Computer Aided Structural Analysis

Saikat Basak

Tips and Tricks for

Computer Aided Structural Analysis

Saikat Basak M.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 preparing this book. These efforts include the development, research and testing of the theories and programs to determine their effectiveness. The author and publisher shall not be liable in any event for the incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. All rights reserved. No part of this book may be reproduced, in any form or by any means, without the permission in writing from the author. 1st Edition 2001 Published on the web

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Computer Aided Structural Analysis

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 .................................................................................................................... 10 DYNAMIC ANALYSIS .......................................................................................................................................... 11 RANDOM VIBRATION......................................................................................................................................... 12 RESPONSE SPECTRUM ANALYSIS....................................................................................................................... 12 TIME HISTORY ANALYSIS.................................................................................................................................. 13 TRANSIENT VIBRATION ANALYSIS .................................................................................................................... 13 VIBRATION ANALYSIS (MODAL ANALYSIS) ...................................................................................................... 14 BUCKLING ANALYSIS......................................................................................................................................... 15 THERMAL ANALYSIS .......................................................................................................................................... 16 BOUNDARY 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 -4-

Computer Aided Structural Analysis 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 ............................................................................................................................ 115 GEOMETRIC 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 ...................................................................................................................... 131 MECHANICAL ENGINEERING PROGRAMS ......................................................................................................... 134 SOME 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 ......................................................................................................................... 144 ANALYSIS PROGRAM RELATED ........................................................................................................................ 145 46.

REFERENCES ..................................................................................................................................... 148

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Computer Aided Structural Analysis Abbreviation

Several abbreviations have been used throughout this book. They have been defined in respective sections, but here is a list of them at a glance. [k] – Stiffness BM – bending moment C - damping CAD – computer aided design/drawing CAE – computer aided engineering CAM – computer aided manufacturing E – modulus of elasticity FE – finite element FEA – finite element analysis FEM – finite element method fy – yield strength of steel G – shear modulus I – 2nd moment of inertia IS – Indian Standard code LRFD – load and resistance factor design LSSA – linear static stress analysis M, m – mass MDF – multi degree freedom MES – mechanical event simulation NLA – non-linear analysis RSA – response spectra analysis SDF – single degree freedom SF – shear force T – time period of vibration THA – time history analysis UDL – uniformly distributed load VRML – virtual reality markup language x – displacement x’ – velocity x” – 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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Computer Aided Structural Analysis

1. Introduction (before you begin…) In this book I shall tell you some practical tips for structural analysis using computer. Most structural engineering books are written to tell you how you will perform the calculation by hand. But even sometimes analysis using computer can be very tricky. You may need to manipulate computer input to solve a problem, which may at first appear to be unsolvable by that program. Finite element programs and structural analysis programs tend to be very expensive. Most small-scale engineering firms keep only one analysis program. Even for a large corporate companies it is seldom possible to maintain more than two standard analysis packages. Therefore it is essential that you use your present analysis program to its full extent. This is not a textbook. I make no attempt to teach you theory of structural analysis to score good marks in the exam! But it can help you to earn more money by enabling you to analyze some structures more easily and accurately, which you were previously thought too difficult to deal with your existing analysis program. Also, I am not going to teach you any particular structural analysis computer program. However, the techniques of analysis discussed here are applicable to most standard analysis packages. I presented the whole thing in an informative yet informal manner. I confined the boring theory and calculation to minimum level. No special knowledge is required to get the most out of this book. Only Bachelor 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 familiar with 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 know that 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 -7-

Computer Aided Structural Analysis

otherwise cannot be produced with a ton of computer output”. You should paste this 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 the market. • Some features I discussed here may not be available in your program. It can even happen that the program you are using has better option to handle a particular problem compared to what I discussed in this book. • I am only providing you some “clues” for more effective use of structural analysis programs. However, every analysis problem is unique depending on type of project, cost, client’s requirement etc. Those specific criteria you have to solve yourself. • Documentation of the program you are using is very important. The program manuals are the best source of help always. The sections of this book are arranged in somewhat haphazard manner deliberately so that you don’t feel bored. The paragraphs are small and to the point. We have often returned to same topics in several sections from different viewpoints. Wherever necessary, numerical examples have been presented. There are also some exercises. Please try to solve them with your structural/FE analysis programs. I like to see your comments and suggestions. You can reach me at www.enselsoftware. com in World Wide Web. I shall be more than happy to answer your queries. Now sit back, relax and enjoy the book. Have a nice reading!

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Computer Aided Structural Analysis

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 is the method of solving your structural analysis problem with the help of computer software. In earlier generation analysis programs, you had to supply the programs the nodal co-ordinates, member incidence (i.e. between what nodal points a particular member is connected), material properties, sectional properties of the all members and the loads (nodal force/moment/distributed member loads etc.). You also had to supply how the structure was supported, fixed, hinged or roller. The program then calculated the member forces, nodal reactions 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 a first 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 models graphically, apply loads and boundary conditions graphically and visualize the shear force, bending moment and even deflected shape diagram graphically. For the first time users, it seems rather like a magic! The availability of these programs has completely changed the way we analyze structures compared to we 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 structures using computers has created many other new problems. First, you must be very familiarize with the programs you are using. You must clearly understand its limitation and assumptions. All programs can’t be applied for analyzing all types of structures. Most programs solve the structures by stiffness method, though solution algorithm may differ from one program to another. What is most 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 this book, you will also learn about some very recent concepts of structural analysis. Bon Voyage!

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Computer Aided Structural Analysis

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 some are 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, the body 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 at the 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 full magnitudes. After reaching their full magnitudes, load will remain constant (i.e. load will not vary against time). This assumption lets us disregard insignificant inertial and damping forces due to negligibly small accelerations and velocities. Time-variant loads that induce considerable inertial and/or damping forces may warrant 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 responses is linear. If you double the magnitude of loads, for example, the response of the model (displacements, strains and stresses) will also double. You can make linearity 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 change while the model is deforming. If the above assumptions are not valid, then we shall have to treat the problem as 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 hence the load distribution in the model will change based on the gap forces generated. - 10 -

Computer Aided Structural Analysis

This functionality offers a linearized solution to a nonlinear problem. Sounds crazy? Calculation of stresses Stress results are first calculated at special points, known ‘Gaussian’ or ‘Quadrature’ points, located inside each element. (See you FEA textbook for details) These points are selected to give optimal results. The program then calculates stresses at the nodes of each element by extrapolating the results available at the ‘Gaussian’ points. After a successful run, multiple results are available at nodes common to two or more elements. These results will not be identical because the finite element method is an approximate method. For example, if a node is common to three elements, there can be three slightly different 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 nodal stresses for each element. In calculating nodal stresses at a node, the program averages the corresponding results from all elements contributing to the stresses at that node. Dynamic analysis

In general, we have to perform dynamic analysis on a structure when the load applied to it varies with time. The most common case of dynamic analysis is the evaluation of responses of a building due to earthquake acceleration at its base. Every structure has a tendency to vibrate at certain frequencies, called natural frequencies. Each natural frequency is associated with a certain shape, called mode 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 its natural frequencies, the structure undergoes large displacements. This phenomenon is known as ‘resonance’. Damping prevents the response of the structures to resonant loads. In reality, a continuous model has an infinite number of natural frequencies. However, a finite element model has a finite number of natural frequencies that is equal to the number of degrees of freedom considered in the model. The first few modes of a model (those with the lowest natural frequencies), are normally important. The natural frequencies and corresponding mode shapes depend on the geometry of the structure, its material properties, as well as its support conditions and static loads. The computation of natural frequencies and mode shapes is known as modal analysis. When building the geometry of a model, you usually create it based on the original (undeformed) shape of the model. Some loading, like a structure’s - 11 -

Computer Aided Structural Analysis

self-weight, is always present and can cause considerable changes in the structure’s original geometry. These geometric changes may have, in some cases, significant impact on the structure’s modal properties. In many cases, this effect can be ignored because the induced deflections are small. This is just a prelude to dynamic analysis. You will find several topics on dynamic analysis later in this book. However, since I shall not discuss theory of structural dynamics here, I strongly recommend that you read a structural dynamic 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 modal analysis are extensions of dynamic analysis. Random Vibration

Engineers use this type of analysis to find out how a device or structure responds 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 the vehicle, such as on-board electronics or cargo of any kind, may need Random Vibration Analysis. The vibration generated in vehicles from the motors, road conditions, etc. is a combination of a great many frequencies from a variety of sources and has a certain "random" nature. Random Vibration Analysis is used by mechanical engineers who design various kinds of transportation equipment. Engineers provide input to the processor in the form of a ‘Power Spectral Density’ (PSD), which is a representation of the vibration frequencies and energy in a statistical form. When an engineer uses Random Vibration he is looking to determine the maximum stresses resulting from the vibration. These stresses are important in determining the lifetime of a structure of a transportation vehicle. Also, it would be important to know if things being transported 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 structure responds to sudden forces or shocks. It is assumed that these shocks or forces occur at boundary points, which are normally fixed. An example would be a building, dam or nuclear reactor when an earthquake strikes. During an earthquake, violent shaking occurs. This shaking transmits into the structure or device at the points where they are attached to the ground (boundary points).

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Computer Aided Structural Analysis

Response spectrum analysis is used extensively by Civil Engineers who must design structures in earthquake-prone areas of the world. The quantities describing many of the great earthquakes of the recent past have been captured with 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 must use response spectrum analysis as well. Such components might include nuclear reactor parts, pumps, valves, piping, condensers, etc. When an engineer uses response spectrum analysis, he is looking for the maximum stresses or acceleration, velocity and displacements that occur after the shock. These in turn lead to maximum stresses. You will find an example of response spectrum analysis later. Time History Analysis

This analysis plots response (displacements, velocities, accelerations, internal forces etc.) of the structure against time due to dynamic excitation applied on the structure. You will find more stuff on this particular type of analysis in later sections. Transient Vibration Analysis

When you strike a guitar string or a tuning fork, it goes from a state of inactivity into a vibration to make a musical tone. This tone seems loudest at first, then gradually dies out. Conditions are changing from the first moment the note is struck. When an electric motor is started up, it eventually reaches a steady state of operation. But to get there, it starts from zero RPM and passes through an infinite number of speeds until it attains the operating speed. Every time you rev the 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 go across it. Very complex vibration patters can occur. Because things are constantly changing, engineers must know what the frequencies and stresses are at all moments in time. Sometimes transient vibrations are extremely violent and short-lived. Imagine a torpedo striking the side of a ship and exploding, or a car 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. In real life, shock is rarely a good thing and almost always unplanned. But shocks occur anyhow. Because of vibration, shock is always more devastating than if the same force were applied gradually.

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Computer Aided Structural Analysis 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 to the vibrations of sound waves. Many times things are made to vibrate for a purpose. For example, a special shaking device is used in foundries to loosen a mold placed in sand. Or, in the food and bulk materials industries, conveyors frequently work by vibration. Usually, however, vibration is bad and frequently unavoidable. It may cause gradual weakening of structures and the deterioration of metals (fatigue) in cars and airplanes. Rotating machines from small electric motors 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 cannot be avoided. For example, tyres and shock absorbers (dampers) help reduce vibration in automobiles. Similarly, flexible couplings help isolate vibrations produced by the engines. Vibration is about frequencies. By its very nature, vibration involves repetitive motion. Each occurrence of a complete motion sequence is called a "cycle." Frequency is defined as so many cycles in a given time period. "Cycles per seconds” or "Hertz”. Individual parts have what engineers call "natural" frequencies. For example, a violin 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 frequency in which the entire string is going back and forth in a simple bow shape. Harmonics and overtones occur because individual sections of the string can vibrate independently within the larger vibration. These various shapes are called "modes". The base frequency is said to vibrate in the first mode, and so on up the ladder. Each mode shape will have an associated frequency. Higher mode 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. This event 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 marching feet align with a natural frequency of the bridge, it could fall down. Engineers must design so that resonance does not occur during regular operation of machines. This is a major purpose of Modal Analysis. Ideally, the first mode has 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 may pass through a resonant frequency. Other vibration processes such as Time History, Response Spectrum, Random Vibration, etc. are used in addition to - 14 -

Computer Aided Structural Analysis

Modal Analysis to deal with this type of more complex situation. These are called Transient Natural Frequency Processors. Buckling analysis

If you press down on an empty soft drink can with your hand, not much will seem to happen. If you put the can on the floor and gradually increase the force by 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 be defined 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 matrix becomes 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 the geometry 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 that buckling usually occurs at relatively low stress values for what the material can withstand. 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 direction buckle under relatively small axial loads. Such structures may fail in buckling while their stresses are far below critical levels. For such structures, the buckling load becomes a critical design factor. Stocky structures, on the other hand, require large loads to buckle, therefore buckling analysis is usually not required. Buckling almost always involves compression. In civil engineering, buckling is to be avoided when designing support columns, load bearing walls and sections of 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 just one local spot of a flange should buckle! In mechanical engineering, designs involving thin parts in flexible structures like airplanes and automobiles are susceptible to buckling. Even though stress can be very low, buckling of local areas can cause the whole structure to collapse by a rapid series of ‘propagating buckling’.

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Computer Aided Structural Analysis

Buckling analysis calculates the smallest (critical) loading required buckling a model. Buckling loads are associated with buckling modes. Designers are usually interested in the lowest mode because it is associated with the lowest critical load. When buckling is the critical design factor, calculating multiple buckling modes helps in locating the weak areas of the model. This may prevent the occurrence of lower buckling modes by simple modifications. Thermal analysis

There are three mechanisms of heat transfer. These mechanisms are Conduction, Convection and Radiation. Thermal analysis calculates the temperature distribution in a body due to some or all of these mechanisms. In all three mechanisms, heat flows from a higher-temperature medium to a lowertemperature one. Heat transfer by conduction and convection requires the presence of an intervening medium while heat transfer by radiation does not. I include a brief discussion on thermal analysis here. You must have read all these in high school. In this book, I shall not discuss anything more about thermal 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 will eventually cool down to the room-temperature mainly by conduction from the coffee 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 conduction through a plane layer of thickness X is proportional to the heat transfer area and the temperature gradient, and inversely proportional to the thickness of the layer. Rate of Heat Conduction = (K) (Area) (Difference in Temperature / Thickness) Convection Convection is the heat transfer mode in which heat transfers between a solid face and an adjacent moving fluid (liquid or gas). Convection involves the combined effects of conduction and the moving fluid. The fluid particles act as carriers of thermal energy. Radiation

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Computer Aided Structural Analysis

Thermal radiation is the thermal energy emitted by bodies in the form of electromagnetic waves because of their temperature. All bodies with temperatures above the absolute zero emit thermal energy. Because electromagnetic waves travel in vacuum, no medium is necessary for radiation to take place. The thermal energy of the sun reaches earth by radiation. Because electromagnetic waves travel at the speed of light, radiation is the fastest heat transfer mechanism. Generally, heat transfer by radiation becomes significant only at high temperatures. Types of Heat Transfer Analysis There are two modes of heat transfer analysis based on whether or not we are interested in the time domain. Steady State Thermal Analysis In this type of analysis, we are only interested in the thermal conditions of the body when it reaches thermal equilibrium, but we are not interested in the time it takes to reach this status. The temperature of each point in the model will remain unchanged until a change occurs in the system. At equilibrium, the thermal energy entering the system is equal to the thermal energy leaving it. Generally, the only material property that is needed for steady state analysis is the thermal conductivity. Transient Thermal Analysis In this type of analysis, we are interested in knowing the thermal status of the model at different instances of time. A thermos designer, for example, knows that the temperature of the fluid inside will eventually be equal to the roomtemperature (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 also need 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 to fixed points in space. Typically this fixity is set with global boundary conditions, in which the fixity is totally rigid. A boundary element, on the other hand, allows for a flexible connection to the fixed space. Boundary elements and boundary points are normally used to simulate the constraints that actually occur when an object is used in the real world. For example, if a coffee cup is - 17 -

Computer Aided Structural Analysis

sitting on the table and a weight is placed on top of the coffee cup, then the table is the boundary. Boundary points would be points on the plane of the table that are defined as being fixed in space and to which nodes of a finite element model of 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 FEA software, boundary elements have an additional capability of imposing and enforced displacement upon a model. The force created by this imposed displacement would be calculated automatically. Additionally, the forces generated at a boundary by forces on the model can be obtained as output using boundary elements. There are another very powerful types of analysis offered by high-end FEA programs, known as Mechanical Event Simulation or Virtual Prototyping. You will find this in section 36.

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Computer Aided Structural Analysis

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 taken as positive and force acting upwards is considered positive. Anti-clockwise moment is taken as positive. This has been shown in following figure for 2D plane. Fy All positive Mz Fx Figure 4-1

Most standard analysis programs follow this sign convention. Although you can use any convention of your own, but I strongly advise you against that. You will always be fine with this convention. Please note that, because of taking y positive 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 somewhat complicate. This is shown below. Y My Fy

All positive Mx X

Fz Z

Fx Mz Figure 4-2

When you see bending moment diagrams, remember that some programs draw them 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 and 4-2), 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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Computer Aided Structural Analysis

When dealing with 3D structures, the program will generally consider y-axis as elevation. This is as expected, because when dealing with 2D structures, you will normally use x-y plane. But it has exceptions as well. Some programs, by default use x-z plane for 2D analysis. Of course you can direct every analysis program to consider z-axis (or even x-axis) as elevation. My main point here to make you understand that co-ordinate system is very flexible. But you must follow same sign convention throughout. Different programs may follow slight different sign conventions. Before using the program, you should be familiar with that program’s sign convention. Solve some basic problems with them first and consult the user guide. As an example, the following figures show how SAP90/2000 describes frame member internal forces.

AXIS 2

T

P

AXIS 1

T AXIS 3 P

Positive Axial Force and Torque Figure 4-3

Compression face V2 Tension face M2

Compression face

V3 Tension face

M3

Positive Moment and Shear [1-2 plane] Positive Moment and Shear [1-3 plane] Figure 4-4 - 20 -

Computer Aided Structural Analysis

Now please solve the following problems using your program and check the result with the answer given. -3 kN/m (case 2) 5 kN (case 1) Node 4

5m

Node 3

6m

6m All members are of 250-mm side square Cross 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 body diagram of each member. Now check the result and the sign with your analysis program. Please note that your program may draw the bending moment diagram on opposite side compared to what shown here! Observe the sign convention. - 21 -

Computer Aided Structural Analysis

Now solve the following truss.

3m

0.208 2.833

All members are made of 3 kN steel (E=200 GPa) with 100-mm side square section -3.642 2.0 2.833 -2 kN

4m

4m Figure 4-7

The axial forces are shown as italics in the above figure. Note that the left end is hinged and right end is roller. It is interesting to know that with some programs, you may need to “tell” the program that the structure is a ‘truss’ by specifying ‘moment releases’ in the truss members. Otherwise, you may wonder why the program result shows bending moment diagram in truss! Different programs have different options for specifying moment (or axial force, torsion etc.) releases. Some programs, which allow you to draw plate elements on screen, you should draw them in counter clockwise fashion. Otherwise you may get awkward result.

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Computer Aided Structural Analysis

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 and member numbers automatically. This default scheme may not always be convenient for you, especially if you are analyzing a multi-story building. Fortunately, most programs offer re-labeling option and you can even use alphanumeric labels. Since it is impractical to re-number hundreds of members manually, you should do it automatically. Generally, beams, columns and slabs are numbered on the story or floor level they reside. In that case, you can direct the program to use X-Z-Y re-labeling pattern (assuming Y-axis is the elevation). You may number all beams in the B5-10 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 other structural members. You can also create ‘group’ for same type of members whose 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 while solving and again display the result in user specified numbering. Wondering what is ‘bandwidth minimization’? It is a technique for assembling global 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. Lower bandwidth means less time necessary for solving equations. For example, in a multistory frame (assuming the height is more than the length); if you number nodes 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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Computer Aided Structural Analysis

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 get help 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 mm oriented as shown in plan. Beams can be specified as 200x300 mm without any problem. But for columns, you have to be careful. Generally, the programs will ask 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 figure 6-2. If you specify the dimension in opposite manner, then you will get wrongly oriented section for the columns. The above figure is taken from real time view of SAP2000. If your program does not offer real time view (i.e. the members should look like in the real structure in 3 dimension) option, you’re out of luck! Many programs, however, have the option for specifying sectional dimension using ‘tx’ and ‘ty’ (or it might be ‘ty’ and ‘tz’ or ‘t2’ and ‘t3’) option. I have tried with various programs this sectional dimension input. In most cases width = 200 and height (or depth) = 400 worked.

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Computer Aided Structural Analysis

Z

Y

X

Figure 6-2

Sometimes you may need to specify inertia directly especially for irregular shaped sections. Normally the programs offer only ‘Iz’ and ‘Iy’ options. The most often used is the ‘Iz’. For the beam discussed above, Iz = 200x3003/12 and Iy = 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 ‘global axes’. 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 your program’s resource files to see how it handles display of local member axes. - 25 -

Computer Aided Structural Analysis

Y Z X Y

X Z Figure 6-4

The above figure shows orientation of local axes for the inclined member. Note that in your program, the orientation for local axes may be slightly different; for example, 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 defining section properties in terms of local axes, even an ‘inclined’ member is considered as ‘straight’. We shall come to local and global axes story again when we discuss interpretation of analysis output.

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Computer Aided Structural Analysis

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 beam without creating a node at that point (the program itself creates a node there internally) where as most programs require that you can assign concentrated loads only at nodes. So, you may need to ‘split’ the member to create intermediate nodes. If your concentrated load is inclined, you better resolve it into horizontal and vertical components yourself and then apply them. Specifying UDL is easy. However, trouble arises, then the load becomes varying. The most common example of varying load is on the beams coming from slabs as shown in figure 7-1. The lengths of the beams are ‘L’. Unfortunately, very few programs will calculate distributed loads form slabs automatically. 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 into three segments (not necessarily equal) and apply triangular loads at end segments 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! But you can avoid trapezoidal loads all together with slight loss of accuracy as shown in figure 7-2. If we equate fixed end moments in two beams (of figure 71 and 7-2), we get 1 – 2α2 + α3 wL2 -------------- WL = ------ or 12 x (1-α) 12

(1 – 2α2 + α3) x W w = ----------------------(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

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… (7.1)

Computer Aided Structural Analysis

α r

0 1

0.2 1.02

0.375 1.05

0.5 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 UDL whose 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 load through equation. The only way out is, split your member into several sections, and specify concentrated loads varying through nodes (or UDL varying through segments). More number of divisions, better the result is. You may wonder whether you can model all the slabs in your building frame using plate elements instead of converting loads to beams as shown in figure 71. Of course you could. But there are several disadvantages! First of all, your analysis program must have ‘plate’ element to do this. Many frame analysis programs don’t have plate element! If you use ‘plates’, then you must ‘mesh’ it before running analysis. If you have, say 100 slabs (i.e. plates) with 10x10 mesh, you’ll have 11x11x100 = 12,100 extra nodes compared to that you’ll have if you transfer the loads on beams. Not only this takes much more time to have your analysis done, but also it will swamp you with tons of output (just count the number of total plate elements – their stress values etc.)! It has been proved 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 earth pressure. If your analysis program has easy method to specify such type of loads, consider yourself really lucky! If you’re applying hydrostatic load on a plate element, apply load before meshing the plate. Sometimes the program allows you to specify separate load at four nodes of the plates (and intermediate values are interpolated) though this is not really necessary in day to day analysis. Hydrostatic load normally takes the shape as of figure 7-4.

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Computer Aided Structural Analysis

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 from plate)? Surprisingly, changing the density into negative works! (Argh!) (I don’t know whether all programs behave in this way, but I found this trick works in Visual 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 fluid level at higher level. Water level

Figure 7-5

Figure 7-6

Figure 7-6 shows another trick where you need to superpose two types of loads to get the desired resultant load distribution. Uniform pressure on plates can easily be applied. While you analyze water tanks, these tricks come handy. When you’re applying distributed load on inclined member, it may act in two different manners as shown in figures 7-7 and 7-8.

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Computer Aided Structural Analysis

Figure 7-7

Figure 7-8

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 analysis programs can handle both types of loading conditions shown. But it’s your responsibility to apply correct method.

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Computer Aided Structural Analysis

8. Column Buckling test Solve the following problem: A steel (E=200 GPa) column of 100x100 mm square cross section (I = 8.33E-6 m4) is 5 m long and fixed at bottom end. A load 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, it won’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 will show you a message, which will say that the program has encountered a negative diagonal term in member stiffness matrix and analysis will terminate. Note this load. This is the minimum buckling load. You are likely to see that even when the column buckles, the deflected shape of the column is drawn straight! 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 you instruct it to perform P-∆ analysis, it performs iteration to find out actual axial forces. Remember if you are using a very cheap program or some noncommercial 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 performing such analysis. I once analyzed a 20 storied 3D frame in VA, which had 10 bays in 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 - 31 -

Computer Aided Structural Analysis

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

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Computer Aided Structural Analysis

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 grossly inaccurate. Just analyze any frame subjected to lateral loading by these methods and then compare the results with exact analysis by computer. You will find as much as 50% to 60% difference of moments and shear forces. If computer is available, you must not use these methods. Even for preliminary analysis, when you do not know the size of the members in the structure, still these methods are not useful. You can do the same easily by using computer.

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Computer Aided Structural Analysis

10.

Deflection of Reinforced Concrete member

Consider a simply supported beam made of reinforced concrete. It is loaded by uniformly 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 gross moment of inertia of the section if applied moment (wL2/8 in this example) exceeds cracking moment capacity. Here w stands for dead + live load. Most computer programs do not take into account the reduced moment of inertia because of cracking. Since sometimes Ie comes equal to 50% of Ig, when you do not calculate Ie, you may just double the deflection as found from computer analysis which takes Ig. Please note that, for all members you may not need to use Ie because for all members calculated moments may not exceed cracking moments. Once you have got Ie, you can use the same analysis program to find out 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. You can achieve this by splitting the beam into two members. Most analysis programs have the option of doing this. 2. Changing I values of some members does not alter moment and shear values which you have got previously using Ig. 3. Ie can be calculated only when you have designed the member i.e. you have specified number and diameter of reinforcement bars. 4. When you are specifying I value explicitly, ensure that you do not define beam 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)hf3/12 + (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 - 34 -

Computer Aided Structural Analysis

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√fck MPa if fck (MPa) is measured as cube compressive strength of concrete; and Ec = 4700√fck MPa if fck 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-term deflection (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 2 way members e.g. slab, things are not as easy. We shall discuss later how to find the deflection of 2-way slab by using finite element analysis. Most codes provide you minimum depth of members if you do not calculate deflection. But these values are always overestimated and thus lead to uneconomical design for multistory buildings. Don’t be lazy. Always calculate deflection, you can save money!

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Computer Aided Structural Analysis

11.

Shear deformation

Most programs do not take into account deformation due to shear force. For normal beams where depth of beams are much less than their lengths, neglecting shear deformation does not lead to erroneous result but where length of beam is very close to depth of beam it can lead to large error. In fact if (Length of beam/Depth of beam) < 2 then the beam is termed as deep beam. There, shear deformation must be taken into account. Consider the following problem. A point load of 1 MN is applied at the free end of 1-m long steel cantilever beam. The cross section of the beam is 400x600 mm. The total deflection is ∆ = PL3/3EI + 6PL/5GA (the equation comes from strain energy theorem) = 0.000234 + 0.00007143 = 0.0003029 m. See what deflection your program shows! Chances are that it will show only 0.000234 m. So, what do you learn? Now make the beam section 600x400 mm and you will find that the total deflection is very near to bending deflection. Some programs offer option for specifying shear area. In that case, they can take into effect of shear deformation. Check whether your program has this option.

Figure 11-1

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Computer Aided Structural Analysis

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 to rotate the local axes of the selected member; then you will specify the joint restraints 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 you can specify a spring at any angle. The spring reaction is the resultant of X and Y components of reaction. In case of roller support you will get the reaction automatically from spring reaction.

Think what you have learnt… How many of following analysis methods you have learnt in the university? – Moment distribution, Slope deflection, Portal, Cantilever, Kani’s rotation contribution, Conjugate beam, Graphical – Funicular polygon & Maxwell diagram – Williot-Mohr diagram, Three moments theorem, 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 started using computer analysis programs? Probably none! Academic people will argue that all the said methods are to be mastered for a better understanding of structural response. Do you think so? I don’t. Well, among the methods listed above, the moment distribution is most popular. This is quite logical, because this method is easy, does not involve solution of simultaneous equations and converges rapidly. We shall discuss substitute frame method later (see section 13) while considering maximum bending moment, shear force etc. in building frames. Did you notice that all these methods are used for frame analysis only? You may like to know that 80% to 90% of all real world structures analyzed are frame structures. Although you have learnt flexibility and stiffness approach while studying computer method of analysis, only stiffness method is used in computer programs. Modern world’s most powerful analysis method – finite element method is also a stiffness method in essence.

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Computer Aided Structural Analysis

13. Maximum bending moment, shear force and reaction in building 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 classical approximate substitute frame with computer analysis. But before that note the following live load distribution criteria. To get this Maximum positive bending moment at center of span Maximum positive bending moment at center of span Maximum negative bending moment at support Maximum column reaction

Do this Load that span and then alternates spans Load adjacent spans and then alternate spans Load adjacent spans and then alternate spans Load adjacent spans and then alternate spans Maximum positive bending moment at Load all spans except adjacent spans support 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 load intensity, then you can load all spans together with dead and live load without any combination. But if you have computer, it is always better to perform the actual combination to get maximum values of force and moment. In classical substitute frame (see figure 13-1), we isolate one single floor with the assumption of columns at top and bottom floors are fixed. Then we apply the combinations described above to get maximum member forces. In case of computer analysis, though you still need to apply the live load in same combination as discussed above, yet you need not isolate one particular floor. Rather, you should just apply the required span load combination in any floor. In case of regular shaped building elevation, result obtained from one floor will be same for other floors. For example, in the (figure 13-2) shown, the load combination stands for maximum negative support moment in first interior column (actually both interior columns since this structure is symmetric) in 2nd floor (bottom most floor, i.e. ground floor is normally denoted by “0” in structural analysis convention). The value obtained for support moment under this condition will also be the maximum support moment for 1st, 3rd and 4th

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Computer Aided Structural Analysis

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

4 3 2 1

Figure 13-2

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Computer Aided Structural Analysis

14.

Support settlement

Take any statically determinate structure, for example a simply supported beam or a simple truss. Apply a settlement in one of its supports. Now analyze the structure. You will see an interesting phenomenon. Though the program will correctly say zero member force, still it will draw a bending moment or axial force diagram! In a statically determinate structure, there should not be any member force developed due to support settlement. Hooray, you have discovered a bug in the program! The reason of this awkward shape can be explained. Although the member force is 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 one of its support settle to an amount and perform the analysis. You should find some member forces in the beam. Well, now take a statically indeterminate truss. The truss should be externally indeterminate. For example, you can take a 2-support truss whose both supports are hinged (pinned) as shown in figure 141. Now apply a downward settlement in any one of its supports and analyze the structure. 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 the support displacement perpendicular to member’s local axis. That’s why you get the wrong answer. But since the members’ length change, there are strains, which would create the axial forces. If you want to know the actual member forces after such support settlement, you need to modify the member stiffness matrix considering the displacement in perpendicular direction as well. Unfortunately, you can’t do it with most available programs.

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Computer Aided Structural Analysis

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 as shown in figure 15-1.

Figure 15-1

You can easily analyze just one plane frame as shown in fig. 15-2. Whenever possible, try to convert 3D structures into 2D in this way. 2D structures are not only easier to model, but also they can be ‘handled’ and analyzed much more easily compared to 3D structures.

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Computer Aided Structural Analysis

Figure 15-2

Look, in fig. 15-1, the loads are towards X direction. If there were additional loads of same type towards Z direction, you could adopt similar 2D frame (on YZ plane) as shown in fig. 15-2. You can then superpose the result as long as it is 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 3D into 2D? I shall discuss this when covering dynamic analysis in detail.

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Computer Aided Structural Analysis

16.

Curved member

Most frame analysis programs do not have curve element. You will need to replace the curved member by a number of straight members. Obviously, more the number of straight members used better the accuracy is. While drawing straight 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 is to figuring the straight members’ nodal co-ordinates in spreadsheet (for example, 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-in commands, you can easily find out y co-ordinates of the curve against each x co-ordinate. Some programs can “copy” and “paste” member and nodal information to and from spreadsheet file. You may like to know that it is theoretically possible to create stiffness matrix of a curved member. Now solve the following two hinged parabolic arch. -50 kN/m (downward) Y 4m

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-line segments, you should get exact answer within 1% accuracy. Note that the theoretical answer has been obtained by (H = ∫ My dx / ∫ y2 dx) formula.

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Computer Aided Structural Analysis

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 this book! To specify a tapered section by yourself, you should ‘break’ the members into a number of parts (more the number, better is the result). Then you should specify various A (areas) and I (inertia) for each segment. This will become clear from the following problem. 1.5

-5000

2

1.5 4

15

15 Figure 17-1

The figure shows a tapered beam. Hinged at left end and fixed at right end. A clockwise moment (hence minus sign) of 5000 has been applied at middle of the beam. 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. Width of the sections is same throughout. But the depth is varied as (from left most section) 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 to theoretical answer of 170. The bending moment at just left of mid-point of beam is – 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 specify any unit or E value of material! You should get same answer whatever unit you use. Although some programs do allow you to specify “linearly” tapered members; you still need to apply this trick for “non-linearly” (e.g. cubic or parabolic) tapered members.

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Computer Aided Structural Analysis

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 by a 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 this member i.e. this spring replacement member should carry axial load only. After analysis, 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

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Computer Aided Structural Analysis

19. Sub-structuring technique and symmetry (break them into pieces…) 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 the results. When an isolated part of a structure is analyzed, it is crucial that the boundary conditions ‘sub-structures’ accurately represent the conditions in the actual structure. As a first simple example, consider the following structure as shown in figure 19-1. You are required to analyze the structure. A = 2002 mm² E = Steel -5 kN/m

10 kN 4m 20 kN 4m 4m

4m

4m

Figure 19-1

If you separate the upper floor and then analyze only that portion, you will get the result as shown below. 10 kN

11.44 kNm

11.44 kNm 5 kN

4.28 kN

5 kN 4.28 kN

Figure 19-2

With the result shown above, the applied loads on the bottom floor of the actual structure will be as shown below. - 46 -

Computer Aided Structural Analysis

11.42 kNm 25 kN 4.28 kN

11.42 kNm

-5 kN/m

5 kN

A

4.28 kN

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 lower floor should be same as that of obtained if you considered the whole structure as shown in figure 19-1. For your check, the ultimate results are as given below for problem figure 19-1. Node A B C D

Fx kN -6.223 -5.99 -11.2 -6.583

Fy kN -14.85 30.54 20.64 3.675

Mz kNm 15.58 15.18 22.08 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, but this 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, this sub-structuring technique is a great time saver. When the structure has one or more planes of symmetry, it is possible to perform the analysis on one-half, onequarter or an even smaller part of the structure, provided that the appropriate boundary conditions are applied at the nodes of the plane(s) of symmetry. Followings are some examples of exploiting symmetry of structures.

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Computer Aided Structural Analysis

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 sign convention is same as described in section 4 earlier. Now consider plane frames with 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 frame with odd number of spans has been shown in figure 19-10. Here you will have to apply boundary condition of X translation and Z rotation prevented in mid points of the middle beams as shown in fig. 19-11. Symmetrical structures are not only easier to analyze but also perform better than unsymmetrical structures in real life!

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Computer Aided Structural Analysis

Figure 19-8

Fixed

Fixed

Fixed

Figure 19-9

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Computer Aided Structural Analysis

Figure 19-10

X translation & Z rotation fixed X translation & Z rotation fixed

X translation & Z rotation fixed

Figure 19-11

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

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Computer Aided Structural Analysis

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 this assumption 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 (including self-weight). The second figure is the approximation of the same staircase as simple beam. The section of concrete staircase may be taken as 1-m width x 150-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. In approximate 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. In this case we have considered the staircase as simply supported. Depending on casting, it may be fixed-fixed or fixed-pinned as the case may be. In fact staircases are more often analyzed as fixed-fixed support condition. From the analysis it is found that maximum mid span moment is almost same in both analyses. Shear forces (reactions) are also more or less equal. This proves that approximate 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. - 51 -

Computer Aided Structural Analysis

Figure 20-2

This analysis was done in Visual Analysis 3.5. An interesting point is to note that, both structures were analyzed as a single file. This is applicable to most analysis programs. You may analyze as many as separate structures in a single file even they are not connected together. Now think about the following paradox. Following three beams are all simply supported (left end pinned and right 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 (say 10 kN/m). Find out what will be the bending moment at mid spans.

10 m

10 m

10 m Figure 20-3 - 52 -

Computer Aided Structural Analysis

What result do you see? The bending moment (and reactions as well) is same in all 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 same irrespective 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 simply supported in both ends. That’s why you got same bending moment! Had they been fixed at ends, the results would not have matched. However, they still would 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 reactions at supports are zero (since we have only loading acting downward). So, moment due to ‘eccentricity’ of geometry is also equal to zero. This will be from following 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 the bending 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 are non-zero. So, the internal forces differ depending on geometry of the beam. When you analyzed two hinged arches as a student you probably used the equation: Arch moment = Beam moment – He. Didn’t you?

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Computer Aided Structural Analysis

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 truss member (which carries axial force only) but additionally you will have to specify that it can take tension only (no compression). Some analysis programs may not have the option of defining a tension only member! Once you have specified cables in this way, the analyses are pretty straightforward. While viewing the result, you should check whether cables’ axial force diagram shows tension only (generally positive number) and no bending moments. That’s all. An example of cable structure is shown in fig 21-1. After performing the analysis, check your answer with exact result as given.

- 54 -

Computer Aided Structural Analysis

Figure 21-1

- 55 -

Computer Aided Structural Analysis

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 = 84 kNm.

Figure 21-2

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

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Computer Aided Structural Analysis

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 yA

θA

θB

yB

PyB PθB

PθA

c P

P Upward UDL

w = 8Pc/L²

c L/2

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

L/2

Actual pre-stressed cable

Equivalent load Figure 22-1

PyB PθB

PyA yA

θA

θB

yB P

PθA 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 to comprehend the following worked out problem. Please note that the θ values are in radians. Note that the yA and yB indicate eccentricity of the cable at supports in upward direction from center of gravity of concrete (cgc) line. Upward distance is positive at supports and downward distance is positive at mid spans for pre-stressed cable profile (majority of standard analysis programs follow this sign convention). If the cable distances are of opposite sense compared to what shown in above figures, then ‘arrows’ of moments will be reversed.

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Computer Aided Structural Analysis

With reference to the figure 22-3, the calculation is shown 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 end and 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 rad And c = (0.5 + 0.4)/2 + 0.6 = 1.05 m So, 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 is not shown).

Figure 22-3

250 kNm 250 kN

250 kNm 150 kN

250 kN

200 kNm 18.67 kN/m

250 kN

Figure 22-4

Now the forces shown in blue color will go to support directly. Moments shown in orange color will cancel each other. So, the remaining forces that will act are - 58 -

Computer Aided Structural Analysis

shown in green color. The ultimate equivalent load will be that of as shown in figure 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 shown above. 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 results as 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 concrete in addition to the bending stress due to pre-stress, dead load and live load. For more information on this subject, please see any standard textbook on prestressed concrete. I have shown here only linear and parabolic cable profile. Although parabolic profile is the most common, there are other types of profiles possible. See your textbook for details.

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Computer Aided Structural Analysis

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 virtually all kinds of structures that human brain ever can imagine! If you have studied finite element before, you may skip this section. Those who did not, I present a very very brief introduction of the subject. There exist more than 1001 books in this subject. But I warn you; the theory of finite element analysis is very complex! What is meant by finite element? The answer is any element, which is not infinite. 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 a model car or house by “Lego” parts? Now consider each part of mechano as “finite element”. A number of mechano elements were needed to build your model 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 in essence where each finite element is a “bar”. Y

X Figure 23-1

This is the longitudinal section of a beam shown. That is, you are viewing a beam from its length side. Observe that here we consider the beam as 2dimensional “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 all previous cases, we treat all beams as “bars” like a “stick”. But in the above figure 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” in the beam. The beam is simply supported, left end pinned, right end roller. It is - 60 -

Computer Aided Structural Analysis

loaded by a uniformly distributed load. We like to find out the stresses at various points of the beam. Please note that analysis of this problem by classical method is close to impossible. So, first we divide the beam into a number of “triangular finite elements”. Then we shall determine the member stiffness matrix [k] of each individual triangular element and ultimately we shall have to combine the member stiffness matrices into “global stiffness matrix” [K]; pretty much the way we did in case of frame analysis. 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 must be converted to appropriate nodal loads by applicable equations. So, our problem can be represented by familiar equation [P] = [K][D]. From this equation we can solve for [D] and then we can find out nodal stresses form equation [σ σ] = [C][εε] where [C] matrix differs in various cases like plane stress, plane strain etc. We are describing this problem as plane stress because we considered 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. Y

X Figure 23-3

If you analyze the beam with both triangular and rectangular elements as shown above, you will see that you get accurate answer when you use rectangular elements. It proves one very fundamental concept of finite element analysis: You must choose proper element for particular problem. You do get correct result with triangular finite element but you must use very fine mesh compared to rectangular element. In general, triangular element is not a good choice. If you are interested to know why triangular element behaves in such way, you should consult any standard finite element analysis textbook. - 61 -

Computer Aided Structural Analysis

Figure 23-4

As a crude rule, when you use triangular you will normally need much finer “mesh” than rectangular elements. The assembly of elements in finite element analysis is called “mesh”. Most powerful finite element programs can generate mesh automatically if you specify the boundary surfaces of the models. If you want to analyze the same beam in 3D then your model will look like as shown in figure 23-5.

Figure 23-5

In this case finite element will be 3D solid element like shown in figure 23-6.

Figure 23-6

This is an 8 noded finite element because it has 8 nodal points. If its each vertex has one additional point in the middle, then it would have been 20 noded finite element. Higher is the number of nodal points in an element better is the accuracy of the solution. But higher noded elements are difficult to calculate even with a computer since total number of nodes increases the size of global stiffness matrix. Whatever element you use, it must be compatible. Compatibility means there must not be any discontinuity or overlapping among the elements when the analysis model deforms under applied load. You can combine more than one kind of element in single structure. You should use more number of elements where you anticipate stress variation is more irregular. There are a lot more other finite elements in addition to basic triangular and rectangular elements discussed above. - 62 -

Computer Aided Structural Analysis

One distinguishing feature of finite element method is that it does not provide “closed form” solution. Every problem in finite element analysis is unique. This probably 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 input and 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 you change the length of the beam, it becomes another new problem because the geometry of the model changes. Of course you can change E or w values or boundary condition without remodeling the whole structure. Another aspect of Finite element analysis is that it almost always produces an approximate result. I used the word “almost” because finite element analysis does produce exact result only when the finite element is “bar” that is in “frame structures”. 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! A real question indeed! You can realize it yourself. Just think of solving a simple beam in finite element method (this is presented just after this section). After you solve this beam by finite element method, you can easily check whether the result 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 must accept the finite element analysis result as exact result! That’s why it is so important that finite element analysis models must be created to simulate the actual structure as much as possible. You must use proper combination of finite elements, sufficiently accurate mesh, proper load and applicable boundary conditions. It is often a common practice to analyzing the structure first with a particular mesh and then repeating the whole analysis after doubling the mesh to see whether the result converges. But this method has drawbacks! Your program cannot analyze the structure if your number of mesh nodal points exceed the program’s capacity. Moreover, it is very difficult to predict beforehand what particular “finite element” will best simulate the structure. This is especially a demanding task for very complex structures. Finite element method is nowadays widely used in all branches of aerospace engineering, bio-medical engineering, mechanical engineering and structural engineering etc. Some manufacturing companies spend millions of dollars every year in finite element analysis! I am concluding finite element introduction here. But you must realize that it is not so easy as it seems. Researchers are still developing new finite elements. - 63 -

Computer Aided Structural Analysis

Sometimes even the most expensive finite element analysis programs produce wrong answer to complex problems. If you feel inclined to know more about this wonderful (?) tool of analysis, I strongly recommend that you to go through some standard finite element method textbooks. One word of advice, many engineers tempt to use finite element analysis everywhere even when it is possible to analyze the particular structure using classical method of analysis. My main aim is to make you realize that finite element analysis is required only when it is absolutely necessary. Remember that finite element analysis programs are very expensive and they also demand great 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’s graphical interface or importing it from a CAD program). 2. Creating the mesh (most programs can automatically generate mesh for best result). 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 uniformly distribute load analyzed by FEA method. This example is for your understanding of the basic concept of FEA only. In practice, this problem should be solved by simple flexure formula of σ = My/I. Remember this!

- 64 -

Computer Aided Structural Analysis

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 and right end is roller. Find deflection at mid point and maximum bending stress in the 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 Y direction. You can convert the uniform load into nodal loads by applying 0.25 kN at extreme nodes and 0.5 kN at intermediate nodes. 2. Solid model analysis. Use standard solid brick or tetrahedral element. Most finite element analysis programs offer these elements.

For 2D analysis, after modeling your structure should look similar to this figure.

Figure 23-7

In case of plane stress model formulation, you should use plate finite element whose thickness will be equal to the depth in Z direction. In this problem, this is equal to width of the beam. After performing the finite element analysis, you should get the answer: mid point deflection 1.95x10-4 m maximum stress 3.75 MPa. Your program may display slight different result due to numerical round off in calculation. The deflected shape should resemble the following figure. Original shape is shown by dotted line. This 2D-beam analysis was performed in Visual Analysis.

- 65 -

Computer Aided Structural Analysis

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 finite element analysis.

Figure 23-9

- 66 -

Computer Aided Structural Analysis

To analyze the beam as 2D, you should not face any difficulty. However, you should take into account many other things when you analyze the same beam as 3D 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 upper surface. Be careful about the load’s direction. Now, comes the main hurdle, the meshing. If you are using a high-end FEA program, it will mesh the model itself. By default, the program will mesh it by using brick elements or tetrahedral elements. The next figure shows the beam with automatically generated tetrahedral mesh. You may note that, such high density meshing is not really required for this very problem. If you manually mesh with 20 numbers (2 elements along depth and 10 elements along length, similar to shown in fig. 23-7) 8-noded solid elements, (as in SAP2000) you will get exact result for this problem. Left end boundary condition is X, Y, Z translation fixed along 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. - 67 -

Computer Aided Structural Analysis

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

- 68 -

Computer Aided Structural Analysis

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 rescue you. Unfortunately, most FEA textbooks do not contain sufficient numerical examples to make the whole thing transparent to the readers. If you study the following numerical example along with your FEA textbook, you might find it easier 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 is taken 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 simple numerical 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 divided into a larger number of elements (triangular or rectangular etc.). Anyway, we now begin solving the problem. I hope you’ve already familiar with ‘shape function’. This is the function, which describes displacement of any point 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].

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Computer Aided Structural Analysis

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 triangular element. We consider element 1 at first. Its nodal points are 1,2 and 3. The co-ordinates are (0,0), (4,0) and (4,4) respectively. If the area of the triangle is A, then we know 1 x1 1 A = 1 x2 2 1 x3

y1 1 0 0 1 y 2 = 1 4 0 = 8...m 2 2 y3 1 4 4

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 ( x 2. y3 − x3. y 2) + x( y 2 − y3) + y ( x3 − x 2) = 1 − 0.25 x 2A ( x3. y1 − x1. y3) + x( y3 − y1) + y ( x1 − x3) N2 = = 0.25 x − 0.25 y 2A ( x1. y 2 − x 2. y1) + x( y1 − y 2) + y ( x 2 − x1) N3 = = 0.25 y 2A N1 =

Now, the ‘strain displacement’ matrix is ∂N 1 ∂x [ B] = ∂[ N ] =

0 ∂N 1 ∂y

0 ∂N 1 ∂y ∂N 1 ∂x

∂N 2 ∂x 0 ∂N 2 ∂y

0 ∂N 2 ∂y ∂N 2 ∂x

∂N 3 ∂x 0 ∂N 3 ∂y

0 ∂N 3 ∂y ∂N 3 ∂x

− 0.25 0 0.25 0 0 0 = 0 0 0 − 0.25 0 0.25 ......m −1 0 − 0.25 − 0.25 0.25 0.25 0

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

- 70 -

Computer Aided Structural Analysis

1 ν 0 200 0 0 E [C ] = 0 = 0 200 0 .10 6...N / m 2 ν 1 2 1 −ν 1 −ν 0 0 100 0 0

ν

We have considered E = 200x106 N/m² and Poisson’s ratio as 0. The stiffness matrix of triangular element is given by [k ] = ∫ [ B]T [C ][ B]dV = [ B]T [C ][ B] At......( for... plane...stress ) V

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

[k ]1 =

1

0

−1

0

0 .5

0 .5

− 0 .5 − 0 .5

0

−1

0 .5

1 .5

− 0 .5 − 0 .5

0

0

0

0

0

− 0 .5 − 0 .5

1 .5

0 .5

−1

0

− 0 .5 − 0 .5

0 .5

0 .5

0

−1

0

1

0

0

0

.10 6......N / m

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

[k ] 2 =

− 0 .5 − 0 .5

0 .5

0

0

0

1

0

0

0

−1

0

0

1

0

−1

0

− .0 .5

0

0

0 .5

0 .5

− 0 .5

− 0 .5

0

−1

0 .5

1 .5

− 0 .5

0 .5

−1

0

− 0 .5 − 0 .5

0 .5

.10 6......N / m

1 .5

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,

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Computer Aided Structural Analysis

[K ] =

1 .5

0

−1

0

1 .5

0 .5

− 0 .5 − 0 .5

0

0

−1

−1

0 .5

1 .5

− 0 .5 − 0 .5

0

0

0

0

− 0 .5 − 0 .5

0

0 .5

0

− 0 .5 − 0 .5

1 .5

0 .5

−1

0

0

0

− 0 .5 − 0 .5

0 .5

1 .5

0

−1

0

− 0 .5

0

0

−1

0

1 .5

0 .5

− 0 .5

− 0 .5

0

0

0

−1

0 .5

1 .5

− 0 .5

0 .5

−1

0

0

0

− 0 .5 − 0 .5

.10 6.......N / m

1 .5

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 a varying distributed force. In FEA, we must transfer distributed loads into equivalent nodal loads. The formula for converting distributed loads into nodal loads is: {re } = ∫ [ N ]T q...dx

For more explanation on this topic, see your favorite (or not so favorite) FEA textbook. q1

q2

L(2q1+q2)/6

L Actual distributed load

L(q1+2q2)/6

L Equivalent nodal load Figure 20-2

Note that the applied load is in DOF 3 and 5 direction. So, force matrix becomes an 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 to adjust both global stiffness and force matrix with DOF 3, 4, 5, 6 only. The force matrix is

- 72 -

Computer Aided Structural Analysis 0 0 8000 [F ] =

0

.....N

4000 0 0 0

After applying boundary condition, global [K] takes the form 1 .5 [K ]B =

− 0 .5 − 0 .5

0

− 0 .5

1 .5

0 .5

−1

− 0 .5

0 .5

1 .5

0

0

−1

0

1 .5

.10 6.........N / m

Force matrix is (for 3, 4, 5, 6 DOF only) 8000 [F ]D =

0 4000

..........N

0

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 7.43 [d ]c =

1.71 4.57

.10 −3..........m

1.14

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

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Computer Aided Structural Analysis

0 0 7.43 [d ] =

1.71

.10 −3........m

4.57 1.14 0 0

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]. 0 0

εx − 0.25 0 0.25 0 0 0 1875 7.43 −3 − 0.25 [ε ] = ε y = 0 0 0 0 0.25 . .10 = − 143 10 − 6 1.71 γ xy − 0.25 − 0.25 0.25 0.25 − 286 0 0 4.57

1.14

The stress is defined by [σ] = [C][ε]. σx 371450 [σ ] = σ y = − 28550 . .......N / m 2 τ xy − 28600

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] d1 d2 ux uy

=

N1

0

N2

0

N3

0

0

N1

0

N2

0

N3

.

d3 d4 d5 d6

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=

Computer Aided Structural Analysis

0 0 0.25

0

0.25

0

0 .5

0

7.43

0

0.25

0

0.25

0

0 .5

1.71

10 −3 =

4.14 1 .0

10 −3........m

4.57 1.14

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 x or y. However, this is the case only for triangular element. That’s why it’s called Constant Strain Triangle. For rectangular and other elements, [B] is a function of x and y (and z for 3D cases). 2. Since strain is constant, stresses at any point within element are also constant. 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 vindicates absolute necessity of computers in FEA. 6. The answer we have got here is not correct. This is as expected because we have considered only two elements. Use your analysis program to generate a mesh and see what will be the exact answer. Next try the problem with four rectangular elements. See your FEA textbook to find out [N] matrix for rectangular elements. The stiffness matrix for rectangular elements is given by b

[k ] =

a

∫ ∫ [ B]

T

[C ][ B].t...dxdy

−b − a

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

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Computer Aided Structural Analysis

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 simply supported. Ignore self-weight of the plate. It is acted by 100-kPa uniformly distributed 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 0.0007 m deflection using your program, it may be considered sufficiently accurate. First hurdle is how you should ‘mesh’ the plate. Most programs will allow you to divide the plate into number of smaller plates within it. So, do it. Expensive programs can create the optimum mesh for you! If you mesh the plate 15x10 elements, it should be enough for this problem. Now comes the boundary condition. Specify two adjacent edges as pinned and the other two adjacent edges as rollers. It will make the plate simply supported. Consider sign convention as described in section 4. Roller

Pinned

Roller Pinned Figure 25-1

When applying surface load, be careful. Different programs have different options for specifying surface loads. Also make sure the load acts ‘downward’. If everything goes ok, you should get the answer. The deflected shape should resemble a saucer. The programs typically show many other stress-components like Von Mises, S11, Max. Principal etc. I shall discuss about them later. Now solve the same problem with some different boundary conditions like all edges fixed and heavier loads etc. Check your answers with theoretical solutions. 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 and concrete floor slabs as ‘plates’. Now why you need to ‘mesh’ the plates? The problem with this is that the plate elements are only connected to the supporting members at the nodes. In the real world structures, the plates are in continuous contact with the supporting members. Modeling the structure in this way can cause larger deflections in some members than if they were modeled with intermediate connections. The walls may also be too stiff when modeled as a single element. Another concern involves load path continuity. If a distributed surface load were applied to the floor plates, the load would be transferred to the corners of the plate and directly down to the columns. The supporting beams would receive no load. To fix these problems the single plate will have to be split into smaller pieces, so that more nodes are provided to allow the real world connection to be more accurately modeled. The more each plate is split the more accurate the model becomes. However, more elements require more time to analyze, this is particularly important with large models. Interpreting the result also becomes complicated. Finally, it is not clear into how many elements you 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, we can 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 point loads. For linear structures, even if you split frame members, you still get same result. But for non-linear structures, you will get different (usually more accurate) 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 only 6x4 mesh. Your mid point deflection should come 5.91x10-4 m. Clearly, this is not equal to theoretical 6.5x10-4 m. So, you need to increase your mesh density i.e. you have to decrease your mesh size. Let’s apply 15x10 mesh (see fig. 252). 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 you won’t know the answer beforehand! So, how do you know that you get exact answer with 15x10 mesh? What you have to do is that, you analyze the plate again but with 30x20 mesh. You will observe that you get same displacement as in case of 15x10 mesh. Now your result is correct. So, we just increased mesh density and see whether our result converges and we stopped when done. This trick normally works in most of the FEA problems, however, there may be situations where increasing mesh density only may not produce good result! We may 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? - 77 -

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In general, stresses in FEA are ‘less’ accurate than displacements (since stresses are calculated from displacements, see the worked out problem of previous section). So, you should aim at displacement convergence. Take stress diagrams of 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 then gradually move to complex stress components. First consider a simple truss member. When it’s in tension, the force acts on it as shown below.

Figure 26-1

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

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 avoid compression member as far as possible. This is current trend in design. All tension member structures like tents etc. are very efficient and cost effective. In case of straight beam, the forces we generally do consider are – bending moment and shear force, although sometimes effect of axial forces may be quite as much. The next figure shows positive direction of the bending moments and shear forces acting on a beam.

Figure 26-3

The bending moment causes bending stress (= My/I) on the beam. The reactions cause 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. - 79 -

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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 τzy

Y

τzx τxz

X Z σz Figure 26-5

All stress components are shown in figure 26-5. I’m sure you’ve seen this figure several times since your first year of university, didn’t you? In straight beam, we have only σx. Please see figure 26-1 and you will appreciate 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 after analysis is finished. Now what does this mean anyway? Von Mises stress is a measurement of ‘distortion’ of the element. This is based on Von Mises – Hencky theory which, predicts that yielding in ductile material occurs when distortion energy per unit volume of the material equals or exceeds the distortion energy per unit volume of the same material when it is subjected to yielding in a tensile test. The theory takes into account the energy associated with changes in the shape of the material. This criterion is used to analyze materials that would fail in a ductile manner. In a nutshell, The Von Mises stress is a measure of stress intensity required for a material (generally a metallic material), to start yielding and become plastic. Before showing you mathematical concept of Von Mises stress, please let me introduce Principal - 80 -

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stresses. Von Mises stress has no implication in brittle materials like concrete and 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 remaining three normal stress components are called principal stresses. The directions associated 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: 1

σ VM

1 2 2 2 = [ ((σ x − σ y ) 2 + (σ y − σ z ) 2 + (σ z − σ x ) 2 ) + 3.(τ xy + τ yz + τ zx )] 2 2

σ VM =

(σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2 2

Hence, the Factor of Safety = σ limit / σ Von Mises Then comes Maximum Shear Stress criterion, which is also known as Tresca yield criterion. According to this theory yielding of material begins when the absolute maximum shear stress reaches the shear stress that causes the same material to yield in a tensile test. This criterion is mostly used to analyze materials 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 Coulomb Stress and Maximum Normal Stress. Mohr Coulomb theory is also known as Internal Friction theory. According to theory, failure occurs when: σ1 >= σ Tensile Limit if σ1 > 0 and σ3 > 0 σ3 >= -σ Compressive Limit if σ1 < 0 and σ3 = 0 and σ3 = 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 for both tension and compression. Please remember that brittle materials do not have 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 code specification. If the ratio is less than 1, then design is same. Ratio more than 1 indicates re-design is necessary. Please see Section the section on Folded Plate for more information on interpreting 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 the following advises. Use quadrilateral elements – in general, try to use quadrilateral elements instead of triangular elements as they give more accurate results. Remember that, the four corners of a quadrilateral element should all lie on the same plane. If this is not possible, use two triangular elements in place of each quadrilateral. Element shape – quadrilateral elements – the greatest accuracy is achieved with a square – 1:1 element. Elements with a base/height ratio up to 1:2 give good results, but elements with a ratio of 1:5 will be unreliable. However, many FEA programs 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 able to generate the mesh successfully at all! Many a times you will be forced to use 1:50 as lowest acceptable aspect ratio value. Yikes! If you do not specify such upper limit for aspect ratio, the program may occasionally churn out elements with 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 should never be used. However, due to geometry of the model, more often than not you will have to use quadrilateral elements. Triangular elements – equilateral triangles will produce the most accurate results. 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 the actual result through the elements is not linear but parabolic, for example, it is obvious that there will be a decrease in the accuracy. In a fine mesh, the result diagram through any one element will always be approximately linear. Increase the number of elements where there is a greater rate of change in the internal forces. For example; around supports (where bending moments increase sharply), openings and large concentrated loads. To decrease the number of elements – use a rough mesh in areas where relatively low results are expected. Remember that the connection to adjacent - 83 -

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elements is through the element end nodes only and so nodes located along an edge of an element between end nodes are ineffective. Use triangular or trapezoidal shaped elements to step between rough and fine quadrilateral meshes. If you have doubts as to the accuracy of the results in a particular area of the model, 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, by default, generate ‘tetrahedral’ mesh for solids. Some programs, however, allow you to specify how you want the mesh to be generated. Figure 27-1 shows how your model looks with a tetrahedral mesh. Figure 27-2 shows same model with combination 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 quality mesh and the lowest number of elements. Standard solid meshing works from the surface inward. It will make 8-node brick elements on and near the surface of the model while making 6, 5 or 4-node transition elements in the center of the model as needed. All 8-Node Bricks – this option should be used only for processors that accept only 8-node bricks. In many cases, these are fluid flow processors (for analysis of pipe network, this topic is not discussed in this book). This option can make 4 to 5 times the number of elements as the "Standard" option. No Pyramids – the "No Pyramids" option builds brick meshes with 8, 6 and 4node 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 8node elements plus 6-node and 4-node elements, but no 5-node pyramid elements. Tetrahedral – the all-tetrahedral option is for generating a nearly equilateral tetrahedral solid mesh from an equilateral triangular surface mesh. This is the most 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 1000 lbf/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 is described in just next section. We shall come to the same ‘yoke’ model later on, when we discuss how to interpret FEA results.

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Figure 27-1

In the above figure only the mesh has been shown. The boundary condition and pressure load for this particular analysis has been shown in next figure. You may use ‘coarser’ or ‘finer’ mesh in your program. Normally, the programs create the mesh using a default mesh density. If can control the mesh size/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 special cases, where you should use particular type of mesh for best result. See the chart in 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 27-1 after 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 Dynamic Stress Analysis Element type 3-D Truss, 2-nodes

Illustration

Description Truss elements are used to provide stiffness between two nodes. These elements transmit compressive and tensile loads along their axis. They do not carry any bending load. Beam elements are used to provide elongational, flexural and rotational stiffness between two nodes. These elements can possess a wide variety of cross-sectional geometries including many standard types Membrane plane stress elements are used to model "fabric-like" structures, such as tents, cots, domed stadiums, etc. They support three translational degrees of freedom and in-plane (membrane) loading. Orthotropic material properties may be temperature dependent. Incompatible modes are available.

3-D Beam, 2-nodes

3-D Membrane Plane Stress, 3nodes

3-D Membrane Plane Stress, 4nodes

Elasticity elements are used for plane strain, plane stress and axisymmetric formulations. They support two translational degrees of freedom. Orthotropic material properties may be temperature dependent. Incompatible modes are available.

2-D Elasticity, 3nodes

2-D Elasticity, 4nodes 3-D Brick, 4-nodes

Brick elements are used to simulate the behavior of solids. They support three translational degrees of freedom as well as incompatible displacement modes. Applications include solid objects, such as wheels, turbine blades, flanges, etc.

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3-D Brick, 5-nodes

Same as above

3-D Brick, 6-nodes

Same as above

3-D Brick, 8-nodes

Same as above

3-D Plate, 3-nodes

Plate elements are used in the design of pressure vessels, automobile body parts, etc. They support three translational and two rotational degrees of freedom as well as orthotropic material properties. An optional rotational stiffness around the perpendicular axis is automatically added to the node of each element. A thin composite plate element is available for use in models such as mechanical equipment, bicycle frames, etc. A thick composite plate element is also available and can be used in models such as honeycomb sandwich structures, aerospace products, etc. Both thin and thick composite plate elements have no limitations regarding orientation or stacking sequence and support the Tsai-Wu, maximum stress and maximum strain failure criteria. Tetrahedral elements are used to model solid objects, such as gears, engine blocks and other unusually shaped objects. They support three translational degrees of freedom. They are also available in higher order formulations (mid-side nodes).

3-D Plate, 4-nodes

Tetrahedral, 4nodes

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Boundary, 2-nodes

Boundary elements are used in conjunction with other elements. A boundary element rigidly or elastically supports a model and enables the extraction of support reactions. Boundary elements are also used to impose a specified rotation or translation. The gap element simulates compression, where deflection makes two nodes touch and transmit force, such as when a ball bearing moves in a joint. Using gap elements, stresses, bending moments and axial forces where the bearing and joint meet can be determined. A cable element simulates tension, where two nodes moving away from each other a specified distance cause the element to become active. It is still a small-deflection, small-strain analysis, but with deflection-sensitive connectivity.

Gap/Cable, 2-nodes

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’ to resist horizontal forces induced by wind or earthquake. For such buildings, if we solely depend on beams and columns for providing lateral stiffness, then the sway movement will be quite disturbing, vibration can easily be felt by the occupants and structural members may develop high stresses in them. Concrete walls, which have high ‘in-plane’ stiffness, are placed at convenient locations in the building to provide necessary resistance to horizontal forces are known as shear walls. Shear walls are often provided surrounding elevator or staircase. Figure 29-1 shows a typical arrangement of shear walls in a building frame. The exterior shear walls are shown red and the interior shear walls surrounding central columns are shown yellow for easy visualization. This is one of the most common arrangements for shear walls. In the figure only a 5-storied building has 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 very complicated task. It involved lots of assumptions and mind-boggling calculations. One common classical method was that of MacLeod. If you want to torment yourself, go and get an Advanced Structural Analysis textbook and try to solve a shear wall problem using classical methods. The advent of FEA has made life much easier for us. Just model the building first. 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 shear walls. 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 is ready for cooking (I mean analyzing)! Run your favorite FEA analyzer, and you will instantly get displacements, shear forces, bending moments, stresses in the beams, 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 already that 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, lesser values 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 structures with various shear wall arrangements to get the most economical yet practical structure. Sometimes, it may be necessary to provide shear walls with openings for windows, doors etc. for architectural reasons. Openings reduce the stiffness of the shear wall. Exercise Assume suitable dimensions for the building frame shown in figure 29-1. Apply some realistic horizontal forces at floor levels and analyze the structure with and without shear walls. See how the result changes with shear walls. Use a different arrangement for shear walls and analyze again. Which arrangement comes 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 single classical method available for general analysis of folded plates. They depend on the shape of folded plates. Some common methods are – Simpson’s, Whiteney’s, Iterative, Three-shear equations etc. However, if you use FEA, then the analysis itself is quite simple and the labor involves only preparing the model correctly and interpreting the result properly. Your analysis programs must have ‘plate’ elements to successfully analyze folded plates. Modeling the folded 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. The front view is shown in the figure and the length of the structures is 12 m as shown in figure 30-2 with mesh for sufficiently accurate result. C

G

D 3m

Thickness 150 mm

B

E

Thickness 300 mm

A

1.5 m

F 3m

4.5 m

3m

Figure 30-1

The frame is loaded with 1.5 kPa uniformly distributed load in addition to the self-weight. The folded plate is made of concrete. If you use your analysis program’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) after the plates are drawn, delete the frame members since they were drawn here only to ease the model (5) now mesh the plates (6) apply boundary condition, in this problem, the sides of the plates are fixed (7) apply plate thickness, material properties 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 appear awkward as shown in figure 30-4. If it happens, delete the particular plate and redraw again in the opposite direction compared to previous case. For this reason, 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 - 94 -

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

Fixed on side Fixed on side

12 m Figure 30-2

Figure 30-3

Figure 30-4

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Are you wondering whether we can take advantage of the symmetry? Of course we do. In fact, only 1/4th of the structure needs to be analyzed. The appropriate boundary conditions are shown in figure 30-5 with respect to the global axes shown in the same figure. all free Z rotation fixed

Y Z

X all fixed Figure 30-5

To assign proper boundary conditions accurately, you must visualize the deflected shape of the structure yourself in your mind before performing the actual analysis. Of course, you may bypass this mental exercise by analyzing the 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). I admit that you need some benchmark problems to compare your analysis output, 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 needs clarification. 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 is actually ‘inclined’ in the real structure. For example, a ‘plate element’ of BC may 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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Computer Aided Structural Analysis

σy

τyx = - τxy τxy

y x

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. Y X Y X Z

Global

Figure 30-7

A different situation comes when we speak stress components in terms of global axes. 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 axes stress components. How do I know which convention it is following? Answer: Difficult to say. Generally, most programs give output with respect to local 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 some other type of ‘elements’! Really confusing! See your programs’ manuals for details. However, you better solve some benchmark problems (with known answer) to check. Question: Shall I provide the reinforcement on the basis of forces on global axes or local axes? Answer: Local axes forces.

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Figure 30-8

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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 figure 30-10.

Figure 30-10

The span of plate (along X axis) is 10 m and thickness of all four plates is 100 mm. 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. Rest assured, I’m not going to swamp you with differential equations. Theoretical background on shell is extensive and if you are interested, you should consult a textbook on shell theory. Here I shall discuss only what you should know for shell analysis using FEA.

20 m

5m 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 every point, not horizontally projected) is acting over it (including self-weight i.e. you need not add self-weight load). The sides of the shell are pinned. You are to analyze the shell. Your first task is to model the shell. Some programs can generate this kind of shell through a model wizard. Then you are lucky. If your program does not have this option, you need to start from scratch. First create the arc. Note that, either you can draw the arc exactly (if your programs graphic editor 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 you break the arc into lines before, copying into suitable interval creates the mesh automatically. The concept may appear garbled if you just are reading, but will be clear if you try to draw the shell model yourself. A 20x80 mesh is sufficient for 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 - 102 -

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shells is same for all shell types. The shell theory is extremely complicated and if you lean too much in that subject you may find a place in lunatic asylum too soon! One advantage of shell is that, because of its curve shape, less material is required 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, because of constructional difficulties, shells are still of limited use. Yes, like the folded plate discussed in previous section, you can again take advantage of symmetry as shown in figure 31-2. X rotation fixed

boundary condition Y Z

X

pinned free Figure 31-2

For other types of shells, if the loading and geometry are symmetrical, you can always take advantage of symmetry. Although you have the right to place the origin at anywhere, I prefer to keep the origin at the center of the shell. It has some ‘psychological’ advantages. The analysis output for σ in local X-axis is shown in figure 31-3. The analysis was done in Algor. You should note some important points. First of all, all FEA programs display output in colorful stress contour range. Normally, you can ‘adjust’ the range yourself. It is always advisable to specify a range of your own. This is because, the highest (and lowest) range generally covers only one or 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 range shown 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.

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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 structural dynamics are very large. Even a decent introduction of this subject would require at least 100 pages. I am just describing some very basic concepts of structural dynamic useful for practicing engineers. If you’ve already studied the theory of structural dynamic then it’s great. If not, I advise that you start reading a textbook of structural dynamic besides this book. It will make you understand the applications of dynamics discussed in this book in an easily understandable manner. At the beginning, let me explain the difference between static and dynamic analysis. In static analysis, the applied force is constant but in dynamic analysis, the applied force varies with time. It is not necessary that dynamic analysis always involves application force only, it may be due to shaking of ground due to earthquake. In civil engineering applications, dynamic analysis mostly involves determination of maximum response (i.e. displacement etc.) of the structure 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 shown below. 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 the arrow. 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 top of model as shown by the filled circle. This means, the mass of the stick, though distributed though out its length, we assume as if it is concentrated at one place as shown by the circle. This model may reflect idealization of a single story building, 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 - 105 -

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'lumped' at the respective floor levels. In that we shall have multi-degree-offreedom (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 that we need not study theory of SDF structures, because many MDF structures can be 'broken' into separate SDF structures and can be easily analyzed rather than analyzing the whole structures! Damping – if a structure is displaced from its equilibrium position by a small amount 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 and after some time, the model will come to rest. This process by which free vibration diminishes is known as ‘damping’. Natural frequency (ωn) – it is the number of cycles per second a structure vibrates. 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 simple static analysis. More displacement means higher values of internal forces (bending moment, shear force etc.) and subsequently higher values of stresses in the members. If the stress reaches the yield strength of the material the structure will collapse! For a time varying applied force, if I apply calculated values of the force at particular instant of time and then perform static analysis with that force, shall I get correct displacement etc.? No, you won’t (even for a linear structure)! The concept of structural dynamics is different from static analysis theory. You must perform exact dynamic analysis. - 106 -

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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 for earthquake analysis. Normally, in low height buildings, dynamic analysis does not produce much different result compared to static analysis. However, for all structures, where vibration is a major factor in design, dynamic analysis must be performed. Designing of machine foundations always requires dynamic analysis. For highly important structures like bridges, dams, nuclear reactors etc. dynamic analyses are very important. Theory of structural dynamics is applicable 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 &} + [C ]{ X&} + [ K ]{ X } = F (t ) [ M ]{ X&

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 ground acceleration 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, then we call undamped motion. However, in real life, [C] is never equal to zero. In the subsequent sections, we shall explore various examples of structural dynamic analyses – simple to complex! By this time I expect that you will also study a few pages of dynamics textbook. Suggested chapters for reading in your textbook 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 dynamic response, equation of motion for MDF systems, modal and response spectrum analyses. If you are interested, you may read all the chapters of the book, but above topics are enough for understanding the calculations presented in this book. You may read Ref. 5, 16, 17 to start with in dynamics.

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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 shall show you how to perform accurate dynamic analysis without any computer program! 1000 kN acceleration m/s2 t2 3m 0

1 time (s)

1m Figure 33-1

All the members of the tower shown are made of steel with 200x200 mm square sides. The base of the tower is given an acceleration of 1-s duration as shown in the 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. Then apply base acceleration input and find out the displacement response curve for entire duration. However, you will soon discover that you can solve it much quickly 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 should look like figure 33-2. Apply a lumped mass of (1000/10)/2 = 50 kNs2/m on the two uppermost nodes as shown (we divide the weight 1000 kN by g = 10 m/s² to get the mass). Note the direction of applied lumped masses because we are interested in calculating the displacement in this direction. Next apply the base acceleration 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 - 108 -

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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 1 mode is used in the analysis. After performing the dynamic analysis we get from the time vs. displacement curve that the displacement at t = 0.5 s is – 7.64E-4 m. 50 kNs²/m

50 kNs²/m

Y X

Figure 33-2

Now we are going to see what we get if we perform the analysis using Duhamel Integral. In this case, after modeling the structure, our first step is to calculate its stiffness. For this we need to apply a force (say 1 kN) in the upper left nodal point. Then run a static analysis and note the displacement of any of the uppermost nodes. Displacement ∆ 1 kN

Figure 33-3

The stiffness of the entire structure is k = F/∆ = 1 kN/3.211E-5 m = 31143 kN/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 = - 109 -

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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 input and 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 and mass m = 100 kNs²/m. The favorite Duhamel’s integral has the well-known form: u (t ) =

1 mϖ d

t

∫ p(τ )e

−ςω n ( t −τ )

sin ω d (t − τ )dτ

0

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

u (0.5) =

− 100 τ 2 e ( −0.05)(17.64 )( 0.5−τ ) sin 17.6(0.5 − τ )dτ = −7.64 E − 4...m 100 x17.6 ∫0

After performing the numerical integration, we get the same answer as that of in exact 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 at several points to get the maximum response. In this particular problem, the input excitation is defined by simple algebraic function, which makes use of Duhamel’s integral feasible. In actual practice, the input excitation is often defined as a set of data at some specified interval (normally at 0.02 s interval for earthquake ground acceleration data). So, there you must use numerical methods like what your analysis programs do. Now probably you have a comprehensive idea of what dynamic analysis involves. 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’s dimension and sectional properties). You also need to specify a suitable value of damping ratio. 3. Apply an input excitation (time varying ground acceleration or time varying force). - 110 -

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4. Specify how many mode shapes you do want (generally you need same number of mode shapes equal to the number of stories in structure, however sometimes you may need to venture more or less number of mode shapes depending on problem type). Normally (but not always) first three mode shapes are sufficient for subsequent calculations. However, for multi-story buildings, contribution of higher modes is quite significant. 5. Some programs may ask you what algorithm you want to follow – eigenvector 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 mode shapes. After finishing it displays the natural time periods (hence frequency therefrom) of the structure for each modes, mode shapes and time vs. displacement (and velocity, acceleration etc.). Some programs can also display other parameters such as base shear etc. One word of caution, the mode 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 the stresses developed in the members due to dynamic analysis. You can do it by noting maximum displacement and then calculating the bending moment developed 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 the deformation of the system, or displacement of the mass relative to moving ground, to which the internal forces are linearly related. Knowing maximum lateral displacement of the structure would be useful in providing enough separation between adjacent buildings to prevent their pounding against each other during an earthquake. Moreover, total acceleration of the structure would be needed if the structure is supporting sensitive equipment and the motion imparted 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 a specified 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 each time instant (discussed later). A typical response spectra for 1940 El Centro earthquake has been shown in figure 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 damping ratio of the structure. Using this, we can directly read the maximum displacement of the structure from the figure as shown. One thing to note is that, you must have response spectra chart for the particular excitation (e.g. El Centro earthquake in this case). If you need to analyze a structure for another earthquake, you need to use response spectra for that earthquake. It means that, someone must have prepared the response spectra chart before you use it in your analysis. The procedure of developing response spectra is discussed in any standard structural dynamics textbook and not discussed here. However, you may note that considerable computational effort is required to generate such charts. So, the modern trend is to perform full time history analysis, which is more versatile and accurate, for structures. If your analysis program offers time history analysis option (unfortunately majority of analysis programs do not), there is no point of going for response spectra 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 time history analysis.

Response spectra for elastic system for the 1940 El Centro earthquake ξ =0, 2, 5, 10, 20% Figure 34-1

For example, assume our structure has natural frequency of 1 Hz. Also assume 10% 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 relative displacement response of our structure is 3.3 inches if it is excited by the 1940 El Centro earthquake ground acceleration. Also seen that horizontal arrow (shown blue) extends to the pseudo velocity axis at 18.5 inch/s. Please note that the axes are in logarithmic scale. Since earthquake can’t be predicted, you may like to analyze your structure for several past earthquake effects. Instead of using response spectra for each earthquake, 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 the structure 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 Working stress, no more than about ½ yield point At or just below yield point

Type and condition of the structure

Damping ratio % Welded steel, pre-stressed concrete, well reinforced concrete 2-3 (only slight cracking) Reinforced concrete with considerable cracking 3-5 Bolted and/or riveted steel, wood structures with nailed or 5-7 bolted joints Welded steel, pre-stressed concrete (without complete loss in 5-7 pre-stress) Pre-stress concrete with no pre-stress left 7-10 Reinforced concrete 7-10 Bolted and/or riveted steel, wood structures with nailed or 10-15 bolted joints Wood structures with nailed joints 15-20 Figure 34-2

Since, damping properties of the materials are still not well established; it is quite a challenging task to determine exact damping of the structures, especially in 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 techniques for this purpose, but the most popular is ‘wavelet’ analysis, which has recently attracted 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 nonlinear 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 stressstrain curve is linear and deformation is small. Right? But we have to consider NLA if any or both of the above assumptions are violated. There are mainly two types 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

σ Et Ei Yield point

E

E ε

(a)

ε (b)

ε (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). The curve is linear (slope E) up to yield point, where from it changes its slope to Et though remain linear again. E is our familiar modulus of elasticity or Young’s modulus. Et is known as ‘strain hardening modulus’. In the range of ‘E’ the material remains elastic, but in the range of ‘Et’ it becomes ‘plastic’. This material 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 the curve. This is also a typical example of material non-linearity. Now how do we incorporate material non-linearity in analysis program? The steps are given below. 1. Divide the total load incrementally i.e. F = Σ(∆Fi) - 115 -

Computer Aided Structural Analysis

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 global stiffness 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 structure using 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 thus global stiffness matrix) remains constant throughout analysis. This stiffness matrix is formed on the basis of co-ordinates of the nodes of the structure. If the deformation 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 and deflected 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 line shows 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 new co-ordinates of the nodes. If the deflection is ‘large’, then [K] will not be equal to [KD]. This is the main theory behind geometric non-linearity. If we increment the load at each time instant and update [K] according to ‘changing’ displaced position of the structure, then it will be a geometric non-linear analysis. So, the main 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 you will have to perform the whole calculation of setting up stiffness matrix at every step! Now imagine what will happen if you need to analyze a large complex structure 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 later sections! But there remains one most important question – when you should consider a deformation large enough for geometric NLA? The answer is not - 117 -

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easy! Often in large deformed state, the structure may behave entirely different manner than that of small-deformed state. This depends on particular type of structure, loading and material properties etc. Some analysis programs have an in-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 not displacements are large. In a very long or very tall structure, inter-story or inter-span deflections can easily be less than 5% of the overall dimension yet large enough to violate the small displacement assumptions. The ultimate decision 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 ‘virtual experiment’. Traditional FEA programs calculate stresses usually at a single instant of time and requires assumption about forces. That means, in linear static stress analysis, you must input the force quantity explicitly. But MES intrinsically calculates loads and stresses as motion takes place at each instant in time throughout the event, facilitating a more efficient design/analysis process since the need to estimate and specify forces is eliminated. The whole thing will become 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 bar of known dimension form a specific height ‘h’. We need to determine the stresses 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, all we need to do is specify the dimension of the bar, height h and dimension and weight 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 at each instant of time. You will see the steps like animation. Did you realize the advantage? You actually performed a virtual experiment of a mechanical event! The example given here may appear too simple, but imagine that this method can be successfully employed in crash test simulation of an automobile! However, there are a few disadvantages as well. The very simple example described above took me 90 minutes to perform for just 1-second simulation in a 233 MHz PII computer. So, when you make a simulation of real world problems 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 super computer 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 can write 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 involves simultaneous analyses of a model for more than one physical effect. Some example of multi-physics analyses are – thermal stress, fluid flow, combined thermal and fluid flow, electricity and electro-magnetism. Let me explain in a lucid way. Suppose you want to analyze fluid flow around an airfoil. (Sounds whacky?) 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 whether the flow at a particular point is laminar or turbulent. The things are really mindboggling. 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 FEA programs, 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 draw the actual structure in the analysis program, particularly 3D frame or complex structures for finite element analysis, drawing them in AutoCAD/Mechanical Desktop/Solid Edge/Solid works etc. are an easy alternative. Since the CAD programs themselves are high-end drawing programs, creating the drawing in them is always easier. But be careful! Not all analysis programs can import all CAD 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 in AutoCAD, be sure to “Export” it into “DXF” format. Please note that DXF files are AutoCAD’s version specific. If your analysis program can’t read AutoCAD R2000 DXF then you should save the AutoCAD drawing file in R14 or R13 DXF format. You can also perform reverse process that is exporting your analysis model from your analysis program to AutoCAD DXF format. Analysis Program Frame Nodes

AutoCAD Line Automatically created at intersections of elements 2D Solid / 3D Face 3D Solid

Plate Shell

Except for simple lines, other AutoCAD entities are often treated differently by various programs. If you are importing DXF file with plate, shell or wire frame elements, you may find discrepancy in geometry upon importing. Therefore, it is always better to check what AutoCAD entities your program can successfully import. If you find your program can’t successfully import AutoCAD drawing (DWG/DXF), then you should draw in the analysis program’s graphical environment. 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. In these programs, you can modify the geometry by just changing the dimensions. An example is given below to explain parameter-defined geometry. - 121 -

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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 as that 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 be updated automatically as shown in fig. 37-3.

Figure 37-1

Figure 37-2

Figure 37-3

Did you see the difference? This kind of relationship is very convenient for solid object modeling. Some of the already named CAD programs have the capability 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 into this format. These files have *.IGES or *.IGS extension. This is actually a text file. Programs write model information in the files. When you import an IGES file, 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 file has an extension of WRL. You can view VRML files using any standard web browser like Internet Explorer. However, you must install first the VRML viewer support files (normally come with your operating system CD). VRML files 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 different materials of the structure. It’s an excellent feature to impress your clients because you can show them what your structure will look like when it would be built in real life. You are probably aware of ‘rendering’ feature of CAD programs. VRML is just like that, but looks more realistic. VRML is actually a text file. The browser reads the data in the file first and then develops the realistic model. But here’s a warning! Finite element solid models, which contain 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 computer may create staggeringly large (> 1 GB) virtual memory in your hard disk and it will 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 linear programming problems using your favorite spreadsheet! Linear programming is often required by engineers to solve certain design problem for example – in pre-stressed concrete section design and various other ‘optimization’ problems etc. Linear programming can be also be solved in programs like MATHCAD or MATLAB etc., however, since these mathematical applications are not very common in design offices, you better bet on versatile spreadsheets. Do you know spreadsheet is the largest selling type of application in the world? The procedure presented here are for Microsoft Excel 97, but if you any other spreadsheet like Lotus 123 or Borland Quattro Pro you will find similar functions in those programs as well. I assume that you are familiar with basic spreadsheet operations. So, I’m directly going to the problem. Please make sure that you have installed Analysis ToolPak add-in in Excel. Consider the equation Z = 3x + 4y subjected to constraints 4x + 3y 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 the cells of the spreadsheet. A B 1 X 0 2 Y 0 3 Z =3*B1+4*B2 4 C1 =4*B1+2*B2 5 C2 =2*B1+5*B2 Figure 39-1

Note that in cell B3, the equation for Z has been input. Here cells B1 and B2 stands for variable x and y respectively. The strings in column A is for understanding 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 to be positive number. These constraints will later be defined using B1 and B2 cells. 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 solver dialog box appears as shown in figure 39-2. - 124 -

Computer Aided Structural Analysis

Figure 39-2

Make the ‘Set Target Cell’ box to cell B3, because this cell contains the definition 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 the range on the worksheet by clicking the red arrows as shown in the figure. Also make sure that the ‘Equal To:’ selection is set to ‘Max’ for this problem. When done, click on ‘Add’ button to specify constraints. When you do so, you will see following dialog box.

Figure 39-3

Specify the Cell Reference and Constraint (for 4x + 3y 0. Click OK when done. After that ‘Solver Parameters’ dialog box should look like figure 39-5.

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Figure 39-5

Remember to use proper ‘’ sign while specifying constraints. If everything seems ok, click the ‘Solve’ button. And that’s all. Excel will solve it within 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 as shown in figure 39-7. A B 1 X 2.5 2 Y 35 3 Z 147.5 4 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 - 126 -

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Find the minimum value of P = a – 3b + 3c subjected to the constraints a, b, c >= 0, 3a – b + 2c = - 12, - 4a + 3b + 8c