Computer Aided Structural Analysis[1]

8/3/2019 Computer Aided Structural Analysis[1]

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Saikat Basak



Tips and Tricks forComputer Aided

Structural Analysis

Saikat BasakM.Eng (Structural), BCE, CIC, AIE (Ind.), A.ASCE

Structural Engineer

PUBLISHED BY ENSEL SOFTWARE



© Saikat Basak

The author and publisher of this book have used their best efforts in preparingthis book. These efforts include the development, research and testing of thetheories and programs to determine their effectiveness. The author and publishershall not be liable in any event for the incidental or consequential damages inconnection with, or arising out of, the furnishing, performance, or use of theseprograms.

All rights reserved. No part of this book may be reproduced, in any form or byany means, without the permission in writing from the author.

1st Edition 2001

Published on the web

Price: FREE for online viewing



Computer Aided Structural Analysis

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CONTENTS

ABBREVIATION.................................................... ............................................................ .................................... 6

1. INTRODUCTION (BEFORE YOU BEGIN…)..................................................... .................................... 7

2. WHAT IS COMPUTER AIDED STRUCTURAL ANALYSIS? ............................................................. 9

3. ANALYSIS TYPES .......................................................... ........................................................... ............... 10

LINEAR STATIC STRESS ANALYSIS .................................................... ........................................................... ..... 10DYNAMIC ANALYSIS...................................................... ........................................................... ......................... 11RANDOM VIBRATION..................................................... ........................................................... ......................... 12RESPONSE SPECTRUM ANALYSIS....................................................... ........................................................... ..... 12TIME HISTORY ANALYSIS........................................................ ........................................................... ............... 13TRANSIENT VIBRATION ANALYSIS .................................................... ........................................................... ..... 13VIBRATION ANALYSIS (MODAL ANALYSIS) .......................................................... ............................................ 14BUCKLING ANALYSIS..................................................... ........................................................... ......................... 15

THERMAL ANALYSIS...................................................... ........................................................... ......................... 16BOUNDARY ELEMENT ................................................... ........................................................... ......................... 17

4. SIGN CONVENTION (MIND YOUR SIGNS).... ........................................................... ......................... 19

5. NUMBERING OF JOINTS AND MEMBERS ...................................................... .................................. 23

6. SPECIFYING MOMENT OF INERTIA ..................................................... ............................................ 24

7. SPECIFYING LOADS..................................................... ........................................................... ............... 27

8. COLUMN BUCKLING TEST ............................................................ ...................................................... 31

9. PORTAL AND CANTILEVER METHOD ........................................................... .................................. 33

10. DEFLECTION OF REINFORCED CONCRETE MEMBER ...................................................... ..... 34

11. SHEAR DEFORMATION ..................................................... ........................................................... ..... 36

12. INCLINED SUPPORT........................................................... ........................................................... ..... 37

13. MAXIMUM BENDING MOMENT, SHEAR FORCE AND REACTION IN BUILDING FRAME:

SUBSTITUTE (EQUIVALENT) FRAME ........................................................... ............................................ 38

14. SUPPORT SETTLEMENT ................................................... ........................................................... ..... 40

15. 2D VERSUS 3D............ ........................................................... ........................................................... ..... 41

16. CURVED MEMBER.................................................... ........................................................... ............... 43

17. TAPERED SECTION .................................................. ........................................................... ............... 44

18. NODES CONNECTED BY A SPRING.............................................................. .................................. 45

19. SUB-STRUCTURING TECHNIQUE AND SYMMETRY (BREAK THEM INTO PIECES…)... 46

20. STAIRCASE ANALYSIS ...................................................... ........................................................... ..... 51

21. CABLES .................................................... ........................................................... ................................... 54

22. PRE-STRESSED CABLE PROFILE ....................................................... ............................................ 57




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23. FINITE ELEMENT ANALYSIS (FEA) METHOD IS APPROACHING… .................................... 60

24. A TYPICAL WORKED OUT PROBLEM OF FEA........... ........................................................... ..... 69

25. PLATES BY FEM ........................................................ ........................................................... ............... 76

26. INTERPRETING FEA RESULT.............................................................. ............................................ 79

27. TIPS FOR CREATING BETTER MESH.......................................................... .................................. 83

28. COMMON FINITE ELEMENTS LIBRARY FOR LINEAR STATIC AND DYNAMIC STRESS

ANALYSIS .................................................. ............................................................ ............................................ 89

29. SHEAR WALL ................................................... ........................................................... ......................... 92

30. FOLDED PLATE ......................................................... ........................................................... ............... 94

31. SHELLS........................ ........................................................... ........................................................... ... 102

32. A FIRST STEP IN STRUCTURAL DYNAMICS ...................................................... ....................... 105

33. AN EXAMPLE OF A SINGLE DEGREE OF FREEDOM PROBLEM............................ ............. 108

34. WHAT DYNAMIC ANALYSIS YOU SHOULD PERFORM? .................................................... ... 112

35. NON-LINEAR ANALYSIS (NLA) – AN INTRODUCTION FOR BEGINNERS.........................115

MATERIAL NON-LINEARITY .................................................... ........................................................... ............. 115GEOMETRIC NON-LINEARITY .................................................. ........................................................... ............. 116

36. MECHANICAL EVENT SIMULATION .......................................................... ................................ 119

37. IMPORTING MODEL FROM CAD PROGRAMS .................................................. ....................... 121

38. VIRTUAL REALITY IN ENGINEERING (VRML) ........................................................... ............. 123

39. LINEAR PROGRAMMING IN SPREADSHEET ..................................................... ....................... 124

40. REINFORCEMENT DETAILING IN CONTINUOUS BEAMS ....................................................128

41. A GUIDE TO SOME STRUCTURAL ENGINEERING & FINITE ELEMENT ANALYSIS

PROGRAMS......................................................... ............................................................ ................................ 130

CIVIL ENGINEERING PROGRAMS ........................................................ ........................................................... ... 131MECHANICAL ENGINEERING PROGRAMS ..................................................... .................................................... 134SOME CAD PROGRAMS… ....................................................... ........................................................... ............. 137

42. HOW TO SELECT THE MOST APPROPRIATE PROGRAM FOR YOUR NEED?.................139

43. HOW TO CHECK THE RESULT FOR ACCURACY? ..................................................... ............. 142

44. FILE NAME EXTENSION GUIDE (FOR SOME CAD/CAE PROGRAMS) ...............................143

45. COMMON ERROR MESSAGES AND SOLUTIONS ........................................................ ............. 144

OPERATING SYSTEM RELATED........................................................... ........................................................... ... 144ANALYSIS PROGRAM RELATED.......................................................... ........................................................... ... 145

46. REFERENCES ................................................... ........................................................... ....................... 148




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Abbreviation

Several abbreviations have been used throughout this book. They have beendefined in respective sections, but here is a list of them at a glance.

[k] – StiffnessBM – bending momentC - dampingCAD – computer aided design/drawingCAE – computer aided engineeringCAM – computer aided manufacturingE – modulus of elasticityFE – finite element

FEA – finite element analysisFEM – finite element methodf y – yield strength of steelG – shear modulusI – 2nd moment of inertiaIS – Indian Standard codeLRFD – load and resistance factor designLSSA – linear static stress analysisM, m – massMDF – multi degree freedomMES – mechanical event simulationNLA – non-linear analysisRSA – response spectra analysisSDF – single degree freedomSF – shear forceT – time period of vibrationTHA – time history analysisUDL – uniformly distributed loadVRML – virtual reality markup language

x – displacementx’ – velocityx” – acceleration

ε – Strain

µ, ν – Poisson’s ratio

σ – Normal stress

τ – Shear stress

ωn – natural frequency of the structure

ξ – Damping ratio




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1. Introduction (before you begin…)

In this book I shall tell you some practical tips for structural analysis usingcomputer. Most structural engineering books are written to tell you how youwill perform the calculation by hand. But even sometimes analysis usingcomputer can be very tricky. You may need to manipulate computer input tosolve a problem, which may at first appear to be unsolvable by that program.

Finite element programs and structural analysis programs tend to be veryexpensive. Most small-scale engineering firms keep only one analysis program.Even for a large corporate companies it is seldom possible to maintain morethan two standard analysis packages. Therefore it is essential that you use yourpresent analysis program to its full extent.

This is not a textbook. I make no attempt to teach you theory of structuralanalysis to score good marks in the exam! But it can help you to earn moremoney by enabling you to analyze some structures more easily and accurately,which you were previously thought too difficult to deal with your existinganalysis program.

Also, I am not going to teach you any particular structural analysis computer

program. However, the techniques of analysis discussed here are applicable tomost standard analysis packages.

I presented the whole thing in an informative yet informal manner. I confinedthe boring theory and calculation to minimum level.

No special knowledge is required to get the most out of this book. OnlyBachelor Degree knowledge in Civil/Mechanical Engineering is assumed.However some parts of the book do discuss some topics which are normally

covered in Master’s degree level in detail. Also, I expect that you are familiarwith at least one standard structural analysis package otherwise you may

find the contents of this book quite terse!

This book does not contain listing of any computer program; because I knowthat most readers will not bother to type them or to even read them.

But remember the most important advice: A structure will not behave as the

computer program tells it should regardless of how accurate the program

seems or how expensive it is! Thus goes the famous proverb “With good engineering judgement you can produce on the back of an envelop that which




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otherwise cannot be produced with a ton of computer output”. You should pastethis in front of your computer so that you see it everyday. (I did it!)

Before you accuse me by complaining that my tips do not work with your

program, I like to mention following important points.

• I did not work with all the structural analysis programs available in themarket.

• Some features I discussed here may not be available in your program. It caneven happen that the program you are using has better option to handle aparticular problem compared to what I discussed in this book.

• I am only providing you some “clues” for more effective use of structuralanalysis programs. However, every analysis problem is unique depending on

type of project, cost, client’s requirement etc. Those specific criteria youhave to solve yourself.

• Documentation of the program you are using is very important. The programmanuals are the best source of help always.

The sections of this book are arranged in somewhat haphazard mannerdeliberately so that you don’t feel bored. The paragraphs are small and to thepoint. We have often returned to same topics in several sections from differentviewpoints. Wherever necessary, numerical examples have been presented.

There are also some exercises. Please try to solve them with your structural/FEanalysis programs.

I like to see your comments and suggestions. You can reach me atwww.enselsoftware. com in World Wide Web. I shall be more than happy toanswer your queries. Now sit back, relax and enjoy the book.

Have a nice reading!




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2. What is Computer Aided Structural Analysis?

This section is a head start for those who are using structural analysis programs

for the first time. As the name suggests, Computer Aided Structural Analysis isthe method of solving your structural analysis problem with the help of computer software. In earlier generation analysis programs, you had to supplythe programs the nodal co-ordinates, member incidence (i.e. between whatnodal points a particular member is connected), material properties, sectionalproperties of the all members and the loads (nodal force/moment/distributedmember loads etc.). You also had to supply how the structure was supported,fixed, hinged or roller. The program then calculated the member forces, nodalreactions and joint displacements and presented in a tabular format. This type of structural analysis programs is still used in junior years in the university as afirst learning tool. However, the commercial structural analysis programs of modern days are far more powerful and easy to use. Here, you can actually‘draw’ your model on screen (as if you’re drawing in a paper with a pencil!)with the mouse and keyboard! Everything is graphical. You draw modelsgraphically, apply loads and boundary conditions graphically and visualize theshear force, bending moment and even deflected shape diagram graphically. Forthe first time users, it seems rather like a magic! The availability of theseprograms has completely changed the way we analyze structures compared towe did the same just a decade ago! Now it is a child’s play to analyze structures

having more than 10,000 degrees of freedom! However, analyzing structuresusing computers has created many other new problems. First, you must be veryfamiliarize with the programs you are using. You must clearly understand itslimitation and assumptions. All programs can’t be applied for analyzing alltypes of structures. Most programs solve the structures by stiffness method,though solution algorithm may differ from one program to another. What ismost important is that you must interpret the output result accurately. This book will show you how to perform quickly, accurately and proper interpretation of data in easiest way. You will also learn to analyze many new kinds of structures

without learning theoretical calculations! Sounds interesting? At the end of thisbook, you will also learn about some very recent concepts of structural analysis.Bon Voyage!




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3. Analysis types

In this section, you will learn various analysis options those are offered by FEA

programs. You are already familiar with most of the types of analyses, and someare new to you. (References 8 and 15 were considered for this section.)

Linear Static Stress Analysis

This is the most common type of analysis. When loads are applied to a body, thebody deforms and the effects of the loads are transmitted throughout the body.To absorb the effect of loads, the body generates internal forces and reactions atthe supports to balance the applied external loads. Linear Static analysis refers

to the calculation of displacements, strains, and stresses under the effect of external loads, based on some assumptions. They are discussed below.

1. All loads are applied slowly and gradually until they reach their fullmagnitudes. After reaching their full magnitudes, load will remain constant (i.e.load will not vary against time). This assumption lets us disregard insignificantinertial and damping forces due to negligibly small accelerations and velocities.Time-variant loads that induce considerable inertial and/or damping forces maywarrant dynamic analysis. Dynamic loads change with time and in many cases

induces considerable inertial and damping forces that cannot be neglected.

2. Linearity assumption: The relationship between loads and resulting responsesis linear. If you double the magnitude of loads, for example, the response of themodel (displacements, strains and stresses) will also double. You can makelinearity assumption if a. All materials in the model comply with Hooke’s Law that is stress is directly

proportional to strain.b. The induced displacements are small enough to ignore the change is stiffness

caused by loading.c. Boundary conditions do not vary during the application of loads. Loads must

be constant in magnitude, direction and distribution. They should not changewhile the model is deforming.

If the above assumptions are not valid, then we shall have to treat the problemas non-linear analysis. I shall devote a few sections on non-linear analysis later.

Some FEA programs offer contact/gap elements. With this option, available

during meshing, contacting mating faces may separate during loading and hencethe load distribution in the model will change based on the gap forces generated.




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This functionality offers a linearized solution to a nonlinear problem. Soundscrazy?

Calculation of stresses

Stress results are first calculated at special points, known ‘Gaussian’ or‘Quadrature’ points, located inside each element. (See you FEA textbook fordetails) These points are selected to give optimal results. The program thencalculates stresses at the nodes of each element by extrapolating the resultsavailable at the ‘Gaussian’ points. After a successful run, multiple results areavailable at nodes common to two or more elements. These results will not beidentical because the finite element method is an approximate method. Forexample, if a node is common to three elements, there can be three slightlydifferent values for every stress component at that node.

During result visualization, you may ask for element stresses or nodal stresses.In calculating element stresses, the program averages the corresponding nodalstresses for each element. In calculating nodal stresses at a node, the programaverages the corresponding results from all elements contributing to the stressesat that node.

Dynamic analysis

In general, we have to perform dynamic analysis on a structure when the loadapplied to it varies with time. The most common case of dynamic analysis is theevaluation of responses of a building due to earthquake acceleration at its base.Every structure has a tendency to vibrate at certain frequencies, called naturalfrequencies. Each natural frequency is associated with a certain shape, calledmode shape that the model tends to assume when vibrating at that frequency.When a structure is excited by a dynamic load that coincides with one of itsnatural frequencies, the structure undergoes large displacements. This

phenomenon is known as ‘resonance’. Damping prevents the response of thestructures to resonant loads. In reality, a continuous model has an infinitenumber of natural frequencies. However, a finite element model has a finitenumber of natural frequencies that is equal to the number of degrees of freedomconsidered in the model. The first few modes of a model (those with the lowestnatural frequencies), are normally important. The natural frequencies andcorresponding mode shapes depend on the geometry of the structure, itsmaterial properties, as well as its support conditions and static loads. Thecomputation of natural frequencies and mode shapes is known as modal

analysis. When building the geometry of a model, you usually create it based onthe original (undeformed) shape of the model. Some loading, like a structure’s




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self-weight, is always present and can cause considerable changes in thestructure’s original geometry. These geometric changes may have, in somecases, significant impact on the structure’s modal properties. In many cases, thiseffect can be ignored because the induced deflections are small.

This is just a prelude to dynamic analysis. You will find several topics ondynamic analysis later in this book. However, since I shall not discuss theory of structural dynamics here, I strongly recommend that you read a structuraldynamic textbook if you haven’t done so already.

The following few topics – Random Vibration, Response Spectrum analysis,Time History analysis, Transient vibration analysis and Vibration modalanalysis are extensions of dynamic analysis.

Random Vibration

Engineers use this type of analysis to find out how a device or structureresponds to steady shaking of the kind you would feel riding in a truck, rail car,rocket (when the motor is on), and so on. Also, things that are riding in thevehicle, such as on-board electronics or cargo of any kind, may need RandomVibration Analysis. The vibration generated in vehicles from the motors, roadconditions, etc. is a combination of a great many frequencies from a variety of

sources and has a certain "random" nature. Random Vibration Analysis is usedby mechanical engineers who design various kinds of transportation equipment.Engineers provide input to the processor in the form of a ‘Power SpectralDensity’ (PSD), which is a representation of the vibration frequencies andenergy in a statistical form. When an engineer uses Random Vibration he islooking to determine the maximum stresses resulting from the vibration. Thesestresses are important in determining the lifetime of a structure of atransportation vehicle. Also, it would be important to know if things beingtransported in vehicles will survive until they reach the destination.

Response Spectrum Analysis

Engineers use this type of analysis to find out how a device or structureresponds to sudden forces or shocks. It is assumed that these shocks or forcesoccur at boundary points, which are normally fixed. An example would be abuilding, dam or nuclear reactor when an earthquake strikes. During anearthquake, violent shaking occurs. This shaking transmits into the structure ordevice at the points where they are attached to the ground (boundary points).




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Response spectrum analysis is used extensively by Civil Engineers who mustdesign structures in earthquake-prone areas of the world. The quantitiesdescribing many of the great earthquakes of the recent past have been capturedwith instruments and can now be fed into a response spectrum program to

determine how a structure would react to a past real-world earthquake.Mechanical engineers who design components for nuclear power plants mustuse response spectrum analysis as well. Such components might include nuclearreactor parts, pumps, valves, piping, condensers, etc. When an engineer usesresponse spectrum analysis, he is looking for the maximum stresses oracceleration, velocity and displacements that occur after the shock. These inturn lead to maximum stresses. You will find an example of response spectrumanalysis later.

Time History Analysis

This analysis plots response (displacements, velocities, accelerations, internalforces etc.) of the structure against time due to dynamic excitation applied onthe structure. You will find more stuff on this particular type of analysis in latersections.

Transient Vibration Analysis

When you strike a guitar string or a tuning fork, it goes from a state of inactivityinto a vibration to make a musical tone. This tone seems loudest at first, thengradually dies out. Conditions are changing from the first moment the note isstruck. When an electric motor is started up, it eventually reaches a steady stateof operation. But to get there, it starts from zero RPM and passes through aninfinite number of speeds until it attains the operating speed. Every time you revthe motor in your car, you are creating transient vibration. When things vibrate,internal stresses are created by the vibration. These stresses can be devastating if resonance occurs between a device producing vibration and a structure

responding to. A bridge may vibrate in the wind or when cars and trucks goacross it. Very complex vibration patters can occur. Because things areconstantly changing, engineers must know what the frequencies and stresses areat all moments in time. Sometimes transient vibrations are extremely violentand short-lived. Imagine a torpedo striking the side of a ship and exploding, or acar slamming into a concrete abutment or dropping a coffeepot on a hard floor.Such vibrations are called "shock, " which is just what you would imagine. Inreal life, shock is rarely a good thing and almost always unplanned. But shocksoccur anyhow. Because of vibration, shock is always more devastating than if the same force were applied gradually.




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Vibration Analysis (Modal Analysis)

All things vibrate. Think of musical instruments, think of riding in a car, think of the tires being out of balance, think of the rattles in an airplane when they are

revving up the engines, or the vibration under your feet when a train goes by.Sometimes vibration is good. Our ears enable us to hear because they respond tothe vibrations of sound waves. Many times things are made to vibrate for apurpose. For example, a special shaking device is used in foundries to loosen amold placed in sand. Or, in the food and bulk materials industries, conveyorsfrequently work by vibration. Usually, however, vibration is bad and frequentlyunavoidable. It may cause gradual weakening of structures and the deteriorationof metals (fatigue) in cars and airplanes. Rotating machines from small electricmotors to giant generators and turbines will self destruct if the parts are not well

balanced. Engineers have to design things to withstand vibration when it cannotbe avoided. For example, tyres and shock absorbers (dampers) help reducevibration in automobiles. Similarly, flexible couplings help isolate vibrationsproduced by the engines. Vibration is about frequencies. By its very nature,vibration involves repetitive motion. Each occurrence of a complete motionsequence is called a "cycle." Frequency is defined as so many cycles in a giventime period. "Cycles per seconds” or "Hertz”.Individual parts have what engineers call "natural" frequencies. For example, aviolin string at a certain tension will vibrate only at a set number of frequencies,which is why you can produce specific musical tones. There is a base frequencyin which the entire string is going back and forth in a simple bow shape.Harmonics and overtones occur because individual sections of the string canvibrate independently within the larger vibration. These various shapes arecalled "modes". The base frequency is said to vibrate in the first mode, and soon up the ladder. Each mode shape will have an associated frequency. Highermode shapes have higher frequencies. The most disastrous kinds of consequences occur when a power-driven device such as a motor for example,produces a frequency at which an attached structure naturally vibrates. Thisevent is called "resonance." If sufficient power is applied, the attached structure

will be destroyed. Note that ancient armies, which normally marched "in step,"were taken out of step when crossing bridges. Should the beat of the marchingfeet align with a natural frequency of the bridge, it could fall down. Engineersmust design so that resonance does not occur during regular operation of machines. This is a major purpose of Modal Analysis. Ideally, the first modehas a frequency higher than any potential driving frequency. Frequently,resonance cannot be avoided, especially for short periods of time. For example,when a motor comes up to speed it produces a variety of frequencies. So it maypass through a resonant frequency. Other vibration processes such as Time

History, Response Spectrum, Random Vibration, etc. are used in addition to




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Modal Analysis to deal with this type of more complex situation. These arecalled Transient Natural Frequency Processors.

Buckling analysis

If you press down on an empty soft drink can with your hand, not much willseem to happen. If you put the can on the floor and gradually increase the forceby stepping down on it with your foot, at some point it will suddenly squash.This sudden scrunching is known as "buckling."

Models with thin parts tend to buckle under axial loading. Buckling can bedefined as the sudden deformation, which occurs when the stored membrane(axial) energy is converted into bending energy with no change in the externally

applied loads. Mathematically, when buckling occurs, the total stiffness matrixbecomes singular (see section 8).

In the normal use of most products, buckling can be catastrophic if it occurs.The failure is not one because of stress but geometric stability. Once thegeometry of the part starts to deform, it can no longer support even a fraction of the force initially applied. The worst part about buckling for engineers is thatbuckling usually occurs at relatively low stress values for what the material canwithstand. So they have to make a separate check to see if a product or part

thereof is okay with respect to buckling.

Slender structures and structures with slender parts loaded in the axial directionbuckle under relatively small axial loads. Such structures may fail in bucklingwhile their stresses are far below critical levels. For such structures, thebuckling load becomes a critical design factor. Stocky structures, on the otherhand, require large loads to buckle, therefore buckling analysis is usually notrequired.

Buckling almost always involves compression. In civil engineering, buckling isto be avoided when designing support columns, load bearing walls and sectionsof bridges which may flex under load. For example an I-beam may be perfectly"safe" when considering only the maximum stress, but fail disastrously if justone local spot of a flange should buckle! In mechanical engineering, designsinvolving thin parts in flexible structures like airplanes and automobiles aresusceptible to buckling. Even though stress can be very low, buckling of localareas can cause the whole structure to collapse by a rapid series of ‘propagatingbuckling’.




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Buckling analysis calculates the smallest (critical) loading required buckling amodel. Buckling loads are associated with buckling modes. Designers areusually interested in the lowest mode because it is associated with the lowestcritical load. When buckling is the critical design factor, calculating multiple

buckling modes helps in locating the weak areas of the model. This may preventthe occurrence of lower buckling modes by simple modifications.

Thermal analysis

There are three mechanisms of heat transfer. These mechanisms areConduction, Convection and Radiation. Thermal analysis calculates thetemperature distribution in a body due to some or all of these mechanisms. In allthree mechanisms, heat flows from a higher-temperature medium to a lower-

temperature one. Heat transfer by conduction and convection requires thepresence of an intervening medium while heat transfer by radiation does not. Iinclude a brief discussion on thermal analysis here. You must have read allthese in high school. In this book, I shall not discuss anything more aboutthermal analysis.

Conduction

Thermal energy transfers from one point to another through the interaction

between the atoms or molecules of the matter. Conduction occurs in solids,liquids, and gasses. For example, a hot cup of coffee on your desk willeventually cool down to the room-temperature mainly by conduction from thecoffee directly to the air and through the body of the cup. There is no bulk motion of matter when heat transfers by conduction. The rate of heat conductionthrough a plane layer of thickness X is proportional to the heat transfer area andthe temperature gradient, and inversely proportional to the thickness of thelayer.

Rate of Heat Conduction = (K) (Area) (Difference in Temperature / Thickness)

Convection

Convection is the heat transfer mode in which heat transfers between a solidface and an adjacent moving fluid (liquid or gas). Convection involves thecombined effects of conduction and the moving fluid. The fluid particles act ascarriers of thermal energy.

Radiation




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Thermal radiation is the thermal energy emitted by bodies in the form of electromagnetic waves because of their temperature. All bodies withtemperatures above the absolute zero emit thermal energy. Becauseelectromagnetic waves travel in vacuum, no medium is necessary for radiation

to take place. The thermal energy of the sun reaches earth by radiation. Becauseelectromagnetic waves travel at the speed of light, radiation is the fastest heattransfer mechanism. Generally, heat transfer by radiation becomes significantonly at high temperatures.

Types of Heat Transfer Analysis

There are two modes of heat transfer analysis based on whether or not we areinterested in the time domain.

Steady State Thermal Analysis

In this type of analysis, we are only interested in the thermal conditions of thebody when it reaches thermal equilibrium, but we are not interested in the timeit takes to reach this status. The temperature of each point in the model willremain unchanged until a change occurs in the system. At equilibrium, thethermal energy entering the system is equal to the thermal energy leaving it.Generally, the only material property that is needed for steady state analysis isthe thermal conductivity.

Transient Thermal Analysis

In this type of analysis, we are interested in knowing the thermal status of themodel at different instances of time. A thermos designer, for example, knowsthat the temperature of the fluid inside will eventually be equal to the room-temperature (steady state), but he is interested in finding out the temperature of the fluid as a function of time. In addition to the thermal conductivity, we alsoneed to specify density, specific heat, initial temperature profile, and the period

of time for which solutions are desired.

Boundary Element

A type of finite element sometimes used to connect the finite element model tofixed points in space. Typically this fixity is set with global boundaryconditions, in which the fixity is totally rigid. A boundary element, on the otherhand, allows for a flexible connection to the fixed space. Boundary elements

and boundary points are normally used to simulate the constraints that actuallyoccur when an object is used in the real world. For example, if a coffee cup is




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sitting on the table and a weight is placed on top of the coffee cup, then the tableis the boundary. Boundary points would be points on the plane of the table thatare defined as being fixed in space and to which nodes of a finite element modelof the coffee cup are attached. If the table has a spongy surface, you might want

to use boundary elements to account for the flexibility. With many FEAsoftware, boundary elements have an additional capability of imposing andenforced displacement upon a model. The force created by this imposeddisplacement would be calculated automatically. Additionally, the forcesgenerated at a boundary by forces on the model can be obtained as output usingboundary elements.

There are another very powerful types of analysis offered by high-end FEAprograms, known as Mechanical Event Simulation or Virtual Prototyping. Youwill find this in section 36.




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4. Sign convention (mind your signs)

In structural analysis, sign convention is very important. You must follow same

sign convention throughout your life! Normally, the force towards right is takenas positive and force acting upwards is considered positive. Anti-clockwisemoment is taken as positive. This has been shown in following figure for 2Dplane.

Fy

All positiveMz

Fx

Figure 4-1

Most standard analysis programs follow this sign convention. Although you canuse any convention of your own, but I strongly advise you against that. You willalways be fine with this convention. Please note that, because of taking ypositive upwards, when specifying gravity loads, you often need to use “minus”sign to do so.

For 3D structures, the sign convention will be of same type but somewhatcomplicate. This is shown below.

YMy All positive

Fy MxX

Fz FxZ Mz

Figure 4-2

When you see bending moment diagrams, remember that some programs drawthem in tension side or some may do the opposite. Also note that the “sign” of bending moment diagrams indicate the “direction” (as shown in figure 4-1 and4-2), they do not indicate whether the bending moment is sagging or hogging.

Axial forces are normally considered positive for tensile forces and negative for

compressive forces.




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When dealing with 3D structures, the program will generally consider y-axis aselevation. This is as expected, because when dealing with 2D structures, youwill normally use x-y plane. But it has exceptions as well. Some programs, bydefault use x-z plane for 2D analysis. Of course you can direct every analysisprogram to consider z-axis (or even x-axis) as elevation. My main point here tomake you understand that co-ordinate system is very flexible. But you mustfollow same sign convention throughout.

Different programs may follow slight different sign conventions. Before usingthe program, you should be familiar with that program’s sign convention. Solvesome basic problems with them first and consult the user guide. As an example,the following figures show how SAP90/2000 describes frame member internalforces.

AXIS 2 T P AXIS 1

TAXIS 3

P

Positive Axial Force and Torque

Figure 4-3

Compression face

V2

Tension face Compression faceM2

V3 M3Tension face

Positive Moment and Shear [1-2 plane]Positive Moment and Shear [1-3 plane]

Figure 4-4




- 21 -

Now please solve the following problems using your program and check theresult with the answer given.

-3 kN/m (case 2)

5 kN (case 1) Node 4 Node 3 5 m6 m 6 m

All members are of 250-mm side squareCross section made of concrete E=20GPa

Node 1 Node 2

fixed hinged

Figure 4-5

Figure 4-6

The above figure shows the bending moment diagram and the free bodydiagram of each member. Now check the result and the sign with your analysisprogram. Please note that your program may draw the bending moment diagram

on opposite side compared to what shown here! Observe the sign convention.




- 22 -

Now solve the following truss.All members are made of

3 kN steel (E=200 GPa) with100-mm side square section

3 m 0.208 -3.642 2.0

2.833 2.833

-2 kN4 m 4 m

Figure 4-7

The axial forces are shown as italics in the above figure. Note that the left end ishinged and right end is roller.

It is interesting to know that with some programs, you may need to “tell” theprogram that the structure is a ‘truss’ by specifying ‘moment releases’ in thetruss members. Otherwise, you may wonder why the program result showsbending moment diagram in truss! Different programs have different options forspecifying moment (or axial force, torsion etc.) releases.

Some programs, which allow you to draw plate elements on screen, you shoulddraw them in counter clockwise fashion. Otherwise you may get awkwardresult.




- 23 -

5. Numbering of joints and members

Proper node and joint numbering is very important for large models. Those

programs, which allow you to “draw” the model on screen, apply joint andmember numbers automatically. This default scheme may not always beconvenient for you, especially if you are analyzing a multi-story building.Fortunately, most programs offer re-labeling option and you can even usealphanumeric labels. Since it is impractical to re-number hundreds of membersmanually, you should do it automatically.

Generally, beams, columns and slabs are numbered on the story or floor levelthey reside. In that case, you can direct the program to use X-Z-Y re-labelingpattern (assuming Y-axis is the elevation). You may number all beams in the B-5-10 or B05010 fashion where “B” indicates beam, next number indicates“floor” and the last number stands for serial number of beam on that floor.Similar procedure may be adopted for numbering columns, slabs and otherstructural members. You can also create ‘group’ for same type of memberswhose design will be same such as all columns in a particular floor.

Improper node numbering may increase bandwidth of global stiffness matrix.However, most programs automatically re-number nodes internally whilesolving and again display the result in user specified numbering.

Wondering what is ‘bandwidth minimization’? It is a technique for assemblingglobal stiffness matrix so that non-zero terms in the matrix tend to become‘closer’ rather than getting ‘dispersed’. Generally, the non-zero elements of global stiffness matrix are limited to a band adjacent to its diagonal. Lowerbandwidth means less time necessary for solving equations. For example, in amultistory frame (assuming the height is more than the length); if you numbernodes row wise (horizontally or more precisely along smaller dimension),bandwidth will be less compared to column wise (vertical i.e. larger dimension)

node numbering.




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6. Specifying moment of inertia

New users of structural analysis programs often find it confusing to define

section properties of the members, particularly for 3D structures. You may gethelp from the following examples.

Y

X

Plan of columns

Figure 6-1

In the above figure, the beams are of 200 x 300 mm and columns 200 x 400 mmoriented as shown in plan. Beams can be specified as 200x300 mm without anyproblem. But for columns, you have to be careful. Generally, the programs willask you to specify ‘depth’ and ‘width’ of the member. If you specify depth =400 mm and width = 200 mm then you will get exact section as shown in figure6-2.

If you specify the dimension in opposite manner, then you will get wronglyoriented section for the columns. The above figure is taken from real time viewof SAP2000. If your program does not offer real time view (i.e. the membersshould look like in the real structure in 3 dimension) option, you’re out of luck!Many programs, however, have the option for specifying sectional dimensionusing ‘tx’ and ‘ty’ (or it might be ‘ty’ and ‘tz’ or ‘t2’ and ‘t3’) option. I havetried with various programs this sectional dimension input. In most cases width= 200 and height (or depth) = 400 worked.




- 25 -

Figure 6-2

Sometimes you may need to specify inertia directly especially for irregularshaped sections. Normally the programs offer only ‘Iz’ and ‘Iy’ options. Themost often used is the ‘Iz’. For the beam discussed above, Iz = 200x3003 /12 andIy = 300x2003 /12. For the column, Iz = 200x4003 /12 and Iy = 400x2003 /12.

One important thing you must understand is the concept of ‘member local axes’.In most analysis programs, the ‘local axes’ settings are different from ‘globalaxes’. Normally, the ‘local axes’ are defined as shown in figure 6-3.

Y

Z

Figure 6-3

Some programs can display ‘local axes’ for all members. Please explore yourprogram’s resource files to see how it handles display of local member axes.

XZY




- 26 -

YZ

X

Y

X

Z

Figure 6-4

The above figure shows orientation of local axes for the inclined member. Notethat in your program, the orientation for local axes may be slightly different; forexample, direction of Z axis may be in opposite direction. The orientation of global axes is also shown in blue color. It is clear that, when you are definingsection properties in terms of local axes, even an ‘inclined’ member isconsidered as ‘straight’. We shall come to local and global axes story againwhen we discuss interpretation of analysis output.




- 27 -

7. Specifying Loads

All programs have the option for specifying concentrated and uniformly

distributed loads. Some programs allow you to assign a point load on beamwithout creating a node at that point (the program itself creates a node thereinternally) where as most programs require that you can assign concentratedloads only at nodes. So, you may need to ‘split’ the member to createintermediate nodes. If your concentrated load is inclined, you better resolve itinto horizontal and vertical components yourself and then apply them.Specifying UDL is easy. However, trouble arises, then the load becomesvarying. The most common example of varying load is on the beams comingfrom slabs as shown in figure 7-1. The lengths of the beams are ‘L’.Unfortunately, very few programs will calculate distributed loads form slabsautomatically. More often than not, you’ll have to specify the slab load yourself.Some programs allow specifying trapezoidal loads on beam members, however,some allow only triangular load. In that case, you need to ‘split’ each beam intothree segments (not necessarily equal) and apply triangular loads at endsegments and UDL on mid segment.

αL Total Load = W (N) UDL = w (N/m)

Figure 7-1 Figure 7-2

Yes, this is somewhat cumbersome if you have, say 200 beam members! Butyou can avoid trapezoidal loads all together with slight loss of accuracy asshown in figure 7-2. If we equate fixed end moments in two beams (of figure 7-1 and 7-2), we get

1 – 2α2 + α3 wL2 (1 – 2α2 + α3) x W

-------------- WL = ------ or w = ----------------------- … (7.1)12 x (1-α) 12 (1 – α) x L

So, the ratio of mid span moment of trapezoidal/uniform r =

[(3-4α2)WL/(24(1-α)] / [(1-2α2+α3)(W/L)(L2 /8) / (1-α)] =

(1 - 1.33α2) / (1 - 2α2 + α3)

If we tabulate α vs. r values as shown below




- 28 -

α 0 0.2 0.375 0.5

r 1 1.02 1.05 1.068

It is seen that maximum difference of mid span moment for figure 7-1 and 7-2 is

6.8%, which is quite small. So, we can safely replace trapezoidal load with UDLwhose magnitude is given by w as shown in (7.1).

Under some circumstances, you may have non-linearly varying (e.g. parabolic)type of loads. Except in high-end FEA programs, you can’t input the loadthrough equation. The only way out is, split your member into several sections,and specify concentrated loads varying through nodes (or UDL varying throughsegments). More number of divisions, better the result is.

You may wonder whether you can model all the slabs in your building frameusing plate elements instead of converting loads to beams as shown in figure 7-1. Of course you could. But there are several disadvantages! First of all, youranalysis program must have ‘plate’ element to do this. Many frame analysisprograms don’t have plate element! If you use ‘plates’, then you must ‘mesh’ itbefore running analysis. If you have, say 100 slabs (i.e. plates) with 10x10mesh, you’ll have 11x11x100 = 12,100 extra nodes compared to that you’llhave if you transfer the loads on beams. Not only this takes much more time tohave your analysis done, but also it will swamp you with tons of output (justcount the number of total plate elements – their stress values etc.)! It has beenproved that with the conventional slab load distribution as shown in figure 7-3,you’re quite correct.

Figure 7-3

Another type of load, which often creates problem, is due to hydrostatic of earthpressure. If your analysis program has easy method to specify such type of loads, consider yourself really lucky! If you’re applying hydrostatic load on aplate element, apply load before meshing the plate. Sometimes the programallows you to specify separate load at four nodes of the plates (and intermediatevalues are interpolated) though this is not really necessary in day to dayanalysis. Hydrostatic load normally takes the shape as of figure 7-4.




- 29 -

Figure 7-4

To specify you generally supply fluid height, axis and density. Now the density

may be tricky. For example, in the above figure, the load acts towards the plate.But what to do if we want make it act in opposite direction (i.e. away fromplate)? Surprisingly, changing the density into negative works! (Argh!) (I don’tknow whether all programs behave in this way, but I found this trick works inVisual Analysis).

Surcharge or earth pressure load can be specified in the same way as that of hydrostatic load. If your load needs to be like figure 7-5, then just place the fluidlevel at higher level.

Water level


Figure 7-6 shows another trick where you need to superpose two types of loadsto get the desired resultant load distribution. Uniform pressure on plates caneasily be applied. While you analyze water tanks, these tricks come handy.

When you’re applying distributed load on inclined member, it may act in twodifferent manners as shown in figures 7-7 and 7-8.




- 30 -


In figure 7-8, the load is projected on horizontal axis. This is the common case.In figure 7-7, the load is acting perpendicular to the member. Most analysisprograms can handle both types of loading conditions shown. But it’s yourresponsibility to apply correct method.




- 31 -

8. Column Buckling test

Solve the following problem: A steel (E=200 GPa) column of 100x100 mmsquare cross section (I = 8.33E-6 m4) is 5 m long and fixed at bottom end. Aload is applied axially to the column. Find the buckling load.

By calculation, buckling load is given by Pcr = π2EI/4L2 = 165 kN

Figure 8-1

First draw the column, define the column properties and then apply any load,which you know that is well below Pcr. Now see if the column buckles! No, itwon’t. To get the correct result you must activate the “Frame instability” or “P-

∆” analysis option yourself to force the computer to make iterations! Now

gradually increase the load and re-analyze. At one instant, the computer willshow you a message, which will say that the program has encountered anegative diagonal term in member stiffness matrix and analysis will terminate.Note this load. This is the minimum buckling load. You are likely to see thateven when the column buckles, the deflected shape of the column is drawnstraight!

You may ask why computer can’t account for buckling in normal analysis.Well, most analysis programs, by default, perform first order analysis. That is, it

sets stiffness matrix, solves it and then calculates axial forces from it. When youinstruct it to perform P-∆ analysis, it performs iteration to find out actual axialforces. Remember if you are using a very cheap program or some non-

commercial program, it may not have P-∆ analysis option! Be careful!

You may wonder, why the computer itself does not choose P-∆ analysis always.Hmm, it would have been nicer. But think of the time required for performingsuch analysis. I once analyzed a 20 storied 3D frame in VA, which had 10 baysin both x and y direction. With 233 MHz, 16 MB RAM computer it took me 20

minutes to perform first order analysis. If you want to perform P-∆ analysis for




- 32 -

such structure using a standard PC and inexpensive program, chances are thatyour system will crash! Check it!




- 33 -

9. Portal and Cantilever method

You may have been taught to use portal and cantilever method for analysis of effect of lateral loads in frames. Both of these methods assume a point of contraflexure at mid point of beams and columns, which is often grosslyinaccurate. Just analyze any frame subjected to lateral loading by these methodsand then compare the results with exact analysis by computer. You will find asmuch as 50% to 60% difference of moments and shear forces. If computer isavailable, you must not use these methods. Even for preliminary analysis, whenyou do not know the size of the members in the structure, still these methods arenot useful. You can do the same easily by using computer.




- 34 -

10. Deflection of Reinforced Concrete member

Consider a simply supported beam made of reinforced concrete. It is loaded byuniformly distributed load. How do you calculate its deflection at midpoint?

You may, of course, use the familiar equation ∆ = 5wL4 /384EI. But remember,here you must use effective moment of inertia of the section and not the grossmoment of inertia of the section if applied moment (wL2 /8 in this example)exceeds cracking moment capacity. Here w stands for dead + live load. Mostcomputer programs do not take into account the reduced moment of inertiabecause of cracking. Since sometimes Ie comes equal to 50% of Ig, when you donot calculate Ie, you may just double the deflection as found from computeranalysis which takes I

g. Please note that, for all members you may not need to

use Ie because for all members calculated moments may not exceed crackingmoments. Once you have got Ie, you can use the same analysis program to findout the deflection of desired members. But you must note following things.

1. To find out deflection at middle of a beam, you must have a node there. Youcan achieve this by splitting the beam into two members. Most analysisprograms have the option of doing this.

2. Changing I values of some members does not alter moment and shear valueswhich you have got previously using Ig.

3. Ie can be calculated only when you have designed the member i.e. you havespecified number and diameter of reinforcement bars.

4. When you are specifying I value explicitly, ensure that you do not definebeam width and depth or radius, otherwise you may get absurd results.

The formulas for calculating cracked moment of inertia are given below (Ref.1).

For rectangular beam reinforced for tension only:Icr = b(kd)3 /3 + nAs(d-kd)2

Where k = ((2ρn + (ρn)2)0.5 – ρn and ρ = As /bd

For a beam with both tension and compression reinforcement:Icr = b(kd)3 /3 + (2n-1)As’(kd-d’) + nAs(d-kd)2

Where k = ((2n(ρ+2 ρ’d’/d) + n2(ρ +2 ρ’)2)0.5 – n(ρ + 2 ρ’), ρ = As /bd and ρ’ =A’s /bd

For a T-beam with k>hf Icr = bw(kd)3 /3 + (b-bw)hf

3 /12 + (b-bw)hf (kd-hf /2)2 + nAs(d-kd)2

Where k = (ρ n + 0.5(hf /d)2)/( ρ n + hf /d) and ρ = As /bd




- 35 -

For a T-beam with k>hf – use same equation as that for a rectangular beam.In all cases n = Es /Ec.

Modulus of elasticity of concrete is given by Ec = 5700√f ck MPa if f ck (MPa) is

measured as cube compressive strength of concrete; and Ec = 4700√f ck MPa if f ck is cylinder compressive strength.

What is said above stands for short-term (immediate) deflection. You must add

long term deflection due to creep and shrinkage as well.

This additional deflection can be obtained by multiplying the short-termdeflection (discussed above) due to dead load (+ live load, if live load remains

in place for extended periods of time) by creep factor ξ = ν /(1+50ρ’)

Where ν = 0.787(months)0.229

but not greater than 2.0 and ρ’ = area of compression steel/gross cross sectional area of the member.

This simple trick works for 1 dimensional member only i.e. for beams. For 2way members e.g. slab, things are not as easy. We shall discuss later how to findthe deflection of 2-way slab by using finite element analysis.

Most codes provide you minimum depth of members if you do not calculatedeflection. But these values are always overestimated and thus lead to

uneconomical design for multistory buildings. Don’t be lazy. Always calculatedeflection, you can save money!






- 37 -

12. Inclined support

If your program supports specifying inclined local axes for a particular member,

you are lucky. In this case you just need to mention in what angle you want torotate the local axes of the selected member; then you will specify the jointrestraints in usual manner and it will be considered as an inclined support. But if your program does not have this feature, you need to try out something else.

You can achieve this by specifying a “spring” of infinite stiffness. Normally youcan specify a spring at any angle. The spring reaction is the resultant of X and Ycomponents of reaction. In case of roller support you will get the reactionautomatically from spring reaction.

Think what you have learnt…

How many of following analysis methods you have learnt in theuniversity? – Moment distribution, Slope deflection, Portal, Cantilever,Kani’s rotation contribution, Conjugate beam, Graphical – Funicularpolygon & Maxwell diagram – Williot-Mohr diagram, Three momentstheorem, Column analogy, Moment area, Substitute frame, Method of

joints & method of section for trusses. Probably you know all or most of the above classical methods of analysis. Now be honest, how many of the above methods you still use to solve structures after you have startedusing computer analysis programs? Probably none! Academic peoplewill argue that all the said methods are to be mastered for a betterunderstanding of structural response. Do you think so? I don’t. Well,among the methods listed above, the moment distribution is mostpopular. This is quite logical, because this method is easy, does notinvolve solution of simultaneous equations and converges rapidly. Weshall discuss substitute frame method later (see section 13) while

considering maximum bending moment, shear force etc. in buildingframes. Did you notice that all these methods are used for frame analysisonly? You may like to know that 80% to 90% of all real world structuresanalyzed are frame structures. Although you have learnt flexibility andstiffness approach while studying computer method of analysis, onlystiffness method is used in computer programs. Modern world’s mostpowerful analysis method – finite element method is also a stiffnessmethod in essence.




- 38 -

13. Maximum bending moment, shear force and reaction inbuilding frame: Substitute (Equivalent) frame

A frame member will not experience maximum bending moment, shear force

and reaction when it is fully loaded. Here we shall combine classicalapproximate substitute frame with computer analysis. But before that note thefollowing live load distribution criteria.

To get this Do this

Maximum positive bending moment atcenter of span

Load that span and then alternatesspans

Maximum positive bending moment atcenter of span

Load adjacent spans and then alternatespans

Maximum negative bending momentat support

Load adjacent spans and then alternatespans

Maximum column reaction Load adjacent spans and then alternatespans

Maximum positive bending moment atsupport

Load all spans except adjacent spans

In all cases, dead load must always be applied over all spans.

Some codes say that if live load intensity does not exceed 75% of dead loadintensity, then you can load all spans together with dead and live load withoutany combination. But if you have computer, it is always better to perform theactual combination to get maximum values of force and moment.

In classical substitute frame (see figure 13-1), we isolate one single floor withthe assumption of columns at top and bottom floors are fixed. Then we applythe combinations described above to get maximum member forces.

In case of computer analysis, though you still need to apply the live load insame combination as discussed above, yet you need not isolate one particularfloor. Rather, you should just apply the required span load combination in anyfloor. In case of regular shaped building elevation, result obtained from onefloor will be same for other floors. For example, in the (figure 13-2) shown, theload combination stands for maximum negative support moment in first interiorcolumn (actually both interior columns since this structure is symmetric) in 2nd floor (bottom most floor, i.e. ground floor is normally denoted by “0” instructural analysis convention). The value obtained for support moment underthis condition will also be the maximum support moment for 1st, 3rd and 4th




- 39 -

floor. (Though it is customary to use reduced live load in roof level). Similarly,other load combinations can be used in same manner.

Figure 13-1

Figure 13-2

4

3

2

1




- 40 -

14. Support settlement

Take any statically determinate structure, for example a simply supported beamor a simple truss. Apply a settlement in one of its supports. Now analyze thestructure. You will see an interesting phenomenon. Though the program willcorrectly say zero member force, still it will draw a bending moment or axialforce diagram! In a statically determinate structure, there should not be anymember force developed due to support settlement. Hooray, you havediscovered a bug in the program!

The reason of this awkward shape can be explained. Although the member forceis zero, the program calculates it as a very small (say 10-100) number. The

graphic code picks up this small but finite number and draws the force diagram.

Now take a statically indeterminate structure. Say a continuous beam. Make oneof its support settle to an amount and perform the analysis. You should findsome member forces in the beam. Well, now take a statically indeterminatetruss. The truss should be externally indeterminate. For example, you can take a2-support truss whose both supports are hinged (pinned) as shown in figure 14-1. Now apply a downward settlement in any one of its supports and analyze thestructure. Most likely, you will see zero force in all members after the analysis.

This is not correct!

Figure 14-1

Most standard analysis package use truss stiffness matrix based on ignoring thesupport displacement perpendicular to member’s local axis. That’s why you getthe wrong answer. But since the members’ length change, there are strains,which would create the axial forces. If you want to know the actual memberforces after such support settlement, you need to modify the member stiffnessmatrix considering the displacement in perpendicular direction as well.Unfortunately, you can’t do it with most available programs.




- 41 -

15. 2D versus 3D

For symmetrical structures, often it is possible to convert the 3D model to 2D

for easier input and analysis. For example, consider the following structure asshown in figure 15-1.

Figure 15-1

You can easily analyze just one plane frame as shown in fig. 15-2. Wheneverpossible, try to convert 3D structures into 2D in this way. 2D structures are notonly easier to model, but also they can be ‘handled’ and analyzed much moreeasily compared to 3D structures.




- 42 -

Figure 15-2

Look, in fig. 15-1, the loads are towards X direction. If there were additionalloads of same type towards Z direction, you could adopt similar 2D frame (onYZ plane) as shown in fig. 15-2. You can then superpose the result as long as itis a linear structure with material and member section properties are the same.

How about dynamic analysis of the frame shown? Is it possible to convert 3Dinto 2D? I shall discuss this when covering dynamic analysis in detail.




- 43 -

16. Curved member

Most frame analysis programs do not have curve element. You will need toreplace the curved member by a number of straight members. Obviously, morethe number of straight members used better the accuracy is. While drawingstraight members for curve elements, it is a good idea to change grid setting into“polar” form instead of normal rectangular setting. Another way of doing this isto figuring the straight members’ nodal co-ordinates in spreadsheet (forexample, Microsoft Excel or Lotus 123). This is useful when the equation of curve is known as y = f(x) e.g. parabolic arch. By using spreadsheet’s built-incommands, you can easily find out y co-ordinates of the curve against each xco-ordinate. Some programs can “copy” and “paste” member and nodalinformation to and from spreadsheet file. You may like to know that it istheoretically possible to create stiffness matrix of a curved member.

Now solve the following two hinged parabolic arch.

-50 kN/m (downward)Y

4 m X

20 m

Figure 16-1

The answer is: left vertical reaction 375 kN ↑, right vertical reaction 125 kN ↑,horizontal reactions are 312.5 kN inward at both ends. With 20 straight-linesegments, you should get exact answer within 1% accuracy. Note that the

theoretical answer has been obtained by (H = ∫ My dx / ∫ y2 dx) formula.




- 44 -

17. Tapered section

Many programs have the option of specifying tapered or variable cross sectional

members. If so, you’re lucky. If not, still you’re lucky as you are reading thisbook! To specify a tapered section by yourself, you should ‘break’ the membersinto a number of parts (more the number, better is the result). Then you shouldspecify various A (areas) and I (inertia) for each segment. This will becomeclear from the following problem.

1.5 -5000 1.5

2 4

15 15

Figure 17-1

The figure shows a tapered beam. Hinged at left end and fixed at right end. Aclockwise moment (hence minus sign) of 5000 has been applied at middle of thebeam. It is required to analyze the beam. The cross sections at both ends have

been shown. Please note that the beam has been ‘divided’ into 8 sections. Widthof the sections is same throughout. But the depth is varied as (from left mostsection) 2, 2.25, 2.5, 3, 3.25, 3.5, 3.75, 4. The calculated reaction at left end is –163 (i.e. downward) and at right end is +163 (i.e. upward) compared totheoretical answer of 170. The bending moment at just left of mid-point of beamis – 2443 (theoretically –2549) and that of right is 2557 (theoretically 2451).You will get more accurate answer if you divide the beam into more number of elements. Note another interesting point that, in this problem, I didn’t specifyany unit or E value of material! You should get same answer whatever unit youuse. Although some programs do allow you to specify “linearly” taperedmembers; you still need to apply this trick for “non-linearly” (e.g. cubic orparabolic) tapered members.




- 45 -

18. Nodes connected by a spring

Many programs allow you to define a spring support, but none will allow you to

connect two nodes by a spring. But you can achieve this! Replace the spring bya member connected between those two nodes where the spring is required.Choose the properties of that member so that stiffness of spring equals AE/L of that connected member. E should be same as that of material of the spring.Choose A and L value properly, keeping L small; because if you choose large L,the member will buckle easily. Also, do not forget to release moment on thismember i.e. this spring replacement member should carry axial load only. Afteranalysis, you must check whether axial load in spring replacement member is

below its buckling load (π2EI/L2). This can be automatically checked if you

activate P-∆

analysis (see section 8) option in your program. This trick works!

Figure 18-1




- 46 -

19. Sub-structuring technique and symmetry (break them intopieces…)

In the analysis of large structures, it is often possible to consider only a part of the structure rather than the whole. This approach is useful to reduce the labor(cost and time) of preparing the data, of computing and of interpretation of theresults. When an isolated part of a structure is analyzed, it is crucial that theboundary conditions ‘sub-structures’ accurately represent the conditions in theactual structure. As a first simple example, consider the following structure asshown in figure 19-1. You are required to analyze the structure.

10 kN A = 2002 mm²4 m E = Steel

-5 kN/m20 kN

4 m

4 m 4 m 4 m

Figure 19-1

If you separate the upper floor and then analyze only that portion, you will getthe result as shown below.

10 kN

11.44 kNm11.44 kNm 5 kN 5 kN

4.28 kN 4.28 kN

Figure 19-2

With the result shown above, the applied loads on the bottom floor of the actualstructure will be as shown below.




- 47 -

11.42 kNm 11.42 kNm -5 kN/m25 kN

4.28 kN 5 kN 4.28 kN

A B C D

Figure 19-3

Observe that on leftmost node, 25 kN loads comes from 20 kN applied at that

node and 5 kN reaction from upper floor. The reactions you will get in the lowerfloor should be same as that of obtained if you considered the whole structure asshown in figure 19-1.

For your check, the ultimate results are as given below for problem figure 19-1.

Node Fx kN Fy kN Mz kNmA -6.223 -14.85 15.58B -5.99 30.54 15.18

C -11.2 20.64 22.08D -6.583 3.675 15.81

From the above example, it is clear that; you need to apply opposite of reactions

as loads on lower floor frames.

The procedure described here seems too meager for this particular structure, butthis method is an absolute must for doing a fine meshed finite element analysis.It may happen that, if you run the whole structure once, it may exceed the

program’s or your computer’s resource limit. That’s why it’s so important to‘break them into pieces’.

Whenever possible, try to design symmetrical structure as much as possible.They behave better than unsymmetrical ones. For symmetrical structures, thissub-structuring technique is a great time saver. When the structure has one ormore planes of symmetry, it is possible to perform the analysis on one-half, one-quarter or an even smaller part of the structure, provided that the appropriateboundary conditions are applied at the nodes of the plane(s) of symmetry.Followings are some examples of exploiting symmetry of structures.




- 48 -

Continuous beams, with even number of spans.Actual beam

Figure 19-4

Symmetry utilized beam

Fixed

Figure 19-5

Continuous beams, with odd number of spans.Actual beam

Figure 19-6

Symmetry utilized beam

Z direction rotation fixed

Figure 19-7

The ‘key’ to utilizing symmetry, is applying proper boundary condition.Remember, in order to take advantage of symmetry, both the structure

(geometry and material) and the applied load must be symmetric. Although,you can still take advantage of symmetry even if the loading is ‘anti-symmetric’(i.e. one half of the loading is similar to other half in magnitude but opposite in

direction), the procedure will be somewhat screwy. In all cases, our signconvention is same as described in section 4 earlier. Now consider plane frameswith even number of bays as shown in fig. 19-8. This frame can be detached,after applying proper boundary condition, as shown in fig. 19-9. Plane framewith odd number of spans has been shown in figure 19-10. Here you will haveto apply boundary condition of X translation and Z rotation prevented in midpoints of the middle beams as shown in fig. 19-11. Symmetrical structures arenot only easier to analyze but also perform better than unsymmetrical structuresin real life!




- 49 -

Figure 19-8

Fixed

Fixed

Fixed

Figure 19-9




- 50 -

Figure 19-10

X translation & Z rotation fixed



Figure 19-11

Exercise

Solve some problems yourself on the basis of above example models. Unlessyou analyze the models and visualize the results, things will not be crystal clearto you. If you face any problem, don’t hesitate to send me an e-mail! You willfind advanced info on 3D structures’ symmetry, plate’s symmetry etc. in latersections.




- 51 -

20. Staircase analysis

A staircase is actually a folded plate structure. But in our traditional simplified

method of analysis, we consider it as a straight beam. How far is thisassumption justified? Consider the figure of the staircase shown below.

Figure 20-1

The first figure shows exact shape of a flight of a staircase with loads (includingself-weight). The second figure is the approximation of the same staircase assimple beam. The section of concrete staircase may be taken as 1-m width x150-mm depth. The length of simple beam equals 1.25 + 2.75 + 1 = 5 m.Theoretically, loading on landing should be less than that of inclined flight. Inapproximate calculation, it is assumed same load is acting through out the span

for conservative result. The results of both analyses are shown in next figure. Inthis case we have considered the staircase as simply supported. Depending oncasting, it may be fixed-fixed or fixed-pinned as the case may be. In factstaircases are more often analyzed as fixed-fixed support condition. From theanalysis it is found that maximum mid span moment is almost same in bothanalyses. Shear forces (reactions) are also more or less equal. This proves thatapproximate analysis of staircase is not really inaccurate! In hand calculation,moment was computed using simple M = wL2 /8 formula.

We shall venture on folded plate analysis in detail in some later section.




- 52 -

Figure 20-2

This analysis was done in Visual Analysis 3.5. An interesting point is to notethat, both structures were analyzed as a single file. This is applicable to mostanalysis programs. You may analyze as many as separate structures in a single

file even they are not connected together. Now think about the followingparadox. Following three beams are all simply supported (left end pinned andright end roller). Their projected length on plan is same in all cases (say 10 m).They are all acted by same uniformly distributed load on ‘projected’ length (say10 kN/m). Find out what will be the bending moment at mid spans.

10 m 10 m

10 m

Figure 20-3




- 53 -

What result do you see? The bending moment (and reactions as well) is same inall cases! If you took w = 10 kN/m and L = 10 m, then Mmax = wL2 /8 = 10*102 /8= 125 kNm. It shows that, for the simple beam, bending moment is sameirrespective of beam’s geometry. This happens because all three beams shown

are statically determinate structures. Now make all the beams fixed at both ends.Now re-analyze them and you will see different bending moments for all cases.The example problem I presented in this section for staircase, was simplysupported in both ends. That’s why you got same bending moment! Had theybeen fixed at ends, the results would not have matched. However, they stillwould not differ appreciably from traditional straight beam calculation.

Still in doubt why you got same result for statically determinate beams? Well,the reason is simple. As the beams were simply supported, horizontal reactionsat supports are zero (since we have only loading acting downward). So, momentdue to ‘eccentricity’ of geometry is also equal to zero. This will be fromfollowing figure.

Internal moment developed

Horizontal reaction H ex = eccentricity

x

Figure 20-4

This internal moment (= Hex at any section of distance x from end) causes thebending moment to differ from the value as in case of straight beams (where ex = 0 at all sections). In case of statically indeterminate beams, both H and ex arenon-zero. So, the internal forces differ depending on geometry of the beam.When you analyzed two hinged arches as a student you probably used theequation: Arch moment = Beam moment – He. Didn’t you?




- 54 -

21. Cables

It is possible to analyze cables with a mere frame analysis program. A cable

carries ‘tension’ only. So, you should define a cable in the same way as trussmember (which carries axial force only) but additionally you will have tospecify that it can take tension only (no compression). Some analysis programsmay not have the option of defining a tension only member!

Once you have specified cables in this way, the analyses are prettystraightforward. While viewing the result, you should check whether cables’axial force diagram shows tension only (generally positive number) and nobending moments. That’s all.An example of cable structure is shown in fig 21-1. After performing theanalysis, check your answer with exact result as given.




- 55 -

Figure 21-1




- 56 -

Vertical reaction at A = 104 kN (down), at B = 250 kN (up). Moments: MAB =0, MBA = 75, MBE = -117, MEB = 41, MEC = -41, MBD = 41, Mid span of EC = 84kNm.

Figure 21-2

It is also possible to analyze the cable shown in figure 21-2. Use suitable valuesfor span, sags and loads. Then find out the tension in cables. This is given as anexercise to you! If the loads are all unequal, the tensions in the cables will bedifferent. Check if equation of static equilibrium is satisfied at each node.




- 57 -

22. Pre-stressed cable profile

Does your program offer specifying pre-stressed cable profile? If yes, then

good. If not then read the following tricks. Observe sign conventions carefully.

PyA PyB

yA θA θB yB PθA PθB c

P P

Upward UDL w = 8Pc/L²c

θA = (4c + yA – yB)/LL/2 L/2 θB = (4c – yA + yB)/L

Actual pre-stressed cable Equivalent load

Figure 22-1

PyA PyB

yA θA θB yB PθA PθB

P P(θA + θB) P

Total length L

Actual pre-stressed cable Equivalent load

Figure 22-2

Observe the figures very carefully. They are really confusing! Try tocomprehend the following worked out problem. Please note that the θ values arein radians. Note that the yA and yB indicate eccentricity of the cable at supportsin upward direction from center of gravity of concrete (cgc) line. Upwarddistance is positive at supports and downward distance is positive at mid spansfor pre-stressed cable profile (majority of standard analysis programs follow thissign convention). If the cable distances are of opposite sense compared to whatshown in above figures, then ‘arrows’ of moments will be reversed.




- 58 -

With reference to the figure 22-3, the calculation isshown below.

For left span, L = 15 m, yA = 0.5 m, yB = 0.5 m.So, θA = (0.5+0.5)/10 = 0.1 rad

And θB = (0.5+0.5)/5 = 0.2 rad

So, moments are 500 x 0.5 = 250 kNm on left endand 500 x 0.5 = 250 kNm on right end of left span.

Concentrated force at 10 m from left span is 500 x(0.1 + 0.2) = 150 kN.

For right span, L = 15 m, yA = 0.5 m, yB = 0.4 m.

So, θA = (4 x 0.6 + 0.5 – 0.4)/15 = 0.17 rad

And θB = (4 x 0.6 – 0.5 + 0.4)/15 = 0.15 radAnd c = (0.5 + 0.4)/2 + 0.6 = 1.05 mSo, equivalent upward load w = 8 x 500 x 1.05/152 =18.67 kN/m. Also, support moment at left end of right span is 250 kNm and on right end is 500 x 0.4= 200 kNm. So, the equivalent forces on the beam

will be of as shown in figure 22-4 (axial force P isnot shown).

Figure 22-3

250 kNm 250 kNm 200 kNm

250 kN 150 kN 250 kN 18.67 kN/m 250 kN

Figure 22-4

Now the forces shown in blue color will go to support directly. Moments shownin orange color will cancel each other. So, the remaining forces that will act are




- 59 -

shown in green color. The ultimate equivalent load will be that of as shown infigure 22-5.

250 kNm 200 kNm

150 kN 18.67 kN/m

Figure 22-5

So, for pre-stressing force, the beam should be analyzed for the loading shownabove. Naturally, the beam will also carry dead load and live load as well.Analyze the beam for these loads as separate cases and then combine the resultsas desired. In actual practice there are always more than one cables. You can

analyze effect of each cable separately and then superpose to get the net result.Also remember that, there is a uniform compressive stress ‘P/A’ in the concretein addition to the bending stress due to pre-stress, dead load and live load. Formore information on this subject, please see any standard textbook on pre-stressed concrete. I have shown here only linear and parabolic cable profile.Although parabolic profile is the most common, there are other types of profilespossible. See your textbook for details.




- 60 -

23. Finite Element Analysis (FEA) Method is approaching…

We now come to the most outstanding and most versatile method of structural

analysis: the Finite Element Method. It has made possible to analyze virtuallyall kinds of structures that human brain ever can imagine! If you have studiedfinite element before, you may skip this section. Those who did not, I present avery very brief introduction of the subject. There exist more than 1001 books inthis subject. But I warn you; the theory of finite element analysis is verycomplex!

What is meant by finite element? The answer is any element, which is notinfinite. Don’t be exasperated; this is the real definition of finite element.

Did you play with mechano when you were a child? Just think how you built amodel car or house by “Lego” parts? Now consider each part of mechano as“finite element”. A number of mechano elements were needed to build yourmodel car or house.

Now consider a frame. It is made of a number of beam/column members or“bars”. Here the “bars” are “finite elements” of the “frame”.

I hope you have probably realized now that the frame analysis, so far what we

have discussed in preceding sections, is actually finite element analysis inessence where each finite element is a “bar”.

Figure 23-1

This is the longitudinal section of a beam shown. That is, you are viewing abeam from its length side. Observe that here we consider the beam as 2-dimensional “Plane stress” structure. Don’t confuse this with 2D or 3D frame.By 2D or 3D frame we actually mean “Plane” and “Space” frame. In allprevious cases, we treat all beams as “bars” like a “stick”. But in the abovefigure we are treating the beam taking into effect of its length as well as depth.That’s why it is 2D. Had we considered the width of beam in the analysis, it

would have been termed as 3D solid. Pretty confusing! Look, there is a “cut” inthe beam. The beam is simply supported, left end pinned, right end roller. It is

X

Y




- 61 -

loaded by a uniformly distributed load. We like to find out the stresses atvarious points of the beam. Please note that analysis of this problem by classicalmethod is close to impossible.

So, first we divide the beam into a number of “triangular finite elements”. Thenwe shall determine the member stiffness matrix [k] of each individual triangularelement and ultimately we shall have to combine the member stiffness matricesinto “global stiffness matrix” [K]; pretty much the way we did in case of frameanalysis. Then we shall have to apply the boundary condition on [K] matrix.

Figure 23-2

After that we need to construct force matrix [P]. For this, distributed load mustbe converted to appropriate nodal loads by applicable equations. So, ourproblem can be represented by familiar equation [P] = [K][D]. From thisequation we can solve for [D] and then we can find out nodal stresses form

equation [σσσσ] = [C][εεεε] where [C] matrix differs in various cases like plane stress,plane strain etc. We are describing this problem as plane stress because weconsidered only 2 dimensions (X and Y) and stress variation along width (Z

direction) has not been taken into account. That means we have taken care of only σx, σy and τxy. In this problem we considered the beam is made of “triangular” finite elements, but we could have also considered it is made of “rectangular” finite elements as shown in figure 23-3.

Figure 23-3

If you analyze the beam with both triangular and rectangular elements as shownabove, you will see that you get accurate answer when you use rectangularelements. It proves one very fundamental concept of finite element analysis:You must choose proper element for particular problem. You do get correctresult with triangular finite element but you must use very fine mesh comparedto rectangular element. In general, triangular element is not a good choice. If

you are interested to know why triangular element behaves in such way, youshould consult any standard finite element analysis textbook.

X

Y






- 63 -

One distinguishing feature of finite element method is that it does not provide“closed form” solution. Every problem in finite element analysis is unique. Thisprobably needs little more explanation. Think of a simple beam. In classical

method of analysis, you can make a program, which takes L, E, I and w as inputand computes deflection at any point by solving the equation of elastic line,which can be easily formulated. But in case of finite element analysis, if youchange the length of the beam, it becomes another new problem because thegeometry of the model changes. Of course you can change E or w values orboundary condition without remodeling the whole structure.

Another aspect of Finite element analysis is that it almost always produces anapproximate result. I used the word “almost” because finite element analysisdoes produce exact result only when the finite element is “bar” that is in “framestructures”.

You may be wondering that if finite element method can solve any structure,then what is the justification of studying classical methods of analysis. Aha! Areal question indeed! You can realize it yourself. Just think of solving a simplebeam in finite element method (this is presented just after this section). Afteryou solve this beam by finite element method, you can easily check whether theresult is correct or not by comparing the answer obtained by classical method.But now imagine the analysis of the fuselage of an airplane or the propeller of a

ship. How do you check the correctness of these analyses? Therefore you mustaccept the finite element analysis result as exact result! That’s why it is soimportant that finite element analysis models must be created to simulate theactual structure as much as possible. You must use proper combination of finiteelements, sufficiently accurate mesh, proper load and applicable boundaryconditions. It is often a common practice to analyzing the structure first with aparticular mesh and then repeating the whole analysis after doubling the mesh tosee whether the result converges. But this method has drawbacks! Yourprogram cannot analyze the structure if your number of mesh nodal points

exceed the program’s capacity. Moreover, it is very difficult to predictbeforehand what particular “finite element” will best simulate the structure. Thisis especially a demanding task for very complex structures.

Finite element method is nowadays widely used in all branches of aerospaceengineering, bio-medical engineering, mechanical engineering and structuralengineering etc. Some manufacturing companies spend millions of dollars everyyear in finite element analysis!

I am concluding finite element introduction here. But you must realize that it isnot so easy as it seems. Researchers are still developing new finite elements.




- 64 -

Sometimes even the most expensive finite element analysis programs producewrong answer to complex problems. If you feel inclined to know more aboutthis wonderful (?) tool of analysis, I strongly recommend that you to go throughsome standard finite element method textbooks.

One word of advice, many engineers tempt to use finite element analysiseverywhere even when it is possible to analyze the particular structure usingclassical method of analysis. My main aim is to make you realize that finiteelement analysis is required only when it is absolutely necessary. Rememberthat finite element analysis programs are very expensive and they also demandgreat part of contribution from you for preparing input and interpreting output.

Typically, a finite element analysis consists of following steps.

1. Defining the model (i.e. drawing it either in the finite element program’sgraphical interface or importing it from a CAD program).

2. Creating the mesh (most programs can automatically generate mesh for bestresult).

3. Defining the boundary conditions.4. Defining the loads.5. Performing analysis (may take hours for complicated models!)6. Interpreting the result (very important).

The steps are pretty straightforward. But there are many glitches!

In next page you will find an exercise of simply supported beam with uniformlydistribute load analyzed by FEA method. This example is for yourunderstanding of the basic concept of FEA only. In practice, this problem

should be solved by simple flexure formula of σ = My/I. Remember this!




- 65 -

Exercise

A 5-m steel (E=200GPa) beam has width 200 mm and depth 500 mm. It is

loaded by 10-kN/m uniformly distributed load. ν = 0.3. Its left end is hinged andright end is roller. Find deflection at mid point and maximum bending stress inthe beam by finite element analysis. Try following modeling:1. Plane stress analysis with 20 rectangular elements, each 0.5x0.25 m size.

That means there are 10 elements in X direction and 2 elements in Ydirection. You can convert the uniform load into nodal loads by applying0.25 kN at extreme nodes and 0.5 kN at intermediate nodes.

2. Solid model analysis. Use standard solid brick or tetrahedral element. Mostfinite element analysis programs offer these elements.

Figure 23-7

In case of plane stress model formulation, you should use plate finite elementwhose thickness will be equal to the depth in Z direction. In this problem, this isequal to width of the beam. After performing the finite element analysis, youshould get the answer: mid point deflection 1.95x10-4 m maximum stress 3.75MPa. Your program may display slight different result due to numerical roundoff in calculation.

The deflected shape should resemble the following figure. Original shape isshown by dotted line. This 2D-beam analysis was performed in Visual Analysis.

For 2D analysis, after modeling yourstructure should look similar to thisfigure.




- 66 -

Figure 23-8

The figure 23-9 shows one of mid plane stresses, local σx distribution.

Your program should have the option to display other stresses e.g. σy , τxy etc.Interpreting the finite element analysis result is very important. It is expected

that you spend equal or more time in interpreting analysis result compared to the time previously spent in preparation of the model .

Later we shall see how finite element analysis can produce incompatible result.There you will realize why it is essential to learn some theory behind the finiteelement analysis.

Figure 23-9




- 67 -

To analyze the beam as 2D, you should not face any difficulty. However, youshould take into account many other things when you analyze the same beam as3D solid. Firstly, what will be the load? Look, here we’ve applied a total load of 10 kN/m x 5 m = 50 kN acting on the upper face area of 5 m x 0.2 m = 1 m. So,

the applied load we have to specify as 50 kN/m² pressure normal to the uppersurface. Be careful about the load’s direction. Now, comes the main hurdle, themeshing. If you are using a high-end FEA program, it will mesh the modelitself. By default, the program will mesh it by using brick elements ortetrahedral elements. The next figure shows the beam with automaticallygenerated tetrahedral mesh. You may note that, such high density meshing isnot really required for this very problem. If you manually mesh with 20numbers (2 elements along depth and 10 elements along length, similar toshown in fig. 23-7) 8-noded solid elements, (as in SAP2000) you will get exactresult for this problem. Left end boundary condition is X, Y, Z translation fixedalong bottom edge and Z translation fixed along bottom edge on right end.

Figure 23-10

The next figure shows stress (σx) diagram on displaced shape.

Figure 23-11

This 3D analysis was performed in Cosmos/Design Star.




- 68 -

Did you see that finite element analysis programs normally give you output inthe form of nodal displacements and stresses. It does not show you bendingmoment or shear force diagram. Why? Well, why do you need bending momentand shear force values? To calculate stresses later, isn’t it? Finite element

analysis programs directly give you the stress values!




- 69 -

24. A typical worked out problem of FEA

I think by this time you have at least tried to open first few pages of a FEA

textbook and probably bogged down by heavy theory. Well, I am here to rescueyou. Unfortunately, most FEA textbooks do not contain sufficient numericalexamples to make the whole thing transparent to the readers. If you study thefollowing numerical example along with your FEA textbook, you might find iteasier to comprehend now. So, let’s start…

Figure 24-1

Figure 24-1 shows a plate divided into two triangular shaped finite elements.This is a plane stress problem. The figure also shows the degrees of freedom(DOF) of the system. Please note that the value of modulus of elasticity E istaken as 200x106 N/m². In fact, the values taken in this problem are not realistic.

My main aim is to present various steps of FEA computation through a simplenumerical problem. The division of the plate into a mere two elements is done

just for illustration purpose only. In actual practice, the plate should be dividedinto a larger number of elements (triangular or rectangular etc.).

Anyway, we now begin solving the problem. I hope you’ve already familiarwith ‘shape function’. This is the function, which describes displacement of anypoint within an element as a function of nodal displacements of the element.The definition will be clear as we solve the problem. Shape function matrix is

normally denoted by [N].




- 70 -

All FEA textbooks describe how to derive shape functions for various elements.So, we shall assume that we already know the shape function for the triangularelement.

We consider element 1 at first. Its nodal points are 1,2 and 3. The co-ordinatesare (0,0), (4,0) and (4,4) respectively. If the area of the triangle is A, then weknow

Where, (x1, y1), (x2, y2) and (x3, y3) are the co-ordinates of node 1, 2 and 3

respectively.

The shape functions for triangular element 1 are

Now, the ‘strain displacement’ matrix is

Now we set the ‘Constitutive matrix’ [C] as (for plane stress only)

2...8

441

041

001

2

1

331

221

111

2

1m

y x

y x

y x

A ===

y

A

x x y y y x y x y x N

y x A


x A


25.0

2

)12()21()1.22.1(

25.025.02

)31()13()3.11.3(

25.012

)23()32()2.33.2(

3

2

1

=−+−+−

=

−=−+−+−

=

−=−+−+−

=

1

332211

321

321

......

025.025.025.025.00

25.0025.0000

00025.0025.0

000

000

][][

−

−−

−

−

=

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂∂

∂

∂

∂

∂

∂∂

∂

∂

∂

∂

∂

=∂=

m

x

N

y

N

x

N

y

N

x

N

y

N

y

N

y

N

y

N x

N

x

N

x

N

N B




- 71 -

We have considered E = 200x106 N/m² and Poisson’s ratio as 0.

The stiffness matrix of triangular element is given by

If we perform the calculation, we shall get stiffness matrix for element 1 as:

Note that, it is a 6x6 matrix. Its DOF are 1, 2, 3, 4, 5 and 6. Thickness t = 0.01m. Following exactly same steps, we can easily find stiffness matrix for element2. This is given as an exercise for you. Calculation is same. Only you have touse x1 = 0, y1 = 0 (point 1), x2 = 4, y2 = 4 (point 3), x3 = 0, y3 = 4 (point 4).

For element 2, DOF are 1, 2, 5, 6, 7 and 8. So, the global stiffness matrix [K]becomes, [K] = [k]1 + [k]2. In matrix format,

26

2 / ...10.

10000

02000

00200

1

00

01

01

1][ m N

E C =

−−=

ν

ν ν

ν

ν

∫ ==V

T T stress plane for At BC BdV BC Bk )............(]][[][]][[][][

m N k / ......10.

101000

05.05.05.05.00

15.05.15.05.00

05.05.05.15.01

05.05.05.05.00

000101

][ 6

1

−

−−

−−−

−−−

−−

−

=

m N k / ......10.

5.15.05.0015.0

5.05.15.0105.0

5.05.05.0005.0.010100

100010

5.05.05.0005.0

][ 62

−−−

−−−

−−−

−

−−

=




- 72 -

Observe that it is an 8x8 matrix since we have 8 DOF in the problem.

Our next step is to calculate applied nodal forces. In the problem, we have avarying distributed force. In FEA, we must transfer distributed loads intoequivalent nodal loads. The formula for converting distributed loads into nodalloads is:

For more explanation on this topic, see your favorite (or not so favorite) FEAtextbook.

q1 q2 L(2q1+q2)/6 L(q1+2q2)/6

L LActual distributed load Equivalent nodal load

Figure 20-2

Note that the applied load is in DOF 3 and 5 direction. So, force matrix becomesan 8x1 matrix. However, still we did not specify the boundary condition. Since,nodes 1 and 4 are pinned, DOF 1, 2 and 7, 8 will be fixed. Therefore, we have toadjust both global stiffness and force matrix with DOF 3, 4, 5, 6 only.

The force matrix is

m N K / .......10.

5.15.05.000015.0

5.05.15.010005.0

5.05.05.101005.0

0105.15.05.05.00

0015.05.15.05.00

0005.05.05.15.01

1005.05.05.05.10

5.05.05.000105.1

][ 6

−−−

−−−

−−−

−−−

−−−

−−−

−−−

−−−

=

∫= dxq N r T

e...][}{




- 73 -

After applying boundary condition, global [K] takes the form

Force matrix is (for 3, 4, 5, 6 DOF only)

Recall very well known stiffness method formula [F] = [K][d]Calculated displacement for DOF 3, 4, 5, 6 is

[d]c = [K]B-1.[F]D or

Global displacement matrix [d] becomes 8x1 matrix as shown next.

N F .....

0

0

0

4000

0

8000

0

0

][=

m N K B / .........10.

5.1010

05.15.05.015.05.15.0

05.05.05.1

][ 6

−

− −−

−−

=

N F D ..........

0

4000

0

8000

][ =

md c ..........10.

14.1

57.4

71.1

43.7

][

3−

=




- 74 -

Our main analysis is complete. Now we shall ‘post process’ our result. Suppose,we like to know the strains and stresses at all three nodes of element 1. The

strain matrix is defined by [ε] = [B][d].

The stress is defined by [σ] = [C][ε].

What if we want to find out the displacement at point x = 3 and y = 2?Displacement at any point within the element [u] = [N][d]

md ........10.

0

0

14.1

57.471.1

43.7

0

0

][ 3−=

63 10

286

143

1875

10.

14.1

57.4

71.1

43.7

0

0

.

025.025.025.025.00

25.0025.0000

00025.0025.0

][ −−

−

−=

−−

−

−

==

xy

y

x

γ

ε

ε

ε

2 / ........

28600

28550

371450

][ m N

xy

y

x

−

−==

τ

σ

σ

σ

==

6

5

4

3

2

1

321

321.

000

000

d

d

d

d

d

d

N N N

N N N

uy

ux




- 75 -

Where, dn indicates nodal displacement of the element.

You should appreciate following interesting points:

1. In this problem, [B] = constant i.e. it does not involve any terms containing xor y. However, this is the case only for triangular element. That’s why it’scalled Constant Strain Triangle. For rectangular and other elements, [B] is afunction of x and y (and z for 3D cases).

2. Since strain is constant, stresses at any point within element are alsoconstant.

3. Distributed load should be converted to equivalent nodal loads.4. The commercial FEA programs basically performs the same operation as

described above. They calculate stress/strain at all points inside the elements

and plots as colorful contour diagram as output.5. Did you realize the labor involved in solving with just 2 elements and 8

DOF. Now imagine what will happen with 100,000 nodes. It vindicatesabsolute necessity of computers in FEA.

6. The answer we have got here is not correct. This is as expected because wehave considered only two elements. Use your analysis program to generate amesh and see what will be the exact answer. Next try the problem with fourrectangular elements. See your FEA textbook to find out [N] matrix forrectangular elements. The stiffness matrix for rectangular elements is given

by

Here you have to integrate ‘numerically’. (Oops!) I think you are already feelingbore. Let’s have a snack break!

m........10

0.1

14.410

14.1

57.471.1

43.7

0

0

5.0025.0025.00

05.0025.0025.033 −− =

∫∫−−

=a

a

T

b

b

dxdyt BC Bk ...].][[][][




- 76 -

25. Plates by FEM

Let’s solve a simple plate by finite element analysis. Our model is 3-m x 2-m

size and 10 mm thick. The plate is made of steel (E = 200 GPa). It is simplysupported. Ignore self-weight of the plate. It is acted by 100-kPa uniformlydistributed load over its surface. Find the deflection at mid point of the plate.The theoretical answer is 0.65-mm i.e. 0.00065 m. If you get 0.0005 or 0.0007m deflection using your program, it may be considered sufficiently accurate.

First hurdle is how you should ‘mesh’ the plate. Most programs will allow youto divide the plate into number of smaller plates within it. So, do it. Expensiveprograms can create the optimum mesh for you! If you mesh the plate 15x10elements, it should be enough for this problem. Now comes the boundarycondition. Specify two adjacent edges as pinned and the other two adjacentedges as rollers. It will make the plate simply supported. Consider signconvention as described in section 4.

Roller

Pinned Roller

Pinned

Figure 25-1

When applying surface load, be careful. Different programs have differentoptions for specifying surface loads. Also make sure the load acts ‘downward’.If everything goes ok, you should get the answer. The deflected shape shouldresemble a saucer. The programs typically show many other stress-componentslike Von Mises, S11, Max. Principal etc. I shall discuss about them later.

Now solve the same problem with some different boundary conditions like alledges fixed and heavier loads etc. Check your answers with theoreticalsolutions. Make a ‘cut’ anywhere in the plate and see what happens.

Why it is necessary to ‘mesh’ plates?




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Suppose you have made a model of moment frame with masonry walls andconcrete floor slabs as ‘plates’. Now why you need to ‘mesh’ the plates? Theproblem with this is that the plate elements are only connected to the supportingmembers at the nodes. In the real world structures, the plates are in continuous

contact with the supporting members. Modeling the structure in this way cancause larger deflections in some members than if they were modeled withintermediate connections. The walls may also be too stiff when modeled as asingle element. Another concern involves load path continuity. If a distributedsurface load were applied to the floor plates, the load would be transferred to thecorners of the plate and directly down to the columns. The supporting beamswould receive no load. To fix these problems the single plate will have to besplit into smaller pieces, so that more nodes are provided to allow the real worldconnection to be more accurately modeled. The more each plate is split themore accurate the model becomes. However, more elements require more timeto analyze, this is particularly important with large models. Interpreting theresult also becomes complicated. Finally, it is not clear into how many elementsyou should split the plates (Ref. 6). You may wonder that we don’t split frame(beam/truss) members. Well, as I already stated that for ‘bar’ finite elements, wecan form the exact stiffness matrix. So we don’t need to split them. Although,occasionally you may need to split frame members too for applying nodal pointloads. For linear structures, even if you split frame members, you still get sameresult. But for non-linear structures, you will get different (usually moreaccurate) result if you split frame members. More on non-linear structures in

some later sections. So, keep going on!

How do I know whether my meshing is accurate enough?

Why not start with this very plate problem? First, analyze the plate with only6x4 mesh. Your mid point deflection should come 5.91x10-4 m. Clearly, this isnot equal to theoretical 6.5x10-4 m. So, you need to increase your mesh densityi.e. you have to decrease your mesh size. Let’s apply 15x10 mesh (see fig. 25-2). As I already mentioned you earlier, in this case you should get exact mid

point deflection. So, this mesh is sufficient. But hey, in actual problem youwon’t know the answer beforehand! So, how do you know that you get exactanswer with 15x10 mesh? What you have to do is that, you analyze the plateagain but with 30x20 mesh. You will observe that you get same displacement asin case of 15x10 mesh. Now your result is correct. So, we just increased meshdensity and see whether our result converges and we stopped when done. Thistrick normally works in most of the FEA problems, however, there may besituations where increasing mesh density only may not produce good result! Wemay need to use more complex elements. For example, instead of 4-noded plate

element we can use 8-noded plate element. I shall discuss more about it later.Another point, should you rely upon convergence of displacement or of stress?




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In general, stresses in FEA are ‘less’ accurate than displacements (since stressesare calculated from displacements, see the worked out problem of previoussection). So, you should aim at displacement convergence. Take stress diagramsof the point when displacements converge (in 30x20 case for this problem).

Figure 25-2




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26. Interpreting FEA result

In this section you will learn how to interpret FEA result. This is one of the

most important tasks. I shall start with very basic beam and truss and thengradually move to complex stress components.

First consider a simple truss member. When it’s in tension, the force acts on itas shown below.

Figure 26-1

The tensile stress developed in the member is simply equal to Force/Area. Untilthe stress exceeds the value of yield stress of the material of the member, themember will not fail. On the other hand, consider a member under compressiveforce.

Figure 26-2

Here the stress is also Force/Area but in this case, buckling may occur. The

maximum stress that the member can withstand depends on the material and the‘slenderness ratio’ (= length/side dimension) of the member. Try to avoidcompression member as far as possible. This is current trend in design. Alltension member structures like tents etc. are very efficient and cost effective.

In case of straight beam, the forces we generally do consider are – bendingmoment and shear force, although sometimes effect of axial forces may be quiteas much. The next figure shows positive direction of the bending moments andshear forces acting on a beam.

Figure 26-3

The bending moment causes bending stress (= My/I) on the beam. The reactionscause shear stress (= VQ/Ib). For beams curved in plan, in addition to these

forces, torsion also comes (see figure 26-4). Things will get really messy if unsymmetrical bending is considered. This is discussed later.




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Figure 26-4

Now consider a most general case of the stresses acting on a solid element.

σy

τyx

τxy

τyz

σx

Y τzy τzx

τxzX

Z

σz

Figure 26-5

All stress components are shown in figure 26-5. I’m sure you’ve seen this figureseveral times since your first year of university, didn’t you?

In straight beam, we have only σx. Please see figure 26-1 and you willappreciate the stress component diagram shown there. Note that in that beam,

you will get values for only σx stress components, which is obvious.

Typically, all FEA programs will show you Von Mises stress by default afteranalysis is finished. Now what does this mean anyway?

Von Mises stress is a measurement of ‘distortion’ of the element. This is basedon Von Mises – Hencky theory which, predicts that yielding in ductile materialoccurs when distortion energy per unit volume of the material equals or exceedsthe distortion energy per unit volume of the same material when it is subjectedto yielding in a tensile test. The theory takes into account the energy associatedwith changes in the shape of the material. This criterion is used to analyzematerials that would fail in a ductile manner. In a nutshell, The Von Misesstress is a measure of stress intensity required for a material (generally a

metallic material), to start yielding and become plastic. Before showing youmathematical concept of Von Mises stress, please let me introduce Principal




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stresses. Von Mises stress has no implication in brittle materials like concreteand soil.

Stress components depend on the directions in which they are calculated. At a

point, for certain co-ordinate axis rotations, shear stresses vanish; the remainingthree normal stress components are called principal stresses. The directionsassociated with principal stresses are called the principal directions. The three

principal stresses are denoted by σ1, σ2 and σ3.

I apologize for the following boring equations!

The Von Mises stress is computed from the six stress components as follows:

Hence, the Factor of Safety = σ limit / σ Von Mises

Then comes Maximum Shear Stress criterion, which is also known as Trescayield criterion. According to this theory yielding of material begins when theabsolute maximum shear stress reaches the shear stress that causes the same

material to yield in a tensile test. This criterion is mostly used to analyzematerials that would fail in a ductile manner.

τ max ƒ (σ limit / 2) where τ max is the maximum of τ 1/2, τ 2/3 or τ 3/1.

τ 1/2 = (σ1 – σ2)/2, τ 2/3 = (σ2 – σ3)/2, τ 3/1 = (σ3 – σ1)/2.

Hence, the Factor of Safety = σ limit / 2τ max

There are two other failure criteria mainly for brittle materials – Mohr CoulombStress and Maximum Normal Stress.

Mohr Coulomb theory is also known as Internal Friction theory. According totheory, failure occurs when:

σ1 >= σ Tensile Limit if σ1 > 0 and σ3 > 0

σ3 >= -σ Compressive Limit if σ1 < 0 and σ3 <0

(σ1/ σ Tensile Limit – σ3/ σ Compressive Limit) < 1 if σ1 >= 0 and σ3 <= 0

Hence, the Factor of Safety = 1/(σ1/ σ Tensile Limit – σ3/ σ Compressive Limit)

2

)()()(

)].(3))()()((21[

2

13

2

32

2

21

2

1

222222

σ σ σ σ σ σ σ

τ τ τ σ σ σ σ σ σ σ

−+−+−=

+++−+−+−=

VM

zx yz xy x z z y y xVM




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This criterion is used for brittle materials whose tensile and compressivestrength properties are different.

According to Maximum Normal Stress criterion, failure of the material occurs

when maximum principal stress reaches the ultimate strength of the materialwhen subjected to simple tension. That means, failure is predicted to occur

when σ1 >= σ limit where σ1 is the maximum principal stress.

Hence, the Factor of Safety = σ limit / σ1

This criterion is used for brittle materials whose ultimate strength is same forboth tension and compression. Please remember that brittle materials do nothave specific yield points. (Ref. 15)

What factor of safety you use is entirely your responsibility. In machine design,it is not uncommon to use a factor of safety value in the range of 10 to 20.

Most frame analysis and design programs show the ratio of (applied stress/ allowable stress) for every member after performing design as per codespecification. If the ratio is less than 1, then design is same. Ratio more than 1indicates re-design is necessary.

Please see Section the section on Folded Plate for more information oninterpreting FEA result.




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27. Tips for creating better mesh

In this section you will learn some tips for creating a better FE mesh. But the

things are really messy! When your model is quite simple, remember thefollowing advises.

Use quadrilateral elements – in general, try to use quadrilateral elements insteadof triangular elements as they give more accurate results. Remember that, thefour corners of a quadrilateral element should all lie on the same plane. If this isnot possible, use two triangular elements in place of each quadrilateral.

Element shape – quadrilateral elements – the greatest accuracy is achieved witha square – 1:1 element. Elements with a base/height ratio up to 1:2 give goodresults, but elements with a ratio of 1:5 will be unreliable. However, many FEAprograms have the options, which allow you to specify maximum aspect ratio.But be careful, if you specify too low aspect ratio, the program may not be ableto generate the mesh successfully at all! Many a times you will be forced to use1:50 as lowest acceptable aspect ratio value. Yikes! If you do not specify suchupper limit for aspect ratio, the program may occasionally churn out elementswith aspect ratio as high as 1:1,000 even if you are using a $20,000 program!

Try to use rectangular (rather than quadrilateral) shaped elements whenever

possible, If not, the internal angles should not vary greatly from 90°. Angles of 30° or 150° will greatly reduce accuracy. Elements with convex angles shouldnever be used. However, due to geometry of the model, more often than not youwill have to use quadrilateral elements.

Triangular elements – equilateral triangles will produce the most accurateresults. However, it is always better to avoid triangular elements.

Mesh density – the mesh density need not be constant throughout the model.

The program assumes a linear result distribution through the element. If theactual result through the elements is not linear but parabolic, for example, it isobvious that there will be a decrease in the accuracy. In a fine mesh, the resultdiagram through any one element will always be approximately linear.

Increase the number of elements where there is a greater rate of change in theinternal forces. For example; around supports (where bending moments increasesharply), openings and large concentrated loads.

To decrease the number of elements – use a rough mesh in areas whererelatively low results are expected. Remember that the connection to adjacent




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elements is through the element end nodes only and so nodes located along anedge of an element between end nodes are ineffective. Use triangular ortrapezoidal shaped elements to step between rough and fine quadrilateralmeshes.

If you have doubts as to the accuracy of the results in a particular area of themodel, rerun the problem with a finer mesh in that area and compare results.The results converge to the exact solution, as the mesh becomes more refined.

Life will be much troublesome if you need to analyze complex solid models.Here you need to consider many other parameters. Most FEA programs, bydefault, generate ‘tetrahedral’ mesh for solids. Some programs, however, allowyou to specify how you want the mesh to be generated. Figure 27-1 shows howyour model looks with a tetrahedral mesh. Figure 27-2 shows same model withcombination of 8-noded brick, tetrahedral, 5 or 6-noded transition elements.Some programs (e.g. Algor) offer following types of mesh generations.

Standard – used for most meshes by default. It gives you the highest qualitymesh and the lowest number of elements. Standard solid meshing works fromthe surface inward. It will make 8-node brick elements on and near the surfaceof the model while making 6, 5 or 4-node transition elements in the center of themodel as needed.

All 8-Node Bricks – this option should be used only for processors that acceptonly 8-node bricks. In many cases, these are fluid flow processors (for analysisof pipe network, this topic is not discussed in this book). This option can make 4to 5 times the number of elements as the "Standard" option.

No Pyramids – the "No Pyramids" option builds brick meshes with 8, 6 and 4-node elements, but no 5-node pyramid elements.

Tetrahedral from Quads – the "Tetrahedral from Quads" option is for generating

a tetrahedral solid mesh from a quadrilateral surface mesh.

Enhanced No Pyramids – this option makes brick meshes with predominantly 8-node elements plus 6-node and 4-node elements, but no 5-node pyramidelements.

Tetrahedral – the all-tetrahedral option is for generating a nearly equilateraltetrahedral solid mesh from an equilateral triangular surface mesh. This is themost common type of mesh for a large number of FEA programs.




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Figure 27-2 shows boundary condition (fixed) and pressure load of 1000lbf/sq.inch (I had to use FPS unit because the YOKE model file was in ‘inch’unit.)

If you’re wondering how brick or tetrahedral elements look like, this isdescribed in just next section.

We shall come to the same ‘yoke’ model later on, when we discuss how tointerpret FEA results.




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Figure 27-1

In the above figure only the mesh has been shown. The boundary condition andpressure load for this particular analysis has been shown in next figure. Youmay use ‘coarser’ or ‘finer’ mesh in your program. Normally, the programscreate the mesh using a default mesh density. If can control the meshsize/density using a slider in the program.

You may wonder whether the result will change depending on what kind of

mesh you are using. Well, not really in general. However, there are specialcases, where you should use particular type of mesh for best result. See the chartin next section.




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Figure 27-2

I apologize if the stuffs seem too boring!

Figure 27-3 shows Von-mises stress (described later) diagram of figure 27-1after analyzing the model in COSMOS/Design Star.

The same model with mesh as of figure 27-2, is analyzed using Algor and

shown in figure 27-4.




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Figure 27-3

Figure 27-4

The stress is shown in ‘psi’ (pound/sq.inch) unit.




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28. Common Finite Elements library for Linear Static and DynamicStress Analysis

Element type Illustration Description3-D Truss, 2-nodes Truss elements are used to providestiffness between two nodes. Theseelements transmit compressive andtensile loads along their axis. They donot carry any bending load.

3-D Beam, 2-nodes Beam elements are used to provideelongational, flexural and rotationalstiffness between two nodes. These

elements can possess a wide variety of cross-sectional geometries includingmany standard types

3-D MembranePlane Stress, 3-nodes

3-D MembranePlane Stress, 4-nodes

Membrane plane stress elements areused to model "fabric-like" structures,such as tents, cots, domed stadiums,etc. They support three translationaldegrees of freedom and in-plane(membrane) loading. Orthotropicmaterial properties may betemperature dependent. Incompatiblemodes are available.

2-D Elasticity, 3-nodes

2-D Elasticity, 4-nodes

Elasticity elements are used for planestrain, plane stress and axisymmetricformulations. They support twotranslational degrees of freedom.Orthotropic material properties may betemperature dependent. Incompatiblemodes are available.

3-D Brick, 4-nodes Brick elements are used to simulatethe behavior of solids. They supportthree translational degrees of freedomas well as incompatible displacementmodes. Applications include solidobjects, such as wheels, turbineblades, flanges, etc.




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3-D Brick, 5-nodes Same as above



3-D Plate, 3-nodes

3-D Plate, 4-nodes

Plate elements are used in the designof pressure vessels, automobile bodyparts, etc. They support threetranslational and two rotationaldegrees of freedom as well asorthotropic material properties. Anoptional rotational stiffness around theperpendicular axis is automaticallyadded to the node of each element.

A thin composite plate element isavailable for use in models such asmechanical equipment, bicycleframes, etc. A thick composite plateelement is also available and can beused in models such as honeycombsandwich structures, aerospaceproducts, etc. Both thin and thick composite plate elements have no

limitations regarding orientation orstacking sequence and support theTsai-Wu, maximum stress andmaximum strain failure criteria.

Tetrahedral, 4-nodes

Tetrahedral elements are used tomodel solid objects, such as gears,engine blocks and other unusuallyshaped objects. They support threetranslational degrees of freedom. They

are also available in higher orderformulations (mid-side nodes).




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Boundary, 2-nodes Boundary elements are used inconjunction with other elements. Aboundary element rigidly or elasticallysupports a model and enables the

extraction of support reactions.Boundary elements are also used toimpose a specified rotation ortranslation.

Gap/Cable, 2-nodes The gap element simulatescompression, where deflection makestwo nodes touch and transmit force,such as when a ball bearing moves in a

joint. Using gap elements, stresses,

bending moments and axial forceswhere the bearing and joint meet canbe determined.A cable element simulates tension,where two nodes moving away fromeach other a specified distance causethe element to become active. It is stilla small-deflection, small-strainanalysis, but with deflection-sensitiveconnectivity.

Courtesy: Algor Inc.




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29. Shear Wall

In normal reinforced concrete framed buildings, we know walls are not

designed to carry any load. Beams, columns and slabs carry all loads. However,in high rise buildings, it is important to ensure adequate ‘lateral stiffness’ toresist horizontal forces induced by wind or earthquake. For such buildings, if wesolely depend on beams and columns for providing lateral stiffness, then thesway movement will be quite disturbing, vibration can easily be felt by theoccupants and structural members may develop high stresses in them. Concretewalls, which have high ‘in-plane’ stiffness, are placed at convenient locations inthe building to provide necessary resistance to horizontal forces are known asshear walls.

Shear walls are often provided surrounding elevator or staircase. Figure 29-1shows a typical arrangement of shear walls in a building frame. The exteriorshear walls are shown red and the interior shear walls surrounding centralcolumns are shown yellow for easy visualization. This is one of the mostcommon arrangements for shear walls. In the figure only a 5-storied buildinghas been shown. In practice, shear walls are generally provided in tall buildings.

Shear walls resist bending. So, reinforcement must be provided inside them.The area and arrangements of steel bars are calculated in usual manner.

However, analysis of shear wall and frame must be done properly. In early days,when computer was not available, the frame-shear wall interaction was a verycomplicated task. It involved lots of assumptions and mind-bogglingcalculations. One common classical method was that of MacLeod. If you wantto torment yourself, go and get an Advanced Structural Analysis textbook andtry to solve a shear wall problem using classical methods.

The advent of FEA has made life much easier for us. Just model the buildingfirst. You can do it easily using your analysis program’s graphics editor (or you

can import it from CAD program) and then add ‘plates’ in the place of shearwalls. Specify thickness and material properties of the walls (i.e. plates). Then‘mesh’ the plates (see Section 25 for how to mesh plates). And your model isready for cooking (I mean analyzing)! Run your favorite FEA analyzer, and youwill instantly get displacements, shear forces, bending moments, stresses in thebeams, columns and walls. Isn’t that easy?




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Figure 29-1

However, actual difficult part comes after the analysis. Look, I told you alreadythat if there were no shear walls, the forces in the beams and columns would

have been much higher compared to those of with shear walls present. So, lesservalues of forces mean smaller dimension of members and less amount of reinforcement. But inclusion of shear walls will raise the expense again. So,there must a trade off at some point. You must analyze several model structureswith various shear wall arrangements to get the most economical yet practicalstructure. Sometimes, it may be necessary to provide shear walls with openingsfor windows, doors etc. for architectural reasons. Openings reduce the stiffnessof the shear wall.

Exercise

Assume suitable dimensions for the building frame shown in figure 29-1. Applysome realistic horizontal forces at floor levels and analyze the structure with andwithout shear walls. See how the result changes with shear walls. Use adifferent arrangement for shear walls and analyze again. Which arrangementcomes out to be the best?




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30. Folded Plate

The folded plate is a very complex structure. Classical methods of analyzing

folded plates involve laborious calculations. Moreover, there is no singleclassical method available for general analysis of folded plates. They depend onthe shape of folded plates. Some common methods are – Simpson’s,Whiteney’s, Iterative, Three-shear equations etc. However, if you use FEA, thenthe analysis itself is quite simple and the labor involves only preparing themodel correctly and interpreting the result properly. Your analysis programsmust have ‘plate’ elements to successfully analyze folded plates. Modeling thefolded plates sometimes can be really tricky. We shall discuss various aspects of folded plate analysis with the following example as shown in figure 30-1. Thefront view is shown in the figure and the length of the structures is 12 m asshown in figure 30-2 with mesh for sufficiently accurate result.

C G D

Thickness 150 mm 3 m

B Thickness 300 mm E 1.5 m

A F3 m 4.5 m 3 m

Figure 30-1

The frame is loaded with 1.5 kPa uniformly distributed load in addition to theself-weight. The folded plate is made of concrete. If you use your analysisprogram’s graphics editor for input, the quickest way to create the model is: (1)draw the model first using frame element so that the shape like as of figure 35-1(2) then copy the frames over 12 m distance (3) now draw the plates (4) afterthe plates are drawn, delete the frame members since they were drawn here onlyto ease the model (5) now mesh the plates (6) apply boundary condition, in thisproblem, the sides of the plates are fixed (7) apply plate thickness, materialproperties etc. (8) apply surface loads on plates BC, CD and DE.

After you apply the loads, visualization should look like as that of figure 30-3.If you draw the plates in wrong orientation, the load direction may appearawkward as shown in figure 30-4. If it happens, delete the particular plate andredraw again in the opposite direction compared to previous case. For thisreason, it is a good practice to mesh the plates, after you have applied the loads

properly. Although, to make your figure change from 30-4 to 30-3, you may be




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tempted to change the load into –1.5 kPa for plate DE, however, it may createproblem while interpreting the output.

Figure 30-2

Figure 30-3

Figure 30-4

Fixed on side12 m

Fixed on side




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Are you wondering whether we can take advantage of the symmetry? Of coursewe do. In fact, only 1/4th of the structure needs to be analyzed. The appropriateboundary conditions are shown in figure 30-5 with respect to the global axesshown in the same figure.

all freeZ rotation fixed

YZ X

all fixed

Figure 30-5

To assign proper boundary conditions accurately, you must visualize thedeflected shape of the structure yourself in your mind before performing theactual analysis. Of course, you may bypass this mental exercise by analyzingthe whole structure rather than taking advantages of the symmetry. You may

wonder why I did not give you theoretical result the above folded plate analysis.Well, theoretical calculations are also based on certain simplified assumptions,which may not completely valid many cases (See Section 3, for example). Iadmit that you need some benchmark problems to compare your analysisoutput, still you should start relying on your FEA output to gain confidence!

After performing the analysis, you will get following stress components – σx,

σy, σz, τxy, τyz, τzx in global axes direction and σx, σy, τxy in local axes

direction along with maximum principal normal stress σ, minimum principal

normal stress σ and maximum principal shear stress τ. The concept needsclarification.

We have already discussed local axes concept. Now, every ‘plate’ element of the above folded plate may be considered ‘lying’ in a ‘2D plane’ even if it isactually ‘inclined’ in the real structure. For example, a ‘plate element’ of BCmay be visualized as shown in figure 30-6. This figure is a specific case of figure 30-5 where stress variation along Z-axis is negligible.




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σy τyx = - τxy

y τxyx

σx

Figure 30-6

The visualization will be more apparent from the figure 30-7, where local axes

for the vertical, inclined and horizontal plate elements’ are shown.

YX

Y

XZ Global

Figure 30-7

A different situation comes when we speak stress components in terms of globalaxes. Here you’ll find all 6-stress components since we now speak in 3D space.

Question: My analysis program doesn’t explicitly show global and local axesstress components. How do I know which convention it is following?Answer: Difficult to say. Generally, most programs give output with respect tolocal axes. But there are some programs, which show the result with respect to

local axes for some type of ‘elements’ and with respect to global axes for someother type of ‘elements’! Really confusing! See your programs’ manuals fordetails. However, you better solve some benchmark problems (with knownanswer) to check.

Question: Shall I provide the reinforcement on the basis of forces on global axesor local axes?Answer: Local axes forces.




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Figure 30-8




- 99 -

Figure 30-9

These analysis outputs are from Visual Analysis.




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As a second example, consider another type of folded plate as shown in figure30-10.

Figure 30-10

The span of plate (along X axis) is 10 m and thickness of all four plates is 100mm. The plates are made of concrete and are acted by 3-kPa downward (i.e.along –Z direction) load perpendicular to the surface of the plates. Sides(leftmost and rightmost edges but not middle edge) of the plates are fixed. The

global X stress (σx in N/m2) after analysis (in Algor) is shown in figure 30-11.




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Figure 30-11

Exercise

Model and analyze the plate yourself and check if you’ve got the same result.




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31. Shells

Analysis of shells involves solving fourth order differential equations. Restassured, I’m not going to swamp you with differential equations. Theoretical

background on shell is extensive and if you are interested, you should consult atextbook on shell theory. Here I shall discuss only what you should know forshell analysis using FEA.

20 m

5 m

Figure 31-1

As a first example consider the translational shell shown above. It is made of 75-mm thick concrete. A uniformly distributed load of 3 kPa inward (i.e.

towards center of the shell, perpendicular to the surface of the plate at everypoint, not horizontally projected) is acting over it (including self-weight i.e. youneed not add self-weight load). The sides of the shell are pinned. You are toanalyze the shell. Your first task is to model the shell. Some programs cangenerate this kind of shell through a model wizard. Then you are lucky. If yourprogram does not have this option, you need to start from scratch. First createthe arc. Note that, either you can draw the arc exactly (if your programs graphiceditor permits) or you need to follow the procedure as outlined in Section 5. If you draw the arc, remember to ‘break’ it into a combination of ‘lines’. Then

copy the ‘lines’ through out the length of the shell (20 m in this example). If youbreak the arc into lines before, copying into suitable interval creates the meshautomatically. The concept may appear garbled if you just are reading, but willbe clear if you try to draw the shell model yourself. A 20x80 mesh is sufficientfor this shell. In most practical cases, your shell will be ‘thin’ i.e. thickness of the shell is much less compared to width or length. So, there will be only three

stress components σx, σy and τxy and other three stress components will be zero.Thin shell theory is also known as ‘membrane’ theory. On the other hand,‘thick’ shells are analyzed using ‘bending’ theory. There are various types of

shells – hemispherical, cycloid, conical, hyperbolic etc. Classical theories of shells often vary with different geometric shapes, although FEA treatment of




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shells is same for all shell types. The shell theory is extremely complicated andif you lean too much in that subject you may find a place in lunatic asylum toosoon! One advantage of shell is that, because of its curve shape, less material isrequired compared to beam or slab to carry same load. In other words, the ratio

of load carried/ amount of material used is higher for shells. However, becauseof constructional difficulties, shells are still of limited use. Yes, like the foldedplate discussed in previous section, you can again take advantage of symmetryas shown in figure 31-2.

X rotation fixed boundary condition

YZ X

pinnedfree

Figure 31-2

For other types of shells, if the loading and geometry are symmetrical, you canalways take advantage of symmetry. Although you have the right to place theorigin at anywhere, I prefer to keep the origin at the center of the shell. It hassome ‘psychological’ advantages.

The analysis output for σ in local X-axis is shown in figure 31-3. The analysiswas done in Algor. You should note some important points. First of all, all FEAprograms display output in colorful stress contour range. Normally, you can‘adjust’ the range yourself. It is always advisable to specify a range of yourown. This is because, the highest (and lowest) range generally covers only oneor two elements in extreme ends of the model due to some numerical round of.In fact, 99% of all elements’ stresses fall within the 80% region of stress rangeshown in the display by default. So, keep on adjusting stress range until you are

convinced that your model shows optimum design stress range. In figure 31-3,the maximum and minimum stresses are respectively 5 kPa and –20 kPa. Check what answer you get in your program.




- 104 -

Figure 31-3




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32. A first step in Structural Dynamics

So far our study was confined within static analysis. In this section we shall

learn some aspects of dynamic analysis. The theory and field of structuraldynamics are very large. Even a decent introduction of this subject wouldrequire at least 100 pages. I am just describing some very basic concepts of structural dynamic useful for practicing engineers. If you’ve already studied thetheory of structural dynamic then it’s great. If not, I advise that you start readinga textbook of structural dynamic besides this book. It will make you understandthe applications of dynamics discussed in this book in an easily understandablemanner.

At the beginning, let me explain the difference between static and dynamicanalysis. In static analysis, the applied force is constant but in dynamic analysis,the applied force varies with time. It is not necessary that dynamic analysisalways involves application force only, it may be due to shaking of ground dueto earthquake. In civil engineering applications, dynamic analysis mostlyinvolves determination of maximum response (i.e. displacement etc.) of thestructure due to some applied ground acceleration.

To understand dynamic analysis properly, we need understand the concepts of some common terms.

Degrees of Freedom – consider the typical spring mass damper system as shownbelow.

Mass m Displacement x(t)

Stiffness k

Figure 32-1

It has mass 'm' and stiffness 'k'. It can move in only the direction shown by thearrow. So, this model has one degree of freedom. We call this single degree of freedom (SDF) system. We also assume that the mass 'm' is 'lumped' at the topof model as shown by the filled circle. This means, the mass of the stick, thoughdistributed though out its length, we assume as if it is concentrated at one placeas shown by the circle. This model may reflect idealization of a single storybuilding, where the roof mass is concentrated as shown in the figure. If there is

more than one story, then floor mass of each story may be considered as




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'lumped' at the respective floor levels. In that we shall have multi-degree-of-freedom (MDF) structures. Things may appear little bit terse at the beginning,but gradually everything will seem comprehensive. I assure you!

In practice most structures are of MDF type. However, that does not imply thatwe need not study theory of SDF structures, because many MDF structures canbe 'broken' into separate SDF structures and can be easily analyzed rather thananalyzing the whole structures!

Damping – if a structure is displaced from its equilibrium position by a smallamount of force, it will vibrate (move from this direction to that direction).Unless there is ‘something’ to prevent vibration, it will go on vibrating forever.But in actual practice, the ‘amplitude’ of vibration will gradually diminish andafter some time, the model will come to rest. This process by which freevibration diminishes is known as ‘damping’.

Natural frequency (ωn) – it is the number of cycles per second a structurevibrates. It’s measured in radian per second. It is related to natural time period

Tn = 2π / ωn. Natural frequency is computed using ωn = (stiffness/mass)0.5. For

damped structures, damped natural frequency ωd = ωn(1- ξ ²)0.5.

Damping ratio (ξ) – it is a measurement of how much damping is there. It has

values in the range of 0 to 1. It is often expressed in %. For example 100%damping means the structure will not vibrate at all. However, for most practical

structures, this value lies in the range of 5% to 20% i.e. ξ = 0.05 to 0.2.

Why bother studying dynamic analysis?

If a dynamic excitation (force or ground acceleration) is applied to a structure,the resulting displacement might be much more than that obtained by simplestatic analysis. More displacement means higher values of internal forces(bending moment, shear force etc.) and subsequently higher values of stresses inthe members. If the stress reaches the yield strength of the material the structurewill collapse!

For a time varying applied force, if I apply calculated values of the force atparticular instant of time and then perform static analysis with that force, shall Iget correct displacement etc.?

No, you won’t (even for a linear structure)! The concept of structural dynamicsis different from static analysis theory. You must perform exact dynamic

analysis.




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Is it mandatory to perform dynamic analysis for all structures?

It depends. Dynamic analysis is especially required for multi-story buildings forearthquake analysis. Normally, in low height buildings, dynamic analysis does

not produce much different result compared to static analysis. However, for allstructures, where vibration is a major factor in design, dynamic analysis must beperformed. Designing of machine foundations always requires dynamicanalysis. For highly important structures like bridges, dams, nuclear reactorsetc. dynamic analyses are very important. Theory of structural dynamics isapplicable equally to buildings as well as automobiles!

How the programs calculate dynamic response?

Nearly all analysis programs calculate dynamic response by numerical methods.

The main governing equation for dynamic analyses is conventionally written as

Where, [M] = mass matrix (kg or Ns²/m), [C] = damping matrix (Ns/m), [K] =stiffness matrix (N/m), X = displacement (m), X’ = velocity (m/s), X” =acceleration (m/s²) and F(t) = Force (N).

The F(t) in above equation will be replaced by –[M]{u(t)”g} if there is groundacceleration instead of nodal forces. Here, u(t)”g denotes ground acceleration(m/s²). If F(t) = 0, then the situation is known as free vibration. If [C] = 0, thenwe call undamped motion. However, in real life, [C] is never equal to zero.

In the subsequent sections, we shall explore various examples of structuraldynamic analyses – simple to complex! By this time I expect that you will alsostudy a few pages of dynamics textbook. Suggested chapters for reading in yourtextbook are – introduction and simple formulation for SDF systems, direct

solution of differential equation of motion, damped and undamped motion,response to harmonic and periodic excitation, numerical evaluation of dynamicresponse, equation of motion for MDF systems, modal and response spectrumanalyses. If you are interested, you may read all the chapters of the book, butabove topics are enough for understanding the calculations presented in thisbook. You may read Ref. 5, 16, 17 to start with in dynamics.

)(}]{[}]{[}]{[ t F X K X C X M =++ &&&




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33. An example of a Single Degree of Freedom problem

I hope that by this time you’ve read a few pages of structural dynamics

textbook. Yes? Great, they you can appreciate following calculation.

Consider the following structure. It resembles a water tank over a tower. I shallshow you how to perform accurate dynamic analysis without any computerprogram!

1000 kNaccelerationm/s2

t2 3 m

0 1time (s)

1 m

Figure 33-1

All the members of the tower shown are made of steel with 200x200 mm squaresides. The base of the tower is given an acceleration of 1-s duration as shown inthe figure. It is desired to calculate the displacement of the structure at t = 0.5 s.Of course, you can model the entire structure using your analysis program. Thenapply base acceleration input and find out the displacement response curve forentire duration. However, you will soon discover that you can solve it muchquickly using simple hand calculation and just static analysis program! Anyway,you need to model the structure first! After all, we need to know the stiffness of the whole structure.

When you are doing exact time history dynamic analysis, your model shouldlook like figure 33-2. Apply a lumped mass of (1000/10)/2 = 50 kNs2 /m on thetwo uppermost nodes as shown (we divide the weight 1000 kN by g = 10 m/s²to get the mass). Note the direction of applied lumped masses because we areinterested in calculating the displacement in this direction. Next apply the baseacceleration data as shown in the figure 33-1. You may generate the data using

spreadsheet and refer the text file in your analysis program. For linear dynamic




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analysis time interval of one tenth of duration of applied force is sufficient. So,here we use 0.1 s interval. Since we are considering the structure as SDF, only 1mode is used in the analysis. After performing the dynamic analysis we getfrom the time vs. displacement curve that the displacement at t = 0.5 s is –

7.64E-4 m.

50 kNs²/m 50 kNs²/m

YX

Figure 33-2

Now we are going to see what we get if we perform the analysis using DuhamelIntegral. In this case, after modeling the structure, our first step is to calculate itsstiffness. For this we need to apply a force (say 1 kN) in the upper left nodalpoint. Then run a static analysis and note the displacement of any of the

uppermost nodes.

Figure 33-3

The stiffness of the entire structure is k = F/ ∆ = 1 kN/3.211E-5 m = 31143kN/m. We treat the whole structure as a SDF with a lumped mass of 100 kNs²/m

and stiffness k. Let’s determine the natural frequency of the structure as ωn =

1 kN

Dis lacement ∆




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(k/m)0.5 = (31143/100)0.5 = 17.64 rad/s. Time period Tn = 2p/ ωn = 0.36 s.

Consider 5% of damping i.e. ζ = 0.05. In Duhamel integral we need to use the

term p(τ) for the applied force. But here we have ‘ground acceleration’ as inputand not the ‘force’. The relationship between the ‘force’ and ‘acceleration’ can

be related by ‘Force = - mass x acceleration’ i.e. p(τ) = - m.a(τ). So, our p(τ)will equal -100τ2 because in this problem, the acceleration is defined as t² andmass m = 100 kNs²/m.

The favorite Duhamel’s integral has the well-known form:

ωd = ωn (1- ζ2)0.5 = 17.6 rad/s. So, the ultimate equation takes the form asshown below.

After performing the numerical integration, we get the same answer as that of inexact dynamic analysis program! However, in dynamic analysis using the

program, you will get the time vs. displacement plot over the entire duration,wherefrom you can get the maximum displacement at once. But in case of analysis using Duhamel’s integral, you need to perform the calculation atseveral points to get the maximum response. In this particular problem, theinput excitation is defined by simple algebraic function, which makes use of Duhamel’s integral feasible. In actual practice, the input excitation is oftendefined as a set of data at some specified interval (normally at 0.02 s interval forearthquake ground acceleration data). So, there you must use numericalmethods like what your analysis programs do.

Now probably you have a comprehensive idea of what dynamic analysisinvolves. The basic steps of dynamic analyses are:

1. Model the structure as usual (you’ll have to do it yourself).2. Define the ‘lumped masses’ (normally your duty, however some smart

programs can calculate lumped masses themselves from structure’sdimension and sectional properties). You also need to specify a suitablevalue of damping ratio.

3. Apply an input excitation (time varying ground acceleration or time varyingforce).

τ τ ω τ ϖ

τ ςω d t e p

mt u d

t

t

d

n )(sin)(1

)()(

0

−= −−∫

∫ −−=−−

= −−5.0

0

)5.0)(64.17)(05.0(2 ...464.7)5.0(6.17sin6.17100

100)5.0( m E d e

xu τ τ τ

τ




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4. Specify how many mode shapes you do want (generally you need samenumber of mode shapes equal to the number of stories in structure, howeversometimes you may need to venture more or less number of mode shapesdepending on problem type). Normally (but not always) first three mode

shapes are sufficient for subsequent calculations. However, for multi-storybuildings, contribution of higher modes is quite significant.5. Some programs may ask you what algorithm you want to follow – eigen-

vector or ritz-vector (more about this later).6. Now it’s time to analyze the structure. Dynamic analysis invariably takes

more time than static analysis because it performs iteration to find out modeshapes. After finishing it displays the natural time periods (hence frequencytherefrom) of the structure for each modes, mode shapes and time vs.displacement (and velocity, acceleration etc.). Some programs can alsodisplay other parameters such as base shear etc. One word of caution, themode shapes are ‘relative’ displacements and not the actual displacements.

7. Now it’s time to interpret the result. Generally it’s your task to find out thestresses developed in the members due to dynamic analysis. You can do it bynoting maximum displacement and then calculating the bending momentdeveloped there from.

I hear and I forget... I see and I remember… I do and I understand. Aha!




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34. What dynamic analysis you should perform?

In this section, we shall discuss the ‘Response Spectra’ and ‘Time History’

dynamic analyses in detail. Of greatest interest in dynamic analysis is thedeformation of the system, or displacement of the mass relative to movingground, to which the internal forces are linearly related. Knowing maximumlateral displacement of the structure would be useful in providing enoughseparation between adjacent buildings to prevent their pounding against eachother during an earthquake. Moreover, total acceleration of the structure wouldbe needed if the structure is supporting sensitive equipment and the motionimparted to the equipment is to be determined.

Response spectra may be defined as – a plot of the maximum response(displacement, velocity, acceleration or any other quantity of interest) to aspecified load function for all possible single degree of freedom systems.

Once the deformation response history has been evaluated by dynamic analysis,the internal forces can be determined by static analysis of the structure at eachtime instant (discussed later).

A typical response spectra for 1940 El Centro earthquake has been shown infigure 34-1. First we need to determine natural period Tn (or frequency f = 1/Tn

in cycles per second) of the structure. Then we’ll have to check the dampingratio of the structure. Using this, we can directly read the maximumdisplacement of the structure from the figure as shown.

One thing to note is that, you must have response spectra chart for the particularexcitation (e.g. El Centro earthquake in this case). If you need to analyze astructure for another earthquake, you need to use response spectra for thatearthquake. It means that, someone must have prepared the response spectrachart before you use it in your analysis. The procedure of developing response

spectra is discussed in any standard structural dynamics textbook and notdiscussed here. However, you may note that considerable computational effortis required to generate such charts. So, the modern trend is to perform full timehistory analysis, which is more versatile and accurate, for structures.

If your analysis program offers time history analysis option (unfortunatelymajority of analysis programs do not), there is no point of going for responsespectra method.




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By now you must have realized that the response spectra analysis is a ‘short cut’to find out maximum response directly from the chart without performing timehistory analysis.

Response spectra for elastic system for the 1940 El Centro earthquake ξξξξ =0, 2, 5, 10, 20%

Figure 34-1

For example, assume our structure has natural frequency of 1 Hz. Also assume10% of critical damping. From the figure 34-1, we see that the vertical arrow(shown blue) drawn from Tn = 1 s, intersects the displacement line for 10%damping at 3.3 inches displacement (shown green). So, the maximum relativedisplacement response of our structure is 3.3 inches if it is excited by the 1940El Centro earthquake ground acceleration. Also seen that horizontal arrow(shown blue) extends to the pseudo velocity axis at 18.5 inch/s. Please note thatthe axes are in logarithmic scale.

Since earthquake can’t be predicted, you may like to analyze your structure forseveral past earthquake effects. Instead of using response spectra for eachearthquake, you can use ‘design response spectrum’, which represents a kind of average response spectrum for design.

The non-linear response spectrum is slightly different (not discussed here).




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During dynamic analysis, the program will calculate the stiffness of thestructure internally. However, you need to supply modal damping ratio yourself.This method is sufficiently accurate for linear structures with classical damping.Figure 34-2 recommends damping values (Ref.5).

Stress level Type and condition of the structure Damping

ratio %

Welded steel, pre-stressed concrete, well reinforced concrete(only slight cracking)

2-3

Reinforced concrete with considerable cracking 3-5

Working stress,no more thanabout ½ yieldpoint Bolted and/or riveted steel, wood structures with nailed or

bolted joints5-7

Welded steel, pre-stressed concrete (without complete loss inpre-stress)

5-7

Pre-stress concrete with no pre-stress left 7-10

Reinforced concrete 7-10Bolted and/or riveted steel, wood structures with nailed orbolted joints

10-15

At or just belowyield point

Wood structures with nailed joints 15-20

Figure 34-2

Since, damping properties of the materials are still not well established; it isquite a challenging task to determine exact damping of the structures, especiallyin non-linear range. In fact, stiffness of the structure also varies with time(Yikes!) due to deterioration of the structure. The change in stiffness is often

used for ‘retrofitting’ the structure. Determining exact stiffness and damping of the structure is known as ‘system identification’. There are several techniquesfor this purpose, but the most popular is ‘wavelet’ analysis, which has recentlyattracted attention of the engineers.




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35. Non-linear analysis (NLA) – an introduction for beginners

Ultimately we come to hottest topic of structural analysis, the non-linear

analysis. Availability of powerful computers and software, more and more non-linear analyses are being done than ever before. But what is NLA all about?Why there is so much hype about it? Let’s the adventure starts!

Recall that the main assumption in linear analysis (LSSA), is that the stress-strain curve is linear and deformation is small. Right? But we have to considerNLA if any or both of the above assumptions are violated. There are mainly twotypes of non-linearity. The first is ‘Material Non-linearity’ and the second is‘Geometric Non-linearity’. We shall study them in detail.

Material Non-linearity

Consider the following three stress-strain curves of any material.

σ σ σEt

Ei Yield point

E E

ε ε ε(a) (b) (c)

Figure 35-1

The figure (a) shows stress-strain curve for perfectly linear material. Figure (b)shows stress-strain curve for ‘bi-linear’ material (typical for carbon steel). Thecurve is linear (slope E) up to yield point, where from it changes its slope to E t

though remain linear again. E is our familiar modulus of elasticity or Young’smodulus. Et is known as ‘strain hardening modulus’. In the range of ‘E’ thematerial remains elastic, but in the range of ‘E t’ it becomes ‘plastic’. Thismaterial behavior is known as material non-linearity. Now consider figure (c).Here the curve is entirely non-linear. E value changes at every point of thecurve. This is also a typical example of material non-linearity.

Now how do we incorporate material non-linearity in analysis program? Thesteps are given below.

1. Divide the total load incrementally i.e. F = Σ(∆Fi)




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2. Apply load ∆Fi at each time instant.

3. Find deflection at that instant of time as ∆Di = ∆Fi /[K]i, where [K]i is globalstiffness matrix of the structure at that instant of time.

4. Find strain ε = ∆L/L. There from, find Ei as shown in figure 35-1 (c). You

will have to find ε and corresponding Ei for every member of the structureusing above equation.

5. Update each element stiffness matrix [k] using this new value of Ei.6. Update global [K]i as [K]i+1.

7. Go to step 2 and apply load ∆Fi+1.

8. Find deflection ∆Di+1 = ∆Fi+1 /[K]i+1 as in step 3 and repeat through step 6.

9. Total deflection D = Σ(∆Di).

Did you realize the labor involved in the calculation for a large structure?

Geometric Non-linearity

In conventional linear analysis, the stiffness matrix for each element (and thusglobal stiffness matrix) remains constant throughout analysis. This stiffnessmatrix is formed on the basis of co-ordinates of the nodes of the structure. If thedeformation of the structure is small, then co-ordinates of the deflected nodes of the structures will not move too much from its original configuration. In that

case, stiffness matrix formed on the basis of original nodal co-ordinates anddeflected nodal co-ordinates will be almost same.

Let’s consider the truss shown in figure 35-2.

Figure 35-2

Its deflected shape will look like as shown in next figure.




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Figure 35-3

Now, the stiffness matrix [K] depends on nodal co-ordinates of nodes N1, N2,

N3 and N4 of the truss. Original geometry is shown by dotted line, solid lineshows deflected shape. Now if the co-ordinates of deflected truss be ND1, ND2,ND3 and ND4, then we can construct another stiffness matrix [KD] from newco-ordinates of the nodes. If the deflection is ‘large’, then [K] will not be equalto [KD]. This is the main theory behind geometric non-linearity. If we incrementthe load at each time instant and update [K] according to ‘changing’ displacedposition of the structure, then it will be a geometric non-linear analysis. So, themain steps for geometric non-linear analysis are summarized below.

1. Divide the total load incrementally i.e. F = Σ(∆Fi)

2. Apply load ∆Fi at each time instant.3. Compute (element and then) global stiffness matrix [K]i depending on

original configuration of the structure.

4. Find deflection ∆Di = ∆Fi /[K]i 5. Update [K] to [K]i+1 on the displaced nodes of the structure.

6. Go to step 1, update load to ∆Fi+1

7. Find deflection ∆Di+1 = ∆Fi+1 /[K]i+1

8. Repeat above steps until ∆Dn = ∆Dn-1

9.

Then total deflection D = Σ(∆Di).

This calculation is more demanding than material non-linearity case, since youwill have to perform the whole calculation of setting up stiffness matrix at everystep! Now imagine what will happen if you need to analyze a large complexstructure with both types of non-linearity. May I add some dynamics as well?Hey buddy, why people learn engineering?

What are you looking for? A numerical example? Well, may be in some latersections! But there remains one most important question – when you shouldconsider a deformation large enough for geometric NLA? The answer is not




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easy! Often in large deformed state, the structure may behave entirely differentmanner than that of small-deformed state. This depends on particular type of structure, loading and material properties etc. Some analysis programs have anin-built option to warn you if deflection comes out to be more than certain

percentage (say 5%) of length of largest dimension of the structure. The %value is by no means to be taken as an absolute in determining whether or notdisplacements are large. In a very long or very tall structure, inter-story orinter-span deflections can easily be less than 5% of the overall dimension yetlarge enough to violate the small displacement assumptions. The ultimatedecision is yours!




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36. Mechanical Event Simulation

Mechanical Event Simulation (MES) is also known as ‘Virtual Prototyping’. As

the name suggests, this is a method by which you can perform a ‘virtualexperiment’. Traditional FEA programs calculate stresses usually at a singleinstant of time and requires assumption about forces. That means, in linear staticstress analysis, you must input the force quantity explicitly. But MESintrinsically calculates loads and stresses as motion takes place at each instant intime throughout the event, facilitating a more efficient design/analysis processsince the need to estimate and specify forces is eliminated. The whole thing willbecome clear if you consider an example.

Mass m

h

dimension L x B x d

Figure 36-1

The figure shows an experiment where a weight of ‘mg’ is dropped over a barof known dimension form a specific height ‘h’. We need to determine thestresses at the bars.

In conventional LSSA, out model will be as shown in next figure.

F F

Figure 36-2




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Look, here we need to specify the force ‘F’ explicitly. But in case of MES, allwe need to do is specify the dimension of the bar, height h and dimension andweight of the falling object, position of the objects and the meshing the model(as shown in fig. 36-1). When we perform MES on the model, it does a time

varying analysis at each instant. The program will display stress and motion ateach instant of time. You will see the steps like animation. Did you realize theadvantage? You actually performed a virtual experiment of a mechanical event!The example given here may appear too simple, but imagine that this methodcan be successfully employed in crash test simulation of an automobile!However, there are a few disadvantages as well. The very simple exampledescribed above took me 90 minutes to perform for just 1-second simulation ina 233 MHz PII computer. So, when you make a simulation of real worldproblems say, crash test simulation, it may take as much as 24 hours of computing! No joking! Anyway, MES is highly sophisticated analysis indeed.In fact, to use these kinds of calculations seriously, you will need a supercomputer rather than a PC!

What is the theory behind MES?

In classical LSSA, we solve familiar equation F = Kd, where F = force, K =stiffness of the structure and d = displacement. But, from Newtons 2nd law of motion, we know, F = ma, where m = mass and a = acceleration. So, we canwrite ma = Kd and this is our governing equation for MES. Later we shall see

how this equation is part of general structural dynamic equation ma + cv + kd =F(t), where c = damping, v = velocity of body and F(t) = time varying force.

A similar method of MES is ‘Multi-physics analysis’ which involvessimultaneous analyses of a model for more than one physical effect. Someexample of multi-physics analyses are – thermal stress, fluid flow, combinedthermal and fluid flow, electricity and electro-magnetism. Let me explain in alucid way. Suppose you want to analyze fluid flow around an airfoil. (Soundswhacky?) With multi-physics analysis, you can visualize pattern of fluid flow

over the surface of the airfoil in real time view. You can verify and see whetherthe flow at a particular point is laminar or turbulent. The things are really mind-boggling. A detailed analysis of these topics is beyond the scope of this book. If you are really inclined this kind of analyses, you better try using the FEAprograms, which offer these features, for example, Algor, Adams, Ansys etc.




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37. Importing model from CAD programs

Most Windows based structural analysis programs do allow you to import

geometry from CAD packages like AutoCAD. If you find it difficult to drawthe actual structure in the analysis program, particularly 3D frame or complexstructures for finite element analysis, drawing them in AutoCAD/MechanicalDesktop/Solid Edge/Solid works etc. are an easy alternative. Since the CADprograms themselves are high-end drawing programs, creating the drawing inthem is always easier. But be careful! Not all analysis programs can import allCAD object types.

“Lines” in AutoCAD are converted to frame members upon importing. “Nodes”are automatically created at the intersections. Sometimes “Plates” (known as“2D Solid” in AutoCAD) can also be imported. Remember to “Explode”rectangles, poly-lines and polygons into “Lines”. After drawing the structure inAutoCAD, be sure to “Export” it into “DXF” format. Please note that DXF filesare AutoCAD’s version specific. If your analysis program can’t read AutoCADR2000 DXF then you should save the AutoCAD drawing file in R14 or R13DXF format. You can also perform reverse process that is exporting youranalysis model from your analysis program to AutoCAD DXF format.

Analysis Program AutoCAD

Frame LineNodes Automatically created at intersections

of elementsPlate 2D Solid / 3D Face

Shell 3D Solid

Except for simple lines, other AutoCAD entities are often treated differently byvarious programs. If you are importing DXF file with plate, shell or wire frameelements, you may find discrepancy in geometry upon importing. Therefore, it

is always better to check what AutoCAD entities your program can successfullyimport. If you find your program can’t successfully import AutoCAD drawing(DWG/DXF), then you should draw in the analysis program’s graphicalenvironment. Note that your analysis programs may or may not keep track of AutoCAD’s ‘layers’ feature in imported drawing.

However, when drawing solid models, it is better to use ‘parametrically defined’drawing programs like Mechanical Desktop, Solid Edge, Solid Works etc. Inthese programs, you can modify the geometry by just changing the dimensions.

An example is given below to explain parameter-defined geometry.




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Suppose you have drawn the following drawing in AutoCAD as shown in fig.37-1. If you change the radius of the right end circle, the drawing will look asthat of fig. 37-2. However, in parameter defined CAD programs, if you ‘tell’ the

program, that the lines will always be tangent to the circles, the drawing will beupdated automatically as shown in fig. 37-3.


Figure 37-3

Did you see the difference? This kind of relationship is very convenient forsolid object modeling. Some of the already named CAD programs have thecapability of converting 2D drawings in to 3D solid models! Wow!

A popular ‘neutral file’ format for solid models is IGES (Initial Graphics

Exchange Specification). Many CAD programs can export solid models intothis format. These files have *.IGES or *.IGS extension. This is actually a textfile. Programs write model information in the files. When you import an IGESfile, the analysis programs read that information and re-generate the model.




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38. Virtual Reality in Engineering (VRML)

Do you know what VRML stands for? Well, it’s Virtual Reality Mark-up

Language. It’s a new wonder of visual display. Fortunately, a large number of structural analysis programs can export your model to VRML format. This filehas an extension of WRL. You can view VRML files using any standard webbrowser like Internet Explorer. However, you must install first the VRMLviewer support files (normally come with your operating system CD). VRMLfiles can show the models in a 3D view like in actual life! You can rotate,enlarge or even ‘walk through’ your model! You can even visualize differentmaterials of the structure. It’s an excellent feature to impress your clientsbecause you can show them what your structure will look like when it would bebuilt in real life. You are probably aware of ‘rendering’ feature of CADprograms. VRML is just like that, but looks more realistic. VRML is actually atext file. The browser reads the data in the file first and then develops therealistic model. But here’s a warning! Finite element solid models, whichcontain thousands of nodes, can result to very large VRML files. Moreover,when you will try to open the same file using your web browser, the computermay create staggeringly large (> 1 GB) virtual memory in your hard disk and itwill take a few minutes to display the model on screen.




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39. Linear Programming in spreadsheet

In this section, I’m going little bit off track. I’ll show you how to solve linearprogramming problems using your favorite spreadsheet! Linear programming isoften required by engineers to solve certain design problem for example – inpre-stressed concrete section design and various other ‘optimization’ problemsetc. Linear programming can be also be solved in programs like MATHCAD orMATLAB etc., however, since these mathematical applications are not verycommon in design offices, you better bet on versatile spreadsheets. Do youknow spreadsheet is the largest selling type of application in the world? Theprocedure presented here are for Microsoft Excel 97, but if you any otherspreadsheet like Lotus 123 or Borland Quattro Pro you will find similarfunctions in those programs as well. I assume that you are familiar with basicspreadsheet operations. So, I’m directly going to the problem. Please make surethat you have installed Analysis ToolPak add-in in Excel.

Consider the equation Z = 3x + 4y subjected to constraints 4x + 3y <= 80, 2x +5y <= 180 and x => 0, y => 0. Our aim is find out the maximum value of ‘Z’subjected to above constraints.

To solve it by Excel, follow the steps illustrated.

Step 1: Define the problem as shown in the figure 39-1, which resembles thecells of the spreadsheet.

A B1 X 02 Y 0

3 Z =3*B1+4*B24 C1 =4*B1+2*B25 C2 =2*B1+5*B2

Figure 39-1

Note that in cell B3, the equation for Z has been input. Here cells B1 and B2stands for variable x and y respectively. The strings in column A is forunderstanding purpose only. The constraints are defined in cells B4 and B5.You may like to note that there are two more constraints that both x and y has tobe positive number. These constraints will later be defined using B1 and B2cells. After you do this, initially all cells in the range B1:B5 will show 0.

Step 2: Click on ‘Tools’ and then ‘Solver…’ from Excel’s menu bar. The solverdialog box appears as shown in figure 39-2.




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Figure 39-2

Make the ‘Set Target Cell’ box to cell B3, because this cell contains thedefinition of our main function Z. Now set the ‘By Changing Cells:’ to$B$1:$B$2. You can either type the cell range yourself or you can select therange on the worksheet by clicking the red arrows as shown in the figure. Alsomake sure that the ‘Equal To:’ selection is set to ‘Max’ for this problem. Whendone, click on ‘Add’ button to specify constraints. When you do so, you will seefollowing dialog box.

Figure 39-3

Specify the Cell Reference and Constraint (for 4x + 3y <= 80) so that the figurelook like as shown below.

Figure 39-4

In the same way specify the other three constraints namely 2x + 5y <= 180, x=> 0, y => 0. Click OK when done. After that ‘Solver Parameters’ dialog boxshould look like figure 39-5.




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Figure 39-5

Remember to use proper ‘<=’ or ‘=>’ sign while specifying constraints. If everything seems ok, click the ‘Solve’ button. And that’s all. Excel will solve itwithin seconds and dumps you another dialog box like figure 39-6.

Figure 39-6

You will of course want to retain solver solution. Now you get the solution asshown in figure 39-7.

A B1 X 2.52 Y 353 Z 147.54 C1 80

5 C2 180

Figure 39-7

Obviously the maximum value of Z comes out to be 147.5 and for x = 2.5 and y= 35. So, our problem is solved.

Exercise




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Find the minimum value of P = a – 3b + 3c subjected to the constraints a, b, c>= 0, 3a – b + 2c <= 7, 2a + 4b >= - 12, - 4a + 3b + 8c <= 10.Answer is Pmin = - 28.6 for a = 6.2, b = 11.6 and c = 0.

Remember spreadsheet is a very powerful as well as useful application. Try toexploit its full potential. Often you’ll find that spreadsheet is your best rescuer.




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40. Reinforcement detailing in continuous beams

Figure 40-1




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When students see the reinforcement bar arrangements in building codes, theynormally have a hard time digesting it. This is because; detailing is often shownin single color, which makes it difficult to follow. Figure 40-1 shows detailingarrangement for a continuous beam as per IS456-1978 and SP34. Note that only

the bending reinforcement is shown. You must provide shear reinforcementwith vertical stirrups as well. Reinforcement arrangement generally followsbending moment pattern in the beam. If interested, you can model the beam as2D FEM (as discussed in Section 23) to visualize what are the highest zones of stresses. Building codes specify lots of guidelines in ‘terse’ languages!Unfortunately, detailing in most analysis programs is often unreliable. This ismainly because, unlike analysis, design is subjective. Many codes allow theengineers to use their ‘judgment’ in detailing. In fact, detailing is a subjectitself. A good designer must not finish his duty only after analysis, he shouldprepare the accurate detailing as well let alone the drawing too if possible. Thegeneral trend in design offices is that the detailing is done by the draftsmanrather than the engineer who performs the analysis. This often creates a‘communication gap’, which is undesirable. Remember, if detailing is wrong,then all the result of a good analysis will go astray! Traditionally, the detailingdrawing is drawn in monochrome (single) color. However, because of availability of low cost color printing, the printing cost has been drasticallyreduced in recent times. So, whenever possible, use colors in your drawings. Itwill make the whole thing much clearer to everybody. The world around us iscolorful, why not your drawings?




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41. A guide to some Structural Engineering & Finite ElementAnalysis Programs

I have discussed my personal views about the programs. You should test them

yourself, before deciding which program will serve your purpose best. Actual

prices may vary.

We shall divide the programs in two main groups – Civil Engineering andMechanical Engineering analysis programs. You must differentiate betweenthem first. A typical civil/structural-engineering program offers beam and trusselements and sometimes plate/shell elements as well. It also often offers designfeatures (concrete/steel/aluminum/wood) for slabs, beams, columns, trusses,footings, base plates etc. according to various country codes. It may includeboth static and dynamic analysis. In contrasts, a typical mechanical engineeringanalysis program contains a large number of element types for solid modeling.This includes brick, tetrahedral etc. elements. In addition, it also includes all theelements used in civil engineering analysis i.e. beams, plates etc. In fact,mechanical engineering programs use finite element analysis in true sense.These programs’ features include static, dynamic, thermal, multi-physics/mechanical event simulation/virtual prototyping etc. Mechanicalengineering analysis programs are normally far more expensive than pure civil

engineering analysis programs. Since, mechanical programs offer versatile finiteelement library, it is possible to solve all types of civil engineering structuralanalysis problems using these programs. However, there are drawbacks as well.Mechanical engineering programs do not offer concrete design facilityaccording to country codes. They mainly concentrate on solid model analysise.g. machine parts, rather than pure frame structures. Also, their output forframe structures is generally much less varied compared to pure civilengineering structures. So, for day to day building or industrial structureanalysis, it is much better to use civil engineering analysis programs. Although

some developers offer product for both civil and mechanical engineeringanalysis, they generally contain separate modules for specific type of analyses.

I only described a brief description of the most used programs only. Note that Ididn’t have the opportunity to use all the programs myself. This is because notall developers offer demo/trial version of their programs. The programs, which Ihave used, I presented the features available, ease of use, reliability etc. Forother programs, I wrote the descriptions from product literature. I again tell youthat my personal views about the programs may differ a lot from your views.

So, judge them yourself before buying.




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Civil Engineering programs

Program name: Visual Analysis, Visual Design, Visual Tools etc.Developer: Integrated Engineering Software

Website: www.iesweb.com Description: 3D frame and 2D plate, P-delta analysis, response spectra dynamicanalysis, concrete/steel/wood/aluminum design as per US codes.Price: USD 700 – 1,500 (Student version USD 50 only!)Demo: 30-day trial full version CD availableMy comment: An excellent program for frame and plate analysis. Simple to useand very user-friendly. Student version has no node or member size limitation!Has excellent customer support.

Program name: SAP2000 Standard/Plus/Non-linearDeveloper: Computer and Structures Inc.Website: www.csiberkeley.com Description (for non-linear version): Static, Dynamic Response Spectrum, TimeHistory, Bridge and Dynamic Nonlinear Time History Analysis with frame,shell, solid, asolid and non-linear link (as external Damping, Base Isolators, Gapand Hook) elements. Concrete and steel design as per ACI, AASTHO,Canadian, British and Euro codes. Its predecessor SAP90 is still used foreducational purposes.Price: USD 5,000 – 7,000Demo: 30-node demo CD availableMy comment: Good overall general-purpose static/dynamic structural analysisprogram. Easy to use. However, non-linear analysis options are difficult tocomprehend. Also, solid elements can’t be drawn graphically.

Program name: SpacegassDeveloper: Integrated Technical SoftwareWebsite: www.spacegass.com Description: Frame analysis only – beam, truss, grillage and cable etc., non-

linear dynamic and concrete/steel design. Supports Australian, British, Euro andUS code.Price: USD 1,000Demo: 30-day full version copy can be downloaded from net. File size 6 MB.My comment: Good program for frame analysis. However, graphics shouldhave been better. It shows different sections in various colors, which makesgeometry creation easier. Some dialog boxes offer too many options. Data canbe entered conveniently by on screen as well as using spreadsheet like format.However, there is no plate element and time history dynamic analysis is not

available.




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Program name: RISADeveloper: Risa TechnologyWebsite: www.risatech.com Description: 3D-frame and 2D-plate analysis program.

Price: Not knownDemo: 150-node 150-member demo can be downloaded from net. File sizearound 6 MB including tutorial.My comment: Graphics should be better.

Program name: STAAD Pro/ Fabricad/ FEMkit etc.Developer: Research Engineers Inc. (recently re-named as Netguru Inc.)Website: www.reiworld.com Description: Frame and finite element analysis, dynamic and non-linearanalysis, design of piles, retaining wall, bolt groups, pre-stress concrete, as perUS, Indian, Canadian, Japanese, Chinese, European etc. codes.Price: USD 5,000 – 6,000/ INR 75,000 – 150,000Demo: 6-member demo CD available.My comment: Overloaded with many features but I found quite difficult to use.Program’s help files are not very comprehensive. Demo version is ridiculouslylimited to do anything good.

Program name: STRAPDeveloper: ATIR

Website: www.atir.com Description: 3D-frame and plate/shell static/dynamic analysis program.Concrete and steel design according to European, Indian, Canadian, British andUS code. Separate bridge and foundation design modules are available.Price: USD 4,000 – 4,500/ INR 75,000 – 150,000 (full capacity student versionINR 25,000)Demo: 12-node demo CD available.My comment: Interface is somewhat complex. Its model wizard can generatevarious types of structures easily. Design feature is quite good. Due to node size

limitation, I couldn’t test it extensively.

Program name: GT-STRUDLDeveloper: Georgia Institute of TechnologyWebsite: www.gtstrudl.gatech.edu/gtstrudl Description: 3D frame, plate, non-linear dynamic, finite element analysis etc.Price: USD 2,000 – 12,500/ INR 60,000 – 80,000Demo: 50-node 200-member Student version CD availableMy comment: Somewhat complicated to use.

Program name: SAFI 2D/3D/TOWER/BRIDGE etc.




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Developer: SAFIWebsite: www.safi.com Description: Frame, plate/shell linear/non-linear static and dynamic analysis.Price: Not known

Demo: 50-node 50-member demo can be downloaded from web site.My comment: I didn’t use it.

Program name: RobotDeveloper: Integrated Structural SoftwareWebsite: www.robot-structures.com http://www.autofea.com/ Description: Frame, truss, cable, plate, shell, grillage non-linear static/dynamicanalysis, concrete/steel design.Price: Not knownDemo: 25-member demo CD available.My comment: I didn’t use it.

Program name: AxisVMDeveloper: Inter CADWebsite: www.axisvm.com Description: Performs static, vibration and buckling analysis on anycombination of truss, beam, rib, membrane, plate, and shell 3D structures, gapand spring elements for nonlinear support modeling. Supports Eurocode.Price: USD 800 – 2,000

Demo: 30-beam/truss, 100-surface element version can be downloaded from thenet. File size around 5 MB.My comment: Graphics is good, but has too many dialog boxes for modelcreation. Some common commands are difficult to find.

Program name: ETABSDeveloper: Computers and Structures Inc.Website: www.csiberkeley.com Description: Specific program for building frame analysis. Similar to SAP2000

in features – static & dynamic analyses – automatic load distribution tomembers, steel, concrete design and optimization, non-linear pushover analysis,US and Euro codes.Price: Not knownDemo: Not offeredMy comment: User friendliness is similar to that of SAP2000. Unless yourdesign job involves only multistory building frames, it is better to use SAP2000,which is more versatile.

Program name: Diana




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Developer: Diana Analysis bv

Website: www.diana.nl Description: finite element analysis – static/dynamic, soil/concrete, solid

modeling, pre-stress, mobile loads, modeling of embedded reinforcement inconcrete, bridge, dam, offshore structure design. Various separatedesign/analysis modules are available. Available on Unix as well.Price: Not knownDemo: Not offeredMy comment: I didn’t use it.

Mechanical Engineering programs

Program name: FEMAPDeveloper: Enterprise Software Products Inc., Structural Dynamic ResearchCorporationWebsite: www.femap.com Description: Finite element pre and post processor only. Does not analyzeanything by itself.Price: USD 3,500 – 5,000Demo: 300-node demo CD available (30-day trial full version available in North

America only) with m-tab (www.sai-mtab.com) stress for analysis.My comment: Draws models, creates mesh and import/export models from/to20 different finite element analysis programs’ formats. Has many advancedoptions but difficult to use. Sometimes cannot translate models successfully tosome specific formats. FEMAP is sometimes bundled with other finite elementanalysis programs.

Program name: NISADeveloper: Engineering Mechanics Research Corporation

Website: www.emrc.com Description: Finite element analysisPrice: Not knownDemo: Available from net, but not clear what file to download.My comment: I didn’t use it.

Program name: ALGORDeveloper: Algor Inc.Website: www.algor.com

Description: Linear and non-linear static and dynamic finite element stressanalysis with motion for physics based mechanical event simulation/virtual




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prototyping. Incorporates beam, truss, plate, 2D, membrane, brick, contact, gap,laminar, composite, sandwich, tetrahedral etc. elements. Can import model fromvarieties of CAD programs. Separate modules are available for CADintegration, heat, electricity, fluid flow, pipe stress etc. analysis.

Price: USD 2,000 – 18,000Demo: 30-day trial full version CD available.My comment: Very powerful programs containing lots of features, but advancedoptions are difficult to use without training. Good buy if you analyze complexfinite element models often. To use this program to its full extent, you shouldhave good knowledge of finite element theory.

Program name: Cosmos/Design Star/Geo Star/N Star etc.Developer: Structural Research & Analysis Corp.Website: www.cosmosm.com Description: Analyzes solid by finite element method; linear stress analysis,dynamic analysis, buckling, heat transfer, can import model from a variety of CAD programs. Separate modules are available for non-linear static/dynamicanalysis, fatigue, optimization, fluid flow and electro-magnetism.Price: USD 5,000 – 20,000Demo: 30-day trial program can be downloaded from net. File size 55 MB. TrialCD also available.My comment: Excellent FEA program. Easy to use. Uses special Fast FE solver,which is many times faster than common FEA solver. Can’t create model by

itself (GeoStar is required to create models, Nstar for non-linear analysis), butcan import model from a variety of CAD programs. Graphics is also very good.Help system is comprehensive. However, Design Star analyzes 3D solid modelsonly and does not have beam/truss elements.

Program name: ABAQUSDeveloper: HKSWebsite: www.hks.com Description: Non-linear/dynamic finite element analysis for mechanical, civil,

structural, biomedical etc. Various other modules are available for wavepropagation, heat, offshore design, CAD/CAE integration.Price: Not knownDemo: Not offered.My comment: I didn’t use it.

Program name: AnsysDeveloper: AnsysWebsite: www.ansys.com




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Description: Finite element analysis. Various other separate analysis/designmodules are available for structural, mechanical, multi-physics, thermal, fluidflow, acoustic, electro-magnetism etc.Price: Not known

Demo: Not offered.My comment: I didn’t use it.

Program name: ADAMSDeveloper: Mechanical Dynamics Inc.Website: www.adams.com Description: Analysis and virtual prototyping (mechanical event simulation) of mechanical systems. Separate modules are available for linear, car, engine, tyre,rail, hydraulics, mechanism etc.Price: Not knownDemo: Not offered.My comment: Finite element application for mechanical engineering. Versatileprogram but complex. To use this program to its full extent, you should havegood knowledge of mechanics and finite element theory. With Adams, you cancreate a whole model of virtually anything – a sports car to a railwaylocomotive with all engineering details and they you can analyze it! You caneven simulate it! It’s a really wonderful program! Like a video game, you candrive a car here and can refine analysis at every stage! Besides this, Adams hasall other features of a typical FEA program. These programs will drive you

mad!

Program name: NastranDeveloper: Macro Industries/NASAWebsite: www.nastran.com Description: Finite element analysis of frame, plate, shell, solid, shear panel,fluid etc.Price: USD 2,750Demo: Version 1.0 is available for downloading. File size 12 MB.

My comment: I didn’t use it.

Program name: PatranDeveloper: Mechanical SolutionsWebsite: www.mechsolutions.com/products/patran Description: Finite element analyses, can read/write to various other FEAprograms’ formats, various modules are available separately.Price: Not knownDemo: Not offered.

My comment: I didn’t use it.




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Program name: AdinaDeveloper: Adina R & D Inc.Website: www.adina.com Description: Finite element analysis of solid, fluid flow etc.

Price: Not knownDemo: Not offeredMy comment: I didn’t use it.

Program name: I-DEASDeveloper: Structural Dynamics Research CorporationWebsite: www.i-deas.com Description: Integrated CAD/CAM/CAE Finite element analysis. Mechanicalevent simulation – virtual prototyping etc.Price: Not knownDemo: Not offeredMy comment: I didn’t use it.

Program name: JL AnalyzerDeveloper: AutoFEA Inc.Website: www.autofea.com http://www.autofea.com/ Description: Static, buckling, frequency, dynamic, spectrum, thermal, electric,seepage and nonlinear static FE analyses.Price: USD 285 – 5,000 (depending on node capacity)

Demo: 300-node demo can be downloaded from net.My comment: Complicated to use.

Some CAD programs…

Some developers provide feature-limited demo. See their web site for details.

Program name: AutoCAD/ Mechanical DesktopWebsite: www.autodesk.com

Program name: Solid EdgeWebsite: www.solid-edge.com

Program name: Solid WorksWebsite: www.solidworks.com

Program name: CadkeyWebsite: www.cadkey.com




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Program name: Turbo CADWebsite: www.turbocad.com

Prices in INR are shown only for those companies, which have dealer/ representative in India and you can buy the product in Indian currency. Somecompanies sell their full capacity student version at reduced cost, but thatversion cannot be used commercially. Some developers do offer universitylicense at very nominal cost.

A useful point on downloading large files from the Internet. Use a downloadutility program such as Go!Zilla (www.gozilla.com) or something else, whichcan resume broken download from where it left. Also, it can be set to disconnectfrom Internet automatically after downloading is complete.




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42. How to select the most appropriate program for your need?

Here are some simple points to check before buying the program.

How easy is the program? How quickly you can create and analyze a modelusing the program after installing? Use the sample problems in this book andsee how quickly you can analyze the structures with your programs.

Try reading the programs’ manuals at first. See if you can analyze modelssuccessfully by just reading the example problems rather than by getting helpfrom the developers.

How good is the interface? Does the program follow standard Windowsconventions? For example, pressing F1 should bring context sensitive help andCtrl+O should show Open File… dialog box.

How good is the visualization? Does the model look similar to actual structure?Does it offer dynamic viewing? Can it show the model in VRML format?

Can it generate complex models automatically? For example, to generate a10x10 bays 20-story building, the program should draw the model by itself. Canit create finite element mesh automatically? Can you control the size of mesh

yourself?

Before buying a program decide what features you do really need. There is no justification of buying a high-end ‘feature overloaded’ expensive finite elementanalysis program if your main purpose is building-frame analysis. Suchprograms are generally much less expensive. In general, most civil

engineering design firms do not need high-performance FEA programs. Only a frame analysis program will serve the purpose in most of the cases.

FEA is still mainly used in mechanical engineering.

Does the program is ‘forgiving’? Can it check model for instability or bucklingby itself? Does it allow ‘snapping’ members to intersection or end points?

How good is the program’s post processing capability? Can it show forcediagrams for any individual member? Can you clearly see the reaction forces inthe result? Can it display bending moment or shear force values in therespective diagram?

How about printing through the program? Will it allow you to change papersize, margin etc. before printing?




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What about generating report of analysis? Does it have the option of displayingtailor made report specific to project requirement or it just dumps 100 pages of numbers without telling you what to do with them?

Check if your program is compatible with other programs in the market. Manyprograms allow copy-paste option with spreadsheet programs. Some offer crossplatform (i.e. different operating system) compatibility. Check if the programcan import/export models from/to CAD or other analysis programs.

Next comes the design part. Some programs do not offer design capabilityaccording to your country code. This may dictate you to choose an inferiorprogram just for the sake of design calculation. Unlike analysis, design is notsomething impossible to do manually. Sometimes, it is possible to ‘tailor made’other codes according to your country codes. For example, in reinforcedconcrete design, Euro code and Indian code give almost identical values of steelarea. You can use one code instead of other without much loss of accuracy. So,the choice is yours.

An important point – how good is the customer support? Some companies mayoffer free technical support by phone/e-mail. Others may insist on annualmaintenance contact (and demands a hefty sum for it). But to use high-end FEAprograms, you really need developers support. There will be many situations

when you will require technical support to know what a particular commanddoes or whether the program does a particular calculation automatically etc.

Lastly, what is your budget? Can you afford buying the program what impressesyou most? Some finite element analysis programs are very fascinating but theirprices often make them affordable by large corporate firms only.

Ask the vendors whether they offer free demo. Some generous companies mayoffer 30-day full product trial. However, most companies offer limited

capability (e.g. 100-node maximum) demos. These limited demos may hide theprograms actual performance from you. You may not realize certain pitfalls of the program unless you use full product for sometime. Therefore, try to selectyour program from those, which offer full-product tryouts. Sometimes even thelow cost commercial versions of the programs do have node or member sizelimitation (e.g. 1500 nodes maximum for SAP2000 Standard version)!

Many structural analysis programs nowadays come with hardware locks(hasp/dongle) as copy protection. It means that if you want to use the programs

in more than one computer, you need to pay more! Although it may be againstthe law to copy the program into another computer, I don’t find it wrong to




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make another copy in your laptop computer. But copy protection will preventyou even doing that. Remember this while selecting your program.

Since these programs are quite expensive it does not make sense to change your

analysis programs frequently. More so, new programs may not open or analyzeyour existing old projects. So, when you will select programs, you have keep inmind that you must use it continuously for a couple of years at least. Ask thedevelopers how they like to upgrade their products in the near future.

My main aim is to make you realize that you must try the programs first beforemaking a final decision on purchase. Every company will boast that theirproducts are the best. They will swamp you with pictures of structures designedusing their programs! Very few companies offer you a money back guaranteewith first few weeks of purchase. Therefore, it is not an easy task to choose themost suitable product for your need. But if you follow the above guidelines youmay by a gainer.




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43. How to check the result for accuracy?

Before performing an analysis, check carefully all input data. Verify loads arein the directions you intended. Make sure co-ordinate directions as per youspecified. Check the member properties and end releases. Verify that you havethe correct support conditions. Take a look at the real time view (if this featureis available in your program) of the structure and visually observe member sizesand locations. Are they correct? You might also use spreadsheet like inputreports to make sure no information have missed.

When you obtain analysis results from analysis programs, you should performsome simple checks to verify that the results are valid. First of all, look at thedeformed shapes that are presented to you. Do they make sense based on theloading and structure?

Secondly, look at the magnitude of the largest displacement shown. Is it large?(Generally, the displaced shape is automatically scaled suitably so that you cansee it!) If displacements are too large, the basic assumption of smalldisplacement is violated and results must be questioned. In that case you shouldperform geometrically non-linear analysis.

Third, look at the member internal forces and stress levels reported in thehighest stressed members (beam/shell etc.) elements. Do these stresses yield,crack, or fail or buckle the material? Again if they do, the response is violatingthe basic assumption of linear-elastic material behavior.

As a fourth check, look for static check. If your loads are correct, yet reactionsdo not balance, check for nearly unstable structures. If you have poor geometryor very flexible members in critical areas, the structure may collapse due toinstability.

Finally use your engineering judgment. If something seems wrong, investigateit carefully. Convince yourself that the results are correct before continuing.Recall that “With good engineering judgment you can produce on the back of an

envelop that which otherwise cannot be produced with a ton of computer

output”.

If you think you have found an error in the program, contact the programdevelopers immediately for clarification.




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44. File name extension guide (for some CAD/CAE programs)

Extension Description

ANS Ansys inputASM Mechanical Desktop/Solid Edge assembly

CDL Cadkey wire frame model

DAT Nastran input

DFT Solid Edge draft

DGXASM Cosmos Design Star Assembly

DGXPRT Cosmos Design Star Part

DWG AutoCAD drawing /Mechanical Desktop model

DXF AutoCAD drawing interchange

ESD Algor inputIGS IGES solid model

INP Abaquas input

MOD Femap input

NAS Nastran input

NEU Patran input

NIS NISA input

PAR Solid Edge part model

PRT Solid Works model

SAT AutoCAD objects representing trimmed NURB surfaces, regions,and solids to an ACIS file in ASCII (SAT) format. Other objects,such as lines and arcs, are ignored.

SDB SAP2000 input

SG Spacegass input

SLDPRT Solid Works

STD STAAD input

STL Stereolithograph Apparatus (SLA) format, in which the solid datais transferred as a faceted mesh representation consisting of a set

of trianglesSTP STEP model

UNV SDRC I-DEAS

VAP Visual Analysis project

WRL VRML file

X_T Parasolid

Note: during runtime, programs create various files with different file name

extensions. If you delete those files, you may not view result without analyzingthe model again.




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45. Common error messages and solutions

Operating system related

1. Not enough memory for calculation/initialization of graphics...

May be misleading message sometimes. Check the program in anothercomputer with same amount of memory (preferably with same hardwareconfiguration as well). If the program runs successfully there, try reinstalling theprogram in your computer. Rarely, the program may demand better qualitygraphics card. (This is more common in computer games rather than structuralanalysis programs, though!) If all fails, stuff some more RAM into yourcomputer and see what happens. If you're lucky everything will go ok, if notcontact the program vendor. Sometimes, simply changing your color settingfrom 256 to 16/24-bit color works!

2. This program has preformed an illegal operation and will be shut down. If the problem persists, contact the program vendor.

This is a common error message for all Windows programs. It can be rectifiedby just rebooting the machine in most cases. However, if you continue to get the

message every time, something is definitely wrong. Contact the developers.

3. This program is set to run in DOS mode. All other programs will shutdown if you proceed. Continue?

Some analysis programs are made in such cumbersome manner, that theyincorporate some Windows programs and some DOS programs. I am not surewhy this happen. It may be due to operating system problem rather than that of

your analysis program. Try reinstalling the program. If it's of no use, then I amafraid that you may need to reformat your hard disk and load all from scratch. Imyself faced this very situation while installing Algor for the first time. Aftererasing everything from my hard disk, and then refilling all, I was finally to runthe program.

4. A program was running fine in last week, but to my surprise, it refuses torun now.

The possibilities of this cause are endless. Did the problem start after installingany new application? Then uninstalling that application might help. There may




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be some jumbled DLL and OCX files. My SAP2000 refused to show dynamicanalysis result after I loaded GT-STRUDL. Ultimately I discovered that anOCX file required by SAP2000 was overwritten by STRUDL with backdatedversion. After restoring that file, both were happy again.

Analysis program related

Analysis programs, which display the specific error message, are shown within

bracket. “Common” means many analysis programs show this

message/problem.

1. The structure is unstable (common)

You must specify proper boundary conditions, so that the structure does preventrigid body motion. That means the global stiffness matrix must not be singular.Also see (8).

2. Your model may not be tied down (Algor)

Same as above. However, some FEA programs can still continue solving theproblem and ultimately you may get result with no force or stress at all. Quiteconfusing!

3. Load case "X" has no load (common)

Some programs will not accept any load case that has no loading specified.Remove that load case or apply some load in that load case.

4. I have analyzed the structure successfully. But it shows all zeroforces/stress in the result (common)

The resulting forces/stress might be too small. Try changing the units, for

example, from kN-m to N-mm and see if you find any value now. Also see (2).

5. Zero stiffness found during assembly for DOF Ry of joint 5 (SAP)

It means that the joint has no stiffness in that direction. It should be connectedto at least one member or support without releases.

6. Excessive loss of accuracy during the solution of equations – the structureis unstable of ill-conditioned (also Lost 6.4 digits of accuracy) (SAP)




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Equation solutions in most analysis programs are done in double precisionarithmetic. This allows for 15 significant digits to be carried in the calculation.An internal check estimates the number of digits of accuracy lost during thereduction of each degree of freedom. A warning is issued when the loss is

estimated to be above 6 digits and the degree of freedom is listed when thisoccurs. Users should check the displacements, reactions and global equilibriumbalances to assure that the digit loss has not unacceptably degraded the results.When the program detects over 11 digits accuracy loss it stops furtherprocessing as the results are suspect. You should check for global instability andlocal instability at or around the degree of freedom reported. See also (13).

7. Rigid body motion is not prevented in Z direction (common)

Somewhat similar to (1) and (2). There is no boundary condition specified toprevent rigid body displacement (translational or rotational) to Z direction.Specify a suitable constraint in that direction.

8. The structure is internally unstable (common)

A really annoying situation! Your structure has become a mechanism. So, anyload will make it fail (collapse). This is where you need to use your engineering

judgement. Add or remove some members until it becomes a stable structure.You will have to perform some trial and error. This problem mainly occurs in

complex trusses.

9. I have applied loads on the model, but I can’t see the loads (common)

Look if there is any load ‘toggle’ menu. Some programs treat nodal load andsurface/pressure loads differently. You may need to specify surface/pressureloads by some separate command. Sometimes there may be a pressure load‘multiplier’. Make sure that it is not set to zero.

10. A smoothed stress tensor may not be meaningful for this model (Algor)

Stress values like σx, σy, τxy etc. is not meaningful for particular analysisresult. It is not so much that a smoothed tensor may not be accurate for a model,but rather that with some element types and element generation methods, thetensor may not provide meaningful results, i.e. a random mesh on a compositeelement type.




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11. I see that my program does not have plate and shell elements separately(common)

Some programs only have ‘shell’ element. In that case, ‘plate’ is considered a

special type of shell element (e.g. SAP2000). In some other programs, onlyplate element is offered and shell is considered a type of ‘plate’ element (e.g.Algor). However, there are many programs (e.g. Visual Analysis, Risa etc.)which offer only ‘plate’ element, which can’t be used for shell analysis.

12. I can’t convert/export my model from one FEA format to another.

Exporting or converting your model from one analysis program’s format toanother’s is always bad. Except for very simple models, it often leads to loss of data and thus unsuccessful translation. I strongly discourage you to exportingwhole FEA model data in this way. However, if you still want to do that, check the following points:

• Have you defined the mesh? Most programs can’t export to another FEAformat unless you create the mesh.

• Be careful about the units used in both programs.

• If full model (i.e. with mesh, boundary conditions, loads etc.) export doesnot work, then try exporting the model with mesh only.

•

Upon importing the model in desired program, check the model carefully.• Suppose you are exporting the model from program ‘A’ to ‘B’s format andyou will open it using another program ‘C’. Then make sure ‘C’ can also‘read’ that ‘version’ of ‘B’s data!

13. The structure is ill-conditioned (common)

Ill-conditioning commonly occurs when frames contain members of widelyvarying stiffness. When a very stiff member is connected to a very flexiblemember and their stiffness matrices are assembled into the structure stiffnessmatrix, some of the stiffness terms of the flexible member can be completelylost due to their insignificance in comparison with the stiffness terms of the stiff member. Hence, the flexible member is not completely represented and illconditioning occurs. If after the analysis, the sum of the reactions equals thesum of the applied loads then it can be assumed that the frame is wellconditioned.




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46. References

1. Schaum’s Outline of Theory and Problems of Reinforced Concrete Design,

3

rd

edition, by Noel J. Everard, McGraw Hill2. Finite Element Analysis: Theory and Programming, by Krishnamoorthy,Tata McGraw Hill

3. Concept and Application of Finite Element Method, by Cook, Malkus,Plesha, John Willey & Sons.

4. Schaum outline series of Finite Element, by Buchanan, McGraw Hill5. Dynamics of Structures, by Chopra, Prentice Hall6. Users’ Guide, Visual Analysis 3.5, IES7. Users’ Guide, SAP2000, CSI8. Algor R12 information brochures, tutorials and newsletters, Algor Inc.9. Reinforced Concrete Design Vol 1 & 2, by Punmia, Laxmi Publication10. Basic Structural Analysis, 2nd edition, by Reddy, Tata McGraw Hill11. Shell Analysis, by Bairagi, Khanna Publisher12. Users’ Guide, Femap13. Users’ Guide, Strap14. User’s Guide, Strudl15. User’s Guide, Cosmos/Design Star, SRAC16. Structural Dynamics, by M. Paz, E & FN Spons.17. Elements of Structural Dynamics, by Glen Berg, Prentice Hall

18. Users’ Guide, Spacegass

The above list is not exhaustive. There are many other better books available.You can find online reviews of a good number of books at www.amazon.com.



FOR THE FIRST TIME

• A PRACTICAL GUIDE• An absolute must for all practicing

Mechanical/Civil/StructuralEngineers, Researchers and Students!

• Use your structural analysis program

to its full extent!• Analyze those structures, which you

previously thought impossible,without going through heavy and

boring theoretical calculation!

Date post:	06-Apr-2018
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Computer Aided Structural Analysis[1]

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