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CONTENTS
ABBREVIATION....................................................................................................................................................6
1. INTRODUCTION (BEFORE YOU BEGIN…).........................................................................................7
2. WHAT IS COMPUTER AIDED STRUCTURAL ANALYSIS? .............................................................9
3. ANALYSIS TYPES ....................................................................................................................................10
LINEAR STATIC STRESS ANALYSIS....................................................................................................................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
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23. FINITE ELEMENT ANALYSIS (FEA) METHOD IS APPROACHING… ....................................60
24. A TYPICAL WORKED OUT PROBLEM OF FEA...........................................................................69
25. PLATES BY FEM ..................................................................................................................................76
26. INTERPRETING FEA RESULT..........................................................................................................79
27. TIPS FOR CREATING BETTER MESH............................................................................................83
28. COMMON FINITE ELEMENTS LIBRARY FOR LINEAR STATIC AND DYNAMIC STRESS
ANALYSIS..........................................................................................................................................................89
29. SHEAR WALL .......................................................................................................................................92
30. FOLDED PLATE ...................................................................................................................................94
31. SHELLS.................................................................................................................................................102
32. A FIRST STEP IN STRUCTURAL DYNAMICS.............................................................................105
33. AN EXAMPLE OF A SINGLE DEGREE OF FREEDOM PROBLEM.........................................108
34. WHAT DYNAMIC ANALYSIS YOU SHOULD PERFORM? .......................................................112
35. NON-LINEAR ANALYSIS (NLA) – AN INTRODUCTION FOR BEGINNERS .........................115
MATERIAL NON-LINEARITY ............................................................................................................................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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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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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
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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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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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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.
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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
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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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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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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
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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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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 lower-
temperature 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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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 room-
temperature (steady state), but he is interested in finding out the temperature of
the fluid as a function of time. In addition to the thermal conductivity, we 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
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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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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 All positive
Fy Mx
X
Fz Fx
Z 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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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 Compression face
M2
V3 M3
Tension face
Positive Moment and Shear [1-2 plane]
Positive Moment and Shear [1-3 plane]
Figure 4-4
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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
Node 3 5 m
6 m 6 m
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.
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Now solve the following truss.
All members are made of
3 kN steel (E=200 GPa) with
100-mm side square section
3 m 0.208 -3.642
2.0
2.833 2.833
-2 kN
4 m 4 m
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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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 B-
5-10 or B05010 fashion where “B” indicates beam, next number indicates
“floor” and the last number stands for serial number of beam on that floor.
Similar procedure may be adopted for numbering columns, slabs and 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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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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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.
XZ
Y
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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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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 7-
1 and 7-2), we get
1 – 2α2
+ α3
wL2
(1 – 2α2
+ α3
) x W
-------------- WL = ------ or w = ----------------------- … (7.1)
12 x (1-α) 12 (1 – α) x L
So, the ratio of mid span moment of trapezoidal/uniform r =
[(3-4α2
)WL/(24(1-α)] / [(1-2α2
+α3
)(W/L)(L2
/8) / (1-α)] =
(1 - 1.33α2
) / (1 - 2α2
+ α3
)
If we tabulate α vs. r values as shown below
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α 0 0.2 0.375 0.5
r 1 1.02 1.05 1.068
It is seen that maximum difference of mid span moment for figure 7-1 and 7-2 is
6.8%, which is quite small. So, we can safely replace trapezoidal load with 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 7-
1. 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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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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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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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 = π2
EI/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 non-
commercial program, it may not have P-∆ analysis option! Be careful!
You may wonder, why the computer itself does not choose P-∆ analysis always.
Hmm, it would have been nicer. But think of the time required for 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
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such structure using a standard PC and inexpensive program, chances are that
your system will crash! Check it!
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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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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)hf
3
/12 + (b-bw)hf(kd-hf/2)2
+ nAs(d-kd)2
Where k = (ρ n + 0.5(hf/d)2
)/( ρ n + hf/d) and ρ = As/bd
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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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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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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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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 Do this
Maximum positive bending moment at
center of span
Load that span and then alternates
spans
Maximum positive bending moment at
center of span
Load adjacent spans and then alternate
spans
Maximum negative bending moment
at support
Load adjacent spans and then alternate
spans
Maximum column reaction Load adjacent spans and then alternate
spans
Maximum positive bending moment at
support
Load all spans except adjacent spans
In all cases, dead load must always be applied over all spans.
Some codes say that if live load intensity does not exceed 75% of dead 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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floor. (Though it is customary to use reduced live load in roof level). Similarly,
other load combinations can be used in same manner.
Figure 13-1
Figure 13-2
4
3
2
1
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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 14-
1. 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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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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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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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
4 m X
20 m
Figure 16-1
The answer is: left vertical reaction 375 kN ↑, right vertical reaction 125 kN ↑,
horizontal reactions are 312.5 kN inward at both ends. With 20 straight-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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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 1.5
2 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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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 (π2
EI/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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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.
10 kN A = 2002
mm²
4 m E = Steel
-5 kN/m
20 kN
4 m
4 m 4 m 4 m
Figure 19-1
If you separate the upper floor and then analyze only that portion, you will get
the result as shown below.
10 kN
11.44 kNm
11.44 kNm 5 kN 5 kN
4.28 kN 4.28 kN
Figure 19-2
With the result shown above, the applied loads on the bottom floor of the actual
structure will be as shown below.
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11.42 kNm 11.42 kNm -5 kN/m
25 kN
4.28 kN 5 kN 4.28 kN
A B C D
Figure 19-3
Observe that on leftmost node, 25 kN loads comes from 20 kN applied at that
node and 5 kN reaction from upper floor. The reactions you will get in the 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 Fx kN Fy kN Mz kNm
A -6.223 -14.85 15.58
B -5.99 30.54 15.18
C -11.2 20.64 22.08
D -6.583 3.675 15.81
From the above example, it is clear that; you need to apply opposite of reactions
as loads on lower floor frames.
The procedure described here seems too meager for this particular structure, 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, one-
quarter 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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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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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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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.
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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
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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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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.
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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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22. Pre-stressed cable profile
Does your program offer specifying pre-stressed cable profile? If yes, then
good. If not then read the following tricks. Observe sign conventions carefully.
PyA PyB
yA θA θB yB PθA PθB
c
P P
Upward UDL w = 8Pc/L²
c
θA = (4c + yA – yB)/L
L/2 L/2 θB = (4c – yA + yB)/L
Actual pre-stressed cable Equivalent load
Figure 22-1
PyA PyB
yA θA θB yB PθA PθB
P P(θA + θB) P
Total length L
Actual pre-stressed cable Equivalent load
Figure 22-2
Observe the figures very carefully. They are really confusing! Try 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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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 kNm 200 kNm
250 kN 150 kN 250 kN 18.67 kN/m 250 kN
Figure 22-4
Now the forces shown in blue color will go to support directly. Moments shown
in orange color will cancel each other. So, the remaining forces that will act are
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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 pre-
stressed concrete. I have shown here only linear and parabolic cable profile.
Although parabolic profile is the most common, there are other types of profiles
possible. See your textbook for details.
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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”.
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 2-
dimensional “Plane stress” structure. Don’t confuse this with 2D or 3D frame.
By 2D or 3D frame we actually mean “Plane” and “Space” frame. In 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
X
Y
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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 [σσσσ] = [C][εεεε] where [C] matrix differs in various cases like plane stress,
plane strain etc. We are describing this problem as plane stress because 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.
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.
X
Y
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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.
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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.
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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!
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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.
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.
For 2D analysis, after modeling your
structure should look similar to this
figure.
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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
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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.
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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!
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