Dept. of Mechanical & Materials Engineering

The University of Western Australia

NOTES ON

DESIGN AND ANALYSIS

OF

MACHINE ELEMENTS

Douglas Wright

February 2001

INTRODUCTION AND EXPLANATION FOR STUDENTS

The current course in mechanical engineering design at UWA spans three years ( II, III, IV) in which it- introduces the concept of design in the engineering environment (II) and provides hands-on experience of the design

process- reviews failure mechanisms under steady loading (II) and examines failure under fluctuating loads (fatigue) and in

unstable situations (buckling, fracture mechanics) (III)- considers (III) the analysis and safe design of various common elements of engineering systems such as pressure ves-

sels, shafts, gears and the like- provides experience (IV) in the design of systems under the guidance of practising professionals.

These Notes form a resource for all years of the course, though they do not cover every topic (eg. fatigue is not yetincluded). Most chapters appear also online at www.mech.uwa.edu.au/DANotes/ The coloured diagrams (and a fewanimations) of the website are generally much easier ro understand than the black and white copies in the printed Notes.

It is presumed that students commencing the course are familiar with the concepts of equilbrium, stresses etc. but have :- very little or no practical background involving engineering hardware and construction methods,- very little or no experience in solving open-ended design problems in which an expressed need must be transformed

into a physical artefact or an action.

For these reasons the course first explains what is meant by design and why we go to the trouble of designing. A proce-dure for creative problem-solving, referred to as the rudimentary design process, is described in depth. This procedure issimple enough to be understood and to be applied successfully by newcomers to design, yet forms a more-than-adequate basis for solving any open-ended problem likely to be encountered by students. The process should be usedfor design projects in the course.

Design is exemplified in particular through students taking part in a Design & Build competition where the taskdemands a creative solution realisable by kitchen table-top materials and methods of construction rather than by sophis-ticated metal working. This experience should convince students that solving real problems demands both creativityand criticism.

In view of students' lack of exposure to machine elements (belt drives, springs etc.) the Notes adopt a simple mathemati-cal approach to explain elements' behaviour and safety - however it should be realised that although computers andmathematical models may help in this regard, their ability to reflect all nuances of real behaviour cannot be guaranteed.Engineers cannot do without sound engineering judgement based on practical knowledge acquired through experience.

Students may wish to supplement the often necessarily brief descriptions of the Notes by consulting the many librarytexts and references. The web is an increasingly useful descriptive resource.

In practice, some components are designed to guidelines laid down in standard Codes of Practice whose implementa-tion could be disastrous if they are treated like recipe books. While there is nothing intrinsically wrong with recipes -provided that they do not replace or inhibit creativity and provided that their limitations are clearly understood - it is afact that Codes are often applied indiscriminately by students. To help avoid this, the Notes provide background toassist intelligent application of some important Codes, as undergraduate texts usually offer little help in this regard.

An extremely important objective of the design course is to prepare students for their subsequent career - not necessarilyas 'designers' but as ingenious solvers of real-life problems ( pronounced ‘engineers’ ! ) - so this course differs somewhatfrom other University subjects in that it does not serve up a host of facts and figures for memorising, with subsequentregurgitation in examinations. Rather students are expected to demonstrate :- an understanding of why a certain approach is used to throw light on a particular component's behaviour,- an appreciation of the general trends of that behaviour, and - an ability to modify the component economically to suit the design problem in hand.

For these reasons :- Lectures will trace out only the broad arguments.- Students are expected to read the Notes in detail, to follow through the development of the theory whilst appreciat-

ing its assumptions - that is, generally, to flesh out the lecture material.- Examinations are open-book and test ability to adapt course material to new situations. Students should therefore

attempt many tutorial examples, to become adept at adaptation. Having answers for some problems to hand in anexamination (and in real life) is useless unless it is known how to modify the solution processes intelligently.

- Detailed answers to all tutorial problems are provided online, but they should not be consulted until the problemshave been tried and the solution steps appreciated.

Students should realise that when they graduate they must be prepared to tackle the difficult, ie. previously unencoun-tered, problems - the easy, mundane ones can be solved by someone less qualified and less expensive to hire than them.

Douglas Wright

NOMENCLATURE

Symbols are defined when they first appear in

each chapter; however the forms shown here are

in general use.

PROGRAM DIRECTORY

The following Mac programs, referred to in the Notes, have been prepared to assist in the design task and may be down-

loaded from the website : www.mech.uwa.edu.au/DANotes/ Copies may be available also on other departmental serv-

ers from time to time.

Pascal source code is freely available on request, provided that authorship credit is retained always with the code.

COMPILED APPLICATIONS FOR COMPONENT DESIGN & ANALYSIS

brakes analyses twin shoe brakes for sensitivity, torque, bearing loads &c

FEM1 analyses two-dimensional linear elastic systems (with sample data file)

fillet welds analyses fillet welded planar joints consisting of a number of straight runs

motors assists selection of a squirrel cage motor for a given duty & acceleration time

springs facilitates the fatigue design of round wire steel compression springs

steel spur gears analyses steel spur gears for safety against strength and wear failure

tooth generator simulates manufacture of an involute gear tooth by rack generation

V-belts selects V-belt drives suitable for a defined duty

TEXT FILES

FEX00 specimen datafile for FEM1

heads.txt data base for pressure vessel design

FRACTURE MECHANICSFatigue of ductiles; stress concentration; linear elastic fracture mechanics; plasticity; yielding fracture mechanics -the R6 technique; fatigue crack growth. Crack growth kinetics.

FINITE ELEMENTSLinear 1-networks; extension to 2- and 3-networks. The Rayleigh-Ritz method. Finite element theory applied toelasticity - equilibrium of the discretised body; element stiffness. Implementation; condensation and bandwidth;discretisation. Appendices - the refining process; 'FEM1' User's Guide.

UNITS, DIMENSIONS AND CONVERSION FACTORS

x ≡ x ≡

x ≡ x ≡ x

xmin lo

x ≡ x ≡ xtilde a~

x ≡ x ≡ xbar m-

max hix~x-

x

x

CONTENTS

Most chapters include a bibliography and problems whose solutions are online at www.mech.uwa.edu.au/DANotes/

DESIGNWhat is Design ?; why do we design ?; how do we design ?; problem statement; generation of ideas; criteria & con-straints; practicalising the candidates; evaluating the candidates; the feasibility study; where do we go from here ?;more advanced considerations.Appendices A - improvement problem; B - tube end problem; C - lessons in frustration; D - analysis of a springdriven vehicle; E - springs as energy stores; F - JCH Roberts on Creativity.

STRESS, STRENGTH AND SAFETYSafety factor; stress concentration; static indeterminacy; elementary load building blocks; stress resolution; strainresolution; failure theories; putting it all together; design equations for static shafts; power transmission shafts. Appendix - indeterminate assemblies of multiple components.

MISCELLANEOUS STRENGTH TOPICSCastigliano's theorem; thin curved beams; thick curved beams; asymmetric bending; contact stresses.

SPRINGSClose coiled round wire helical compression springs; the spring characteristic; stresses & stiffness; buckling; wirematerials; presetting; fatigue loading; spring design. Appendix - presetting a torsion bar.

THREADED FASTENERSThread geometry; screw thread mechanics; static failure; loads in an elastic bolted assembly; preload and its con-trol; fluid pressurised joints; bolt fatigue; non-uniformly loaded bolt groups.

WELDED JOINTSFillet welded joints; geometric properties of lines; traditional analysis; throat stresses and joint safety; unified anal-ysis; resolution. Appendices - the compliant lap joint; extract from AWRA, Technical Note No. 8.

CYLINDERSAxial stress; thin cylinders; thick cylinders; design equations; thin cylinder errors; strains; autofrettage; compoundcylinders; torsional loading.

PRESSURE VESSELSCorrosion; welded joint efficiency; thin shells of revolution - heads; compensation; pipes and flanges; inspectionopenings; supports; design.

SQUIRREL CAGE MOTORSCharacteristics of a steady load and of a motor; matching a motor to a given steady load; periodic loading; acceler-ation; hydraulic couplings. Appendix - integration in practice.

V-BELT DRIVESOverall geometry; kinetics; fatigue; effectiveness; drive selection; approximate solutions; V-flat and pivoted motordrives. Traction mechanics. Appendix - commercial selection tables.

BRAKESSystem dynamics; linings; brake shoe analysis - short translational shoe; long translational shoe; short rotationalshoe; shoe figures of merit; long rigid shoe; long hinged shoes; twin shoe brakes. The braked wheel; braking ofvehicles; wheel lock - vehicle characteristic; brake control characteristic.

SPUR GEARSOverall kinetics of a gear pair; epicyclic trains; conjugate tooth action; the involute tooth; the generation process -tooth systems and profile shift; gear meshing. Gear failure - reliability; tooth forces; bending strength; pitting resis-tance; periodic duty. Appendices - continued fractions; geometry of the involute gear tooth.

BUCKLINGBuckling of thin walled structures; stability of equilibrium; effects of imperfections; submerged pipelines; practicalcolumns - design equations.

Design 1

much thought. This is true enough - if the solution can be based on direct experience. However weshall soon come to realise that without experience such a thoughtless solution usually comes to griefsooner or later - the more involved the problem and the more folk affected by the solution, the morelikely is the solution going to fall in a heap.

Any old solution will not do - we must strive for the optimum solution.

We expect that the design process, if properly carried out, will show a high probability of disclosinga solution which is optimum or close-to-optimum, if indeed a unique optimum exists.

The prime aim of this chapter is to develop a structured approach to design - an approach whichwill promote confidence in effectively solving real life problems. We shall focus on problems involv-ing engineering hardware - particularly for Design and Build (D&B) Competitions - however theapproach is perfectly general and applicable to problems arising from a marketing sortie or a labourwrangle for example. The approach is thus very relevant to managers for example - not just to 'hard-ware designers'.

Before presenting the method however, let us look briefly at why we go to the trouble of designing.

Why do we design ?

In a nutshell . . . . TO SURVIVE.

Most people these days exist by providing 'things' to others; in the case of engineers these 'things'are technical muscle-power or know-how, or physical artefacts - that is solutions to buyers' or hirers'particular problems. If these clients are not completely satisfied with the 'thing' provided then theywill dismiss the provider, go somewhere else for their next 'thing', and tell everyone about the pro-vider's unsatisfactory 'things'. If this happens often enough to a particular provider then he willcease to exist as a market force - nobody will want to know.

So clearly, if 'things' are not designed with care and attention to clients' needs then the provider willhave problems - just like Jane and John . . . .

• Jane worked as an engineer for a firm of consulting engineers, one of a number of such firmsspecialising in minerals processing.

A certain mining company intended to develop a new deposit and therefore required plant toprocess the mineral. It called for 'tenders' that is for plans and cost estimates for constructionand operation of the plant. The various consulting firms simultaneously each set about design-ing the plant - ie. solving the client mining company's particular problem - and then reportedto the client outlining its proposed optimum solution. Consultants receive no remuneration forthis service.The client reviewed the solutions submitted by the various consulting firms, and awarded thecontract for ongoing project management to the firm which had best satisfied its perceivedneeds. The successful consulting firm therefore had ongoing work for a year or two.Jane's firm was not successful in this instance. But this was unexceptional - consultants do notnormally expect to win every contract which is put out to tender.But Jane's firm did not win the next job to come up either.Or the next . . . . . . Or the next !So what happened eventually ? Predictably, with no successful designs and with no moneycoming in, Jane's firm folded and Jane is now out looking for work. There could of course bereasons aplenty for Jane's firm sinking - but it could not hope to exist with designs which were

DESIGN

The word 'design' means different things to different people - a wallpaper pattern, a fashionabledress, the appearance of a racing car and so on. We therefore start by defining what we mean by'design' in the present context - ie. What design is all about.

This understanding will lead to an examination of• Why we need to 'design', particularly in an engineering environment, and• How we might best go about 'designing'.

What is design ?The Concise Oxford Dictionary explains design as 'a mental plan, a scheme of attack, end in view,adaptation of means to ends, . . . preliminary sketch for picture, . . . invention.' Evidently there is alot more to design than mere visual aspects, and design is not restricted to engineering. Key compo-nents of this explanation are as follows :-

• Means to ends implies that we design not for the abstract mental exercise, but with a definitegoal in view - some action or some physical object (artefact) will result from the design.

• Mental suggests that design is a thinking process. When we design we deal primarily withideas, with abstractions rather than with numbers - and computers for example cannot do thejob for us, though they can help in certain tasks. No matter what we design, it is vital that we develop and apply our imagination to visualiserealistically the future form of the artefact or action, how it will eventually come into beingand most importantly how it will thereafter interact with people and other artefacts or actions.

• Plan, scheme infers that design is distinct from implementation. Designers especially in engi-neering seldom execute their plans, but rather communicate them to others - either by word ofmouth, or visually (sketches, engineering drawings, computer simulations &c), or through thewritten word. Again, note the lack of emphasis on numbers.

• Invention means just that, we are coming up with something NEW - at least partly. Creativityis crucial as we shall see later.

So, can we now define design? No ! and neither do we need to. A rigid definition implies a rigid pro-cess, and design is anything but that. We shall adopt the following interpretation as it incorporatesthe above concepts and conveys a reasonably clear idea of what design is all about -

Design is the application of creativity to planning the optimum solution

of a given problem and the communication of that plan to others.

Apart from the communication aspect therefore, we understand the essence of design to be prob-lem-solving, though the type of problem encountered in design is not like a typical textbook mathe-matics problem for example in which the unique 'correct' solution is guaranteed by following,automaton-like, a series of learned solution steps. A design problem on the other hand is a real-lifeproblem with many solutions, some of which meet the problem requirements better, some worse,and where the process of discovering the solutions is not mechanistic.

Some problems might appear not to need 'design' as a solution can be cobbled together without

Design 2

A problem is not a problem if it has been solved successfully in the past - it is trivial. Conversely ifthe solution to a problem is not known prior to design, then the problem is new and the solutionalso must be new.

The necessity for novelty in design is obvious where a number of competing providers of the same'thing' coexist by continually providing new 'things'. Computer-'things' are a case in point - providerA first launches a completely new type of memory, provider B counters by making it half the size,provider C attacks via a drastic price cut enabled by a novel manufacturing technique, provider Dedges ahead with a much faster operating system, and so on. Nobody can afford to stand still;nobody can exist by slavish copying; novelty is a necessity for good design, for survival.

Survival = Good design = Creativity

This does not imply that all aspects of a successful design have to be novel; you need not re-inventthe wheel.

It is useful to view design in the context of a typical artefact which evolves from initial conception,through the distinct stages shown below, to eventual obsolescence. A planned action undergoes ananalogous sequence, however we shall concentrate on hardware.

• A need is recognised, ie. a problem is posed, so• a certain artefact is designed to meet the problem - thereafter• the artefact is manufactured, and• sold/delivered to the user . . . .• . . . . who operates it, causing wear and• requiring maintenance to restore its effectiveness, until• eventually it reaches the end of its economic life and is retired.

Various people are involved in the various stages - the designers, the manufacturers, the salespeo-ple, the operators, the maintainers and the eventual dismantlers of an artefact are all completely dif-ferent folk carrying out completely different tasks.

Feasibility Study

Operations Research

Detail Design

Research & Development

Industrial Design, Ergonomics

Industrial Relations

&c &c

Murphy's Law

Tech. Specs.Safety

STAGES IN THE LIFE OFA TYPICAL ARTEFACT

CreativityEconomics

Conception ( need )

DESIGN

Manufacture

Distribution, Sale

Operation

Maintenance

Retiral

feed

bac

k o

f an

tici

pat

ory

idea

s

demonstrably not competitive. Jane now realises that, while a score of 80% in a Universityexamination might be regarded as excellent, in real life there are no marks whatsoever forcoming second.

• John's firm, which makes and installs large industrial ovens, was approached by a client whowanted to install such an oven in its existing factory. John was delegated to look after the con-tract, so he examined thoroughly the myriad technical issues, including the most suitablechoice of . . . .- energy source necessary to raise the oven's temperature,- location for the oven in the factory with regard to minimising transport of products from/to

other manufacturing operations in the factory,- control mechanism needed to ensure that the oven's temperature stays within bounds,- fail-safe safety procedures which prevent any employee from being inadvertently locked in

the oven,- insulation thickness to optimally balance the first and ongoing costs of insulation,- chimney dimensions for projecting the exhaust gases high enough to ensure clean air for the

surrounding environment- . . . . and so on, there were lots of other aspects to consider.

John was technically competent. He carried out all his sums correctly - though we need notworry about the details at this stage. He was very satisfied when his recommended optimumsolution (a gas-fired oven) was accepted and the oven was designed in detail, built and putinto service.All went well until reports fil-tered in that the client's officestaff were reporting headachesdue to vibration of the officestructure. Expensive investiga-tion proved that the culprit wasthe oven's fan which drew airfor gas combustion through anintake duct crossing the officeceiling. After further work Johneventually had the duct re-routed from the adjacent wallas shown in the factory plan. John's firm had to pay for the investigation, for the modifications, for a number of medicalbills, and for losses in production while the oven was out of service being modified.Again all went well until the gatekeepers started to complain bitterly about noise from the re-positioned air inlet next their hut. So it was back to the drawing board once again for John . . . .he was not popular !

Lessons that John learned from this experience included :- There is a lot more to design than mere technical calculations.- An incomplete design which does not take everyone's viewpoint into consideration is a

recipe for trouble.- It is the designer's reponsibility to seek out these viewpoints.- A solution must be close-to-optimum to start with, as retrospective fixes are never wholly

satisfactory.

officefactoryfloor

ORIGINAL PLANWITH AIR INTAKE DUCT THROUGH OFFICE CEILING

factoryfloor

oven

office

MODIFIEDARRANGEMENT OFAIR INTAKE DUCT

hut

oven intake

inta

ke

Design 3

If something can go wrong then it will go wrong - and at the worst possible time

Murphy cannot be ignored - there is no excuse for designers throwing up their hands andexclaiming 'How were we to foresee that happening ?' . . . . but they must foresee it (whateverit might be) and make allowance at the design stage to minimise its deleterious effects.Murphy is especially hard on beginning designers who have yet to learn that Nature does notalways follow simple theoretical predictions. But Murphy is no respecter of persons, and manyan experienced designer has suffered at his hands !

We wrap this section up by drawing attention to two articles reproduced below from the technicalpress which throw further light on why we design :

• Seymour in 'Competition Analysis' lists many criteria which are commonly used by clients tocompare the products of competing providers. Designers must be aware of all these criteriaand design accordingly, and not focus solely on the 'technical specifications'.

• Somerville reports that 'Woodside Critical of Aust Suppliers' following construction of the NWShelf LNG facilities. Local industry is lambasted for poor performance : 'Some 70% of the tend-ers (ie. feasibility studies) received from Australian companies were technically inadequate'and '. . related to lack of effort in preparation . .' This is certainly an indictment that designersignore at their peril. However it’s the very last sentence in this article which is particularlydamning - Why ?

Having emphasized the importance of design, it is now time to look at how we go about it . . . .

How do we design ?

Newcomers to design often feel unsure of themselvesbecause the problems encountered are unlike those pre-viously solved successfully in other units such asDynamics, Strength of Materials &c. Although there isno mechanistic series of steps leading to the 'correct'solution of a design problem, there are techniques whichmay be learned for tackling design with confidence andwith reasonable expectation of achieving a 'close-to-optimum' solution.

The illustrated Rudimentary Design Process is one suchtechnique, and will form the model for design through-out this course - however the success of the process, likethat of any human endeavour, depends largely on theattitude, skill and effort of the practitioner(s).

The rudimentary model is the engine lying at the veryheart of all professional engineering design processessuch as the Pahl & Beitz model (illustrated later) typify-ing Continental practice, or the SEED model common inthe UK. Despite the complexity of these, the rudimen-tary model is itself sufficiently simple to be used effec-tively for problem solving by the rawest tyro.

It requires the designer(s) to carry out sequentially five

BANK OF SOLUTION CANDIDATESuncritical - quantity not quality

OPTIMUMSOLUTION

communicate

maybe

RUDIMENTARYDESIGN

PROCESS

AVAILABLEKNOWLEDGE

GENERATEIDEAS

methods

STATE THEPROBLEM

broad, complete

SPECIFY CRITERIA &

CONSTRAINTS

RENDERPRACTICABLE

EVALUATECANDIDATES

Design is the springboard for all subsequent stages, and so it is at the design stage that the later sat-isfaction of each and every one of these folk is, or is not, effectively set in stone. That is why the'feedback of anticipatory ideas' is highlighted in the sketch, as it is vital that designers foresee - inevery last detail - the interaction of the planned artefact with all these people, and endeavour to ful-fill their wishlists. A designer must put herself in other folks' shoes, close her eyes and realisticallyimagine their interactions with the artefact.

Do not get carried away by technicalities. Remember always that it is people who make decisions topurchase; it is people who have to live with your design. A designer's primary goal is the satisfac-tion of people, not of elegant mathematical expressions.

Design is keeping everybody happy.

. . . or at least as happy as possible. Sometimes it may be nigh on impossible to please everyone, butyou'll never-never know if you never-never have a go at trying to please them. That is why wedesign ! We'll see later how to factor in conflicting criteria and different agenda. If you are not surewhat these are likely to be in a particular case, then don't be like John - find out.

The importance of good design is underlined by the fact that in Australian manufacturing industryaround 70% of product costs are defined at the design stage. As the average profit is only some 7% itwill be appreciated that indifferent design is commercially intolerable, as Jane's employers discov-ered to their cost. The life stages sketch emphasises the importance of creativity and economics indesign, and of the technical specifications and safety in operation. We shall return to this criticalsafety issue later.

Also shown in the sketch are some facets of the design process which it is useful to introduce here :

• A feasibility study is a report describing in broad but realisable terms the optimum solution.An important component of a real life feasibility study is the solution's cost, but detailed cost-ing is generally not expected in this course.

• Operations research is the name given to the branch of mathematics which models industrialand commercial processes such as queuing, distribution, scheduling &c.

• Detail design completes all details necessary for the next stage, manufacture, details which areomitted in the deliberately broad-brush treatment of the feasibility study. In practice a solutionmust first be confirmed as feasible and the decision made to proceed with it, before detailingcommences.

• If a design lies at the cutting edge of known practice or science then it may not be possible toaccurately model certain aspects of its behaviour. Further research and development (R&D)involving experimentation must then be conducted before these aspects of the design can befinalised with confidence.

• Industrial design deals with artefacts' aesthetics, safety and ergonomics among other things.The principles of ergonomics are used to optimise human-machine interaction when designingeg. the controls of a bobcat (a mini bulldozer) so that the operator and the bobcat are essentiallyseamless with the operator's eyes, two feet and hands integral non-fatigued components of thecontrol loops for turning, accelerating, reversing, braking, blade lifting, blade orienting and soon.

• Industrial relations together with occupational health and safety are obvious and importantconsiderations in design - they are just facets of 'keeping everyone happy'.

• Murphy's Law states that :

range of piping materials requiredand in many instances could not meetthe required schedule," he said.

Overall Mittertreiner estimates $226miliion worth of sales were "lost" byAustralian industry through lack ofcapability and experience. Essentialitems purchased overseas includecryogenic heat exchangers, gas tur-bine generators, gas turbine and elec-tric drive compressors and cryogenicpumps.

Other factors pushing up the cost,he claims, are decisions to reject thelowest overseas price offered and theacceptance of Australian tenders.

Rankling Mittertreiner is the choiceof Australian suppliers for transform-ers, air fin coolers for the LNG trains,and power cables.

These purchases came after discus-sion with the State and Federal Gov-ernments through the National Liai-son Group (NLG), a body set up tomonitor and boost Australian partici-pation in the project.

In the case of the power transform-ers it appeared to the NLG, after rep-resentations from the WA Govern-ment, manufacturing groups and thetrade unions - who threatened indus-trial action if nothing was done tostop the tender going offshore - thatan anti-dumping inquiry would beundertaken into the price submittedby the originally successful foreigntenderer once the transformerslanded in Australia.

A spokesman for Woodside said theoverseas tenderer withdrew from thecontract rather than face costly litiga-tion and Westralian Transformers, aWA-based Westinghouse subsidiary,picked it up.

The general manager for WestralianTransformers, Beavan Oakes, said theoriginal winner was substantiallybelow the 4 other tenderers.

"On the information we had it wasquite clearly a dumping case. Afterduty the overseas price was 20%below the next tendered price," hesaid. Oakes is not convinced that Mit-tertreiner is correct in claiming Aus-tralian participation pushed up thecost of the project.

"Sure Australian costs are high butWoodside chose the site so far fromcivilisation and imposed the stringentrequirements," he said.

- Paul Somerville

Many Australian suppliers to theNorthwest Shelf liquid natural gasproject have difficulty meeting thehigh quality standards set down forthe LNG plant, are often behindschedule and are sometimes not pricecompetitive with overseas suppliers,said Woodside's LNG project man-ager, Frans Mittertreiner.

He made the remarks when review-ing the performance of Australiancompanies on the $3 billion project ata briefing in Perth, staged by Wood-side Offshore Petroleum Pty Ltd atthe end of last year.

Mittertreiner said the review ofindustry's performance had been atthe request of the WA State Govern-ment, the Federal Government andAustralian industry.

By October 1987 $2240 million, or75% of the total estimated cost of theLNG plant, had been committed.

Of the $638 million worth of equip-ment and materials ordered for theLNG plant, 54% was acquired locally.

Overall, he said, the performance ofAustralian industry had been goodbut required improvement in specificareas.

Despite some problems the plant isexpected to be completed on timewith the first shipment of LNG head-ing for Japan in October 1989.

According to Mittertreiner Austra-lian companies had problems main-taining consistently high quality.

"In particular difficulties in produc-ing acceptable castings for both valvesand pumps caused numerous repairsand recastings, which resulted in seri-ous delivery delays," he said.

He went on to say that the mostcommon problem was the relativelyhigh incidence of dimensional errorsand fabrication misalignments in Aus-tralian shops as evidenced by the rela-tively large number of concessionrequests.

He pointed to a lack of understand-ing by vendors of the specificationsand quality assurance and quality con-trol requirements within orders. This,he said, was related to a lack of effortin the bid preparation stage and tend-ers received from Australian compa-nies were generally below the stan-dard required and poor in comparisonto overseas tenders.

"Some 70% of the tenders receivedfrom Australian companies were tech-

nically inadequate" he said.Some local suppliers, he observed,

placed more emphasis on quantityrather than quality and often lackedattention to detail.

Getting Australian made or sup-plied items to the plant site on timepresented Woodside with a few head-aches.

"Australian vendors have not per-

formed as well on equipment deliver-ies as overseas vendors. While 75% ofoverseas equipment items were deliv-ered within 3 months of the promiseddate, only 52% of the Australiansourced items were delivered in thesame time. For Australian vendorsmore than 26% of deliveries weremore than 6 months late" said Mitter-treiner.

All overseas vessels were deliveredon time while only 18% of vesselsfrom Australia were delivered ontime. For columns, 60% were deliv-ered within 3 months of promiseddeliveries for overseas vendors, whileno columns from Australia weredelivered in the same period," Mitter-treiner said.

He also noted the need to stay costcompetitive with overseas vendors.

"Where orders [for columns andvessels] were won by overseas manu-facturers, the lowest Australian ten-dered prices were between 25% and110% higher," he said.

The prices for locally made piping,flanges and process valves were some100% to 200% higher than those ofsimilar overseas items, he said. So farlocal industry has picked up 41% byvalue of potential orders.

For the purchase of items such aspipes and valves $67 million worthcould not be purchased here.

Australian industry did not havethe capability to produce the pipingand flanges in many of the sizes andtypes required, he said.

"In general Australian industrycould only produce a small part of the

"Some 70% of thetenders received fromAustralian companies

were technicallyinadequate"

NORTHWEST SHELF

Engineers Australia February 5th 1988 17

First, it is essential to define and identify competingproducts. Usually, these will be products that either oper-ate by the same mechanisms, or by utilising differentmechanisms achievc the same end result. The latter is par-ticularly influential today with the rapid advances in tech-nology. In consequence, has your company's market intel-ligence isolated a competitor developing a new process ortechnology which may greatly affect your product's mar-ketability?

For each competing product, it is necessary to identify(1) the technical advantages and disadvantages, bearing

in mind the function of the product(2) the range of applications(3) those products protected by patents, and the advan-

tages created by the patents(4) the extent of adherence to British Standards(5) the relative superiority on performance, reliability,

quality, finish and serviceability(6) the reasons for any modifications which have been

carried out in the past four years(7) those which must be used in conjunction with

another product, and may be dependent on factorsoutside the control of the manufacturer.

For a product to sell successfully, it must appeal to theneeds of the potential customer. In order to orientate thedesign and marketing effort, those factors consideredimportant by the potential customer in making a purchasedecision should be identifïed, eg :

pricedeliveryafter-sales servicemaintenance cost including sparesproduct and company reputationbrand loyaltypersonal contactssales historyintra-company trading restrictionstechnical specificationspolitical considerationsnumber of purchase outletsproduct range.In anticipating the potential growth of each competitor,

it is necessary to identify which- have excess production capacity- have facilities for factory expansion- are well placed to obtain capital.Marketing not only involves selling products in the most

effective manner, it demands feedback to various func-tions within a company, such as design. Hence it is advisa-ble for engineers to be aware of the commercial position ofcompetitors.

Increasing sales in a rapidly growing market is obvi-ously more straightforward than in a mature, slow growth

CME January 1981 39

market. It should be appreciated, however, that maturemarkets can often provide a stable source of return. What-ever the dynamic position, the market type and structuremust be determined to enable marketing plans and bud-gets to be developed. Therefore the following informationshould be collected

(1) the competing companies(2) the market size in value and volume, by product(3) the competitors' market shares by value and volume,

by product(4) the market growth rate over the past four years,

together with the anticipated growth rate over thenext three years

(5) the changes in competitors' market shares in relationto market size alterations

(6) the percentage of sales exported, by product, bycompetitor

(7) the major competitors which dominate the market,and any changes in leadership over the last fouryears

(8) the number of companies entering and leaving themarket over the last four years, and the reasons forthose movements.

This detail should give a picture of the market structure.It is now necessary to to analyse the marketing operationsof competitors. When conducting this, always make acomparison to your own company. You should include

- distribution systems and effectiveness- disribution agreements and margins- discount and credit facilities- stocking policy, at plant, distributors and users- sales promotion, the mix of techniques and budget- number of salesmen- basis for salesmen's remuneration- basis for territory allocation- licensing and franchise agreements- tendering policies.

The above will detail major similarities and differencesto your own company. It may also illustrate whether theproduct type offered by your company is considered pri-mary or secondary to your competitors. Identifying theproduct which provides each competitor with its greatestprofit will give some insight into the possible reactions ofthose competitors to technological improvements orchanges in marketing strategy by your company. Which ofthese companies take competition and market researchseriously, and hence will be able to react quickly to anymarket changes?

Answering the above questions will indicate whatstrengths, currently enjoyed by the market leader, can bedeveloped by your company.

Competition Analysisby S Seymour

This article is a checklist of the major factorsthat should be considered in conducting ananalysis of competitors and their products.

Woodside critical of Aust suppliers

Design 5

ties repeated for the whole initial bank of candidates.

It is important to complete each activity over all the candidates before starting the next activity.

The outcome of the design process is the feasibility study, which in the present context describessuccintly the major decisions made whilst arriving at the optimum solution.You, the designer, have to sell your ideas. You will have to convince others that your optimum solu-tion is in fact THE optimum, and this requires justification of all significant design decisions. Justifi-cation does not mean unsubstantiated opinions - you must be prepared to defend your argumentswith demonstrable facts.

When you have acquired the necessary engineering judgement you will be able to shortcut steps inthe rudimentary design process - for example you might not need to consciously practicalise all can-didates or to fully justify each and every decision, relying instead upon experience. However if yourengineering judgement is still at the toddling stage then you should carry out all the activitiesexactly as described.

Design & build competitions provide practice at applying the design process by those whose knowl-edge of practical solutions, manufacturing techniques and the like is generally limited. The competi-tions thus demand creative solutions, and enable student designers to appreciate the implications oftheir designs when it comes to the later stages of manufacture, of operation and of maintenance.Details of the design process will be explained with particular emphasis on D&B competitions.

Let us now look at the individual activities in more detail, starting with the problem statement.

Problem statement

Factors relevant to the problem statement include the following :

• Understand the problem - communicate.Problems are often specified rather imprecisely because the client does not understand exactlywhat's required, or because complete clarification requires initial exploratory work. An impre-cise specification should not prevent a start being made on the problem, but it is important thatdesigner and client communicate at the earliest opportunity - and continue to communicatethroughout the design process - to make sure that they are both on the same wavelengthregarding what has to be done. It is all too easy for a poorly briefed designer to go off at a tangent. It is the designer's responsi-bility - not the client's - to initiate communication and to clear up any misunderstanding.Specifications, for D&B competitions in particular, are often misinterpreted simply becausethey are not read with enough attention to detail. There is no place for loose interpretation.Each and every clause must be isolated, put under the microscope and examined critically todeduce its ongoing implications. Especial care must be taken to . . . .

• Avoid artificial constraints.The solution space for a problem must always be finite andfurther constrained by the designer's limited knowledge.Extreme care must be taken to avoid tighter artificial con-straints. These are fictitious restrictions which are not intrin-sic to the problem but which are introduced incorrectly andusually unwittingly by the designer. - In the SLAMDUNK competition, students had to design

and build a device to pick up a ball from a rest position on the ground, transport it to a verti-cal pipe 3 m away, and drop it into the end of the pipe 1 m above ground - the accent was on

real boundary

fictitiousrestrictions

your limitedknowledge

solutionspace

solutionspace

discrete activities. Each of these activities will be explained in depth later but it is useful first todescribe broadly what each consists of and how they fit together :-

• It is not enough to unthinkingly accept the problem as given, instead the designer mustamplify and state the problem in terms which are both :- broad, in not being constrained unnecessarily, and- complete, in identifying everyone and everything likely to be affected by any solution.The problem statement contains information relevant to the next two activities :

• The designer, assisted by methods which enhance creativity, generates ideas for possible solu-tions to the problem. This activity is inventive and often comprises a free association of ideasbased upon the designer's knowledge of solutions to analogous problems. Ideas are recordedbut not criticised in any way - the activity must be totally non-judgemental since criticism crip-ples creativity.When idea generation is exhausted, the result of this activity is a bank of many potential solu-tion candidates - some are later found to be good, some bad, some ugly.

• The designer must then specify the various :- constraints (such as 'must be shorter than 1m') to which every candidate must conform if it is

to be a viable solution, and- criteria ( such as 'cost' and 'degree of safety' ) by means of which candidates may later be

compared with one another.The constraints and criteria reflect the wishlists of all those identified as having a stake in theproblem, so this very detailed activity is really one of completing the problem in unequivocalterms.This activity is left until after ideation has finished, thereby avoiding the stifling of ideation byprior nit-picking detail.

• A practical solution satisfies all the problem's constraints - it can be manufactured, it will oper-ate, and so on. 'Practicalisation' is the process of rendering a candidate practical - of fleshingout or transforming a raw idea, possibly recorded during the ideation activity only by a key-word or rough sketch, into a practical solution to the problem.During practicalisation the designer continually criticises each candidate, moulding it to con-form to the constraints and to satisfy the criteria as much as possible. Practicalisation invari-ably uncovers fresh secondary problems, some of which may be trivial while others must besolved by further application of the rudimentary process.Since ideas were generated non-judgementally, it will not be possible to practicalise every ideato comply with all constraints - these ideas must therefore be trashed. But impractical ideasoften trigger other worthwhile ideas - that is why trashing is delayed until this late stage of thedesign process. Trashing is due to the inability to meet constraints, not criteria.On completion of this activity, every candidate must have been either :- rendered entirely practical, or- scrapped . . .. . . . . . . there are no half measures.

• The designer finally evaluates the remaining, now fully practical candidates to see how theycompare with one another in meeting the complete set of criteria defined earlier. The optimumsolution to the problem is the candidate which best satisfies ALL the criteria, not just one ortwo.The evaluation activity may indicate that some of the criteria or constraints are inappropriate.If this should occur then the criteria and constraints must be updated and the last three activi-

Design 6

etc

allow insurance

to lapse

petrolconsumption

ignoremaintenance

exchangeRolls for pre-owned Lada

how to reduce car's running costs ?

or, more broadly :

MY POSER

etc

walk bike publictransport

car

how to travel between home and work at minimum cost ?

or, more broadly :

etc

I go to workworkcomesto me

webinterface

tele-conferencing

how to interface me and work at minimum cost ?

or, more broadly :

etchow to make money ?

play thegee-gee's

work atUniversity

retire on aGovernment

pension

send myfamily out

to work!!

or, more broadly :

I am feeling the pinch financially, so I pose the problem . . .

how to reduce my car's petrol consumption ?etc

improve driving

habits& narrower

checktyre

pressures

removedinghy from

roof

tune engine

minimising the run-time from pickup to drop. Most studentsdivided this task into two components :

i carry the ball 3 m horizontally, andii lift the ball 1 m vertically.

This subdivision was perfectly legitimate because a separate source of energy was appropri-ate for each of the two motion components. But many students artificially bounded the prob-lem by separating also the kinematics - eg. the horizontal component was completed first, atrigger thereafter activating the vertical. If the motion components had been carried outsimultaneously then a quicker run could have resulted.

- A pressure vessel is used to store a pressurised fluid. It usually consists of a cylinder with endclosures and with attached pipes through which fluid enters and leaves the vessel. Studentswere asked to design such a vessel with a 'pipe diameter of 500 mm'. They took this to meanthat the pipe's bore (inside diameter) was 500 mm, but in fact this dimension referred to thepipe's 'nominal diameter' - a different thing entirely. By introducing this artificial constraintthe students' optimum vessel was in fact some 25% unnecessarily expensive and would nothave gotten past first base in a real-life tendering situation.The detailed logic of why this fictitious bound led inexorably to a cost blow-out need notconcern us here, but the point which must be made is that slack interpretation of the specifi-cation was the root cause of the blow-out - exactly the sort of thing that Somerville (above)warned against.

One insidious artificial constraint involves accepting the problem as presented without consid-ering whether the solution to a broader problem might in fact give greater satisfaction. So youshould try to . . . .

• Broaden the problem.Try to define a less specific problem which encompasses the given problem, thereby increasingthe potential advantages of a successful solution. My Poser shown opposite is a case in point -rather than pussy-foot around adjusting my car's timing in an effort to reduce fuel costs andimprove my finances, I am likely to be substantially more satisfied by posing and solving abroader problem.Broadening is not always so attractive as it was in this example: it depends on the context. Theclient may not wish it broadened; you may be given a very specific design task to carry out - nofrills, no fancy ideas, just solve the problem! While you do not have much option under thesecircumstances, attempts to broaden the issue - if only for your own enlightenment - will assistyour understanding and solution of the specific problem.

• Complete the problem.Completion is a very detailed exercise, and in order to prevent its inhibiting creativity it shouldgenerally be left until after the bank of solution candidates has been generated.During the problem statement therefore, completion should consist only of identifying folkwho are likely to be affected by, and who must be satisfied by the design. Their agenda will beexamined in detail under 'constraints and criteria' below.

• Consider subdividing the problem.Sometimes a problem lends itself to splitting into sub-problems which are each easier to solvethan the original problem. For example if the problem were to collect a pile of grain at groundlevel and deliver it to a bin 3 m high and some metres distant horizontally, then separate sub-problems might be posed :

i how to collect the grain and lift it off the ground ?

Design 7

Generation of ideas

This inventive phase of design is unlike other activities in the design process - and indeed differsfrom other subjects in the engineering course - in being absolutely non-critical. You may find it diffi-cult to switch off your analytical faculties, but it is vital that you completely separate creation fromcriticism, since nothing inhibits creativity more than does nit-picking criticism. There is plenty ofscope for criticism during later activities in the design process.

Avoid ALL criticism while creating

Don't be disheartened because you think you lack any powers of invention. Krick op cit concludesthat inventiveness depends upon :

• Inherited qualities : we cannot all inherit Leonardo's genes, but read on . . . .• Method : you can employ proven techniques to increase your inventive prowess• Attitude : you must be positive, you can invent !• Knowledge : your understanding of how related problems are solved can be increased• Effort : Edison (inventor of the light bulb) was spot-on when he observed that

‘Invention is 1% inspiration and 99% perspiration’

Apart from the first, these are skills, and like all skills they develop only with practice. This coursecannot make you a good designer, it can only point you in the right direction.

Ideas cannot emerge from a complete vacuum. Our minds generate ideas, however indirectly, fromour store of knowledge - moulding, modifying, and trying on for size the myriad items buried awayin our subconscious which might conceivably be relevant to the problem in hand. This necessaryknowledge base is indicated in the rudimentary design process - the deeper and wider the base, themore useful ideas are likely to be generated. Student designers are therefore advised to continuallyincrease their general, scientific and technical knowledge by taking advantage of the resources avail-able in university libraries eg. Some interesting and not-too-heavy periodicals are cited in the Bibli-ography at the end of this chapter.

There are numerous techniques to enhance creativity - Synectics, Morphological Analysis, and theTheory of Inventive Problem Solving (TIPS), to name a few. Some techniques are more structured thanothers; some are more hit-and-miss - but they all suggest ways of modifying and/or combiningexisting partial or related solutions in order to devise new solutions. Which is all very well, pro-vided the inventor knows about existing solutions. The knowledge component is therefore crucialwhen ideating.

Structured techniques are suited to more bounded problems for which detailed knowledge ofrelated solutions in a specialised field is particularly important. For example the designers of a newindustrial vacuum cleaner would need to know about the fluid dynamics of multi-stage fans, theintricacies of manufacture with various plastic materials, the latest trends in visual appeal, what sortof jobs the cleaner might tackle (slurp up liquids? back-pack transportability?), cost/performance ofpotential competitors, ease of emptying, etc etc.

The present course is aimed at jolting the tyro designer away from ingrained analytical habits, andso the problems introduced here - typified by D&B tasks - are amenable to less structured methodssuch as brainstorming, based on general knowledge of how similar problems are solved by Naturefor example. Brainstorming is essentially a free association of ideas where each idea is recordedwhen first thought of, and actively encouraged to beget further ideas. A designer can brainstormalone, or preferably with others who have different interests, different experiences . . . . ie. essentially

ii how to elevate the grain to a height of 3 m ?iii how to transport the grain some metres horizontally ?iv how to energise these various transportation components ?v how to control these various transportation components ?vi from what materials should a device be constructed ? etc. etc.

Although this divide-and-rule philosophy can lead to a more tractable solution process, it isnot a universal panacea because :- If you think about the problem too critically with a view to subdivision then your mind may

be pre-configured into the critical mode, to the detriment of subsequent ideation.- If subdivision leads to a plethora of problems then you might unconsciously throw in the

towel - the prospect of having to solve so many problems is just too daunting.- Subdivision of the problem may obscure unified solutions. For example if each grain

transporter sub-problem were individually addressed, then it would be more difficult tocome up with solutions in the form of a large industrial vacuum cleaner (answering sub-problems i-v in one fell swoop), or a couple of guys equipped with wheelbarrows and shov-els (answering all the above).

- Sub-problems are seldom of equal difficulty. You must learn to distinguish between thosewhich are relatively difficult and those which are not, putting most of your energy into solv-ing the former. For example in D&B competitions the question of materials is often quiteminor compared to the difficulty of figuring out a mechanism to carry out the required task -so the choice of materials might advisedly be left until later detailing.

• Beware the 'improvement' brief.Look out when you are asked to improve an existing solution! An improvement is often soughtbecause someone else's design is less than perfect, and by sticking to the letter of your instruc-tions you may automatically retain drawbacks which are inherent in the existing solution -recall John's oven and the inevitable limitations of retro-fixes.Another class of 'improvement' brief arises when an existing solution must be adapted to con-ditions which differ from those for which it was designed in the first place - conditions forwhich it may be totally unsuited. Appendix A describes an unfortunate request for improve-ments to a computer workstation.

As a problem-solver, you cannot afford to accept uncritically whatever information is handed toyou. In many cases you have to figure out the best problem before figuring out the best solution !And for this . . . .

You must ask the right questions.

It wasn't until after John had stated and solved his oven problem that he appreciated the need to askaround. And by then of course it was too late.

Before progressing to the next stage of the design process therefore, the problem statement must beas broad and as complete as you can make it. Let's now consider this next stage - ideation.

Design 8

( 1 )

( 2 )

( 3 )

( 4 )

( 5 )

( 6 )

( 7 )( 8 )

( 9 )

( 10 )( 11 )

( 12 )

( 13 ) ( 14 )

fixed

diaphragm

different spans of knowledge.

Any hint of criticism during ideation is strictly taboo - there is no such thing as a 'crazy' idea whilstgenerating ideas. This is the key. If you think that an idea is unworkable or laughable, then thethought MUST be suppressed. Any idea which gives birth to other ideas cannot be wholly daft, andas this spawning is not weighed up until post-ideation criticism, there cannot be any justificationduring ideation for concluding that the idea is crazy or not. There is plenty of scope for criticismafter the ideation activity has ceased.

Roberts, in Appendix F, presents a number of idea-generation techniques. You should read thisextract to obtain an overview of the various methods available for enhancing creativity - if nothingelse it indicates that the topic is sufficiently regarded to have attracted a lot of interest over the years.We shall concentrate on brainstorming as described by Roberts - you should practice this whereverpossible to gain proficiency.

Some hints which are generally applicable to the ideating phase and which have not yet been men-tioned include the following :

• Two activities follow the problem statement in the rudimentary design process :- ideating to create the bank of solution candidates, and- completing the problem by defining the constraints and criteria.It is suggested strongly that ideation be carried out first, before one's thoughtsturn critically to constraints and criteria. The reason for this recommendationmay be seen from the fluid filled pipe - "How can fluid be prevented from escaping from the end of the tube ?"What is the effect of defining the constraints and criteria before ideation ? App-endix B illustrates the benefits from delaying critical thinking till after ideation.

• Ideas in engineering are often but not always recorded more succintly by sketches than bykeywords.

• Although the more ideas you generate the better, you must recognisewhen to stop inventing. Design costs money; so if you find that yourcreativity is drying up, then it may be time to quit.

• Further candidate solutions often pop spontaneously into one's headsome days after a deliberate brainstorming session has been termi-nated. If possible, time should be allowed for this subconscious gesta-tion as it can't be hurried.

• Finally, remember that it takes effort to search for new ideas, and, as later criticism invariablythrows up lesser problems, you must be prepared to confront these lesser problems. We'll lookat this more closely when we come to the practicalisation activity.

EXAMPLE Generate ideas for the operating principle of a mechanical device to raise and retain a1 kg unopened can of baked beans as high as possible from rest. Prior to raising, the device must fitinside a cubic envelope of 0.4 m side.

Possible mechanisms are sketched opposite- the descriptions are amplified by comments (in italics)which were added after ideation had been completed :

( 1) large spring like a Jack-in-the-box - conical for stability, possibly using inter-coil anti-topplingrestraints

( 2) spirally wound compression spring - from flat stock for enhanced inherent stability

fluid

time

cumulative number of ideas generated

rate of ideageneration

QU

IT T

IME ?

Design 9

Problem completion - constraints & criteria

The problem has been treated broadly up till this stage in order to derive as many candidate solu-tions as possible. It is now time to complete the problem, to focus on exactly what is required ofsolutions by way of constraints and criteria.

• ConstraintsA constraint is a bound, a limit with which every candidate must comply if it is to be a validsolution. Problems are usually characterised by a number of constraints.Satisfaction of a constraint by a candidate is binary - the candidate either does or does not sat-isfy the constraint, there are no ifs, no buts, no maybes and no 85% about it.Typical constraints are :

i “To solve my transport problem I need a vehicle which must cost less than $10000"ii “Our device for the D&B competition has to fit inside a 400 mm cube”

iii “Any solution to the greenhouse gas problem must not put any of our people out of work”There are constraints - often implied and not spelled out - which are obvious and particularlyimportant to D&B competition groups. Each group designs, constructs, operates and repairs itsown solution device for the competition. The group therefore must be able to build the device,the device must operate as intended, and so on.

• CriteriaA criterion is a yardstick by which the suitability of candidates may be judged - some candi-dates may satisfy the criterion well, some poorly. Problems are usually characterised by anumber of criteria.The degree to which the candidate i satisfies the criterion j may be expressed by the utility,ui.j - often a real number between 0 and 100%, though other scales are used :- A high utility indicates that the candidate satisfies the criterion to a high degree.- A low utility signifies that the candidate satisfies the criterion poorly.- A utility of 100% means that the candidate satisfies the criterion perfectly. - Zero utility means that the candidate is completely useless - but only as far as the relevant

criterion is concerned.NB : A zero utility does NOT imply that the candidate is useless as a whole, so a zero utilityfor one candidate with respect to one criterion must NOT be cause for trashing the candidate.Typical criteria are :iv “To solve my transport problem a vehicle will have to be fast, cheap to run, manoeuvera-

ble, safe and commodious”v “Simple manufacture without powertools” will be a requirement for any device we adopt

to solve the D&B problem“vi “Any solution to the greenhouse gas problem must minimise the number of our people put

out of work”

A constraint is more limiting than an equivalent criterion - the $10000 limit of constraint ‘i’ aboveleaves less scope for a solution than a criterion along the lines of 'cheap as possible' or 'minimumcost'. You should therefore try to enlarge the solution space wherever possible, by converting con-straints into equivalent criteria. For example if Australia had adopted the criterion ‘vi’ above ratherthan the constraint ‘iii’ during the 1997 Kyoto talks then its stance would likely have attracted lesscriticism - ie. more folk might have been kept happy. Unfortunately many political decisions appearto be argued on the basis of constraints rather than criteria, thus antagonising whole sections of thecommunity.

In this completion activity we are not interested in whether a candidate complies with a constraint

( 3) constant force spring - very thin steel tape pre-formed to circular cross-section becomes a tube whenunwound flattened from a roll (similar to a steel tape-measure but substantially more pre-forming)

( 4) party blow-up - or more practically, fluid pressurised coiled hose( 5) fluid pressurised telescopic tubes - need careful practicalising to render leakproof with negligible

friction( 6) sectionalised extending mechanism - eg. wire braced for lightness, extended like fireman's ladder

#8 below( 7) rolling diaphragm instead of sliding seal to stop leakage between telescopic tubes (practicalising

detail)( 8) fireman's ladder - operated by pulling on a single cord (not sketched)( 9) roll of 'tank tracks' (one-way bending chain) or flat belts - probably arranged as in #14 for

stability(10) unwinding arm like elephant's trunk, possibly straightened by pulling a cord (how are mus-

cles/tendons arranged to enable elbow bending for example?)(11) cord-operated straightening arm - consisting of identical pin-jointed bars with pulleys at the joints(12) lazy tongs - operated by tension springs or elastic bands(13) foaming agent expanded through a nozzle becomes rigid when exposed to air(14) triangular arrangement of three effectors such as #3 (or 4 or 9) for mutual support against buckling(15) thermal expansion of vapourising fluid used instead of pressurised air(16) very flexible fishing rod(s) bent and coiled initially within a box(17) air-operated bellows similar to a bamboo/tissue paper Chinese lantern that folds flat(18) screw jack eg. pump-action screwdriver(19) project the can - with packed parachute if necessary (depending upon the elevated retention time

desired)(20) project can upwards, trailing cords which solidify on exposure to air - note similarity with pre-

vious idea(21) ditto, but trailing chains whose specially shaped links lock when aligned thus supporting

compression - note how this followed directly from the previous impractical idea(22) suspend from a balloon(23) utilise magnetic repulsion/levitation(24) plant a seed of Jack's QuikGro Beanstalk under the can, water it and stand clear - not all crazy

ideas bear further fruit, however . . .(25) train a snake to balance a can on its nose like a seal - is crazy too, but triggers the following

which isn't . . .(26) feed coiled wet thick rope vertically out of a container above which dry ice freezes rope, ren-

dering it rigid - and so on.

A common failing on the part of budding designers is to abandon ideation before sufficient candi-dates have been identified - during ideation you should always aim to

Generate as many ideas as possible - quality doesn't matter at this stage

If a candidate bank is not large enough to start with, then a few unsuccessful attempts at practicalis-ing could see you with nothing left to work with.

Having generated a bank of solution candidates, let's see how we go about setting up the problem'sconstraints & criteria.

Design 10

The practicalisation activity takes the form of an assessment and development of the candidate bythe designer, who must modify the candidate to surmount if possible all barriers to a practical solu-tion. The activity is seldom instantaneous but rather requires the designer to repeatedly traverse thepracticalisation loop :

• proactively searching for and identifying ALL subsidiary problems, and• solving these subsidiary problems (if possible).

Practicalisation is not mere criticism, it is positive remedial action

A problem is trivial - it is no problem - if it has previ-ously been encountered and solved satisfactorily. For real problems . . .

. . . if we adopt an existing solution then there is noproblem, but if we try a novel solution then furtherproblems are likely to arise . . . for each of which . . .

. . . if we adopt an existing solution then there is noproblem, but if we try a novel solution then furtherproblems are likely to arise . . . for each of which . . .

. . . and so on - the problems becoming smaller andsmaller until they reach triviality.

Practicalisation is not so exhausting as might be infer-red from this description, but it must be exhaustive. One major lesson learned by students in D &Bcompetitions is that real life does not forgive “practicalisation” which is not thorough and complete.

One example of problems leading to other problems involves the primary problem "to design a new fac-tory for producing a given chemical" . There may be a number of different solutions to this, each with itsown set of raw materials, chemical reactions, economics and so on. If one of these solutions is adopted then the various necessary heat exchangers, pressure vessels(containers for pressurised fluids), pipes, pumps, cooling towers etc. each forms a secondary prob-lem - some of which may be trivial as the items may be procured off-the-shelf. The solution to one of these non-trivial secondary problems - let's say a pressure vessel - wouldthrow up tertiary problems involving the choice of material, of dimensions, lagging to reduce heatloss, etc. etc. leading eventually into detail design.

Large problems like this require considerable resources involving large design teams for their solu-tion. But whether the problem is large or little - if it is not known that the solution is practical beforecomparing it with others, then obviously the comparison is an utter waste of time as it could resultin an 'optimum' which can't be made or which won't work !

Practicalisation does not equate to detail design. For example, two parts may have to be joinedtogether demountably. Looking into the means of joining, it might be concluded that a set screw isperfectly feasible as the loads are unremarkable, access for tightening and loosening is not restricted,and so on. This is practicalisation. Further analysis involving fatigue loading, materials and safety might lead to a solution in the formof an M10x1.5 class 10.9 socket-headed set screw, length overall 80 mm, length of thread 25 mm,unlubricated and torqued to 90% proof. This is detail design ; it has no place in this introductorychapter on design.

During practicalisation the designer must foresee and must overcome (if possible) all drawbacks to

secondary

primary

PROBLEM

SOLUTION SOLUTION

PROBLEM PROBLEM

SOLUTION SOLUTION

PROBLEM

tertiary

PROBLEM

SOLUTION

PROBLEM

SOLUTION

PROBLEM

or not, neither are we interested in how well a candidate satisfies a criterion - all we are doing isidentifying and recording the constraints and criteria, so that they might be applied to candidatesduring later activities in the design process.

One of the most crucial tasks in the whole design process is to :

Ensure that ALL constraints and criteria are identified.

The constraints and criteria are determined by the wishlists of everyone who, and everything whichwill interact with the solution. A designer must therefore realistically visualise the solution as itprogresses through its life stages, and anticipate these interactions together with the correspond-ing constraints and criteria. If other sources can assist with this task then clearly they should be con-sulted. A designer's incomplete knowledge of manufacturing processes for example is good reasonfor talking with a fitter and turner, for further reading, or for experimenting personally.

One of the reasons for mounting D&B competitions is that members of student design groups areinvolved with all life stages of their solution, and so must themselves establish the completion wish-list. Although other folks' agenda are not addressed, students still have to foresee the solution'sfuture and to set constraints and criteria accordingly. These should not have to be inferred duringlater activities, but must be written down in black and white in the present completion activity.

One common criterion which is often overlooked is the need for simplicity. Appendix C provides acouple of graphic illustrations where complexity equates to frustration.

Keep it as simple as possible

If you don't identify and write down all constraints ("we must be able to make it") and criteria ("it's gotto be as simple as possible") then you'll most probably overlook them when it comes to the next step,practicalisation, resulting in a less than optimum solution.

Practicalisation - rendering candidates practical

A solution candidate is practical if it complies with all the problem constraints and well satisfies theproblem criteria which have been identified previously. Practicalisation is the activity in which anembryo candidate solution (in all probability described only by a very hazy sketch or a few key-words) is transformed into a practical solution - if this is possible - before candidates are comparedwith one another and certainly before any manufacture is contemplated.

The outcome of practicali-sation for a particular can-didate is that :

• either the candidateis practical, that iscomplies with allconstraints,

• or the candidate isnot practical as itcannot meet everyconstraint. . . . there are no halfmeasures.

YES YES

NO

Are youcertain that candidate

cannot comply with a l l constraints

?

practical solution to evaluation activity

impracticalcandidate to trash

candidate solution from bank

Solve secondaryproblems, modify

Are youcertain that candidate

can comply with a l l constraints

?

NO

PRACTICALISATION

Design 11

former might provide adequate reinforcement.

(d) Tubes could be centrifugally cast by quickly rotating a tubular mould into which a settingplastic eg. is poured. Centrifugal force causes the liquid to form a uniform film which setsinside the mould. A tube so formed could become the mould for the next smaller tube.

(e) Tubes might be made from rubber hose, allowing radial expansion to prevent air leakage.

(f) The large radial gap between adjacent sizes of tube available in the shops - which led to theproposed scrapping of the telescoping tube idea - might be put togood use as a mould cavity in which to cast an intermediatetube. There are many casting plastics and latex rubbers availa-ble on the market. The fishing line reinforcement mentionedabove might be incorporated again here.

And so on. The foregoing doesn't pretend to be an exhaustive list of ideas, and all the secondaryproblems have not yet been resolved - ie. telescoping tubes have not yet been fully practicalised -but certainly we now have sufficient confidence to retain the underlying idea and to invest in thebuilding and development of prototype tubes.

EXAMPLE Here is another example of what can be done with even lower-tech materialsand manufacture. Students were asked to design and build a stair climber. One common resultof initial ideation was a vehicle equipped with caterpillar tracks like those on military tanks.The tracks were perceived as rubber belts with treads, but because nobody could convenientlylay their hands on such peculiar components the whole idea was trashed - students didn't tryto practicalise . . . .

How may caterpillar tracks be made from scratch using commonly available materials ?

Some ideas are sketched :-

Both these examples are typical of D&B problems in which practicalisation is incomplete becauseour knowledge/experience cannot predict every detail - we don't know if we can manufacture atube from wound fishing line or a tank track from bent cardboard. However the above ideation hasindicated some possible solutions to the secondary problem of manufacture, and importantly hasdemonstrated that manufacture is not necessarily impossible. Note the glimmer of hope here, com-pared to students' knee-jerk reaction to trash both telescoping tubes and tank tracks because theycouldn't immediately lay their hands on them.

Practicalisation can be completed only by acquiring the necessary knowledge - with D&B devicesthis is usually best done by direct experimentation . The investment of time and effort in experimen-

in operation

as cast

bent cardboard cut-outpaper clip hinge pins

match glued on for traction

stapled upholstery webbing

lengths of plastic tube, riveted on

narrow strip of flexible carpet forms track,

or attached around wheel periphery

form chain track from bent wire

the artefact's practical realisation. Secondary problems which the idea's novelty has introduced mustbe solved satisfactorily. This again is time-consuming. You have to work at it. You will have to startthe problem-solving process all over again - "How can this drawback be overcome so that this candidatecan be rendered practicable ?" You will have to consider the manufacture and operation of the candi-date IN DETAIL. You will need to know something about materials and how you yourself can fash-ion them. You may have to set up mathematical models of the device's operation, and so on.

With D&B competitions for example, the major constraints relate to the two significant life stagesafter design - the constraints concerned with manufacture and the constraints relative to operation.

We shall demonstrate practicalisation first with respect to manufacture.

EXAMPLE Candidate #5 of the above can raising device involves telescoping tubes. Fromwhere might we obtain these ? Are we able to manufacture them ? Practicalise this, given thatthe pressure of any fluid used must not exceed 100 kPa for safety reasons.

An all too common approach here is to visit the nearest hardware shop on the lookout for plas-tic or light alloy tubes of different diameters and thicknesses which may nest inside oneanother, and on being unable to unearth suitable tubes to scrap the candidate as impracticable.This lack of effort is deplorable. What one might do is . . . .

A simple p = F/A demonstrates that a pressure of 100 kPa acting over a circular area of only 11mm diameter will support a mass of 1 kg. Alternatively, if the whole 0.16 m2 plan area of thedevice is available to lift the can then the necessary air pressure is a paltry 60 Pa (6 mm H2O).These limits indicate that intermediate pressures and areas could be used successfully, and thatmetal tubes may not be necessary to withstand the operating pressure. So, recalling that theenvironment is dry ambient, we further practicalise by asking what other materials/manufacture might be used?

(a) We might glue cardboard or paper or plastic, winding up tubes helicallylike the support tubes for toilet rolls eg., using the next smaller tube as amandrel. Each tube might consist of multiple layers of differing hand,built up on a temporary innermost layer which is later removed to ensure clearancebetween adjacent tubes. The thickness of each tube would then be easily adjusted to with-stand the internal pressure.

(b) A very light tissue paper tube could be close wound with nylon fishing line eg. - the formercomponent ensures a leakproof tube, the latter provides reinforcement to withstand the burst-ing effect of the internal pressure. This idea might be criticised becauseof the difficulty of a leakproof sliding seal outside the tube due to thecorrugations formed by the fishing line. But further practicalisationmight reveal that the interstices could be filled with smoothed settingplastic - or they might be put to good use as a reservoir for a honey-like substance doing double duty as a speed retarder and an air seal (due to its surface tensionand viscosity).

(c) Composite tubes could be formed in the same manner as Saturn fuel tanks, by winding rein-forcing fibres helically around a mandrel (possibly using a lathe for uniform pitch) thenimpregnating/spraying them with a solidifying plastic. Subsequent machining for a smoothexternal surface might be necessary. A silk stocking or a long knitted sock stretched over a

composite tubehalf-section of

Design 12

a device before building, it may be possible to assess operation by means of a mathematical model -though it must be emphasised that mathematical modelling is no substitute for direct experimenta-tion if that is possible. Students generally seem reluctant to study hardware by mathematical modelsof their own devising - they're excellent at analysing existing models, but this ability is completelyuseless if no reasonably accurate model of the device exists.

It is not suggested that mathematical modelling always be attempted - the simple party blow-up(candidate #4 of the can-raising device above) is far too complex theoretically - but rigid bodydynamics eg. is often very useful in enabling prediction of operation. Students should practise theconstruction of such models - ensuring that free bodies are correct, a common source of error in stu-dents' work. A few illustrative mathematical models are now given.

EXAMPLE An idea proposed for the stair climber involves motorised wheels equippedwith lobes which engage with the steps to allow the device to progress smoothly from one stepto the next. What shape of lobes should be used?

The 'Sherpa' model shown overleaf considers lobe geometry based on the involute for smoothascent. Although this model may be practical from the point of view of kinematics, it says nothingabout forces or ascent speed, so the idea has not been practicalised fully. Rather than extend themodel to include kinetics and strength, the kinematic model may provide sufficient confidence tobuild a device, test it and develop it physically to full performance capability.

EXAMPLE What size of spring should be used in the lazy tongs, candidate #12 for the canlifter above.

The model of a rudimentary light lazy tongs is illustrated and analysed here. The expressionrelating the load's acceleration 'a' to the arm's inclination 'θ' is found to comprise three terms,

only the first of which is positive. So unless the inclination is large enough to yield a positiveacceleration, it doesn't matter what size of spring is used - the device won't work!

This conclusion was found out the hard way by students who built the device then found thatthey had to incorporate a compression spring perpendicular to the spring sketched in order tostart the device. Given the springs and dimensions, the model enables the compressive load Pto be found as a function of inclination, if required for later detail design to avoid buckling andfailure of members.

EXAMPLE This concerns practicalisation of the can lifter candidate#11, the cord-operated straightening arm in which the load 'W' is ele-vated through the distance 'h' by means of a cord pulled by tension 'F'through the distance 's'.

W

S

S

a

eight W raised by light mechanism of strut length c, with tension spring of stiffness k and free length L o :-

pring force : F = k ( L – Lo ) = k ( 2 c cos θ – Lo )For light wheel : F – P cos θ = 0For rising mass : 2 P sin θ – W = m a

o acceleration : ag = 4kc

W sinθ – 2kLo

W tanθ – 1

nd velocity : v2

2gc = 2kLo

W cosθ – kc

W cos2θ – sin θ + const.

c x

P

F

W

a

P

P

c

θ

k

N (≥W/2)

W

device

W

F

s h

tation has been shown to be justifiable in the above examples, and would concentrate only on theunknown aspects of manufacture, not necessarily on the complete build (recall that our main aimduring practicalisation is to become certain that we can or we cannot manufacture the candidate).

Let's now consider practicalising from the point of view of operation.

The principle of operation has been established during the previous ideation activity; the principlenow has to be fleshed out during practicalisation - again sufficiently to satisfy ourselves that the can-didate either can or cannot operate. By far the most satisfactory basis for assessing operation isdirect experience of this operation. While we can't experience all aspects of operation before the arte-fact is completed, in most cases it is possible to quickly and cheaply experiment with certain compo-nents or sub-systems whose behaviour is particularly problematical, and to modify these to obtainthe operation required. You must be prepared to try it !

EXAMPLE A D&B project required collection of as many dried split peas from a pile as pos-sible, using a device which could be battery powered but which had to be as light as possible.The pile was situated 1.5m away from the device's initial position, and the peas had to be deliv-ered to a collection container.

The initial problem could be broken down into sub-problems such as "how to transport device tovicinity of pile?", "how to pick up peas?", "how to deliver peas to collector?" , "how to power the device -electric motor(s), clockwork motor(s), springs, compressed gas cylinders etc.?".

Concentrating here on the pick-up phase, there are many possible candidate solutions - rotat-ing/sweeping brush, conveyor belt, bulldozer/scraper, vacuum, air jets and so on. Thevacuum was popular with contestants as a cheap vacuum unit for cleaning the interior of a carusing the car's electric system was available.

Some contestants dismissed this solution on the grounds that "it needed batteries which were tooheavy" (Note the unforgiveable error of trashing a candidate on the basis of a criterion), "itssuction was insufficient to pick up peas" etc. - without any justification whatsoever. They did notactually test the unit, but based their conclusions on their preconceptions, which - not beingbased on past experience - couldn't have been more wrong.

It would have been so easy to have actually tried the unit out, to have experimented with it byitself. How close to the peas would the vacuum's nozzle have to be in order to pick peas up?Could a snow-plough blade be attached to the moving nozzle to pick up more peas? What arethe lightest batteries required to give the desired period of operation? How can the unit be inte-grated with other sub-systems such as transportation?

All these sorts of questions could have been answered definitely, and all the necessary modifi-cations completed, ie. the candidate substantially practicalised by direct experimentationbefore comparing it with other candidates or integrating it into the complete device. Even if agroup couldn’t afford to purchase one of these units without some confidence in its potential, itmight have been possible to borrow one, or to try one out in the shop with a plate of split peas.

This project was edifying as some competitors rejected the vacuum candidate without proper practi-calisation, whereas other competitors produced successful devices based on it. Conversely, somecompetitors rejected conveyors on the basis of mere arguments - mere hot air with absolutely no jus-tification, no basis in the real world. The winning device was a conveyor.

Lacking the opportunity for direct experimental appreciation of the performance of a sub-system or

Design 13

This demonstrates that the cord displacement 's' increases in proportion to the bar inclination'θ', the constant of proportionality being the pulley diameter 'd'. Thus for small forces in themechanism we need a large pulley diameter.

It is not difficult to extend this analysis to cover changing forces and buckling proclivities inthe device as bar inclination changes during the arm-straightening process. Again, we gain anunderstanding of performance before building starts.

Students were asked to design and build a vehicle powered by a supplied rubber band to travel asfar as possible along a straight horizontal track. In a common device the rubber band, modelled as a

spring, was connected to a cord wrapped around a drum attached toa large driving wheel. The spring was first wound up by rotating thewheel by hand, the vehicle was then released on the track and trav-elled due to the cord unwinding off the drum.

Students made the drum diameter small in an effort to achieve alarge travel from a given spring displacement - ie. they examined only the overall geometry ofmotion. But the ground traction force accelerating the vehicle was proportional to the drum diame-ter, so a small drum resulted also in a small accelerating force. This force was so swamped by fric-tion (which students hadn't allowed for) that there was no acceleration whatsoever. The vehiclerefused to budge and caused much embarassment - or hilarity, depending on the point of view !

This inter-dependence of kinematics and forces occurs in all devices.Beware the dangers of a kinematic analysis without looking also at the forces.

A more complete analysis involving the kinetics of this vehicle is presented in Appendix D, and pro-vides insight into the interaction between the major design parameters such as weight, overallgeometry, spring characteristics, etc. and the resulting performance - all vital stuff when assessingwhether or not the device will operate.

A brief overview of springs as energy storage devices is given in Appendix E to illustrate the kind ofinformation available in the literature - a few simple measurements enable the stiffness of a givenspring to be estimated, taking for steel E = 207 GPa, G = 80 GPa. The later chapter on Springs shouldbe consulted for further information.

An even more realistic model of such a device must recognise further consequences of the choice ofspring with a given energy storage - either :

• a stiff (large k) spring requires a heavy vehicle with massive membersto withstand the buckling effect of large spring forces, and a large inef-ficient speed increaser to amplify the small spring deflections, or

• a compliant (low k) spring which needs a vehicle of large dimensionsto accomodate the large spring deflections, though compactness maybe achieved by a clock spring rather than the tension spring foreseen.

Extension of the model to include these effects is not particularly difficult, but there often comes atime when the preparation of a realistic mathematical model is too demanding given the significanceof the problem. Then is the time to experiment with hardware.

Mathematical models are not restricted to mechanics, as the following demonstrates.

EXAMPLE Some effort has been devoted already towards practicalising manufacture of the

k

khi

klo

deflection

force Springs storingequal amounts

of energy

In first assessing feasibility we consider an ideal mechanism for which 'F' is constant. Work-energy thus requires Fs = Wh so that if the geometry/kinematics require that the cord's dis-placement 's' is small, then a correspondingly large tension 'F' is required in the cord, with con-sequent implications on bar buckling etc.

So, what are the kinematics? They are examined in the box below for a single bar of the device.

Bar of length c carries two pulleys of diameter d around which acord is wrapped. To find the relation between cord length x-x andbar inclination to the horizontal θ.

Inclination of cord to bar : γ = arcsin(d/c) which is constant

From geometry : π = φ + ( π/2 – γ ) + θ and so

cord length between points 'x' = 2 ( 1/2 d.φ + √( (c/2)2 – (d/2)2 ) )= d ( π/2 + γ – θ ) + √( c2 – d2 )= constant – d. θ

γ

φ

x

xθ

c

Design 14

lel tongs in the device . . . . and so on.

If you don't ask these sorts of questions and answer them satisfactorily then don't be surprised when- not if - Murphy appears on the scene! So, once again . . . . YOU MUST ASK THE RIGHT QUES-TIONS.

We have illustrated practicalisation by means of devices built and operated by the designers - clearlymanufacture and operation are the most important life stages. But the same thoroughness and atten-tion to detail are necessary in other aspects of more usual problems involving other folk. This is whyit's so important for the designer to approach any problem from the points of view of all those likelyto be affected by the solution - the lathe operator, the user, the sales person, the maintainer etc. Onlyby visualising the step-by-step actions of these people can the designer appreciate the subtleties oftheir interaction with the solution candidate.

This section concludes by emphasising the need for definite knowledge of a candidate's practicabil-ity after the practicalisation activity. Theoretically, an accurate comparison (evaluation) cannot beundertaken until all candidates have been designed completely. In choosing a new car for example,all candidates are physically available - however this is hardly a design problem, it's purely a matterof selection.

In the design context we cannot afford the luxury of designing in all detail every likely looking can-didate in order to select a single 'best' solution. At the other extreme, what confidence can we havein a choice between one candidate which we don't know will work and a second which we don'tknow how to make? So the designer must continue around the practicalisation loop until the ulti-mate practicality or uselessness of each candidate is known with some certainty. Failure to do so isone of the most common shortcomings of students' designs. All decisions must be justifiable, andthey can't be if they are based on incomplete knowledge.

Candidates must be practicalised before evaluation is attempted.

Let's now see what this evaluation activity entails.

Evaluation - choosing the optimum candidate

At the conclusion of the practicalisation activity, we have a number of candidate solutions whichmeet all the constraints. These now have to be evaluated - that is compared with one another on thebasis of the problem criteria - in order to select the optimum solution.

Evaluation consists of three distinct steps, carried out by the designer :1. the relative importance of the various criteria is defined2. the degree to which each candidate satisfies each criterion is established3. the degrees to which the candidates satisfy the overall problem are finally worked out.

Let's examine these steps individually.

1. Relative importance of the criteria

Different sets of criteria are associated with different problems. One particular criterion which maybe common to different problems usually assumes different significance in these problems.

The following sketch illustrates the importance of five criteria common to two different artefacts. In the design of a domestic vacuum cleaner, cost is probably the most important criterion sincepotential purchasers' first thought is the effect on their pockets. Undiscerning consumers are not so

telescoping tubes, candidate #5 of the can raising device. The device is conceived as an air res-ervoir at an initial high pressure, to which the unexpanded tube of cross-sectional area 'A' isconnected. On being released, the tube expands, raising the 1 kg can to a height 'H'. Whatheight may be expected?

The air undergoes an expansion process ( p1 + p0 ).V1n = ( p2 + p0 ).V2

n where p0 is atmos-

pheric pressure and other pressures are gauge.

The constraints are : - V1 (the reservoir initial volume) ≤ (0.4 m)3

- p1 ≤ 00 kPa- V2 = V1 + HA- p2 ≥ Wcan/A.

Assuming an adiabatic process ( n = γ ), this may be solved to obtain an idea of the height Hachievable - the feasibility of the device may thus be assessed.

Once a device has been built, a mathematical model can be a useful aid to understanding any unex-pected behaviour which testing uncovers - eg. a vehicle will not start, or it flips over and kicks itswheels in the air, or, if propellor driven, it rotates wildly while the propellor remains stationary, andso on! Real life Research and Development entails the testing of physical models when availablemathematical models lack realism. Clearly the sophistication of any modelling, whether it be mathe-matical or physical, must be in keeping with the importance and sophistication of the project.

Don't get carried away by mathematical or computer modelling, remember that a mathematicalmodel is a means to an end, not an end in itself.

An extremely important task in the practicalisation exercise is to thwart Murphy by foreseeing allhis worst tricks and sabotaging them. Again this requires effort on the part of the designer to visual-ise each life stage step by step, and to ask questions about what could go wrong with each. Thus inthe case of the lobed stair climber "What would be the effect on climber operation if the stairs were imper-fect?", eg. stairs provided with an anti-slip bead, or built with a tolerance of ± 5 mm on tread dimen-sions, or covered with a fluffy carpet, or copiously treated with slippery polish by the cleaner, andso on. Or again, "If overall stability requires two identical lobed wheels on a single driving axle, what wouldhappen if the axle became misaligned to the stair treads?"

What could go wrong with the lazy tongs, candidate #12 of the can lifter? Elas-tic bands would probably be used instead of the linear springs envisaged in themathematical model; the bands would introduce severe non-linear and hyster-itic behaviour which would render useless any quantitative deductions fromthe model - though qualitative findings would still be very useful.

If the tongs' struts were not 'identical' then they could bind, causing unexpectedfriction in the mechanism. Manufacture of the struts by drilling the three holes in each through anaccurately pre-drilled jig would ensure adequate dimensional similarity.

Or, "Murphy would try to tip over the extended can lifter - what steps would minimise overturning tenden-cies?" This might lead to practicalising ideas such as :

- a heavy wide base together with light struts to impart overall stability- a speed retarder to slow the ascent and minimise dynamic effects- close fits on all rotating joints to avoid excessive play in the mechanism- use of lubricant to minimise friction together with the correct disposition of springs to improve

uniformity of forces internal to the device- the possible use of cords within the mechanism to ensure ascent synchronism of the two paral-

2

A

H

1

V1

deflection

force

Design 15

The ith candidate's degree of satisfaction of the problem as a whole is its overall utility Ui which in

its simplest form is represented by the mean weighted utility :

U i = Σ w j . u i , j / Σ w j

If utilities are not independent then this simple linear weighting may be inaccurate and call for morecomplex formulation of the overall utility.

The optimum candidate is that with the highest overall utility, as this reflects a candidate's fitnessfor the criteria-defined problem.

EXAMPLE The accepted method of laying claim tonew mineral deposits in the bush involves stickingsurvey markers (or 'pegs') into the ground around theperiphery of the claimed area. A marker usually con-sists of a coloured plastic flag attached to a metal wiremast which is thick enough to permit insertion into theground without buckling.

Markers are manufactured in their thousands by a shel-tered workshop for disabled people who currently cut uprolls of coloured plastic cloth into rectangles, paint PVAadhesive onto an edge of each rectangle with a paintbrush, then fold the edge over to secure a pre-cut lengthof straightened wire as illustrated.

We pick up the story as the workshop re-assesses itsoperations in an effort to speed up the current manual processes by supplementing them withlabour-saving devices. The particular sub-problem of interest is "How to apply adhesive to the flag?" -other problems such as "How to fold the flag flap?" are not addressed here.

The following candidates for applying adhesive have been proposed and practicalised elsewhere· brush - a hand held paint brush (the current method)· coated wire - the wire is dipped into adhesive then rolled over flag to distribute it· dip - the flag edge is dipped completely into a pool of adhesive· drip - a pressurised pipe equipped with nozzles is positioned above the flag edge and drips

adhesive when a tap is operated, like multiple pipettes· extrude - a bead of adhesive is extruded onto the flag edge rather like toothpaste· fixed roller - the flag edge is pressed against a roller which is partially immersed in a bath of

adhesive (similar to postage stamp wetters in Post Offices)· gravity - the flag edge is transported under an opening in the bottom of an adhesive container· guided brush - as current method, but the brush is guided to assist accurate location of film· paint roller - a paint roller is substituted for the paint brush above· stamp - an adhesive-retaining stamp is guided by a

linkage alternately between a bath of adhesive andthe flag edge as shown :

· spray - adhesive is sprayed from a gun onto the flagedge which may require protection by a stencildepending upon the spray nozzle-to-flag distance

· wheel - a multiple wheel retains adhesive around its periphery by viscosity/surface tension, and transfers adhesive to the flag when rolled over it.

flag & wirespray paintcoloured cairn

stamp

position wireapply adhesive

clothes pegssew flag to wireglue wire in flag

coated wirepaintbrush

fold flap

delineate mineral claim

prepareflag clothmaterial

applyadhesiveto an edge

positionwire mast

fold overand press flap down

150 10

0

flagadhesivereservoir

stamp

aware of reliability issues - the importance of the relia-bility criterion to the designer might therefore beslight, though he has to remember that poor reliabil-ity can be expensive to the manufacturer by having tohonour guarantees.The reliability of a military aircraft on the other handis of prime importance since air-readiness is a vitalround-the-clock necessity. The visual aesthetics - theappearance criterion - of the aircraft is relatively trivial,whereas appearance is important in making a profit from vacuum cleaner sales.

The relative importance of the jth criterion in a problem is termed the weight, wj (or weighting or

weight factor) of that criterion. It is the relative, rather than the absolute, values of the criterias'weights which are significant.

From the sketch for example, wappearance might be 75% for a domestic vacuum cleaner but only 25%for a military aircraft. Different people often attach different degrees of importance to the same cri-terion for the same problem - thus the manufacturer of a vacuum cleaner might assign wmass = 25%,whereas the person who has to clean a lot of stairs might assign 90% to this weighting. Again, it's upto the designer to find out who are involved with the solution, and what their agenda are likely tobe. Only then can realistic weights be assigned.

If the weighting for a certain criterion differs markedly between folk then it's probably best to con-sider the two extremes in further work. A certain candidate may be so much better than the othersthat a perturbation in one or two weightings may not affect the outcome of evaluation. Conversely ifthe optimum is sensitive to certain weights then a more rigorous examination is called for.

2. Satisfaction of the criteria - utility

In general the various candidate solutions for a particular problem satisfy the various criteria to vari-ous degrees. The degree to which the ith candidate satisfies the jth criterion is characterised by the

utility, ui,j - often a real number normalised such that 0 <= ui,j <= 1 or 100%.

If there are M candidates and N criteria then it is up to the designer to assign the MxN utility values.The utilities must be logical and consistent - it's no good setting u4,9 = 50% and u7,9= 25% unless thedesigner can justify that candidate 4 satisfies criterion 9 approximately twice as well as candidate 7.

Setting utility values must be carried out properly or not at all

It is recommended that when assigning utilities the designer i. focuses on one criterion at a time,

ii. sorts the candidates in ascending or descending degree of satisfaction of that criterion(probably by comparing combinations of two candidates on a one-to-one basis) and finally

iii. assigns corresponding utility values for that criterion.

If manufacturing costs can be predicted accurately, then assigning utilitites for the cost criterion isfairly straightforward, but when the criterion is abstruse then assignment necessarily is more subjec-tive. Sorting prior to assignment (ii. then iii. above) helps to maintain consistency. Software is availa-ble to assist in assigning statistically consistent utilities - the interested reader is refered to Tabuca-non eg. cited in the Bibliography.

3. Overall utility

The ith candidate's degree of satisfaction of the jth criterion is the utility ui,j

ease ofinspection

cost

mass

appearance

reliability

DOMESTICVACUUMCLEANER

MILITARYAIRCRAFTcriterion

Design 16

problems are foreseen when the operator attempts to minimise spillage - so the utility is lessthan the maximum.

5. Stamp and fixed roller are both easily operated / controlled, so attract the maximum utility.

Utilities should not be expressed as either 'bad' or 'good' - binary utilities are too coarse . The binary0/1 is too easily misinterpreted as reject/accept (rather than bad/good) which leads to the incorrectrejection of a candidate whose satisfaction of a single criterion is poor.

The utilities for the remaining criteria areassigned in a similar manner and result inthe utility matrix illustrated above. Theoverall utilities are obtained from thelinear weighting formula above, andexpressed as percentages of the maximumutility which is '5' here.

On the basis of the assigned weights andutilities, the stamp candidate is clearly theoptimum, while solutions based upon roll-ers or wheels are also suitable.

No matter what the scenario - alwayscheck your assumptions if you're givenhalf a chance. In the present context itwould be advisable to re-assess the utilitymatrix with amended values of weightsand utilities, to see if the problem is sensi-tive or insensitive to the values assumed.You usually cannot be confident in a one-off result until you have also assessed sen-sitivity.

The evaluation of utility for a given criterion in more sophisticated problems may be assisted by autility function which enables the utility to be quickly evaluated in terms of candidate attributes.

For example, supposing you were about to buy a new car and you were interested in the criterioncheapness ie. running costs per month. You might set up theutility function illustrated, assigning a utility of 100% to run-ning costs which approached zero, and a utility of effectivelyzero for costs over $600. The function between these limitswould be established only by detailed knowledge of the effectof costs on your personal financial state, and clearly reflectsvery subjective utilities. Setting up the function is time-consuming, but again there are computer programs to assist indefining the statistically 'best' function.

Once the function is set up however, it is a simple matter to read off the utilities corresponding tothe running costs of the various candidates. The candidates are practical - they're on the showroomfloor - and their running costs can be calculated in terms of known attributes such as

· initial cost (and corresponding monthly loan repayments)· fuel economy (km/l) together with petrol prices and average distance travelled

UTILITYMATRIXFORAPPLYINGADHESIVE

5 2 1 5 3 2 2 5 5 633 1 2 5 4 1 4 5 5 643 4 2 5 0 4 4 5 5 724 4 3 3 3 2 1 3 2 574 3 3 3 4 3 2 3 3 612 5 3 4 4 4 3 4 3 762 5 4 3 1 3 1 2 1 544 3 3 4 3 2 2 4 3 624 4 3 4 4 4 3 4 4 764 3 3 3 3 3 1 3 2 564 5 4 4 4 5 4 4 2 834 4 3 4 4 4 3 4 3 74

weight 1 3 2 3 2 2 2 2 1

com

pat

ibil

ity

op

erat

or

dem

and

s

film

to

lera

nce

rob

ust

nes

s

adh

esiv

e ec

on

om

y

pre

mat

ure

dry

ing

clea

nab

ilit

y

rep

aira

bil

ity

init

ial c

ost

ov

eral

l uti

lity

%

paint brushcoated wiredip edgedripextrudefixed rollergravityguided brushpaint rollerspraystampwheel

running cost, $/month0

1

0200 400 600 800

utility

The safety constraint is particularly relevant to handicapped operators.

Criteria relevant to the adhesive application process are :· compatibility - adhesive application should be compatible with other processes such as folding

the flag edge over the wire· operator - the dexterity and effort demanded of the operator must be minimised whilst main-

taining the need for some concentration (eg. pressing a button on a machine which does every-thing automatically is hardly a rewarding exercise)

· tolerance - the extent and the thickness of the adhesive film should be controlled so that - the glued joint is sufficiently strong for the eventual duty of the marker- mess due to excessive or poorly positioned adhesive is avoided

· robust - the applicator must be robust to withstand heavy-handed uncoordinated operators· economy - adhesive should not be wasted as its cost is significant· drying - potential problems due to premature drying of the adhesive must be minimised (an

aspect of the tolerance criterion which merits special attention)· cleaning - the applicator should be easily cleaned between shifts to prevent accumulation of

solidified adhesive· repair - if necessary the applicator should be capable of easy repair and adjustment by a rela-

tively unskilled person, though not by the operator· cost - devices' initial cost must be negligible as the enterprise is not commercial.

The requirement of corrosion resistance is not mentioned because all candidates will need to beequally resistant. Corrosion resistance is really a constraint here, and will be assured by detaildesign.

There is no point in using numerically refined weights here (eg. 47%) because the nature of the prob-lem cannot justify such refinement - integer weights in the range [ 0: 3] will be quite sufficient. Of theabove criteria, operator demands are clearly extremely important in this sensitive application, whilerobustness is required for trouble-free operation. Weight factors of '3' will be assigned to these crite-ria. (Although robustness depends to some extent on later detail design, candidates based on smallorifices eg. are thought to increase maintenance requirements and this high weighting will militateagainst these candidates.)

Compatibility with other operations is relatively insignificant as it is not seen to be difficult tomarry other operations to adhesive application (unless this is very complicated). The initial cost isnot highly weighted either, mainly because all candidates are unsophisticated and cheap. So aweight of '1' will be adopted for both these criteria.

Nothing much separates the remaining criteria so they will be assigned weights of '2'.

The criterion of operator demands will be used to demonstrate brief arguments for the choice of util-ities. The candidates are first sorted in order, then the utilities in the range [ 0: 5 ] are assigned (againthere is no justification for using numerically more refined values).

1. Coated wire - after dipping the wire, the operator has to position the wire on the flag andmove/rotate it to ensure uniform adhesive film. This is difficult for folk with poor concentra-tion or motor skills, so a low utility of '1' seems appropriate.

2. Paint brush - again a uniform film is highly dependant on operator skills. A guided brush willbe less demanding on operators, so its utility could be 3.

3. Extrude - it's easier to place a bead of (thick) adhesive than a film. A spray will place aboutthe same demands on operators, so its utility will also be taken as 3 for this criterion.

4. Paint roller - will be easier than the guided brush as film uniformity is more assured. Theoperations of dipping and dripping will also be straightforward for the operator, however

Design 17

cess. It is at first, if it's to be done well. But like all skills - practice makes perfect, practice makes lessonerous, practice makes more interesting.

The optimum solution has now been deduced. It only remains to complete the design by explainingit in a feasibility study.

The feasibility study

A feasibility study presents the argument for the optimum solution to a problem.

For the purposes of this course, it outlines the designer's solution activities whilst following the rudi-mentary design process, and describes :

· the problem statement (including any broadening considered by the designer)· the bank of solution candidates generated· all the constraints and criteria deemed relevant· how the candidates were practicalised· the decisions made during candidate evaluation· the optimum solution (without painting it rosier than it actually is)· any development of the solution carried out after manufacture.

Feasibility studies should not be regarded as mere exercises in essay writing

A professional feasibility study would usually not mention all the ideas considered for example, butwould concentrate only upon the optimum solution, including detailed costing. Undergraduates'reports however are expected to spell out everything ( in brief ! ) - though students' inability to fore-cast accurate manufacturing cost for example calls for care when costing ( if costing is required ).

Writing such a report is just another occasion for 'putting yourself in the other guy's shoes' . Any fea-ture of your report which prevents the reader from assimilating it as briefly, as completely, and aspainlessly as possible is guaranteed to raise hackles. You must try to visualise a reader's reaction toyour submission.

Essential attributes of feasibility studies which are sought in this course appear overleaf:

Where do we go from here ?

The rudimentary design process is a more than adequate prescription for the problems you'll meetin your course and in most of your professional life. Design processes in industry are usually moreformal than this, since the design function must be systematised, managed and integrated withother functions such as manufacture. Nonetheless, all professional design processes are based onthe rudimentary design process, and the activities which you have learned here are just as vital inmore elaborate design schemes. Obviously the sophistication and cost of any design process willreflect the sophistication and cost of the product.

Commercial design processes require continual review and documentation, and so the progressionspecification-to-concept-to-preliminary layout-to-definitive layout is brought up short after each stage toallow confirmation before proceeding to the next. Leading models of the design activity include :

• The Pahl and Beitz op cit model illustrated overleaf is typical of Continental practice and sub-divides the design phase into three major areas :- concept design: search for adaptable solution principles (this is essentially the ideation activity

above)- embodiment design: find suitable hardware within technical and economic constraints

(which is just practicalisation)

GUIDELINES WHEN EVALUATING

Evaluation is a complete waste of time unless all candidates are practical. The most commonfaults with tyros’ designs include insufficient candidates and inadequate practicalisation, so

don't waste effort evaluating at the expense of these activities.

Weight factors and utilities must be justifiable (logical & self-consistent) and applied objectively.

Constraints and criteria once identified must not be ignored.

If you have partially completed an evaluation and you realise that you have omitted a criterion,then you must start evaluation once again with this new criterion incorporated.

Evaluation must be based on ALL criteria. Candidates should be trashed on the basis of con-straints not of criteria. A very common mistake is to trash on the basis of a single criterion - for

example metal A is rejected because it is too heavy, metal B is passed over because it is difficult toobtain . . . and so on. It is easy to kill off all candidates by selecting a single criterion for each -not that one would deliberately set out to do this, but misapplication has just the same effect

whether it is inadvertent or deliberate.

Another failing is to reject all candidates except one, which by default must then be adoptedwith inadequate criticism - the solution then depends to a large extent upon the order in which

candidates are addressed. This approach can lead to dreadful difficulties later on, when theinherent deficiencies of the poorly assessed 'optimum' become all too apparent.

ALL CRITERIA MUST BE APPLIED CONSISTENTLY TO ALL CANDIDATES

Guard against making up your mind, perhaps unwittingly, that a particular solution is opti-mum, then tailoring arguments (again perhaps unwittingly) to bolster this preconception.

Do not vacillate when making a decision - your time is precious.

· insurance requirements· estimated cost of periodic servicing . . . and so on.

Another criterion in new car selection is maneuverability which depends upon attributes such asturning radius, vehicle length, steering wheel turns lock-to-lock, all round visibility whilst parking,effort necessary to turn steering wheel and so on. Setting up a utility function for this criterion isobviously very difficult - establishing values of the attributes is easy enough, but ascribing a uniquelogical utility value to an arbitrary combination of them is not.

Evaluation is like any other activity - the end justifies the means. The effort devoted to evaluationshould be in keeping with the importance of the problem. Utility functions are unnecessary for thescope of problem encountered in this course. But whatever the evaluation technique used - basic orsophisticated - the guidelines below must be followed.

It may help your appreciation of the evaluation activity if you consider it being carried out on you -one of the candidates who sits an examination in order to answer the problem "Is this candidate fit toproceed to the next unit ?" You would object strongly if you were barred from proceeding just becauseof a low utility in one of the criteria - ie. just because you achieved 2/10 for one question in theexam. Evaluation must be based on the overall utility (your mark for the exam as a whole) and noton an individual utility.

The evaluation activity - like many facets of design - seems at first meeting to be a long-winded pro-

Design 18

Identify essential problemsEstablish function structuresSearch for solution principlesCombine and firm up into concept variantsEvaluate against technical and economic criteria

Develop preliminary layouts and form designsSelect best preliminary layoutsRefine and evaluate against technical and economic criteria

Finalise detailsComplete detail drawings and production documentsCheck all documents

Optimise and complete form designsCheck for errors and cost effectivenessPrepare preliminary parts list and production documents

Clarify the taskElaborate the specification

Specification

Concept

Preliminary layout

Definitive layout

Documentation

Upg

rade

and

impr

ove

Task

Solution

Cla

rific

atio

nof

the

task

Con

cept

ual d

esig

nE

mbo

dim

ent d

esig

nD

etai

l des

ign

Opt

imis

atio

n of

the

prin

cipl

e

Opt

imis

atio

n of

layo

ut a

nd fo

rms

Info

rmat

ion:

ada

pt th

e sp

ecifi

catio

n

Specific engineering info.

Electronic devices

Mechanical devices

Engineering science

Control

Materials

Manufacture

Costing

Stress analysis

Decision making

Synthesis

Market analysis

Total Activity

Market

Specification

Concept design

Detail design

Sell

Manufacture

Planned Organised

elements ofspecification

specificationformulation

conceptualdesign equatesto specification

detail design equates to

specification

design completely in balance with specification

flow

main

desi

gn

AB

GDE

HF C

GD

ECB

AH F

GF

EBD

AC H

BC

DEF

GH A

Information Technique

Design core bounded by product specification

order ofimportance

Pahl & Beitz Design Model SEED Design Model

Design 19

tion, of which design is but one facet. The authors' conclusions with regard to industry in the UShave the same implications in Australia.

• "without products and services to offer, a company has no reason to exist"• "the US-made share of global consumer electronics products declined from almost 100% to a

mere 5% over 30 years"• "US product development efforts are too slow, too expensive, and too often fail to create prod-

ucts with the features, performance and quality that customers want"• "world class product delivery will be the key enabler in the battle where quality, cost and

speed will determine the survivors"• "customers purchase a firm's products only when they find those products to be the most effec-

tive in meeting their needs"• "new products are truly the lifeblood of a company's long-term economic existence"• "better firms create two and a half times more new products on average than lagging firms".

You will recognise some key elements higlighted earlier - survival, the customer, novelty, etc.

After investigating the product development process in many US companies, the authors identifyfour elements of the process which are essential to success :

1. control by a single team (collaboration, communication)2. vision of the future product (look ahead; anticipate actions, interactions and reactions)3. a convergence of information from marketing (the firm's antenna to the outside world) engi-

neering and manufacturing (design must integrate with other functions); and4. a continuity in the information collected about critical product characteristics (documenta-

tion, feedback &c).

The elements of the rudimentary design process are thus seen to be fundamental (if not sufficient)components of successful design and development processes.

We conclude our examination of design by presenting some more advanced considerations whichare useful to know about when embarking on professional design.

More advanced considerations

So far we have concentrated upon design applied to elementary D&B projects. We now look at someprinciples which will be found relevant to the design of more advanced artefacts. The principles arerelevant when practicalising and optimising later stages of an artefact's life, so we'll consider themunder "designing for optimum manufacture", "designing for optimum use" - for the most part they arejust aspects of seeing the problem through others' eyes.

When designing for manufacture (or better - for the manufacturer) . . . .

• Specify components to be made as roughly as possible, consistent with their ability to performtheir duty. Thus, if you estimate that a certain 25 mm dia shaft can work satisfactorily with a tolerance of ±0.1 mm (ie. any diameter between 24.9 and 25.1 mm is workable) then a tolerance of ± 0.01 mmshould not be specified because the increased accuracy requires - a more skilled operative, - a more sophisticated machining process,- a more accurate measuring device,- more time spent in checking size during manufacture, etc

. . . . . all of which lead to unnecessary expense. The same philosophy applies to surface finish - surfaces should be as rough as possible consis-

- detail design: finalise the form, dimensions, materials etc. of individual components.

• The SEED model (Cross op cit) which is prominent in the UK, envisages Design at the core ofproduct development, surrounded by the requirements of the product's specification andembedded in the constraints of engineering science, economics etc.

These models are somewhat flawed as they emphasise engineering aspects, and do not stress suffi-ciently the people aspect and the necessity for a designer to find out about all potential reactions tothe solution. In spite of their tendency towards regimentation, the models clearly reflect the iterativenature of design.

This is probably a suitable point to put the foregoing lessons into context by quoting Wilson, Ken-nedy & Trammell op cit who address product development - the whole idea-to-product transforma-

FEASIBILITY STUDY ESSENTIALS

SHOW CONSIDERATION FOR READERNeat and legible; succint; unified; preferably word-processed

Introduction/synopsis to put the reader on trackLayout (clear section headings might parallel design steps)

Table of contents with page or section indexingMultiple calculations, sketches and programs relegated to appendices

Report bound in a folder and LH pages blankEnglish must be impeccable, for example :-

· spelling - “it's wagging its tail” ( it’s = it is, it’s not possessive )· absence of colloquialisms - "go for", "basically" &c; · impersonality - "the writer" rather than "I".

PROVIDE EVIDENCE OF COMPLETE PROBLEM DEFINITIONWhat criteria are relevant apart from the obvious ones in the (appended) Specification?

Has the Specification been interpreted too constrictively (fictitious bounds)?

PROVIDE EVIDENCE OF INVENTIONBreakdown of total problem into sub-problems if appropriate

Techniques; sketches; ingenuity; creative effort and determinationMore than just a couple of ideas

PROVIDE EVIDENCE OF PRACTICALISATION ATTEMPTSPracticalisation before attempting to choose optimum solution

Remedial action rather than mere criticism; perseverancePossible theoretical models of certain features

Testing and physical development if appropriate

PROVIDE EVIDENCE OF FORESIGHTWhat problems are foreseen and how can they be corrected? Outsmarting Murphy

PROVIDE EVIDENCE OF LOGICAL ARGUMENTJustification - not unsubstantiated opinions. Beware ". . the writer thought . ." if you're wrong!

Consistent use of the Evaluating Guidelines - all criteria uniformly applied across all candidateswhen selecting the most suitable solution

Arguments must not be tailored in an attempt to justify preconceptions

PROOF READ EFFECTIVELY BEFORE PRESENTATION

Design 20

prior to final machining in way of adjacent components - the shaft and the belt, Fig F.A designer who was not conversant with this process might have specified the design of Fig G which will be more expensive as it has the following disadvantages :- there is no draft which allows pattern withdrawl from the sand without damage, Fig C- the thin web solidifies first due to heat transfer, and will be torn by massive sections which

solidify and contract later, unless special conductors are inserted; this tendency is exacer-bated by large blending radii here

- the non-flat parting surface requires awkward consolidation of sand around the pattern inthe cope, rather than easily on a flat surface, Fig D. . . . and so on - there are many texts devoted to the details of casting design.

Students were asked to design a sheet-metal domestic letterbox to be made in reasonable quan-tities by a jobbing shop, using conventional manual techniques - mass production processes,such as heavy presswork, were not available. A typical design, involving slanting sides and acurved top for aesthetics and a hinged rear panel for access, is illustrated below.

You should criticise this design from the point of view of the manufacturer and users. Howmany different shortcomings can you identify ?

When designing for sale . . . .

• You should refer again to Seymour's article on Competition Analysis and put yourself in thebuyer's shoes. The economics in particular must be attractive. Remember always the ubiquityof the cost equation :

total costs = first costs + running costsin which costs are typically discounted cash flows.

First costs include interest accrued by borrowing for the initial purchase, or loss of potentialearning power if existing capital is used, costs associated with the purchase (special founda-tions or buildings), and so on. Running costs include fuel, labour, outage due to breakdown,

holes forpivot BACK

BASEROOF

holes formountingscrews

bend

FRONT

LETTERBOXWITH CURVED ROOF

tent with duty, with fatigue life, with sales appeal and so on.Use the least skilled labour and the least sophisticated processes possible - though this outlookdoes not apply when mass production is contemplated.

• Maximise the use of stock sizes and existing components. If the size of a certain shaft to satisfy fatigue strength, lubrication and other necessities is calcu-lated to be 23.4 mm, then the use of 25 mm dia bar stock should be considered - provided thesurface finish of off-the-shelf bright stock is adequate, and the larger size does not compromiseother components. The resulting 'waste' of material is usually more than offset by savings inmanufacturing time. If a design calls for a component which is very similar to the existing part of another assem-blage then you should examine the possibility of changing the design to adapt that part, per-haps modified slightly. This Lego/Meccano outlook is referred to as Value Analysis - a tool forreducing machining and the variety of parts which must be held in stock for later assemblyand/or spares. Stock and spares represent an investment which is just sitting there, requiringinterest to be paid on the investment, and not earning profits.

• Pay attention to details. If you are not aware of what these could be, find out. For example, the steps in casting a flatbelt pulley are essentially as follows. A wooden or plastic pattern is surrounded by a flask asshown in cross-section, Fig A above. This is filled with moulding sand which is consolidatedby ramming before the lid is fitted, Fig B. The flask is then inverted and the pattern carefully removed, Fig C, to form the lower part ofthe mould cavity - the drag- (in practice, a separate core would probably be used for the centralhole). The cope or top portion of the mould, whose sand has a plane parting surface here, ismade separately and mounted on the drag. Holes for the flow of molten metal are cut in thecope, Fig D. Metal is poured into the cavity through the feeder until seen in the riser afterwhich it is allowed to solidify. The casting is removed when solid, Fig E, and the sprues cut off

cope

drag

riserfeeder

machined surfacessprues

partingsurface

woodenpattern draft

flask

A B

C D

E F G

sand

Design 21

machine on the basis of its rating (a single figure) the load cycle should be understood clearlyso that the effects of part- and over-load operation can be quantified.

A final instance of economic reasoning concerns the supply of power to theelectricity grid by a number of steam-powered turbo-alternators, some ofwhich are large, some old, some out of service for (hopefully routine) mainte-nance, and so on. A major problem is in deciding which turbo-alternatorsshould be generating at any one time for minimum overall running costs,given the expected daily variation ofdemand in the grid as shown. The newest,largest and most efficient turbo-alternators

are driven constantly at their rated load, supplying the grid'sbase load. The load on the smaller less efficient machinesvaries, as dictated by demand peaks. Other machines are onhot standby, ready for unexpected demand or breakdown ofother alternators.

• Do not ignore the contribution which industrial design can make towards aesthetics and salesappeal. For example, the functional design of a drill press, Fig A below, is not in tune withmodern fashion which accents simplicity in form (the expensive computer lookalike).

A sheet metal cowl, Fig B, is therefore added to give an uncluttered illusion whilst still allow-ing adequate motor ventilation and permitting the belt to be easily swapped between pulleysto alter the drilling speed. This modification also enhances the design's safety by reducing thechance of fingers being accidentally trapped in the moving belt. (Consideration of industrialsafety is sometimes regarded as the prerogative of industrial designers, but safety is so vitalthat everyone must be involved, not just a select few.) Further styling might conclude that thecowl makes the drill seem top-heavy, so the base is enlarged and rendered in a contrastingcolour to appeal through an enhanced impression of stability, Fig C.

When designing for use . . . .

Remember that your design will not act in isolation, but in concert with other MEN, MATERI-ALS & MACHINES - plan accordingly. Other considerations include :

• The required technical specifications must be met with confidence - the solution must be practi-cal in other words. This calls for detailed application of theory from other subject areas(Dynamics, Thermodynamics etc).

base base peak peak peak

hot hot coldroutinemaint'ce

standbystandby

TURBO-ALTERNATORS

base

peak

0 12 24time hour

load

cost of maintenance, etc. Although the assignation of insurance, depreciation and the like - either to first costs or to run-ning costs - is essentially an accounting exercise, the implications of these costs must not be for-gotten at the design stage. Minimisation of total costs thus requires design optimisation in allareas - optimum 'design for manufacture' for example, primarily affects first costs, whereas opti-mum 'design for use' reflects principally in running costs.Subtle interplay between cost components often arises - for example better 'quality of manu-facture' may increase first costs but decrease running costs through improved and/or morereliable performance. Suppose for example a buyer requires a water heater for a certainapplication, and the choice lies between gas-fired and solar heaters.Knowing the initial prices of various sizes of each, the price of gas,the expected solar insolation, and making some allowance for fore-seeable maintenance etc, the buyer can build up a picture as shownhere. The solar heater will come into its own in the larger sizes asthe 'fuel' costs are negligible. From this picture buyers can deter-mine that, if their installation is below the cross-over size then theyshould select a gas-fired heater, otherwise they should select a solar. Buyers will try to assesswhat they are buying before parting with hard earned cash, so designers must weigh up care-fully what they are trying to sell and how buyers will view it.A further example of economic analysis is shown; it concerns arange of automatic machine tools for mass production. As themodels become more sophisticated the initial price rises but the timespent in machining a given batch of components decreases, leadingto a reduction in running costs. The total costs for machining a par-ticular part exhibit a clear minimum, so the model corresponding tothis degree of sophistication should be bought to machine that part -though other non-economic arguments would have to be consideredof course.Further examples of cost curves similar to these abound. Consider the choice of a heatexchanger. If the exchanger is small, so is its first cost. So also is its thermodynamic effective-ness - which leads to high running costs. Conversely a large exchanger implies a high first costwith lesser running costs. So the above graph applies again with exchanger size as the abscissa.

The larger the machine or item of plant, the more efficient is it likely to beThis is a general truism - that's why gas turbines, though acceptable in large sizes for airplanes,have not caught on for automotive applications - the efficiency in the smaller sizes is unaccept-ably low.

Another general trend is illustrated here. The size or capacity of amachine is usually indicated by a figure which represents the ratedcontinuous load which the machine can deliver or handle - thus a"150 kW" engine means that the engine can deliver a steady 150 kWcontinuously. This is also known as "full load".The efficiency (or effectiveness) of a machine increases fairly stead-ily as the load on the machine is increased, until a maximum effi-ciency is reached at or close to the rated load. Further load increasecauses a sharp drop in efficiency as shown.It is rare for the load on a machine to remain constant, so although a designer might choose a

size of water heater

tota

l cos

ts

cross-oversize

solar

gas-fired

sophistication of machine tool

cost

first running

total optimum

load

effic

ienc

y

rate

d lo

ad

design point

Design 22

A commonly accepted approximation is life * loadindex = constant in which the positive

index also is constant, both constants being characteristics of the machine under consideration.Typical indicative values of the index are 3 for ball bearings, 7 for ore-hauling railways, and 12for elastomeric V-belts. The magnitude of life reductions caused by increased loads, in view ofthese large indices, can readily be appreciated. The characteristic constants can often be pre-dicted from fatigue or crack propagation analyses (based on material properties from tests) orfrom maintenance records.

Safety is at all times of paramount importance.

Failure of hardware, through fracture or buckling for example, must be prevented by correctdetail design along the lines of the following chapters. Designers must be satisfied that anytheory which they apply is in fact applicable to the case in hand, and that no variations in load-ing for example render the theory irrelevant. Liberty ships operated satisfactorily in WW2, asdid Comet aircraft in the postwar years - both designed by the elastic theory which we still usetoday. But they still fell to bits (literally) for no reason which was understood at the time. Elas-tic theory has since been shown to be inadequate by itself for certain combinations of manufac-ture, materials, environment and load (see the later chapter on fracture mechanics).Avoidance of potential danger to life and limb is more significant than hardware breakdown.Damages for injury commonly reach six figures, and the courts seem to accept little frailty onthe part of designers, who can be found liable for compensation. Guards around dangerousmachines are usually specified by Codes and are therefore mandatory. But they may not beenough. People must be guarded against themselves, since they can be bloody stupid - and thisadjective is used advisedly. Loss of fingers resulted from circular saw guards being deliber-ately immobilised in a railway woodworking shop.

You have to foresee all possible interactions between your design and people, rational and idi-otic, and plan accordingly. Peoples' safety awareness must be fostered.

• Ergonomics is the study of the human/machine interface (Bibliography) with a view to optim-ising the effectiveness of both human and machine. A typical application is the provision of aworking environment for a bulldozer driver who, for maximum effectiveness, should :- be located in a cabin to minimise fatigue from airborne noise, dust and heat- sit on a sprung cushioned seat to minimise vibration and shock transmission- have excellent all-round vision for tractor manuevering and bladework- have all controls - for blade height, angle etc - adjacent to fingertips

. . . . and so on.

These requirements would be obvious to any designer who cared to observe what the averagebulldozer driver experiences and does from one second to the next. Unfortunately, ergonomicsoften plays second fiddle to economics in such instances. The fact that it is probably false econ-omy is often not recognised - not only are the delayed effect of prolonged exposure to a harm-ful environment legally actionable (eg. mesothelioma), but enhanced comfort often leads toimproved efficiency - ie. decreased running costs despite increased first cost.

When designing for maintenance . . . .

Again, it is a case of imagining yourself at the coalface. If a repairer's fingers are continuallybeing bruised through lack of room to wield a spanner, then the resulting frequent trips to firstaid will earn the overseer's wrath and lead to an determination not to order another designfrom the same stable.

• One factor which is often overlooked is the inherent variability of supposedly constant datagiven to the designer as part of the problem statement. The variation may be significant, or it may not, but the designer has got to know one way orthe other. Students who designed a stair climber often ignored variations in tread height fromone step to the next. In many cases this did not matter because the climbers' modes of propul-sion were insensitive to this parameter - the students were lucky. But if they had used a Sherpa-type of device then variations in nominally constant dimensions would have had a profoundeffect on operation.Typical occurrences of non-constant loading include :- Fatigue loading of components : for example, the bending moment on a rotating shaft may

be constant, but the stress on an element of the shaft must vary from tensile to compressive.- Busy road bridges are provided with peak lanes (analogous to the electricity grid above).- Manufacturing and mineral processing plants have stockpiles between the various operations

to cater for differing operating rates and outage of individual machines.

These examples are widely recognised and, where relevant, may be analysed by the techniquesof operations research - the name given to the branch of mathematics which deals with industrialproblems - queuing, routing and sequencing, minimising and optimising, learning and the like(Bibliography). But mathematical models are no substitute for common sense.

Other examples may not be so obvious; therefore it is important to plan for untoward varia-tions. Is there any chance of localised buckling, of vibration resonance, of imperfection interac-tion, or of foundation settlement ? What are the consequences of applying elastic theory whenthe material behaves plastically, and so on ? Altering the position of a heat exchanger baffle ledto the establishment of an extremely intense local vortex, which went undetected until highvelocity liquid (like a cyclone) had eroded its way right through the 50 mm thick steel cover.Designers must visualise fluid and stress flows; designers have to be subtle. Again - design isnot merely a matter of applying a few formulae.

A certain mineral processing plant is operating at just below itsrated throughput when a strike occurs, halting production in alloperations. On resolution of the labour difficulties, management - whichhas little empathy with things engineering - decrees that all pro-cessing machines will operate flat out until production hascaught up with the backlog. The result ? Breakdown of machines. No.2 crusher is the first towilt under the strain of being asked to do more than it was designed to do. Stockpiles mayallow the rest of the plant to continue to operate for a limited period while this machine isrepaired, as suggested by the sketch, but further breakdowns can obviously be expected. The moral here ? This scenario should have been foreseen by the designer(s). Stockpiles offerlittle help here; the only hope of protecting the machines from abuse lies in education. In otherwords, users must be provided with sufficient information to allow them to predict the resultsof their actions. If users want the ability to increase production, then they must be prepared tobuy larger items of plant. And this brings us on to the next point . . . .

The life of a component, machine or item of plant depends upon its load - the greater the load, the shorter the life.

rated continuousoutput

strikeou

tpu

t

time

no.2 crusherbreakdown

Design 23

processing plant. A fossil fuelled power station for example excretes large quantities of hotwater, gases and dust, whose effective dispersal does not come cheaply.

It is realised that you may not understand all the ramifications of this section, but the argumentsshould have made you aware that the art of engineering design is not merely a series of calculations.You must plan ahead.

Conclusions

Well, that’s it ! You should now have the fundamentals of designing under your belt. But recall thatknowing the fundamentals of anything is pretty useless unless you can apply them advantageously.

Design is no different - you’ve got to practice application of the fundamentals, particularly when theproblem doesn’t seem to require any design process, rudimentary or otherwise.

Remember always that when you’ve got to solve a real life problem, it’s the optimum solution thatyou’re looking for. Good luck in your search !

Maintenance is rather like safety - the designer's task is to foster an awareness of the need andto make things as easy as possible, not just for carrying out maintenance, but also for monitor-ing to reveal the need for maintenance - testability and maintainability are mandatoryrequirements of most defence equipment for example.If continuous monitoring of performance is impractical, then intermittent monitoring at leastshould be carried out to expose incipient malfunction. Preventative maintenance can then beplanned and executed as with the turbo-alternators mentioned above. This approach is muchsuperior to waiting until something actually breaks down before fixing it. Obviously, the morecritical a machine in the overall scheme of things, the closer must be the watch over it. Suchthinking leads to the provision, as a matter of course, of thermometer pockets in heatexchanger pipework to check for the build-up of deposits over a period of time. Even littleexchangers are often so equipped, since their failure could trigger a disastrous chain reaction -though if they are vital links then they are usually replicated.

• Providing spare sub-assemblies as opposed to individual spare components may be an attrac-tive proposition.

For example the water pump on a car engine shown below usually fails because the seal wears,allowing water to leak past it to corrode the shaft bearings, which then seize. Depending on themanufacturer, it is often cheaper to replace the complete pump rather than just the seal andbearings, because - the level of skill required to bolt on the complete replace-

ment pump is much less than that needed to dismantle thepump and rebuild it with new components

- if one component has reached the end of its useful lifethen other components are probably not far behind

- the turnaround time is much less.Extending this concept, the designer should investigate theadvisability of the manufacturer or his specialist agent assum-ing the responsibility for all maintenance. This becomes amatter of necessity rather than choice when the equipmentbecomes too sophisticated for the user to repair, as with com-puters and domestic refrigerators.

The provision of complete reconditioned machines is another option along the same lines. Throw-away razors and throw-away cigarette lighters are evidently commercially feasible. . . . whatnext?

When designing for retiral . . . .

· Open cut mining operations in environmentally sensitive areas usually begin with removingand storing the topsoil for eventual replacement and revegetation when mining has ceased.The carcases of other industries may not be so easily retired - nuclear power stations and off-shore oil rigs are cases in point.

· There is probably not much general advice to give here, except once again, to think of the reti-ral phase, to think of those who must effect it. There are many instances of successful venturesbased on recycling retired products - car tyres transformed into non-slip sports surfaces,marine life nuclei and so on.

· The disposal of waste products in an acceptable fashion must be solved during the design of

seal

bearing

beltpulley

water flow

Design 24

Appendix A - "Improvements" to a computer workstation

The 'CompuChamp' is an easily moved workstation for micro-computers. It consists essentially of two load-bearing inclineduprights rigidly attached to a base equipped with castors, togetherwith various tables and trays for monitor, keyboard, printerpaper, desktop and other equipment. Tables are attached tosleeves which can slide up and down the uprights, and which are

locked by hand-operated screws tightened tobear on the uprights. A user selects the tablesrequired, adjusts their height, and locks them inposition before finally placing equipment onthem. The workstation can then be moved easily around as a complete unit.

The need for improvement arose from an increasing awareness of repetitive straininjury (RSI) in the workforce at the time, which led to a market requirement for table heights to beadjustable under load, ie. after equipment had been placed on the tables. Without improvement it isvirtually impossible for an unaided person to adjust the height of a loaded table - even although theload by itself is within the strength capabilities of that person, the necessity for supporting the loadwhilst simultaneously releasing two locking screws, adjusting the table to the desired height, andthen relocking the screws, renders the whole procedure too dangerous (imagine yourself in theadjuster's shoes!). So the CompuChamp's manufacturer approached students to put forward ideas -ie. to prepare feasibility studies - for improving the workstation by permitting on-load table heightadjustment. Students' responses were not accepted - admittedly it was not an easy brief (but thenthey never are !); it is instructive to learn by their mistakes.

It is very poor design to rely on friction as the major load support mechanism. As sure as fate, some-one somewhere will at sometime leave a screw somewhat untightened; a slight knock could theneasily cause the table to slip and a falling computer to inflict injury. This is Murphy again - thingsare fine for 99.99% of the time, but it is not difficult to foresee what can happen and design it outbefore being landed with a hefty damages claim - it is thedesigner's responsibility. To drive this point home, consider theadjustable office chair whose mechanism's essentials are sketchedin both the loaded position (when someone is sitting on the seat)and released to enable seat height adjustment. Not shown are var-ious springs &c which cause the lobes to automatically springback into engagement with the corrugations when the operatingrod is deactivated after adjustment has been completed and thechair is again ready to support an occupant. The mechanism isingenious as operation of the multiple lobed support wedge is notaffected by heavier users - it is perfectly secure once properlyengaged in the corrugations.

However one day a particular chair collapsed when someone saton it after adjustment and received serious and permanent spinaldamage. Collapse was due to a number of factors which pre-vented the lobes from re-engaging the corrugations when theoccupant deactivated the operating rod, so the support wedgeremained in the released mode and the mechanism descended too

table

lockinghandscrew

sleeve

upright

corrug-ated tubemountedon base

multiplelobed

supportwedge offlexibleplastic

supportwedge

operatingrod

seatsupport

tubetaperedat lower

end

LOADED RELEASED

Appendix A - “Improvements” to a computer workstation - 1

quickly under the occupant's weight for the lobes to engage lower down. A rare occurrence perhaps,but certainly foreseeable - it wasn't fail-safe.

Returning to the 'CompuChamp', its reliance on friction though poor enough when table heights areadjusted once only prior to loading, could be disastrous if adjustments were carried out on a routinebasis. But there are further features which prevent easy improvement as desired.

The side sectional view of an unlocked sleeve with cantilevered load 2w issketched. The geometry is characterised by the single parameter q - theratio between the load offset and half the sleeve depth, b. If the sleeve is toremain in equilibrium then the upright must exert force components w andq.w on it at each of the two contact points as shown. If the coefficient of fric-tion between sleeve and upright is µ then equilibrium is possible and slipavoided provided µq ≥ 1, no matter how big the load - indeed it is possibleto sit on a 'CompuChamp's table with the locking screws completely removed ! But of course ifwhile sitting, one were to move to reduce the load offset q, then the table would plunge - and thesame capricious behaviour could occur if equipment were to be moved aroundon an improperly locked table. This principle is used to advantage in an adjusta-ble clamp which can be adjusted before tightening, and where q is large andfixed. It is distinctly disadvantageous in the 'CompuChamp' however, becausejamming will occur if the load is offset by more than b/µ. This behaviour meansthat support of a load in a manner which promotes free sliding can occur reliablyonly if the reaction is approximately coaxial with the load (ie. q < 1/µ) - anyother location leads to a couple and potential jamming. It may be appreciatedthat it is difficult to provide a height-adjusting drive to a table which is coaxialwith the load rather than along the uprights.

Jamming is promoted also by the twin uprights. The frontview shows a table's two sleeves in contact with the uprightsat two discrete points due to an asymmetric table-tipping load.A couple similar to that above is created and leads to binding.The table can be moved only through synchronised equal dis-placements of the two sleeves - a mechanism to carry this out,though not particularly complex in itself, could be expected to

be awkward to incorporate retrospectively into the existing design. Binding on the two uprights isexacerbated by inaccurate manufacture when a number of tables with slightly different sleeve spac-ing are assembled onto a pair of uprights.

The students who tackled this 'improvement' problem missed the above points. They carried outpaper exercises without paying enough (any?) attention to the real world, though they were pro-vided with a 'CompuChamp', on which to experiment had they wished. The conclusions here werethat the 'CompuChamp', though commercially successful when tables are loaded only after adjust-ment, is not suitable for modifications which permit routine on-load adjustment.

The moral here ? Beware the improvement brief :

'Improve the present design'really means

'Select the optimum solution to suit criteria which differfrom those for which the present design is optimum'

qb2w

wqw

2b

qw

w

contact points

load

Appendix A - “Improvements” to a computer workstation - 2

Design 25

Appendix B - Tube end problem

The benefits of delaying problem completion until after ideation is complete aredemonstrated by the problem :

"How to prevent fluid escaping from the end of the tube ?"

We keep the problem as broad as possible by not being more specific, and brainstorm the followingcandidate solutions :

- manually close an upstream valve - screw cap over end- weld flange and bolt cover to it - balance a book on the end- crimp the tube (eg. squeeze it flat) - explosive ditto- liquid trap - rely on surface tension- stick your thumb in the end - push cork or squash ball into end- the reverse - stick end into cork - blob of Blu.Tack or chewing gum- drain contents then no more can escape - clothes peg to squeeze end shut- automatic ball-valve like toilet cistern - slip a toy balloon over end- reduce pressure in tube - weld up the end- increase the pressure outside the end - melt the end & fuse into sealing blob- freeze fluid - tie a knot in the tube- surround end with magnet - gladwrap, bandaid or other sticky tape- ignite any escaping fluid - insert a flat soccer ball then inflate it- turn upside down & stick into ground - jet another fluid into end- weighted piston inside tube - set up standing pressure wave- cover end with a pile of earth - electrostatic precipitator (smoke)- &c &c &c

You don't need a degree in quantum physics to generate ideas like these - but good general knowl-edge clearly helps.

Having generated the bank of solution candidates, we would then complete the problem by defininga set of appropriate constraints and criteria involving typical attributes as follows :

- fluid : air / water / steam / liquid chlorine / blood / conc sulphuric acid / person's breath /sewage &c

- hazard : toxic / benign / environmentally sensitive &c- tube : steel/ glass/ artery / plastic / paper / earthenware / wooden &c- size : 2 m dia Goldfields pipeline / capillary / 40 mm dia car radiator hose / garden sprinkler

/ 500 mm dia air-conditioning duct &c - effectiveness : absolutely leakproof / just prevent some escaping / the odd drip doesn't matter- pressure : 1 mm H2O / high pressure (10 MPa) / water mains / sound pressure amplitude &c- closure : releasable / permanent / manually controllable (bath tap) / only for 5 minutes / auto-

matic &c- cost : of no consequence (emergency) / quantity production &c- &c &c &c

The complete problem might be one of the following, for example :- a flexible armoured high pressure 20 mm dia steam hose- a 1500 mm dia asbestos stormwater culvert during a flood- a 2 mm dia stainless pipe with liquid chlorine adjacent to a schoolroom- a hypodermic needle containing HIV positive blood- a fractured underground pipe conveying avgas under a busy air terminal

Appendix B - Tube end problem - 1

- a windmill-filled water pipe to a stock trough way out on the range- and so on, the combinations are infinite.

How many of the above solutions would we have dreamed up if the complete problem had been known beforehand ?

Precious few !

Of course many if not most of the solutions are completely impractical for a particular problem sce-nario, but the point is that once solution candidates have been generated it is relatively easy to criti-cise in order to weed out the crazy candidates - and, as we have noted before, crazy candidates oftentrigger, or can be transformed into, practical solutions. If you don't try for novel competitive solu-tions by ideating in this manner, then you're heading for the same fate as Jane !

Besides, what at first glance may appear crazy, could turn out to be perfectly viable after practicalis-ing. Let's demonstrate by following the design process a little further here. We shall try to practical-ise the toy balloon candidate for example, on the presumption that the scenario is not extreme, ie.tube diameter around 20 mm, non-hazardous liquid at a pressure P well below domestic water pres-sure, and so on.

Assuming that the neck can be forced over the tube end, the situation mightappear as in sketch a here. The balloon's grip around the tube - determined byelastic properties and initial stretching - manifests itself as a contact pressurebetween neck and tube exterior surface. Provided the fluid pressure is not toogreat, its tendency to blow the balloon off the end may be resisted by friction cor-responding to the pressure over the contact area.

Depending upon the relative magnitude of contact and fluid pressures, fluidmight find its way through the contact region between neck and tube, thusdestroying friction and causing immediate blow-off.

Recognising this, the basic idea might be modified as in sketch bby replacing the balloon by a more robust but still flexible cover,the idea being to increase the contact pressure. The cover couldbe moulded from a high density polymer.

The enlargement indicates fluid pressure on the cover if a leak-age path should develop - the fluid pressure decreases from P atthe interior to zero at the outside of the leakage path. The fluid pressure tends to bend the cover'swall (like a beam) away from the tube, with a corresponding reduction in equilibrating friction andin cover retention capabilities.

In an effort to improve retention, the arrangement of sketch cis proposed, the external cover being replaced by a flexibleinternal bung. Clearly the higher the fluid pressure, thehigher will be the contact pressure and friction, as the fluidpressure tends to push the flexible wall of the bung againstthe tube's interior surface.

If a leakage path should develop, as indicated by the enlarge-ment, then the fluid pressure outside the wall must be less than the full fluid pressure inside - thenet tendency being to force the bung wall against the tube thereby closing off the path. Note theadvantages of this arrangement compared to a solid bung or cork.

Practicalisation could continue with detailed mathematical modelling based on presumed pressure-

Appendix B - Tube end problem - 2

fluid

P

a

P

P

b

PP Pc

Design 26

deformation characteristics of the bung wall and tube, probable coefficient of friction, etc etc. Theanalysis would have to allow for the non-elastic behaviour of the plastic bung, the inevitability ofmanufacturing inaccuracies which could drastically affect frictional "grip", and other non-ideal beha-viour.

Recognising that any mathematical model which avoided expensive complexity wasnot going to be very accurate, practicalising emphasis might switch to R&D. An arte-fact would therefore be manufactured, tested and modified until its service behaviourcould be predicted with confidence. It may prove advisable eg. to incorporate groovesaround the bung wall's exterior, sketch d, which in service would contain adhesive toenhance the bung's holding power.

Appendix B - Tube end problem - 3

Appendix C - Lessons in frustration

1 : Devices for the PEP D&B competitionThe track for the Potential Energy Propulsion Warman D&B competitioninvolved a false ceiling 3 m above the floor, a 0.5 m square hole in thisceiling, and a target on the floor 7.5 m distant from the hole centre. Thetask was to transport a can of cool drink from above the ceiling,through the hole to a rest position as close to the target as possible. Thesole means of powering the transporter was potential energy. Speed wasirrelevant, the competition score being reckoned by the final distanceof the can from the target.

It was just natural for Tom and Jerry ( T & J ) to form a stu-dent design group as they had been close friends for years,living just around the corner from each other. Tom's fatheroperated a machining workshop - Tom had cut his teeth ona lathe and was a competent if unqualified tradesman. Jerryalso was experienced in fitting, having worked part-timewith Tom's father during vacations.

Like other groups they had six weeks in which to designand build a device - theirs consisted of two major compo-nents :

- A four wheeled cart which carried the cool drink canwas made from aluminium alloy. Each of its wheelswas turned, drilled for lightness, equipped with a rub-ber tyre and attached to the cart by six counterbored

setscrews. Underneath the cart (not visiblein the photograph) was a small drumaround which thin fishing line waswrapped, to be unwound during the runto limit the cart's travel to 7.5m.

- A frame riveted up from aluminium alloysquare section was mounted on four verti-cal retractable legs and carried a curvedramp along which the cart travelled. Apulley was mounted atop each leg, and alinear bearing allowed the leg to slide upand down. When any leg was pushed upit turned a common central shaft via a sys-tem of pulleys and thick fishing line, theleg travel being retarded by an adjustabledisc brake on the shaft.

In operation the cart was mounted at the top ofthe frame's ramp and retained there by a latch. The assembly was then dropped through the ceilinghole and free fell to the ground where the shock was absorbed by the disc brake. Simultaneously, oncontact with the ground the latch was released by an automatic trigger, allowing the cart to roll

Appendix C - Lessons in frustration - 1

d

3m

7.5m

target

Design 27

down the ramp and travel along the ground until brought up short by the unwinding thin fishingline.

Like other groups, T & J tested their device exhaustively to ensure that Murphy didn't get a look in,but the fiddly manufacture took so long that the device was not finished in time for the event, so thegroup received no credit for competing.

Undeterred, T & J were ready a day or so later to salvage some marks for completing a successfuldevice. The finished model was so beautifully crafted that half the workshop and academic staffturned out to see it through its paces. A few last minute adjustments and at last all was set - thedevice was released, fell with a resounding crump, leaned gracefully on its side, and sent the cart offat right angles to the track . . . . .

A moment of stunned disbelief before it was realised that one of the shock absorber legs had buckledbeyond repair. A disaster! A rerun was out of the question. And this after such extensive testing.Needless to say T & J were distraught that the fruits of so much labour came to naught.

The group received very little credit for this. Why? Because the design was far and away too com-plex - the more bits and pieces there are, the more likely will something go wrong. In this case it wasnot difficult to predict that there was an outside but still non-zero chance of the device landing awk-wardly on one leg so that sliding through the bearing was impossible - behaviour identical to theCompuChamp table sleeves of appendix A which had been brought to the attention of T & J.

As an article of craftsmanship the device was excellent; as a design it was a shocker.

The group adopted criteria to exhibit (or indulge in ?) its members' skills (vide the six setscrews ineach wheel, the lovely finish on the turned pieces &c) rather than to maximise simplicity. Don't belike T & J - learn the lesson before you waste a lot of time and effort. Youare not expected to be as skilled in manufacturing as this group - and forsuccess in D&B competitions you don't need to be.

For example and as a contrast, Peter, Paul & Mary based their PEP deviceon three rolls of computer paper, shown here at its initial and final posi-tions. Two fixed rolls A & B rotated freely in bearings attached to a roughwooden frame positioned over the hole in the ceiling.

The ends of these fixed rolls were attached to the underside of a curvedramp which was built up from pieces of expanded polystyrene gluedtogether. One end of a third travelling roll C was taped to the cool drinkcan then the roll was wrapped around the can. The free end of this roll wasattached to the top of the ramp and the roll as a whole supported at the topof the ramp by a releasable latch. Not shown on the sketch are :

- a cord wrapped around the two fixed rolls to synchronise unwinding,- a disc, attached to the shaft of one of the fixed rolls, dipping into a tin of

honey to prevent unwinding too quickly, - a trigger made from a wire coat hanger, protruding under the ramp

to release the travelling roll when the ramp hit the ground.The assembly was located above the hole in the ceiling, and theramp released. It fell in a controlled fashion due to theunwinding synchronised fixed rolls and the honeyretarder. When the ramp reached the ground, thetrigger released the travelling roll and allowedit to roll down the ramp, unwinding as it did

Appendix C - Lessons in frustration - 2

so. Motion eventually came to a halt when this roll had unwound completely and revealed the can.

A remarkable feature of this device, apart from its simplicity, was its relatively slow controlledmotion and resulting accuracy. It won the local competition, and would have triumphed in theNational Final had the track not been positioned right in front of an air conditioning diffuser whichtook up the whole wall. Although the air velocity from this was so low that it could not be felt, itwas sufficient to throw the extremely light device slightly off course. This effect was not discovereduntil after the first of the two Finals runs - it was rectified in time for the second run, but unfortu-nately for Peter, Paul and Mary by that time it was too late. Murphy had the last laugh yet again !

Note the similarities and differences between these two devices. They both involved a descendingcurved ramp which initiated a second transportation component. The first device was very complex,was made from metal and therefore required relatively complex and sophisticated manufacture. Thesecond was a simple, elegant and creative solution made from household/office materials whichdemanded little manufacturing expertise. Although it was made from plastic and paper, it solvedthe problem - whereas the first device failed, despite being made of metal. Which all goes to showthat there's no substitute for creativity !

2 : Accelerator cable gland in the Citroën CX 2200 saloon

In motor cars without fuel injection the accelerator pedal is usually connected to the carburettor by aBowden cable - just like a bicycle's brake cable only more rugged. The cable passes through a hole inthe sheet metal bulkhead between passenger space and engine compartment, and requires someform of a seal between cable and hole edge in order to carry out indefinitely the duties :

- protect passengers from dust, noise, fumes and possibly fire under the bonnet,- permit the cable sheath to move axially through the hole as its core tightens and relaxes

slightly during operation,- prevent the cable from bending excessively or from abrading on the edge of the hole in the

sheet metal.

Citroën solve this sealingproblem by using a glandas follows. A circularhole with two lobes (i) iscut (how?) in the sheetmetal bulkhead. Thecable is free to slidethrough a close-fittingplastic gland cylinder, shown in section (ii), in which there are no sharp corners on which the cablemight chafe. Two lugs on the cylinder match the hole's lobes.

Assembly commences by poking the cable and gland cylinder through the bulkhead hole from theengine compartment, as shown at (iii) below - the cable being omitted from this sketch for clarity.Once through the hole, the cylinder is rotated through 90o to prevent it falling back into the enginecompartment. A circular plastic collar equipped with a couple of fingers (iv), is then slipped over thecylinder with the fingers protruding through the hole lobes thus preventing the cylinder from acci-dentally rotating and coming out if the bulkhead hole (v).

Unwanted disengagement of the collar by vibration is avoided by a compression spring forcing it upagainst the bulkhead (vi), the spring reacting against a thrust washer retained by a circlip which

Appendix C - Lessons in frustration - 3

ramp

roll C

4

roll A

roll Bframe

( i )

( ii )

Design 28

engages with a groove around the outside of the cylinder.

Citroën's solution certainly works, but has a couple of serious drawbacks :- There are five gland components which have to be sourced individually and assembled on

the cable for sale as a unit - a costly exercise, given the gland's mundane duty.- Without special tools, assembly into the bulkhead is an exercise in frustration, as :

- The operator cannot bring both hands to bear on the assembly because the space adja-cent to the hole is severely constricted by other features such as the wheel arch.

- Compressing the spring and collar prior to fitting the circlip into its groove is very diffi-cult due to the spring's appreciable stiffness and the lack of anything for the operator toreact against. (It's a bit like trying to pump up a bicycle tyre using only one hand!)

The conclusion reached by one bruised and extremely frustrated installer was that this solution is fartoo complicated for the problem posed, though admittedly something more elaborate than a blob ofBluTack is necessary. Little if any thought had been given to the wishlist of non-specialist folkengaged in maintenance, an accusation which has been levelled at Citroën in the past . . . . . but ofcourse Citroën cars have other attractions !

Appendix C - Lessons in frustration - 4

Appendix D - Analysis of a spring driven vehicle

The vehicle of mass m consists of a light double driving wheel -centre O and outside radius r - a chassis and two small trailingwheels for lateral stability. It is powered by a light tensionspring of stiffness k, attached to a cord wrapped around thesmall diameter portion of the driving wheel of radius ρr, where0 ≤ ρ ≤ 1. In the sketch, the vehicle travels rightwards due to therelaxing spring. Realistically, motion will be opposed by friction

at all contacts - at the ground, with the surrounding air, at the wheel/shaft bearings etc - and theopposition will vary; but for the purposes of this simplified analysis, all these frictional effects arelumped into a single drag force D which is constant. This drag could include a gravitycomponent if the vehicle were climbing a slope.

The vehicle's displacement is x, as sketched, from some arbitrary fixed datum. Thespring force and deflection are P and δ respectively, the spring being wound initially to P i- less than the limiting P dictated by the material's strength. In certain situations, geomet-rical constraints may also limit δ to δ.

The analysis seeks to determine the effect of build parameters on the displacement-time relationship,and it must consider three broad aspects - kinematics (compatibility, geometric necessities), kinetics(forces, accelerations), and the spring's constitutive law (characteristic). It is assumed initially thatthere is no wheel slip or flip-over; this will be checked later.

Compatibility. If the double wheel rotates through a small clockwise increment, dθ, asshown, then the spring's deflection, δ, would reduce by ρr.dθ while, in the absence ofslip, the vehicle's displacement, x, would increase by r.dθ. So, for an appreciable rota-tion, θ, from the initial position :

(1) δ = δi – ρr.θ = δi – ρx

Kinetics. For the vehicle as a whole :

(2) ΣF→ = F – D = mx"

and, for the light uniform driving wheel :

(3) ΣMO = P.ρr – F.r = Iα = 0 ; ie F = ρP

Note the general effect of ρ for a given range of spring deformation. For a desirably large vehicle dis-placement a small value of ρ is indicated by (1); for a desirably large tractive effort a large value of ρis indicated by (3) - ρ affects both displacement and force.

(4) P = kδ the spring's Constitutive Law.

Solution. There are four equations in the four time-dependent variables P, δ, F, x; eliminating thefirst three leads to a differential equation with constant coefficients :

(5) m.x'' + ρ2k.x = ( Fi – D ) where the RHS also is constant.

Fi = ρkδi is the initial tractive effort - obviously F i must exceed D before any motion is possible. Theprominence of ρ should again be noted - its square arises from its relevance to force and displace-ment, both of which considerations are embodied in equation (5). Solving this with x = x' = 0 at timet = 0 leads to the relation sought :

(6) x = xo ( 1 – cos t/τ ) where the constants xo and τ depend on the vehicle characteris-tics, thus : xo = ( Fi – D )/ρ2k and ρτ = √( m/k ).

Having found the displacement-time relation, (6), the kinematics can be examined to evaluate thetotal distance travelled, x, and the corresponding period of motion, t .

Appendix D - Analysis of spring-driven vehicle - 1

( v ) ( vi )

( iv )( iii )

enginecompartment

bulkhead

Design 29

Kinematics. Two alternative travel modes can be foreseen before the vehicle halts :-A - drag overcomes tractive effort in halting the vehicle before the spring relaxes fully, orB - the spring relaxes fully before halt, implying a period of deceleration due to drag alone.As Mode A does not fully utilise the spring's energy, it seems less satisfactory than Mode B. Theratio of initial traction to drag is crucial; it is symbolised in its own right : f ≡ Fi/D.

Mode A As there is no discontinuity in the spring force over the journey, (6) applies throughout. From it,the velocity is x' = (xo/τ) sin t/τ, being zero when t = 0 and when t = πτ, in which latter case thetotal travel is found from (6) to be x = 2xo. If this mode is valid, the final minimum spring deflection, δ = δi – ρx from (1), must be > 0. Substi-tuting from above, this requires that Fi ≤ 2D. So, summarising for this mode :

(7a) If 1 ≤ f ≤ 2 then x /xo = 2 and t/τ = π where xo and τ are given by (6).

Mode BThe total displacement may be found from work-energy. Since the initial and final speeds arezero, the initial elastic potential of the spring, 1/2kδi

2, is dissipated completely in overcoming theconstant drag, D, over the distance x. So x = kδi

2/2D.The time at the end of the powered phase, tδ=0, when the spring has run down fully, is given by (6)

and (1) as tδ=0/τ = arccos( 1 – δi/ρxo ) noting that 1 – δi/ρxo = 1/( 1–f ). The corresponding veloc-ity at that instant is x'δ=0 = (xo/τ) √[ 1 – (1 – δi/ρxo)2 ].Applying v = u+at over the second phase of constant decelera-tion, D/m, yields 0 = x'δ=0 – (D/m)( t – tδ=0) from which fol-lows t . Summarising for the mode :

(7b) If 2 ≤ f then x /xo = f2/2(f–1) and t /τ = π + g – arctan(g) where g ≡ √( (f–1)2 –1 ).

The normalised total distance and period given by (7) are plottedhere against f, in the expected range of interest; it is apparent thatneither of these vary markedly.

Wheel Slip is most likely at vehicle release when the tractive effort is a maximum, Fi. It will occuronly if Fi → µN, µ being the coefficient of friction between driving wheel and track. The vehicle'swheelbase is L. Proportions (h,v) which define the mass centre, G, also affect the slip tendency andare shown in the free body, together with the drag, D, which is assumed to act through G. Momentsabout the point q are taken to eliminate the unwanted reaction, Q :

ΣMq = IGα + ma*(perpendicular from q to a)N.L + F.vL – W.hL = 0 that is N = hW – vF

If therefore, F ≤ µN initially, it follows that :

(8) Fi ≤ µhW/( 1+µv ) to avoid wheel slip

Vehicle Flip-over is not expected to be a problem since the vehicle isfront-wheel drive.

This typical mathematical model is approximate as it assumes constant drag, an inertia-less wheeletc. Accuracy of results from it cannot exceed that of data input to it. Nevertheless it gives a ball-park 'feel' for the problem - it enables assessment of feasibility, appreciation of sensitivity to parame-ter variations, choice of close-to-optimum parameters before building, and so on. But it is still onlythe kinetics side of the story - the safety aspects (strength and buckling) would also have to be mod-elled, to assess the effect of a beefy big spring eg.

Use of the model is exemplified overleaf.

Appendix D - Analysis of spring-driven vehicle - 2

Example

Details of a vehicle, which must travel 1m, together with measurements of a handy spring, are givenbelow. Is the spring suitable for driving the vehicle, and if it is, what wheel ratio, ρ, and spring initialwind-up, Pi, would give the shortest travel time ?

Vehicle : weight, W = 20 N mass centre normalised coordinates (h,v) = (0.75, 0.25) drag, D = 2 N friction coefficient, µ = 0.5⇒ peak traction, F = µhW/( 1+µv ) = 0.5*0.75*20/(1+0.5*0.25) = 6.67 N eq (8)

Spring : wire diameter, d = 2 mmcoil mean diameter, D = 20 mm number of turns, n = 30steel - modulus, G = 80 GPa; design stress, say Ss = 400 MPa 1MPa ≡ 1N/mm2

⇒ stiffness, k = d4G/8nD3 = 24*80*103/8*30*203 = 0.667 N/mm⇒ capacity, P = πd3Ss/8D = π*23*400/8*20 = 62.8 N check units

There are limitations on both the traction and wind-up initially, and as these forces are mutuallydependent on one another through the double wheel proportions ρ and equilibrium equation (3), it isadvisable to first obtain an idea of the limiting forces which are feasible, as functions of ρ. Concen-trating on the wind-up, these limits (in N) are :

motion - if the vehicle is to move, then F > D, and as F = ρP, then P = drag/ρ = 2/ρ slip - if slip is to be avoided, then F < F, and as F = ρP, then P = F /ρ = 6.67/ρstrength - if the spring is to remain intact, then, from above, P = 62.8

Thus, for any ratio, ρ, the initial wind-up must lie above themotion limit P, and below the lesser of the two values of P, corre-sponding to slip and strength limits - it must lie in the unshadedband sketched at left. A reasonable wind-up for any ρ-ratio is taken to be the maximumlimit less one quarter of the band height, ie

Pi = P – ( P – P )/4 = ( 3*P + P )/4The range corresponding to this wind-up varies with wheel ratioas shown below. Typically, for ρ = 0.05 :

P = 2/0.05= 40 N P = minimum(6.67/0.05, 62.8) = 62.8 NPi = (3*62.8+40)/4 = 57.1 NFi = 57.1*0.05 = 2.86 Nxo = (Fi – D)/ρ2k = (2.86–2)/0.052

*0.667 = 514 mm eq (6)

τ = √(m/k)/ρ = √(20/9810*0.667)/0.05 = 1.11 sf = Fi/D = 2.86/2 = 1.43 < 2 not preferred Mode Bx = 2*xo = 1.03m eq (7a)

t = πτ = 3.47 s

The spring is unsuitable as it allows no margin for range error -eg, under what conditions can we be sure that drag will not exc-

ced 2 N? The weight of this spring is about 25% of the all-up weight, so weights of heftier trialsprings cannot be ignored. The Pascal program Spring Driven Vehicle enables rapid evaluation along the foregoing lines. Butcomputing and mathematical modelling should never be overdone.

Appendix D - Analysis of spring-driven vehicle - 3

Design 30

Appendix E - Springs as energy storesThe performance of a spring is embodied in its load-deflection characteristic,deflection being reckoned from the free, unloaded state. The characteristic of anelastic, hysterisis-free spring is a straight line - whose slope is known as the stiff-ness of the spring - together with limits on load to avoid material failure, andpossibly on deflection too, if geometric constraints such as solidification occur.The deflection limit is more critical than the load limit in the case illustratedhere.

The manner in which a spring performs is either :-linear deflection , δ (m); force load, P (N), & stiffness, k = P/δ (N/m), orangular deflection , θ (rad); torque load, T (Nm), & stiffness, k = T/θ (Nm/rad)

The energy stored by a spring at any state is the 'area' under the characteristic between the originand that state. Thus in the sketch, the energy stored at state 1 is :-

energy = shaded 'area' = 1/2*(deflection)1*(load)1 = 1/2*(deflection)12*k = 1/2*(load)1

2/k

Metallic springs having appreciable deflections (ie not tension bars) are loaded in either :-bending : material modulus, E; load limited by maximum normal stress, Sn (MPa), ortorsion : material modulus, G; load limited by maximum shear stress, Ss (MPa),

Neither of these promotes compact energy storage since stresses are not uniform. Fatigue is a com-mon failure mechanism. Some common springs are illustrated below; the equations cited for elasticsprings are approximations which neglect stress concentration for example.

Appendix E - Springs as energy stores

Tension bar

k = bdE /LP = bdSn

Buckle-prone in compression.Deflections small, if metallic.

Cantilever spring

k = bd3E /4L3

P = bd2Sn /6LLeaf springs are similar.

Helical compression spring

k = d4G /8nD3

P = πd3Ss /8D n: number of turnsTensile need special ends.

Torsion bar

k = πd4G /32LT = πd3Ss /16

Helical torsion springSimilar to helicalcompression spring,but coil ends shapedto enable transfer oftorque.

k = d4E /64nDT = πd3Sn /32

Spiral torsion (clock) spring

k = bd3E /12L T = bd2Sn /6 on shaft.L: total length of spiral

Constant force spring

Horizontal characteristic.Large deflections possible.

Bellville washer

Often highly non-linear; stiff-ness may be negative.

Non-metallic springsMany forms exist, including

fluid spring(piston-cylinder), therubber vib-ration mountshown, etc etc.

The following selection provides a general background to engineering design and creative problem-solving :

Roberts JCH, Creativity Guide, Design Engineering Series, Morgan-Grampian 1960Glegg GL, The Science of Design, The Selection of Design & The Design of Design, Cambridge

1969-72Krick EV, An Introduction to Engineering, Wiley, 1976 - excellent introduction to engineering French MJ, Conceptual Design for Engineers, Springer 1985 - highly recommended Petroski H, To Engineer is Human - the Role of Failure in Successful Design, St Martins 1985Maunder L, Machines in Motion, Cambridge 1986French MJ, Invention and Evolution: Design in Nature and Engineering, Cambridge 1988Fabian J, Creative Thinking & Problem Solving, Lewis 1990Hollins W & Pugh S, Successful Product Design, Butterworths 1990 - rather more advanced French MJ, Form Structure & Mechanism - a Design Primer, MacMillan 1991De Bono E, Serious Creativity, Harper Collins 1992 - interesting; non-engineering focus Cross N, Engineering Design Methods, Wiley 1994Roozenburg NFM & Eekels J, Product Design: Fundamentals and Methods, Wiley 1995Wilson CC, Kennedy ME & Trammell CJ, Superior Product Development, Blackwell 1996Pahl G & Beitz W, Engineering Design, Springer 1996 - comprehensive, highly recommended

More specific/advanced texts, cited mainly to show the topics available :

Wagner HM, Principles of Operations Research, Prentice Hall 1975Mildren KW, Use of Engineering Literature, Butterworths 1976Gordon JE, Structures - or Why Things Don't Fall Down, Penguin 1978 - light enlightenment Flurscheim C ed, Industrial Design in Engineering, Springer 1983Tabucanon MT, Multiple Criteria Decision Making in Industry, Elsevier 1988Shigley JE & Mischke CR, Mechanical Engineering Design, McGraw-Hill 1989 - component detailsJensen PW, Classical & Modern Mechanisms for Engineers & Inventors, Dekker 1991Chironis N, Mechanisms & Mechanical Devices Sourcebook, McGraw-Hill 1991 - browseworthyCushman WH, Human Factors in Product Design, Elsevier 1991 - man-machine interactionHubka V & Eder WE, Design Science, Springer 1995 - more advanced systematic design methods

Periodicals (vital for increasing awareness of current issues, techniques & hardware) :

Engineering, Design Engineering, Machine Design , New Scientist, Scientific American , Engi-neers Australia, Professional Engineering, etc

Learned societies' journals (often rather advanced for the emerging designer) :

Trans. American Society of Mechanical Engineers, Proc. Institution of Mechanical Engineers, etc

Standards and Codes of Practice :

VDI 2221: Systematic Approach to the Design of Technical Systems and Products

Videos :

Warman Design & Build Competition - a number of videos of recent heats and finals is available

Design for Manufacture: The Case of a Pump Body, Monash Teaching Services Unit 1992

JCH Roberts, Creativity Design 31

iv redesign?d. Are there any other problems?e. What relationship exists between the main prob-

lem and the sub-problem?f. What effect has time on the problem?g. Is the problem really worth solving?

4. Hypothesis. Purpose - To enable every conceivableaspect of the problem to be imaginatively consideredand without prior evaluation. The individual relaxeshis judgment and deliberately allows his imaginationto wander, at the same time keeping the problemclearly in focus. He tries to get away from conven-tional possibilities by imagining new combinations,features that are opposite to existing ones etc.

5. Incubation. Purpose - To provide opportunity for theindividual's subconscious to unravel a connectingthread between his mental 'storehouse' and the prob-lem needs. This stage is a subliminal repetition of thedeliberate hypothesising of the previous stage - theEureka factor. No conscious action is required thougha period of time must be allowed for gestation.

6. Synthesis. Purpose - To reassemble the componentparts of the problem. The ideas developed during ear-lier stages are carefully evaluated from different view-points, i.e. feasibility, practicability, availability, cost,time etc.

7. Verification. Purpose - To evaluate the resultantideas and ensure that the final solution will really workand is acceptable to prevailing conditions.

The process of accumulating tentative ideas is an indis-pensable part of any problem solving project, whether itbe creating a new household appliance or in solving acomplex production problem. Almost always it is neces-sary to think up a number of unusable ideas in order toarrive at one that may work.

Similarly, the process of analysing these ideas is impor-tant. Often the mere breaking down of the problem mayreveal the answer, or show that the real problem is otherthan the one originally considered.

ROAD-BLOCKS TO PROBLEM APPRECIATION

Success in creative problem solving partly depends upona clear understanding of the actual problem involved.Often the definition of a problem is more essential thanits solution, which may be merely a matter of mathemati-cal or experimental skill.

Inability to understand a particular problem clearly isoften due to one or more of the following perceptualblocks:

1. Difficulty in isolating the problem. The individual isunable to separate the real problem from related prob-lems. Too often he is not tackling the real problem atall.

2. Difficulty caused by over-narrowing the problem. Littleor no attention is paid to the environment surroundingthe problem. The individual is thus unable to see theproblem in any other terms than those presented in it.For example, try and solve this problem. (You can doanything you like with the 'nines'.) How can you makefour nines equal one hundred?

3. Inability to define terms. Most problems are describedby means of words and these can, and often do,cause confusion. The manner in which words areused often makes a problem more difficult to solve.The more words we use to describe the problem andwhich bring stereotyped or particular images to mind,the more our thinking is likely to be influenced alongthese particular lines. For example: 'Two Indiansstood on a hill. One is the father of the other's son!What is their relationship?' Nine out of ten who seethis simple problem for the first time find it difficult,and invariably produce the wrong answer. If, however,the problem is rephrased, the correct answerbecomes immediately obvious, e.g. 'Two indiansstood on a hill. He is the father of her son. What istheir relationship?'

4. Difficulty in seeing remote relationships. This is theability to form or transfer concepts. It necessitates theability to look at different objects or situations, or evenproblems, so as to determine what they have incommon. The common factor may not be a physicallikeness. It may be one of relationship, having func-tional similarities in one situation which apply inanother.

5. Failure to use all senses in observing. Ordinarily, wethink of observations as seeing only. However, it oftenhelps in solving problems to use the complementarysenses of smell, taste, hearing, feeling, etc. For exam-ple, the maintenance department of one well-knownfirm uses tape-recorded noises of their productionmachinery to diagnose machine faults because theyfound that watching a machine in operation proveddistracting and audible indications to possible faultswent by unheard.

6. Difficulty in not investigating the obvious. Once anindividual becomes accustomed to looking at certainsituations and problems in a particular way, itbecomes increasingly difficult to depart from that rou-tine. Because of familiarity he really ceases to 'see'them in their fullest detail.The designer's first reaction to the task of designing anew part is to look for components which are physi-cally similar, and provide the same function ratherthan ask himself: 'What better, simpler or cheapermethod could be used to achieve the same endresult?'All too often designers, engineers, managers, etc.search for the complicated solution when the simplewill suffice; they develop complex designs and pro-cesses because they think they are more in keepingwith modern technological standards when, in fact,

Businesses concerned with the manufacture or selling ofcommodities must look closely at five basic problems:

1. How to manufacture products more cheaply.2. How to get more money for them.3. How to enlarge the uses of the products and thereby

improve sales.4. How to improve profitability.5. How to find new products.

The successful solution of these problems determinesthe profitability of the business.

Unfortunately, most business executives are untrained inproblem solving techniques and therefore tackle theirdaily problems in a non-creative manner.

WHAT IS A PROBLEM?

Basically, there are two types of problem (a) Analyticaland (b) Creative. Both can be further defined as follows:

a. The analytical problem. This is defined exactly andwith a clue to the probable answer, e.g. A boy says 'Ihave as many brothers as sisters.' His older sisterreplies 'I have twice as many brothers as sisters.' Howmany brothers and sisters constitute the family? Thequestion is solved by the individual applying one ormore mathematical skills, by direct experimentation,or by means of 'trial and error' techniques. Thereexists only one correct answer to the stated problem(see last page).

b. The creative problem. In general terms this can bedefined as a situation demanding the positive applica-tion of an individual's creative abilities. Usually, theproblem is stated in very general terms, e.g. 'Whatwould happen if we changed the material?' or 'Howcan we improve reliability, or reduce costs' etc. Nohint of the answer is given. Unlike (a) the Creativeproblem can be solved by using many different crea-tive approaches which, again unlike the Analyticalproblem, will produce several possible solutions, allperhaps equally good.

TWO APPROACHES TO SOLVINGCREATIVE PROBLEMS

Broadly, there are two different approaches to solvingproblems creatively, namely:

1. Imagineering - a process involving a very free use ofthe imagination, by which the problem-solver virtuallycatapults his thinking into the unknown and thenworks - or engineers - his way back to reality. Thisnecessitates considerable control of the user's evalu-ating and censoring attitudes.

2. Organised thinking - a system of rules and proce-dures developed so that tested creative techniquescan be applied to problems on a step-by-step basis.

The Organised Thinking approach is more applicable toindustry in that engineers, managers and others canquickly be taught to use the techniques involved in acreative manner. Three separate processes are involved,each containing a number of individual steps. Ideally theprocess comprises the following procedures, each ofwhich, regardless of sequence, calls for deliberate effortand creative imagination on the part of the thinker:

a. Fact-finding - involves locating and defining the spe-cific problem then collecting and analysing pertinentinformation.

b. Idea-finding - involves thinking up tentative ideas aspossible pointers to acceptable solutions and thenselecting the most promising ideas, adding others,modifying them or arranging them in different combi-nations.

c. Solution-finding - involves verifying tentative solutionsby tests and other means before deciding upon, andimplementing, the final solutions.

In a sense, this 'approach', coupled with use of variouscreative devices, provides a pattern for attacking a prob-lem effectively, and one that enables the user to see itclearly and in its simplest terms.

STANDARD PROBLEM SOLVING DEVICES

For effectiveness the 'Organised Thinking' approach toproblem solving is divided into seven progressive stagesof attack:

1. Orientation. Purpose - To ensure that the individualis thoroughly familiar with the problem and, havingdecided whether it is analytical or creative, can adjusthis method of treatment accordingly.

2. Preparation. Purpose - To ensure that the individual'smental 'storehouse' is suitably equipped to tackle theproblem by delving into the background of the matterand absorbing maximum factual information concern-ing the problem and its history.

3. Analysis. Purpose - Here the individual systemati-cally evaluates the problem so that he can decidewhether or not there are any sub-problems to accom-plish. He may ask himself the following questions :a. Why is there a problem?b. What effects or consequences are directly and

indirectly attributable to the problem?c. Are the troublesome items essential? Can they be

dispensed without creating further problems. Canthe problem by solved by :i substituting a different part?ii replacing a part?iii repair ?

4. How to solve problems creatively

JCH Roberts, Creativity Design 32

duce the largest number of different answers. Avoid useof image-provoking words when preparing final descrip-tion)

6. How can the float be repaired quickly?7. How can further loss of buoyancy be prevented?8. How can buoyancy be restored and maintained?9. How can the valve be eliminated?

On the basis of the foregoing, the problem can bedefined exactly as: 'How can the float's buoyancy berestored and maintained?'

Stage C. Consider the problem generally. (Have anyaspects been overlooked?)

10. Is there a 'spare' float?11 What other part can be used to reliably replace the

float?12. How can the hole be enlarged to facilitate discov-

ery?13. Can the float be modified - with the water in it - to

nullify the effect of the hole?14. What other device can be used with the float to com-

pensate for the hole's effect?15. What liquids would penetrate the hole more easily

than water?16. What else can help overcome the loss of buoyancy?

Stage D. Dream-up all possible ideas relevant to theproblem but do not evaluate.

17. Use a microscope to look for the hole?18. Cover the float with fine white chalk powder?19. Replace the float with a block of wood?20. Make an identical float out of plastics material?21. X-ray the float to find the hole?22. Drill another hole in the float and pump it full of col-

oured liquid?23. Heat float and watch for release of steam from hole.24. Submerge float in water and fill with compressed air,

watch for bubbles indicating hole.25. Apply adhesive rubber material to entire surface of

float. Magnetise whole surface of float, then coverwith powdered iron. Put float in vacuum tank andwatch for seepage.

26. Put float inside tight-fitting rubber balloon to seal-offleak.

Stage E and F. Evaluate ideas produced and refine byconsidering such factors as feasibility, practicability, cost,time, who does what with what etc. in respect of eachidea.

FINAL SOLUTION

Play the flame from an ordinary welding torch over thebottom surface of the float until the water boils andcreates steam. Roll the float until the leak is under waterand watch for emission of pressured water. Then weld uprevealed hole. Note: When the described method wasused in practice the steam pressure drove out a thin jet

of water a foot long.

Now try creatively solving this problem:

Problem 1. The youngsters in a large city delighted instealing electric light bulbs from the underground railwaycarriages, often leaving a train in semi or total darknessby the end of the journey. The railway company could notafford to employ more personnel to watch for bulb-snatchers. Neither could the police department afford tostation men in underground trains.

How can the pilferage be halted? (Answers on last page)

these are not needed and represent areas of unnec-essary cost.

7. Failure to distinguish between 'cause' and 'effect'.Most people believe quite confidently that they knowthe difference between 'cause' and 'effect'. Too oftenhowever, the distinctions are not clear-cut and confu-sion results.

FOURTEEN RULES FOR SOLVINGPROBLEMS CREATIVELY

From the foregoing certain relatively simple rules can bedeveloped which, when used in an orderly progressivemanner, can prove invaluable aids for effective problem-solving.

Rule 1. Understand the problem clearly - To understandthe problem so that its purpose is clear, visualisethe problem as a whole, clearly and vividly. For themoment, ignore confusing details and side issues.

Rule 2. Define the problem exactly - If the problem isclearly understood it should be possible to state itclearly in ten words or less. Make sure that thewords used communicate a crystal clear picture ofthe problem and nothing else.

Rule 3. Think about the problem generally - illuminate thewhole problem by thinking about all its aspects.Vital angles of a problem are often so obscure thatthey escape notice. Emphasise different parts,examine different details, note the same detailsrepeatedly but in different ways. Combine thedetails differently, approach them from differentsides, different viewpoints, etc.

Rule 4. Isolate the principal parts of the problem - List theprincipal parts of the problern. Consider themsingly, in turn, in various combinations, and byrelating each detail to other details and to thewhole of the problem.

Rule 5. Think up what information might help to solve theproblem - In general terms consider the variouskinds of information and material that might provehelpful in solving the problem.

Rule 6. List the most likely sources of information -Having considered the various types of informationmost likely to be of help, list them and decide inwhat order they are to be researched.

Rule 7. Dream up all possible ideas that keep to theproblem - Apply imagination creatively to produceas many ideas as possible freely. Do not, at thisstage, evaluate ideas. Remember different wordsproduce different mental images and, therefore,different ideas. Try using different words todescribe the problem and see what ideas are pro-duced.

Rule 8. Change direction and let your subconscious takethe problem over - Once you seem to havestopped producing ideas, change direction byswitching to some other type of work completelyunassociated with the problem, but which will fullyoccupy thinking. After a while, return to the prob-lem and see what ideas result from your subcon-

scious mind's attack on the problem.Rule 9. Select the ideas most likely to lead to an accepta-

ble solution - Evaluate each idea by judicial think-ing and so remove valueless ideas.

Rule 10. Think up all possible ways to test the solution -Use creative thinking techniques to produce possi-ble ways for testing the proposed solution.

Rule 11. Choose the soundest ways to test the final solu-tion - List these ways in the form of a check-listagainst which the solution can be reliably tested.

Rule 12. Imagine all possible contingencies - Eventhough the final solution has been corroborated byexperiment, it is helpful to envisage what mighthappen as a result of its adoption. For example, animprovement to new products must also be pro-cessed on the basis of competitors' actions tocounteract the advantages the improvementbrings.

Rule 13. Decide on the final answer - Carefully weigh upall the pros and cons to ensure that the solutionfully satisfies the problem.

Rule 14. Activate the solution - Expedite whatever actionis essential for the successful implementation ofthe solution.

EXAMPLE OF CREATIVE PROBLEMSOLVING TECHNIQUES APPLIEDTO A SPECIFIC PROBLEM

Problem as stated to the works engineer of a chemicalworks:

A stainless steel, cylindrical, one gallon capacity valve-control float, specially made by a German firm, had verygradually lost buoyancy. This was detected just in time toavoid serious production difficulties.

The works engineer instructed the maintenance foremanto have the float examined immediately. On shaking thefloat, the foreman estimated that about a quart of waterremained inside the float - but none came out of thestainless steel shell. He reported his findings to the worksengineer and although both carefully examined the float,they could see no leak.

SOLVING THE PROBLEM

Stage A. Specify the problem (by listing all the creativetype of questions the problem suggests)

1. How can the leak be found?2. How can the hole be sealed?3. What material can be used to coat the entire float

without having to find the hole?4. How can the hole be kept at the top of the float, so

that the water will not seep in?5. How can I seal the hole up?

Stage B. Define the problem exactly. (What is the realproblem? What is the basic objective? What preciselyhas to be accomplished? Apply questions that will pro-

JCH Roberts, Creativity Design 33

PREPARATION FOR GROUP BRAINSTORMING

The size of a brainstorming group depends upon theextent and type of organisation it is to serve. Ideally, thegroup should consist of a Chairman, a Recorder (other-wise known as the 'idea collector'), six regular 'core'members and about six guests. Certain qualificationrequirements govern the selection of a group's members,namely :

1. Group Chairman. The Group Chairman, or Leader,should be trained in advance of his function. Preferablyhe should have taken a course in creative thinking andhave participated, as a group member, in various brain-storming sessions. He should NOT be a senior memberof the organisation's management.

2. Recorder. The only contribution required of theRecorder is that he or she be able to record all ideasquickly and reliably, and without interruption of the idea-tion proceedings. Often a tape-recorder can be usedeffectively for this function.

3. 'Core' Members. As the 'core' members are thegroup's pace-setters, they must be people who haverepeatedly demonstrated their ability to produce originalideas or suggestions.

4. Guests. The guests should be invited from variousdepartments of either the organisation or its associateconcerns. A different group of guests should be invitedfor each brainstorming session. This rotation helps tospread a creative spirit throughout an organisation, andprevents the development of a rigid pattern of thinking,such as would occur if the same guests were invited toeach and every session.

Often it is helpful to include among the chosen guests atleast one, or possibly two, people who know nothing atall about the problem under discussion. People withoutexperience in a particular field bring a new, often valu-able viewpoint to the problem. Several organisationshave found it helpful to include women in each brain-storm group. The women try to 'out-ideate' the men, andvice versa. This tends to induce an extra spirit of rivalrywhich stimulates the flow of ideas.

5, Pre-conditioning. Since guests who have neverbefore participated in brainstorming sessions are unfamil-iar with the various creative techniques involved, a rela-tively thorough orientation is recommended. Ideally, thisshould be accomplished in a 30 minute briefing lecturewhich covers the basic principles of ideation as well asbrainstorming procedures.

6. Attendance of Top Management. Experience indi-cates that brainstorming sessions tend to be less produc-tive when a high-ranking member of the organisation'smanagement is present. They tend by facial expression,or otherwise, to induce an inferiority complex on the partof the remaining members of the group, and thus discou-

rage 'free-wheeling '.

7. Size of Group. Although larger sized groups havebeen used for brainstorming sessions, experience indi-cates that the optimum size of a group is about a dozen.

HOW TO CONDUCT ABRAINSTORMING SESSION

Essential to the success of any group brainstorming ses-sion is an atmosphere that will allow each member of thegroup to depart freely from his logical and conformingmental control, and to assume the framework establishedby the Group Chairman.

(a) Preliminary procedures

1. Simplify the problem - The first task of the ChairmanIn organlsing a brainstorming session is to ensure thatthe problem to be brainstormed is presented in a mannerbest calculated to produce the largest number of alterna-tive ideas. This he does by carefully examining the prob-lem with the individual who submitted it. His objective isto ensure that the problem is stated simply and specifi-cally. It should not be an all-encompassing problem, suchas 'how to reduce manufacturing costs?' Such a problemmust first be analysed into its numerous, more specificsub-problems like 'Ideas for reducing material costs';'Ideas for reducing machinery costs'; 'Ideas for reducingdrilling costs' etc., which may be more amenable to thebrainstorming process.

2. Highlight the problem's background - Once a singlespecific problem has been selected and its definitionagreed upon, the individual who submitted it must pro-vide the informative background necessary to familiarisethe group's participants with the problem. This he doesby providing the material for a one-page memorandum,to be written by the group Chairman, which high-lightsthe background of the problem, states the problem in itssimplest terms, and gives at least two examples of thetype of ideas which are sought (as shown opposite).

3. Select and invite participants - The Chairman thenselects his group participants in the light of the nature ofthe problem to be solved. At least two days in advance ofthe session the participants are invited by telephone orotherwise and are simultaneously supplied with copies ofthe 'background' memorandum, so as to allow the partici-pants to 'sleep on the problem' thus allowing incubationto enhance the working of free association of ideas.

Similarly, the Chairman should develop in advance hisown list of possible solutions to the problem so that, ifand when the session slows down or gets off track, he isin a position to reactivate the flow of ideas by contributingsome of his own. The same objective can also beachieved by the Chairman introducing idea-spurringquestions.

The formation of any business begins with someone pro-ducing the initial idea for the project. The continued suc-cess of an established business depends upon thenumber and quality of the ideas fed into it. Without a con-tinual flow of new ideas, a business cannot function prof-itably or expand successfully and must, therefore eventu-ally fade into total obscurity.

Ideas for a new business project, a new product, ameans of reducing manufacturing costs, or for solvingindustrial labour problems, begin in the human mind.Most people conceive their ideas unconsciously, andbecause they are unaware of the mental mechanics thatcaused the 'idea' to be produced, they cannot repeat theideation process to produce further profitable ideas atwill.

Fortunately, there are available established creative tech-niques which, when used correctly, do enable a personto produce a large number of first-class ideas at will. Onesuch creative technique, and probably the most widelyused in American industry, is 'brainstorming'.

WHAT IS 'BRAINSTORMING'?

This unique technique is used for the deliberate produc-tion of a large number of ideas in the shortest possibletime.

Developed some 28 years ago by Alex F Osborn, co-founder of one of America's most successful advertisingagencies, 'brainstorming' originally referred only to delib-erate ideation by groups. As a result of practical usage,modification and research, the term has evolved untilnow it refers to the basic principle of suspended judg-ment which scientific research has proved to be highlyproductive in individual creative effort as well as in groupeffort.

Deferred judgment functions on the basis of the deliber-ate alteration of normal thought processes. Instead oftrying to think critically and imaginatively at one and thesame time, the creative thinker uses his creative mindand his judicial mind separately. In other words judgmentis not allowed to jam creative imagination.

To test this principle a group of ideators jointly brain-stormed a particular problem while, simultaneously, anequal number of ideators individually attacked the sameproblem in the conventional way - but without deferringjudgment. The results showed that, in the same amountof time, the groups which applied the deferred judgmentprinciple produced 75 per cent more good ideas than didthe groups who judged each idea separately before pro-ducing another.

GENERAL PRINCIPLES OF BRAINSTORMING

Although the detailed procedures used in applying the'brainstorming' technique tend to vary in line with the spe-cial needs of the user, in almost all applications the fol-lowing general rules apply when using the techniques.

Rule 1. Considerably more ideas will be produced if criti-cal judgment is entirely eliminated during the ideaproduction process.

Because education and experience have trained mostpeople to think judicially (i.e. critically) rather than imagi-natively, the flow of ideas they are capable of producingis impeded because they apply their critical evaluativefaculties too soon. They are more concerned withassessing the value of individual ideas than with creatinga large number of alternative ideas. By deferring judg-ment during the idea-producing process, however, alter-native ideas can be produced for a longer period, andtherefore, a considerably larger number of ideas areavailable for evaluation at the end of the ideation period.

Rule 2. Group ideation can add to an individual's ideaoutput.

Usually a person's experience of joint thinking is gainedfrom his attendance at the traditional kind of conferenceor lecture, where original ideas are neither asked for norencouraged.

The principal value of group brainstorming lies in the factthat a brainstorming session, when properly conducted,can produce far more good ideas than a conventionalconference - and in much less time. A striking example ofthis can be instanced by a brainstorming session held bythe American Cyanamid Company which produced 92ideas in a single 15 minute session - more than 6 ideasper minute, and an average of over 8 ideas per personattending the session.

Group brainstorming procedures call for individual idea-tion both before and after each session. Since a combi-nation of these two methods of approach to 'brainstorm-ing' usually produces maximum results, an alternationbetween group ideation and individual ideation is recom-mended as the best means of obtaining really effectiveresults.

Rule 3. The more ideas that can be created, the betterthe overall results.

Characteristic of brainstorming is the fact that by drivingfor a few more ideas, you get far more. Ideas create stillmore ideas. First you get 30 and you want to get 60.Then when you get 60 you want to get 75. Probablyseven of those last 15 ideas are first rate, and wouldnever have been conceived had the quantity of ideasrequired been, say, limited to only 30. The more sugges-tions produced during a brainstorming session, thegreater the chances of producing first-class ideas.

5. Brainstorming positively develops creative ideas

JCH Roberts, Creativity Design 34

9. Post-session activities - The following day, thegroup's Recorder either sees or phones the participantsto obtain from them any further suggestions they mayhave to offer.

Because each of the participants has 'slept' on the prob-lem, some of the most valuable ideas are likely to be pro-duced as a result of individual ideation pursuant to thesession.

Many companies using brainstorming adopt a multipleprogramme for gathering post-session ideas by :

a. Sending typed minutes of each session to each partici-pant, with plenty of blank space provided so that they canadd further suggestions that have since occurred tothem.b. Circulating a round-robin idea-folder to be passed toeach participant within a definite time, and with instruc-tions that new ideas are to be added in handwriting.

Once all ideas have been gathered from participants, theRecorder finally prepares a triple spaced typewritten listof all the ideas suggesting during the brainstorm session.Afterwards the group Chairman edits the list, making surethat each idea is correctly and properly described. Healso classifies the ideas into logical categories. Undereach classification he gives the numbers of the individualideas (as taken from the original list) which should beincluded in that section. Once this task has been accom-plished, the ideas accumulated are then ready for evalua-tion and final selection.

10. Evaluation of tdeas - Evaluating the final selectionof ideas is generally done by the individual responsiblefor submitting the problem that has been brainstormed.He selects those ideas which seem to be the most prom-ising, at the same time seeking to develop more andbetter ideas by combining one idea with another, or aseries of ideas, or by reprocessing the ideas throughcombination, elaboration or other creative means. Waryof rejecting apparently farfetched ideas too quickly, heconsiders the seemingly silliest idea from all possibleviewpoints. As a result, the most unpracticable idea isoften developed into a first-rate suggestion.

To facilitate the evaluation of ideas, it is often helpful tocreate a check-list of the various criteria to which therequired ideas should conform and so as to ensure thateach idea is considered from various essential view-points. The following examples illustrate the (a) type ofcheck-list that could be used to evaluate ideas for a newproduct and (b) that could be applied to ideas for improv-ing the manufacture of a particular product.

a. Criteria for evaluating ideas for a new product

1. Is the idea simple enough?2. Is it compatible with human nature?3. Is it timely?4. Is it feasible?5. Can it be duplicated by competition?

6. Is its application limited?7. Is it costly to produce?8. Is it safe? etc.

b. Check-list of criteria for ideas relating to improving themanufacture of a particular product.

1. Will it increase production - or improve quality?2. Is it an improvement over the present tools and

machinery?3. Does it improve methods of operation, maintenance

or construction?4. Does it permit a more effective utilisation of man-

power?5. Does it prevent waste or conserve materials?6. Does it eliminate unnecessary work?7. Does it improve present methods?8. Does it reduce costs?9. Will it improve working conditions?10 Does it improve safety?11 Does it improve labour relations?

Pertinent crlteria can be added to each check-list and asdictated by the nature of the ideas to be evaluated.

11. The presentation of selected ideas - Havingdlscarded all worthless ideas, the valuer then prepares,in order of merit, a short list of what he considers to bethe ideas most likely to solve the problem satisfactorily,and which he is prepared to recommend to his manage-ment.

Problem 1.

A small sized village church congregation of limited finan-cial means faced the problem of repainting its church hallin time for its annual anniversary and village fete. Toreduce costs, the men of the congregation decided to dothe painting during their spare time.

Two weeks before the day of the fete most of the hall stillremained unpainted. Most of the male volunteersseemed disinterested in honouring their promises and itbecame evident to the worried 'Repair Chairman' that hewould have to act quickly, and creatively, if the hall wasto be painted in time for the much publicised anniversary.Studying the expanse to be painted, he soon hit upon anidea for ensuring that the work was completed in timeand at no extra cost.

Putting yourself in the 'Repair Chairman's' shoes, whatwould you do?

(b) Group session procedure

4. State the basic rules - Once the group is assembledthe Chairman outlines the four basic rules to be adheredto throughout the session, namely :

Rule 1. Judicial judgment of ideas is not allowed - Criti-cism of ideas produced MUST be withheld untillater. (Someone, or another group, will evaluatethe ideas later. )

Rule 2. 'Free-Wheeling' is welcomed - The wilder theidea, the better. It is easier to moderate than toproduce.

Rule 3. Quantity of ideas is wanted - The greater thenumber of ideas, the more likelihood of good ones.

Rule 4. Combination and improvement are sought - Inaddition to contributing their own ideas, eachmember suggests how ideas contributed by otherscan be improved, or how two or more ideas could,with advantage, be combined.

A large placard on the wall - positioned so that it can beeasily seen by each member of the group - states theserules. In front of the Chairman is a handbell which he willring whenever any member of the group violates a rule.

5. The Practice Session - Before seriously brainstorm-ing the given problem, a 10 minute practice, or 'warm-up'session is provided. This exercise should consist ofbrainstorming some simple problem, such as: 'In whatways could office chairs be improved?'

6. Brainstorming the problem - At the end of 10 min-utes the Chairman calls a halt to the practice session. Hethen repeats the rules of the session, presents the prob-lem simply and then calls for suggestions in solution ofthe stated problem.

When several hands are raised simultaneously there isthe danger that in absorbing ideas propounded, the laterparticipants may forget their ideas. It is recommended,therefore that each member be provided with a scratch-pad so that he can jot down details of ideas he plans tooffer when his turn comes.

Participants are not allowed to bring written ideas into themeeting. Only one idea should be offered at a time byany participant, otherwise the pace of the session will beimpeded.

7. Recording the ideas produced - The job of theRecorder is to list every idea suggested during the brain-storm session. Ideas should not be taken down word forword, but briefly reported. The Recorder should beseated next to the Chairman so that he, or she, is indirect line of communication with him, as well as with theother members of the group.

Some firms tape-record their brainstorm sessions, thusenabling the Recorder to re-check the list of ideas whichhave been reported during the session.

Regardless of how the ideas are recorded, it is helpful towrite a few on a black-board during the progress of thesession in order to provide visual stimulation to the partic-ipants. This enables the Chairman to refer the group tothese listed ideas in case of lulls, or to encourage idea-combinations by free-association of ideas.

Each idea recorded should be numbered so as to enablethe Chairman to know how many ideas have been pro-duced up to any point of time in the session. If he haspreviously decided upon a particular quota of ideas forthe session, he is then able to know at any time howclose he is to his target.

No idea should be identified with the participant responsi-ble for suggesting it. Possibly, the very same idea mayhave been previously thought up by another participant.Or the idea may have resulted directly from an earliersuggestion made by someone else.

8. Terminating the sesslon - Experience indicates thatthe optimum length for a group brainstorming session isabout 40 minutes. If more time is needed, it is a goodidea to break up the problem into smaller problems, eachof which can be handled satisfactorily in a 40 minute ses-sion.

When closing the session the Chairman expresses hisappreciation and requests the participants to keep theproblem in their minds until the next day when they willbe asked for their afterthoughts.

To: Mr. John Tagmarch,Section Leader - Airframe Design Office.

You are invited to participate in a brainstorming session onTuesday, May 14th at 3.00 p.m. in the Staff Conference Room.

The problem as stated by our Technical Director is as follows:"What ways and means can you think up for removing ice from

aircraft and helicopters that are parked in the open".Ice formation has been a peril and hindrance in the field of

aviation not only to airborne aircraft, but in some regions even toaircraft that are parked in the open. In such regions, many air-craft have been rendered useless while awaiting a signal to repelan impending attack or to depart on a mission of mercy. So anyideas for new equipment or protective procedures will be appre-ciated.

Exa mples of the kind of ideas we are seeking are as follows :1. Spray aircraft with liquid which will quickly dissolve ice.2. Invent electrical apparatus shaped like a mattress which may

be placed under aircraft to heat it, thus melting ice.3. Invent ice extinguishers containing a gas which would react

with ice to dissolve it.4. Coat aircraft with ice resistant paint which would allow any

ice that formed to be quickly chipped off.5. Park aircraft over grids that allow heated air to be blown

upwards and over total external surface of parked aircraft.6. Put hot water bottles on aircraft to melt ice.

FRANK DOBSON Brainstorm Panel Chairman

Example of a suitable 'background' memorandum

JCH Roberts, Creativity Design 35

If Attribute listing is to be used effectively, the followingfour progressive steps must be applied:

Stage 1 Isolate, and list all the principal attributes of theproduct, process, or item under review,

Stage 2 Carefully consider each 'attribute' in turn, ingroups, and then in various combinations bychanging the combination in every conceivablemanner. Relate each particular characteristic tothe other attributes and each to the whole of theproblem in order to see how it can be changedor improved.

Stage 3 List all the ideas produced.Stage 4 Evaluate each idea in the light of the limitations

of the problem and list - in order of merit - thoseideas worthy of serious consideration.

Example

Suggest design improvements for the common tele-phone. The telephone's main attributes can be listed asfollows:

1. Coloured, usually black, cream, or grey.2. Comprises two components, a handset (incorporating

a microphone and speaker) and base.3. Circular dial for communicating number required.4. Constructed mainly of plastics.5. Electrically operated.

Carefully and creative consideration of the product'spresent attributes suggest the following changes:

Present Attribute Possible Change1. Coloured, usually 1. Use red or other colours.

black, cream or 2. Could be two-toned.grey 3. Could incorporate visual

signs (e.g. polka dot, abstract art, personalised designs, owner's name or initials). 4. Could be self-illuminating (for use at night).2. Comprises hand- 1. Design so that handset fits

set and base comfortably to user's fingers.

2. Incorporate device to allow user to hold speaker to ear without use of hands.

3. Make base square, oval, round, higher, or lower.

4. Make handset and base a single, self-contained unit.

5. Incorporate TV screen so that user can see speaker.

6. Incorporate separate recording and playback mechanism.

7. Incorporate independentand variable volume control.

3. Circular com- 1. Replace with press-buttonmunicating dial system.

2. Use pictures instead of

numbers for speedy contactwith emergency services.

3. Incorporate a simple comp-uter to interpret vocal com-mands.

4. Incorporate automatic faultindicator.

4. Constructed 1. Could be made of metal,mainly of plastics glass, wood, compressed

paper, hard rubber, or otherplastics.

5. Electrically 1. Redesign to operate onoperated independent power supply.

A quick evaluation suggests that the conventionaldomestic telephone could be considerably improved byredesigning it as a single, self-contained, easily-handledunit that is self-illuminating and which incorporates thefollowing additional useful features:

Small TV screen.Separate recording and playback device.Independent volume control.Press-button dialing system with pictures to indicateemergency services or computer to interpret vocalcommands.

(b) The input-output method

This method generally operates when the product, orproblem, under review involves the use of one or moreforms of energy. Basically, the method involves present-ing the problem in the form of a dynamic system compris-ing input, output, and specifications. These are general-ised, and questions are asked about all three divisions insuch a way as to suggest variant approaches to possibleanswers. For example, if the problem is the discovery ofmeans for improving an electric clothes dryer, the INPUTis electric heat, and the OUTPUT covers the dry clothes,or, by finer analysis, evaporated water or water vapour.

The specifications are determined by the objectiverequired, which might be a faster machine.

Often, when using this method, the 'output' feature canbe used as a new 'input' aimed toward the specifiedobjective and the solution of the given problem. Forinstance, in the case of the electric clothes dryer theevolved water vapour could be conducted over an indica-tor which switches the machine off when the vapourreaches a predetermined low concentration, thus utilisingthe 'output' as a new 'input'

The progressive steps involved in applying this particularmethod to a problem are as follows:

Step 1- Establish the desired end result, i.e. the 'output'or solution required.

Step 2- Determine the 'input', i.e. the various forms ofenergy available and which are likely to be ofuse in providing the desired 'output '.

Step 3- Decide the problem's limiting conditions.Step 4- Analyse the 'input' in order to decide how the

Finding a good new product idea is the cornerstone ofsuccessful product development. Without good ideas todevelop, there can be no successful new products. Busi-ness firms who fail because their products are out-dated,too costly, or because they have been unable to developnew products, do so mainly because their key employeeshave fallen down on ideation techniques for the produc-tion of new product ideas.

ANALYTICAL TECHNIQUES

Today, not even a new-product development specialistcan afford to depend upon the accidental discovery ofideas. He must know, and be able to apply, the variouscreative techniques available for producing good ideasdeliberately and in large numbers.

These techniques are divided into three distinct catego-ries, namely:

1. Analytical techniques.2. Free association techniques.3. Forced relations techniques.

Each category contains a number of creative 'tools' ormethods each designed specially to stimulate creativethinking in the individual and thereby help him producegood ideas deliberately.

The most popular analytical methods, or creative 'tools ',are:

(a)Attribute listing - developed by Professor L. Crawfordof the University of Nebraska.

(b) Input-output - developed by the Creative EngineeringDepartment of the General Electric Co., USA.

(c) Morphological analysis - developed by Dr. F. Zwickyof the California Institute of Technology's CreativeEngineering Department.

Each of the above methods employs a logical step-by-step approach to the creative solution of the problem towhich it is applied, and forces the individual concerned tothink in a creative manner.

Each method involves the serious application of twobasic rules, namely:

1. All ideas conceived, no matter how impractical theymay appear to be, MUST be seriously considered.

2. Prior judgment and evaluation of individual ideas isNOT permitted whilst ideas are being produced.

Although the basic rules for each of the three techniquesare the same, their approach to the problem is entirelydifferent.

(a) Attribute listing

Generally, this method is employed when new ideas arerequired to improve an existing product, process or fieldof usage. The underlying principle is that a 'new' idea is acombination of existing ideas to which various new attrib-utes have been applied.

First, however, it is necessary to have a clear under-standing of what constitutes an 'attribute'. Attributes aredefined along three dimensions:

1. The material of which the particular item is made (i.e.type, physical properties, etc.).

2. The method(s) of manufacture (i.e. stamped, mould-ed, welded, etc. ).

3. The specific way(s) in which the item is used (i.e.what it does, what function it performs, how it is used,etc.).

'Attributes' are NOT opinions concerning the item undercreative review, e.g. 'difficult to hold', 'too heavy'. 'tooslow', 'too dangerous', etc. These are opinions and notattributes. It is imperative, when using this method toensure that the various attributes selected for creativeattack relate to the immediate physical characteristics,manufacturing methods, and usage purposes of the par-ticular items under review, Usually unique attributes tendto produce the greatest opportunity for original ideas,whilst those attributes common to most products (i.e.colour, shape, size, weight, etc. ) tend to yield opportuni-ties for fewer extraordinary or original ideas.

For further clarification of what is meant by 'Attribute List-ing', and how it is used to improve an existing product, letus consider the old-fashioned, conventional woodenscrewdriver. Its main attributes are:

1. A round, steel shank.2. A wooden handle riveted to the shank.3. A steel, wedge shaped, end for engaging the slot in

the screw-head.4. It is manually operated in that the screw driving torque

is provided with a physical, twisting action.

Over the years each attribute has been changed manytimes to provide supposedly, a new and better screw-driver. The round steel shank has been changed to ahexagonal shape, so that a spanner could be applied to itand so increase the applied torque. The wooden handlehas been replaced by a moulded rubber or plasticshandle, thus reducing breakage hazards and dangerfrom electrical shock.The wedge-shaped end has been modified to fit all kindsof screw heads. Pneumatic or electric power has beensubstituted for physical power. Other modificationsinclude detachable ends, containers built into the handle,reversible, push-action manual drives, etc.

6. Practical techniques for producing profitable ideas

JCH Roberts, Creativity Design 36

wheels, rollers.

4(c) Source of motive power.Subdivide to give - atomic energy, electricity, air,water, steam, gas, cable or endless belt, IC engineor turbine.

Let us assume that the above three variables will com-pletely describe some device for removing somethingfrom one place to another.

This may seem an over-simplification, but it helps to clar-ify the function of the chart. Now think of the chart as afiling cabinet with drawers opening in all three directions,and each containing a different 'idea'. The 'idea' contentsof each of the drawers will be defined by ONE of the vari-ations of each of the three independent variables.

For example, if the contents of drawer lettered 'A' (Fig. 1)were examined, they would suggest a bed-type vehiclemoving over a hard surface and powered by an InternalCombustion Engine. Similarly, the contents of the drawerlettered 'B' would suggest a sling-type vehicle moving inair and powered by a moving cable. Again, the contentsof drawer 'C' would suggest a chair-type vehicle travellingon wheels powered by electricity.

Using the subdivisions chosen for the three independentvariables, the morphological chart constructed in theabove example will, by using the various combinations,produce 330 ideas. By increasing the number of vari-ables along each axis the number of idea-combinationscan be increased.

To evaluate the large number of ideas yielded by thechart, each idea would be screened against a carefullycompiled Specification which outlines the particular fea-tures required of the final solution, e.g. low cost, high

degree of safety, speed and manœuvrability, fuel con-sumption, acceleration, seating, and storage accommo-dation, etc.

Problem 1Look at a simple screw driver (metal blade - woodenhandle). Jot down all the ways by which this screw drivercould be further improved.

Problem 2Name three ways in which current television sets mightbe improved.

Problem 3What features in your house might be improved if theywere curved instead of straight?

AIR

WATER

OIL

HARD SURFACE

RAILS

ROLLERS

WHEELS

BIC

YC

LES

LIN

GS

HIP

BE

DW

AG

ON

CH

AIR

IC E

NG

INE

EN

DL

ES

S B

EL

T GA

S

WA

TE

R AIR

ELE

CT

RIC

ITY

AT

OM

IC E

NE

RG

Y

ST

EA

M

B

C

A

Figure 1. Morphological chart

various forms of energy produced by it can bestbe used to achieve the desired 'output' or solu-tion.

Example

Problem - To design an effective fire-warning system foruse in a factory.

1. Establish desired 'output' = warning of the presence offire.

2. Determine 'input' = fire.3. Decide, and specify the problem's limiting conditions:

Desired specifications - The warning must bereceived one mile from the factory premises andwithin 15 seconds after the fire starts. The cost of thesystem must be within £800. The system must beeasy to install, relatively trouble-free, and simple tomaintain. It must be capable of operating efficiently 24hours a day and on a seven-day a week basis.

4. Analyse the input to decide how the various forms ofenergy produced can best be used to achieve thedesired solution.

(a) What 'outputs' are directly produced by thestated 'input' ?

Answer - Fire, heat, light. Also, the gaseous and solidproducts of combustion and smoke.

(b) Can any of these features be used to producethe desired output directly?

Answer - No. Although the combination of light andsmoke would provide its own warning signal,the fire producing this combination would, bythat time, be almost uncontrollable and havedamaged much of the premises.

(c) What reactions are caused by heat?Answer - Light, smoke, fumes, the expansion of metals,

glass, liquids, and gases. Intense heat canalso change the composition of certain chemi-cals. Similarly, light and smoke can cause vari-ous chemical and physical actions and reac-tions.

(d) Can any of these reactions be used to achievethe desired 'output'?

Answer - Yes. The difference in the rate of the thermalexpansion of glass and metal could be used tooperate a water or chemical 'sprinkler' system,or to break an electric current; a liquid thatexceeds a predetermined level of expansion ata specified temperature could be used to trig-ger a valve; an alloy which quickly melts at atemperature below the boiling point of freshwater could be used to break an electric circuitor to trigger open a mechanised valve; variouschemicals can be used to detect light; certainchemicals, when exposed to heat, can quicklyproduce a non-inflammable, non-toxic, highdensity gas or foam which can be used tosmother a fire.

The problem is finally solved by carefully evaluating eachsuggestion in the light of the previously specified limitingrequirements.

(c) Morphological analysis

Morphological Analysis is a creative approach to prob-lem-solving that makes use of a matrix to relate all con-ceivable features or functions required of the solution toall possible means of obtaining such features or func-tions. The technique is especially useful for developing anew product. It has the effect of gathering together alarge volume of relevant information before applying criti-cal judgment in the form of specified criteria for the rejec-tion of impractical solutions. The steps used are as fol-lows:

Step 1- Prepare a broad, and general, statement of theproblem.

Step 2- Define as broadly and as completely as possibleall the various independent variables to be con-sidered, then sub-divide them.

Step 3- Prepare a matrix, or 'morphological' chart, usingeach variable as an axis (see Fig. 1), the size ofwhich is determined by the number of variablesto be considered.

Step 4- List all idea combinations suggested by the'chart '.

Step 5- Evaluate listed ideas by subjecting each idea toa previously prepared specification or compari-son 'check list'.

Thus, when using this method to develop a new product,the designer would first list the chart's vertical parame-ters such as the form, size, material, weight, etc., of thedesired product. He would then extend against each ahorizontal list of the various means by which thesedesired features might be achieved, or of various descrip-tions of the required form, size, material, weight, etc.When both lists have been extended to the limit of hisimagination, a useful field of possible solutions is availa-ble in which the designer can readily combine one sug-gestion with another, and in various combinations. Thenext task is to prepare a Specification against which toaccept or reject each possible solution both in quantitiveand qualitative terms. This Specification, prepared byconsulting the original problem statement, is used toreduce the field of possible solutions.

The following simplified example illustrates the use of thismethod for producing a large number of ideas.

Example

1. Problem (stated in broadest possible terms) -To remove something from one place to another bymeans of a powered vehicle.

2. Independent Variables to be considered, (a) Type of vehicle.

Subdivide to give - ship, wagon, chair, bed, sling,bicycle (numerous others could be included).

3(b) Media on which vehicles operate. Subdivide to give - air, water, oil, hard surface, rails,

JCH Roberts, Creativity Design 37

in restricted spaces'. Someone might thereupon mentionhow bees store their honey, or ants their food. Possiblythis would provide the key idea to the undisclosed prob-lem. Someone else might mention that an object con-sumes less space when stood on its end; that sausagesand onions are sometimes stored in a suspended posi-tion to save space. At this point, the Chairman wouldredefine the problem as 'storing motor cars'. This mayresult in suggestions that a device be developed enablingcars to be parked on their noses rather than on fourwheels.

New fields are opened up for discussion as the problemis gradually defined more exactly. Finally, when theChairman senses that the group is close to providing asound solution, he reveals the exact nature of the prob-lem. The Group's competitive spirit is now keenly devel-oped and is highly creative so that the principle behindthe problems solution is immediately clear to them, andthey begin to develop new ideas in detail.

For best results full attention must be given to the choiceof the discussion group. Participants should be selectedon the basis of their experience of the problem asdefined in its very broadest sense. For example, if theproblem were 'to invent something which fills a realneed', the participants chosen to comprise the Groupwould probably include:- a production specialist- an artist with experience in electronics and industrial

engineering- a chemical engineer and sculptor- a mechanical design engineer- a private inventor with a knowledge of machine-shop

practice, and possibly- a housewife.

Irnagination-Undeveloped-Resource (IUR)

Two main criticisms of the 'Gordon Method' of appliedcreativity are:(1)

an opportunity for an incubation of ideas is deniedand

(2) an opportunity to re-structure the problem is scarcelyavailable when only one member of the Group knowswhat the problem is.

The 'Imagination-Undeveloped-Resource' method ofcreative thinking compensates for this lack of a suitableincubation period by dividing the problem session intotwo parts.

Although generally following the same problem generali-zation procedures as those used in the 'Gordon Method',IUR differs in that:

(a) The Group Chairman does not disclose the actualproblem until just before the end of the session.

(b)A second session is held a few days later.(c) At the commencement of the second session a tape-

recording of the previous session is played to the

assembled group. The previous problem is then pre-cisely stated and the group start to develop possiblesituations for use in solving it.

The 'Hypothetical Situation' method

Developed by Professor Arnold the purpose of this tech-nique is two-fold:

1. to free the user ol routine thinking habits and the ever-present influence of his environment,

2. to stimulate the user's imagination to produce newideas.

The method is simple to apply, and can be used either byan individual or a group. Although it can and must beregarded as a creative 'tool' in industry, it is more gener-ally employed as a means of training designers and engi-neers to be more creative.

The method involves creating a hypothetical situationrelated to a particular area of activity, or interest. Theindividual, or group participants are then asked to pro-duce ideas for solving the described situation.

For profitable results it is essential that the 'hypotheticalsituation' conforms to the following rules:

1. It must be one that provides a means of freeing theindividual from his present limitations.

2. It should be made as unique as possible.

Similarly, if there is to be a chance of applying the resultsproduced to existing circumstances, the hypothetical situ-ation must:

3. Ensure that thelr ldeas for new designs or productsare based on present sound engineering principles.No miracle substances, materials or perpetual motiondevices should be considered.

4. Assume that the conditions described in the givenhypothetical situation actually exist, and make a deter-mined effort to solve the problems within these condi-tions.

Some examples of the kind of hypothetical situationwhich could be used to stimulate a design engineer'simagination to produce profitable ideas are:

a. In the year 2500, the motor car instrument panel willbe so complicated that the information presentedmust make use of senses other than sight. Produceideas for the design and selection of instruments withsound indications, vibration indications, etc.

b. Assume that several manufacturers of machine toolsand process equipment, now intend to produce ultra-sonic equipment also. To get a firm foothold in theproduct manufacturers market, they are offering theirnewly produced ultrasonic equipment at a third of thecost of the equivalent conventional equipment. At thesame time your own company, also a machine toolmanufacturer, is planning a factory expansion of 100,000 square feet, involving the purchase of necessary

More and more far-seeing, profit-minded members ofindustry's top management are developing an acuteawareness that their company's profitability is closelyrelated to the ability of its key personnel to develop newproducts, efficient manufacturing methods, and success-ful marketing techniques. They are also aware that therapidity of technological advancement and the ever-dynamic competitive response, compress the time scaleof a product's life-cycle to such an extent that the newproduct ideas must be produced at maximum rate. If con-tinued profitability is to be assured, key personnel mustbe purposely trained to use every established creativetechnique now available for deliberately producing a con-tinuous supply of new ideas that will enhance productionspeed.

As explained in Part 6, these techniques are divided intothree distinct categories, namely:1. Analytical Techniques (described in Part 6)2. Free Association Techniques.3. Forced Relations Techniques.

FREE ASSOCIATION TECHNIQUES

Scientific investigation has established that the averageindividual's ability to produce good ideas is deliberatelyincreased by the application of two basic creative princi-ples:

(a) Deferrnent of judgement

Twice as many good ideas can be produced (in the sameamount of time) providing judgment of each idea as it isproduced, is deliberately withheld.

(b) Quantity improves quality

The overall quality of the ideas generated improves withthe quantity of ideas produced.

Free Association techniques are largely based on theprinciples described above and include:

i. Brainstormingii. The Gordon Methodiii. Imagination- Underdeveloped- Resource (IUR)iv. The 'Hypothetical Situation' Methodv. Buzz Session

vii. Reverse Brainstormingvii. Slip Writing

The advantages of 'Brainstorming'. the application ofwhich has been previously described (see Part 5), are:

1. Less inhibitionism and defeatism. The rapid fireexchange of ideas by a Brainstorm Group quicklyexplodes the myth that the individual cannot think of anew and different solution to a problem.

2. Building upon the ideas of others. What may seem anabsurd idea to one member of the group may stimu-late another to produce a new and useful idea.

3. Contagion of enthusiasm.4. Development of a competitive 'creative' spirit as a

result of which each participant wants to top theother's idea.

The 'Gordon' Method

Devised by William J. Gordon of Arthur D. Little, Inc., anAmerican firm of professional creative thinkers who spe-cialise in producing ideas for new products for otherfirms, the 'Gordon' Method of creative thinking bears acertain similarity to 'Brainstorming' in that it:

(a)is a Group procedure,(b) encourages unevaluated, free-flowing creative discus-

sion,(c) prior criticism of ideas is ruled out.

Unlike 'Brainstorming', the purpose of the 'GordonMethod' is to produce, not many ideas but one or tworadically new ideas. At the outset. only the Group Chair-man knows the exact nature of the problem under con-sideration.

The initial discussions focus on a very broad subjectrelated to the problem only in a very general way. Forexample if the problem was to invent a new type of lawnmower, the Group Chairman would simply define theproblem as 'cutting' or perhaps 'separation'. As particularareas of thought became exhausted, the Chairman inter-jects further limited facts which, in a progressive manner,are a little more explicit to the problem. In the examplegiven the problem would be progressed by consideringthe nature of blades or the value of rotary motion, thusopening up new, related fields to explore as the discus-sion continues.

The 'Gordon Method' of applied creativity is largely usedfor producing new ideas for products. Its success is con-trolled by the creative leadership of the Group Chairmanand his expertise in isolating, and presenting variousrelated aspects of the problem under consideration. Toregard another totally different example. Suppose theproblem is to find a new way to park motor cars in acrowded city, the Group Chairman would describe thesubject for discussion as 'storing things'. The sessionwould probably start with a discussion on what 'storing'means. This, in turn, might lead to discourse upon theparticular features desirable in a good storage system,considering features such as, for example, a minimum oflow-cost space, ready availability when wanted, thiefproofing etc. Next the discussion might turn to methodsused for storing things in nature, in the home, aboard air-craft in ships, or in industry etc. At this stage, the Chair-man, would narrow the problem by providing a rathermore specific definition of the problem e.g. 'Storing things

7. Practical techniques for producing new ideas

JCH Roberts, Creativity Design 38

cant and impact it into the wooden launchways. The con-ventional method used to clean launch-ways involves asquad of 6 or 8 men systematically scraping theimpacted grease from the launch-way with shovels andmetal scrapers. This method is both costly and time-consuming.

Source for possible solution: Forced relationship betweenship sliding down launchways and curved metal snow-dozer blade.

Ideas produced: Fit snow-dozer blade to bow of shipabout to be launched so that blade curves towards sternof vessel and the blade edges slide on launchway. Asship travels down launchway, snow-dozer blade willsimultaneously remove hard impacted grease cleanlyfrom launchways thus eliminating necessity to cleanlaunchways afterwards.(Note: Patent applied for by Author).

B. Systematic listing

List a number of objects, or ideas, all associated with ageneral subject. After the items have been listed, giveeach a number. Now consider each item in turn with theother items in various combinations, to produce possiblenew ideas.

Example: A manufacturer of office furniture might con-sider the following objects: a desk, chair, desk lamp, filingcabinet, bookcase. He would consider such combinationsas chair and desk, chair and lamp, chair and filing cabi-net, chair and bookcase, etc., as possible profitable newproduct ideas.

As with all other creative techniques, no attempt is madeto judge any of the ideas produced until all possible ideashave been collected.

C. Focused object method

This method differs from the two methods previouslydescribed in that the particular elements in the forcedrelationship are preselected with a definite purpose inmind. To apply the 'Focused Object' method.

1. Select the fixed element in the forced relationship.(This may be a product, an idea, or a problem state-ment.)

2. Deliberately focus attention on some other element -usually something in the immediate vicinity. Once thefixed element and the element selected at randomhave been chosen, the forced, or unnatural, relation-ship has been established. This is now used as thebasis for a flowing chain of 'free associations' fromwhich are produced new and original ideas. Usually,the first ideas will come from a simple transfer of theattributes of the random to the fixed element.

Example:

A furniture manufacturer might choose a chair as the'fixed' element and an electric light bulb as the 'random'

element. The initial idea suggested might be a glasschair, a thinner chair, bulb- shaped chair, screw-plugbase construction, an electrically operated reclining chair,or a chair with a built-in reading light, etc. Usually themost profitable ideas evolve from the second-level ideasi.e. those developed from free association of the initial orfirst-level ideas.

D. Modification method

This is an individual operational technique which consistsessentially of attempting to modify or twist a given idea inas many ways as possible so as to produce further ideavariations. Usually a specific goal, consisting of a certainnumber of modifications is set. For some people it ishelpful to specify a requirement of 30 or more modifica-tions of one basic idea.

E. 'The Fresh Eye' method

An ordinary object is selected and examined with a largeamount of concentration. All senses are applied toenable the familiar object to be seen in a 'fresh' light, andthus provide a source of new ideas. When using thismethod it is often helpful to feature the object against anunusual background

PROBLEMS

Problem 1

Suppose you were a manufacturer over-stocked withtooth-brushes. For what other uses (other than cleaningteeth) might you try to sell your surplus stock?

Problem 2

Large flocks of starlings annually create a public nui-sance m many cities and towns. Suggest six, or more,possible solutions to this problem.

production equipment. Produce ideas involving theapplication of ultrasonics either to a new or existingproduct which will help combat this situation.

c. Assume that the overseas markets for your Com-pany's products have suddenly rocketed. To meetdemands, your present production must be increasedtenfold.Your company has no additional floor space availablenor any sizeable quantity of manpower. Produceideas for re-arranging existing processes, premisesand labour that will increase your present productiveoutput tenfold.

d. Assume that the present materials used in the manu-facture of your Company's product(s) are suddenly nolonger available. Produce ideas for redesigning theproduct so that it can be made, say, wholly from plas-tics, aluminium or some other NEW material.

Similar 'hypothetical' situations can easily be designed sothat the individual's imagination is activated.

The 'Buzz' session

The Buzz session approach to profitable creative thinkingwas developed by J. Donald Phillips, President of Hills-dale College, Michigan, USA. It is an extension of theBrainstorming technique in that the large BrainstormingGroup is divided into a series of small Groups each com-prising 5 participants headed by a Chairman. The sameproblem is given to each Group and is Brainstormed for aperiod of 20-30 minutes, or longer if needed. Each Groupthen selects its best idea for presentation and final evalu-ation at a joint session of all the individual Groups.

The main objective of the technique is:1. To get all possible ideas.2. To encourage maximum participation.3. To instruct large Groups of people at one time on the

technique of Brainstorming and its possibilities.

The overall effectiveness of the 'Buzz' session is duelargely to the fact that:

4. Participants are often more willing to speak within asmall Group than as members of a large-sizedassembly.

5. It provides adequate time for each participant to statehis ideas. In a large-size Brainstorming Group slowerthinking, less forceful participants are robbed of thechance to state their ideas.

'Buzz' sessions can also be effectively used to developquestions in a 'question-and-answer' session after a lec-ture, talk or technical presentation.

Reverse brainstorming

Developed by the American Hotpoint Company, this tech-nique works in the opposite way to 'Brainstorming' inas-much as the product, idea or problem statement is sub-jected to a barrage of ideas as to why and how they areinadequate. The object is for the Group to think up allpossible limitations, shortcomings, and failings of a given

product or idea so as to produce a series of new, previ-ously undisclosed problems. These are then subjected tovarious creative problem-solving methods.The same Company profitably employs two other inter-esting brainstorming variants, which can be described as:

(a) The 'Waste Not' Method. The Group participants areshown a discarded product, or item of large-volumewaste material (e.g. a small plastic container, the wastematerial from a particular machining or cold-pressingoperation). The Group is then asked to Brainstorm forprofitable uses.

(b) The 'And-Also' Method. A practical suggestion con-cerning a possible use, improvement or modification for aproduct is made. Each participant in the Group attemptsto cap the previous suggestion with a better one of hisown.

Slip writing

This method is more accurately described as 'Brain-storming' limited to a single person. The individualapplies the technique by:

1. Formulating his own problem statement, and2. Writing down, or tape-recording, all the ideas on the

problem that occur to him. As with other methods, theobject is to produce a large number of ideas. Thequality of each idea so produced is evaluated in aseparate, later session.

FORCED RELATIONS TECHNIQUES

The underlying principle of 'Forced Relations' Techniquesis to generate further ideas from the use of an estab-lished relationship between one product, idea or problemstatement, and another. In most instances, the forcedrelationship is established arbitrarily - often by mechani-cal means. Among the methods used are:

A. Catalogue review method

Open a manufacturer's catalogue (or series of unrelatedcatalogues) or any other source of printed information,preferabIy illustrated. Select any item or subject, or evena single descrlptive word (noun and/or adjective) atrandom, without conscious effort. Now choose a seconditem, subject or word in the same arbitrary manner. Con-sider these elements together (i. e. force-fit their relation-ship) and attempt to create original ideas based upon theestablished forced relationship.

Example:

Problem statement: Means to degrease a ship's launch-ways cheaply and quickly.

Background to problem: After a ship is launched the lubri-cant applied over the top surface of the launchways mustbe removed in readiness for the next launch. The com-bined effect of friction and weight tend to 'fire' the lubri-

JCH Roberts, Creativity Design 39

3. Cleaning awkward-to-get-at mechanical parts.4. Making designs in stucco.5. Pattern paintings.6. As a dolls hair brush.7. As a brush for eyebrows, moustaches, etc.8. Cleaning fingernails and fingers.9. Cleaning dentures.10. Removing dirt from venetlan blinds.

Problem 2

1. Set poison or traps near nest sites.2. Find out what attracts them and eliminate it (e.g.

garbage, crumbs, warmth of chimneys, etc.).3. Install fans where birds congregate and discourage

them with blasts of air.4. Get the fire services to hose nests regularly.5. Paint nesting surfaces with potassium chromate.6. Paint nesting surfaces with slippery substance.7. Encourage local boys to shoot at the birds with cata-

pults or air rifles.8. Place stuffed owls near nesting sites. 9. Fit photocell operated electronic noise making appli-

ances near nests.10. Spray the nests with a chemical which is offensive to

the birds.11. Encourage ferocious cats to habitate trees near

nests.

PART 4

Problem 1

The standard lightbulb is turned counter-clockwise forremoval. So the railway company simply installed sock-ets and bulbs which turned in the opposite direction.Thus youngsters attempting to twist bulbs out of theirsockets merely screwed them tighter instead, sometimescutting their hands if they turned a bulb too hard. Within avery short time, the theft of light bulbs dropped to almostnothing. Other ideas that could well have proved equallyeffective are: 1. Fit a wire guard over each bulb.2. Manufacture light-bulbs with thorny surfaces.3. Fit a loud-sounding alarm activated by the removal

of a bulb, the sound of which increases in volume asmore bulbs are removed.

Problem on boy's sister:

Answer: The boy has three brothers and the girl has twosisters.Solution: Boy has 3 brothers making 4 boys in the family.Girl says she has twice as many brothers as sisterstherefore, she has 4/2=2 sisters making three girls in thefamily. Thus, the boy's statement that he has as manybrothers as sisters is correct

How to make four nines equal one hundred : 999/9 =

100

PART 5

The 'Repair Chairman' studied the external area of theChurch to be painted and divided it into 7 approximatelyequal parts. In each of the 7 parts he painted, in boldwhite lettering, the full name of the member of the con-gregation who had promised to help paint the church andadded after each name the words 'PROMISED to paintthis area' and so that the message read: 'Mr. . . . PROM-ISED to paint this area'. Once the various male volun-teers saw their names in print, and realised that everyoneelse could see that they had not kept their promises, theyquickly set to and completed the promised paintwork.

PART 6

Problem 1

1. Groove the handle with finger-tips as on a car driv-ing wheel.

2. Coat handle with luminuous paint, to make screwdriver easy to find.

3. Drill a hole at top of handle, or add an eye, so thatscrew driver could be hung on tool rack.

4. Provide an assortment of angled blades to allow

user to reach into hard-to-get-at places.5. Fit small light at base of handle to illuminate screw

location in dark areas.6. Provide magnetised blades to pick up screws.7. Supply battery for driving blades.

Problem 2

1. Provide a single control for ALL purposes.2. Provide photo-electric eye for automatic control of

picture brightness.3. Set designed so that it could be built into a wall.4. Design set so that chassis is independent of picture

screen and so that the screen could be hung as aconventional picture on a wall.

5. Provide a control which could be set in advance soas to limit selection of programmes solely to thosemost suitable for children.

6. Provide control that automatically shuts sound, and/or picture, off during commercials, or political broad-casts.

7. Provide tiltable screen of adjustable height for view-ing from the prone position.

8. Provide arrangement that would enable viewer towatch two different programmes at the same time.

9. Provide control that automatically switches toanother programme at a pre-determined time.

10. Provide set with picture doors for decoration whenset not in use during day-time.

11. Provide automatic Fault Indicator at rear of setwhich lights up to indicate particular named fault .

Problem 3

1. Legs of beds and tables to curve inwardly, so thatthey will be less in the way.

2. Rounded tables and dressers in Children's rooms -to prevent bumps.

3. Rounded play-pen for same purpose.4. Rounded rooms, so that dust won't collect in the cor-

ners.5. Rounded roofs to provide more head-room in attics.6. Rounded ('cinerama') TV screens so curved that

people on all sides of the room can watch.7. Curved windows.8. Curved bottom of bath for more comfortable contour 9. Desk with its lower drawers curved inwardly, to pro-

vide more knee room.10. Refrigerator door curved outwardly to allow more

shelf room inside the door.11. Edges of stair steps curved to prevent wear.

PART 7

Problem 1

1. Cleaning the edges of shoe soles.2. Staining shoes.

Answers to problems

Stress, Strength and Safety 1

tion ) of any component and derive the equation for a round shaft - one of the most common compo-nents in mechanical engineering. Static failure only is considered in the first instance, however thesame approach will be implemented later for other failure mechanisms such as fatigue and fracture.

Safety Factor

A component subjected to a solitary load will be considered in the first instance. This load is inter-preted in the context of the component's nature and duty - thus load usually implies a transverseforce in the case of a beam component, or a longitudinal compressive force in a column, or atorque in the case of a shaft, or a pressure in a fluid containment vessel, and so on.

There are two completely different manifestations of the load, which have important consequencesfor the component :

- the extrinsic actual load is the load exerted on the component by its surrounds, and- the intrinsic maximum load is the largest load that the component can withstand without fail-

ure; the maximum load is a property of the component, a function of dimensions and material.

Each of these two loads is expressed usually by its nominal value. ( "Nominal" means "in name only";a nominal value of something is merely a convenient label. For example, 500 mm pipes are made with an out-side diameter of 508 mm and various wall thicknesses - and hence come in various internal diameters - butthey are all referred to by the convenient rounded "500" nominal diameter as a descriptor of size. )

There are many reasons why the actual load on a particular component at a particular time willdiffer from the nominal value, including -

- the inherent variability of the load ( eg. in practice the mass of a "ten tonne truck" will dependon the load it's carrying ),

- static indeterminacy ( when components share the load in proportion to their elastic responses -more of this anon ),

- dynamic ( or shock ) effects - for example if a weight W is dropped from aheight h onto an elastic component of stiffness k, then the peak force in the com-ponent is :- Fdyn = dynamic magnification factor ( dmf ) * W.Elementary energy methods give :- dmf = [ 1+ ( 1 +2hk/W)1/2 ] for this typeof loading, demonstrating that releasing a load W suddenly even from zeroheight induces a peak force of at least twice that arising from gradual release -the effective actual load is at least twice its nominal value,

- . . . . and so on.

Similarly the maximum load which a particular component can withstand at a particular moment intime will differ from the nominal maximum load due to many factors including -

- dimensions differing from their nominal or expected values - eg. we might look at our creek-bridging branch, calling it 50 mm and figuring the nominal maximum load on that basis; butthe diameter may in fact be 47 mm, or 54 mm, or more likely vary along the branch;

- material strength differing from its nominal value due in turn to variations in material compo-sition or heat treatment, to unsuspected flaws ( is there an undetected split or other anomaly inour bridging branch? ),

- . . . . and so on.

Deviations may be highly significant or they may be insignificant - it's important that we have anidea of their magnitudes. Sometimes we haven't much of a clue and therefore have to fear the worstand make suitable allowance for our ignorance.

Clearly a component is safe only if the actual load applied to the component does not exceed the

W

k

h

STRESS, STRENGTH AND SAFETY

You are walking in the countryside when you encounter a creek which is too wide to jump. Problem:How can you cross to the far side without a soaking? Applying creativity, you may come up with thefollowing possible solutions, depending upon the circumstances :-

- roll up your jeans and wade across;- dam the creek;- adapt a couple of tent-poles as stilts to walk dry-shod through the creek;- bend a sapling, secure it with a rope, climb the sapling, cut the rope and catapult across;- construct a bridge from a convenient uprooted tree trunk;- use a broken branch to pole-vault across;- position piles of boulders to use as stepping stones;- attach a rope to a tree, lassoo another tree on the far side, and swing across hand-over-hand;- build a raft and float across . . . . and so on.

These and other solutions, apart from the first, require implements - ropes, trees and the like - whosefailure may be through either fracture ( they break ) or excessive deformation ( they bend or stretchtoo much ). We may examine an implement’s safety (absence of failure) by one of three approaches :-

- Try it with care, an approach which is not possible if failure could be hazardous or cause signif-icant financial loss, or

- Analyse it via a mathematical model of the existing implement, from which the stresses anddeflections may be calculated for the known loading and the implement’s size (dimensions)and material (strength, modulus), thus indicating whether the implement is safe or not; or

- Synthesise it, ie. analyse in reverse, where the minimum dimensions necessary to avoid failureare first calculated before the implement is selected or later made to suit.

Analysis and synthesis involve four major aspects of theimplement, as shown :-

An implement in mechanical engineering is more complexthan exemplified above, and requires careful design to ensurethat everyone who is associated with it is satisfied with it. Awell designed artefact is cheap to manufacture, and is easyand safe to use and to maintain, among other things.

Although safety is just one aspect of design, it is a fundamental necessity for all designs

Implement safety can of course be assured by building a physical model and testing it, but this isusually uneconomical and so one of the major aspects of this course is to demonstrate the formationof mathematical models of various mechanical components - bits and pieces such as shafts, brakes,welds, bearings and the like, which are assembled into machines for transforming mechanical poweroutside the human performance envelope. These models may be analysed to predict the prototypes'behaviour and safety before they are built, and in conjunction with sketches enable componentdesign to be carried out. It must be appreciated that the techniques of ( mathematical ) model build-ing which are introduced in the context of one particular component are usually applicable to manyother components which cannot be considered in the course, and therefore the course emphasis is asmuch on how we arrive at a result as on what that result is.

In this chapter we trace the steps necessary for setting up the mathematical model ( the design equa-

MATERIAL

SIZE

LOADS

SAFETY

synthesis

analysis

Stress, Strength and Safety 2

The stress in a brittle material, Fig.D, cannot exceed the ultimate strength, Su.

Ductiles also display such an upper limit, Fig.E, but the yield strength, S y, is often more relevant as itforms a bound above which plasticity and excessive deformation may occur. Most modern ductilesdo not possess a distinct yield so in this case an artificial value is usually defined - the offset yield,Fig.F - based upon some maximum acceptable permanent deformation (eg 0.2%) remaining afterload release. Yields and ultimates are material properties; representative values for many materialsare cited in the literature.

Let us derive a design equation for the simplest of all components - thetensile bar. The bar's cross-sectional area is A, it is subjected to a tensileforce P and the strength of its material is S - which may be the ultimate iffracture is important, or the yield if the material is ductile and excessive deformation is relevant.

Assuming uniform stress across the cross-section, the maximum load that the component can sus-tain occurs when the stress σ reaches the material strength, and is P = A S - this expression whencombined with ( 1) leads to the design equation for direct normal stress, namely :-

( 1b ) n P = A S

The design equation embodies conveniently and directly the four aspects mentioned previously -safety, load, dimensions and material (strength). The equation may be used either :

- for analysis to determine the degree of safety n = A S /P for given S, A, P, or - for synthesis (design) to ascertain the dimensions required A ≥ n P/S to withstand a given P

with material of strength S and specified degree of safety n. Codes often cite a material’s allow-able or design stress, σd = S/n as a characteristic in its own right, rather than stipulating boththe strength and the necessary safety factor. The required area would then be A ≥ P/σd.

Equation ( 1b) applies also to compressive loading, however components in compression are usuallymore likely to fail by buckling than by any strength limitation being exceeded. Buckling (ie. geomet-ric instability) is dictated as much by the component's overall geometry as by the inherent strengthof the component's material. A cardboard tube can withstand a larger compressive load than a steelwire. The design equation for buckling is derived in a later chapter.

Suitable nominal factors of safety for use in elementary design work are tabulated below.

The table illustrates clearly the need to increase safety factors when designing with loads eitheractual or maximum which are not known with certainty. In order to avoid waste of material to caterfor ignorance, we try to forecast loads as accurately as possible. For example, rather than use a largedesign factor to allow for unknown shock effects, we may obtain a realistic dynamic magnificationfactor from the literature citing other folks' experience with similar components, and then increasethe nominal actual load by this dmf. Statically determinate assemblies are generally preferred to inde-terminate because component loads are predictable from simple statics and do not rely on a complexinteraction between the components.

PσΑ

strain ε

stre

ss σ

E

Sufracture

ultimate

BRITTLE(eg. cast iron)

strain εstre

ss σ

E

Su

Sy

fracture

excessive deformationyield

ultimate

DUCTILE (eg. steel)

( D ) ( E ) ( F )

Sy

parallel

specifiedpermanent

set eg. 0.2 %

OFFSETYIELD

component's inherent maximum sustainable load. The degree of safety is usually expressed by thesafety factor n :-

( 1 ) n = maximum load / actual load = F / F

Thus if n = 1 then the component is on the point of failure< 1 then the component is in a failed state> 1 then the component is safe.

The safety factor is usually expressed as a ratioof nominal loads. A higher value of the safetyfactor seems to indicate a safer component -however this is not necessarily the case as theinevitable variations must be kept in mind. In Fig.A, n = 1.25 based on nominal values,but because of the relatively large variations inboth actual and maximum loads, there is a sig-nificant probability of the actual load exceed-ing the maximum, and hence of failure.For a negligible failure probability with theselevels of variations, the nominal safety factormust be increased by eg. reducing the actualload applied to the component, as indicated inFig.B - alternatively the maximum load couldbe increased, by increasing the material'sstrength or the component's dimensions.Fig.C illustrates how the probability of failuremay be decreased also by reducing the varia-tions of the actual and/or maximum loads.This can be accomplished by a better under-standing of what's going on - we'll amplify this later.

When an assemblage of components is subjected to a single load, the assembly's safety factor is thesmallest of the component safety factors - 'a chain is only as strong as its weakest link'. If a componentor an assemblage is subjected to a number of different simultaneous loads, then the concept of asingle safety factor may be inappropriate - but nevertheless all potential failure mechanisms must beinvestigated when deciding whether an implement is safe to use or not.

IF - and only IF - the stress in a component is proportional to the actual load on the component, thenthe safety factor may be interpreted also as a stress ratio :-

( 1a) n = S / σ [ provided σ ∝ F ]

where S is the strength of the component’s material and σ is the stress in the componentdue to the actual load, F.

Such an interpretation is evidently inapplicable to assemblages. Stresses are proportional to load forthe majority of practical elements, so we usually assume that the stress ratio interpretation ( 1a)applies. But this is NOT the case with some not uncommon failure mechanisms (examined later)such as buckling and Hertzian contact - in these cases we must fall back on the fundamental loadinterpretation ( 1).

The strength of a material is the maximum stress it can withstand without failure; it is obtained froma tensile test on a specimen. Common metals follow stress-strain relations of the forms illustrated :

fre

qu

en

cy

load

many failures indicated by large area of overlap

maximum load

nominal actual load, F maxF , nominalmax load

actual load showing statistical variations about nominal value

some but acceptably few failures due to separation of F, F (large safety factor)max

some but acceptably few failures due to reduced variability of both actual and maximum loads (small safety factor)

n = 2.25

n = 1.25

( A )

( B )

( C )

Stress, Strength and Safety 3

compression ), shear, bending and torsion - though others will be encountered. If the memberis itself indeterminate then the three-pronged attack mentioned above must be applied to ele-ments within the member - it will be recalled that the elementary equations for bending andtorsional stresses were deduced in this manner based on an assumed deformation geometry.

3 Resolution of the stress components into principal stresses, using either analytic techniques orMohr's circles. We shall not consider tensor resolution of the general 3-dimensional state - it issomewhat complex and seldom encountered in routine design.

4 Implementation of an appropriate failure theory whereby the biaxial or triaxial stresses at theelement may be correlated with material strengths, which are derived from uniaxial tests, toevaluate the safety factor.

These four steps will be examined individually in the following pages, and will be exemplified insetting up the design equation for a very common component - a shaft of circular cross-section sub-jected to simultaneous bending and torsion. But before embarking on this and writing down stressequations such as σ = P/A, we’ve got to appreciate the limitations of such mathematical models -we must be able to visualise qualitatively how stresses vary throughout a body and to appreciate theconcept of stress concentration.

For example consider equation ( 1b) applied to the assembly illustrated compris-ing two short coaxial bars, 1 and 2, of cross-sectional areas 100 and 200 mm2

and strengths 600 and 400 MPa respectively (recall that 1 MPa is equiva-lent to 1 N/mm2). If the assembly is compressed by 50 kN distributed uniformly overeach end, is the assembly safe ?[ Work this out, recalling that nassembly = minimum ( n1 , n2 ) - then check with your tutor. ]

Stress Concentration

We shall use a tightened nut and bolt, in which the detailed stress variation is extremely compli-cated, to illustrate some of the simplifying assumptions which are usually made in routine analysisand design.

The leftmost sketch below shows an assembly comprising a nut and bolt which fasten two compo-nents together. The nut and bolt pair can be regarded as a sub-assembly or component in its ownright. Before any identification of the stresses can be attempted, all external effects on this sub-assembly must be known. A free body of the nut and bolt demonstrates that the external load onthem is the force P due to contact over annular areas with the two fastened components.

The flow analogy is useful when visualising how stress is transmitted through a loaded component.In the analogy, lines of force (or force paths) in the component are likened to streamlines in a fluid

stressconcentrations

assemblydrawing

free body of nut & bolt the flow analogy

free body of bolt portion

p

p

contactpressure

here

here

chan

nel

in

out

fluid"lines of

force"

p

A

σ = P/A

12

For the same reason we employ mathematical models which describe real behaviour as accurately aspossible - rather than use simplistic models which are known to be poor predictors. Don't confuse amodel's descriptive accuracy with its numerical accuracy - a mathematical model might give ananswer to 53 decimal places, but it may describe a component's actual behaviour so simplisticallythat only one decimal place is significant.

Design factors are increased also when the consequences of failure - economic, social, environmentalor political - are serious. For example, a local manufacturer of processing machinery destined for theheadwaters of the Yangtze River, doubled the size of every motor predicted by the design processbecause of the delay in providing replacements if failure should occur due to overload. This isanother example of the designer having to foresee the whole future life of an artefact.

The tensile component is the simplest of all machine members. Investigation into the safety of a morecomplex component involves searching over all elements of the component ( ie. over all locations inthe component ) to find the weakest link - the element with the smallest safety factor. In practice fora given component only a few elements are potentially critical, so an intelligent search is not neces-sarily protracted. At each element, safety is assessed by means of the following general approach :-

1 Find the loads on the component, given the loads on the machine or structure ( assemblage ). Ifthe assemblage is statically determinate then free bodies will give the components' loads imme-diately, but if it is statically indeterminate then a three-pronged attack based on equilibrium,compatibility and the constitutive laws of the assemblage's individual members must be car-ried out.

2 Ascertain the stress components ( often Cartesian ) at the element in question again by the useof free bodies. This is essentially a superposition of load building blocks such as tension ( or

SUGGESTED SAFETY (DESIGN) FACTORS FOR ELEMENTARY WORKbased on yield strength - according to Juvinall & Marshek op cit.

1. 1.25 - 1.5 for exceptionally reliable materials used under controllable conditions and subjected to loadsand stresses that can be determined with certainty - used almost invariably wherer low weightis a particularly important consideration

2. 1.5 - 2 for well-known materials under reasonably constant environmental conditions, subjected toloads and stresses that can be determined readily.

3. 2 - 2.5 for average materials operated in ordinary environments and subjected to loads and stressesthat can be determined.

4. 2.5 - 3 for less tried materials or for brittle materials under average conditions of environment, loadand stress.

5. 3 - 4 for untried materials used under average conditions of environment, load and stress.

6. 3 - 4 should also be used with better-known materials that are to be used in uncertain environmentsor subject to uncertain stresses.

7. Repeated loads : the factors established in items 1 to 6 are acceptable but must be applied to theendurance limit (ie. a fatigue strength ) rather than to the yield strength of the material.

8. Impact forces : the factors given in items 3 to 6 are acceptable, but an impact factor (the dynamicfactor above) should be included.

9. Brittle materials : where the ultimate strength is used as the theoretical maximum, the factorspresented in items 1 to 6 should be approximately doubled.

10. Where higher factors might appear desirable, a more thorough analysis of the problem shouldbe undertaken before deciding on their use.

Stress, Strength and Safety 4

where δ is the extension of the component whose length is L and cross-sectional area is A, andon which the tensile load is P; the axial stiffness of the component is k = AE/L. If the load iscompressive or if the deformation is a shortening then the signs of the various terms of ( 2a)will have to be changed by inspection or by the positive tensile convention.If a component is bent then its stiffness (load/deflection) follows from beam theory.

The technique is applied to an assemblage of uniaxially loaded components in the example below.

Two- or three-component assemblies are examined here, but statically indeterminate assemblieswith many components are by no means uncommon. The Appendix considers one class of suchassemblages in which two loaded members are joined by a number of identical connectors in paral-lel, the problem being to ascertain the proportion of the total load transferred through each connec-tor. Examples of this class include :

- a chain drive in which the chain tension is transferred into the sprocket wheel through anumber of identical sprocket teeth, and

- a threaded joint where the bolt load is transferred into the nut by a number of thread turns.

This latter case may be appreciated qualitatively by imagining a rubber bolt screwed into a steel nut.When the bolt is pulled it stretches noticeably, but the relatively rigid nut prevents the bolt stretch-ing inside the nut . . . and if the bolt suffers no strain then it suffers no stress - that is, no load is trans-ferred through the bolt in way of the nut. It is concluded therefore that all the load is transferred inthe first thread and the threads are not loaded equally. The implications of this on the safety of thethreads is obvious. Although most assemblages do not contain rubber components, load distributionnevertheless depends markedly upon the relative stiffnesses of the various components. That's whywe generally much prefer statically determinate assemblages - the loads depend solely on statics; weknow where we are with the loads and don't need large factors to account for our ignorance.

channel whose shape is similar to that of the component - fluid enters and leaves the channel at loca-tions which correspond to the areas where the external loads are applied to the component.

- A complicated streamline pattern infers a complex stress situation.- Stresses are low where the streamlines are widely spaced. - Stresses are high where the streamlines are bunched together due to geometric shape varia-

tions. The more sudden these variations, the higher the local stresses. This last is known asstress concentration. Geometric irregularities give rise to non-uniform stresses.

A free body of portion of the bolt which includes the bolt head shows that the external load P is equi-librated by stress σ in the shank. If the free body boundary cuts the shank in the middle then thestress is uniform across the shank cross-section of area A - that is σ = P/A. But if the boundary liesclose to the head or to the thread and nut, then stress concentration occurs and the maximum stressis greater than P/A.

If it is required to calculate the elastic extension, δ, of a bolt whose length L is relativey large com-pared to its shank diameter, then a first approximation neglects stress concentration to obtain δ = Lε= L(σ/E) = LP/AE = P/k where the stiffness, k, of the bolt is defined as k = AE/L.

If the accent is on bolt safety rather than on deflection, then the core area Ac should be usedfor stress calculations - this minimum area in the force transmission path is formed by atransverse plane cutting the bolt thread between the plain shank and the nut's contact face.

The rightmost sketch above employs similar reasoning to illustrate details of the complex transfer offorce across the thread into the bolt, together with the high stress concentration in way of the threadroot, which may lead to localised yielding.

Static Indeterminacy

The external effects on each member of an assembly - the loads and deformations - must be resolvedbefore stresses and strains internal to the member can be examined. This resolution is carried out bya three-pronged attack :-

a. Equilibrium of all the members must be assured, so free bodies of the individual members aredrawn, complete with known and unknown loads consistent with the physical contact betweenthe various members. The equilibrium equations are written down.

b. Compatibility considers all the members' deformations, and the constraints between these def-ormations which are necessary to preserve the geometric integrity of the assemblage. This stepis carried out by sketching the assembly in the free condition and in the loaded condition withmember deformations greatly exaggerated for clarity; the compatibility equations which statethe integrity constraints are then written down by inspection.

A very common error is not to pay sufficient attention to compatibility.

c. Definition of the constitutive law for each of the members which undergoes significant defor-mation enables resolution of the indeterminacy. The constitutive law for each member inter-relates the member's external load and deformation. If the member is elastic, the constitutivelaw is based upon :-

( 2 ) εx = [ σx - ν ( σy + σz ) ] / E + α t ; γxy = τxy / G etc.which includes a thermal expansion term due to the temperature rise t. If a component isloaded uniaxially and uniformly then, like the bolt above, equations ( 2) degenerate to :-

( 2a) δ = P / k + L α t

Ac

Stress, Strength and Safety 5

F12 (or P + F31 from ( ii)) represents the unknown tension in the (great majority of the) eyebolt 1

F12 (or F23 from ( iii)) represents the unknown compression of the cylindrical sleeve 2.

If it is not clear from free body (c) that the eyebolt is tensioned by F12 and not by F31, then the flow analogy

should be applied or free bodies drawn cutting the shank of (c) and embracing either end of the bolt.

Constitutive Laws. Equation ( 2a) is applied to each deformable component, ensuring that the signs in the

constitutive equations tally with the senses of loads and deformations used in the previous two steps.

Eyebolt 1 From above, the eyebolt is subjected to a tensile load F12 and suffers a tensile deformation δ1, so theconstitutive law ( 2a) consistent with compatibility, equilibrium and constant temperature is :-

( iv) δ1 = F12 / k1 ; k1 = ( A E / L )1Sleeve 2 From above, the sleeve is subjected to a compressive load F 12 and a temperature rise t, and suffers a

tensile deformation δ2, so by inspection or by adhering to a positive tensile convention, the form of(2a) consistent with compatibility, equilibrium and temperature rise of this component is :-

( v) δ2 = - F12 / k2 + L2 α2 t ; k2 = ( A E / L )2

Solution Resolving the indetermincay by solving the five equations for the three unknown forces and two

unknown deformations gives :-

( vi) F12 = Fo where the constant initial load is defined as Fo = ke ( ∆ + L2 α2 t )in which the equivalent series stiffness is 1/ke = 1/k1 + 1/k2

Fo is called the initial load because it is due to nut tightening and heating before the external load P is applied.

This result seems ridiculous since it has F12 (the only failure agent in both 1 and 2) completely independent of the

applied load P - ie. the load

may be raised out-of-hand

with no adverse effects on

bolt or sleeve !!! What's

wrong with the analysis ???

The channel for the flow anal-

ogy is shown at (f); the

restrictive annular areas are

arrowed. Two distinct force

paths are suggested in (g) -

that of P from the eye directly

into the support and byepass-

ing the bolt shank and sleeve,

and that of the self-equilibrating initial load Fo in the bolt shank and sleeve. These make physical sense as no

external support is necessary to cater for a temperature rise of the sleeve or for a tightening of the nut - these

effects, embodied in Fo, are resisted internally in the assemblage. Note that the force paths have been greatly sim-

plified here to avoid confusion in the sketch.

The two forces F12 and F31 are graphed against increasing external load P in (h). F12 is constant at Fo from ( vi )

while from ( ii) F31 drops off from this value in proportion to P, until it becomes zero when P reaches a value of

Fo. But F31 is the force between eyebolt and support; it cannot be negative (there is no adhesive to enable the eye-

bolt to pull down on the support). The physical interpretation of F31 reaching zero is that contact between eyebolt

and support is lost as indicated by ( j). This then is the reason for the apparent anomaly in the foregoing analysis.

The analysis is perfectly correct - but only within the limitations of its assumption, implicit in the free bodies (c)

and (e), that contact between 1 and 3 is maintained, and F31 > 0. This assumption should not be confused with the

unrelated deformation assumption of (b).

If it is desired to increase the load P at which contact is lost, then (h) indicates that Fo must be raised, probably by

further nut tightening, ∆. After contact is lost (P > Fo) the arrangement is statically determinate and componentloads can be found from equilibrium only. Setting F31 = 0 and F12 = P in the free bodies (c), (d) and (e) demon-

strates that all components transmit the external load P. This is confirmed physically by the force path ( j).

Having found the force F12 transmitted through the bolt eg. the bolt's safety can be assessed from the tensile

0

contact no contact

P

Fo

F12

F31

P

Fo

P

contact lost when P > Fo

( j )( h )( g )( f )

EXAMPLE An eyebolt 1 and cylindrical sleeve 2 are supported by the plate 3 as in (a), (b) below.

The nut is just finger tight ('snubbed' - there are no gaps in, and no appreciable loads on the assembly). Subse-

quently the nut is tightened by screwing it a known distance ∆ along the thread, then the temperature of thesleeve only is increased by t, and finally the load P is suspended from the eye.

Given the axial stiffnesses k1 and k2, resolve the indeterminacy in the final loaded state.

The washer and support plate 3 are thin - ie. very short axially compared to the bolt and sleeve - and therefore rel-

atively rigid (k ∝ 1/L). It follows that the only deformations of significance are those of the bolt 1 and sleeve 2.

The sequence of load application (heat, tightening, external load) is irrelevant in this elastic analysis - what mat-

ters is the final loaded state after all three have been applied. The significant deformations (δ1, δ2) in this finalloaded state are assumed to be as shown in (b) in which deformations are grossly exaggerated in order to clarify

compatibility constraints. Other deformation assumptions may be equally valid - the result should not be affected

by the assumption provided any subsequent analysis is consistent with the assumption. More of this anon.

Compatibility. By the assumptions of (b) :

δ1 represents the unknown extension of the (great majority of the) eyebolt 1

δ2 represents the unknown extension of the cylindrical sleeve 2, and

∆ is the known nut movement along the thread.

For the components to remain in contact - ie. for compatibility - the geometry of (a) and (b) requires that :-

( i) δ1 - δ2 = ∆Equilibrium. Before the components are examined individually, all externals on the assembly must be known.

The given external load P applied to the eye is supported by the plate 3, hence a known upward load P appears on

the free body (b) of the assembly. The support must also exert a bending moment on the plate to equilibrate the P-

couple, but this is neglected here as having little bearing on resolution of the indeterminacy - eg. if the cantilever 3

droops then other components undergo rigid body motion which doesn’t affect the relative motion between 1 & 2.

Free bodies of the separate components are now prepared - (c), (d), (e) - in which the unknown action/reaction at

each contact is identified conveniently by the indices of the two components in contact. No components are stuck

together so the action/reactions must be 'pushes' on the contacting components - hence F12 is shown in free body

(c) pushing on component 1, and on free body (d) pushing on component 2. This force is the resultant of (presum-

ably uniform) pressure over the annular contact area - the washer's purpose is to distribute this pressure evenly.

There are three contacts 1-2, 2-3, 3-1 between the three components, hence the three contact resultants F12, F23,

F31 appear on the three free bodies. Equilibrium of all three free bodies requires that :-

( ii) F12 - F31 = P

( iii) F12 = F23

It is noted particularly from the free bodies (c), (d) that :

F23

F31

( b )

δ2

∆ δ1

imaginary mark on bolt which after tightening liesin way ofnut thrustface

P

( a ) (c ) ( d ) ( e )

L1

L 2

washer

muchexaggerateddeflections!

P

P P

F12

A 1

F31

F12

F23

A 21

2

3

3

2

1

Stress, Strength and Safety 6

Elementary Load Building Blocks

We refer to tension, shear, bending and torsion as elementary load building blocks because the load-ing which is applied to most components is a combination (superposition) of these four - for exam-ple, a rotating shaft which transmits power is subjected not only to torsion, but usually also to bend-ing and to direct shear. The figure opposite illustrates the similarities and differences between thefour elementary loads whose use is illustrated by this example inv olving a bent cantilever :

This example illustrates an important general conclusion :if bending is present then direct normal stresses are usually insignificant if torsion is present then direct shear stresses are usually insignificant - though like all general

stress resultant

property of mem

ber's cross-sectional geom

etry

material

property(elastic

modulus)

measure of

deformation

( strain )

stress in mem

ber ( at

distance from neutral

axis, if linear )

==

xT

he practical unit of stress is neither 10 N/m

nor 10 Pa, but M

Pa

(equivalent to N/m

m ).

Stress conversion factor : 6.895 kP

a per lbf/in

62

62

2

ELEMENTARY LOAD BUILDING BLOCKS

σ

τ= = G

J Tr τ

L θ

= = E I M

y σR 1

FORCE RESULTANTUNIFORM STRESS

MOMENT RESULTANTLINEARLY VARYING STRESS

τ σT

EN

SILE

OR

CO

MP

RE

SSIVE

FO

RC

E - P

SHE

AR

FO

RC

E - V

BE

ND

ING

MO

ME

NT

- M

TO

RSIO

NA

LM

OM

EN

T - T

(TO

RQ

UE

OF

RO

UN

D S

HA

FT

)

P

V

σ

0

τ

0σy

n.p.

τr

n.a.

neutral plane

Stress is uniform

in transverse direction across the beam

side view

Stress is uniform

in circumfer-

ential direction around the shaft

Tneutral

axis

Shear stress is uniform

to a first approximation . . . but m

ore realistically

end view

σ = = E = E

εPA

L δ

sδL

VAτ = = G

= G γγ

sδ

LC

omplem

entary shearusually requires a non-linear stress distribution

eg. τ = V Q

/ b I

Lδ

L

Rθ

Note the analogy

between bending and

torsional deformations :

1/R =

θ/L

Torsional deform

ation θ/L is usually constant along a m

ember w

hereas bending deform

ation varies with

bending mom

ent and so requires integration for deflections

Geom

etric instability(buckling) is often crucial in com

pression

u = σ ε12

u = τ γ12

u : specific strain energy ( N

m/m

)3

U =

2EI

M2

∫dL

0 L

U : total strain

energy ( Nm

)

∫U

=dL

0 L2GJ

T2

STR

ESS R

ESU

LT

AN

TST

RE

SS DIST

RIB

UT

ION

EL

AST

IC C

ON

STIT

UT

ION

EN

ER

GY

M

strain

stress

EXAMPLE The bent cantilever, shown at ( a) below, is made from Ø 60mm rod. Determine the stress compo-

nents at the surface element A lying in the horizontal plane of the cantilever’s centreline.

The stress resultants at the cross-section which contains A are readily established by the free bodies shown in ( b).

Evaluating the stresses from the building block equations :-

Tension P = 4 kN

σ = P / A = 4E3 * 4 / π * 602 = 1.4 MPa Note, the practical unit of stress is MPa

Shear V = 3 kN . . . . . since we know the distribution of shear stress (see any Mechanics of Materials text

under bending shear ) we shall use this in preference to the simplistic uniform model. As A lies in

the neutral plane it will be subjected to the maximum shear stress, which for a circular cross-

section is :-

τ = 4V / 3A = 4 * 3E3 * 4 / 3 π * 602 = 1.4 MPa Don't forget to monitor units

Bending Mz = 1.2 kNm

Since A lies on the neutral plane there is no stress at A attributable to this moment component.

Bending My = 2 kNm

σ = M z / I = 2E6 * 30 * 64 /π * 604 = 94.3 MPa compressive at A by inspection.

Torsion Tx = 1.5 kNm

τ = T r / J = 1.5E6 * 30 *32 /π * 604 = 35.4 MPa downward as sketched in (b) by inspection.

Alternatively, the components of the equilibrating moment at A might have been found by vector algebra :

M = - r x F = - [ 0.4 0 -0.5 ]' x [ 4 -3 0 ]' kN.m = [ 1.5 2 1.2 ]' kNm = [ Mx My Mz ]'

where Mx here is in fact the torque Tx identified above. Care should be exercised in using such a

vector approach not to lose sight of the physical significance of the numbers and senses.

The stress components from the building blocks are sketched and then combined below.

4 kN

3 kN

A

x

y

z

A

3 kN

4 kN

3 kN A

( a )4 kN

4 kN

A

400

M = 2 kNmyz

x

y

3 kN

A

3 kNz

3 kN

M = 1.2 kNmz

A4 kN

3 kN

4 kN

y

x

T = 1.5 kNmx

( b )

3 kN

2 kNm

4 kN

1.5 kNm

1.2 kNm

x

y

zA

4 kN500

A

300

700

94.3 35.4 93

34

1.4 1.4

Tension Shear Bending Torsion Combined

x

y MPa

Stress, Strength and Safety 7

EXAMPLE Select suitable dimensions for the dismantlable cot ter joint illustrated, which transmits a tensile load,

F. The allowable stresses, St in tension and Ss in shear, are known for each ductile component.

The first step is to evaluate the external effects on each component - then, for each component, the force paths can

be traced, the possible modes of failure identified and the corresponding design equations applied. To illustrate

the technique, we shall consider only the left component; the right component is very similar. The joint is statically

determinate, so the free body of the component with simplified lines of force will be as shown in Figure A below.

Some of the failure modes are :-

Tension of rod end Area = πd2/4 σ = 4F/πd2 ≤ St d ≥ √(4F/πSt)

Tension in way of hole, Fig B. Area = b(a-e) σ = F/b(a-e) ≤ St b(a-e) ≥ F/StNote that this is the most critical tensile location in the left component's body.

Shear of end ligament, Fig C. Area = b(a-e) approx. τ = F/b(a-e) ≤ Ss b(a-e) ≥ F/Ss

In practice, we expect interaction between tension and shear (and contact phenomena, see below)

around the end of the component, which, together with stress concentration, would lead to a more

complex stress state, failure mechanism and fracture surface. However the simple approach used

here is acceptable, provided suitably large factors of safety (ie. of ignorance) are incorporated in the

allowable stresses used in design. Such an elementary approach usually leads to wasteful overdesign

of components, and it is to avoid this that we try to employ more realistic mathematical models.

Bearing in hole, Fig D. If the two contacting members are rigid then the contact area is zero and the

stresses infinite. In reality both contacting components deform slightly resulting in a small contact

area; actual stresses are therefore finite but large as evidenced by the bunched force paths in Fig E.

We shall examine contact stresses (another load building block) in more detail later; for the time

being we use the simplified concept of a critical crushing or bearing pressure, pc, which is taken to be

uniform over the equivalent projected area Fig F. This critical pressure is obviously very much less

than the material strengths, however the simple approach is reasonable provided we know from sep-

arate experiments the critical pressure which the weaker of the two materials is capable of sustaining.

We then have :- Area = eb (projected) p = F/eb ≤ pc (known) e b ≥ F/pc

. . . . and so on. When all failure modes have been investigated (and these would include buckling, excessive

deflection, crack growth rate etc, if relevant, not just strength as above) then a set of dimensions which satisfies all

the inequalities would be chosen as a design basis. Some dimensions may have to be increased for other practical

reasons - a part might be too flimsy for clamping during manufacture, or too prone to damage during transporta-

tion, unless dimensions were greater than necessary for safety under load.

F

b

cc

Ø e Ø d

F F

F

a

FFσ

τwidth b

Fig A Fig B Fig C Fig D Fig FFig E

= b(a–e)stress area

= b(a–e)stress area

FF

conclusions this has to be applied with some discretion. It will not apply near a beam's simplesupport for example - however this region of the beam is unlikely to be critical.

The load building blocks which occur in a particular component are not always immediately obvi-ous. The cotter joint example opposite, in the realm of design rather than analysis, illustrates how togo about discovering blocks’ occurrence. It also introduces another load building block - crushing orbearing due to direct contact between two bodies, and emphasizes the need to identify ALL potentialfailure modes. After all, the designer must define all dimensions . . . . and materials, and tolerances,and finishes, etc. . . . . in fact the designer is responsible for everything.

In the cotter joint example we assume that the failure mechanisms are separate -tension, shear &c occur at disparate points. The more general state of affairs istypified by the bent cantilever in which a number of load building blocks occursimultaneously at the element of interest. The example demonstrated how thestresses from the various load building blocks are combined to define the stressstate at the element :

So far, so good - but if we want to determine the degree of safety at this element (given the materialstrength) we face problems. It was easy to weigh up the safety of the tensile bar for example (equa-tion ( 1b)) because both the bar and the specimen which provided the material strength (Sy or Su)were loaded purely in tension. But that's not the case here - while the tensile specimen is loaded inpure tension, the component certainly is not. Shear loading also is present.

There is no theoretical rationale which we can adopt to correlate these 'different' loadings and assesssafety - instead, we have to employ a suitable empirical failure theory.

Suppose we were analysing a body on which loading was not known completely; we might ascertainthat the stress components in the x- and y- directions were zero. Would we be justified in deducingfrom this that there was no load on the body and therefore that the safety factor was infinite?

Of course not! as a quick appraisal of a tensile bar loaded in the z-direction willconfirm. Degree of safety must be intrinsic to the loaded body and completelyindependent of whichever Cartesian axes we happen to choose for the analysis.

So before we can look at failure theories we must be able to assess the stress components as the Car-tesian system is rotated, and to evaluate the principal stresses on which safety issues depend.

xy

z

93

34 MPa

xy

z

Stress, Strength and Safety 8

We now examine the variation of normal and shear stress components as the inclination of the faceon which they occur changes.

Consider the elemental unit cube ( size 1x1x1 ) under the known positive 2-dimensional stress com-ponents shown in ( i) below - the third dimension (z) is principal.

Rotational equilibrium requires complementary shear, that is τyx must equal -τxy. This necessity hasbeen incorporated into ( ii), from which it is apparent that the three components ( σx, σy, τxy )together with the third principal are necessary to define the stress state. The three components arecalled the Cartesian triad.

We wish to evaluate - in terms of the Cartesian triad - the stress components ( σ, τ) in the generaldirection θ, so we consider force equilibrium of the wedge element ( iii), one of whose faces isinclined at θ. The height of the wedge remains 1 unit, however the dimension in the x-sense becomes1.tanθ and the length of the hypotenuse is 1.secθ.

The force components on each face of the wedge are the stress components multiplied by the facearea - these are shown in ( iv). For force equilibrium of the wedge :

in the σ-sense : σ.secθ - ( σx + τxy.tanθ ) cosθ - ( σy.tanθ + τxy ) sinθ = 0in the τ-sense : τ.secθ + ( σx + τxy.tanθ ) sinθ - ( σy.tanθ + τxy ) cosθ = 0

which on simplification give the required resolution equations :σ = 1/2 ( σx + σy ) + 1/2 ( σx - σy ) cos 2θ + τxy sin 2θτ = - 1/2 ( σx - σy ) sin 2θ + τxy cos 2θ

Similar equations but with different signs are encountered in the literature - sign differences arisefrom positive conventions other than the above.

The simultaneous occurence of sine and cosine terms in these equations makes it difficult to visual-ise how the resolved components ( σ, τ) vary as the direction θ changes. More easily interpretedequations result if the stress state is defined by the basic triad ( σ, σ, θp) rather than by the Cartesiantriad. θp is the inclination of the plane of maximum principal stress with respect to the x-reference.

The basic triad is related formally to the Cartesian as follows :-

( 3a) σ = 1/2 ( σx + σy )σ cos 2θp = 1/2 ( σx - σy ) that is σ = √[ { ( σx - σy )/2 }2 + τxy

2 ]σ sin 2θp = τxy θp = 1/2 arctan [ 2 τxy/( σx - σy ) ]. . . . though these equations seldom need to be implemented.

Making these substitutions leads to resolution equations in the more meaningful form :-

( 4a) σ = σ + σ cos 2( θp - θ ) ; τ = σ sin 2( θp - θ )

It is apparent that normal and shear stress components vary sinusoidally with direction θ (notunlike vector components) however the variation is second harmonic - that is stress components arethe same along axes which lie at 180o to one another. The two sinusoids are of the same amplitude σ

xy

yxτ

xσxσ

yσ

yσ

xyτ

yxτxyτ

= -yxτ xyτ

xσ

xyτ

θ σ.secθ

xyτ .tanθ

yσ .tanθ

τ.secθ

FORCE =STRESS x AREA

( ii )( i )

xσxσ

yσ

yσ

xyτ

xyτ

STRESS

( iii )

στ

xyτ

θθ

yσ

xσ

( iv )

Stress Resolution

An element's stress state is essentially three-dimensional, generally with both normal andshear components in each of the three dimen-sions. The components are usually deducedby superposition of load building blocks as inthe bent cantilever example of the previoussection. The two- and one-dimensional casesillustrated here are particularisations of whatis essentially three-dimensional.

The element may be rotated about all threeaxes into a unique principal orientation inwhich all shear stresses vanish. The corre-sponding normal stresses in this principal orientation are termed the principal stresses. A stressstate is characterised most succintly by the principal stresses, say ( σ1, σ2, σ3 ) and failure theories -the next step in failure assessment - are expressed in terms of principal stresses. It is therefore neces-sary to examine how principal stresses are derived from Cartesian components.

Stress is a tensor entity, so complex tensor arithmetic must be applied in the three-dimensional caseto evaluate stress components as the element rotates. We consider here only the simpler two dimen-sional case in which one direction is known to be principal, and resolution consists of element rota-tion only about that principal axis. Thus in the 2-dimensional sketch above, the z-axis is principaland x-y stress components vary as the element rotates about that axis. Fortunately the great majority

of practical cases are two -dimensional since the relatively simple load-ing in conjunction with the natural choice of axes leads to one of theaxes being principal automatically. For example there can be no stresson the free surface of a body, so the surface normal is an obvious choicefor one of the three Cartesian axes - as there can be no shear on this sur-face, the axis is automatically a principal axis.

Although we consider only "two-dimensional stresses" , it is important to remember always that stressstates are essentially three dimensional.

In this course the positive convention adopted for the orientation of a plane (characterised by itsnormal) is shown at (a) below, and for normal and shear stress and strain at (b) - positive shear iscounter-clockwise. The two sketches (c) illustrate the consistency of the positive senses for shearstresses and strains; the total shear strain (distortion) is γ = 2( γ/2).

Using these conventions, the plane stress state at 'A' in the bent cantilever of the previous section is :σx = -93, σy = 0, τxy = +34 MPa

2σ

1σ

1 - dimension 2 - dimensions 3 - dimensions

UNIAXIAL GENERALPLANE STRESS σ = 0PLANE STRAIN ε = 0

z

z

BIAXIAL TRIAXIAL

Rotationintoprincipaldirections

xσx

2σ

1σ

3σ

σy τxy

xσx

y

τxy

τxz

xσ

y

zx

τγ

γ2

( c )

( a ) ( b )

ORIENTATIONOF PLANE

anticlockwisefrom reference

STRESS

NORMAL:SHEAR :

tensileanticlockwise

STRAIN

elongationclockwise

θx

τ

σ

εγ 2

no shear stress on this free

surface

principal axis

Stress, Strength and Safety 9

EXAMPLE An element is stressed as in Fig A below. All stresses are in MPa.

Sketch the element oriented principally.

Employing the conventions outlined above, one technique for constructing Mohr's circle is as follows :-

Fig B By the convention, the stress state defined by Cartesian components is σx = -420, σy = -280, τxy = -240

MPa; so the points X (σx, τxy) and Y (σy, τyx = -τxy = 240 MPa) are plotted in τ-σ space.

Fig C Since the line X-Y is the diameter of the circle, the trigonometry of the circle requires :

circle centre : σ = ( -280 -420)/2 = -350 MPa

circle radius : σ = ( 702 +2402 )1/2 = 250 MPa

inclination of X-Y diameter to vertical = arctan( 70/240) = of 16.2o

Fig D The circle is completed, noting that the two principals from ( 5) are σ = -100, σ = -600 MPa. We choose

here to define the principal orientation by reckoning σ at 73.8o clockwise from Y in the circle, so . . .

Fig E . . . σ on the element lies at 73.8/2 = 36.9o anticlockwise from y, which enables completion of the sketch

of the principally oriented element. Alternatively the minimum might be reckoned from y, from x, etc.

The variations of the normal and shear stresses with inclination θ, calculated from ( 4a), are plotted below together

with the Mohr's circles corresponding to the inclinations of principals, of maximum shear etc. The reader should

appreciate the trends indicated.

The foregoing should not be construed as being the only way for resolving a two dimensional stress state.

Fig E

280240

420

xy

MPa

Fig A

( –280, 240 )

( –420, –240 )

X

Y

σ

τ

0

Fig B

σ = –350

σ = 250

70

240

X

σ

τ

0

Y

arctan 70/240 =16.2

o

Fig C

600y

100

36.9o

σ

τ

0

X

Y

σ = –100σ = –600

16.2 o

73.8 o

Fig D

0 30 60 90 120 150 180

250

0

-250

-500

-750rotation θ, degst

ress

on

incl

ined

fac

e, M

Pa

36

.9

81

.9

-350

12

6.9

17

1.9

shear

normal

350 250600 100

x420

240

240280

420240

X

Y

X

Y

X

Y

X

Y

X

Y

X

Y

X

Y

250350

and out of phase by 45o. σ is the constant component of the normal stress.

The principal and maximum shear stresses follow immediately as :-

( 5 ) σ = σ + σ ( at θp )σ = σ - σ ( at θp - π/2 ) τ = σ ( at θp - π/4 )

These relations are often expressed graphically via Mohr's stress circle,in which σ and σ represent the circle's centre location and radius respec-tively. The conventions require that angles in the circle, reckoned fromthe X-radius, are double the corresponding angles on the element(which are measured from the x-reference), and in the opposite sense.The reader should confirm that this construction satisfies equation ( 4a).

The example opposite demonstrates typical stress resolution using the simple trigonometry ofMohr's circle rather than the formal resolution equations derived above. The example also showsclearly the variation of stress components with orientation, θ, as predicted by ( 4a).

It is important to remember that a stress-strain state is always essentially three-dimensional,involving three principals . We have addressed only resolution in two dimensions to obtain a singleMohr's circle involving the principals in these two dimensions - however two other circles mustexist relating each of these two principals with the third.

As noted above the third principal is usually deduced from the nature of the problem - ie. its valuefollows from inspection Two common situations arise when the state is either one of :

plane stress : If there is no stress orthogonal to the 1-2 resolution plane then σ3 = 0

plane strain : If there is no strain orthogonal to the 1-2 resolution plane then ε3 = 0 and itfollows from ( 2) that σ3 = ν ( σ1 + σ2 ).

Fig F completes the Mohr's circles for theexample opposite (Fig D) assuming that theelement is loaded in plane stress, that is thethree principals are ( -600, -100, 0 ) MPa.

Fig G illustrates the three principals andthree Mohr's circles for a completely unre-lated stress state where two-dimensionalresolution happens to relate to the largest(3-2) circle and σ1 is the principal stressorthogonal to the resolution plane.

The outcome of stress resolution at an element must be a set of three principals - all three must beknown before the element's safety can be assessed by application of an appropriate failure theory.

X

σ

τ

στ

2θ2θP

σσ

σσ

σ

τ

–100–600 0

X

Y

Fig F Fig G

τ

σ12 3

X

Y

12 3σ < 0 < σ < σ

Stress, Strength and Safety 10

VARIATION OF STRESS COMPONENTS WITH ORIENTATION

Graphical ApproachSelect a strain scale of, say $ε = 1 E-5 per mm, in which case the points X ( 20, -30 mm ) and Y ( 100, 30 mm )

define the ends of Mohr's circle diameter. The strain circle is then drawn as shown below, it being found that the

centre lies at Cε = 60 mm from the shear strain axis. Converting to the stress circle via ( 7) gives $σ = 1.6 MPa/

mm, Cσ = 100 mm, which latter figure enables the shear stress axis to be located with respect to the circle centre.

Locate T at 2*30o = 60o clockwise from X, hence T ( 54.0, 19.6 mm ) by measurement.

Scaling these stress components : σ = 54.0*1.6 = 86 MPa; τ = 19.6*1.6 = 31 MPa (graphics → two sigfigs)

Finally - and this is what is significant, no matter which convention or technique is used - the stresses on the

oriented element should be drawn.

P2X T Q1 Y P1 Q2 –X300

200

100

0

–1000 30 60 120 150 180

τ

σ= 80σ

=160σ

= 80σ

θ =

18.

4o

θ =

63.

4o

θ =

153

.4o

θ =

108

.4o

Inclination, θ deg

Str

ess,

MP

a

P2

X

T

Q1

Y

P1

Q2

40 mm 60 mm 50 mm

2 θ = 216.8o

P

τ γ/2

σ, ε

Scales -

ε : 10 / mm

σ : 1.6 MPa/mm

–6

2 θ = 60o

T

y

x

X

Y224

48 96

θ = 30To

T31.4

86.4

P2

P1θ = 108.4

oP

240

80

Strain Resolution

Resolution of strain is generally unnecessary when assessing the safety of common engineering com-ponents, however the following description is given for completeness.

If a material behaves in a linear elastic manner then the directions of principal strains are identical tothe directions of principal stresses, and all the preceding equations, and Mohr's circles, may beexpressed in strain terms - provided that everywhere in the stress equations :

normal stress, σ, is replaced by normal strain, ε, and shear stress, τ, is replaced by half the shear strain, ie. by γ/2.

The strain analogues of the foregoing are therefore :-( 3b) ε = 1/2( εx+ εy) ; ε = 1/2√[ ( εx - εy)2 + γxy

2 ] ; θp = 1/2 arctan [ γxy/( εx - εy)] ( 4b) ε = ε + ε cos 2( θp - θ ) ; γ /2 = ε sin 2( θp - θ )

Experimentally, surface stresses are found from strain gauges attached to accessible surfaces. It isuseful therefore to be able to quickly interrelate stresses and strains for a plane stress state, forwhich, since the principal stress and strain directions coincide, it follows from ( 2) and ( 3) that :-( 6 ) E ε = ( 1 - ν ) σ ; E ε = ( 1 + ν ) σ

which relate the centres and radii of the two circles. If these circles are drawn to scale, and it isrequired that their circumferences coincide for ease of drawing, then it may be shown that :( 7 ) E $ε = ( 1 + ν ) $σ ; ( 1 + ν ) Cε = ( 1 - ν ) Cσ

where $ε is the scale of the strain circle ( strain units/mm )$σ is the scale of the stress circle ( MPa/mm )Cε is the distance (mm) of the strain circle centre from the shear axisCσ is the distance (mm) of the stress circle centre from the shear axis

Although the use of scaled Mohr's circles is not necessarily advocated, it is strongly recommendedthat the circles are at least sketched free-hand as an aid to interpretation of the equations. It is impor-tant that skill is developed in visualising the interplay between components and principals.

The following example demonstrates application of strain-to-stress transformation.

EXAMPLE Given the surface strain state : εx = 2 E-4 (ie. 2*10-4), εy = 10 E-4 , γxy = - 6 E-4 , deter-

mine the principal stresses, the planes in which they occur, and the stress components on the t-plane inclined at

30o from the x-reference. Take E = 200 GPa, ν = 0.25 (ie. the material is not steel).

Refer to sketches below.

Analytic ApproachTransforming the given strain Cartesian components into the basic strain triad via ( 3b):-

ε = 6 E-4 , ε = 5 E-4 , θp = 108.4o

Transforming from strain to stress via ( 6) :-

σ = 160 MPa , σ = 80 MPa, (θp = 108.4o )So the principal stresses are :-

σ = σ + σ = 240 MPa at θp = 108.4o

σÿ = σ - σ = 80 MPa at θp ± 90o = 18.4o

and, from ( 4a) the components at the general angle θo are :-

σ = 160 + 80 cos 2( 108.4 - θ )o ;

τ = 80 sin 2( 108.4 - θ )o which are plotted below.

When θ = 30o in the t-direction, then σ = 86.4 ; τ = 31.4 MPa.

Stress, Strength and Safety 11

DUCTILE BRITTLE

(A) PLASTIC BEHAVIOUR Significant Negligible

(B) NORMAL STRESS-STRAIN

Superposition of results of separate tension andcompression tests

Sut

–Suc

comp'n

tension

[ ≡ S ]ySytSyc ≈ SutSuc >>

σ

ε

Syt

–Syc

compression

tension

σ

ε

(C) CRITICAL STRENGTH(S)

Superposition of Mohr's circles for separate tensionand compression tests.

σ

τ

Sut–Suc

SucSut ,

–Sy Sy

σ

τSy

Maximum shear stress aka. TrescaDistortion energy aka. von Mises

( Maximum normal stress )( Internal friction aka. Mohr ) Modified Mohr

(G) TENABLE FAILURE THEORIES

(F) HYDROSTATIC FAILURE

SySh >> SucSh ≈Three circlesdegenerate

to a point

–Sy

–Sh στ

–Suc

–Sh στ

Sh

ShSh

(E) FAILURE MODES ( indicates failure plane )

Tension(normal stress)

Torsion(shear stress)

Failure sense maximum shear sense maximum normal sense

xyτx

y

xσ x

x

x

x

σ

τB

D

X

σB

DX

Superposition ofMohr's circles forseparate tensionand torsion tests.

(D) SHEAR CRITICAL

σ

τ

Sut

Susnormal

limitσ

τSys

Sy

shear limit

Sus Sut≈Sys Sy/0.5 < < 0.6

OBSERVED BEHAVIOURAL DIFFERENCES - DUCTILES v BRITTLESFailure Theories

We have seen that failure of a tensile member occurs when the stress caused by the actual loadreaches the stress limit - the strength - of the member's material. Correlation of the actual stress withthe maximum stress (strength) is straightforward in this case because they are both uniaxial. Buthow can we correlate the triaxial stress state in a component - whose material strength(s) is meas-ured in uniaxial tests - to assess failure tendency?

Unfortunately there is at present no fundamental rationale for any such correlation. We therefore postulate some attribute of the stress state as a descriptor of that state - an attributesuch as the maximum stress or the specific energy - and then compare the values of this attributefor the given component triaxial state on the one hand and the uniaxial test state on the other. Thispostulate is the failure theory based upon the particular attribute selected; it is a useful theory onlyif its predictions are confirmed by experiment.

Currently no universal attribute has been identified which enables prediction of failure of both duc-tiles and brittles to an acceptable degree of accuracy. To appreciate why this is so, it is useful torecall the differences in behaviour of ductiles and brittles, as outlined in the sketch opposite.

- Under normal stress, ductiles' compressive characteristics are approximately the same as theirtensile characteristics (B,C) whereas brittles are significantly weaker in tension - eg. Suc/Sut ≈3.5 to 4 for the cast irons ( Suc & Sut are the compressive & tensile ultimates ).

- A ductile's shear strength is about half its tensile strength, whereas shear and tensile strengthsare about the same in the case of a brittle. Superposition of Mohr's circles of sizes correspond-ing to these proportions (D) suggests that a shear-based attribute is relevant to ductiles,whereas a normal-based attribute to brittles. (Note: the two superimposed circles correspond to com-pletely different stress states - they have been superimposed only to highlight the circles' relative sizes.)This conclusion is reinforced when we consider the directions of the fracture surfaces under

normal and shear loadings (E). Thus in tension, ductiles fail at 45o to the load sense, which isthe direction of maximum shear. This same 45o fracture sense is found with brittles under tor-sion, however in this case the Mohr's circle is centred on the origin thus indicating a maximumnormal attribute for brittles.

- A final qualitative argument involves strength under hydrostatic pressure, for which the threeMohr's circles degenerate to a single point (F). Brittles fail when the pressure correspondsapproximately to the compressive strength, but it is difficult to assess when ductiles fail as thehydrostatic strength is so large - the absence of failure may be equated to the absence of shear.

The upshot of these observations is the adoption of the failure theories indicated - using - shear-based attributes for ductiles- normal-based for the brittles.

These theories are examined below.

We refer to the equivalent stress, σe, for a ductile material only, as the uniaxial stress which has thesame failure tendency as a given triaxial state - on the basis of whichever attribute is relevant. Thesafety factor then follows from ( 1a) as n = S/σe.

Distortion Energy Theory ( aka. von Mises Theory )It may be shown that the specific distortion (or shear strain) energy for a linear material under thetriaxial state (σ1, σ2, σ3 ) is proportional to [ ( σ1 -σ2 )2 + ( σ2 -σ3 )2 + ( σ3 -σ1 )2 ].

For a uniaxial stress ( σe, 0, 0 ), or merely σe, this expression becomes 2σe2. So the uniaxial stress is

equivalent to the triaxial system - ie. gives rise to the same distortion energy - if :-

Stress, Strength and Safety 12

There is no numerical difference between the equivalent stresses predicted by the theories when theintermediate coincides with either extreme; the maximum difference is 13.4% when the intermediatelies midway between the extremes. The maximum shear stress theory is always conservative.

Both theories have their drawbacks in use - the distortion energy theory introduces non-linearitiesinto otherwise linear problems, whereas the maximum shear stress theory requires the ordering ofthe principals - which is easy enough when these are enumerated but most awkward when the prin-cipals are expressed algebraically, as ordering leads to multiple possible expressions for the equiva-lent stress. The particularisation of the equations to plane stress is left to the reader (Problem #16).

Modified Mohr TheoryThis theory derives from the maximum normal stress criterion which historically was the first fail-ure theory proposed for brittles, and is similar to the maximum shear stress theory in that the inter-mediate principal plays no part in the failure process. The concept of equivalent stress is not appli-cable to brittles owing to their differing tensile and compressive characteristics (250 MPa tensile hasa greater failure tendency than 250 MPa compressive) so the safety factor must be obtained from theprincipals and strengths directly, either graphically or numerically.

The maximum normal stress criterion failure locus is shown in σ/σ space at (a) below. The locus isdefined by the two lines σ = St the tensile ultimate, and σÿ = -Sc the compressive ultimate; the twolines meet in the corner x. The pure shear line cuts the failure locus at the shear ultimate Ss - so thistheory has Ss = St which we have seen is typical for brittles. Above the line from origin throughcorner x, failure is dictated by tensile considerations; to the left of this line the compressive mecha-nism is more critical - so, to cater for both tension and compression :-

( i) 1/n = maximum ( σ/St , - σ/Sc )

Experimental failure points are shown; the theory is evidently dangerous to use for design in thefourth quadrant. Mohr's internal friction theory (b) overcompensates as it is unduly conservative inthe fourth quadrant and unrealistic in prediciting a shear strength substantially less than the tensile,however the modified Mohr theory represented by the three linear segments of (c) describes experi-ments well. The third segment is defined empirically to pass through the points ( St, -S t ) & ( 0, -Sc ),thus ensuring equal tensile and shear strengths; the addition of this third segment to ( i) leads to :-

The modified Mohr theory will be used in this course for brittle materials.

It should be emphasised that all the theories examined here relate to the fundamental static failure

EQUALITY

PURE SHEARS t

-Sc

S s

FAILURE

n = 4

n = 2

n = 4/3

( b ) Internal Friction (Mohr) ( c ) Modified Mohr

EQUALITY

PURE SHEARS t

-Sc

S s

FAILURE

n = 4

n = 2

n = 4/3EQUALITY

PURE SHEARS t

S s

σ

-Sc

FAILURE

n = 4

n = 2

n = 4/3

x

EXPERIMENT

( a ) Maximum Normal Stress

σ σ σ

σ σ

( 8 ) 2 σe2 = ( σ1 - σ2 )2 + ( σ2 - σ3 )2 + ( σ3 - σ1 )2

The equivalent stress is often referred to as the von Mises stress, after the propounder.

Maximum Shear Stress Theory ( aka. Tresca Theory )This theory postulates that if a uniaxial stress σe, and a triaxial stress state give rise to the same max-imum shear stress, then the failure tendencies of the two states are identical; that is, σe is equivalentto the triaxial state. So, for the same diameter of the largest Mohr's circle :-

( 9 ) σe = σ - σ where σ = max ( σ1, σ2, σ3 )

σ = min ( σ1, σ2, σ3 )

The intermediate principal - whichhappens to be σ2 in the sketch - hasno effect on failure tendency accord-ing to this theory.

The theory appears as a series of lines at 45o when the minimum principal is plotted against the max-imum principal as shown below. Each point on this graph corresponds to a unique largest Mohr'scircle and hence to an infinite number of stress states when the intermediate principal is considered.

The 45o lines are each asso-ciated with a particularvalue of stress difference, ie.of equivalent stress from (9),and hence of safety factor.No point can appear to theleft of the equality line onthis graph, as this wouldimply a reversal of the mini-mum/maximum roles. Theequivalent stress varies fromzero on the equality line,through to the strength, S,on the failure locus (n=1).The safety factor may beinterpreted graphically as

the ratio of intercepts on the axis of maximum principal, n = S/σe. The safe band extends to infinityin both directions as foreshadowed by the hydrostatic results.Pure shear is represented by the line through the origin, perpendicular to the equality line, cuttingthe failure locus at a shear strength Ss which is half the tensile strength. This tallies with experimen-tal findings as noted above.

Either of the foregoing theories may be applied to ductiles -distortion energy represents experimental results some-what better than does maximum shear stress. The smallnumerical difference between the two theories may beappreciated by defining the intermediate principal's valuein relation to σ, σÿ by the parameter λ ( 0 ≤ λ ≤ 1 ) as shownin the sketch, substituting into ( 8) and invoking ( 9) to give

σe DE /σe MSS

= √( 1 - λ + λ2 ) ≤ 1 which is plotted here.

σ

τ

τ

σE

σ

σ

τ

1 2 3

MAXIMUM SHEAR STRESS THEORY

σ σ

S

-S

Ss

σe-

to ∞

to ∞

σ σ

σ σ

σ σ

σeIMPOSSIBLE

FAIL

SAFE

σ

σ

Failureσσ – = S

Equalityσσ – = 0

Pure Shearσσ = 0 +

eσσ

Constant Safety– = = S / n σ

σ σ

τ

σλ=0 λ=1

normalised intermediate stress, λ

1.0

0

0.866

1.00

max

imum

she

ar s

tres

s σ

e

dist

ortio

n en

ergy

σe

λ + (1−λ) σσ

Stress, Strength and Safety 13

section as it is subjected to the maximum bending tensile stress σb and also to the maximum tor-sional shear stress τt. The diametrically opposite element, for which the bending stress is compres-sive, would be equally critical if the material were ductile, but would be less critical if the materialwere a compressively strong brittle.

Elemental loading is plane stress, with σy = σz = 0, and, from the load building block equations :

bending : σx = σb = M 1/2D / I = M / Z where Z ≡ 2 I / D = π ( D4 - d4 ) / 32 Dtorsion : τxy = τt = T 1/2 D / J = T / 2 Z since J = 2 I for a circle.

Principal StressesResolving in the x-y plane using ( 3a) or Mohr's circle :-

σ = M / 2 Zσ = √ [ ( M /2Z )2 + ( T /2Z )2 ] = √ ( M2 + T2 ) /2Zσ = σ + σ = + [ √ ( M2 + T2 ) + M ] / 2 Z > 0σ = σ - σ = - [ √ ( M2 + T2 ) - M ] / 2 Z ≤ 0

The following argument is important : The maximum principal must be ten-sile; the minimum cannot be tensile; the intermediate must be zero, corre-sponding to the z-component. The x-y circle must therefore be the largest.

σ

X

σσ

M/Z T/2Z

Y

Z

σ

τ

σ

EXAMPLE The cantilevered shaft of fig (a) is loaded only by the 18.2 kN force. Ascertain the factor of safety at

the hatched cross-section if the shaft is ( i) ductile with a yield of 350 MPa, or

(ii) brittle with tensile and compressive ultimates of 300 and 750 MPa.

Find all loads on member, resolving any static indeterminacies if necessary The 18.2kN force is the sole load here.

Identify load building blocks and potentially critical element(s); evaluate corresponding stress components

Constructing a free body embracing the load(s) and the required cross-section, the load building blocks are

found here to include bending, torsion and direct shear - fig (b). The cross-sectional element of fig (c) is

identified as the most critically loaded; the stress components on it are :

bending σx = My/I = 16380*50/π/64 (1004 -524) = 180 MPa

torsion τxy = Tr/J = 21840*50/π/32 (1004 -524) = 120 MPaThe direct shear stress maximises in the centroidal x-y plane at 4V/3A

= 4*18.2E3/3*π/4 (1002 -522) = 4 MPa but vanishes at the element in

question; its neglect is justified. Other stresses on the element are zero.

Resolve stresses to ascertain the three principals

Stresses on the x-y faces of the element are resolved by Mohr's circle fig

(d) to give the principals in the x-y plane. Since there are no z-stresses,

the three principals are 240, 0, -60 MPa.

Apply the relevant failure theory

Ductile - Distortion energy ( 8) σe = √[[ (240-0)2 +(0-(-60))2 +(-60-240)2 ]/2] = 275 MPa ; n = 1.27

Max. shear stress ( 9) σe = ( 240 - (-60) ) = 300 MPa ; n = Sy /σe = 1.17Either theory may be applied to ductiles; the maximum shear stress is clearly more conservative.

Brittle - Modified Mohr ( 10) 1/n= max( 240/300 , 60/750 , 240/300 - 180/750 ) = 0.8 ; n = 1.25

18.2kN

dimensions in mm

Ø 52

Ø 100

1200 900

(b)

16 380 Nm

21 840 Nm

MPa

(c)x

yz

σ = 0y

σ = 0z

τ = 120xy σ = 180

x

(a)

18.2 kN

(d)

2400-60

90

120

150

90

150

τ

σZ

(180,120)X

Y

( 0,-120)

mechanism. Other common mechanisms, including fatigue and fracture of ductile materials in a brit-tle fashion, will be examined later in the course.

Putting It All Together

We have examined the individual steps in assessing safety of any element of a member :- resolving indeterminacies if necessary to discover loads on the component of which the ele-

ment is a part,- application of load building blocks to establish stresses on the element,- resolution of stresses to determine principal stresses, and- application of a failure theory which is relevant to the element's material.

These steps are fundamental and applicable to members of any form - for example we apply themopposite to a particular cross-section of a hollow circular shaft.

This example concerns a certain hatched cross-section. But what about the other sections, and in par-ticular the most severely loaded section? To establish the minimum safety factor for the shaft, asearch must be conducted over all likely looking cross-sections, undertaking all the above steps foreach. However we avoid unnecessary repetition by formulating a shaft design equation - an alge-braic equation based on the above steps - which may then be applied directly and immediately toeach cross-section to evaluate the safety factor.

Design Equations for (Static) Round Shafts

The foregoing steps are strung together (algebraically, rather than numerically as in the foregoingexample) to derive the design equation for a shaft's circular cross-section of outer diameter D andinner diameter (or ‘bore’) d. The equation will interrelate for the cross-section the four aspects identi-fied previously :- load (the stress resultant) consisting of a torque T and a bending moment

M; direct stresses are presumed negligible compared to bending and tor-sional stresses as in the above example

- dimensions, characterised by the section modulus, Z ≡ 2 I / D- material, of tensile strength S (yield or ultimate as appropriate)- safety factor, n, at the cross-section under consideration.

Load Building BlocksThe surface element lying on the z-axis is the most critical in the cross-

T M

xy

^tτb

σ

τ xyσx

criticalelement

z

EXAMPLE A component is made from a brittle material of tensile

and compressive strengths 200 and 600 MPa respectively.

The table illustrates how the safety factor is evaluated at each of

three different points (a), (b) and (c) on the component, the stress

state at each point being defined by the three given principals which

are tabulated.

principals σ σ 1/n from ( 10) factor

(a) ( 100, 75, 50) 100 50 max( 1/2, -1/12, 1/4) = 1/2 2.0

(b) ( -200, 10, 50) 50 -200 max( 1/4, 1/3, 1/2) = 1/2 2.0

(c) ( -200, -300, -100) -100 -300 max( -1/2, 1/2, 1/6) = 1/2 2.0

The selected points happen to lie on the n = 2 locus as illustrated.

S t

bc

a

σ

–Sc

S = 600 MPac

S = 200 MPat

n = 2

n = 1

MODIFIEDMOHRTHEORY

σ

Stress, Strength and Safety 14

50

250a

b

c

d

Ø 25

Ø 22

1 kN

Ø 30

e

200

EXAMPLE The torque wrench is made from a 350 MPa yield steel and loaded

by 1 kN. Ascertain the safety factor based on the maximum shear stress theory.

First identify those cross-sections which appear likely to be critical - a, b and c

are all subjected to a torque of 200 Nm. c is of the same diameter as b and the

bending moment there is less than that at b, so c may be neglected as being

less critical than b.

Although the bending moment at b is less than at a, the diameter is also less, so it is

not immediately clear which of the two sections is the more critical. At d, although there

is no torque, the diameter is again different so the degree of safety is not obvious. Cross-

sections a, b and d must therefore all be included in the search for the weakest link.

cross-section a b d

M ( Nm ) 300 250 200

T ( Nm ) 200 200 -

Me ( Nm ) = √( M2 + T2) 361 320 200diameter, D ( mm ) 30 25 22 From this table, it is evident that cross-section b

Z ( mm3) = π D3 / 32 2650 1535 1045 is the most critical, since the safety factor is

σe ( MPa) = Me / Z 136 208 191 smallest at that location.

n = 350/σe 2.6 1.7 1.8 The shaft safety factor is min ( 2.6, 1.7, 1.8) = 1.7

grams will help visualisation.- The wrench has been designed quite well as the safety factors for the components - the shaft and

the handle - are about the same. That is, there is no markedly weakest link.- Two significant figures in the answer are more than adequate, since the yield is given only to this

accuracy, and stress concentrations in way of the diameter change at b have been neglected, ashave the contact stresses in the hole for the handle, etc, etc.

Avoid excessive significant figures - they usually give a completely erroneousimpression of a mathematical model's inherent approximation of reality.

Having derived the design equation for shafts, it's appropriate to now apply it to the design ofpower transmission shafting, noting that at this stage we're obliged to employ large design factors toaccount for the fact that the static theory on which ( 11) are based can only describe fatigue effectsvery approximately. We shall apply fatigue theory later in the course, and derive in a manner simi-lar to the above a more practical design equation than ( 11) specifically for rotating shafts.

Failure Theories

Distortion Energy - equation ( 8) gives :-σe = √ ( σ 2 - σ σ + σ 2 ) and, since n = S/σe it follows that

( 11a) Z S = n √ ( M2 + 3/4 T2 )

Maximum Shear Stress - from equation ( 9) :-σe = σ - σ = √ ( M2 + T2 ) / Z so that

( 11b) Z S = n √ ( M2 + T2 )

Modified Mohr - equation ( 10) with the convenient substitution Q ≡ √( M2 +T2 ) ≥ M becomes :-2Z/n = maximum ( ( Q+M)/St , ( Q-M)/Sc , ( Q+M)/St - 2M/Sc ) The first term must always be the largest of the three as its numerator is larger than thesecond's and its denominator smaller; it is clearly larger than the third term. So :-

( 11c) Z St = n [ √ ( M2 + T2 ) + M ] / 2 ; the compressive strength is evidently irrelevant.

Equations ( 11) are the design equations according to whichever failure theory is being employed.Building blocks, resolution and failure theories are essential steps in any safety evaluation, but thedesign equations, being based on these steps, obviate the need to carry them out in detail every timea shaft is examined - provided the loading assumed in the derivation of the equations applies.

The limitations of the design equations here must be clearly understood :- Direct and contact stresses have been ignored (see above), as have stress concentrations.- The failure theories are static theories, therefore ( 11) are strictly applicable only to stationary

shafts. Power transmission shaftsrotate and are subject to bending,so every element is subject toalternating tension and compres-sion. This reversal of stress ismuch more damaging than staticstress of the same magnitude and will later be examined more thoroughly using fatigue theory.However it is permissible to employ ( 11) for the elementary design of rotating shafts, providedsuitably large factors of safety/ignorance are incorporated.

It is sometimes useful to consider an equivalent bending moment Me, which, acting alone, is equiva-lent failure-wise to a given combination of bending and torsion. The design equations then all takethe form: Z S = n Me where Me is given by the right hand sides of ( 11) appropriate to the theory inuse, thus eg. : Me = √ ( M2 + T2 ) for the maximum shear stress theory.

The cross-section with the lowest safety factor is the most criticaland defines the safety factor for the whole shaft

Equations (11) must therefore be applied at all likely cross-sections in a search for the most critical. Itis not enough to consider only the cross-section with the largest M, nor is it enough in a shaft withvarying diameters to consider only the section with the largest Me. This is demonstrated in theexample opposite.

The following points should be noted about this example :-- The specified loading is rather uncommon; an upward reaction is usually provided at c-d.- Once the stress resultant is evaluated at each cross-section, the design equation is applied directly,

without having to consider the general steps of load building blocks, principal stresses . . . .- The 200 Nm bending moment in the handle at d, becomes a torque of the same amount when the

corner is turned, in the shaft at c. If stress resultants are difficult to appreciate, clear free body dia-

compression

tension

half-a-turn later

exaggerated deflections !

Stress, Strength and Safety 15

or pitch cylinder of diameter D. The load on the spur geararises from inter-tooth contact with its mating gear and com-prises two components - the useful tangential component Ft

and the unwanted but unavoidable radial component Fr (com-monly 0.36Ft ). Gear forms other than spur (see Spur Gears fron-tispiece) give rise also to a load component parallel to the shaftaxis - but for all gears, shifting the offset force as above, T = Ft.D/2.

A belt, being flexible, cannot withstand compression - the pulley is therefore subjected to two strandtensions F and F both of which must be non-zero. The net torque T = ( F - F ) D/2 is clockwise here.The tension ratio F/F must lie between a lower limit of unity (corresponding to no torque transfer)and an upper limit dictated by incipient belt slip around the pulley. ( This may be demonstrated byequally pulling the ends of a length of rope wrapped around a post, then progressively increasingthe pull at one end.) A chain sprocket is similar though the minimum tension may drop to zero dueto the positive drive not relying on friction.

The mounting of a half-coupling and the like on a shaft is, locally, intrinsically indeterminate. This isexemplified by the pulley boss illustrated which extends from A to C along the stepped shaft. Theactual distributions of bending and torque in the shaft are shown. Tothe left of A and to the right of C there are no immediate transverseloads, so the bending moment is linear. But within AC the boss contrib-utes to the shaft's stiffness and also distributes the pulley load graduallyinto the shaft, so the bending moment variation is non-linear. Similarlythe torque is some constant value (zero here) in the shaft to the left of Aand another constant value T to the right of C - within AC the torqueincreases in a complex indeterminate fashion.

No attempt is made to resolve this bending/torsional indeterminacy -the actual distributions are instead replaced by approximate distribu-tions based on external loads concentrated at the mid-point B. Thisleads to a bending moment diagram characterised by straight segmentsand a torque diagram which increases stepwise at B. The approximationis justified in view of unknown stiffness details and stress concentrationdue to shoulders (used for positive longitudinal location) and mount-ings (square keys, shrink fits &c). In this example a cross-sectionbetween B and C is critical since the diameter is the smaller DA (neglecting the blending radii whichreduce stress concentration at the shoulder), the torque is the fully developed T, and the bendingmoment is to all intents and purposes the peak MB.

Shafts are supported in two bearings (slid-ing or rolling) which allow the shafts toturn freely - there is no appreciable torqueexerted by the bearings. A sliding bearingneeds a lubricant film in the clearancespace between shaft and bearing bush; inthe fully hydrodynamic bearing illustrated,oil is dragged into the wedge-shaped gapcausing a pressure build-up (similar to thatin hydroplaning) which supports the shaft without metal-to-metal contact and little friction.

A rolling bearing (of which the ball bearing illustrated is but one example) comprises two hardened

DA D

C

C

ACTUAL

moment

0

A

B

APPROX.

moment

torque

shaft

boss

torque0

T

T

Fr

spur gear pitch cylinders

D

F

F

beltstrands

DFt

clearance fit

oil film, thicknessexaggerated

bush

press fit

high

SLIDING BEARING

lowpressure

innerrace

outerrace

ball

ROLLING BEARING

Power Transmission Shafts

Continuous mechanical power is usually transmitted along and between rotating shafts. The transferbetween shafts is accomplished by gears, belts, chains or other similar means for matching thetorque/speed characteristics of the interconnected shafts - eg. a car needs gears between the enginecrankshaft and drive wheel half-shafts. Matching will be examined more closely in later chapters(Squirrel Cage Motors, V-Belt Drives, Spur Gears), here we focus on shafts.

Shafts rotating only at constant speed n (rev/s) are considered here, and as shafts are usually stati-cally determinate they may be examined by the techniques of elementary statics. Also, since

power = force ( N) * linear velocity ( m/s) in translational applications andpower = torque ( Nm) * angular velocity ( = 2πn rad/s) in rotational applications,

then it follows that torque is a major load component in power transmitting rotating shafts.

Torque may be transferred to or from the end of one shaft by a secondcoaxial shaft - this is a pure torque, a twist about the shaft axis. The transfer is accomplished by a shaft coupling (illustrated inSquirrel Cage Motors frontispiece) which consists of two oftenidentical half-couplings which are each mounted on one of theshafts before being connected together non-permanently.

The mounting of a half-coupling on a shaft must prevent relative rotationwhen torque is applied - this is achieved by a positively interlocking ele-ment exemplified by a key (or a derivative of this such as a spline) or by friction. Shown below is thefree body of a shaft keyed to a coupling half - the torque on the shaft is equilibrated by an equal andopposite torque on the half-coupling.Free bodies of the individual elementsclarify how the key's faces give rise toan equilibrating couple on both theshaft and the half-coupling (refer to thetutorial problems for more details).

The typical friction-based mounting below comprises a split tapered bush which is forced by screwsinto a mating taper machined in the bore of the half-coupling. The bush contracts onto the shaftcausing highpressures andcorrespondinghigh surface fric-tion which equili-brates the torque -provided tighteningis sufficient.

Torque may be transferred also at any point along a shaft by a gear,belt pulley, or chain sprocket for example, mounted on the shaft asdescribed above. These common elements apply forces offset fromthe shaft axis, and therefore useful torque T must be accompaniedby a corresponding radial load with consequential bending - this iswhy the generalised shaft design model ( 11) incorporates bothtorque and bending moment load components.

A spur gear and a belt pulley are sketched below, each subjected to loading tangential to its effective

shaft 1

T

shaft 2

shaftcoupling

FBD of shaftT

FBD of half-couplinghalf-coupling

T

Tshaft

key

T

FBD of shaft

TT

FBD of half-coupling

3 set-screwsT

splittaperbush

T

force on pulley or gear &c

equivalenteffect onshaft

F

T = Fr

Fr

Stress, Strength and Safety 16

introduced when the gear's tangential force is shifted to the axis is Ft*0.05 Nm, assuming Funits are N. For rotational equilibrium therefore 0.05 Ft = 120, ie. Ft = 2400 N. Finally Fr maybe taken as 0.36 Ft = 864 N. All external loads are now known, so they may be resolved intoorthogonal planes, figure (c) prior to calculating bending moments. Note that figure (f) is not a complete free body, however the bearing reactions which are notshown do not contribute to rotational equilibrium.

- As shaft loading is concentrated, it is usually easier to calcu-late bending moments from standard results rather than toprepare a complete bending moment diagram from firstprinciples. Thus a body of length L between simple supportsis loaded by a transverse force F located at p from one sup-port. The bending moment at a point Q distant q from theother support is derived in the sketch as

( 12) MQ = F pq/L a result which can be applied tosuperpose the bending effects of multiple loads, though careis needed when load senses differ or loads overhang the sup-ports - it is suggested that the steps shown in the sketch be retraced for these cases.

- Tutorial problem #17 (very important) proves that the maximum equivalent loading Me cannotoccur between concentrated loads, so it is necessary to search for the critical cross-section onlyamongst those cross-sections where (or close to where) loads or supports occur.

The following example demonstrates a typical rudimentary design procedure.

= F

FQ

F Fp

L

Fp

LM

Q

pq

L

qL

p

races between which a number of identical hardened elements roll. Not shown are various necessarydetails such as means for preventing axial movement of races, for dust sealing and for lubricantretention.

There are many different embodiments of these two bearing classes. A bearing supports a shaft overa finite axial length and so is intrinsically indeterminate no matter what its form - however it isapproximated like the boss ABC above by a concentrated reaction at its mid-point.

The transverse forces on a shaft are seldom coplanar, so it is usually necessary to resolve forces intotwo convenient orthogonal longitudinal planes and to separately consider bending in each. Forexample the horizontal shaft ABCD, figure (a), is simply supported in bearings at A, D and carriesgears or pulleys at B, C.

The offset gear/belt forces are shifted to the shaft axis as above, giving rise to the shaft bendingloads, figure (a). The corresponding torque diagram figure (b) shows that torque is transmittedbetween B and C (there is no torque sink outside BC as the bearings can offer no appreciable tor-sional resistance). The bending loads are resolved into vertical and horizontal planes, figure (c).Treating each plane separately, the bearing reactions are computed and thebending moment diagram completed, resulting in two component BMDs, figure(d). Since moments are vectors, the resultant bending moment at any cross-section along the shaft can be found easily from the two components by Pythag-oras as M = √( MH

2 +MV2 ), recalling that it is the moment magnitude only which

is relevant in ( 11).

Finally, some practical tips for shaft analysis :-- Equilibrium in the axial view is usually examined first, to figure out the magnitude of the

torque(s) and the locations between which torque is transmitted. Thus the shaft, figure (e)similar to ABCD above, is freely supported inbearings and equipped with- a 150 mm diameter belt pulley with strand

tensions given as 2100 and 500 N- a spur gear of 100 mm pitch diameter with

unknown tangential and radial forces.Each force is shifted if necessary to passthrough the shaft axis whilst introducing thecorresponding torque, figure (f). Thus, starting

with the two known belt forces, the net clockwise torque due to their shift is ( 2100 - 500 )*0.075= 120 Nm. This is the torque T transmitted between belt and gear. The anticlockwise torque

A

B

C

DFB

FC

TBC

A B

C

D

( a )

( b )

M BH

M CH

M CV

M BV

A

B

C

D

( d )

vertical

horizontalA

D

( c )

FBH

FCH

FBV

FCV

B

C

M H

M V

M

ø 100

ø 150

500 N

2100 N

( e )

FrT = 0.05 Ft

500 N

2100 N

120 Nm( f )

bearing supportreactions not shown

Ft

Fr

Ft

Stress, Strength and Safety 17

Section c is straightforward as there is no torque transfer in its vicinity -ie. any torque passes right through the section. So a table is set up forthese two sections, and the distortion energy failure theory eg. isapplied :-

This simplified analysis indicates 47 mm as being suitable for bothdiameters - there is little difference between these because the commontorque is the dominating load component. Considerations other thansafety might require diameters different from but not less than 47mmbecause :- further shoulders may be necessary to afford positive axial location of other components not shown- ø 50mm stock might be available off-the-shelf and require no machining prior to use as the intermediate shaft- bearings of the required load capacity might only come in diameters of 48 mm, 50 mm, etc- an existing pulley, suitable apart from being bored out to ø 50 mm, may yield the cheapest solution, and so on.

The raised diameter of the shaft to the left of c has not been considered - as long as it is greater than 47mm then itis safe. This diameter would be selected on grounds other than safety, eg. it must be large enough for positiveabutment of the boss and RH bearing's inner race (considering assembly and the possibility of possible stray longi-tudinal forces in operation), but not too large to interfere with the bearing's outer race.

This shaft has worked out very short for its diameter, so shear effects are likely to be significant. In practice, amore rigorous analysis acknowledging fatigue would therefore be carried out for complete confidence in thedesign.

M = 2552*75*135 / 210 - 4443*60*75 / 210 = 28 NmM = 2552*75*60 / 210 - 4443*60*150 / 210 = -136 Nm

b

c

M = 3580*75*135 / 210 - 2314*60*75 / 210 = 123 NmM = 3580*75*60 / 210 - 2314*60*150 / 210 = -22 Nm

b

c

371 N

2552 N

4443 N

2262 N

75 mm

60 mm75 mm

ab

cd

28 Nm

136 Nm

ab c

d

Fig D - Vertical

75 mm

60 mm75 mm

a

b c

d

1640 N

3580 N

2314 N

374 N

123 Nm

22 Nm

ab c

dFig E - Horizontal

M H

M V

section b c

MV (Nm) 28 136

MH (Nm) 123 22

T (Nm) 255 255

Me ( 11a) (Nm) 254 260

Z = Me/σd (mm3) 10160 10400

D = 3√[32 Z/π] (mm) 47 47

Bibliography

ASM Handbook v20: Materials Selection and Design, ASM 1997Jones DRH, Materials Failure Analysis: Case Studies and Design Implications, Pergammon 1993Juvinall RC & Marshek KM, Fundamentals of Machine Component Design, Wiley 2000

EXAMPLE The layout of a two-stage 20 kW V-belt speed reduction is shown below. The first stage consists of

a φ 125 mm engine-driven input pulley driving a φ 400 mm pulley mounted on an intermediate shaft. The φ 200

mm small pulley of the second stage is also mounted on this intermediate shaft, and drives the φ 630 mm outputpulley connected to the load. Tensions obtained from belt analyses are indicated on the sketch.

The intermediate shaft abcd has to be designed; the preliminary layout illustrated is based upon known widths of

the pulleys and estimated widths of the bearing housings. The second stage small pulley is placed between the

bearings as the belt loads are higher than those of the first stage, whose lesser loaded pulley is overhung.

Select suitable diameters D 1 and D2 for the intermediate shaft given that the ductile material’s design stress ( σd =S/n) is 25 MPa, a very low value to allow for unknown effects of fatigue, stress concentration due to keyways &c.

The shaft design equation may be applied since there is (more than) enough information to determine the shaftloading. Fig A illustrates the belt tensions - pulley weights are negligible in comparison. As outlined above, eachstrand's tension force, F, is shifted perpendicular to its line of action to act through the shaft axis (Fig B) thus intro-ducing a torque of magnitude F*r where r is the strand's pulley radius. When viewed from end a, the net effect of

this is a 255 Nm anticlockwise torque input at pulley b, equilibrated by an equal and opposite torque extracted atthe other pulley d. The corresponding power transmitted along the shaft from d to b, from first stage to second, is :

P = ωT = [2π*(125/400)*40]*255 = 20 kW

In Fig C the forces are resolved vertically and horizontally - usually this is sufficient for the calculation of bendingmoments via (12). However because the shaft is not supported at its extremities here, as is (12)'s model, the bear-ing reaction at c is first calculated then treated as a load in the application of (12), with the results tabulated on theright above. The complete loading and bending moment diagrams are given for interest in Figs D and E.

The cross-section b here is analogous to section B of the stepped shaft examined above, and is analysed similarly.

3535 N

990 N

547 N

1820 N

Ø 630

Ø 125

Ø 200

Ø 400

45o

17.1o

16.6o

500 mm

752

cent

res

belt tension

SECOND STAGE: 5B2870

FIRST STAGE: 4SPZ1800

467

cent

res

7575 60

a dcb

D1 D2

INTERMEDIATE SHAFTPRELIMINARY LAYOUT

Not to scaleDimensions in mm

17.1o17.1o

28.4 o

990 N

547 N 1820 N

3535 N

255 Nm

255 Nm

28.4 o

Fig B

bTTd

2552 N

255 Nm

255 Nm

d

3580 N

374N

2262 N

ab c

255 Nm

axial view a-b-c-d

Fig C

547 N1820 N

3535 N

a

cd

ø 400

Fig A

990 N

b

ø 200

Stress, Strength and Safety 18

can be seen qualitatively by imagining A and B to be made of steel and the connectors to take the form of rubber blocks.

Since all connectors deform in an identical manner they must transmit the same force, 1/6 or 16.7% of the external load

on the assemblage, F0.

The compliance (flexibility) of A increases in the three other distributions illustrated above. The area under the distribu-

tions remains the same (F0), but it is apparent that the connectors at the right hand end of the assembly (where the

unconnected A meets the connection assemblage) progressively take more of the load, until as cA/cC → ∞ the first con-nector at the right hand end takes all 100% of the external load. This is the case argued qualitatively earlier when a flexi-

ble rubber bolt in a steel nut with steel connecting threads, stretches such that all the load is taken by the first engaged

thread adjacent to the pulled bolt head.

Some further results from (iv), in which B becomes progressively more flexible with respect to A, are also presented.

As may be seen, the effect of B's increasing flexibility is to place more load on the connectors at the left hand end of the

connection assembly adjacent to the unconnected B, until, when the compliances of A and B are equal as in the rightmost

graph, the load distribution is symmetric. If B's compliance were to further increase with respect to that of A, then the

distribution would become skewed with the larger peak at the left hand end.

Although this simple model is somewhat unrealistic - for example in its assuming A and B are in tension only - the

resulting load distributions are certainly realistic and confirm the general conclusion that the load distribution between

components of a statically indeterminate assembly depends upon the relative stiffness of the components.

A similar situation arises when the connector is continuous as in the case of a continuous weld C between A and B. The

non-uniform load transfer is often referred to as shear lag and will be examined in the chapter on Welded Joints.

cA

cC

/ =cB

cC

/ =0

016.7 %

29 %

62 %

92 %

cA

cC

/ =cB

cC

/ =0.1

0

cA

cC

/ =cB

cC

/ =1

0

cA

cC

/ =cB

cC

/ =10

0

LOAD SHARING BETWEEN SIX CONNECTORSWITH COMPONENT 'B' RIGID

79 %

13 %

66 %

26 %

53 %

40 % 40 %cA

cC

/ =cB

cC

/ =3

0

cA

cC

/ =cB

cC

/ =2.5

0.5

cA

cC

/ =cB

cC

/ =2

1

cA

cC

/ =cB

cC

/ =1.5

1.5

EFFECT OF VARYING RELATIVE STIFFNESSOF BOTH TENSILE COMPONENTS

APPENDIX : INDETERMINATE ASSEMBLIES WITH MULTIPLE COMPONENTS

The model consists of two members A and B which are

joined by a number of discrete identical connectors C - the

assembly being subjected to an external tensile load F0. A

and B are assumed to be in tension everywhere. Between the

connectors, each segment of A and of B has an unloaded

length L, and a compliance (reciprocal stiffness) of cA and cBrespectively. The precise manner in which each connector deforms is irrelevant, however its compliance cC is known,

and so the deflection of a connector is given by δC = cCF, where F is the force external to the connector exerted by A & B.

We want to find out how the load transfer between A and B is shared between the various

connectors - A could represent a bolt, B a nut and C a thread. The example illustrated will

be examined - there are N = 5 inter-connector segments of A and of B, and N+1 = 6 connec-

tors. Results for other numbers of connectors may be deduced by inspection.

COMPATIBILITY

The assemblage is shown deformed as a result of the

external tensile load F0. The extensions of succesive

inter-connector segments of A are δA1, δA

2 . . . etc,

and of B δB1, δB

2 . . . etc. The deflections of succes-

sive connectors are δC1, δC

2 . . . and so on asshown. From the diagram, compatibility requires

( i) δCn+1 + δA

n = δCn + δB

n ; 1 ≤ n ≤ N = 5

EQUILIBRIUM

Let the forces in the inter-connector segments of A be

F1, F2 . . . etc, while for equilibrium (of the RHS

free body eg) the corresponding forces in B must

be F0-F1, F0-F2 . . . and so on, as sketched.

A free body of the typical n'th connector illustrates

that the A/B force which is transferred through

the connector is Fn-1-Fn.

CONSTITUTIVE LAWS

For A and B in tension :

(ii) δAn = cA Fn and δB

n = cB (F0-Fn) ; 1 ≤ n ≤ N = 5

and for the connectors :

(iii) δCn = cC (Fn-1-Fn) ; 1 ≤ n ≤ N+1 = 6 ; F6 = 0

SOLUTION

Inserting the 3N+1 constitutive laws into the N compatibility equations leads to the final N equations from which the N

unknown forces may be found :

( iv) M F = Fo where F = [ F1 F2 F3 F4 F5 ]' is the unknown force vector

while Fo = F0 cB/cC [ ( 1 + cC/cB ) 1 1 1 1 ]' is known,

as is M = m -1 . . . -1 m -1 . . . -1 m -1 . in which : . . -1 m -1 m = 2 + (cA+ cB)/cC . . . -1 m

The equations for N other than 5 may evidently be written down immediately.

The fraction of the total load which passes through each of the N+1 =6 connectors, as found from solution of (iv), is

shown in the distributions below for B rigid and for various ratios cA/cC .

In the leftmost illustration both A and B are rigid therefore all connectors experience the same deformation - a fact that

123F

F2 F1

0F –F2 0F –F1

04

F4

0F –F4

F3

0F –F3

123456

F5 F4 F3 F2 F1

F00F –F5 0F –F4 0F –F3 0F –F2 0F –F1

F0

L

123456

L L L L

C

B

AF0

F0

F

δC

FFF

δC

L+δB5

δ C6

L+δB4

L+δA4

L+δB3

L+δA3

L+δB2

L+δA2

L+δB1

L+δA1L+δA

5

δ C5 δ C

4 δ C3 δ C

2 δ C1F0

F0

Stress, Strength and Safety 19

rary load Fo, and the concrete poured into a mould surrounding it. When the concrete has set,the mould and the load Fo are removed. Since there is no relative longitudinal displacementbetween steel and concrete, the steel relaxes somewhat and compresses the concrete - this pre-stressing occurs before external loading of the composite.An external tensile load, F, is finally applied to the composite member.Ascertain the tensile force in the concrete matrix when the composite member is loaded, in termsof the loads Fo and F, together with the longitudinal stiffness of the steel rod, ks, and of the con-crete matrix, kc. [ Fconcrete = ( F - Fo )/( 1 + ks/kc ) ]

6 A press consists of the central screw 1 threaded through thebeam 2 which is connected to the base by the two identical tie-rods 3. All components are steel; their effective dimensions are- the pitch of the central screw is 3 mm, its diameter is 20 mm

and its length 250 mm;- the beam is 30 mm wide, 60 mm deep, and spans 250 mm;- the 250 mm long tie-rods are each 15 mm diameter.The screw is rotated manually until it bottoms snugly on thebase. What is the stress in the tie rods when the screw is there-after rotated by quarter of a turn? Neglect base deflections andbuckling effects. [ 209 MPa ]

7 (More difficult) The annular disc, of radii ri and ro and thickness b, issupported around its outer periphery. A load is transmitted uniformly toits inner periphery by shear. Assuming that the shear in the disc atradius r is uniform, estimate the stiffness experienced by the load if it is :- ( a ) an axial force, F [ 2 π b G / ln ( ro/ri ) ]

( b ) a torque, T [ 4 π b G /( 1/ri2 - 1/ro

2 ) ]

8 A hollow plastic tube is bonded to two identi-cal circular rubber support collars. The allowa-ble shear stress in the adhesive is 100 kPa. Anaxial force and a torque are applied to one endof the tube as shown.Use the results of the previous problem to deduce the bond's safety factor. Values of E, ν are : for plastic 2 GPa, 0.35; for rubber 2 MPa, 0.5. [1.05]

9 A plastic ball is inflated to produce membrane stresses σx = σy = 2 MPa. The radial thickness ofthe material is 0.7 mm before inflation. By how much will the thickness change after inflation ifthe moduli of elasticity and of rigidity are 3.5 and 1.4 GPa respectively ? [ 0.2 µm ]

10 A tensile test at 125MPa is conducted ona specimen of a component's material.Strain gauges attached to the specimenand to the component in service indicatethe strains shown. Find the principalstresses in the loaded component.

[ -80, 120 MPa, -70o ]

250

250

60

30

1

2

3 3

base

b

F

T

500 N

20 Ø 100

Ø 60

3

600

10 Nm

–177

604

specimen component

–410

–40

655 x125 MPa

strains x1 0 6

Grade A B C DYield (MPa) 320 360 410 450Relative cost 1.00 1.09 1.14 1.20

STRESS, STRENGTH AND SAFETY - PROBLEMS

Take E = 207 GPa, ν = 0.29, ρ = 7.8 t/m3 for steel, unless indicated otherwise.Sketches are not to scale; dimensions are in mm.

1 A tensile test on a steel specimen, of 12.5mm diameter and 50 mm gauge length, gave the follow-ing results :

Load (kN) 26 36 46.5 54.5 71 75 80.5 85Elongation (mm) 0.05 0.07 0.09 0.11 0.15 0.20 0.31 0.41

Estimate the 0.2% yield strength and the elastic modulus. [ 640 MPa, 207 GPa ]

2 A rod of length 800 mm has to withstand a tensile force of2 kN without its length increasing by more than 1 mm.The safety factor is laid down as 2.5 minimum.Steel rod is available off-the-shelf in any one of the fourgrades shown, and in the diameters : 2 2.5 3 4 5 6 8 10 12 16 20 24 and 30 mm.Select a suitable grade and rod diameter for this duty.

3 This example introduces concepts which will be used later in the treatment offlanged joints. The φ 18 mm eyebolt (1) is assembled through the 20 mm borehole in the 35 mm O.D. sleeve (2), with the nut just fitting snugly. The nut isthen tightened to produce an initial force in the assembly, and the load, P,finally applied. The maximum allowable stresses are 550 and 80 MPa for thebolt and sleeve respectively, and the corresponding moduli of elasticity are 207and 50 GPa. What is the maximum load that the assembly can withstand with-out loss of contact, and what initial force is necessary ?Hint : Plot vs load the magnitudes of the forces internal to the components.

[ 136, 52 kN ] 4 Three 0.5 m long bars which are nominally identical and

made from a 250 MPa yield steel, are connected by twopins and loaded by a force of 15 kN. The bars weredesigned on the assumption of equal load sharing and asafety factor of 2.5. However manufacturing inaccuracieslead to the central bar's length differing by 0.2 mm fromthe outers' length, thus requiring one of the pins to be forced in. Neglecting buckling, determinethe actual safety factor of the assemblage if ( a ) the central bar is the longest [ 2.0 ] ( b ) the central bar is the shortest [ 1.6 ]

5 The sketch illustrates the essentials of a reinforced concrete composite member which may beloaded longitudinally either in tension, as indicated, or in compression - depending upon loads on the adjacent structure. The member consists of a steel rodembedded in a concrete matrix. Byitself, the rod would be weak in com-pression due to buckling; how-ever the compressively strong concrete prevents any buckling. Conversely, concrete by itselfwould be weak in tension, but the steel component protects it from undue tensile effects. Thetwo components of the composite are thus seen to complement one another.The member is manufactured and prestressed as follows. The steel rod is stretched by a tempo-

steel rod

concrete matrix

F F}{

2501

2

P

15

kN

Stress, Strength and Safety 20

Plot the failure locus in (σ1, σ2) space and superimpose the failure locus of the maximum shearstress theory for plane stress.

17 (important) The two variables y1 and y2 are each linearly dependent upon x in the interval 0 ≤ x≤ L. The variable z is defined as z = √( y1

2 + λ2 y22 ) where λ is a constant - all parameters being

real. Prove that any maximum of z must occur either at x = 0 or at x = L - a result which will befound very useful.

18 A uniform solid shaft ABCDE is supported in bear-ings at A and D, and rotates at 900 rev/min (rpm).50 kW are delivered to the shaftthrough the φ 560 mm belt pulleyat C. 30 kW are extracted via theφ 280 mm pulley at B, and 20kW at the φ 210 mm pulleyat E. At each pulley, thetwo belt strands are paralleland the ratio of tensions in them is 3:1. Determine the minimum acceptable shaft diameter if the design stress of 100 MPa incorporatesan allowance for fatigue. [ φ 40 mm ]

19 The drill brace a-b-c-d-e-f-g is made from abent rod of 450 MPa yield steel, and isloaded as shown by the operator. Thesimple support at g also provides thetorque reaction necessary for equilibrium. What is the factor of safety? [ maximum shear stress 1.21; or distortion energy 1.22 ]

20 The horizontal shaft ABCD is mounted in bearings at B and D as shown. A belt passes around the 250 mm diameterpulley fixed to the shaft at A, and a gear pinion of 150mm pitch diameter is mounted on the shaft at C. Shaftdiameters and axial disposition of the components areas sketched.The belt strand tensions are horizontal and in the ratioF1/F2 = 4, while the vertical reaction on the pinion, P,acts tangentially to the pinion's pitch circle.Ascertain the shaft's safety factor when transferring 20kW from belt to pinion at a steady 7.5 Hz, taking theyield strength of the ductile shaft material to be 500 MPa.Fatigue and stress concentration are neglected here - alarge safety factor caters (?) for this neglect. [ maximum shear stress 14.5; or distortion energy 15.6 ]

C

DE

500A

B

B

45o

150

150

350

a

b

cde

fg

250 N

500 N

1208080

80

80

Ø 12

xy

z

200 N

A B C D80 120 100

Ø 250

Ø 150

Ø 60 Ø 55

P

F1

F2

11 When a solid shaft of circular cross-section is subjected to a uniformpressure, p (due to the interference fit of a pulley boss for example) theradial and circumferential stresses in the shaft are compressive, ofmagnitude p, as sketched for a typical element.Using the maximum shear stress failure theory, derive the design equation for ashaft cross-section of section modulus Z, loaded by such a pressure in addition to abending moment, M, and a torque, T. [ n√{ (M/Z + p)2 + (T/Z)2 } = S ]

12 The components of the stress resultant at a particular cross-section of a 50 mm diameter ductile rod are as shown - a tensileforce of 120 kN, a vertical shear force of 120 kN, a bendingmoment of 0.5 kNm and a torque of 1.5 kNm.What is the maximum equivalent stress at this cross-section? [ 292 MPa ]

13A key's rectangular cross-section is wxh, its length is L,and it is made from a ductile material whose designtensile stress is S.The key is provided to transfer the load torque Tbetween the shaft and the collar (eg. the boss of apulley or gear). The key is fitted closely into slotsmachined in these components, whose commoninterface diameter is D, where D » h.Identify the modes by which the key may fail, and derivethe corresponding design inequalities which must be satisfied by the dimensions.

14 Determine for each of the following two-dimensional stress states (MPa) the principal stressesand the orientation of the maximum principal. Sketch the principally-oriented elements.( a ) σx = 80 σy = 170 τxy = 60 c.w. [ 50, 200 MPa, 116.6o ]( b ) σx = -220 σy = -70 τxy = 180 c.c.w. [ -340, 50 MPa, 56.3o ]( c ) σx = -205 σy = -445 τxy = 35 c.w. [ -450, -200 MPa, -8.1o ]

15 Find the safety factors for the two-dimensional states of problem 14, if the following apply : ( a ) plane stress, ductile material with yield strength of 500 MPa,

via distortion energy theory, or [ 2.77, 1.36, 1.28 ]via maximum shear stress theory [ 2.50, 1.28, 1.11 ]

( b ) as (a), but plane strain ( ν = 0.3 )via distortion energy theory, or [ 3.59, 1.46, 1.98 ]via maximum shear stress theory [ 3.33, 1.28, 1.96 ]

( c ) plane stress, brittle, with 250 and 800 MPa ultimates [ 1.25, 1.78, 1.78 ]( d ) as (c), but plane strain ( ν = 0.2 ) [ 1.25, 1.78, 1.78 ]

16 Show that the distortion energy failure locus degenerates to the following alternate forms for aplane stress state :

( a ) σe2 = σ1

2 - σ1 σ2 + σ22 σ1 and σ2 are the biaxial principals, or

( b ) = σ 2 + 3 σ 2 in terms of the 'basic' components, or( c ) = σx

2 - σx σy + σy2 + 3 τxy

2 in terms of the Cartesian components.What ratio between shear and tensile strengths does this predict ? [ 0.577 ]

Ø D

T

T

key

= h

w

=

L

ppp

120 kN

1.5 kNm

0.5 kNm

x

y

z

120 kN

Miscellaneous Strength Topics 1

EXAMPLE What is the slope of the above cantilever in way of the force Q?

If a displacement is needed for which there is no corresponding actual load, then a fictitious dummy load must be

applied to the system. The value of such a dummy load is zero, but it must be included algebraicallly in order to

deduce partial derivatives ( for use in ( ii) ) - for although a dummy load is actually zero, partial derivatives

involving it generally are not zero. Once these partial derivatives have been ascertained, the dummy load is set to

zero when working out the integral in ( ii).

Since the rotational displacement of the end at Q is required here, the dummy rotational load Mo is defined at that

end as shown. This external moment will appear on each of the free bodies

above, so that the bending moments at s become :

(a) M = Mo + Qs; ∂M/∂Mo = 1 ; 0 ≤ s ≤ L/2

(b) M = Mo + Qs + P(s – L/2); ∂M/∂Mo = 1 ; L/2 ≤ s ≤ L

Having evaluated the partial derivative of the bending moments with respect to the dummy load, the dummy load is

zeroed in the energy integral. So, with Mo = 0 :

δMo= ∂U/∂Mo = 1/EI ∫0

L M ∂M/∂Mo

do not confuse the bending moment M with the external load Mo

EI δMo= ∫

0

L/2 Qs.1 ds + ∫

L/2

L { Qs + P(s – L/2) }.1 ds = ( Q/2 + P/8 ) L2 so end slope = ( Q/2 +P/8 ) L2/EI

Thin Curved Beams

Many practical beams are curved rather than straight. We consider only beams in the form of circu-lar arcs in the loading plane, with cross-sections symmetric about that plane. These beams may besubdivided into two categories :

- A thin curved beam is characterised by a beam depth which is small compared to the radius ofcurvature, and as a result curvature effects and stress concentration in the cross-section are negligible, leading to stresses which are essentially linear like thoseof a straight beam. With stress concentration not an issue, attention is focusedon beam deflections, since the familiar formulae used for straight beam deflections cannot beused for curved beams.

- A thick curved beam is characterised by a beam depth which is of the same orderas the radius of curvature and so stresses are non-linear, increasing towards theinside of the bend where curvature effects are more pronounced. We examine thisstress concentration later, however let’s first look at thin curved beams.

A thin curved beam is typified by the quadran-tal cantilever illustrated. Its radius is R and it issubjected to two components of end load, V &H. The loaded end’s deflection is sought.

Castigliano’s theorem may be applied directly,although the angle θ is usually a more conven-ient independent variable than the distance swhen it comes to curved beams. Thus in thesketch, at the general free body’s cut enddefined by θ, the bending moment required for equilibreium is M = HR ( 1–cosθ ) + VR sinθ. The strain energy of direct tensile and shear forces is negligible, so the forces need not be evaluated.Clearly in ( ii) s = Rθ so that all terms are functions of θ, enabling the integral to be evaluated.

V

H

Rθ

s

MV

H

R θ

P

s

Q

Mo

R >> d

d

R =O(d)

d

MISCELLANEOUS STRENGTH TOPICS

Some elementary load building blocks were reviewed in the previous chapter 'Stress, Strength andSafety’; readers should be familiar also with the theory of straight beams including energy methods,shear flow and shear centre.

We now consider further building blocks which are often found in machines :– bending of beams which are initiallly not straight but deliberately curved– asymmetric (or ‘unsymmetrical’) bending where we use a matrix approach which will be later

found indispensable in the analysis of fillet welded joints– detailed stress concentration in way of elastic contacting bodies such as the round pin and hole

which in the previous chapter we were obliged to approximate using projected areas.

Castigliano’s TheoremThis theorem is a useful tool for determining displacements of an elastic system. Thetheorem states that the displacement δ

G at/of any load G acting upon such a system is

given by :

( i) δG

= ∂U/∂G

in which U is the elastic strain energy of the system due to G and to all other loads. Theload G may be a translational force in which case the displacement δ

G is linear; alternatively the load

G may be a rotational moment or torque in which case the displacement δG

also is rotational. Thedisplacement and the load are reckoned positive in the same sense - eg. NNE or clockwise as shown.

For a system in which bending strain only is significant, that is for a beam-system :

δG

= ∂/∂G [ ∫length

( M2/2EI ) ds ] or, changing the order of operations,

( ii) δG

= ∫length

M/EI ( ∂M/∂G ) ds where s is the distance along the beam.

The deflection under a particular load is thus found by expressing M (and EI if it’s not constant)algebraically in terms of distance along the beam, s, then applying ( ii). Castigliano is particularlyuseful for curved members however we demonstrate its application first to straight beams.

EXAMPLE What is the displacement of the cantilever under the force Q?

Use Castigliano in bending ( ii) with Q as the load and EI constant :

δQ = ∂U/∂Q = 1/EI ∫0

L M ∂M/∂Q ds

where M = function(s)

There are two spans, ie. two regions of integration. From a free body of each, the bend-

ing moment may be found in terms of the known external loads :

(a) M = Qs; ∂M/∂Q = s ; 0 ≤ s ≤ L/2

(b) M = Qs + P( s – L/2); ∂M/∂Q = s ; L/2 ≤ s ≤ L

So EI δQ = ∫0

L/2 Qs.s ds + ∫

L/2

L { Qs + P( s – L/2) }.s ds = ( Q/3 + 5P/48 ) L3

The displacement of the cantilever under the load Q is therefore ( Q/3 + 5P/48 ) L3/EI

δG

G

δG

G

s

M Q

( a )

s

M P Q

( b ) L2/

s

P Q

L2/

L2/

Miscellaneous Strength Topics 2

Thick Curved Beams

The cross-sectional depth of a thick curvedbeam is comparable to its radius of curva-ture; we are interested now in stressesrather than in deflections because of poten-tially critical stress concentration. Exact anal-ysis is complex but the simple approximationwhich follows - based on plane cross-sectionsremaining plane under load - is adequate formost practical beams typified by the curved portionof the G-clamp illustrated.

The generalised sector of a thick curved beam with centre of curvature at O and small angular extentΦ appears below. The radii of interest are identified in the LH sketch. Two radii are particularly sig-nificant - the mean (or centroidal) radius rc, and the mean reciprocal radius rd. They are defined by :

( 1 ) A * rc = ∫ri

ro dA * r ; A / rd = ∫ri

ro dA / r

Like ri & ro, both rc & rd are intrinsic to the known geometry [ b = f(r) ] of the cross-section.

As the beam is curved, the radius defining the unstressed neutral axis, rn, is not immediately obvi-ous as it is in pure bending of a straight beam - it depends on the load and is unknown initially.

The RH sketch shows the assumed positive senses of the two components of the resultant of stressesacting on the RH cross-section

- a tensile force, F, which is chosen to pass through the centre of curvature, and- a corresponding bending moment, M, which tends to sharpen the curvature.

These positive conventions must be adhered to for the theory developed below to be applicable.

In the normal situation the resultants may be ascertained easily since they equilibrate other loads,which are usually known. Shear is neglected as in simple bending of a straight beam.

Application of M increases the curvature, so the angle subtended by the sector increases to Φ' underload in the RH sketch. Assuming that plane cross-sections remain plane during loading, the increasein length of the typical fibre at radius r is proportional to the distance from the unstrained neutralaxis, ( r–rn). Since the original length of the fibre is proportional to the radius r, then :

fibre strain = extension/free length ∝ ( r – rn )/r = 1 – rn/r

Presuming elastic behaviour, the stress σ in the fibre at radius r is proportional to this strain, so

40

20

10

20

15

A

A

5

OO

P Pnot toscale

r

rird

rnrc

roneutral axis

centroidal axis

fibreconsidered

σ

centre of curvatureO

ΦΦ

Φ'F

M

O

N.A.

fibre

n( r – r )

plane

given symmetriccross-sectonal

shape

axis ofsymmetry

r

EXAMPLE A proving ring of radius R and rec-

tangular cross-section b*d is subjected to a force,

F, as shown. The elastic modulus and design

stress of the material are E and S respectively.

Determine :

- the ring deflection under the load,

- ditto, transverse to the load, and

- the maximum allowable load on the ring.

Consider only one quadrant since the system is symmetric - say the top left quadrant. This is modelled as a canti-

lever ( which automatically preserves the lower end vertical ) and loaded with

- a vertical force V equalling half the total load F (the top RH quadrant supports the other half)

- a dummy load, H, which is necessary for determining the requested horizontal deflection, and

- an external moment, Mo. This is not a dummy load. If there were

no moment at the end then the end would rotate as shown at (c).

But the end remains horizontal due to symmetry of the complete

ring, therefore there must be a moment Mo exerted by the RH mir-

ror quadrant to ensure compatibility δMo = 0 as shown at (d). Mo

is in fact a bending moment in the complete ring, however it shows

up as an (unknown) external moment on the FBD of the LH quadrant.

The free body of the end of the cantilever at (b) enables correlation of the stress resultants at θ with the external

loads. The bending moment only needs evaluation since, in general, strain energy due to direct normal and shear

stresses will be insignificant compared to that due to bending. For end equilibrium :-

Σ Mcut at θ = M + Mo – V R sinθ – H R ( 1– cosθ ) = 0 that is

M = V R sinθ + H R ( 1 – cosθ ) – Mo → V R sinθ – Mo when dummy H = 0

∂M/∂V = R sinθ ; ∂M/∂H = R ( 1 – cosθ ) ; ∂M/∂Mo = –1

Inserting these into ( ii) with EI constant gives the various deflection components as :-

δV = 1/EI ∫0

L M.∂M/∂V.ds = 1/EI ∫0

π/2(V R sinθ – Mo).R sinθ.R dθ = ( π/4 VR – Mo) R

2/EI

δH = 1/EI ∫0

L M.∂M/∂H.ds = 1/EI ∫0

π/2(V R sinθ – Mo).R(1–cosθ).R dθ = (1/2 VR – ( π/2 –1) Mo) R

2/EI

δMo= 1/EI ∫0

L M.∂M/∂Mo

.ds = 1/EI ∫0

π/2(V R sinθ – Mo).(–1).R dθ = (–VR + π/2 Mo)

R/EI

But there is no end rotation, so setting δMo= 0 in the last equation resolves the indeterminacy giving Mo = 2/π VR.

Substituting this value of Mo into the first two equations : δV = ( π/4 –2/π )

VR3/EI and δH = ( 2/π –

1/2 ) VR3

/EI.

So, for the complete ring, as V = F/2 then :

diametral deflection, coaxial with load = 2 δV = ( π/4 –2/π )

FR3/EI

diametral deflection, transverse to load = 2 δH = ( 2/π –1/2 )

FR3/EI

From the bending moment equation above, the maximum moment occurs at θ = 0 and is Mmax = Mo = 2/πVR.

Since bending is the only significant load building block, the maximum stress is

σ = Mmax ymax/I = ( 2/πVR )( d/2 ) /( bd3/12 )

The maximum allowable load on the ring occurs when σ reaches the design stress, S, and is therefore Fmax =

πbd2S / 6R.

MV

H

R θ

Mo

R

F x

x

db

magnifiedsection

x - x

( a ) ( b )

VV

Mo

( c ) ( d )

Miscellaneous Strength Topics 3

which the integrals in ( 1) are calculable, then applying ( 1) in discrete form :-

( 1a) A = Σ j=1 ( δA )j ; A*rc = Σ j=1 ( δA*rc )j ; A/rd = Σ j=1 ( δA/rd )j

Consider the T-section shown, consisting of two rectangular elements :-

elem- radiient width inner outer δA δA*rc δA/rd

1 bi ri rm bi ( rm– ri )1/2 bi ( rm

2 – ri2 ) bi ln (rm/ri )

2 bo rm ro bo( ro– rm )1/2 bo( ro

2 – rm2 ) bo ln (ro/rm)

Σ j=1 ( δA )j Σ j=1 ( δA*rc )j Σ j=1 ( δA/rd )j

So A = bi ( rm– ri ) + bo( ro– rm ) using ( 1a)

A*rc = 1/2 [ bi ( rm2 – ri

2 ) + bo( ro2 – rm

2 ) ] whence rc for the section . . .

A/rd = bi ln (rm/ri ) + bo ln (ro/rm ) . . . and rd

EXAMPLE The G-clamp illustrated above is made from a steel whose design stress is 50 MPa.

It is supposed to sustain a force, P, of 750 N. Can it ?

The most heavily stressed cross-section is assumed to be just to the left of A-A as shown, since

- it is the furthest cross-section from the load, so it is the cross-section subjected to the largest bending stresses

- it lies in a ‘thick curved beam' portion of the clamp and so is subjected to stress concentration.

Section properties : From the above T-section results with ri = 20, rm = 30, ro = 50, bi = 15, bo = 5 mm :

A = 15 (30–20) +5 (50–30) = 250 mm2 watch units!

A*rc = [ 15 (302 –202 ) +5 (502 –302 ) ]/2 = 7750 mm3 whence rc = 31 mm

A/rd = 15 ln 30/20 + 5 ln 50/30 = 8.636 mm whence rd = 28.95 mm

The stress resultants, F and M, are shown on the free body in the conventional positive

senses used in developing the theory above. For equilibrium of the free body, F = P and

M = – Pa. Inserting these into ( 3) gives :-

rn = rd ( a + rc )/( a + rd ) = 28.95 ( 40 + 31 )/( 40 + 28.95 ) = 29.81 mm

σm = –P ( a + rd )/A ( rc – rd ) = –750 ( 40 + 28.95 )/250 ( 31 – 28.95 ) = – 101 MPa

The stress distribution is therefore :- σ = – 101 ( 1 – 29.81/r ) MPa where r is in mm.

The stress distribution is plotted and yields extreme values of 49 MPa tensile at the inside (20 mm

radius), and 41 MPa compressive at the outside. As the stress nowhere

exceeds 50 MPa, the claimed capacity seems vindicated.

But, for the straight portion :-

I = 15 ( 103/12 +10*62) + 5 ( 203/12 +20*92) = 18083 mm4, so

σ = σd + σb = P/A + My/I

= 750 /250 – 750(40+31). ( r –31)/18083 = 3 – 2.945 ( r –31) MPa

This stress distribution also is plotted and indicates a tensile stress of 35

MPa at the inside, and 53 MPa compressive at the outside. The clamp is

therefore some 6 % under-designed. But it is physically impossible for

two distinct stress distributions to occur at an infinitessimal distance

apart across the interface A-A. The actual distribution at A-A will there-

fore lie somewhere between those found - however this does not invali-

date the conclusions regarding clamp inadequacy . . . . why ?

a=40

A

OA

F

M

P

dr =28.95

cr =31.00

compressive tensile 50050

STRESS (MPa)20

30

40

50

RAD

IUS

(mm

)

curvedstraight

r =29.81n

ri

ro

OO

rmbi

bo

1

2

( 2 ) σ ∝ 1 – rn/r = σm ( 1 – rn/r )

where σm and rn are constants, as yet unknown.

This establishes the form of the stress variation across the cross-section; it is clear that the distribu-tion is non-linear, giving rise to high stress magnitudes at small radii - that is, stress is concentratedwhere curvature is sharpest ( r ⇒ 0 ). Determination of the two constants follows from the necessityfor the resultants of the stress distribution ( 2) to be the known F and M noted above. Thus, forequivalence of the distribution and the resultants, using ( 1) and ( 2) :-

F = ∫A σ dA = σm ∫A

( 1 – rn /r ) dA = σm A ( 1 – rn /rd )

M = ∫A σ r dA = σm ∫A

( 1 – rn /r ) r dA = σm A ( rc – rn )

Solving these for the two unknown constants :-

( 3 ) rn = rd ( M – F rc )/( M – F rd ) ; σm = ( M – F rd )/A ( rc – rd )

It is noticed that these two constants - and hence the stress distribution ( 2) - depend only upon theloading ( F & M) and the cross-sectional geometry ( A, rc & rd ).

Two particular locations of the neutral axis occur when - M = 0 whereupon rn = rc, and - F = 0 then rn = rd - the pure bending case most commonly encountered in the literature.

Summarising - application of the foregoing theory involves the following steps :- compute the salient radii rc and rd from ( 1) and known cross-sectional geometry (see below),- determine the stress resultants ( F & M) necessary to equilibrate other given loads, with posi-

tive senses to the convention noted above, - evaluate the constants from ( 3), hence completely define the stress distribution by ( 2).

The conspicuous radii, rc & rd, for common symmetric cross-sections are found from ( 1) as follows.

Trapezoid - defined by dimensions ri, ro, bi and bo as sketched.

The variables r and dA must be expressed as functions of the same variable before the integrals in(1) can be evaluated. The radius r itself is the logical choice for independentvariable, since, from the sketch dA = b.dr, where b is evidently linearlydependent on r, say

b = α + β r where constants α and β are evaluated at ri and ro :α = ( bi ro – bori )/( ro – ri ) ; β = ( bo – bi )/( ro – ri )

So dA = b dr = ( α + β r ) dr and therefore, from ( 1) :-

A = ∫A dA = ∫riro ( α +βr )dr = 1/2 ( ro – ri ) ( bi + bo )

A*rc = ∫A dA*r = ∫riro ( α +βr ) r dr = 1/6 ( ro– ri ) [ bo( 2ro+ri ) + bi( 2ri +ro ) ]

A/rd = ∫A dA/r = ∫riro ( α/r +β )dr = bo – bi + [ ( biro– bori )/( ro– ri ) ] ln ( ro/ri )

Rectangle bo = bi = b ( constant ), so particularising the trapezoidal results :-

A = b ( ro– ri ) ; A*rc = 1/2 b ( ro2 – ri

2 ) ; A/rd = b ln ( ro/ri )

Circle Integrating as above :- 2 rc = ro + ri ; 2 √rd = √ro

+ √ri

Built-up Sections may be analysed by dividing them into a number of simple shapes for each of

OOri

ro

bi

bo

ri

ro

r

drb

dA

OO

Miscellaneous Strength Topics 4

transverse to the longitudinal z-axis of the beam.The origin of any convenient Cartesian system islocated at the centroid of the cross-section, G,with the x and y axes lying in the plane of thesection. The stress resultant at the cross-sectionis M = [ Mx My Mz ]' where Mx and My arebending moment components and Mz is atorque - all taken positive by the right hand rule.There is no force resultant in pure bending.

A representative small element of cross-sectional area, δA, is identified at r = [ x y z ]' in theCartesian system shown in the right hand sketch, where z = 0 here since all cross-sectional area ele-ments lie wholly in the x-y plane. The centroid is defined by :-

(iii) ∫area r dA = 0 - ie. as ( iiia) above, ∫area x.dA = 0; ∫area y.dA = 0; ∫area z.dA = 0 concisely.

Due to application of the turning effect, M, the cross-section will rotate about some centre of rota-tion, C, whose position though fixed at rc is not (yet) known. The radius vector from the centre ofrotation to δA is

( iv) s = r – rc

This rotation - and correlation of compatibility with the symmetric case - requires that the (small)strain at the element will be orthogonal to and linear in this s-vector. So, presuming elastic behavi-our, the stress on δA, σ = [ σx σy σz ]' is also orthogonal to and linear in s, as indicated by thesketch. This may be expressed as :-

( v) σ = b x s – const = b x r – c using ( iv)

in which the constant b = [ bx by bz ]' incorporates the necessary linearity while the cross productensures orthogonality. c = [ cx cy cz ]' is also a constant for all elements in the cross-section under consideration.

Equation ( v) establishes the form of the stress distribution; the two constants b and c may be evalu-ated from the necessary equivalence of the known stress resultants ( F & M ) and the stress distribu-tion integrated over the cross-section. Thus, since there is no force resultant here :-

F ≡ ∫area σ dA = ∫area ( b x r – c ) dA integrating by parts and using ( v)

= b x ( ∫area r dA ) – c A ≡ 0

Since this last integral vanishes as a result of ( iii), it is concluded that the constant, c, also vanishes.Similarly for the moment resultant :-

M ≡ ∫area r x ( σ dA ) = ∫area r x ( b x r ) dA via ( v) with c = 0

Expanding out the triple product and integrating term-by-term leads to :-

Since the loading M is known and the second area moments I of the given cross-section can be com-puted, it follows that the constant b necessary for equivalence must satisfy ( vi) - ie. b = I -1 M,though inversion is unnecessary due to the null elements of I. The example below reviews the tech-nique for determining second moments of a plane area.

The constants b and c having been found, the stress at any point r in the cross-section follows imme-

( vi) M = I b where I ≡ I xx –I xy 0 ; I = ∫ y . dAxx–I xy I yy 0 ; I = ∫ x y . dAxy 0 0 Jzz ; J = ∫ ( x + y ) dAzz

2A

A

A2 2

G

Cs

σ

rc

δAr

G

My

Mx

Mz

x

y

zA

Asymmetric Bending

The transverse cross-section of a typical prismatic beamis illustrated. It is rectangular with x and y axes chosenparallel to the axes of symmetry of the cross-section, and thez-axis longitudinal to the beam. Events in the x-z plane underload are completely divorced from events in the y-z plane, so that bend-ing stresses for example may be found in each plane separately before being com-bined (superposed) to give the stresses due to simultaneous loading in bothplanes. Deflections may be treated likewise.

When the cross-section is not symmetrichowever - or when it is, and loadingplanes other than those containing the axesof symmetry are chosen - thencoupling between events in orthogonalplanes will occur.

Thus consider the I- and Z-beams below,each loaded in the y-z plane of the web. The bending stress resultants may be approximated byflange forces only, acting at the centroids of the flanges. The side views of the y-z loading planes areunremarkable - the two beams seem identical. However plan views of the transverse x-z planes

illustrate the basic problem with asymmetric sections like the Z-beam here - a bending moment existsin the plane transverse to the loading plane, thus demonstrating the coupling phenomenon.

We shall adopt a vector approach to examine the stress distribution in general asymmetric bending.A vector here is essentially a vertical list of scalars, shown for compactness by its horizontal trans-pose (denoted by ‘ ). A similar approach will be implemented later in the context of fillet welds,however before we delve into this, let’s recapitulate how the centroid of a known area is located.

To ascertain the centroidal abscissa ( XG), an arbitrary reference axis isfirst set up, from which the position, X, of the general area element δAcan be reckoned. Varignon's Theorem (aka. the Theorem of Moments)defines X G from : XG ∫area dA = ∫area X dA - and since the disposition ofthe area is known, the integrals may be evaluated and XG calculated.

Alternatively, transposing terms to the RHS of the equation, the require-ment for G to be the centroid becomes ∫area ( X –XG ) dA = 0; or, if thecoordinate x = X –XG is reckoned from the centroid, then the centroid must be such that :

( iiia) ∫area x dA = 0

Similar centroidal requirements apply in the y and z directions.

The LH sketch below shows the cross-section of a prismatic beam of known geometry which lies

loadingplane y-z

transverseplane x-z

yx

z

free body face

xy

z

σG

20

-2010

50

-50-10 MPa

30

-30 MPa

y-z plane x-z plane superposed

xy

z

G

X

xXG

δA

centroidal axis

refe

renc

e ax

is

y x

z

loadingplane y-z

transverseplane x-z

Miscellaneous Strength Topics 5

Contact Stresses

We have seen how contact stresses may be treated approxi-mately under the heading of 'bearing' or 'crushing', using theconcept of projected area. We now examine these stresses inmore detail.

In the contact between theoretically rigid bodies such as theplane with wedge, cylinder or sphere, stresses are infinite since thecontact area is zero. In real life however neither the base (b) nor the cylin-der (c) is completely rigid - rather both contactingelements must deform as indicatedat (d). Elasticity requires gradualtransition of both members fromtheir free boundaries to the com-mon interface area (e).

These last sketches are of coursegrossly exaggerated - contact areasbetween metal components are small but non-zero, and give rise to stresses which are finite butlarge.

It is possible to analyse certain simple contacting body geometries - simple in that end effects areignored for example. The analysis is still complex, so we shall quote rather than derive the results inorder to obtain an appreciation of the behaviour for common geometries.

Shown are two spheres, or cylinders of length L in end view. ACartesian system is defined with the y-axis normal to the sketchplane. Both components deform; the generally non-planecontact area extends a distance 'a' from the z-axis of symme-try, and the contact pressure and loading severity are max-ima on this axis.

The equivalent modulus E* and the equivalent diameter d* aredefined by :

1/E* = 1/E1' + 1/E2' where E' = E/( 1–ν2 )

1/d* = 1/d1 + 1/d2 where |d2| >|d1| and is negative if internal.

The principal compressive stresses are then given by (Shigley op cit) :

( 5a) Spheres Contact zone size : a3 = 3 F d* / 8 E*

Maximum contact pressure : pmax = 3 F / 2 π a2

Normalised principal stresses sx = sy = ( 1 + ν )( 1 – φ arccot φ ) – 1/( 2 ( 1 + φ2)) ( s = σ/ pmax @ φ = z/a ) : sz = 1/( 1 + φ2)

( 5b) Cylinders Contact zone size: a2 = 2 F d* / π E* L ; L is the cylinders' contact lengthMaximum contact pressure : pmax = 2 F / π a LNormalised principal stresses sx = 2 ν ( √( 1 + φ2 ) – φ ) ; sz = 1/ √( 1 + φ2 ) ( s = σ/ pmax @ φ = z/a ) : sy = √( 1 + φ2 ) ( 2 – 1/( 1 + φ2 ) ) – 2 φ

These stresse ratios are plotted below for Poisson's ratio ν = 0.3, together with the equivalent stressfrom the distortion energy failure theory. Evidently for both spheres and cylinders the maximumequivalent stress occurs just below the surface and may be taken to be 60% of the peak pressure.

F

F

F

( a ) ( b ) ( c ) ( d ) ( e )

z

F F

x

2a

d1

maxp

p

O

d2

EXAMPLE Determine the second area moments of the 12*8*1 cm angle :

Set up convenient X,Y axes and divide the area into elements (two here) of simple shape whose individual

I's can be directly calculated. Tabulate, and find first moments of area to define the centroid using ( iii).

dA X Y X dA Y dA x = X –XG y = Y –YG

1 11 1/2 11/2 11/2 121/2 –1.47 –2.53

2 8 4 23/2 32 92 2.03 3.47

19 75/2 305/2

XG Σ δA = Σ X δA XG = 75/2*19 = 1.97 cm

YG Σ δA = Σ Y δA YG = 305/2*19 = 8.03 cm

Hence set up centroidal x-y axes, as shown, and calculate :

Ixx = ( 1*113/12 + 1*11 (–2.53)2 ) +

( 8*13/12 +8*1 (3.47)2 ) = 278.4 cm4

Iyy = ( 11*13/12 + 11*1 (–1.47)2 ) + ( 1*83/12 + 1*8 (2.03)2 ) = 100.3 cm4

Ixy = ( 0 + 1*11 (–2.53) (–1.47) ) + ( 0 + 8*1 (3.47) (2.03) ) = 97.2 cm4

Note that both elements contribute to the product second moment since their centroids lie in the first and

third quadrants and hence δA.x.y is positive for both.

2

1

O

12

8

1X

Y

1

G

1.47

2.03

3.47

2.53

y

xG1

G2

diately from ( v); thus for the case of pure bending where the torque Mz = 0, the vector σ degener-ates to the scalar σ(z), given by ( v) with ( vi) as :-

( 4 ) σ = ( ( Mx Iyy + My Ixy ) y – ( Mx Ixy + My Ixx ) x ) / Io2 where Io2 = Ixx Iyy – Ixy

2

Some particular cases of ( 4) include the following :

( a) Ixy = 0 - ie the cross-section is symmetric - then from ( 4) :

σ = ( Mx / Ixx ) y – ( My / Iyy ) x that is the loading planes are uncoupled as in theheading sketch for the rectangular beam.

( b) My = 0 ( Ixy ≠ 0 so coupling exists ) - then from ( 4) :

σ = ( Iyy y – Ixy x ) Mx / Io2 that is, the neutral axis is NOT the x-axis, but is

defined by σ = 0, ie. by y/x = Ixy /Iyy - which implies curvature components in boththe x-z and y-z planes.

Asymmetric bending may be tackled also by resolution of the loading into the principal area senses.The minimum principal axis passes close to the centroids of the area elements.

Miscellaneous Strength Topics 6

enlargedsection x-x

50 rad

x

x8rad

The general case in which two contacting bodies 1 & 2 are each curvedthree-dimensionally is too complicated for closed-form evaluation ofthe stresses. Young (op cit) presents a numerical solution for the maxi-mum contact pressure which may be approximated as a guide tosafety. The minimum and maximum radii of curvature at the point ofcontact are r1 and R 1 for body 1, and r2 and R2 for body 2. The recipro-cals 1/r1

and 1/R1 are known as the principal curvatures of body 1, and

1/r2 and 1/R2 of body 2, and in each body the principal curvatures are

mutually perpendicular. The plane containing curvature 1/r1 in body 1

makes an angle φ with the plane containing curvature 1/r2 in body 2.

The equivalent curvature radius r* and the parameter λ are defined by :1/r* = 1/r1

+ 1/R1 + 1/r2

+ 1/R2

λ2 = (2/π) arccos { r* √ [ ( 1/r1 – 1/R1

)2 + ( 1/r2 – 1/R2

)2 + 2( 1/r1 – 1/R1

)( 1/r2 – 1/R2

).cos 2φ ] }

Then, for λ > 0, the elliptical contact area Ac and the maximum contact pressure pmax are :

( 5c) Ac ≈ π ( 3 Fr* /2λE* )2/3 pmax = 3F /2Ac

In the case of ductiles, it is reasonable to take the maximum equivalent stress to be 0.6 pmax analo-gously to spheres and cylinders, since this figure does not appear to be affected significantly by thegeometry. The approximation ( 5c) for the contact area can be 7-8% high, so that the predicted equiv-alent stress may be underestimated by this percentage.

EXAMPLE The diameter of a ball in the steel bearing illustrated is 15 mm and

the groove radius of both inner and outer races is 8mm. If a single ball

were to transmit the whole radial load of 5 kN, what maximum

equivalent stress would result?

Let the ball be body 1 and the inner race body 2. Then r1 = R1 = 7.5 mm;

r2 = –8 mm (since concave) and R2 = 50 mm.

From the above, r*= 6.19 mm, φ = 0 and so λ = 0.540. Since E*= 114 GPa for steel-on-steel,

from ( 5c) Ac = 2.60 mm2, pmax = 2.88 GPa and σ* = 0.6pmax = 1.73 GPa.

Repeating for the outer race, r1 = R1 = 7.5 mm; r2 = –8 mm and R2 = –65 mm (since concave).

From the above, r* = 7.92 mm, φ = 0 and so λ = 0.575; from ( 5c) A c= 2.94 mm2, pmax = 2.55 GPa and σ*= 1.53 GPa

For the bearing as a whole, the maximum equivalent stress is 1.73 GPa.

Bibliography

Burr AH & Cheatham JB, Mechanical Analysis and Design , Prentice Hall 2ed 1995 refers to rotatingdiscs, plasticity etc

Faupel JH & Fisher FE, Engineering Design, Wiley 1981Shigley JE & Mischke CR, Standard Handbook of Mechanical Design, McGraw-Hill 1986Shigley JE & Mitchell LD, Mechanical Engineering Design, McGraw-Hill, latest metric ed.Ugural AC & Fenster SK, Advanced Strength and Applied Elasticity, Elsevier 1987Young WC, Roark's Formulas for Stress and Strain, McGraw-Hill 1989

R1

R2

r1

r2

r1planeof

r2planeof

φ

Stresses are proportional to the peak con-tact pressure, which in turn depends on therelative size of the two bodies. If a refer-ence body (cylinder or sphere) is contactedby an auxiliary body whose diameter var-ies ( with other parameters constant ) thenthe peak pressure will change as shown.The advantage of close conformity inreducing stress levels is apparent - ballbearings capitalise on this by using toroidalrather than cylindrical race surfaces.

EXAMPLE Ascertain the peak pressure between a flat cast iron surface and a 100 mm diameter steel wheel, 5

mm wide and carrying a load of 500 N. If the cast iron's compressive ultimate is 750 MPa and the steel's yield is

350 MPa, estimate the factor of safety.

E1' = [ E/( 1 – ν2 ) ]1 = 207/( 1 – 0.292 ) = 226 GPa d1 = 100 mm

E2' = [ E/( 1 – ν2 ) ]2 = 100/( 1 – 0.212 ) = 105 GPa d2 = ∞E* = 1/( 1/E1' + 1/E2' ) = 1/( 1/226 + 1/105 ) = 71.7 GPa d* = 100 mm

Then, from the cylindrical equations (5b) with L = 5 mm :

a2 = 2 Fd* / π E* L = 2 * 500 * 100 / π* 71.7E3 * 5 whence a = 0.30 mm

pmax = 2F / πaL = 2 * 500 / π * 0.30 * 5 = 213 MPa

It will be recalled that a safety factor is essentially a load criterion - it is a stress criterion only when stress is pro-

portional to load, which is NOT the case here. From the above pmax ∝ F /a where a ∝ √F, so that pmax ∝ √F.

For the steel, the 500 N load gives rise to a maximum equivalent stress of 0.6 pmax = 128 MPa, so the load neces-

sary to produce yield will be 500 (350/128)2 = 3740 N. That is, the steel's safety factor is 3740/500 = 7.5.

Cast iron failure will occur when the maximum pressure, ie the maximum surface compressive stress, reaches the

compressive ultimate. Now the 500 N load above gives rise to pmax of 213 MPa, so the load necessary to induce

fracture will be 500 (750/213)2 = 6200 N. That is, the cast iron's safety factor is 6200/500 = 12.4.

The overall factor of safety is therefore minimum( 7.5, 12.4) = 7.5

EFFECT OF RELATIVE CURVATURE ONCONTACT PRESSURE BETWEEN

CYLINDERS AND SPHERES

5

4

3

2

1

0rela

tive

pres

sure

on

refe

renc

e bo

dy

0.1 1 10diameter ratio: auxiliary body / reference body

spherescylinders

auxiliary

reference

norm

alis

ed s

tres

s, s

= s

tres

s / p

norm

alis

ed s

tres

s, s

= s

tres

s / p

1.0

0.0

normalised distance from contact, φ = z / a0 1 2 3

1.0

0.0

normalised distance from contact, φ = z / a0 1 2 3

xs

zs

ys

zs

STRESSES BELOW SURFACE ON

CONTACT AXIS OF CYLINDERS

STRESSES BELOW SURFACE ON

CONTACT AXIS OF SPHERES

ys

xs ,

se s

e

Miscellaneous Strength Topics 7

9 Re-work the derivation of equations ( vi) & ( 4) showing all steps clearly.

10 Calculate the centroidal second moments of area of the Z-beam's cross-section. If the components of bending moment at the section are 50 and10 kNm about the x- and y-axes respectively, ascertain the safety factorat the section. The material's yield is 250 MPa. [ 1.52 ]

11 The beam ABCD, simply supported at A and D and other-wise identical to that above, is loaded by 75 kN at B, in linewith the web, and by 30 kN at C parallel to the flanges(AB=BC=CD=1m). Determine the safety factor. [1.12]

12 The 10mm diameter ball end of the steel link is socketed into the10.1mm diameter spherical seating of the valve-operating linkage.Determine the maximum allowable linkage force, F, if :-(a) the socket material is bronze and failure occurs due to lubrication

breakdown occasioned by the contact pressure exceeding 200MPa. [ 1.70 kN ]

(b) failure is due to material strength limitations; and [ 13.6 kN ](c) as (b), but the socket material is cast iron. [ 11.0 kN ]Properties of the particular materials used in the joint are :

steel bronze cast ironelastic modulus ( GPa ) 207 110 100Poisson's ratio 0.29 0.30 0.21design stress ( MPa ) 400 240 100(t), 350(c)

13 A rod passes through a hole of length L and is loaded by a radial force F. Both rod and hole areof nominal diameter d, and the small diametral clearance between them is c*d where c is a frac-tion. Show that the peak contact pressure is √( 2cFE*/ πdL ).

14 A wrench, required to produce a torque of 300 Nm, is made from atube of 50 mm mean diameter and wall thickness t, through whichtwo diametrically opposite holes are drilled. A detachable solidhandle, of nominal diameter d, fits through these holes with a dia-metral clearance of 1%. The torque is generated by a force appliedat some distance along the handle. Both components are of steelwith design (tensile) stresses of 200 and 300 MPa for tube and han-dle respectively.Suggest suitable dimensions for the wrench.Indicate the effect on the selected dimensions if the diametral clearance is changed to 5%, tofacilitate handle engagement.

L

P

tØ D

Ø d

F

60

48

120

36

24

90

x

y

yxz

A

B

CD

75 kN30 kN

MISCELLANEOUS STRENGTH TOPICS - PROBLEMS

1 A spring is formed by bending a rod of diameter d into a U-shape ofradius R and leg length L, where R, L >> d. Show that the spring's compli-ance (reciprocal stiffness) is 32( 4L3 +6πRL2 +24R2L +3πR3 )/ 3πEd4.

2 A spring is made by curving a strip of rectangular cross-section intofour identical arcs of radius R and included angle α, as shown.Neglecting friction, show that the deflection, δ, experienced by the loadP is :- δ = ( PR3/4EI )[ 3( 2α +sin2α ) - 4α cos2α - 8 sinα ]

3 The flexible pipe loop may be placed between a steam boiler and pro-cess plant to reduce the forces caused by thermal expansion of the pipe.Determine the force and moment developed by a temperature rise of Tover that at installation. Hint: allow the pipe to expand freely then load it tonegate the expansion. [ 0.228 EI αT/R2 ]

A hook, of uniform rectangular cross-section and madefrom a steel whose design stress is 100 MPa, may supporta load, P, in either of the two positions shown. Whatmaximum load can the hook sustain ? [ 12.6 kN ]

5 Find the stresses at the inner and outer surfaces ofthe trapezoidal section A-A of the riveting frame fora load P of 2 kN. [ 119 MPa (t), 83 MPa (c) ]

6 Show that, for a thick curved beam of circular cross-section 2√rd = √ri + √ro.

7 The end of a straight round rod of radius r is bent to a circular arc ofmean radius R. The severity of curvature is characterised by the ratioλ = r/R. The rod supports a force P which acts through the centre of curvature C,normal to the interface i-i between straight and curved portions.If σd = P/A is the direct stress, show that the maximum stress is

σd . λ ( 1 – λ )

. 1 + √( 1 – λ2 ) 1 – √( 1 – λ2 ) in a cross-section just to the left of i-i as sketched, and

σd . ( 1 + 4 / λ ) in a cross-section just to the right of i-i.Plot these expressions versus λ ( 0 < λ < 1 ), comment on the trends displayed, and evaluate thecurvature-induced stress concentration for λ = 0.05, 0.5, 0.95. [ 1.04, 1.55, 6.96 ]

8 A fish hook is forged from wire having a yield strength of2.3 GPa. Assuming a straight pull from the leader throughthe point, estimate the load which would cause the hook toyield. [ 2.2 N ]

RαP

R

45o

section

x

50 rad

20

100

x

sectionx - x

50 radP

P

AA

100

50P

20

38

6

sectionA-A

Ø 0.384.2 5.2

4

L

R d P

i

ir

RC

P

Springs 1

SPRINGS

Springs are unlike other machine/structure components in thatthey undergo significant deformation when loaded - their com-pliance enables them to store readily recoverable mechanicalenergy. In a vehicle suspension, when the wheel meets an obsta-cle, the springing allows movement of the wheel over the obstacle and thereafter returns the wheelto its normal position. Another common duty is in cam follower return - rather than complicate thecam to provide positive drive in both directions, positive drive is provided in one sense only, andthe spring is used to return the follower to its original position. Springs are common also in force-displacement transducers, eg. in weighing scales, where an easily discerned displacement is a meas-ure of a change in force.

The simplest spring is the tension bar. This is an efficient energy store since all itselements are stressed identically, but its deformation is small if it is madeof metal. Bicycle wheel spokes are the only common applications whichcome to mind.

Beams form the essence of many springs. The deflection δ of the load F onthe end of the cantilever can be appreciable - itdepends upon the cantilever's geometry andelastic modulus, as predicted by elementary

beam theory. Stresses are approximately constant along the leafspring shown on the right, since the linear increase of bending

moment from an end is matched by the beam's widening - not by its deep-ening, as longitudinal shear cannot be transmitted between the leaves.

The shortcoming of most metal springs is that they rely on either bending ortorsion to obtain significant deformations; the stress therefore varies throughout

the material so that the material does not all contribute uni-formly to energy storage. The wires of helical compressionsprings (left) are loaded mainly in torsion and are thereforeusually of circular cross-section. This type of spring is themost common. The (ex)tension spring below is similar to thecompression spring though generally requires special endsto permit application of the load - these ends are a potentialsource of weakness which is not present in compressionsprings. Rigorous duties thus

usually call for compression rather than tension springs. The vastmajority of springs are loaded either in tension or in compression -springs loaded in both senses are restricted to light duty.

Belleville washers (shown in part section on the left) are characterised by short axiallengths and relatively small deformations. They are often used in stacks containingmany individual washers in series/parallel. Their geometry can be engineered toproduce highly non-linear characteristics which may display negative stiffness.

All the foregoing springs are essentially translatory in that forces and linear deflec-tions are involved. Rotary springs involve torque and angular deflection. The sim-

FF

F

F

plest of these is the torsion bar (left) in which loading is pure torque;analysis is based upon the simple torsion equation. Torsion bars aregenerally stiff compared to other forms of rotary springs however they

do have many practical applications including vehicle suspensions.

Torsion springs which are more compliant than the torsionbar include clock- or spiral torsion springs(left) and helical torsion springs (right). Theserely on bending for their action, as a simplefree body will quickly demonstrate. The heli-cal torsion spring is similar to the helical ten-sion spring in requiring special ends to trans-mit the load.

The constant force spring is not unlike a self-retracting tape measure and is used where large rela-tive displacements are required such as for the spring motors used in sliding door closers. Thereexists also a large variety of non-metallic springs often applied to shock absorption and basd onrubber blocks loaded in shear. Springs utilising gas compressibility also find some use.

Close-coiled Round Wire Helical Compression Springs

This type of spring is the most frequently encountered, and italone is examined here. It is made from wire of diameter d woundinto a helix of mean diameter D, helix angle α, pitch p, and totalnumber of turns nt. This last is the number of wire coils prior toend treatment (see Table 1 below) - in the spring illustrated nt ≈ 81/2.

Close-coiled implies a small helix angle, say α ≤ 12o, where tan α = p/πD from the developed helix.

Various wire diameters are obtainable, but the availability of new springs is enhanced by specifyingwire diameters from the R20 series of AS 2338, whose most frequently used decade is . . . 0.8 0.9 1.0 1.12 1.25 1.4 1.6 1.8 2 2.24 2.5 2.8 3.15 3.55 4 4.5 5 5.6 6.3 7.1 8 9 10 11.2 12.5 . . . mm

The ratio of mean coil diameter to wire diameter is the spring index, C = D/d. Por-tions of two springs which have the same mean coil diameter but different wire diam-eters and hence different spring indices are compared here. It is clear that low indicesresult in difficulty with spring manufacture and in stress concentrations induced bycurvature. Springs in the range 5 ≤ C≤ 10 are preferred; C < 3 is impracticable.

Loads are transferred into a spring bymeans of platens, which are usually justflat surfaces bearing on the spring ends.Various end treatments are shown inTable 1. Plain ends - when the wire isjust cropped off to length - are suitableonly for large index, light duty applica-tions unless shaped platens areemployed, because spring ends bearunevenly on flat platens, leading tobending of the spring and uncertain per-formance. Ground ends distribute theload into a spring more uniformly than

C=12

C= 3

p α

D

d nt

T T

EFFECT OFEND TREATMENT

TABLE 1

plain squaredor closed

ground squared& ground

n - 2t

n t

n - 1t

n - 2t

active coils, na

(n +1)dt

n dt

n dt

(n +1)dt

solid length, L s

n ptn p+3da n p+2dan p+ dafree length, L o

Springs 2

do plain ends, but the contact region on a flat platen will be very much less than 360o which is idealfor concentricity of load bearing surface and spring axis. One or more turns at the ends of a springmay be wound with zero pitch, this is called a squared or closed end. Subsequent grinding producesa seating best suited for uniform load transfer, and so squared and ground ends are usually specifiedwhen the duty is appreciable. Grinding the ends becomes difficult when the spring index exceeds10, and is obviously inappropriate for small wire sizes - say under 0.5mm.

The active turns na are the coils which actually deform when the spring is loaded, as opposed toinactive turns at each end which are in contact with the platen and therefore do not deform thoughthey may move bodily with the platen. The free length Lo of a compression spring is the spring’smaximum length when lying freely prior to assembly into its operating position and prior to load-ing. The solid length Ls of a compression spring is its minimum length when the load is sufficientlylarge to close all gaps between the coils.

Table 1 indicates how na, L o and Ls depend upon wire diameter, total turns, pitch and end treat-ment. Because of variability in the squaring and/or grinding operations, Table 1 should be regardedas a set of guideline rather than of rules - especially if there are less than seven turns.

The springs illustrated here are right handed, but left hand lays are just as common. The lay usuallyhas no bearing on performance, except when springs are nested inside one another in which case thetwo lays must differ to avoid interference. Springs with closed ends do not become entangled whenjumbled in a container, which is sometimes an issue in assembly.

The Spring Characteristic

The performance of a compression spring is characterised by the rela-tion between the load F applied to it and the deflection δ which results.The deflection is reckoned from the unloaded free end as shown here.

The F-δ characteristic is approximately linear provided the spring isclose-coiled and the material elastic. The slope of the characteristic isknown as the stiffness of the spring k = F/d (aka. spring constant, rate,scale or gradient). The spring’s geometry and material dictate its freelength Lo, solid length Ls (Table 1) and stiffness (as will be shown).

The maximum point on the characteristic is dictated by either one of :-- The geometric solidity limit occurs when the spring solidifies

with a deflection of δs (= Lo-Ls) and load of Fs (= kδs). Althoughthe spring’s load may exceed Fs, its deflection cannot exceed δs.

- The yield limit occurs when the maximum stress (see below) in thewire material reaches yield causing non-linear behaviour, thoughsometimes to improve performance a spring is yielded (pre-set) ina controlled fashion during manufacture.

A spring should be designed so that the solidity limit is over-riding assketched. This avoids all possibility of yielding even if the spring isinadvertently solidified during assemby prior to operation.

A spring undergoes a working cycle between a minimum operationalstate (F, δ) and a maximum state (F, δ) as shown. If the number of cyclesis small (eg. < 104 ) then loading may be treated as static, otherwise fatigue must be considered.

The largest working length should be appreciably less than the free length to avoid all possibility of

δ

solidlength

sδδ

δ

δ

F

F

Fs

Flo

Ls

workingrange

load, F

clashallw'ce

F

Fs

0

k

Lo

free

yield limit

solid

ity

limit

s

δ

defl-ection

freelength

hi

contact being lost between spring and platen, with consequent shock when contact is re-established.In high frequency applications this may be satisfied by the design constraint F/F ≤ 3.

As the spring approaches solidity, small pitch differences between coils will lead to progressive coil-to-coil contact rather than to sudden contact between all coils simultaneously - solidification is agradual process. To avoid contact - with attendant impact, surface deterioration and increase in stiff-ness - a minimum clash allowance of 10% of the maximum deflection is stipulated - ie. δs - δ ≥ 0.1δthough this allowance might need to be increased in the presence of high speeds and/or inertias.

Stresses and Stiffness

The free body ( a) of the lower endof a spring of coil diameter D :

- embraces the known upwardload F applied externally andaxially to the end coil of thespring, and

- cuts the wire transversely ata location which is remotefrom the irregularities associated with the end coil and where the wire stress resultant consistsof an equilibrating force F and an equilibrating rotational moment FD/2.

The wire axis is inclined at the helix angle α at the free body boundary in the side view ( b) (Notethat this is first angle projection). An enlarged view of the wire cut conceptually at this boundary ( c)includes the force and moment triangles which show that the stress resultant on this cross-sectionconsists of four components - a shear force (F cosα), a compressive force (F sinα), a bending moment(1/2FD.sinα) and a torque (1/2FD.cosα).

Assuming the helix inclination α to be small for close-coiled springs then sinα ≈ 0, cosα ≈ 1 and thesignificant loading reduces to torsion plus direct shear. The maximum shear stress at the inside ofthe coil will be the sum of these two component shears :

τ = τtorsion + τdirect = Tr/J + F/A

= (FD/2) (d/2) /(πd4/32) + F/(πd2/4) = (1+d/2D) 8FD/πd3 . . . . . that is

( 1 ) τ = K 8FC /πd2 in which the stress factor K assumes one of three values :

K → 1 when torsional stresses only are significant - ie the spring behavesessentially as a torsion bar, or

K → Ks ≈ 1 + 0.5/C which accounts approximately for the relativelysmall direct shear component noted above, and is used in staticapplications where the effects of stress concentration can beneglected, or

K → Kh ≈ ( C+0.6)/( C–0.67) which accounts for direct shear and alsothe effect of curvature-induced stress concentration on the inside ofthe coil (similar to that in curved beams). Kh should be used infatigue applications; it is an approximation for the Henrici factorwhich follows from a more complex elastic analysis as reported inWahl op cit. It is often approximated by the Wahl factor Kw = (4C-1)/

(4C-4) + 0.615/C.

The factors increase with decreasing index C as shown in this plot :-

F

F

DFD2

F

F

FD2α

Fcompression

shear

bend-ing

FD/2

torsionα

( a) ( b) ( c)

torsion

+ direct

+ conc'n

τ

Springs 3

3 6 9 12

1.6

1.4

1.2

1.0

STRESSFACTORS

Ks

Kh

spring index, C = D/d

The deflection δ of the load F follows from Castigliano's theorem.Neglecting small direct shear effects in the presence of torsion :

δ = ∂U/∂F = ∂/∂F ∫length

T2/2GJ ds with T = FD/2

= ∫length

T/GJ.∂T/∂F ds = (T/GJ) (D/2)*(wire length)

= (FD/2GJ) (D/2) naπD

( 2 ) k = F/δ = Gd / 8naC3

The number of useful or active coils na will be less than the total number of coils since the end coilsare rendered more or less ineffective by contact with the loading platens - refer to Table 1. Despitemany simplifying assumptions, equation ( 2) tallies well with experiment provided that the correctvalue of rigidity modulus is incorporated, eg. G = 79 GPa for cold drawn carbon steel.

Standard tolerance on wire diameters less than 0.8 mm is ±0.01 mm, so the error of theoretical pre-dictions for springs with small wires can be large due to the high exponents which appear in theequations. It must be appreciated also that flexible components such as springs cannot be manufac-tured to the tight tolerances normally associated with machined rigid components. The springdesigner must allow for these peculiarities. Variations in length and number of active turns can beexpected, so critical springs are often specified with a tolerance on stiffness rather than on coil diam-eter. The reader is referred to BS 1726 or AE-11 op cit for practical advice on tolerances.

Buckling

Compression springs are no different from other members subject to compression in that they willbuckle if the deflection (ie. the load) exceeds some critical value δc which depends upon the slender-ness ratio Lo/D rather like Euler buckling of columns, thus :

( 3a) c1 δc /Lo = 1 – [ 1 – ( c2D /λLo )2 ]1/2 in which the constants are defined as

c1 = (1+2ν)/(1+ν) = 1.23 for steel ; c2 = π √[ (1+2ν)/(2+ν) ] = 2.62 for steel

The end support parameter λ reflects the methodof support. If both ends are guided axially butare free to rotate (like a hinged column) then λ =1. If both ends are guided and prevented fromrotating then λ = 1/2. Other cases are covered inthe literature. The plot of the critical deflection isvery similar to that for Euler columns.

A rearrangement of ( 3a) suitable for evaluatingthe critical free length for a given deflection is :

( 3b) Lo.crit = [ 1 + ( c2D/c1λδ )2 ] c1 δ /2

Static analysis of a typical compression spring is illustrated by the example opposite.

UNSTABLESPRINGS

STABLESPRINGS

absolute conditional

1 2 30

1

0

c D2λ Lo

c δ1 Lo

λ = 1.0

λ = 0.5

Wire Materials

The bulk of springs are cold wound from colddrawn carbon steel. The marked increase oftensile strength Sut with decreasing wire diam-eter d is plotted here for three common steelwires - the corresponding expressions in Table2 overleaf which relate strength to diameter areregressions to data in the Spring DesignManual op cit which may be consulted for dataon other materials.

The strength ratios in Table 2 were preparedfrom graphs presented by Godfrey showingallowable shear stress limits for different wirediameters and materials. Strengths are :

Sus ultimate strength in shearSys yield strength in shear - permanent set results if this is exceededSes fatigue strength in reversed shear corresponding to 10 Mc (107 cycles).

Godfrey includes plots similar to those mentioned but for lives other than 10 Mc, from which the

0 4 8 12 16wire diameter, d (mm)

stre

ngth

, S

(

GPa

)ut

3

2

1

MINIMUM TENSILESTRENGTH OF

STEEL WIRE

A228 musicwire

A227hard drawn wire

A229 oil temperedwire

EXAMPLE Estimate the stiffness and maximum operating stress of the close

coiled steel spring with squared and ground ends illustrated.

The wire diameter d = 4 mm and the external diameter Do = 30 mm, so the

mean coil diameter D =Do–d = 26 mm and the index is C = D/d = 6.5.

Stress factors, from ( 1), are Ks = 1+0.5/C = 1.08; Kh = (C+0.6)/(C–0.67) = 1.22

The total number of turns is now counted. This is a somewhat inaccurate process if the spring cannot be inspected

physically. The leftmost coil here ends in a feather edge at the top, and so the end of the wire (imagined before

grinding) must coincide with the vertical plane which contains the spring axis. Starting to count from this vertical

plane - following the wire around towards the observer, then downwards, around the back and upwards to meet

the plane again at the point 'a' illustrated - 'a' therefore corresponds to one complete turn. Repeating this, we

arrive at point 'b' after 11 complete turns. Continuing from 'b', the bottom is reached (another half turn) without

the wire yet ground to a feather edge. So the wire must end at about 3/4 turn after 'b', therefore nt ≈ 113/4 turns.

Ends are not ground right down to feather edges in practice.

From Table 1 the active turns are na = nt–2 = 93/4, the solid length is Ls = ntd = 47 mm, and the pitch is p = (Lo–

2d)/na = (85-8)/93/4 = 7.9 mm. The corresponding helix angle is α = arctan(p/πD) = 5.5o - the spring is certainlyclose- coiled.

From ( 2) : k = Gd /8naC3 = 79E3*4 / 8*9.75*6.53 = 14.8 N/mm.

The solid deflection δs = Lo-Ls = 38 mm, so the solidification load is Fs = k δs = 14.8*38 = 560 N. Assuming that

solidification is infrequent and outside the working range, then this load will be treated as static so that the direct

shear stress factor Ks applies in ( 1). The corresponding solid shear stress from ( 1) is therefore :

τs = Ks 8FsC / πd2 = 1.08*8*560*6.5 / π*42 = 625 MPa.

Assume that at the maximum working state the above-mentioned 10% minimum clash allowance so that δs –δ ≥0.1*δ ie. δ ≤ δs/1.1 = 38/1.1 say δ = 34mm. Proceeding as with the solid state, F = k δ = 14.8*34 = 505N and so :

τ = Ks 8 F C / πd2 = 1.08*8*505*6.5 / π*42 = 563 MPa - static conditions still assumed.

Considering stability at this maximum state and assuming hinged ends (λ = 1), then λLo/c2D = 1*85/2.62*26 =

1.25. Since this is >1 buckling could occur, so investigate via ( 3a) : δc = ( 1 – √( 1 – 1/1.252))*85/1.23 = 27 mm and

since the maximum deflection δ > δc then clearly buckling will occur unless the spring is supported by a rod orsurrounding cylinder. Alternatively if end rotation is prevented (λ = 0.5) then λLo/c2D < 1 which automatically

guarantees stability.

Ø 4.0

85

Ø 30

a b

Springs 4

corresponding fatigue strengths may be deduced.

Fatigue applications such as engine valves demand a high surface finish of the wire, so the choice ofmaterial is based largely on this characteristic which is a result of the method of wire manufacture.Expensive care is taken to maintain the surface of valve spring quality (VSQ) wire free from defects.

Presetting and Shot-peening

Presetting (or scragging) and shot-peening are two manufacturing operations which may be carriedout after coiling and heat treatment, but before putting a spring into service. They each confer afavourable residual stress distribution in the wire which permits higher operating loads.

The stresses in a torsion bar made from an elastic-perfectly plastic material during and after preset-ting are sketched here. A relatively large presetting load (torque) is applied to cause a band of mate-rial under the surface to become plastic while the core of the wire remains elastic ( b).

When the presetting load is removed the bar recovers elastically but does not return to its originalposition since part of the wire cross-section has yielded - the bar remains permanently deformed.The corresponding residual stresses ( c) are negative at the surface - negative in the sense of thestress induced by the applied load. When the bar is subsequently put into service these residualstresses are elastically subtractive from the stresses attributable to the load ( e), so that surfacestresses are less than those which would have occurred in the absence of preset. A more detailedexplanation of presetting is given in the Appendix.

Presetting a spring, as opposed to a torsion bar, involves making the spring sufficiently long so thatportion of the cross-section yields when the spring is compressed solid during preset. When the pre-setting load is removed the spring recovers elastically to a shorter free length. The residual stressesare similar to those in a torsion bar, but more complex due to direct shear and curvature-induced

material hard drawn wire music wire oil tempered wiredesignation ASTM A227/SAE J113 ASTM A228/SAE J178 ASTM A229/SAE J315nominal carbon, % 0.5 - 0.8 0.7 - 1.0 0.6 - 0.8common sizes (d) mm 0.5 - 12.5 0.2 - 5 0.8 - 16general attributes cheapest; not for fatigue best quality; especially general purpose but

due to surface defects for smaller springs fatigue life limitedtensile ultimate (Sut) MPa 2470 +d( 2910 +40d) 3370 +d( 6560 –230d) 2630 +d( 2180 +56d)(empirical approximation) 1 +d( 2 +0.1d) 1 +3.5d 1 +d( 1.6 +0.08d)

Sus / Sut 0.52 0.50 0.63Ses / Sut 0.13 0.15 0.13Sys / Sut 0.43 0.43 0.48

( a) ( b) ( c) ( d) ( e)

Sys

elastic presettingload applied untilyield at surface

load increased -skin at yield whilecore still elastic

preset releasedleaving residualstress and set

linear service stresssuperposed elastically

with residual stress

stress concentration. Presetting usually reduces the stiffness by 5-10%.

Shot peening involves bombarding the wire with high velocity pellets to impart a surface compres-sion. Residual stresses are more surface-localised than those induced by presetting. Peening is par-ticularly beneficial for fatigue in the presence of surface flaws (die marks, pits and seams), but isappropriate only for wire diameters exceeding 1.5 mm or thereabouts.

Springs destined for arduous duty are invariably preset and/or peened, however in the interests ofsimplicity we shall not make use of the significantly higher design stresses which these operationsallow. The material strengths tabulated above refer to springs without preset or peening.

Fatigue Loading

If a compression spring is designed with theyield limit above the solidity limit and manu-factured correctly then the only way it can failis through fatigue. This photograph of a typi-cal fatigue failure surface reveals a cracksource near he inner most highly stressed sur-face of the wire. If poor manufacture or corro-sion causes stress raisers then fracture is likely to emanate from these thereby reducing fatigue life.

In the absence of stress raisers, correlation of the repeated loading on a spring with its material'sproperties in order to ascertain safety is usually carried out via a Goodman analysis. Given thespring material and wire diameter, the shear ultimate Sus and fatigue strength in reversed shear Sesmay be found from the literature, or as explainedabove. The conservatively straight Goodman fail-ure locus is thus defined in τ - τ space, as shownon the diagram (A).

Spring loading is almost invariably unidirec-tional, so τ ≤ τ and loading states occur only tothe right of the 45o stress equality line. One suchload state is illustrated, lying on a line which cor-responds to a safety factor of n - ie. the line joinsthe points (Sus/n, 0), (0, Ses/n) and so lies parallelto the failure locus. From similar triangles theGoodman fatigue equation is :

( 4a ) τ / Sus + τ / Ses = 1 / n in which the stress components are given by

τ = Ks 8 F C /πd2 where F = 1/2 (F + F ) and

τ = Kh 8 F C /πd2 where F = 1/2 (F – F )

The full stress concentration factor Kh is applied to the alternating component τ in ( 4a). Stress redis-tribution is presumed to follow localised yielding, with the result that stress concentration is usuallyignored in figuring the steady component of fatigue loading. The mean stress τ is therefore based onthe factor Ks, which accounts only for direct stresses in the torsion equation.

The safe operating window of the Goodman diagram is completed by the yield limit. If ny is thefactor of safety against the yield being exceeded, then :

( 4b ) τ + τ = Sys / ny as ( τ + τ ) represents the maximum stress.

τ

τ

loadstate

ττ

SusSys

Ses

n

45o45o

==0 t

yieldlimit

failurelocus

0

τ = Ses

t

τ

τ = τ0 t

τ

τττ

τ < τ

SAFE

( A )

Springs 5

The traditional method of presenting allowable fatigue stresses is to plot the maximum shear stress τagainst the minimum τ, rather than the amplitude versus the mean as above. Although strictly notcorrect, the full stress concentration factor Kh is applied conservatively to both components whichare inserted into the Goodman expression ( 4a) to give the failure locus:

( 4c ) τ = [ 2 SusSes /( Sus +Ses ) ] + [ ( Sus –Ses ) /( Sus +Ses ) ] τ where

τ = Kh 8 C F /πd2 and τ = Kh 8 F C /πd2

Thus for the wire of the above example with stresses in MPa :- Goodman failure locus ( 4a) τ / 888 + τ / 183 = 1 (i)- yield failure criterion ( 4b) τ = 677 (ii)- modified Goodman locus ( 4c) τ = 303 + 0.658 τ (iii)

Though (i) and (iii) appear identical, since τ = τ + τ and τ = τ – τ, thefour stress components are defined with different stress factors in ( 4a)and ( 4c). The safe operating window from the last two relations isshown here. States below the equality line are impossible as this wouldmean reversal of the minimum/maximum roles. The materialstrengths of Table 2 were deduced via ( 4c) from plots such as this in Godfrey.

The fatigue strength in repeated (one-way) shear S'es forms the basis of a Goodman approach - dia-gram (B) - which is less conservative than thatbased on reversed shear, diagram (A). If the teststress varies between zero and S'es, then τ = τ =S'es/2, and the failure line joins the points (Sus, 0),(S'es/2, S'es/2 ). The resulting design equation is :

( 4d ) τ /Sus + τ ( 2/S'es – 1/Sus ) = 1/n

Since it is rare that the fatigue strength in repeatedloading is available, we will henceforth employ(4a) and ( 4b) with the fatigue strength corresponding to whatever life is of interest.

800

800

00 τ

τ Goodman

yield

(MPa)

4mm OILTEMPERED

WIRE

equality

τ

τ

SusSys

==0 t

S'es

S'es12

S'es12

45o

ττ =

( B )

EXAMPLE The spring of the preceding example is made from ASTM A229 oil tempered wire. It is proposed

for a non-critical application in which the load varies between a maximum of F and a minimum of 0.4F. What

maximum load can the spring withstand indefinitely ?

Material properties for 4 mm diameter oil tempered wire are, from Table 2 :

Sut = (2630 +4*(2180 +56*4)) /(1 +4*(1.6 +0.08*4)) = 1410 MPa

Sus = 0.63Sut = 888MPa; Ses = 0.13Sut = 183MPa; Sys = 0.48Sut = 677MPa

The solid stress of 625 MPa, evaluated previously, does not exceed this yield, so the spring will not be

permanently deformed when solidified during assembly.

Operating stress components from ( 4a) with F in Newtons :

F = ( F +0.4F)/2 = 0.7 F ; τ = 1.08*8*0.7F*6.5/π*42 = 0.782 F MPa

F = ( F –0.4F)/2 = 0.3 F ; τ = 1.22*8*0.3F*6.5/π*42 = 0.379 F MPa

Insert these into the Goodman design equation ( 4a) with a safety factor of say 1.1 :

0.782 F /888 + 0.379 F /183 = 1/ 1.1 F = 310 N is the maximum load capacity.

The small safety factor should be noted. It is justified because it relates to material properties which have been

derived from lower bounds of a great amount of data which has not needed any theoretical correction for size or

for surface finish - in other words, there is considerable confidence in the material properties. An additional argu-

ment for the small safety factor here is the non-critical application.

This example might have been tackled by first assuming a dummy load, then scaling the answer appropriately.

Spring Design

Spring design may be based upon charts, computer programs, nomograms or other similar aids - thesuitability of any approach depends to a large extent upon the information available :

- material properties and how they are presented, eg. test results on actual springs may be avail-able which include automatically the effects of size and surface finish; alternatively, tensile testresults on laboratory quality wire may be the only knowledge of the material

- the spring specification may be rudimentary in the extreme, or comprehensive in defining bothoperating limits, tolerances on performance, surface treatment etc.

The approach set out below is particularly suitable to material properties as given by Table 2, but adesigner must be capable of modifiying this (or any) approach to suit the problem in hand.

Whatever approach is used, it is important that the desired spring characteristic is known as fully aspossible before starting on detailed calculations. This may necessitate preliminary assumptions,since specification information is often sparse. In a certain fatigue application the maximum operat-ing force may be specified but nothing is stipulated about the minimum, so we might adopt acommon rule of thumb which has the minimum force no less that one third of the maximum, toavoid all possibility of spring bounce.

It is usually most expedient to search for a suitable combination of material and wire diameter -together with index C rather than coil diameter D. Wire size must be selected from the availablerange. The search for suitable index (say 5 ≤ C ≤ 10) is more bounded that a search for mean diame-ter D. However a space constraint is common - thus the spring may haveto fit over a rod, diameter Dr, in which case a space constraint is Di = D-d= (C-1)d > Dr. Or the spring may have to fit inside a hole of bore Dh, so,realising that a compressed spring may expand diametrically by some 3-4%, the design must satisfy Do = D+d = (C+1)d < 0.96Dh.

Selection of the major spring parameters for static applications (say ≤104

cycles) in which the maximum load is F, might proceed thus. The materialis first selected, and hence the ratio of shear yield strength to tensile ulti-mate strength (Sys/Sut) obtained from Table 2 or similar. The safety factor,n, in this static application may be found from (1) :

τ = Ks 8F C /πd2 = Sys/n

When this is normalised by the tensile ultimate Sut, the term π/4d2 Sut

emerges, and is recognised as the ultimate tensile load on the wire, Fut.Since Sut is a function solely of the wire diameter (Table 2), then Fut also isdependent only upon diameter, as given in Table 3. The foregoing equa-tion may then be written as :

( 5a ) n = Fut /Fe ; Fut = π/4d2Sut ; Fe = 2F.CKs/(Sys/Sut)

Fe is the tensile load on the WIRE which is equivalent failurewise to theactual maximum load on the SPRING. While F ut depends on wire diame-ter for a given material, Fe on the other hand depends only on springindex, C, for a predefined load. Separation of d- and C-effects in thismanner greatly assists manual spring design.

In fatigue applications, the Goodman design equation ( 4a) may be manip-ulated into the same form as ( 5a), however the force Fe is now interpretedas the steady tensile WIRE load which is equivalent failurewise to theSPRING fatigue load - components F and F - and is given by :

TABLE 3 WIRE ULTIMATE LOAD

size A228 A229d(mm) Fut (kN) Fut (kN)

0.2 0.0860.224 0.1070.25 0.1310.28 0.1610.315 0.2010.355 0.2500.4 0.3120.45 0.3880.5 0.4710.56 0.5800.63 0.7210.71 0.8990.8 1.12 0.950.9 1.39 1.181 1.69 1.431.12 2.09 1.751.25 2.56 2.141.4 3.16 2.621.6 4.04 3.341.8 5.03 4.142 6.11 5.012.24 7.54 6.162.5 9.23 7.522.8 11.4 9.253.15 14.1 11.53.55 17.5 14.34 21.7 17.74.5 26.8 22.05 32.3 26.65.6 32.86.3 40.67.1 50.68 62.89 77.910 94.311.2 11612.5 14214 17416 221

Springs 6

( 5b ) Fe = 2F CKs /(Sus/Sut) + 2F CKh /(Ses/Sut)

It is noticeable that this fatigue equivalent load, like the equivalent load in the static case, dependsonly on the spring index, for given load and material.

An important consideration in high-speed cyclic applications is the avoidance of surging, wherewaves of relatively large inter-coil displacement travel along the spring leading to high operatingstresses, clashing and impact. Surging arises from resonance between the operating and natural fre-quencies of the spring, whose mass distribution is significant at the accelerations involved. Thespring's fundamental natural frequency, fn, is given by :

( 6 ) fn = (d/π naD2) √(G/8ρ) and, for steel with density ρ = 76.5 kN/m3 this becomes= 358 / naCD kHz where D is in mm.

To avoid surging, this fundamental frequency must be higher than the operating frequency's highestsignificant harmonic - usually taken to be the 12th harmonic, though this will depend upon theshape of the load cycle.

The example overleaf demonstrates the application of ( 5) to a typical design problem.

The program 'Springs' simplifies spring checking (rating) and design - the dialogue for candidate Cis shown overleaf. To use the program, a problem is first specified in terms of the minimum andmaximum working loads together with the maximum deflection. On this basis the program pre-pares a contoured map of spring index C versus wire diameter d - any point on this map is a candi-date solution resulting in a unique combination of fatigue safety from ( 5b), total turns, coil diameterand stability in addition to index and wire diameter. The user aided by the map thereafter specifies acandidate to the program which prompty rates it. Since map reading is not very accurate this laststep usually has to be repeated on a trial-and-error basis to converge accurately on the candidatesolution.

All five solutions to the specimen problem have been manually overlaid on the map. The specimendialogue corresponds to the converged rating of candidate C.

Study of the map contours reveals general trends of geometric parameter combinations necessary tomeet a specific duty - knowledge of these trends is useful in manual spring design. Clearly the lociof constant coil mean diameters are hyperbolae (Cd=D). For a certain wire diameter, decreasing theindex leads to increased fatigue safety, markedly increased necessary turns, and increased lengthwith a consequent increased tendency to buckle. Buckling propensity is not much affected by wirediameter. Both wire diameter and index have to increase/decrease together to maintain a constantlevel of safety.

The solution space in the C-d plane is wedge-shaped - figure (a) below - bounded by fatigue safety(n=1) and conditional stability. In practice, this space is constrained further by design factors exceed-ing unity and by stability which is preferably absolute. Every dynamic spring is subject to these con-straints. Instability, even whilst the spring is guided, can lead to stiffness fluctuations, scuffing, andother undesirables.

Contours of spring diametral limits appear in figure (b) - unlike the general constraints of figure (a),these diametral limits are relevant only to certain springs. Simultaneous ID and OD limits are unu-sual.

The requirement for a fundamental natural frequency of at least twelve times the running frequency,results in a bound similar to that of figure (c) - eg. 12*400 cpm = 80 Hz. High speed springs oftencannot meet this 'limit', so it should not be regarded as immutable as were the safety/stability limitsof figure (a). Preset and/or peened springs can withstand higher stresses than the untreated springs

examined here - for the sameduty they would thereforebe lighter, possess a highernatural frequency, andwould be more suitable forhigh speed applications.

As a result of all the limits,the problem's solution spaceappears as in figure (d). Theform of this space is quitegeneral, though clearly theactual shape in any particu-lar case will depend uponthe relative significance ofthe various bounds. In thepresent problem for example, the ID limit is over-ridden by the stability bound.

Spring design specifications evidently do not always admit practical solution.

Bibliography

AE-21, Spring Design Manual, SAE 1996 comprehensive coverage of different types of springsAS 1472-1991, Steel Wire for SpringsAS 2338-1980, Preferred Dimensions of Wrought Metal ProductsBS 1726: 1987, Coil Springs. Part 1: Guide for the Design of Helical Compression SpringsGodfrey L, 'Steel Springs', Metals Handbook v1, ASM 10ed 1990Richmond TD, Piston Return Springs for Hydrostatic Machinery, Joint Convention on Marine

Applications of Fluid Power, NEL 1966Wahl AM, Mechanical Springs, McGraw-Hill 1963 definitive Item 65005 - Elastic Stresses and Deflections of Helical Compression Springs of Round Wire,

Engineering Sciences Data Unit 1965Item 65006 - Notes on the Design of Helical Compression Springs of Round Wire, Engineering Sci-

ences Data Unit 1965

fatigue n = 1 OD = 62 mm

ID = 15 mmconditional stability

frequency = 80 Hz

a b

c d

Springs 7

EXAMPLE A return spring of infinite life (ie. 107 cycles) is required for a cam follower which moves through25 mm, 400 times a minute. The spring must fit over a φ 15 mm shaft and inside a φ 65 mm hole, and it must exert

a force which varies between 300 and 600 N. Design a suitable spring with closed and ground ends, made from

ASTM A229.

Although the material specified is not optimum for fatigue applications, the solution to the problem will be a

useful benchmark for other solutions. A small safety factor of 1.1 will be selected for reasons given previously and

presuming the consequences of failure are not severe.

Completing the characteristic, given δ – δ = 25 mm, F = 300 N and F = 600 N,

it follows that the spring stiffness must be k = (600-300)/25 = 12 N/mm, and

further δ = 25 mm, δ = 50 mm. Assuming a 12% clash allowance, then δs–δ ≥0.12δ, so take δs = 56 mm and hence the solid load is Fs = kδs = 672 N. This

completes the characteristic as sketched.

The mean and alternating load components are F = 450 N and F = 150 N, so,

from ( 5b) with Sus/Sut = 0.63 and Ses/Sut = 0.13 for A229 from Table 2, and

taking forces in N :

( a ) Fe = (2*450/0.63).CKs + (2*150/0.13).CKh

= C ( 1430 Ks + 2310 Kh ) - a function of C only.

As a first trial assume C=7.5 which lies midway in the usual recommended range. From ( a) it follows that Fe =

7.5*(1430*1.07 +2310*1.19) = 32.1 kN, so that the tensile capacity of a wire suitable for this fatigue application is Fut= nFe = 1.1*32.1 = 35.3 kN.

From Table 3, a suitable wire diameter lies somewhere between 5 and 6.3mm; so the table opposite is prepared for

candidate solutions based on wire diameters in this range from the R20 series of AS 2338. Succeeding rows of the

table assess the practicability of each candidate wire diameter by addressing :

- Calculation of the maximum spring index via ( a) corresponding to the selected safety factor (1.1 in fatigue

here), with subsequent checks for spring overall geometry - candidates A and E do not meet the present

geometric constraints, so these candidates are dropped.

- Calculation of the corresponding number of turns necessary for the target spring stiffness (12 N/mm here).

- Verification that the candidate is close-coiled (helix angle α ≤ 12o) - all remaining candidates conform here.- Determination of whether the candidate is absolutely stable, conditionally stable or unstable. Candidate B

is unstable here, so is passed over.

- Computation of the candidate's fundamental natural frequency which ideally should be at least 12 times

the running frequency. In the present case, the ratio of natural/running frequencies for the two remaining

candidates is over 14, so neither candidate should lead to resonance problems.

- Assessing whether the candidate is likely to yield if solidified during assembly. The shear stress of both

candidates when solid is much less than their shear yield, so yielding is unlikely - though the effect of toler-

ance on both wire and coil diameters would have to be taken into account in a real design.

Two candidates solutions emerge here - C and D - however D is rejected in favour of C as it is less compact ( π/4Do

2 Lo ) and because its OD, being so close to the limit, may cause problems when practical manufacturing toler-ances are allowed for. Manufacturing dimensions of the chosen solution are therefore :

wire diameter 5.6 mm, mean coil diameter 38.7 mm, total number of turns 16.0, free length 146 mm, ends S&G.

25

lo

672

600

300

50 56

F (N)

(mm)

δ0

solidhi

travel

clashallow'ce

Candidate (dimensions mm, forces N) A B C D E trial wire diameter, d 4.5 5 5.6 6.3 7.1

Compute corresponding spring index to satisfy chosen fatigue safety factor of 1.1, from (a)Fut from Table 3 22000 26600 32800 40600 50600initial trial C 7.5 7.5 7.5 7.5 7.5updating C from (a) : 4.69 5.67 6.99 8.65 10.8

C = Fut /1.1*( 1430.Ks +2310.Kh ) 4.33 5.43 6.92 8.81 11.2where K's based on last C-update 4.25 5.39 6.91 8.83 11.3

final converged maximum C 4.22 5.38 6.91 8.83 11.3D = Cd 19.0 26.9 38.7 55.6 80.0Di = D -d = (C -1)d > 15 ? too small 21.9 33.1 49.3 73.1Do= D+d = (C+1)d < 0.96*65 ? 31.9 44.3 61.9 too big

Calculate corresponding number of turns to give required stiffness of 12 N/mm, from ( 2)

na = Gd/8kC3 = 79E3*d/8*12*C3 26.4 14.0 7.5Finalise solid and free lengths, and verify close-coiled assumption

total turns n t = na+2 (Table 1) 28.4 16.0 9.5solid length Ls = ntd (Table 1) 142 90 60free length Lo = Ls +δs 198 146 116pitch p = (Lo–2d)/na (Table 1) 7.1 9.6 13.8

helix angle α = arctan(p/πD) 4.8o

4.5o

4.5o

Check buckling at maximum deflexion of 50mm, assuming guided ends ( λ=0.5)λLo/c2D (absolutely stable if ≤ 1 ) 1.40 0.72 0.40δc from ( 3a) (conditionally stable if > 50 mm) 48 - -conclusion re stability unstable OK OK

Check for probability of resonance, from ( 6)fund'l natural frequency, fn = 358e3/naCD Hz 96 97ratio natural/running frequencies = fn*60/400 14.4 14.6

Check for yielding when solidifiedtensile ultimate, Sut (Table 2) MPa 1330 1305shear yield Sys = 0.48 Sut (Table 2) 640 625

solid stress, τs = Ks 8FsC/πd2 MPa 403 404

7060

50

10

1520

30

1.0

30

40

5

7

0.8

2.01.51.2

12

11

10

9

8

7

6

5

4

34.54.0 5.0 5.6 6.3 7.1 7.9

A

B

C

D

E

coildiameters

fatiguesafety factors

totalturns

ODlimit

IDlimit

abs

con

wire diameter, mm

sprin

g in

dex

stabilitylimits

300 600

50

Sample dialogue with 'Springs'

Springs 8

APPENDIX : PRESETTING A TORSION BAR

We shall first derive some general results for a homogeneous solid tor-

sion bar of radius ro and length L, made from an elastic-perfectly plas-

tic material whose shear yield and modulus are τy (ie Sys above) and G

respectively.

When torque T is applied to the ends of the bar, one end rotates

through an angle θ with respect to the other. Cross-sections remain

plane and any radial line in a cross-section remains straight, no matter

whether behaviour is elastic or plastic. From the sketch, compatibility requires the shear strain γ at

radius r to satisfy :

Lγ = rθ or

( i ) γ = θ'.r where θ' ≡ θ /L is constant for all radii in the cross-section.

From the constitutive law, the corresponding stress is :

(ii) τ = Gγ = Gθ'.r from (i) if elastic, or

= τy if plastic; ie τ = minimum ( Gθ'.r, τy )

The torque T is the resultant of shear stress over the cross-section, ie from the sketch :

T = ∫0

ro( 2πr.dr ) τ .r = 2π ∫0

roτ r2 dr where τ = f (r).

A piece-wise linear variation is defined by a known stress τi at some interior radius ri, together with a stress τo at the

outer surface, radius ro. From the foregoing, the torque which corresponds to this stress pattern is :

T = 2π ∫0

ri ( τi

r/ri ) r2 dr + 2π ∫ri

ro{ τi + ( τo – τi ).( r – ri )/( ro – ri ) } r

2 dr ie

(iii) Tro /J = τo + ( τi – ρτo ) ( 1–ρ3 ) /3 ( 1–ρ ) in which ρ ≡ ri /ro and J =

π/2 ro

4

The above results will now be applied to presetting and loading of the torsion bar, which involve the following steps.

Under increasing load the torsion bar responds elastically until the maximum

stress (at the outer surface, radius ro) reaches the yield stress, τy. The linear por-

tion of the bar's normalised torque-twist characteristic is terminated by the point y

where the torque Ty and the twist θy' correspond to τy occurring at ro. From the

above, or from the familiar elastic equations, these critical parameters are related

through :

(iv) τy /ro = Ty /J = G θy'

The torque and angle of twist are further increased with progressive yielding of

the bar's outer skin and shrinking of its elastic core. At any particular load the

stress in the core varies linearly from zero at the axis to τy at r i = ρro, the elastic-

plastic interface, while the stress in the skin is constant at yield. From (ii) applied

to yield at the interface, and from (iii) with τi = τo = τy, the variations of torque T

and twist θ' ≥ θy' with core size ρ are :

( v) roT /Jτy = 1/3 ( 4 – ρ3 ) ; Groθ'/τy = 1/ρ

which results in the non-linear portion of the torque-twist characteristic.

This completes the loading phase of the presetting operation, but if loading were

to continue then the limit load Tp would be reached when plasticity had extended

right throughout the bar cross-section - whereupon, from (v) with ρ = 0 : roTp /

Jτy = 4/3. That is the plastic collapse torque Tp is 4/3 the torque at which yieldingfirst commences, Ty.

The preload T of the previous step which caused yielding of the skin to radius ri =

ρro is now removed completely. All the material, including that which has reached

yield, unloads elastically as suggested by the straight unloading line of the torque-

twist characteristic above. However although there is zero net torque after unlo-

ing, residual stresses and a permanent set have been induced, as sketched here.

To find this permanent twist, θr', together with the residual stresses τri and τro, we note the compatibility requirements

(i) : γri = ri θr' and γro = ro θr'. Since the interface lies on the elastic line, τri = Gγri, and considering the linear

unloading process at the outer surface from the previous stress/strain state above, τro = τy – G ( τy /ρG – γro ).

From these and from (iii) with zero net torque we obtain :

( vi) Groθr'/τy = ( 3 –4ρ +ρ4 )/3ρ ; τri /τy = ( 3 –4ρ +ρ4 )/3 ; τro /τy = ( ρ3 –1) /3 ≤ 0

Plotting these states above demonstrates the negative shear on the outer surface which will go some way towards coun-

teracting any subsequent positive shear due to an applied load.

The bar, having been preset, is now put into operation where it sustains the service torque Ts. Assuming the bar to

behave elastically still (we shall check this in a minute), the stress distribution will be a superposition of the elastic load

stress τ/r = Ts/J, and the residual stress given by (vi) whose torque resultant is zero. Thus :

(vii) τo = roTs /J + ( ρ3 –1) τy /3

τi = ρ.roTs /J + ( 3 –4ρ +ρ4 ) τy /3 which are plotted here.

The maximum service torque Ts.hi will be that value of Ts which causes τo or τi to just

reach yield. This is easily found from (vii) to be :

(viii) roTs.hi /Jτy = ( 4 –ρ3 )/3 - exactly the same result as (ii).

It is evident therefore that this elastic torque capacity of the preset bar is the same as the plastic torque which was used

to preset the bar initially, and is higher than the torque which just started to yield the bar, roTy /Jτy = 1 from (iv).

Thus loading after presetting involves elastic return up the unloading line shown in the torque-twist characteristic

above.

Presetting thus increases the torque capacity of a bar by decreasing the stresses in the highly stressed outer skin, whilst

compensating by an increase of stress level in the lightly loaded core.

Springs 9

SPRINGS - PROBLEMS

Springs are steel helical compression springs with ends squared and ground, unless otherwise specified.

1 Identical springs are to be designed to support a machine in order to isolate vibrations. The static load on each spring is 7 kN under which the deflection should be not less than 50mm, and the outside coil diameter must not exceed 150 mm. The shear yield strength of thematerial selected is 690 MPa, which may be assumed independent of wire size.Determine the springs' manufacturing dimensions based upon a design factor of 1.5.

2 A spring has a mean coil diameter of 76 mm, a wire diameter of 12.5 mm, and 5 active coilswith a freelength of 140 mm. The material is ASTM A229 oil tempered wire.(a) Determine the spring's stiffness [ 110 N/mm](b) Check the solid stress [ 619 MPa](c) Calculate the factor of safety for a load which fluctuates between 2 and 3 kN. [ 1.3]

3 A spring from ASTM A229 oil tempered wire is required which will fit inside a guide tube of70 mm bore, exert a minimum force of 1 kN, and when compressed a further 30 mm exert aload not exceeding 1.5 kN. Determine the manufacturing dimensions of a suitable spring to last indefinitely.

[ wire dia. 8 mm, mean coil dia. 51 mm, total turns 20.5, free length 263 mm ]

4 Is there a feasible solution to the previous problem if the guide tube's bore is 40 mm?

5 A helical compression spring is required to have infinite life when subjected to a load whichfluctuates between 250 and 500 N. Wire having Sut=1250 Sus=1000, Sys=680 MPa and a zero-to-maximum torsional endurance limit of 550 MPa will be used. These values apply to the sizerange and surface finish of the spring.What wire size should be used if the spring index is 7 ? [ 4 mm ]

6 A cam rotates at 10 Hz and imparts a sinusoidal oscillation to its follower. Follower total lift is20 mm and the mass of the oscillating parts is 9 kg.A return spring is required to overcome the inertia of the oscillating parts and to maintain con-tact between follower and cam. To fit the space available, the spring inside diameter must be atleast 25 mm and the outside diameter no more than 50 mm. Ascertain the manufacturing dimensions of a spring made from music wire and with indefinitelife.

Threaded Fasteners 1

- a ‘blind’ tapped hole (as opposed to a tapped hole passing rightthrough the component) extending deeper than the stud andending in a conical point of 120o approximately;

- a stud's depth of engagemeni typically 11/4 to 11/2 times its size;- a threaded length sufficient for the nut to be tightened, leaving a

couple of threads ‘exposed’ (ie. not engaged) to cater for variationsin thickness of the assembly - though too much exposed thread should be avoided;

- a thin locknut which may be jammed against the ordinary nut to assist assembly or to prevent loosening under severe vibration, thoughresistance to inadvertent loosening is usuallyeffected by a thread locking fluid, a lock washer ora self-locking nut such as shown here :-

Screws may be supplied complete with captive lock washer to ease assembly - they are thenknown as ‘sems’ and come in many forms including self-tapping screws for joining sheet metal.

Salient geometric features of the thread areillustrated in ( v). The distance between simi-lar points on adjacent threads is the thread’spitch. The load on the bolt Fb passes fromthe nut gradually through the engagedthreads into the bolt, however the wholeload must pass through the transverse cross-sections X-X at the exposed threads outsidethe nut. Neglecting stress concentration, thetensile stress in way of the exposed threads is therefore :

( 1 ) σ = Fb / As where As is the stress area, a function of thread size and geometry.

Since the stress area is less than the cross-sectional area of a normal (non-reduced) shank, theexposed threads usually are the most critically loaded part of the assembly - this is why failure ofthreaded joints occurs most commonly close to the nut face.

A thread can be likenedto a piece of stringwound in a tight helixaround a cylinder - or aconical frustum in thecase of pipe threaddesigned to eliminatefluid leakage. When anut on a screw isrotated by one turn, ittravels along the screw a distance known as the lead L. Developing one turn of the thread at themean diameter dm (the average of major and minor diameters) gives the lead angle (or helixangle) λ as tanλ = L/πdm.

Power screws may employ multiple strings, or starts, so L = p*number of starts as illustrated. Fas-teners on the other hand are almost invariably single start (L=p); they are also right handed toavoid confusion in tightening though LH screws appear commonly in turnbuckles and in cer-tain parts where the prevailing torque would tend to loosen RH fasteners.

tappedhole

locknut

clearancehole

( iv)

dia

met

erm

ino

r

ma

jor

exposedthreads

X

X stress area

As

section X-X throughexposed threads

( v)

nut

pitch

p L p L p L

single start, L = p double start, L = 2p triple start, L = 3p

dm

λ

L

π dm

THREADED FASTENERS

The fundamental operation in manufacture is the creation of shape - this includes assemblywhere a number of components are fastened or joined together either permanently by weldingfor example or detachably by screws, nuts and bolts and so on. Since there is such a variety ofshapes in engineering to be assembled, it is hardly surprising that there is more variety indemountable fasteners than in any other machine element. Fasteners based upon screw threadsare the most common, so it is important that their performance is understood, and the limita-tions of the fastened assemblies appreciated.

There are two distinct uses for screw threads, and they usually demand different behaviourfrom the threads :

- a power screw such as a lathe leadscrew or the screw in a car lifting jack which transformsrotary motion into substantial linear motion (or vice versa in certain applications), and

- a threaded fastener similar to a nut and bolt which joins a number of components togetheragain by transforming rotary motion into linear motion, though in this case the translationis small.

A typical hexagon headed bolt and nut are shown at ( i). Thediameter of the bolt shank is usually the same as the outsidediameter - the major diameter or briefly size - of the thread. Theradiused fillet at the junction of shank and head reduces stressconcentration.

The shank diameter of a ‘waisted’ bolt ( ii) is less than the threaddiameter thus allowing a radiused thread runout which reducesstress concentration - beneficial in fatigue applications. Theassembly illustrated incorporates a washer under the nut whichpromotes uniformity of contact - minimising damage to theunderlying parts and again lessening stress concentration. Thebolt head may be equipped with an optional integral washer face. A bolt's grip is the thickness of the fastened parts.

A screw ( iii) is similar to a bolt - the names are often loosely interchanged- though strictly a bolt is equipped with a nut which is rotated to tightenthe assembly, whereas a screw is itself rotated and engages with a threaded(or ‘tapped’) hole in a stationary component such as an engine block cast-ing. The screw illustrated has no shank, being threaded right up to thehead. There is a great variety of screw head forms available.

The ubiquitous socket headed setscrew shown here is tightened by a hexagonwrench rather than by a spanner.

A stud ( iv) has no head and is threaded at both ends. The ends are not necessar-ily identical. One end is screwed into one of the components usually before the secondcomponent is assembled. The sketch ( iv) illustrates also :

- a clearance hole through a component, typically 5-10% larger than the bolt/stud size to facil-itate assembly and to clear any shank/head fillet;

- a tapped hole which is drilled smaller than the root or minor diameter of the thread - seethe enlargement ( v) below;

threadlength

headradius

shankthread runout

nut

nominallength( i)

washer

reduced shankradius

grip

washer faceradius

( ii)

widthacross

flats

( iii)

Threaded Fasteners 2

threads is based upon the average of theminor and mean diameters, ie. : ds = d – 3/4 p

Screw Thread Mechanics

There are always three major components in practical applications of a screw thread mechanism- the screw - a generic name applied to a setscrew, leadscrew, bolt, stud or other component

equipped with an external thread- the nut - refers to any component whose internal thread engages the screw, such as the nut

of a nut & bolt or a large stationary casting with a tapped hole into which a stud is screwed - the thrust bearing - that is the contact surface between two components which rotate with

respect to one another - examples of this include :a) the under-surface of a screw head which is being

tightened by a spanner, andb) the spherical seating of a G-clamp screw in the sta-

tionary self-aligning anvil.

A nut can spin and move freely along a screw without con-tacting another component, ie. without the need for anythrust bearing, but a thrust bearing comes into existence immediately contact occurs and themechanism is put to practical use.

Clearly there is relative motion in the thrust bearing, and alsobetween the nut and the screw - and where there is relativemotion there is friction. We now examine the role of friction sinceit dominates the behaviour of the mechanism unless special ( read'expensive' ) means are taken to minimise its effects. When con-sidering friction it doesn't matter which component rotates andwhich is stationary - it's the relative motion which is important.We shall therefore analyse the jack shown here to deduce the gen-eral effect of friction on screw thread behaviour.

The jack's screw is fixed; the nut is rotated by a spanner and trans-lates vertically. The thrust collar's only motion is vertical transla-tion as it is prevented from rotating by contact with the load, onecorner only of which is pictured. Since there is relative rotation between contacting nut andcollar, the contacting surface assumes the role of thrust bearing.

The nut shown here in plan is in contact with three bodies :- the spanner exerts the torque T which tends to raise the load (analo-

gous to tightening a nut and bolt)- the screw thread which exerts the frictional torque Tt , and- the thrust bearing which exerts the frictional torque Tb .

We are interested in the tightening torque T, and, if the nut is in equilibrium then

( i ) T = Tt + Tb

from which T follows once Tt and Tb are found individually.

Consider the thrust bearing first. The contact surface of area A is assumedto be a narrow annulus of mean radius rb on which the uniform pressureis W/A, where W is the load supported by the mechanism. If the coeffi-

p

d

p2

p2 =

=

α=29o

SQUARE ACME

( a)

( b)

load

trans-latingthrustcollar

thrustbearing

nut

fixed screw

TtTb

T

δN

δAµ δNb

rb

Thread Geometry

A thread system is a set of basic thread proportions which is scaled to different screw sizes todefine the thread geometry. Whitworth, Sellers, British Standard Pipe (BSP) are just three ofmany systems which proliferated before the adoption of the ISO Metric thread system. Since thislast is now universal, it alone is examined here.

The basic profile of ISO Metric threads isbuilt up from contiguous equiangular tri-angles of height h disposed symmetri-cally about a pitch line which becomesthe pitch cylinder of diameter d2 whenthe system is rotated about the axis toform the thread. The distance betweenadjacent triangle - the pitch - is p = 2/√3 h. The tips of the triangles are truncated by h/8 to formthe major diameter (size) d of the thread, and the bases are truncated by h/4 to form the minordiameter d1. It follows that d1 = d - 5/4h = d - 1.08p. This leads to a rule of thimb for suitable tap-ping size drills in normal materials : dtapping = d - p.

The basic profile becomes a maximum material profile for external threads (on screws) and inter-nal threads (in nuts) through the use of suitable radii and tolerances, so that adequate clearance

is assured when internal and external threads engage. The relatively large radii at the minordiameter of external threads tends to equalise the strengths of external and internal threads. AS1721 sets out comprehensive geometric data including toler-ances, however knowledge of these many details is not neces-sary here.

At its most fundamental, a thread definition comprises a com-bination of size and pitch. Thus M14x1.25 refers to a Metricthread whose major diameter d is 14 mm and whose pitch p is1.25 mm. The stress area of an external thread corresponds to adiameter ds = d - 13/12 h that is As = π/4( d - 0.9382p)2. Othersalient features follow from the underlying geometry.

Most threaded fasteners in general engineering are manufac-tured to the ISO Metric Coarse Pitch (First Preference) Seriesoutlined in Table 1. Fine pitch and constant pitch series areused for special purposes such as IC engine spark plugs andexternally threaded thin-walled pipes.

The 60o thread form is not suitable for power screws whichtransform motion and which therefore must have high effi-ciency. The square thread illustrated opposite offers the bestefficiency but is generally impractical. The Acme thread form isthe best compromise between efficiency and ease of manufac-ture, assembly and wear take-up. The stress area of Acme

TABLE 1ISO METRIC COARSE PITCHFIRST PREFERENCE SERIES

nom. pitch stress hexagonsize area a/c flats

d p As sm m m m m m2 m m

1.6 0.35 1.27 3.22 0.4 2.07 4

2.5 0.45 3.39 53 0.5 5.03 5.54 0.7 8.78 7

5 0.8 14.2 86 1 20.1 108 1.25 36.6 13

10 1.5 58.0 1612 1.75 84.3 18

16 2 157 2420 2.5 245 3024 3 353 3630 3.5 561 4636 4 817 55

internal

external d

p

d

h/8

d2 d1

α = 60o

to axis

h =

=

h/4

pitch line

Threaded Fasteners 3

applied ( Tt = 0 ) the nut is at rest, point c. The torque is thereafter increased until at d the nutbreaks away and rotates up the thread at constant velocity. If the torque were to increase furtherthen the nut would accelerate.

If however at c the torque were to increase in a negative sense - to 'undo' the nut, assisted by theload - then when point e was reached the nut would break away and move down the thread atuniform speed. Any further increase of torque in the negative sense would lead to downwardacceleration.

The point b is significant. It corresponds to the critical coefficient of friction for the lead con-cerned, µc = tan λ, at which the nut is on the point of moving down the thread of its own voli-tion without any torque applie ( Tt = 0 ). Further, if . . .

µ > µc then the thread is self-locking in that the nut cannot undo by itself, it needs to beunscrewed by a definite negative torque; or

µ < µc then the thread is overhauling in that the nut will unscrew by itself under theaction of the load unless prevented by a positive tightening torque.

Clearly self-locking behaviour is essential for threaded fasteners. Car lifting jacks would not beof much use if the load fell as soon as the operating handle was released - so they too demandself-locking characteristics.

Some applications of power screws require over-hauling behaviour. The Archimedean drill (above)and pump action screwdriver (below) incorporatevery large lead angles which guarantee that thecritical friction coefficients are larger than any actual coefficient likely to be encountered. Sensi-

tive linear actuators may incorporate recirculating ballscrews (shown in part-section, left) to reduce thread frictionto levels which go hand-in-hand with overhauling.

Efficiency is important in rotary/linear motion transfor-mation such as provided by power screws. Whilst acting

against a load, the efficiency of a screw thread alone ( ie. neglecting the thrust bearing ) is :

η = Wfor a given Tt /Wfor the same Tt without friction

= tan λ / tan ( λ + φ ) from ( iii)

The effect of lead and friction on thread efficiencyis plotted here - as expected, high efficienciesdemand very low friction. Also shown is the lineof critical friction.

Coefficients of friction around 0.1 to 0.2 may beexpected for common materials under conditionsof ordinary service and lubrication - steel on steelfor fasteners, and steel on bronze or cast iron inthe case of power screws. Special coatings and sur-face treatments can reduce friction significantly.Vibration also reduces friction and can cause fas-teners to unloosen unless locked as mentionedabove.

Lead angles around 3o are typical of single start power screws which are therefore somewhat

inefficient - this is one reason for multiple start screws with their associated large leads.

efficiency,η (%)

0.2 0.4 0.6 0.80

0.2

0.3

0.4

0.5

0

0.1

lead angle, λ (radians)

coef

fici

ent

of

fric

tio

n,

µ

EFFECT OFFRICTION &

LEAD ANGLE ON THREADEFFICIENCY

ove

rha

uli

ng

40

60

self

-lo

ckin

g

20

80

9095

cient of friction in the bearing is µb then the torque exerted by the frictional force on area ele-ment δA is δTb = µb δN rb = µb rb (W/A) δA. Integrating over the whole contact area :

( ii ) Tb = W µb rb

Consider now the thread which is square, of mean radius rm and lead angle λ. The nut engagesthe screw with friction coefficient µ corresponding to a friction angle φ = arctan µ. The static andkinetic coefficients of friction are taken to be essentially equal for this preliminary analysis.

We wish to find the torque Tt which must be exerted on the nut to offset thread friction andmaintain the load W in equilibrium - that is either static or moving at constant speed. A torquewhich tends to raise the load is reckoned positive; a negative torque is one which tends to lowerthe load.

A small element of the nut is shown below sliding or about to slide on the inclined ramp of thethread, this motion being either up or down the thread. Both motion directions are sketched. δPis the small force on the element due to the torque, and the element supports a small part of theload δW. The contact force components are the normal and friction forces, δN and δF - the latteropposing motion or motion tendency.

The nut element free bodyrequires for equilibrium . . .

normal to plane : δN = δP sinλ + δW cosλ parallel to plane : δF = δP cosλ - δW sinλ δF = - δP cosλ + δW sinλ

If the element is moving, or on the point of motion, then δF = µ δN and so eliminating δF, δN δP = δW tan ( λ + φ ) δP = δW tan ( λ − φ )

The total torque Tt is the sum Σ rm δP over all the identical nut elements, ie.( iii ) Tt = W rm tan ( λ + φ ) Tt = W rm tan ( λ − φ )

. . . if motion (tendency) is upwards downwards

When raising a load the thread inclination λ is thus effectively steepened by the thread frictionangle φ.

The upward tightening torque from ( iii) is plot-ted here against friction coeffficient for a leadangle, λ = 0.25 ( 14.3o).

When there is negligible thread friction ( φ = 0 )then Tt/Wrm= tan λ = 0.255, the point a.

The equilibrium line a-d corresponds to incipi-ent motion up the thread; the equilibrium line a-b-e corresponds to incipient motion down thethread. The area between these lines representsstasis ( δF < µ.δN ); the areas beyond the lines cor-respond to non-equilibrium, ie. to acceleration -though it should be realised that only thread fric-tion is being considered at this point.

Consider what happens if the coefficient of fric-tion is 0.5 for example. Before any torque is

δWδP

δNδF

λ

motion (tendency)

δWδP

δN

δF

λ

motion (tendency)

no

rmal

ised

to

rqu

e,

T

/Wr

tm

0.2 0.4 0.6 0.80

1.5

1.0

0.5

0

-0.5

friction coefficient, µ = tan φ

accelerationupward

stasis

1.0-1.0

ab c

d

eOVER-

HAULINGSELF-LOCKING

acceleration downward

EFFECTON NUT OF

UPWARD TIGHTENING TORQUE FOR

A LEAD ANGLEλ = 0.25

tan( λ - φ )

tan( λ + φ )

Threaded Fasteners 4

We have noted that the elastic behaviour of connected members has a significant effect uponthe distribution of load between their connectors. In an elastic bolt and nut connected by anumber of thread turns, the first 'thread' takes a disproportionate frac-tion of the total tensile load - Craddock op cit reports theload distributed among the threads as sketched here. It isclear that there is little benefit from an engagement lengthexceeding half-a-dozen threads. Special nuts which allevi-ate non-uniform load distribution are available. The variation ofnormal stress σ over a cross- section adjacent to the nut face ( assketched ) is dominated by stress concentration induced by thethread root, so it is hardly surprising that bolt fractures usually occur in theexposed threads close to the nut face.

Another failure mechanism is stripping of the nut threads, which is essentially shear failure ofthe nut material on the cylindrical surface at the thread majordiameter. Stripping of the bolt threads is a similar shear of thebolt material at the minor diameter - but this is rare. Other possi-ble failure mechanisms, such as crushing of the nut bearing sur-face and dilation of a thin nut due to its riding up the threadflanks, are not critical in themselves, but contribute to other modes.

If any failure is to occur, then bolt fracture is the preferred mode. Bolt fracture is clearly discerna-ble and often occurs whilst tightening, when torsional shear stresses ( which dissipate quicklyafter tightening ) are superimposed upon tensile stresses. So the operator replaces the bolt andlearns to exercise care in subsequent tightening. Thread stripping on the other hand is insidiousand progressive - the first thread fails putting more load onto the remaining threads, hence thesecond succumbs . . . and so on. When threads strip it is often difficult to separate componentsfor fastener replacement, whereas a broken bolt requires no further separation. The conse-quences of a tapped hole in a car engine block being stripped by over- zealous spanner wieldingis a case in point.

For a given load, bolt fracture tendency will clearly be reduced by a larger stress area ( ie. a largerbolt size ) and/or a higher class of material. Given these parameters ( load, size and bolt class ),nut stripping will be reduced by a longer nut and hence an increased cylindrical shear area, by athicker walled nut to decrease dilation, and/or by a superior nut (property) class.

The first stage of fastener design is bolt selection - size, class and other geometric attributes. ISOMetric nuts have been dimensioned to bias any failure towards bolt fracture rather than nutstripping, on the tacit presumption that the nut class is equal to or greater than the bolt class - ie.the material of the chosen nut is at least as strong as the preselected bolt's material. It is for suchreasons that regular nut lengths are about 80% of bolt size, and of the hexagonal dimensionsshown in Table 1 - though there are various styles of nut which differ from these figures.

Loads in an Elastic Bolted Assembly

The safety of a threaded fastener depends upon the actual load on it and the maximum load itcan withstand. The latter has been considered above. Since the fastener is part of an indetermi-nate assembly it is now necessary to analyse such an assembly to find out just what is the actualload on the fastener itself. The analysis is very similar to that carried out for an indeterminateassembly in the earlier chapter ‘Stress, strength and safety’.

34

2316

11 9 7

% load

σ

From ( i), ( ii) and ( iii) the total torque necessary to raise a load W whilst overcoming thread andthrust bearing friction is :

( 2 ) T = W [ rm tan( λ + φ ) + µb rb ]

The foregoing analysis has been developed for square threads ( α = 0 ), but ( iii) and ( 2) may beapplied with little error to Acme threads ( α = 29o ) and also to ISO Metric fasteners ( α = 60o )provided the effect of wedging ( analogous to V-belts ) is included - thus the effective coefficientof thread friction to be used in ( 2) is µ.sec( arctan( cosλ.tanα/2 )), which tends to µ.secα/2 for

small lead angles.

Static failure

We shall concentrate on steel fasteners, which,because of their strength and cheapness, constituteover 90% of all fasteners used. Steels for commer-cial fasteners are graded into the property classes ofTable 2. Each class number consists of two figures

- the first figure is Su/100 where Su is the steel's nominal tensile ultimate (MPa)- the second figure is the ratio Sy /Su where Sy is the nominal 0.2% offset yield strength.

Minimum material strengths are not less than nominal values.

It is difficult to determine the yield of a full size threaded connector ( as opposed to a cylindricaltest piece ) because of the different strain rates of shank, thread and runout. For this reason theproof stress rather than the yield stress is used as a criterion for failure assessment - the proofstress Sp is the largest stress which does not lead to any permanent set. The proof load of a screwmade from a particular material is the maximum load the screw can withstand without perma-nent deformation, and is given by ( 1) as the product of stress area (Table 1) and proof stress(Table 2). Selection of class 12.9 should not be undertaken lightly - its high strength begets otherpotential problems.

TABLE 2 STRENGTH OF STEEL BOLTS

class no. 4.6 5.8 8.8 9.8 10.9 12.9Su MPa 400 500 800 900 1000 1200Sy MPa 240 400 640 720 900 1080Sp MPa 225 380 590 650 830 970elong'n % 22 20 12 10 9 8

EXAMPLE A screw jack similar to the one illustrated above raises a load of 5 kN. The screw is 24 mmdouble start Acme with 4 mm pitch, and the thrust bearing's mean diameter is 35 mm. Coefficients of run-ning friction in thread and bearing are estimated to be 0.12 and 0.09 respectively.

Find the starting torque for raising and for lowering the load, and calculate the jack's efficiency whilst lifting.

Mean diameter of thread dm = ( d + d1 )/2 = ( d + ( d-p))/2 = d -p/2 = 24 -4/2 = 22 mm

so r m = 11 mm

Lead angle ( 2-start ) λ = arctan( L /πdm) = arctan( 2x4 /22π) = 6.60o

Take static coefficients of friction to be about a third greater than kinetic values, ie. µthread,static = 0.16, µbear-ing,static = 0.12. Assume that the square thread equations ( 2) are sufficiently accurate for the Acme here.

Raising, starting φ = arctan µ = arctan( 0.16) = 9.09o so from ( 2)

T = 5 ( 11 tan( 6.60o + 9.09 o) +17.5x0.12) = 26 Nm

Lowering, starting T = 5 ( 11 tan( 6.60o - 9.09o) - 17.5x0.12) = -13 Nmie. 13 Nm loosening

Raising, in motion φ = arctan µ = arctan( 0.12) = 6.84o

so from ( 2) the load raised by 1 Nm W = 1 / ( 11 tan( 6.60o +6.84o ) +17.5x0.09) = 238 N

while the load raised by an ideal jack W o = 1 / ( 11 tan( 6.60o + 0 ) +17.5x0.0 ) = 785 N

and so the efficiency is η = W/Wo = 238/785 = 30%

Threaded Fasteners 5

the bolt load Fb, however for the case considered, the majority of these components, and allintermediate components such as j3, are subjected to the compressive load Fj only. Weshall here assume that all joint components are wholly compressed by Fj - this approxima-tion should be reviewed for other arrangements.

Constitutive Laws Bolt - Since in the previous steps the bolt is loaded by the tensile force Fb and undergoes thetensile deformation δb, then, if kb is the axial stiffness of the bolt :

Fb = kb δb Joint - Since in the previous steps the joint is loaded by the compressive force Fj and under-goes the compressive deformation δj, then, if kj is the stiffness of the joint - that is of thethree joint-elements in series 1/kj = 1/kj1 + 1/kj2 + 1/kj3 :

Fj = kj δj

Solving the four equations above resolves the indeterminacy and yields component loads :

( 3a ) Fb = Fi + P.ke/kj where 1/ke

= 1/kb + 1/kj

; F i = ke∆Fj = Fi – P.ke/kb

ke is the equivalent stiffness of the assembly ( ke < kb & kj ). Evidently the bolt is in series withthe joint, ie. all four components are in series. Fi is the initial load (or preload) in the assemblydue to tightening the nut by an amount ∆.

The bolt load Fb and the joint load Fj from ( 3a) are plotted in ( vii)versus the external load P, with ke < kb < kj.

When the external load is zero then Fb = Fj = Fi and the load pathis a closed loop through all elements in series. As the externalload increases then the joint-elements become less tightly com-pressed - Fj decreases - while the bolt load increases. For the caseconsidered, the rate of bolt load increase ke/kj

is less than the rateof joint load decrease ke/kb

because kb < kj.

The graph indicates that the joint force becomes zero when the external load reaches some criti-cal value P*. Since Fj cannot be tensile (negative by the present convention) it follows that j3must lose contact with the other joint components. Joint separation occurs and the bolt takes allthe external load, as may be confirmed by free bodies. So, after separation Fj =0 and Fb =P.

In the arrangement ( viii)the assembly is required tosustain the external loadPo without separation, butthe preload and stiffnessesare such that Po exceedsthe separation load P*. Therequirements cannot bemet - a larger separationload is necessary.

One way of increasing the separation load is to increase the preload Fi as in ( ix). The slopes ofthe bolt and joint load characteristics are not altered by this extra tightening. Another approach(x) uses the same preload as ( viii) but with an increased stiffness ratio kb/kj. Clearly it is easier toalter the preload (within the strength capabilities of the bolt and joint components) than it is toalter stiffnesses.

( viii)

P

P*00

Fi

F

Po

( ix)

0

Fi

P*

P

0

F

Po

ke

kbk j

P

P*00

Fi

F

Po

ke

kbk j

( x)

( vii)

P

P*00

Fi

Fb

Fj

F

jointsepar-

ation

ke

kb

450

kek j

The nut and bolt b of figure ( i) are used to connect together three joint-members j1, j2 and j3(ii). After tightening the nut by a known amount ∆, the external load P is applied axially to j1and j2, and tends to separate the joint members. Joint separation must usually be avoided, soexamination of the assembly has two major goals :

- to determine the assembly's tendency towards separation, and- to ascertain the safety of all components of the assembly.

The load on each component is a function of the initial tightening and the external load, but thecomponent loads cannot be found immediately because the assembly is statically indeterminate- so this indeterminacy must first be resolved before separation or safety can be addressed. Reso-lution requires consideration of compatibility, equilibrium and the constitutive laws of all com-ponents, whose behaviour here is assumed to be elastic, frictionless and at room temperature.

Compatibility Shown in section ( iii) is the free assembly with the nut snubbed (or ‘finger tight’)that is the nut has been tightened to close up all gaps in the assembly, without inducingappreciable loads. An imaginary mark is situated on the thread a distance ∆ from the nutunderside. After tightening, this mark will coincide with the nut face, so ∆ represents thedistance moved by the nut along the bolt thread during the tightening operation. Aftertightening and loading by the external force, the disposition of the assembly elements isshown much exaggerated in ( iv). The extension of the bolt is δb while the compression ofthe joint δj is the sum of its elements' contractions, ie. δj = δj1 + δj2 + δj3From the geometry of ( iii) and ( iv), compatibility necessitates :

∆ = δb + δj

Equilibrium The free body of the bolt and nut ( v) illustrates the unknown forces at the bolt's twocontacts. The upper Fb is the axial resultant of the uniformly distributed pressure exerted bythe top surface of j1 over the annular area under the bolt head The other contact - theupper surface of the nut bearing on the underside of j2 - yields a similar pressure resultantwhose magnitude must also be Fb for equilibrium of the bolt.The free body of the joint element j3 in ( vi) shows the unknown pressure resultant Fj dueto annular contact with its neighbours. The reactions to the aforementioned Fb and Fjappear on the remaining free bodies j1 and j2 in ( vi) from either of which results :

P = Fb – FjEvidently the bolt is under tension Fb.

Parts of the joint components j1 and j2 in contact with the nut and bolt are also subjected to

( i)

nut &bolt

section ofsnubbed

assemblytightened& loaded

free bodiesof loaded

components

( ii) ( iii) ( iv) ( v) ( vi)

P

P

P

∆δb

δ j

j1

j2

j3

bFb

Fb

b

PFj

Fb

j1Fj

Fj

j3

P

Fb

Fj

j2

Threaded Fasteners 6

Nut factorsSteel Thread Condition Kas received, stainless on mild or alloy 0.30as received, mild or alloy on same 0.20cadmium plated 0.18molybdenum-disulphide grease 0.14PTFE lubrication 0.12

ally specified between 75% to 100% of proof, with 90% being common. Thus for an M16x2 class 8.8screw the proof load is AsSp = 157 *590 = 93 kN and so a suitable preload is 0.9*93 = 83 kN. Which isall very easy . . . . but how do we ensure that the nut is tightened to produce 83 kN in the field?What measuring devices can be used to control preload? Bickford op cit puts it neatly thus : When we tighten a bolt,

( a) we apply torque to the nut,( b) the nut turns, ( c) the bolt stretches, ( d) creating preload.

Ideally we would like to measure preload directly ( d) and advise the spanner operator when therequired preload is reached. However in practice, feedback is usually from an earlier step, and someof the difficulties which this causes are now examined.

Torque wrenches are commonly used to monitor the torque applied to a nut being tightened - step(a) of Bickford's sequence. We may employ ( 2) to obtain an indication of the spanner torque T nec-essary to induce the preload Fi.

The pitch of ISO Metric fasteners in the sizes of most practical interest lies between 0.12d and 0.16d(Table 1) and correspond to helix angles λ around 2-3o. The thread mean diameter is dm = ( d+d1)/2= d–5h/8 = d–0.5413p so, assuming p ≈ 0.14d, then the mean thread radius is rm ≈ 0.46d. The annu-lar contact area under the nut hexagon plays the role of thrust bearing. Its OD is about 1.5d as notedabove, so the mean bearing radius is rb ≈ (d+1.5d)/4 = 0.63d. Assuming the same coefficient of fric-tion µ in nut bearing and in threads, ( 2) on expansion with the neglect of small helix angles becomes

( 2a ) T = Fi ( rm( p/πdm + µ.sec30o) +µrb ) = Fi d ( 0.022 +0.53µ +0.63µ) = K Fi d

The three terms in the penultimate RHS here represent three portions of the tightening torque :- the torque useful in 'raising the load' ie. in generating preload ( 0.022 or 10% )- the reaction torque dissipated in thread friction ( 0.53µ or 40% )- the torque used in overcoming nut bearing friction ( 0.63µ or 50% )

The percentages quoted refer to an average friction coefficient around 0.15. Only some 10% of theapplied torque generates preload - and herein lies the cause of most joint problems. Preload iswhat's left of the tightening torque after friction has absorbed the lion's share, since preload cannotbe induced until motion has occurred. Anticipated variations in the coefficient of friction will thushave a marked effect upon the preload, and careful calibration of the torque wrench will not muchimprove preload accuracy.

Mean values of reported nut factors K in ( 2a) aretabulated here to give an appreciation of the benefi-cial effect of proper lubrication of thread and nut faceon preload. Variations of ±25% in K are routinelyexperienced; larger variations are not unexpected.Ideally, each application should be tested to deter-mine its own peculiar nut factor since the factorincludes the effects of everything that affects the torque-to-preload transformation, not just frictionbut also the consequences of local deformations, interrupted tightening, inaccurate thread form, mis-aligned components, relaxation due to tightening other bolts in the joint, whether the fastener is newor is being reused, initially bent bolts, speed of tightening etc etc - all of which affect preload repro-ducibility. It is concluded that torque wrenches may be quite adequate for everyday non-criticaltightening operations such as those on vehicles, but the measurement of torque, step ( a), is by itselfinadequate when preload accuracy is needed.

In a fatigue situation with given alternating load on the assembly, P, theamplitude of the damaging fatigue load on the bolt Fb is reduced as the slopeof the Fb-P characteristic is decreased ( xi). Comparing the arrangement ( ix)with ( x) it is apparent that when designing for bolt fatigue, the low slopearrangement ( ix) is preferable - that is a relatively low bolt stiffness kb andhigh preload Fi are desirable.

The stiffness of an elastic component which is subjected uniformly to tension or compression isk = F/δ = AE/L

A bolt may be reckoned as two elements in series, each with a con-stant cross-sectional area - the shank, and the exposed thread whosearea is the stress area As. The lengths of these should include halfthe head and nut thicknesses respectively, to allow for local defor-mation. The stiffness of short bolts is less predictable than that of long bolts.

A gasket is a thin compliant sheet rather similar to j3 above, which is sandwiched between otherjoint components to fill up microscopic irregularities thus preventing fluid leakage. If the fastenedcomponents include a gasket, then determination of their stiffness can pose problems - we shall con-sider gasketed joints separately below.

In a metal-to-metal (non-gasketed) joint, calculation of afastened component's stiffness is straightforward if it is athin tube and the external load is applied uniformly overthe annular end area A as at ( xii) - but this is unusual.More commonly the component is similar to ( xiii).Known are its thickness L and the fact that there isextended reactive contact over the face which is not incontact with the bolt head or nut face. The componentmay be modelled as a conical frustum ( xiv) of diameterDo at the small end and cone angle γ, bored cylindricallywith diameter Di to clear the bolt. The deformation dδ of a small element, thickness dx, will be :

dδ = Fj/kx = Fj( L/EA )x = Fj dx / E π/4 [ ( Do+2x.tanγ)2 –Di

2]

Integrating, the total compressive deformation δ over length L is found to be :δ = ( Fj /πEDi.tanγ ) ln[ ( Do–Di+2L.tanγ )( Do+Di ) / ( Do+Di+2L.tanγ )( Do–Di ) ]

It is usual to particularise this for some assumed cone angle such as 30o, but a general result hasbeen obtained by Wileman op cit using finite elements on the basis of Di ≈ d (small clearances) andDo ≈ 1.5d - ie. the outer diameter of the annular bearing area under the hexagon equals the standardwidth across flats, Table 1. Wileman's results may be correlated for steel by :

( 4 ) k = Fj/δ ≈ Ed ( 0.702 +0.654 d/L ) / ( 1 –0.12 d/L ) ; d/L ≤ 2

Use of ( 4) presumes that there is sufficient material present to allow the lines of force to developunfettered by finite boundaries.

Preload and Its Control

The benefits of high preload were apparent in ( ix) above - tendency towards separation (or leakagein a pipe joint) is reduced, fastener loosening is minimised and so on. Furthermore it is better to usehighly stressed small bolts rather than lightly loaded large bolts because the stiffness of the smallbolts is less. In general, preload should be as high as components can withstand. Bolt preload is usu-

P

Fb

Fi

( xi)

As shankFb

Di

L

Fj

Do

x

dx

γ

d

A

( xiii) ( xiv)( xii)

Threaded Fasteners 7

O-rings are suitable where the environment is non-aggresssive, temperatures are sensibly ambient,and relatively high manufacturing/assembly costs can be tolerated - otherwise gaskets are used.Pipes are frequently connected by welding similar flanges to each and bolting the flanges togetherwith a gasket and multiple bolts as shown below. Gaskets are common also in heat exchanger tubeplates and the cylinder heads of i/c engines, however we shall explain gaskets in the context offlanged pipe joints. Dimensions of flanges suitable for various combinations of pipe size, fluid pres-sure and temperature are standardised (AS 2129).

Gasket materials range from cork to copper - each material best suiting a particular combination offluid corrosiveness, pressure, temperature, leakage impications, cost, and so on. Most gaskets arenon-metallic, highly compliant compared to other joint components, and non-linear. Their low stiff-ness means that other joint components may usually be neglected when figuring joint overall stiff-ness : 1/kj = 1/kgasket + Σ 1/kother large - that is the joint overall stiffness is essentially the same asthe gasket stiffness, which has far-reaching implications in fatigue since it may not be possible to

flange gasket full face gasket raised face flange

pipe

bolts

P

AgAg

EXAMPLE The cast iron cylinder and head of a pump are attached by a number

of M16x2 class 8.8 bolts. How many bolts are required to resist a pressure load of 200 kN

on the head? A design factor of 2 is appropriate.

First find component stiffnesses to relate the bolt load to the external load via ( 3a).

Bolt The threaded length is not given, but from the sketch the runout coincides

with the joint faces - though this exposed thread length is really excessive.

Take half head and nut lengths to be 0.5*0.9*16 = 7 mm.

Shank length = 20+7 = 27 mm area = π/4*162 = 201 mm2

Thread length = 24+7 = 31 mm area = As = 157 mm2 (Table 1)1/kb = Σ L/AE = (27/201 +31/157)/207 kb = 624 kN/mm . . . watch units !

Flange Assume lengths of effective conical frusta are each (20+24)/2 = 22 mm to

equalise reactive areas as shown.

Assume that ( 4) is applicable to cast iron with little error, so for one frustum :

kf = 100*16 ( 0.702 +0.654*16/22) / ( 1 –0.12*16/22) = 2065 kN/mm

So, for the two flange frusta in series

1/kj = 2 ( 1/2065 ) kj = 1030 kN/mm

The equivalent stiffness is therefore 1/ke = 1/624 +1/1030 ; so ke = 389 kN/mm

The proof load of the bolts is Fp = AsSp = 157*590 = 93 kN.

Assume that the bolts are tightened initially to three-quarters of their proof load Fi = 0.75*93 = 70 kN.

So applying ( 3a) with a design factor of 2 on the external load, to 'z' bolts presumably equally loaded :

Fb = 70 + 2*200*389 / 1030z and ≤ Fp = 93 kN from which z = 6.3

Either 8 bolts would be used (7 are difficult to make equidistant) or 6 bolts might be employed with preload

reduced to 73% rather t han 75% of proof. But we have seen that it is impossible to expect 2% accuracy - so select 6

bolts. This decision is bolstered by the knowledge that since the joint is self-energising, a high preload is unneces-

sary from a joint operational point of view.

20 24

22

The measurement of nut rotation θ - Bickford's sequence step ( b) - suggests itself as an alternativeguide to the preload attained, since the distance ∆ moved by a nut along its (single start) thread ispitch*nut rotation, ie with ( 3a), Fi = ke∆ = kepθ. However this rotation-preload correlation requireslinear behaviour of the joint (ke) - and this usually does not happen. Instead, a characteristic alongthe lines of the sketch can be expected. At low loads A-B, non-linearities arisechiefly from warp straightening and embedment of contact surface peaks. Inthe middle range B-C, linear elastic behaviour predominates, as suggested bythe foregoing theory. Finally at high loads, progressive yielding of the vari-ous components produces the non-linearities C-D. It is evident therefore thatcounting turns is also an unsatisfactory preload monitor, unless one knowsthe Fi-θ relationship - the measurement of which rather begs the question.

Measuring bolt stretch - step ( c) - would seem to be more direct than measuring torque or turns.Ultrasonic measurement of bolt length changes whilst tightening are routinely carried out - thereader is referred to Brown op cit. Heat exchanger joints are typical subjects for this, but theyemploy many bolts and it is impossible to apportion preload uniformly between the bolts. Thus onebolt might be tightened with the correct stretch and preload, but tightening subsequent bolts causesthat preload to change. So complex tightening sequences have to be resorted to - and even then pre-load variation between individual bolts can be very large.

Industrial bolt tighteners may be computer controlled and often monitor both torque and turn. Theyfirst apply a specified torque ∆T to overcome initial warpage and embedment, thereby ensuring thebolt is taken into the elastic zone. The nut is then rotated through a specified ∆θ to ensure the boltyields - this results in a preload which is relatively predictable, ie.insensitive to the vagaries of tightening. Although the bolt hasyielded, eventual dissipation of torsional tightening stresses togetherwith bolt ductility ensures that it is still capable of sustaining signifi-cant loads - though careful joint design is required. Computer con-trolled tightening using this technique can be programmed to auto-matically sense and warn of stripped or crossed threads, holes whichare blind or inadequatedly tapped etc.

The most direct control of preload is of course to measure preload itself, Bickford step ( d). Specialcrush washers (akin to wavy washers) may be designed to flatten at the desired preload so that con-trol is exercised by visual inspection. But these also have drawbacks - while they indicate when acertain preload has been attained, they cannot advise by how much it has been exceeded.

Fluid Pressurised Joints

A common application of bolted joints is for detachable connec-tions which contain pressurised fluid. These may be rendered leak-proof either by O-rings or by gaskets. An O-ring is an endless ringof circular cross-section as sketched, which is relatively compliantand usually elastomeric. It is lightly nipped when assembled, sub-sequent application of fluid pressure extrudes the ring to plug all potential leakage paths. Suchjoints are called self-energising as they do not rely on preload for their effectiveness.

The example opposite illustrates a typical static analysis, however the problem concerns a pump’scylinder and head so the pump is presumably reciprocating and calls for a fatigue rather than astatic analysis. Bolt fatigue is considered below.

nutrotation, θ

pre

load

A

B

C D

assembled pressurised

rotation, θ

∆θ ∆T

torque, T

Threaded Fasteners 8

deflection. A material can be characterised by three constant properties - the slope and intercept ofthe initial loading regression line (log-log), together with the point through which all subsequentlinear (un)loading trajectories pass. These results have yet to be incorporated into Codes.

It is clear from the PVRC findings that non-metallic gaskets do not behave elastically.Table 4 gives the range of stiffnesses thatmay be experienced - stiffness beingexpressed as gasket stress (force/Ag) perunit deflection, ie. modulus/thickness. Cal-culation of load sharing from ( 3a) is thusfraught with uncertainty if the gasket stiff-ness is not known accurately, however loadsharing may be characterised by :( 3b ) Fb = Fi + C P ; Fj = Fi – ( 1 – C ) P

in which the joint factor C is a typical measureof the fraction of the total external load P seen by thebolts. Although comparison with ( 3a) indicates that C =ke/kj

= kb/(kb+kj) ≤ 1, the joint factor should be

regarded more as an overall approximate figure ofmerit, which incorporates all vagaries of behaviour andwhich is experimentally determined in its own right,rather than as a ratio of elastic stiffnesses. Experiencesuggests the values of Table 5.

Flange rotation refers to the initially disc-shaped flange assuming a flat conical form under fluidpressure loading, and is significant when a narrow gasket is used. The cause of rotation may beappreciated from these sketches. At ( a) is shown part of the flange with the bolt load Fb equilibrat-

ing the external load P and the joint load Fj, as above. Accordingly Fb is split into its superimposedcomponents ( b), which are shown separately at ( c) and (d). It is clear that both the P- and F j-couplestend to rotate the flange as sketched. This rotation is exacerbated by the fluid pressure p tending toexpand the pipe connected to the flange, however the flange is relatively rigid radially and so pre-vents pipe wall local expansion as suggested by ( e). Free bodies of the separate pipe and flangeappear at ( f) and ( g). The interface remains essentially rigid whilst rotation occurs - that is thedeformed pipe and deformed flange remain perpendicular to one another. This requires the action/reaction sketched - the radial force Fi and the bending moment Mi. Clearly Mi contributes to the rota-tion of the flange ( g).

It is common practice to allow for rotation empirically by considering an effective gasket widthsomewhat less than the actual width, as suggested by ( h) - further details appear in AS1210.

Operation at high temperatures leads to further difficulties with creep, ratcheting and so on.

TABLE 4 Gasket Material Stiffnesswidth thick- stiffness

gasket material mm ness MPa/mmcompressed asbestos 12 3 330-950 ditto. 24 1.6 600-3260compressed aramid fibre (asbestos substitute) 12 1.6 220-490spiral wound, asbestos filled 18 4.6 270-1800non-asbestos laminated head gasket 1.1 1140-2630

TABLE 5 Joint FactorsType of joint Csoft gasket held by studs 1.00soft gasket with through fasteners 0.75asbestos gasket 0.60soft copper & long through bolts 0.50metal-to-metal with long bolts 0.00

P

Fj

Fb

Fi

Mi

P

Fj

P Fj PP

Fj

Fj p p

Fi

Mi

effective width

( a) ( b) ( c) ( d) ( e) ( f) ( g) ( h)

ensure that kb << kj as desired for low bolt load amplitude.

Soft gaskets usually extend over the whole flange (full face), whereas gaskets made of stiffer materi-als are narrow and extend out only to the bolt holes. It is assumed that multiple bolts are loadedequally, and that the gasket stress - ie. the average contact pressure between gasket and flange - isuniform over the gasket area Ag. Clearly the gasket stress must be higher than fluid pressure if leak-age is to be avoided, and, for a given gasket stress, a raised face flange with a narrow gasket willrequire less preload and hence less flange strength than a full face gasket. So the tendency is for highpressure gaskets to be narrow.

In service, the pipe acts as a closed cylinder with axial load P in the wall due to fluid pressure actingover the internal area. This loading - a preload due to initial tightening, together with an externalservice load - is typical of joints in general.

Codes traditionally classify gasket materials by two property values - the 'y' factor and the 'm' factor,Table 3 - these are necessary minimum values for the avoidance of leaks :

y - the initial gasket stress (ie. preloaded prior to pressurisation), andm - the ratio of gasket stress to fluid pressure (after pressurisation).

Theses limits are sketched on the joint characteris-tic below in which the external axial load P equalsthe fluid pressure p times the internal transversearea Ai. The initial gasket stress induced by thetightening load Fi must be less than the gasketcrushing stress pcrush and greater than the 'y'appropriate to the gasket material. Loading ini-tially is the most severe since the joint loaddecreases thereafter as

the external load is applied - but the external load should not be solarge that the gasket stress becomes less than 'm' times the fluid pres-sure. Failure of a joint - as opposed to leakage - will be most likely at ini-tial tightening, so joint design (selection of bolting, flange thickness etc.)is based upon that state.

There has been a deal of debate on the suggested values of 'm' and 'y'which appear in the Pressure Vessel Codes. While the values have been found satisfactory bases forsafe joint design, they are but poor indicators of leakproof operation. The Pressure Vessel ResearchCommittee (PVRC) of the Welding Research Council instituted comprehensive tests to clarify thebehaviour of gaskets, with results shown at left below. Initial tightening proceeds along a-b; as theexternal load is applied the gasket unloads along b-c. Further cycling of the external load leads tothe gasket loading/unloading along the same b-c-b trajectory. If the bolts are further tightened thenthe gasket (un)loads along the d-e-d locus. Gaskets cannot be absolutely leak free when containing

gas; locii of constantleakage rates are shownsuperimposed upon thecharacteristic. Behaviourmay be described mostconveniently by graphssimilar to the one on theright, in which a leakageparameter replaces

F

FiFj

P = pAi

gmpAgy A

gp Acrush

0

TABLE 3 Gasket Materials Properties (from AS 1210) thick- y mgasket material ness mm MPa -asbestos with binder 1 45 3.5 ditto. 1.5 25 2.75 ditto. 3 11 2.0vegetable fibre 7.5 1.75spiral wound, asbestos filled 69 3.0

gasket deflection

gask

et s

tres

s

a

b

c

d

e

constantleakagerate

decreasingleakage

unloadingreloading

initialload-ing

log tightness parameter

log

gask

et s

tres

s initial loading

concurrent

unloadingreloading

Threaded Fasteners 9

plot here, corresponding to a joint with asingle bolt. For the particular bolt mate-rial, initial tightening, and magnitude ofexternal load fluctuation on this joint, thefatigue safety factor is given by ( 6b ) as : nC = As(mm)/127It is apparent that for a given bolt size, thesafety factor is inversely proportional tothe joint factor. Thus for an M8x1.25 bolthere, if the joint factor is assessed fordesign purposes as 0.1, then the designwould be reckoned perfectly acceptablewith a safety factor of 3. But if, due tosome overlooked detail, the joint factor is in fact 0.3, then there is no margin for error as the safetyfactor would only be 1. The implications of a poorly assessed joint factor on aspects other than boltfatigue - eg. leakage - must also be borne in mind.

A typical fatigue analysis appears overleaf.

Non-uniformly Loaded Bolt Groups

So far we have looked at groups of threaded fasteners in which the fasteners' loads are tensile andapproximately the same. We now consider briefly joints in which the loads on individual fastenersdiffer and in which shear and friction can be significant.

Shown at ( a) is a single bolted lap joint in which the bolt is more orless centrally located in the clearance holes in the two components.Provided it is not too large, the external load P is transmitted by fric-tion due to pressure between the components, caused in turn by thebolt's initial tightening load. If P is such as to cause the components toslip then the components bear against the bolt shank ( b). Furtherincrease of the external load could lead to shear failure across the boltshank ( c), however failure in other modes might have occurredbefore P reached this level - failure analysis is similar to that of thecotter joint examined in an earlier chapter. It should be noted that thebolt's shear plane ( c) should lie in the shank and not in way of theexposed threads - bearing on an excessive length of exposed threadmust be avoided too.

The example below demonstrates the approach applicable to a typical bolted con-nection in which tension and shear both occur. The example’s assumptions arerather unrealistic - a completely rigid bracket, accurately located bolt holes ensuring theoretical shar-ing of shear, and so on. We would probably use larger bolts which 'looked right' !

Many of the connections which fall into this category are covered by Codes, which obviously mustbe followed if relevant. Gorenc op cit provides useful information with regard to steel structures,which is applicable to many mechanical situations. Shear is often best resisted by dowels ratherthan by threaded fasteners. Dowels are pins (often tapered) inserted intoclose-fitting holes which are drilled and reamed through the componentsafter assembly to ensure accurate registration.

2.5 3 4 5 6 8 10 1 2 1 6 20 24

10

1

0.1

fati

gu

e sa

fety

fa

cto

r

SAFETY FACTORFOR EXTERNALLOAD ON ABOLTEDJOINT

class 10.9 rolled thread tightened to 85% proof maximum load 10 kN

0.050.1

0.20.4

0.61.0joint factor

bolt nominal size, mm

P

P

P

P

( a)

( b)

( c)

Bolt Fatigue

The external load on a joint assembly may vary over many cycles during the joint's life - the cylinderhead joint of an i.c. engine is a typical case. The bolts of such an assembly undergo fluctuating ten-sion and so may fail in fatigue.

The traditional Goodman analysis may be employed to assessbolt safety in a particular joint assembly. Failure is most likelyin the exposed threads close to the nut face where a stress con-centration factor Kf models the damaging effect of the threadroot notch. A rolled thread is less prone than a cut thread tofatigue damage, due to better grain orientation and surfacework hardening conferred by manufacture. Values of Kf are given in Table 6.

Shigley op cit reports the correlation between steels' ten-sile endurance limits Se' and tensile ultimates plottedhere - the fluctuating tension varying between zero andthe maximum plotted. The endurance limit in threads Seis little affected by surface finish or size, so, employingKf as a fatigue strength reduction factor, Se = Se'/Kf.

In a typical joint assembly, the initial load is Fi and theexternal load varies between zero and P as shown infigure ( xi) above. The bolt load therefore varies betweena minimum of Fi and a maximum of (Fi +CP) where thejoint factor C subscribes either to the stiffness ratio in ( 3a) if the stiffnesses are known accurately, orto Table 5. The corresponding alternating and mean stress components are therefore :

( 5 ) σ = CP/ 2As ; σ = σi + σ ; where σi = Fi /As

in which σ is proportional to the external load P, and σi reflects the initial load Fi.

A Goodman analysis isthereafter straightforward;the stress components ( 5)appearing on the load line

emanating at 45o from theinitial stress on the meanstress axis.

The 'load' upon which thesafety factor is based may be interpreted in one of two ways, either

- the load may refer to the load on the bolt itself (Fb), the Goodman diagram ( i) above then gives( 6a) σi/ Su + σ ( 1/Su + 1/Se ) = 1/n or

- the load may be the external load on the assembly (P), the Goodman diagram ( ii) yielding( 6b) σi/ Su + n σ ( 1/Su + 1/Se ) = 1

It can be seen that the essential difference between these interpretations lies in the inclusion or exclu-sion of the initial load into safety considerations. Although the second interpretation is perhaps themore correct, the imprecision with which the initial load is known may call for the conservatism ofthe first interpretation. A reliability or fatigue life analysis is really preferable to this traditionalsafety factor approach.

The importance in design of assessing the joint factor as accurately as possible is brought out by the

TABLE 6 Stress Concentra-tion Factor Kf

material rolled cutclass thread thread5.8 and below 2.2 2.88.8 and above 3.0 3.8

tensile ultimate, S ( GPa)u

0 1 20

0.4

0.8

axialendur-ance

limitS '(GPa)

e

S ' = (0.55 -0.088 S ) Se u u

AXIAL ENDURANCELIMIT FOR STEELS(data of Landgraf)

σ_

σi

σ~

Su

Se

failurelocus

externalload line

σ~nσ~

( ii)

σ~ σ_

σi

σ~

σ_

S /nu

S /ne

Su

Se

failurelocus

externalload line

( i)

Threaded Fasteners 10

EXAMPLE The bracket is fastened to a vertical support plane by three identi-

cal class 9.8 through-bolts, and supports the eccentric 10 kN steady load. Determine a

suitable bolt size for a safety factor of 5.

Design load P = 5*10 = 50 kN

We shall examine this by means of different free bodies and related assumptions.

Free Body # 1 - the simplest FB, envisaging the bracket as a particle in side view.

Contact with the vertical support plane is represented by the normal

and friction components, N and Ff respectively. If breakaway is

imminent and the bracket is equilibrated solely by friction and bolt

tension, then :

Pbreakaway = Ff = µN = µ 3Fi where Fi is the initial tightening load.Assuming a typical friction coefficient of 0.35 and an initial tightening load of 90% proof then

Pbreakaway = 50 kN = 0.35*3 (0.9*As*650) ⇒ As = 81 mm2 ie. M12x1.75 bolts are adequate.

Free Body # 2 - side view

A more complex FBD of the side view recognises the non-particle nature of the bracket, which is

assumed to be rigid and to rotate about the point Q where the support plane's concentrated

reaction is R. The bolts deform (δb) beyond the initial state now that rotational separation hasoccurred, so the forces in each of the two upper bolts F2 will be greater than that in the lower

bolt F1. Since the small opening angle = δ1/60 = δ2/240 it follows that δ2 = 4δ1 so that from thebolt characteristic (F2–Fi ) = 4(F1–Fi). For rotational equilibrium about the point Q :

60F1 + 240*2F2 = 200*P which, using the above, gives 99F2 + 9Fi = 40P = 2000 kN

Assuming that maximum bolt force F2 equals the proof load Fp at the design load P, and that

initial tightening load is 90% proof, it follows that (99+9*0.9)As*360 = 2E6 N ⇒ As = 52mm2

So this model is by itself less critical than that of FB#1 in which total support is by friction - but

this is not surprising because FB#2 says nothing about vertical equilibrium, assuming it to be

achieved by the friction component of R at Q. Furthermore, the eccentricity of the load in the front

view has not been addressed.

Free Body # 3 (front view)

In this model the bracket is again presumed to be rigid, and equilibrium is taken to be

achieved by shear of the bolt shanks, in a manner similar to sketch ( c) above. The centroid

of the three equal shear areas A is located at G as shown. Since G does not lie on the line of

action of the external load, a rotational tendency (torsion) exists.

Analysis is similar to the traditional analysis of fillet welds in that force equilibrium is

achieved by uniform direct shear, while torsional equilibrium arises from secondary shear

whose sense is perpendicular to the radius from the centroid to the location of the secon-

dary shear, and which is proportional to that radius, given by the torsion equation

Ft = Aτ = A.Tr/J = A.Tr/ΣAr2 and since all A's are the same Ft = Tr/Σr2.

So the torsional shear forces are Ft1 = (P*75)*120/(1202+2(602+802)) [note units check] = 90P/344 while Ft2 = Ft3 = Ft1*100/120 = 75P/344.

The three equal direct shear forces equilibrate the force P, ie F1d =F2d = F3d = P/3.

The bolt 2 is most critical since primary and secondary most nearly coincide in it. Combining these vectorially, the

total shear force on the critical shank is F2 = 0.5243P = 26.2 kN. Assuming the distortion energy failure theory, the

proof strength in shear is 650/√3 = 375 MPa, so the required shank diameter is √(4*26.2E3/375π) = 9.4 mm ie. M10

is adequate. But of course this neglects friction and the prying action recognised by FB #2.

Free Bodies # 2 & 3

Combining the above, a conservative design approach assumes the support plane to be completely smooth, so

that vertical equilibrium is achieved by shank shear, and R at Q is normal to the support plane.

From FB #2 assuming Fi ≈ F2 : tensile F2 = 2000/(99+9) = 18.5 kN and so σ = 18.5/A

From FB #3 : shear F2 = 26.2 kN and so τ = 26.2/A

Applying Mohr's circle and the maximum shear stress theory to this state (σ,τ), the equivalent

normal stress is σe = 2√( (σ/2)2 +τ2 ) = 55.6/A from above, and = Sp at failure. So shank size = √(4*55.6E3/650π) = 10.4 mm. So use M12x1.75 bolts. It is fortuitous that the solution from this model is the same as

that from FB #1.

P

F1

2F2

R Q

Fi

Fp

δb

Fb

P

F1d

G

A

F2dF3d

F1t

F2t

F3t

60

120

80

200 150 75

8010 kN

400

24060

3F

F

N

Pi

f

(σ,τ)

EXAMPLE A φ 100 mm nominal pipe is

equipped with raised face flanges and a spirally

wound asbestos filled gasket, 4 mm thick and of

the dimensions sketched. It is proposed to use 8

M20 x2.5 class 5.8 bolts for the joint, which is sub-

jected to a fluid design pressure varying between

zero and 7.5 MPa. Will this class be suitable?

The whole operati on of the joint will need to be checked - leakage tendency as indicated by 'm' and 'y' factors, fatigue of

the bolts, and so on.

Dimensions & Initial Loading

The width of the gasket here is (150–125)/2 = 12.5 mm. AS 1210 table 3.21.6.4(B) states that the effective width, allow-

ing for rotation etc, is 2.52√(width/2) = 6.3 mm.

Gasket OD is 150 mm, so gasket effective ID is 150–2*6.3 = 137.4 mm, and gasket effective area is Ag = π( 1502–

137.42)/4 = 2844 mm2.

The effective area inside the gasket which is subjected to fluid pressure is Ai = 137.42 π/4 = 14 830 mm2.Bolts' proof load Fp = 245*380 = 93 kN

Assume tightening to 75% proof, then total initial tightening load is Fi = 8*0.75*Fp = 560 kN.

Stiffnesses & Joint Factor On the basis of a single bolt and the gasket associated with it . . . .

Bolt Grip = 2*35 +4 = 74 mm. Assume two exposed threads and head/nut height = 0.9*20 = 18 mm

1/kb = Σ L/AE = [ (74 –2*2.5 +0.5*18)/π/4*202 + (2*2.5 +0.5*18)/245 ]/207 ⇒ kb = 682 kN/mm

Joint Gasket area associated with one bolt Ag = 1/8 of 2844 = 356 mm2

Table 4 quotes limits for the stiffness of the gasket material 270 to 1800 MPa/mm, or, in units of force (multiplying

by Ag= 356mm2) 96 ≤ kg ≤ 640 kN/mm.It is anticipated that a larger value will be more appropriate, but we'll persist with two extreme values using {} to

show results for the smaller stiffness. So select estimated gasket stiffness of kg = 600 kN/mm { 100 kN/mm }.

The portion of each flange in direct compression will be modelled as a 35 mm long conical frustum using ( 4)

kf = 207*20*(0.702 +0.654*20/35)/(1 –0.12*20/35) = 4780 kN/mm

This is so much greater than kg that the approximation inherent in applying the frustum model in this case should

not carry over into the overall joint stiffness. So for the two flanges and the gasket in series in the joint :

1/kj = 2/kf +1/kg = 2/4780 +1/600 ⇒ kj = 480 kN/mm { 96 kN/mm }

The joint factor follows as :

C = ke/kj = kb /( kb +kj ) = 0.59 { 0.88 }

Gasket Stress Limitations Based upon whole gasket . . . .

From Table 3, y = 69 MPa and m= 3.0 for a spiral wound asbestos filled gasket.

The initial gasket stress is Fi/Ag = 560/2844 = 197 MPa > y, so OK.

Check crushing - lacking any recommendations, presume a limiting stress of 2y = 138 MPa (based on total not effec-

tive gasket area). Based upon this area, the initial (ie. maximum) stress on the gasket is 560*4/π( 150 2–1252) = 104MPa < 2y and so crushing should not be a problem.

Now consider gasket stress when full fluid pressure of 7.5MPa is applied. From ( 3b) with external fluid load (or

hydrostatic load) of P = pfluid Ai = 7.5*14830 = 111 kN

Fj = Fi – ( 1 –C ) P = 560 –(1–0.59)*111 = 514 kN { 547 kN }

which corresponds to a gasket stress of pg = Fj/Ag = 514/2844 = 181 MPa { 192 MPa }

Ratio pg /pfluid = 181/7.5 > m = 3.0 so this criterion is satisfied and the gasket should not leak.

Bolt Fatigue Based upon a single bolt . . . .

From Table 2, Su = 0.5 GPa, so Se' = ( 0.55 –0.088*0.5 )*0.5 = 253 MPa

Assuming rolled threads, Kf = 2.2 (Table 6) and so Se = Se'/Kf = 253/2.2 = 115 MPa

σi = Fi/As = 560/8*245 = 286 MPa; σ = CP/2As = 0.59*111/2*8*245 = 17 MPa { 25 MPa }

Applying ( 6a), the safety factor for the load on the bolts themselves (Fb) is

1/n = σi /Su +σ (1/Su + 1/Se ) = 286/500 +17*(1/500+1/115) ⇒ n = 1.33 { 1.19 - note effect of small kj }

Alternatively, applying ( 6b) for the safety factor on fluid pressure loading (P) against bolt fatigue failure

1 = σi /Su + n σ (1/Su + 1/Se ) = 286/500 + n*17*(1/500+1/115) ⇒ n = 2.35 { 1.60 }

All-in-all, the joint is satisfactory as it stands. The bolt class is therefore suitable.

Ø 1

14.3

Ø 1

25

Ø 1

50

6.335

M20 x 2.5

Threaded Fasteners 11

THREADED FASTENERS - PROBLEMS

Two significant figures at most are justified - more are given here to aid checking.

1 A 30 mm single start Acme power screw, of 6 mm pitch, has a thrust bearing of 40 mm meandiameter. Thread and bearing coefficients of kinetic friction are 0.15 and 0.1 respectively.(a) Estimate the torque necessary to raise a load of 100 kN. [ 501 Nm](b) If the screw rotates at 1 Hz ascertain the power input to the screw, and the combined effi-

ciency of screw and thrust bearing. [ 3.15 kW 19%](c) If bearing friction is eliminated by a rolling thrust element, show that the screw is self-

locking and determine the torque necessary to lower the load. [ 106 Nm](d) The thread is further lubricated to halve the friction coefficient. Does this result in over-

hauling behaviour? What is the effect of lubrication on performance here?

2 Arrangements ( a) and ( b) are identical except forthe spring washer's position. The washer is theonly element of the assembly having appreciablecompliance - the bolt and joint components areessentially rigid. The bolt is initially tightened toFi, then the external load P is applied. If Fw is the washer load, Fb is the bolt load, and Fj is the load transferred between the joint com-ponents 1 and 2 prior to separation, plot for each arrangement Fb and Fj versus the externalload, P. Identify the separation load P*. Hint : note that prior to separation Fw must be constant atFi, so equilibrium alone gives the information required.

Evaluate the bolt load when P is 2 kN and Fi is 10 kN. [ 12 kN; 10 kN]

3 The cover of a pressurised cylinder is secured by ten bolts whoseaggregate stiffness is a quarter of the total joint stiffness. Each bolt istightened initially to 5 kN, then a constant external load of 20 kN isapplied to the cover by the pressurised contents of the cylinder.Plot the variation of bolt and joint loads against external load, and evaluate the maximum loadon each bolt, the minimum total joint load, and the separation load. [ 5.4, 34, 62.5 kN]

4 The steel bracket is bolted to steel ceiling joists by two bolts of class 8.8and 48 mm grip length. What tightening torque should be used, and whatis the corresponding load in each bolt when the 40 kN external load issubsequently applied? [ 480 Nm; 125 kN]

5 The cover of a pressurised cylinder is attached by a self-energising sealand a number of identical bolts. The fluid pressure is essentially con-stant at 6 MPa. Select suitable class 8.8 bolts for this duty given that theratio of bolt spacing (around the ø 180 mm pitch circle) to bolt sizeshould not exceed 10 to minimise flange bending, and should not beless than 5 in order to allow spanner access. A safety factor of three isapplicable to the external load on the assembly.

6 An electric motor weighing 20 kN is provided with a steel eye bolt for lifting. Determine a suit-able size of class 8.8 bolt, explaining your choice of safety factor. [ M30x3.5]If the strength of the motor housing (into which the bolt is screwed) is only half thatof the bolt, what minimum number of threads should be engaged? Hint : considercylindrical area in nut (housing) such that shear failure here is equally likely to tensile failurein way of the exposed threads. [ 7 threads]

2

P2

P2

1

2

P2

P2

1

spring washer

40 kN

M20x2.5

50

Ø 180Ø 120

Ø 100

Bibliography

AS 1111 ISO Metric Hexagon Commercial Bolts and Screws

AS 1721 General Purpose Metric Screw Threads

AS 2129 Flanges and Bolting for Pipes Valves and Fittings

Basics of Design Engineering 1991, 'Machine Design' Reference Issue

Bickford JH, Introduction to the Design and Behaviour of Bolted Joints, Dekker 1995 definitive

Blake A, Threaded Fasteners, Dekker 1986

Brown IW, Design and Behaviour of Bolted Joints, Hons Thesis, The University of WA, 1992

Craddock DH, Introduction to Fastening Systems, OUP 1974

Czernik DE, Gaskets: Design, Selection & Testing, McGraw-Hill 1996

Gorenc BE & Tinyou R, Steel Designers Handbook, NSWU 1981

Industrial Fastener Reference Manual, FJ Sweetman & Co 1997 overview of the industrial range

Juvinall RC & Marshek KM, Fundamentals of Machine Component Design, Wiley 1991

Sealing with Security, Richard Klinger Pty Ltd users guide to selection of jointing materials

Shigley JE, Mechanical Engineering Design, McGraw-Hill (first metric ed) 1986

South DW & Mancuso JR, Mechanical Power Transmission Components, Dekker 1994

Thompson G, An Engineer’s Guide to Pipe Joints, PEP 1998

Wileman J et al, 'Computation of Member Stiffness in Bolted Connections', Trans ASME: Journalof Mechanical Design v113 December 1991

Threaded Fasteners 12

11 Transform equation ( 6b) into the form suitable for initial design :As = β n C P / Se

where n is the safety factor corresponding to the external load PAs is the bolting stress area required to support P

and β = 1/2 ( 1 + Se /Su ) / ( 1 – σi /Su ) and lies generally in the range 1.3 <β <2.1for rolled threads, steel classes 5.8 through 12.9, and an initial load of between75 and 90% proof.

Select a suitable size and number of class 9.8 bolts for a narrow face flanged joint to containfluid whose pressure fluctuates continuously between 0 and 150 kPa. The metal jacketed asbes-tos filled gasket has y= 50 MPa, m= 3.5 and stiffness 1.5 GPa/mm, with diameters 1840 and1865 mm, and thickness 3 mm. Bolt spacing between 5 and 10 diameters, and a design factor of5 on the fluid pressure load are appropriate. Assume that a flange thickness of around 50 mmis sufficient to prevent appreciable rotation.

7 The big-end cap of a steel connecting rod is secured by two class 8.8 M12x1.25 (fine pitch) steelbolts. A reversing load of 20 kN amplitude is transmitted between the connecting rod and thecrankshaft journal via the big-end bearing. The portion of the connecting rod which surroundseach bolt and is elastically compressed may be taken to be cylindrical of annular area 300 mm2.Determine the safety factor appropriate to the reversing load, with(a) zero initial bolt load, [ 2.0](b) initial bolt load just sufficient to prevent separation, [ 6.8](c) bolts tightened initially to 70% proof. [ 3.6]Estimate the tightening torque necessary for ( c). [ 91 Nm]

8

A connecting link consists of two circular sections of forged aluminium alloy (E=71 GPa) spi-gotted and joined by six equally spaced steel studs, M8x1.25 class 10.9, having shanks ofreduced diameter with transition radii of 20 mm and polished surfaces.(a) Explain briefly but clearly why the studs are provided with reduced shank sections.(b) To what minimum diameter can the shanks be reduced without compromising the fatigue

strength of standard studs with unreduced shanks? [ 5 mm](c) Studs with 6mm diameter shanks are used, tightened to half their yield strength (caution:

occurring where?). What is the factor of safety when an alternating axial load of 60 kNamplitude is applied? (caution : evaluation of the bolt load needs care) [ 0.92]

9 The components of an hydraulic actuator are of steel - thecylinder is of bore D= 100 mm, wall thickness t= 10 mmand length L= 300 mm. The thickness of the end bracketsis w= 20 mm, and they are connected together by fiveM12x 1.75 grade 5.8 bolts, tightened to 75% proof. In oper-ation the cylinder is pressurised between 0 and 4 MPa.(a) Determine the stiffnesses of the bolts and of the joint

assuming that the cylinder is compressed uniformly andthat the end brackets are rigid. [ 344, 2240 kN/mm]

(b) Find the mean and alternating stresses in the bolts. [ 289, 4.7 MPa](c) Estimate the endurance limit of the bolts based upon 50% reliability. [ 115 MPa](d) What factors of safety guard against fatigue failure and against static failure? [ 8.3, 9.8]

10 A pipe joint incorporates flanges of width w= 12 mmtogether with a full face gasket of bore Di= 150 mm,outside diameter Do= 250 mm and thickness t= 2 mm.The gasket material has a stiffness of 100 MPa/mmwith 'm'= 1.5 and 'y'= 2 MPa.Neglecting rotation, assess the joint's suitability towithstand fluid pressure fluctuating between 0 and 1MPa, if six M10x1.5 class 5.8 steel bolts are used.

wL

D

tw

viewx-x x

x

rod

jointcap

Ø 8

0Ø

110

part section x-x

x

x

100

spigot

R 20

Di

Do

w

t

Welded Joints 1

largely by the joined components, the only means of varying the load capacity for a given electrodematerial is to vary the size w.

Successful welds depend as much on practical issues as on on theory. It is not the present intentionto dwell on practicalities - the reader is referred instead to the Bibliography for advice on practicalnecessities. AWRA Technical Note #8 illustrates many important considerations and is required

background to the serious design of welded joints. Howeverthere is one simple issue which should be appreciated at theoutset - the cost of a weld is approximately proportional to theamount of electrode used, other things being equal. The weldson the right here are obviously more expensive than those onthe left - but by how much in relative terms ?

So, the size of a fillet weld must be sufficiently large for safety,without being so large that unnecessary expense is incurred.Clearly a rational design procedure for such joints is needed -

one based upon the steps already identified : resolution of indeterminacies, load building blocks,stress resolution and implementation of an appropriate failure theory. The derivation and applica-tion of such a procedure is the main thrust of this chapter.

Fillet Welded Joints

A typical fillet welded joint is illustrated. It connects two com-ponents, one of which is conveniently regarded as the loadedmember - as all loads on it are known - the other is the sup-port or reaction member. Clearly the loads are transmittedthrough the joint before being absorbed in the support.

A run may be three-dimensional however the majority of practical runs are two-dimensional and liein a weld plane like the cantilever's joint here. We consider only such two-dimensional runs, the cen-troids of which must also lie in the weld plane. It is convenient to erect a Cartesian system at the cen-troid G, and to designate the x-y plane as the weld plane as shown at ( a) below.

In general the resultant load on the joint is a force F = [ Fx Fy Fz ]' through the centroid of the linearrun, together with a moment M = [ Mx My Mz ]' whose components are given by the RH Rule, ( b).For the cantilever above, this resultant would be found by moving the sole force to act through thecentroid, and introducing the moment corresponding to the force multiplied by the distance trans-verse to the force's line of action between the point of load application and the centroid.

This load is equilibrated by a force distributed along the length L of the run as indicated in ( c). Byvirtue of the stresses in the weld, each element of run length δL contributes an elemental force q.δLtowards equilibrium. q is a force intensity; it is a vector force-per-unit-length and except in simplecases varies in magnitude and direc-tion around the run.

Conceptually, force intensity is nottoo different from stress, which is aforce-per-unit-area, that is δF = q.δL= σ.δA. Force intensity is also similarto the bending moment in a beam:both are stress resultants - of stressesin the weld throat and in the beam's

G

My

Mx

Mz

FyFx

Fz( a ) ( b )

G

M

x

y

z

F

run

L

( c )

q.δL

δL

support member

w fillet weldedjoint run

known load(s)

loadedmember

60o

60o

510

( a )

( b )

WELDED JOINTS

The strongest and most common method of permanently joining steel components together is bywelding. Of the many welding techniques available, arc welding is the most important since it isadaptable to various manufacturing environments and is relatively cheap.

In arc welding an electric arc at the extremity of a travelling consumable electrode maintains a poolof molten metal in which the components and electrode material coalesce, forming a homogeneouswhole (ideally) when the pool later resolidifies. But no joint can be perfect, and it is the designer'sjob to allow for practicalities and to ensure that the joint is adequate and economical.

The materials of components and electrode must be compatible from the point of view of strength,ductility and metallurgy - this last being most important in view of thermal effects arising from theusual uncontrolled localised cooling. The various Codes lay down electrodes which should be usedfor given steels - AS 1554 eg. cites electrode classification E41xx as first preference and E48xx as sec-ond, when welding steel grades 250 through 350 - the number in the classification being one tenth ofthe deposited weld metal's ultimate tensile strength (MPa).

The form of a welded joint is dictated largely by the layout ofthe joined components; the two most common forms are thebutt and fillet joints illustrated.

A butt weld aims usually for full penetration with no voids inthe completed joint, so edge preparation to allow electrodeinsertion is required for all but the thinnest weldable compo-nents. Inhomogeneities may be minimised by double welding (ie. welding from both sides assketched), by ultrasonic examination with subsequent repair if necessary, or by other means. In non-critical applications, correctly fashioned butt joints carried out by competent welders are often takento be just as strong as the joined components - that is, provided the electrode is correctly chosen andthe welding technique is satisfactory then joint safety does not have to be separately addressed. Butthe occurrence of imperfections in potentially hazardous joints must be recognised and allowed forin design, as will be seen later in the context of Pressure Vessels.

The majority of the deposited material in a fillet weld lies outside the silhouette of the joined mem-bers; penetration is not complete and a crack is inherent. Theweakening effect of cracks are examined under FractureMechanics, but it is immediately obvious that the geometric sin-gularity at the end of a crack could not be more extreme and sosevere stress concentration in the fillet is inevitable. The utmostcare is therefore necessary when specifying fillet welds withtheir intrinsic cracks to withstand substantial fatigue loads.

While the dimensions of a butt joint are governed in the main by thethickness of the joined components, the size of a fillet weld is not lim-ited by component thickness except for reasons of heat transfer andresulting weld quality. But do either of the twofillets illustrated here look practicable ?

The weld cross-section is idealised as a 45o fillet whose size is characterised by

the dimension w - only welds of constant size are considered here. The locus offillet centroids (or roots, approximately) traces out the weld run of length L.The run may be continuous or discontinuous. Since the run's shape is dictated

butt

fillet

'crack'

safe ?

which???

economic ?

root

toe

toe

w

w

throatplane

Welded Joints 2

Traditional Analysis

The traditional approach is limited to planar joint runs, and is commonly restricted to runs havingstraight segments. It is stated here without proof as it is a particularisation of the unified approachwhich is derived from first principles below.

A typical planar joint run comprising two segments AB and BC is shown at ( d) below. A convenientCartesian system is erected at the run's centroid with the run lying in the x-y weld plane. In thesketch, the external loads on the loaded member have been transferred to the run centroid to consti-tute the known load F, M on the joint - it is this load which is equilibrated by the force intensity qalong the run.

The traditional approach treats force intensity along a weld run analogously to stresses in a beam,resolving q into :

- The primary intensity qp is uniform along the run and equilibrates the centroidal force compo-

qy

Myqx

Mxqz

Mz

rF

qp

z

xy

G

A

C

B

M xy

Iyyq =

y

M yx

I xxq =

x

M rz

Jzzq =

z

F

Lq =

p

( d ) ( e ) ( f ) ( g ) ( h )

magnitude :-

PRIMARY( FORCES )

in-planetorsion

out-of-plane bending

SECONDARY ( MOMENTS )

zzJ = I + Ixx yy

b d /4L2I xy20 0 0 0 0

2 3L 6 12Lb d b b b( b+2d ) ( b+3d ) ( b+4d )322 3

I3πRyy 0

612 12 12L6( 4b+d )d d d d d( 6b+d ) ( 3b+d )3 3 2 2 3

I3πRxx

b+dL d 2d 2b+d 2( b+d ) 2πR

b

d d d d d

b b b

x

y

G

b /L2 b /2L2

d /2L2

R

3SECOND MOMENTS OF JOINT RUNS ( length )

centroid G lies on symmetry axes where applicable

cross-section respectively, and both vary in general along a linear path - the weld run and beam axis.

For the majority of beams the bending moment is easily found in terms of the loads using statics. Inthe case of a fillet weld however, correlating the intensity q with the load F, M is less straightfor-ward since the arrangement is statically indeterminate.

Two techniques for this correlation (having the same theoretical foundation) are presented below.The first traditional approach is based on recasting the building block stress equations for bendingetc. in terms of force intensity rather than of stress. This approach though simple has limitationswhich in some situations requires the more general second technique, the unified approach.

Geometric Properties of Lines

It is useful to review the concepts of centroids and second moments of linear though not necessarilystraight runs, ie. of lines rather than of areas. Two dimensions only are detailed, however the under-lying concepts may be extended to three dimensions.

The geometry of the planar line ABC shown here is known.

The line's centroid is found by first erecting any convenient (X,Y) Cartesian system in the plane, thenapplying first moments (Varignon's Theorem) to all line elements δL. The centroid's abscissa XG maybe defined in alternative ways :

( i) Σ XG δL = Σ X δL more usually seen as XG = ( Σ X δL )/L or, collecting termsΣ (X–XG) δL = 0 or, shifting the Cartesian origin to the centroid setting x = X-XGΣ x δL = 0 which in the limit becomes ∫

L x dL = 0;that is, the centroid is that point about which the first moment of length vanishes.

Extending this argument to three dimensions leads to the conclusion that the centroid is that uniquepoint for which all components of first moment vanish : Σx.δL = 0, Σy.δL = 0, Σz.δL = 0 or, in brief Σr δL → ∫L r dL = 0 where r = [ x y z ]'. The run sketched lies in the z = 0 weld plane.

The sum-of-increments approach is extended to higher moments, thus the second moment of a lineabout a defined axis is :

Iaxis = Σover all elements { (length δL of each element)*(the element's distance from axis)2 }→∫L (distance of dL from axis)2 dL

Integral forms are suitable for continuous line segments for which the geometry f(x,y,z) = 0 isknown; discrete forms are suitable in conjunction with the Parallel Axes Theorem for a line made upof a number of segments whose individual centroids and second moments are known from tablesfor example, or from the useful results of Problem 6. Properties of some common run geometries aretabulated here. It should be noted that : O [ second moment of length ] = length*distance2 = length3.

It is apparent that the process of finding centroids and second moments of lines is identical to thatfor areas - essentially it's a matter merely of interchanging δL and δA in the appropriate formulae.

Y = d - d /2(b+d)

d ( b+d/2)

b /2(b+d)

Y

XGA

B Cb

d1

2

XO

Y

x

y

G

G/2

2element δL X Y X.δL Y.δL

2 b 2b/ b

2

1 d 0d

2/d

2/0

d bd

L = Σ δL = b+d ΣX.δL = b 2/2

Σ Y δL = Σ Y.δL soG G2

Varignon :

and

Σ X δL = Σ X.δL soG G2X =

= ΣY.δL

Welded Joints 3

EXAMPLE Determine the intensity components at the points

A, B and C in the 6mm fillet weld illustrated.

Geometric properties of the run :-

b = 120 d = 180 L = 300 mm

The centroid is located via the table of run properties, B lies at ( –

24, 54, 0 ) mm in the centroidal system shown, and

Ixx = d3 (4b+d)/12L = 1.0692 E6 mm3 - note units

Iyy = b3 (4d+b)/12L = 0.4032 E6 mm3 - so Jzz = 1.4724 E6 mm3

Ixy = b2 d2/4L = 0.3888 E6 mm3 - but note that this is not used in the traditional

approach

Centroidal loading

The known loads on the loaded member are transferred to the run centroid, with introduction of the requisite

moments. It is clear by inspection here that the resultant centroidal force is F = [ –4 3 12 ]' kN using the

given Cartesian system. The resultant moment may be found either from the three orthogonal views below

which each show one component of the loads’ moment about the centroid, or by vector algebra.

Vector check M = r x F = [ –24 54 400 ]' x [ –4 3 12 ]' kN.mm = [ –552 –1312 144 ]' Nm

Intensity components at line segment end-points for equilibrium : -

12 120

180

x

z

y

3

4

400

A

CB

support

6 forcesin kN

loadedmember

M r / Jq =z z zzA A

yAM x / Iq =y yy

AM y / Iq =x x xxA A

24

B C

x

y

A

54

96

126144 Nm

qzA

qzB

qzC

A

54

y

z552 Nm

126

xqA

xqB,C

B,C

x

24z

1312 Nm

yqC

96

yqA,BA,B

C

view in positive x-direction view in negative y-direction view in negative z-direction

out-of-plane bendingout-of-plane bending in-plane torsion

= - 552 E3x1261.0692 E6

= - 65.1

q = + 27.9 N/mmxC

q = + 27.9 N/mmxB

= 1312 E3x240.4032 E6

= + 78.1 q zxA = -

144 E3x1261.4724 E6

= - 12.3

q zyA =

144 E3x241.4724 E6

= + 2.4

q = -312.4 N/mmyC

q = + 78.1 N/mmyB q = 5.3 N/mmzx

B q = 2.4 N/mmzyB

q = 5.3 N/mmzxC q = -9.4 N/mmzy

C

F = 3 ; F = -4 ; F = -4 ;x z x yy z F = 3 kNF = 12 kN F = 12 kN

M = 4x54 - 3x24 = 144 NmM = 12x24 - 4x400 = -1312 NmyM = 12x54 - 3x400 = -552 Nmx z

view in positive x-direction view in negative y-direction view in negative z-direction

A,B

C

4

12

x

z

400

24yM B C4

x

y

24

zM

3

A

54

A

B,C

3

12

y

z

400

54

xM

beam

nent of the load : qp = - F/L where L is the total length of the run. Thus the sense of qp is oppo-site to that of F as indicated in ( e).It is often more straightforward to resolve F and qp into components qpx = Fx/L and so on.Primary intensity is analogous to the direct stress in a beam's cross-section which also is uni-form - eg. σd = F/A.

- The secondary intensity varies linearly along the run and equilibrates the moment load M. Thesecondary intensity is split into two components which are each analogous to a basic loadbuilding block :- The in-plane or torsional component of secondary intensity, qz of Fig ( f), equilibrates the

moment which is orthogonal to the weld plane, Mz. This moment is analogous to the torqueT in a shaft or beam, so Mz is often referred to as the "torque" on the joint. At any point on the run the torsional intensity, like torsional shear stress in a shaft cross-section :- lies in the weld plane,- is directed perpendicularly to the radius r from the centroid to the point in question, and

in a sense opposing Mz, and- is of magnitude given by qz = Mz r /Jzz ie. the torque load building block with appropri-

ate units (N/mm).- The out-of-plane or bending component of secondary intensity equilibrates the moment

lying in the weld plane - in the general case this latter consists of two Cartesian components,eg. Mx and My in Figs ( g), ( h) above.At any point on the run the bending intensity - say qx of ( g) - like normal bending stress in abeam cross-section :- is orthogonal to the weld plane,- is directed perpendicularly to the coordinate y from the centroid to the point in question,

and in a sense opposing Mx, and- is of magnitude given by qx = Mx y /Ixx, ie. the bending load building block with appro-

priate units (N/mm).

The total force intensity q is the vector sum of primary and secondary components. Vector summa-tion senses must be deduced by inspection, which can be rather tedious. The procedure is demon-strated in the example opposite.

The example is not pursued to its logical conclusion - to ascertain safety for example - because thenumerical values are WRONG ! - the traditional approach does not cater for asymmetric bend-ing which occurs in this example run. Although Ixy is evaluated, it is not used.

The traditional approach is suitable only for :- joint runs which are planar, made up of straight segments, and symmetric;- loading which is simple, eg. consisting of a single force and moment component, because oth-

erwise vector recomposition like that of the example can become very tedious.

The unified approach on the other hand is completely general and suited to asymmetric runs. It is avector treatment which does not distinguish between in- and out-of-plane loading, and which iswell suited for computer implementation. However before this is examined it is necessary to com-plete the foregoing by correlating the resultant intensity at any point along the run with the materialstrength to evaluate the safety factor at that point. To do this, the stresses in the throat plane must beconsidered.

Welded Joints 4

The double subscript notation has the first subscript representing the axis about which the secondary intensity

Thus for the in-plane intensity at the point A :

r = √ ( 242 + 1262 ) = 128.3 mm therefore qz = Mz r /Jzz = 144 E3 x 128.3 / 1.4724 E6 = 12.5 N/mm

Resolving this into x, y components by similar force and dimensional triangles with signs by inspection :

qzx = -12.5 x 126/128.3 = -12.3 N/mm ; qzy = +12.5 x 24/128.3 = 2.4 N/mm as tabulated above

Resultants

The resultant intensity at each point is the sum of all components, primary and secondary :-

qA = [( 13.3–12.3 ) (–10.0+2.4 ) ( –40.0–65.1+ 78.1 )]' = [ 1.0 –7.6 –27.0 ]' N/mm

qB = [( 13.3 + 5.3 ) (–10.0+2.4 ) ( –40.0+27.9+ 78.1 )]' = [ 18.6 –7.6 66.0 ]' N/mm

qC = [( 13.3 + 5.3 ) (–10.0–9.4 ) ( –40.0+27.9–312.4)]' = [ 18.6 –19.4 –324.5 ]' N/mm

Throat Stresses and Joint Safety

The following assumptions are made :-- failure of joints in practice occurs most commonly in the throat so any safety analysis need con-

sider only the throat- in spite of the extreme stress concentration at the root geometric singularity, the stresses in the

critical throat plane may be considered uniform.These gross simplifications are justified as they enable designs to be compared to actual weld testsusing the same fillet loading characterisation.

The resultant intensity in global terms q = [ qx qy qz ]' has been evaluated as above for a certainweld element of length δL. The global intensity must be resolved into components local to the ele-ment q = [ q1 q2 q3 ]' in Fig ( j) below - the 1-direction being longitudinal ie. along the run, and the2- and 3-directions normal to the fillet faces.

For the point C of the preceding example, the 1-2-3 directions local to the joint at C happen to corre-spond to the global x-y-z senses, so :

qC ≡ [ q1 q2 q3 ]' = [ qx qy qz ]' = [ 18.6 –19.4 –324.5 ]' N/mm By inspection at the point A on the other hand :

qA ≡ [ q1 q2 q3 ]' = [ qy –qx qz ]' = [ –7.6 –1.0 –27.0 ]' N/mm

The equilibrating components on the loaded member face of the element in ( j) demonstrate that q1

is essentially different to q2 and q3 as these latter appear as shear and normal effects on adjacent fil-let faces whereas q1 manifests itself as shear on both faces.

To correlate throat stresses with the known intensity of ( j), we consider equilibrium of a free bodyof the top half of the fillet, one of whose boundaries is the throat, Fig( k). The uniform stress over

( j )

q2

q2

q1

udinal directionview in longit-

q1

q3

q3

throatplane

udinal directionview in longit-

q1

τt

w

a =w

√ 2

45o

σ

q2

q3

τl

throatplane

στt

q2

δL

filletoutline

q1q

3

of top halffree body

τl

τ

throat

( k )

q1

q2

q1

q3

q3

q2

longi-tudinal

direction

support

load

δL

the throat consists of a component σ normal to the throat plane, together with two mutually perpen-dicular shear components lying in the throat plane - the longitudinal shear component τl and thetransverse shear component τt.

As δF = q.δL = σ.δA and the throat width a = w/√2, force equilibrium of the free body ( k) requires :

( ii) τl . aδL = q1 δL ; τt . aδL = ( q2 + q3 ) δL /√2 ; σ. aδL = ( q2 – q3 ) δL /√2

Applying the distortion energy failure theory to this plane stress state, theequivalent uniaxial stress in the throat of the fillet element, σe, is given by :-

σe2 = σ2 + 3 τ2 = σ2 + 3 ( τl

2 + τt2 ) using Pythagoras.

Substituting for the stresses from ( ii) and introducing the concept of the sca-lar force intensity qe which is equivalent failurewise to the given q yields :-

( 1 ) w σe = 2 qe where qe ≡ √( 3/2 q12 + q2

2 + q2 q3 + q32 ) ; σe = S/n

in which S is the strength of the weld material and n is the safety factor at the fillet.Note that ( 1) contains the four indispensible ingredients of a design equation - safety (n), dimen-sions (the constant fillet size, w), material (characterised by strength S) and loading characterised bythe intensity (q).

Like the beam, the final analysis step involves a search along the run to determine the most criticallyloaded fillet and hence the minimum safety factor of the joint as a whole.

Unified Analysis

The theory will be developed in three dimensions, and later parti-cularised to two. A general three-dimensional run of length L isillustrated, in which a typical element δL is located at r in a con-venient centroidal Cartesian system, the centroid being such that

( iii) ∫L r dL = 0 ; r = [ x y z ]' see ( i) above.

The centroidal loading F, M causes the loaded member to rotate withrespect to the support about a fixed centre C located at rc. The radius s from C to the element is :

( iv) s = r – rc

As a result of this rotation, compatibility requires that the strain in the joint at the element is bothorthogonal to and linear in s. Presuming elastic behaviour, the stress resultant at the element - ie. theintensity q - is also orthogonal to and linear in s. This may be expressed as :

( 2 ) q = [ qx qy qz ]' = b x s = b x r – c using (iv)

where b = [ bx by bz ]' is a vector constant commensurate with linearity,the cross product ensures q - s orthogonality, and c = b x rc = [ cx cy cz ]' is another constant vector common to all elements.

Equation ( 2) establishes the form of the intensity variation with position r along the run. It remainsto evaluate the two constants b and c, to define completely the dependence of q upon r. This evalua-tion follows from equilibrium :

For force equilibrium, using ( 2) F = [ Fx Fy Fz ]' = – ∫L q dL = – ∫L ( b x r – c ) dL

= – b x ( ∫L r dL ) + c ∫L dL ; the first integral vanishes as a result of ( iii), leaving

( 3 ) F = L c that is c = F / L - the traditional method’s reversed primary intensity.

rs

r cG

q

xy

z

runC

δL

( 0, -τ )

( σ, τ )

Welded Joints 5

The longitudinal sense may be remembered bythe requirement for the run to be traversed clock-wise when the weld plane is viewed normally, assketched here.

Summary

The unified approach to analysing a fillet welded joint - whose run consists of co-planar straight linesegments - involves the following steps :

- erect a convenient X-Y system in the run plane and calculate the centroidal coordinates- erect a parallel centroidal x-y system as the basis for further work

- determine the centroidal second moments of run length- define the joint loading at the centroid, probably using a free body- calculate the constants b and c from ( 4a) and ( 3) respectively.

- for each straight line run segment :- determine the positive longitudinal sense, the slope θ and the t-matrix from ( 5a)- for each of the two end points of the line in turn :

- determine the radius vector, r, from the centroid- compute q in global Cartesian terms from ( 2)- ascertain the local q components from ( 5) and- calculate the equivalent intensity using ( 1)

- search over all end points for the maximum value of qe (corresponding to minimum safety)then apply the design equation ( 1) to find either the safety factor n in the analysis situation orthe necessary fillet size w in the design situation.

The foregoing example carried out inappropriately by the traditional technique is repeated belowvia the unified approach, and highlights how neglect of asymmetry can lead to a dangerous situa-tion. Thus if asymmetric bending had been incorrectly neglected here (ie. zero Ixy used in calcula-tions) then a fillet size of 2.8 mm would have resulted, which, if used in the joint, would haveyielded an actual safety factor of only 2.8*410 /2*464 = 1.24 - a significant and dangerous reductionfrom the target of 1.73.

Conclusion

The foregoing analyses rest upon rather simplistic foundations despite the apparent mathematicalniceties. Three significant and questionable assumptions are the following :

- The connected members are assumed to be rigid, the weld metal itself being the only compli-ance in the connection - however the compliant lap joint modelled in the Appendix demon-strates that stress concentration must occur when all components in the connection undergodeformation, as would happen to some extent in any practical fillet welded joint.

- Stresses in the throat plane are assumed to be uniform over that plane. It is possible to define a

0

cosθcosθsinθ

– sin θ00

0 1

( 5a) t = q2

qy q q

1

qx

θ

x

longit-udinal

longi-tudinalsense

x

y

z

For moment equilibrium, again using ( 2)

M = [ Mx My Mz ]' = – ∫L r x ( q dL ) = – ∫L r x ( b x r – c ) dL = – ∫L r x ( b x r ) dL + ( ∫L r dL ) x c

in which the second integral vanishes, again as a result of ( iii).

Expanding out the products in the first integral and integrating term by term (left as an exer-cise for the reader) :-

( 4 ) M = – I b ie. b = – I -1 M ;

- this matrix of second line-moments can be computed from known geometry.

The constants b and c may thus be determined from known loading and run geometry, via ( 3) and (4), enabling q to be found at any point r from ( 2).

In the case of a two-dimensional joint lying wholly in the x-y plane (the only case examined atlength here), z vanishes everywhere on the run, r degenerates to [ x y 0 ]', and ( 4) becomes :-

which may be solved for b without the necessity for inversion - thus for planarwelds lying in the x-y plane :-

It may be shown that equations ( 2), ( 3) and ( 4a), with Ixy = 0, reduce to the equations cited previ-ously in the traditional approach. However ( 4a) correctly recognises x-y coupling, as may be seenby comparing it with the equations used for the asymmetric bending of beams.

Resolution

The intensity q at a point in the joint run is related to the known load in global terms [ qx qy qz ]', via( 2). Safety at that point in turn depends, via ( 1), on q expressed in terms of local components [ q1q2 q3 ]' whose directions are dictated by the orientation of the fillet locally at the point. To avoid theneed for visual inspection to carry out the global-to-local transformation - inspection which can beawkward as the above example showed - we define the matrix operation :-

( 5 ) qlocal = t q global ie. [ q1 q2 q3 ]' = t [ qx qy qz ]'

in which the 3*3 transformation matrix, t, is a pure geometric entity involving thedirection cosines of the run tangent at the point in question, together with the roll angle of the fillet.We consider resolution only for the particular case of a run lying in the x-y support plane with the z-and 3-senses identical (as above). The fillet does not roll. If θ is the inclination of the longitudinalsense (tangent to the run) in the x-y system - θ being reckoned positive anti-clockwise from the x tothe longitudinal sense - then :

I Jxx

Jyy

Jzz

-Iyz

xz-Ixy-I

sym

Jxx; = ∫ ( y + z ) dL2 2L

Ixy = ∫ x y dLL

&c

=where

= - ; Ixx = ∫ y dL2L

&c

0Ixx

Iyy

Jzz

0xy-I

xy-I

00

b

b

b

x

y

z

M

M

M

x

y

z

M =

b M / J

( M I + M I - I )) / ( I I

( M I + M I - I )) / ( I I( 4a ) b = b

bx

y

z

yyx xyy xx yy xy2

xyx xxy xx yy xy2

zzz

= -

Welded Joints 6

more realistic stress distribution which satisfies equi-librium, compatibility and linear elastic behavioureverywhere within the fillet. The variations of thestress components on the fillet legs predicted by onesuch model are given here for q3 loading. Significantnormal 'bending' stresses evidently occur in the hori-zontal leg, even although their force resultant is zero.The high stresses in way of the geometric singularityat the root are noteworthy. Similar stress concentra-tion occurs with q1 loading, however such models(again) neglect compliance of the joined components - currently the effect of this can only beexamined numerically, by finite elements for example.

- It will be seen in the context of Fracture Mechanics that the stress concen-tration ahead of any crack tip is extreme due to the geometric singularity.A crack is inherent with filleted joints - the tip coinciding with the filletroot - so it is inevitable that at low loads the material will yield and behavenon-linearly over a small region with some consequent stress redistribu-tion.

These assumptions notwithstanding, the theory works in practice because it describes fairly welljoints' tendencies to failure - so provided it is used with experimental safety constraints which alsoare characterised by the theory, then safety assessment is perfectly valid.

This chapter cannot conclude without reiterating forcefully that successful welded joint design reliesas much if not more on common sense and practical experience than on mathematics - no matterhow appealing this last might be.

Bibliography

AS 1250 - SAA Steel Structures Code, SAA (Standards Association of Australia)

AS 1554 - SAA Structural Welding Code, Part 1, SAA

MA 1-4 - Steel Structures - Part 4 : Connections, SAA

Connor LP ed, Welding Handbook, 5 vols, American Welding Society 1987

Lay MG, Source Book for the Australian Steel Structures Code, AISC

Niemi E ed, Stress Determination for Fatigue Analysis of Welded Components, Abington 1995

Zhao X-L & Packer JA, Fatigue Design Procedure for Welded Hollow Section Joints, Woodhead2000

The following are essential background to practical welded joint design :-

Economic Design of Weldments, Technical Note No.8, AWRA (now the WTIA)

Blodgett OW, 'Weld Failures : They Could be the Result of Violating Simple Design Principles',Welding Journal, March & April 1982.

q3

q3

crack

EXAMPLE The joint of the previous problem is made with an E41xx electrode to AS 1250 in which the mini-

mum safety factor is laid down as √3. What size of fillet should be used ?

Recognising the essential asymmetry of the run, analysis will be based on the unified technique.

Run Geometry

Employing the same axes as in the previous example, the properties are unchanged. However it should

be noted that Ixy, though calculated, was not previously employed. This is now rectified.

Centroidal Loading

The force of [ –4 3 12 ]' kN is applied at the point [ –24 54 400 ]' mm from the centroid. Transferring

this force to the centroid gives rise to the centroidal moment of :

M = r x F = [ –24 54 400 ]' x [ –4 3 12 ]' = [ –552 –1312 144 ]' kN.mm

b-Vector

Using ( 4a) for this planar case :-b = – ( –552*0.4032 – 1312*0.3888 )/( 1.0692*0.4032 – 0.38882 ) E–3 = 2.6173 MPa

( –552*0.3888 – 1312*1.0692 )/( 1.0692*0.4032 – 0.38882 ) 5.7778 ( note 144/ ( 1.0692 + 0.4032 ) -0.0978 units )

Line A-B Search - from ( 5a) the t-matrix may be evaluated since :-

sin θ = ( yB – yA ) / L

AB = 1 ; cos θ = ( xB – xA ) /L

AB = 0

Point A : r = [ –24 –126 0 ]' mm, so, from ( 2)

q = [ 2.6173 5.7778 -0.0978 ]'x [ -24 -126 0 ]' – [ -4 3 12 ]' * 103/300

= [ 1.0 –7.7 –231.1 ]' N/mm ; resolving via ( 5) with ( 5a) :q = t q = 0 1 0 1.0 = – 7.7 N/mm

–1 0 0 – 7.7 – 1.0 0 0 1 – 231.1 – 231.1

qe = 231.8 N/mm . . . from ( 1).

Point B : r = [ –24 54 0 ]' mm

q = [ 18.6 –7.7 240.0 ]' N/mm, so, as above for point A :

= t q = [ –7.7 –18.6 240.0 ]' N/mm

qe = 231.5 N/mm

Line B-C Search - From the same relations as for line A-B above :- sin θ = 0 ; cos θ = 1

Point B : q = [ 18.6 –7.7 240.0 ]' . . . . exactly the same as for point B on AB

= t q = [ 18.6 –7.7 240.0 ]' N/mm ;

qe = 237.3 N/mm

Point C : in a similar manner -

qe = 463.9 N/mm

It should be noted that any turning value of qe in a straight segment is assumed to be a minimum, and so the two

end points only of each straight run need to be examined when searching for maximum qe. This follows from sim-

ilar conclusions which were proved in the context of linearly varying bending moments (see Stress, Strength and

Safety, problem 17 ).

So, for this run as a whole, the maximum value of qe is 464 N/mm. Applying the design equation ( 1), with a

maximum equivalent normal stress of 410/√3 = 237 MPa, gives a required fillet size of 2*464/237 = 3.9, say 4 mm.

Four significant figures are quoted here purely to help readers check their calculations.

The program Fillet Welds employs the unified approach - a dialogue with the program for this example illustrates

how the unified approach can obviate traditional difficulties with vector recomposition.

Welded Joints 7

(a) q = dFA/dx = – dFB/dx

Constitutive Laws - Assuming elastic behaviour :-

εA = ( σ/E )A = ( F/AE )A ; εB = ( F/AE )B ; γ = τ / GC = q / bGC

Eliminating the three strains and two deflections from the constitutive laws and compatibility; and defining the stiff-

nesses kA = (AE)A/L ; kB = (AE)B/L ; kC = bLGC/t leads to :-

(b) dq/dx = ( FA/kA – FB /kB ) kC / L2

Solving the three simultaneous differential equations (a) and (b) for q, FA and FB, with boundary conditions FA = 0 at x

= 0 and FA = P at x = L :-

q L/P = [ µ cosh λφ + ( 1–µ ) cosh λ( 1–φ ) ] λ / sinh λFA/P = 1 – FB/P = 1 – µ + [ µ sinh λφ – ( 1–µ ) sinh λ( 1–φ ) ] / sinh λ

in which φ = x/L ( 0 ≤ φ ≤ 1 ) is the normalised location along the joint, and both

λ = √( kC/kA + kC/kB ) and µ = 1/( 1 + kA/kB ) are constant parameters which are characteristic of the joint and determine its behaviour.

These results are plotted below for two different arrangements - on the left, one member is much less compliant than the

other; on the right, the connected members are identical. The upper graphs show how the force intensity q, and hence

shear stress in the connector τ, vary along the joint. The lower figures illustrate the tension variation along member A.Although this model joint is a gross simplification, the trends portrayed by it are quite realistic and tally with the quali-

tative conclusions above - if the joined members are rigid and the connector C alone is compliant, then there is no stress

concentration. The traditional/unified fillet models examined above are assumed to behave in just this ideal manner, so

the models will underestimate the stresses which occur in practice.

A dialogue with the program ‘Fillet Welds’ for the above worked example.

************* Analysis of FilletFILLET WELDS Welded Planar Joints************* - version 1f

enter - title : example of Notes (kN,mm) number of line end-points : 3 and their coordinates point 1 : 0 -180 point 2 : 0 0 point 3 : 120 0 number of straight runs : 0centroid at ( 24,-54) weld length = 300 Ixx= 1.069e+6 Iyy= 4.032e+5 Ixy= 3.888e+5

enter the components of force : -4 3 12 and of moment : -1200 -1600 0elements of the b-vector :- bx = 2.617e-3 by = 5.778e-3 bz =-9.780e-5

end-point intensity - components & equivalentline number 1 ( 0,-180) -7.65e-3 -1.01e-3 -2.31e-1 2.32e-1 ( 0, 0) -7.65e-3 -1.86e-2 2.40e-1 2.31e-1 line number 2 ( 0, 0) 1.86e-2 -7.65e-3 2.40e-1 2.37e-1 ( 120, 0) 1.86e-2 -1.94e-2 -4.53e-1 4.64e-1

maximum equivalent force intensity is 4.639e-1 ********

op : s 's'ave option; 'h'elp option lists the options availabledialog saved in : DCW Hard Disk:welded joints:fillets:eko

op : q 'q'uit optionend of program

qemax, kN/mm

User selects convenient Cartesian system - point 2is origin here (the point B of earlier sketches) - and usesthis system consistently for all input.Zero enforces sequential lines between points 1&2, 2&3

kNkN.mm, about dummy origin - programauto converts to centroidal moment.kN/sq.mm ie. GPa

Consistent units for all input and corresponding output

APPENDIX : THE COMPLIANT LAP JOINT

The force P is transmitted from member A to member B through the connector C,

which may be an adhesive or longitudinal fillet welds for example. Examination of

the force transfer through the connector is a necessary precursor to the determina-

tion of the joint's safety factor. A simple model of the system is considered in which

the members A and B are taken to be in tension only, and C is in shear only. Load transfer between the members and

the connector is by shear. There are two extremes of behaviour which may be appreciated qualitatively :-

- if the members are rigid and the connector is flexible, then the force is transferred uniformly, the connector is

sheared uniformly and connector stresses are the lowest possible;

- if the members are very flexible and the connector is rigid, then neither member will deform substantially in way

of the connector . This means that nearly all the force is transferred between A and C at x = L, and between B and

C at x =0, leading to high stress concentrations at the ends of the joint.

The problem is a limiting continuous case of the multiple discrete connector assemblies examined previously in the

chapter on Stress, Strength and Safety . It is statically indeterminate and so compatibility, equilibrium and the individual

constitutive laws have to be examined.

Compatibility - Let uA and uB be the deflections of A and B

from their unloaded positions at location x along the

joint, and let γ be the shear strain in C. Geometryrequires :

εA = duA /dx ; εB = duB /dx ;

γ = ( uA –uB ) /t

Equilibrium - If q is the force intensity (force per unit length)

on element δx at x, then equilibrium requires that:

BC

A

P

PL

x

t width b

t

uA

uB

γ

COMPATIBILITY EQUILIBRIUM

FA F +δFA A

q δx

q δx

FB F +δFB B

x δx

Welded Joints 8

8 A b x b equal angle is fillet welded orthogonally to a support andloaded by the axial force, F, located at the corner as shown. Derivethe joint design equation : w b S = 4 n F where n is the safety factor,and S is the strength of the fillet, whose size is w.

9 The cantilever, of circular cross-section, is fillet welded to asupport using an E48xx rod, and loaded by a force of magni-tude F, inclined at [ 4 –3 –12 ]' as shown. What is the maxi-mum value of F for a safety factor of 1.5 ? [ 19.7 kN]

The horizontal cantilever of triangular cross-section is fillet welded to avertical wall and supports a weight of 15 kN as sketched. What size offillet should be laid down if the allowable tensile stress in the joint is250 MPa? Neglect the fillet not being everywhere right-angled. [ 4 mm]

11 The Z-beam is joined obliquely to the plane support by two identi-cal fillet welds, one on each flange, and is loaded by a 1400 Nmcouple, M, whose axis is indicated. If the design normal stress inthe joint is 250 MPa, what size of fillet is necessary? [ 6 mm]

w b

bF

Ø 200

10

500

3F

12F z

4F3F

x

y

15 kN145

50

120

75

50

w

10

x

60

90

45o

z

y

M

M

FILLET WELDED JOINTS - PROBLEMS

Traditional methods should be used in problems 1-4 since the loading is simple and bending symmetric.

1 Which lap joint is the more effective, the transverse or longitudinal, and by how much ?[ Transverse, 22% ]

2 The design normal stress for the welds is 240MPa. Determine the maximum permissible load,F, in each case. [ 100, 35.3, 14.5, 10.3 kN ]

3 The two beams are each weldedto a fixed support as shown.Find the maximum equivalentnormal stress in each joint. [ 17.5, 104 MPa]

4 A force of 7.5 kN acts on the bracket. What is the maximumequivalent normal stress in the fillet ? [ 48.0 MPa]

5 Show that ∫L r x ( b x r ) dL = I b ; the symbols being as definedin the above text.

6 A straight line extends between the two points [ x1 y1 z1]' and [ x 2 y2 z2]'. Derive the follow-ing expressions for the second moments Jxx and Ixy of this line, in terms of the end point coordi-nates and the line length, L12 :-

Jxx = 1/3 L12 ( ( y12 + y1y2 + y2

2 ) + ( z12 + z1z2 + z2

2 ) )Ixy = 1/6 L12 ( 2 ( x1y1 + x2y2 ) + x1y2+ x2y1 )

7 The centroidal loading on a circular filletweld of radius R is F & M. Prove that thelocal intensity at a point P defined by theangle φ on the fillet, is given by the vectorshown, where m ≡ M/R.

L/2

L2

F

F

F

F

LongitudinalTransverse

6 6

6

( a )

F

( b ) ( c ) ( d )

60

50

6

30

60F

50

60

F100

6

6

30

F100

60

60 300 50

15

5

250 N tube Ø32 x 6

250

850 N5

6

1206

60

120

7.5kN

45o

– F sin φz yx + F cos φm

– m sin φ ) – F zy x2 ( m cos φx

yφ

P q = 1

2πR – F sin φyx – F cos φ

Cylinders 1

ously adequately supported as in figure ( a) below.

Axial Stress

Cylinders are classed as being either :- open cylinders, in which there is no axial component of wall stress, or- closed cylinders in which an axial stress must exist to equilibrate the fluid pressure.

Open cylinders are typified by interference-fitted bushes figures ( a) and ( b), in which there are nolongitudinal pressures and so no axial stresses. A fluid container sealed by a piston is open ( c) - inthis case an external axial force Fa is mandatory for equilibrium of the overall piston-and-cylinderassembly. If Ai is the internal circular area then a free body of either piston or cylinder end ( c) musthave Fa = piAi and there is no need for axial wall stresses to equilibrate the fluid pressure.

If an axial stress does exist then it is uniform across the cylinder wall, no matter whether the cylin-der is thin or thick. The stress may be found easily from equilibrium in the axial direction, aided bya free body viewed in side sectional elevation. Considering the free body of one end of an internallypressurised closed cylinder ( d) the fluid pressure resultant piAi is equilibrated by the wall stressresultant σaAa - where the annular wall area is Aa = Ao –Ai in which Ao is the outside circular area.In the more general case where an external fluid pressure also exists :

( i ) σa = ( pi Ai – po Ao ) / Aa

A free body ( e) of part of a pipe connecting two vessels (not shown) might be thought open, sinceno ends are evident and the fluid pressure is self-equilibrating across the free body. However eachconnected vessel acts as a pipe closure, so the pipe is in fact closed and axial stresses must occur.

Thin Cylinders

The tangential stress is uniform across the wall; so, from equilibrium of thefree body illustrated : σt = ( pi Di – po Do ) / 2 tSince Di/t >> 1, this equation with ( i) yields the following :

( 1 ) σt = ∆p Di / 2t where ∆p = pi – po σa = ∆p Di / 4t (closed) or σa = 0 (open)

The radial stress is zero, the tangential stress is always the principal ofgreatest magnitude, and the axial stress is either zero in the case of

open thin cylinders or half the tangential stress in closed thin cyl-inders. The Mohr's circles thus appear as shown. Derivation of thefollowing design equations is left as an exercise for the reader :-

A a

A i A o

OPEN

p iOPEN

po

CLOSEDσ

a

p i

p iσ

a

CLOSED

p ip i

σa

σa

σa

σa

OPEN

p iFa Fa

Fa

p i

( a ) ( b ) ( c ) ( d ) ( e )

pi σ

tσ

t

po

σ

τ

σ

τ

ATR

CLOSEDOPEN

TR A

CYLINDERS

Cylinders find many applications, two of the most common being :- fluid containers such as pipes and pressure vessels as illustrated in the frontispiece, and- interference-fitted bearing bushes, sleeves and the like.

Cylinder loading consists of internal and external ( gauge ) pressures pi and po due to adjacentfluids or to contacting cylindrical surfaces. The following notes examine cylinder safety and deriveappropriate design equations. Since the analysis presupposes an elastic material, superposition maybe applied if other loading mechanisms such as bending or torsionoccur. The treatment ignores end effects and the axial variation ofpressures and stresses, so cylinder length is of no consequence. Secon-dary stresses arising from incompatibility between cylinder and endclosure are examined in the chapter on “Pressure Vessels”.

The stress state in the wall is essentially triaxial and initial analysisgives the principal components without need for resolution :

- axial ( or longitudinal ) stress, σa- tangential ( or hoop or circumferential ) stress, σt- radial stress, σr . . . . all of which are taken to be positive tensile.

Stresses are identical in cylinders which have geometrically similar cross-sections and which areloaded by the same pressures - absolute size is irrelevant insofar as stresses are concerned. Geomet-ric similarity is expressed by a single ratio between two of the three dimensions - inside diameter (orbore ) D i, outside diameter Do and wall thickness t. Cylinders are usually considered to be eitherthick - in which stress concentration due to relative curvature is significant - or thin in which stressconcentration is negligible. Other differences are :

Thin ThickLimiting proportions (approx) Di/t > 25 γo ≡ (Do/Di )2 > 1.15Analytic treatment simple approximation accurateStatically determinate yes noStress state membrane - ie. biaxial triaxialStress - radial, σr zero varies with radius

- tangential, σt uniform varies with radius- axial, σa uniform uniform

The limiting proportions - the dividing line between thin and thick - is not hard and fast. Whetherwe choose to treat a particular cylinder as thin or as thick is a matter only of the accuracy we seekfrom the mathematical model together with the effort required to solve the model. For example, ifin a particular design the thin model results in a cylinder with Di/t = 7 then we would either - repeat the calculations using the thick model if an accurate solution was sought, or - accept the result of the thin model knowing that it was in significant error due to the neglect of

curvature-induced stress concentration.We shall address the error associated with the thin cylinders below. The Pressure Vessel Code adoptsthe latter approach as the Code incorporates large design factors to cater for other imponderables.

Physically, thin cylinders are not suitable when the external fluid pressure is much greater than theinternal, unless the cylinder is supported or stiffened against local buckling. We shall not considerthin cylinders with net external pressures sufficient to cause buckling, unless the cylinders are obvi-

t

D iDo

σa σ

r

σt

ip

op

Cylinders 2

The stress variations given by ( 2) and ( 3) for a particular closedcylinder are sketched here, and the Mohr's circles correspondingto the bore ( γ=1 ) and to some other location in the wall ( γ >1 )appear below - the similarity between the Mohr's circles for thickand thin cylinders is noticeable. The tangential /radial stress dis-

tribution is seen to take the form of ahorn disposed symmetrically about σand of opening radius σ at the bore,confirming the general conclusionthat the bore is the critical location -as expected since curvature andstress concentration are most severeat the bore ( see Mohr’s circles ).

If this cylinder were ductile and themaximum shear stress failure criterion is applied, then the equivalentstress is 180 MPa at the bore and only 20 MPa outside - evidently thematerial in a thick cylinder is not used effectively throughout the wall.

The advantages of the normalised treatment used here may be appreciated by comparing ( 2) and (3)with the corresponding equations in any textbook. The pressures or overall proportions ( γo ) maynot be known initially, in which case equations ( 3) are usually applied in the context of the problem,with σ and σ retained as unknown constants whose solution is sought first.

For the particular case of pressure po external to a solid shaft, ri = 0, both γo and γtend to infinity, in which case ( 2) and ( 3) degenerate to :

( 3a) σt = σr = –po these stresses are uniformly compressive.

Design Equations

The following equations have been derived from ( 2) and ( 3) with the appropriate failure theory.The detailed proofs, left to the reader, exemplify the difficulties associated with the application logicof piecewise-defined theories such as the maximum shear stress and modified Mohr.

For DUCTILE materials the concept of equivalent stress, σe, is relevant. If the distortion energy fail-ure theory is applied, then using ( 3) :

σe = √[ ( ( σt – σr )2 + ( σr – σa )2 + ( σa – σt )2 ) / 2 ]= √[ ( σ – σa )2 + 3 ( σ /γ )2 ] where the appropriate σa is given by ( 3).

The maximum equivalent stress, defined as σ* ≡ ( σe)max, occurs when γ is a minimum ie. at the borewhere γ = 1. Resulting design equations are therefore :

( vii ) Distortion Energy

OPEN : σ* = √[ ( pi – γo po )2 + 3 γo2 ( pi – po )2 ] / ( γo – 1 )

CLOSED : σ* = √3 | pi – po | γo / ( γo – 1 )

If, on the other hand, the maximum shear stress theory is applied, it may again be shown that themaximum equivalent stress occurs at the bore, so the principals from ( 2) and ( 3) are :

σt = [ (γo+1)pi – 2γopo ]/(γo–1) ; σr = – pi ( sic ! ) ; σa = (pi – γopo)/(γo–1) or 0Application of the theory requires selection of the minimum and maximum principal stresses, withneglect of the intermediate, and it is apparent that σt may or may not exceed either σr or σa. It is nec-

= 9 CLOSED

p = 120 MPap = 40 MPa

γi

o

o

σ~ = 90

σ t

σr

–20

–40

60

–120

–30

stre

ss (

MP

a)

radius 600

20

σ–σa=

R A T σ

A TR 0

σ

τ

σ–

σ~σ~== σ γ

~

γ > 1

γ = 1

po

po

( ii ) n ∆p Di/t = 2 Sy [ maximum shear stress; distortion energy theory (open)]= 4/√3 Sy [ distortion energy failure theory (closed) ]

= 2 Sut [ modified Mohr failure theory ]

Thick Cylinders

In the absence of bending and direct shear, the radial and tangential stresses are circumferentiallysymmetric - independent of θ - but vary with radius r. It is convenient to normalise the radius by theconstant internal radius, setting :

( iii ) γ ≡ ( r / ri )2 ; 1 ≤ γ ≤ γo = ( ro / ri )2 = ( Do /Di )2

where γo is a constant, representing the proportions of the cylinder.

Consider the element, defined by r, δr and δθ, and of unit width per-pendicular to the sketch plane. For radial equilibrium :

σr r δθ – ( σr + δσr )( r + δr ) δθ + 2 σt δr sin(δθ/2) = 0which, on expanding with neglect of second order terms, gives :

( iv ) ( σt – σr ) δr = r δσr

As σt and σr cannot be found individually from this equilibrium equa-tion, the problem is statically indeterminate - so compatibility and the constitutive law must beinvoked. The compatibility requirement, first postulated by Lamé, that initially plane transversecross-sections remain plane when loaded, implies that theaxial strain, εa, is constant across the cross-section - ie. is inde-pendent of radius. With the elastic constitutive law, this maybe expressed as : E εa = σa – ν ( σt + σr ) = constant ≠ function (r)and since σa is itself constant ( ie. uniform, and not a function of radius ) it follows that :

( v ) σt + σr = constant = 2 σ where σ is some constant, as yet undefined.

This latter compatibility/constitutive equation enables resolution of the indeterminacy. Thus, elimi-nating σt between ( iv ) and (v ) :

δσr / ( σr – σ ) = – 2 δr / r and integrating with γ given by ( iii)( vi ) σr = σ – constant / r2 = σ – σ / γ where σ is an integration constant.

The form of the stress distribution is thus established but the twounknown constants σ and σ must be evaluated before the distribu-tion is defined completely. From ( vi) and from equilibrium of ele-ments at the inner and at the outer surfaces :

inside σr = σ – σ = – pioutside σr = σ – σ /γo = – powhich may be solved for the unknown constants in terms of the known pressures :

( 2 ) σ = ( pi – γo po ) / ( γo – 1 ) ; σ = σ + pi and = ( pi – po ) γo / ( γo – 1 )

Combining ( v) and ( vi), and recalling ( i), yields the final stress distribution equations :

( 3 ) σt = σ + σ / γσr = σ – σ / γ ≤ 0σa = σ (closed) or σa = 0 (open)

by which all stress components are expressible in terms of location, r ( ie. γ ), once the constants havebeen evaluated from ( 2) in terms of proportions and loading.

δr

σ t

σ t

θσr

r

roσ + δσr r

ir

δθ

po

σr σrpi

γ = γo

γ = 1

Cylinders 3

other proportions areof similar shape. Theabsence of upperlimits for ductileclosed cylinders is dueto triaxiality - approxi-mately equal princi-pals with small shearanalogous to hydro-static loading

Error of Thin Cylinder Approximations

The error which arises from the use of the thin cylinder equa-tions will now be evaluated for ductile cylinders subjected tointernal pressure and subscribing to the distortion energy fail-ure theory. Setting w = t/Di, and employing ( ii) & ( vii), theerror in the maximum equivalent stress is :

Error = 1 – σe (thin) / σ*(thick) = 1 – 2 ( 1 + w ) /√[ 1 + 3( 1 + 2w )4 ] Open= 1 – ( 1 + w ) /( 1 + 2w )2 Closed

The plot of these errors indicates that a thin/thick transition at say w ≈ 0.04 corresponds to an errorof about 10 %, and that the thin cylinder equations always underestimate the equivalent stress.While there may be little difference between theories in analysis situations - when equivalentstresses or safety factors are being calculated - the equations' forms often lead to substantial differ-ences in the dimensions predicted by the theories in synthesis situations. Thus note the large differ-ence in the diameters predicted by the thin and thick theories in part (a) of the example below, andby the maximum shear stress and distortion energy ductile failure theories in parts (b) and (c).

Stress concentration at the bore of thick cylinders can lead to very wasteful use of material - see theexample above with internal and external pressures of 120 and 40 MPa. It is not uncommon to befaced with a design which at first glance seems impossible - thus in the selection of a fixed bore cyl-inder, increasing the OD in an effort to “beef it up” merely increases γo and stress concntration, sopreventing a solution from being obtained. It may be possible to choose a stronger material, how-ever a more satisfactory solution - especially if weight is an issue - is to use a compound cylinderwhich we now lead up to.

Strains

Cylinder strains are of fundamental importance when resolving statically indeterminate assem-blages of cylinders such as compound cylinders. Normal strain components are related to normalstress components through the elastic three-dimensional constitutive law. Thus, from ( 3) :

( x ) εt = [ σt – ν ( σr + σa ) ]/E = [ ( 1 – ν ) σ – ν σa + ( 1 + ν ) σ /γ ]/Eεr = [ σr – ν ( σt + σa ) ]/E = [ ( 1 – ν ) σ – ν σa – ( 1 + ν ) σ /γ ]/Eεa = [ σa – ν ( σt + σr ) ]/E = ( σa – 2 ν σ )/E ie. γ-independent as Lamé requires.

If temperature effects occur then α.δT must be added to each strain component.

30

20

10

0.05 0.10

THIN CYLINDERAPPROXIMATION

Err

or %

w = t / Di

CLOSED OPEN

0.15

40

distortionenergy

maximumshearstress

modifiedMohr( m = 3.5 )

1.0

1.0i

pn /Suci

p / σ * &

n /Suco

po

p / σ *

distortionenergy

maximumshear stress

1.0

1.0i

p / σ *

op / σ *

OPEN

γ = 4oCLOSED

γ = 4o

SAFE

FAIL

SAFE

FAIL

essary therefore to consider all feasible combinations of the principals, with the results :

( viii ) Maximum Shear Stress

OPEN ordered application limits, po/pi values of

principals min max ( γo–1 ) σ∗

σr ≤ σa ≤ σt 0 (1+1/γo)/2 2 γo ( pi – po )σr ≤ σt ≤ σa ( 1+1/γo )/2 1 ( γo – 1 ) piσt ≤ σr ≤ σa 1 infinity – ( γo + 1) pi + 2 γo po

CLOSED σ* = 2 | pi – po | γo / ( γo – 1 )

BRITTLE cylinders subscribe to the modified Mohr failure theory for which the safety factor isgiven by the maximum of three terms, each of which is linear in the principal stresses :

1/n = max ( σ /St , – σ /Sc

, σ /St – ( σ + σ ) /Sc

) where σ, σ = max, min ( σt, σr, σa )

The implications of this piecewise defined theory may be appreciated with reference to a particularCLOSED cylinder of geometry γo = 3 in a material whose strengths are St = 200, Sc = 700 MPa. Thecylinder is subjected to a con-stant internal pressure of 10 MPawhile the external pressureincreases from 0 to 30 MPa. Thesketch traces the three principalstresses at the critical bore - thetwo extremes are the tangentialand the radial (which is con-stant). Also shown are the threeterms of the modified Mohrequation - the maximum of thesecomprises four linear segmentsa-b-c-d-e.

The tangential extreme, which istensile initially at a, becomes progressively more compressive as the external pressure rises until at dwhen the pressures are identical and the stresses are all compressively equal to the pressures. In a-dthe varying tangential principal is maximum while the constant radial is minimum. These roles arereversed in d-e. The three modified Mohr terms change accordingly - in a-b the first term is the max-imum of the three, in b-c the third term dominates, in c-d-e the second term. As a result of the inter-action between the individually simple terms the safety factor changes in a complex fashion, increas-ing from 10 to 70 then remaining constant before falling monotonically with further rise in externalpressure. Repeating this analysis algebraically (problem #8) leads to a four-piece design equation.

An identical approach for brittle OPEN cylinders results in a similar design equation :

( ix ) Modified Mohr values ofOPEN ordered application limits, po/pi ( γo –1 )Suc/n

principals min max where m = Suc /Sut

σr ≤ σa ≤ σt 0 1/γo [ ( γo +1) pi – 2 γo po ] m ,, 1/γo ( 1+1/γo )/2 [ ( γo+1)m – 2] pi – 2(m–1) γo poσr ≤ σt ≤ σa ( 1+1/γo )/2 1 ( γo –1) piσt ≤ σr ≤ σa 1 infinity – ( γo +1) pi + 2 γo po

The failure loci corresponding to ( vii), (viii), and ( ix), are illustrated for the case of γo= 4; loci for

d

0 10 20 30External Pressure, MPa

0.15

0.10

0.05

0.00

-0.05

-0.10

Ter

ms

of R

ecip

roca

l Saf

ety

Fac

tor

Str

ess

at B

ore,

MP

a

25

0

-25

-50

-75

-100

a

b

c

e

second

third

first

tangential

axial

radial

SAFETY FACTOR

TERMS

STRESSES

70

10

maximum term heavily lined

BRITTLE CYLINDERCLOSED

Cylinders 4

objection may be overcome by providing a residual compressive stress at the bore prior to puttingthe cylinder into service. Autofrettage is a process, the last in manufacture, in which a controlledinternal pressure is applied, high enough to yield the material adjacent to the bore. When this pres-sure is subsequently removed, a permanent tensile set exists around the bore which is resisted bythe surrounding elastic material - with the result that the bore material is in compression. The nettensile stress when the cylinder is eventually put into service is therefore reduced.

Shot-peening, used for establishing a compressive stress in the surface layers of a shaft for example,is generally unsuited to cylinders since the effect is too surface-localised.

Compound Cylinders

Another method of pre-stressing is touse compound (or composite) cylinders- two or more cylinders which areassembled with an interference fit.The analysis which follows is elasticsince the method does not generallyinvolve yielding - and may be applied tosleeves pressed onto shafts, etc.

Two open cylinders only are considered. They are shown in exaggerated fashion both before assem-bly (individually completely free), and after assembly and pressurising - ie. after all load compo-nents have been applied. The bore of the outer cylinder and the outside diameter of the inner cylin-der are made to the same nominal common diameter Dc - however there is a known small diametralinterference, ∆ << Dc. The cylinders are then assembled concentrically using heat or force, andloaded by pressures internal and external to the assembly - pi and po respectively.

Before safety of either cylinder can be addressed, the common interface pressure pc must be known.But the problem is statically indeterminate since statics reveals only that if pc exists as a contact pres-sure internal to the outer cylinder, then the inner cylinder is pressurised externally by the sameamount. Geometric compatibility and the cylinders' constitutive laws ( xi) must be invoked. If thecommon diameter Dc increases by δi measured on the inner cylinder as sketched, and by δo meas-ured on the outer, then compatibility requires that :

∆ = δo – δiCombination of the equilibrium and constitutive considerations proceeds as follows, in which γo, Eand ν refer to the appropriate cylinder :-

Cylinder INNER OUTER

Internal press. pi pc

External press. pc po

σ - eq ( 2) ( pi – γo pc ) / ( γo – 1 ) ( pc – γo po ) / ( γo – 1 )

σ - eq ( 2) ( pi – pc ) γo / ( γo – 1 ) ( pc – po ) γo / ( γo – 1 )

Strain at Dc at γ = γo at γ = 1

εd - eq (xi) [ (1–ν)σ + (1+ν)σ /γo ]/E [ (1–ν)σ + (1+ν)σ ]/E

ie. δi/Dc = { 2pi -[(1-ν)γo+1+ν]pc }/E(γo-1) δo/Dc = { [(1+ν)γo+1-ν]pc -2γopo}/E(γo-1)

Inserting these last expressions for δi and δo into the compatibility equation above, and solving forpc, leads to the desired result :-

∆

Dcpi

po

outer as made inner as made assembled and loaded

δo δ i

EXAMPLE An open ductile cylinder of 60 mm bore is made from a material whose design stress is 400 MPa.

(a) Use the maximum shear stress failure theory to determine an outside diameter suitable to withstand simultane-

ous internal and external pressures of 125 and 250 MPa.

(b) Repeat (a) with internal and external pressures of 360 and 180 MPa respectively.

(c) Repeat (b) but use the distortion energy failure theory.

(a) First try thin cylinder theory to find wall thickness. From the design equation ( ii) with Sy/n = σd = 400 MPa

t = ∆p Di / 2 σd = | 125 – 250 |*60 / 2*400 = 9.4 mmSince the corresponding t/D i is 0.156, it may be seen from the graph above that the error due to the thin cylin-

der approximation approaches 30%. This is unacceptably rough, so repeat with thick cylinder theory.

Using thick cylinder theory, from ( 2) with as yet unknown proportions, γo :

σ = (125 – 250γo )/( γo–1 ) = – 125 ( 2γo– 1 )/( γo– 1 ) MPa

σ = (125 – 250 ) γo /( γo–1 ) = – 125 γo /( γo– 1 ) MPa

At the critical bore (γ=1), from ( 3) :

σt = σ + σ = – 125 ( 3γo– 1 )/( γo– 1 ) = σ σr = σ – σ = – 125 (of course; it must equal the negative of the internal pressure )

σa = 0 (open) = σApplying the maximum shear stress criterion :

σe = σ – σ = 0 – { – 125 ( 3γo– 1 )/( γo– 1 ) } and = 400 MPa

Solving, γo = 11 and so from the definition of γo, Do = √11 Di = 199 mm

(b) The design equation ( viii) will be used, to demonstrate its directness compared to the method of part (a).

Considering the application limits of ( viii) po/pi = 180/360 = 0.5 and < (1+1/γo)/2 for all γo.

So, for this range of applicability, (γo–1) σ* = 2γo ( pi–po) ie. 400 (γo–1) = 2γo (360–180) MPa

Solving, γo = 10 and so from the definition of γo, Do = √10 Di = 190 mmThe advantages of the design equation should be noted - although the correct application range cannot always

be chosen a priori as here, and a couple of trials may be necessary.

(c) The distortion energy criterion is not piecewise defined like the maximum shear stress theory, so in principle

the design equation ( vii) is more immediately applicable than ( viii). Inserting values into ( vii) :

(γo– 1)2 σ*2 = ( pi – γopo )2 + 3 γo

2 ( pi – po )2 becomes (γo– 1)2 4002 = ( 360 – 180γo )

2 + 3 γo2 (360–180 )

2

which leads to the quadratic 19γo2 – 119γo

+ 19 = 0, from which γo = 6.10 and Do = √6.10 Di = 148 mmAs can be seen the second order distortion energy equation poses its own problems.

The physical interpretation of εr and εt is shown in the sketch, inwhich εr = δ∆/∆ is the radial component of strain. When a cylinderis loaded, we are interested in how it 'breathes', how its diameterchanges as pressures are varied. This is measured by the diametralstrain, εd. Circumference is always proportional to diameter, there-fore circumferential strain (which is just the tangential strain compo-nent, εt) is always identical to diametral strain - so from ( x) :

( xi ) εd = δD/D = εt= [ ( 1 – ν ) σ – ν σa + ( 1 + ν ) σ /γ ]/E

Note the distinction between the rather confusing terminology here - diametral strain is the strain ofa diameter, but radial strain is NOT the strain of a radius, it is merely the component of strain in theradial direction, as noted above.

Autofrettage

Thick cylinders use material very inefficiently - referring to the cylinder whose stresses are plottedabove for example, the material at the outer surface is stressed only to 1/9 of that at the bore.Assuming, as is commonly the case, that the tangential stress at the bore is tensile in service, this

∆ + δ ∆ = ∆ ( 1 + ε )r

r + δ r= r ( 1 + ε )t= r ( 1 + ε )d

∆r

LOADED

FREE

Cylinders 5

member τrt must be zero, and at the interface it must equal the friction force developed (which willnot necessarily correspond to friction breakaway, ie. ≠ f pc δA). Geometric compatibility must there-fore be examined - that is equality of the integrated angular displacements of the members at theinterface - which in turn depends upon the relative torsional stiffnesses. The coupling is a rotationalanalogue of the indeterminate tensile connector examined in an earlier chapter.

These considerations lead to the conclusion that interface pressure, stresses and strains all vary inthe axial direction, and this, in conjunction with the presence of both τat and τrt, renders the problemessentially three-dimensional and too complex for analysis at this stage. In practice, the situation isusually further complicated by bending stresses, which, being circumferentially assymetric, lead tofatigue. Shaft design Codes cater for these complexities by means of factors which are based uponexperimental findings and experience.

It is possible to carry out a limited analysis however, if the axial variations noted above and theshear component, τrt, are all presumed negligible - in which case the elastic superposition of con-stant contact pressure and constant torque loadings is feasible. The torsional shear stress at somelocation in the wall is :

( xiii ) τat = Tr/J = τ √γ in which the constant τ ≡ 16 T / π Di3 ( γo

2 – 1 )

The Mohr's circles will be as indicated, the normal stress components being givenby ( 3) and open cylinders presumed. It is evident that the radial direction isprincipal and that resolution in the a-t plane is possible, yielding the principals :

σ1,2 = σt /2 ± √ [ (σt /2 )2 + τat2 ] ; σ3 = σr

Employing the distortion energy theory for example, and expanding these princi-pals via ( 3) and ( xiii), leads to an expression for the equivalent stress :

σe2 = σ

2 + 3 [ ( σ/γ )2 + τ 2 γ ]

It may be shown that, if any turning value of σe exists in the wall ( 1 ≤ γ ≤ γo ), then it must be a mini-mum. The maximum equivalent stress, σ*, must therefore occur either at the inner surface, whenpressure effects prevail, or at the external surface, when torsional effects are dominant. Hence thedesign equation is :

( xiv ) σ*2 = σ 2 + 3 max [ σ

2 + τ 2 , ( σ/γo )2 + τ 2 γo ]

Bibliography

There are no texts known which employ the normalised stress approach as above.

Burr AH & Cheatham JB, Mechanical Analysis and Design, Prentice Hall 2ed 1995 refers to rotatingdiscs, plasticity etcAS 1835, Seamless Steel Tubes for Pressure PurposesAS 1210, SAA Unfired Pressure Vessel Code

R

A

T

12

σ

τ

Thus, knowing the parameters for each cylinder ( γo, E, ν ), the initial interference ( ∆/Dc ) and theloading ( pi , po ), this equation enables solution for the common interface pressure, pc. The staticindeterminacy having been thus resolved, the safety of each cylinder may then be examined.

The stresses in a particular compound cylinder are shown below, after initial assembly and subse-quent internal pressurisation. The stresses are significantly lower than those in a single cylinder ofthe same overall propor-tions and load. The tan-gential stress in the innercylinder is compressive asa result of the as-assembled interference fit,but becomes tensile afterthe internal pressure israised to 80 MPa. Quotedequivalent stresses arefrom the maximum shearstress failure theory.

Note that most texts give an equation for calculating pc only upon initial assembly - such an equa-tion corresponds to ( 4) with the pressures pi and po set to zero.

Torsional Loading

A common application of compound cylinders is in torque transmissionbetween a shaft and the hub of a pulley or sprocket through an inter-ference fit or a proprietary radially-expanding coupler.

The principle of torque transfer may be appreciated from the sketchwhich shows an element δA of the contact area π DcL, on which the con-tact pressure, pc, is assumed to be uniform. The maximum torque capacity ofthe element, δT, may be ascertained if it is assumed that friction breakaway is imminent and hencefriction force = coefficient of friction, f, times the normal reaction. The torque arising from the areaelement is therefore δT = ( f pc δA ) Dc /2 so that the torque capacity T of the whole joint is :

( xii ) T = ∫ A δT = π f pc Dc2 L / 2

It should be realised however, that this is only approximate asthe stress-strain state is much more complex than that onwhich ( 4) is based. The axial cross-section of a coupling illus-

trates that torque transfer is gradualalong the length of the interface. Thisimplies that the torsional shear stress, τat,varies with the axial coordinate z, inaddition to the radial variation. Further-more, the shear component, τrt, must vary through the wall of each member,and the radial direction is no longer principal. At the free surface of each

σt

σt

σrσr

σe σe

60

–60–40

60

00

2040

–20

σtσt

σrσr

σe

σe

120 120

40

80

4040

–80

–40

0σr

σt

σe

180

–80

20

0

100 stresses MPa

COMPOUND CYLINDERSINGLECYLINDER- LOADED

LOADEDASSEMBLED

= = 3γo inner

γo outer

z

TB

TA

BA

T T

T

0

T

L

Dc

f p δAc

p δAc

δA

+1γγo

o–1 + ν

E outer

– ν

E inner+

+1γγo

o–1[ ] Dc

∆( 4) p

c=

2pi

E ( γo–1 ) inner

+γo

2po

E ( γo–1 ) outer

+

T

τr t

τa t

tr

a

Cylinders 6

ø 100

liner

block

9 The underlined figures in the table are known for each steel cylinder. Verify the remaining tab-ulated values. Fa is an axial compressive force applied externally to a cylinder's ends.

========== Inside ========== ======== Outside =======Cylinder Fa d p σt σr εD σ* d p σt σr εD

kN mm MPa MPa MPa *104 MPa mm MPa MPa MPa *104

a open 0 40 120 60 –120 4.6 180 120 40 –20 –40. –0.4b open 0 20 190 125 –190 8.8 315 60 50 –15 –50. 0c open 0 15 120 40 –120 3.7 160 60 45 –35 –45. –1.0d closed 0 60 75 125 –75 6.76 200 120 0 50 0. 2.1e closed 600 60 0 –59 0 –1.4 100 120 22 –37 –22. 0

10 An open cylinder of 50 mm bore is required to withstand internal and external pressures of 55and 45 MPa respectively. Determine the necessary OD if the safety factor is 4 and :(a) the material's yield is 200 MPa; use the distortion energy theory [ 69 mm ](b) ditto - but using the maximum shear stress criterion(c) the brittle material has tensile and compressive ultimates of 200 and 700 MPa. [ 56 mm ]

11 The surface diameters of an open, steel compound cylinder are 100, 180 and 270 mm. The com-ponents are assembled with an interference of 0.05 mm prior to loading with an internal pres-sure of 50 MPa. Plot the tangential, radial and equivalent stresses across the wall of each com-ponent after initial assembly and after loading. [ σ*= 82 MPa ]If a single cylinder had been used in this application (a) what pressure could it withstand if it were the same overall size and the equivalent stress

was limited to 82 MPa? [ 35 MPa ](b) what outside diameter would it need to be to contain the pressure of 50 MPa with a maxi-

mum equivalent stress of 82 MPa?

12 The wall of an air compressor's cylinder liner is 5 mm thick. Whencold, the liner sits in the block with a diametral clearance of 0.05 mm.The temperatures of the liner and the block rise by 150 and 80oCrespectively when on load. If the mean pressure in the cylinder isthen 1 MPa, what is the contact pressure between liner and block -which are both of steel? [ 6.2 MPa ]

13 Two coaxial hollow shafts are to be coupled together by means of an interference fit, and arerequired to transmit a torque of 7.3 kNm. Both shafts are made from a 250 MPa yield steel. Theinner shaft is 100*130 mm (bore*OD), the outer shaft is 130*155, and the effective axial length ofthe interference is 125 mm. Assuming a friction coefficient of 1/3, and neglecting axial non-uniformity :(a) determine the necessary diametral interference [ 0.04 mm ](b) plot the variation of equivalent stress through both walls(c) ascertain the safety factor. [ 4.9 ]

14 The overall diameter ratio of an internally pressurised open steel compound cylinder is Γ =(Do/Di )2. Show that :

(a) the optimum choice of interface diameter is such that : (γo)inner = (γo)outer = √ Γ = 1 /( 1 – npi/Sy )(b) the corresponding optimum interference is : ∆opt = pi Dc /E

CYLINDERS - PROBLEMS

For steel, take E = 207 GPa, ν = 0.3, α = 10.8e-6 peroC, and assume the maximum shear stress criterionunless indicated otherwise.

1 Derive the thin cylinder design equations ( ii).

2 It is required to choose a pipe from the following availability list, from AS 1835-1976, for apipeline pressurised to 20 MPa. Use a design stress of 110 MPa in conjunction with the thin cyl-inder equations to select a suitable size.

Wall thicknessOD 3.6 4.0 4.5 4.9 5.4 5.9 6.3 8.0 9.560.3 X X X X X

76.1 X X X X X X

88.9 X X X X X X

101.6 X X X

114.3 X X X X X

139.7 X X X X X

165.1 X X X X X

3 It has been noted that the thin cylinder equations ( 1), which use the internal diameter, Di, onthe numerator of the stress equations, underestimate the stresses. The equations are sometimesmodified by substituting, instead of Di, either :- the mean diameter Dm = ( Di + Do )/2, as used in AS 1210, or- the outside diameter Do (giving the so-called 'Barlow's formula')(a) Calculate the error using Di, Dm and Do in turn for a closed steel cylinder with Di/t = 25.(b) Derive the basic AS 1210 design formula for internally pressurised cylinders, namely :

t = Di /( 2 σd/pi – 1 ) where σd is the allowable (design) stress.(c) A pressure vessel to contain steam at 2 MPa is designed to AS 1210 with a design stress of

80 MPa. The vessel is 780 mm bore, 2.5 m long and simply supported over a 2 m span. Isbending significant when the vessel is filled with water for an hydraulic test?

4 A tube is made from an aluminium alloy with properties E = 70 GPa, ν = 0.33. The tube ODand wall thickness are 90 and 2 mm. Determine the principal strains on the exterior surfacewhen internally pressurised to 12 MPa. By how much does the OD increase under load?

[ 0.33 mm open; 0.28 mm closed ]

5 A thin-walled steel pipeline of 84 mm OD contains pressurised hot oil.Since it is feared that expansion is giving rise to bending, three straingauges are mounted at the critical location and indicate the strains shown

(in 10-6 units). What is the maximum equivalent stress at the cross-section? [ 208 MPa ]

6 A closed steel cylinder, having bore and OD of 80 and 240 mm, is pressur-ised to 60 MPa internally and 20 MPa externally. Plot the stress variationthrough the wall and calculate the maximum equivalent stress.

[ σ* = 90 MPa]7 Verify the design equations ( viii).

8 Derive design equations, analogous to ( ix), for a brittle closed cylinder. Plot the failure locusfor γo = 4 and m = 3.5 (a typical value for cast iron).

A = 800

C = 500

B = –200

y

x

µ-strain

Pressure Vessels 1

σ lσcLongitudinaljoint subjectedto circumferentialstress- usuallythe MOREcritical

Circumferentialjoint subjectedto longitudinal

stress- usually

the LESScritical

Welded Joint Efficiency

Welded joints are not as strong as the parent plate unless welds are thoroughly inspected and, ifflawed, repaired during manufacture - all of which is expensive. This strength reduction is charac-terised by the weld or joint efficiency η = joint strength/parent strength - which varies from 100%for a perfect weld (ie. virtually seamless) through 75-85% for a tolerably good weld.

Consider the three bars shown below which are of the same width b and material strength S, andwhich are loaded by the tensile load P. The first bar is seamless of thickness ts, and the secondwelded with joint efficiency η and thickness tw locally at the joint. The safety factor in this secondbar is Sbts/P away from the joint and (ηS)btw/P at the joint. If in design these safety factors are to beidentical (ie. the bar is not to be weakened by the joint) then tw must equal ts/η.

However it is most unusual for the economics to justify increase of bar thickness locally in way ofthe joint - the whole bar must therefore be of thickness tw to cater for an isolated joint inhomogene-ity, as shown in the right sketch. The implications of joint efficiency upon cost of raw plate materialsare thus readily apparent, however it should be appreciated that away from the joint region thematerial really needs only to be as thick as a seamless bar.

Pressure vessels are usually made from flat platesrolled into cylinders then welded along longitudi-nal joints. Circumferential joints are used toattach end closures (dished ends or heads) to thecylinder, and to weld together rolled plates for along vessel if plate size availability or rollingmachine capacity are restricted.

Weld types and efficiencies usually differ for longitudi-nal and circumferential joints, and therefore the jointstresses in a vessel must satisfy both the requirements :-

longitudinal joint ( efficiency ηl ), circumferential stress σc = pD / 2t ≤ ηl Scircumferential joint ( efficiency ηc ), longitudinal stress σl = pD / 4t ≤ ηc S

In the design context, having selected the material and hence the design stress from ( i), these ine-qualities may be solved for the plate necessary minimum or calculation thickness, t :-

( 1 ) t = maximum ( pD/2ηlS , pD/4ηcS )

In this formula, AS 1210 takes the diameter to be the mean at the wall mid-surface (D =Di+t) thoughit quotes the equation conveniently in terms of Di. As the circumferential stress is twice the longitu-dinal it follows that the first of these inequalities is usually controlling - provided that ηc is greaterthan half ηl, which is commonly the case. A plate thickness, T such that T–c ≥ t, would be chosenfrom a commercially available range such as the BHP plate size schedule, cited in Problem 3 below.

The Code defines four classes of vessel manufacture - 1, 2 (A&B) and 3 - and provides details of theweld types and efficiencies which are permissible in each class. Some of the major differences

Pt w

S

ηS

P

Sb

t s

St w

t s

P

ηS

η = 1 η < 1 η < 1( i ) ( ii ) ( iii )

PRESSURE VESSELS

Air receivers and LNG tanks are two common examples of pressurevessels. They are usually cylindrical in form and hence employ cylin-der theory in their design, but there are numerous practical aspects which transcend the basic theory- end closures have to be fixed to the cylinder, holes have to be cut and inlet/outlet pipes ( branchesor nozzles) attached, weld flaw probabilities have to be acknowledged, and so on. Some of theseaspects are examined below.

We consider only internally pressurised, welded steel unfired vessels operating at room tempera-ture and above, thus avoiding the complexities associated with

- buckling due to external pressure and manufacturing inaccuracies, and- the dangerous partnership of low temperatures and welding which always requires special

precautions in design and construction.Unfired implies the absence of any energy source in the vessel.

AS 1210 Code for Unfired Pressure Vessels (the Code) forms the legal background to most contractsbetween manufacturer and purchaser; these notes aim also to provide a brief introduction to thisCode - sufficient for the outline design of a simple vessel. One must be careful not to regard AS 1210as a design cookbook - engineering judgement remains a sine qua non, and pitfalls abound for theindiscriminate user. Some of the equations developed below are amended slightly in the Code, andthe nomenclature also differs.

AS 1210 adopts a simple approach to vessel design based on thin cylnder theory and the neglect ofsecondary complexities such as the essential incompatibility between intersecting cylinders. Thisapproach requires substantial safety factors, despite the secondary effects being minor within theCode’s scope. Appendix A stipulates that the design tensile stress of the plate material, S, shall be :-

( i ) S = minimum ( Sy/1.6, Su/4, SyT/1.5, Scr/1.6 )

in which Sy and Su are the tensile yield and ultimate at room temperature, and thelast two terms are applicable only at design temperatures above 50oC, SyT being the yield strengthand Scr the 100 000 hr creep rupture strength at the design temperature. Minimum mechanical prop-erties are laid down in AS 1548 Steel Plates for Boilers and Unfired Pressure Vessels, a steel beingreferenced typically as 'AS1548-2-430-H' in which the '2' refers to Section 2 of AS1548 which dealswith a particular type of steel, '430' denotes the minimum ultimate (MPa), and 'H', if it appears, indi-cates that properties have been verified by high temperature testing rather than being inferred fromroom temperature properties. Properties are slightly thickness-dependent.

Corrosion

In design, corrosion which occurs over the life of a vessel is catered for by a corrosion allowance, c,whose design value depends upon the vessel duty - 1 mm is typical for air receivers. It is importantto realise that when dimensions in any formula refer to a corrodable surface, then the dimensionsinserted into the formula are the dimensions at the end of the vessel's life, when all the corrosionallowance has been eaten away. So, if a plate is of nominal thickness T now, and is subject to corro-sion on one side, then (T–c) must be substituted whenever nominal thickness appears in an equa-tion. Similarly a tube of current bore Di which corrodes will have a bore of ( Di+2c ) at the end of itslife.

Pressure Vessels 2

and δθ. The local surface normal cuts the z-axis at the point B, ABbeing defined as the radius rθ. The centre of curvature lies at C onthe normal, AC being the instantaneous radius of curvature of themeridian, rφ.

The components of the pressure and stress resultants along theoutward normal are :

pressure : p δA = p ( r δθ ) rφ.δφ meridional stress : – 2 σφ ( t r δθ ) sin δφ/2

circumferential stress : – 2 σθ ( t rφ δφ ) sin δθ/2 sin φ

Taking limits and noting that r = rθ sinφ, then equilibrium of the element requires that :-( ii ) σθ /rθ + σφ /rφ = p/t - the membrane equation.

Furthermore, for equilibrium of the dish area above the hoop :-( iii ) π r2 p = 2 π r t σφ sin φ

Solving equations ( ii) and ( iii) gives the stress components in termsof rθ and rφ, which are in turn functions solely of the meridionalgeometry - its shape and location with respect to the rotation axis :

( 2 ) σφ = ( p/2t ) rθ ; σθ = σφ ( 2 – rθ/rφ )

Some specialisations of ( 2) are as follows :-

Cylinder rθ = D/2 ; rφ → ∞ therefore from ( 2)( iv ) σθ = 2σφ = pD/2t ie. the thin cylinder equations once again.

Sphere rθ = rφ = D/2 and so from (2)( v ) σθ = σφ = pD/4t

The sphere is an ideal end closure since the stresses are less than those in other shapes, how-ever the degree of forming necessary renders it impractical except for very high pressureswhen the manufacturing cost may be justified.

Ellipsoid An elliptical meridian ( overleaf ) of semi-major and -minor axes a, b and eccentricity ε= √[1–(b/a)2], is rotated about the minor axis to form the head of the cylinder whose diameteris D = 2a as sketched below. The location of an element on the ellipse is defined by radius rfrom the rotation axis. The geometry of the ellipse may be invoked to derive the radii of inter-est in terms of r, thus defining the variable u for convenience :-

rφ = (a2/b) u3 ; rθ = (a2/b) u where u ≡ √[1 – ε2 (r/a)2] , b/a ≤ u ≤ 1 The stresses at the r-element follow immediately from ( 2) as :-

( vi ) σφ = ( pa2/2bt) u ; σθ = σφ (2 –1/u2) where u = f(r) as defined above.

The stresses are graphed overleaf for a = 2b, the most common proportions for practical ellip-soidal ends. The prominent feature of this stress pattern is the tensile-to-compressive transitionof the hoop stress at about 80%D. The consequences of compressive behaviour include a pro-pensity for local buckling ( see chapter on “Buckling” ) and an increase in the equivalent stress -since the third principal stress ( the radial ) is always zero. But the major outcome is a signifi-cant incompatibility between cylinder and head, due to different senses of the hoop ( ie. diame-tral ) strain - under pressure the cylinder tends to expand, and the ellipsoidal end to contractdiametrically at the junction. Bending moments in the walls of the cylinder and head are there-fore set up at the junction, with corresponding bending stresses. These relatively minor secon-

B

r

z δθ

φ

σθσφ

p

centre of curvature of meridian at element

elementnormal

intersection of element normal

and z rotation axis

A

C

r = ACφr = ABθ

δφ

σφφ

2rp

between the classes are summarised in the table above; other require-ments such as certification of welders' skills need not be gone intohere. The sketch indicates the area of applicability of the classes - themost expensive Class 1 vessels are mandatory if failure is potentiallylethal, no matter what the thickness. Class 2 vessels having a wallthickness greater than 32 mm are inadmissible, and so on. Steel vessels arehydrostatically tested after manufacture to 150% of the design pressure.

It is therefore apparent that in the design process, the selection of class (if scopefor choice exists) and of weld type involves an economic balance between welding costs ( includinginspection and repairing flaws during manufacture, if carried out ) and the cost of plate - a categori-cal decision cannot be made until the relative cost of all potential candidates can be compared.

Welds should preferably not be situated adjacent to anomalies such as stress concentrations or inho-mogeneities due to other welds, especially major structural welds. Hence the staggering of the twolongitudinal welds in the sketch above.

Thin Shells of Revolution - Heads

The stresses in pressurised thin axisymmetric shells of revolution arenow considered so that the behaviour of dished ends may be appre-ciated.

A shell is formed conceptually by rotating the meridian, a curved lineof selected shape lying in the r-z meridional plane , aboutthe z-axis. The resulting surface of revolution is clothed bya small, symmetrically disposed thickness,t, and the resulting shell loaded byinternal pressure, p. If themeridian were a straight linefor example, parallel to the z-axis and distant D/2 from it,then the shell would be a cyl-inder of diameter D.

These shells are similar to thin cyl-inders in that radial stresses are negligibleand the membrane stresses

- the circumferential or hoop stress σθ (that is σt of cylinder theory) and - the meridional stress σφ (analogous to σa of cylinder theory)

can be found from equilibrium since they, and the loading, are axisymmetric.

Consider the element located at the point A in the r-z plane as shown below, and defined by φ, δφ

0 12 32vessel maximumthickness, mm

conse-quencesof failure

minor

major class 1

2

3

σθσφ

z

r

θ

meridionalplane

hoopcircle

meridian

datum

element normal

φ

Class of vessel 1 2A 2B 3Plate thickness, T mm no limit T ≤ 32 T ≤ 32 T ≤ 12ηl (double butt mandatory) 1.0 0.85 0.75 0.65ηc -------- depends on weld type, usually less than ηl --------Heat treatment required ----------------- not required ------------------Radiographic examination required spot ------- not required -------

Pressure Vessels 3

- dimensions being reckoned in the fully corroded condition. The stress-transmitting area removedis A =Dbt where the calculation (necessary minimum) thickness t is given by ( 1).

The figure on the right demonstrates compensation for area removal by providing equal area foralternate force paths in otherwise unused material of cylinder and branch. Not all the branch wallcan be devoted to compensation since the branch is a cylinder in its own right, with calculation andnominal thicknesses, tb and Tb, determined in a manner identical to the main shell.

Provided that the longitudinal welds in both shell and branch do not lie in the critical longitudinalplane then - from a compensation point of view - both t and tb would be reckoned from ( 1) with η =1. The thickness differences (T–t) and (Tb –tb ) contribute to compensation - though reinforcement isineffective beyond the limits Ln normal to the vessel wall, and Lp from the branch centreline parallelto the wall, as shown for the set-in branch protruding through the wall :-

AS 1210 gives the limits as :-

( ix ) Ln = maximum [ 0.8√(DbTb) +Tr , minimum ( 2.5T, 2.5Tb+Tr ) ]or √( DbT) for a flanged-in head

( x ) Lp = maximum [ Db , Db/2 +T +Tb +2c ]

Usually the first of the Lp limits, Db, controls; however a compensating area cannot contribute tomore than one branch, so if the spacing of two branches Db1 and Db2 is less than (Db1+Db2), then byproportion Lp1 = Db1*spacing/(Db1+Db2).

Furthermore, if the branch is attached to a dished end, then no compensation area is effective if itlies outside the aforementioned 80% limits. If the head is torispherical, the hole should lie in thespherical portion and 't' will be given by ( v). If the head is ellipsoidal, then AS 1210 defines anequivalent sphere for the application of ( v), since the hole will not lie close to the rim region ofsharp curvature which dictates the head thickness via the stress concentration factor in ( vii).

material nototherwise necessary which

may be used for compensating

Tt

σc

Db

necessary material lostwhen hole is bored -to be compensated

material necessary to transmit circum-ferential stress of magnitude S

additional reinforc-ing ring may or may not be necessary

compensating material

4

Tbnecessary for transmission of pressure-induced hoop stress

compensatedmaterial

Section in Critical Longitudinal Plane

bt

tT

A2

nL

nL

Tr

1

2

3

5

D2

b effective region

pLlongitudinal welds outsidecritical longitudinal plane

dary stresses cannot be explainedby membrane theory, but otherwork shows that they becomeinsignificant at a distance of aboutfive times the wall thickness fromthe junction.

Tensile strains in the cylinder atthe junction tend to relieve thecompression in the dished endand for this reason the stress con-centration factors K cited in AS1210 are less than might beinferred from the above graph -for example :-

( vii ) S ≥ K pD/2tη ; K = [ 2 +(a/b)2 ]/6 - ie. K = 1 and not the graphed 1.5 for a/b = 2.

Torisphere Another common shape for a dished end is torispherical.This consists of a spherical central portion of radius R and a tor-oidal knuckle of radius r, where R/r is often 12 or thereabouts, andR is about 95 % of the cylinder diameter. Junctions of the torus withboth sphere and cylinder give rise to geometric singularities andhence to secondary bending stresses as discussed above. The AS1210 stress concentration factor, M, reflects this behaviour :-

( viii ) S ≥ M pR/2tη ; M = [ 3 +√(R/r) ]/4

- the greater the deviation from a sphere (R/r = 1), the larger the factor. Noticeably, thehighly stressed region again extends outwards from about 80% D. Torispherical ends are oftenpreferred to ellipsoidal since the depth of drawing is less and hence they are slightly cheaper -about 10 % on average - but this is often outweighed by their higher stress concentration andconsequent lesser allowable pressure.

Dimensions are taken to the wall mid-surface in ( vii) and ( viii). If a dished end is seamless, as isusually the case, then theoretically η = 1. However the pressure ratings in one commercial brochureare based on η = 0.875 for thicknesses ≥ 25 mm and 0.85 below, to allow for the thinning whichresults from drawing. Heads are always provided with a short flat, ie. the heads are manufacturedwith a short integral cylinder, to avoid the junction weld coinciding with stress concentration due tothe head-to-cylinder incompatibility and the small head radii of curvature.

Flat plate end closures are not suitable in larger sizes, though often used for doors - being flat, thereare no membrane stresses and pressure is resisted solely by plate bending. We have noted alreadythat, for a given load, bending stresses are generally much larger than direct (eg. membrane)stresses. Flat plates therefore have to be much thicker than dished ends for similar duty.

Compensation

Compensation ( or reinforcement ) is the provision of extra stress-transmitting area in the wall of acylinder when some area is removed by boring a hole for branch attachment, as sketched opposite.

The left sketch shows part of a cylinder's longitudinal section; the major circumferential stress actsacross the critical longitudinal plane. The nominal thickness is T, and a hole of diameter Db is bored

r

R

r D2

10 0.2 0.4 0.6 0.8 1.0

1.0

0.5

0

-0.5

crow

n

tensile

compressive

cylin

der

junc

tion

σe

σθ

σφ

σφ

σe

&

STRESSES IN A 2:1 (a:b)SEMI-ELLIPSOIDAL HEAD

σpD/2t

D

b

r

rφ

a rθ

0.816

Pressure Vessels 4

branch

bridge

stud

gasket

fluid

spigot

cylinder shell or head

door

INSIDE OUTSIDE

details are given in the chapter on “Threaded Fasteners”.

The higher a fluid’s pressure and temperature the more robust must be the pipes, flanges and bolt-ing to contain it. AS2129 Flanges and Bolting for Pipes, Valves and Fittings contains tables ofdimensions for flanges and other pipe fittings. Each Table ( D, E, F . . . T ) lists for the various nomi-nal pipe sizes the minimum dimensions which are suitable for a certain limiting combination offluid pressure and temperature. Thus for steels in the range -50 to 232 oC the maximum pressure pFT

permitted by each flange Table is :

Table D E F H J K R S TpFT (MPa) 0.69 1.38 2.07 3.45 4.83 6.20 8.27 12.4 19.3

These limits are halved linearly from 232 to 427 oC and drop tozero at 532 oC as sketched. So, when steel fittings have to beselected for a design pressure p and temperature T, their size isdictated by a Table for which :

( xiii ) pFT ≥ p * maximum ( 1 , 390/( 622–T) , 210/( 532–T) ) ; T (oC) < 532

A small bore pipe is often demountably attached to a vessel by means of a flange butt-welded to thepipe and mating with a pad which is welded into or formed on the vessel wall, ( c ) above. Clearlythe pad dimensions must match the flange. Studs are screwed into holes tapped in the pad, sincethrough-bolts like those of ( b) would allow leakage. Acceptable pad forms are defined in AS 1210.Pipe wall thicknesses, if required, may be found from ( 1) using design stresses for pipe materialscited in AS 1210, in conjunction with manufacturers' lists or AS 1835.

Inspection Openings

Openings are required to monitor the condition of the vessel's interior if subject to corrosion, andmay be necessary also for manufacture. The size and disposition of the opening(s) depend upon theduty and size of the vessel - in a small vessel a single handhole or a flanged-in inspection openingmay be adequate whereas a large vessel requires an elliptical manhole, often with a reinforcement/seating ring, though heads may be flanged inwards (reverse knuckle) to provide a seating surface.The minor axis of an elliptical opening in a cylindrical shell should lie parallel to the longitudinalaxis of the shell.

The opening is sealed by aninternal door, a gasket and oneor two bridges and studs. Thedoor is elliptical to permit itsremoval if necessary for remach-ining a damaged gasket seatingsurface.

The studs provide the initial seal-ing force, ie. the initial seatingpressure on the gasket facebefore the fluid is pressurised.When the fluid pressure laterrises, the door tends to be self-sealing as the pressure load on the door increases the gasket contactpressure. The load on the studs therefore decreases, however the Code specifies that the door mustwithstand simultaneously bending by both fluid pressure and maximum possible stud (or bolt)

p

532–50 232 427temperature T ( C)o

acceptable

region

AS 2129 FLANGETABLE LIMITS

pressure, p

p2

Within the Ln , Lp limits, compensation requires that :-

( xi ) A1 + A2 + A3 + A4 + A5 ≥ A = Db tThe inward protrusion '3' is subjected to corrosion on three surfaces but there is no pressure diffe-rential across it; it will not exist for a set-on branch. The sketch indicates that:-

( xii ) A1 = ( 2Lp –Db –2tb ) ( T – t ) ; A2 = 2 Ln ( Tb – tb ) etc.

These equations should be compared carefully with the corresponding equations in AS 1210 to clar-ify the restrictions implicit in the Code.

Compensation should be disposed symmetrically about the hole and as close to the hole as possible.It is usually more economical to increase the branch thickness than to provide a separate reinforcingring, however such an increase should not be excessive.

The principle of compensation is very simple and ignores inevitable stress concentrations - hence thesubstantial safety factors mentioned above. AS 1210 lays down branch size upper limits beyondwhich this simplistic approach is no longer permissible, namely :-

Maximum branch bore = minimum ( Di /2, 500 mm) if Di ≤ 1500 mm, or= minimum ( Di /3, 1000 mm) if Di > 1500 mm

Strictly, reinforcement ought to be checked in all planes which contain the branch centreline; AS1210 defines A = FDbt where F is a factor which varies between unity for the longitudinal plane con-sidered above, and 1/2 ( ie. σa/σt ) for the transverse plane. Generally such a check is unnecessary.

A single opening 75 mm diameter or less, in plate thicker than 6mm does not require compensation. Short pipes of this size aresometimes welded straight into the shell as in the photograph,however not only is the pipe prone to damage in transport but ifthe pipe thickness differs appreciably from the shell thicknessthen welding could be awkward.

Pipes and Flanges

Long pipe runs which convey fluids between items of plant such as pressure vessels are usuallymade up from short lengths of pipe welded together, however the pipes are not welded permanen-etly into the vessels but rather attached in a manner which permits easy separation. One of the sim-plest demountable attachments is the screwed connection ( a) though these are suitable only for rela-tively low pressures due to the difficulty of making good a leaking thread.

The most common demountable connection involves welding a flange to each of the pipes, then con-necting the flanges by nuts and bolts ( b). A gasket of some relatively soft material is interposedbetween the flanges to fill any imperfections in their faces thus preventing fluid leakage - further

pipe welded to AS 2129 welding neck flange

pad / disc matching welding neck flange welded into shell wall

studs screwed into blind tapped holes in pad

gasket

pipes butt-welded to AS 2129 welding neck flanges

nuts & bolts gasket

( b )( c )

( a )

internally threaded coupling

pipes with externally threaded ends

Pressure Vessels 5

The design of a complete pressure vessel is complex, especially if the vessel is a heat exchangerwhere wide scope exists for the choice of pass and tube layouts. For a straightforward containerhowever the scope is more restricted and the design procedure simpler. If it is not clear where tostart a design (of pressure vessel or anything else) then a start can be made anywhere, realising thatthe first pass through the design process will be something of a trial run. The skeleton procedure setout overleaf for a container-type pressure vessel is recommended as a starting point.

There are two extreme approaches to the use of such a design procedure. The first approach ismechanistic, suitable for computer implementation, and involves the mindless selection of a numberof alternatives for each parameter. Thus, following the procedure, one might select both extremes ofdesign pressures, all four classes, all available heads etc. Each of the resulting large number of candi-date solutions would incorporate a unique combination of all parameters. Only a few of these candi-dates would on later analysis be found practicable. Of these, the optimum solution would be thecheapest - within the usual constraints of safety over the vessel's life, consumer acceptance &c.

The apparent drawback to this approach - apart from stifling any individuality on the part of thedesigner, and the large bank of candidates - is the difficulty in obtaining realistic cost figures for allthe alternatives; normal commercial confidentiality precludes it. Head costs are known to be approx-imately proportional to thickness and to (diameter)7/4, but this is of little use as heads are only a partof the total vessel. Graphs in Sinnott op cit suggests that in very general trms, cost ∝ diameter1.05

length0.85; however one should not read too much into this as it does not refer to one fixed pressure.Such approximate generalisations notwithstanding, it is possible to formulate simple realistic quan-tifiable relative cost utilities - on the basis of known trends in relative welding costs for example.Thus it is possible to compare relatively but realistically the costs of the two tabled candidates incor-porating Rheem semi-ellipsoidal heads for a 14.5m3 capacity class 2B vessel whose plate widths arelimited by the rolling machine’s 2 m capacity. What criteria would you adopt for this comparison?

head cylindercandidate bore thickness length thickness

1 2134 25 3200 282 1372 16 9200 20

A second design approach comprises a single pass through the procedure, at each step making a sin-gle choice of the corresponding parameter, and leading to a single solution. As the design unfoldshowever, the designer continually questions the effects of earlier decisions, and if a change appearslikely to be beneficial then the designer must double back to the earlier stage, confirm whether ornot the change is indeed beneficial, and if it is then effect the change. In the hands of an experienceddesigner such an iterative approach can lead to a close-to-optimum solution, however the novicedesigner often cannot weigh up the probable outcome of an altered earlier decision, so the novicedoes not iterate adequately and ends up with a far-from-optimum solution. With experience theneed for iterative review lessens and the order of the procedural steps may profitably be altered.

An inadequate search is a common failing with novices' designs - not just of pressure vessels. Forexample the required thickness of a shell may work out to be 12.4 mm, so novices select the nexthighest available plate - 16 mm - rather than seeing whether an earlier decision can be altered so thatthe required thickness is 12.0 mm, allowing selection of standard 12 mm plate. The search and alter-ation to an earlier decision saves not 0.4 mm but ten times this - 25 %.

The approach recommended is not unlike the first approach, however the choice of parameters ateach step is both intelligent and limited. Thus at the procedural step for selecting the parameterclass, the designer might know from the vessels' general specification that class 1 is too sophisticatedand expensive, and that class 3 is impractical owing to the vessel's likely diameter and hence thick-ness over 12 mm. So the limited range of 2A and 2B might be selected as candidate attributes,

tightening. The flat door calculation thickness t is thus given by :-

( xiv ) ( C1 * fluid pressure * door area + C2 * bolt stress * bolt area ) / t2 ≤ S ; C1, C2 constant

The door is equipped with a locating spigot to aid its engagement when closing. If the door is heavythen provision must be made for supporting it during opening or closing - any such support mustnot interfere with even take-up of the gasket, nor must it hinder easy access to the vessel. Thedesigner of the door support must visualise the door's detailed operation.

The choice of gasket material depends upon the vessel duty - fluid, temperature and pressure - andthe flanges' surface finish and rigidity. The stiffer the gasket, the greater must be the initial seatingforce and hence door thickness.

Allowable stresses for studding materials are quoted in AS 1210, which stipulates that the core areasand not the effective stress areas should be used in stress calculations :

Screw size (mm) M8 M10 M12 M16 M20 M24 M30 M36Core area (mm2) 32.8 52.3 76.2 144 225 324 519 759

Bridges, which must be weaker than the studs, are designed as simple beams with cross-sectionalproportions of depth/width ≥ 3 if rectangular.

Supports

A horizontal pressure vessel (length L, diameter D mm) is commonly mounted on two saddle sup-ports - more would result in static indeterminacy and difficulty in predicting the load distribution inthe event of foundation settlement. Each support should extend at least 120o around and approxi-mately √(30D) along the vessel [BS 5500] in order to transmit the reaction gradually into the shellwall. One support is attached to the vessel to prevent axial movement, the other is not attached butmerely supports the vessel's weight, thus permitting free longitudinal expansion of the vessel whenthermal strains occur.

The safety of any artefact must be verified under all possible circumstances - not just in normal dutybut also during manufacture, erection, test, aberrant service and so on. Under hydrostatic test forexample, a pressure vessel is subjected to a superposition of loads - internal pressure plus bendingdue to the distributed weight w of shell and water charge. A simply supported beam model of thevessel indicates that the supports are optimally located at 0.207L from the ends, corresponding tobending moment magnitudes of 0.0214wL2 at both the centre and at the supports - but the modelneglects possible distortion which may occur when concentrated loads are applied to the relativelythin shell of a pressure vessel. To lessen this possibility, supports should be situated within D/4 ofthe ends to take advantage of the stiffening afforded by the heads, although this location will lead tobending stresses larger than those arising from the optimum location.

Design

The lessons presented earlier in these Notes about design generally are relevant when embroiled inthe details of pressure vessel design. It should be re-emphasized in the present context that :

- Design is the search for the optimum solution, not the verification of any old solution.- Design is undertaken by stating the problem completely, creating a bank of solution candi-

dates, setting up clear criteria and constraints, practicalising the candidates and finally evaluat-ing them.

This applies not just to the design of the overall vessel but also to the design of details like the doorgasket of the inspection opening, the attachment of branches to the main shell, and so on.

Pressure Vessels 6

instead of all four classes as in the first approach. Clearly this leads to a more manageable bank ofcandidates than that resulting from the first approach. Using the cheapest suitable material, simpleweld preparation and the like will all promote the optimum solution. Selection of the optimum solu-tion will consequently also define the design pressure, the class etc. because these parameters areintrinsic attributes of any candidate.

Don't finalise anything until ALL (available?) evidence is to hand.

It should be realised that one's decisions are always subject to criticism by superiors or by those whohold the purse-strings, so any decisions must be capable of rational justification. Designers worththeir salt should always question the input specification for the design process, since actual con-straints may in fact not be so rigid as they first appear.

In the absence of a specialised computer program, spreadsheets can assist greatly in the design taskbecause they foster consideration of a bank of solution candidates more comprehensive than couldever be attempted by hand. It is vital to accompany each spreadsheet with one complete set of clearmanual calculations, both as a check for the designer on the spreadsheet's correctness and as a guidefor the reader on the logic and equations used. A spreadsheet appearing in a report should be editedto remove the bulk of confusing irrelevant data especially unnecessary/downright misleading insig-nificant figures, and formatted so that like tabulated values appear with the same number of deci-mal places. It must be stressed that spreadsheets do not absolve the designer from the need to think.

Design is an exercise in trying to keep everybody happy - the welder, the user, the production man-ager, the salesman, the cleaner, the repairer and so on - in fact everybody who is or who might beaffected by the design. This includes the reader of the design report. Designers must put them-selves in these peoples' shoes and pay due attention to their wishlists.

With regard to the presentation of a design report, the prospective reader (who is probably busierand more senior than the designer) should be offered an easily assimilable document - a report mustbe short & sweet. The following layout is suggested for a pressure vessel report :

- the title page, including at least the addressee, the author(s) and the date;- the Executive Summary appears on the next page: this outlines in a paragraph what the report is

all about, and tabulates the major features of the chosen solution;- the list of contents with correct pagination appears on the next page;- then follow word processed pages which summarise the design process and attendant logic;- the drawings if bound with the report come next, noting that :

- all drawings must be to accepted standards, and sketches must be approximately to scale- the general arrangement (GA) is of particular interest to the client and must show all major

dimensions so that the vessels' integration with other plant can be quickly appreciated;- the report is concluded by appendices, each containing longhand calculations on a major

aspect of the design, with significant decisions in the RH margin ( see attached sample calcula-tion sheet ) to assist both the reader and the designer(s).

Results of similar calculations for a number of candidates should be tabulated, with a single speci-men calculation only being provided. Code equations must be used wherever relevant.Sophisticated maths should not be over-used (unless absolutely necessary); it is better to demon-strate a practical grasp of the problem using simple approximations rather than to indulge in mathe-matical complexities.

All members of a vessel design group should contribute to the major task, that of selecting the ves-sel's overall parameters such as bore, class etc. Individual members of the group may consider sep-arate features such as inspection openings, compensation and the like - remembering always theneed to question earlier decisions.

SKELETON PROCEDURE

for

PRESSURE VESSEL DESIGN

SPECIFICATION

Duty ( fluid, steady operating temperature and pressure )Capacity, and possibly limits for diameter and lengthSize of branchesLife

PROCEDURE

1 Select design pressure, say 5-10 % in excess of operating pressure2 Select material - generally the cheapest most readily available steel (eg AS

1548-1-430) consistent with dutyAdopt a suitable corrosion allowanceDetermine design stress, equation (i)

3 Select class if choice exists (fluid) - generally opt for cheapest class 3 if possibleDetermine longitudinal joint efficiency

4 Select circumferential weld type for head connection - butt or fillet5 Select head shape and size from available range

Determine head calculation thickness, equation (vii) or (viii)Determine vessel bore consistent with head and end joint

Check that diameter lies within specification limitsCheck branch-to-shell bore limits

Determine calculation thickness of shell, equation ( 1)Determine vessel length from capacity

Check that length lies within specification limitsCheck that length/bore ratio is feasible, say between 1.5 and 6,

though higher values may apply to heat exchangersCheck plate size availability and bending machine capacity

6 Select form, size and location of inspection openings7 Select thickness of shell, heads and branches from available ranges consistent

with calculation thicknesses and corrosion allowanceCheck that class limits are not violated

Determine compensation needs, if any8 Select details of major welds9 Select details of small bore pipe attachments, supports etc

10 Determine total cost of solution

THE UNIVERSITY OF WESTERN AUSTRALIA project final pageDept. of Mechanical and Materials Engineering temp pagecomponent item comps by

date

conclusions

BibliographyAustralian Standards (SAA: Standards Association of Australia) :

AS 1210 : Code for Unfired Pressure VesselsAS 1275 : Metric Screw Threads for FastenersAS 1548 : Steel Plates for Boilers and Unfired Pressure VesselsAS 1835 : Seamless Steel Tubes for Pressure PurposesAS 2129 : Flanges and Bolting for Pipes, Valves and FitttingsHB1 : Technical Drawing for Students

Manufacturers' Information :Heads, Rheem Australia LtdSize Schedules for Plates and Wide Slabs, The Broken Hill Pty Ltd Sealing with Security, Richard Klinger Pty Ltd gasket materials

Further Reading (most design handbooks are too advanced for undergraduate work) :Bednar HH, Pressure Vessel Design Handbook, Krieger 1991BS 5500:1991 Unfired Fusion Welded Pressure Vessels, British Standards Institution 1991Burr AH & Cheatham JB, Mechanical Analysis and Design, Prentice Hall 2ed 1995 - general intro-

ductory text includes some discussion on shells, etcChilds P, 'Take the Guesswork out of Gaskets', CME, November 1981, 23-29Czernik DE, Gaskets: Design, Selection andTesting, McGraw-Hill 1996 - comprehensiveFlitney RK et al, The Seal User's Handbook, BHRA 1984 - gives an introduction to seals, both static

(eg gaskets) and dynamic (eg face seals for rotating shafts)Goodall PM ed, The Efficient Use of Steam, IPC Science & Technology Press 1980 - a very useful

and readable general reference Harvey JF, Theory and Design of Pressure Vessels, van Nostrand 1991Hewitt GF ed, Hemisphere Handbook of Heat Exchanger Design, Taylor & Francis 1990McKetta JJ ed, Piping Design Handbook, Dekker 1992Moss DR, Pressure Vessel Design Manual, Gulf 1987Nichols RW ed, Developments in Pressure Vessel Technology, vols 1-5, Applied Science 1979-87

- vol 5 is a useful commentary on world-wide codes Singh K & Soler A, Mechanical Design of Heat Exchangers, Arcturus 1984Sinnott RK, Chemical Engineering , vol 6, Pergamon 1985 - includes generally useful information on

heat transfer equipment, pressure vessels, piping etc.South DW & Mancuso JR ed, Mechanical Power Transmission Components, Dekker 1994 - useful

overview of gaskets and shaft sealsSpence J & Tooth AS ed, Pressure Vessel Design: Concepts & Principles, Spon 1994 - provides use-

ful background to codes (UK particularly)Stephens RI ed, STP 1250: Case Studies for Fatigue Education, ASTM 1994 - invaluable insight into

practical application of fatigue/fracture techniques to a variety of componentsThompson G, An Engineer’s Guide to Pipe Joints, PEP 1998 - a mainly descriptive introduction

Texts - more advanced classics - on shells, flat plates and the like :Dym CL, Introduction to the Theory of Shells, Hemisphere 1990Flügge W, Stresses in Shells, Springer-Verlag 1973Ford H & Alexander JM, Advanced Mechanics of Materials, Wiley 1977Novozhilov VV, The Theory of Thin Shells, Goningen 1959Timoshenko S & Woinowsky-Krieger S, Theory of Plates and Shells, McGraw Hill-Kogakusha 1959Urugal AC, Streses in Plates and Shells, McGraw-Hill 1999 includes pressure vessels applications

Pressure Vessels 8

PRESSURE VESSELS - PROBLEMS In the following, reference to Australian Standards implies use of the extracts appearing above.

1 The equation of a parabola in the meridional plane is 2cz = r2 where c is a constant. Show thatthe membrane stresses in a paraboloid of revolution, of wall thickness t and subjected to apressure differential p, are :-σφ = k ( r2 + c2 )1/2 ; σθ = k ( 2r2 + c2 )/( r2 + c2 )1/2 where k = p/2tAn observation dome for an undersea vehicle is made of glass with an allowable compressivestrength of 35 MPa. If the dome geometry is defined by c = 60 mm and rmax = 150 mm, howthick should it be if the vessel is to operate at depths up to 200 m? The density of sea water

may be taken as 10 kN/m3. [ 8.6 mm]

2 A conical tank is supported circumferentially around its periphery and filled with a liquid ofdensity ρ to a height h. Derive the membrane stress equations :-σφ = k( 3h–2z)z ; σθ = 6k( h–z)z ; where k = ρ sinα /6t cos2αnoting that ( iii) must be modified for liquid weight in the apex.A steel tank of this form and semi-angle α = 45o contains water to adepth of 6.5 m. If the design stress is 65 MPa based on the distortionenergy failure theory, calculate the wall thickness necessary and thediametral strain at the water level. [ 2 mm; –71 µstrain]

3 An unfired pressure vessl of 920 mm bore and 2.5 MPa operating pressure has to be manufac-tured to AS 1210. A design pressure is chosen 10% in excess of the operating pressure and theinternal corrosion allowance is 2 mm. The design stress of the steel selected is 108 MPa and thematerial is available in plate thicknesses :- 3 4 5 6 8 10 12 16 20 25 28 32 36 40 45 50 55 60 70 80 mm(a) Determine a suitable plate thickness, given that the longitudinal and circumferential joint

efficiencies are 85 and 75% respectively. [ 16 mm](b) What should be the thickness of a seamless ellipsoidal head (a/b = 2) to suit ? [ 16 mm](c) Calculate the increase in bore at the operating pressure, measured separately on the cylin-

der and on the head, assuming each is not constrained by the other. [ 0.28, –0.37 mm](d) Repeat (b) with a torispherical head of common proportions. [ 25 mm](e) Design a 250 mm bore nozzle set into the cylinder wall, and reinforce if necessary.(f) Repeat (e), but the branch is set onto the head of (d).

(g) Which flange Table from AS 2129 should be used for pad design if the design temperature

of the vessel is ( i) 350oC (ii) 450oC ? [ J; R](h) A 330 *230 mm elliptical opening in the shell is closed by a flat door whose gasket is effec-

tively 7 mm wide. The minimum seating pressure (the 'y-factor') is 10 MPa. If the designstress for the studding material is 100 MPa, determine the size and number of studsrequired. [ 2off M24]What plate thickness should be used for the door if the constants in ( xiv) are C1 = 0.4 & C2

= 0.8 ? [ 32 mm](j) Design bridges suitable for the above opening.

h zα

σφ

4 The table below presents alternative solutions which are proposed by a pressure vessel designprogram, for a 6.8 m3 pressure vessel with design pressure and stress of 1 and 95 MPa respec-tively; incorporating 1 mm corrosion allowance and 350 mm nominal size branches which donot require additional reinforcing rings. The ends of class 3 vessels are attached by double filletwelds. Verify a few of the proposals at random, checking in particular the adequacy of intrinsicreinforcement.

Solution Class RHEEM Head Nominal thickness

diameter form head shell branch

1 3 1118 i.d. s.e. 12 12 52 " 1137 i.d. tor. " " " 3 " 1143 o.d. tor. 10 " " 4 " " " " 12 " " 5 1 1219 o.d. tor. " 8 12

6 2A " " " " 10 107 2B " " " " " " 8 3 " " " " 12 89 1 1219 i.d. s.e. 10 8 12

10 2A " " " " 10 10

11 2B " " " " " " 12 3 " " " " 12 813 " " " " 12 " "

Squirrel Cage Motors 1

Squirrel cage motors are not without their drawbacks, notably during starting when currentdrain is high. Mechanical aspects will be stressed here - we are inter-ested mainly in how to select a motor to drive a given mechanicalload - but it is essential to appreciate in broad terms the thermal-electrical behaviour when selecting a motor, since the windingtemperature dictates the life that results, as suggested by thesketch. It is the motor's heat dissipation capability dictated byheat transfer and the integral fan which to a large extent deter-mines the motor's maximum continuous mechanical powerrating for winding temperatures commensurate with anacceptable life.

Characteristic of a Steady Load and of a Motor

A load needs a torque T to drive it at speed nL. A load which is predomi-nately frictional in nature requires a constant ( speed-independent ) torque;conversely a hydrodynamic machine such as a centrifugal fan or pumpneeds a torque which varies approximately as the square of the speed. Before an economicmotor can be selected to drive a load, the load's steady state torque-speed characteristic should beknown - this may be generalised by :

( 1a ) T = T0 + T1 (nL/no) + T2 (nL/no)2 + . . . etc

in which no is some convenient referencespeed and T0, T1 and T2 are constant properties of theload, dictated by the type and size of the load. We willneglect third and higher orders here. Most practicalloads are modelled by two terms at most in ( 1a), thus acharacteristic with non-zero T2 and with T0 = T1 = 0represents an axial flow fan's approximate behaviour.

More often than not a load is driven through a speedreducer based on gears, or belt & pulleys, chain & sprockets or other mechanism. A speed reduceris usually provided not to reduce speed but to increase torque. Thus a car's gears eg. are necessaryto amplify engine torque and so provide high torque at the road wheels to accelerate the car or tomove it uphill against gravity - if speed reduction were the only goal then it could be achievedwithout a speed reducer simply by easing up on the accelerator.

If a load is driven through an ideal speed reducer of ratio R ( ≥ 1)then the speed of the reducer's input shaft is n = nL*R, whilethe torque on this shaft necessary to drive the load is TL = T/Rsince the power fed into the ideal reducer ( ∝ nTL ) must equalthe output power ( ∝ nLT ) transferred from the reducer to theload. The steady state torque -speed characteristic of the loadreferred to the reducer's input shaft is therefore, from ( 1a) :

( 1b ) TL = { T0 + T1 ( n/Rno ) + T2 ( n/Rno ) 2 } / R

The losses from practical speed reducers are normally around 5%, though gear efficiency canexceed 99% whereas the efficiency of a hydrostatic speed reduction may not reach 70%. Losses areneglected in ( 1b). The effects of cyclic load variation are examined below.

Turning now to the motor - a squirrel cage motor may be regarded simplistically as a stator

electricalpower input

useful

LIFE

mechanicaloutput

electricalI R losses

ie. heat

temperaturerise of

windings

heatrejected tosurrounds

2

torq

ue, T

speed, nL

on

Ln

1T

0T

referenceLOAD CHARACTERISTIC

2T

n , TL

input

speed reducer

load

T , nL

R

T , nL

load

SQUIRREL CAGE MOTORS

Before the widespread industrial use of electric-ity, the individual machines in a manufactoryfor example were powered by flat leather beltsfrom a roof -mounted network of rotating shaftsand pulleys, driven in turn by a centralisedpower source - perhaps a waterwheel or a steamengine. The whole power distribution systemwas mechanical. Napier & Sons engineeringworks shown in this photograph circa 1910 wastypical in resembling a forest of belts whichwere inefficient, unreliable, dangerous and lim-ited in the power which could be transferred toa single machine.

We take it for granted nowadays that we can power a machine by its own dedicated electricmotor, often energised from the AC mains. Indeed it is difficult to imagine how modern indus-trial plant - say an extensive conveyor system - could operate satisfactorily if powered through-out by mechanical energy from a centralised engine.

This chapter on squirrel cage electric motors, together with the following chapters dealing withV-belts, gears and brakes, concentrates on the problem of safely powering machines. The electri-cal theory which underpins the motors is hardly mentioned - we concentrate on the mechanicalaspects.

Squirrel cage motors get their name from theappearance of early rotors. They are the mostcommon type of industrial AC electric motor,being rugged and requiring neither a separateDC power source nor slip-rings. They are essen-tially constant speed devices when supplied byconstant frequency AC, however elec-tronic speed control is available.

The majority of industrial squirrel cagemotors are foot mounted Totally EnclosedFan Cooled (TEFC) as shown here, inwhich the motor internals are isolated from the sur-rounding environment by shaft seals etc, thus minimising

the ingress of dust and moisture. The inevitable heat generated by I2R losses in the internalwindings is transferred to the surrounds by air circulation within the casing together with cool-ing air blown by an integral fan along fins on the casing's exterior.

A variety of forms other than TEFC meet specialised needs- motors may be dust protected, ignition proof etc. A flange-rather than foot- mounted motor with shaft extension isillustrated. Integral brakes or gearboxes also are available -suppliers' brochures should be consulted for full details.

Squirrel Cage Motors 2

rated full full torque ratios full rotormotormotor cont. load load start break load inertiatype frame output speed torque -ing -down eff'y J

size kW rpm Nm Ts/Tf Tb/Tf % kg.m 2

MT 63 B 0.18 1370 1.25 2.5 3.0 54 2.8E-4MT 71 A 0.25 1400 1.7 2.5 3.0 60 7.3E-4MT 71 B 0.37 1400 2.5 2.5 3.0 68 9.8E-4MT 80 A 0.55 1410 3.7 2.5 3.0 73 1.7E-3MT 80 B 0.75 1410 5.0 2.5 3.0 75 2.1E-3MT 90 S 1.1 1410 7.5 2.5 3.0 75 3.2E-3MT 90 L 1.5 1420 10 2.7 3.2 79 4.3E-3MT 100 LA 2.2 1430 15 2.7 3.2 81 6.9E-3MT 100 LB 3.0 1430 20 2.8 3.3 82 8.2E-3

MBT 112 M 4.0 1440 27 3.0 3.3 84 0.012MBT 132 S 5.5 1440 37 3.0 3.5 85 0.018MBT 132 M 7.5 1440 50 3.0 3.4 87 0.023MBT 132 MD 9.0 1450 60 2.2 2.7 88 0.036MBT 160 M 11 1450 72 2.8 3.0 88 0.073MBT 160 L 15 1455 98 2.9 3.2 90 0.094MBT 180 M 18.5 1470 120 2.3 2.9 91 0.16MBT 180 L 22 1470 143 2.3 2.9 91.5 0.19

M2BA 200 MLA 30 1475 194 2.8 3.0 93.0 0.30M2BA 225 SMA 37 1482 238 2.5 2.7 93.7 0.40M2BA 225 SMB 45 1482 290 2.8 3.0 94.2 0.45M2BA 250 SMA 55 1482 354 2.6 3.0 94.7 0.77M2BA 280 SMA 75 1485 482 2.5 3.0 95.0 1.25M2BA 280 SMB 90 1485 579 2.6 2.9 95.3 1.5M2BA 315 SMA 110 1488 706 2.7 3.1 95.6 2.3M2BA 315 SMB 132 1488 847 2.7 3.2 95.8 2.6

M2BA 315 SMC 160 1488 1027 2.8 3.2 96.0 2.9M2BA 315 MLA 200 1487 1284 2.7 3.2 96.3 3.5M2BA 355 S 250 1488 1605 2.5 3.0 96.5 6.5M2BA 355 SMA 315 1489 2020 2.7 3.2 96.7 8.2M2BA 355 SMB 355 1487 2280 2.4 2.8 96.7 8.2M2BA 355 MLA 400 1490 2584 1.7 3.0 96.8 10.0M2BA 355 MLB 450 1490 2884 1.6 3.2 96.8 10.0M2BA 355 MLC 500 1490 3205 1.4 3.2 96.8 10.5

winding through which AC flows thus causing an EMF to rotate at synchronousspeed, ns = 2 * frequency/number of poles. The number of poles depends uponthe electrical build - two, four, six or eight pole motors are available, howeverthe most common machines are four pole, with a synchronous speed of 1500rev/min when supplied at 50 Hz.

If this simplistic motor were ideal and there was no external drag on the rotor then the rotorwould be pulled around by the EMF also at synchronous speed, however when an external brakeis applied to the rotor shaft it slows down to some speed n which is less than synchronous. Theproportional drop in speed is called the slip, s = ( ns - n )/ns. Clearly the greater the brakingtorque applied externally to the rotor (ie. the torque developed by the motor), the greater the slipand the lower the rotor speed. If there is no slip then there is no torque developed by the motor.

The complete torque-speed characteristic for a typicalsquirrel cage motor shown here demonstrates thisincrease in torque as the slip increases from zero - ie. asthe motor speed decreases from synchronous. At startingthe speed is zero, the slip is unity and the starting torqueis Ts. The torque of a small motor may decrease mono-tonically from starting, without a distinct minimum ormaximum.

Full load refers to the maximum continuous torque Tfthat a motor can generate without overheating - a motorcan operate continuously only at points on the character-istic between full load and synchronism.

The data given opposite for ABB motors is typical of the information published by manufactur-ers. Some notable features of the data include :

- Motor torque is approximately proportional to the square of the supply voltage. - The general truism that "efficiency increases with size" is reflected by the high efficiency and

small full load slip achieved by the large motors listed here.- The system inertia (of which the tabulated rotor

inertia is a part) is relevant to starting, as may beappreciated by the variation of current in a typicalmotor illustrated here. The starting current isaround seven times the full load current, so the I2Rlosses at starting are some fifty times the full loadlosses - and the motor's cooling sytem can handleonly the full load losses over an extended periodwithout overheating. It is therefore necessary whenselecting a motor to ensure that the time of accelera-tion does not exceed the manufacturer's tabulatedlimit - the DOL time, when the motor is connectedDirect On Line (as opposed to other current-reducing starting techniques).

- The maximum shaft load tabulated refers to a radial load right at the end of the shaft and a40 khr L10 ball bearing life; a load closer to the drive-end bearing can exceed this limit, androller bearings may be fitted to some motors to increase capacity.

- Motor dimensional information also is available.

MOTORCHARACTERISTIC

speed, n

sT

uT

bT

pull-up

synch

ronous,

n s

torq

ue, T M

Tf

slip, s s=0s=1

starting

breakdown (pull-out)

fullloadnf

0 500 1000 1500

full load

torque( Nm )

300

200

100

0

current( A )

speed, rev/min

400

rotorsn

Under test at 500 rpm (rev/min), a centrifugal pump delivers a flow of 0.05

m3/s against a pressure difference of 350 kPa with overall efficiency of 70%.

Assuming that rotodynamic machines’ torque-speed characteristics are approx-imately quadratic, select a motor and speed reducer to operate this pump con-tinuously at 600 rpm.

First consider the given test state. The power delivered by the pump to the fluid is

Pfluid = flow rate * pressure difference, ∆p = Q ( pout – pin ) = 0.05 (m3/s) * 350(kN/m2) = 17.5 kW

The overall efficiency of the pump is ηoverall = output/input = Pfluid /Pshaft and sothe shaft power necessary to drive the pump must be

Pshaft = Pfluid /ηoverall = 17.5 / 0.7 = 25 kW corresponding to a shaft torque

T = P / ω = 25*103 (Nm/s) / { 2π * 500/60 (rad/s) } = 477 Nm

At the operating speed of 600 rpm, from ( 1a) the torque necessary to drive the pump is

T = 477 * ( 600/500 )2 = 687 Nm

One might be tempted to choose a motor on this basis, eg. from the above table the 315 SMA with a full load torque of706 > 687 Nm, but this would be WRONG because it ignores the torque amplification of the speed reducer - choosing amotor on the basis of power rather than on torque is generally less prone to error.

The corresponding power necessary to drive the pump is

P = T ω = 687 (Nm) * { 2π * 600/60 (rad/s) } = 43.2 kW

Alternatively, since T ∝ speed2 it follows that P = Tω ∝ speed3 for arotodynamic machine, ie. the power necessary to drive the pump at the

operating speed is 25*(600/500)3 = 43.2 kW

But the speed reducer will incur losses which must be supplied bythe motor. If the reducer's efficiency is around 95% (see eg. the tablein next chapter) then the required minimum motor power will be

Pmotor = 43.2 / 0.95 = 45.4 kW

The choice of motor thus appears to lie between the 225 SMB with a rated continuous output of 45 kW andthe 250 SMA at 55 kW. Apparently the smaller motor would be overloaded by 1%; it's worth reviewing previ-ous work to see if this can be ignored or eliminated. Factors affecting the choice of motor and reducer include :

- The foregoing numbers should not be treated as gospel, since :- Published motor characteristics might be expected to err on the conservative side, so the 225 SMB

motor utilised in this installation might handle the 45.4 kW quite happily on a continuous basis. Canone risk it? The question of rating tolerance is covered by AS 1359.

- The scaling laws are not exact; the effect of pump speed changes in important installations should beconfirmed by manufacturers. The efficiency at 600 rpm would probably not be the 70% at 500 rpm.

- The speed reducer's 95% efficiency is a guesstimate - even when the reducer is selected, it's doubtful ifits efficiency could be ascertained to two significant figures without testing. If it were 96% then themotor power required would be 43.2/0.96 = 45.0 kW and the smaller 225 SMB motor is indicated.

- A pump is analogous to a motor; its operating point 'O' wheninstalled in an hydraulic load circuit is dictated by the intersection

of the hydraulic (Q, ∆p) characteristics of pump and load assketched. The approximate effect of speed on the pump characteris-

tic is : flow ∝ speed, pressure difference ∝ speed2 , so the point '2'on the 600 rpm characteristic is related to the point '1' on the 500

rpm characteristic by Q2 = 1.2 Q1, ∆p2 = 1.22 ∆p1. The load characteristic is concave upwards as it consists of two pres-sure difference components one of which (the static head) is con-stant while the other (the velocity or dynamic head) varies approxi-mately as the square of the flow. These components depend uponphysical attributes such as elevation differences, pipe diameters and lengths.It follows that if the pump speed is less then 600 rpm then the flow through the hydraulic circuit wouldbe reduced from that at 600 rpm. So the question of 'What speed is tolerable?' is really one of 'Whatflow is tolerable?' and we would need to know a lot more about the installation before such a questioncould be answered. A similar argument usually applies to ANY load on a motor, not just a pump.

pump losses

Pfluid

Pshaft

speed reducer

R687 Nm

600 rpm

43.2 kWto load

frommotor

losses

Q

pin

pump

pout

600 rpm

500 rpm

flow, Qpre

ssure

diffe

rence

, ∆

p hydraulicload

pump

1

2

O

The determination of motor acceleration requires the complete torque-speed characteristic simi-lar to the curve above - the tabulated performance data is not sufficient. The complete character-istic is usually available from the motor supplier on request, but lacking this the followingtorque-speed approximation is useful :-

( 2 ) TM = Tb / { 1 + ( sb – s ) 2 ( a /s – b s2 ) } ; s = 1 – n /ns

in which sb and Tb ( the slip and torque at break-down ie. at maximum torque ), aand b are all constant properties of the motor. The tabulated values have been chosen to fit ( 2)to the manufacturer's performance data together with an estimated pull-up ( minimum ) torqueof Ts - ( Tb -Ts )/2. Elementary theory predicts b = 0.

Matching a Motor to a Given Steady Load

We now consider a motor connected to a load. The load may incorporatea speed reducer, or it may not (R=1) - in either case the load characteristicis given by ( 1b). When energised, the motor torque TM must alwaysexceed the torque absorbed by the load TL since the excess torque Tnet = TM - TL is necessary toaccelerate the system of motor-plus-load until the system settles at a steady running speed nrwhere torque equilibrium obtains, ie. TM = TL = Tr , and the two characteristics ( 1b) and ( 2) inter-sect. It is unlikely that an accurate solution for the running speed will be necessary since Codespecifications allow for a tolerance of ± 20% on the full load slip - if a precise running speed wererequired then a motor/control system more sophisti-cated than a DOL started squirrel cage would be used.

As noted previously, full load refers to the maximumtorque which a motor can develop on a continuousbasis; it is dictated largely by the motor's inbuilt cool-ing system. A running point above full load is possibleover restricted running periods ( see below ) however ifduty is continuous then Tr must be less than Tf asillustrated to avoid gradual build-up of winding tem-perature and consequent insulation breakdown.

At its most basic therefore, the task of selecting a motor to drive a given continuous load is oneof choosing a motor whose rated continuous output (ie. full load power) exceeds the loaddemand and transmission losses. However in many cases the optmum selection will requiremore thorough investigation, as the example opposite illustrates.

Squirrel cage motors are termed 'constant speed' because their steep characteristic in the operatingregion around full load results in large excursions in torque demand being accomodated bysmall variations of motor speed. Since the approximate motor characteristic ( 2) is usually fittedto full- and no-load ( synchronism) it is a fair approximation in the continuous operating regionbelow full load.

Theoretically, driving a load at low speed may be carried out equally well by a directly coupledlow speed motor or by a high speed motor and speed reducer. In practice the high speed motor/reducer option is very often the cheaper for a number of reasons :

- A motor's mass is approximately proportional to the torque it develops as suggested by thetrends graphed below - for example the mass of a 200 kW four pole machine is about halfthat of an eight pole machine which generates the same power at half the speed. Costsobviously are related to mass.

T = T - TLMnet

speed, n

torq

ue, T

Tr

rn

Tf

fullload

runningpoint

Lload, T

Mmotor, T

load

motor

n

Squirrel Cage Motors 4

motor is then tightened down and the half couplings are finally connected together by a dis-mountable flexible element.

The choice of coupling depends on the speed and energy level of the connected machines, onthe degree of misalignment which the coupling can tolerate, on the torsional vibration levelpresent, and so on. Thus the flexible spider coupling might be expected to cater for greater misa-lignment than the gear coupling, and also to provide more damping - but the torque capacity ofthe gear coupling would be much greater than that of the flexible spider because the metal gearteeth are much stronger than the rubber spider.

Shaft alignment is important also when driving through belts. The plan view here shows amotor equipped with a belt pulley and mounted on slide rails which arefixed. During installation the motor can slide on the rails and is posi-tioned by the two adjusting screws so that the belt is correctly tensionedand the motor axis is perpendicular to the belt length (if it isn't perpen-dicular then the belt might run off the pulley). When adjustment iscomplete the motor is secured to the rails. The orientation of the adjust-ing screws opposes the torque exerted on the motor by the belt tension.

Periodic Loading

When the load torque is not constant, but varies cyclicallyin some manner with period to as shown, it is usual todefine an equivalent constant load torque, Te, which hasthe same life reduction effect on the motor as has thecycle, thus :-

( 3 ) Tem to = ∫0

to Tm dt

where m is a constant, noting that, as m → ∞ then Te → TmaxSince motor speed is essentially constant, power may be substituted for torque here.

The rationale behind ( 3) is the approximate relation : life * loadm ≈ constant, which many

components and machines are found to follow. Thus m ≈ 3 for rolling element bearings, m ≈ 12for V-belts, and so on. Electric motor life depends on a complex interaction of electrical and ther-mal phenomena, as has been seen, and there is thus no real justification for adopting the tradi-tional RMS value of m = 2 - though the concept of RMS current rather than load may be appro-priate. Rather, manufacturer's data indirectly suggests a value of m ≈ 5.

If the torque consists of a sinusoid, of amplitude T superimposed on amean component T as sketched, where T ≤ T , then ( 3) gives theequivalent load Te as :-

( Te )m = ( 1/2π ) ∫

0

2π ( T )m dθ

= ( 1/2π ) ∫0

2π ( T + T sinθ )m dθ or, normalising

( 3a) ( Te / T )m = ( 1/2π ) ∫0

2π ( 1 + λ.sinθ )m dθ where λ = T/ T

= Σ q=0

m div 2

[ m! (λ/2)2q ] / [ (m-2q)! (q!)2 ] for integer m, and

= 1 + 5λ2 + 15/8 λ4 for m = 5 for example.

The choice of a motor suitable to drive a cyclic load is based on the motor's full load torqueequalling or exceeding the load's equivalent constant torque Te, whilst ensuring that the break-

period, to time, t

torq

ue, T Te

T~

Ttime, tto

rque, T

- An electric motor is much more com-plex than a gearbox for example, and itspower density (kW/m3) is less since it

relies on EMF rather than on highlystressed metal teeth for torque genera-tion. So it makes economic sense todevelop torque in a mechanical trans-former rather than in an electro-mechanical transducer such as a motor.However there is a limit to this argue-ment ( why are two pole machines notmore common ? ) and the real costs of various solutioncandidates must be compared before any optimum can beselected.

It is seldom prudent to rigidly connect together nominally coax-ial power transmitting shafts - of motor and load or of anythingelse - since real misalignment gives rise to untoward bendingstresses and vibration. Misalignment to some extent is inevitabledue to tolerances in initial assembly, to load- or thermally-induced support movement, to shaftbending under load, and so on.

Misalignment is allowed for by connect-ing the shafts by a shaft coupling - some ofthe many types of coupling are shownhere. Each half-coupling is connected rig-idly to one of the two shafts by key, con-tracting friction bush, or by other means.The motor and drive assembly is thenadjusted to minimise the initial misa-lignment (see eg. tutorial problem #9) the

- The full load speed of the 225 SMB motor is 1482 rpm, indicating a required speed reduction of 1482/600= 2.47. The full load speed of the 250 SMA is the same, but it will be runnng light; so interpolating line-arly between full load and synchronism, the speed when delivering 45.4 kW will be about 1500 - (1500-

1482) * 45.4/45 = 1484 rpm which corresponds to a requred speed ratio of 1485/600 ≈ 2.47 also. So it is con-cluded that a 2.47:1 speed reducer is necessary with either motor. But it would be fortuitous indeed if such an odd ratio were available off-the-shelf. If a more likely 2.5:1unit were used then the motor demand would be reduced - as the pump speed would be 1482/2.5 = 593

rpm and from the scaling laws Pmotor = 45.4*(593/600)3 = 43.8 < 45 kW, so the smaller 225 SMB suffices.

It is unlikely that any of the preceding arguments can be firmed up by striving for greater precision in pumpand reducer efficiencies, more exact scaling laws etc. We have to resign ourselves to arguments based uponimprecise/incomplete knowledge. It is for this reason that we may incorporate appropriate design factors (ie.factors of ignorance) to ensure that a critical installation is not compromised if operational parameters assumevalues at the dangerous end of their tolerance band. If the present installation is critical then we might select a20% over-design and specify a required motor power of 1.2*45.4 = 54.5 kW, necessitating the 250 SMA motor.

Good design requires accurate knowledge of the the physical units and processes, enabling minimisation offactors of ignorance and hence minimisation of over-design. But does the effort outweigh the cost?

Concluding and pending further investigation, adopt the smaller 225 SMB motor with a 2.5:1 speed reducer.

outp

ut

kW

400

300

200

100

1000 1500 2000motor weight, kg

ABB SQUIRREL CAGE MOTORS3 phase 415v 50 Hz

315 SMC315 MLA

355 S

355 SMA355 MLA

frame size8 pole 750 rpm

4 pole 1500 rpm

6 pole 1000 rpm

end float

mradial

practically misaligned shafts (much exaggerated) with float

ideally coaxial shafts

mangular

Squirrel Cage Motors 5

( 4 ) ∆t = 2 π J ∫onr ( 1/ Tnet ) dn

Although this period tends theoretically to infinity, a slight reduction of the upper limit enablesrealisitic integration to be carried out, possibly by the graphical techniques of the Appendix. Thereduction is excusable in view of the full-load current being so much lower than starting, andthe questionable applicability of the motor's steady state characteristic to a transient analysis.

The starting period so calculated must not exceed the maximum allowable starting time for aDOL-started motor laid down by the manufacturer - refer to the table above. Although star-deltastarters allow about three times the starting period permitted with DOL, torques are reduced. Ifthe starting time for an intended motor is excessive - that is the system inertia is relatively large- then consideration should be given to a clutch or hydraulic coupling, rather than opting for alarger motor. As will be seen, these devices isolate the motor from high inertia loads thus allow-ing the motor to accelerate quickly.

Hydraulic Couplings

A hydraulic coupling consists of an impeller (pump) anda turbine, each equipped with multiple vanes and eachmounted on a separate shaft in a casing which is partiallyfilled with oil. The impeller is mounted on the motorshaft, the turbine on the coaxial load shaft as shown. Thedevice operates on the principle of oil being flung out-wards by the vanes of the rotating impeller; the oil thenrecirculates around the shroud and gives up its angularmomentum to drive the turbine. Conceptually the deviceis identical to a motor-driven fan blowing over a wind-mill which is connected to the load. The motor is thuseffectively isolated from high load inertias, and also from load transients (ie. from shock).

On startup, the motor accelerates quicklysince the inertia of the impeller is substan-tially less than the load inertia, but when anappreciable speed has been attained, hydro-dynamic coupling between impeller andturbine occurs and the latter gradually accel-erates the load.

It is evident from a free body of the coupling running at steady speed, that the torque must beconstant across it. Since the coupling efficiency must be less than 100%, and power = torque *speed, it follows that the steady state speed of the turbine/load shaft, ωt, must be less than thespeed of the impeller/motor shaft, ωi. The proportional speed loss is referred to as the hydraulicslip, s, of the coupling - not to be confused with the electrical slip of the motor. This state ofaffairs is the reverse of that in a 1:1 gearbox in which speed constancy is assured by the positivedrive of the gear teeth, and the inevitable power loss is reflected in a torque loss across the unit.

An approximate model of a coupling will be examined in order to estimate the steady statetorque developed. The coupling considered is one of a series of geometrically similar modelseach characterised by its 'size', D - eg. the runner diameter.

Fluid momentum theory predicts the torque on either shaft due to oil flow ( Q m3/s) over the

blades to be :

motor-impeller

time

spee

d

spee

d

time

direct drive hydraulic coupling

turbine-loadmotor & load

down torque substantially exceeds the peak torque of the load cycle. The effect of system inertiain reducing speed fluctuations, and hence the torque variations experienced by the motor, isdemonstrated in the Problems which follow.

AS 1359 classifies eight duty types, S1 through S8, three of which are shown :-S1 - continuous maximum duty

This has already been discussed.S2 - short time duty

The motor is loaded by a uniformtorque T for a short period t* =10, 30, or 60 minutes and thereaf-ter it is shut down until its tem-perature returns to ambient.ABB recommends that :if t* = 30 minutes, then Te = T/1.2if t * = 60 minutes, then Te = T/1.1

S3 - intermittent periodic dutyThe cyclic duration factor (cdf) is defined as t*/to in which the period to is 10 minutes.Again the motor is de-energised when off-loaded.ABB recommendations correspond to m ≈ 16/3 in ( 3) ie. Te /T = ( cdf )1/m; it is for thisreason that m ≈ 5 is suggested above - at least for preliminary estimation purposes pendingconfirmation from the manufacturer - though S3 is not a cycle as conceived in ( 3) since themotor is shut down during part of the cycle and there is no fan cooling while shut down.

Clearly when a load subscribes to one of the standard duty types then the recommended motorselection procedure should be followed. However many loads fall outside the scope of the stan-dard, and ( 3) may be the only feasible selection rationale.

When a motor is started frequently, in a situation similar to S3 but with a period less than 10minutes, the temperature build-up due to frequent acceleration may be significant evenalthough the eventual load is negligible. ABB suggests that the load be increased by the factor :-

( 3b ) ( 1 – ( 1 + JL / JM ) ( x / xmax ) ) –1/2 ≥ 1

where x is the number of starts per hour, xmax is the maximum allowablenumber of starts per hour and JM is the motor inertia (both of which are tabulated above), andJL is the load inertia referred to the motor shaft (see below).

Acceleration

Squirrel cage motors suffer from drawing high starting current - typically some 6-8 times fullload current - so I2R losses at starting are much greater than the full load losses which the cool-

ing system can handle on a continuous basis. Lengthy starting periods thus causeearly failure. Star delta starting reduces initial current but is generally not recom-mended, for quadratic loads at least; direct-on-line (DOL) starting is usually themore favoured. The starting period for a motor -driven load may be foundfrom the equation of motion of the system referred to the motor shaft :-

Tnet = TM – TL = J dω/dt ; ω = 2 π n

where J is the system inertia referred to the motor shaft ie. J = Jmotor + Jload /R2 etc.

Integrating this yields the acceleration time, ∆t, to reach running speed, nr, from rest :-

T T Tt*

timeload

losses

temper- ature

S1 - cont-inuous

S2 - shorttime

S3 - intermittent periodic

dutytype

t* to

JT

Ln , T

M

Squirrel Cage Motors 6

APPENDIX : INTEGRATION IN PRACTICEClosed form evaluation of the integral z = ∫A

B f(x) dx is feasible only for a very limited class of simple applications;

in practice, numerical or graphical techniques must usually be employed.

You might prepare a numerical integration procedure for your calculator or computer. Such a procedure could be

based on recursive subdivision using Simpson's method. To increase robustness it might embody the transforma-

tion x = m y ( 3 - y2 ) + n where m, n are constants chosen so that the newly introduced independent variable, y,

assumes a minimum value of -1 when x = A; and a maximum value of +1 when x = B; ie m = (B-A)/4; n = (B

+A)/2. In terms of y, the integral then assumes the form :-

z = 3m ∫-11 ( 1 - y2 ) function ( m y ( 3 - y2 ) + n ) dy

The transformed integrand automatically vanishes at the two limits, so it does not need to be evaluated there and

the transformation avoids potential difficulties due to the fairly common occurrence of a singularity at either limit.

This transformation is incorporated into the Romberg integration procedure in the program Motors. Lacking such

procedures, functions may be graphed and the following applied.

Graphical Integration

To evaluate graphically the integral

( i ) z = ∫AB

y dx where x and y may be temperature, or stress, or length, or whatever.

The y-x relation is plotted as a Y-X graph, both X and Y being lengths in Figure 1 eg. drawn to known scales :

( ii ) Sx = x/X ( eg - units of 'x' per mm ) ; Sy = y/Y ( units of 'y' per mm )

The following construction gives the integral, Z, also as a dimension, to some scale, Sz, as yet unknown.

Choose any line Q parallel to the Y-axis, and a pole P, with ordinate Yp and distant p (mm) from Q. In Figure 2, P

has been taken with the same ordinate as A, but this is not necessary.

Divide the curve into a number of vertical strips, Figure 3, and for each strip :

- select the mid-point

- construct a line parallel to the X-axis from the mid-point to the line Q and thence to the pole, P

- parallel to this polar line, draw a line across the strip from the similar line in the previous strip.

This construction would start from any point such as B' having the same abscissa as one of the limits, and yields

TC = oil mass flowrate * angular momentum change = ρ Q ( ωi r22 – ωt r1

2 )where ρ is the oil’s mass density.

If dimensions are proportional to size, and it is assumed that Q is pro-portional to :

- tangential velocity, ie. ∝ ωi r2 ∝ ni D- slip, s = 1 – nt/ni ( ie. if slip were zero, solid body rotation

would result in no flow )- flow area which is proportional to D2, then :

TC ∝ ( ni D3 s ) ( kni - nt ) D2 ∝ D5 ni

2 s ( k + s) ; k is constant.

A more realistic formula allowing for losses due to excessive turbu-lence at large slips is :

( 5 ) TC = C D5 n i2 s ( k – s ) . . . . the constants k & C applying over the whole size range.

The variation of torque with slip, predicted by ( 5) with steady motor/impeller speed is shownon the left here for a particular series of couplings, for which k = 1.5.

Couplings normally operate with 3-5 % slip. The choice of a particular size of coupling to suit agiven motor is based on this - eg. solution of ( 5) for D, with Tc and ni corresponding to motorfull load and 3 ≤ s ≤ 5 %. The solution may be carried out graphically, as sketched on the right fora motor with 1400 rpm full load speed. Running parameters are found from equality of TC, TM,and TL, with impeller speed = motor speed, and turbine speed = load speed; again, an accuratesolution is rarely necessary.

Centrifugal and magnetic clutches are two of many other devices which allow motors to acceler-ate quickly. Various forms of compliant couplings are available for isolating motors from shockloads. A special form of hydraulic coupling, with an adjustable scoop for altering the amount ofoil, is used as a variable speed device.

D

shroud

vaner2

r1

Squirrel Cage Motors 7

Hence the M/EI diagram is plotted, Figure 2, to scales :

Sx = 200 ( mm ) / 48 ( mm ) = 4.17 mm / mm

SM/EI = 40 ( mrad/m ) / 5 ( mm) = 2.67 mrad/m.mm

The pole Pθ is selected at zero ordinate and at distance pθ = 11 mm

as shown in Figure 2. Graphical integration then follows, yielding

the slope curve ( θ-x : Figure 3 ) to a scale, from ( iii) of :

Sθ = 11(mm)*4.17*2.67 (mrad/m.mm) = 122 mrad /mm

Although the zero axis of the slope curve is undetermined at this

stage, it is known to be horizontal since the pole had been chosen

at zero ordinate in Figure 2. Integration of the slope yields the

deflection, v, so a pole Pv is chosen at a distance pv = 9 mm in

Figure 3, and the v-x curve constructed, Figure 4, whose scale,

from ( iii ) is :

Sv = 9 ( mm ) * 4.17 * 122 ( mrad/mm) = 4.59 mm /mm

Since the ordinate of the pole was not zero in Figure 3, the v = 0

axis will be inclined. It is known that there can be no deflection of

the shaft supports, A & E, so the line a-e must represent the unde-

flected axis. A line parallel to a-e and tangent to the curve identi-

fies the maximum deflection, which occurs at the point g. As this

point is also the point of zero slope, the zero axis in Figure 3 can

now be defined from the v-x curve.

From the deflection curve Figure 4 the scaled maximum deflec-

tion is measured as Vmax = 11 mm.

The maximum deflection is therefore :

vmax = 11 ( mm ) * 4.59 ( mm / mm ) = 50 mm

In this example, two integrations are necessary before any integration constants can be evaluated. Analytic

approaches to the bending of simply supported beams demonstrate this same necessity.

Accuracy will obviously be improved by graphs larger than the above, and by measurements better than to the near-

est millimetre.

BibliographyAS 1359.30 : General Requirements for Rotating Electrical Machines, SAA 1974

ABB Three Phase Squirrel Cage Alloy Motors, CMG Electric Motors, Publication ABBS9212

Burr AH & Cheatham JB, Mechanical Analysis and Design, Prentice Hall 2ed 1995

Calistrat MM, Flexible Couplings: Their Design Selection and Use, Caroline 1994

Constant Filling Fluid Couplings, Fluidrive Catalogue B1-1 (1976)

Finney D, Variable Frequency AC Motor Drive Systems, Peter Peregrinus 1988

Hurst K ed, Rotary Power Transmission Design, McGraw-Hill 1994

The Induction Motor, ABB, Pamphlet A 10-2004 E 1990-04

Nesbitt B ed, Guide to European Pumps and Pumping, PEP 2000 - a comprehensive overview

South DW & Mancuso JR, Mechanical Power Transmission Components, Dekker 1994

Totally-Enclosed Squirrel-Cage Three-Phase Motors, ABB, Brochure FIMOT/M2BA 1 GB 93-06

the Z-curve, from which the integral follows. Thus from similar triangles in Figure 4 :

( Y -Yp ) /p = ( ( Z + δZ ) - Z ) / δX ; δZ = ( ( Y -Yp ) / p ) δX or, integrating

Z = 1/p ( ∫ Y dX - Yp X + constant ) - and using ( ii)

z /Sz = ( ∫ y dx - yp x - zo ) /p SxSy - where zo is constant; then setting . . . .

( iii) Sz = p Sx Sy ( mm * ( units of x per mm )* ( units of y per mm ) ) . . . . leads to

( iv) ∫ y dx = z + zo + ypx - which, with limits, yields

∫AB y dx = zB - zA + yp ( xB - xA ) . . . . the desired result.

If P is chosen at Y = 0, then the last term of ( iv) vanishes and ( i) is satisfied identically. If this be the case, and fur-

thermore the integral is known to vanish at some point C, then the plot of the integral may be completed by the

addition of the X-axis as shown in Figure 5. But this is not always immediately possible - sometimes a second inte-

gration is necessary to simultaneously establish two integration constants; the closed form integration of the sim-

ple beam's elastic curve is a case in point. If Yp<> 0 then the integral must be reckoned from the inclined line

inferred by (iv) and indicated in Figure 6.

EXAMPLE 1 A load is characterised by constant, linear and

quadratic components of 55, 80 and 60 Nm respectively, at 1500

rev/min. The load's inertia is 25 kg m2 and it is driven directly

by a now superseded ASEA squirrel cage motor, type MBN 200

L. Estimate the acceleration time graphically.

The motor and load torques are plotted against speed, Figure 1,

using a speed scale of :

Sn = 25 (Hz) / 47 (mm)

The net torque is computed at various speeds and its reciprocal

plotted, Figure 2, with a scale of :

ST = 0.01 (Nm)-1 / 23 (mm)

A pole, P, and reference line, Q, are conveniently selected on

Figure 2, with pole distance p = 16 mm.

The integral is then drawn, Figure 3, whose value up to run-

ning speed is measured as Zr = 26 mm.

The acceleration time is therefore :

∆t = 2π J ∫0nr ( 1/Tnet ) dn = 2π J zr

= 2π J ( Sz Zr ) = 2π J ( p Sn ST ) Z r= 2π ( 25+0.35 ) 16 ( 25 / 47 ) ( 0.01 / 23 ) 26 * 1

rad kg m2 mm rev 1 mm N s2

rev s mm Nm mm kg m

= 15 s

This value is comparable with the period calculated by the program Motors, though drawing inaccuracies become

relatively large as the running speed is approached, and the program is based upon current motor characteristics.

Note that the Z = 0 axis is horizontal here as the pole is selected on the zero-ordinate axis.

EXAMPLE 2 Determine graphically the maximum

deflection of the steel shaft illustrated.

Static analysis yields the bending moment diagram of

Figure 1. Preparing the M/EI diagram :

EIABC = ( π /64 ) * 0.0304 * 207E9 = 8230 Nm2

EICD = ( π/64 ) * 0.0404 * 207E9 = 26012 Nm2

EIDE = ( π /64 ) * 0.0364 * 207E9 = 17067 Nm2

Squirrel Cage Motors 8

SQUIRREL CAGE MOTORS - PROBLEMS

Characteristics refer to ABB totally enclosed, 4 pole, three phase motors - started direct-on-line at 415 v. Loadsare coupled directly to motors unless stated otherwise. The program 'Motors' may be used to aid motor selec-tion.

1 A load fluctuates cyclically as shown. Determine themean effective load and select a motor suitable to driveit, using an exponent of (a) 2; (b) 5 in the equivalentload equation. [ 100 LB; 112 M]

2 Select a motor to drive a load whose inertia is 0.2 kg.m2

and which varies periodically as shown. [ 132 S]

3 A fan delivers 10 m3/s against a head of 120 mm watergauge, when driven at 1450 rpm. The fan's efficiency is 78%, its rotor's inertia is 3.5 kg.m2,and bearing friction accounts for a constant torque of 10 Nm. Select a motor to drive the fanand estimate the running speed and acceleration time. Check your results with 'Motors'. [ 180M, 1472 rpm, 2.7s ]

4 The inertia of a dewatering drum is 400 kg.m2. Friction and windage are estimated to con-tribute equally to the drum's power requirement of 6 kW at its nominal operating speed of250 rpm. Select a motor to drive the drum through a 5.75 : 1 reduction gearbox. [ 160 M ]

5 This problem concerns the effect of system inertia, J, in reducing the torque fluctuationsexperienced by a motor which drives a cyclically varying load consisting of a sinusoid of

period to and amplitude T superimposed on a mean torque of T ( where T ≥ T ). That is, theload is of the form T + T sin 2πt/to.Assuming the motor torque-speed characteristic at the mean torque, T, to be approximatelylinear with negative slope of 'k' (Nms/rad), show that the amplitude of the motor torque

excursions, Tv, about the mean, is given by : Tv / T = ( 1 + ( 2π J/kto )2 )-1/2 ie. < 1.

6 The duty cycle of an automatic punch may be represented by a sinusoid of period 5s,between the limits of 10 and 210 Nm. It is proposed to drive the punch by a 180M motor in

conjunction with a flywheel, such that the system inertia is 20 kg.m2. The punch will notbe loaded until the motor has run up to speed. Discuss the feasibility of the proposal usingthe results of the previous problem in conjunction with an exponent of 2 in the equivalentload equation.

7 Select a motor to drive a fairground machine whichconsists of a ø5m rotating drum. The drum loadedwith people accelerates for 30s; it then runs at constantspeed before slowing to a stop for passenger exchangeprior to the next cycle. The motor runs continuously, even while the drum is stationary.The motor drives the drum through an hydraulic coupling and a 70:1 speed reducer whose

efficiency is 75%. The loaded drum's inertia is 10t.m2. Its power consumption due to fric-tion is negligible, ie. it needs appreciable power only for acceleration. [ 90 S ]

0 10 60

7.5

1.5

time (s)

load (

kW)

0 20 40time (s)

25

load (

Nm

)

drum speed

0 0.5 5 10time (min)0

brake

load/unload

run

Acceleration example of AppendixTorque increment 100.0 Nm

A Dialogue with 'Motors'

Squirrel Cage Motors 9

8 A family of geometrically similar hydraulic couplings includes the sizes :- 180, 205, 235, 265,290, 325, 370, 415, 450, 510, 585, 660, 735, 815, 915 and 1040 mm, for which the torque charac-teristic at small slips is given by : T ( kNm ) = 0.9 d5 n2 s ( 1.5 - s ) where d is the size(m) and n (rev/s) the impeller speed. Select a coupling for use with a 132M motor in thedrum drive of Problem 4 and determine the speed of the drum. Note that the coupling isinterposed between motor and gearbox. [ 265 mm, 242 rpm ]

9 A flexible shaft coupling can operate satisfactorily only when the shaft misalignment is lessthan a certain limiting value. This problem concerns the measurement and subsequentminimisation of misalignment prior to coupling assembly.

Misalignment may be measured by the arrangement of Fig (b). An arm is mounted on themotor shaft eg. by means of the boss B. Two dial gauges are rigidly attached to the arm,their styli bear upon the coupling half, C :- gauge R measures radial runout εr resulting from shaft radial misalignment, Fig (a), and- gauge A measures axial runout εa resulting from shaft angular misalignment.Radial and angular misalignments are independent of one another, as are εr and εa.Gauge readings are recorded against rotation angle θ, Fig (d), as the shafts are rotatedtogether slowly by hand. Neglecting any inconsequential mean value, the variation of each(small) runout is sinusoidal ε = εo cos ( θ – θo ) the amplitude εo and phase θo of whichenable correction of the misalignment, radial or angular as the case may be, by means ofadjustments to the motor mounting position.For a particular motor the locations of the shaftand measuring plane in relation to the fourmounting bolt positions are sketched. The fol-lowing readings [1 gauge division = 0.01mm] areobtained as above, gauge A readings being takenat the diameter D = 400 mm, Fig (b) : angle from vertical, θ (deg) 0 45 90 135 180 225 270 315reading (divisions) of radial gauge R 61 65 43 7 –20 –22 –3 31reading (divisions) of axial gauge A 37 26 29 46 69 80 79 61What displacements of the motor mounts are necessary to minimise misalignment ?

[ front 0.77 up, 0.85 right; rear 1.33 up, 1.77 right (mm) ]

θA

R

C

(a) (c) (d)

C

B

R arm

DA

(b)

εr εa

motorshaft

left right right

500 750

420

mea

surin

g

plan

e

700

rightleftfront rear

V-Belt Drives 1

r2 1rFr

Ft

Ft

T1

T2r2 r1

F^

F

T2

T1

r2 r1

ω1ω2

v

r2 1rω2

ω1v

P /P < 12 2 2 1 1 1 2 1ω T < ω T = P ; η =P = power - for gears and for belt :

T /r2 2 1

^1

T /r = ( F -F) = T /r < F < T /r (friction)2 2 1 1t

< ω r (creep)2 2 1 1ω r < v2 2 1 1

= ω rω r = v

SPUR GEARS BELT

velocity

freebodies

torque

1 - driveR2 - driveN

contact - or more particularly by the tangential component of this force Ft whose moment about thecentre of each free body equilibrates the shaft torque T ( assuming constant velocity ). Friction playsonly a minor role - inescapable and deleterious but subsidiary nonetheless.

The transfer of power in a belt drive on the other hand relies critically on friction. The tensions F & Fin the two strands ( the nominally straight parts of the belt not in contact with the pulleys ) cause anormal pressure over each belt-pulley contact, and it is the corresponding distributed friction whosemoment about the pulley centre equilbrates the shaft torque T - provided gross slip of the belt on thepulley surface does not occur due to friction breakaway.

Ideally, for gearsand for belts, thespeed reductionratio and the torqueamplification ratioare each equal to theradius ratio, so thatthe output powerequals the inputpower and the effi-ciency is 100%. Thespeed ratio across areal pair of gearsalways equals theideal ratio becauseof the positive drive,however sliding fric-tion results in a torque ratio which is less than ideal. A real belt drive is just the opposite - the torqueratio equals the ideal ratio ( as may be seen from the free bodies), but creep results in the speed ratiobeing less than ideal. Creep - not to be confused with gross slip - is due to belt elements changinglength as they travel between F and F , and since a pulley is rigid then there must be relative motionbetween belt element and pulley.

Since power equals the product of torque and (angular) speed, the consequence of the foregoing isthat efficiencies of real gears and belts are less than 100%.

Some of the many forms of belt are introduced below.

Historically, flat belts made from joined hides were first on the scene,however modern flat belts are of composite construction with cordreinforcement. They are particularly suitable for high speeds.

Classical banded ( ie. covered ) V-belts comprise cord tensile memberslocated at the pitchline, embedded in a relatively soft matrix which is

encased in a wear resistant cover. The wedging action of a V-beltin a pulley groove results in a drive which is more compact than a

flat belt drive, but short centre V-belt drives are not conducive to shock absorption.

Wedge belts are narrower and hence lighter than V-belts -centrifugal effects which reduce belt-pulley contact pressureand hence frictional torque are therefore not so deleterious in

wedge belt drives as they are in V-belt drives.

V-BELT DRIVES

The great majority of mechanicalpower transmission applicationsinvolve rotating shafts, since rotationis continuous and the shafts/mountings are cheap relative to othermeans of power transmission.Matching a prime-mover to a loadthus involves transformation ofpower between shafts - usually froma high speed/low torque drive shaft,through a speed reducer of ratio R ≥1, to a low speed/high torque loadshaft. So far we have considered onlyelectric motor prime movers, withindustrial load speeds of the order of10 Hz - however the range of torquesand speeds encountered in practice ismuch wider than this viewpoint, asthis diagram from Palmgren op cit suggests.

As has been noted, speed reducers are employed almost invariably to amplify torque rather than toreduce speed. The two most common speed reduction mechanisms in industry are belts ( usually V-belts ) and gears - though chains, hydrostatic transmissions or other drives may be used. The majorfactors other than cost which must be borne in mind when choosing a reducer are listed below how-ever shock absorption capacity, distance between shaft centres, accuracy required of shafts andmountings, tolerable vibration levels and so on may also need to be considered.

The efficiencies of belts are generally less than those of gears - that is why belts are not found in themain drive train of road vehicles where fuel economy is critical.

The diagram below compares the kinematics and kinetics of a pair of mating spur gears with thoseof a belt wrapped around two pulleys. The gears are represented by their pitch cylinders which rollwithout slip on one another, slip being prevented by the positive drive - ie. by the meshing teeth.

The transfer of power between the gears is enabled by the normal action/reaction force at the tooth

CHARACTERISTICS OF SPEED REDUCERS gears chain timing V-belt flat hydrosta-

belt belt tic trans.drive mechanism positive positive positive friction friction -max capacity, kW 10 000 500 500 1 000 1 000 1 000max torque, Nm 108 106 104 104 104 108

max speed, m/s 50 15 60 30 50 -speed ratio accuracy high high high moderate moderate moderatefull load efficiency, % 94-98 94-98 93-98 92-97 92-97 84-92lubrication required yes yes no no no -

AB

C

DE

FG

H

IJ K L

M

N

O

P Q

rotational speed ( rev/s )

10-4 10–2 100 102 104

10–4

10–2

10–1

101

102

104

105

106

107

108

103

100

10–3

torque ( Nm )

10–9

10–6

10–3

100

103

power ( kW )106 TORQUE-SPEED REGIMESOF VARIOUS MACHINES

A rotary cement kilnB container ship gas turbineC electric power generatorD diesel engine in ferryE caterpillarF windmillG refrigeration compressorH Volvo 340 carI washing machineJ windscreen wiper motorK hand power toolL truck turbochargerM timer clockN electric razorO gyroscopeP recording cylinderQ dentist's drill

V-Belt Drives 2

A typical single V-belt drive is illustrated. The effec-tive or pitch diameter of the small driveR pulley isD1; that of the large driveN pulley is D2.

Before the drive starts to rotate and transmit power,an initial tension Fo is produced in both belt strandsby the shafts being pulled apart and then locked ( eg.by a motor on slide rails ) or by other means.

Drive commences by the power source applying aclockwise ( as sketched ) torque T1 to the shaft of thesmall driveR pulley, causing it to rotate clockwise at a steady speed n1 ( rev/s ). The tension in thetight upper straight strand will then exceed Fo while the tension in the slack lower strand willbecome less than Fo - this tension difference applies a torque to the driveN load pulley, equilibratingthe load torque T2 while the pulley rotates at uniform speed n2, also clockwise. Neglecting creep:

( 1 ) v = π D1 n1 = π D2 n2 ; R = n1 / n2 = D2 / D1 ≥ 1

where the belt speed, v, is limited to 30 m/s for the usual cast iron pulley material, thoughhigher speeds can be achieved with more expensive builds. V-belts are designed for optimum per-formance at speeds of around 20 m/s.

In the most common V-belt drive design problem, the transmitted power and the speed of the smallpulley are stipulated, together with a specified range of the speed ratio and possibly acceptableranges of shaft centre distance and of drive life - though if the drive is designed by the methods out-lined by the Code(s) then its life, though presumably commercially acceptable, cannot be evaluated.

The required drive is specified by a suitable size ( B, or SPA &c), number and length of the belt(s),and by the two pulley diameters. We therefore must consider two aspects of V-belt drives :

- the drive overall geometry, ie. the inter-relationship between centre distance and belt length- the fatigue life of the belts as dictated by the loading on them, which is in turn affected by the

drive kinetics and the power transmitted through the drive.

100 200 400 20001000 4000speed of faster shaft, rpm

200

400600

100

1000

20

4060

10

2

46

1

desi

gn p

ower

, kW

100 200 400 20001000 4000speed of faster shaft, rpm

200

400600

100

1000

20

4060

10

2

46

1

desi

gn p

ower

, kW

SPC

SPB

SPA

SPZ

WEDGE BELTS

100 200 400 20001000 4000speed of faster shaft, rpm

200

400600

100

1000

20

4060

10

2

46

1

desi

gn p

ower

, kW

D

A

B

C

VEE BELTS

n2

tight

slack

driveN, Ø D2 driveR, Ø D1

n1T2

T1

v

cover

tensilemembers

gap

matrix

pitchdiameter

2β

Modern materials allow cut belts to dispense with a separate cover. The belt illus-trated also incorporates slots on the underside known as cogging which alleviatedeleterious bending stresses as the belt is forced to conform to pulley curvature.

The synchronous or timing belt drive on the left is apositive rather than friction drive as it relies on gear-like teeth on pulley and belt enabled by modern materials and manufac-turing methods. It is mentioned here only for completeness - we shallnot examine it further.

If a single V-belt is inadequate for power transmission then multiple beltsand corresponding multi-grooved pulleys are necessary - this pulley isequipped with a tapered bush for axle clamping without the stress concentra-tion associated with a key.

The rather extreme short centredrive on the right illustrates a prob-lem with multiple belts - how toensure equitable load sharing

between flexible belts whose as-manufactured dimensionaltolerances are significantly looser than those of machinedcomponents for example. Two types of belt for avoidingmismatched lengths are shown :

Each component of a V-belt performs a particular function. The mainload- carrying elements are the tensile members, often in the form of lon-gitudinally stiff rayon cords located near the centroidal axis of the belt'scross-section, embedded in a relatively soft elastomeric matrix whosemain purpose is to channel the load from the contacts with the groovesides into the tensile members.

The groove semi-angle lies usually in the range 17o ≤ β ≤ 19o. It should benoted that there is a gap ie. no contact at the bottom of the groove. Flat beltsmay be regarded as particular cases of V-belts in which β = 90o.

V-belts are available in a number of standard cross-sectional sizes, designated in order of increasingsize A, B &c while wedge belts are designated variously as SPA, SPB &c (or α, β &c in the US). Eachsize is suitable for a particular power range as suggested by the carpet diagrams below. The applica-bility regions of the various sizes in these diagrams overlap substantially.

As the belts are endless, only certain discrete standard pitch lengths are manufactured, as tabulatedbelow. Discrete dimensions apply also to off-the-shelf pulleys, which are available only with the tab-ulated recommended pitch diameters and number of grooves. A special pulley may be manufacturedof course - but would cost more than a mass-produced commercial product. A pulley is referred toby its pitch diameter - other dimensions including its OD are available from suppliers' manualswhich should be consulted also for local availability.

V-Belt Drives 3

Overall Geometry

V-belt drives are essentially short centre drives. If indrive design the centre distance C is not specified,then it should be set at around 2D1√( R+1) thoughpreferably not less than D2. Since the diameters andbelt length are discrete variables so also is the theoret-ical centre distance, though in the absence of idlersthe nominally fixed centre distance must be capableof slight variation by motor slide rails for example, toallow for belt installation and subsequent take-up ( initial tightening ) before rotation commences.This capability also allows for manufacturing tolerances on belt length, L. From the geometry :-

( 2a) L = π/2 ( D1 +D2 ) + 2C ( γ sinγ + cosγ ) where sinγ = ( D2 –D1 )/2C

This is used to find the belt length, L, for given centre distance, C ( and pulley diameters ).Conversely, to find the centre distance corresponding to a certain belt length, ( 2a) must be solvediteratively - a very close first approximation is given by :-

( 2b) 4 C = √[ g2 – 2( D2 –D1 )2 ] + g ; g = L – π/2 ( D2 +D1 )

The wrap angle ( or arc of contact ) on each pulley is evidently :-

( 2c) θ1 = π – 2γ ; θ2 = π + 2γ ; θ1 ≤ θ2

The power transmission capability of a drive is usually limited by the arc of contacton the small pulley, and so is reduced by large speed ratios and by short centres - eg.θ1 is only about 100o here.

Any particular cross-section of the belt traverses alternately the slack and tight strands and is subjectto bending when in way of a pulley, so it is clear that cyclic loading and fatigue are prevalent. Beforewe can look at fatigue however, we have to know the belt forces and stresses. These will depend onthe belt load and speed - so belt kinetics must now be examined.

Kinetics

When a rope is wrapped around a stationary cylinder, it can remain in equilibrium when its endsare pulled provided one end is not pulled excessively. If the pull is sufficient to overcome frictionthen gross slip occurs and the rope slides around the cylinder. We now consider similar behaviourof a V-belt/pulley contact in order to find out the friction-limited torque capacity.

We consider a belt wrapped around a pulley of radius r as shown at ( a), the contact extending fromφ = 0 where the slack side tension is F, to φ = θ where the tight side tension is F. The uniform beltspeed is v, say as shown ( ie. the pulley is the driveR ).

In order to clarify how the belt tension increases from F to F we consider the belt element δφ lying

n

φ

δφ

T

φ = θ

φ = 0

v

t

F

F

2 δN sinβF + δF

F

δφ /2 2 δFf

free body of element δφ at φ

end view of element with resolved normal reaction

at each groove contact( a ) ( b ) ( c )

δN sinβ

δN

β

C

γ

D2

D1θ2 θ1

1

2 3

4

STANDARD PITCH LENGTHS ( mm )Y Z A B C D SPZ SPA SPB SPC

200 405 630 930 1560 2740 630 800 1250 2000224 475 700 1000 1760 3130 710 900 1400 2240

250 530 790 1100 1950 3330 800 1000 1600 2500 280 625 890 1210 2190 3730 900 1120 1800 2800 315 700 990 1370 2340 4080 1000 1250 2000 3150

355 780 1100 1560 2490 4620 1120 1400 2240 3550 400 920 1250 1690 2720 5400 1250 1600 2500 4000 450 1080 1430 1760 2880 6100 1400 1800 2800 4500 500 1550 1950 3080 6840 1600 2000 3150 5000

1750 2180 3310 7620 1800 2240 3550 5600

1940 2300 3520 8410 2000 2500 4000 63002050 2500 4060 9140 2240 2800 4500 71002200 2700 4600 10700 2500 3150 5000 80002300 2870 5380 12200 2800 3550 5600 90002480 3200 6100 13700 3150 4000 6300 10000

2570 3600 6860 15200 3550 4500 7100 112002700 4060 7600 8000 125002910 4430 91003080 4820 107003290 5370

3540 6070

RECOMMENDED PULLEY PITCH DIAMETERS ( mm )Y Z A B C D SPZ SPA SPB SPC

20 50 75 125 200 355 67 100 160 22422.4 53 80 132 212 375 71 106 170 236

25 56 85 140 224 400 75 112 180 250 28 60 90 150 236 425 80 118 190 265

31.5 63 95 160 250 450 85 125 200 280

35.5 67 100 170 265 475 90 132 212 315 40 71 106 180 280 500 95 140 224 355 45 75 112 190 315 530 100 150 236 400 50 80 118 200 355 560 112 160 250 450 56 85 125 212 400 630 125 180 280 500

63 90 132 224 450 710 140 200 315 560 71 95 140 236 500 800 150 250 355 630 80 100 150 250 560 900 160 315 400 800 90 112 160 280 630 1000 180 400 500 1000

100 125 180 315 800 1250 200 500 630 1250

112 140 200 355 1000 1600 250 630 800 125 150 250 400 1250 2000 315 800 1000

160 315 500 400 1000 1250180 400 630 500200 500 800 630

250 630 1000 800315 800 1250400 1000500630

800

V-Belt Drives 4

to increase drive capacity by increasing the initial tension, but if belts are overtightened then theirfatigue lives will suffer, though by how much we cannot say, since the foregoing model ignores ini-tial tension except insofar as "normal" tightening practice is assumed. Whilst large pulleys reduceloads and hence the cost of mounting shafts and bearings, the initial cost of the pulleys themselvesincreases with size. So the design of a power transmission train in practice needs to be carefully bal-anced from an economic point of view. The manufacturers of electric motors recommend minimumpulley diameters for acceptable motor bearing lives ( see tutorial problems ).

Every cross-section of the belt is subjected alternately to the strand tensions F and F and thereforeundergoes fatigue. These tensions at least must be detemined before fatigue can be addressed. If thepower transmitted by the drive P is shared equally between z belts in parallel then since the torqueper belt always equals ( F – F ) * radius :-

( 4a) P = z ( F – F ) v ie. F – F = P/zv

This relation alone is insufficient for determining F, F individually for a given power-per-belt ( P/z )and velocity (v). If however we focus on the maximum power transmissible without gross slip, thenequation ( 3) applies. Eliminating F between ( 3) and ( 4a), the tight side tension F per belt for agiven slip-limited power P is :-

( 4b) F = P/ z kθv + ρv2 where the drive property kθ ≡ 1 – e–(fθ)min by definition.

It must be emphasised that although ( 4a) is always relevant, equation ( 4b) applies only when slipis imminent , and relates to the maximum power capacity of the drive, dictated by slip on the moresusceptible pulley. At part load, when slip is not limiting, it is common practice to deduce approxi-mate tensions from the full-load ratio given by ( 3), together with the tension difference correspond-ing to the actual part-load torque from ( 4a) ( see the tutorial problems ).

Fatigue

In addition to the direct stress in the belt due to the peak tension, F, belt wrap around a pulleycauses bending stresses. Invoking elastic bending theory, σbending max = ymaxE/R where R is thebending radius here. Bending stress is thus inversely proportional to pulley diameter D - andalthough the belt will probably not behave elastically, this proportionality is a reasonable measureof damaging bending effects. The overall severity of belt loading on each of the two pulleys maythus be expressed as an equivalent damaging force, F* :-

( ii ) F1* = F + M/D1 = P/ z kθv + M/D1 + ρv2 using ( 4b)

F2* = F + M/D2 = P/ z kθv + M/D2 + ρv2

where M is a constant of proportionality whose experimentally obtainedvalue will depend only upon the build of the belt, ie. it is a property of the belt just like the lineardensity, ρ - see Table 1 below.

The sketch here illustrates the quali-tative variation of equivalent forceon a belt element as it passes thepoints 1-2-3-4 of the above geometricsketch.

An element in the slack side straightencounters the large pulley at 1, thetension in the element thereafterincreases due to friction over the arc 1 2 3 4

one revolution of beltlocation

equ

ival

ent

forc

e,

F*

F

useful

M/D2

M/D11

F*

2F*

D θ2 22

1

D θ12

11

F

at φ. The tension in the belt increases from F to F+δF across the element as shown in the enlargedfree body ( b).

The element contacts the two inclined sides of the groove, giving rise to a normal reaction δNshown in ( c) together with a friction force δFf at each contact. Since the two normal reaction com-ponents lying parallel to the pulley shaft in ( c) equilibrate one another, the resultant of the two con-tacts appear as in ( b).

The equations of motion for the elemental free body travelling at constant belt speed v, from ( b) are

tangentially : ( F + δF ) cos1/2δφ – F cos1/2δφ – 2 δFf = 0 as there is no angular accelerationnormally : ( F + δF ) sin1/2δφ + F sin1/2δφ – 2 δN sinβ = δm.an = ( ρ.r δφ ).( v2/r ) = ρv2.δφ

where ρ is the belt's linear density ( kg/m), a belt property - see Table 1 below.

In the limit these two equations yield dFf/dN = dF.sinβ /( F – ρv2 ).dφ and, furthermoreif gross slip is to be avoided then dFf/dN ≤ µ the coefficient of belt/groove friction.

Eliminating dFf/dN here, separating variables, and integrating over the contact arc 0 ≤ φ ≤ θ :-

( i ) ( F – ρv2 ) / ( F – ρv2 ) ≤ e (µ.cosec β) θ

This relation specifies the maximum ratio of belt tensions which a given belt/pulley interface cansupport without gross slip. The ρv2 term, often conveniently if erroneously referred to as the centrif-ugal tension, detracts from the interface's useful tension ratio capabilities. We define f ≡ µ.cosecβ asthe effective coefficient of friction which reflects the amplification of the actual coefficient µ bywedging action in the groove of semi-angle β.

A drive comprises two pulleys, potentially with different values of µ, β and θ - although the slackand tight strand tensions are common to both pulleys. The maximum tension ratio which the drivecan support without slip on either of the two pulleys is therefore :

( 3 ) ( F – ρv2 ) / ( F – ρv2 ) ≤ e( fθ )min where f ≡ µ.cosec β and ( fθ )min ≡ minimum [ ( fθ )1, ( fθ )2 ]

If f is the same for both pulleys, as is usual when both pulleys are grooved, then the smaller pulleywill be limiting since θ1 <θ2 , but this does not necessarily apply to V-flat drives ( see below ).

A representative coefficient of friction is 1/6, and this together with a wedge semi-angle β = 19o and

a wrap angle of 180o in a 1:1 drive, corresponds to a static tension ratio F /F = e0.512π = 5.0

The effects of tensionratio and pulley diame-ter on loads generallymay be appreciated bythe three pulleys in 1:1drives illustrated here,all of whose torquecapacities are 2 kNm.The second pulley isone fifth the diameterof the first, so the forces have to be five times those of the first for the same torque - clearly this hassevere repercussions not just on belt fatigue but also on shaft bending, bearing reactions etc. Thethird pulley is the same as the first, however the tension ratio is much smaller due to a reduced RHSof ( 3) - again larger forces occur right throughout the drive train.

Large pulleys and tension ratios are clearly beneficial, however tension ratios are limited by the fric-tion coefficients and wrap angles which are encountered in practice. Attempts are sometimes made

5

1

25

5

24

20

necessary belttensions, kN( ρv negligible )2

pitch diameter, m 1 0.2 1

tension ratio 5 1.25

total reaction, kN 6 30 44

V-Belt Drives 5

sketch here by distortion of cross-sections due to pulley contact and the ten-sile members

- the complete disregard of initial tension and tension variation between belts- non-metallic materials may not follow the stress/strength theories applicable

to metals, and will probably not behave elastically- the inexactness inherent in any fatigue analysis, and in particular, the ques-

tionable accuracy of Miner's rule- probable operation of the drive at less than its gross-slip-limited capacity- the belt is not ideal - it is not perfectly circular on one side of a tangent point with the pulley,

and perfectly straight an infinitessimal distance away on the other side of the tangent point- the belt tends to ride out of the groove as its tension reduces, so increasing the pulley's effective

pitch diameter- variations in the coefficient of friction- neglect of transients such as high starting torques in the fatigue asssessment- the composite nature of the belt, with components having different load-life relations, etc.

Despite these reservations, experiments show that the design equation ( 5a) describes the generaltrend of results. Its solution for drive life T whilst transmitting power P is straightforward, but theinverse solution for the power capacity for a given life must be carried out iteratively, except in theparticular case of a single belt ( z=1) in a 1:1 drive ( D1 = D2 = D), whereupon ( 5a) may be trans-posed to give :-

( 5b) Pπ = kπ v [ F ( L/2vT )1/m – M/D – ρv2 ] ; where kπ ≡ 1 – e–fπ

The variation of capacity with velocity is plotted below for two typical V-belts - A1750 and D6100.Effectiveness is explained in the next section.

The capacity is evidently a fatigue-limited value which depends upon the belt size and whichdecreases as the fatigue life increases ( ie. the first term on the RHS of ( 5b)) - this value beingreduced by bending and by centrifugal effects ( the second and last terms on the RHS). Large pul-leys clearly improve belt capacity, though the reciprocal nature of the bending term means thatincreasing the diameter of a pulley which is already large may have little effect on belt capacity.Centrifugal effects cause the capacity to peak and eventually to decrease with increasing speed.

Application of the design equation ( 5a) to deduce the life of a wedge belt drive is illustrated by theexample below.

SPZ373034.04

0.0728312.8

1592-

SPA623587.480.1287

13.02278

-

SPB9057182.00.2037

13.43204

-

SPC16585555.10.4123

13.85070

-

Y318

0.5870.03962

11.11466

6

Z734

3.0450.04678

11.11780

8

A321623.93

0.0968211.111717

11

B553562.720.166611.112266

14

C9842174.40.295811.113653

19

D20080618.50.603511.116112

24

SectionF (N)M (Nm) (kg/m)mL (mm) (mm)

ρ

∆

TABLE 1 - BELT PROPERTIES ( corresponding to f = 0.512 )

o

of contact θ2, the equivalent damaging force in the element is increased also by M/D2. When the ele-ment leaves the pulley at 2 the bending term no longer applies and the element's load is merely thetight side tension F. The reverse process occurs as the element passes around the smaller pulley, theelement tension returning to F after 4.

Each element evidently undergoes a double-peaked fatigue cycle during one revolution of the belt.Since D1 ≤ D2 it follows that the F1

* peak cannot be less than the F1* peak. A

snub idler, that is an idler located on the outside of the belt to increase the arcof contact on the small driveR, should normally be avoided since it introducesreversed bending and hence increases the stress range markedly.

It is required to determine the belt life under this double-peaked fatigue cycle.

For many cyclically loaded components, including belts, there exists an approximate log-log linearrelationship between :

- the load on the component, characterised by the equivalent damaging force F* in the case of abelt, and

- the life of the component N* at failure ( expressed in cycles of load application ), viz :-life * loadindex = constant or, for a belt N* = ( F / F* )m

in which both the index m and the reference load F are constant properties of the belt, Table 1.

A belt subjected only to the force varying between zero and Fi* willfail after Ni* cycles as sketched. At a certain point in time prior to fail-ure it will have undergone Ni cycles, so the cumulative fatigue dam-age at that instant may be taken as N i/Ni*. Thus when the load is firstapplied Ni ≈ 0 and no appreciable damage has occurred; when Ni ≈Ni* damage is almost 100% complete and so failure is imminent.

In a drive situation every belt element undergoes a cycle character-ised by two peaks of magnitude F1* and F2*. If the life of the belt underthese combined loads is Nc then Miner's rule states that failure occurswhen the cumulative damage via both mechanisms reaches 100%, ie. :N1/N1* + N2/N2* = 1 where by the nature of loading : N1 = N2 = Nc

The effects of further pulleys could be incorporated if appropriate atthis juncture.

The elapsed time for one such belt cycle is L/v, so the drive life T in units of time ( not to be con-fused with torque ) is T = Nc L/v. Substituting for the lives Ni* via the load-life equation and for theloads Fi* from ( ii) leads to the desired result :-

( 5a) ( P/ z kθv + M/D1 + ρv2 )m + ( P/ z kθv + M/D2 + ρv2 )m = Fm ( L/vT )

This is the final fatigue design equation for the drive - like all design equations it relates the fourmajor aspects :

- load power per belt ( P/z), velocity ( v)- dimensions pulley diameters (D1, D2 ), belt length (L) and friction/arc factor ( kθ)- material ie. belt section and corresponding properties ( F, M, ρ, m) from Table 1- degree of safety or more appropriately in a fatigue situation, the drive life ( T)

Properties found by curve fitting to belt capacities cited in the Code, are given in Table 1 opposite.

The design equation appears to be a relatively sophisticated model, but it is important to realise thatmost of the sophisitication lies in the arithmetic since the model is based upon some rather grossassumptions, including :-

- neglect of significant 3-D stressing, apart from bending and direct tension, as indicated in the

snubidler

N* = (F/F*)m

loadF*

Ni N*i

F*i

life, N

F*1F*2

timeone cycle of the belt

V-Belt Drives 6

Effectiveness

The design equation ( 5a) may be normalised by the right hand side to give :

( 5c ) ( p + b1 + c )m + ( p + b2 + c )m = 2 ; where p = ( P/Fkθvz ) ( 2vT/L )1/m

The three terms in each bracket of this equation represent the fractional consumption of finitefatigue life by useful power transmission, and by deleterious bending and centrifugal effects respec-tively. The first term, p, is interpreted as the effectiveness of life usage. If bending and centrifugaleffects are negligible then the majority of the life is devoted to power transmission and the effective-ness approaches 100%. If, on the other hand, bending or centrifugal effects predominate, then thebelt is not being used effectively and a low value of p results. Effectiveness should not be confusedwith efficiency.

The effectiveness of the two belts graphed above illustrates a monotonic drop-off with increasingspeed, and effectiveness is reduced markedly by increased bending over small pulleys. Lackingcosts, effectiveness may be used to assist in the selection of the optimum drive for a given duty.Drives cited in manufacturers' brochures correspond to values between about 30 and 70%.

Drive Selection

Input to the V-belt drive selection process is the drive specification which defines :- the power capacity of the drive and the small pulley's shaft speed (rev/s)- acceptable limits of the speed ratio R - though it is common for only a single nominal speed

ratio to be specified, around which the designer must set bounds commensurate with the duty- limits of pulley centre distance - once again the designer may have to define a tolerable band

around a sole nominal value or, lacking even this, around 2D1√ R+1)- limits of acceptable belt fatigue

life: it is rare for these to be statedexplicitly, and the designer mayhave to adopt limits in line withthe duty ( see the table here ) andeconomic belt replacement fre-quency, or with the usual com-mercial expectation of 26 kh.

Drive selection involves choice of size(B, SPA &c) number and pitch length ofthe belt(s), and diameters of the twopulleys. As the design equations cannotbe solved in closed form, and sincemany parameters are discrete, a trial-and-error selection process must beadopted. The steps illustrated belowform a useful framework for this, andshould be read from top to bottom.

The process starts with the choice of a suitable belt size - aided but not prescribed by a carpet dia-gram - with belt properties from Table 1 and with corresponding lists of readily available beltlengths and pulley diameters. The smaller pulley diameter is next selected, and is usually taken assmall as possible for compactness consistent with acceptable belt life and effectivness. Belt speedand bounds on large pulley diameter follow from the desired speed ratio limits and equations ( 1) ie.

NORMAL BELT LIFE REQUIREMENTS

Industrial khequipment with long operating hours or continuous

operation eg. fans, pumps, conveyors 12 – 26hand tools, office equipment 7.5 – 20equipment with intermittent or occasional operation 6 – 12

Agriculturalstationary equipment with long operating hours or

continuous operation eg. fans, pumps, conveyors 6 – 12stationary equipment with intermittent operation 2 – 6mobile equipment eg. harvesters, sowing machines,

manure spreaders, hay balers 0.5 – 1

Automotivepassenger cars, vans 1 – 3lorries, buses, tractors, road construction machines 5 – 10

Home appliancesheating, ventilation, air-conditioning 5 – 10washing machines, tumbler dryers, dish washers 1.5 – 2sewing machines, lawn mowers, hand tools 0.2 – 1

A1750

pulleydiameterD (mm)

fatiguelife, T (kh)

0.22

20

0 30 40 502010

75

75

75

300

300

300

15

10

5

0

speed v (m/s)

capa

city

, P

(kW

)π

5

limit

0 30 40 502010

400

100

80

60

40

20

0

400

400

1600

1600

1600

speed v (m/s)

fatiguelife, T (kh)

0.22

20

capa

city

, P

(kW

)π

pulleydiameter D (mm)

limit

fatiguelife, T (kh)

0.22

20

0 30 40 502010

effe

ctiv

enes

s, p

(%

)

100

40

20

0

160060

80

400

400

400

1600

speed v (m/s)

pulleydiameter D (mm)

limit

fatiguelife, T (kh)

0.22

20

0 30 40 502010

100

40

20

0

75

75

75

300

300

60

80

speed v (m/s)

effe

ctiv

enes

s, p

(%

)

pulleydiameter D (mm)

limit

D6100

D6100

A1750

EXAMPLE 45 kW are transmitted by 2 m long wedge belts which operate at full capacity. The diame-ter of the motor pulley is 180 mm, and it rotates at 1440 rpm; the output pulley is 400 mm diameter.Determine the effect of belt number on life if the belts are (a) SPA; (b) SPB.

Given P = 45 kW ; n1 = 24 Hz ; so, from (1) : v = πD1n1 = π (0.18)(24) = 13.6 m/s

Geometry : D1 = 180 mm ; D2 = 400 mm ; L = 2000 mm so, from (2a) (2b) and (2c) :-C = 533 mm ; γ = 0.208 rad; θ1 = θmin = 2.726 rad

Since the belts operate at full capacity, slip is imminent and ( 5a) applies. Assume f = 0.512, so

kθ = 1 – e–fθmin = 1 – e–0.512 * 2.726 = 0.752

and, in ( 5a) P/kθv = 45*103/0.752*13.6 = 4400 N

Properties of SPA from Table 1 : F = 6235 N ; M = 87.48 Nm ; ρ = 0.1287 kg/m ; m = 13.0

and, in ( 5a) ρv2 = 0.1287*13.62 = 23.8 N

( P/kθvz +M/D1 + ρv2 )/F = ( 4400/z +87.48/0.180 +23.8 )/ 6235 = 0.7057/z + 0.0818( P/kθvz +M/D2 + ρv2 )/F = ( 4400/z +87.48/0.400 +23.8 )/ 6235 = 0.7057/z + 0.0389

So, from ( 5a) with z = 6 belts for example :-

L/vT = ( 0.7057/6 + 0.0818 )13.0 + ( 0.7057/6 + 0.0389 )13.0 = 8.23 E-10 ∴ T = L / 8.23 E-10 v = 2.0 / 8.23 E-10*13.6 = 1.79 E8 s = 50 k.hours ( kh)

Repeating these latter steps with different numbers of belts, and then repeating all calculations in a similarmanner for SPB belts, illustrates the marked effect of number of belts upon the resulting belt life, thus :

Number of belts 1 2 3 4 5 6 8 10 12SPA life ( kh ) 6.1E-7 1.6E-3 0.11 1.6 11 50 400 1700 4800 [ 26kh ≡SPB life ( kh ) 3.0E-5 0.035 1.1 9.5 40 120 500 1300 2600 10 years]

Only one significant figure is really justifiable when quoting these lives, since the analysis is so approximateas mentioned. The general trend of the results is surprising - at normal lives and as expected, the larger SPBbelts live longer than SPA for a certain duty, or, alternatively, fewer SPB belts are required for a given life.However the reverse is indicated when loads are heavy and large numbers of belts are necessary. This is dueto the changing relative significance of the power transmission, bending and centrifugal terms in the designequation. In fact, when loads are heavy and large numbers of small belts are indicated, then fewer largebelts are usually more economical.

V-Belt Drives 7

Codes and belt suppliers' literature do not employ the design equation ( 5a) directly, but presentinstead a selection procedure based on charts and factors which are derived from approximations tothe design equation. Since we cannot always have software like “V-belts” handy it is advisable to beconversant with these approximate solutions.

Approximate Solutions

Codes and commercial literature employ a tabular approach to the solution of ( 5a) for the powercapacity. This is based upon the following approximations for large index, m :

( iv ) ( 1 + x/m )m ≈ ex ; x1/m ≈ 1 + 1/m ln x

Without going into unnecessary detail, the technique depends upon the definition of base, or refer-ence, values of :

- life, To, corresponding to 10 years at 50 weeks/year, 6.5 days/week and 8 hours/day - ie. 26kh - and is common to all belt sizes;

- length, Lo, which is a function of belt size ( refer to Table 1).

The basic rating , Po, is defined as the power capacity of a single belt of a particular size in a 1:1drive, whose length is the reference length and whose life is the reference life. Thus, from ( 5b) :

Po = kπ v [ F ( Lo /2vTo )1/m – M/D – ρv2 ] ; and using ( iv) :-

( v ) Po ≈ v [ K1 – K2/D – K3v2 – K4 log(v) ] in which K1 through K4 are size-

dependent constants, ie. belt properties. The basic rating is thus evidently a function of belt speedand pulley diameter only - see the above graphs and the belt tables in commercial literature.

The steps in designing with typical belt tables are as follows :-1 Determine the : design power = actual power * service factor where the tabulated service

factor is a combination of the above duty factor and a life factor which caters for lives otherthan the reference life.

2 Select an appropriate belt size, guided by the above carpet diagram of design power versussmall pulley speed. This diagram is indicative, not prescriptive.

3 Ascertain the power rating of a single belt of that size; this is a tabulated function of belt speedand small pulley diameter as noted above, the tabulation also making provision for the largepulley diameter by approximations similar to ( iv).

4 The arc factor, again tabulated, caters for contact arcs other than 180o on the small pulley; it is

just the ratio kθ/kπ.5 A length factor reflects the effect of a belt length other than the reference length; again the fac-

tor is tablulated.6 The number of necessary belts of that particular size follows from :

belt number ≥ design power / ( belt power rating * arc factor * length factor )

The tabular technique was devised long before the advent of personal calculators and computers,when calculations were necessarily and laboriously carried out by slide rules. The technique is easierto use than equation ( 5a) if software such as “V-belts” is not available, however it gives no indica-tion of the lives which result from the choice of a belt number different from that given by the tech-nique. Whatever approach is employed, selection depends on costs. Belts in a drive with large pul-leys may live longer than those in a drive with small pulleys, but do the savings accruing from lessfrequent belt replacement outweigh the cost of the larger pulleys? Is the net present worth of a fewbelts requiring frequent replacement greater than that of a larger number of belts which seldomneed maintenance?

STEPS IN V-BELT DRIVE SELECTIONinput choice result

BELT SIZE properties (Table 1)

diameters & lengths

SMALL DIA.

small pulley speed belt velocity (1)

ratio limits large diameter limits (1)

LARGE DIA.

centre limits length limits (2a)

LENGTH centres (2b), wrap angles (2c)

capacity, life limits number limits (5a)

NUMBER life (5a)

RminD1 ≤ D2 ≤ RmaxD1.

If feasible, a large pulley D2 is nextchosen from the diameters availa-ble within this range, and beltpitch length limits Lmin, L max cor-responding to the stated centredistance limits Cmin, Cmax workedout from ( 2a). Again if feasible, abelt length Lmin ≤ L ≤ Lmax is nextchosen from the lengths available,and the geometry finalised ( 2b,c).

The sole design parameter whichnow remains to be chosen is the number of belts, z. Rather than follow the schema by working outlimits to the number of belts consistent with stated life constraints, it is probably easier just to choosevarious belt numbers on a trial-and-error basis and evaluate belt life from ( 5a) until a suitable lifeeventuates.

This process is repeated with different initial choices of belt size and small pulley diameter, to pro-vide a bank of solution candidates with acceptable lives. Selection of the 'best' solution is basedthereafter upon economic considerations. The drive-selection program “V-belts” operates in thisway, with nested loops corresponding to belt size, small pulley diameter etc, as in the schema above.Users should check that belt lengths and pulley diameters cited by it are available locally. “V-belts”is capable of handling V-flat drives ( see the dialogue appended ) and alerts if the aforementioned speedlimits are exceeded or if dynamic balancing of the pulleys is necessary.

The design power input to the selection process must incorporate a duty factor to allow for shock,high starting torques and other expected non-uniformities :-

( iii ) design power = actual nominal power * duty factor ( from the table below eg.)

Note that a duty factor of 1.5 implies a life reduction of approximately 1.512 = 130.

DUTY FACTORS Driving machinery 'soft starts' 'heavy starts' Electric motors, a.c. star-delta start direct-on-line start

Consult Codes for Electric motors, d.c. shunt wound series & compound woundabnormal applications Int. comb. engines four or more cylinders less than four cylinders

All prime movers with fluid,

Driven machinery dry or centrifugal couplings

Agitators(uniform density); blowers, fans & exhausters(up to 75kW); centrifugal compressors & pumps; belt 1.0 1.1conveyors(uniformly loaded).

Agitators and mixers ( variable density ); blowers, fans &exhausters(over 75kW); rotary compressors & pumps otherthan centrifugal; belt conveyors(non-uniformly loaded); 1.2 1.2generators & exciters; laundry, printing & woodworkingmachinery; lineshafts; machine tools; rotary screens.

Textile, rubber & brick machinery; reciprocating pumps& compressors; bucket elevators, hoists & heavy duty 1.2 1.4conveyors; hammer mills; vibrating screens, pulverisers & quarry plant; punches, presses & shears.

Crushers (gyratory, jaw or roll); mills (ball, rod or tube). 1.3 1.5

V-Belt Drives 8

EXAMPLE A pivoted motor drive is envisaged for a launderer which absorbs 7 kW at its design speed of 500rpm. The launderer is equipped with a 500 mm diameter flywheel which may be utilised in a V-flat drive.

(A) Select a suitable motor and belts.(B) Select a suitable pivot position given that the motor should lie vertically underneath the launderer pul-

ley for overall compactness.

(A) Assuming a drive efficiency of around 95%, the motor capacity must be at least 7/0.95 = 7.4 kW.

Select, for example, an ABB MBT 132M squirrel cage motor, whose rating is 7.5 kW at 1440 rpm.

The duty factor will be 1.2 from the table above. Since it is unlikely that the drive can be chosen to power thelaunderer at exactly 500 rpm, and since the launderer's torque-speed behaviour is unknown, the belt driveshould be capable of handling the full motor output. The design power will thus be 1.2 * 7.5 = 9 kW.

The carpet diagram indicates that B belts are suitable. From Table 1, ∆ =14 mm so D2 = 500+14 = 514 mm

The motor pulley must be around 514 * 500/1440 = 179 mm. Take nearest standard : D1 = 180 mm so thatthe reduction is R = 514/180 = 2.86 and the belt speed is v = πD1n1 = π * 0.18 * 1440/60 = 13.6 m/s

Geometry : Lacking anything definite regarding layout, take C ≈ 2*180√( 2.86+1) = 707 mm

From ( 2a) L ≈ 2544 mm, so select nearest standard L = 2500 mm

It follows from ( 2b), ( 2c) that C = 685 mm and 2γ = 0.493 rad.

The belts will slip first on that pulley with the lesser fθ product. From ( 3) and ( 4b) with µ = 1/6

for the small pulley (f θ)1 = ( π – 0.493 )*(1/6) cosec19o = 1.356

for the large pulley (f θ)2 = ( π + 0.493 )*(1/6) cosec90o = 0.606

. . . . that is, the large flat pulley is limiting and kθ = 1 – e–(fθ)min = 1 – e–0.606 = 0.454

Applying ( 5a) with say z=3 belts as a trial : P/ kθvz = 9 E3 /0.454 * 13.6 * 3 = 487 N

ρv2 = 0.1666 * 13.62 = 31 N

(L/vT) = ( (487+62.72/0.17+31) / 5535 )1/0.09+ ( (487+62.72/0.514+31) / 5535 )1/0.09 = 1.15 E-9 no units

∴ T = 2.5/13.6 * 1.15 E-9 = 44 kh which is acceptable.

So the choice of 3 B-belts is vindicated. This solution candidate is one of many which the drive design pro-gram “V-Belts” is capable of generating; the corresponding dialogue appears below.

(B) The salient dimensions of the motor selected are as shown here :-

A hinge location at (100, –180) seems appropriate and so is selected.

From ( vi), with φ = 90o : s = 137 mm; t = 57 mm.

In terms of normalised tensions, we have from ( 6), ( 4a) and ( 3) :

( a) 57 f + 137 f = 100 mm

( b) f – f = P / W v

( c) f / f ≤ e (fθ)min = 1.83

The mass of the motor itself is 50 kg. Allowing around 15% extra for pulley and base, and presuming springaugmentation to be unnecessary, then Wv = 7.6 kW in ( b).

The variation of the normalised tensions, as the powertransmitted through the drive increases, is plotted here -values being calculated from ( a) and ( b) simultaneously.Full load corresponds to the power reaching 7 kW, where-upon the RHS of ( b) is 7/7.6 = 0.92.

It is clear from the graph that this value cannot beattained, because gross slip will occur when the ratio oftensions reaches a value of 1.83 - that is when P/Wv =0.344 and the transmitted power is 0.344*7.6 = 2.6 kW. Thediagram thus really terminates at the 0.344 abscissa.

Thus the drive is unsuitable because the arbitrary choiceof the hinge point means that slip will prevent attainmentof full load. The choice of suitable hinge points is left asan exercise for the reader: a problem at the end of thechapter is devoted to this.

0 0.5 1.0 1.5

3

2

1

0

slip limit1.83

0.9

2fu

ll lo

ad0.3

44

f^

f

f^

f

P/Wv

250

132

100

180

H

tight

V-Flat and Pivoted-Motor Drives

A V-flat drive incorporates an ungrooved, or flat,large pulley in the form of an existing flywheel forexample, or a relatively cheap flat-belt pulley withits belt-tracking crown removed. V-belts only, andnot wedge belts can be used in this way. As thebelts contact the pulley on their flat bases ( β = 90o )

rather than in any grooves, the contact on the largepulley will probably limit the drive - both fθ prod-ucts must be evaluated so that the minimum may be used in ( 3) et seq.

The effective pitch diameter of the larger flat pulley, D2, is greater than its outside diameter by theamount ∆ ( refer to Table 1 ).

In the typical pivoted-motor drive shown at ( a) below, the base onto which the motor is bolted isfree to pivot about the fixed hinge H, adjacent usually to the tight strand, thus allowing the motor'sweight to tension the belt strands. Sometimes it is necessary to augment the motor's weight by aspring similar to the compression spring illustrated - however a spring is not always needed. Suchan arrangement is particularly suited to short centre drives where limited belt compliance wouldreduce shock tolerance and part-load life if the centres were fixed - the pivot enables the belt strandtensions and bearing reactions to vary beneficially with load.

The strand tensions F, F in each of the z identical belts will clearly depend on the motor's weightand the layout's geometry - in order to find the tensions a Cartesian system is erected at the motorshaft axis O as shown enlarged at ( b). The hinge H is defined by ( hx, hy ), and the shafts' centrelineis inclined at φ to the x-axis. The moment arms about the hinge of the slack and tight side tensionsare respectively :

( vi ) s = hx sin( φ + γ) − hy cos( φ + γ) + D1/2 where γ is as given by ( 2a)t = hx sin( φ − γ) − hy cos( φ − γ) − D1/2

Angular impulse/momentum theory is then applied about the hinge for the system of sketch ( c),over the small time period δt during which identical elements of belt, mass δm, leave and enter thesystem with speed v. If the effective weight of motor ( with spring augmentation if appropriate ),pulley and base plate is W acting through the motor shaft axis, then, assuming speed and forces tobe constant and taking clockwise to be positive above :

moment of impulse about H = final moment of momentum about H less initial ditto

crownexaggerated

V-pulleyflat pulley"de-crowned"

∆2

D2

drivenpulley

spring

motor hingeH

base

line ofcentres

hx

h y

x

ys

γ

φ

D1ø

O

H

( a) ( b)

Hs

t

O

δm v

δm v

W

z F

z F

vv

h x

^

system

( c)

V-Belt Drives 9

this limiting case still requires adhesion locally at z.

A relatively small torque load is applied to each of thepulleys in ( c), setting up unequal strand tensions F1and F2 as shown at ( d). The pulleys are mirror imagesof one another, but the torque destroys any symmetryabout the line of centres. The model assumes that theadhesion/slip zone system on each pulley does notchange significantly under low load, apart from rotat-ing bodily through the angle φ1 as shown in ( B)below. z now divides the total contact arc into two dis-similar regions in each of which the friction acts in asingle sense, as shown at ( d). In the larger or positiveregion adjacent to the tight strand, friction acts on thebelt in the same sense as the load torque, whereas inthe negative region adjacent to the slack strand, fric-tion on the belt is opposite to the load sense. The loadis evidently transmitted by the net friction resultant. Itfollows that each region in general consists of an adhe-sion zone and a slip zone, and that the minimum ten-sion in the belt is the tension at z, not the slack strandtension.

Upon increasing the load, the zone system rotatescausing the positive slip zone to enlarge and the nega-tive slip zone to shrink and eventually disappear whenthe interface i between adhesion and slip zones coin-cides with the slack strand tangent. Eventually thenegative adhesion zone also vanishes, whereupon zcoincides with the slack strand tangent, and all frictionis in the positive sense contributing to torque transfer -ie. all cross-sections are rotated in the same sense as sketched in ( e). As the torque finallyapproaches full-load, the positive adhesion zone shrinks in turn until finally at full-load, adhesiongives way to imminent slip at the slack side contact point - gross slip is imminent and further loadincrease is impossible.

The relative motion of the beltover the pulley surface whichthe model allows is permissi-ble, even at no-load, becausegross slip is always preventedby adhesion somewhere - untilfull-load is reached and slip orimminent slip is universal, cor-responding to the slip-limited classical theory. When the model is further developed and applied topractical belts, it is expected that they will be found to be relatively rigid in shear and the extent ofthe adhesion zones will be negligible. However positive and negative friction regions are still pre-dicted, with the latter shrinking and vanishing when the load reaches its limiting value.

Development of the theory to enable evaluation of the part-load tension variation is unnecessary

T

F2

F1

F0

F0

F1

F2

TT

z

zz

z

z

r

rp

z

positive

negative

frictionsenses :-

line of symmetry

line of centres

( a) wrapping

( b) initial tension: no-load

( c) part-load

( d) part- load

( e) approaching full-load torque

F0

F0

adhe-sion

slip

adhe-sion

slip

z

i

i

( A)no load

F2

F1

adhe-sion

adhe-sion

slip

slip z

i

i

φ z

T( B)low load

( z F t + z F s - W hx ) δτ = δm v s − ( − δm v t ) where δm = z ρ v δτRearranging this leads to the necessary inter-relationship between the two strand tensions of a piv-oted motor drive :

( 6 ) ( F − ρv2 ) t + ( F − ρv2 ) s = W hx /z or, normalising the tensions

f .t + f .s = hx where f = ( F −ρv2) z/W; f = ( F −ρv2) z/W

As the power transferred in a pivoted motor drive increases, the two strand tensions vary, beinggiven by ( 4a) simultaneously with ( 6). The ratio of the two tensions also increases until it reachesthe slip-limited value, given by ( 3), thus setting an upper limit to the power that can be transmittedby the drive. It is clear from ( 6) that a hinge situated close to the motor weight's line of action ( hx→0 ) nullifies the weight's influence, and is therefore not recommended.

The following synthesis example relates to a typical V-flat pivoted-motor installation. It illustrateshow the hinge location must be selected with care.

Traction MechanicsIt has been pointed out that the classical thin belt, whose analysis led to the design equation ( 5a),models only full-load, when gross slip is imminent - it ignores adhesion as a possible mechanism fortraction between belt and pulley. Some aspects of a model ( Wright, Bibliography) which includesboth the adhesion and slip traction mechanisms, and which helps to explain the effects of initial ten-sion and of creep will now be described. For simplicity the model is developed initially only for aflat belt in a 1:1 drive for which dynamic effects ( ρv2) are insignificant. The belt's thickness is notnegligible as was assumed in the classical theory, but is sufficient for shear deflections to be signifi-cant, thus allowing tangential friction between belt and pulley to be sustained by slip and adhesion.

Assembly and loading of the model are illustrated opposite. At ( a) the belt is wrapped looselyaround the pulleys whose radius is rp. Edge views of belt cross-sections are shown - these are radialin way of the pulleys since the belt is taken to be wholly compliant in flexure. The pattern of cross-sections is symmetric about the line of centres and about the transverse line of symmetry lying mid-way between the pulleys.

At ( b) an initial tension, Fo - acting through the neutral axis of the cross-section at radius r - isinduced in both strands by the pulleys being forced apart and locked in that position. The extensionof the straight strands and of the belt in way of the pulleys gives rise to friction and causes shear ofthe belt - so cross-sections ( which are assumed to remain always straight ) rotate from their previ-ous positions in ( a). By symmetry, the two cross-sections z which lie on the line of centres cannot besheared, and therefore remain radial after initial tensioning. Symmetry also requires the sense of thefriction forces to reverse across z, so there can be no friction force at z. Slip is thus impossible at zand the belt must adhere to the pulleys at z.

Tangential friction forces arise at cross-sections other than z. Depending on the belt and pulleyproperties and the local surface pressure, these local friction forces may or may not reach breakawayvalues. If breakaway is not attained then adhesion prevails; if breakaway is reached then the beltmoves locally relative to the pulley, ie. slip prevails. The tendency towards slip movement mustincrease the further a cross-section is removed from z on a pulley, figure ( b). The zero shear loca-tion, z is thus surrounded by two adhesion zones, each of which abuts a slip zone which leads intothe corresponding straight strand as shown at ( A) below. The maximum circumferential extent ofeach adhesion zone, θA, and the size of the slip zones, will depend upon the belt and surface charac-teristics, however symmetry about both the line of centres and the transverse line is maintained. Ifthe belt is thin and relatively rigid in shear then the adhesion zones will be insignificant, however

V-Belt Drives 10

Bibliography

AS 2784-1985 : Endless Wedge Belt and V-belt Drives, SAA BS 3790-1981 is virtually identical

Firbank TC, On Forces Between Belt and Driving Pulley etc, ASME paper 77-DET-161

Freakey PK & Payne AR, Theory and Practice of Engineering with Rubber, Applied Science 1978Not relevant to belts, but a useful general reference for rubber behaviour, with applications such asvibration mountings, fluid seals etc.

Oliver LR, Johnson CO & Breig WF, V-Belt Life Prediction and Power Rating, Trans ASME, Journalof Engineering for Industry, Feb 1976, pp 340-347

Palmgren H, The V-belt Handbook, Studentlitteratur/Chartwell-Bratt 1986

Shigley JE & Mitchell LD, Mechanical Engineering Design, McGraw-Hill 1983

South DW & Mancuso JR ed, Mechanical Power Transmission Components, Dekker 1994

Wedge Belt, V-belt, Timing Belt and Chain Drives, Fenner Pty Ltd - catalogue ( latest edition )

Wright DC, Adhesion and Slip in Flat Belt Traction Mechanics, Department of Mechanical &Materials Engineering, University of Western Australia 1995

Sample Dialogue with “V-Belts”

here - the system is statically indeterminate and must therefore be tackled using equilibrium, com-patiblity and the belt constitutive laws in both tension and shear. The model has been mentionedonly to give some understanding of the role of initial tension in a fixed centre drive. Clearly, oncethe tension variation around the pulley has been found, it is possible to integrate the resultingstrains around the contact, and to find the total length of the belt. In a fixed centre drive this lengthremains constant as the load changes - extensions in one part of the belt due to local increase in ten-sion must be offset by a decrease in another part.

Typical results from the model are shown here; these cor-respond to a fixed centre drive for which the coefficientof friction is 0.5, the maximum extent of an adhesionzone θA is 60o, and the slip-limited tension ratio F1/F2

is 4.05. The interesting aspect is the essentially lineardependence of the strand tensions with load. Thisapproximate linearity was found for all cases studied viathe model, including thin belts with friction coefficientsas low as 0.1. It was found that, for the most part, theratio of the slopes of the two tension variations fellwithin the bounds :

( 7 ) 0.66 ≤ ( F1 – Fo )/( Fo – F2 ) ≤ 0.80

- that is Fo's increase to F1 is less than itsdecrease to F2. It is noticed that this linear behaviour issimilar to that of the pivoted motor drive ( 6).

Apart from giving a clearer understanding of the mechanics involved in power transfer by belts themodel enables part- and no-load tensions to be evaluated, and more economic drives chosen.

The example utilised a maximum tension ratio of 5 which is characteristic of V- and wedge-belts,however it must be pointed out that the model has yet to be proved suitable for such belts.

150

120

90

60

30

0

1.5

1.0

0.5

0.00 20 40 60 10080

load, %

F0

F1

F2F0

φz

( deg )

EXAMPLE A belt which behaves according to the above traction model must be selected to transmit 3 kW

continuously, however it might be called upon to transmit 4 kW in an emergency. The belt speed is 20 m/s but

centrifugal effects are negligible. The maximum possible friction-dictated tension ratio is 5. The speed ratio is 1:1.

Gross slip occurs in peak emergency conditions, so from ( 4a) ( 3) :

F1 – F2 = 4 E3/20 = 200 N, F1/F2 = 5 solving F1 = 250 N; F2 = 50 N

Under normal continuous conditions at 3 kW, from ( 4a) ( 7) :

F1 – F2 = 3 E3/20 = 150 N,

0.66 ≤ ( 250 – F1 ) /( F2 – 50 ) ≤ 0.80 from similar triangles :-

Solving gives mid-range values F1 = 230 N, F2 = 80 N, Fo = 165 N

The belt would be selected on the basis of these last figures and the belt's known life-

load relation - clearly a more rational selection process than any based on the classi-

cal theory.

250

50

F0

F1

F2

150

loadten

sio

n

V-Belt Drives 11

9 A squirrel cage motor is usually equipped with deep groove ball bear-ings, but life considerations might require replacement of the drive-endbearing by a larger capacity roller bearing when :- the shaft load is heavy due to a small belt pulley for example, or- the load overhang (x in the diagram) is large. An ABB motor type M2BA 280 SMA delivers 75 kW at 1485 rpm viaa fully loaded belt drive comprising 5 SPB 3150 belts on Ø 212 and 630 mm pulleys at 889 mmcentres. From catalogues [ABB, Fenner] the pulley width w = 102 mm, the motor shaft length E= 140 mm, and shaft radial loads F corresponding to two different bearing lives at two loadpositions along the shaft are : life (kh) 40 63

F0 at x=0 (kN) 7.32 6.29FE at x=E (kN) 6.18 5.31

Will the drive-end ball bearing last for the target life of 25 kh or is a roller bearing needed?

10 This concerns part-load belt tensions and components with different load-life equations.Select a squirrel cage motor and fixed centre belts to drive an agitator at about 700 rpm for 1 khper annum. The power demand varies cyclically as shown and a 3-year belt replacementperiod is acceptable.The drive should be compact, but not to the extent that the motor'susual ball bearings have to be replaced by roller bearings. For the pur-poses of this problem, motors' maximum shaft loads tabulated in theMotors chapter refer to loads at the end of the shaft ( FE of the previousproblem ), to bearing lives of 40 kh, and to a load-life index of n = 2.5.

E

xFw

power, kW

time,min

328

14

0 15

V-BELT DRIVE - PROBLEMS

AS 2784 Endless Wedge Belt and V-belt Drives is referred to as 'the Code'.

1 A belt drive incorporates a small pulley of 100 mm diameter and a belt whose length is 1100mm. For speed ratios of (a) 1.5:1 (b) 2:1 (c) 3.15:1, calculate the theoretical shaft centre distanceand angle of wrap on the small pulley. [ 353, 310, 193 mm]

2 (a) Use the belt propertiesmof Table 1 to calculate the basic rating of an A-section V-belt withtwo Ø 100 mm pulleys rotating at 4200 rpm. Check, using the Code tables. [ 3.07 kW]

(b) If the pitch length of the above belt is 3080 mm, what then is the rating and what is the corre-sponding power correction factor for pitch length? Check this latter value with the Code. [ 3.50 kW, 1.14]

(c) If the drive in (a) is required to last for only 10 khr, by what percentage is the above capacityincreased? [ 24%]

3 Plot rating versus belt speed, similar to the curves on page 6, for an A-section belt. Use pulleydiameters of 75, 132, 250, 500 and 1000 mm. Superimpose upon this, trajectories of constanteffectiveness : 20, 40, 60, 80 and 100%. Comment upon the effect of pulley diameter on rating asthe diameter increases.

4 A V-belt drive employs a single B belt of length 2300 mm, together with 200 and 400 mm diame-ter pulleys. The smaller pulley rotates at 1440 rpm.(a) What is the capacity of this drive using the Code method ? [ 6.04 kW](b) What is the life of the drive when transmitting 6.04 kW ? [ 31 khr](c) Repeat (a), but use ( 5a) with the standard life of 26 khr. [ 6.19 kW](d) Check this last result using the program “V-Belts”(e) A multi-strand drive, otherwise identical to the above, is required to transmit 12 kW with a

duty factor of 1.3. Use ( 5a) to determine the number of belts required. [ 2.5](f) What life may be expected, if 2, or if 3 belts are used ? [ 5.0, 82 khr]

5 Two 1750 mm long A-section belts are incorporated into thedrive whose layout is sketched. The wrap angle on the 150 mmdiameter motor pulley “1" is 118o and the pulley rotates at 2880rpm. The 400 mm diameter driven pulley “2" absorbs the designpower of 10 kW. Pulley “3", of 80 mm diameter, is an idler andabsorbs no appreciable power. Estimate the life of the belts if . . . . .(a) the pulleys rotate clockwise, or(b) they rotate counterclockwise, or(c) the idler is removed and the centre distance between “1" and “2" increased accordingly.

[ 6, 0.6, 14 khr ]6 A 7.5 kW 1445 rpm squirrel cage motor, started direct-on-line, is required to drive a machine

tool at a speed of around 860 to 870 rpm. Duty is expected to be 7 hr/day, 5 days/week, 49weeks/year with 4 years between belt replacements. The centre distance should lie within therange 280 to 320 mm. Select a drive for this duty.

7 Select a suitable hinge location for the pivoted motor drive of the foregoing worked example inwhich an ABB MBT 132M motor transmits 7 kW to a launderer through three 2500 B belts on180 and 514 mm diameter pulleys, the motor lying vertically under the launderer pulley.

8 A blower absorbs 3.5 kW at its design speed of 650 rpm, and is equipped with a 260 mm diame-ter, 90 mm wide flat pulley. It is proposed to drive it by a pivoted motor, V-flat arrangement.Select a squirrel cage motor and finalise the drive, including pivot location.

1 2

3

Brakes 1

These are external rigid shoebrakes - rigid because the shoeswith attached linings are rigidlyconnected to the pivoted posts;external because they lie outsidethe rotating drum. An actuationlinkage distributes the actuationforce to the posts thereby causingthem both to rotate towards thedrum - the linings thus contractaround the drum and develop a frictional braking torque. The RH brake features improved hingelocations and integral posts/shoes.

The two hydraulically actuated rigid shoebrakes here are located internal to thedrum. The LH brake incorporates a rotat-ing cam which causes the shoes to expandand the linings to bear on the surroundingdrum. The RH brake features two leadingshoes, enabled by an individual (andmore expensive) hydraulic cylinder andpiston for each shoe. The terms leadingand trailing are explained below.

These rigid external shoe brakes act on the rope drumof a mine (cage) winder. The arrangement is fail-safe asan electric solenoid disengages the brake to allowmotion, but in the event of power failure the brakes areengaged automatically by the large springs visible atthe sides of the drum. Visible in the photograph are theactuation mechanisms with wear-compensating turn-buckles, the electric drive motor on the right, and thecage level indicator on the left.

In the hingedshoe brake atleft the shoes are hinged to the posts. As wear proceeds theextra degree of freedom allows the linings to conform moreclosely to the drum than would be the case with rigid shoes.This permits the linings to act more effectively and reducesthe need for wear adjustment. The commercial unit comescomplete with actuating solenoid.

About 5% of the heat generated at the sliding interface of a friction brake must be transferredthrough the lining to the surrounding environment without allowing the lining to reach excessivetemperatures, since high temperatures lead to hot spots and distortion, to fade (the fall-off in fric-tion coefficient) or, worse, to degradation and charring of the lining which often incoporates organicconstituents. Thorough design of a brake therefore requires a detailed transient thermal analysis ofthe interplay between heat generated by friction, heat transferred through the lining via the sur-rounding metalwork to the environment, and the instantaneous temperature of the lining surface.

BRAKES

A brake decelerates a system by transferring power from it. A clutch both accelerates and deceler-ates a system by transferring power to/from it. The two devices in rotary applications are thus verysimilar as they both transmit torque whilst supporting a varying speed difference across them.

Brakes take a number of forms - for example a system may drive a pump or electric generator, so thepump or generator acts as a brake on the system. However the most common brakes employ frictionto transform the braked system's mechanical energy irreversibly into heat which is then transferredto the surrounding environment. The friction mechanism is convenient since it allows force andtorque to be developed between surfaces which slide over one another due to their different speeds.One of the sliding surfaces is usually metal, the other a special friction material - the lining - whichis sacrificial. Wear (ie. material loss) of the lining must be catered for, and the lining usually needs tobe renewed periodically.

We examine only friction brakes in anydetail here - some common embodie-ments are first described . . . . .

This hydraulically actuated clutch com-prises a number of discs faced with liningmaterial which are connected alternatelyto input and output shafts by torquetransferring splines. The clutch is engagedby high pressure oil applied to an annularpiston pressing the discs against oneanother while they rotate at the different speeds of the two shafts. The normal pressure betweendiscs enable them to exert friction torque on one another which tends to equalise the two speeds.

A hydraulically activated disc brake comprises two opposingpistons each faced with a pad of lining material. When thehydraulic pressure is increased the pads are forced against therotating metal friction disc, exerting a normal force at each con-tact. The two normal forces cancel one another axially but causeadditive tangential friction forces which oppose the disc's motionand decelerate it.

The band brake below consists of a flexible band faced withfriction material bearing on the periphery of a drum which mayrotate in either direction. The actuation force P isapplied to the band's extremities through an actu-ation linkage such as the cranked lever illus-trated. Tension build-up in the band is identicalto that in a stationary flat belt.

The band cross-section shows lining materialriveted to the band. Allowance for lining wearis provided - when the rivets start to rub on thedrum they are drilled out prior to new liningsbeing riveted to the band.

seal

sliding

bushing

alignmentchamber

oil

piston

discs

splines

oil ways

P

wear allowance

cross-section of band

r i v e t l ining

caliper

fluid pressure actuation

pad

hydrauliccylinder

friction disc

Brakes 2

consist of three components - metal fibres for strength, modifiers toimprove heat conduction, and a phenolic matrix to bind everythingtogether. The characteristics of Ferodo AM 2, a typical mouldedasbestos, are illustrated. The coefficient of friction, which may betaken as 0.39 for design purposes, is not much affected by pressureor by velocity - which should not exceed 18 m/s. The maximumallowable temperature is 400 oC.

Linings are attached to shoes either by soft countersunk rivets or bybonding, though set-screws and proprietary fixings may be used inthe larger sizes. In order to withstand the inherent abrasion, matingsurfaces should be ferrous with a hardness of at least 150 BHN, or200 if the duty is heavy. Fine grain high tensile pearlitic iron is generally suitable. The interestedreader should refer to manufacturers' publications for further information.

Having ascertained the braking requirements from the system dynamics, we now wish to formsome idea of the leading dimensions of a suitable brake.

Practically achievable power density limitations apply to brakes as they do to other mechanicalplant such as engines and heat exchangers. For a given size of brake there is a limit to the mechani-cal power that can be transformed into heat and dissipated without lining temperatures reachingdamaging levels. Brake size is characterised by lining contact area, A, so denoting this maximumsafe power density as Rp we have, for a reasonable lining life :-

( 4 ) Pm /A ≤ ( Pm /A )critical ≡ Rp (kW/m2)

Experience suggests the following values of Rp for various types of brake in different applications.

The table indicates that the improved heat transfer capabilities of disc brakes compared to othertypes enables them to handle greater power densities - per unit area of lining, not necessarily perunit volume of brake. All tabulated power densities should be treated as indicative rather than abso-lute maxima; their use with ( 3) and ( 4) enables reasonable estimation of required lining areas - opti-mum designs would have to consider thermal analyses, which is beyond the present scope.

Coe

ffici

ent o

f Fric

tion,

µ

0.5

Spe

cific

Wea

r R

ate,

Rw

Temperature ( C )o0 4000

FerodoAM 2

TYPICAL VALUES OF Rp ( kW/m2 ) spot plate

type of duty cooling typical drum disc brakes & cone conditions applications brakes brakes clutches clutches

Intermittent duty Time between applications Emergency andor infrequent full permits assembly to cool to safety brakes, 1800 6000 600 800duty applications ambient prior to actuation. safety and torque-

limiting clutches.

Normal Some cooling between applic- All general dutyintermittent ations, but temperature applications - 600 2400 240 400duty builds up to a moderate level winding engines,

over a period of time. cranes, winchesand lifts.

Heavy frequent Frequency of applications too Presses, drop duty where life high to permit appreciable stamps, excavators 300 1200 120 240is critical. cooling between applications. and haulage gear.

Typical lining pressure range ( kPa ) 70- 350- 70- 70-700 1750 350 350

Brake design investigations generally proceed along the following lines. - The braked system is first examined to find out the required brake capacity, that is the torque

and average power developed over the braking period. - The brake is then either selected from a commercially available range or designed from scratch.

In the latter case, conservative rather than optimum brake sizing may be based upon powerdensities which experience has shown to be acceptable, thus avoiding the difficulties associatedwith heat transfer appraisal.

- Analysis of the actuating mechanism is necessary to disclose the actuation requirements, brakesensitivity, bearing loads and the like.

The following notes consider these aspects for rotating drum brakes only, and go on to introduce theeffect of a road vehicle's braking control system on the vehicle's stability.

System Dynamics

The braked system must be analysed to throw light on its braking requirements. Analysis requiresknowledge of :

- the system's total energy (comprising eg. kinetic, gravitational and elastic potential) initially, ie.before braking

- the system's final total energy ie. after braking- the initial and final velocities of the brake drum- the desired braking period ∆t, or alternatively the corresponding rotation of the drum ∆θ- the life of the brake lining would also be specified or estimated by the designer.

During deceleration, the system is subjected to the essentially constanttorque T exerted by the brake, and in the usual situation this constancyimplies constant deceleration too. The elementary equations of constantrotational deceleration apply, thus when the brake drum is brought torest from an initial speed ωo :-

( 1 ) deceleration = ωo2/ 2 ∆θ = ωo/∆t ; ωm = ωo/2 = ∆θ/∆t

where ωm is the mean drum speed over the deceleration period.

Application of the work/energy principle to the system enables the torque exerted by the brake andthe work done by the brake, U, to be calculated from :-

( 2 ) U = ∆Ε = Τ.∆θ

where ∆Ε is the loss of system total energy which is absorbed by the brake during decelera-tion, transformed into heat, and eventually dissipated.

The mean rate of power transformation by the brake over the braking period is :-

( 3 ) Pm = T ωm = U /∆t

which forms a basis for the selection or the design of the necessary brake.

Linings

The choice of lining material for a given application is based upon criteria such as the expected coef-ficient of friction, fade resistance, wear resistance, ease of attachment, rigidity/formability, cost,abrasive tendencies on drum, etc.

Linings traditionally were made from asbestos fibres bound in an organic matrix, however the car-cinogenic tendencies of asbestos have led to the decline of its use. Non-asbestos linings generally

drum speed, ω

time, t

ωo

ωm

0 ∆ t0

Brakes 3

Brake shoe analysis

Before examining a practical twin-shoe brake we must understand the behaviour of a single shoe.

Various shoe con-figurations areillustrated. Eachconsists of a bodywhose motion isbraked togetherwith a shoewhich can swingfreely about afixed hinge H. Alining is attachedto the shoe andcontacts thebraked body. Theactuation force Papplied to the shoe gives rise to a normal pressure and corresponding braking friction distributedover the area A of contact between lining and braked body.

This illustration explains this contact in more detail. An elementdA of the lining is shown with the braked body moving past atvelocity v. Since sliding must occur the force resultant on dAcomprises the elemental component dN = p.dA orthogonal to dAtogether with the elemental friction component dFf = µp.dA coplanar with dA and in the sense of v.

Shoes are classed as being either short or long. A short shoe is one whose lining dimension in thedirection of motion is so small that contact pressure variation is negligible, ie. the pressure is every-where uniform, at pm say. The contact resultant therefore consists of the normal reaction N = Σ dN= Σ p dA → ∫ p dA = p m ∫ dA = pmA together with the friction force Σ µ.dN = µ.pmA. Shortshoes are statically determinate.

In a long shoe the variation of contact pressure in the direction of motion is not negligible and inte-grals of the above form ( ∫ p dA ) cannot be evaluated unless the p-variation is known. A long shoeis thus statically indeterminate and must be appraised by the three-pronged attack used for welds,cylinders, thick curved beams and so on - a plausible spatial distribution of lining deformation ( geo-metric compatibility ) is first assumed, then the lining constitutive law ( presumed elastic ) andfinally equilibrium are applied to establish completely the pressure variation for a given actuation.Long shoes are more complex than short.

The pivot G of a hinged shoe introduces a second degree of freedom permitting the lining to adopt aposition which tends to spread wear more uniformly over the lining thus increasing lining life.

The braked body may translate or it may rotate about a fixed shaft axis O. External rotational shoesonly are sketched above - the treatment of internal shoes is identical to that of external.

All these configurations behave qualitatively in identical fashion, and they are all analysed in a simi-lar manner - it's only their differing geometries which introduce complications. We shall lead up tothe analysis of practical rotational long shoes by first considering the following simpler cases

- translational shoes - although these are uncommon since mechanical power is usually trans-mitted rotationally, they are geometrically simpler than rotational shoes, and

P

H

ωO

G

P

H

ωOP

H

ωO

H P

G

v

H P

v

H P

v

trans-lating

rotating

short shoeuniform pressure

long rigid shoevarying pressure

long hinged shoevarying pressure

l ining

braked body

shoe

v

dAdN = p.dA

dF = µ.dN = µp.dAf

An alternative brake rating procedure is based upon the product of average pressure, pm , over thelining contact area and the mean rubbing speed, vm , during deceleration. This procedure requiresknowledge of the coefficient of friction, µ , so it is less useful than ( 4) and is mentioned only becauseit is commonly used - we shall persevere with ( 4). However if F is the lining contact resultant then :-( i ) pmvm = ( Fnormal /A ) vm = ( Ftangential /µA ) vm = Pm/µA = Rp /µ

which may be used analogously to ( 4) to determine the minimum lining area necessaryto dissipate a given power if µ is known.

If a drum brake has to be designed for a particular system (rather than chosen from an availablerange) then the salient brake dimensions may be estimated from the necessary lining area, A,together with a drum diameter-to-lining width ratio somewhere between 3:1 and 10:1, and an angu-lar extent of 100o say for each of the two shoe.

The lining is sacrificial - it is worn away. The necessary thickness of the lining is therefore dictatedby the volume of material lost - this in turn is the product of the total energy dissipated by the liningthroughout its life, and the specific wear rate Rw (volume sacrificed per unit energy dissipated)which is a material property and strongly temperature dependent as may be seen from the graphabove for Ferodo AM 2. This temperature dependence may be expressed as :-

( ii ) Rw ≈ Rwo exp ( (lining temperature oC / To ) n ) where Rwo , To and n are constant material properties.

The example below demonstrates how a brake and its lining may be sized for a given duty when thelining properties and temperature are known, however in practice wear rate is difficult to assesssince a thermal analysis is needed to predict the lining temperature.

The final aspect of drum brakes which we will investigate involves the analysis of a given brake inorder to correlate the input actuation with the output braking torque. The analysis will also evaluateloads throughout the brake - loads such as the drum reaction, the knowledge of which is necessaryfor the design of drum shaft and bearings.

EXAMPLE Select salient dimensions for a drum brake whose lining temperature is estimated to be 200oC. Thelining wear constants are Rwo = 8.0 mm3/MJ ; To = 185 oC ; n = 1.78 and the normal intermittent duty is :-

energy dissipated : 7.5 kJ /application (which includes a design factor)

average speed : 240 rpm (during braking) ( rpm ≡ rev/min )braking time : 0.25 s / applicationrequired lining life : 1.6 E6 applications

For one application : mean power dissipation rate ( 3) : 7.5/0.25 = 30 kW

Rp from the above table : 600 kW/m2

so, minimum req'd lining area ( 4) : 30/600 = 0.05 m2

Over the whole life : total energy dissipated : 7.5 E-3 * 1.6 E6 = 12 E3 MJ

Rw from ( ii) : 8.0 exp( (200/185)1.78 ) = 25 mm3/MJ

so volume lost over the life : 12 E3 * 25 = 300 E3 mm3

and thickness = volume/area : 300 E3/0.05 E6 = 6.1 mm

This is increased by 100% to allow for possible uneven wear and fixing with rivets - so required thickness is 12 mm.

Note that if the temperature were to increase by only 60 oC, then the thickness would have to be doubled.Over the braking period : displacement of drum from ( 1) : (240/60) * 0.25 = 1 rev

braking torque from ( 2) : 7.5/2π = 1.2 kNm

The drum diameter D and lining width w may be estimated as above. Thus with two 100o shoes :-A = w ( radius * lning extent ) = w D * 5π/9 and from the above-recommended proportions :-

3 ≤ D/w ≤ 5π D2 / 9A ≤ 10 leads to 0.29 ≤ D ≤ 0.53 say D = 400 mm.

The mean pressure pmmay be estimated from ( i), assuming µ = 0.35 say. vm = ( πDn)m = π * 0.40 * (240/60) = 5.0 m/s as always, keep track of units !

pm = 600/50 * 0.35 = 340 kPa which is acceptable by the above table.

Brakes 4

Mathematically, the negative sign in the denominator of ( iii) causes a significantly larger contactresultant for a given actuation than does the positive sign.

Once the normal and friction forces have been found via ( iii) for a given actuation, the hinge reac-tion, RH, may be ascertained easily from force equilibrium of the shoe free body. All forces are thenknown, allowing safe design to be carried out.

Long Translational Shoe

The lining thickness and compliance of this translational longshoe are grossly exaggerated to demonstrate how liningdeflection varies in response to changing actuation. For agiven actuation the deflection at any point along the lining is approximately proportional to thepoint's distance x from the hinge, with the maximum deflection occuring at the end of the liningfurthest from the hinge. Assuming the lining to behave elastically, the compressive stress and thecontact pressure p are also proportional to x. This proportionality is a reasonable postulate for prac-tical linings which are not wholly elastic.

A long shoe is analysed by dividing it up conceptually into infinitessimal elements - each of whosecontact with the translating body is identical to that of the short shoe above - then summing (inte-grating) the equilibrating effect of all the elements.

The free body of the long shoe is similar to that of the short inhaving three contacts, however the contact with the brakedbody is distributed. Consider such a trailing shoe whose lin-ing, of constant width w normal to the sketch plane, extendsfrom xmin to xmax as sketched. The lining pressure is p onthe typical element of length dx located at x from the hinge,so the normal force on the element is dN = p.dA = p w dx,while the corresponding friction force will be µ p w dx.

Moment equilibrium of the assembly of elemental short shoes about the hinge (to eliminate the asyet unwanted hinge reaction) requires :-

M - Σ dN.x - Σ µdN.h = 0 or in the limit on substituting for dN : M = w ∫ p ( x + µh ) dx

Integration is possible only if the pressure distribution p(x) is known, so the above-noted propor-tionality between pressure p and distance x is formalised as p = pmax ( x /xmax ) - the severity ofcontact loading being characterised by the peak pressure pmax which depends upon the actuation.Inserting this into the equilibrium equation and integrating between the lining limits xmin and xmaxleads to :-

( iv ) pmax = ( M xmax / w ) / [ ( x3max - x3

min )/3 + µh ( x2max - x2

min )/2 ]

The relation ( iv) for a long shoe is similar to ( iii) for a short shoe as it expresses the essential pro-portionality between actuation and the resulting contact effects - if the actuation is doubled then thepeak pressure pmax and the pressures at all points of the lining are also doubled. The total brakingeffect, hinge reaction etc. produced by a certain actuation follow from ( iv) - for example :-

total braking effect = Σ µpw dx = ∫ µ [ pmax ( x/xmax )] w dx

= µ w pmax ( x2max - x2

min ) / ( 2 xmax ) dx or substituting for pmax from ( iv) :

= M / [ h + 2 ( x3max - x3

min) / ( 3 µ ( x2max - x2

min) ) ]

The details of this final result notwithstanding, it is clear that the braking effect is again proportionalto actuation since only constant geometric and frictional terms occur.

x

dN = p w dx

dA = w dx

M

RH

µ p w dxx

x

x p

- short shoes - although these are generally impractical because of their small lining area ( recallthe 'Linings' section above ), they are simpler than long shoes as contact pressure is uniform.

A shoe behaves very differently depending upon whether it trails or leads. If the braked body dragsthe lining away from the hinge then the lining trails behind the hinge and the shoe as a whole is said'to trail’. The shoes in the lower row of the above sketch all trail while the shoes in the upper row alllead for the motion senses drawn.

In each of the following shoe analyses, the shoe geometry is defined and the coefficient of friction µbetween lining and braked body is known. In the first instance it is required to find the resultant ofthe lining/braked-body contact for a given actuation. Determination of the braking effect, shoe per-formance parameters, and hinge & shaft reactions is thereafter straightforward.

It is useful to realise that effect is always proportional to cause for constant geometry and frictioncoefficient - thus if a shoe's actuation is doubled then the braking effect, lining pressure, hinge reac-tion etc. all double.

Short Translational Shoe

The geometry of a short translational shoe is sketched. The uniformpressure over the contact area A is pm. The braked body may movein either direction at a fixed distance, h, from the hinge axis.

Free bodies of the shoe are shown for the two possible directions -evidently the shoe in the left configuration trails while that in theright leads. The three contacts which each free body makes with its surrounds are :-

- the actuation force P- the braked body characterised by the nor-

mal reaction N = pmA (see previous section),together with the friction force µN in thesense of the braked body's motion, and

- the hinge support reaction RH.

It is the reaction to the friction force µN sketchedwhich brakes the translating body. The normalreaction and friction force are correlated with agiven actuation via the equation for moment equi-librium about the hinge (thus eliminating the asyet irrelevant hinge reaction) - the equationappears under each free body.

The normal reaction depends upon the moment Pe rather than on P itself. The product Pe istermed the actuation moment, M, so the two equilibrium equations may be written :-

( iii ) N = pmA = M / ( a ± µh ) - the sign depends on whether the shoe trails or leads.

All terms on the RHS of this equation are geometric - apart from the actuation and the (presumablyconstant) friction coefficient - so the normal reaction and the frictional braking force are directly pro-portional to the actuation. The difference between the two behaviours is as follows :-

- In the LH sketch the moment of the friction force opposes the actuation moment; friction tendsto throw the shoe off the braked body, to reduce the actuation - the shoe is said to be counter-actuating.

- In the RH sketch the moment of the friction force assists the actuating moment; friction tendsto drag the shoe onto the braked body, to augment the actuation - the shoe is self-actuating.

h

eP

fixedhingeH

lining

shoe

braked body

a

Pe – Na – µNh = 0

N = Pe / ( a + µh )

counter-actuating

P

R

NtrailingµNv

H

Pe – Na + µNh = 0

N = Pe / ( a – µh )

self-actuating

P

NleadingµNv

RH

Brakes 5

Next the drum's net effect on the shoe sketch ( c), is replaced by the equiva-lent effect in Cartesian components at the drum centre, sketch ( d). If thearrangements of sketches ( c) and ( d) are identical then :-

Fx = ∆N cos θm – δµN sin θm ; Fy = ∆N sin θm + δµN cos θm ;T = µNr - or briefly :-

pm is the uniform pressure over the short lining whose contact area is A. The reaction to the torque Tis the braking effect exerted by the shoe on the drum.

Taking moments about the hinge for the free body ( d) - [ ( c) serves equally well ] -

( vi ) ∆M = aFy – δT = N [ a ( δµ cos θm + ∆ sinθm ) – δµ r ] - and solving for N :-

( 6a ) N = M /[ a sinθm - δ∆µ ( r - a cosθm) ] - a generalisation of (v) applicable to all layouts

= M / [ µr ( m - δ∆ n ) ] in which m ≡ 1/µ.a/r. sinθm and n ≡ 1 – a/r. cosθmare constant dimensionless characteristics of the shoe.

Summarising the analysis steps for a short shoe of known geometry and friction coefficient :- - evaluate the shoe constants, m and n, from ( 6a)- for a given actuation M ascertain the corresponding normal reaction N from ( 6a) with δ, ∆

appropriate to the shoe's configuration- determine contact resultants including braking torque from ( 5a), then other forces if desired

from force equilibrium.

So much for the mechanics of analysis - let's now examine the physical implications . . . .

Shoe Figures of Merit

The performance of a single shoe or of a complete brake is described by either of two dimensionlessfigures of merit.

The mechanical advantage ( or "brake factor"), η :-( vii) η = output/input = braking torque/corresponding actuation necessary

The sensitivity, S, reflects the proportional variation of braking torque with fixed actuation as thefriction coefficient varies, thus :-( viii) S = µ/T ( ∂T/∂µ )M

= µ/η . dη/dµ

For a single shoe, use of ( 5a, 6a) leads to :-( 8 ) η = T/M = 1 / ( m – δ∆.n ) and S = m η

These are plotted below for representative short shoes of typical proportions. The curves demon-strate that the same trends apply for both figures of merit, namely :-

δ∆ = –1External leading and internal trailing shoes of the given proportions are counter-actuating.Both mechanical advantage and sensitivity are low and the shoes are inherently stable.

δ∆ = +1External trailing and internal leading shoes of the given proportions are self-actuating. Bothmechanical advantage and sensitivity can become high - giving rise to instability, to vibrationand squeal, to judder if the assembly is flexible, or even more dangerous, to grabbing of the

( d)

δT

∆M

Fx

Fy

( 5a)

Fy

Fx ∆ −δµ∆δµ θsin m

θcos m= N

T = µ N r where

and

mN = p A

Short Rotational Shoe

Turning now to drum brakes, we first con-sider the external, trailing short shoesketched at ( a). The drum of radius r is cen-tred at O which forms the origin of aCartesian system whose x-axis passesthrough the shoe hinge H distant a from O.The y-axis is directed generally towards theshort lining - whether this system is right- orleft-handed is irrelevant. The lining is located atangle θm from the x-axis.

The shoe free body is shown at ( b) with the actuation moment M = Pe and the effect of the brakedbody (drum) contact on the lining expressed by the normal reaction N and the friction force µN inthe sense of braked body motion - just like the short translational shoe.

For rotational equilibrium about the hinge to eliminate the as yet unwanted hinge reaction :-

( v ) N = pmA = M / ( a sin θm - µ ( r - a cos θm ) )

This is analogous to ( iii) in expressing the essential proportionality between the normal reactionand the actuation - other forces follow from force equilibrium.

This is really all there is to short shoe brake analysis - however it is useful to introduce two artificesto simplify later work :-

- ( v) applies only to an external trailing shoe. Rather than derive other relations for internaland/or leading shoes, we shall employ a "double delta" notation and re-derive a general form of( v) which will describe all possible configurations simply by inserting appropriate ±1 valuesfor the deltas.

- When short shoe results are extended to long shoes it will be seen that the normal and frictioncomponents vary in both magnitude and direction around the lining contact. It is much easierto handle these if they are moved to the shaft centre, O. It will be recalled from basic Staticsthat a force may be moved transverse to its line of action without altering the equilibrium of abody on which it acts provided a corresponding moment is introduced equal to the force multi-plied by the transverse distance moved.

The double delta notation utilises :-- upper case ∆ to designate shoe position - if a shoe is external then ∆ = +1; if the shoe is inter-

nal then ∆ = -1- lower case δ to designate shoe sense - if a shoe is trailing then δ = +1; if the shoe is leading

then δ = -1

The forces and torques of sketch ( b) which enter into the hinge momentequation are repeated with the δ−∆ terminology in sketch ( c), though this isnot a complete free body of the shoe. It should be noted that :-

- if the shoe leads rather than trails then the friction force on the shoe µNwill act in the SE direction, ie. negatively in the NW sense sketchedsince δ = -1 for the leading shoe;

- if the shoe is internal to the drum rather than external, then the normalreaction N will act radially inwards on the lining while, to rotate theshoe clockwise against the surrounding drum, the actuation M must be

clockwise ie. negative in the senses sketched since ∆ = -1 for the internal shoe.

liningr

a

O

H

ω

θ m

e

P

x

y

θ m

P

O

M

NµN

RH( a) ( b)

( c) ∆M

∆NδµN

θ m

O

Brakes 6

common shoe geometry - refer to short shoe sketch ( a) above

r = 150 mm ; a = 240 mm ; e = 150 mm ; θm = 70o ; so from ( 6a)

m = 1/0.5 x 240/150 x cos 70o = 3.007 ; and n = 1 - 240/150 x sin 70o = 0.4528

The "significant figures" quoted here for checking purposes are anything but - the friction coefficient accuracy is twosigfigs at the very most !

individual shoe parameters LEFT SHOE RIGHT SHOE

External shoes, ∆ = +1 leading so δ = -1 trailing so δ = +1

P (N) 800 600

M (Nm) 0.15x800 = 120 0.15x600 = 90

( 6a) N = M/µr( m– δ∆n) (N) 120/0.5x0.15( 3.007+0.453) = 462 90/0.5x0.15( 3.007-0.453) = 470

(vii) η = 1/( m– δ∆n) 1/( 3.007+0.453) = 0.289 1/( 3.007-0.453) = 0.392

(viii) S = mη 3.007x0.289 =0.869 3.007x0.392 = 1.177

(5a) Fx = N( ∆ cosθm- δµ sinθm) (N) 462(0.342+0.5x0.940)=375 470(0.342-0.5x0.940)= -60

(5a) Fy = N( δµ cosθm+ ∆ sinθm) (N) 462(-0.5x0.342+0.940)=355 470(0.5x0.342+0.940)=522

(5a) T = µNr or = ηM (Nm) 0.5x462x0.15 = 34.7 0.5x470x0.15 = 35.2

So for the brake as a whole, To = TL + TR = 34.7 + 35.2 = 69.9 Nm

Shoe free bodies with actuation,drum contact and hinge reactioncomponents

RHx from Σ Fx = 0 (N) 800 cos40o – 375 = 238 600 cos40o +60 = 520

RHy from Σ Fy = 0 (N) 800 sin40o – 355 = 159 600 sin40o – 522 = –136

RH by Pythagoras (N) 286 537

Drum free body with reactions to above shoe-drum contacts at drum centre, and horizontal and verticalcomponents of drum shaft support

ROh from Σ Fhoriz = 0 (N) (522-355) cos30o - (60+375) sin30o = -73

ROv from Σ Fvert = 0 (N) (522+355) cos60o + (60-375) sin60o = 166

RO from Pythagoras (N) 181

If the brake is to be practicable then the average lining pressure pm = N/A ( 5a) can-not exceed a value which experience has shown to be feasible - around 700 kPa for adrum brake ( see earlier Rp table). The pressure resultant N is higher for the right shoehere so the highest average lining pressure in the brake is pm = 470/100 = 4.7 >> 0.7MPa. The brake is clearly not practicable unless actuations and resulting braking torque arereduced to 0.7/4.7 = 15% of their values above. Short shoe brakes just don't have enough lining area to give usefulbraking effects with sustainable lining pressures, however they may be applicable to one-off emergency situations.

The left shoe is counter-actuating hence requires a larger actuation than the self-actuating right shoe for approxi-mately the same braking torque.

If the left shoe's LH coordinate system causes difficulty in comprehension, then the brake should be viewed fromthe rear - the left shoe in the front view becomes the right in the rear view.

We shall consider later the practical means for applying different actuation forces to the two shoes from a singlecontrol signal (eg. a car driver's stamping on the brake pedal).

The brake illustrates how the two opposed linings tend to balance one another so that the net load on drum shaftand bearings is small.

800 N

x

y

375 N

355 N

34.7Nm120

Nm

RHyRHx

600 N

90 Nm

522 N

-60 N35.2 Nm

y

x

RHy RHx

522 N

60 N375 N

355 N

34.7Nm

35.2 Nm

ROv

brake. Physically this is due to friction not merely assisting actuation but swamping it.

Self-actuation occursif m and n in ( 6a, 8)are approximatelyequal due to design,or to the shoe becom-ing out of adjustmentbecause of hingewear, or if the coeffi-cient of frictionincreases after theingress of dirt, drumscoring, or other envi-ronmental changes.While significant self-actuation may be advantageous for an emergency brake, generally it should be avoided.

Experience has shown that shoes should be designed with a maximum sensitivity of 2 under heavyduty conditions, or 2.5 for moderate duty.

The following example analyses a brake with two separately actuated short shoes, and considersbraking torque, sensitivity and bearing loads. The consequences of common actuation will be exam-ined after long shoes have been analysed.

0 0.2 0.4 0.6 0.8coefficient of friction

0 0.2 0.4 0.6 0.8coefficient of friction

4

3

2

1

0

mec

hani

cal a

dvan

tage

4

3

2

1

0

sens

itiv

ity

θ = 90a/r = 1.2 externala/r = 0.8 internal

mo

internalleading

externaltrailing

external leading

externaltrailing

internalleading

internal trailing

θ = 90a/r = 1.2 externala/r = 0.8 internal

mo

internal trailing

external leading

EXAMPLE A brake consists of two identical short shoes arranged andactuated as shown. The distance between drum and hinge centres is 240 mm.Assuming a friction coefficient of 0.5, determine :

- the total braking torque on the drum, To- the sensitivity S of each shoe- the load RH on each shoe hinge- the load supported by the drum shaft, RO- the brake's practicability given that the contact area of each

lining is 100 mm2. 20 o

70 o

60 o

150

Ø 300

600 N800 N

Brakes 7

The similarity between ( 5b) and the short shoe ( 5a) is notable, however the wr product is not anactual area so No is not a pressure resultant as was N in the short shoe case - rather No is a pres-sure parameter, a convenient group of terms which reflects the pressure level arising from the actua-tion present.

Using ( vi) again, the actuating moment from sketch ( d) is :-

∆M = a Fy – δT = No [ a ( δµ Isc + ∆ Iss ) – δµ r Is ] - and solving for No

( 6b ) No = M / [ µr Is ( m - δ∆ n ) ] in which m ≡ 1/µ. a/r. Iss/Is and n ≡ 1 – a/r. Isc/Isare constant dimensionless characteristics of the shoe.

It is apparent that ashort shoe is just a lim-iting case of a longshoe - the geometricintegrals in ( 5b, 6b)become simple circularfunctions in the shortshoe equations ( 5a,6a). The general trendsof mechanical advan-tage and sensitivity forthe two cases are thusvery similar, as may beseen by comparing thegraphs here for typical long shoes with the corresponding graphs for short shoes.

Long Hinged Shoe

The hinged shoe is connected to the actuated post by the hinge, G,which introduces another degree of freedom - so the shoe tends toassume an optimum position in which the pressure distribution overit is less peaked than in a rigid shoe. The extra expense of providinganother hinge is thus justified on the grounds of more uniform liningwear. The pressure consists of two components; one, proportional tosinθ, due to rotation of the post-and-shoe about H; the second, propor-tional to sinφ, due to rotation of the shoe alone about G. Since φ = θ - θG we maytake the pressure variation as :-

p = ps sinθ + pc cosθ where ps and pc are proportionality constants ( akin to pmaxof the rigid shoe ) whose combined action leads to equilibrium of the shoe.

The x-component of the contact force resultant becomes :-

Fx = ∆ ∫1

2 cosθ ( ps sinθ + pc cosθ ) w r dθ – δ µ ∫12 sinθ ( ps sinθ + pc cosθ ) w.r.dθ

= ∆ ( Ns Isc + Nc Iss ) – δ µ ( Ns Iss + Nc Isc ) from ( 10), and

where by definition : Ns ≡ w r ps and Nc ≡ w r pc

The y-component and the torque may be found in a similar manner, to yield :-

0 0.2 0.4 0.6 0.8coefficient of friction

0 0.2 0.4 0.6 0.8coefficient of friction

4

3

2

1

0

mec

hani

cal a

dvan

tage

4

3

2

1

0

sens

itiv

ity

internalleading

externaltrailing

external leading

externaltrailing

internalleading

internal trailinginternal trailing

external leading

a/r = 1.25 externala/r = 0.80 internalθ = 45θ = 1352

o1 o

a/r = 1.25 externala/r = 0.80 internalθ = 45θ = 1352

o1 o

G

H

Oθ

φ

θG

OH = aOG = b

x

T = µ r ( I N + I N )

( 5c)

Fy

Fx ∆ −δµ∆δµ I ss

I sc and=

s

N sI sc

Icc N c+

s c c

Long Rigid Shoe

The long rigid rotational shoe is analogous to the long translationalshoe examined above. In both cases the location of a point along thelining is reckoned from the hinge, however in the present cylindricalembodiment the angle θ rather than x is the obvious independentgeometric variable. The lining of constant width w normal to thesketch plane extends between θ1 and θ2.

The lining illustrated below is impractical in extending from 0o to 180o

with grossly exaggerated thickness and compliance to hilite the crescent-shaped dis-tribution of lining deformation arising from changing actuation. By inspec-tion for a given actuation the deformation is zero at θ = 0o, 180o and a

maximum at 90o. Assuming the lining to behave elastically, the com-pressive stress and the contact pressure p also vary in this man-ner. A sinusoidal pressure variation is thus a reasonable postulatefor practical linings which are not wholly elastic, thus :-

( 9 ) p = pmax sinθ ; θ1 ≤ θ ≤ θ2

pm = ∫1

2 p dθ / ( θ2 – θ1 ) = pmax Is / ( θ2 – θ1 )

in which pmax is the pressure at θ = π/2 . For acertain actuation there is a corresponding pmax and hence a corre-sponding pressure distribution via ( 9).

pm is the average pressure over the whole lining, and the integral Istogether with other relevant integrals are defined as :-

( 10 ) Is = ∫1

2 sin θ dθ = cos θ1 – cos θ2

Iss = ∫1

2 sin2 θ dθ = ( 2θ2 – 2θ1 – sin 2θ2 + sin 2θ1 )/4

Isc = ∫1

2 sin θ cos θ dθ = ( cos 2θ1 – cos 2θ2 )/4

Ic = ∫1

2 cos θ dθ = sin θ2 – sin θ1

Icc = ∫1

2 cos2 θ dθ = ( 2θ2 – 2θ1 + sin 2θ2 – sin 2θ1 )/4

A long shoe is analysed by dividing it up conceptually into infinitessimal elements - each of whosecontact with the braked body is identical to that of the short shoe above - then summing ( integrat-ing ) the equilibrating effect of all the elements. Thus the element of lining dθ at θ in the sketchbelow is analogous to the short lining of sketch ( c). The element's contribution to the contact resul-tant at the shaft centre O is :-

dFx = ∆.dN cosθ – δ.µ.dN sinθ ; dFy = ∆.dN sinθ + δ.µ.dN cosθ ; dT = µ.dN r

The pressure p is essentially uniform over the short element, so the normalreaction on the element is dN = p dA = pmax sinθ.wr dθ using ( 9). The x-component of the resultant contact force, sketch (d), is therefore :-

Fx = pmax wr ∫1

2 ( ∆ cosθ – δµ sinθ ) sinθ dθ= pmax wr [ ∆ Isc – δµ Iss ] using ( 10)

Proceeding in like manner for the y-component and for the torque leads to :-

H

O

x

θ

θ1

θ 2

O

x

θ

∆.dN

δ.µ.dN

dθ

H

oT = µ N r I where

( 5b)

Fy

Fx ∆ −δµ∆δµ I ss

I sc and

max

= No

o s N ≡ p wr by definition

Brakes 8

EXAMPLE The two shoes of the internal brake are identical.

The lining is 40 mm wide and the coefficient of friction is estimated to be

0.35. The brake is hydraulically operated - the maximum available

hydraulic pressure is 8 MPa, and the piston diameters are 18 and 10 mm

for left and right shoes respectively. Find the torque capacity of the

brake if the lining mean pressure is limited to 400 kPa.

What are the corresponding bearing reactions and brake sensitivity ?

common shoe geometry

inclination of x-axis to vertical, φ = arctan 30/72 = 22.6o

P-inclination θP = 90 – φ = 67.4o

lining limits θ1 = 45 – φ = 22.4o; θ2 = 180 – 35 – φ = 122.4o

from ( 10) : Is = 1.460 ; Iss = 1.275 ; Isc = 0.284

OH = a = √( 722 + 302 ) = 78 mm ; a/r = 0.78

from ( 6b) : m = (1/µ) (a/r) Iss / Is = 1.946 and n = 1 – (a/r) Isc / Is = 0.848

φ

35o

100

72

75O

H

hydraulic cylinder

45o

30

If rotation is essentially unidirectional, as in a car, the two shoes may be arranged so that they areboth self-actuating. Although a more elaborate actuating linkage will be required, hydraulics domake this easier to accomplish.

Despite the apparent complexity of the hinged shoe equations, proportionality is inherent betweenthe three parameters of most interest - actuation torque, lining mean pressure, and braking torque -for all shoe types considered. The simplest way of tackling many brake problems, if not the most ele-gant, is therefore to assume any convenient reference actuation source value, and work out all shoeparameters on that basis using ( 12), ( 5) and ( 6). A final scaling of all values would then be carriedout as shown by the following example which refers to a twin rigid shoe brake.

The operation of brakes particularly for industrial applications has been explained above. Vehiclebrakes are of interest since we use them all the time on our cars and bicycles. The topic is complexdue to the many factors which affect vehicle dynamics - the ground- tyre interface, vehicle suspen-sion and cornering dynamics, the brake control system, driver behaviour and so on. In order to gainsome understanding of the subject, we now examine the effect of the weight distribution and brakecontrol system on the stability of vehicle braking during straight forward motion.

Equation ( vi) still applies to the free body of the post-and-shoe, giving :-

( 6c ) M = α ( Js Ns + Jc Nc ) where

α = [ δ∆µa a – δ∆µr ] ; Js = [ Isc Iss Is ]' ; Jc = [ Icc Isc Ic ]'

The actuating force P is applied to the post HG so the shoe itself is subject to two contacts only - the(ideal) pin at G and the distributed contact with the drum. For equilibrium of the shoe :-

Σ MG = T + Fx by – Fy bx = 0 where bx = b cosθG ; by = b sinθG

which, with ( 5c), requires that :-

( 11 ) β ( Js Ns + Jc Nc ) = 0 where β = [ ( δ∆µbx – by ) ( δ∆µby + bx ) ( – δ∆µr ) ]

This relation lays down the necessary inter-dependence of Ns and Nc, that is of ps and pc, for equi-librium. The forces and torques as given above are therefore all expressible in terms of a single pres-sure parameter, and equations ( 5c, 6c) take the same form as ( 5a,b & 6a,b). Thereafter the treat-ments of the various cases are very similar.

Twin Shoe Brakes

Shoe behaviour has been discussedat length. Two such shoes are com-bined into a complete practicalbrake unit, two being used to mini-mise the unbalanced forces on thedrum, shaft and bearings, andbecause, as has been seen, liningsbecome increasingly ineffective ifthey extend much beyond 90o to 110o. The shoes will be designated 1 and 2.

The brake torque of the complete brake To is the sum of the torque contributions of the two shoes.The shoes are operated by a single brake actuation source Po which may be a force in a brake rod oran hydraulic pressure for example. This source is converted into the individual shoe actuations bysome actuating linkage, examples of which are shown at the beginning of the chapter. Since the twoshoes usually behave differently - one leading while the other trails - the actuating linkage isarranged to have different transformation ratios between the source and the shoes so that the lin-ings' peak pressures and lives are not too different. Expressing this mathematically :-

( 12 ) M1 = λ1 Po ; M2 = λ2 Po

where λ1 and λ2 are geometric actuation constants which reflect the actuationlinkage's mechanical advantages. The dimensions of λ will be [length3 ] if Po is an hydraulic pres-sure, or [length] if Po is a force - thus for the linkage sketched above, λ1 = P1e1/Po and λ2 = P2e2/Poboth depend upon the geometry of the bell crank. Clearly in designing such brakes, the economicsand reliability of more complex mechanisms must be considered.

For geometrically similar shoes, the shoe with the largest λη product will be subjected to the great-est lining pressure and will contribute most to the total braking torque To. The contributions of thetwo shoes to the mechanical advantage ηo and sensitivity So of the brake as a whole may beexpressed as :-

( 13 ) ηo = To /Po = ∑ λ η ; So = ∑ λ η S / ηo

which lie somewhere between the values for the individual shoes ( see tutorial problem #2 ).

Po

P1

P2Po

e1

bellcrank

e2

Brakes 9

The Braked Wheel

We now investigate the stability of road vehicleswhilst braking during straight line motion, and con-sider first a typical single non-driven wheel equippedwith a drum brake of the type examined above. Thereare three contacts between the wheel/brake drum/tyre system and its surrounds :

- the ground, characterised by a tyre/ground adhesion coefficient k, and represented on thefree body by a normal force component N and a friction component F whose sense opposesvehicle velocity vo and whose magnitude cannot exceed the adhesion limit Fmax = kN whenthe wheel slides without rotation on the ground, a condition known as wheel lock

- the two brake shoes, which give rise to the braking torque T and the force resultant R whichare proportional to the brake actuation Po as discussed above - provided the wheel is turning,since friction force = µ*normal reaction at the lining-drum contact only when there is relativemotion between lining and drum

- the wheel bearing at O, taken to be essentially frictionless so that the sole effect here is the bear-ing reaction RO.

The equation of angular motion of this system about its centre of gravity O is Frw – T = I dω/dt, and

if the system's inertia is negligible then :

( x ) F = T/rw under all conditions.

The vehicle decelerating force F and the braking torque T are plotted against brake actuation Po.

In the absence of wheel lock, the torque increases in proportion toactuation, until a limit dictated eg. by a relief valve is reached -this actuation limit is shown as the point A. During this period ofbrake application, wheel rotation occurs and the decelerating forceF also is proportional to Po via ( x).

If a sufficiently small value of either k or N should occur then theadhesion limit is reached at point B, below the actuation limit. Thetorque produced by the brake is also limited since the wheel islocked and does not rotate - and from ( x) Tmax = Fmaxrw = kNrw.Such a situation occurs when sliding on a gravel road with thebrake pedal frantically pressed to the floor - no matter how hardthe brake is applied, wheel rotation has ceased and vehicle deceleration is dictated by the (lack of)ground adhesion and not by any limitations of the brake itself or the brake's source of actuation.

Braking of Vehicles

The front wheels of the upper vehicle here are locked by thebrakes and slip on the ground. The braking force, F, is thusgenerated in the main by the rear wheels and its momentabout the mass centre tends to correct any yaw, leading tostable braking. Conversely the rear wheels of the lower vehi-cle are locked so the major braking force is due to the frontwheels and tends to increase yaw, rendering braking unsta-ble.

The effect on a single wheel when reduced normal ground

F

O

rwT

N

RRoω

vo

adhesionlimit

actu

atio

nlim

it

rotation lock

brake actuation, Po

F = T/rw

brake torque, T

ground friction, F

A

B

motion

F

frontlock -stable

F

rearlock -unstable

8 MPa

400 kPa

lining meanpressure

hydraulicpressure

right

left

individual shoe parameters

As the hydraulic actuation pressure rises so does the mean pressure

on each lining. It is not clear which pressure bound will limit the

brake's performance - the hydraulic pressure or the mean contact

pressure on left or right lining - so evaluate pm on basis of a conven-

ient reference hydraulic pressure of 1 MPa say, then scale as necessary.

The left column of figures below for each shoe is based on this reference;

the right column comprises the scaled value. Results are given at length

to facilitate checking; two figures at most are significant.

LEFT RIGHT

Both shoes are internal, so ∆ = –1 trailing δ = +1 leading δ = –1P = ( p A )hydraulic 255 1858 79 573 N

M = P e ( e = 75 +72 mm) 37.4 273 11.6 84 Nm

( 12) λ = M/brake actuation(pressure) 37.4 11.6 Nm/MPa

( 6b) No = M / µ r Is ( m – δ ∆ n ) 262 1912 206 1503 N( 5b) pmax = No / w r 65.5 478 51.5 376 kPa

**( 9 ) pm = pmax Is / ( θ2 – θ1 ) 54.8 400 43.1 314 kPa

( 5b) Fx = No ( ∆ Isc – δ µ Iss ) –191 –1396 33 244 N

( 5b) Fy = No ( δ µ Isc + ∆ Iss ) –308 –2248 –283 –2065 N

( 5b) T = µ r Is No 13.4 98 10.5 77 Nm

( 8) η = 1 / ( m - δ∆ n ) 0.358 0.358 0.911 0.911

( 8) S = m / ( m - δ∆ n ) 0.696 0.696 1.772 1.772 ** On the basis of 1 MPa hydraulic pressure, the greater mean lining pressure is 54.8 kPa, so an hydraulic pres-

sure of 1*400/54.8 = 7.30 MPa will correspond to the maximum allowable lining pressure and hence to

maximum torque capacity. The hydraulic pressure capacity is adequate, ie. torque capacity is not limited by

it, and a relief valve will be necessary in the circuit, set at 7.3 MPa, to ensure that the lining pressure does

not exceed 400 kPa. Multiplying all results by 7.3 gives the final underlined figures in the right columns -

though the mechanical advantage and the sensitivity are not affected by the actuation.

The mechanical advantage of the brake - the overall transformation of pressure to braking torque - is :-

ηo = 13.4 +10.5 = 23.9 Nm/MPa [ ie. (98+77)/7.3 Nm/MPa ]

In this problem the actuating linkage factors λ represent (brake actuation pressure)-to-(shoe actuation moment)transformations (Nm/MPa), so from ( 13) the brake overall sensitivity is :

So = ∑ λ η S / ∑ λ η = ( 37.4x0.358x0.696 + 11.6x0.911x1.722 ) /( 37.4x0.358 + 11.6x0.911 ) = 1.171 Hinge and drum shaft reactions

may be found from free bodies

complete with the shoe/drum

contacts calculated above from (

5b). Torques are not shown as

only force equilibrium is being

considered. Here the contact Fx,

Fy are drawn on the shoes in

their correct senses from above;

the reactions to these appear on

the drum free body.

For force equilibrium of these free bodies :

Left : ΣFx = RHx – 1396 + 1858sinφ = 0 RHx = 681

ΣFy = RHy – 2248 + 1858cosφ = 0 RHy = 533 RH.left = 865 N

Right : ΣFx = RHx + 244 + 573sinφ = 0 RHx = -464

ΣFy = RHy – 2065 + 573cosφ = 0 RHy = 1536 RH.right = 1605 N

Drum: ΣFv = ROv + ( 1396 – 244 ) cosφ – ( 2248 +2065) sinφ = 0 ROv = –595

ΣFh = ROh + ( 2065 –2248) cosφ – ( 1396 + 244) sinφ = 0 ROh = –800 RO = 997 N

summary Actuating pressure : 7.3 MPa Braking torque : 175 Nm

Sensitivity : 1.17 Drum shaft reaction : 997 N

Shoe hinge reactions : 865 N (left), 1605 N (right)

LEFT RIGHT

1396

φ2248

2442065

φ

OvR

OhR

1858

1396

2248

x

y

φ

HxR

HyR

573

244

2065

x

y

φ

HxR

HyR

Brakes 10

Wheel Lock - the Vehicle Characteristic

If the front wheels should lock, then, as above : FF = kNF. Similarly if the rear wheels lock, then FR =kNR. Combining these with ( 21a) and ( 21b) in turn, and defining the normalised wheel friction/braking force as φ = F/W, results in either . . . . .

( 22a) ( 1/k – h ) φF – h φR = cR front wheel lock - stable, or . . . . .

( 22b) h φF + ( 1/k + h ) φR = cF rear wheel lock - unstable

These linear relationsdescribe how the totalbraking effort is appor-tioned between front andrear wheels, as eitherwheel set locks. They areplotted for a class of vehi-cles whose mass centresare located proportionallyat cF = 0.4, h = 0.3.

Also shown on this dia-gram are the loci of con-stant deceleration, ie.from ( 20) :-

( 23 ) φF + φR = z

The lines which represent front and rear wheel lock meet in a curve, which therefore corresponds tolock of both front and rear wheels, and which, from ( 22), is given by :-

( 24 ) h ( φF + φR )2 = cF φF – cR φR ; cF + cR = 1

This curve - defined uniquely by the location of the mass centre - is the vehicle characteristic. Thedeceleration at any point on a z-locus, such as point A, is evidently less than the maximum possibledeceleration which is achievable on the characteristic - eg. zmax = 0.3 = k for the point A. As thecharacteristic represents the locus of maximum possible decelerations, it is sometimes called theideal control locus, though it is hardly ideal from the driver's point of view.

The Brake Control Characteristic

The proportional braking mentioned above appears on the diagram as a line of constant slope :

( 25 ) FR / FF = φR / φF = constant ≤ ( φR / φF )max = cF / cR

The maximum slope is the slope of the vehicle characteristic at the origin, which would result inunstable braking at all times, if implemented. The sketch overleaf illustrates a slope of half this, ie.φR /φF = 1/3 for the vehicle under consideration, and shows how such a plot may be used to assistbrake selection for a particular vehicle. Point B represents a modest deceleration of 0.2g, which isstable provided that road conditions allow it, ie provided that k ≥ 0.23.

If on the other hand, the adhesion coefficient were 0.3 and the brakes further actuated to give greaterdeceleration, then point C would be reached, corresponding to a deceleration of 0.27g. At this pointthe front wheels would start to lock and the proportional braking trajectory BC would give way tothe front wheel adhesion locus CD. However since braking is still stable, the point D could bereached - mainly due to an increase in the rear wheels' contribution. At D of course, the limiting

0.1 0.2

0.40.5 0.6

0.1

0.1

0.1

0.2

0.3

0.3

0.4

0.4

k = 0.50.6

0.7

0.5 0.6

0.7

AdhesionCoefficient

NormalisedDeceleration

REAR LOCKUNSTABLE

00

FRONT LOCKSTABLE

0.2

0.2

z = 0.3

A

Rea

r N

orm

alis

ed F

rict

ion

Fo

rce,

φ R

Front Normalised Friction Force, φF

c = 0.4; h = 0.3F

reaction N lowers the adhesion limit Fmax has been noted. When a multi-wheeled vehicle deceler-ates, then the normal reaction on the front wheels increases from its static value, while that on therear wheels decreases - the amount of increase/decrease depends upon the magnitude of the decel-eration and upon the location of the vehicle's centre of gravity. The increase in front wheel reactionincreases also the adhesion limit and renders the front brakes relatively more effective (more able todecelerate without front wheel lock), as anyone who has flown over the handlebars of a bicycle willtestify!

Motor vehicles are thus provided with front brakes which are more powerful than the rear -achieved by larger lining areas and/or hydraulic cylinders - but if this is overdone then the rearwheels may lock rendering the vehicle unstable as pointed out above. The brake control system for atypical truck is sketched - it consists of separ-ate air and hydraulic fluid circuits. The air ispressurised by an engine driven compressorand driver controlled by a foot pedal oper-ated servo-valve. The air acts over the largepiston of the front brake pressure intensifierwhose small piston causes the hydraulicfluid circuit to reach very high pressure. Aseparate intensifier and hydraulic circuit isprovided for the rear brakes. This system transforms the driver's braking effort Po into hydraulicpressure in the brake cylinders of front and rear wheels. Thus, provided locking of neither front norrear wheels occurs, the ratios of front-to-rear braking torques and decelerating friction forces areconstant and braking is said to be proportional. Such a high performance braking system is notneeded in a car - a single fluid system is adequate.

Although proportional control is straightforward, it is also potentially dangerous if road conditionsand deceleration lead to unstable rear wheel lock. The following paragraphs examine this possiblityby considering the interaction between :

- the vehicle, characterised by the location of its mass centre- the nature of the tyre/road contact, described by the adhesion coefficient k- the decleration, and- the installed brake control system, which must ensure stable braking, in straight forward travel

at least, under as wide a range of road conditions as possible.

The dimensions which locate the mass centre G of a vehicle are normalised by the wheelbase L. Themass centre is thus defined by cF ( or cR as cF + cR =1 ) and h. The deceleration normalised by g is z.

Neglecting as small any pitching tendency, theequations of motion are :-

( 20 ) Σ F → = FF + FR = (W/g) zg = W z

( 21a) Σ MpF= W cF – NR = W z h

( 21b) Σ MpR= W cR – NF = -W z h

For a given deceleration and build - ie. cF, h and weight W - these equations are insufficient fromwhich to find the ground contact components. Another physical condition is necessary - either :

- the braking control system, which dictates the front-to-rear braking distribution in the absenceof wheel lock, or

- locking of either the front wheels, or the rear wheels, or both front and rear simultaneously.

brake servo-valveair

compr-essor

pressureintensifier

Pohigh pressure air

air receiver

very high pressure hydraulic fluid

RNh L

c LF c LR

L

G zg

W

FFNF

FRPF

PR

Brakes 11

Bibliography

Manufacturers' publications.

Friction Materials for Engineers, Ferodo Ltd (UK), 1973

Industrial Brakes and Clutches, Hardie-Ferodo Pty Ltd, NSWUP 1976

Texts

Ageikin IS, Off-the-Road Wheeled and Combined Traction Devices, AA Balkema 1987

Baker AK, Industrial Brake & Clutch Design, Pentech 1992

Burr AH & Cheatham JB, Mechanical Analysis and Design, Prentice Hall 2ed 1995

Neal MJ ed, Component Failures, Maintenance & Repair, Butterworth-Heinemann 1995 usefulguide to failure of bearings, brakes, gears and other elements

Shigley JE & Mitchell LD, Mechanical Engineering Design, McGraw-Hill 1983

South DW & Mancuso JR eds, Mechanical Power Transmission Components, Dekker 1994

Conference proceedings

Braking of Road Vehicles, IMechE 1976, 1993

Apetaur M ed, The Dynamics of Vehicles on Roads and on Tracks, Swets & Zeitlinger 1988

Limpert R ed, Brake Design & Safety, SAE 2ed 1999 very comprehensive but restricted to road vehicles

Sample Dialogue with 'Brakes'

† eg. P = 1 (MPa) 182 (mm ) = 0.255 kN per unit pressure actuationπ/4**left2

†

# this angle is used to determine the resultant drum contacts and hence shaft reaction

#

* any two of these four duty parameters (except the combination of torque and actuation) must be specified

*

- scaled to suit pressure actuation

inputnull on

inputnull on

deceleration is attained ( z= k = 0.3 ). This sequenceis acceptable as the driverexperiences feedback ofincipient loss of control inthe segment CD.

Suppose however thatroad conditions improvedsuch that k = 0.5 (pointE). If the brakes wereapplied severely, allwheels would lock simul-taneously - there wouldbe no warning. It wouldalso be impossible toachieve decelerations greater than 0.5g if road conditions were to further improve (ie. k > 0.5). Sosimple proportional control is deficient in both these regards, unless the slope is unacceptably small.

One way of overcoming these drawbacks at little cost is to incorporate an actuation limit such as apressure relief device in the hydraulic circuit of the rear brakes only, to limit the braking torque andbraking effort on the rear treads. Such a brake control characteristic is called proportional-limiting.An example of this is shown below, set to give a maximum rear braking force of 0.095*vehicleweight, and installed in four different vehicles.

Application of the fore-going arguments dem-onstrates that such acontrol characteristicwill be stable if it liesunder the vehicle char-acteristic for all roadconditions likely to beencountered.

The characteristic of thevehicle previously con-sidered is shown at 1.Characteristic 2 corresponds to a mass centre height increase of L/20. In 3 the mass centre has beenmoved forward by L/20, whilst both these displacements give rise to the characteristic 4. From thisit may be seen that if the original vehicle 1, equipped with the proportional-limiting control schemeillustrated, were loaded in a manner that gave rise to either of these latter two mass centre shifts,then the vehicle would be a potential death trap. A control limitation which is set too low in anattempt to guarantee stability will have an adverse effect in unnecessarily constraining the decelera-tion which is attainable for any given road surface.

Plotting the vehicle and control characteristics in the above manner enables the effects of differentbrake control philosophies to be appreciated quickly. Although vehicle brake design in practice iscomplicated by aspects of real behaviour neglected in the preceding analysis (eg. the inertia of thewheel-tyre-drum system may be significant), and by requirements such as braking stability whilstcornering or jack-knife avoidance in the case of articulated vehicles, such an analysis is a necessaryfirst step.

0.1 0.2

0.40.5

0.1

0.1

0.1

0.2

0.3

0.3

0.4

0.4

k = 0.50.6

0.5 0.600

0.2

0.2

z = 0.3

BR

ear

No

rmal

ised

Fri

ctio

n F

orc

e,

φ R

Front Normalised Friction Force, φF

c = 0.4; h = 0.3F

C

DE

vehicle 1 2 3 4 c 0.4 0.4 0.35 0.35

h 0.3 0.35 0.3 0.35F

Front Normalised Friction Force, φF

0.1 0.2

0.1

0.3 0.4 0.5 0.600R

ear

No

rmal

ised

Fri

ctio

n

F

orc

e,

φ R

1

23

4

proportional-limitingcontrol characteristic

φ = 0.095R max

Brakes 12

so arranged that in normal opera-tion the brake is disengaged by anelectric solenoid. In the event of anelectrical power failure however,the brake is automatically engagedby a steel compression spring.Determine the initial compressionof the spring necessary to arrestthe loaded skip, falling at 2 m/s,within 2 s of brake engagement.The inertia of the pulley/drum is0.5 t.m2 ( 1 t ≡ 1 tonne ) the coeffi-cient of friction for the brake may be taken as 0.3, and the spring, which is made from 10 mmdiameter stock, has 10 active turns of 60 mm mean coil diameter. [ 27mm]

9 At first glance, a sprag type of over-running, uni-directional clutch looks rather like a cylindri-cal roller bearing, in that it consists of two concentric circular rings. However instead of havingcylinders between these, there is a series of closely spaced sprags or cams,similar to the one sketched. These fit loosely for one direction of relativering rotation allowing free-wheeling. Light springs keep the sprags in touchwith the rings. A reversal of relative rotation causes a rocking of the spragsand a tightening-up so that a high torque may be carried. The two linesdrawn from the centre O of the rings to the centres of curvature O1 and O2of the sprag surfaces at the contact points, make a small angle α with one another.

Draw the free body of a sprag, perhaps exaggerating the angle α for clarity, and obtain graphi-cally the relative magnitudes and directions of the two contact forces with their normal andtangential components. Derive approximate equations for the angles betwen the forces and their normal componentsas functions of α and the ring contact radii r1 and r2, and find the maximum value of α in termsof the coefficient of friction µ if slipping is not to occur. [ αmax ≈ µ ( 1– r1/r2) ]

10 The centre of mass of a vehicle lies at cF = h = 1/3.( a) Plot the vehicle braking characteristic along with representative loci of constant adhesion

coefficient and of constant decelerations;( b) The vehicle is equipped with proportional braking of 1:4 (rear:front). What is the maximum

deceleration that can be achieved safely, and the corresponding necessary adhesion coeffi-cient ? [ 0.4]

( c) Repeat ( b) if the normalised rear braking force is limited to 0.06. [ 0.765]

11 The centre of mass of a 1.2 t vehicle lies midway between frontand rear axles, at a height above road level of one quarter of thewheel base. The vehicle is equipped with hydraulically operated,symmetric brakes as shown, the front and rear sets being identi-cal except for lining width and hydraulic cylinder diameter.Further details are :- Front Rear

hydraulic cylinder diameter, mm 29 20.5maximum hydraulic pressure, MPa 5.5 4

The friction coefficient between lining and drum is 0.4 and the tyres

1.5 t

0.5 t

loaded skip

empty skip

1200

head pulleybrake drum

2m/s

390 180

150 150

tosolenoid

50

350

360

180

brakeadjustingscrew

140o Ø 600

O1

r1

r2

α

O2

30 o

90 o

30 o45

160

R 125

BRAKES - PROBLEMS

The program Brakes may be used to check answers.

1 A motor whose inertia is 0.3 kg.m2 drives the rope drum of a hoist through a 5:1 gear reduc-tion. The average diameter, radial thickness and face width of the larger gear's rim are 360, 16and 50 mm. The mass of the rope drum is 120 kg, its radius of gyration is 110 mm, and it isequipped with grooves of 250 mm pitch diameter, on which is wrapped the hoisting ropewhose mass is 0.5 kg/m. The maximum extended length of the rope is 60 m. A brake is incorporated into the motor shaft. Determine the brake torque and average powerover the braking period when stopping within 1m, a load of 1t being lowered at 3 m/s.

[ 435 Nm, 26.1 kW ]2 Derive equations ( 13).

For each brake, determine the sensitivity, the hinge and drum shaft reactions, and the parame-ters which are not defined in the duty statements. The brakes are symmetric, except for themechanism of Problems 5 & 6; the geometry of Problem 6 is identical to that of Problem 5, how-ever the drum rotation is clockwise in 5, counterclockwise in 6. Duty statements are ( psi ≡ lbf/in2; 1000 lbf/kip ) :problem 3 4 5 6friction coefficient 0.32 0.24 0.3 0.3braking torque - - - 25 kip inactuation force 1.2 kN - 400 lbf -lining mean pressure limit - 1.0 MPa 150 psi -lining width 28 mm 75 mm - 1.7 in

[ 3: T= 231 Nm p = 543 kPa S= 1.337 RHl = 462 N RHr = 2716 N RO = 2323 N ][ 4: T= 2646 Nm P = 8604 N S= 1.117 RH = 8900 N RO = 0 ][ 5: T= 2150 lbf.ft w= 1.7 in S= 1.143 RHl = 6540 lbf RHr = 4130 lbf RO = 2510 lbf ][ 6: P= 404 lbf p = 125 psi S= 1.101 RHl = 4720 lbf RHr = 5925 lbf RO = 1290 lbf ]

7 Design an external rigid shoe brake for the hoist of Problem 1, given that braking occurs twice

a minute, that the lining is expected to reach a temperature of 300oC, and that the required lin-ing life is 5 khr. A design factor of 1.2 should be applied to the system energy loss of Problem 1.

8 A skip hoist for lifting bulk material, consists of two identical buckets or skips connected by awire rope which passes around a motor-driven head pulley. The loaded skip is partially coun-terbalanced by the empty skip. The head pulley is equipped with the brake illustrated, which is

50 150

F

16510 o

200F

15 o

pivot

Problem 3 Problem 4

F

ø 250

30 o

30 o

R 100

Problems 5,6

F

30o

20o

5 16

4

12

3

Ø 2012

3

5

dimen-sions ininches

3, 4, 5, 6

Brakes 13

are 640mm diameter.( a) Criticise the safety of this braking arrangement. ( b) Determine the maximum vehicle deceleration that may be obtained, and the tyre-road

adhesion coefficient necessary to achieve it. [ 7.3 m/s2, 0.79]( c) Calculate the maximum value of the average lining pressure if the front and rear lining

widths are 60 and 40 mm respectively. [ 1.34 MPa]

12 Repeat Problem 5 with the shoes pivoted to the posts at b = 11 in, θG = 80o.The sensitivity expression for a pivoted shoe is more complex than that of a rigid shoe - the m& n parameters are no longer appropriate. One way to proceed is to denote ' ≡ d/dµ, where-upon α' and β' follow from ( 6c) and ( 11). Call the scalar ratio between the pressure compo-nents ν and evaluate it from ( 11) as ν = Nc/Ns = pc/ps = – β Js / β Jc. Thereafter form the vectorJo as Jo = Js + ν Jc and hence ν' = – δ∆.β' Jo / β Jc. It follows that η = µr Jo[3] / α Jo. Differentiatingthis last expression and inserting into ( vii) leads to the expression for sensitivity.

[ T = 2160 lbf.ft w = 1.67 in S = 1.148 RHl = 3610 lbf RHr = 1400 lbf RO = 2475 lbf]

Spur Gears 1

- compactness - shafts are short and simply supported where practicable, with gears located asclose as possible to bearings in order to minimise shaft bending

- sturdiness increasing from input through output - the sizes of input & output shafts, and sec-ond & third stage gear teeth, should be compared.

A pair of meshing gears is a power transformer, a coupler or interface which marries the speed andtorque characteristics of a power source and a power sink (load). A single pair may be inadequatefor certain sources and loads, in which case more complex combinations known as gear trains arenecessary - eg. the above gearbox. In the vast majority of applications such a device acts as a speedreducer in which the power source drives the device through the high speed low torque input shaft,while power is fed from the device to the load through the low speed high torque output shaft.

Speed reducers are much more common than speed-up drives not so much because they reducespeed, but rather because they amplify torque. Thus gears are used to accelerate a car from rest, notto provide the initial low speeds ( which could be accomplished by easing up on the acceleratorpedal ) but to increase the torque at the wheels which is necessary to accelerate the vehicle. Torqueamplification is the reason for the gearbox's increasing sturdiness mentioned above.

These notes will consider the following aspects of spur gearing :-- overall kinetics of a gear pair (for cases only of steady speeds and loads)- tooth geometry requirements for a constant velocity ratio (eg. size and conjugate action)- detailed geometry of the involute tooth and meshing gears- the consequences of power transfer on the fatigue life of the components, and hence- the essentials of gear design.

Some of the main features of spur gear teethare illustrated. The teeth extend from theroot, or dedendum cylinder ( or colloquially,"circle" ) to the tip, or addendum circle: boththese circles can be measured. The useful por-tion of the tooth is the flank ( or “face” ), it isthis surface which contacts the mating gear.The fillet in the root region is kinematicallyirrelevant since there is no contact there, butit is important insofar as fatigue is concerned.

Overall Kinetics of a Gear Pair

Analysis of gears follows along familiar lines in that we con-sider kinetics of the overall assembly first, before examininginternal details such as individual gear teeth.

The free body of a typical single stage gearbox is shown. Thepower source applies the torque T1 to the input shaft, drivingit at speed ω1 in the sense of the torque ( clockwise here ). Fora single pair of gears the output shaft rotates at speed ω2 inthe opposite sense to the input shaft, and the torque T2 sup-plied by the gearbox drives the load in the sense of ω2. Thereaction to this latter torque is shown on the free body of the gearbox - apparently the output torqueT2 must act on the gearbox in the same sense as that of the input torque T1.

The gears appear in more detail at ( i) below: O1 and O2 are the centres of the pinion and wheel

root fillet

tip

circularpitch, p

facewidth, fflank

dedendumcircle, r

addendum, a

dedendum, b

a

b

pitchcircle, R

addendumcircle, r

T2

T2

powerflow

output toload

input frompower source

GEARBOXFREE BODY

ω1T1

ω2

ω2

A pair of spur gears for mount-ing on parallel shafts. The 10

teeth of the smaller pinion andthe 20 teeth of the wheel lie

parallel to the shaft axes

A rack and pinion. Thestraight rack translates recti-

linearly and may be regardedas part of a wheel of infinite

diameter

Like spur gears, helicalgears connect parallel shafts

however the teeth are not par-allel to the shaft axes but liealong helices about the axes

SPUR GEARS

Gears are used to transmit power between shafts rotating usually at different speeds. Some of themany types of gears are illustrated below.

Straight bevel gears for shaftswhose axes intersect

Hypoid gears - one of a num-ber of gear types for offset

shafts

A worm and wormwheelgives a large speed ratio but

with significant sliding

In order to demonstrate briefly the development of gear drives, from firstprinciples through to safety implications, we consider hereonly spur gears. Knowledge of these is fundamental to under-standing the behaviour of geometrically more complex types,including helical gears which are generally preferredto spurs since they are more compact and smoother inoperation, thus permitting higher speeds.

A typical commercial gearbox is shown here with itscover removed. It demonstrates that it is usually moreattractive economically to split a larger speed ratio into anumber of stages (pairs of gears) rather than to effect itwith a single pair. There are three stages here - thefirst spiral bevel pair is followed by two helical pairs.

A couple of features of the box are immediatelyapparent :

Spur Gears 2

above gearbox is used when the desired speed ratio cannot be achieved economically by a singlepair. Applying ( 1) to each stage in turn, the overall speed ratio of a compound train is found to bethe product of the speed ratios for the individual stages.

Selecting suitable integral tooth numbers to provide a specifiedspeed ratio can be awkward if the speed tolerance is tight and therange of available tooth numbers is limited. Until the advent ofcomputers allowed such problems to be solved by iterative trials,techniques based on continued fractions were used. Appendix Ais provided to illustrate the concepts and advantages of contin-ued fractions and attendant Padé approximations - this is for gen-eral interest, not just for gears.

Unlike the above gearbox, the input and output shafts are coaxial in the train illustrated here; this israther an unusual feature but necessary in certain change speed boxes and the like.

Epicyclic Gear Trains

An epicyclic train is often suitable when a large torque/speed ratio is required in a compact enve-lope. It is made up of a number of elements which are interconnected to form the train. Each elementconsists of the three components illustrated below :

- a central gear ( c) which rotates at angular velocity ωc about the fixed axis O-O of the element,under the action of the torque Tc applied to the central gear's integral shaft; this central gearmay be either an external gear (also referred to as a sun gear) Fig 1a, or an internal gear, Fig 1b

- an arm ( a) which rotates at angular velocity ωa about the same O-O axis under the action of thetorque, Ta; an axle A is rigidly attached to the end of the arm

- a planet gear ( p) which rotates freely on the axle A at angular velocity ωp, meshing with thecentral gear at the pitch point P - the torque Tp acts on the planet gear itself, not on its axle, A.

The epicyclic gear photographed overleaf without its arms consists of two elements. The central gearof one element is an external gear; the central gear of the other element is an internal gear. The three

P Ft

Ft

Ft

Ft

Rp

Rc

Tp

Tc

A

O

Ft

Ft

Ra

Ta

O

A

planet gear (p)

ωa

Ta

ωc

Tc

ωp

Tp

arm (a)

external central gear (c)

O

A

Fig 1a

Fig 2a

ωa

Ta

ωc

Tc

ωp

Tp

planet gear (p)

internal central gear (c)

arm (a)

Fig 1b

PFt

Rc

Tc

O Ft

Fig 2b

O

A

Ta

Rp

Ft T

p

A

P

Ft

Ra

O

A

input output

a

bc

d

shafts respectively. We may regard the gears as equivalent pitch cylinders ( ii) which roll togetherwithout slip - the requirements for preventing slip by the positive drive provided by the meshingteeth is examined below. Unlike the addendum and dedendum cylinders, pitch cylinders cannot bemeasured directly; they are notional and must be inferred from other measurements.

One essential for correct meshing of the gears is that the 'size' of the teeth on the pinion is the sameas the size of teeth on the wheel. One measure of size is the circular pitch, p, the distance betweenadjacent teeth around the pitch circle ( ii); thus p = πD/z where z is the number of teeth on a gear ofpitch diameter D. The SI measure of size is the module, m = p/π - which should not be confusedwith the SI abbreviation for metre. So the geometry of pinion 1 and wheel 2 must be such that :

D1 /z1 = D2 /z2 = p / π = m

. . . . that is the module must be common to both gears. For the rack illustrated above,both the diameter and tooth number tend to infinity, but their quotient remains the finite module.

The pitch circles contact one another at the pitch point , P, figure ( iii), which is also notional. Sincethe positive drive precludes slip between the pitch cylinders, the pitch line tangential velocity, v,must be the same for both pinion and wheel :

v = ω1 R1 = ω2 R2 ; R = D/2

Separate free bodies of pinion and wheel appear in ( iv). Ft is the tangential component of action-reaction at the pitch point due to contact between the gears. The corresponding radial componentplays no part in power transfer and is therefore not shown on the bodies. Ideal gears only are con-sidered initially, so the friction due to sliding contact is omitted also. The free bodies show that themagnitude of the shaft reactions must be Ft, and that for equilibrium in the absence of friction :

Ft = T1 /R1 = T2 /R2 ; R = D/2

The preceding concepts may be combined conveniently into :-

( 1 ) ω1 /ω2 = T2 /T1 = D2 /D1 = z2 /z1 ; D = 2R = m z

That is, gears reduce speed and amplify torque in proportion to their teeth numbers. In practice,rotational speed is described by N ( rpm ie. rev/min, or Hz ) rather than by ω ( rad/s ).

The only way that the input and output shafts of a gear pair can be made to rotatein the same sense is by interposition of an odd number of intermediate gears asshown - these do not affect the speed ratio between input and output shafts. Sucha gear train is called a simple train. If there is no power flow through the shaft ofan intermediate gear then it is an idler gear.

A gear train comprising two or more pairs is termed compound when the wheelof one stage drives the pinion of the next stage. A compound train as in the input

output

idler

p

p

T2

T1

O1

O2

Ft F

t

Ft

Ft

P

ω1

ω2

2D

P v

1DO1

O2

ω2

ω1T1

T2

O1

O2

z = 91

z = 122

( i ) ( ii ) ( iii ) ( iv )

Spur Gears 3

= Tc { ( ωc – ωa ) zc + ( ωp – ωa ) zp }/zc = 0confirming that energy is conserved in the ideal element.

In practice, a number of identical planets are employed for balance andshaft load minimisation. Since ( 2) deal only with effects external to theelement, this multiplicity of planets is analytically irrelevant provided Tp

is interpreted as the total torque on all the planets, as suggested by thesketch here. The reason for the “sun-and-planet” terminology is obvious;the arm is often referred to as the spider or planet carrier.

Application of the element relations to a complete train is carried out as shownin the example below. More complex epicyclic trains may be analysed in a similar manner, but thetechnique is not of much assistance when the problem is one of gear train design - the interesteddesigner is referred to the Bibliography.

Conjugate Tooth Action

We have seen that one essential for correctly meshing gears is that the size of the the teeth ( the mod-ule ) must be the same for the two gears. We now examine another requirement - the shape of theteeth necessary for the speed ratio to remain constant during an increment of rotation; this behavi-our of the contacting surfaces ( ie. the teeth flanks ) is known as conjuate action.

Consider the two rigid bodies 1 and 2 overleaf which rotate about fixed centres, O, with angularvelocities ω. The bodies touch at the contact point, C, through which the common tangent and nor-mal are drawn.

Tp /6Tp /6

Tp /6Tp /6

Tp /6 Tp /6Ta

EXAMPLE An epicyclic train consists of two three-component elements of the kind

examined above. The first element comprises the external sun gear 1 and planet 2; the

second comprises the planet 3 and internal ring gear 4. The planets 2 and 3 are com-

pounded together on the common arm axles. Determine the relationships between the

kinetic variables external to the train in terms of the tooth numbers z1, z2, z3 & z4.

The train is analysed via equations ( 2) applied to the two elements in turn, together

with the appropriate equations which set out the velocity and torque constraints

across the interface between the two elements 1-2-arm and 3-4-arm.

1-2-arm : ( ω1 – ωa ) z1 + ( ω2 – ωa ) z2 = 0 from ( 2a)

T1/z1 = T2 /z2 = –Ta2 /( z1 + z2 ) from ( 2b)

3-4-arm : ( ω4 – ωa ) (–z4 ) + ( ω3 – ωa ) z3 = 0 in which z4 is a positive integer as central gear is internal

T4/(–z4 )= T3 /z3 = –Ta3 /( –z4 + z3 )

Ta2 and Ta3 are the parts of the total external torque on the arm, Ta, which are applied individually

to the two elements : 1-2-arm and 3-4-arm.

Interface : ω3 = ω2 since the planets 2 & 3 are coupled

T3 = – T2 since the planets 2 & 3 are coupled (action/reaction)

Ta = Ta2 + Ta3 as the arm is common to both elements 1-2-a and 3-4-a

Solution : The basic speed ratio, io, of an epicyclic train is defined as the ratio of input to output speeds when the

arm is held stationary. Neither input nor output is defined here - indeed this teminology is rather confusing with

multiple degrees of freedom - so select gear 1 as input, gear 4 as output. It follows that io = ( ω1/ω4 )ωa=0.Solving the three velocity equations and the six torque equations leads to the desired relations :

Velocities : ( ω1 – ωa ) = io ( ω4 – ωa ) where io = – z2z4 /z1z3Torques : T1 = –T4 /io = Ta /( io – 1 )

Evidently this train possesses the same degrees of freedom as an individual element.

1

2 3

4

arm

identical planets of one element are compounded with ( joinedto ) those of the second element.

We shall examine first the angular velocities and torques in asingle three-component element as they relate to the tooth num-bers of central and planet gears, zc and zp respectively. Thekinetic relations for a complete epicyclic train consisting of twoor more elements may then be deduced by combining appropri-ately the relations for the individual elements.

All angular velocities, ω, are absolute and constant, and thetorques, T, are external to the three-component element; for convenience all these variables are takenpositive in one particular sense, say anticlockwise as here. Friction is presumed negligible, ie. thesystem is ideal.

Separate free bodies of each of the three components - including the torques which are applied oneto each component - are illustrated in Figs 2a and 2b for the external and internal central geararrangements respectively. Also shown are the shaft centre O and axle A, the radii Rc & Rp of thecentral and planet pitch cylinders, the radius of the arm Ra.

There are two contacts between the components :- the planet engages with the central gear at the pitch point P where the action/reaction due to

tooth contact is the tangential force Ft, the radial component being irrelevant;- the free rotary contact between planet gear and axle A requires a radial force action/reaction;

the magnitude of this force at A must also be Ft as sketched, for equilibrium of the planet.

With velocities taken to be positive leftwards for example, we have for the external central gear :geometry from Fig 2a : Ra = Rc + Rp

velocity of P : vP = vA + vPA so with the given senses : ωcRc = ωaRa – ωpRp

torques from Fig 2a : Ft = –Tc/Rc = –Tp/Rp = Ta/Ra

and for the internal central gear :geometry from Fig 2b : Ra = Rc - Rp

velocity of P : vP = vA + vPA so with the given senses : ωcRc = ωaRa + ωpRp

torques from Fig 2b : Ft = –Tc/Rc = Tp/Rp = Ta/Ra

Substituting for Ra from the geometric equations into the respective velocity and torque equations,and noting that Rc/Rp = zc/zp, leads to the same result for both internal and external central geararrangements. These are the desired relations for the three-component element:

(2a) ( ωc – ωa ) zc + ( ωp – ωa ) zp = 0 where zc is taken to be a positive integer for an external central(2b) Tc /zc = Tp /zp = –Ta /( zc + zp ) gear, and a negative integer for an internal central gear

It is apparent that the element has one degree of kinetic (torque) freedom since only one of the threetorques may be arbitrarily defined, the other two following from the two equations ( 2b). On theother hand the element possesses two degrees of kinematic freedom, as any two of the three veloci-ties may be arbitrarily chosen, the third being dictated by the single equation ( 2a).

From ( 2b) the net external torque on the three-component element as a whole is : ΣT = Tc + Tp + Ta = Tc { 1 + zp /zc – ( zc + zp )/zc } = 0

which indicates that equilibrium of the element is assured.

Energy is supplied to the element through any component whose torque and velocity senses areidentical. From ( 2) the total external power being fed into the three-component element is : ΣP = Pc + Pp + Pa = ωcTc + ωpTp + ωaTa = Tc { ωc + ωpzp /zc – ωa( zc + zp )/zc }

Spur Gears 4

cutting the fixed reference at the fixed pitch point P through which the pitch cylinder passes. Theline of action is inclined to the pitch point tangent at the pressure angle, α. The knot C always movesalong the line of action, tracing out an involute with respect to the rotating cylinder. The relationbetween the base and pitch circle radii is evidently :-

( 5 ) Ro = R cos α

Extending this to two cylinders - representing meshing gears,1 & 2 Fig C - the taut cord T1-P-C-T2 winds off one base cylin-der and onto the other to form the line of action inclined at thepressure angle α. The knot C on the mating involutes coin-cides with the contact point and moves along the line of actionas the gears and base cylinders rotate.

The pitch cylinders extend to the pitch point P situated at theintersection of the lines of action and of centres.

Since the cord remains taut, the gears rotate at a constantvelocity ratio which equals the ratio of base cylinder radii. Itfollows from ( 5) that the pitch cylinders rotate without slip. Evidently the distance between the cyl-inders does not affect the speed ratio since the base cylinder diameters are fixed.

The distance between knots - ie. between tooth flanks along the line of action, Fig C - is the basepitch, po, given by :-

( 6 ) po = π Do / z = p cos α = π m cos α . . . . . from ( 1)

For continuous motion transfer, at least two pairs of teeth must be in contact as one of the pairscomes into or leaves mesh. The teeth in Fig C are truncated in practice to permit rotation.

Involute generation by knotted cord is all very well conceptually, but useless as a basis for manufac-ture. There are many methods of gear manufacture in practice, however we shall examine only therack generation technique which is fundamental to the understanding of gear behaviour.

The Generation Process: Tooth Systems & Profile Shift

A tooth system is described by a unique set of tooth proportions andpressure angle, characterised by the system’s basic rack, Fig D.Various systems find application in particular specialised sectors ofindustry, but by far the most widespread - and the only system con-sidered here - is the 20° full depth system.. This incorporates a 20°pressure angle ( α) and the proportions :-

half-width h = π/4 addendum a = 1dedendum b = 1.25 fillet radius e = 0.38

that is the tooth occupies half the pitch (π) measured along the reference line.

A circular gear of specified tooth system and module is generated by a rack cutterwhich takes the form of the shaded mask of Fig D. The rack’s profile is a cutting edgewhose straight sides and fillets machine respectively the active flanks and the roots ofthe gear’s teeth. The cutter is set out about the reference and centre lines, with abso-lute dimensions derived from the system’s proportions multiplied by the specified module.Cutters are available commercially to the following modules :-

m = 1 1.25 1.5 2 2.5 3 4 5 6 8 10 12 16 20 25 32 40 50 mm

α

T1

line of centres

P

Fig C

T22

1

line of action

C

base pitch p

o

reference line

Fig D centre line

α

v

uO

erad

π / 2π / 2

a

bh

The absolute velocity v of the contact point reckoned as a point on either body, is perpendicular tothe radius from that body's centre O. For the bodies to remain in contact, there must be no compo-nent of relative motion along the common normal, so that from the velocity triangles :-

v2 cos θ2 = v1 cos θ1 where v1 = ω1.O1C ; v2 = ω2.O2CNote that the tangential components of velocity are generally different, so sliding must occur.

For the speed ratio to be constant therefore, from the above and similar triangles :-

( 3 ) ω2/ω1= v2.O1C/v1.O2C = O1C.cos θ1/O2C.cos θ2= O1C1 / O2C2 = O1P / O2P ie. this ratio also must be constant.

This indicates that, since the centres are fixed,the point P is fixed too. In general therefore,whatever the shapes of the bodies, the contactpoint C will move along some locus as rota-tion proceeds; but if the action is to be conju-gate then the body geometry must be suchthat the common normal at the contact pointpasses always through one unique pointlying on the line of centres - this point is thepitch point referred to above, and the pitchcircles' radii are O1P and O2P .

There exists a host of shapes which ensure conjugacy - indeed it is possible, within certain restric-tions, to arbitrarily choose the shape of one body then determine the shape of the second necessaryfor conjugacy. But by far the most common gear geometry which satisfies conjugacy is based on theinvolute, in which case both gears are similar in form, and the contact point's locus is a simplestraight line - the line of action.

The Involute Tooth

One method of generating an involuteis shown in Fig A. A generating cord, inwhich there is a knot C, is wrappedaround a fixed cylinder - the base cylin-der ( idiomatically “circle” ) of radiusRo. When the taut cord is subsequentlyunwound, the knot traces out an invo-lute whose polar coordinates may beexpressed implicitly in terms of the var-iable generating angle ψ, reckoned fromthe radius through the initial knot posi-tion, C'. The coordinate origin is taken at the circle centre, O, with a fixed reference direction definedat some constant angle γ, also from the initial radius. The tangent, TC, is normal to the involute at C,and since the tangent length TC is equal to the arc TC', the polar coordinates of C (r,θ) are

( 4 ) r = Ro√( 1 + ψ2 ) ; θ = γ – ψ + arctan ψ

In order to see how the involute leads to gear teeth and conjugate action, we alter this cord-and-cylinder model slightly. The cord is wrapped around the base cylinder which in Fig B is now free torotate about its centre as the cord is pulled off in a fixed direction. This fixed cord direction formsthe line of action, tangent to the base cylinder at the fixed point T, and clearly satisfies conjugacy by

Fig A

T

unwindinggeneratingcord

θ

ψ

ψRο

Rο

C

refe

renc

e

γ

involute

fixedbasecircle

rC'

translatinggeneratingcord (lineof action) P

R

rotatingbase circle

pitch circle

Fig B

T

Rο

Cre

fere

nce

involute

pressureangle, α

P

1

2

v1

v2

O1

O2

C1

C

C2

θ1

θ 2ω1

ω2

Which body is thedriver here ?

Spur Gears 5

Gear proportions and absolute dimensions are often used interchangeably - this should not causeconfusion provided the absence or presence of units is noted carefully. Thus the dedendum of a 4mm module gear may appear alternately as 1.25 (no units) or as 5 mm (ie. 1.25 multiplied by 4 mm).If this gear possessed 20 teeth then its pitch radius could be quoted as 40 mm ( 1) or as 10 (no units).

Generated gears of the same system which have the same number of teeth are geometrically similarto one another. One cutter only is necessary for each module, no matter how many teeth are in thegear being cut - though obviously gears become increasingly impractical as their tooth numbersdecrease (imagine trying to achieve conjugacy when a gear with two teeth engages a gear having three teeth :try sketching this!)

Once the cutter has been prepared, generation commences by aligning the rack's reference line withthe pitch circle of the gear blank about to be cut - Fig E shows this for unit module.

During machining of the gear blank by the rack, the cutter undergoes two distinct simple motions :- it reciprocates perpendicularly to the sketch plane Fig E, shaving material off the blank as

shown in the photograph of a partially cut cut gear at left below, and- it translates parallel to its reference line while the circular gear blank rotates, an external mech-

anism being provided so that there is no slip between these feed movements.

Generation may be carried out by a hob in the form of a rotating worm-wheel (at right above) ratherthan by a reciprocating rack.

Before cutting commences, the rack may be displaced transverse to its reference line by an amount s(or m*s dimensionally) where s is the so-called profile shift coefficient (also known as the addendummodification coefficient) Fig F, which is introduced to lower the practical tooth number limitation.Profile shift is reckoned positive if the rack is moved away from the blank as shown above; negativeif the cutter is moved towards the gear blank’s centre.

A gear’s pitch circle represents the notional cylinder which rolls without slip on the pitch circle ofits mating gear. The introduction of profile shift makes it necessary to distinguish between three dif-ferent pitch circles :-

- The standard pitch circle (Fig E) is defined by the gear’s tooth number only, thus from ( 1) thiscircle’s proportional radius is R = z/2, or in dimensional terms R = mz/2.

- When profile shift s is present then the pitch circle is referred to as the extended pitch circle;its radius is R + s proportionally as may be seen from Fig F.

- The (actual) pitch circles , R ’1 and R ’2, at which two involute gears operate depend on the dis-tance between their centres, as will be explained in the Gear Meshing section below.

The Macintosh program ‘Tooth Generator’ simulates the rack generation process, duplicating the rela-tive motion of a unit module rack cutting a gear blank whilst holding the blank stationary to betterillustrate formation of tooth flank and fillet. The program and the animation http://www.mech.uwa.edu.au/DANotes/gears/generation/generation.gif illustrate how the cutter shaves a portion of thetooth during each reciprocation, only a few of which are plotted.

The potential advantage to be gained by using profile shift is exemplified below by the program’soutput for three 20o full depth 9-tooth gears which are identical apart from being generated with dif-ferent profile shifts. The base, standard pitch, addendum and dedendum circles appear in each plot -since the gears have the same number of teeth they share the same standard pitch and base circles -

and the rack reference origin is highlighted. The useful involute flank exists only outside the basecircle (recall the generating cord above): the root fillet is trochoidal.

- With no profile shift as shown on the leftmost sketch above, the rack reference origin lies onthe standard pitch circle and the tooth is severely undercut, seriously weakening the cantile-vered tooth in the region of highest bending stress.

- A profile shift of 0.6 in the centre sketch removes this undercut but the reduced tip width isdangerous as the tip is liable to be easily broken off, especially if the teeth are hardened.

- A larger shift in the rightmost tooth leads to extremepointing; the tooth does not extend to the theoreticaladdendum circle.

The effect of profile shift is further exemplified by thismontage of five different gears. They all have 10 teeth sothey share common base and standard pitch circles - theirtooth flanks are therefore all part of the same involute.However the gears’ profile shifts differ and so the usefulflanks are based on different portions of that involute.

For a particular number of teeth there is a critical lowerprofile shift below which the teeth become undercut, and acritical upper profile shift above which the tooth tipsbecome pointed. It is unnecessary to consider tooth geom-etry further here - full details of the involute flank and the

undercut pointed

1.00.6

z = 9s = 0.0

z = 9s = 0.6

z = 9s = 1.0

base standard pitch

0.0

0.5

1.0

1.5– 0.5

s =

extended pitch addendum

basestandard

pitch

dedendum

z = 10s = 0.5s = 0.6ˆˇ

Fig F

π

Fig E standard pitch circle, Rextendedpitch circle

rack reference line

s

Spur Gears 6

Number of teeth, z

Pro

file

shift

, s

Fig G105 15 20 25 30 35

1.0

0.5

0.0

- 0.5

theorypractice

POINTED

UNDERCUT

Gear Meshing

Involute gears have the invaluable ability of providingconjugate action when the gears' centre distance isvaried either deliberately or involuntarily due tomanufacturing and/or mounting errors.

The exaggerated effect of this is shown in Fig H forgears manufactured without profile shift to pressureangle α. On the left, the gears are mounted at thestandard centre distance (the sum of the two stan-dard radii given by ( 1)) with the “cord” line ofaction wrapped around the base circles with radii given by( 5).

On the right of Fig H the gears’ centre distance is increased - but since the same generating cord iswrapped around the same base circles then it follows that the speed ratio is unaltered and the sameinvolutes ( and hence teeth ) are involved. Clearly there are practical limits to centre distance varia-tions - eg. the gears may lose contact completely - however provided these limits are not reachedthen the pitch point and resulting pitch circles are defined by intersection of the lines of action andof centres, exactly as occurred in Fig C above. From the similar triangles of Fig H :-

( 8 ) R’i = D’i /2 = C zi /Σz = Roi sec α’ ( i = 1,2 ) using ( 5)

There is an infinite number of possible centre distances for a given pair of profile shifted gears, how-ever we consider only the particular case known as the extended centre distance which correspondsto mutual tangency of the two extended pitch circles mentioned above, that is :-

( 9 ) C = ( R1 + s1 ) + ( R2 + s2 ) = 1/2 Σ z + Σ s proportionally, using ( 1), or

= m ( 1/2 Σ z + Σ s ) dimensionally.

Without profile shift, the choice of centre distance is limited by modules available from the standardlist and by integral tooth numbers, however the profile shift sum in ( 9) provides flexibility in thechoice of (extended) centre distance - invaluable in a coaxial reduction for example.

The meshing of typical gears at extended centres is further detailed in Figs I and J which are frames

EXAMPLE A pair of 20o full depth gears mesh at extended centres - the 12 tooth pinion’s profile shift is 0.5, the

18 tooth wheel’s is 0.4. What are the pitch and addendum diameters of the gears and what is the pressure angle ?

No absolute dimensions are given so it is proportions which are relevant in this example.

From ( 1) the diameters of the standard pitch circles are D1 = 12

D2 = 18. Corresponding base circle diameters are Do1 = 12*cos20o = 11.28, Do2 = 18*cos20o = 16.92

The pinion’s extended pitch radius is R1 +s1 = 12/2 + 0.5 = 6.5

and similarly the wheel’s is 18/2 + 0.4 = 9.4

The extended centre distance is thus (9) 6.5 +9.4 = 15.9

From (8) the actual pitch diameters are D’1 = 2*15.9*12/(12+18)

= 12.72 and D’2 = 2*15.9*18/(12+18) = 19.08

Note that the actual pitch circles preserve the velocity ratio :

19.08/12.72 = 16.92/11.28 = 18/12 = 1.5

Also from the geometry and ( 8) the addendum diameters are Da1 = 2( 6.5 +1.0) = 15.0, Da2 = 2( 9.4 +1.0) = 20.8,

while the actual pressure angle is α’ = arcos( Doi /D’i ) = arcos( 16.92/19.08) = 27.5o

s

a

b

R R o

rack reference line

addendumcircle

standardpitch

base circlededendum

circle

extendedpitch

pitchcircle

Ro2

Ro1

α R'2

R'1

α'

standardpitchcircle

basecircle

R1

R2

s = s = 01 2

line ofaction

C

Fig H

trochoidal fillet are given in Appendix B. Fig G illustratesthe profile shift limits for the 20o full depth systemrevealed by that theory, together with the usual practicallimits suggested by ISO TR 4467, namely :-

( 7 ) max [ – 0.5, (30–z)/40 ] ≤ s ≤ 0.6

Desirable compactness necessitates the smallest toothnumbers possible, but these tooth form limits becomemore constraining as tooth numbers decrease - eg. 20 isthe minimum which theoretically avoids undercutting inthe absence of profile shift. Depending upon the drive asa whole, spur gear tooth numbers of 12 are possible with suitable profile shift. Despite the indica-tions of Fig G, less than 12 teeth are seldom used - except for lightly-loaded essentially kinematicapplications - due to other considerations such as the weakness of non-pointed but thin tooth tipsand the need for two pairs of teeth to be in contact simultaneously as the drive is transferred fromone pair to the next, manufacturing and/or mounting inaccuracies notwithstanding (see contactratio below).

The specialised business of gear manufacture and inspection is complex. The example below givessome insight into the necessity for the involute geometry of Appendix B. Not infrequently the stan-dardised 20o full depth system is altered subtlely to improve operation.

Helical gears ( see photograph page 1) are preferred to spur gears in high duty applications since toothengagement is more gradual and operation smoother. Helical gears may have smaller tooth num-

EXAMPLE Show that a 20o full depth spur gear is a body of con-stant width w hen measured as shown across tooth flanks, the difference

between two such measurements giving the module of the gear.

With what standard module rack would this gear have been generated?

If there are 13 teeth on the gear, what is the profile shift ?

The measurement C1C2 is nor-

mal to the teeth flanks and so

coincides with the tangent to the base circle at T. In a similar manner, the

measurement C1'C2

' coincides with the tangent at T '. But the arc length

TT ' must equal the differences in generating cord lengths TC1 – T ' C1'

etc. so the measurements are equal.

Radii are drawn to the starts of the flank involutes on the base circle; these radii are inclined at γ to the tooth centre

line, and corresponding radii on adjacent teeth subtend 2π/z at the centre, O. Equality of

the generating cord and arc lengths requires that the measurement across n

teeth is Ln = Ro ( 2γ + ( n – 1) 2 π/z ), so that the difference in measurements

between n and n+1 teeth is evidently Doπ/z, or, using ( 1, 5), π m cos α.

Assuming a 20o pressure angle here, m = (62.14 – 38.52)/ π cos 20o = 8 mm. A 13-tooth gear therefore has a pitch diameter of D = mz = 104 mm and

from the Ln equation above :-

γ = L2/ D cos α – π/z = 38.52/104 cos 20o – π/13 = 0.1525 rad

We now need to evaluate γ as a function of s. Figures D and F indicate that the

rack space half-width on the gear rolling (pitch) circle is ( h + s tan α ), and this must

equal the arc subtended by the gear half-tooth around the pitch circle, ie. rθ, where r and

θ are given by ( 4) at r = R. From this we have :-

γ = ( h + s tan α ) / R + tan α – α or, inserting known values

0.1525 = ( π/4 + s tan 20o ) 8 / 52 + tan 20o – π/9 whence s = 0.30

TT'C'1

C'2

C1C2

base circle

γ γ2πz

2πz

O

Ro

62.14

38.52

EXAMPLE Outline a procedure for selecting a spur gear pair on kinematic/geometric grounds, for the front

end of a general design program whose later segments could examine fatigue and reliability.

Input data are - minimum and maximum bounds of speed ratio ( ≥ 1 ) and of extended centre distance ( mm );

while output is - module ( mm), with corresponding tooth numbers, z1 & z2 ( say 12 ≤ z1 ≤ 25), and profile shifts,

s1 & s2 , to give a speed ratio and extended centre distance falling within the specified bounds,

and a contact ratio no less than some stipulated critical value ( eg. 1.2 ).

The procedure might take the form of a number of nested loops, each assigning a trial value to one of the design

variables. For example the outermost loop would select a module from the standard list and the next would

assume a trial pinion tooth number between the limits suggested above. The wheel tooth number would be

assigned in the next inner loop, corresponding to the pinion teeth and a speed ratio between the specified limits.

The innermost loop would consider both profile shifts simultaneously and is best understood through a graph of

s2 versus s1, realising that the module and tooth numbers have been settled in outer loops prior to this aspect being

being examined. The tooth numbers will define practical limits to the profile shifts ( 7), hence solutions must lie

within the known rectangular region of the graph.

In addition to these limitations, there are further bounds, Σsc-minand Σsc-max, corresponding to the centre distance limits from ( 9) -these appear as the straight boundaries superimposed on the graph;

and a high bound, Σse-max, corresponding to the critical contactratio via ( 10), which appears as the curved boundary. The actual

disposition of all these boundaries on the graph - which may or may

not enclose a viable solution space - depends of course on the prob-

lem in hand and the values of module and tooth numbers assigned

in outer loops. Having defined any such solution space, the proce-

dure could output any solution lying within it - any point on the line between the two extreme points 'lo' and 'hi'

would appear to be a suitable compromise - or alternatively ( 11) might be implemented.

hi

lo

solution

space

s1

s2

2s

1s

Σ cs

2s

Σ csˆΣ es

1s

Continuous motion transfer requires two pairs of teeth in contact at the ends of the path of contact,though there is only one pair in contact in the middle of the path, as in Fig J. The average number ofteeth in contact is an important parameter - if it is too low due to the use of inappropriate profileshifts or to an excessive centre distance for example, then manufacturing inaccuracies may lead toloss of kinematic continuity - that is to impact, vibration and noise. The average number of teeth incontact is also a guide to load sharing between teeth; it is termed the contact ratio, εγ, given by :-

εγ = length of path of contact / distance between teeth along the line of action

= Q2PQ1 / base pitch, po and for extended centres with ( 6) for the 20o system :

( 10 ) ( 2 π cosα ) εγ = Σ i = 1,2 √[ ( z i + 2( 1+s i ))2 - ( z i cosα )2 ] - √[ ( Σz + 2 Σs )2 - ( Σz cosα )2 ]

Gears having a contact ratio below about 1.2 are not nor-mally recommended as the gears themselves, their shaftsand bearings would all require especial care in design andmanufacture to preserve conjugacy. The effect of toothnumber on contact ratio is shown in Fig K for two identicalgears with a profile shift which varies between the practicallimits of Fig G. Evidently it becomes more difficult toachieve an acceptable contact ratio as tooth numbersdecrease - hence the statement above that 12 teeth are theusual minimum. The plot further indicates that contact ratioincreases as profile shift decreases.

In design, Σs is often taken as zero; or, if the centre distance is closely specified, by the selection of Σsfor use in ( 9) - the individual profile shifts may be estimated from a specified Σs as follows :-

( 11 ) s1 = [ λz2 + ( Σs - λ ) z1 ] /Σz ; s2 = Σs - s1 ; where 0.5 ≤ λ ≤ 0.75

which tends to balance the strengths of the gears.

0.0 1.0Profile Shift

2.0

1.0

Con

tact

Rat

io

1.2

number of teeth inidentical gears

50

30

201512

FIG K

from the animation http://www.mech.uwa.edu.au/DANotes/gears/meshing/gearAnimation.gif ofthe above 12:18 drive example .

Fig I shows the initial no-load situation. The gears are mounted so that the extended pitch circles (ofradii 6.5 and 9.4 in the example) are mutually tangential with the pinion tooth symmetrically dis-posed with respect to the wheel inter-tooth space. Tangency of the extended pitch circles impliesthat the generating rack (shown dotted) is simultaneously tangent to the tooth profiles of both gears,and leads to a gap - an absence of contact - between the pinion tooth and both adjacent wheel teeth.

A subsequent on-load state of affairs when the pinion rotates clockwise is shown in Fig J. The tautgenerating cord (line of action) extends between the tangent points T1 and T2 on the base circles. Thecord’s intersection with the line of centres defines the actual pitch point P - exactly as in Fig H - andin turn the actual pitch circles whose diameters in the particular case are 12.72 and 19.08.

A pinion tooth touches a wheel tooth at the contact point C (the knot) which moves up the line ofaction and along the teeth faces as rotation proceeds. Since contact cannot occur outside the teeth, ittakes place along the line of action only between the points Q2 and Q1 inside both addendum circles.The line segment Q2Q1 is named the path of contact.

The animation, and to a lesser extent Figures I and J, shows clearly :- the contact point marching along the line of action- the path of contact bounded by the two addenda- the orthogonality between the line of action and involute tooth flanks at the contact point- how load is transferred from one pair of contacting teeth to the next as rotation proceeds- relative sliding between the teeth - particularly noticable at the beginning and end of contact- guaranteed tooth tip clearance due to the dedendum exceeding the addendum- a significant gap between the non-drive face of a pinion tooth and the adjacent wheel tooth

Gears formed by a milling cutter instead of being generated by a rack cutter as above, may exhibitundercutting or interference (which prevents complete rotation of the two gears due to teeth bind-ing) - these faults result from Q2 for example lying to the left of T1 in Fig J.

The gap between the non-drive face of the pinion tooth and the adjacent wheel tooth is known asbacklash. If the rotational sense of the pinion were to reverse, then a period of unrestrained pinionmotion would take place until the backlash gap closed and contact with the wheel tooth re-established impulsively. Shock in a torsionally vibrating drive is exacerbated by significant backlash,though a small amount of backlash is provided in all drives to prevent binding due to manufactur-ing or mounting inaccuracies and to facilitate lubrication. Backlash may be reduced by subtle altera-tions to tooth profile or by shortening the centre distance from the extended value, however we con-sider gears meshing only at the extended centre distance.

P

T1

T2

Q1

9.54 rad

C

Fig J 6.36 rad

Q2

9.4 rad

z =18 s =0.42 2

Fig I 6.5 radz =12 s =0.51 1

Spur Gears 8

fails to reach the life given by the curve for a particular load.We shall consider only steel materials (carbon steels induc-tion- or through-hardened, nitrided alloy steels etc). Ratherthan provide the complete load-life diagram for each steel, theAGMA chooses a reference point on the curve correspondingto 99% survival rate after 107 unidirectional loading cycles,and cites the corresponding allowable stress, S, as a represen-tative property of the steel. So, for any given load (σ) the lifeand survival rate (reliability) may be correlated through :

( 12 ) σ = S * life factor ( KL or CL ) where :

reliability factor ( KR

) - The reliability factor, KR, caters for survival

rates other than 99%. Since the survival con-tours are essentially parallel to one another on alogarithmic scale, then simple multiplying fac-tors enable load correlation, as tabled :-

- The life factor ( KL for bending, CL for pitting) caters for lives other than 107 cycles. Since theload-life diagrams for all the steels considered are of the same shape essentially, normalisingby the allowable stress will result in a unique KL (or CL) -versus- life curve for all steels; thiscurve will be presented later.

These aspects when combined will result in gear tooth design equations, one for bending strengthand one for pitting resistance, which embrace load (transmitted power), dimensions (primarily sizecharacterised by module), material (allowable stress) and life rather than safety.

Tooth Forces

It will be recalled that the free body of a gear pair necessitates identical senses of the torques on pin-ion and wheel - Fig L below repeats the essentials of Fig J with clockwise torques. Two teeth are incontact at the point C on the line of action, which is tangential to the two base circles and inclined atthe pressure angle, α. ( Strictly, parameters corresponding to extended centres should be consideredhere, but this sophistication is not necessary for the present introduction to gear reliability. )

On the right of Fig L, separate incomplete free bod-ies of the gears are shown to highlight the equal-and-opposite contact forces, F, acting on thegears at the contact point. Presuming negligi-ble friction, these forces are normal to theinvolutes ie. along the line of action. Usingthe principle of transmissibility, the contactforce on the lower gear has been shownmoved along the line of action to act at thepitch point, P, where it is resolved into a tangential com-ponent Ft and a radial component Fr. The same can be done with theupper gear - in both cases the torque is equilibrated by the contact force, so using ( 5) :-

( 13 ) Ti = F R oi = F ( R i cos α ) = ( F cos α ) R i = Ft R i ; i = 1,2

where Ft is the useful tangential component of the contact force at the pitch point.

reference point

survivalrate (%)

9099

99.9

99.99

107

log

load

, σ

allowablestress, S

log life (cycles)

RELIABILITY FACTOR survival KR %

fewer than one failure in 10 000 99.99 1.50

fewer than one failure in 1 000 99.9 1.25

fewer than one failure in 100 99 1.00

fewer than one failure in 10 90 0.85

T2

1T

α

T1

P

Fig L

P

Fr

T2

C FF

αFt

C

Gear Failure - Reliability

Photoelasticity is used to estimate the stresses in a loaded element.Polarised light is transmitted through the transparent model of a sliceof the element. The model deforms under load, causing the transmittedbeam to interfere with a reference beam. The resulting interference pat-tern consists of a series of bands, each representing an area of constantprototype stress whose value can be predicted by calibration and bycounting bands from unloaded areas. Qualitatively, the closer thesebands are bunched up in the model. the higher are the protoypicalstresses.The photoelastic results of contacting gear teeth shown herehave important implications for tooth safety :

- high stresses occur where the teeth contact one another- bending stresses occur at the toot of the lower cantilevered tooth whose fillet radius is large- much higher bending stresses occur at the root of the upper tooth whose fillet radius is small :

this is yet another manifestation of stress concentration in way of geometric singularities.

These stresses are alternating since a particular tooth is loaded only briefly during one rotation of agear. Failure is therefore very much a case of fatigue, though a one-off static overload obviously maycause failure if sufficiently large.

The following treatment of gear reliability is a gross simplification of the AGMA ( American GearManufacturers Association ) code of practice, AGMA 2001. The treatment aims to demonstrate oneapproach to fatigue design, rather than to transform the reader into an expert on gears.

Apart from one-off overloads, there are three common modes of tooth failure :- bending fatigue leading to root cracking,- surface contact fatigue leading to flank pitting, and- lubrication breakdown leading to scuffing.

We shall not examine the last of these - it involves the relativevelocity and pressure between the teeth together with thelubricant viscosity which is affected by the complex interac-tion between the heat generated due to the gears' inefficiency,the thermal inertia and dissipative properties of the completegearbox, and the temperature of the surrounds.

Since fatigue is prevalent in the other two failure modes, the AGMA employs a reliability approachrather than a safety factor approach in assessing a tooth's tolerance of damage - that is its bendingstrength and pitting resistance. In a reliability analysis, knowledge of a component's load-life (S-N)relationship enables a load to be considered from the point of view of its effect on component liferather than whether it leads to total failure or to total non-failure. A smaller load increases life, alarger load reduces life; whether the safety factor is greater or less than one is irrelevant - indeed thewhole concept of safety factor is inappropriate in the context of reliability.

In the succeeding sections therefore, we first determine the effective damaging tooth force F* due tothe power transmitted and the shock/vibration intrinsic to the gears' manufacture and operationalenvironment. The stress, σ, due to this force is then deduced from bending theory, or from Hertziancontact theory in the case of pitting, and the resulting life follows from the load-life curve appropri-ate to bending or to contact fatigue as the case may be.

It will be recalled that the load-life diagram for a particular material and given type of loading isgenerally of the form illustrated, with curves corresponding to different survival rates being approx-imately parallel to one another - the 99% survival curve eg. implies that one sample in a hundred

flank pitting -surface fatigue

root cracking -bending fatigue

Spur Gears 9

Kv does not account for vibration induced byrunning near shaft critical speeds or by otherresonances - these should be avoided or separ-ately catered for. Note that the AGMA definesKv as the reciprocal of the Kv used here.

- Km is a load distribution factor which reflectsthe non-uniformity of tooth loading over theface width of the teeth, arising from gear andmounting inaccuracy, elastic deformation ofshafting and the like.

For a face width f (mm) the AGMA proposes theempiricisms :

Km = 1 + Cpf + Cma where

Cpf is a pinion proportion factor which reflects misalignment due to load induced elasticdeformations of the pinion; it may be evaluated (with dimensions in mm) from :

Cpf = 0.1 max( 0.5, f/D1 ) + max ( 0, f - 25 )/2000 - 0.025 graphed below.

Cma is a mesh alignment factor which accounts for misalignment due to causes other thanelastic deformation such as inaccurate location of shaft bearings. It is approximated :

Cma = A + f/B - (f/C)2 in which the constants A, B & C for variousclasses of gears - open, commercial and precision enclosed - are shown below.

The face width should be reasonably proportioned to other gear dimensions. If a tooth is too wide itmay bend excessively across its width, if it is too narrow then an uneconomically large diametermust be provided to compensate for lack of width. Proportions may be expressed as :-

( 16 ) f = β m where, usually, 9 ≤ β ≤ 15 for economic gears - these limits should notbe regarded as inviolable, but costs should be expected to escalate ifthey are exceeded.

0.247 1500 2900

0.0675 2000 2600

0.127 1600 2400

A B C (mm) (mm)

face width, f (mm)0 100 200 300 400

0.5

0.4

0.3

0.2

0.1

0

Cm

a

precisioncommercial

open

Cpf

face width, f (mm)0 100 200 300 400

0.5

0.4

0.3

0.2

0.1

0

fD_

1

≤0.51.0

1.52.0

0 10 20 30 40 50pitch line speed, v (m/s)

2.0

1.8

1.6

1.4

1.2

1.0

dyna

mic

fact

or,

Kv

≤ 5

6

7

8

9

10

11

≥ 12

accuracylevel, Q

The radial component, Fr, is useless and tends only to separate the gears (or pitch cylinders) - notethat ( 13) tallies with ( 1) which neglected Fr.

The pressure angle should more correctly be the extended value, and correspondingly ( 13) shouldreally be cast in terms of the actual pitch radius. We neglect such niceties in the present generaldevelopment.

The steady power transferred, P, is :-

( 14 ) P = ωi Ti ; i = 1, 2 the power transfer in rotational terms equals

= ωi Ft R i = Ft v the power transfer in translational terms, from ( 1)

Athough the positive drive ensures that there is no speed loss, there will be torque and power lossesdue to sliding. Spur gear efficiencies exceeding 99% have been reported, however a value of around96% is more appropriate for run-of-the-mill design - this allows for bearing and aerodynamic lossesin addition to tooth friction.

Gear analysis or design usually starts with a known time-averaged (ie. steady) power transfer -these last equations enable the uniform transmitted load Ft to be found. The corresponding failure-producing (ie. damaging) load F* will be greater than Ft because of shock, vibration caused by less-than-perfect tooth profiles and mounting rigidity, and so on. We write :

( 15 ) F* = Ka Kv Km Ft

in which the various empirical K-factors, all ≥1, each reflects the extra damage caused bya particular, separately identifiable practical non-uniformity, as follows :-

- Ka is an application factor to allow for the non-uniformity of input and/or output torqueinherent in the machinery connected to the gears. Typical values are :-

- Kv is a dynamic factor which accounts for internally generated tooth loads induced by non-conjugate action of the teeth, by gear mesh stiffness variation, by gear imbalance and the like.

The AGMA code defines a transmission accuracy level number, Q, which reflects these latter(high Q implies low excitation, a low Q refers to low accuracy and high vibration) and suggeststhe Kv factors plotted below, expressing them empirically as follows (v in m/s) :-

Q ≤ 5 Kv = 1 + √v/3.6 gears of this class are limited to speeds under 13m/s.

6 ≤ Q ≤ 11 Kv = ( 1 + √v / ( 7.6 - 4B ) )B where B = 0.25*( 12 - Q )2/3

Q ≥ 12 Kv = 1 - this corresponds to ultra precision gearing, or to situations wheredynamic loads have been accurately forecast and separately allowed for.

The selection of a suitably realistic accuracy level requires considerable experience.

generator, belt conveyor, light elevator,electrichoist, machine tool feed drive, ventilator, turbo-blower, turbocompressor, mixer (constant density)

press, shear, rolling mill drive, heavy centrifuge,heavy feed pump, pug mill, power shovel, rotarydrilling apparatus, briquette press

machine tool main drive, heavy elevator, craneturning gears, mine ventilator, mixer (variable den-sity), multicylinder piston pump, feed pump

uniform

heavyshocks

mediumshocks

uniformelectric motor,steam turbine,

gas turbine

1.00

1.25

1.75

light shocksmulti-cylinder

combustionengine

1.25

1.5

2.00or higher

medium shockssingle cylinder

combustionengine

1.5

1.75

2.25or higher

Driven Machinery

DrivingMachinery

APPLICATION FACTOR K aFOR REDUCTION GEARS

(x1.1 if speed-up)

Spur Gears 10

The first bracketed term of ( 19) is common to both gears, noting that N1 z1 = N2 z2. In the analysissituation, the two bending life factors KL and hence the lives of the two gears may be evaluated indi-vidually from the equations ( 19). In design, a trial module or face width is often selected whichmust satisfy the inequalities ( 19) for both gears, whose minimum life would be specified - theweaker gear, that with the lesser J.Sb.KL product, dictates the minimum module acceptable.

As an indication of the information available in AGMA2001, the maximum allowable bending stresses for twogrades of through-hardened carbon steel gears areshown on the left as functions of Brinell hardness - thedifferences between the grades are metallurgical asexplained in AGMA 2001, which gives also allowablestresses for other materials. These stresses are the result

of extensive gear testing; attempts todeduce them from other material proper-ties such as the reversed bending endu-rance limit are ill advised.

AGMA proposals for the bending life fac-tor, KL, of steel gears (except nitrided) areshown also - however values are influencedby factors other than the Brinell Hardness.

Pitting Resistance

Flank pitting is caused by alternating normal pressure on the contactsurfaces of the teeth. It is found to occur most frequently at the pitchcircle - where relative sliding of the teeth is zero and the hydrody-namic lubricant film tends to break down - so attention is focussed onthis location. From previous work on Hertzian contact stresses, themaximum normal pressure, p, between contacting cylinders of equiv-alent diameters de1 and de2 and length L, and on which the normalforce is F, is :-

( 20 ) p2 = ( 2 E* F / π L ) ( 1/de1 +1/de2 )

where 1/E* = ( 1 - ν12 )/E1 + ( 1 - ν2

2 )/E2

ie. E* = E / 2 ( 1 - ν2 ) if the materials are identical, and

= 113 GPa if the materials are both steel.

This simplified model is adapted to involute teeth contacting at the pitch point as follows :-- The radii of the equivalent cylinders are the radii of curvature of the contacting teeth at the

pitch point - the distances T1P and T2P of Fig J (in which the contact point C has moved to thepitch point). So de = 2 TP = 2 R sin α = D sin α = m z sin α

- The model cylinders' common contact length is the gears' face width, ie. L = f = βm using ( 16)- The maximum surface stress, σ, in the gear materials at the contact is proportional to the maxi-

mum pressure on the model cylinders’ surface, ie. p ∝ σ.

S = B ( 1.2 – B/1150 ) + 43b

S = B ( 1.15 – B/955 ) – 2b

Grade 2 maximum

Grade 1 maximum

200 300 400Brinell hardness, B

400

300

200

Sb

(MP

a)al

low

able

ben

ding

str

ess

101010 810 610 410 2

number of load cycles

3.4

2.7

2.4

1.6

bend

ing

life

fact

or,

KL

4.0

3.0

2.5

2.0

1.5

1.25

1.0

0.8

1.04

3 1

0*

6

carburised &case hardened

400 BHN250 BHN

160 BHN

throughhardened

BENDINGLIFE

FACTOR

T1

T2

C

d12 e1

d12 e2

Bending Strength

As a first approximation the tooth is modelled as a cantilever of rectangularcross-section and length L and subjected to the damaging load, F*. The maxi-mum bending stress at the critical failure region is :-

σ = M y / I = ( F* L ) ( t /2 ) / ( f t 3 /12 ) and

∝ F* /m f since, for geometrically similar teeth, both t and L are proportional to

the 'size', ie. to the module, m. So :

( 17 ) σ = F* /m f J

in which the AGMA proportionality constant, J, is known as the bendingstrength geometry factor and is evaluated by photo-elastic or finite ele-ment methods, and correlates the stress in the simple cantilever modelwith the actual maximum stress in a gear tooth, which is influencedby stress concentration, by variation in load direction and momentarm length, and by load sharing between teeth as the contact pointmoves along the tooth as suggested by the sketch - though the twostatically indeterminate tooth pairs prevent the sudden stepssketched.

The factor's value depends upon the number of teeth and profile shift of both gears and is obtainablefrom AGMA 2001, or calculated by the Macintosh program 'Steel Spur Gears' which is based on AS2938. Its variation is graphed below when both profile shifts are zero and when they are both at themiddle of the practical range (Fig G).

The graphs demonstrate that without profile shifts, J-factors decrease and stresses ( 17) increasemarkedly with decreasing tooth numbers - however with small numbers of teeth, significantincreases of J-factor and reductions of stress result from incorporating suitable profile shifts. Whenboth profile shifts are mid-range the J-factor may be approximated by :

( 18 ) J ≈ { 0.535 + ( - 3.030 + 14.5 / zb ) /zb } + { - 0.904 + ( 28.43 - 147.0 / zb ) /zb } /za

in which za is the number of teeth in the gear whose J-factor is required and zb is thenumber of teeth in the mating gear.

Combining ( 12) ( 15) ( 16) and ( 17) leads to the final design equation for tooth bending strengthwhich is applicable to each of two mating gears :

( 19 ) ( π m3 β N1 z1/ Ka Kv Km KR P ) . ( J Sb KL )i ≥ 1 ; i = 1,2

where Sb is the allowable bending stress for the gear tooth.

10 20 30 40 50 60 70 80 90 100

0.6

0.5

0.4

0.3

0.2

number of teeth in gear whose factor is required, za

number of teeth inmating gear, zb

bend

ing

stre

ngth

geo

met

ry fa

ctor

, J

5020

10

1000100

BENDING STRENGTHGEOMETRY FACTOR

WITH ZERO PROFILE SHIFTS

highest point of

single tooth

contact

one tooth in contact

two teeth in contact

failureregion

L

tf

F*

number of teeth in mating gear, zb

BENDING STRENGTHGEOMETRY FACTOR

WITH MID-RANGE PROFILE SHIFTS

20 30 40 50 60 70 80 90 10010

0.6

0.5

0.4

0.3

0.2

number of teeth in gear whose factor is required, za

bend

ing

stre

ngth

geo

met

ry fa

ctor

, J

50

2010

100

14

30

1000

Spur Gears 11

Both fatigue design equations ( 19), ( 21) indicate that the capacity (the maximum power transferra-ble) is proportional to the cube of the module - ie. a slight increase in module enables a large increasein capacity. The examples below illustrate application of the equations to analysis and to design.

The analysis example reveals the general truism - pitting of steel gears is much more damaging thanbending, ie. the life of the gearset is more often than not limited by contact fatigue. A gear's capacityis proportional to the square of its allowable surface contact stress ( 21) so that surface hardening ofthe tooth flanks (by case- or flame-hardening, by nitriding or carburising &c) leads to a better bal-anced solution in which the contact-limited fatigue life may be increased to approach the bending-limited fatigue life. AGMA 2001 cites properties for surface-hardened steel gears - allowable contactstresses exceeding 2 GPa are possible.

The design example adopts flame hardened teeth on the pinion, however the life of the chosen can-didate is also dictated by contact. Clearly different materials and surface treatments are design vari-ables which, together with different tooth numbers, characterise solution candidates in a real prob-lem - and the optimum choice is mainly a question of economics. To put lives in perspective, a life of1000 khr represents more than a century of continuous running - all day, every day!

The graph shows the results of repeating the design example in an effort to obtain the most compactbox. Trial pinion tooth numbers increasing from 9 to 25 lead to modules which decrease montoni-cally from 20 to 8 mm, pinion diameters (D =m*z) which generally increase and face-widths which generally decrease. The sur-prising outcome of this study is that pinion(and hence gearbox) volume is essentiallyconstant over the range studied - that issmall pinion tooth numbers are not neces-sary for compactness. However the analysisrecognises only two fatigue failure modes; itignores eg. the possibility of tooth yield dueto a single large overload - a possibilitywhich would be reduced by the larger teeth associated with small tooth numbers. AGMA 2001 ismuch more comprehensive than the fatigue overview given above; it caters for overloads and othereventualities, and introduces a number of additional factors which we need not examine here.

It should be stressed that gear design is an art. Gear geometry and metallurgy in particular can besubtle. As AGMA 2001 puts it "This Standard is intended for use by the experienced gear designer capableof selecting reasonable values for the factors. It is not intended for use by the engineering public at large."However there is no reason why intelligent application of the foregoing should not be used for non-critical gears.

Periodic Duty

If a component is subjected to a repeating stress of constant magnitude,σi , then it will fail after n*

ii application cycles - that is its life is n*i cycles.

At some stage prior to failure it will have been subjected to ni cycles, sothe cumulative damage at that stage ( the fraction of its life used up ) isni/n*

i . These lives are shown here on a stress-life diagram which, in thecase of steel gears and when normalised by the allowable contact stressSc , is just the contact life factor CL examined above. It follows that for steel gears :

n i

σin*ilo

g st

ress

, σ λ

1Sc

107

fail

log cycles, n

pinion volume

10 15 20 2511 16 219 12 17 2213 18 2314 19 24

MIXER GEARBOX DESIGN EXAMPLEWITH MID-RANGE PROFILE SHIFTS

module, mm

pinion teeth

pinion diameter, mm180

160

208

168

204

240

200

2016

12 10 8

180

- The contact force on the cylinders, F, corresponds to the normal force between the teeth, Fig L.Since it is the effective damaging tangential force, F*, which is relevant above, the correspond-ing damaging normal force is F = F*sec α from the triangle of forces, Fig L.

Making these adjustments to ( 20), invoking ( 12) ( 15) ( 16) and introducing a proportionality con-stant - to correlate the simple model with the actual tooth - leads to the final design equation for pit-ting resistance :

( 21 ) ( π2 m3 β N1 z12 I / Ka Kv Km KR

2 E* P ) . ( Sc CL )i2 ≥ 1 ; i = 1,2

in which Sc is the allowable contact stress for the gear material. Values are cited inAGMA 2001 - maximum values for two grades of through-hardened steels are shown here.

CL is a gear's contact life factor - the normalised load-life curve under repeated contact loading as explainedabove. The AGMA relationship for through-hardenedsteels is plotted below. The implications of the slope ofthe load-life plot must be appreciated - a load increaseof only 2% leads to a life reduction of 30%.

The contact design equation ( 21) is appliedto the two gears individually in the samemanner as is the bending design equation (19) - again, the first bracketed term is com-mon to both gears. In applying ( 21) todesign therefore, the gearset will be limitedby the weaker gear - that which has thelesser ScCL product.

The remaining parameter in ( 21) as yet unexplained is the pitting resistance geometry factor, I,which correlates the stresses between the simple Hertzian model and actuality. Like J its valuewhich depends upon tooth numbers and profile shifts can be obtained from AGMA 2001 or com-puted from AS 2938. The value is the same for both gears and tends to sin2α/4(1+z1/z2) as the num-ber of pinion teeth z1 becomes large. It is plotted below corresponding to the profile shifts of bothgears being either zero or in the middle of the practical range (Fig G). When profile shifts are mid-range the I-factor is essentially dependent only upon the gear ratio ρ, and may be approximated by :

( 22 ) I ≈ ( 0.0404 + 0.1127 ρ ) / ( 1 + 0.70 ρ ) in which ρ = z2/z1 ≥ 1.

20 30 40 50 60 70 80 90 10010

0.15

0.125

0.10

0.075

0.05

pitti

ng r

esis

tanc

e ge

omet

ry fa

ctor

, I

PITTING RESISTANCE GEOMETRY FACTOR WITH ZERO PROFILE SHIFTS

gear ratio, z / z2 1

10

5

3

2

1.4

1

number of teeth in pinion, z1

20 30 40 50 60 70 80 90 10010

0.15

0.125

0.10

0.075

0.05

number of teeth in pinion, z1

pitti

ng r

esis

tanc

e ge

omet

ry fa

ctor

, I

PITTING RESISTANCEGEOMETRY FACTOR

WITH MID-RANGE PROFILE SHIFTS

10

5

3

2

1.41

gear ratio, z / z2 1

S = 2.51 B + 186c

S = 2.25 B + 179c

Grade 2 maximum

Grade 1 maximum

Sc

(MP

a)

allo

wab

le c

onta

ct s

tres

s

1000

800

600200 300 400

Brinell hardness, B

1200

101010 810 610 410 2

number of load cycles

cont

act l

ife fa

ctor

, C

L2.0

1.6

1.4

1.2

1.0

0.8

0.6

1.47

17.93

1

CONTACTLIFE

FACTOR

Spur Gears 12

ble more certainty in this choice.

From the lessons learned in the previous example, both gears will be made from grade 2 steel - the pinion will be

flame hardened to 52 HRC while the wheel will be through hardened to 360 BHN. The pinion's allowable stresses

must be obtained directly from AGMA 2001 - they are Sc = 1320 MPa, Sb = 380 MPa.

As contact rather than bending could be critical, the gears will be designed for contact, then checked in bending.

tooth number 18 65

allowable contact stress, Sc (grade 2) MPa 1320 1090

life (16 khr * rotational speed) cycles 1.9 E8 5.3 E7

life factor corresponding to this life, from graph, CL 0.849 0.912

product ScCL to deduce the weaker gear MPa 1120 994

The wheel is weaker than the pinion in contact fatigue - it might pay to use a somewhat less sophisticated pinion

material/heat treatment, but any decision will be delayed until all results (eg. bending too) are to hand.

The contact design equation (21) will now be solved to obtain a rough idea of the module - and to this end, middle-

of-the-road values will be assumed for those parameters which depend upon module for evaluation. Thus take β =12, Kv = 1.1 since the pinion rotates fairly slowly at 200 rev/min and so the pitch speed will be low; Km = 1.2 since

the mesh alignment factor cannot be less than 0.13 for commercial enclosed gears. It is known that KR = 1, corre-

sponding to a reliability of 99% and the contact geometry factor I = 0.127 from (22) corresponding to a ratio of 65/

18. So applying (21) :-

m3 = Ka Kv Km KR2 E* P / π2 β N1 z1

2 I ( Sc CL )2

1.25*1.1*1.2*12*113 E9*1 E5 / π2

*12*(200/60)*182*0.127*(994 E6)2 whence m = 10.5 mm

N/m2 Nm/s s m4/N2 A standard module of either 10 or 12 mm is indicated; a trial candidate m = 12 mm will be used - this will enable

evaluation of some of the factors more closely, thus :

- Kv D1 = m.z1 = 12*18 = 216 mm; v = πD1N1 = π*0.216*(200/60) = 2.26 m/s (ie. very low as predicted)

Assuming a transmission accuracy level of Q = 8, then, from the graph, Kv = 1.26

- Km Assuming an average value of β = 9 (on the low side to offset the choice of module on the high side ofthe range around 10.5mm). Then the facewidth f = 12*9 = 108, say 110 mm which leads to Cpf = 0.068;

Cma = 0.194 and so Km = 1 + 0.068 + 0.194 = 1.26

So, solving (21) for the facewidth ratio, β, and for the weaker gear :

β = Ka Kv Km KR2 E* P / π2 m3 N1 z1

2 I ( Sc CL )2

1.25*1.26*1.26*1.02*113 E9*1 E5 / π2

*0.0123*(200/60)*182

*0.127*(994 E6)2 whence β = 9.7

This is within the usual economic range 9 ≤ β ≤ 15 so looks like a practical candidate. The face width is f = 9.7*12 =116 mm. Take f = 120 mm - sufficiently close to 110mm to obviate need to update Km for refined calculations.

The chosen parameters so far (m = 12mm, f = 120 mm) will now be checked via the bending equation ( 19), from

which the bending lives of the two gears will be deduced, thus :

( J Sb KL )i = Ka Kv Km KR P / π m2 f N1 z1

1.25*1.26*1.26*1.0*1 E5 / π*0.0122*0.120*(200/60)*18 ie. the product ( J Sb KL ) is 60.9 MPa

Nm/s 1/m2 1/m sproduct J Sb KL MPa 60.9 60.9

bending strength - Sb, pinion from AGMA 2001; wheel corr. to 360 BHN MPa 380 360

bending strength geometry factor, J, from (18) 0.464 0.415

KL = ( J Sb KL ) / J Sb 0.345 0.408

It may be seen from the graph that the lives corresponding to these KL factors are very much greater than those

specified, ie. bending again is less critical than contact.

Summarising, the candidate considered utilises a module of 12 mm, a facewidth of 120 mm, gears of 18 and 65

teeth with mid-range profile shifts and materials as specified above. In practice, further design candidates with

various steels and heat treatments etc. would be carried out to aid an economic assessment. The program 'Steel

Spur Gears' is based on the foregoing theory and may be used to assist in the design task - the dialogue for these

examples appear in Appendix C below.

EXAMPLE - ANALYSIS A crane traversing gearset is driven at 425 rev/min by an electric motor through pri-

mary gearing. The open gearset transmits 75 kW. The gears are manufactured to a transmission accuracy level of 5,

with 8 mm module and 110 mm facewidth from through-hardened grade 1 steel, the 20 tooth pinion to 250 BHN

and the 53 tooth wheel to 180 BHN. What life can be expected of the gearset at 99.9% reliability ?

Evaluating the various common factors first :

- Ka Referring to the table of Application Factors, the input is uniform and the output rather less damaging

than medium shock, so assume a value of Ka = 1.15

- Kv D1 = m.z1 = 8*20 = 160 mm; the pitch line speed is v = πDN = π 0.160*425/60 = 3.56 m/s. Since Q =5

the corresponding dynamic factor is Kv = 1 + √v /3.6 = 1 + √3.56 /3.6 = 1.52 - tallies with the Kv graph.

- Km f/D1 = 110/160 = 0.688 which gives a pinion proportion factor Cpf = 0.1*max( 0.5,0.688) +max( 0, 110 -

25)/2000 - 0.025 = 0.086. The mesh alignment factor for open gearing of this facewidth is Cma = 0.247

+110/1500 - (110/2900)2 = 0.319. So the load distribution factor is Km = 1 +0.086 +0.319= 1.41

- KR corresponding to a reliability of 99.9% is 1.25

- I assuming both gears’ profile shifts are mid-range, then for a ratio of 53/20 = 2.65, I = 0.119 from (22)

Apply the bending design equation (19) to find life factors and hence pinion wheel

lives of the two gears. Calculations are shown for pinion only.

tooth number 20 53

bending strength geometry factor, J, with mid-range shifts, from (18) 0.462 0.423

Brinell hardness number 250 180

allowable bending stress, Sb, from diagram (grade 1) MPa 220 171

KL = Ka Kv Km KR P / π m2 f N1 z1 J Sb

1.15*1.52*1.41*1.25*75 E3 / π 0.0082*0.11*(425/60)*20*0.462*220E6 0.726 1.020

Nm /s 1/m2 1/m s m2/N always check units carefully

corresponding life from KL plot Mc 205 000 5.4

dividing by rotational speed for life in units of time khr 8020 !! 0.56

Apply contact design equation ( 21) in a similar manner to obtain life factors, CL, and hence the lives of the two

gears in units of time. The gearset life will then be the least of the four computed lives.

allowable contact stress, Sc, from diagram (grade 1) MPa 740 585

CL = √( Ka Kv Km E* P / m2 f N1 I ) KR /π z1 Sc

√(1.15*1.52*1.41*113 E9*75 E3 / 0.0082*0.11* (425/60)*0.119)*1.25/π 20*740 E6 1.60 2.02

√( N/m2 Nm/s 1/m2 1/m s ) m2/N

corresponding life from CL plot Mc ? ?

The maximum possible contact life factor is 1.47 (refer to the CL graph); since the calculated values of CL for both

pinion and wheel exceed this, there is no way that the desired reliabilty can be achieved with a reasonable life.

Note that CL∝KR above, so a substantially reduced probability of survival would result in gears with non-zerolives. But the probability of the gears surviving 75 kW with an economic life is low.

EXAMPLE - DESIGN Select suitable gears for the final stage of a commercial quality 100 kW gearbox whose

input and output shafts rotate at 200 and at about 55 rev/min respectively. The box will be interposed between an

electric motor and a variable density mixer, and should have a life of 16 khr with 99% reliability.

Design candidates would be characterised by the number of teeth in the pinion - this should be small for overall

compactness of the box. We shall exemplify the fatigue aspects of design using (19) and (21) for the candidate z1 =

18 ( z2 = 65) and mid-range profile shifts.

An application factor of 1.25 seems appropriate, though in practice further details of the mixer's charge wouldena-

Spur Gears 13

APPENDIX A : Continued Fractions and Padé Approximations

Truncating infinite power series in order to use the resulting polynomials as approximating functions is a well known

technique. While some infinite series lend themselves to this, others do not - needing many terms and significant com-

puting time to reach the desired accuracy. Some may indeed diverge. It is useful to realise that infinite series are not the

only bases for numerical approximations, and it is for this reason that we now introduce the topic of continued fractions

which often provide a much more efficient approximation basis than series. Continued fractions and power series are

analogous to one another ( compare ( iii), ( iv) below ). Padé approximations are truncations of continued fractions, in

the same way as polynomials are truncations of infinite power series.

The concept of a ‘convergent’ arises from the study of continued fractions. Convergents find many applications - notably

in finite elements - however their immediate relevance is as the basis of a procedure for selecting gear tooth numbers

from a limited range to approximate irrational speed ratios. This procedure is carried out mechanistically however its

grounding in convergents and continued fractions is reason enough to introduce these latter at this juncture.

We show, for interest, how a continued fraction may be derived from a power series, and the advantages of the resulting

Padé expression over the corresponding polynomial as an approximation for a particular function. The reader is referred

to Wall HS, Continued Fractions, van Nostrand 1948, for further details.

A typical example of the advantages of Padé approximations is the following.

The real-time dynamic simulation of the air-operated brakes of an ore train consisting of hundreds of cars requires thou-

sands of iterations over the whole train. During each iteration the compressible flow function { p-2/γ - p- (1+1/γ )} has to

be evaluated a number of times over each car to determine the flows through the various pipes and brake cylinder

restrictions in the car - p being the variable pressure ratio across a restriction ( 1 ≤ p ≤ pcrit ), and γ the constant ratio of

specific heats for air. The flow function is plotted here.

The two exponentiations in the many functional evaluations

took up such an inordinate amount of computing time that

real-time simulation was jeopardised. To circumvent this the

flow function was approximated by polynomials - three

were required - these weren’t particularly accurate but

more significantly they gave rise to further computing con-

vergence problems due to their piecewise nature.

A much superior approximation applicable over the whole

p-range is of the form ( p -1 )( a +b.p )/( 1 +c.p ) in which a, b & c are carefully chosen constants. This is a Padé approxi-

mation and is indistinguishable from the the correct function at the scale of the plot - the accuracy of the approximation

is for the most part within 0.01%, which is much superior to the polynomials.

In the context of gears, the foregoing expressions ( 18) ( 22) for tooth geometry factors are further examples where ratios

of polynomials may be set up as Padé approximations to functions - in this case derived from finite element and photoe-

lastic analyses.

A function of the single variable x may be written in the various forms :-

( i ) f ( x ) = a1 x + a2 x2 + a3 x3 + a4 x4 + . . . . ai constant

( ii) = x ( a1 + x ( a2 + x ( a3 + x ( a4 + . . . . ))))

( iii) = c1 x * ( 1 + c2 x * ( 1 + c3 x * ( 1 + c4 x * ( 1 + . . . . )))) ci constant

( iv) = b1 x /( 1 + b2 x /( 1 + b3 x /( 1 + b4 x /( 1 + . . . . )))) bi constant

( v ) = b1 x 1 + b2 x

1 + b3 x 1 + b4 x

1 + . . . .

1 2

0.3

0.2

0.1

0

0.01

-0.01

0

erro

r %

pressure ratio, p

errorflowfunction

flow

func

tion

σ1 σ2

σ3

σ4

n1 n2 n3 n4

( 23) n*i = 10 ( Sc/σci

)λ =17.93 Mc ; σci ≤ 1.47 Sc - provided failure is due to pitting.

The concept may be extended to a component under different stresses according to the spectrumshown below which comprises a block of n1 cycles at stress magnitude σ1, a block of n2 cycles at σ2,and so on. The life of the component if σ1 were applied alone is n*

1, itslife under σ2 alone is n*

2 etc. It follows that the fraction of life used up

in the first block is n1/n*1, the fraction used up in the second is n2/n*

2etc. Assuming that the stress blocks may be treated independently,failure of the component occurs when the sum of the life-fractionsused up by the various blocks (the cumulative damage ) reaches 100%.That is, at failure :

∑i=1

ni /n*i = 1 Miner's Rule (strictly the Palmgren-Miner Rule)

This convenient but simplistic view implies that the damage caused by a stress block is constant irre-spective of whether block occurred at the start of loading or just before failure. This is not the case inpractice so Miner's Rule can be rather inexact, but is preferred in many non-critical applications tomore complex/expensive techniques for handling variable load fatigue such as the 'RainflowMethod' popular in aeronautical circles.

Many components other than gears follow approximate logarithmically linear stress-life relation-

ships : σλi.n*

i = constant. Inserting this into Miner's Rule leads to :

∑i=1

σλi . ni = constant, and = σλ

e ∑i=1ni

in which σe is the equivalent stress - the single stress level which causes the samedamage as the actual stress spectrum over the period of interest. If,WOWOWOWOWOWOWOOWfurther, stress is proportional to load as is very common, then load may be substituted for stress in

the foregoing, leading to the ubiquitous relationship : loadλ * life = constant.

This proportionality usually does not apply to gears however, as failure is most often caused by pit-ting rather than by bending - and contact stresses are proportional to the square root of the tangen-

tial force and hence of the load (power or torque). The life of a gear n*i due to the power in the i’th

block of a periodic load spectrum can be computed using ( 21) as if the power were constant and notpart of a spectrum.

All potential failure modes must be considered when applying Miner's Rule.

EXAMPLE Estimate the life at 99% reliability of a steel σci GPa 1.63 1.54 1.45 1.37

gear under the given stress spectrum. Failure is through Ni rev/min 65 85 125 14

pitting - the allowable contact stress Sc being 1.55GPa. ti hr 1 2 3 4

If the gear’s life is L (khr) then the total number of cycles

in each stress block over that life is : ni = ( ti /∑ti ) L * Ni ni Mc 0.39L 1.02L 2.25L 0.34L

The life n*i corresponds to the stress σci in each stress block

acting alone, and follows from ( 23) : n*i Mc 4.1 11.2 33.1 91.5

Applying Miner's Rule : L ( 0.39/4.1 +1.02/11.2 +2.25/33.1 +0.34/91.5) = 1 whence L = 3.9 khr.

Spur Gears 14

C2 = A2 /B2 = 2 x / ( 2 + x ) -2 0.6667 1 1.4286

C3 = A3 /B3 = x ( 6 + x ) / ( 6 + 4 x ) -2.5 0.7 1.1429 2.1154

C4 = A4 /B4 = x ( 6 + 3 x )/( 6 + 6 x + x2 ) -3 0.6923 1.0909 1.7213

C5 = A5 /B5 = x ( 30 + 21 x + x2 )/( 30 + 36 x + 9 x2 ) -3.3333 0.6933 1.1014 1.8391

C6 = A6 /B6 = x ( 60 + 60 x + 11 x2 )/( 60 + 90 x + 36 x2 + 3 x3 ) -3.6667 0.6931 1.0980 1.7787

etc.

This exemplifies the superiority of an approximation based on a continued fraction over one based on a power series -

for this particular function at least. It should be noted how successive “convergents” are alternately greater and less than

the correct value ( except for the infinite case ), and converge upon it - hence their name.

The ability of continued fractions to identify simple ratio approximations, may be put to use for deriving gear trains for

otherwise awkward ratios. The technique used is simpler than the above and is best shown by example. Suppose we

require a ratio of 0.3711, correct to the fourth significant figure. We first express the fraction rationally - 3711 / 10000 -

then factorise both numerator and denominator in an attempt to implement a compound train - 3 * 1237 / ( 2 * 5)4 -

which gets us nowhere as 1237 is prime and far too large for a practical gear; we will have to use a simpler approxima-

tion. An extended division is therefore carried out below; initially dividing the numerator into the denominator and

thereafter dividing the remainder into the previous divisor. The quotients thus found ( ie. the bn ) form a continued frac-

tion ( of form different to ( v)), whose convergents yield approximations to the original fraction.

3711 ) 10000 ( 2 ≡ b1 7422

2578 ) 3711 ( 1 ≡ b2 1 C4 = 1 = 10 / 272578 2 + 1 2 + 1

1133 )2578 ( 2 ≡ b3 1 + 1 1 + 1 2266 2 + 1 2 + 1

312 )1133 ( 3 ≡ b4 3 + 1 3 936 1 + 1

197 )312 ( 1 ≡ b5 1 + 1 197 1 + 1

115 ) 197 ( 1 ≡ b6 2 + 1 115 2 + 1

82 ) 115 ( 1 ≡ b7 etc 82

33 ) 82 ( 2 ≡ b866

16 ) 33 ( 2 ≡ b932 etc

The convergents may be found as C4 above - or by using the scheme below, in which C0 is a seed, C1 is the reciprocal of

b1, and thereafter the numerator and denominator follow from the same recurrence relation : An = bnAn-1 +An-2; Bn =

bnBn-1 +Bn-2. The seventh convergent would evidently give the desired results -

Conv'gt order, n 0 1 2 3 4 5 6 7 8 9

Quotient, bn - 2 1 2 3 1 1 1 2 2

Numerator, An 0 1 1 3 10 13 23 36 95 226Denominator, Bn 1 b1=2 3 8 27 35 62 97 256 609

Cn = An /Bn - 0.5 0.333 0.375 0.3704 0.37143 0.37097 0.37113 0.37109 0.37110

Proportional error - -3.5e-1 1.0e-1 -1.1e-2 2.0e-3 -8.9e-4 3.6e-4 -9.2e-5 1.7e-5 -4.4e-7

but it is impractical since 97 is prime and unavailable in our selection of change wheels. However the eighth is appropri-

ate as 95/256 = 5 * 19/28, and a compound train of 19/32 and 20/32 could be used.

Since all convergents are approximations to the same target value, their combinations may be more suitable than the

convergents themselves - eg. ( C8 +C6 )/2 = 59/159 = 0.3711, and so on.

Admittedly, this technique has been largely superseded by cut-and-try methods on a computer.

Form ( i) is the standard infinite power series representation ( neglecting any constant term) which must be truncated

when used for computations - in which case it is often expressed more conveniently as ( ii). Form (iii) is an alternate rep-

resentation in which the c-coefficients may easily be derived from the a-coefficients; this form is included here only to

show the similarity between the power series and the continued fraction (iv). Thus the multiplications which appear in

the power series ( iii) are replaced by divisions in the continued fraction ( iv), which is often seen in expanded form ( v).

If we set b4 = 0 in ( v), then the continued fraction is truncated after the third term - this is referred to as the third conver-

gent :- C3 = b1 x ( 1 + b3 x ) / ( 1 + ( b2 + b3 ) x ) ≡ A3 / B3 and demonstrates the general form of the n'th conver-

gent as a quotient between a numerator, An, and a denominator, Bn, both of which are polynomials. The convergent is

an approximation of the original function - referred to as a Padé approximation.

A tabular procedure, involving only two simple recursive operations, may be used to derive the b-coefficients from the

a-coefficients; that is to derive the continued fraction from the power series, since the latter approximations are widely

available. The technique is exemplified below for a particular function, the natural logarithm, which is first expressed as

an infinite power series of which six terms only ( for example ) are considered :-

f ( x ) = ln ( 1 + x ) = x - x2/2 + x3/3 - x4/4 + x5/5 - x6/6 + etc.

These a-coefficients are entered into the second row of the table below, the first row of which is a seed. Operations on

succeeding rows are as follows.

An odd row is formed by dividing the preceding even row by its leading ( leftmost ) term - this term

becoming the latest b-coefficient.

An even row is formed by subtracting terms in the preceding odd row from terms in the odd row

before that, and moving the terms one column leftwards when entering them into the even row.

1 1 0 0 0 0 0 2 1 -1/2 1/3 -1/4 1/5 -1/6

3 b1 = 1 1 -1/2 1/3 -1/4 1/5 -1/64 1/2 -1/3 1/4 -1/5 1/6

5 b2 = 1/2 1 -2/3 1/2 -2/5 1/3 6 1/6 -1/6 3/20 -2/15

7 b3 = 1/6 1 -1 9/10 -4/5 8 1/3 -2/5 2/5

9 b4 = 1/3 1 -6/5 6/510 1/5 -3/10

11 b5 = 1/5 1 -3/212 3/10

13 b6 = 3/10 1 etc

Recurrence relations for the numerator ( An) and denominator ( Bn) of the n'th convergent are as follows :-

A0 = 0 B0 = 1

A1 = x b1 = x B1 = 1

A2 = A1 + x b2 A0 = x B2 = B1 + x b2 B0 = 1 + x/2

A3 = A2 + x b3 A1 = x + x2/6 B3 = B2 + x b3 B1 = 1 + 2x/3

A4 = A3 + x b4 A2 = x + x2/2 B4 = B3 + x b4 B2 = 1 + x + x2/6

A5 = A4 + x b5 A3 = x + 7x2/10 + x3/30 B5 = B4 + x b5 B3 = 1 + 6x/5 + 3x2/10

A6 = A5 + x b6 A4 = x + x2 + 11 x3/60 B6 = B5 + x b6 B4 = 1 + 3x/2 + 3x2/5 + x3/20

etc etc

The convergents (Cn ) are finally expressed as ratios of polynomials. Approximations calculated therefrom, for some rep-

resentative values of x, are tabulated :-

Trial variable : x -1 1 2 5

Correct value : ln ( 1 + x ) - ∞ 0.6931 1.0986 1.7918

Trunc. series : x - x2/2 + x3/3 - x4/4 + x5/5 - x6/6 -2.1167 0.9500 15.73 3107.1

C1 = A1 /B1 = x -1 1 2 5

Spur Gears 15

( iv ) ψ = φ + α + γ

In order to determine the tooth constant half-angle, γ, the Cartesian coordinates of the contact point, calculated from (iii)

in the symmetric position with ψ = α + γ, are inserted into the equation for the rack flank derived from the geometry of

Figure D as : x = h + ( R + s – y ) tan α. This yields the necessity :-

( v ) γ = ( h + s tan α ) / R + inv α ; inv α ≡ tan α – α . . . . the involute function

Checking for tip-pointing. The maximum generation angle, at the addendum circle, is, from ( iii) :-

( vi ) ψmax = √ ( ( ra / Ro )2 – 1 ) ; ra = R + s + a . . . . the addendum radius

The teeth are pointed if the corresponding θmax <= 0, due in turn to the profile shift exceeding some critical shi which

may be found, from ( iii) and ( v), to be :-

( vii ) shi = ( R ( ψmax – arctan ψmax – inv α ) – h ) cot αnoting that this must be solved by trial, since ψmax is a function of shi here, via ( vi).

On the other hand, the minimum generation angle, ψmin, corresponds to the tangent point T1 being tangent also to the

involute, Fig B-5, provided the tooth is not undercut, that is provided that T1 lies outside the base circle since the invo-

lute is not defined inside this circle. Comparing the ordinates of the fillet centre, C, with respect to the involute and to

the rack :-

yC = Ro cos α + ( Ro ψmin + e ) sin α = R + s – b + e from which

(viii ) ψmin = tan α + ( s + e ( 1 – sin α ) – b ) / Ro sin α

and the corresponding lower limit for the trochoid, φmin, follows from ( iv).

Setting ψmin = 0 in ( viii) gives the critical profile shift, slo , below which undercutting occurs, thus :-

( ix ) slo = b – R sin2α – e ( 1 – sin α )

The Trochoidal Fillet

It may be appreciated from the sketches of tooth generation that the rack radius scours out the trochoidal tooth fillet,

prior to the involute tooth flank being formed by the straight side of the rack. If C is the centre of the fillet circle, Fig B-6,

then the coordinates of the corresponding point on the trochoidal fillet envelope are, from the geometry of Fig B-7 & ( ii):

( x ) x = xC + e dy/ds = λ ( R φ + uC ) cos φ – ( R + λ ( s + vC ) ) sin φ ( uC , vC ) from ( i)

y = yC – e dx/ds= λ ( R φ + uC ) sin φ + ( R + λ ( s + vC ) ) cos φwhere λ = 1 + e / √( ( R φ + uC )2 + ( s + vC )2 )

The roll angle which defines the upper limit of the trochoid ( φmax ) corresponds to the point T2, Figure D, lying on the

dedendum circle, radius rb = R + s – b. Inserting coordinates of T2 into ( ii) leads to :-

( xi ) φmax = – uT2 / R = – ( h + e sec α + ( b – e ) tan α ) / R

Undercutting

If undercutting is indicated by ( ix), then the intersection between involute and trochoid must be found by simultaneous

solution of the intersection coordinates, via ( iii) at ψmin for the involute, and ( x) at φmin for the trochoid. Note that φminand ψmin are not interrelated through ( iv) in this instance.

Calculation Sequence

Given the system parameters ( α, a, b, e, h ); compute coordinates of C from ( i).

Given the gear parameters ( R, s ); compute Ro, slo and shi via ( ix) and ( vii) respectively.

Set up the limits for the various profile segments :-

If s < slo then the tooth is undercut; calculate φmin from ( x) and ψmin from ( iii) simultaneously

else determine ψmin from ( viii) and corresponding φmin from ( iv)

If s >= shi then the tooth is pointed; calculate ψmax from ( iii) corresponding to θ = 0

else determine ψmax from ( vi)

Ascertain φmax from ( xi)

Compute the various profile segments :-

The dedendum circle, radius rb, from the tooth boundary at θ = π / z, to the start of the trochoid; the trochoid

from φmax to φmin; the involute from ψmin to ψmax; and finally, if not pointed, the addendum circle, radius

ra, to θ = 0.

APPENDIX B : Geometry of the Involute Gear ToothThe equations which describe the profile of a rack-generated gear tooth, including both the involute and the fillet tro-

choid, are now derived. These are useful not only for power transmission gearing, but also for hydrostatic power trans-

formation in gear pumps and so on, where fluid sealing and inter-tooth volume are important.

Coordinates of a Point on the Rolling Rack Any point such as Q, Fig B-1, is defined by coordinates ( u, v ) on the rack

whose reference point, O, is offset by the profile shift, s, from the gear rolling circle at P. Thus for the fillet centre of

Figure D :-

( i ) vC = e – b ; uC = h + e sec α – vC tan α

It is required to determine the location of the Q relative to the gear, after a roll angle, φ, is undergone.

In Fig B-2, the coordinates ( X, Y ) of the point with respect to the gear centre are :

X = R φ + u ; R = z / 2 z being the number of teeth on the gear, and

Y = R + s + v which is constant

Transforming these to axes ( x, y ) fixed to the gear, Fig B-3, they become :-

( ii ) x = r sin ( θ – φ ) = X cos φ – Y sin φ = ( R φ + u ) cos φ – ( R + s + v ) sin φ

y = r cos ( θ – φ ) = X sin φ + Y cos φ = ( R φ + u ) sin φ + ( R + s + v ) cos φ

So, the system parameters ( α, a, b, e and h ) having been laid down, and the gear parameters ( R and s ) specified, the

coordinates of any point ( u, v ) may be found from ( ii) for any roll angle, φ.

The Involute Flank

An involute is generated most simply by a point attached to a taught cord which unwinds from a base circle of radius

Ro = R cos α, as shown in Fig B-4.

The polar coordinates of a point on the profile corresponding to the generation angle ψ, are :-

( iii ) r = Ro √ ( 1 + ψ 2 ) ; θ = γ – ψ + arctan ψ

With the rack in the symmetric position ( φ = 0 ), the cord generator must be inclined at α to the horizontal, Fig B-5, since

it is normal to the non-rotated rack at the contact point. It follows that the relation between the generation angle ( ψ ) and

the rack roll angle ( φ ) is :-

Spur Gears 16

SPUR GEARS - PROBLEMS

In the following problems, assume that :- gears with any tooth number up to 120 are procurable ( constraints are more severe in practice )- all gears are of steel, to the 20 deg full depth system unless otherwise indicated- mid-range profile shifts apply, where relevant.The program 'Steel Spur Gears' should be used to assist solution of asterisked problems, and may be used tocheck longhand solution of other fatigue problems.

1 Tooth numbers of certain gears in the epicyclic train are indicated;all gears are of the same module. Gear A rotates at 1000 rev/minclockwise while E rotates anticlockwise at 500 rev/min.Determine the speed and direction of rotation of the ring-gear Dand of the arm shaft F. If the power output through each of D andF is 1 kW, what are the power transfers through A and E?[ 371 rev/min anticlockwise; 40 rev/min clockwise; 8.77 kWinput; 6.77 kW output ]

2 The arm of the epicyclic train is driven clockwise at 1450 rev/min bya 5 kW motor. What torque is necessary to lock the 33 teeth gear?What is the speed of the 31 tooth gear? Note the reduction! [ 16.3 kNm clockwise, 2.92 rev/min clockwise ]

The sun wheels A and D are integral with the input shaft ofthe compound epicyclic gear illustrated, and the annularwheel C is fixed. The planet wheel B rotates freely on an axlecarried by the annular wheel F, and the planet E on an axlemounted on the output shaft's arm. Given the tooth numbersindicated, find the speed of the output shaft when the inputshaft rotates at 1000 rev/min. [ 524 rev/min ]

4 In the epicyclic train illustrated, the gear C is fixed and thecompound planet BD revolves freely on a spindle which iscoaxial with the input and output shafts. (a) Show that, if zb ze > zc zd then input and output shafts

rotate in the same direction.(b) 7.5 kW is fed into the input shaft at 500 rev/min, losses

are negligible, and tooth numbers are sketched.Determine the torque on the output shaft. [15.5kNm ]

5 Select spur gears suitable for speed ratios of (i) 1/√2, and (ii) π, to four significant figures.

6 Determine the practical limits of profile shift on a 6 mm module gear with 19 teeth. If a profileshift of 0.4 is implemented, what are the dedendum, base, pitch, extended pitch and addendumcircle diameters of the gear ? Evaluate the base pitch and the angle γ of Fig A.

[ 103.8, 107.1, 114, 118.8 and 130.8 mm. 17.7 mm, 6.47o ]

C

D E

FA

B

inputoutput

A = 17 B = 60C = 75 D = 19 E = 25

AD

E

C

B

16T

33T

15T

31T

A D

C F

input output

A = 30B = 20D = 32E = 18 B E

APPENDIX C - Program 'STEEL SPUR GEARS'

This program is based upon the theory above - the fatigue analysis of AGMA 2001-B88 and the geometric factors defined

in AS 2938-1987. The program's solutions (shown below) to the above analysis and synthesis problems are more accurate

than the previous solutions because they employ correct extended centre geometry and incorporate correct geometry

factors - not the approximations ( 18) & ( 22) for mid-range profile shifts. The interactive nature of the program may be

employed to appreciate quickly the effect on life or power capacity of varying parameters such as tooth number, profile

shift and allowable contact stress.

The program requires values of maximum bending life factor to be input only if the gears are limited by strength rather

than by contact, and the bending life factors exceed the changeover value of 1.04 ( see KL graph above).

Spur Gears 17

15 A commercial gear pair having a transmission accuracy level of 8 is required to transmit 100 kWin a shockfree application with 99% reliability. The speeds of pinion and wheel are 1450 andapproximately 470 rev/min. Allowable stresses for contact and for bending of the pinion are1450 and 400 MPa respectively; for the wheel 1300 and 350 MPa.Select suitable tooth numbers and profile shifts, along with a corresponding module and face-width for a compact pair with a design life of 20 khr.

16 Estimate the life of a gear whose allowable contact stress is 1.2 GPa and which undergoes thestress spectrum : contact stress, σc (GPa) 1.0 1.1 0.9

speed, N (rev/min) 500 400 300duration, t (hours) 2 1 3 [ 7.9 khr]

17 A pair of 8 mm module, 100 mm facewidth commercial gears is manufactured to a transmissionaccuracy level of 7 and employed in a periodic duty of 1.25 application factor. The 23 tooth pin-ion’s allowable contact stress is 1.2 GPa at 99% reliability, the 47 tooth wheel’s is 1.1 GPa. If power is transmitted to the following cycle, what life may be expected of the pair ?

power, P (kW) 60 45 35pinion speed, N1 (rev/min) 200 150 100duration, t (min) 10 20 30 [ 13 khr ]

BibliographyANSI/AGMA 2001-B88, Fundamental Rating Factors & Calculation Methods for Involute Spur

and Helical Gear Teeth, 1988 mandatory if serious design is contemplated

AS 2938-1993, Gears - Spur and Helical - Guide to Specification and Rating relies on AGMA

Colbourne JR, The Geometry of Involute Gears, Springer-Verlag 1987 rather theoretical

Drago RJ, Fundamentals of Gear Design, Butterworths 1988

Dudley DW, Handbook of Practical Gear Design, McGraw-Hill 1984 superseded by Townsend

Erdman AG & Sandor GN, Mechanism Design: Analysis and Synthesis , Prentice Hall 1997 includesepicyclic gear trains

ISO/TR 4467-1982 (E), Addendum Modification of the Teeth of Cylindrical Gears . . ., 1982

Jensen PW, Classical and Modern Mechanisms for Engineers and Inventors, Dekker 1991

Merritt HE, Gear Engineering, Pitman 1971 most comprehensive geometric text

Mitchiner RG & Mabie HH, The Determination of the Lewis Factor and the AGMA GeometryFactor for External Spur Gear Teeth, Trans ASME, Journal of Mech. Design, January 1982

Müller HW, Epicyclic Drive Trains, Wayne State University Press 1982

SAE AE-15, Gear Design Manufacture and Inspection Manual, 1990 automotive oriented

Seireg AA, Friction and Lubrication in Mechanical Design, Dekker 1998 contact, fluid film bearings,gear contact

South DW & Mancuso JR ed, Mechanical Power Transmission Components, Dekker 1994

Townsend DP (ed), Dudley's Gear Handbook, McGraw-Hill 1991 useful

7 Derive equation ( 10) from which the contact ratio may be calculated (20o full depth system).

8 What is the practical range of centre distance for a pair of 4 mm module spur gears with 19 and35 teeth ? If they are manufactured with profile shifts of 1.5 mm and 2 mm respectively, evaluatethe extended pressure angle and the contact ratio. [ 108.6≤C≤112.8mm, 24.47o, 1.42 ]

9 Use the design procedure outlined in the Notes to determine gears suitable for a speed ratio of√2 ± 0.5 % and a centre distance of 200±1 mm. [ 6 mm module, with 27 and 38 teeth, and profile shifts of 0.45 say, for pinion and correspondingly 0.38 for wheel ]

10 Evaluate the contact ratio and the fatigue geometric factors I and J for each of the following :(a) the pairs 13:35, 23:62 and 36:97 (all of which approximate the ratio 0.3711 to within 0.1%)

taking profile shifts which lie midway between the practical limits for each gear;(b)* 23:62 teeth, assuming the minimum practical profile shifts for both gears; (c)* repeat (b) but use the maximum practical profile shifts.Comment upon the trends suggested by these results.

11 The transmission accuracy level number of a pair of open gears is 6. Further particulars of the 25mm module 300 mm facewidth gears are as follows :

number allowable stresses, MPa speed,of teeth contact bending rev/min

pinion 25 1100 290 150wheel 55 1000 280 -

What life may be expected of the gears whilst transmitting 1 MW uniformly? [ 39 khr ]Shock loading of the foregoing drive results from unsuspected torsional vibration. If the effec-tive application factor is in fact 1.25, what life may now be expected? [ 5.4 khr ]

12* Two mating gears of commercial quality are to hand with 18 and 56 teeth. Their common face-width is measured as 50 mm and their addendum diameters as 83.2 and 233.6 mm.Metallurgical analysis reveals that the expected contact and bending stresses of the gears' com-mon material are 1100 and 300 MPa respectively. Estimate the pair's capacity (kW) for a 10 khr life in a shock-free application in which the pinionspeed is 300 rev/min. The transmission accuracy level number is 6. [ 9.1 kW ]

13 A gear pair transmits 75 kW with an application factor of 1.5 and reliability of 99%. Particularsof the commercial 6-accuracy level gears are :

number allowable stresses, MPa speed,of teeth contact bending rev/min

pinion 20 1300 180 90wheel 37 1250 175 -

Select a suitable module and facewidth for a life of 15 khr. [ 16, 144 mm ]

14 Details of a pair of commercial gears having a transmission accuracy level of 8 are as follows :number allowable stresses, MPa speed,of teeth contact bending rev/min

pinion 10 1320 380 200wheel 36 1100 360 -

Select a suitable module and facewidth for a design life of 16 khr whilst transmitting a uniform125 kW with a reliability of 99%. [ 16, 187 mm ]

Buckling 1

The Crymlyn Viaduct over the Ebbw Alley opened in 1857 asWelsh coal mining expanded. It was constructed of wrought andcast iron, and remained the highest railway viaduct in the UK untilits closure in 1964 due to increased locomotive weights ( 1908photo ). The advance from masonary to the slender metal compres-sive members which make up each column requires substantialbracing to prevent buckling.

The Humber Road Bridge, opened in 1981, comprises a continu-ously welded closed box road deck suspended from catenarycables supported on reinforced concrete towers. Suspensionbridges eliminate the need for struts other than the two towers,however avoiding buckles in other slender components becomesan issue.

The dangers associated with over-slender build were tragicallydriven home by the collapse of the Tacoma Narrows road bridgeover the Puget Sound in 1940. Although this failure was appar-ently due to wind-structure aerodynamic coupling rather thanbuckling as such, the film from which this frame is taken demon-strates graphically the ability of large structures to undergo signifi-cant elastic deflections.

Buckling of Thin-Walled Structures

A thin-walled structure is made from a material whose thickness is much less than other structuraldimensions. Into this category fall plate assemblies, common hot- and cold-formed structural sec-tions, tubes and cylinders, and many bridge and aeroplane structures.

Cold-formed sections such as those illustrated are increasingly sup-planting traditional hot-rolled I-beams and channels. They are partic-ularly prone to buckling and in general must be designed againstseveral different types of buckling.

It is not difficult to visualise what can happen if a beam is made fromsuch a cold-rolled channel section. One flange is in substantial com-pression and may therefore buckle locally at a low stress ( ie. much

less than yield ) thus reducing the load capacity of the beam as a whole. Buckling rather thanstrength considerations thus dictate the performance.

Let us now look at typical examples of buckling.

The slender elastic pin-ended column is the protoype formost buckling studies. It was examined first by Euler inthe 18th century. The model assumes perfection - the col-umn is perfectly straight prior to loading, and the loadwhen applied is perfectly coaxial with the column.

The behaviour of a buckling system is reflected in theshape of its load-displacement curve - referred to as theequilibrium path. The lateral or 'out-of-plane' displace-ment, δ, is preferred to the load displacement, q, in thiscontext since it is more descriptive of buckling.

out-of-planedisplacement

q

P

δ

load dis-placement

P

Pc

P

δprimary path

bifurcation

post-bucklingsecondary path

directexaggerated

shorteningPc

P

q

criticalload

BUCKLING

When a structure ( subjected usually to compression ) undergoes visibly large displacements trans-verse to the load then it is said to buckle. Buckling may be demonstrated by pressing the oppositeedges of a flat sheet of cardboard towards one another. For small loads the process is elastic sincebuckling displacements disappear when the load is removed.

Local buckling of plates or shells is indicated by the growth of bulges, waves or ripples, and is com-monly encountered in the component plates of thin structural members.

Buckling proceeds in manner which may be either :- stable, in which case displacements increase in a controlled fashion as loads are increased, ie.

the structure's ability to sustain loads is maintained, or- unstable, in which case deformations increase instantaneously, the load carrying capacity

nose-dives and the structure collapses catastrophically.Neutral equilibrium is also a theoretical possibility during buckling - this is characterised by defor-mation increase without change in load.

Buckling and bending are similar in that they both involve bendingmoments. In bending these moments are substantially independent of theresulting deflections, whereas in buckling the moments and deflections aremutually inter-dependent - so moments, deflections and stresses are not pro-portional to loads.

If buckling deflections become too large then the structure fails - this is a geo-metric consideration, completely divorced from any material strength con-sideration. If a component or part thereof is prone to buckling then its designmust satisfy both strength and buckling safety constraints - that is why wenow examine the subject of buckling.

Buckling has become more of a problem in recent times since the use of high strength materialrequires less material for load support - structures and components have become generally moreslender and buckle-prone. This trend has continued throughout technological history, as is demon-strated by bridges in the following sequence :

The Pont du Gard in Provence was completed by the Romans inthe first century AD as part of a 50 km aqueduct to convey waterfrom a spring at Uzès to the garrison town of Nemausus (Nimes).The bridge is constructed from limestone blocks fitted togetherwithout mortar and secured with iron clamps. The three tieredstructure avoids the need for long compressive members.

The Royal Border Bridge, Berwick upon Tweed, was built byRobert Stephenson whose father George built the Stockton andDarlington Railway ( the first public railway ) in 1825. Opened in1850, the bridge continues today as an important link in the busyKing's Cross ( London ) - Edinburgh line. The increased slender-ness of the columns compared to the Pont du Gard reflect techno-logical improvements over many centuries.

P

M=Px

x

bending

M=Pδ

P

δ

buckling

Buckling 2

structure again snaps instantaneously - this time back to the point 3 on the primary path.

Clearly this behaviour makes it imperative in design to apply large safety factors to the theoreticalbuckling loads of compressed cylinders.

It has been noted that a pressure vessel head is sub-jected to a compressive hoop stress at its junctionwith the cylinder. The two photographs here ( fromRamm op cit ) show both inward and outwardbuckles arising from this compression in the toris-pherical heads of internally pressurised 3 m diame-ter stainless steel vessels.

Longitudinal stresses in a vertical cyl-inder may also promote buckling asthese two photographs illustrate ( fromRhodes & Walker op cit ).

Warning of impending failure of the7.3 m diameter vitreous enamelled siloon the left is provided by the visiblebuckles. Grain pours out of the buck-

led bin on the right - the ladder gives an idea of the bin size.

Torsional buckling of columns can arise when a section under compression is very weak in torsion,and leads to the column rotating about the force axis.

More commonly, where the section does not possess two axes of symmetry as in the case of an anglesection, this rotation is accompanied by bending and is known as flexural torsional buckling.Lateral buckling of beams is possible when a beam is stiff in the bending plane but weak in thetransverse plane and in torsion, as is the I-beam of the sketch.

It often happens that a system is prone to buckling in various modes. These usually interact toreduce the load capacity of the system compared to that under the buckling modes individually. Anexample of mode interaction is the thin box section which develops local buckles at an early stage ofloading, as shown greatly exaggerated here.

The behaviour of the column is influenced by these local buckles, and gross column buckle willoccur at a load much less than the ideal Euler load. The Steel Structures Code, AS 1250 op cit sets

before

afterP

P

Torsional buckling

before

after

Flexuraltorsionalbuckling

P

beforeafter

Lateralbucklingof anI-beam

overallbuckling

displace-ment

local buck-ling

enlarged section A-A

A

A

Local and overall buckling interactionin a hollow box column

P P

Nothing is visible when the load on a perfect column first increases from zero - the column is stable,there is no buckling, and no out-of-plane displacement. The δ equilibrium path is thus characterisedby a vertical segment - the primary path - which lasts until the increasing load reaches the criticalEuler load Pc = π2 EImin/L2 - a constant characteristic of the column ( for a derivation of this, see below

or Timoshenko & Gere op cit for example ).

When the load reaches the Euler load, buckling suddenly takes place without any further loadincrease, and lateral deflections δ grow instantaneously in either equally probable direction. Afterbuckling therefore, the equilibrium path bifurcates into two symmetric secondary paths as illus-trated. Clearly the critical Euler load limits the column's safe load capacity.

Local buckling of an edge-supported thin plate doesnot necessarily lead to total collapse as in the case ofcolumns, since plates can generally withstand loadsgreater than critical. However the P-q curve illustratesplates' greatly reduced stiffness after buckling, so platescannot be used in the post- buckling region unless thebehaviour in that region is known with confidence.

It is emphasised that the knee in the P-q curve is unre-lated to any elastic-plastic yield transition; the systemsbeing discussed are totally elastic. The knee is an effectof overall geometric rather than material instability.

This photograph illus-trates local buckling of a model box girder constructed from thinplates, not unlike the road deck of the Humber bridge above.

Inclined striations are caused by shear loading in the web of a beamor in a torqued tube giving rise to compressive buckling stresses at45o to the longitudinal direction as predicted by Mohr's circle below.

The behaviour of a compressed shell after buckling is quite differentto that of a plate; in this case an unstable ( negative ) stiffness isaccompanied by a sudden reduction of load capacity.

Since the displacements are uncontrolled in most practical systems,shells behave in a snap-buckling mode - ie. as an increasing loadreaches the bifurcationpoint, the cylinder mustundergo an instantaneous

increase in deflection ( "snap" ) to the point 1 in order toaccomodate the increasing load. A subsequent decrease inload is accomodated by a corresponding decrease in buck-ling deflection until the point 2 is reached whereupon the

stable post-buckling path

Pc

P

δ

P

Pc

q

qP

δ

edge

s su

ppor

ted

TT

VV

rolled hollow sectionin bending shear

torqued tube

τ

σX

τxy

x

ypureshear

compr-ession

at 45 o

P

Pc

q

Pc

P

δ

unstable post-buckling path (snap)

1

23

Buckling 3

b

k

A

B

b

b

b

k

a aC

P

1 2

y

θ

( a ) ( b )

EXAMPLE The perfect mechanism of Figure ( a) is pinned at A, B and C. The members are rigid, apart from

the two identical springs of stiffness k which can withstand both tension and compression.

Initially A, B and C are collinear and the springs are free.

The equilibrium of the system has to be examined when it is com-

pressed by a load at A.

It is obvious that the system is in equilibrium as it stands in ( a). If the

load had been tensile then no other deformed configuration would be

possible; however since the load is compressive the system may not be

stable in this undeformed position and other positions must be consid-

ered - the loaded configuration of Figure ( b) and its mirror image.

The angle θ is selected as the most convenient measure of displace-ment in the loaded position ( b). It is necessary therefore to express, in

terms of θ, the PE gain of the deformed system over that of the freeconfiguration ( a) in which the elastic members are unstrained.

From ( b) the lengths of the springs 1 and 2 are :

L1 = 2( b cos θ + a sin θ ) = 2b ( cos θ + α sin θ )

L2 = 2( b cos θ – a sin θ ) = 2b ( cos θ – α sin θ )

where the constant α = a/b reflects the characteristic proportions of the mechanism.

The corresponding unstrained lengths Lo are found by substituting θ = 0, and so Lo1 = Lo2 = 2b. The spring deformations are therefore :

δ1 = L1 – Lo1 = 2b ( cos θ + α sin θ –1 )

δ2 = L2 – Lo2 = 2b ( cos θ – α sin θ –1 )

and hence the strain energy gain of the two springs ( k1 = k2 ≡ k ) is :Uk = 1/2 k1 δ1

2 + 1/2 k2 δ22 = 4 k b2 [ ( 1 – cos θ )2 + ( α sin θ )2 ]

For the load, with the base reckoned as the PE datum, the height above datum is : y = 4 b cos θ. The correspond-

ing height in the unstrained mechanism follows with θ = 0 as yo = 4 b. So the gain in potential energy of theload is :

UP = P ( y – yo ) = 4 P b ( cos θ – 1 ) - actually a loss.

The total potential of the system as a whole is therefore :

U = Uk + UP = 4 k b2 [ ( 1 – cos θ )2 + ( α sin θ )2 ] + 4 P b ( cos θ – 1 )

Now we are interested solely in the vanishing first derivative of U, and in the sign of its second derivative. Its

absolute size is irrelevant, so U is non-dimensionalised by 8 k b2 to give the normalised energy, u :u = U/8kb2 = 1/2 [ ( 1 – cos θ )2 + ( α sin θ )2 ] + p ( cos θ – 1 )

where p = P/2bk is the load factor - since 2bk is a known constant, then P may always be found for

any given p, and vice versa. Normalising is optional; it is carried out purely to simplify later algebra.

Differentiating :

( A) u' = sin θ [ 1 – p + ( α2 – 1 ) cos θ ]

( B) u" = ( 1 – p ) cos θ + ( α2 – 1 ) cos 2θ

The equilibrium paths, from ( 1a) and ( A) are :

how this is governed by the energy principles embodied in equations ( 1). We shall therefore applythe principles to various mechanisms to illustrate how the response may be stable or unstable - andsignificantly how imperfections affect behaviour. The conclusions reached for mechanisms aredirectly applicable also to plate-like components.

These energy equations are applied first to a typical perfect mechanism which behaves in a stable,neutral or unstable manner depending upon the mechanism’s geometry.

out rules for the avoidance of mode interaction in large components, and its guidelines should befollowed in design.

Buckling has mixed blessings in automotive applications.

The photograph on the left illustrates how local buck-ling of a car's thin-walled A-pillar dramaticallyreduces passenger cell integrity in the event of roll-over.

Conversely, the energy absorbed by plastic bucklingcan reduce significantly the injuries suffered by a vehi-cle's occupants in the event of a crash. The energyabsorption capability of thin-walled sections is demon-strated clearly by the experiment photographed on theright ( from Murray op cit ).

The detailed analysis of most practical buckle-prone structures is too complex mathematically toattempt here. We therefore examine instead some mechanisms which demonstrate (un)stable beha-viour similar to that of structures. The mechanisms allow us to appreciate buckling behaviour andthe tools used to analyse it, and to introduce the concept of imperfections which must occur in prac-tical components and which have a relatively large effect on buckling behaviour and safety.

This work leads to the derivation of a design equation for practical columns, in which the twin fail-ure modes of strength and geometric instability invariably interact. This interaction is apparent alsoin the behaviour of cracks - the subject of a later chapter.

Prediction of the plastic collapse of sub-sea pipelines is also addressed.

Stability of Equilibrium

As noted, the buckling propensities of a system are reflected in the shape of the equilibrium path, ie.the load-deflection curve. If a system is in equilibrium then its total potential energy, U, has a turn-ing value since an infinitessimal disturbance, dδ, of the systemfrom its equilibrium position does not change the potentialenergy (PE) - δ is any convenient characterising displacement.Expressed mathematically, the equilibrium path is thus definedby :

( 1a ) U' ≡ dU/dδ = 0

The type of equilibrium which exists at any point on the path - stable, neutral or unstable - is also animportant consideration; it may be deduced by taking the second derivative :-

( 1b ) U" ≡ d2U/dδ2 > 0 ; PE - minimum ; stable equilibrium= 0 ; PE - zero slope ; neutral equilibrium< 0 ; PE - maximum ; unstable equlibrium

It will be appreciated from the introduction that, compared to the elementary load building blocks,the plate-like components of thin-walled structures are complex in their deflections and in the math-ematical decriptions thereof. We shall therefore not attempt to analyse these components - the inter-ested reader is instead referred to the ample literature on plate and shells.

It is important however to appreciate the general physical behaviour of plate-like components and

U'' > 0stable

U'' < 0unstable

U'' = 0neutral

Buckling 4

EXAMPLE Repeat the example above if there is an initial out-of-straightness, θo, when the springs are free.

The analysis is similar to the above except that the free position is defined by θ = θo rather than by θ = 0 - so, asthe initial deflections must increase :

δ1 = 2b [ ( cos θ + α sin θ ) – ( cos θo + α sin θo ) ] ; |θ| ≥ |θo|δ2 = 2b [ ( cos θ – α sin θ ) – ( cos θo – α sin θo ) ]

y – yo = 4b ( cos θ – cos θo )

For the system therefore - normalising and differentiating as before :-

U = Σ k δ2/2 + P ( y – yo )

u = U/8kb2 = 1/2 [ ( cos θ – cos θo )2 + α2 ( sin θ – sin θo )2 ] + p ( cos θ – cos θo )

( C) u' = ( cos θo – cos θ ) sin θ + α2 ( sin θ – sin θo ) cos θ – p sin θ( D) u" = ( cos θo cos θ – cos 2 θ ) + α2 ( sin θo sin θ + cos 2 θ ) – p cos θ

The equilibrium path, from ( 1a) and ( C) is evidently :

( E) p = cos θo – cos θ + α2 ( 1 – sin θo / sin θ ) cos θ ; |θ| ≥ |θo|which is plotted below for the same three mechanism proportions as above, for the ideal case

(θo=0), and for positive and negative out-of-straightnesses |θo| = 5o and 10o.

Considering stability, inserting ( E) into ( D) yields :-

u" = sin2 θ [ 1 + α2 ( sin θo / sin3 θ – 1 ) ]

Thus, for the positive branches, stability is lost if

( F) u" ≤ 0 i.e. sin θ ≥ [ α2 sin θo / ( α2 – 1) ]1/3

and can only occur if α > 1; the corresponding maximum load capacity, from ( F), being :

p* = cos θo + [ ( α2 – 1)2/3 – ( α2 sin θo )2/3 ] 3/2 ; α > 1

Some values of p* are shown for α = 2.

These examples, though dealing with one particular mechanism, confirm the following conclusionswhich are quite general for imperfect buckle-prone structures :

- Imperfections lead to deflections growing from the start of loading, and eliminate any sharpcritical bifurcation point. Collapse is therefore not catastrophic; growing deflections warn ofimminent failure.

- As deflections grow, the imperfect structure's behaviour tends to that of the ideal structure.- Imperfections reduce the load carrying capability of a structure for a given deflection, and also

the maximum capacity ( p* above ) when the secondary path is unstable.- The effects of imperfections are much more pronounced when the post-buckling path is unsta-

ble - for example α = 2 above. - For structures with neutral or stable post-buckling paths, the main effect of imperfections is

usually to introduce high bending stresses at an early stage of loading thus leading to failurethrough strength limitations rather than geometric collapse. This conclusion influences the

Effects of Imperfections

Perfect structures are figments of the imagination. If the mechanism above were manufactured, itwould be impossible to ensure that initially A, B, and C all lay on the load's line of action, or thatboth springs were exactly free in the configuration of Figure ( a).

Small imperfections can have a marked effect on buckling behaviouras this next example demonstrates.

either sin θ = 0, ie θ = 0, in which case, from ( B) :

u" = ( 1–p) + (α2–1) = α2 – p - so, if p < α2 then u" > 0 and the path is stable;

if p = α2 then u" = 0 and the path is neutral;

if p > α2 then u" < 0 and the path is unstable.

This is the primary (undeformed) path between no load p = 0 and the critical load (factor) pc = α2.

or p = 1 + ( α2 – 1 ) cos θ, in which case, from (B):

u" = ( 1 – α2 ) sin2θ - so, if α < 1 then u" > 0 and the path is stable;

if α = 1 then u" = 0 and the path is neutral;

if α > 1 then u" < 0 and the path is unstable.

The stability of these secondary ( deformed ) paths is thus dictated solely by the proportions, α.

These trajectories, and their mirror images are plotted for three different proportions α = a/b = 0.5, 1 & 2.

The three mechanisms display the same primary path shapes, but differ in the value of critical load pc = α2 - thus

for a = b/2 then Pc = 2bk.pc = 2bk.(1/2)2 = bk/2. The primary path extends into the post-buckling region, but isunstable.

The secondary post-buckling paths are stable for the mechanism for which a = b/2. This behaviour is analogous

to that of plates and plate structures.

If a = b then the critical load is 2bk and the secondary paths are neutral - which is the performance noted for

Euler columns. Finally for a = 2b, the critical load is 8bk and the secondary paths are unstable - similar to those of

cylindrical shells and similar systems.

In practice, interference would prohibit realisation of the full range of theoretical displacements.

To verify the stabilitiy of the secondary paths we may find the P-q rela-

tions, q being the load displacement. From the geometry and above :

q = yo – y = 4 b ( 1 – cos θ ) Eliminating q between this and the secondary path equation :

p = P / 2 bk = α2 + ( 1 – α2 ) ( q / 4 b ) which plots as :-

This confirms prior reasoning - the stable path ( α < 1 ) has positive

slope, the neutral path ( α = 1 ) zero slope, and the unstable path ( α > 1 )negative slope.

load

fac

tor

p

-90 -60 -30 0 30 9060

3

2

1

displacement, θo

α = 2 4

stable primary path p < α2

-90 -60 -30 0 30 9060

displacement, θo

α = 1

neutral secondary paths p = α2

unstable primary path p > α2

p critical buckling loadcr

3

2

4

load

fac

tor

p

α = 12/

-90 -60 -30 0 30 9060

3

2

4

1

displacement, θo

1/4

α = 2

α = 1

α = 1/2

4

3

2

1

010 displacement, q/4b

load

fac

tor,

p inter-

ference??

Buckling 5

of each stress block is a force Sy t/2 ( per unit length ). The hinge therefore develops a constantbending moment My = Sy t

2/4 per unit length as collapse proceeds ( iv).

The work-energy principle may now be applied to the propagation pressure collapsing the plastichinged model.

Work done by external pressure, per unit length= pressure * change in volume ; or, per unit length := pc * decrease in cross-sectional area = pc * square area a-b-c-d = pc D

2 /2Strain energy gain of four plastic hinges, per unit length

= 4 My * rotation of each hinge over the process = Sy t2 π /2Equating work and energy leads to an estimate of the propagation pressure : pc /Sy = π (t/D)2

The model underestimates the propagation pressuremainly because it neglects strain hardening and thefinite extent of the plastic hinges - this last may beappreciated by comparing the actual buckled shapewith the collapsed model. A more realistic empiricalexpression fitted to experimental results is shown;this tends to the model as t/D → 0.

The ability of this simple model (without empiri-cisms ) to provide a ball park estimate of the col-lapse load should be appreciated. 0 10 20 30 40 50

25

20

15

10

5

0D/t

p /

S

10

cy

3*

p = π ( 1+30 t/D ) ( t/D ) Sc y 2

model

treatment of imperfect columns below.- Different forms of imperfection - for example out-

of-straightness compared to load eccentricity -give rise to similar behaviour. Thus if the soleimperfection in the mechanism of Figure( a) werea load eccentricity e shown here, then the result-ing performance is very similar to that deducedpreviously for out-of-straightness - the readershould confirm this.

- An important corollary of this last conclusion isthe impossibility of categorising imperfectionssolely from the observed behaviour of a structure

which is nominally ideal. Indeed this categorisation is unnecessary in most cases, as, when allsaid and done, it is the behaviour itself which is important.

Submerged Pipelines

With the gradual depletion of easilyexploitable gas and oil fields, less access-ible sub-sea reserves are being increas-ingly tapped. The distribution pipeworkof these often lie at depths of around 200m and is thus subjected to substantial external pressures. It is imperative that pipe buckling is pre-vented as it propagates catastrophically, being driven by a component of the water pressure at thespeed of sound in the metal wall. Cross-sections taken from vari-ous positions along a partially collapsed pipe are shown here :-

The minimum pressure which maintains buckle travel in a particu-lar pipe is the critical propagation pressure, pc, for the pipe, andwe turn now to prediction of this ‘buckling load’.

The buckled shape indicates gross plastic deformation - indeed elastic effects pale into insignificanceand the pipe material assumed to be perfectly plastic, with yield strength Sy. As a result of this non-conservatism, the energy methods of the previous sections cannot be used; however the work-energy principle is applicable.

The adopted model consists of unit length of thethin pipe ( t/D → 0 ) illustrated here, with fourplastic hinges - a, b, c and d. Deformation occursonly at the hinges, so the four lobes of the cross-section rotate about these hinges without other-wise deforming, thus forming the collapsed shapea’-b’-c’-d’.

The material at a plastic hinge is taken to be perfectly plastic - ie. stresses can only be either zero oryield as shown at ( i). If pure bending occurs across the thickness t then tensile yield occurs across

half the thickness ( ii) andcompressive yield extendsacross the remaining halfthickness. The resultant ( iii)

buckle travel

pressureundeformed buckled

a'

d' c' b'D

a

b

c

d

collapsed

plastic hinge, My

t

Sy

SyS t/2y

t/2M = S t /4y y

2

Sy

σ

εplastic hinge

i ii iii iv

0 30 9060

3

2

4

1

load

fac

tor

p

displacement, θo

p*

α = 2

3.072.69

0

0.2

0.1

e/4b

e

b

P

k k

A

B

1 + ( α –1) cosθ2

1 + ( e/4b ) cotθp =

Buckling 6

Although such non-Eulerian initial conditions may be quite deliberate, as in the case of a walkingstick, we shall for convenience refer to them as 'imperfections'.

Imperfect columns are not unlike beams in that bending is a major failure mechanism, howeverthere are two significant differences between columns and beams :-

- Beam loading geometry is essentially invariant when loads are applied ( deflections are smalland do not affect bending moments to a significant degree ) whereas the bending moments inan imperfect column are affected by bending deflections. This last phenomenon causes non-linear behaviour ( stresses and deflections are NOT proportional to load ) and geometric insta-bility.

- Beams are usually subjected only to bending, with direct stresses for the most part negligible.Columns are loaded in bending plus direct compression - one of the few situations wheredirect effects cannot be neglected in the presence of bending. The maximum stress in an imper-fect column is a combination of direct and bending stresses, and the column is assumed to beon the point of failure when this maximum stress reaches the strength of the material.

The variables which enter into the analysis are reviewed in this schema :

The variables are condensed conveniently into three dimensionless groups : - The eccentricity ratio, η, reflects the relative severity of the imperfections .- The strength parameter, θ, is the ratio of the direct compressive stress to the maximum com-

pressive stress ( direct + bending ) which equals the strength S of the material at failure. Thestrength parameter is a measure of tendency towards strength failure - the parameter cannotexceed unity.

- The stability parameter, ψ, is the ratio of the direct compressive stress to the critical buckling (Euler) stress. The stability parameter is a measure of tendency towards stability failure - theparameter cannot exceed unity.

As the three dimensionless groups embody load, strength, dimensional and safety aspects, thedesign equation sought will be of the form : function ( η, θ, ψ ) = 0. The equation is derived frombeam deflection theory rather than by energy methods.

A horizontal column is symmetric about its mid-section x = 0. A free body of its right hand end is

Initially straight column, offset load Initially bent column

e e

unloaded column axis

load axis

z

z

A

y

σσ

dθ =

slenderness ratio

Column Material

Column Dimensions

External Load

Induced MaximumCompressive Stress

elastic modulus

equivalent lengthmax. fibre distancemin. area momentcross-sect'l area

magnitudeinitial max. ecc'y

E

L

yIA

Fe

^

z

σ

Izr=√A

σ = F/Ad

radius of gyration

direct compressivestress

ρ = L/r

σ = E ( )c

πρ 2

Euler criticalstress σ

dσ

c

stabilityparameter

ψ =

strengthparameter

e y

r2η =

eccen-tricity

ratio

Practical Columns - Design Equations

An ideal column is perfectly straight initially, supported by pinned ends which are constrained tomove axially, and loaded by a coaxial compressive force F as shown in sketch ( a) below.

Illustrated also are other support arrange-ments which may be modelled by theconfiguration ( a) with an equivalentlength L related to the actual length ofthe arrangement by a factor k whichreflects the degree of end fixing.

The behaviour of an ideal column hav-ing a particular slenderness is illustratedat left below. The load may be increasedundramatically until it reaches the Euler sta-bility limit Fc whereupon deflections instantaneously increase out of hand and the column col-lapses. This instability is geometric in nature and unlike yielding or fracture has got little to do withmaterial - for example a cardboard tube can sustain a higher compressive load than a steel wire.

If a squat barrel is compressed then buckling is impossible, and failure occurs when the load-induced stresses reach the strength S of the material. This additional failure mechanism is indicatedin the Two Criteria plot above in which the ordinates of the LH graph have been reduced by thecross-sectional area, A, that is, transformed into direct compressive stress, σd = F/A. The materialstrength S acts as a bound additional to the Euler stability limit σc = Fc /A, however experimentalfailure points lie below both limits. The plot shows clearly that σd cannot exceed either σc or S.

Historically, columns have been classed as either :- short, which fail when the strength limit S is reached ( a material criterion ), or - long, which fail when the stability limit σc is reached ( a geometric criterion )

. . . often described by empirical design equations such as the 'parabolic' or 'JB Johnson'formulae, however such a distinction will be found unnecessary with the rational approach adoptedhere which recognises the interaction between the two criteria - stability and strength. Theapproach is based on the concept of imperfect columns.

Perfect columns, like perfect mechanisms, are theoretical idealities. Imperfections must arise in prac-tice - and the effect of these is invariably to reduce the load carrying capacity predicted by elemen-tary idealised theory. We examine a practical column whose imperfections prior to load applicationmay be modelled by either of the two arrangements illustrated below, in which the imperfection ischaracterised by the maximum eccentricity e max .

slenderness ratio ρ = L/r

load

, F

F ≤ Fc

Euler stability limit

F = π EI/Lc= AE (π/ρ)

2 2

2

σ = F /A c c2= E (π/ρ)

typical experi-mental points

short, S<σc

strengthlimit, S

dir

ect

stre

ss,

σ =

F/A

d

long, σ <Sc

stability limit

slenderness ratio ρ = L/r

TWOCRITERIA

( 2 )

2 0.7

( )/12

( )/1√2

0.85

( 1 )

1L

F

( a) ( b) ( c) ( d)

k theoreticalk recommended

L = k Lactual

Buckling 7

clusion that an imperfection'sactual form is of lesser signifi-cance than its magnitude. It isjustifiable therefore to adoptthe simplest equation here - forthe column initially bent sinu-soidally - to represent all formsof imperfection.

This also has the added advan-tage of enabling closed formsolutions.

So, substituting for ψ, θ in therightmost equation ( h) :-

( j) ( σ – σd ) ( σc – σd ) = η σd σc ; σd ≤ σ , σc

from which the maximum stress may be found for a given load ( implicit in the direct stress σd ) act-ing on an imperfect column which is characterised by a certain critical Euler limit σc and eccentricityratio η.

If the load safety factor, n, corresponds to a certain actual load, F, then the direct stress at failure, σd

= Ffailure/A = nF/A, corresponds to σ reaching the material's compressive strength, S. Applicationof ( j) to this failure situation yields the design equation :-

( 2 ) σ2d – σd [ S + ( 1 + η ) σc ] + S σc = 0 ; σd = n F/A ≤ σc, S

The direct stress from ( 2) isplotted here for a particularsteel. The graph demon-strates the generality that :

the larger the imperfections,the smaller the load capacity

The intercepts on the directstress axis are S/(1+η), sincefor a short squat column ρ →0, instability tendencies arenegligible, σc → ∞, and so theintercepts follow from ( 2) ormore immediately from thesketch of the short column :-

If the column is brittle then the ten-sile stress on the convex side, eval-uated along the lines of ( g), may bemore significant than the largercompressive stress which must bethe more critical for ductiles.

Safety factor is a factor on LOAD - ALWAYS

When stress is proportional to load then the safety factor applies equally well to stresses. But col-umn stresses are NOT proportional to load as the following example demonstrates :

stability parameter, ψ = σ /σd c 10

1

stre

ngth

par

amet

er,

θ =

σ

/ σ

d^ 0.01

0.050.1

0.2

1

2

5

eccentricityη = 0.5

perfect η = 0

perfect η = 0

.01.05

0.2

0.5

1

2

5

η

perfect η = 0

y

STEELS = 350 MPaE = 207 GPa

slenderness ratio ρ = L/r0 50 100 150 200 250

dire

ct c

om

pres

sive

str

ess 400

300

200

100

0

σ

= n

F/A

(M

Pa)

d

0.1

F

F M=Fe

F

e

^σ = +F Fe y

A I= σ ( 1+η )

^d

^

σ =d ( 1+η )

S∴

shown, with boundary at x = φL where –1/2 ≤ φ ≤1/2. The stress resultant at the free body’s LHboundary consists of the axial force, F (whichequals the column load) and the bending moment,M. The eccentricity of the unloaded column axisfrom the load's line of action is e - the relationshipbetween e and x will depend upon the form of ini-tial imperfection which exists.

Three different cases each characterised by maximum eccentricity e are examined in parallel - an ini-tially straight column, a column initially bent parabolically and a column initially bent sinusoidally :

CASE 1 2 3straight: offset load bent parabolically bent sinusoidally

( a) e at φ e : constant e ( 1 – 4φ2 ) e cos π φ

For free body equilibrium, letting v be the additional deflection at φ when the load F is applied( b) M = F ( e + v ) - for equilibrium of the free body, and

= – E I d2v/dx2 - from bending theory

Equating these right hand sides, substituting x = φL, and noting that FL2/EI = π2ψ where ψ is aconstant common to all cross-sections, yields :( c) d2v/dφ2 + π2ψ v = – π2ψ e whose solution is of the form( d) v = C cos( π√ψ . φ ) + D sin( π√ψ . φ ) + particular integral

in which C and D are integration constants, and the particular integralrefers to the relevant equation ( a) for e on the RHS of ( c). The integration constant D must vanishby symmetry about φ = 0, and C must satisfy zero deflection at the ends ( v = 0 at φ = ± 1/2 ).

Inserting these boundary conditions, the elastic curves for the various cases are :

( e) v / e sec(π/2√ψ) cos(π√ψ.φ ) -1 (8/π2 ψ) sec(π/2√ψ) cos(π√ψ.φ) [ ψ/( 1 - ψ) ] cos π φ

- 1 - ( 8/π2 ψ ) + 4 φ2

The relevant maximum deflection at the centreline, φ = 0, follows as :-

( f) v / e sec(π/2√ψ) - 1 (8/π2 ψ) [sec(π/2√ψ) -1] - 1 ψ/ ( 1 – ψ)

The maximum compressive stress, on the concave side of the column at φ = 0, is :-( g) σ = σdirect + σbending max - which, with ( b), becomes

= σd + F ( e + v ) y /Iz - or, normalising by σd1/θ = 1 + η ( v / e + 1 )

Finally, substituting the ratios v/e from ( f) gives the desired relationship for each of the three cases

( h) 1/θ 1 + η sec (π/2√ψ) 1 + η (8/π2ψ) [ sec(π/2√ψ) - 1] 1 + η / ( 1 - ψ)

where θ = σd/σ ; ψ = σd/σc and both θ, ψ ≤ 1

The strength-stability interaction equations ( h) are plotted below as loci corresponding to constantvalues of eccentricity ratio.

The leftmost equation ( h) corresponding to an initially straight column with offset load is known asthe ‘Secant Equation’ and appears as dashed loci in the plot.

The full loci refer to both the initially bent columns - there is no difference distinguishable at thescale of the graph between parabolic and sinusoidal imperfections. This reinforces the previous con-

L/2

colu

mn

mid

-sec

tion

F

M

F

e

v

x = φ L

columnloaded axis

free axis

load axis

Buckling 8

least. They may be partially allowed for when applying the above theory, by reducing the designyield strength accordingly.

The causes of scatter may be regarded as equivalent imperfections , and the Code's recommendationsare presumably based on this all-embracing view.

Bibliography

AS 1250: SAA Steel Structures Code, Standards Assocation of Australia

AS 1835: Seamless Steel Tubes for Pressure Purposes, Standards Assocation of Australia

Croll JGA & Walker AC, Elements of Structural Stability, Macmillan 1972

Dowling PJ et al eds, Steel Plated Structures, Granada 1977

Elishakoff I et al eds, Buckling of Structures: Theory and Experiment, Elsevier 1988

Godoy LA, Thin-Walled Structures with Structural Imperfections, Pergammon 1996

Gorenc BE & Tinyou R, Steel Designers Handbook, NSWUP 1981 - a commentary on AS 1250

Kenny JP & Partners, Buckling of Offshore Structures, Granada 1984

Murray NW, When it Comes to the Crunch: the Mechanics of Car Collisions, World Scientific 1995

Ramm E ed, Buckling of Shells, Springer 1982

Rhodes JR & Walker AC eds, Thin Walled Structures, Granada 1980

Rhodes J ed, Design of Cold Formed Steel Members, Elsevier 1991

Steel WJM & Spence J, 'On Propagating Buckles and Their Arrest in Sub-sea Pipelines', Proc IMechE,v197A, April 1983

Timoshenko SP & Gere JM, Theory of Elastic Stability, McGraw-Hill 1961 - more advanced

Tooth AS & Spence J eds, Applied Solid Mechanics - 2, Elsevier 1988

Trahair NS & Bradford MA, The Behaviour & Design of Steel Structures, Chapman & Hall 1988

Young BW, 'Steel Column Design', Structural Engineer, v51, n9, Sept 1973

EXAMPLE Particulars of a column made from a 250 MPa yield steel are :

L = 1 m A = 103 mm2 I = 105 mm4 η = 0.3

What load can it support if a safety factor of 2 is specified ?

Properties : r = √(I/A) = 10 mm ρ = L/r = 100 σc = E(π/ρ)2 = 204 MPa

From ( 2) with S = 250 MPa, the direct stress at failure is σd = 134 MPa ( note < both S and σc ).

The failure load is therefore this mean stress times the area, A, ie. 134 kN.

So if the safety factor is 2, the permissible load will be 134/2 = 67 kN.

Let us see what wo uld happen if this had been carried out incorrectly with safety factor applied to strength

rather than to load.

From ( 2) with S = Sy/n = 250/2 = 125 MPa, the direct stress is found to be 83 MPa. Multiplying by the area, the

corresponding load is 83 kN. Hence the incorrect conclusion is that the column could withstand a load of 83 kN

with a safety factor of 2. But 2*83 = 166 kN which exceeds the 134 kN critical load - ie the safety factor on load is in

fact less than 2.

Column design is more complex than this example suggests. Cross-sectional geometric propertiesoccur in all terms of ( 2) except the material strength, so design requires a trial-and-error approach.A more fundamental difficulty in applying ( 2) however, is deciding on what value of eccentricity touse when imperfections are nominally absent - the column is perceived as perfect but manufactureand assembly must introduce some imperfections. We would suspect that the longer and thinner thecolumn, the greater the difficulty in ensuring perfection, so :-

maximum eccentricity e or eccentricity ratio η = some increasing function of ( length L or slenderness ratio ρ )

Various proposals (based upon observations of course) have appeared in the literature, however therecommendations of the Steel Structures Code will be adopted here. This Code implements equation(2) with a safety factor of 5/3, and stipulates that :

( k) η ∝ ρ2 = 3 * 10-5 ρ2 ie. = 3 * 10-5 ( π2 E/σc ) ≡ ση / σc

. . . . where the equivalent imperfection stress ση ≡ 3 * 10 -5 π2 E is evidently a materialcharacteristic, say 60 MPa for steel. A lower value is probably permissible for small non-critical com-ponents in mechanical engineering.

Equations ( 2) and ( k) lead to the more useful design equation applicable to nominally perfect col-umns, namely :-

( 3 ) σ2d – σd ( S + σc + ση ) + S σc = 0 where σd = n F/A ≤ σc , S

The direct stress at failure ( n = 1 ) according to thisequation, forms a lower bound to experimentalresults from tests on ostensibly perfect columns.The scatter of these results is due to a number ofcauses which the foregoing treatment neglects,among which are plastic behaviour, non-ideal sup-ports, buckling mode interaction ( especially localbuckling ) and residual stresses.

Residual stresses arise when a complex sectioncools differentially after manufacture by hot roll-ing, forging, welding, etc, and may typically reach 30-40% of yield. They are thus far from negligi-ble, and heat treatment aimed at reducing them is usually impracticable, for structural sections at

0 50 100 150 200 250

dire

ct c

ompr

essi

ve

slenderness ratio, ρ

400

300

200

100

0

stre

ss,

σ

(MPa

)d

equation ( 3)

STEELS = 350 MPaE = 207 GPa

y

ησ = 60 MPa

Buckling 9

7 A 1 m long pin-ended column is made from a 250 MPa yield, equal angle section 80*80*10 mm.Assuming an equivalent eccentricity of L/1000, determine the maximum allowable load with asafety factor of 2. [ 153 kN]Repeat this if the column is nominally perfect with ση = 60 MPa. [ 156 kN]

8 (a) The critical Euler stress of a column is σc and the eccentricity ratio is η. It is made from abrittle material with compressive and tensile strengths Sc and St respectively. Prove the follow-ing algorithm from which the direct compressive stress σd may be calculated :

If 1/( Sc – St ) – η/( Sc + St ) < 1/2σc

then σ2d + σd [ St – ( 1 – η ) σc ] – St σc = 0

else σ2d – σd [ Sc + ( 1 + η ) σc ] + Sc σc = 0

Hint : Apply equation (g) to the convex side where bending stresses are tensile(b) A 400 mm long end-fixed column is made from a cast iron whose compressive strength is765 MPa. The cross-section is square, 20*20 mm, and due to fears of casting inaccuracies, aneccentricity ratio of 1 is assumed for estimating purposes. What maximum load can the columnwithstand on this basis ? [ 85 kN]

9 Trial-and-error methods must be employed when applying equation ( 3) to the design of duc-tile columns. Input constraints necessary for the design process include the column equivalentlength, L, the actual load, F, and desired safety factor, n. A range of sections is chosen of a material whose known properties include its elastic modulus,E, and yield strength, S. A value for the imperfection stress, ση, also results from this choice.The trial-and-error process commences by selecting one particular section from the range, ofarea A, and minimum second area moment I. Equation ( 3) enables calculation of the corre-sponding safety factor. This process is repeated until the safety factor is just larger than thedesired value.(a) Prepare a calculation sequence using consistent units, to assist in the above task.(b) A 2.4 m long, nominally straight pin-ended column has to support a 300 kN load with asafety factor of 2. A range of 250 MPa yield steel sections is available for which the imperfec-tion stress, ση, is 60 MPa. What size of section should be used if the sections are :-

( i) geometrically similar, with I/A2 = 0.15 ? [ 4200 mm2]( ii) tubes with an outside diameter-to-thickness ratio of 15 ? [ φ 120 mm](iii) channels whose properties are listed in the textbook ? [ 229*89 mm]

10 The sketch illustrates the right half of an ideal pin-ended column which is subjected to a trans-verse load Q at its centre, in addition to the compressive load P. Using the nomenclature of thepreceding notes, show that the elastic curve is :

v/L = (Q/P) [ sin (π√ψ.φ)/( π√ψ.cos(π/2√ψ) ) - φ ]

Hence derive the design equation for this loading,analogous to equation ( 2) above :

σd = S – σQ.tan( π/2√ψ )/( π/2√ψ )

where σd = nF/A and σQ = 1/2QLymax/I is themaximum bending stress due to the transverse load Q acting alone. Plot σd versus slendernessratio for some representative values of the parameter σQ - say 0, 5, 20, 50, 100, 200 MPa - andcompare the behaviour with that of imperfect columns with eccentricity ratio as the parameter.

L/2P

vvx = φL

Q

BUCKLING - PROBLEMS

In the following problems buckling mode interaction is not considered, the material is steel unless otherwisestated, and section properties are as cited in the Text.

1 Show that the post-buckling paths for the perfect mechanism are :p = 2P/kL = cos θ. Plot the p-θ paths, and ascertain whether equi-librium is stable, neutral or unstable.If the mechanism is imperfect, with an initial out-of-straightnessθo when the springs are free, show that the maximum load capac-ity is :- p* = ( 1 – ( sin θo )2/3 )3/2

Plot p* versus θo, and also the equilibrium paths for some repre-sentative θo.

2 The links of the mechanism are connected by a torsion spring of stiffness k(Nm/rad) whose strain energy is 1/2k∆2, where ∆ is the spring's angulardeflection. The mechanism may embody either one of two imperfections :

( i) The load is not coaxial with the mechanism's axis, but is offset by themoment arm : e = εL, measured perpendicular to the upper link.

( ii) The mechanism is not straight initially, the links being inclined at θo - inwhich configuration the spring is undeformed.

Plot the equilibrium paths for the perfect mechanism and- for the imperfect mechanism ( i) with θo = 0 and ε = 0, 0.05, 0.1, 0.2, 0.5- for the imperfect mechanism (ii) with ε = 0 and θo = 0, 2.5, 5, 10, 20o

Comment on the paths' shapes. Show that the mechanism is always stable.

3 The mechanism is inclined initially at angle α, when the spring is free.Application of the load P causes displacement through the angle θ. Show that the equilibrium path isp = P/2kL = sin φ ( 1 – cos α / cos φ ) ; φ = α –θ and plot this for α = 22.5o in the range –10o ≤ θ ≤ 50o. Prove that the local maximum/minimum loads are p* = ( 1 – ( cos α )2/3 )3/2, andstate clearly what happens when P increases from zero.

4 A sub-sea pipeline, of nominal size 100 mm and 400 MPa yield strength, lies at 200m. Using a safety factor of 2, select a suitable pipe wall thickness from AS 1835 towithstand buckling. A corrosion allowance of 1 mm is applicable both inside andoutside the pipe. [ 6.3 mm ]

5 (a) A straight steel rod, 20 mm diameter and 200 mm long, is fixed at one end. The free end issubjected to a compressive load offset from the rod's centreline by 1 mm. What load ispermissible if the maximum stress is limited to 200 MPa ? [ 37.1 kN]

(b) Repeat (a) but with rod diameter halved. [ 4.2 kN](c) Repeat (a) but with rod length doubled. [ 19.6 kN](d) Repeat (a) but with load eccentricity doubled. [ 28.6 kN]

6 A column is characterised by S = 200 MPa and σc = 150 MPa. Determine the maximum per-missible compressive stress predicted by each of the three imperfecet column models - equa-tions ( h) in turn, for η = 0.1 and for η = 1.0. Comment on the significance of the form and mag-nitude of the imperfections demonstrated by these values. [ 121.7, 124.6, 125.0; 66.7, 69.3, 69.7 MPa]

L/2

P

k k

θL/2

L/2

P

k

θL/2

e

L/2

k

P

θα

L/2

Fracture Mechanics 1

explained by conventional steady state safety appraisals or fatigue arguments :- Ocean-going vessels have been known to break in two since time immemorial - but not while

they were berthed! In January 1943 the one-day old T2 tanker SS Schenectady had just completed successful seatrials and returned to harbour in calm cool weather when . . ."Without warning and with a report which was heard for atleast a mile, the deck and sides of the vessel fractured just aftof the bridge superstructure. The fracture extended almostinstantaneously to the turn of the bilge port and starboard.The deck side shell, longitudinal bulkhead and bottom girdersfractured. Only the bottom plating held. The vessel jack-knifed and the center portion rose so that no water entered.The bow and stern settled into the silt of the river bottom."The ship was successfully repaired.There were nearly five hundred T2 tankers built in US yards between 1942 and 1945, with aver-age production time around 70 days.

Liberty Ships were lightly armed cargo vessels built inthe US for transporting desperately needed suppliesacross the U-boat infested Atlantic to a beleagueredEurope in WWII. Some 2700 vessels were built from1942 until the end of the war. Such huge numbers werepossible only through prefabricated all-welded con-struction to a standard design, together with a massiveinvestment of capital, materials and workers. Towardsthe end of the programme one vessel was completed inless than five days.At the start of the programme some 30% of LibertyShips suffered catastrophic fracture, though not neces-sarily so dramatically as the Schenectady. The stern ofSS John P Gaines is pictured here after the vessel split intwo off the Aleutians in 1943. Later design changesreduced the fracture rate to 5%.

The general technological fraternity was unaware of Fracture Mechanics principles when these ships weredesigned, and the reason for the disastrous fractures was a mystery since conventional safetyassessments were unremarkable and the extremely short lives ruled out fatigue as the culprit.It later became clear that the failures could be attributed to :- the all-welded construction which eliminated crack-arresting plate boundaries which are

present in riveted joints- the presence of crack-like flaws in welded joints performed by inexperienced operators

pressed into service by the exigencies of the programme- the use of materials whose low resistance to crack advance ( toughness ) was further

reduced by low temperatures.

Recent recoveries from the Titanic suggest that poor steels in association with low tempera-tures might have contributed to that disaster too, although this vessel was riveted throughout.Closer to home, the SS Bridgewater though not a Liberty Ship broke in two in the Indian Ocean(c 1960), only the stern half making it back to Fremantle harbour. A close-up of portion of the

Fatigue of Ductiles

Fatigue is recognised as a mechanism of crack growth terminated by catastrophic fracture - hence theS-N diagram which may be used to predict failure.

If we can understand, and formalise mathematically, the fundamentals of the crack growth processand the interaction between the factors which affect it, then the onset of catastrophe may be pre-dicted more confidently - rather than having to rely upon a somewhat contrived correlation betweenS-N curves, notch sensitivity, stress concentration and the like.

But this is not the only advantage ! There are many well-documented catastrophies which cannot be

10 4

Eventual brittlefracture surface

Slow advance ofcrack front - 'treerings', 'beach marks'

Crack initiator

10 5106

10 4 10 5 106Cycles to

failure, N

possibleendurance limit

Strength, S

S-N Diagram

Sut

Se

1

FRACTURE MECHANICS

Fracture Mechanics is the study of cracks and crack-like defects - such as might occur in welds forexample - with a view to understanding and predicting the cracks' growth tendencies. Such growthmay be either stable - that is relatively slow and safe - or unstable, virtually instantaneous and catas-trophic.

Let us first recapitulate with some familiar topics from Strength of Materials, to set the scene for Frac-ture Mechanics concepts.

Ductiles versus Brittles - their relative behaviour under the conventional, slow tensile test.

brittles

Fast catastrophic failure with no warning.Transverse granular cleavage surface - noshear lip.

Low energy absorption capacity. No significant yielding so no reduction ofhigh stresses. Triaxial stresses cause failure, unlikeductiles.

ductiles

Slow controlled extension; final instabilitydue to gross area reduction. Cup-and-cone failure surface characterised by 45o

shear lip.

General ability to absorb energy: 'tough'.Localised yielding at high stress concen-trations redistributes stresses advanta-geously. Shear causes failure, triaxiality isrelatively benign.

shear lip

cup cone

transverse fracturesurface

fra

ctu

re

stre

ss

s train

fra

ctu

re

stre

ss

s train

Fracture Mechanics 2

In both the slow tensile test and in the fatigue of a ductile material, the final stage of failure is brittle(ie. unstable ) since the crack speed is so high that the material's toughness has been reduced to avalue which is more characteristic of a brittle material.

This is partly the reason for the 'unzipping' of sub-sea pressurised pipelines, with disastrous conse-quences. Although these ductile pipes are designed via cylinder theory with the usual factors ofsafety against yield, the materials behaves in a brittle fashion at the propagation speeds involved.Furthermore, since :

- the crack advances at the speed of sound in the metal wall ( around 2 km/s ), and- a pressure pulse travels only at the speed of sound in the fluid, ie. less than the above

. . . . then it follows that the crack front is subjected always to the original pressure of the fluidand a leak is NOT self-relieving.

The Column Analogy

The behaviour of a crack of characteristic length 'a' and loaded by a background stress 'σ', is similarto that of a column of length 'L', loaded by the force 'F'.

columnIf ideal loading is wholly elastic then the load may beincreased, slowly and with complete safety, until it reachesthe critical Euler load, Fc, governed by elastic instability,whereupon catastrophe is immediate. If ideal loading iswholly plastic on the other hand then the material's yieldstrength limits the load that can be applied. Failure in practiceis governed by elastic-plastic interaction indicated by the dashed locus of the sketch for example.

crackIf ideal loading is wholly elastic then the load may beincreased, slowly and with complete safety, until it reachesthe critical LEFM load, σc (see below), governed by elasticinstability, whereupon catastrophe is immediate. If ideal load-ing is wholly plastic on the other hand then the material'syield strength limits the load that can be applied. Failure inpractice is governed by elastic-plastic interaction indicated by the dashed locus of the sketch eg.Additionally in the case of a crack the failure locus may be reached by various paths - for exam-ple the load may remain constant while the crack length increases in a controlled manner, untileventually a critical length is reached and the crack front advances catastrophically.

With the preceding background in mind, we can summarise what Fracture Mechanics does. It . . .. . . . presupposes the existence of cracks in the material, which may be

- microscopic ( grain cleavage or rough surface for example ), or- large ( casting or weld defects for example )- due to manufacture, to corrosion, to fatigue, etc, etc

. . . . correlates three parameters quantitatively- load - the background stress, 'σ', for example- geometry - the crack size, 'a' ( and to a lesser extent, shape )- material - its resistance to cracking, ie. its fracture toughness, Kc, measured by special tests

. . . . and predicts, amongst other things- degree of safety, or imminence of catastrophic ( brittle ) fracture

Eulerstability

l imit

COLUMN

yield limit

Lo

ad (

F )

L

Fc

L

F

F

LEFMstability

l imit

Loa

d (

σ )

yield limit

CRACKa

σc

a

σ

σ

transverse fracture suggests strongly that the deck plating fracturedinstantaneously in a brittle fashion with none of the ductile tearingwhich is evident elsewhere.

- The de Havilland Comet was the first jet-propelled airliner. It enteredcommercial service with BOAC in 1952 and was successful until aseries of unexplained mid-air disintegrations forced its grounding in1954 well before approaching the end of its expected fatigue life. Subse-quent examination of wreckage from the bottom of the Mediterraneanand fatigue pressurisation of a full size fuselage confirmed that explo-sive decompression was due to catastrophic advance of fatigue cracksin escape hatch surrounds.The state-of-the-art technical leader turned out to be a commercial disaster.

Fracture Mechanics currently offers the only satisfactory route to understanding these and similarproblems.

Stress Concentration

The effect of plasticity in way ofhigh stress concentrations in aductile tensile member is nowreviewed. The bar a-b-c-d-e, sub-jected to a uniform tensile loadover its ends, can be expected tofail first at e as the load is increased. Why? Because it is here that the geometric singularities are themost severe, and hence stress concentrations are highest.

Regarding nomenclature, if the load at some stage is 1 kN, then, in c, d, e :-σa = 103 / 5 = 200 MPa is referred to as the 'background' stressσb = 103 / 4 = 250 MPa is referred to as the average 'ligament' stress

Consider the notional stress distribution across the ligament of e under this load if the material iselastic/perfectly plastic with a yield strength of 400 MPa. The simplistic model of uniform stress isuntenable due to the stress concentra-tion at the root of the crack. If the mate-rial had been linear elastic then thisstress concentration could lead to apeak stress of say 500 MPa, as shown :-

However the material cannot supportstresses greater than yield, so the actualdistribution will be as indicated. The fact that some material has reached yield is not particularly del-eterious as it offers load support and is backed up by adjacent sub-yield elastic material.

If this material had been brittle with an ultimate strength of 400 MPa, then the peak stresses wouldhave corresponded to the linear elastic case and the bar would have broken at a load of 800N. 'Notchsensitivity' is a measure of a material's inability to shrug off localised stress concentrations.

Temperature / Rate Effects

These effects have a marked influence on a material's ability to withstand loads. Cracks and low tem-peratures are most unhappy bedfellows - as evidenced by many weld failures at low temperatures.The pressure vessel Code eg. makes special provision for welded vessels at low temperatures.

4 m

m2

5 m

m2

( a ) ( b ) ( c ) ( d ) ( e )

notch crack

UNIFORMSIMPLISTIC

LINEARELASTIC

ACTUAL

240 say 245 say

σ = 500 MPa σ = S = 400y

σ =

σ =

25

0b

^^

Fracture Mechanics 3

dominate, ie. σy → σ ; σx, τxy → 0.

- The stress components ( 1) are of the general form :- stress component = constant ( K ) * function of element location ( r, θ ) The constant is termed the stress intensity factor and embodies the geometry and loading. Korefers to the particular centrally cracked infinite plate arrangement considered above. K is anamplitude, a scale factor, a measure of the intensity of the singularity and the severity of thestress field immediately adjacent to the crack tip - and it is the near-tip field which directlyinfluences crack advance. If the stress intensity factor were to double, due to doubling of the load for example, then allstress components on all elements would also double. Note the implicit linearity here.

- A compressive background stress tends to close the crack and is therefore not deleterious ( aslong as it is steady ).

- Provided the loading is tensile, mode I, then forgeometries other than the infinite plate with cen-tral through-crack, and background loads otherthan a uniform tensile stress, the stress field closeto the crack tip is identical in form to ( 1) - sincethe near-tip region is not directly affected by dis-tant events. However the field strength, that is the stress intensity, KΙ corresponding to mode I,will reflect the particular geometry and loading, and will thus differ from Ko. A configuration factor 'Y'= KΙ /Ko reflects this difference; values of 'Y' for various loadings andgeometries are charted in the literature. Defining the normalised crack size as α = a/w,approximations for some common cases and for α ≤ 0.7 are shown here :-

- The above analysis would be completely useless if the aim were to calculate the maximumequivalent stress (in order to correlate with the yield to find the safety factor for example)because all stress components theoretically tend to infinity as r → 0. Stress redistribution pre-vents this happening in practice of course, but this is a plastic phenomenon, irrelevant to elasticanalyses. So, rather than using stress magnitude as a measure of loading severity ( and hencea basis from which to reckon safety ) we use stress intensity instead. The stress intensityfactor is one of a number of measures whose values describe the damage due to the crack -other common parameters are the 'crack opening displacement' ( COD, see below ), and the 'J-

mode III - KIIImode II - KIImode I - KI

P

aw

4w

-0.88)1 - 0.92Y =

1.12+ α (1.30αα

( a) Long strip, central crack, tensile stress

aw

σ σ

σ σ

2a2w

aw

M M

σa2wσ

( e) Finite plate, edge crack, bending ( SEN )( b) Long strip, edge crack, tensile stress

( d) Long strip, edge crack, pure bending

( f) Long rod, circumf'l crack, tensile stress( c) Finite plate, edge crack, tensile force (CTS)

K =I Y√πaσ

K =I Y√πaσ

K =I Y√πaσ

K =I Y√πa( P/bw )

K =I Y√πa( 6M/bw )2

K =I Y√πa( 6P/bw )

Y =1.12 + α ( 2.91 - 0.64)α

1 - 0.93 α

Y =5.23 + α ( 5.16 - 5.88)α

1 - 1.07 α

Y =1.12 + α ( 3.43 - 1.89)α

1 - 0.55 α

Y =1.12 + α ( 2.62 -1.59)α

1 - 0.7 αY = √sec( πα/2)

0.6w 0.6w

awP P

- crack growth rate whilst advancing in a controlled manner- component life which remains.

The application of Fracture Mechanics can thus be regarded assomewhat similar to the conventionaluse of design equations, viz :-

function ( material, load, geometry, “degree of safety” ) = 0however it should be understood that, at its present level of development, Fracture Mechan-

ics is less precise than conventional stress-strength-safety analyses. The predictions mentioned abovecannot therefore be viewed with quite the same confidence afforded conventional safety factors.

As noted previously, temperature and rate effects are both significant parameters in the crack propa-gation process. They affect the material's fracture toughness generally in the manner sketched.

The effect of a change of loading/crack growth rate is to move bodilythe fracture toughness curve paral-lel to the temperature axis. This,together with the shape of thecurve, means that a higher rate or alower temperature will decrease thematerial's fracture toughness.

Also indicated on the sketch are the regions in which the various failure mechanisms dominate.When the material's resistance to crack growth is low, elastic crack instability is the major failuremechanism; plastic effects are negligible and the material behaves in a brittle, linear elastic manneranalogous to a long Euler column. Fracture mechanics in the absence of plastic effects is known asLinear Elastic Fracture Mechanics ( LEFM ).

Conversely, if fracture toughness is high then failure is dominated by plastic yield - elastic instabilityis relatively insignificant as in the analogous case of a short column.

As with column buckling we shall start by examining linear elastic instability (LEFM) before intro-ducing the effects of plasticity then demonstrating simple techniques for handling mixed mode ( elas-tic-plastic ) fracture.

Linear Elastic Fracture Mechanics

A crack of length 2a extends right through an infiniteelastic flat plate of small thickness b. The plate is loadedby a background stress, σ, normal to the crack. Fromequilibrium, compatibility and the linear elastic constitu-tive law, the stresses in an element located at ( r, θ) closeto the crack tip, may be shown to be :-

The following points should be noted :

- The stress field ( 1) arises from the crack itself and not directly from the background stress. Ifthe element were well removed from the crack tip vicinity, then the background stress would

2ab

σ

σ

x

y

r

θσx

σy τx y

origin at leading edge of crack

σ

τ

σx

y

xy

θ/2sinθ/2sinθ/2sin

θ/2sinθ/2sinθ/2cos

3

3

31 +

1 –θ/2cos

√ 2πr= Ko σ √ πa+ . . . . ; K =o( 1 )

Temperature, T

high KNon-linear plastic,ductile behaviour

c

low KLinear elastic,brittle behaviour

c

Elastic-plastic(mixed mode)

Fra

ctu

re t

ou

gh

nes

s

Kc SLOW FAST

increasing rate

Fracture Mechanics 4

material - ie, its fracture toughness - is reached, whereupon catastrophic fracture occurs. This expla-nation and the above example neglect the real effects of plasticity in the ductile material, so plasticityeffects must now be examined.

Plasticity

Specimens of a given ductile, having standard proportionsbut different absolute size ( characterised by thickness ) giverise to different measured fracture toughnesses, as indicatedhere. Fracture toughness is constant for thicknesses exceed-ing some critical dimension, bo, and is referred to as theplane strain fracture toughness, KΙc. It is a true materialproperty, independent of size. As with materials’ other mechanical properties, fracture toughness istabulated in the literature though not so extensively as is yield strength for example.

Some typical values are tabulated here - from which increasing yield strength implies decreasingfracture toughness is evidently a general trend.

Below bo, toughness increases with decreasingthickness (within limits) and is termed the planestress fracture toughness, Kc. This behaviour maybe explained by the following argument.

Material within the crack tip stress field situatedclose to a free surface can expand laterally - in thez-direction of figure ( g) above. The state of stressthus tends to biaxiality and the material fracturesin a characteristic ductile manner, with a 45o

shear lip being formed at each free surface.

Material at the centre of the component however, is not free to deform laterally as it is constrained bythe surrounding material. The stress state thus tends to plane strain ( triaxiality ) and fracture in this

material yield toughnessMPa MPa√m

Aluminium 2024-T851 455 26 7075-T651 495 24

Titanium Ti-6Al-4V 910 115

*Ti-6Al-4V 1035 55

Steel 4340 860 99 *4340 1515 60 52100 2070 14

* heat treated for higher strength

Specimen Thickness, b

True material property,independent of thickness

bo

K Ic

Mea

sure

d T

ou

gh

nes

s, K

c

EXAMPLE The long rectangular bar is made from a material

whose fracture toughness is 60 MPa √m. During routine maintenance,

a 20 mm deep edge crack is found. Assuming LEFM, is it safe to

return the bar to service without repair ?

The load is not central here and we do not have the Y-factor for this

geometry/loading. However we note that tension and bending are

present, and that for each of these we do have the configuration factor. So, exploiting the inherent linearity between

stress (load) and stress intensity, we use superposition and calculate the stress intensities for the tensile and bend-

ing components separately, then combine ( add ) them :-

The computed stress intensity is

greater than the fracture toughness -

though theoretically it can't be, but

such an inaccuracy is typical of Frac-

ture Mechanics. Nevertheless, it is

clear that the bar was inspected just in

time and will have to be repaired.

20

20

60

100

250 N

+

tension σ = 125 MPa

case ( b)

Y = 1.36K = 42.7I

=

case ( d)

bending M = 2.5 kNm

Y = 1.05K = 19.8I

combined

α = 20/100 = 0.2K = 62.5IMPa √m

integral' ( an energy integral around a path enclosing the crack ).The material’s fracture toughness ( Kc ) can be recognised as just the critical maximum stressintensity ( K ) which the material can withstand without catastrophic crack propagation. Anadditional subscript ‘I’ refers to the most common tensile mode I loading. The distinctionbetween stress intensity , KΙ , a crack characterising parameter and fracture toughness, KΙc , amaterial parameter should be clearly understood.

The appraisal of safety in fracture mechanics is similar in principle to the determination of safetyfactor in traditional Strength of Materials contexts - a parameter expressing loading severity is com-pared to the maximum severity which the material can withstand ie. a material property. The twoappraisal approaches are compared here, though it must be pointed out that in Fracture Mechanics it

is more usual to cite the fatigue life which remains, rather than a factor of safety per se. Life aspectsare considered below.

There are two common forms of fracture toughness test specimen, the 'single edge notch' (SEN, case(e) above ) and the 'compact tensile specimen' ( CTS, case ( c) ). In both of these a crack is started bymachining a chevron notch in a specimen of standard proportions ( thickness b = w/2 ) then fatigu-ing the specimen to obtain a sharp crack front. The load - bending or tension respectively - is thenapplied and increased until catastrophe occurs. In principle, the maximum load in conjunction withthe configuration factor ( Y) corresponding to the final crack size enables the fracture toughness to becalculated. Testing in practice is not quite so straightforward as this might indicate, eg. we shall seethat size effects also have to be reckoned with.

The example below illustrates the appraisal of safety in thecontext of LEFM.

It is now possible to appreciate how fatigue occurs.

On the stress/crack-size diagram here, loci of constant stressintensity appear as hyperbolae ( σ√a = constant ) and a typi-cal fatigue trajectory under constant stress is horizontal. Aninitial crack will thus grow with corresponding increase instress intensity ( 1) until the critical stress intensity of the

aP

P

unstable

KIc

K I

aP

P

Sy

σ (P)

ε (δ)

unstable

TRADITIONAL SAFETY APPRAISAL

FRACTURE MECHANICSAPPRAISAL

Behaviour of specimen in standard test,and . . . .

derived material property S = ( σ ) criticaly K = ( K )criticalIc I

2a 2w load, σComponent under load, and . . . .

loading severity

load, FA

σ = F/A K = σ √πa YI

Safety estimate n = S /σy n = K / K IIc

fractureinitial crack

K increasingI

Kc

Ba

ckg

rou

nd

Str

ess,

σ

Crack Size, a

σ√a constantK constant ie.I

Fracture Mechanics 5

stress fracture toughness for use in the safety determination - but this toughness is not a materialproperty.

It is clear from ( 2) that the plastic zone size in front of a given crack must increase as the loadincreases, until eventually it extends right through the ligament as suggested below, so an upper limit

of the load which can be withstood by a cracked component will correspond to gross plastic collapseacross the ligament. Plastic copplapse loads can usually be found easily from equilibrium withoutrecourse to any fracture mechanics considerations, since for a perfectly plastic material any stressmust equal the yield stress, either tensile or compressive. The approach is demonstrated here :

We now examine Yielding Fracture Mechanics to see how we might handle the inevitable interactionbetween the two failure mechanisms of LEFM and plastic collapse.

Yielding Fracture Mechanics

We have examined the effect of load and crack size when the material is either perfectly elastic orperfectly plastic - but how does a real ductile behave ? There are various approaches to answering

EXAMPLE Estimate the critical loading on a long, 20 mm wide by 2 mm thick

strip in which an 8 mm central crack occurs. Properties are S y = 1.4 GPa, K Ι c = 50

MPa√m.

First estimate the minimum dimension necessary for plastic effects to be ignored.

From ( 2a) bo = 2.5 (50/1400)2*103 = 3.2 mm. Although this is less than the ligament

length of 6 mm, it exceeds the thickness of 2 mm, so plastic effects are not necessar-

ily negligible and we would expect both failure mechanisms - elastic fracture and

plastic collapse - to interact.

Letting the normalised crack length be α = a/w = 4/10 = 0.4, the critical load arising from each of the mechanisms

acting independently is as follows :

Plastic Collapse. Let σP be the critical load if plastic effects dominate; then from the free body above :

σP = Sy ( 1 - α ) = 1400*0.6 = 840 MPa

Elastic Fracture. Let σE be the critical load if elastic effects dominate ( ie. plane strain ), then :

K Ιc = σE Y √( πa ) where for case ( a) :- Y = √( sec α π /2 ) = 1.11 and so

σE = 50/( 1.11 √( 0.004 π) ) = 401 MPa don’t forget the units !

It would appear that elastic fracture plays the major role and that the critical load is somewhat less than 401 MPa -

though not much less because 840 >> 401. But we cannot be more definite at this stage.

8 20σ σ

Sy

Sy

σP

increasing load

Plastic collapse of the ligament - yield stress extends right across it

Sy

region is brittle. ( It will be recalled that shear failure tendency in ductiles is negligible under triaxialhydrostatic loading. )

If the size of a shear lip is assumed to be approximately constant for a given material, then the differ-ing behaviour of thin and thick components follows from there being insufficient material at thecentre of the thin component ( h) to exhibit significant constrained behaviour - so the material seemsductile with high toughness. As the thickness increases ( j) the shear lips occupy a decreasing propor-tion of the cross-section and so the material acts more and more like a brittle, with reducing tough-ness. Eventually the lips' effect is insignificant and the toughness assumes a constant minimum value- the plane strain fracture toughness, KΙ c.

Evidently the accurate determination of the plane strain fracture toughness requires a specimenwhose thickness exceeds the critical thickness, bo, and the question now arises - what is this criticalthickness for a particular material ?

To answer this, we first recognise that a plas-tic zone must exist in front of a crack, sincestresses are high - theoretically tending toinfinity at the crack tip. Setting the equiva-lent stress derived from the stress field ( 1) toequal the yield strength, the polar coordi-nates of the plastic zone boundary are foundto be :-

( 2 ) r = ro( (1-k)2 +3 sin2 θ/2 ) cos2 θ/2 where ro = 1/2π ( KΙ /Sy )2 = a/2 ( σ/Sy )2 in which

the constant k = 0 ( plane stress ) or k = 2ν ( plane strain ).

The proof of this is left as an exercise for the reader. The yield zones ( r,θ) plot out as shown on theright above, from which characteristic zone sizes are :-

( i ) ry ≈ ro ( plane stress ) and ry ≈ ro/3 ( plane strain, ν = 0.3 ).

Despite the questionable use of an elastic field to deduce the size and shape of a plastic zone, experi-ments confirm the above general findings. The derivation assumed perfectly plastic behaviour - tomake some allowance for strain hardening, Sy is often taken as the average of the yield and ultimatestrengths, whereupon it is referred to as the 'flow stress'.

For plastic effects to be negligible, other dimensions such as component width or ligament size mustbe at least 50 times the plane strain plastic zone size, so :-

( 2a) bo ≈ 50 ro / 3 ≈ 2.5 ( K Ιc / Sy )2

In testing for plane strain fracture toughness therefore, the toughness must first be estimated toenable prediction of the minimum specimen dimensions, via ( 2a). The specimen is then made andtested. If the measured toughness is less than that assumed initially then the measured value is valid.Otherwise a larger specimen must be tested. It sometimes happens that the required specimen size orforces exceed the capacity of available testing machines, however it is usually possible, and perfectlyvalid, to test a specimen of the same thickness as the eventual component - ie. to measure the plane

Shear lips occupy smallproportion of thickness.Brittle behaviour.low toughness.Triaxial stress state.Plane stress ductile failure

free to move laterally, σ = 0Shear lip - surface layers

z

Plane strain brittle failureCore constrained, ε = 0zFinal fracture

surface

xy

zb

Shear lips occupymost of thickness.Ductile behaviour.high toughness.Biaxial stress state.

( h) THIN ( j) THICK

( g)

Equivalent stressahead of crack

tip from ( 1)

plastic zone

σe

ry

Sy

∞

plane stressν = 0.3

planestrain

0.83

1.25

0 0.16 0.37 1.0x / ro

y / ro

Fracture Mechanics 6

Predicting the failure load in this way via the COD equation, is knownas the 'two criteria' or ‘CEGB R6' approach to failure assessment - thefailure locus from ( 3) plots as shown. Although ( 3) was deduced onthe basis of a centrally cracked infinite plate, it is applicable to any con-figuration provided the axes are interpreted as load ratios rather thanmerely stress ratios.

This approach is just one of many possible interaction models. Recall-ing the Column Analogy above and adapting the nomenclature usedwith imperfect columns, it is apparent that the R6 locus is a particularcase of the general interaction model shown below, which requiresθ → 1 as ψ → 0 and θ → 0 as ψ → 1. :-

Various empirical correlations such as ( 1/θ - 1) * ( 1/ψ - 1) = con-stant have been proposed - some of which describe particularexperimental results better than others. One model which finds wideacceptance is the 'circular' failure locus θ2 + ψ2 = 1, that is :

( 4 ) 1/σF2 = 1/σE

2 + 1/σP2

For an infinite plate KΙc = σE√( π a ) and σP = Sy as above, so thismay be written :

( 4a) σF = Sy / √( 1 + a/ a* ) in which

a* = 1/π ( K Ιc /Sy )2 is a material property like bo ( 2a).

The plotted trajectory ( 4a) is seen to be rather conservative :-

The examples below illustrate application of the foregoing interactionmodels to deduce approximately real elastic-plastic behaviour.

The interaction approach does not provide detailed understanding ofthe failure mechanism - for this one must revert to more complex teachniques - however it givessome insight into the role of plasticity in fracture mechanics.

The implication of critical crack size and limitations of crack size detecction during inspection mustbe appreciated. If a crack of the critical size remains undetected by the inspection process then noamount of inspection will avert potential catastrophe.

Fatigue Crack Growth

Steady loads only have been considered so far, for which we havenoted the effect of increasing crack size on failure tendency. We nowconsider crack growth under alternating loads, ie. the fatigue process.

A typical crack history under a varying load of constant amplitude,∆σ, is sketched. A crack of size a1 exists initially, and grows in astable, controlled manner until the critical crack size is approached -when the crack growth rate increases out of hand and disaster strikes.

For a given material, the instantaneous rate of crack growth, the slopeda/dN, is found to depend mainly upon the stress intensity range,∆K, as expected, since

- it is the near tip field ( characterised by K ) which affects crack advance, and

number of cycles, N

cra

ck s

ize,

a

fati

gu

e l

ife

fastfracture

a1

∆σ constant

aa*

σSy

0

yS√2

SAFE

σF

0 1

1

σ /σE

σ /σP

FAIL

SAFE

1

1

0

FAIL

loadψ = elastic failure load

loa

dθ

= p

last

ic f

ail

ure

lo

ad

this, ranging from simple interaction models between the two perfect mechanisms which we havealready examined, through to the 'J-integral' technique involving an energy integral around the cracktip, and three-dimensional finite element models with elastic-plastic constitutive laws. Let's see howwe might tackle the problem without the complexity ( or accuracy ) of these last methods.

In practice, the inevitable yield zone ahead of the crack leadsto a blunt crack tip. Thus, LEFM predicts a sharp tip to thecrack of length 'a' as indicated in figure ( k) here. Furtherconsiderations require a yield zone, of size 'ry' as shown inthe previous section. In actuality ( l), the tip is stretched byan amount 'δ' - the crack tip opening displacement or COD -which may be confirmed and measured experimentally.

Plastic effects are sometimes approximately allowed for, byapplying LEFM with an ‘equivalent crack size’, a ' figure ( m),equal to the actual crack size plus some proportion of the plastic zone size. This 'small scale yieldingcorrection’ is inadequate when the background stress becomes an appreciable fraction of yield.

δ, being strain-based, is a more suitable crack characterising parameter in the post-yield region, thanis K which is stress-based. It may be shown that, for a centrally cracked infinite plate, the crack open-ing displacement and the plastic zone size are :-

( ii ) δ = a 8/π Sy/E' ln sec ( π/2

σ/Sy ) ; where E' = E plane stress

ry = a ( sec ( π/2 σ/Sy ) - 1 ) or E' = E/(1-ν2) plane strain

Crack opening displacement is similar to stress intensity in that it is a function of crack length andload ( eg. background stress ), however it also takes material properties - particularly yield - intoaccount. The ratio of background stress to yield stress is particularly relevant as the following resultsfrom ( ii) illustrate :-

σ → Sy Tending towards plastic collapse; both δ and ry → ∞.

σ = 2/3 Sy The plastic zone and crack sizes are equal.

σ → 0 Tending towards LEFM (negligible plasticity) and enabling binomial expansion of ( ii):-

( iii) δ → π a σ2 / E' Sy and since this refers to LEFM, then from ( 1) :

→ KΙ2 / E' Sy

The advantages of COD as a crack characterising parameter should now be apparent - it is applicableover the whole range of loading, from linear elastic fracture to plastic collapse. One may expand onthe COD theme as was done for stress intensity - configuration factors for non-infinite shapes, a criti-cal maximum COD ( δc ) being a material property, and so on - however we shall restrict ourselvesto the adaptation of the COD equation to provide a simple interaction formula for correlating the twoextreme mechanisms noted above.

We take σE as the failure load when the mechanism is perfectly elastic ( LEFM ), and σP when themechanism is perfectly plastic ( plastic collapse ) - both of these can be computed by the methods ofprevious sections. To find σF , the failure load when both elastic fracture and plastic collapse occur,we proceed as follows :-

δc = a 8/π Sy/E' ln sec ( π/2

σF/Sy ) is the critical COD from ( ii) under failure load σF

= KΙc2 / E' Sy on equating critical properties from ( iii).

But for an infinite plate, KΙc = σE√( π a ) and σP = Sy, so the interaction is :-

( 3 ) π/2 ( σF / σP ) = arccos ( exp [ - 1/2 { π/2 ( σE / σP ) }2 ] )

( k)

( l )

( m)

δ

a

a'

ry

Fracture Mechanics 7

per cycle ( 4 E-7 mm/cycle ); growth requires the stress intensityrange to exceed some threshold value; influenced greatly by envi-ronment. This stage is byepassed if a crack exists prior to loading(eg. the point 1 sketched may be the initial state ).

II Stable Propagation. This is the most important stage, dealingwith identifiable cracks (say > 0.1 mm ) growing in a stablemanner. The direction of propagation is less random than in stageI and the material behaves more homogeneously. The materialcharacteristic crack growth rate vs intensity range is approxi-mately log-log linear throughout stage II and so we may write :-

( 5 ) da/dN = C ∆Kn - the so-called Paris equation 'C' and the index 'n' are constant material properties. Selected values appear in the liter-ature, it being customary to cite 'C' indirectly by the stress intensity range, ∆Ko, which corre-sponds to a certain crack growth rate, (da/dN)o - often 1 mm/Mc - thus C = (da/dN)o / ∆Ko

n

III Instability. Although important, this stage exists only for a very small fraction of the compo-nent's life, since the instability is catastrophic. The onset of stage III is dictated by the criticalcrack size being approached, that is by Kmax tending to Kc - 2 in the sketch. One empirical modification of ( 5) which caters for both stages II and III, is :

( 5a) da/dN = C ∆Kn / { 1 - ( Kmax/Kc )n }

Thus if Kmax<< Kc, then the RHS denominator → 1 corresponding to stage II; alternatively ifKmax ≈ Kc then da/dN → ∞ ( stage III ).

Integration of ( 5a) yields the component life ( ∆N12, cycles ) which elapses during crack growth froma1 to a2, thus, in normalised form :

( 5b) { (da/dN)o ∆N12/w } { ∆σ√( πw)/∆Ko }n = ∫12( Y√α)-n dα - ( Yc√αc)-n ( α2 - α1)

in which αc is the normalised critical crack size corresponding to Kc, and α2 ≤ αc.

The Pascal program Crack Growth, exemplified below, enables integration along these lines; itassumes that the configuration factors, both elastic and plastic, are single functions of crack size.

When plane strain conditions are assured, then Kc ≡ KΙ c in the foregoing; otherwise it is suggestedthat the Kc predicted by the R6 or other elastic-plastic technique should be used.

It is usually the initial, rather than the final state of affairs which has largest bearing on componentlife, as is demonstrated by the next example.

Crack Growth Kinetics

It was stated that Fracture Mechanics leads to a better understanding of the crack propagation mech-anism than conventional fatigue arguments can provide, and without the need for the latter's empiri-cism. The preceding discussion has borne this out, but some aspects remain which were notexplained satisfactorily :

- the rate and temperature dependency of fracture toughness- the sigmoidal shape of the fatigue velocity curve which requires Paris' empiricism - the effect of corrosion in exacerbating crack formation and early growth . . . etc.

Greater appreciation of crack growth will follow from examination at the atomic scale, as, when allsaid and done, it is here that temperature effects for instance make themselves felt.

The simplest model at the atomic level involves an isolated pair of interacting atoms which behavesas a non-linear spring. The force between the atoms, F, is the net effect of an attraction and short

I

II

III

thre

sho

ld

fast

fra

ctu

re

log

da/

dN

log ∆K

1

2

1n

K → K c

- fatigue is known to be greatly dependent upon the range of stress and of load ( the 'S' of the S-N diagram ) which is proportional to the range of stress intensity, as suggested here :

Growth rate da/dN is affected also by the mean component of intensity - typically characterised bythe load ratio R = Kmin/Kmax - but this is largely a reflection of Kmax approaching the critical Kc , ie.of impending fracture.

Crack closure, if and when the load becomes compressive, is usually allowed for analytically bysimply ignoring any sub-zero compressive excursions. Further evidence suggests that as the loadchanges from tensile to compressive, the crack closes while the load is still tensile - a result of resid-ual stresses; however we shall neglect such complications here. So, accepting that intensity range isthe main contributor to growth rate, the relationship is found to be as follows - this being a uniquesigmoidal curve for each particular material. There are three stages :-

I Initiation. Relates to cleavage along grain boundaries at rates of the order of one lattice spacing

∆σ

σ

σ timestre

ss (

load

)

t ime

∆KK

K

K c

stre

ss i

nte

nsi

ty All K-parametersincrease withcrack size :

K = ∆σ Y √( πa)∆

K = σ Y √( πa)

K = σ Y √( πa)

EXAMPLE Complete the last example - find the failure load given the previous results σE = 401, σP = 840 MPa

Inserting these two values ino the interacton models and solving for the failure load gives :-

CEGB R6 model - from ( 3 ) σF = 382 MPa.

‘circular’ model - from ( 4 ) σF = 362 MPa - evidently conservative cf. R6

EXAMPLE Using the same 20*2mm long strip, find the critical crack length for a load of 250 MPa.

This is the reverse problem to the above - a closed form solution is impossible, so a graphical or trial-and-error

approach must be adopted - thus :-

Trial normalised crack size α 0.1 0.2 0.3 0.4 0.5 0.6 0.63 0.7

σE = K Ι c/( Y√( πwα)) corresponding to α 887 615 486 401 335 279 263 227

σP = Sy ( 1- α ) corresponding to α 1260 1120 980 840 700 560 518 420

σF from ( 3) corresponding to σE and σP 800 578 462 382 320 265 250 214

The stress ratios computed from this table are plotted to illustrate the fatigue

locus as crack size increases.

The safety factor at any point, p, on the locus is the ratio 0p'/0p.

Failure occurs when σF /σP = 250/518 = 0.483 here, corresponding to α = 0.63

- the critical crack size is therefore 0.63*20 = 12.6 mm

Repeating this for comparison purposes using the more conservative ‘circular’

model . . . . . Despite this model’s equation being simpler than the R6, a closed

form solution is impossible and the above table must be repeated using ( 4)

rather than ( 3). Thus at α = 0.6 from the table and ( 4) σF = 1 /√( 1/2792 +

1/5602) = 250 MPa

The critical crack size predicted by the ‘circular’ interaction modesl is therefore 0.6*20 = 12 mm - not significantly

different from the R6 conclusion.

SAFE

p

p'

increasingcrack size

0.483σ

F

0 1

1

σ /σE

σ /σP

FAIL

Fracture Mechanics 8

2γ = Uo/so2 = (1/so

2)∫so

∞ F ds = so ∫0

∞ σ dε ≈ 2soσmax

2/E from this sinusoid.

Rearranging this equation gives the fracture stress according to the model as σmax = √( γE/so ) - ofthe order of E/10 since γ is around Eso/100 for many materials. Experimental values are often acouple of orders of magnitude below this, indicating severe shortcomings in this simple model.

In an attempt to explain this disparity, Griffith argued from an energy viewpoint, recognising thatthe impetus for crack extension is the energy balance between that absorbed in creating new surfacesand the elastic potential released by a larger crack. Thus, if a through-crack of length2a occurs in a large plate of unit thickness, transverse to a background load, σ, thenlines of force suggest that two triangular areas become stress-free, as indicated by thesketch. Letting the triangles' height be βa, then for half the crack :

gain in surface energy = surface tension * crack surface area = γ.2aloss in elastic energy = specific elastic energy * stress free area = (σ2/2E).βa2

A more sophisticated analysis reveals that β = π in plane stress. The netenergy available for driving the crack is just the difference between theseenergy transfers - which depend on crack size as shown. The net energyexhibits an unstable equilibrium at a certain critical crack size, ac, given by :

d( 2γa - πσ2a2/2E )/da = 2γ - πσ2ac/E = 0 ie. πσ2ac = 2γE.

Cracks larger than the critical will grow out of hand, to reduce the systemenergy. The critical stress predicted by this equation is σmax = √( 2γE/πac )which is much more realistic than that of the static lattice model above, sinceac >> so. Despite the neglect of plasticity in this Griffith model ( though plastic energy may be incor-porated at the cost of extra complication ), its focus on energy, rather than on 'elastic' phenomena atthe crack tip, is of great significance.

Further modelling realism can be gained by reverting to the lattice above, but recognising that it isnot static since all atoms are in continuous motion with random vibrations about their mean latticepositions. Concentrating on an isolated atom pair at the crack tip for simplicity, we have :

( 6 ) U = H + W where H is random vibrational energy which is perceived macroscopically as thermal energy, W is energy due to work done by external loading on the pair, and U is the total energy of the pair - clearly this also varies randomly.

As changes in vibrational amplitude sweep through the material with wavelike motion, sooner orlater the atoms at the crack tip will be host to a crest which is large enough to result in U attaining avalue of Uo, whereupon the bond breaks and the crack advances byone step, that is, by so. Diagrammatically, for the crack tip region,crack advance by one step needs sufficient energy to be input to hoistthe total energy over the hill of height Uo into the next stable energyvalley, where the energy level is increased by the energy inherent inthe newly created free surfaces, ∆Us.

One fundamental aspect of atomic behaviour is the possible reversalof processes, given appropriate directions and magnitudes of the random vibrations. Thus a pair ofwell separated atoms may be juxtaposed to create a bond and the crack is partially healed. Whilstthis rebonding occurs on an atomic scale, macroscopic crack healing is uncommon (for reasons givenbelow) though not unknown under controlled conditions.

These concepts may now be extended from the simple, essentially one-dimensional near-tip systemconsisting of a few isolated atom pairs, to the real three-dimensional lattice. It is possible to describe

range repulsion, while the interaction energy, U, is represented by thearea under the F-s characteristic ( dU = F.ds ). The force and energy varywith atom separation, s, as shown. At the equilibrium spacing, so, the netforce vanishes and the energy exhibits a stable minimum, Uo - the workrequired to break the bond between the atoms, ie. to separate them toinfinity.

A very simple model of a material's microstructure is a static lattice ofsuch atom pairs, in which the influence of adjacent atoms only is recog-nised. Crack advance demands progressive breaking of the inter-atomicbonds and the formation of two

free surfaces. An atom on such a surface has no neighbours onone side, and, as bond energies are negative, its energy exceedsthat of interior atoms by Uo.

A sinusoidal approximation to the F-s characteristic near equi-librium is indicated on the sketch above. Since the stress andstrain in the loaded material are σ = F/so

2 and ε = (s-so)/so, thestress/strain behaviour may be taken as : σ = σmax sin (Eε/σmax) to ensure an initial slope of E. The energy density on eachof the two surfaces of a crack - the surface tension, γ - is given by :

EXAMPLE The 20*2 mm centre-cracked strip of the previous examples is loaded by a stress which varies cycli-

cally between 150 and 250 MPa. The material's Paris index is 4.4, and a stress intensity range of 5.7 MPa√m causes a

crack growth rate of 1 mm/Mc.

Integrate the Paris equation ( 5) to determine the number of cycles required for a crack to grow from 0.2 to 2 mm,

and from 2 to 20 mm. Repeat, with a more realistic provision for instability.

The integral of the basic Paris equation ( 5) is ( 5b) with the last term negligible ( αc → ∞ ).

With normalised crack size limits of 0.01 and 0.1, the integral ( 5b) may be evaluated numerically, noting that w =

10 mm or 0.01 m and that each term in { } is dimensionless :-

{ 1 * ∆N12 / 10 }* { (250-150)* √( 0.010π ) / 5.7 }4.4 = ∫0.010.1( ( cos πα/2 )/α )2.2 dα

{ ( mm/Mc) Mc /mm } { MPa . √(m) / MPa√m } = 195.7 → ∆N12 = 13 Mc

In the second case, with limits of 0.1 and 1, the integral is = 10.25 → ∆N12 = 0.70 Mc

Despite advancing 18 mm in the latter case, compared to only 1.8 mm in the former, the period of the advance has

been reduced by a factor of about 20. This reflects the accelerating growth rate.

A prior R6 analysis of this model, with a peak stress of 250 MPa, has shown that the critical normalised crack size is

0.63, with corresponding elastic configuration factor of 1.35.

So, applying ( 5b) to the first case, the RHS becomes : 195.7 - ( 1.35 √0.63)-4.4*( 0.1 -0.01).

Clearly the integral and hence the period is unaffected by the modification - because the period lies wholly within

stage II.

For the second case, the upper limit must be reduced to the critical 0.63. The RHS of ( 5b) becomes :

∫0.10.63 ( ( cos πα/2 )/α )2.2 dα - ( 1.35 √0.63)-4.4

* (0.63 -0.1) = 10.2 -0.39 → ∆N12 = 0.67 Mc

The introduction of the more realistic upper bound has not made much difference to the period - simply because

the advance is intrinsically so rapid in the last phase of failure.

In practice, closed form integration is usually impossible; graphical techniques may be used.

Fracture Mechanics 9

plots, as shown here for the two separate processes, 1 and 2.

If the two processes are not independent, but interact in promoting growth, then their net effectdepends upon whether they act simultaneously (parallel) or sequentially (serial). Thus, given theindividual process velocities v1 = λ1/τ1, v2 = λ2/τ2 the velocity, v, when they act in concert is :

Simultaneous if the waiting period is τ = τ1 = τ2, then the step size is λ = λ1 + λ2 = v1.τ + v2.τ and = v.τ ; so

( 8a) v = Σ vi

Sequential if the step size is λ = λ1 = λ2, then the waiting period is τ = τ1 + τ2 = λ/v1 + λ/v2 and = λ/v ; so

( 8b) 1/v = Σ 1/vi

Plots of the velocity due to two interacting processes indicate that thelarger of the two velocities controls when the processes are simultaneous; the lesser of the two veloci-ties controls when the processes are sequential. Crack growth mechanisms involving an arbitrarynumber of known processes, simultaneous and/or sequential, can be described in a similar manner.

Crack growth processes only have just been considered - but what happens when breaking and heal-ing processes act together ? Noting that these processes counteract one another, then, using (7), thenet velocity is :

v = vb -vh = λ( 1/τb - 1/τh ) = λ (kT/h) { exp( - ( Ub -Wb )/kT ) - exp( - ( Uh +Wh )/kT ) }

An unloaded crack will not grow. Since it is unloaded, Wb = Wh = 0, and the absence of macroscopicgrowth requires Ub = Uh so that v = 0. This does not imply that atomic bond breaking does not occur;it does, but bond healing progresses at the same pace, so there is no discernable growth. The sketch

illustrates such a situation, with load characterised by stress intensity. Asthe 'load' grows, by increases in background stress or crack length, theeffect of the healing process rapidly diminishes, and the net effectapproaches that of a single bond breaking process. Such behaviour is typ-ical at threshold.

An example of the application of fracturekinetics is shown. It involves isothermal fatigue test results of a certainsteel in a corrosive environment, at various loading rates. The fatiguemechanism here apparently involves three processes, I, II and III, with Iand III being unaffected by loading rate, whilst process II is rate-dependent. Processes I and II appear to be sequential; while processes IIand III are simultaneous. The velocity due to the three can thus be for-mulated variously via ( 8) as :

v = 1/( 1/vI + 1/vII ) + v III or= 1/( 1/vI + 1/( vII + vIII ) ) ; vII = function( frequency)

where each process velocity is of the form vi = exp( ai + bi.∆K ).

This example demonstrates the practical relevance of Fracture Kinetics - by its use it is possible toexamine the individual processes involved in any given crack propagation mechanism, and to pre-dict their combined effect. Fracture Kinetics is not merely a curve-fitting exercise, however the rea-soned application of the technique will not be explored further here - the literature should be con-sulted for fuller details.

In conclusion, it has been said before, and is particularly relevant to Fracture Mechanics - a mathe-matical model is all very well, but engineering judgement remains a sine qua non. Non-mathematical

such a large multi-body system only in probabilistic terms, ie. via statistical mechanics.

Rate theory, as used to describe transport processes - heat, mass and momentum transfer for example- is applicable also in the present context. The following account makes no claims to rigour, for thisthe literature must be consulted.

A crack advances by a series of steps; its average velocity may be expressed as :

v= λ/τ where λ is the average step length, an integral number of atomic distances, so

τ is the average waiting time between successive steps - in the literature, its inverse isreferred to as a 'rate constant'.

One of the prime aims of fracture kinetics is to identify and quantify the individual processes whichcontrol any particular crack's behaviour. This is accomplished by evaluating the velocity peculiar toeach process - one velocity might appertain to crack growth in a certain corrosive environment,another to a particular crack advance mechanism, and so on - then combining the velocities in amanner appropriate to the sequence of the processes, be they simultaneous, or sequential, or what-ever. Since energy fluctuations govern any transport process at the atomic level, the correspondingwaiting time must be dependent on and described statistically in terms of thermal energy, thus forthe i'th process :

( 7 ) τi = ( h/kT ) exp( Hi / kT ) where k, h are Boltzmann's, Planck's constants ( 1.38 * 10-23 J/K, 6.62 * 10-34 Js ) respectively,T is the absolute temperature ( K ), andHi is the thermal energy ( J ) level necessary to activate the i'th process.

It is apparent that temperature effects, which have been noted previously as playing a major role infracture, are automatically included in ( 7). Further analysis demonstrates that temperature, in theusual range encountered, is significant only in the exponential term - variations of the (h/kT) pre-multiplier in ( 7) are swamped - and the equation takes Arrhenius' form. But it should be stressedthat while Arrhenius equations are empirical, the equations developed here are based upon theknown behaviour of real processes.

For bond breaking, ie for the crack growth process, from ( 6) :Hb = Ub - Wb

in which the 'b' subscript refers to the breaking process. For the heal-ing process (subscript 'h'), external work which tends to open thecrack will oppose this process, so :

Hh = Uh + Wh ( ie. W is negative in ( 6) )

The RHS of these last two equations replace Hi in ( 7) applied to the appropriate process. Ui takes theform of a constant energy magnitude representing the total energy crest - again appropriate to theprocess. Wi expresses the overall loading; in LEFM, this term would be a function of stress intensity,K, while plastic behaviour of the crack would necessitate a function of COD or the J-integral. Thebreaking and healing processes are shown diagrammatically above. Due to the senses of the workterms and consequent energy crests, breaking is more likely than healing.

Neglecting variations in the premultiplier of ( 7), and taking stress inten-sity as a measure of loading for example, the waiting time for the i'th pro-cess is of the form :

τi = exp( ai + bi K ) ; ai, bi being temperature dependent.

Within the context of LEFM therefore, the effects of isothermal crackgrowth promulgating processes are linear log(velocity) vs stress intensity

Fracture Mechanics 10

Bibliography

Blake A, Practical Fracture Mechanics in Design, Dekker 1996

Cruse TA (ed), Fracture Mechanics XIXth National Symposium, ASTM 1988

Ellyin F, Fatigue Damage Crack Growth and Life Prediction, Chapman Hall 1997

Gray TGF & Spence J, Rational Welding Design, Butterworths 1982

Knott JF, Fundamentals of Fracture Mechanics, Butterworths 1973

Knott J & Withey P, Fracture Mechanics Worked Examples, Inst Materials 1993

Krausz AS & Krausz K, Fracture Kinetics of Crack Growth, Kluwer 1988

Larsson LH (ed), Elastic-Plastic Fracture Mechanics, Kluwer Academic Press 1985

Meguid SA, Engineering Fracture Mechanics, Elsevier 1989 a useful extension of the NotesS

Rooke DP & Cartwright DJ, Compendium of Stress Intensity Factors, HMSO 1976

Solin J et al ed, Fatigue Design, MEP 1993

Stephens RI ed, STP 1250: Case Studies for Fatigue Education, ASTM 1994 invaluable insight into prac-tical application of fatigue/fracture techniques to a variety of components

The following are required reading before trying to apply Fracture Mechanics in practice :-

Failure Analysis and Prevention, Metals Handbook v 11, 9ed, ASM 1986

Hertzberg RW, Deformation and Fracture Mechanics of Engineering Materials, Wiley 1989

A typical dialogue with the Pascal program Crack Growth into which the user has incorporated con-figuration factors relevant to the problem in hand :

************ Component lifeCRACK GROWTH from a modified************ Paris equationTitle : Example from Fracture Mechanics NotesEnter, for the material -

yield strength : 1400 Note consistent MPa, m unitsplane strain toughness : 50Paris index : 4.4reference growth rate : 0 Default 1 mm/Mccorresponding K-range : 5.7

and, for the configuration -component reference width : 0.010cyclic stress limits : 150 250

so, critical normalised size 0.629normalised crack limits : 0.1 1.0

so, component life is 6.65 Ee+5 cyclesEnd of program

considerations which have not been mentioned here include the very significant implications of met-allurgy; the very real possibility of changing crack direction, or of mode or flaw interaction; the limi-tations of non-destructive testing; the difficulties associated with cumulative fatigue, and so on. Forexample how would one tackle a situation in which a crack spends its early days in the shadow of astress concentration due to some geometric feature of the component; but on maturing, the crack's tipemerges from the shade ?

Fracture Mechanics 11

10 Welded plates, 10 mm thick, are subjected to bending as shown. Crude manufacture leads tothe expectation of 2 mm cracks extending right along the weld root. Multiple service failures occur when the deposition properties are as (b) below. Would a change to (a) or to (c) alleviate the problem ?

deposition (a) (b) (c)yield ( MPa ) 600 800 1000toughness ( MPa√m ) 120 90 60

11 A pressure vessel, of bore 850 mm and wall thickness, w =24mm, is designed with a safety factor of 2.5 based on theyield of 500 MPa. The material's fracture toughness is 50MPa√m. A semi-elliptical longitudinal fatigue crack ( a= 10,b= 20 mm) is discovered at the bore during routine inspection.What is the actual safety factor of the flawed vessel. [ 1.26 ]

If the toughness were to drop to 35 MPa√m due to a drop inambient temperature, what then would be the safety factor ?Assume configuration factors - plastic as graphed, and elastic'Y' thus : Y = C1( 1 + C2 ( 1 -cos πα ))

where C1 = 2.24 /(1+exp πβ/4) ; C2 = 2.34/(1+exp 4√β) ; β = a/b

12 Derive equation ( 5b) from ( 5a).

13 An aluminium shaft, 50 mm diameter and rotating at 3000 rpm, is subjected to a reversedbending moment of 200 Nm. A crack, 0.1 mm deep, extends radially from the surface. Estimatethe crack depth after 100 hours operation, assuming a Paris exponent of 2.7 and a stress inten-sity range of 1.6 MPa√m corresponding to a growth rate of 1 mm/Mc.The configuration factor may be approximated by that of case (d). [ 11 mm ]

14 An axially stressed component of width 20 mm is made from a material which obeys the Parisequation with an index of 4 and a crack growth rate of 1 mm/Mc corresponding to a stressintensity range of 6 MPa√m. The configuration factor may be approximated by Y = 0.84/( 1 -α).

Neglecting instability, determine the number of cycles necessary for a 5 mm crack to grow to 15mm, if the component is subjected to a cyclically varying stress of :(a) 0 to 40 MPa ; (b) 100 to 120 MPa. [ 2.1, 34 Mc ]

15 Repeat the previous problem with KΙ c = 60 MPa√m. [ 2.1, 32 Mc ]

16 Estimate the life of the component of the previous problem with an initial crack size of 5 mm, ifthe material yield is 250 MPa. Assume a plastic configuration factor of ( 1 - α ). [ 2.1, 27 Mc ]

0.5 1.00

0.5

1.0

1.0

00.20.6

a/w

a/b

σPSy

a

w

2b

a

M

M

FRACTURE MECHANICS - PROBLEMS

1 A large sheet containing a 50 mm long crack fractures when loaded to 500 MPa. Determine thefracture load of a similar sheet with a 100 mm crack. [ 354 MPa ]

2 Rocket motor casings may be fabricated from either :(a) low alloy steel : yield 1.2 GPa, toughness 70 MPa√m, or(b) maraging steel : yield 1.8 GPa, toughness 50 MPa√mThe relevant Code specifies a design stress of yield/1.5.Calculate the minimum defect size which will cause brittle fracture in service for each material, and comment on the result (this is important). [ 4.9, 1.1 mm]

3 The bar of 100 *20 mm rectangular cross-section is loadedby a force of 250 kN as shown. Determine the critical cracklength if the toughness is 50 MPa√m. [ 14 mm ]

4 Derive equation ( 2 ) by applying the distortion energy fail-ure criterion to the stress components of ( 1 ).

5 The CTS testpiece is from a 1.2 GPa steel. If the failure load is10 kN, what fracture toughness is indicated ? Is the result valid ?Note that width and crack size are reckoned from the load'sline of action. [ 80 MPa√m ]

6 The toughness of a 700 MPa yield structural steel is estimated to be 140 MPa√m. What size andmass of SEN bend test specimen is necessary, and what capacity of testing machine would berequired ? Assume fracture at α = 0.5. [ 126 kg, 590 kN ]

7 The long strip may be made from either of the two materials :(a) tough, weak : yield 700 MPa, plane strain toughness 100 MPa√m(b) brittle, strong : yield 1400 MPa, plane strain toughness 50 MPa√mA central crack extends through the strip. Plot, as a function of crack length,the failure stress for each material due to the separate mechanisms of elasticfracture and plastic collapse. Comment on the trends of these graphs.

8 The bar of rectangular cross-section, w *b, is edge-cracked and loaded by a tensile force, N, and bendingmoment, M. Consider the equilibrated distribution ofyield stress across the ligament and hence show thatplastic collapse may be caused by any combination ofM and N which satisfies :- m + n ( n +2α ) = ( 1- α )2 where α = a/w ; n = N/bwSy ; m = 4M/bw2Sy

9 A long, 50 mm diameter rod is manufactured from a material of 700 MPa yield and 40 MPa√mtoughness. The rod is circumferentially cracked, case ( f), whilst tensioned by a force 'P'.What is the maximum safe load if the crack depth is 2 mm ? [ 770 kN ]If the load is 200 kN, what crack depth is tolerable ? [ 12 mm ]

M

N N

M =

=

b

aw

52a

20

σ

30

9

10

60

100

20

250kN

a

Finite Elements 1

FINITE ELEMENTS

The vast majority of shapes which occur in Engineering - whether they be chunks of stressedmetal or volumes of flowing gases - are complex three-dimensional continua which cannot berepresented adequately by the simple closed-form mathematical models so beloved of engineer-ing students. These theoretical models also cannot portray the appreciable non-linear or aniso-tropic material characteristics which are often met with in practice. For example, although com-plex mathematical transformations enable the flow of ideal fluids around certain aerofoils to beanalysed, the very real effects of viscosity may render such analyses impracticably inaccurate.

In such situations therefore, we must resort to numerical methods in which the prototypicalcontinuum of infinitesimal material particles is represented by an approximately equivalentassembly of inter-connected discrete elements which are each so simple that they can be treatedindividually as mathematical continua. There are a number of methods whereby such networkscan be analysed - numerical solution of differential equations, finite differences, finite elements,boundary elements, relaxation techniques, and so on. We have chosen to demonstrate theFinite Element Method (FEM) as a typical powerful approach which can handle equilibrium,eigenvalue and propagation problems - though we shall restrict our considerations to equilib-rium applications in linear elasticity.

Finite Elements involve a particular type of network. A network is an arrangement of 'ele-ments' interconnected between 'nodes', such as an electrical system in which the elements areresistors, batteries, etc, or a fluid network where the elements may be pumps or fans, pipes,valves or mine airway resistances, and so on. Networks abound also in non-engineering disci-plines - in economics, in political and the social sciences, and in decision theory to name but afew. This ubiquity is sufficient reason to first examine networks in their own right to see howbest to analyse them - this examination will lead naturally into the concept of Finite Elements.

Linear 1-Networks

The sketch illustrates portion of a net which consists of two-noded elements, 'M' in number and labelled 1, 2 . . m . . . M,arranged between 'N' nodes which are indexed 1, 2 . . i, j . . . N.The typical m'th element, lying between the i'th and j'thnodes, is also shown. The net variables are :-

'element' or 'through' variables, 'y' say, exemplified by the cur-rent or flow, ym, through the m'th element, and

'node' or 'across' variables, 'x' say, typified by the voltages orheads, xi and x j, at the i'th and j'th nodes respectively.

Analysis of the net as a whole requires determination of the complete vectors :-

( 1 ) x = [ x1 x2 . . . . xi xj. . . xN ]' ; y = [ y1 y2 . . . . ym . . . yM ]'

The solution involves three major and quite discrete aspects :-

A - TOPOLOGY of the net, which in turn involves :DEFINITION of the net - the labelling of the nodes and elements, and the statement ofwhich elements are connected between what nodes.ORIENTATION of the elements - the nomination of a positive flow sense through eachelement; as a result of this, one of the two nodes will be the inlet node to the element,the other becomes the exit.

Finite Elements 2

CONSTRAINTS - the definition of the operable network laws; eg Kirchoff's Laws :

Σ ym = 0 at any node, eg flow continuity - or its dualΣ ( xi - xj ) = 0 around any closed loop.

Some observations about these laws include the following :- one set only is necessary; we shall use only the continuity equations here- the laws are intrinsically linear, eg no squares of the variables appear- the formulation of the equations depends only on the topology, eg the equations for

electric and fluid systems are identical.

B - CONSTITUTIVE LAWS which relate the across- and through-variables for the elements. These are known, application-dependent equations of the form :-

xi - xj = function ( ym )

Eg, an electrical resistance : v i - vj = Rm im Linearor an hydraulic resistance : h i - hj ≈ Rm qm

2 Non-linear

The constitutive laws may not be algebraic equations, but differential or integral equa-tions as occur in dynamic systems or inductor/capacitor nets. Note the consequences of orien-tation when the constitutive laws are non-linear - what happens if hi <hj ?

C - SOLUTION OF EQUATIONS which result from combination of the constitutive and net-work laws. Although this might be thought the easiest step, especially if the network islinear, the desired accuracy may be difficult to achieve economically if the net is large.

Network analysis will be demonstrated only for linear systems of the type shown above, inwhich the elements are two-noded - that is the net can be regarded as a series of interconnected1-dimensional lines. Such a net is known as a 1-net.

Analysis will first be carried out for a representative net using a familiar algebraic technique.This will give the background for a more general matrix approach which is the theoretical basisof a computer-oriented method for assembling the net equations - the Direct Assembly method.

EXAMPLE Determine the across-variables, x, and the through-variables, y, for the typical network illustrated at(a) below. Constitutive laws are :-

Passives ym = am ( xi - xj ) where the a-constants have the values shown in (a)Active y = 2 ( xi - xj ) + 70 a pump or battery with orientation as sketched ( xi ≤ xj ); also see (b).

The specified value of the across-variable is xo = 5 at the specified datum node.

Firstly the nodes, and then the elements, are numbered sequentially in any order and a positive sense nominatedfor the through-variable in each passive, ie the orientation of the net elements is completed, sketch (c).

The constitutive laws for all elements are written in an ordered manner, equations (A) below. Although the speci-fied datum value x4 = 5 is known, the value's substitution is delayed.

Finite Elements 3

The continuity equations, (B),are written for all the nodes, pre-serving the same order of nodesand of elements as in the previousstep.

Any one of these equations (B) isredundant, but its deletion isdelayed.

The constitutive equations (A)are inserted into the continuityequations (B), producing (C),whose across-variable terms arecollected to yield the symmetricset of 5 continuity equations (D)in the 5 nodal variables.

The known datum, x4 = 5, is sub-stituted in (D) and the columntaken to the RHS, to give the set(E), from which the superfluous

continuity equation corresponding to thedatum index ( the fourth here ) is elimi-nated, leaving the symmetric set of ( N - 1 )independent equations (F).

This is solved for the across-variables, yield-ing :-

x1 = 11 x2 = 9 x3 = 4 ( x4 = 5) x5 = 1

Inserting these values into the constitutive laws(A) for the through-variables gives :-

y1 = 10 y2 = 28 y3 = 12 y4 = 42y5 = 50 y6 = –4 y7 = 22 y8 = 8

The initially assumed sense of the through-variable in element #6 was wrong; this is correctedin the sketch of the final solution below :-

Finite Elements 4

This example is now repeated using matrix notation, assuming that the first step involvingannotation and orientation has already been completed.

The constitutive laws of the M=8 elements, involving the N=5 nodes, are as follows. Zeromatrix elements are represented by periods for legibility.

or, in brief :

( A') y = a t x + yo

in which the unknown vectors, x and y, are as defined in (1), and the knownmatrices a and t and the vector yo are defined to be :-

- a is an M*M square diagonal matrix of admittance constants; all off-diagonal matrix elementsare zero. The ordering of the constants in a is the same as that of the net elements in y.

- t is an M*N topology incidence matrix, which identifies those two elements of the across-variable x which are relevant to each net element; so each row of t consists of a single +1 anda single -1, all remaining columns being zero.

- yo is an M-vector of specified through-variables, with non-zero elements corresponding to thenetwork active elements, or sources.

The N = 5 nodal continuity equations are

( B') t' y = 0 or, in full :-

in which the y pre-multiplier is just the trans-pose of the topology matrix t, due to the equa-tions being written down in an orderly man-ner. Combining the two sets of equations as was done algebraically above, by inserting (A') into (B') leads to : t' ( a t x + yo ) = 0 or, in short :

( 2 ) A x = b where A = t' a t and b = - t' yo

This matrix equation is identical to the algebraic equations (D) above. Once it is reduced by thedatum across-variable and the corresponding redundant continuity equation, it can be solved bystandard procedures, since both A and b are known.

Strict matrix algebra is seldom used in practical applications since it demands too much unnec-essary storage, and computations take too long with the sparse matrices which usually occur.The matrix approach is useful however in demonstrating how the transpose triple product A,and b, may be assembled directly in the computer. Thus, consider the contribution to A and to b

Finite Elements 5

of the m'th element, for which the following are known ( being held typically in a data file ) :

- the inlet node index 'i', and outlet node index 'j',- the admittance constant 'a', and- the specified through-variable, yo ( zero for a passive element )

It can be seen that the overall result of these detailed matrix multiplications is equivalent tomerely adding/subtracting the element data to/from the system equations.

The direct assembly process is therefore :- null all elements of the N*N A-matrix and of the N-vector, b;- for each network element in turn, read the above data from file and :

- add the a-constant to the ( i, i )th and ( j, j )th diagonal elements of A, and- subtract it from the ( i, j )th and ( j, i )th off-diagonal elements of A;- subtract yo from the ith element of b, and add it to the jth element.

Carrying out this process for the example above :-

Reduction to the final equations here consists of substituting for the datum value and transpos-ing the corresponding terms to the RHS, then, rather than eliminating the relevant continuityequation, it is more simply overwritten by the expression for the datum across-variable. Thispreserves symmetry and requires storage of half the A matrix only. It is noticeable that the ele-ment input data does not have to be stored but is used immediately it is read from file.

The output of the Pascal program 'Linet' for analysing linear networks appears below, togetherwith the data file for the foregoing example. The program uses the direct assembly technique,incorporates the simultaneous equation solver 'Simeq', automatically accumulates the total

Finite Elements 6

number of elements and ofnodes, avoids internal storageof element data but echoes ittogether with the problem titleas a visual check, orients cor-rectly the node indices on out-put (eg element #6) andincludes as a check on numeri-cal accuracy the theoretically-zero nodal continuity excesses.

Network analysis usually involves a transpose triple productakin to (2); non-linear nets can often be linearised into this form.Direct assembly as above is a useful technique for such situationsbut it is not always the most efficient, since in large systems thebandwidth requirements may be crucial, as will be shown later.

Extension to 2- and 3-Networks

In order to generalise the foregoing to more complex networks,we first redefine y as a nodal variable akin to x. That is, x and yare dual or complementary sets of field variables measured at thenodes of the network, and the concept of element orientation becomes irrelevant. If, in the flownet above, y is reckoned as a nodal flow towards the element in question, the constitutive lawfor the m'th element may be written in matrix form as below :

Direct assembly is thus seen to be the addition of the relevant sub-matrices of the element's a-matrix to the corresponding locations in the system's A-matrix.

In the nets above, there is one degree of freedom at each node ( ie both 'x' and 'y' are scalar ) andso the sub-matrices of the element a-matrix are also scalar.

These ideas may easily be expanded to more complex networks in which the concept of y as aflow is unnecessarily restrictive and will now be dropped - x and y are just a pair of complemen-tary nodal variables. Thus x may be interpreted as nodal displacement and y as nodal force, inwhich case a and A correspond to element and system stiffnesses respectively, and (2) are equa-tions of equilibrium - we shall concentrate on this interpretation in what follows; however, it ispossible to reverse the roles of x and y, in which case (2) are compatibility equations and A and aare flexibilities. This conjugality between equilibrium and compatibility, like that already notedbetween the two Kirchoff's Laws, is a consequence of the fundamental duality of networks; anygiven net can be transformed into its dual net by interchanging nodes with elements.

Consider the triangular element in two dimensions ( z1, z2 ) illustrated below, the nodes beinglabelled 'i', 'j', 'k'. There are two components of both 'x' and 'y' at each node, so the constitutivelaw for the m'th element may be written variously as shown - in the full 6*6 scalar form ( i), or

Finite Elements 7

in shorthand vector notation (iii)where the a-matrix consists of 3*3sub-matrices corresponding to thethree nodes, each of the nine sub-matrices being itself 2*2 (ii) corre-sponding to the two degrees of freedom. Evaluation of the various a-scalars is an application-dependent detail, however no matter what the application ( provided it involves the transposetriple product ) the A-assembly process is identical to that outlined above.

b on the RHS of (2) is just a vector of specified y-values - that is, since we are interpreting y asforce, of defined concentrated nodal loads. Assembly of b is now simpler than it was in the caseof flow nets; in that case, sources were added to one element of b and subtracted from another -a specified-y affected two nodes. In the present case however, (2) are equations of nodal equilib-rium, the specified-y's are known external forces at single nodes of the system, and orientationas considered previously is irrelevant. So the external loads are merely inserted individuallyinto b. Thus¡ :

EXAMPLE The triangular lamina is located atthe corner of the assemblage illustrated, and itsstiffness matrix is quoted in MN/mm units.Investigate its contribution to the equilibriumequations of the assemblage.

The element's stiffness sub-matrices are assem-bled directly into the system A-matrix in thesketch below, remembering that all terms are additive to existing coefficients and do not overwrite them.

The specified force components at node 11 are simply substituted into the b-vector.

When all elements have been similarly assembled into the system equations, any specified displacements, x, areevaluated and extracted to the RHS, and thecorresponding equilibrium equation overwrit-ten by the expressions for the specified values -exactly as was done for the 1-networks above.

Support reactions would thus be defined aspoints having no displacement. Substitution of

Finite Elements 8

x10,1 = 0.01 mm is shown in the matrix here,neglecting for demonstration simplicity thecontributions to A of elements other than thepresent one of interest.

Once the equilibrium equations have beensolved, the nodal displacements are substitutedback into the elements' constitutive laws -again, just like the simple 1-nets above - todetermine nodal forces. We shall see later howstresses also can be derived.

Extension to three-dimensions evi-dently requires 3*3 sub-matrices of aand A - their assembly is readily visual-ised - but how do we determine an ele-ment stiffness matrix to assemble in this manner ? Before we demonstrate this, we shall intro-duce the Rayleigh-Ritz method - a powerful general technique for structural analysis, notrestricted to Finite Elements.

The Rayleigh-Ritz Method

The edge-supported buckled rectangular plate is a typical practi-cal component of simple unloaded shape where it is impossibleto deduce the true, closed-form stresses. To obtain some idea of the stress levels therefore, anapproximation to the deformations is first postulated. This mathematical function, the 'dis-placement model', reflects the form or general shape of the deformations, and it is usually aseries incorporating a number of undetermined coefficients. Although the inter-relationbetween these is initially chosen so that the model satisfies compatibility ( eg the geometricboundary conditions ), this necessity is insufficient for the complete evaluation of the coeffi-cients - they must be quantified later to complete the model.

The out-of-plane model displacement of the above plate, u ( scalar ), might here be taken as atruncation of the Fourier series :-

u { x,y } = Σm =1 am sin mπ x/Lx. sin π y/Ly where the constant am's are as yet unknown.

This displacement model automatically satisfies the boundary condition around the edges, nomatter what values are assumed by the a-coefficients, or at what harmonic the model is trun-cated. Truncated power series - again with undetermined constant coefficients - are superior toFourier series for certain other components. Obviously, the greater the number of terms, the bet-ter the model's accuracy ( provided the exact solution is not contained in the series. )

Having laid down the form of the model, the undetermined coefficients are evaluated after-wards by considering equilibrium. We have employed this technique in the past for stresses inshafts, in beams, both straight and curved, and in fillet welds - however in these cases theassumed deformations were the correct ones and so the equilibrium equations also were exact. Ifdeformations are approximate however, the equilibrium equations which are obtained fromthem ( by differentiation ) are also approximate - in fact since differentiation is an error-magnifying process, derived reactions and stresses are less accurate than the displacements. Inview of the approximate nature of the equilibrium equations, they are usually gotten from thePrinciple of Minimum Potential Energy rather than from elementary Statics.

The method may be demonstrated by the simple beam shown overleaf, whose exact solutionmay be used as a yardstick. A candidate model for the deflection might be either :

Finite Elements 9

u = a1 sin πx/L + a2 sin 2πx/L + a3 sin 3πx/L + . . . . . oru = b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + . . . . .

where the a's and b's are constants, as yet unknown.

A single Fourier term will be chosen here for example; it is evident that a1 is the maximum dis-placement and that the model meets the geometric boundary conditions automatically :-

u = umax sin πx/L ( satisfying :- u = 0 @ x = 0, L and u = umax @ x = L/2 )

It may be shown that the strain energy of a linear elastic beam may be expressed variously as :-

Πbeam = 1/2EI ∫L M2 dx or = EI/2 ∫L ( u")2 dx the second formulation is relevant hereso the potential of the system becomes

Π = Πbeam + Πload = π4 EI u2max / 4 L3 - P.umax

ie, a function of the undetermined coefficients, of which there is only one here, umax.

The Principle of Minimum Potential Energy stipulates that, for equilibrium :-

∂Π/∂umax = 0 ie, here ( π4 EI/2L3 ) umax - P = 0

This last equation is clearly an equilibrium equation, of the form K.umax = P, where K is the(approximate) stiffness of the beam as seen by the load, P. The equation predicts that umax = 2PL3

/π4 EI = PL3 /48.7 EI, which compares favourably with the exact expression (48 in the denomina-tor). Evaluation of the unknown displacement coefficient(s) by the above technique enablescompletion of the displacement model, u. The stress resultant follows, from beam theory

M = - EI u" = EI umax (π/L)2 sin πx/L = (2/π2) PL sin πx/L = 0.203 PL sin πx/LAs pointed out above, this approximation is not as satisfactory as that of the displacements - theexact variation is linear with a central moment of 0.25 PL.

Application of the Rayleigh-Ritz method to complex unloaded geometries isoften impractical - the displacement model for a holed plate for examplehas to account for the discontinuity at the hole's circumference andalso has to include a large number of terms to adequately describethe gross non-linearities. This is not impossible, since geometrictransformations are available to map the plate into a rectangle - butthe resulting model is far too cumbersome for routine work. This is where the concept of FiniteElements comes into its own. The body is divided into a number of contiguous elements whichare laid out to describe its approximate geometry, as in the frontispiece; body integrals become

Ubody = ∫over the whole body = Σover all elements in the body ( ∫over one element )

The displacement model now only needs to be applied to the individual elements - whoseshapes are simple - and the model itself can be relatively crude since parameters which varysteeply over the body do not vary so markedly over each individual element. Accuracy mayapparently be enhanced by elements which are either more numerous or more complex, butgenerally all elements in an assemblage should be based on the same displacement model.

The following example shows some further features of the FEM. It is required to estimate thevalue of π by approximating the area of a circle of radius 'r', by triangles. The area is subdividedrationally into 'M' sectors as shown at (a) below for M=8. Refining the net by doubling the num-ber of elements is straightforward, unlike the haphazard net of (b) which contains also four- andfive-sided elements - nets should be set up with ease of possible subdivision in mind.

The sectors of (a) are approximated by triangles, either internal as shown at (c), or external as at(d), and the total area of the triangles evaluated in each case. The graph of these results (e) shows

Finite Elements

some trends which are generally true of FEM analyses :-- As the network is refined, ie as the number of elements increases, the results become moreaccurate - converge to the true value - until swamped by computational inaccuracies when theequation set becomes large.- When the net is reducible, and the other techniques which are used are correct, then we maybe fairly certain that results are bounds to the true value. A reducible net is one which is con-tained completely by the previous coarser net as refinement proceeds. Bounding is importanttheoretically, although not routinely used for practical problems since confidence in the FEMhas been established. It can be achieved by using the dual ( again ! ) energy theorems, for exam-ple; or the dual stiffness and flexibility methods of analysis.

As noted, accuracy may be improved by using elements which are more complex individually,rather than more numerous. Thus if we had attempted the estimation above based upon circum-ference rather than on area, then, for a given result with a certain number of straight line ele-ments, we might have achieved the same accuracy using fewer second order curved line ele-ments ( if we had them ). If a circular element were available, then of course only one would berequired for the 'correct' solution. Analysis economics in practice thus requires a trade-offbetween many simple elements on the one hand, where computation time is devoted mainly toassembling and solving a large number of simple equations, or relatively few elements havingcomplex constitutive laws on the other, in which case the constitutive laws of the individualelements form the bulk of computation costs. The cost of setting up a model for a given proto-type will increase with the number of elements.

Finite Element Theory - Equilibrium of the Discretised Body

In presenting an outline of the finite element method in thissection, no attempt is made to be rigorous or to explain everydetail; the purpose is solely to give a general broad-brush appre-ciation of the method.

An elastic body of known geometry is defined in three-dimensional z-space and subjected to thebody force X throughout its volume, V. Prescribed tractions T ( ie loads ) are applied over part ofthe surface S of the body's boundary, and a further portion of the boundary is subjected to pre-scribed displacements ( eg supported ).

The body is discretised into a system of 'M' contig-uous elements, interconnected at their nodes. Thedisplacement of the body is described by the vectorof discrete nodal displacments, q = [ . . q i q j . . . ]'.

The continuous displacement field, u { z } = [ u1 u2u3 ]', over any particular element must be compati-ble with the corresponding nodal displacements of

Finite Elements 11

that element - ie with a sub-vector of q. The displacement model for the element in questioncan thus be expressed as :-

(3) u = N q where the 'shape function' N { z }, is a continuous function of location.

We shall examine the implications of this in more detail in the next section, but in order tofix our ideas for the time being, consider a one-dimensional element, located betweennodes 'i' and 'j' of the assemblage. Displacement is scalar, and we may postulate the secondorder model : u = a0 +a1z +a2z2 involving three undetermined coefficients. But compati-bility requires that u = q i at z i and u = q j at z j , so that the model must be along the lines :

u = q i ( z j - z )/( z j - z i ) + q j ( z - z i )/( z j - z i ) + a ( z - z i )/( z j - z ) in which 'a' is the solecoefficient open to arbitrary choice. It is evident that the number of coefficients must not be less than the totaldegrees of freedom of 'external' nodal displacements, two in this case, otherwise compatibility cannot possibly beachieved. External nodes connect an element to the rest of the assemblage; it is quite in order to define 'internal'nodes which are intrinsic to the element and are not connected to anything else.

If 'a' is chosen to be zero for the element in question, then the linear model is :

u = q i ( z j - z )/( z j - z i ) + q j ( z - z i )/( z j - z i ) = ( 1/( z j - z i ) ) [ ( z j - z ) ( z - z1 ) ] [ q i q j ] '

which is of the form (3) with the continuous shape function : N = ( 1/( z j - z i ) ) [ ( z j - z ) ( z - z1 ) ]

Stress and strain will be characterised simply here by principals, thus σ = [ σ1 σ2 σ3 ]' and ε = [ ε1ε2 ε3 ]'; a more rigourous treatment incorporates shear in a similar manner. From elementaryelasticity, the continuous geometric entities strain and displacement are inter-related via :

ε1 = ∂u1/ ∂z1 and similarly for other components. Or, in brief :ε = ∂ u where ∂ is a matrix of partial differential operators.

From the above it follows that the strain also must depend upon the nodal displacement vector:

ε = ∂ N q = B q where the continuous strain-nodal displacement matrix, B = ∂ N.

Furthermore, since the body is elastic, stresses are linearly related to strains since :

ε1 = ( σ1 - νσ2 - νσ3 )/E etc, and so, solving for the principal stresses :σ1 = ( ( 1- ν )ε1 + νε2 + νε3 ) E / ( 1 + ν ) ( 1 -2ν ) etc; or, briefly :

σ = C ε = C B q in which C is a matrix of material elastic constants.

Once an element's displacement model has been defined, N and B may be computed; then,when q is evaluated, these constitutive laws above may be recalled to ascertain stress and strain.

The work/energy terms relevant to the system consisting of an element and its associated loadsare as follows :

- The gain in strain energy of the element, ΔΠε, over the element's volume.Characterising stress and strain in the volume δV of the element by their prin-cipals, then ΔΠε = ∫V 1/2 ( ε1σ1 + ε2σ2 + ε3σ3 ) dV = ∫V 1/2 ε' σ dV.

- The work done by body forces such as weight, ΔΠX, over the element's volume. If the dis-placement and ( constant ) body force at the volume δV of the element are u = [ u1 u2 u3 ]'and X = [ X1 X2 X3 ]' respectively, then ΔΠX = ∫V u' X dV.

- The work done by tractions, ΔΠT, over the element's surface area. In a manner analogousto body force work, ΔΠT = ∫S u' T dS. If the element's surface is not part of the assembly'ssurface which is subject to traction, then this work term will vanish.

- The work done by other elements' contacts at the element's nodes, ΔΠN.

The gain in potential of the element and its loads is therefore :

Finite Elements

ΔΠ = ΔΠε - ΔΠX - ΔΠT - ΔΠN

= ∫V 1/2 ε' σ dV - ∫V u' X dV - ∫S u' T dS - ΔΠNor, substituting from above for u, ε and σ :

= ∫V 1/2 q' B' C B q dV - ∫V q' N' X dV - ∫S q' N' T dS - ΔΠN

Summing this over all elements in the assemblage, the potential gain of the body and its asso-ciated loading is :

Π = ΣM { ∫V 1/2 q' B' C B q dV - ∫V q' N' X dV - ∫S q' N' T dS }

Note that the overall effect of ΣM ΔΠN must be zero, due to equal-and-opposite action/reactioncontributions at the inter-element nodes. Since q is a vector of discrete nodal displacements,common to all elements and not subject to integration, it is extracted from the integrals to give :

Π = q' ΣM { ( ∫V 1/2 B' C B dV ) q - ∫V N' X dV - ∫S N' T dS }

Applying the Principle of Minimum Potential Energy, equilibrium of the body requires that∂Π/∂q = 0, or :

{ ΣM ( ∫V B' C B dV ) } q = ΣM ( ∫V N' X dV + ∫S N' T dS ) ie

(4a) ( ΣM k ) q = ΣM pwhere k = ∫V B' C B dV is the stiffness of an element, and

p = ∫V N' X dV+∫S N' T dS is the external force on an element.

Simplifying this still further, the equilibrium equations for the discretised body are :

(4b) Kq = Pwhere K = ΣM k is the stiffness of the discretised body system, and

P = ΣM p is the external force vector acting on it.

The equilibrium equations (4b) are analagous to the network equations (2), and may be assem-bled identically since (4a) demonstrates that elemental stiffness involves the transpose tripleproduct B' C B. The equations only have to be reduced by the prescribed displacements beforebeing solved for q in the same way as (2) were solved previously for x. Having found q, thesolution is completed by calculating stresses and strains from the constitutive laws above.

Finite Element Theory - Element Stiffness

We have seen that knowledge of the strain-displacement matrix, B, is afundamental necessity for evaluating element stiffnesses from (4a). Weshall now illustrate typical approaches to finding B, and particularise on asimple element - the constant strain triangle used in the package, 'FEM1'.

The first step is the choice of a suitable shape function, N - that is, an equation which approxi-mates the displacement, u, at any point in the element under consideration, in terms of :

- the co-ordinates of the point in question, and- the displacement of the element's nodes, q, eg in the case of the 3-node, 2-dof element

shown, either q = [ u1 v1 u2 v2 u3 v3 ]' or q = [ u1 u2 u3 v1 v2 v3 ]'

There are two distinct methods of establishing this function - using either 'generalised coordi-nates' or 'natural coordinates'.

Generalised Co-ordinatesConsider a one-dimensional element of length 'b', with twoexternal nodes, '1' & '2', and a central internal node '3'. There isone degree of freedom in (transverse) displacement, u. We take the displacement model in

Finite Elements 13

power series form with constant coefficients, αi :-

u = α1 + α2 z + α3 z2 or, in brief (a) u = φ' α

where φ = [ 1 z z2 ]' is a vector of powers of the coordinate(s) of the point, andα = [ α1 α2 α3 ]' is a vector of constant coefficients, the 'generalised coordinates'.

In this way, the displacement 'u', which is a scalar here, may be approximated over the region ofthe element. The generalised coordinates cannot be chosen at random however, since compati-bility necessitates that (a) must apply also to the nodes, whose displacements are given as :-

u1 = α1 + α2 z1 + α3 z12 ⎡ u1⎤ ⎡ 1 z1 z1

2⎤u2 = α1 + α2 z2 + α3 z2

2 or q = ⎢ u2⎥ = Φ α where Φ = ⎢ 1 z2 z22⎥

u3 = α1 + α2 z3 + α3 z32 ⎣ u3⎦ ⎣ 1 z3 z3

2⎦

It follows that, for compatibility, the generalised coordinates must satisfy :-

α = Φ-1 q so, from (a) u = φ' α = φ' Φ-1 q = N q that is N = φ' Φ-1

Inserting some values, to show the form of these variables, ⎡ 1 4 16 ⎤suppose an element to extend from z1 = 4 to z2 = 6. Therefore Φ = ⎢ 1 6 36 ⎥

⎣ 1 5 25 ⎦

From the above :- ⎡ 15 10 -24 ⎤⎢ -11/2 -9/2 10⎥

N = φ' Φ-1 = [ 1 z z2 ] ⎣ 1/2 1/2 -1 ⎦ = 1/2 [ (30-11z+z2) (20-9z+z2) (-48+20z-2z 2) ]

N having thus been found for the element, the continuous variations of displacement, for twopossible sets of nodal displacements are as follows :-

q = [ 2 4 0 ]' q = [ 2 -1 3 ]'u = N q = 70 - 29z +3z2 ( 4 ≤ z ≤ 6 ) u = N q = 1/2 ( - 104 + 47z - 5z2 )

The form of N should be noted particularly; the reason for its name is apparent - it shapes a'curve' of the chosen order ( 2nd here ) between the nodal displacements, whatever these mighthappen to be. Note that the number of generalised displacements ( three here ) must tally withthe total nodal degrees of freedom, so that their solution is possible. Thus, in two dimensions(x,y) with two degrees of freedom ( u = [ u v ]' ) and say complete second order, we might have :

u = α1 + α2 x + α3 y + α4 x2 + α5 x y + α6 y2

v = α7 + α8 x + α9 y + α10x2 + α11x y + α12y2 eg. six nodes are necessary.

If the plane element for which this model is chosen has less than six external nodes, then theinternal nodes may be condensed out using the principle of minimum potential energy ( seebelow for the meaning of condensation ). Alternatively, terms may be dropped whilst preserv-ing symmetry of the co-ordinate powers, to tally with the number of external nodes, thus :-

5 nodes : u = α1 + α2 x + α3 y + α4 x2 + α5 y2 or, if there are not 5 nodes,4 nodes : u = α1 + α2 x + α3 y + α4 x y and similarly for 'v'.

Finite Elements

Natural CoordinatesIn the one-dimensional element above, the location of a point was defined by global coordinates(z). Natural coordinates on the other hand, specify a point within an element by a set of dimen-sionless numbers which assume unit value, for example, when the point coincides with anexternal node. Thus the point is located with respect to the element's nodes, independently of

how the element is sized or positioned in the global system. Naturalco-ordinates are intrinsic to the element. Thus, in the example above,suppose a natural co-ordinate 'L' is chosen so that L= –1 at node '1' andL= +1 at node '2' as shown, so the linear mapping of 'x' into 'L' is :

L = ( 2z - ( z1 + z2 )) / b or conversely : z = ( b L + ( z1 + z2 )) / 2

It follows that L= 0 at the central internal node '3'. Using this mapping, differentiation and inte-gration may be expressed as :-

∂/∂z = ( 2/b ) ∂/∂L and, for integer index 'p', the integral over the length 'b'

∫b Lp dl = ∫z1z2 Lp dz = b/2 ∫-1

+1 Lp dL = b/2 Lp+1/(p+1) ⎟ -1+1

= 0 if p is odd, or = b/(p+1) if p is even - a very simple result.

We now set up, corresponding to each node, an 'interpolation function, f', which assumes avalue of unity at the node, and zero at other nodes. Thus, by inspection :-

'1' : f1 = 1/2 L ( L - 1 ) ie f1 = 1 @ '1' when L = -1 and f1 = 0 @ '2' & '3''2' : f2 = 1/2 L ( L + 1 ) ie f2 = 1 @ '2' when L = +1 and f2 = 0 @ '3' & '1''3' : f3 = ( 1 - L2 ) ie f3 = 1 @ '3' when L = 0 and f3 = 0 @ '1' & '2'

These three interpolation functions are plotted below :

We thus have, in general for this element :

z = [ f1 f2 f3 ] [ z1 z2 z3 ]' = f1z1 + f2z2 + f3 z3= [ 1/2 L( L-1) 1/2 L( L+1) ( 1 -L2) ] [ z1 z2 z3 ]'

We can then use the same shape factor and say immediately that :

u = [ 1/2 L( L-1) 1/2 L( L+1) ( 1 -L2) ] [ u1 u2 u3 ]' eg u = u1 when L = -1 &c

This is just the equation : u = N q again, however N is nowexpressed in terms of the natural coordinate, L, rather than aspreviously through the global Cartesian z. By way of illustra-tion, suppose that the vector of nodal displacements here hap-pens to be q = [ 2 -1 3 ]'. Inserting this into the above yields u= ( 6 -3L -5L2 )/2, which plots out exactly as before :-

Elements which employ the identical shape function to define the location and displacement ofthe general point are termed 'isoparametric' elements.

When dealing with triangular elements, the most straightforward natural coordinates withwhich to define the general point 'P', are the area ratios 'Li' sketched below; only two of theseare independent :-

L = [ L1 L2 L3 ]' where Li = Ai /A ( 0 ≤ Li ≤ 1 ) ; A = Σ Ai ; i = 1 ,2, 3

Finite Elements 15

The mapping of the global and natural coordinates may be written :-

⎡ 1 ⎤ ⎡ 1 1 1 ⎤ ⎡ L1 ⎤⎢ x ⎥ = ⎢ x1 x2 x3 ⎥ ⎢ L2 ⎥ eg L1 + L 2 + L3 = 1 and⎣ y ⎦ ⎣ y1 y2 y3 ⎦ ⎣ L3 ⎦ x = x1L1 + x2L2 + x3L3 or, conversely :-

⎡ L1 ⎤ ⎡ 2A23 b1 a1 ⎤ ⎡ 1 ⎤ where ai = - xj + xk ; bi = yj - yk⎢ L2 ⎥ = ( 1 / 2A ) ⎢ 2A31 b2 a2 ⎥ ⎢ x ⎥ and 2 A i j = xi yj - xj yi⎣ L3 ⎦ ⎣ 2A12 b3 a3 ⎦ ⎣ y ⎦ with cyclic permutation.

Differentiation and integration, analagous to the one-dimensional equations, are

∂/∂x = Σi=13 ( ∂Li/∂x ) ∂/∂Li = ( 1/2A) Σi=1

3 bi ∂/∂Li ; ∂/∂y = ( 1/2A) Σi=13 ai ∂/∂Li

∫A L1p L2

q L3r dA = 2A p! q! r! / ( p +q +r +2 )! again, a simple result.

The Strain-Displacement Matrix for the Constant Strain Triangle

Using triangular natural coordinates, the B matrix for the constant strain triangle is as follows.

In two dimensions, the kinematic relationship between strain [ εx εy γxy ]' at a point [ x y ]', andthe displacement [ u v ]' at the point is :-

⎡ εx ⎤ ⎡ ∂/∂x 0 ⎤ [ u v ]'⎢ εy ⎥ = ⎢ 0 ∂/∂y ⎥⎣ γxy ⎦ ⎣ ∂/∂y ∂/∂x ⎦ that is the equation : ε = ∂ u above, in complete detail.

For constant strain, a linear displacement model is appropriate - ie a point's displacement com-ponents are directly proportional to its coordinates. Using the isoparametric concept in conjunc-tion with the triangular natural coordinates above, we have immediately that :-

⎡ u ⎤ = ⎡ L1 L2 L3 0 0 0 ⎤ [ u1 u2 u3 v1 v2 v3 ]' ie u = N q⎣ v ⎦ ⎣ 0 0 0 L1 L2 L3 ⎦

From the above :-

εx = ∂u/∂x = ∂/∂x ( L1 u1 + L 2 u2 + L 3 u3 ) and so, using the chain rule= ( b1 u1 + b2 u2 + b3 u3 ) / 2A

Carrying out similar steps for the other strain components leads to :-

⎡ εx ⎤ ⎡ b1 b2 b3 0 0 0 ⎤ (b) ⎢ εy ⎥ = 1/2A ⎢ 0 0 0 a1 a2 a3 ⎥ q that is, ε = B q

⎣ γxy ⎦ ⎣ a1 a2 a3 b1 b2 b3 ⎦

The strain-nodal displacement matrix, B, can thus be evaluated in terms of the element's nodalcoordinates which are embodied in the 'ai' and 'bi' terms. B, and the strains are constant overthis element ( of thickness 'h' ) so integration for the element's stiffness is trivial :

k = ∫ B' C B dV = hA B' C B.

Finite Elements

Finite Elements - Implementation

We have seen that, overall, application of the FEM requires the following steps :- Discretise the continuum (define the topology) into a network of membrane elements, or axi-

symmetric thick shell elements - or whatever element is appropriate to the body's shape- Select the displacement model (the order of the virtual displacement-coordinate equation)

and hence the complexity of the elements' constitutive laws- Derive the element stiffness matrices, as exemplified above- Assemble the equations, using direct assembly for example, and substitute values for the spec-

ified displacements and loads- Solve the equations for the vector of nodal displacements, q- Compute element stresses and strains from the constitutive laws.

We shall refer to the elementary demonstration package 'FEM1' which is typical in that the useris required to carry out the first step only. The package is restricted to two-dimensional linearelastic problems; its basic elements are quadrilateral and triangular laminae with a node at eachcorner. These are subdivided by the program into constant strain triangles whose B-matrix hasbeen derived in (b) above.

The user communicates with the package via a data file. The appended 'FEM1 User's Guide'should be consulted for full details of file preparation, but essentially the prototype is first subdi-vided into quadrilaterals ( and/or triangles ) by the user, who then enters the coordinates of theresulting nodes into the data file, together with the node indices which define each element.Element thickness, material properties and prescribed loads and displacements are also enteredinto the file.

The program reads the data file element by element, reading each element's geometry, thenevaluating its stiffness matrix from this data, and finally assembling the submatrices directlyinto the system equations before moving on to the next element. Which is all very similar tothe linear 1-net procedure above.

Before the discretisation process itself is examined, it is necessary to consider a couple of topicswhich are relevant to it.

Condensation and Bandwidth

Condensation is an elimination process applied during the solution of a set of simultaneous equa-tions, in which a selected variable is eliminated between one particular equation and the others,the coefficients of that equation being stored temporarily out of core, on disc for example. This isexemplified for x4 :-

The reduced set of equations is then further processed, and eventually solved in core for thevariables not so selected. The equation coefficients may later be retrieved and the selected vari-able evaluated.

A number of variables may be eliminated in this way, thus increasing the size of system whichcan be handled in a restricted core; for example the daughter sub-system of (A) below is con-

Finite Elements 17

nected to the larger mother only via the five nodes 77-102-265-319-341. The daughter's matrices canbe assembled initially as separate entities, then all the displacement variables except q77, q102 etccondensed out. These five remaining equations are then assembled as part of the mother'smatrices, which are then solved.

This avoids having to assemble in core the much larger mother-plus-daughter system. Themother might represent the skin of a ship's hull, the various bulkheads being treated as off-spring, and so on.

Another application for condensation concerns the basic elements used in FEM1 - the quadrilat-eral consists of four constant strain triangles. The 2*2 stiffness sub-matrices for a single constantstrain triangle are contained algebraically in the source code and can be evaluated once the mate-rial and geometric data are supplied - the B matrix from equation (b) above, and the C matrixcorresponding to the element's material.

The triangles' stiffness sub-matrices are assembled after evaluation by the following process, typ-ified by the element defined globally by 31-25-34-66 at (B) above. The program assigns local nodeindices 1-2-3-4, computes the centroid and defines it as the central node 5 - which is isolated fromthe rest of the net - thus splitting the element into four triangles (C). The stiffness sub-matricesof each triangle are computed in turn and assembled into the quadrilateral's 5*5 local matrix, asshown at (D) for triangle II. Once this local matrix is complete, the displacement of the centralnode is condensed out (E), and the resulting 4*4 matrix assembled into the system matrix; eg theshaded sub-matrix is added to the ( 25, 34) locations of the system matrix. All this is transparent tothe user and is aimed at increasing the network's verisimilitude ( degrees of freedom ), withsimple elements and without excessive core demand. The triangular element is similar, beingsubdivided into three constant strain triangles.

The sub-matrices of the element appear in the system stiffness matrix as shown at (F) below, inwhich the shaded area has been set aside in core for the problem. Due to symmetry, the upperhalf only needs be stored, as indicated at (G) in which the dominant diagonal sub-matrices arerelocated in the first column of the skewed band.

The number of sub-matrix columns required by any element is ( D +1 ) where D is the differencebetween the maximum and minimum node indices for the element.

As the size of the sub-matrices equals the number of dimensions, two here, the system as a

Finite Elements

whole requires a semi-bandwidth, or number of columns of scalars :

B = ( Dmax +1 ) * number of degrees of freedom ;

where Dmax is the maximum difference of elements' nodal indices encountered in the net - andobviously depends upon how the nodes are numbered by the user, as indicated by the nets (H)and (I).

It is necessary to construct and index meshes so that the number of nodes and the semi-bandwidth are not excessive, for although the user of FEM1 can pre-declare the maximum num-ber of rows and columns to cover the demands of any reasonably sized problem, large arrays willlead to inefficient and inaccurate solution. This construction of the mesh requires ingenuity, forat the same time it is usually desirable to incorporate as many elements as possible to obtain anaccurate representation of stress variations in way of high stress gradients.

The frontal method and diakoptics are solution schemes which are not subject to such restrictions,but are rather too complex for our present purposes.

Discretisation

Some further points which should be borne in mind whilst preparing a net will now be exem-plified via the prototype cantilever illustrated at (J) below. This consists of an inclined plate ofknown material and geometry, and two different thicknesses. A concentrated force is applied atone corner, one point on the edge is subjected to a specified displacement, and there are two lin-early distributed loads which the program replaces by equivalent concentrated nodal forces.

Nodes must be placed (K) :-at b, d, f, h - the discontinuities of the boundaryon line c-g - line of discontinuity of material property ( thickness )on line f-h - line of discontinuity between body and supportsat a, b - points of application of specified loads or displacementsat c, d, e - end points of distributed loads

Based upon this reasoning the mesh of (L) is drawn up and the nodes indexed. A Cartesian sys-tem is set up for the definition of the node coordinates and the senses of the force/displacementcomponents. One support node (#1) is fixed to prevent rigid body motion; other support nodesare arranged on the assumption that, in this particular example, the physical support is stiff onlyin the vertical direction. The elements are annotated in (M) - there are five quadrilaterals and asingle triangle. Element 3 for example, is defined by nodes 8-4-3-7.

This completes the discretisation; there are 6 elements, 11 nodes and the semibandwidth, B, is12. A data file is then prepared to the format explained in the 'User's Guide', for reading byFEM1. The stresses output by the program will be rather meaningless here because the demon-stration mesh is far too coarse.

Finite Elements 19

As a general guide, a mesh should be based upon orthogonality, with elements as well-conditioned as possible to avoid mathematical ill-conditioning and a consequent tendencytowards inaccuracy - thus ideally quadrilaterals should be square and triangles equiangular. Themesh should be refined in way of expected stress gradients and concentrations. Some techniquesfor local refinement are shown here to illustrate the use of triangles to minimise ill-

conditioning. The degree of refinement, and hence the number of nodes and bandwidth, will ofcourse depend upon the problem - the stress gradients which occur, the accuracy sought and thecapacity of the implementation. Some of the consequences of refining are dealt with in theAppendix. Full advantage should be taken of any symmetry implicit in the prototype.

Consider half a beam in pure bending, supported at its left end as shown opposite. Instead of thissupport, another linear traction, equal and opposite to the load on the right, might be applied tothe left end. Equilibrium is thus theoreticaly assured - BUT this may give rise to numerical prob-lems with FEM1, because round-off errors lead to loss of equilibrium, and although net forcesmight be small, rigid body motion results. So loads and reactions must not be over-specifiedwhen defining the model.

Bibliography

A whole literature exists on this subject; the following are particularly recommended :

Desai CS & Abel JF, Introduction to the Finite Element Method, van Nostrand 1972OC Zienkiewicz, The Finite Element Method, McGraw Hill 1977Irons B & Ahmad S, Techniques of Finite Elements, Ellis Horwood 1980

See also :

Kardestuncer H (ed), Finite Element Handbook, McGraw-Hill 1987Barn NM, Finite Element Analysis on Microcomputers, McGraw-Hill 1988

Finite Elements

Finite Elements 21

APPENDIX : THE REFINING PROCESSJim's Example, consisting of a dual-thickness plate loaded in-plane, has been used to demon-strate the more important aspects of setting up a two-dimensional mesh, and to illustrate how atypical data file is prepared for the linear elastic program FEM1. The demonstration mesh wasfar too coarse to be considered a realistic approximation to the prototype; the present intention isto progressively subdivide - ie 'refine' - that mesh in a bid for greater realism, and at the sametime to highlight some typical results which might be expected of such a refining process.The undeformed meshes are illustrated opposite with a few node and element indices, the latterbeing italicised. Some overall particulars of the meshes are :

mesh subdivision step 0 1 2 3number of nodes 11 31 101 361number of elements 6 22 85 328semi-bandwidth of assemblage equations 12 18 30 54program - maximum nodes*semi-bandwidth 250*72 250*72 250*72 361*54solution time, compiled (sec) - Macintosh SE 20 82 427 3098

- Macintosh SE/30 4 17 88 621number of pages of output 1/3 1 3 10

An ideal square mesh forms a useful basis for estimating, as refining proceeds, the increase inthe number of nodes and in other parameters. For such a square mesh :

node number element number semi-bandwidthif, at a certain stage of subdivision n e wthen, after a further subdivision (2√n -1)2 → 4n 4e 2( w-3) → 2w

The 'area' required for the assemblage equations in memory is proportional to the product ofnode number*semi-bandwidth, and thus increases by a factor of 8 approximately over each sub-division of the square mesh. Jim's mesh is seen to be very similar to this. It will be noticed thatthe ratio of solution times also approaches this multiple.The standard program has been compiled to cater for maxima of 250 nodes and 72 semi-bandwidth - this was extended readily to cater for the last stage of refinement. Clearly, furthersubdivision is rendered impractical on the basic SE by time constraints as much as by memorylimitations - though the speed of the SE/30 nullifies this argument. Solution time using inter-preted code is about 5% longer than that above; solution on a Macintosh Plus takes only a fewseconds longer than on an SE.The outputs from these refinements have not been included because the vast majority is ofabsolutely no interest whatsoever - a suitable post-processor is really a necessity - however thefollowing table compares some facets of the solutions :

mesh subdivision step 0 1 2 3error in reconstituted (6,-9)kN force, N 0, 0 0, 0 0.01, 0.02 0.01, 0deflection at (6,-9) kN force, mm (0.33,-0.48) (0.41,-0.62) (0.46,-0.67) (0.52,-0.77)force at node with specified 0.01mm deflection, kN 17.4 15.7 14.6 13.7peak equivalent stress, MPa 126 237 342 503

The round-off errors, as reflected in the reconstituted concentrated (6,-9)kN force, are in all casesnegligible. The arithmetic procedures of FEM1 would thus seem adequate for the scope of prob-lems allowed by other program limitations.Neither the deflection at the node where the concentrated force is applied, nor the force at thenode where the specified horizontal deflection is 0.01mm, demonstrate the asymptotic tendencywhich one would expect to follow from the refining process. The reason for this unusual beha-viour lies in the example, not in the method or program. This may be appreciated from thedeformed meshes, shown overleaf, the most refined mesh in particular demonstrates large localdeformations at the concentrated effects - the sharp change in slope where the concentrated force

Finite Elements

Finite Elements 23

limits the nodal deflection to 0.01mm, and the deformation of the element where the (6,-9)kNconcentrated force is applied. In practice such concentrated effects would not occur; for examplethe force might be applied through a two-force member welded to the plate, or via a pin bearingupon the interior of a hole in the plate, and so on - ie effects are distributed over the plate. Jim'sexample is therefore not realistic. (It was never meant to be realistic; it started life purely as anexample for data file layout.) Despite the impractical quirks of Jim's plate, it is quite clear thatincreased refinement enables the finite element model to better approach realism. In other words -the more degrees of freedom, the more realism.The unusual behaviour of Jim's plate is evident also in the stresses which occur in the finiteelement model. The table above includes values of the peak equivalent stress, which increase asrefinement proceeds without displaying the asymptotic tendency expected from a 'practical'model. Further aspects of stressing may be appreciated from the stress patterns. The patternoption of FEM1 allows the user to specify a maximum stress contour, and each element is pat-terned according to the ratio which its equivalent stress bears to this maximum :

ratio equivalent stress/maximum stress contour < 1/3 ≥ 1/3 ≥ 2/3 ≥ 1element pattern light gray gray dark gray black

The four mesh patterns are shown with a common maximum stress contour of 275MPa, a valuewhich was chosen somewhat arbitrarily to show desired features - it happens to be about thegeometric mean of the four peak equivalent stresses.The plate's behaviour is sensibly that of a cantilever, with high stresses on the LH and RH faces.It is interesting that the maximum stress occurs at the base of the original mesh and of the firsttwo refinements, but the peak stress of the most refined mesh occurs at the point of applicationof the concentrated (6,-9)kN force. This is another quirk of Jim's unrealistic plate - though a geo-metric singularity like a crack front would lead to similar results.There is not much more of any immediate interest in the above.Subdivision 4 has been devised to duplicate the fine mesh of subdivision 3 in way of the stressconcentrations and large deflections along the RHS of the plate, without the attendant disadvan-tages of the more refined meshes regarding memory usage and execution time. Subdivision 4was achieved by adopting a fine mesh in the areas of interest, and a coarse mesh in the areas oflittle interest. The mesh is shown together with the deformed shape of the plate and the stresscontour patterns.Comparing the output for Subdivision 4 with the above, there are 68 nodes, 73 elements, a semi-bandwidth of 52, and just over 2 pages of output; the solution took 466 s and so in many respectsthe mesh is similar to Subdivision 2. However the selective refining means that the perfor-mance is more in line with Subdivision 3, as the force at the 0.01mm deflection node is 13.9kN,the deflection at the (6,-9)kN load is (0.48,-0.71)mm, and the peak stress is 499MPa - the highlystresses areas are seen to be very similar to those of Subdivision 3. The similarity betweenSubdivisions 3 and 4 is further borne out by the superposition of their deflected shapes.The conclusion from this last step is that it is possible to model realistically localised stress con-centrations by means of meshes which have been selectively refined in way of the high stress/strain gradients, without incurring the memory demand and execution time penalties asso-ciated with globally refined meshes.

Finite Elements

Finite Elements 25 Finite Elements

Finite Elements 27

'FEM1' USER'S GUIDE : version k

The application 'FEM1' carries out finite element static analyses of two-dimensional linearlyelastic objects in a state of plane stress or strain, and enables the objects to be plotted. It is basedon the FORTRAN code appearing in 'Introduction to the Finite Element Method', Desai & Abel (vanNostrand 1971), with major enhancements which include object plotting, subdivision of all ele-ments (not just quadrilaterals) into constant strain triangles, use of a more efficient equationsolver, nodal load reconstitution, support of tractions normal to element edges, and so on. Thesubdivision of an element into constant strain triangles - with a common centroidal nodewhich is statically condensed before assembly of the element equations into the network - allowsclosed-form, as opposed to numerical, integration to be carried out.

Loading of the assembly is by prescribed displacements (zero in the case of supports) and concen-trated forces at the nodes, by body forces such as weight, and by linearly varying inter-node trac-tions which the program replaces by equivalent concentrated nodal forces.

In order to optimise memory usage, the Pascal source organises data dynamically into handleand pointer lists rather than by the fixed vectors so characteristic of FORTRAN - however avail-able memory is still limited. The program defines two constants - 'maxnodes' the maximumnumber of nodes, and 'maxsemibandwidth' the maximum semi-bandwidth of the assemblyequations. These are set at 250 and 72 respectively in the complied application, but may bealtered (within the memory's capabilities) in the Pascal code which must then be recompiled.

The program pays scant regard to execution time, preferring to use easily understandable coding- so if one tries to further subdivide a network in an attempt to improve the accuracy of thefinite element approximation, then round-off errors and time constraints could be just as signif-icant considerations as the node and bandwidth limitations mentioned above.

The program consists of three major suites of procedures :INPUT This unit reads assembly-defining data from a user-prepared file and checks the data

for consistency, whilst setting up pointer lists of material, node and loading records,together with a scratch file of element records. If desired, node and element input data maybe echoed to assist data file debugging. Gross errors are picked up but more subtle mistakeslike mis-assigning a certain material to a particular element, cannot be detected.

SOLVE Each element is subdivided into constant strain triangles, the centroidal node is stati-cally condensed and the eliminated equations written to scratch for eventual determina-tion of the centroidal stress; the element stiffness is then assembled into the system equilib-rium equations which are stored in banded form via handles. When assembly is complete,the equations are reduced by the specified nodal loads and displacements, then solved forthe nodal displacements. These results are output, together with the nodal loads which arereconstituted from them to indicate the numerical accuracy of the solution. In addition tothese nodal results the stress components at the elements' centroids are output.

DRAW The two-dimensional network including supports and loads may be drawn in theundeformed state after INPUT has executed - drawing may be slowed down if desired to aiddebugging. Following completion of SOLVE, the network may either be drawn in thedeformed state with a user-selected displacement multiplier, or the elements shadedaccording to their centroidal equivalent stresses - an element whose centroidal stress isgreater than the user-specified maximum contour is shaded black.

The program requires the user to first subdivide the two-dimensional object into an assembly ofthree- and/or four-noded elements, ensuring that the number of nodes and the semi-bandwidth ofthe assembly do not exceed the maxima defined in the program as mentioned above. The user then

Finite Elements

writes the resulting network data - material, geometric and loading - to a data file which is sub-sequently read by the program. File details are as follows.

Data File PreparationAn existing data file such as FEX00 should be copied to the user's area, renamed and altered todescribe the object using a text editor such as Edit or SimpleText exactly to the following layout :

Consistent units must be employed throughout the data file

first line - the problem title ( including the units used for the data, for future information )next line - either 'plane stress' or 'plane strain' ( 12 characters ), as the case may beone line for each material :

- material index, elastic modulus, Poisson's ratio, weight density, thicknessone line - '0' : this single zero terminates the lines of material propertiesone line for each node, in ascending sequence 1,2,3 etc; or interpolate ( see below ) :

- node index, x-coordinate, y-coordinateone line - '0' : this single zero terminates the lines of nodal parametersone line for each element, in ascending sequence 1,2,3 etc; or generate ( see below ) :

- element index, material index, three or four nodal indices (no blank at end of line)one line - '0' : this single zero terminates the lines of elemental definitionsone line for each specified nodal displacment, or load, or inter-node traction (no leading blank) :

- type identifier ( see below ), node index/indices, specified value/valuesone line - '0' : this single zero signifies the end of transmissible data; further notes may follow

this marker, but are ignored by the program.Notes :Coordinates and vectors described in a user-defined Cartesian system (counter-clockwise x-to-y).Usually a separate line is required for each node, however if a group of sequential nodes is col-

linear and equi-spaced then only the first and last nodes of the group need be included inthe file; the program interpolates intermediate nodes.

An element is defined by three or four nodal indices, counter-clockwise around it.A separate line is usually necessary for each element, however if a group of elements is such

that the node indices each increase by one within the group (eg. 5-2-13-9, 6-3-14-10, 7-4-15-11) then only the first element of the group needs be included in the file; the program willautomatically generate the other elements of the group, assigning the material of the firstmember to them. The last element of the assemblage must always be supplied however.

Specified nodal displacements and loads are identified by a two-character 'type' code :First character, one of . . . . . Second character, one of . . . . .'d' specified nodal displacement 'x' in the x-direction' f ' specified nodal concentrated force 'y' in the y-direction' t ' specified inter-node traction 'n' normal to the edge, towards the body

( force per unit edge length, ie ( this code is relevant only to an inter-thickness*pressure or *stress ) node traction )

A fixed support is thus represented by two lines, type 'dx' and 'dy', at the same node and ofzero value. Tractions are specified only between neighbouring nodes.

Jim's ExampleThis example is presented solely as an illustration of data file preparation - its mesh is far toocoarse for accurate representation of the prototype.The geometry of the cantilevered flat plate, Figure 1 below, is known. The plate is made in twothicknesses from a material with elastic modulus 50 GPa and Poisson's ratio 0.25. The density is

Finite Elements 29

unspecified, so presumably weight forces are negligible.The plate is discretised into the 11-noded, 6-element mesh of Figure 2, in which the edges 1-2and 2-3 are of equal length, and the four elements 2,3,5,6 are identical parallelograms. The non-uniform pressures of the prototype are transformed into the equivalent edge tractions shown inFigure 2. Node 10 suffers a prescibed displacement, and node 11 is loaded by an inclined forcewhose components are sketched. The sliding supports at nodes 5 and 9 are justified by knowl-edge of the support of the prototype.

The data file for this problem, FEX00, appears as follows with italicised comments added :

Jim's Example ( N, mm ) Units stated for later ease of interpreting stress resultsplane stress1 50000 0.25 0.0 9 Notice the use of consistent units2 50000 0.25 0.0 70 Zero material-list-terminator1 0 0 Nodal index and coordinates4 48 102 Nodes 2 and 3 will be interpolated . . . .5 32 06 40 17 . . . . as will node 7 . . . .8 72 859 64 0 . . . . and node 10

11 96 680 Zero node-list-terminator1 1 1 5 6 2 Element index, material and counterclockwise node indices4 2 5 9 6 Elements 2 and 3 will be generated - see output5 2 6 9 10 7 Ensure that there is no trailing blank at end of an element-line6 2 7 10 11 8 The last element of the assembly must be included0 Zero element-list-terminator

dx 1 0 Specified nodal value : type, node and value - the first twody 1 0 lines refer to the fixed supporttx 2 3 0 108 Specified inter-nodal edge traction : type, indices of the neigh-tx 3 4 108 216 bouring node pair and the corresponding values of the ty 4 8 0 -270 traction at the two nodes (no leading blanks)

dy 5 0dy 9 0dx 10 0.01 Specified nodal displacement . . . .fx 11 6000 . . . . and concentrated force componentsfy 11 -90000 Zero end-of-data indicator

Care should be exercised to ensure that the data file is open only to one application at a time.

Finite Elements

Program Operation

Initiating the program leads to a call for selection of the input data file. The user is then asked todefine an output filename on a write-enabled disc - this name becomes the basis for :

- a text echo file, which saves the contents of the text window,- two trashable scratch files, which are empty on leaving the application, and - three files for saving drawings of the undeformed, deformed and patterned mesh.

The input data is checked for consistency by INPUT, after which the DRAW or SOLVE optionsmay be selected; a typical dialogue runs as follows, user-responses being underlined :

******** 17: 3:97FEM1k Finite Element Analysis in Two-dimensional Linear Elasticity******** with 250*72 maximum nodes*semibandwidth

FEX00 : Jim's Example ( N,mm ) Stress units will subsequently be N/mm2, ie. MPa

Echo nodes & elements ? (y/n) : n 'y' enables data verification, stepped through with button

Plane stress - with 11 nodes, 6 elements, and a semi-bandwidth of 12

op ? : d Draws the undeformed net, Figure 3, and saves to file Enter drawing retardation 0..4 : 2 Slowing down drawing execution aids checking for data errors

op ? : s Solves the network, and writes results to text echo file

Node Location Displacement Forcex y x y x y

1 0.0 0.0 0.0000 0.0000 3239.97 -9576.942 16.0 34.0 0.0484 0.0657 676.38 -0.003 32.0 68.0 0.2771 0.0616 4058.26 0.00 Depress button4 48.0 102.0 0.5832 -0.0459 3381.90 -1323.49 to suppress5 32.0 0.0 -0.0198 0.0000 0.00 -4277.10 screen output6 40.0 17.0 -0.0192 -0.0054 -0.00 0.00 of results though7 56.0 51.0 0.1453 -0.0861 0.00 -0.00 they are still8 72.0 85.0 0.4413 -0.2410 -0.01 -2646.98 written to text9 64.0 0.0 -0.0455 0.0000 -0.00 26824.51 echo file

10 80.0 34.0 0.0100 -0.2318 -17356.52 -0.0111 96.0 68.0 0.3267 -0.4787 6000.00 -9000.00

Element Centroid Stress - components - and principals equi-

& mat'l x y x y shear max min deg valent1 :1 22.00 12.75 -39.17 60.49 9.66 61.42 -40.10 84.5 88.62 :1 36.00 42.50 6.66 35.34 34.41 58.27 -16.27 56.3 67.93 :1 52.00 76.50 17.88 -14.15 24.98 31.53 -27.81 28.7 51.44 :2 45.33 5.67 -35.22 -20.78 6.22 -18.47 -37.53 69.6 32.55 :2 60.00 25.50 -75.18 -139.55 -20.36 -69.28 -145.45 -16.2 126.06 :2 76.00 59.50 25.58 -74.20 -10.66 26.70 -75.33 -6.0 91.6

op ? : d Draws and saves the deformed network, Figure 4, nodal Enter displacement multiplier : 25 displacements being magnified by the multiplier.

op ? : p Patterns the undeformed elements according to centroidal von Maximum equivalent stress is : 126 Mises equivalent stress (not shown here) - black if greater

than Enter desired maximum contour : 100 the maximum contour, then dark grey, etc. The plot is saved.

op ? : q Quit ( Option 'h' lists the options available ).

Clicking the mouse button enables exit from the drawing window to the text window.

UNITS, DIMENSIONS&CONVERSION FACTORS

UnitAll meaningful measurements in engineering science con-sist of at least two parts - a magnitude and a unit. Thus themeasurement 3 metres consists of the magnitude or num-ber, 3, and the unit, in this case the metre.

EntityA unit is measure of a certain physical 'entity', for examplethe centimetre unit is a measure of the entity length, thekilometre/hour a unit of the entity speed, and so on.

Entity Inter-Relationship - DimensionsFor any particular physical system under review, it is con-venient to class the entities which occur in it as either'fundamental' or 'derived'. The fundamental entities arethose which are selected - purely on convenience grounds -as basic buiding blocks of the system. The derived entitiesare then expressible as combinations of the fundamentalsraised to certain powers - we say that the derived entities'have certain dimensions' in the fundamentals.Suppose for example a geometric system was being exam-ined - a system comprising the entities length (denoted by[L]), area [A], volume [V] and angle [Ø]. Suppose also thatlength is chosen to be the fundamental entity, in which casearea may be conceived as a length multiplied by a length -this is written dimensionally as [A] = [L]*[L] = [L2] - that is'the derived entity area has the dimensions of lengthsquared'. Similarly [V] = [A]*[L] = [L3] and [Ø] = [L]/[L] =[L0] = [ ] since an angle may be thought of as the arc lengthof a circular sector divided by its radius. Entities in whichthe powers of the fundamentals are zero are termed'dimensionless', written [ ].Alternatively, if volume had been selected as fundamental,then [L] = [V1/3], [A] = [V2/3] while [Ø] remains dimension-less in terms of any fundamental.

Homogeneity of EquationsThe equation : 1 kilometre + 2 litres = 3 seconds is obvi-ously meaningless, and is so because the terms are of dif-ferent entities. An equation must be 'entity homogeneous' -ie. the terms of the equation must all be of the same entity -before the arithmetic 1 + 2 = ? can be carried out.Entity homogeneity is not always so obvious as in theabove example. Suppose we set out to calculate the area 'A'

of the Earth's surface which is visible from an aeroplaneflying at height 'h' above the surface, the Earth beingassumed spherical of radius 'r'. Suppose we came up withthe expression : A = 2π h r3 / (r+h). Choosing length [L] asthe fundamental entity, then the dimensions of the RHS ofthis equation are : [Lo]*[L1]*[L3]/([L]+[L]) = [L4]/[L1] = [L3]since 2π may be regarded either as a pure number or as anangle, both of which are dimensionless.The LHS of the equation, being an area, has the dimensions[L2], so the dimensions of the two sides of the equation aredifferent - the equation is not dimensionally homogeneousand must therefore be wrong - we can spot the error imme-diately. The necessity for entity homogeneity leads to apowerful technique known as 'Dimensional Analysis' foractually setting up the form of the equations governing anyphysical system, as opposed to the mere checking above.We shall not examine the technique further here.Consider now the equation : 1 centimetre + 2 metres = 3kilometres. This too is obviously incorrect even although itis homogeneous in the entity length. It is wrong becausethe units of the various additive terms are not identical -the equation is not 'unit homogeneous' and so 1 + 2 ≠ 3.The process of homogenising this simple equation will besecond nature to the reader, however the underlying prin-ciples of homogenisation still apply, and must be spelt outin detail, when dealing with more complicated functions.The technique is outlined below.

Conversion FactorsConversion factors are dimensionless numbers whichinter-relate, or convert, different units of the same entity.Thus '100 centimetres/metre' and '60 second/minute' arefamiliar conversion factors - their dimensions are respec-tively [L]/[L] = [ ] and [T]/[T] = [ ] where 'T' is the entitytime.Considering the above equation : 1 cm + 2 m = ? again, thehomogenising procedure is :

1 * 1 * 1 + 2 * 1 = 10–5 + 2*10–3 = 2.01*10

–3 km102 103 103

cm m km m km km kmcm m m

Note that before any addition is carried out, each term isbrought to a common unit by successive application ofconversion factors. Whether these multiply or divide isseen by inspection of the units - in the present case they areapplied to eliminate 'cm' and 'm' in turn. The resulting

common unit of the additive terms is 'km' here, though 'm'is much preferred as it eliminates any power of 10 - ie. themost succint answer is 2.01 m.

Introduction to Unit SystemsAlthough the Systeme International d'Unites (SI) is nowfirmly established as the standard system of units in thiscountry, the US system - which is similar to the old Impe-rial System - is still very common in the mineral and chem-ical processing fields, and looks likely to remain so forsome time. This, and the necessity to interpret overseastexts, means that engineers must be fluent in systems otherthan the SI. We shall examine the SI and the Imperialbelow, with a view to illustrating the techniques of con-verting from any one system to any other.Three fundamental entities only are a necessary foundationfor Mechanics (though Thermodynamics and Electricityeach requires an additional fundamental). Those usuallyselected - on the basis of simplicity - are length [L], mass[M] and time [T], and their units are defined arbitrarily.The units of the derived entities are then based on thesefundamental units.

Newton's Second LawThis law, which states that :

force = mass * accelerationis the foundation of Mechanics, and must be homo-

geneous in both entities and units. Acceleration is 'timerate of change of velocity', that is a velocity incrementdivided by a time increment - velocity itself being displace-ment per unit time. So [acceleration] = ( [L]/[T] )/[T] =[LT-2] where L and T are fundamental. For entity homoge-neity therefore, Newton's Law requires that the dimen-sions of the derived entity force, F, are given by :

[F] = [M]*[LT-2] = [MLT-2].For unit homogeneity on the other hand :

force units = mass units * length units / (time units)2and to ensure this the equation is written :

force = mass * acceleration / gcwhere gc is a dimensionless conversion factor

inserted to preserve unit homogeneity in whatever systemis being used - that is each system is characterised by itsown unique gc.It will be seen below that in some systems the magnitude ofgc equals the magnitude of standard gravitational accelera-tion - this results from the manner in which the unit of the

derived entity force is defined in the system and is NOTdue to any supposed direct dependence of gc on gravity -gc has got nothing to do with gravity - it is simply a con-version factor.

The Systeme InternationalIn this system the standard units of the fundamental enti-ties are defined as follows :mass - kilogram (kg) - the mass of an arbitrary lump of

platinumlength - metre (m) - an arbitrary number of wavelengths

of atomic krypton radiation (nominally an evenfraction of the Earth's equatorial circumference)

time - second (s) - an arbitrary fraction of the year 1900.The unit of the derived entity force is called the Newton(N) and is defined as the force necessary to produce anacceleration of 1 m/s2 in a mass of 1 kg. The correspondingvalue of gc follows from the necessary homogeneity ofNewton's Law :

gc = ma/F = 1 {kg} * 1 {m/s2} / 1 {N} = 1 kg.m/N.s2The weight, W, of a mass of 5 kg for example, is the forceexerted by the Earth on the mass, accelerating the mass ifnot equilibrated at the acceleration of gravity (for whichthe standard value is 9.81 m/s2). So:

W = ma/gc = 5 {kg} *9.81 {m/s2}/ 1 {kg.m/N.s2} = 49NThe reader is advised to follow through the cancelling ofunits here, in a manner similar to the '1+2=?' equation onthe previous page. Systems like the SI in which the magni-tude of gc is unity are called 'absolute' systems.

The Imperial SystemThe unit of mass, the 'pound' (lb), and the unit of length,the 'foot' (ft), are currently defined in terms of SI units -conversion factors of 2.205 lb/kg and 0.3048 m/ft apply -while the unit of time is identical to the SI second.The unit of force, the 'pound force' ( lbf ), is defined as theweight of a pound mass under standard gravity conditionswhen the acceleration is 32.174 ft/s2. The value of gc for theImperial system is found in a manner identical to the abovegc = ma/F = 1{lb}*32.174 {ft/s2}/1 {lbf} = 32.174 lb.ft/lbf.s2The weight of a 5 lb mass under standard gravity condi-tions is therefore :W = ma/gc = 5 {lb}*32.174 {ft/s2}/32.174 {lb.ft/lbf.s2} = 5 lbfAgain, the reader should confirm manipulation of the units.

Systems like the Imperial in which a mass's weight is equalnumerically to the mass itself are called 'gravita- tional sys-tems'. US notation differs from the Imperial in using theabbreviations 'lbm' for pound mass, and 'lb' for poundforce.Suppose it is required to find the weight of a body, WM, ata point on the Moon where the gravitational acceleration is5.2 ft/s2, the body weighing 3 lbf under standard Earthgravity conditions. From the above, we see immediatelythat the mass of the body is 3 lb, so, using the last equationWM = ma/gc = 3{lb}*5.2{ft/s2}/32.174 {lb.ft/lbf.s2} = 0.48 lbfConversion of force from one system to another is usuallymost easily carried out by invoking the gc factors of thetwo systems. Thus to find the Imperial equivalent of 10 N,we have :

F = 10 * 1 * 2.205*( 1/0.3048 )*( 1/32.174 ) = 2.25 lbf

N kg.m lb ft lbf.s2N.s2 kg m lb.ft

Proceeding from the left, the derived unit in one system (eg.N here) is first reduced to fundamental units in that (SI)system, then inter-system fundamental conversion factorsare used - in numerator or denominator by inspection - toconvert t o fundamental units i n t he second system(Imperial here) before final reduction to the derived unitsin the second system via that system's gc. This approach isquite general and may be used for derived entities otherthan force.

EnergyAll forms of energy - be they mechanical, thermal, electri-cal, nuclear etc. - are equivalent, that is they have the samedimensions. The two forms of most immediate interest tous are thermal energy or 'heat', Q, and mechanical energyor 'work', W (whose symbol should not be confused withthat of weight). Their equivalence is expressed in the FirstLaw of Thermodynamics : W = Q (briefly, this will be for-mulated more rigorously in Thermodynamics).Work, a derived entity, is conceived as that which is donewhen a force moves its point of application, and is theproduct of the force and the distance moved in the direc-tion of the force, so :[energy] = [force]*[distance] = [MLT-2]*[L] = [ML2T-2]

The unit of work, energy and quantity of heat in the SI isthe 'Joule' { J } which is defined as the work done when aforce of 1 N moves its point of application through a dis-

tance of 1 m - ie. a conversion factor of 1 J/Nm applies.In the Imperial system, different units are used for thermaland mechanical energies, so a conversion factor 'J' is neces-sary in the First Law statement, thus : W = J Q.J is sometimes referred to as 'the mechanical equivalent ofheat', however it just another conversion factor - analogousto Newton's gc but for the entity energy/work/heat. TheImperial unit of work is the foot-pound force { ft.lbf }, andthe unit of heat is the British Thermal Unit { BTU }. Usingthese units, J has the value of 778 ft.lbf/BTU (approx). So,from the First Law, the amount of work which is equiva-lent to 10 BTU of heat is :

W = 778 { ft.lbf/BTU } * 10 { BTU } = 7780 ft.lbfWhen inserting values into algebraic equations, the readeris strongly advised to adopt the technique used above, inwhich :

- the units associated with each number are writtenunderneath the number

- a running check of the units is kept so that placementof conversion factors, in numerator or denominator,can quickly be determined.