A Handbook 0f Building Structure = Allan Hodgkinson

8/8/2019 A Handbook 0f Building Structure = Allan Hodgkinson

1/47

AJ Handbook ofBuilding StructureEDITED BY Allan Hodgkinson

The Architectural Press, London- .


2/47

AJ Handbook of Building Structure Introduction 4 . I

Allan HodgkinsonConsultant editor and authorsThe consultant editor for the Handbook i s Allan HodgkinsonMEng, FICE, FIStrUCtE, MCOnSE,, Principal of Allan Hodgkin-son & Associates, consulting civil and struc tura l engineers.Allan Hodgkinson has been the AJ consultant for structuraldesign since 1951;he is a frequent AJ contributor and is theauthor of various sections of this handbook.The authors of Lach section will be cre dited at the star t ofthe section of the Handbook in which their material appears.The original Architeck Journal articles wero edited byEsmond Reid, BArCh, and John RlcKean, BArCh, MA, ARIBA,ACIA, ARIAS.The frontispiece illttstralaon shows one o f th e mostmagnificent builrlitq structures from the era of the EiffelTower , the Forth Bridge and the great rai lway stations.The Palais des Machines for the Paris Exhibition of 1889(Contarnin, Pierron (e: Charton, engineers) was a pioneerexample of three hinged arches.Preface to the second editionThere have been considerable changes in some BritishStandards, Codes of Practice, and Building Regulationssinco 1974; nd unlike the reprints of 1976 and 1977, hi s isa substantially revised and updated re-issue of the nowwell-established AJ Handbook of Building Structure.The principal changes are i n the sections on Masonry (r e-written to take account of the 197F Building Regnln ons,and the new BS 5628 limit state code of pract ice); and onTimber (substantially revised to take account of the newtimber gradings).Steel handbooks have been replaced for a11 types of struc-tural sections; and technical study Steel 3 has thereforebeen revised accordingly.In gcnernl, t he new limit state npproach bo design is tlis-cussed (eg in the seciion on JIasonry); but in VICW of thercjcction of tho limit state Codes and draft Codcs in theirpresent form, by the majority of practical designcrs, it hasbeen thought prudent to retain the allowable stress methodsof design as the basis of the handbook.Finally, it should be mentioned that tho opportunity hasboon taken to bring all rcfercnccs i n this Handbook up todato; and to correcta number of misprints of th e first edition.I S BN 0 5139 273 3 (paperbound)First published in book form in 1974 byThe Architec;ural Press Limited LondonReprinted 1976, 1077Second edition 1980, 1982, 1983Printed i n Grea t Britain byMackays of Chatham Ltd

This handbookScopeThoro arc two underlying themos in this new handbook onbuilding structure. First, the architect and engineer hnvocomplementary roles which cannot be separatod. A mainobjoct of this handbook is to allow tho architect to talkintelligontly to his engineer, to apprcciatc his skills and toundorstand tho reasons for his decisions. Socond, thubuilding mus t always bo scon as a whole, wheru the succcss-fill conclusion is thc result of optimiscd decisions. A balanccof planning, struc ture o r services docisions may no t ncccs-sarily provide tlio cheapest or best solution from any oftheso soparato standpoints, b ut the wholo building shouldprovido tho right solution within both tho clients brief andhis budget.Tho handbook provides a reviow of tho wholo structuralfield. It includes soctions on movemont in buildings, fireprotection, and structural lcgislation, where philosophy ofdosign is discusssed from the firm base of practical cxperi-once. Foundations an d spocific structura l matorials are alsocovered, while sufficient guidance on analysis and designis givun for the archi tect t o deal with simplo structure shimself.ArrangementTho handbook doals with its subject in two broad parts.The first deals with building stru cturo generally, tho secondwith t he main structural materials individually.Tho history of the structural designer and a generalsurvey of his field tod ay is followed by a section on basicstructural analysis. The general part of tho handbookconcludes with scctions on structural safoty-includingdeformation, f i e and legislation-and on the sub-structure:foundations and retaining structures.Having discussed th e overall structure, t he sections in th esecond part of tho handbook discuss concrete, steelwork,timber and masonry in much greater detail. Finally themare sections on composite structures and on new andinnovatory forms of structuro.PresentationInformatio n is presentod in three kinds of format: technicalstudies, information sheets an d a design guide. Th e technicalstudies are intended to givo background understanding.They summarisc goncral principles a nd include informationth at is too general for direct application. Information sheetsaro intended to give spocific data that can bo applicddirectly by the dosigncr.Keywords aro usod for idontifying and niimboring technicalstudios and inforination shoots: thus, tochnical studySTRUCTURE 1, information shoot FOUSDATIONS 3, an d so on.The design guido is intondcd to roniind designers of thopropcr sequence in which decisions rcquirod in the designprocess should be takon. It contains concise advice anclroforences to detailed information at each stage. This mightseem tho normal starting p oint , bu t the guide is published atthe ond of the handbook as it can be employcd only whenthe dosigner fully understands what. ha s been discussedearlier.Th e gciwral piittern of use, then , is first to read tho relevanttoclinical studies, to undorstand t.ho design aims, t.hoprobleins involved an d tho rango of nvailablo solutioiie.,Tho inforinntion shocts then may be r:s~d s a design aid , tLsoiirco of dtittx nnd design information. Tho design guide,acting also as n check list, ensures th nt docisions are t.akenin tho righ t, sequenco and th at nothing is left out.


3/47


4/47

AJ Handbook of Building structure Introduction: Section 2

Section 2Structural analysisScopeThe later sections on foundations and structural materialsare presented in such a way that complex mathematicsere not essential to an understanding of structural design.They are intended to give readers a structural awareness,and an appreciation of structural form; the detailed designwork being carried out by a consulting engineer.However, information is given on the calculation of simplestructu ral mombors, and the purpose of this present sectionis to provide an clementary knowledge of terms and theirmathematical basis. Detailed examples on the sizing ofmembers is dealt with unde r each structura l material. Thissection covers olomentary statics, the internal conditionsand strengths of materials, beam and stiht theory, andanalysis of some typcs of structure .

AuthorThe author for Section 2 is David Adler BSC DIC MICE, acivil engineer with varied experience in both consultingand contracting.

i

David Adler

28

The cover illustration which appears on the previous page:ompares suapenaion by a single rope, bearing a total lo&d,withhe load borne b y two ropes using a pulley. The engraving isdapted from Leonis English translation (1755) f AlbertisTe n books on architecture, and illustrates book VI, chapter VIIuhich wncerns simple mechanics.

/- - ..


5/47

29 Technical study Analysis1 para 1.01 to 2.08

Technical studyAnalysis 1 Section 2 Structural analysis

Statics and strengthsof mater a s

1 Introduction1.01 The majority of architects would regard structuralanalysis as bcing entirely pure and applied mathematics.This is not stri ctly correct. Mathematics is an imp ortan t toolbut not the only one. The testing of existing structures,mock-ups and models, and analysis of the results, as wellas photo-elastic methods, provide an alternative approach.1.02 In addition, it should be appreciated that the mostabstruse mathematics and the most rigorous analysis canproduce no better answer than the correctness of the basicassumptions made. Mathematics cannot give all theanswers about a problem any more than an architecturalmodel will exactly reproduce the appearance of the finishedbuilding. Mathematical analysis, in fact, does construct anabstract model of the structure. The limitations of theparticular model in each case must always be rememberedby t he problem-solver, jus t as the architect remembers thatthe brickwork of his model is actually perspex.2 StaticsForce2.01 Force is almost indefinable. It is known mainly by itseffects: acceleration, strain and so on.2.02 A force has three characteristics: magnitude, direction,an d point of application. I n these respects it can be repre-sented by a straight line, its length proportional to themagnitude of the force 1.2.03 Any two or more forces acting at the same point canbe replaced by one combined force called their resultant.The magnitude and direction of the resultant can beobtained by the method known as the parallelogram offorces 2 .

1 Force represented bystraight line. Length isproportional to magnitude 2 Parallelogram of forces

This w he first of two technical studies by DAVID ADLER,which provide a simplified mathematical background tosome of the terms used in the later sections on strucluralmaterials, and on foundations and retaining w alls

. .2.04 Conversely, any force can be resolved into separate,smaller forces called components in given directions. Assu-ming forces are acting in one plane only, a force P at anangle 4with the horizontal xx axis 3, can be resolved intocomponents P cos 4 for the xx axis and P sin 4 or the YYaxis in th e direction of these axes 4.2.05 Resultants and components of forces are replacements.Replacements must not be confused with equilibriants whichare described in para 2.12.

X

Y P sin q5 P

3 Diagram showingconv ention of horizon tal xxaxis and vertical YY axis4 Resolution of forces. ForceP is resolved into itahorizontal and verticalcomponents

Moment2.06 To understand the concept of moment, first considera plane system of forces.This s a system which is confinedto two dimensions, and can be represented on a piece ofpaper by lines as described in para 2.02. No forces act outof, or into th e plane of the paper.2.07 The moment of a force about a point is found bymultiplying the magnitude of the force by the distance ofits line of action. Thus in 5, the moment M of force P aboutpoint A, is the product of P, the magnitude of the force, andthe distance d of its line of action from A measured perpen-dicular to t he direction of the line of action.Thus Moment = Force x distance, or M = P x d2.08 But, more usually, forces act in three dimensions, inwhich case point A becomes an axis perpendicular to th eplane of the pap er, bu t th e same definition applies.

,

\;Moment of a force.Moment is qual to forceP x distanced


6/47

Technicalstudy Analysis 1 para 2.09 to 2.17

Couples2.09 A couple is formed by two forces, equal in magnitudeP, acting in parallel but opposite directions distance dapart. The moments M of those forces from any point A,distance x from one of the forces, are given in the formulabelow and shown in 6.

D /

6 A couple. Tw oforces equalin magnitude, parallel and

I P opposite in directionMoment = force x distance x + force (distance d -,

distance x)= force x distanced

or M = P x x + P ( d - x ) = P x dThis value is independent of the position of the point aboutwhich moments are taken. T he two forces form a couple orpure moment. The value of a couple in a given plane isconstant everywhere in that plane.Equilibrium2.10 A force P acting on a free body of mass m in space, willresult in that body moving with acceleration -. If a bodydoes not move (or moves with constant speed in a rictionlessenvironment), then forces acting on t ha t body are said to bein equilibrium. That is, th e resultant of all the forces is zero.2.11 Firs t consider a very small body, small enough, in fact,to be a point. This point body is in equilibrium, so that allthe forces acting on it must have zero resultant 7a. Insteadof using the parallelogram of forces, described in para 2.03,the polygon of forces is used 7b. Alternatively, the com-ponents of each force can be considered in the x and Ydirections. These components also have zero resultant, andIM they all act in the same direction, the arithmetical sumof all the components is zero. in each direction 8. Findingthe components in direction x and summing to zero isreferred to as resolving i n direction x.

Pm

b wrresponding polygon ofJorces. Magnitude anddirection of forces representedb y length and direction of

l a Forces acting at a pointin equilibrium, thereforeresultant is zero: lines2.12 If th e point body is not in equilibrium, then the forceson it have a resultant. A force equal in magnitude to thisresultant, but exactly opposite in direction, will cause thesystem to be in equilibrium, so this force is called theepuilibriant. For example, a brick weighing 45 N resting ona evel table-top has a downwards force of 45 N acting onit 9. It does not move, so the table is supplying an equili-briant of 45 N upwards. This type of equilibriant is usually

30called a reaction. This proves t he first law of statics which is:Action and reaction are equal and opposite.2.13 Now consider a perfectly rigid body of larger dimen-sions, in space. Forces &FB acting on this body, but unlikethe point body considered earlier, these forces are not allapplied a t one point, but at a number of positions around,and possibly even inside, the body. These forces can beresolved in the X,Y, nd z directions*, and the componentsin these directions will still sum to zero if the forces are inequilibrium. This means that the body will not move up,down, or sideways. There is, however, a further mode ofmovement not yet investigated: th e body could rotate.2.14 If the body does not rotate, the moments of the forceson that body must also be in equilibrium. In a plane body(of only two dimensions) moments can be taken about anypoint, a nd th e sum of the moments of all the forces will bezero. In three dimensions, moments can bo taken about an yaxis and the same will be true.

YII P i?

3- -.

-X

8 Forces at a point withweiqht 45N resolution into components

along the xx and YY axeS9 Ac!ion and reaction areequal and opposite (first lawof statics)reaction 45N

92.15 Four equations apply to a plane body in equilibrium:1 and 2: Resolution of forces in two different directions3 and 4: Moments taken about two different points.Only three of these four equations are independent. By thelaws of algebra the values of only three unknown quantitiescan be discovored. If this is not sufficient to find out all theforces on the body, the system is said to be staticallyindeterminate.2.16 When considering a body in three dimensions it ispossible to determine:1 resolution of forces in three directions2 moments about three axes.Of these six quantit ies, any five will give all the informationit is feasible to obtain in thi s way.Example2.1 7 In the example of the plane body shown in 10 , discoverthe forces necessary to maintain equilibrium horizontally:P - 100 = 0 (mus t have zero resultant to maintainequilibrium): P = 100*As prevlously explained in fig 3 Xx and YY are the horizontal axis andvertical axi s respectively in one plane only. The zz axis is the third dimensionaxls, acting perpendlcular to the other two


7/47

31Now take moments about point X, he intersection of forcesP an d Q :R x 6 - 100 x 3 = 0.. R = 50Finally, resolve vertically:.. Q = ItBut R = 50.. Q = 50In this examplo there wero throe unknown forces, P, Q andR so that three equations wore sufficient to find theirmag nit udes.

R - Q = O

Forces and moments: Summary2.18 Three facts are self-evident and import ant to remember:1 Two forces in equilibrium must be equal in magnitudeand opposite in direction at tho same point of application 112 Three forcos in equilibrium must pass through a point(if tho lines of action are extended far enough). Theirmagnitudes must conform to t he triangle of forces, 123 Any force P acting a t a given point can be replaced by aforce of equal magnitude acting at any other point in aparallel direction distant d from the original line plus acouple of.magnitude Pd , 13 .

A bI50

Q10 Equilibrium. Example inpara 2.17 shows how tocalculate forces P , Q and R

11 Two orces in equilibriummust be equal in magnitudeand opposite in direction

trianqle offorces

12 IP - 100 100t ,/--,500 t 12 Three, rces in

r t I equilibrium. Alagnitudesand directions represented in13 Force replaced byforce + couple. ForceP = 100. Couple = P x

I6 triangle of forcesI

13 d = 100 x 5 = 500Rigid body mechanics2.19 A rigid body i n the rod \vorld h n s mass. It is thcrcforoacted on by gmvity, which iii:poscs n downward forcecalled its weight. I f th e bocly is susponded by n string, i t willrotatc until th e tcnsion in tho str ing is i n th e same line as thcline of action of the wcight. The linos of act ion of the weightalways pass through on e particular point i n tho bocly, evenwhen the position of tlic point wliere the string is attache d ischanged 14 . This point is called the centre of gravity, and i f q

Technical study Analysis 1 para 2.17 to 3.01the mass of the body werc concentrated at this point, itsbohaviour under outside forces would be the same.Example2.20 A convenient type of rigid body to use as an exampleis a retaining wall. Assume th at thc a.ctua1wall is quite long,but that a slicc is taken from the middle, of unit length 15 .

2.21 There a re only three forces on this slice of wall:1 its own weight W, acting at the centre of gravity of theslice2 pressure of the earth behind the wall P : soil mechanicstheory indicatcs tha t this pressure is likely to act a s shownin 15.3 the reaction under the foot of the wall R.2.22 These three forces must pass through one point; thereaction must thereforc pass through the intersection of theweight W and thc earth pressure P. The magnitude anddirection of this reaction is obtained from the triangle offorces (see also 12). If th is reaction as drawn does not passthrough th e base of th c wall, bu t falls outside it, th e wallwill fall over.2.23 It will be shown later that the wall can still fall overeven when tho reaction passes through the base, if it doesnot pass through th e middle third of tha t base.3 Strength of materials3.01 The previous section on statics considered forcesacting on the structure as a whole. The later section ontheory of structures will deal with the forces acting withinthe structure. This present section is concerned with theforces acting within the materials of which these membersare composed. T TA A

1W +W14 Find ing the centre ofgravity. The body issuspended f ro m each of it scorners ( T ) .Where the

verticals d u e to its weight( W ) onverge, there is it scentre of gravity or centroid

15 Forces o n a retainingwall. Earth pressure P andweight o j wall W catme areaction. R under the f o o t ofthe wall. Its magnitude anddirection is obtained fr omthe triangle of forces (right).

trianqle offorcesProcedure: plot P and W toscale and inclination in thetrian.gle of forces. J o i n endsto obtain R whosemagnitude can be scaled anddirection determii? d- -


8/47

Technical study Analysis 1para3.02 o3.08

Structure3.02 A structure is any body of material acting in a waywhich changcs the magnitudo, position, or direction ofnatural forces to t he advantage of the user 16.3.03 A structure is usually coinposed of structural mcmbers.A structural mcmber is nearly always of ono, or at the mosttwo homogenous matcrials, is prismatic (SCC later) andconnects two nodes, or points where other membors converge17 . The line joining tho nodes is callcd a longitudinal axis,and the section of the inombcr is the plane figuro producedby cutting t he mcmber at right-angles to th is axis. A mem-ber is said to be prismatic whcn the scction dcos not varyalong its length 18 .3.04 A structure can bo composed of an amorphous mass ofmiscellaneous content, or of non-prismatic mombers ofchanging composition, but the vast majority of structuresfall into th e normal category. The analysis of other types ofstruc turo is not within t he scope of this section.

16a 16ba -----, 16 Examples of structures:a reducing the magnitude of

a force; b altering theposition of a force; andc altering the direction of a

5 --.,73 ----,

1 i force16c4D P E

A H17a

B

17bB

17 Typical structures: abuilding frame; b roof truss;

c cable-stayed bridge. Nodesare lettered. Members arenumbered (see para 3 . 0 3 )

32

Stress3.05 When studying the forces acting 011 tho section of amcmber it IS ncccssary t o un derstand thc concopt of s t r e s s .If a very sinall area of thc section IS consiciered, it can beessumod tha t th c forcc on it IS evenly distribu ted. Thc stresson that small are% 6A is then cqual to thc force 6P on the

6Parea, divided by the area (see 19): stress = -SA3.06 Usually thc stress wil l ac t at scme angle to t he planeof thc section. For conveniencc, it is rcsolvcd into com-ponents (sec 2 0 ) :1 direct stress (f) ,acts perpendicular, or normal to th c planeof t he section2 shear strcss (s ) acts parallel to, or in the plane of thesection.

18 Ezample of a prismaticmember which has aconstant cross section. Thisone is member 7 from 17a

/9a

19b

20

19a Forces acting on asection; b forces on a smallarea 6A20 Resolution of stress intodirect ( f )and shear ( 8 )stresses. (Shear stress isshown on diagramsthroughout this section b ysingle headed arrow)

3.07 Direct stress can bo compressive or tensile depending onwhether it is tending to shorten or lengthen t he member.A compressivo stress is usually referred to as positive, anda tensile stress as negative 21a.3.08 It is important to distinguish betwcen th e stress actingon th e section, and t he stress which is the equal but oppositereaction to it (para 2.12).As defined in par a 3.03, the sectionis prodticod whcn the member is cut. This reveals twoopposite faces, each face acting on the ot her with the equaland opposite stress. In this context opposite meansopposite in direction. For example, a left-to-right directstress on the right-hand face implies a right-to-left directstress on the left-hand face 21b. However, both of thesestresses would tend to shorten thc member, and they are


9/47

.. .. .

3 3 Technical study Analysis 1 para3.08 to3.18

therefore both compressive. A downward-acting shearstress on the left-hand face would mean t ha t there was a nupward-acting shear stress on the right-hand face 21C.3.09 It is usually most appropriate to consider the stress asacting on the section, rather tha n emanating out of it. If allthe little bits of direct stress acting on the section in thisway are added up, the total direct force P on the seotion isobtained.

cornpre ssive stress tensile stress21a

21 b

21 c21a Conventions for showingcompressive and tensilestress, b compressive stress;each force acts on the otherwith equal and opposite

stress. c shear stress;downward acting stress onleft hand face balanced b ysimilar upward acting stresson the right h n d face

22 Summation of directstresses on a section intototal direct force P

3.10 The little b it of force on each area is 6P which is equal tof x 6A from para 3.05 above. Adding all the little bitstogether can bo written as follows:

The symbol x means the sum of all suc?~. uantities . . , an dBA means a little bit of A .3.11 The shear forces can also be summod 23. As these willvary in direction all over the plane of thc section, it is usualto resolve these a econd time into the vertical an d horizontaldirections Y and x thus:Sx =x(sxx 6A ) an d S, = x ( s y x 6A). (S = total shear)3.12 All the small forces on the section have now beenrecLced to three forces at three mutually perpendiculardirections: P, S and S (see 24). However, the points ofapplication of these forces are not known as hese depend onthe actual distribution of the strosscs across the section.To progress further it is necessary to investigate thegeometry of tha t section.

P = C ( f x 6A).

Section geometry3.13 Consider the section in 25 . First eatablish x an d Y axesso that any point on the section can be referred to by itsco-ordinates. The aroa of this point is 6 8 .3.14 If all the littl e areas are added togother the area of the

iy

23 Resolulion of shearstresses on a section into vertical and horizontalcomponents

24 Forces acting on asection; P = direct force,S , and S, = shear forcespi s, I

5 A ( a t 1,3), ' first moment about x axis- y I 6 A - 3 ~ d A

second moment about x axis- Y ' X b A - 9 s d A

- - _ -X25 How tojind t hejrs t andsecond moments of area of asectionIY

section A is obtained:A = Z 6 A3.15 The value of the moment of the small area about theY axis is x x SA. All these moments added together giveth e jirst moment of area of the section, G.Thus G, = Z (x x 6A) and G, = 2' y x 6A)

Centre of area3.16 If the section is symmetrical about the x an d Y axes,each little area will be balanoed by an identical area on theother side of the axis. The values of G, an d G, will both bezero. Even if the section is not symmetrical, it is alwayspossible to choose x and Y so that the first moments of areaabout them ere zero. The origin of these axes (their inter-section) is called the centre of area of the section. If the shapeof the soction were cut out in cardboard, the centre ofgravity of the body would be the same point as the centreof area. It follows that if G is zero about two axes through apoint, it will always be zero for any other axis at any anglethrough the same point.3.17 The longitudinal axis of a symmetrical structuralmember is usually assumed to run through the centres ofareas of the sections of the member.3.18 A simple horizontal structural member is shown in 26 .A longitudinal axis passes through t he cen tre of area of thesection. An x axis, which is horizontal, and a Y axis, whichis vertical, pass through the same centre of area. - -


10/47

Technical study Analysis 1 para 3.19 to 3.30

26b I26a Horizontal structuralmember subjected to directforce P ; b force P replacedby replacement force acting

at centre of area creatingtwo couples My and Mx,called bending moments

Bending moments 3.19.In para 3:12.and 24, itwtwseen hat the forces on.the

section could be reduced to three forces in mutually perpen-dicular directions: P, S, an d S ,. The shear forces S, and S,will be considered later; in this section the direct force Pwill be examined in detail.3.20 When all the little elements of force on the sectionwere added together, the magnitude of their resultant was,found to be P 22 . The position of the point of applicationof this resultant was not discovered. Assume that this forceacts at a point on the section with co-ordinates X an d Y26a.3.21 It was shown in para 2.18 that any force at a givenpoint could be replaced by an equal force through anotherpoint plus a couple. In this way, the force P at X, Y can bereplaced by a orce P a t 0 , O ie at the centre of th e section)plus couples PX and PY acting in the horizontal andvertical planes 26b.3.22 All the little elements of direct stress on the sectionhave now been replaced by a force P through the centre ofarea of the section, and two couples. These couples are calledbending moments and are represented My or the moment inthe horizontal plane, and M, for the moment in the verticalplane.3.23 Each little element of stress makes a contribution toeach of these quantities. The contributions can be separatedout :fa is the part of the stress that adds up to make P acting atthe centre of area of the sectionf,, is the part th at adds up to make M,f,, is the part th at adds up to make M y.

34These separate parts are each stresses acting at the samepoints, so that3.24 From the definitions in the paragraph above, theFollowing equations can be writ ten (see also 27):f = f a + fb x fb yP = Z f x SAM,= Z f x y x SAM y = f x x x SA

+ 4M x - moment resultant in

M y - moment resultanthorizontal plane- I k ~ y 6 A t I ~ x d Aj x 6 A+ + -

27 Diagram to illustrateequations. described i n para3.26,,showing direct forceond,resulhnt moments3.25, The proofs OK , he following assumptions are given inp&a,3.47. It wou1d:not be appropriate to give them here,as they rely on the concept of strain which has not yet beenreached. At this point, these assumptions may simply beaccepted:fa is cons tant in value over the whole section.The value of. fbr is p r o p tional, to the distance from thex axis of. he element ixiquestionie: f,, = k x y where k is a constantsimilarly the value of fby s proportional t o the distancefrom the Y axisie: fby= j x x,where j is a second constant.3.26 If these xaluw are.now put into the equation in para3.23 it is seen tha t:f = f a + k x y + j x xand this may be substituted into the three equations inpara 3.24 to give:P = f a Z S A + k Z y x S A + j Z x x SAM, = fa Z*(y,x SA) + k Z(ya x SA) + j Z(x x y x SA)My = fa Z ( x x SA) + k Z ( x x y x SA) + j Z(xa x SA)These equations are illustrated in 27 .3 .27. In p$a 3.16 i t was demonstrated that if the origin ofth e x an d Y axes was the centre of area of the section, thenZ (x x SA) and Z x SA were both equal to zero.3.28 The same is not true of Z (x2 x SA) and Z (y2 x SA).These are called the second moments of area of the section,or sometimes the moments of inertia; and are representedI, and I,.3.29 The quantity Z(x x y x SA) is called the product ofinertia and is shown as I,,.3.30 The equations in para 3.26 can therefore be writtenas follows:P = fa x AM, = k x I, + j x I,,My = k x I,, + j x I,These are the fundamental equations of bending for thegeneral case. In most practical examples of sections, thevalue of I,, is zero. This is because most sections are sym-metrical about at least one axis: and this means that anelement of positive x x y always has a correspondingnegative value to balance it. However, a section like theone in 32c is anti-symmetric about a diagonal axis, and for(


11/47


12/47

Technical study Analysis 1 para 3.40 to 3.524 the section of the member always remains plane5 the material of the section is homogenous and elastic.Strain3.41 A thin metal rod under tension will stretch. Thismovement under a force is called strain, and strain isdefined as the extension or contraction of unit length of themember 29.

1 - L - eI- -Tstrain- QL

29 Strain i s extension orwntraction of urvit length. formL = length of member;

e = extension; 1 = teneile

Elastic strain3.42 One form of strain is called elastic. Elastic strain haatwo properties:1 Hoakes Law applies. This states that stress is propor-bional to strain, ie- constant (called Youngsmodulus)2 Removal of the stress causes the member to return to itsoriginal state.3.43 Not all materials behave elastically. Some are likemodelling clay and deform increasingly under a constantstress, not returning to their original shape after the stressis removed. This behaviour is called plastic deformation.3.44 Most structural materials behave in a manner similarto that shown on the stress strain curve in 30. For low valuesof stress the material behaves elastically until the elasticlimit is reached at point B. The behaviour is then more orless plaatic until the breaking point is reached at C. If thestress is reduced during the non-elastic deformation, thematerial does not return to its original form but retains apermanent deformation as shown by the dotted line.

stressatram

stresC.breaks

strain30 Stwm-8tmA diagramshowing elastio limit whereHookes Low (stress isproportdonal to strain) nolonger operates. Dotted line

shows dejomnation causedwhen stress is reduced afterelastic limit and materdaldoes not return to i t r ,original form

Bending stress and strain3.45 Having introduced the concept of strain, the behaviourof a member under bending can be examined more closely.3.46 Fig 31 illustrates a member of elastic material underbending. Any section that was plane before bending tookplace is assumed to remain plane during bending. Twosections distance z apar t will then subtend a n angle 4 t thecentre of bending such tha t z = 4 x R. (R is the radius ofourvature).

36

31 Strain i n bending.Diagram shwa a memberunder bending with twoseotwna distance Z apartsubtending an angle 4. R Lsradiw of ourvatwe. Fibreconsidered is d i s t am Yfrom neutral axis (seepam3.41)3.47 Consider a fibre above the longitudinal axis of themember through the centre of area of the section (calledthe neutral axis). If this fibre is distance y from the neutralaxis, it will be extended during bending to a length of4 R + y) ; the extension is therefore 4 x y and the strain:6 x R R+or X Y -

stressstrainNow- E (Youngs constant) for an elastic materialE

R(see para 3.42). Therefore stress = - x y = k x y asassumed in para 3.25.Shear3.48 The detailed theory of shear stress is more complicatedthan for direct stress. Therefore only an outline of theresults is given here rather than a rigorous analysis of themethod.3.49 Shear stresses at each point on the section are resolvedinto vertical and horizontal components (see para 3.11).The resultants of all these components comprise thevertical and horizontal shear forces on the section S, andS,.As for axial stresses, it can be shown that if these forcespass through a certain point some special conditionsapply: there wil l be no torsion or twist of the member. Ifthe resultant of all the shear stresses does not p m hroughthis point, the moment of that resultant about thepoint is the torsion on the section.3.50 In the case of shear forces the point is called the shearcentre. This is not always the same point a& the centre ofarea. I n fact, i t is the same point only when there axe twoaxes of symmetry or anti-symmetry 32. A very commoncase of misconception is the channel shown in 33. The shearcentre is a t point S, and a oad of 20 N applied at the middleof the flange M will cause a torsion of 1 - 3 1 Nm on thesection(ie 28 mm +- x 20N = 1310 Nmm75 mm2

or 1 - 3 1 Nm )Distribution of shear stress3.51 For a symmetrical section with no horizontal shear, atheory can be derived to find the maximum shear stress andits position.3.52 Consider the section in 34a. This section is assumed tobe subject to bending about the x axis only, with no directforce. An arbitrary shape is chosen to develop a generalisedtheory before considering special cases like I-sections orrectangular shapes. II


13/47

37

a b32 Sectionqvwit?&?Learcentres at centre of a,ren

2ON! -+-I

centre of/area--

-t 28 4 - 7 5 4

C Y

33 Chatme1 section withnhear centre S not in thename position, as centre ofarea A

3.53 A narro \v horizontr .Btrip of section has ami 6A. Thedirect stress on this strip is f = k x ( y ) , where (y)repre-nents the distance of the strip from the neutral axis (thehorizontal axis through the centre of area of the section),aind k is the constan t (from para 3.25).3.54 Consider the area By n 34b. This arca extends from aline distant y from the neutral axis to the extremity of th esection. The to tal force on this area 18given by Z x (y) x6A over this area or k Z (y) x SA.Now Z (y) x 6A is thefirst moment of area of the area By bout the neutral axis,or By yb if y b is the distance of its centrc of area fromt,hat axis.'rhus: P = k x By ybBut k x yb is the value of the stress tit t ,he ccmtrtx of arm ofBywhich can be called f b .40 P = f b x By

i- -t

34b34 Diagram to illuetrateshear theory as developed i~para 3.52 et seq

Technical study Analysis1para3.53 to 3.57

3.55 No\\. consider a thin slicc of the bea.ni of thickness 6z.This is shown i n 35 which is not a section of the beam butan elevation. The part of the slice corresponding to thearea H, of the section has been separated from the rest ofthe slice. It has a force P on each face, and a shear forces x s x 6z on its base from th e rest of the slice. Theseforces must be in equilibrium, so by horizontal resolution:P I - P , = s x s x 6z: n, (fb,- b2 )= 8 x s x

Y b By(M, M,)3.56 By niet,hods outside the scope of these notes, i t can beshown that when 6z is allowed to get very small, the valueof IS equal to the vertical shear force S, on thewhole section (see 34b).

MI INO\\.. f = - SO that s x x x 6~ =-M, - M , .

62

Hetice s = By-b x S )I x s4 r a x x b 2

neutralaxis

35 Elevation of beamdeaoribed in para 3.55 (seealso 34)

3.57 By the formula above, the value of the longitudinalshear stress in the beam can be determined. Th at thi s stresscxists can be demonstrated by using two planks to span awide gap. If the two planks are simply placed one on theother , they sag when the load is applied: each sliding on theother. If the two planks are nailed together this slip cannot.occur and th e load-carrying capacity of the pair & greatlyincreased 36.

a

b

C

36 Two planks spanning awide gap; a beforeapplication of point load,b after appl ication of pointLoad with no jixing betweenand c effect of nailing plankstogether


14/47


3.58 However, there is also vertical shear stress on the faceof the section. I f this is shown on a figure similar to th at in35 it becomes a hown in 37. From the argument in para3.65, these two shear stresses are equal in value. The formulaabove, therefore, is also used to find the vertical shearstress at a point on the section, as demonstrated in thefollowing example.

tbrl

neutral -37 Diagram to illustrateeffect of vertical shear on

-

I . beam shown in 35ExampleUsing the section shown in 28, find the shcar stress at thejoin between parts I an d 11, when the shear force on thesection is 20 kN. (All dimensions i n mm).By = area of part I = 2.125 x 103yb = centre of area of part I from neutral axis = 154.5I, = 127.1 x l oo (from table I )x = brcadth of section at point of ca1c;ilation = 20S, = shear on section = 20 x 103so that:

2125 X 154-5S= x 10-6 x 20 x 103127.1 x 20S= 2.58 N/mni2.Shear on a rectangular section3.59 Consider a rectangular section depth d and breadth b(see 38).By= (i - y) x b

32b x d b x d 3

aection (.we para 3.59)+- b-t

38

3.60 The value of this is obviously greatest when y = 0(tha t is a t the neutral axis) and then:3 s

2 b ds = - x -or the maximum shear stress in a rectangular section is 14times the average shear stress, 39 .3.61 Fig 39 shows the approximate distribution of stress onthe rectangular section in graph form. It will be seen thatwhen y = - ha t is, on the edge of the section, the stress iszero,a ight be expected.

d,2

rectonqularsection7-i39 Shear stress in arectangular section

Shear on an I-section3.62 The distribution of shear across an I-section is shownin 40 . It can be seen tha t th e distribution is almost even onthe web, and th at the influence of the flanges is negligible.The shcar stress on the web of the scction in thc cxample inpara 3.58 can thercfore be calculatccl as:20 000

250 x 20 = 4 N/mm2--I-section-2 max--opproxavero eshear rtrerr $web aloneca rr ie r rk a r40 Shear stress on anI-section

Stress relationships3.63 Stresses do not exist in isolation. Each form of stressis interdependent on other forms of stress.Poissons ratio3.64 A block of hard rubber if squeezed together will resultin a barrel-shape 41 . Similarly, a ubber band when stretchedgets thinner. This type of phenomenon happens with allmaterials: if the strain in the x direction is e, and thestrain in the Y direction e,0,- = U (sigma, or Poissons ratio)e*As Hookos Law also applies it is obvious that, in a similarway- - afxf Y

deformor oni------,P - PI I- _ _ _ e -41 Effeot of squeezing ablock of rubber


15/47

39

Shear balance3.65 In 42a is shown a small block of material taken fromthe middle of a structural member, length a, breadth b,and thickness c. On onc end there is a shear stress 8 , or aforce s x bc. If resolved vertically, ano ther shear stress s onthe other end of the block will prevent vertical movementof the block 42b, but this will produce a couple of values x abc. To balance this couple, shear forces s' at the topand base of the block are introduced, producing a couplein the othe r direction s' x abc, 4 2 ~ . ence s = s'; a verticalshear stress will thus always induce a correspondinghorizontal shear stress of equal magnitudc.

42 Shear balance: a shearstress on one end; b verticaland opposite force on otherend prevents verticalmovement but creates acouple; c wuple balanced bycouple in horizontaldirection s'

bDirect stress induced by shear3.66 Diagram 43 can be used to discover how shear strcssproduces direct stress. It is a triangular slice of the blockshown in 42 . The length of the hypotenuse is 1. Assume aninduced direct stress of f and shear stress of s on thishypotenuse. By resolution in the direction o f f f x lb =s' x lb x cos 4 x sin 4 .+ s' x lb x sin $ x cos 4and resolving in the direction of s:s x lb = s' x lb x sin24 - ' x lb x cosp43.67 The interesting result of this exercise becomes evidentwhen 4 = 4 5" because then s = 0 and f = s', that is, ashear stress produces a direct stress of equal value across aplane at 4 5 " to t he shear plane. This, of course, is commonknowledge 44.3.68 The converse is also true: a direct stress induces ashear stress in R plane at 4 5 " t,o ita p lane 45 .

: f = 8 sin 24

: s = 3 cos 24

Principal stresses

43 Diagram to show howshear stress produces directstress (see para 3 .6 6 )

3.69 In para 3.67 it \vas soen that the plane when 4 = 4 5 "had no shear stress across it . By the theorem in para 3.65 i tfollows tha t the plane at right-angles to this plane wouldalso be a plane of zero shear. The direct stresses across theseplanes are known a8 principal stresses, and one of these willbe a maximum direct stress, the other a minimum. Thedirections of these stresses arc called principal axes, and if


they are drawn on , say, an elevation of the member, theyform stress trajectorics.

44 Tension failure due toshear force. Shear stress planepmdwea direct stress of

equal value across a 45'-_A .A

45 Shear failure du e todirect force. Converse of 44

Stress trajectories3.70 Stress trajectories in the side elevation of a cantilevurare shown in 46 . The full lines indicate the directions ofprincipal tensile stresses, the dotted lines the directions ofprincipal compressive stresses. These linos cross, as des-cribed, a t right-angles. The stress magnitudes along theselines do not necessarily stay constant.

r -IIIIIIIIL -

-+

FI 'compression 46 Stress trajectories in acantilever

Photo -elasticity3.71 The directions of principal stresses can also be investi-gated by means of the curious phenomenon known asphoto-elasticity. This results from the fact that wheneverpolarised light passes through certain plastic-basodmaterials, the planes of polarisation are rotated. Theamount of rotation has been found to vary with the condi-tions of stress within the plastic material, so that a modelof a struc ture made of this plastic will show a pattern underpolarised light when subjected to forces simulating theloading on the structure 47 . A similar offoct is seen whenlooking at a toughened glass car windscreen throiighpolarised sunglasses.

I ! 47 Photo-elasticpallerns i n a beaiic.iVote th e region OJpure bending in the

I

T middle, th e shear nearI a the ends, and the

stress wncentrationvtcndersthe j o w loadsf--- -- -'


16/47

Technical study Analysis 2 para 1.01 to 1-08 40

Technical studyAnalysis 2 Section 2 Structural analysis

Structural types

1 Beams1.01 Beams are either statically determinate or staticallyindeterminate. For statically determinate beams bendingiiioments and shears can be determined using statics only.This type of beam is considered first; statically indeter-minate beams are dealt with in para 1.24.1.02 Statically determinate beams are either simply sup-ported or cantilever.Simply supported beams1.03 A simply supp orted beam is shown in 1a. It is assumedtha t the support at each end is a perfect hinge, and th at oneend is also frec to move horizontally. The same beam, con-ventionally drawn, is shown in 1 b. These perfect conditionsare never achieved, but assumptions are necessary to pro-duce a workable theory.

I lood uniformly distributed olonqlenqth of beam II

I Il b1 Simply supported beam: a support at each en d assumed tobe perfect hinge, right end can, move horizontally;b conventional way of showincl beam a with uniformlydistributed ( u n ) oad1.04 The beam has a span L measured between centres ofsupports, and has one or more loads on it. I n theory, loadsare of thre e types:1 point loads (P )2 uniformly distr ibuted loads (W or w)3 non-uniformly distrib uted loads1.05 Point loads are assumed to set at a point, but i l lpractice thcy always spread a short distance. An exampleof a point load is tt column resting on a beam 2a . This isrepresented conventionally in 2b.1.06 !l!he forces i n 2b a t A and C are reactions, and areusually shown R, nd R,. Taking moments about A for thesystem :20 000 x :3 - It, x G = 0 (must have zero resultant, to: It, = 10 000 N or lOkN maintain equilibrium)~

This i s the second of two technical studies by D ~ V I D D mnwhich together give a brief mathematical and desoriptivebackground to the later sections of the handbook. This studydeals with beams and struts and varioua types of surface andskeletal structures

simply supported beam/'

2a 1 2OkNA c-+m

BM D

2c

2b2a Simply supported beam with point load in centre of span:b diagrammatic representation of beam with reactions atA and C and load of 20 kN.at B, bending moment diagram( B M D ) and shear force diagram (SFD) shown below;c dimensions used in calculating bending moment underpoint load1.07 No w imagine th at the beam is cut immediately to th eright of B, where the load is applied. The forces on th e halfof the beam between this cut an d C are:reaction R, = 10 kNshear force at R = S,bonding momen t at B = M,1.08 These forces on the pa rt of thc beam betwcen th e cutand C must be in equilibrium:: 8, - It, = 0 by ver tical resolution (2b)and M, - R, x 3 = 0 by momcnts about B (2c).ThusS, - 10 = 0.'. S, = 10 kNan d M, - 10 x 3 = 0: M, = 30 lcNm


17/47

41

1.09 The bending moment and shear force diagrams (BMDand SFD) for the beam can then be drawn (2b).1.10 The beam shown in 3 haa a uniformly +stributed loadof 1 - 5 kN/m over a span of 10 m. Using the same analysis asbefore, or simply-bysymmetry:

1;s- x i o = 7 . 5 k N2, = R, =1.11 Bending moment Mb a t the centre of the span is foundin the same way:(The effective resultant of the unifoim load over half thespan acts at the centre of gravity of the half-load, ie thequarter-span point, or 2 . 5 m from C): Mb = 3 7 . 5 - 1 8 . 7 5 = 1 8 . 7 5 kNm1.12 Shear force at the centre is Sb= R, - 1 5 x 5 = 0 orThe BM and SE diagrams are shown in 3a.1.13 These examples give an indication of methods forsolving problems from first principles. Normally formulaeobtained from tables are used (see Information sheetWL wL2ANALYSIS 1)an d 2 an d 3 would be solved using- nd-4 8respectively, where W or w = load and L = span.

M b - R, x 6 + 1 . 5 x 5 x 2 . 5 = 0 (3b)

s b = 7 . 6 - 7.6 = 0 .

Cantilevers1.14 The cantilever beam (in 4) and the loads on it aresupported at the left-hand end A, where there is a reactionR, and an end fixity moment M,. End A is often describedaa erboaatre or built-in.1.15 It is very simple to calculate R, and M, by statics.In the example in 4:Vertical resolutionR, - 4 x 3 - 10 = 0

.*. R, = 22 kNMoments abou t AM, - 4 x 3 x 1.5 - 10 X 3 = 0: M, = 48kNmDeflection of beams1.16 As well aa calculating stresses in structural material, i tmay be necessary to predict how much a structure will move.Then the movement of individual members is assessed;each component can move in three ways:1 It can lengthen or shorten. This is strain due to directstress and is readily calculated:movement = length x strain2 It can bend. The longitudinal axis can move at right-angles to itself.3 It can twist. A simple calculation if torsion on the memberis known.1.17 Bending deflection is most difficult to calculate,mainly because the bending stresses are usually not constantalong the length of the member an d the theo ry requires theuse of calculus. [B ut for simple problems there are twotheorems tha t avoid higher mat,hematics; their proofs,however, will have to be taken on trust.

= length x stress/E (Youngs modulus)

THEOREM 11.18 The change i i i the slope of ~t eam over H. given lciigthis q u a l to the area of the- iagram which equalsME1Bending moment

011 t,h>it, ength 5a.YouI>gs modulus x moment of inertiaTHEOREM d1.19 In 5b the vortical (listmicey of -4 clow t,lic twtrgent t i t

Technical study Analysis 2 para 1.09 to 1.19I .SkN/m

B

I--- I o mII 3b

I-4 jSFD3a3a Sim ply supported beam with uni,formly distributed (WD)load, showing bending moment and shear force di ag ra m ;b dimensions used in calculating maximum bending momentat centre of span

4a

I

I-SFDb

OkN

-4-IlOkNI1-

-44c4a Cantilever beam with UD oad plus point load at the end;b bending moment and shear force diag vaw s; c imensionsw e d i?ccalczdatiiig ~ D l n nd SED


18/47

Technical study Analysis 2 para 1.I to 1.25-span or pa r t o f span.

dopedope,i

5a

t-;Ic -

,.beam os deflected

BA

centre of area

ib .I x' 45 Change of slope and deflection: a illustrates theorem 1 andshows either complete span or part span where change ofslope = area of - iagram; b illustrates theorem 2 and showsonly part of a span where y = area of - iagram x x'B is equal to the first moment of area of the - iagram onAB about A .

ME l

ME l

ME1

1.20 In most cases, the material (a nd therefore E ) and t hcsection of the member (an d so I ) s constant over the lengthof the member. Thereforc i t is only necessary to consider

Mthe moment variation when using these theorems. The -E1diagram becomes the &I or bending moment diagram, theE1 factor being used when the slope or deflection has to becalculated.1.21 To show how these theorems are used, consider thecantilever in 6. The M (bending moment) diagram is atriangle of area- ts centre of area- L from the free endA. The beam is horizontal a t B, so that the tangent at B isalso horizontal. Thus the deflcction of A (y.) is equal to the

Mfirst moment of area of the - diagram about A:E 1

PLS 22 3

PL2 2 PLS2EI 3 3EIy n = - x - L = -1.22 In practice, cantilevers are often placed at the ends of

continuous beam runs. Then the beam is not horizontal at Band the deflection of A will be greater by the slope at B(which must be calculated) x length L .1.23 The deflections of beams are not usually calculatedfrom first principles, becausc there are tables coveringcommon loadings. The .real use of slope and deflectioncalculations is in the design of structures that are notstatically determinate.

4 2Encastr6 beams*1.24 The simplest form of statically indeterminate structureis the beam built-in (encastre') t both supports 7. A staticallydoterminate beam can have two simple supports, or onebuilt-in support. With one built-i n support and one simplesupport the beam has one degree of redundancy, becauseif the bending moment on the enmtre support wereremoved, the beam would be statically determinate.Tho doubly built-in beam has two degrees of redundancy.

L 1.- i 1

1 centre of area

If.BMD

PL33 E IDejledion in o oantilever, y a is equal to- ie the first

moment of areu of the- iagram about A )ME I

7a

7b I I

7a Beam built in (encastre) at both supports; b momentdiagram deacribed in para 1.25. M , = ree support moment,

wL2M , = end fixity moment =-121.25 The bcnding moment on a statically indeterminatebeam has two components:1 the free support moment M, hat would occur on theequivalent statically determi nate beam2 the end-fixed moment M,, which has a straight-linodistribution along the beam. This is limited by values ofthe moments a t each end.From thcorem 1, since there is no change of slope along thisMbetun (i t is zcro at each enoastre end), the area of t h o -ET.diagram must be zero.The area of the free support bcnding moment diagram isequal and opposite in sign ( + - to the area of thc cnd-fixity moment diagram.

*Hem eMaslrd i i i e m s beams rigidly tlxetl nt both ends and not memly built.into brlckwnrk


19/47

43Because t.his beam is xymmetrical; the Inoineiits at each endare equal, and the area of the end-fixity mointwt diagramis MF L.'rhus:

LMFL = - M,L (The I L ~ C Rof (I parabola, i n this case MsL, ist\\o -t,h irds of it s enclosing rectang le)

2 wL2 wL23 8 12

3

... M, = - x - -Example : co mplete analysis of a propped cantilever1.26 The beam is shown in 8.1.27 First find th e value of MFThe t,angent at A is horizontal. Hrcansc 13 is propped thereMis no deflection there and so the first moment of t,he -E1diagram about B must be zero.

L 2 2 wL2 LM F 1 - x - L = - x- L x -2 3 3 8 2'. .whereM, ,v - S the are a of the triangle formed bet\vecn R Aan dR , (shown dott ed on t he BM D in 8)2-L ts th e distance of its centre of gravity from It,32 wL23L Y- x L is two- thirds of th e enclosing rectangle ofthe parabola, which equals the ar ea of t he parabola

L2

8

L- is t.he distance of its centre of gravity from It,.2wL2 2 x 5 - 5 28 = 7 . 5 6 kNm8: M, =- =

I B M D rnox BMI It-.5 8 7It--

ISFD8 Propped cantilever used i n analysis described i n paras1.26 to 1.33. Maximum B M occurs at point of zero shear1.28 The vertical reactions RAand Re t each end of thebeam are calculated by taking moments about each end inturn.Moments about ARE x L + MF -- -= 0 (Must have zero resultant to

maintain equilibrium)wL22

Technical study .Analysis 2 para 1.25 to 1.33

2 x 5 . 5 7 . 5 62 5 . 5

= 5 . 5 - 1 - 3 7 = 4 . 1 3 kN... K., = -- -Moments abou t B

wL22R A x L - M F - - = o.'. R A= 5 . 5 -t 1 . 3 7 = 6 . 8 7 kN

Check by vertical resolution1.29 Calculation of the maximum positive moment in thespan of the beam is easier becauso the maximum positivemoment occurs at the point of zero shear. (Proof of thisinvolves calculus and is outside the scope of these notes.)1.30 Assume that the point of zero shear is x metres from13 towards A. Resolving vertically on XB :w x x - R , - S x = 0 (Where S, = the shear force at X)

R,But S, = 0 as X is th e point of zero shear, so that x =-

4 . 1 3 + 6 . 8 7 - 2 x 5 . 5 = 0

W4 . 1 3

2 = 2.065m-'Caking moments at B for XB :

wx22M, - - - - 0 (Where M x = the bending moment at X )

2 x 2 . 0 6 P2 = 4 - 2 6 kNm (approx).. Mx =

1.31 Now calculate the maximum stresses in the boam.Assume a beam section of 250 x 7 5 mm rectangular.Area of the section is 250 x 7 5 = 1 8 7 5 0 mm2The second moment of area I about the neutral axis isbd3 7 5 - X 5 2 0 3From the general equations of bending developed earlier*M f- = -(where f = stress, and y = distance of extremefibres from neutral axis)

= 98 x 10' mm4.=1 2 1 2

I Y

: f = -yMIThe maximum value of M is MF, an d the maximum value ofy is at the top an d bottom of the section.In term s of th e units by which these values have so far been

M (in kNm)1 (in mm4)easured, f = x y (in mm)

To make them comparable, M must be multiplied by1000 (or 103) ,so that all lengths are in mm, and by a further1000 (a total of 10e) to reduce kN t o N, rn t he h a 1 answeris bctter expressed in tha t form. (Conversion to like unit s isused th roughout this section.)

7 . 5 6 x 10'98 x 10' x 125 = 9.64N/mm2f,,, =

1.32 The maximum shear force is R A = 6 . 8 7 kN (see 8).The maximum shear stress on a rectangular section is 1.5times the average stress?

RA.*. S m a x = x 1 .5area (b x d)6 . 8 7 x 103 x 1 . 5 = 0.55N/mm218 750s,,, =1.33 The calculntion of maximum dcflcction under loadWL'

18 5 E1could be very difficult, bu t t he formula- s availablefrom tablcs and t he deflection under load can be computedprovided the Young's modulus (E) or t he materiF1 is known.

*see technicnl study ANALYSIS 1 para 3.30'see technical study ANALYSIS 1 para 3.60


20/47

Technical study Analysis 2 para 2.m to 2.11

I

2 Struts2.01 A strut is any structural member that is subjected tomainly compressive forces. Often the word is reserved fornon-vertical members; vertical struts are described ascolumn stanchions or piers, the use depending to someextent on the material (stanchions are mainly iron andsteel, and piers brickwork or masonry).2.02 The theory of struts involves the concept of unstableequilibrium. A system of forces in unstable equilibrium isshown in 9; this complies with all the requirements forequilibrium, but the smallest change in any of the forceswould destroy the system. Systems in unstable equilibriummust be avoided in structural work!5\N 5kN

I

lOkN9 Unstable equilibrium, any change would destroy system2.03 Each end of the structural member in 10a is a pin-jointthat can only move in the direction of the length of themember. As long as the member is perfectly straight, theonly stress in the member will be pure compression, ofmagnitude: A( 'z).f the member were slightlybowed

I BMD Ilo b10 Preswre on a strut: a pin-jointed member with appliedpressures resulting i n pure compression; b strut dejlected bypressures. Maximum B M at centre &? P y2.04 A rigorous analysis of the deflection requires calculusbut the approximate method used here gives a similaranswer. If the deflection of the st ru t is parabolic the momentdiagram follows the same parabola. The moment at anypoint x is Py. From theorem 2, para 1.19, the deflection of Babove the tangent at C (the centre of the member) is equalto the first moment of the- iagram on CB about B.ME1

516(note -L s distance from B of centre of gravity of that

section of the parabola)NOW c= P x y5 PL2so that y = - x - y48 E1

44

This is a rath er curious result: eithery = 0 and - x - an have any value,

5 PL2or - x - 1 and y can have any value.48 E1Thus the re will be no appreciable bending of the s trut until

48 E1P, the compression on the strut, approaches- henbLa *the deflection of the centre of the s trut increases rapidly, andthe str ut buckles. (- = 9 .6 . A rigorous analysis gives acoefficient of approximately 9.87) .2.05 When designing strut s it is therefore important not toexceed the P calculated above. This load is called the Ederload. Normally a oad factor of at least two is used so thatfor safety a working load of half the Euler load should no t

5 PL248 E1

485

AQ AQA" A"be exceeded. Thus in theorem 2, a igure of half - r-5 5 x 2is used.2.06 If the compressive stress is f, then P = f, x A so t h t48 If, max =- E X -5 x 2 AL2I

Anow - = r2 (r is the radius of gyration of the section):. f, max = 4 . 8 E -(r)PL- is called the slenderness ratw of the member.r2.07 For many materials, values of fc m ax are tabulatedagainst values of the slenderness ratio so that calculationis easier. Of course, the maximum stress must not exceedthe safe compressive stress of the m aterial of the memberwhatever th e value of the slenderness ratio.2.08 The theory of struts is not confined to members inpure compression. The compression flanges of membersunder bending can also be unstable, and th e magnitudes ofthe compressive stresses on such mem6ers are limited bythe theory. Here instability depends on a number offactors, and tables are used to determine the safe com-pressive stresses.Bending and compression2.09 Columns in building frames, which are struts, arecommonly subjected to both compressive and bendingstresses (produced b y eccentricities of loading on to thecolumns). It is therefore important to understand thetheory of combined bending a nd compression.2.10 The rectangular section ( l l a ) 460 x 260 mm under aload of 300 kN and subject to a moment of 4 kNm can betreated in two ways:METHOD 12.11 First calculate the stress due to pure compressiveloadingP 300 x 103f a = - = = 2 67 N/mm2(P s multiplied by 1 0 3

to translate IrN o N)450 X 250Then work out the tensile and compressive stresses due tobending

M bd3 df, = f - - y where I =- nd y = -' - 1 12 24 x 106

4503250 x-12:. f, = f - X 225 = 0.47 NJ-91 -The maximum compressive stress is therefore 2 67 + 0 47= 3.14 N/mm2 and th e stress on th e opposite edge of thesection is 2.67 - 0.47 = 2 . 2 N/mm*.(See l b ) .

f -


21/47

45METHOD 22.12 Convert the compression force and moment into aneccentric force. If e is the eccentricity:M 4 x 103e = - = - - - 13-3mmP 3002.1 3 The distribution of stresses on the rectangular sectionof the column is shown in 12a. The area of this diagramequals the compressive force 300 kN, and the centre of areacoincides with the point of action of the,force. Thus thestresses on each edge of the section can be found.This method is import ant because it covers the behaviourof the stru ctur e under changing conditions.2.14 If t he load is increased without changing its eccentricity12b, the stresses would all increase in proportion to theincrease in load wi thout changing thei r distribution.2.15 If t he eccentricity of a constant load were to increase13, then the stress on the left-hand edge of the sectionwould decrease 13a, and th e stress on the right-hand edgeincrease. When the stress on the left-hand edge becomeszero 13b, the stress diagram becomes a triangle, its centreof area one-th ird of the base length from the righ t-handedge, or one-sixth of the base length from the centre-axis.This middle third zone of the column is extremely important-if the load does not leave thi s zone, there is no tension onthe section, if the load 1s outside the middle thir d, there istension on part of the section 1 3 ~ . his is not always verysignificant, but some materials have very small tensilestrength, and then there is a fundamental change in thebehaviour of the section. This will be described in th e late rsections on structural materials.21 It is possible for the load to have so great an eccentricityas to lie outside the section. Then bendingis more significantthan compression, and the member should be treated aa abeam. 0 4 k N r n

3 0 0 k N

- - - 4 5 0 - - 4cross section

a I

0.47,

-t-1--

2 2 0I-t-1b

l l a Rectangular section under load with a moment on thesection (method 1, see para 2.11); b (top) tress due towmpression loading, (middle)stress due to bending,(below) wmbined stress


Ia I

position ofcentre ofarea of stressblock is underI load

at same eccentricityb

12 Rectangular section as 11 but with load and momentsombined to form eccentric load (method 2 , see para 2.12):a distribution of stresses; b effect of i n ~ r ~ i n goad but notchanging eccentricity

C I13 Effect on beam (shown in 11 ) of increasing eccentricity:a stress on left-hand edge decreases until it reaches b, atrkngle where load f a h within middle third, andjinallyc when tension i s produced because load l k utside middlethird

Ties2.17 Members subject mainly to tension are fairly easyto analyse, but if the member also bends the treatment issimilar to that for combined bending and compression. Asthere is no case of instability, tension members can bevery slender; wire or thin strip is often used.rhe area of a tension member in the stress calculations is:he minimum section. Often this coincides with a hole in


22/47


the member 14 . For a threaded bolt the minimum sectionoccurs at the root of the t hread.3Types of structure3.01 Having looked a t stresses in structura l materials a ndthe design of structural members methods of analysingstructural systems can now be examined.3.02 These systems are classified:1 Skeletal structures: Pin-jointed (eg roof-truss 15a) Rigid-jointed (eg Vierendeel frame 15b)2 Surface structures (eg shell roof 15d)3.03 There are also systems which combine elements of eachtype-thus a continuous beam system is partly rigid-jointed and partly pin-jointed ( at the supports) 15e.3.04 Mathematical analysis of structures is a 'kind ofmodel-making; the structural systems' analysed are modelsys ems.3.05 Real structures are, of course, three-dimensional, butfor simplicity they are usually split into a number of planarsystems. The building frame in 15 f is an example of suchtreatment.3.06 Surface structures are often analysed as if they wereskeletal struc tures. The multi-storey flat slab system in 15 gis analysed as two separate but interlocking rigid-jointedskeletal systems.3.0'1 Rigid-jointed skeletal structures are often analysed asif they were pin-jointed. A typical example is the commonroof truss 15c where the rafters are large continuousmembers.3.08 The successful solution of a structural problem dependson th e correct choice of model system. There is a particularmethod of analysis for each model.

IILT 1 - o

'm~nimm iect:on14 Tension member- tie; minimum section area is useain calculations

a 1

C15

1

Y1I

e

6c

46

b frame5like thir 4 fromeslike thisg15 Types of structure: a pin-jointed roof truss (skeletonstructure); b rigid-jointed Vierendeel frame (skeletonstructure); c common roof truss (rigid-jointedskeletalstructure often analysed as though pin-jointed); d Shell roof(surface structure); e contiicuous beam (combined skeletal andsurface); C building frame (analysed as series of planarsystems); g multi-storey flat slab system (surface structureanalysed as thcmgh skeletal)


23/47

47 Technicalstudy A116lySi6 2 psra 4.01 to 7.04

4 Isostatic truss4.01 I n an isostatic or statically determinate truss all forcescan be determined without consideration of the size ormaterial of the member. It is the archetype of the pin-jointed skeletal system.4.02 The members of this truss are all straight, joints are allpinned, and all loads and reactions act at the joints. Thusthere can be no bending moments in any of the members.4.03 Forces on the members are purely axial, and can befound by resolution a t each joint in turn.Alternatively the method of section can be used-the trussis assumed to be c ut in two and th e forces on each part canbe resolved, so that the in ternal forces in the cut membersare established.4.04 For most cases the quickest and easiest way is agraphical method using the polygon of forces+-Bowsnotations. (See 16.)4.05 A letter is given to each of the spaces enclosed by themembers and applied forces, 16a.

3kNI

I4.5kNa

b

A I4.5kN/Ib

16 aaphioa. method of determining forces in a roof trwrs:a t ruas with spaces lettered to identqy forces and members;b Bows notation (a special form of the polygon of forces)which, (f drawn .a scale, will enable force% in members to bemeasure& directly (see para 4.04 to 4.10). Direction awowson triangle abc relate to directions of forces around left-handsupport in a. These directions indicate whether the memberie in tension (t ie)or compression (strut)4.06 The Bows notation diagram 16b is built up startingfrom the left-hand support. The reaction of 4.5 kN is anupward force dividing space A from space B, going round thejoint in a clockwise direction. Therefore, draw ab verticallyupwards of length proportional to 4 . 5 kN. Continuinground the joint clockwise, the rafter dividing B from C isreached. The force in this member is compressive, actingdown the slope towards the joint. Line bc can be drawnparallel to this force in the same direction, although theposition of c is still unknown. The next member at thisjoint is the tie CA-its tensile force acting away from thejoint. Drawing ca parallel to th is tie establishes the positionof c and hence the lengths of bc and ca. The magnitudes of*see technical study ANALYSIS para 2.11

the forces in BC and CA can therefore be established bymeasuring lines bc and ca respectively.4.07 The joint in the middle of the rafter is considered nextmd the procedure is repeated working.clockwisearound th ejoint. cb is already drawn, so this acts as the base for thisjoint. The magnitude of the load BD is 3 kN, so the positionof d can be established. de and ec are drawn parallel to t helines of members DE and EC to establish point e.4.08 Each joint is treated in this way to complete 16b.Notice that c and j turn out to be the same point in theliagram. The line bd-dh -hk is the same line in reverse aska-ab, and represents the external loads on the truss.4.09 The real truss differs considerably from the model.For example, the compression in rafter BC is found to be9 kN. The load from the roof is not applied at the joint, bu trtt a number of points along the rafter, and the bendingmoment due to these loads must be found. The combinedbending and axial stresses must also be examined to ensurethat they are less than the permitted maxima.0.10 Thus analysis of the model system is not the end of th ecalculation, but only one of the stages.5 Space frames5.01 To a certain extent th e popularity of the space frame isdue to the advent of the computer. Nearly all space framesttre rigid-jointed, and the analysis of axial, shear, bendingand torsion forces in the members and on the joints wouldbe beyond normal hand-methods. I n practice the usualassumptions for pin-joints, which are so useful in planeframes, are both dangerous and costly when applied tothree-dimensional problems.5.02 Most space frames are designed by specialist COIU3UltantEo r contractors who have access to t he complicated computerprograms required.6 Rigid-jointed frames6.01 The analysis of a large multi-storey building framewith rigid joints (17a) is also a job for the computer. Butthere is a subdivision process that allows manual methodsto be used.6.02 First, the beams at each floor level are analysed ascontinuous beams supported on pin joints a t each columnposition. The beams can be designed using results from th isanalysis 17b.6.03 Next, the bending moments imposed on the externalcolumns are estimated using empirical formulae. Thisinformation and the axial loads (estimated from the loads ofthe building fabric and superimposed loading 1 7 ~ ) reneeded when designing the columns.6.04 Finally, the wind moments in the columns can becalculated by one of a number of semi-empirical methods.This checks that the allowable increase in stress for forcesinduced by wind is not exceeded 17d.7 Continuous beams7.01 The most common problem in statically indeterminatestructures is the continuous beam. This s a combination ofa rigid and a pin-jointed skeletal system.7.02 Various ihethods exist for solving the problem-thetheorem of three moments, virtual work, influence co-efficients, and so on-these can be referred to in standardtextbooks. There are also tables that solve the problem forvarious span ratios and loading conditions.7.03 The method recommended for analysing continuousbeams is the Hardy-Cross or moment distribution method.7.04 Assume that all the support joints in a run of con-


24/47


X

C

17th Typical multi-storey frame with rigid joints; b beam-to-column junction at Y ; C beam analysis of typical beam X ;d typical empirical wind analysis of frame in 17atinuous beams 18a, are locked so tha t th e ends of each spanare horizontal and encastre.7.05 The end moments in each span are calculated.7.06 Then each joint is unlocked, allowed to take up its freeposition, and locked again. Each time that a joint is un-locked, the clockwise and anticlockwise moments from theend moments of the adjacent beams are balanced out 18b.7.07 There is also a carry-over of moment from the unlockedjoint to the locked joints on either side. This must becalculated an d allowed for. As a esult, each time a joint isunlocked, the previously balanced neighbouring joint hassome out-of-balance moment reimposed on it. Thus thedistribution method consists of several repetitions oriterations of the locking and unlocking, until the resjdualout-of-balance moment at each joint is very small 1 8 ~ .Theory of mom ent distribution method7.08 Apply a moment M A to the left-hand end of the beam

48

a I

b I

C18aContinmm beam with encastre ends showing loading;b beam when joints are unlocked; C final &-of -balancemoment at emh joint is quite small (see paras 7.04 to 7.07)with fixed ends 19a by rotating it through an angle 4.Amoment M , is induced at the right-hand end (see the

Mmoment diagram 19b). From para 1.18 the area of th e -E1diagram is equal to the change in th e slope of the beam ie

4 x (M A fE1=7.09 Also, from para 1.19 the vertical distance of A below

the tangent at B is equal to t he first moment of area of th eM- diagram about AE1(MA f 2MB) x La

BE1=hence M e = - MAsubstituti ng this in the equation above

L4EI= M A X -

4EILr M A =- t P

4EIL.10 The quantity- s th e stiffness of the beam. Stiffness

atMAb '-19a Beam with fixed ends, with moment applied to left-handend; b moment diagram

kEITable I Beam stiffnnesses.h'tiffnws =-LContinuous both endsstiffness factor k = 4carry-over 4

~~

Continuous one endstiffness factor k = 3carry-over 0Symmetricalstiffness factor k = 2carry-over 0

I Anti-symmetricalstiffness factor k = 6qn n, carry-over 0


25/47

49

+ 1 2 3

is defined as the moment required to achieve unit rotat ion.Table 1 gives the values of stiffness for a few useful beamconditions.7.11 Now M , = - #MA, tha t is, the carry-over moment ishalf the magnitude of the applied moment. The negativesign indicates that a hogging applied moment produces asagging carry-over (see 21). This is not the usual signconvention, and it will be seen later t ha t t he carry-over hasthe same sign as the applied moment.

-123 + 56 -56 + 9 7 - 97 1

Continuous beam calculation7.12 A continuous beam system is shown in 20.7.13 The stiflness ratios for each pair of spans is calculatedusing

kEIStiffness =-LAny quantity (E in this example) that is common to allspans can be ignored because it cancels out in the ratios.Table 11 Stiffness ratios for the beam in 20Span k I L Stiffness B C D

each x E x 10'AB 3 600 x l o o 7000 256 0.35

0.41 0.57BC 4 600 x 10' 5000 480CD 4 500 x 10 e 6000 333DE 3 500 x 10' 6000 250 0. 43

0.65 0.59

583

central point load3 0 k N uniform1 distributedI /load 2lkN/m

' ~ 6 0 0 0 ~ 6 0 0 0 ~beams of same material, E is constanta+M moment7 -ve momentT)

b20a Beam loading for example described i n paras 7 . 1 2 to7.28; b sign oonventionfor moment distribution7.14 The calculation in the second half of table 11 can beconaidered in the form:Stiffness of AB = 256Stiffness of BC = 480

256256 + 4 80Therefore, stiffness ratio of AB = = 0 . 3 5480the stiffness ratio of BC = -736

= 0 . 6 57.15 Next calculate the end-fixity moments for each span,assuming that it is encastre at the supports. The endsupports are assumed pinned, both in the calqulation ofend-fixity moments, and also, as above, in the calculationof span stiffness.At B, for span AB

wL2 25 x 7 2MF =- From table I ) =- 153 kNm8 8at B, in span BC, for uniformly distributed loadingwLa 25 x12 12MF=- =-- - 5 2 kNmand for point load


PL 30 x 5MF =- =- 19kNm8 8Total = 71 kNm

at C, in span BC, the same as at B by symmetry = 71 kNmat C an d D in span CD

wLa 25 x 62 = 75 kNm- - = -12 12-at D in span DEwLa 25 x 6aMF =- - 1 1 2 k N m8 8Sign oonventwn7.1 6 When putting values into the moment distributiontable, the correct sign has to be applied to each end-fiuitymoment. Considering the moment acting on the end of thespan, clockwise moments are positive, anti-clockwisemoments are negative 20b.Table 111 ilrloment distribution tableA B C D E0.35 10.65 0.59 10.41 0 . 57 10.43

-1-153 - 71 i-71 -75 + 7 5 -112- 29 - 53 + 2 2 +XI2:1 -26 + 1 0- 1+ 4- 1 - 3 - 2 - 1

Free span bending moments7.20 Before the bending moment diagram can be drawn, itis necessary to calculate the free span bending moments:in AB wL2Ms =- from table I) = 153 kNm8in BC, from distributed load

wL2 25 x 5 2M, =- - 78 kNmfrom point load8 8P L 30 x 5M, = - = -= 384 4

Total = 116 kNmin CD and DE

wL28Ms =- 1 12 kNm

Bending moment diagram7.21 For the bending moment diagram 21, the normalbending moment sign convention is used-a saggingmoment is positive, and a hogging moment negative 22.In practice, most midspan moments are positive, andsupport moments are negative.


26/47


A 8 C D E

-123

21 Bending moment diagram for beam ahown in 20

+ve moment condition -ve moment condition

--_ 4'+ve moment condition -ve moment condition

--_ 4'22 Normal e n onvention. Bending moment diagrams aregenerally drawn to reflect th e &fleot& profile of the beam,k with poaative BM below the line and negative BM abovethe lineCalculations of shears and reactions7.22 Calculation of shears and reactions Consider span AB,and take moments for it about B: (see 23)

I-7-------*7 00023 Calculation of ahew and reaction of apan A B

1237

= 25 x 7 x 0 . 5 - -= 88 - 18 = 70kN7.23 Similarly, taking moments about A

MBSBA = 4wL +123. = 2 5 x 7 ~ 0 . 5 + -7

= 88 + 18 = 106kNAnd for span BC, moments about C

M B - MCS B C = +wL + 8P +123-56

5= 25 x 5 x 0 . 5 + 30 x 0 . 5 +-62 + 15 + 13 = 9OkNBy this method, all the shears and reactions on the beam

c an be found.RB = SBA+ SBc = 106 + 90 = 196 kNMethod used for a-symmetrical arrangement of beams7.24 I n many cases the structure is symmetrical, but theloads upon it are not. A typical example is shown in 24.7.25 Any system of loads can be built up of a gombinationof symmetrical and anti-symmetrical systems. This arrange-ment is split as shown in 25.

/F,, ,D-,-4ooo-1---9 000- - 4 4 ooo-l-24 Structure i s symmetrical but loada are not

50

2 8 k N / m 2 8 k N l m

a2 8 k N l m

b25a Symmetrical loading ayatem; b anti-symmetrical loadingTable IV Swan atiffnesa-tvmmetrioalw e"" YSpan k E I L Stiffness RatiosA B b C D 3 - - 0'75}om 0.77.2 30c 2 - - 9 0.22

7.26 End fixity moment:At B in span AB M, =- - 56 kNmWLZ 28 x .4z8 8

+1 3 I -13Table v Span stiffness- anti-wmmetrical caaeSpan k E I L Stlffness RatiosA B b C D 3 - - 0'75}~ . 42 0 . 53.47BC 6 - - 9 0.67

7.27 End-fixity moment as before = 56 kNmA B C

0 .53 1 0.47

+26 I -267.28 Reverting to 'normal' sign convention, and combiningthe load cases to give the loading in figures:MB = ( -13) + ( -26) = -39kNmMc = ( -13) + (+26) = +13kNm8 Portal frames8.01 The continuous beam type of structure may be con-sidered as a one-dimensional structure, although the loadsand reactions on it are in a second dimension. Bu t as soonas the stru ctur e becomes two-dimensional, as in the portalframe, the problem is immediately more complex. There a r evarious methods for dealing with two- and three-dimen-sional skeletal frames:1 Moment distribution (as in th e previous section)2 Influence coefficients3 Slope-deflection4 Kleinlogels tables5 Package computer programsThe last two are not strictly methods of analysis but theresults of other methods, and can be used to solve a imitednumber of problems.8.02 Kleinlogels books of formulae and tables for con-tinuous beams and various portal and multibay frames canbe useful, but can involve a ot of work, so engineers tend touse package computer programs. These are availablethrough computer bureaux but the engineer who frequentlymeets this problem will find a computer terminal installed inhis own office, giving access to such programs, will save himmuch time and effort. If neither of these methods isavailable one of the first three procedures is used. 7I


27/47

51

-.

8.03 Moment distribution is an iterative method. In somecases the number of iterations can be excessive and inextreme cases the method fails to reach a balanced solution.A separate distribution is needed for almost every type ofloading but no simultaneous equations are generated andcalculations do not have to be taken to many decimal places.8.04 The influence coeficients method is very much morepowerful, and can be used on virtually any two- or three-dimensional skeletal structure. Once the flexibility matrixhas been found and inverted, the result may be used tocalculate any number of loading cases without too muchextr a work. However, it is not always possible to invert t heflexibility matrix by hand. Usually the results depend onsmall differences in large quantities, and so sufficientdecimal places to ensure a high level of accuracy are needed.8.05 Slope-deflection is useful for small structures, but likeinjZuence coeficients it generally leads to simultaneousequations.8.06 To demonstrate the analysis of a statically indeter-minate frame structure, and give some idea of the workinvolved in quite simple cases, shortened versions of theanalysis of the fixed-foot portal frame 26 by both themoment distribution method and the influence coeficientsmethod are given below. This example is deliberately non-symmetrical. If either the portal or the load on it are notsymmetrical the frame sways 27 and the direction of swayis not always obvious. Particularly in the moment distribu-tion method, it is important to consider sway.

I

Moment distribution method8.07 The frame is first translated into a continuous beam 28.The stiffness of each arm (method para 7.14) and the end-fixity moments in BC (method para 7.16) are determined.

I 2 0

HA6 Portal frame example which is analysed by momentdistribution and b y influence coeficients methods

HD

' I\ I

27 Non-symmetrioal portal frame sways

28 Portal translated into wntinuoua beam

Technical study Analysis 2 para 8.03 to 8.08Table VI Moment distributionA B C D

0.67 0.33 0.38 0.62-60+20 +40 +2 0- - 1..$+ 3.8 + 7 . & + 3 .0- 1. 9+ 1.3 + 0. 60 .6 - 0. 4+ 0 . 2 + 0 .3 + 0.1

-- + 60-22.8 -37.2 - 18.6+1 0 - 3.13.8 - 6.2+ 1 . 9 - 0. 60 .7 - 1.2+ 0 .3 - 0. 10 . 1 - 0.2i -24 .6 +49.2 -49.2 +44.8 -44.8 - 22.48.08 These results, table VI, are used to draw the bendingmoment diagram 29 (B an d C are assumed fixed in space;a force H, is needed to achieve this). Divide the structureinto its component members 30. For BC, taking momentsabout, C:V, x 6 - 120 x 3 + 49 . 2 - 44.8: V, = 60-73kN an d V, = 59.27For CD, moments about D:Hc 5 = 44.8 + 22 . 4.'. Hc = 13.45= H,For AB, moments about A:: HE= 1 . 3 1 and HA= 14.76(HE+ Hc) x 5 = 49 . 2 + 24.6

A 2 4 6 2 2 4 D29 Bending moment diagram for portal frame

44.0nHc IVD

1,DU22.410 Bending moment diagram divided into componentnembers -\


28/47

Technical study Analysis 2 para 8.09 t o 8.198.09 The force at B to maintain the frame in its originalposition is 1 * 31 kN to th e left and th e imposed load at thispoint is 35 kN to the right. Thus the force applied a t this&ed point is 33 69 kN. If all the other loads on the frameare now removed, and the restraints at B and C released,the frame will sway to the right a distance d. The rigidity,in the rotational sense, of the joints at B and C is main-tained when the restraints on the frame are released.8.10 Both ends of the member in 31 are encastre; B hasbeen deflected through a vertical distance d. From Theorem1 (para 1.18) he area of the - iagram is zero, thereforeME1MA = - Mg.From Theorem 2 (para 1.19)

231 d = 3L x ( M A + Mg) - L - Mg x L x &L

MAL%65-

8.11 These results are used in a second moment distribu-tion to find the end-fixity moments. d is unknown and thu scan be given any value at this stage, so put 6EI d = 1.The results of this distribution are shown in 32. But themoments at the feet of the portal legs are no longer halfthe moments a t the heads because of the sway of the frame.8.12 As before, the external forces on the frame are foundby statics:V, = - VA = 1.84 x lO-'kNH D = 2.7 x lO-'kNHA = 3 .1 x 10-6kNH, = 6 8 x 1 0 - 6 k N8.13 This establishes t,hat a horizontal force of 5 . 8 x 10-6kN at B produces a sway of 1/6E m and moments as shownin 32.

31 Diagram showing bending moment of encastre member(see 7a)

32 Second moment distribution diagram8.14 If these moments are multiplied by theresult will be the moments produced by the applied horizon-tal force corrected by the sway due t o t he non-symmetryof the frame. Table IX allows comparison of the values of theunknowns obtained by various methods.

33.695.8 x 10-6'

'

52

nfluence coefficients method1.15 This method produces a series of simultaneous equa-ions which, for. anything other t han an extremely simpletructure, have to be solved on a computer. The theoryBehind the method is not really necessary for its under-tanding, and is beyond the scope of these notes.1.16 The equation controlling the method is written:C, G an d U are not single numbers, b ut represent arrays oflumbers called matrices. (A brief outline of the theory ofnatrices is given in a footnote after 10.)i.17 X is a column matrix (or vector) representing thewtraints; G is the flexibility matrix for the structure, andCT is the particular solution for the given imposed loading.5.18 The portal frame in 26 is statically indeterminate in,he third degree because it has th ree redundant restraints.rhe first operation is to make the frame statically deter-ninate by reducing the restraints by three. Methods forioing this are illustrated in 33 but for this example the wayihown in 34 is used. The resulting structure is called theeeleased structure.5.19 In 34 a hinge is introduced at A , an d a hinge and t%moller support a t D. The removed restraint s are moments at

C = - G - l x U

Fable VII Graphical repreeentution of releaseeSign Name Transm its

cut nothing

rollers momentaxial force

ax ia l forceinge__o_ shearsleeve moment-- shear

r-3 Various forms of released structures. Graphicalconventionsfor these frame8 are illustrated i n table VIIIA t&+a

C

a34 Forms of released structure and restraints chosen foruse in the example


29/47

53

G;-1 =

A and D, an d a hori

Date post:	10-Apr-2018
Category:	Documents
Upload:	niques-dawson
View:	226 times
Download:	1 times

A Handbook 0f Building Structure = Allan Hodgkinson

Documents