ELEMENTARY FINITE
ELEMENT
METHOD C.S.
DESAI
B&l
D45e 2035801 Desai element finite Lementary 15.624
jthod
Mi
u
o
Ou.
Q library public AMD AU*N COUNT*, SORT \WAYN6
INO.
/
ALLEN COUNTY PUBLIC LIBRARY
4JHB|||a|i|_.
ELEMENTARY FINITE ELEMENT METHOD
CHANDRAKANT S. DESAI Professor, Virginia Polytechnic Institute
and State
University, Blacksburg, Virginia
PRENTICE-HALL,
INC., Englewood
Cliffs,
New Jersey 07632
Library of Congress Cataloging
Desai,
Chandrakant
Elementary
finite
(Civil engineering
Bibliography;
in
Publication Data
S.
element method.
and engineering mechanics
series)
p.
Includes index. 1.
Finite element method.
TA347.F5D47
I.
Title.
78-10389
624'. 171
ISBN 0-13-256636-2
Civil Engineering
N. M.
and Engineering Mechanics
Newmark and W.
V»
J.
Series
Hall, Editors
w.
©1979 by Prentice-Hall, Inc., Englewood Cliffs, N.J. 07632
All rights reserved.
may
No
part of this
book
be reproduced in any form or
by any means without permission from the publisher.
in writing
Printed in the United States of America
10
987654321
Prentice-Hall Prentice-Hall Prentice-Hall Prentice-Hall Prentice-Hall Prentice-Hall
International, Inc., London of Australia Pty. Limited, Sydney of Canada, Ltd., Toronto of India Private Limited, New Delhi of Japan, Inc., Tokyo of Southeast Asia Pte. Ltd., Singapore
Whitehall Books Limited,
Wellington,
New Zealand
2G3S8Q1 To
My
Parents,
Maya and
Sanjay
Digitized by the Internet Archive in
2012
http://archive.org/details/elenrientaryfiniteOOdesa
CONTENTS
PREFACE
1.
XI
INTRODUCTION Basic Concept
l
Process of Discretization
and Laws 13 Cause and Effect 14 Review Assignments 14
Principles
References
2.
16
STEPS IN THE FINITE ELEMENT Introduction
17
General Idea References
34
17
METHOD
17
Contents
vi
3.
ONE-DIMENSIONAL STRESS DEFORMATION Introduction
35
35
Explanation of Global and Local Coordinates
36
Local and Global Coordinate System for the
One-Dimensional Problem Interpolation Functions
38
41
Relation Between Local and Global Coordinates
Requirements for Approximation Functions Stress-Strain Relation
Principle of
Minimum
Expansion of Terms Integration
46 Potential Energy
47
52
53
Approach (for assembly) Method 58
Potential Energy
Direct Stiffness
Boundary Conditions 59 Strains and Stresses 64 Formulation by Galerkin's Method Computer Implementation 76 Other Procedures for Formulation
76
Complementary Energy Approach Mixed Approach 79 Bounds 83
76
References
55
67
Advantages of the Finite Element Method Problems 85
4.
85
91
ONE-DIMENSIONAL FLOW Problems
93
102
Bibliography
102
ONE-DIMENSIONAL TIME-DEPENDENT FLOW (Introduction to Uncoupled and Coupled Problems)
Uncoupled Case 103 Time-Dependent Problems 108 One-Dimensional Consolidation Computer Code 124 Problems 130 References
42
43
132
[8, 9]
121
103
Contents
6.
COMBINED COMPUTER CODE FOR ONEDIMENSIONAL DEFORMATION, FLOW, AND TEMPERATURE/CONSOLIDATION Philosophy of Codes Stages
Listing
134
144
and Samples of Input/Output
Beam-Column
177 189
Other Procedures for Formulation Problems 197 References
192
200
ONE-DIMENSIONAL MASS TRANSPORT Introduction
References
Bibliography
202
202
Finite Element Formulation
202
209 209
ONE-DIMENSIONAL OVERLAND FLOW Introduction
References
211
211
Approximation for Overland and Channel Flows Finite Element Formulation 213
10.
172
172
Physical Models
9.
145
BEAM BENDING AND BEAM-COLUMN Introduction
8.
133
134
Problems
7.
vii
213
222
ONE-DIMENSIONAL STRESS WAVE PROPAGATION Introduction
224
Element Formulation Problems 234 Finite
References
Bibliography
235 235
225
224
Contents
viii
11.
TORSION
237
Introduction
237
Triangular Finite Elements
238
Finite Element Formulation
240
Comparisons of Numerical Predictions and Closed
Form
Solutions
252
Approach 254 Review and Comments Hybrid Approach 269 Mixed Approach 284 Stress
Static Condensation Problems 293
References
12.
268
291
297
OTHER FIELD PROBLEMS: POTENTIAL, THERMAL, AND FLUID FLOW Introduction Potential
299
299
Flow
300
Element Formulation 302 Stream Function Formulation 314 Thermal or Heat Flow Problem 321 Seepage 323 Finite
Electromagnetic Problems
327
Computer Code FIELD-2DFE Problems References
332
Bibliography
13.
328
328
332
TWO-DIMENSIONAL STRESS-DEFORMATION ANALYSIS Introduction
333 Plane Deformation Finite
333
Element Formulation
Computer Code Problems References
368 371
354
338
333
Contents
14.
MULTICOMPONENT SYSTEMS: BUILDING FRAME AND FOUNDATION Introduction
Computer Code
373
382
Transformation of Coordinates Problems 392
15.
389
392
PRELUDE TO ADVANCED STUDY AND APPLICATIONS Theoretical Aspects
Bibliography
394
395
396
VARIOUS NUMERICAL PROCEDURES; SOLUTION TO BEAM BENDING PROBLEM
399
SOLUTION OF SIMULTANEOUS EQUATIONS
400
3.
PHYSICAL MODELS
419
4.
COMPUTER CODES
424
Appendix
1.
Appendix
2.
Appendix
Appendix
INDEX
372
372
Various Components
References
jx
429
PREFACE
The finite element method has gained tremendous attention and popularThe method is now taught at most universities and colleges, is researched extensively, and is used by the practicing engineer, industry, and government agencies. The teaching of the method has essentially been concentrated at the postgraduate level. In view of the growth and wide use of the method, however, it becomes highly desirable and necessary to teach it at the underity.
graduate
level.
There are a number of books and publications available on the finite element method. It appears that almost all of them are suitable for the advanced student and require a number of prerequisites such as theories of constitutive or stress-strain laws, mechanics,
and variational
calculus.
Some
of the introductory treatments have presented the method as an extension of matrix methods of structural analysis. This viewpoint necessary, since the finite element
maturity and generality.
It
method has reached a
may no
longer be
significant level
of
has acquired a sound theoretical basis, and in
has been established as a general procedure relevant to engineering and mathematical physics. These developments permit its teaching and use as a general technique from which applications to topics such as mechanics, itself
structures, geomechanics, hydraulics,
as special cases. It
is
and environmental engineering method be treated
therefore essential that the
arise
as a
general procedure and taught as such.
This book sufficiently
is
intended mainly for the undergraduate.
elementary so that
it
Its
approach
is
can be introduced with the background of xi
Preface
xii
essentially
undergraduate subjects. At the same time, the treatment
enough so
that the reader or the teacher interested in various topics such as
broad
is
stress-deformation analysis, fluid and heat flow, overland flow, potential flow,
time-dependent problems, diffusion, torsion, and wave propagation can use
and teach from
it.
of the method and provides a distinct and element method at an elementary
The book brings out the
intrinsic nature
that permits confluence of various disciplines
rather novel approach for teaching the finite level.
Although the book
is
intended mainly for the undergraduate,
it
can
be used for the fresh graduate and the beginner with no prior exposure to the finite element method.
The
prerequisites for understanding the material
be undergraduate mathematics, strength of materials and undergraduate
will
courses in structures, hydraulics, geomechanics and matrix algebra. Intro-
ductory knowledge of computer programming
The
text
written in such a
is
way
that
is
desirable but not necessary.
no prior knowledge of variational
The derivations are presented through the use of Over a period of the last five years or so, the author has taught, based on these prerequisites, an undergraduate course and a course for user groups composed of beginners. This experience has shown that principles
is
necessary.
differential calculus.
undergraduates or beginners equipped with these prerequisites, available to
them
in the
undergraduate curricula at most academic institutions, can under-
stand and use the material presented in this book.
The
first
chapter presents a rather philosophical discussion of the
finite
element method and often defines various terms on the basis of eastern and western concepts from antiquity. The second chapter gives a description of the eight basic steps. Chapters 3-5 cover one-dimensional problems in stress-
deformation analysis and steady and time-dependent flow of heat and
The fundamental
mon
method
fluids.
by showing the comcharacteristics of the formulation for these topics and by indicating the
fact that their
generality of the
is
illustrated
governing equations are essentially similar. The generality
further established by including a
computer code
in
is
chapter 6 that can solve
the three problems in chapters 3-5.
Understanding and using the the use of the computer.
teaching of the method
It is
may
finite
element method
is
closely linked with
the belief of the author that strictly theoretical
not give the student an idea of the details and
the ranges of applicability of the technique. Consequently this text endeavors
and simultaneously with the theoretical and understanding of computer codes. The code in chapter 6 is thoroughly documented and detailed so that it can be used and understood without difficulty. Moreover, a number of rather simple codes are introduced in the later chapters and their applications are included. Details of these codes, designed for the beginner, are given in appendix 4. It is recommended that these or other available codes be used by the student while to introduce the student, gradually
teaching, to the use
learning various topics in chapters 7-14.
Preface
xiii
Chapter 7 introduces the idea of higher-order approximation for the problem of beam bending and beam-column. One-dimensional problems in mass transport (diffusion-convection), overland flow due to rainfall, and wave propagation are covered in chapters 8, 9 and 10, respectively. These problems illustrate, by following the general procedure, formulations for different categories of time-dependent problems.
Chapters 11-14 enter into the realm of two-dimensional problems. The chapter on Torsion (chapter 11) and Other Flow Problems (chapter 12) have
been chosen because they involve only one degree-of-freedom
at a point.
Chapters 13 and 14 cover two-dimensional stress-deformation problems involving two and higher degrees-of-freedom at a point.
The text presents the finite element method by using simple problems. It must be understood, however, that it is for the sake of easy introduction that we have used relatively simple problems. The main thrust of the method, on the other hand, is for solving complex problems that cannot be easily solved by the conventional procedures. In order to emphasize this and to show the reader what kind of complex factors can be handled, chapter 1 5 includes a rather qualitative description of the advanced study and applications of the method. Here, a number of factors and aspects that are not covered in chapters 1-14 are stated and references are given for a detailed study. For a thorough understanding of the finite element method, it is essential that the student perform hand calculations. With this in mind, most chapters include a number of problems to be solved by hand calculations. They also include problems for home assignments and self-study. The formulations have been presented by using both the energy and residual procedures. In the former, the potential, complementary, hybrid, and mixed procedures have been discussed. In the residual procedures, main attention has been given to Galerkin's method. A number of other residual methods are also becoming popular. They are described, therefore, in appendix I, which gives descriptions, solutions and comparisons for a problem by using a number of methods: Closed form, Galerkin, collocation, subdomain, least squares, Ritz, finite difference, and finite element. Formulations by the finite element method usually result in algebraic simultaneous equations. Detailed description of these methods is beyond the scope of this book. Included in appendix to the
commonly used
direct
and
iterative
2,
however, are brief introductions
procedures for solution of algebraic
simultaneous equations. Physical models can help significantly in the understanding of various concepts of the method. Appendix 3 gives the descriptions of some physical
models. Appendix 4 presents details of a number of computer codes relevant to various topics in the text.
one or two undergraduate courses. The second course may overlap with or be an introductory graduate course. Although a
The book can be used
for
Preface
xiv
number of topics have been covered
in the
book, a semester or quarter course
could include a selected number of topics. For instance, a quarter course can
and then one or two topics from the remaining chapters. mechanics and stress-deformation analyses, the topics can be Beam Bending and Beam-Column (chapter 7), and Two-Dimensional Stress Deformation (chapter 13). If time is available (in the case of a semester course), chapter 10 on One-Dimensional Wave Propagation, chapter 11 on Torsion and/or chapter 14 on Multicomponent Systems can be added. A class oriented toward field problems and hydraulics can choose one or more of chapters 8, 9, 11, and 12 in addition to chapters 1-6. Thanks are due to Y. Yamada, University of Tokyo; Peter Hoadley, Vanderbilt University; William J. Hall, University of Illinois at Urbana; E. L. Wilson, University of California at Berkeley; John F. Abel, Cornell University and my colleagues S. Sture, T. Kuppusamy and D. N. Contractor for reading the manuscript and for offering useful comments and suggestions. A number of my students helped in solutions of some of the problems I would like to express special appreciation to John Lightner for his assistance in implementing some of the computer solutions. I realize that it is not easy to write at an elementary level for the finite element method with so many auxiliary disciplines. The judgment of this book cover chapters
For a
1-6,
class interested in
;
;
is
better left to the reader.
All natural systems are essentially continuous or interconnected,
influenced by a large
number of parameters.
we must understand all we make approximations, by
and are
In order to understand
such a system,
the parameters. Since this
possible
selecting only the significant of
is
not
them and neglecting the others. Such a procedure allows understanding of the entire system by comprehending its components taken one at a time. These approximations or models obviously involve errors, and we strive continuously to improve the models and reduce the errors.
Chandrakant
S.
Desai
INTRODUCTION
BASIC CONCEPT current form the finite element (FE)
method was formalized by civil The method was proposed and formulated previously in different manifestations by mathematicians and physicists. The basic concept underlying the finite element method is not new The principle of discretization is used in most forms of human endeavor. Perhaps In
its
engineers.
:
the necessity of discretizing, or dividing a thing into smaller manageable
from a fundamental limitation of human beings in that they cannot see or perceive things surrounding them in the universe in their entirety things, arises
or totality. Even to see things immediately surrounding us, several turns to obtain
we
a.
we must make
jointed mental picture of our surroundings. In other
around us into small segments, and the final one that simulates the real continuous surroundings. Usually such jointed views contain an element of error. In perhaps the first act toward a rational process of discretization, man
words,
discretize the space
assemblage that we visualize
is
divided the matter of the universe into five interconnected basic essences
(Panchmahabhutd), namely, sky or vacuum,
air,
added to them perhaps the most important of all,
water, earth, and time,
by singing
Time created beings, sky, earth, Time burns the sun and time will bring What is to come. Time is the master of everything
^he number
[l].
1
within brackets indicates references at the end of the chapter.
fire,
and
Chapter 1
Introduction
<**J-W
Figure 1-1 'Discretization' of universe, (a) 'Discretized' galaxy. (From Smithsonian, Nov. 1976, reproduced by permission of Mr. Clifton
(From Smithsonian, Aug. 1976, by
Line.) (b) 'Discretized' earth.
permission of Dr. Athelstan Spilhaus.)
We finite)
conceived the universe to be composed of an innumerable (perhaps
number of
solar systems, each system
composed of
its
own
stars,
our solar system we divided the planet earth into interconnected continents and oceans. The plate of earth we live
planets,
on
is
and galaxies
[Fig. 1-1 (a)]. In
composed of interconnected
When man
finite plates [Fig. l-l(b)].
started counting, the
To compute
numeral system evolved.
drew and outside
the circumference or area of a circle, early thinkers
polygons of progressively increasing and decreasing the circle, respectively, and found the value of
n
size inside
to
a high degree of
made of
blocks or elements
accuracy.
In
(civil)
engineering
(Fig. 1-2). Figure 1-3
we
started buildings
shows how early constructions by man, made of mud,
naturally cracked into blocks or elements
building clue.
;
thus nature provided us with the
1
Chapter 1
Introduction
1
1
\_
IN!
_/
r
T_
r
.44
r r
r r
+-
r r
Block or element
r r r r~
1
Figure 1-2 Building column composed of blocks or elements.
When engineers surveyed tracts of land, the track was divided into
smaller
was surveyed individually [Fig. l-4(a)]. The connecting of the individual surveys provided an "approximate" survey of the entire tract. Depending on the accuracy of the survey performed, a "closing error" would be involved. In aerial photography a survey of the total area is obtained by matching or patching together a number of photographs. The tracts,
and each small
tract
aberrations or discontinuities at the interfaces or junctions of the individual
photographs are shown in Fig.
For
stress analysis
l-4(b).
of modern framed structures in
civil
engineering clas-
methods such as slope deflection and moment distribution were used. The structure was divided into component elements, each component was examined separately, and (stiffness) properties were established (Fig. 1-5). The parts were assembled so that the laws of equilibrium and physical condition of continuity at the junctions were enforced. Although a system or a thing could be discretized in smaller systems, components, or finite elements, we must realize that the original system itself is indeed a whole. Our final aim is to combine the understandings of individual components and obtain an understanding of the wholeness or continuous sically,
nature of the system. In a general sense, as the
modern
scientific
thinking
Chapter 1
Introduction
Figure 1-3 'Discretized' hut due to cracking. {From Smithsonian, Aug. 1976, by permission of Dr. Athe Is tan Spilhaus.)
recognizes,
which the Eastern philosophical and metaphysical concepts had
recognized in the past,
all
systems or things are but parts of the ultimate
continuity in the universe!
The foregoing
many
activities
of
abstract
man
and engineering examples make us aware of the on discretization.
that are based
Chapter
Introduction
1
Closure error
Survey of smaller areas
(a)
^^5":
# /£>'
#
:
-
(b) Figure 1-4 'Discretization' in surveying,
(a)
Closure error in survey
of subdivided plot, (b) Patching of aerial photographs to obtain
assemblage of a survey.
Figure 1-5 'Discretization' of engineering structure, (a) Actual structure, (b) 'Discretized' structure, (c) Idealized
(a)
(b)
one-dimensional model.
(c)
PROCESS OF DISCRETIZATION Discretization implies approximation of the real
and the continuous.
We
use
a number of terms to process the scheme of discretization such as subdivision, continuity,
potential,
compatibility,
minimum
convergence, upper and lower bounds, stationary
residual,
and
error.
As we
shall see later,
although these
terms have specific meanings in engineering applications, their conception
we discuss some of these and aspects have been adopted from Russell [2].
has deep roots in man's thinking. In the following terms; a
number of
figures
Subdivision
Zeno argued that space is finite and infinitely divisible and that for things must have magnitudes. Figure l-6(a) shows the concept of finite
to exist they
Figure 1-6 Finiteness and nitely divisible triangle.
divisibility,
{From Ref.
[2],
(a) Infinite space,
Limited, London.)
(a)
(b) Infi-
by permission of Aldus Books
(b)
Chapter 1
Introduction
8
space If the earth were contained in space, what contained the space in turn :
[2]? Figure l-6(b) illustrates this idea for the divisibility
number of component
of a triangle into a
triangles.
Continuity is made up of divisible elements. between any two points in a line, and there exist other moments between two moments in a period of time. Therefore, space and time are continuous and infinitely divisible [2], and things
Aristotle said that a continuous quantity
For
instance, there exist other points
are consecutive, contiguous,
1
i
and continuous
(Fig. 1-7).
3 (b)
Figure 1-7 Concepts of continuity, (a) Consecutive, (b) Contiguous, (c)
Continuous. {From Ref.
[2],
by permission of Aldus Books
Limited, London.)
These ideas of finiteness, divisibility, and continuity allow us to divide continuous things into smaller components, units, or elements. Convergence
For evaluating the approximate value of n, or the area of a circle, we can draw polygons within [Fig. l-8(a)] and around [Fig. l-8(b)] the circle. As we make a polygon, say, the outside one, smaller and smaller, with a
number of sides, we approach the circumference or the area of the This process of successively moving toward the exact or correct solution can be termed convergence.
greater circle.
!
Chapter 1
The
Introduction
idea
is
analogous to what Eudoxus and Archimedes called the method
of exhaustion. This concept was used to find areas bounded by curves; the available space was filled with simpler figures whose areas could be easily
Archimedes employed the method of exhaustion for the parabola by inscribing an infinite sequence of smaller and smaller one can find the exact numerical formula for the parabola. Indeed, practitioner of the finite element method soon discovers that the
calculated.
(Fig. 1-9); here triangles,
an active
pursuit of convergence of a numerical procedure
is
indeed fraught with
exhaustion Figure 1-8 Convergence and bounds for approximate area of circle, (a)
Polygons inside
circle, (b)
gence for approximate area of
Polygons outside
circle, (c)
Conver-
circle,
A(5) = 1.322
A(4) = 1.125 units
A(6) = 1.462
A(7) = 1.540
A(8) = 1.591
A(9) = 1.628
(a)
— Chapter 1
Introduction
10
1
'^^
y
^X
l\ 1
y
J\
\ :>^-
1
1
\l
/
i
A(4) = 2.250
A(5) = 2.043
A(6) = 1.945
A(7) = 1.896
A(9) = 1.843
A(8) = 1.864
(b)
Figure 1-8 (Continued)
In the case of the circle (Fig. or outside polygon
is
1-8),
convergence implies that as the inside number of sides, we
assigned an increasingly greater
approach or converge to the area of the circle. Figure l-8(c) shows the plots of successive improvement in the values of the area of the circle from the two procedures polygons of greater sides drawn inside and outside. We can see that as the number of sides of the polygons is increased, the approximate :
areas converge or approach or tend toward the exact area.
11
Inside polygons
1.0
0.0
8
7
Number
9
10
11
12
13
of sides
(c)
Figure 1-8 (Continued)
Physical models:
The student can prepare
board or
models to
plastic)
illustrate
pictorial or physical (card-
convergence from the example of
the area of the circle.
Bounds
Depending on the course of action that we take from within or from without, we approach the exact solution of the area of the circle. However, the value from each method will be different. Figure l-8(c) shows that convergence from within the circle gives a value lower than the exact, while that from outside gives a value higher. The former yields the lower bound and the latter,
the upper bound. Figure 1-10 depicts the process of convergence to
upper and lower bound solutions.
12
Chapter 1
Introduction
Parabola
Figure 1-9 Concept of convergence or exhaustion. {From Ref.
by permission of Aldus Books Limited, London.)
Figure 1-10 Concept of bounds.
Upper bound
Exact solution-
Lower bound
[2],
Error It
should be apparent that discretization involves approximation. Consewhat we obtain is not the exact solution but an approximation to
quently,
that solution.
The amount by which we
differ
can be termed the
error.
For
example, the areas (or perimeters) of the polygons inscribed in the circle (Fig. 1-8) are
always
less
than the area (or perimeter) of the
circle,
and the areas
(or perimeters) of the circumscribed polygons are always greater than the
area (or perimeter) of the circle.
and the exact perimeter the
number of
is
sides of the
The
difference between the approximation
the error, which becomes smaller and smaller as
polygons increases.
We
can express error in the
area as
A*
= the exact area, A =
where A*
- A = e,
(1-1)
approximate area, and e
= error.
PRINCIPLES AND LAWS
To
describe the behavior of things or systems around us,
laws based on principles.
we need
to establish
A law can be a statement or can be expressed by a
mathematical formula. Principles have often been proposed by intuition, hypothesized, and then proved.
Newton's second law
states
F= ma
(l-2a)
F=mp
d-2b)
or
where
F=
force,
m=
mass, and a
displacement u with respect to time
body is
in
t.
= acceleration The
principle
or second derivative of
is
that at a given time the
dynamic equilibrium and a measure of energy contained
in the
body
assumes a stationary value.
A
simple statically loaded linear elastic column [Fig. 1-1
1(a)]
follows the
principle that at equilibrium, under a given load and boundary constraints, the potential energy of the system assumes a stationary (minimum) value and
that the equation governing displacement
v
where v
= length
=
is
= ^>
displacement in the vertical y direction,
of the column,
A =
(1-3)
AE
cross-sectional area,
P = applied load, L E = modulus of
and
elasticity.
13
14
Chapter
Introduction
A
1
= area of
cross section
4TP
tU^
Y
(b)
(a)
Figure 1-11 Structures subjected to loads (causes), (a) Column, (b)
Body
or structure.
CAUSE AND EFFECT The essence of all investigations is the examination and understanding of causes and their effects. The effect of work is tiredness and that of too much work is fatigue or stress. The effect of load on a structure [Fig. 1-1 1(b)] is to cause deformations, strains, and stresses too much load causes fatigue and ;
failure.
When effect
studying
finite
element methods, our main concern
is
the cause
and
of forcing functions (loads) on engineering systems.
The foregoing
offers a rather abstract description
process of discretization, inherent in almost
all
of ideas underlying the
human
endeavors. Compre-
hending these ideas significantly helps us to understand and extend the element concept to engineering; that is the goal of this text.
finite
IMPORTANT COMMENT Although we have presented the descriptions in this book by using simple problems, we should keep in mind that the finite element method is powerful and popular because it allows solution of complex problems in engineering and mathematical physics. The complexities arise due to factors such as irregular geometries, nonhomogeneities, nonlinear behavior, and arbitrary loading conditions. Hence, after learning the method through simple examples, computations, and derivations, our ultimate goal will be to apply it to complex and challenging problems for which conventional procedures are not available or are very
difficult.
REVIEW ASSIGNMENTS In the beginning stages of the study, student
homework
it
may prove
that requires review of
very useful to assign the
some of the
basic laws, principles,
::
Chapter
Introduction
1
and equations. This
will facilitate
understanding the method and also reduce
the necessity of reviews by the teacher.
assignments that cover topics from
Home 1.
Assignment
15
The following
are
many undergraduate
two suggested home curricula.
1
Define: (a)
Stress at a point
(b) Strain (c)
2.
Hooke's law
Define (a) Principal stresses
and
(b) Invariants of stresses 3.
(a)
strains
and
strains
Define potential energy as a
sum
of strain energy and potential of applied
loads, (b)
Give examples for analysis in
(civil)
engineering in which you have used
the concept of potential energy. 4.
Define (a)
Darcy's law and coefficient of permeability
(b) Coefficient of thermal conductivity
and
coefficient of thermal
expansion
5.
Derive the Laplace equation for steady-state flow.
6.
Derive the fourth-order differential equation governing beam bending,
wA EI ir*)= p - k w °
where
w =
displacement,
spring constant,
coordinate along
E=
p
=
axis.
= support reaction, k = = moment of inertia, and x =
applied load, and k s w
modulus of
beam
>
elasticity,
/
See Fig. 1-12.
1— Figure 1-12
Home 1.
Assignment 2
Define: (a)
(b) (c)
s
Determinant Row, column, and rectangular matrices Matrix addition and subtraction
x
Chapter 1
Introduction
16
(d)
Matrix multiplication
(e)
Inverse of a matrix
Transpose of a matrix (g) Symmetric matrix (h) Sparsely populated and banded matrices (f )
2.
(a)
Define a
set
of algebraic simultaneous equations, (b) Describe Gaussian
elimination with respect to the following equations,
2*i
4*!
and 3.
find the value of
x and x 2 x
+ +
3* 2
5* 2
= =
14, 10,
.
Define: (a)
Total derivative
(b) Partial derivative for
one variable and two variables and the chain rule of
differentiation
REFERENCES [1]
[2]
—
The Upnishads Praise of Time. See available translations of The Upnishads from Sanskrit to English, e.g., The Upnishads by Swami Nikhilananda, Harper Torchbooks, New York, 1964. Russell, Bertrand, Wisdom of the West, Crescent Books,
Books
Ltd.,
London, 1959.
Inc.,
Rathbone
THE FINITE ELEMENT METHOD STEPS IN
INTRODUCTION Formulation and application of the
finite
element method are considered to
consist of eight basic steps. These steps are stated in this chapter in a very
The main aim of this general description is to prepare for complete and detailed consideration of each of these steps in this and the subsequent chapters. At this stage, the reader may find the very general general sense.
description of the basic steps in this chapter
However, when these
somewhat overwhelming.
steps are followed in detail with simple illustrations in
the subsequent chapters, the ideas
and concepts
will
become
clear.
GENERAL IDEA Engineers are interested in evaluating effects such as deformations, stresses, temperature, fluid pressure, and fluid velocities caused by forces such as
The nature of disbody depends on the characteristics of the force system and of the body itself. Our aim is to find this distribution of the effects. For convenience, we shall often use displacements or deformations u (Fig. 2-1) in place of effects. Subsequently, when other problems such as heat and fluid flow are discussed they will involve distribution of temperature and fluid heads and their gradients. applied loads or pressures and thermal and fluid fluxes.
tribution of the effects (deformations) in a
17
18
Steps
in the Finite
Element Method
Chapter 2
Distribution of u(x, y) for entire body
u e (x, y)
Nodal
line
Additional
node Finite element
Corner or primary node
(b)
(a)
Figure 2-1 Distribution of displacement
head
(p.
(a) Discretization
tion of Ue over a generic element
We assume that tional
it is
u,
temperature T, or fluid
of two-dimensional body, (b) Distribue.
difficult to find the distribution
methods and decide to use the
finite
the concept of discretization, as explained in Chapter into a
number of
of u by using conven-
element method, which 1
.
We
based on
body
smaller regions [Fig. 2- 1(a)] called finite elements [1,2].
consequence of such subdivision
is
that the distribution of displacement
also discretized into corresponding subzones [Fig. 2- 1(b)].
elements are
is
divide the
now
easier to
distribution of u over
examine as compared to the
A is
The subdivided entire body and
it.
For stress-deformation
analysis of a
body
in equilibrium
under external
loading, the examination of the elements involves derivation of the stiffness-
load relationship.
To
derive such a relationship,
we make
use of the laws and
principles governing the behavior of the body. Since our primary concern to find the distribution of u,
terms of
u.
we
contrive to express the laws and
is
principles in
We do this by making an advance choice of the pattern,
shape, or
outline of the distribution of u over an element. In choosing a shape,
we
follow certain rules dictated by the laws and principles. For example, one law says that the loaded body, to be reliable
breaks anywhere in tinuous. Let us
now
its
and
functional, cannot experience
regime. In other words, the
body must remain con-
describe in detail various steps involved in the foregoing
qualitative statements.
Step
1.
Discretize and Select Element Configuration
This step involves subdividing the body into a suitable number of "small" bodies, called finite elements.
The
intersections of the sides of the elements
and the
are called nodes or nodal points,
interfaces between the elements are and nodal planes. Often we may need to introduce additional node points along the nodal lines and planes [Fig. 2- 1(b)]. An immediate question that arises is, How small should the elements chosen be? In other words, how many elements would approximate the continuous medium as closely as possible? This depends on a number of factors, which we shall discuss. What type of element should be used ? This will depend on the characteristics of the continuum and the idealization that we may choose to use. For instance, if a structure or a body is idealized as a one-dimensional line, the element we use is a "line" element [Fig. 2-2(a)]. For two-dimensional bodies, we use triangles and quadrilaterals [Fig. 2-2(b)]; for three-dimensional idealization, a hexahedron with different specializations [Fig. 2-2(c)] can be
called nodal lines
used.
Although we could subdivide the body into regular-shaped elements in we may have to make special provisions if the boundary irregular. For many cases, the irregular boundary can be approximated by
the interior (Fig. 2-3), is
a number of straight lines (Fig. problems,
it
may be
2-3).
On
the other hand, for
many
other
necessary to use mathematical functions of sufficient
order to approximate the boundary. For example,
if
the boundary shape
is
approximate that
we can use a second-order quadratic function to boundary. The concept of isoparametric elements that we
shall discuss later
makes use of
similar to a parabolic curve,
this idea. It
may
be noted that inclusion of
irregular boundaries in a finite element formulation poses
Step
2.
Select Approximation
no great
difficulty.
Models or Functions
we choose
a pattern or shape for the distribution (Fig. 2-1) quantity that can be a displacement and/or stress for stressdeformation problems, temperature in heat flow problems, fluid pressures
In this step,
of the
unknown
and/or velocity for fluid flow problems, and both temperature (fluid pressure) and displacement for coupled problems involving effects of both flow and deformation.
The nodal points of the element provide strategic points for writing mathematical functions to describe the shape of the distribution of the unknown quantity over the domain of the element. A number of mathematical functions such as polynomials and trigonometric series can be used for this purpose, especially
polynomials because of the ease and simplification 19
20
Steps
in the Finite
Element Method
Chapter 2
r
A
i
'
'
One-dimensional body
Line elements
(a)
2
unknowns each node
at
Quadrilateral
Triangle
Two-dimensional body
Quadrilateral and triangular elements
(b)
Hexahedron element
Three-dimensional body
(0 Figure 2-2 Different types of elements, (a) One-dimensional ele-
ments,
(b)
elements.
Two-dimensional
elements,
(c)
Three-dimensional
Chapter 2
Steps
in the Finite
Element Method
\
/
21
body
Original
Discretized
body
Figure 2-3 Discretization for irregular boundary.
they provide in the
finite
element formulation. If we denote u as the unknown,
the polynomial interpolation function can be expressed as
= N lUl + N u + N
u
Here u u u 2l u 3 , and u 2
N N
we
,
.
.
2
.
.
.
.
,
2
3
u3
+
•
•
•
+ Nm u m
,um are the values of the unknowns
Nm are the
shall give details
(2-1)
.
at the nodal points
interpolation functions; in subsequent chapters
of these functions. For example, in the case of the
we can have u and u 2 as unknowns or degrees of freedom and for the triangle [Fig. 2-2(b)] we can have u u u2 u 6 as unknowns or degrees offreedom if we are dealing with line
element with two end nodes [Fig. 2-2(a)]
x
, . . . ,
a plane deformation problem where there are two displacements at each node.
A degree of freedom can be defined as an independent (unknown) displacement that can occur at a point. For instance, for the problem of onedimensional deformation in a column [Fig. 2-2(a)], there is only one way in which a point is free to move, that is, in the uniaxial direction. Then a point has one degree of freedom. For a two-dimensional problem [Fig. 2-2(b)], if deformations can occur only in the plane of the body (and bending effects are ignored), a point is free to move only in two independent coordinate directions; thus a point has two degrees of freedom. In Chapter 6, when bending considered, it will be necessary to consider rotations or slopes as independent degrees of freedom. We note here that after all the steps of the finite element method are
is
accomplished,
we
shall find the solution as the values of the
unknowns u
at
u m To initiate action toward obtaining the u l9 u2> solution, however, we have assumed a priori or in advance a shape or pattern that we hope will satisfy the conditions, laws, and principles of the problem
all
the nodes, that
is,
.
.
.
,
.
at hand.
The reader should realize that the solution obtained unknowns only at the nodal points. This is one of discretization process. Figure 2-4
shows that the
will
be
in
terms of the
the outcomes of the
final solution is
a combina-
22
Steps
in the Finite
Element Method
Chapter 2
u(x, y)
u 3 (x, y)
Common
boundary
(b)
Finite element approximation
Exact
Common
boundary
(0 Figure 2-4 Approximate solution as patchwork of solutions over elements, (a) Assemblage, (b) Neighboring elements, (c) Section
along A-A.
tion of solutions in each element patched together at the
This
is
common boundaries.
further illustrated by sketching a cross section along A-A. It can be
seen that the computed solution
is
not necessarily the same as the exact
continuous solution shown by the solid curve. The statement in Chapter that discretization yields approximate solutions can be visualized
schematic representation (Fig. solution; that
Step
3.
is,
2-4).
Obviously,
computed solution the error is a minimum.
tion to be such that the
is
we would
from
1
this
like the discretiza-
as close as possible to the exact
Define Strain (Gradient)-Displacement (Unknown)
and Stress-Strain (Constitutive) Relationships
To proceed to the next step, which uses a principle (say, the principle of minimum potential energy) for deriving equations for the element, we must define appropriate quantities that appear in the principle.
For
stress-defor-
Chapter 2
Steps
Element Method
in the Finite
23
mation problems one such quantity is the strain (or gradient) of displacement. For instance, in the case of deformation occurring only in one direction
y
assumed to be
[Fig. 2-5(a)], the strain e , y
small,
is
given by
dv €y
where v
is
the deformation in the
direction, such a relation
is
(2-2)
dy
y
For the case of fluid flow
direction.
the gradient g x of fluid head [Fig. 2-5(b)] gx
in
one
:
(2-3)
dx
Here cp is the fluid head or potential and g x is the gradient of
with respect to the x coordinate.
Variation of
is,
rate of
g x gradient or slope
v?
of
<£
I e
y
gradient
or slope of v x,
v?
(b)
Variation
of v
Figure 2-5 Problems idealized as one-dimensional, (a) One-dimensional stress-deformation, (b) One-dimensional flow.
In addition to the strain or gradient,
we must
quantity, the stress or velocity; usually, this
ship with the strain. Such a relation alized sense,
it
is
is
is
also define an additional done by expressing its relation-
called a stress-strain law. In a gener-
a constitutive law and describes the response or effect
(displacement, strain) in a system due to applied cause (force). strain
law
is
one of the most
vital parts
The
stress-
of finite element analysis. Unless
it is
defined to reflect precisely the behavior of the material or the system, the results
from the analysis can be of very little significance. As an elementary Hooke's law, which defines the relationship of stress to in a solid body:
illustration, consider
strain
ay where a y
=
elasticity. If
= Eye
(2-4a)
y,
stress in the vertical direction
we
substitute e y
from Eq.
and
Ey =
Young's modulus of we have the
(2-2) into Eq. (2-4a),
expression for stress in terms of displacements as
~ dv dy
(2-4b)
:
24
Steps
One of
:
in the Finite
Element Method
the other simple linear constitutive laws
is
Chapter 2
Darcy's law for fluid
flow through porous media vx
= coefficient of permeability,
where k x
electrical engineering the
Step
= -k x gx
4.
(2-4c)
,
vx
= velocity,
corresponding law
is
Ohm's
and gx
= gradient.
In
law.
Derive Element Equations
By invoking
and principles, we obtain equations governing The equations here are obtained in general terms
available laws
the behavior of the element.
and hence can be used for
all
elements in the discretized body.
A
number of alternatives are possible for the derivation of element equations. The two most commonly used are the energy methods and the residual methods.
Use of the energy procedures
At
this stage
of our study of the
requires knowledge of variational calculus. finite
element method, we shall postpone
detailed consideration of variational calculus
manner introduce
and
in a
somewhat
less
rigorous
the ideas simply through the use of differential calculus.
ENERGY METHODS These procedures are based on the idea of finding consistent
states
of
bodies or structures associated with stationary values of a scalar quantity
assumed by the loaded bodies. In engineering, usually this quantity is a measure of energy or work. The process of finding stationary values of energy requires use of the mathematical disciplines called calculus of variations involving use of variational principles. In this introductory book, it is not considered necessary to elaborate on this subject.
As noted above, we
shall
introduce and use the energy methods through the familiar topic of differential calculus.
Within the realm of energy methods, there are a number of methods and variational principles, e.g., the principle of stationary potential
and com-
plementary energies, Reissner's mixed principle, and hybrid formulations,
which are commonly used
in finite element applications [3-6].
STATIONARY VALUE
maximum, minimum, or Under certain conditions, the
In simple words, the term stationary can imply a saddle point of a function F(x) (Fig. 2-6).
function
may
minimum or a maximum value. To we equate the derivative of F to zero
simply assume a
point of a stationary value,
find the
Chapter 2
Steps
in the Finite
Element Method
25
F(x)
Maximum Neutral
Minimum
Figure 2-6 Stationary values of a function.
POTENTIAL ENERGY In the case of stress-deformation analysis, the function
F is
often repre-
sented by one of the energy functions stated previously. For instance, define
Fto
simple column under the given support conditions [Fig. 2-5(a)], elastic
and
minimum
we can
be the potential energy in a body under load. If the body, say, a if it is in
equilibrium,
it
To comply
potential energy.
is
can be shown that the column with the
np
denote potential energy by the symbol
and assume
linear
will
commonly used notation we
where the subscript denotes
,
potential energy.
The
potential energy
is
defined as the
sum of the
W
internal strain energy
U
and the potential of the external loads the latter term denotes the p capacity of load P to perform work through a deformation v of the column. \
Therefore, II P
When we
= U+
apply the principle of
take the derivative (or variation) of
W
p
(2-6a)
.
minimum
potential energy,
U p and equate
it
to zero.
the load remains constant while taking the derivative
m The symbol 3 denotes
p
= 8u
sw„
;
we can
essentially
then
= o.
(2-6b)
variation of the potential energy n^.
subsequently in Eq. (2-9),
we
We assume that
interpret the variation or
As
indicated
change as com-
posed of a series of partial differentiation of 11^. Here we use the relation between the variation in potential of external loads and in work done by the loads as
dW Note
5W
(2-6c)
t
that the negative sign in Eqs. (2-6b)
potential of the external loads in Eq. (2-6a)
the external loads.
is
and
(2-6c) arises because the
lost
through the work done by
:
Steps
26
The
Element Method
Chapter 2
fact that for linear, elastic bodies in equilibrium the value of
minimum can be of n^,
in the Finite
verified
greater than zero
is
S2
Up
is
a
by showing that the second derivative or variation that
;
is,
U P =3 U-S WP 2
2
>0.
(2-7)
Proof of Eq. (2-7) can be found in advance treatments on energy methods and is not included in this text. The symbol S is a compact symbol used to denote variation or a series of partial differentiations. For our purpose, we shall interpret it simply as a symbol that denotes derivatives of 11^ with respect to the independent coordinates or expressed.
For example,
np where u ,u 2 1
3T1 P
=
,
.
.
.
,
unknowns
in terms of
which
it is
if
=np(u
w„ are the total
19
u2 ,...,uj,
(2-8)
number of unknowns
then
(at the nodes),
implies
*L=o, £L=o, du 2
(2-9)
Here n
=
total
number of unknowns. we shall
In subsequent chapters stationary energy
and other energy
illustrate the
use of the principle of
principles for finite element formulations
of various problems.
METHOD OF WEIGHTED RESIDUALS One of the two major alternatives method
is
are employed under the least squares,
for formulating the finite element
(MWR). A number of schemes among which are collocation, subdomain, methods [3, 7]. For many problems with certain
method of weighted
the
residuals
MWR,
and Galerkin's
later chapters), Galerkin's method from variational procedures and is closely related to them. Galerkin's method has been the most commonly used residual method for finite element applications. The is based on minimization of the residual left after an approximate or trial solution is substituted into the differential equations governing a
mathematical characteristics (discussed in yields results identical to those
MWR
problem. As a simple
illustration, let us consider the following differential
equation
££-£='<*>•
<2
-
10a >
Chapter 2
where w*
Steps in the Finite Element
unknown, x
the
is
is
Method
the coordinate,
27
the time,
t is
forcing function. In mathematical notation, Eq. (2-10a)
is
and f(x)
is
the
written as
Lu* =/,
(2-10b)
where
- dx is
2
dt
the differential operator.
We
are seeking an approximate solution to Eq. (2-10) and denote an
approximate or
function u for u* as
trial
n
ft
=
+
cp
(2-11)
a^j
+
+
•
•
•
+
a n
.
Here p t q> 2
,
Eq. (2-11)
.
.
,
.
written as
is
u in
which a
x
=
and
1
into Eq. (2-10),
we
are
l
=
left
(p Q
.
=±* If the
(2-12)
t
approximate solution u
is
R(x)=Lu-f
(2-13)
i
which
is
zero
In the
if
u
=
u*.
method of weighted
residuals, the
aim
to find an approximate
is
solution u for w* such that the residual R{x) in Eq. (2-13) possible or
is
substituted
with a residual
is
made
as small as
minimized. In other words the error between the approximate
and the exact solution u is minimized. A number of schemes aim of minimization of R(x) details of some of the major schemes collocation, subdomain, least squares, and Galerkin methods are given in Appendix 1. Moreover, in the subsequent chapters, Galerkin's method is used to derive finite element equations for a number of simple problems. For the sake of completeness, only brief and general statements for these methods are given below. solution w*
are possible to achieve the
;
—
—
Mathematically the idea of minimization can be expressed as
f
R(x)W (x)dx t
= 0,
I
=
1, 2,
.
.
.
,
n,
(2-14)
where D denotes the domain of a structure or body under consideration. For the one-dimensional column problem, the domain is simply the linear extent of the column.
:
28
Steps in the Finite Element
Method
Chapter 2
W
In Eq. (2-14) the denote weighting functions. Various residual schemes such as collocation, subdomain, and Galerkin use different weighting t
For
functions.
instance, in the case of the collocation
method,
W=
1.
{
The
expression in Eq. (2-14) implies that the weighted value of R(x) over the
domain of a
structure vanishes. Figure 2-7 shows a schematic representation of Eq. (2-14). The shaded areas in Fig. 2-7(b) denote error between the
approximate and the exact solution u*
D
R(x) over
—
u over the domain D.
[Fig. 2-7(a)] is related to the error
D
integral of R(x) over
u*
—
w,
The residual and the sum or the
minimized.
is
u
i
R(x)= Lu-f
(a)
Figure 2-7 Schematic representation of residual, (a) Integration of
R
over D. (b) Error u*
As a simple
—
u over D.
illustration, let us consider the following
second-order
differ-
ential equation that governs the problems of one-dimensional stress deforma-
column and flow
tion in a
in
Chapters 3 and
d 2v dy
where v*
the
is
unknown
4, respectively
3
(2-1 5a)
f,
2
(deformation), y
is
the coordinate axis, c denotes a
material property, and /is the forcing function. For a column, c the applied load, area.
Assume
E is
that
the
= EA, /is
modulus of elasticity, and A is the cross-sectional 10 units of load; then Eq. (2-1 5a) 1 and
/=
EA =
specializes to
d 2 v* dy
An
(2-15b)
10.
:
approximate solution for v* can be written as a special case of
Eq. (2-11) as v
where
t
=
satisfy the
= ai +
cc 2
y
+
(x 3
y
2
= 2 *t9t,
(2
"
16)
2 This function should be chosen so as to 1,
Chapter 2
Steps in the Finite Element
Method
29
given by
R(y)
According to the
MWR, f*
=
W~ 1
I0
(2 " 17a)
'
Eq. (2-14) leads to
Riy)W {y)dy )
= 0,
i
=
(2-17b)
1, 2, 3,
Jo
or
^(^-10)w,(y)dy = or
j^d^y w
y
(y)dy
0,
(2-17C)
=
(2 . 17d)
Here L
is the length of the column which represents the domain D. Now, we can substitute for the second derivative in Eq. (2-17) by differentiating v in Eq. (2-16) twice. The final results will yield three simultaneous equations in
au
cc 2
,
and a 3
as
[[^^-^]w
[[^^y
2
(y)dy
[ -^-10]w d
which are solved for % u
substituted into Eq. (2-16),
At
this stage,
to proceed in the
it is
from Eq.
and a 3
,
we
.
3
y)dy
i(
=
0,
= 0,
(y)dy
When
(2-18)
=
0,
these values of
a„ a 2 and a 3 ,
are
obtain the approximate solution for v*.
not necessary to go into the details of the steps required (2-17) to Eq. (2-18); they are given in
Appendix
1
and
subsequent chapters.
ELEMENT EQUATIONS Use of either of the two foregoing methods will lead to equations ing the behavior of an element, which are commonly expressed as [k]{q]
= {Q},
describ-
(2-19)
=
=
vector of unknowns at the [k] element property matrix, {q} element nodes, and {Q} vector of element nodal forcing parameters. Equation (2-19) is expressed in a general sense; for the specific problem of vector of nodal displacements, stress analysis, [k] stiffness matrix, {q}
where
=
and {Q}
= = = vector of nodal forces. Details of the matrices in Eq. (2-19) will be
developed and described fully in subsequent chapters.
Step
Assemble Element Equations to Obtain Global
5.
or Assemblage Equations and Introduce Boundary
Conditions
Our
final
aim
to obtain equations for the entire
is
body
approximately the behavior of the entire body or structure. In
that define
fact, as will
discussed in various chapters, use of the variational or residual procedure
body
relevant to the entire
;
it is
for simplicity that
we view
be is
the procedure in
Step 4 as having been applied to a single element.
Once element,
the element equations, Eq. (2-19), are established for a generic
we
are ready to generate equations recursively for other elements by
using Eq. (2-19) again and again. Then equations. This assembling process continuity (Chapter
1). It
is
we add them
together to find global
based on the law of compatibility or
requires that the
body remain continuous; that
is,
the neighboring points should remain in the neighborhood of each other after
the load
is
applied (Fig. 2-4). In other words, the displacements of two
adjacent or consecutive points must have identical values [Fig. 2-8(a)].
Depending on the type and nature of the problem, we may need to enforce the continuity conditions more severely. For instance, for deformations occurring in a plane,
it
may be
sufficient to enforce continuity
of the dis-
Figure 2-8 Interelement compatibility, (a) Compatibility for plane
problems, (b) Compatibility for bending-type problems. Slopes
may not
be
equal
Elements
Equal displacements (a)
Slope
Equal slopes or gradients
Equal displacements (b)
30
Chapter 2
Steps
On
placements only.
in the Finite
Element Method
31
the other hand, for bending problems, the physical
properties of the deformed
body under the load
the continuity of displacements
we
requires that in addition to
ensure that the slopes or the
first
deriva-
of displacements are also continuous or compatible at adjacent nodes [Fig. 2-8(b)]. Often it may become necessary to satisfy compatibility of the curvatures or the second derivative also.
tive
Finally,
we
obtain the assemblage equations, which are expressed in
matrix notation as [Ktfr}
= {R},
(2-20)
where [K] = assemblage property matrix, {r} = assemblage vector of nodal unknowns, and {Q} = assemblage vector of nodal forcing parameters. For stress-deformation problems, these quantities are the assemblage stiffness matrix, nodal displacement vector, and nodal load vector, respectively.
BOUNDARY CONDITIONS Until now we have considered Equation (2-20)
[Fig. 2-9(a)].
only the properties of a body or structure
tells
us about the capabilities of the body to
is an engineer. How he depend on the surroundings and the problems he faces; these aspects can be called constraints. In the case of engineering bodies, the surroundings or the constraints are the boundary conditions. Only when we introduce these conditions can we decide how the body will perform. Boundary conditions, then, are the physical constraints or supports [Fig. 2-9(b)] that must exist so that the structure or body can stand in space uniquely. These conditions are commonly specified in terms of known values of the unknowns on a part of the surface or boundary S and/or gradients or derivatives of the unknowns on S 2 Figure 2- 10(a) depicts a beam. In the case of the simply supported beam, the boundary S is the two end points where
withstand applied forces. will
perform
It is just like
saying that one
his engineering duties will
x
.
t
the displacements are given. This type of constraint expressed in terms of
displacements
is
often called the essential, forced, or geometric boundary
conditions.
At
the simple supports of the beam, the bending
moment
zero
that
is,
the second derivative of displacement vanishes. This type of constraint
is
is
;
often called a natural boundary condition.
Figure 2- 10(b) shows a cylinder through which fluid or temperature On the boundary S^ temperature or fluid head is known; this is the essential boundary condition. The right end is impervious to water, or
flows.
insulated against heat; then the
heat flux, which
is
boundary condition
proportional to the
first
is
specified as fluid or
derivative of fluid
head or tem-
perature. This is the natural boundary condition. To reflect the boundary conditions in the finite element approximation of the body represented by Eq. (2-20), it is usually necessary to modify these
[K]{r}= {R}
(a)
[K](7}=
{R]
Constraints
(b)
Figure 2-9 Boundary conditions or constraints, (a) constraints (b)
Body with
Figure 2-10 Examples
boundary conditions,
Body without
constraints.
of boundary conditions, (b) Pipe flow with
(a)
s,
"TT (a)
$
\ Flow (b)
Beam
with
boundary conditions.
S,
Chapter 2
Steps
in the Finite
Element Method
33
equations only for the geometric boundary conditions. Further details and procedure for such modification are given in subsequent chapters. The final
modified assemblage equations are expressed by inserting overbars as [K]{f}
Step
6.
Solve for the Primary
Equation (2-21)
is
a
set
= {R}.
(2-21)
Unknowns
of linear (or nonlinear) simultaneous algebraic
equations, which can be written in standard familiar form as
K K
ll
rl
2l r t
+ K 12 r + + K22 r +
•
2
•
2
•
•
+K +K
ln r n
2n rn
= Ru = Ri> (2-22)
K s + Km2 r + n
2
x
•
• •
+K
nn rn
=R
r
These equations can be solved by using the well-known Gaussian elimination or iterative methods. Detailed coverage of these methods is beyond the scope of this book; however, later when individual problems are considered, some of these methods will be illustrated. Moreover, brief descriptions of these in Appendix 2. At the end of this step, we have unknowns (displacements) r l9 r2 , rn These are called primary unknowns because they appear as the first quantities sought in the basic Eq. (2-22). The designation of the word primary will change depending on the unknown quantity that appears in Eq. (2-22). For instance, if the
methods are given
solution
solved for the
problem
is
,
formulated by using stresses as unknowns, the stresses
called the primary quantities.
be the
fluid or velocity
Step
7.
.
will
be
For the flow problem the primary quantity can
head or potential.
Solve for Derived or Secondary Quantities
Very often additional or secondary quantities must be computed from the primary quantities. In the case of stress-deformation problems such quantities can be
strains, stresses,
moments, and shear forces for the flow problem they ;
can be velocities and discharges. It is relatively straightforward to find the secondary quantities once the primary quantities are known, since we can make use of the relations between the strain and displacement and stress
and
strain that are defined in Step 3.
Step
8.
Interpretation of Results
final and the important aim is to reduce the results from the use of the element procedure to a form that can be readily used for analysis and design. The results are usually obtained in the form of printed output from
The
finite
the computer.
We then
select critical sections
of the body and plot the values
Steps
34
in the Finite
Element Method
of displacement and stresses along them, or often very convenient
and
less
we can
Chapter 2
tabulate the results. It
is
time consuming to use (available) routines and
ask the computer to plot or tabulate the results.
REFERENCES [1]
Turner, M.
J.,
Clough, R. W., Martin, H. C, and Topp, L. C, "Stiffness and Complex Structures," /. Aero. Sci., Vol. 23, No. 9,
Deflection Analysis of Sept. 1956. [2]
Clough, R. W., "The Finite Element Method in Plane Stress Analysis," Proc. 2nd Conf. on Electronic Computation, ASCE, Pittsburgh, Sept. 1960.
[3]
Crandall,
[4]
Argyris,
S.
J.
H., Engineering Analysis, McGraw-Hill,
New
in
York, 1956.
H., Energy Theorems and Structural Analysis, Butterworth's,
London, 1960. [5]
[6]
Desai, C. S., and Abel, J. F., Introduction Nostrand Reinhold, New York, 1972.
Element Method,
Vah
Pian, T. H.*H„ and Tong, P., "Finite Element Methods in Continuum .
Mechanic^" York, 1972. [7]
to the Finite
Finlayson,
Academic
in
B
.
Advances
A
Press,
.
,
in
Applied Mechanics, Vol.
12,
Academic
Press,
New
The Method of Weighted Residuals and Variational Principles, York, 1972.
New
ONE-DIMENSIONAL STRESS DEFORMATION
203SQ01 INTRODUCTION From
we shall consider engineering problems idealThe main motive for treating these simple problems
here through Chapter 10
ized as one-dimensional.
to introduce the reader to the details of various steps so that basic concepts can be understood and assimilated thoroughly without undue complex and is
lengthy derivations. An advantage to this approach is that hand calculations can be performed for two- and three-dimensional problems this can become ;
increasingly difficult.
Although simple problems are treated (Chapters 3-10), we introduce concepts and terms that are general and relevant to advanced theory and applications. These concepts are explained and defined in easy terms, often with intuitive and physical explanations. It may be mentioned that
many
one-dimensional idealizations permit not only simple derivations but often provide satisfactory solutions for
As our
first
many
practical problems.
problem we consider the case of a column,
strut,
or bar of
uniform cross section subjected to purely axial loading [Fig. 3- 1(a)]. Under these conditions we can assume that the deformations will occur only in one, vertical direction. Consequently, we can further assume that the column can be replaced by a [Fig. 3- 1(b)].
line
Now we
with the axial
stiffness
EA lumped
consider derivations of the
the step-by-step procedure as described in Chapter
finite
at the centerline
element method in
2.
35
One- Dimensional Stress Deformation
36
Chapter 3
ic
s Nodes
Finite
elements
S EA zMx^
777Z
(0
(b)
(a)
Figure 3-1 Axially loaded column, (a) Actual column, (b) One-
dimensional idealization,
Step
1.
(c)
Discretization.
Discretization and Choice of Element Configuration
Before we proceed further it is necessary to describe the coordinates or geometry of the column by using a convenient coordinate system. In the onedimensional approximation, it is necessary to use only one coordinate along the vertical direction. is
We
y axis. Because this coordinate system column (or structure), it can be called the
call this the
defined to describe the entire
global coordinate system.
We now
discretize the
column
into
an arbitrary number of smaller units
that are called finite elements [Fig. 3-l(c)].
The
intersections of elements are
called nodes or nodal points.
At
this stage, before Step 2,
it is
useful to introduce the concept of a local
or element coordinate system. There are a
number of advantages
local system for deriving element equations (Steps 2-4)
for multidimensional problems,
makes the required
tions extremely simple to handle. Indeed,
it is
;
its
derivatives
and integra-
possible to obtain
by using the global system, but use of the local system subsequently) facilitates the derivatives and is economical.
tions
to using a
use, particularly
(as
all
we
deriva-
shall see
EXPLANATION OF GLOBAL AND LOCAL COORDINATES For a simple explanation, let us consider an example, Fig. 3-2(a); here we need to define or survey a plot of land and relate it to a standard or global point or benchmark A. Surveying the plot could entail locating each point and establishing its distance from point A. Let us assume that point A away from the plot and that it is difficult to establish a direct relation with A. A local point B is available, and its distance from A is known, howin is
it
far
Chapter 3
One- Dimensional Stress Deformation
37
Plot of land
"^S B
Local station
"Global" station
(a)
1
r > 1
®i
>s =
©• .L-+1-
1
A
A(Dfo
j
7
-1 s
Y2
=
y2 Y3 Vi
Vi I
Global y
v/^W s
Global y
= Local coordinate = y//
^^
77777,
L =
(y-y 3 )/(//2)
(b)
(c)
Figure 3-2 Global and local coordinates, (a) Concept of global and local coordinate systems, (b) Local coordinate
point
ever. Point
1. (c)
B is
measured from node
Local coordinate measured from midnode
accessible
from
all
points in the plot.
distances of each point in the plot
3.
We
can
first
define the
from the point B; then, knowing the
distance x AB from point A to point B, it is possible to define the distances of the points in the plot with respect to point A. For example, if the distance
of any point P from
B is x B
,
then
*a
its
distance from point
= x* +
*ab-
A is (3-la)
Here we can call the measurements with respect to point B local coordiand those from point A, global coordinates. We can see that the mea-
nates,
One- Dimensional Stress Deformation
38
surements with respect to point B, which of the plot
much
is
in the vicinity,
Chapter 3
make
the definition
simpler than those with respect to point A, which
is
more
There can be a number of possible points like B with respect to which we can define the local coordinate system such choices will depend on the nature of the problem and the convenience and ease of difficult to reach.
;
measurements.
The basic idea of the use of local coordinate system(s) ment method is very similar to the foregoing concept.
in the finite ele-
LOCAL AND GLOBAL COORDINATE SYSTEM FOR THE ONE-DIMENSIONAL PROBLEM As mentioned above,
there are a
number of ways
in
which we can define local
coordinate systems for the one-dimensional problem (Fig. here two local coordinate systems, Figs. 3-2(b) and In the
first
case,
we measure
3-1).
We
consider
(c).
from the node point 1 is measured from analogous to point B and that
the local coordinate
of a generic element e [Fig. 3-2(b)]. The global coordinate the base of the column. the base
is
Note
that
A
analogous to point
node
1
is
in Fig. 3-2(a).
y then the global coordinate of any point ;
y=y+y
t
We
call the local
in the element
coordinate
is
(3-lb)
;
hence
=y- yi-
y Often
it is
0- lc)
more convenient to express the local coordinate as a nondi-
mensional number; such a procedure can considerably facilitate the integrations and differentiations involved in the subsequent computation. Here we nondimensionalize by dividing^ by the length of the element; thus o
y -y\ _ yi-yx
_
y
(3-ld)
.
i
where s = nondimensionalized local coordinate, / = length of the element, and Vi and y 2 = global coordinates of nodes 1 and 2, respectively. Note that because we nondimensionalized the coordinate by dividing by the length of the element, I = y 2 — J>i, the value of s varies from zero at node 1 to unity at node 2. In the second alternative, we can attach the origin of the local system at an intermediate point in the element, say, at the midpoint [Fig. 3-2(c)]. Here the local coordinate
is
written as
L The values of L range from
—
1
=
^-
at point
1
to
(3-2)
at point 3 to
1
at point 2.
One- Dimensional Stress Deformation
Chapter 3
An
important property of these local coordinates
39
is
that they are in
nondimensionalized form, and their values are expressed as numbers and often lie
between zero and unity.
property that imparts simplicity to the
It is this
subsequent derivations. Step
2.
Select Approximation for the
Unknown
One of the main grasp at this stage
Model
or Function
(Displacement)
ideas in the finite element is
method
that the reader should
the a priori or in advance selection of mathematical
functions to represent the deformed shape of an element under loading. This implies that since
it is
a closed form or exact solution,
difficult to find
we
guess a solution shape or distribution of displacement by using an appropriate this function, we must follow the laws, and constraints or boundary conditions inherent in the problem. The most common functions used are polynomials. In the initial stages of the finite element method, the polynomials used were expressed in terms of generalized coordinates however, now most finite element work is done by using interpolation functions, which can often be considered as transformed generalized coordinate functions.
mathematical function. In choosing principles,
;
Generalized Coordinates
The
we can use
simplest polynomial that
is
the one that gives linear
variation of displacements within the element [Fig. 3-3(b)],
v
=a +
oc 2
1
(3-3a)
y,
or in matrix notation. l
[1
y]\
\
<;)
(3-3b)
or
W= where o^ and a 2 in the element,
y and y2 To show x
- 3c
)
= generalized coordinates, y = the coordinate of any point and v = displacement at any point in the element. The a's
contain the displacements at nodes v .
(3
[+]{«}.
this,
we
first
x
and v 2 and the coordinates of nodes 1 and 2 by substituting
evaluate v for points
fory, Vl
=«,
+a y
v2
=
+
<%!
7
l9
a 2j 2
,
(3-4a)
(3-4b)
or in matrix notation,
N=P "IN
(3-40
= [A]{«}.
(3-4d)
or {q}
One- Dimensional Stress Deformation
40
©
H
v
©
.
©
y3
Yi
Y2
h
'1
^
V
Chapter 3
(a)
V =
OLy
+
OL
2
Y
(b)
N, =/ 2 //=1-sorN 1 = 1(1
-
L)
Equal
© A ©© A © (e)
Figure 3-3 Linear interpolation functions and interelement compatibility.
Here
r
{q}
=
[v x
v2]
is
the vector of nodal displacements. Second,
we
solve
for {a} as
{a}
= [A]-i{q},
where
(3-5a)
[A]-
-yi~
yi
1
_-i
l/l
~ 1
yi
-
(3-5b)
i
yi
-y\ _-i
-yC l
(3-5c)
.
Chapter 3
Here
One- Dimensional Stress Deformation
= Jacobian
|/|
element.
=y —y =
determinant
2
I
l
41
equals the length of the
Thus
\[y
fai]
/
x
(2
-y
2
(3-5d) 1
1
x
(2
1)
2)
(2
v2 \ x 1)
Therefore
(3-5e)
which shows that
a's are functions of
and made up of y u y 2i v l9 and v 2 Note .
that a's are related to but are not explicit functions of nodal displacements.
This
is
one of the reasons we call them generalized coordinates. can substitute {a} into Eq. (3-3) to express v in terms of the
Now we
nodal displacements: v
=
y 2v
i
-y,v 2
+
-»«+*> (
(3-6a)
,,
)
-(Z ^)*.-(ZL f 2 )^ =
fv,
=
tf,*,
l
+ fv = l
2
(l
-
y.) Vl
(3
+
j-v 2
+Nv = — s, N = y/l = s,
6b >
(3-6c)
(3-6d)
2i
2
"
where iVj = 1 — y/l 1 and y === /, [Figs. 3-2(b) and 2 3-3]. Equation (3-6d) leads us to the concept of interpolation function models. In this equation, N and N2 are called interpolation, shape, or fozs/s functions. The displacement v at any point in the element can now be expressed as x
v=[N,
N ]H
(3-7a)
2
= [N]{q}, where [N]
is
(3-7b)
called the matrix of interpolation, shape, or basis functions.
property of the interpolation functions
example, in the above,
N +N = t
2
is
that their
sum
A
equals unity. For
1
INTERPOLATION FUNCTIONS Since our aim in the finite element analysis
is
to find nodal displacements
v x and v 2 we can see the advantage of using approximation models of the type in Eq. (3-7). In contrast to Eq. (3-3a), here the displacement v is expressed directly in terms of nodal displacements. Also, the use of interpola,
42
One- Dimensional Stress Deformation
tion functions
makes
it
Chapter 3
quite easy to perform the differentiations
and
integra-
tions required in the finite element formulations.
In simple words, an interpolation function
of unity at the node point to which
it
is
pertains
a function that bears a value
and a value of zero
nodes. For example, Figs. 3-3(c) and (d) show distributions of
along the element; the function point
JV,
pertains to point
1,
and
N
2
N
x
at other
N
and
2
pertains to
2.
For a given element it is possible to devise and use different types of local coordinates and interpolation functions N Let us consider the second coordinate system [Fig. 3-2(c)] described in Step 2 and in Eq. (3-2), and t
.
express the v as
v=i(l-L) Vl +i(l+L)v
= N v, + N v 2
x
(3-8a)
2
(3-8b)
2
= [tfi N ]l Vl \
(3-8c)
=
(3-8d)
2
Note
that
N
t
(i
=
1,
[N]{q].
2) in
Eq. (3-8b) are different from
N
t
(i
=
1,
2) in Eq.
same linear variation within the element. Alternative coordinate systems based on local measurement from other points on the element are possible; for example, see Prob. 3-1. (3-6d); however, they yield the
RELATION BETWEEN LOCAL AND GLOBAL COORDINATES
An
important point to consider is that a one-to-one correspondence exists between the local coordinate s or L and the global coordinate y of a point in the element. For example, for L in Eq. (3-2) we have
y
= i(l - L)y + ^(1 + L)y = #iJ>i + N y = [N]{y„], t
2
where
T {y„]
=
[y
t
y2 ]
is
(3-9a)
2
(3-9b)
2
(3-9c)
the vector of nodal coordinates.
In fact an explanation of the concept of isoparametric elements, which
is
can be given at this stage. A comparison of Eq. (3-8) and (3-9) shows that both the displacement v and the coordinate y at a point in the element are expressed by using the same (iso) the most
common
procedure
interpolation functions.
An
now
in use,
element formulation where
similar) functions for describing the deformations in
(or geometry) of an element
This
is
is
we use the same (or and the coordinates
called the isoparametric element concept
rather an elementary example;
we
and more general isoparametric elements.
shall subsequently
[1].
look at other
VARIATION OF ELEMENT PROPERTIES We often tacitly assume that the material properties such as cross-sectional area A and the elastic modulus E are constant within the element. It is not necessary to assume that they are constant. We can introduce required variation, linear or higher order, for these quantities.
For instance, they can
be expressed as linear functions:
E = N,E + N E = [NtfEJ, A = N A + N A = [N]{A„}, E and {A„} r = [A A are the x
T
where [E n } of E and A
= at
[E
x
X
2
2
(3-10)
t
2
2
(3-11)
2]
nodes
1
2]
x
and
vectors of nodal values
2, respectively.
REQUIREMENTS FOR APPROXIMATION FUNCTIONS As we have
stated before, the choice of an approximation function is guided by laws and principles governing a given problem. Thus an approximation function should satisfy certain requirements in order to be acceptable. For general use these requirements are expressed in mathematical language. However, in this introductory treatment, we shall discuss them in rather
simple words.
An
approximation function should be continuous within an element. The
linear function for v [Eqs. (3-7)
and (3-8)]
is
indeed continuous. In other words,
it
does not yield a discontinuous value of v but rather a smooth variation of
v,
and the variation does not involve openings, overlaps, or jumps.
up
The approximation function should provide interelement compatibility by the problem. For instance, for the column prob-
to a degree required
lem involving
axial deformations,
it
is
necessary to ensure interelement
compatibility at least for displacements of adjacent nodes. That
is,
the
approximation function should be such that the nodal displacements between adjacent nodes are the same. This is shown in Fig. 3-3(e). Note that for this case, the higher derivatives such as the first derivatives may not be compatible.
The displacement
at node 2 of element 1 should be equal to the displacement node 1 of element 2. For the case of the one-dimensional element, the linear approximation function satisfies this condition automatically. As indicated in Chapter 2 (Fig. 2-8), satisfaction of displacement compatibility by the linear function does not necessarily fulfill compatibility of first derivative of displacement, that is, slope. For axial deformations, however, if we provide for the compatibility up to only the displacement, we can still expect to obtain reliable and convergent solutions. Often, this condition is at
tied in with the highest order of derivative in the energy function such as the
potential energy. derivative dv/dy
For example,
= €,
is
1
;
in
Eq. (3-21) below, the highest order of
hence, the interelement compatibility should 43
:
One- Dimensional Stress Deformation
44
include order of v at least
up
to
(zero), that
is,
Chapter 3
displacement
v.
In general,
should provide interelement compatibility up to order
the formulation
—
n 1, where n is the highest order of derivative in the energy function. Approximation functions that satisfy the condition of compatibility can be called conformable.
The
other and important requirement
is
that the approximation function
should be complete; fulfillment of this requirement will assure monotonic convergence. Monotonic convergence can be explained in simple terms as a process in which the successive approximate solutions approach the exact solution consistently without changing sign or direction.
For
instance, in
approximate areas approach the exact area in such a way that each successive value of the area is smaller or greater than the previous value of area for upper and lower bound solutions, respectively. Completeness can be defined in a number of ways. One of the ways is to Fig. l-8(c) the
relate
it
to the characteristics of the chosen approximation function. If the
function for displacement approximation allows for rigid body displacements
(motions) and constant states of strains (gradients), then the function can be
considered to be complete.
mode For
A
rigid
body motion represents a displacement
that the element can experience without development of stresses in
it.
instance, consider the general polynomial for v as
v
=a + 2
cc 2
y
In Eqs. (3-3a) and (3-6)
|
+
a 3y 2
+
a 4 j> 3
we have chosen
the general polynomial as
+
•
•
•
+
n cc n+
(3-3d)
iy
a linear polynomial by truncating
shown by the
vertical
dashed
line.
The
linear
approximation contains the constant term a, which allows for the rigid body displacement mode. In other words, during this mode, the element remains
and does not experience any strain or stress, that is, a 2 y = 0. The requirement of constant state of strain (e y) for the one-dimensional column deformation is fulfilled by the linear model, Eq. (3-3a) because of the existence of term a 2 y. This condition implies that as the mesh is refined, that in each element is, the elements become smaller and smaller, the strain e y
rigid
approaches a constant value. In the case of one-dimensional plane deformations in the column, the
condition of constant state includes only e y
may
— the
first
derivative or gradient
and more general problems such as beam and plate bending. In such cases, it will be necessary to satisfy the constant strain state requirement for all such generalized strain or gradients of the unknowns involved, e.g., see Chapters 7, 11-14. In addition to monotonic convergence, we may be interested in the rate of
v.
Additional constant strain states
exist in other
of convergence. This aspect is often tied in with the completeness of the polynomial expansion used for the problem. For instance, for the onedimensional column problem, completeness of the approximation function requires that a linear function, that
is,
a polynomial of order («) equal to one
Chapter 3
is
One- Dimensional Stress Deformation
needed. Completeness of the polynomial expansion requires that
including and up to the satisfied for the linear
from Eq.
first
order should be included. This
model, Eq. (3-3a) since
(3-3d). In the case of
cubic approximation function is
45
all
terms up to n
beam bending, Chapter is
6,
to
and includes the order
It
may happen
not include satisfy the
all
we
all
it
terms
automatically
=
1
are chosen
shall see that a
Then
required to satisfy completeness.
necessary to choose a polynomial expansion such that
up
is
includes
all
it
terms
3.
that an approximation function of a certain order (n)
may
terms from the polynomial expansion, Eq. (3-3d), and
still
requirements of rigid body motion and constant states of
As an example of a complete approximation
satisfying rigid
strain.
body motion and
constant state of strain in a two-dimensional problem, but not complete
of the polynomial expansion, see Chapter 12, Eq. (12-9). For two-dimensional problems, the requirement of completeness of polynomial expansion can be explained through the polynomial expansion represented by using Pascal's triangle, see Chaps. 11-13. Here we have given rather an elementary explanation of the requirements for approximation functions. The subject is wide in scope and the reader interested in advanced analysis of finite element method will encounter the subject quite often. For example, the completeness requirement can be further explained by using the so-called "patch test" developed by Irons [2]; it is discussed in refs. [3] and [4]. Moreover, the approximation model should satisfy the requirements of isotropy or geometric invariance [5]. These topics are considered beyond the scope of this text. in the sense
Step
3.
Define Strain-Displacement and Stress-Strain Relations
For stress-deformation problems, the actions or causes (Chapter 2) are and the effects or responses become strains, deformations, and stresses. The basic parameter is the strain or rate of change of deformation. The link connecting the action and response is the stress-strain or constitutive law of the material. It is necessary to define relations between strains and displaceforces,
ments and stresses and strains for the derivation of element equations in Step 4. Hence, in this step we consider these two relations. We note at this stage that although we use familiar laws from strength of materials and elasticity for the stress-deformation
problem,
the relations relevant to specific topics.
flow problem (Chapter
4),
the relation
in later chapters
we
shall use
For instance, in the case of the fluid between gradient and fluid head and
Darcy's law will be used.
Returning to the axial deformation of the column element, the displacement relation, assuming small strains, can be expressed as ey
= ^> ay
strain-
(3-12a)
One- Dimensional Stress Deformation
46
Chapter 3
where e y = axial strain. Since we have chosen to use the local coordinate L and since our aim is to find dv/dy in the global system, we can use the chain rule of differentiation as
€
Now, from Eq.
(3-2)
dv
,~ 101 v (3 " 12b)
= d_(l^M)=\ dy\
dy
dL
dL dv -ayTL'
we have d_L
and from Eq.
dv = Ty =
>
(3-12c)
1/2
(3-8)
= ^[±(1 -
L)v t
+ i(l +
=
L)v 2 ]
1]N.
±[-1
(3-12d)
Substitution of Eqs. (3-12b)-(3-12d) into Eq. (3-12a) leads to
= -}-[-
e,
1
1]H
(3-13a)
or
=
{€,} (1
where
[B]
=
(1//)
x
x [— 1
1)
x
{q} 2) (2
1)
is
a one-dimensional problem, the strain vector
contains only one term, and the matrix [B]
{e y}
(3-13b)
,
x
can be called the strain-displacement transforma-
1]
tion matrix. Because this
[B] (1
is
only a row vector.
retain this terminology for multidimensional problems
;
We
shall
however, with multi-
dimensional problems these matrices will have higher orders.
The student can
easily see that Eq. (3-1 2d) indicates constant value
strain within the element; this
the displacement.
is
because
we have chosen
of
linear variation for
We can then call this element a constant-strain-line element.
STRESS-STRAIN RELATION For
simplicity,
we assume
elastic (Fig. 3-4).
that the material of the
column element
is
linearly
This assumption permits use of the well-known Hooke's
law,
ay
— Ey € y
(3- 14a)
,
or in matrix notation, {«,} (1
where
[C]
is
X
1)
=
[C] (1
X
{€,}
1)(1
X
,
(3-14b)
1)
the stress-strain matrix. Here, for the one-dimensional case,
matrices in Eq. (3- 14b) consist of simply one scalar term.
One- Dimensional Stress Deformation
Chapter 3
47
i\
A^ = U cl u
-
III
Figure 3-4 Linear elastic constitutive or stress-strain (Hooke's) law.
Substitution of Eq. (3-13) into Eq. (3-14b)
now
allows us to express {a y}
in terms of {q} as
{«y } Step
A
4.
(3-15)
Derive Element Equations
number of procedures
Among
= [C][B]{qJ.
are available for deriving element equations.
and residual methods. Principles based on and complementary energies and hybrid and mixed methods are used within the framework of variational methods. As described in Chapter 2 and in Appendix 1, a number of schemes such as Galerkin, collocation, and least squares fall under the category of residual methods. We shall use some of these methods in this chapter and subsequently in other chapters. these are the variational
potential
PRINCIPLE OF In simple words,
MINIMUM POTENTIAL ENERGY if
a loaded elastic body
is
in equilibrium
under given geo-
metric constraints or boundary conditions, the potential energy of the
deformed body assumes a stationary value. In the case of linear elastic bodies is a minimum; since most problems we consider involve this specialization, for convenience we shall use the term minimum. Figure 3-5 shows a simple axial member represented by a linear spring in equilibrium, the value
with spring constant k(F/L). Under a load P, the spring experiences a displacement equal to v. The potential or the potential energy n p of the spring is composed of two load (see Chapter 2) parts, strain energy U and potential p of the external
n,
The
strain energy
U
W u+ W
:
B.
(3-16)
can be interpreted as the area under the stress-strain when we minimize II,, we differentiate or
curve (Fig. 3-4). Mathematically,
:
One- Dimensional Stress Deformation
48
Chapter 3
IP
I
n
w^% Figure 3-5 Idealized linear spring.
take variations of Tl p with respect to the displacement v. While doing this we assume that the force remains constant, and we can relate variation of work
W by the load and the potential of the load as
done
SW
6Wr
(3-17)
where 8 denotes arbitrary change, variation, or perturbation. For our purpose
we can consider
it
to imply a series of partial differentiations.
sign in Eq. (3-17) occurs because the potential
into is
work by
these loads,
W
p
The negative
of external loads
is
lost
W. Then the principle of minimum potential energy
expressed as
sn B = su + 8wD =
su-sw =
(3-18)
o.
There are two ways that we can determine the minimum of II,: manual and mathematical. Both involve essentially an examination of the function represented by lip until we find a minimum point. For simple understanding, we first consider the manual procedure and write the potential energy for the spring (Fig. 3-5), assuming undeformed state of the spring as the datum for potentials, as
n
-Pv Ikv - Pv, 2
(3-19a)
where kv = force in the spring and %(kv)v denotes strain energy as the area under the load-displacement curve (Fig. 3-5). Since the load in the spring goes from to kv, we have to use average strain energy. The term Pv denotes the potential of load P; since we have assumed P to be constant, this term does not include £. We further assume that P = 10 units and k = 10 units per unit deformation. Then
U p = \\0v 2
Now we
search for the
;
a positive v
is
=
5v 2
-
minimum by examining
various values of deformations table
lOv
assumed
v.
The
(3- 19b)
10v.
values of the potential Il p for
results are
shown
in the following
to act in the direction of the applied load
One- Dimensional Stress Deformation
Chapter 3
V
n,
-2.000 -1.000
+40.000
49
+ 15.000
0.000
0.000
0.125
-1.1719 -2.1875 -3.7500 -5.0000
0.250 0.500 1.000
2.000
0.0000
3.000
15.0000
4.000
40.0000
5.000
75.0000
etc.
Figure 3-6 shows a plot of lip versus value at v will
=
1.
deform by
On
Hence, under 1
known
is
can be seen that lip has a
when
minimum
in equilibrium,
we can perform the procedure of "going to and minimum by using mathematics. It is
11, to find its
that a function assumes a
derivative
v. It
10 units, the spring,
unit.
the other hand,
on the function
P=
zero.
Applying
minimum
\kvbv
Pdv
well
value at a point(s) where
this principle to II, in
SU,
fro"
Eq.
(3- 19a),
we
its
obtain
=0
or (kv
- P)Sv = 0.
(3-20a)
Figure 3-6 Variation of potential energy.
/
i
40
Is** 30 20
\
\ \ 10
\
\ -3
-1
^^" '2
3
Minimum
point
1
-10 -20 -30
4
5
v
One- Dimensional Stress Deformation
50
Since Sv
is
arbitrary, the term in parentheses
Chapter 3
must vanish. Therefore,
kv-P = or
=P,
kv
(3-20b)
which is the equation of equilibrium for the spring. STl p = in Eq. (3-20) is analogous to equating d\\ p \dv = 0, which will result in the same equilibrium equation (3-20b). Substitution of the numerical values gives \0v
Therefore, v
We method
=
1.0 unit, the
=
10.
same answer
as before.
note here that in most problems solved by using the
Hp
is
a function of a large
ments. Consequently,
it
is
number of parameters
most economical and
matical methods because the manual procedure
finite
element
or nodal displace-
direct to use the is
mathe-
cumbersome and
often
impossible.
The mathematical procedure involved minimization of lip. For simplicity, we may view the process as simply taking derivatives of Tl p In general, how.
ever, the minimization will involve calculus of variations. In
treatment in this book
we
Now we return to the column element (Fig.
n> =
most of the
shall use the simple differentiation concept.
3-7)
and write FL
as
[5]
M
c e dv iff ^ ' >
~
iff
fvdv
~
if
T'vds
S P„v„
Figure 3-7 Generic column element with loads.
jp„
© U\®
©n
% \
^
(3-2 la)
One- Dimensional Stress Deformation
Chapter 3
=
where \o y e y
strain energy per unit
Ty =
(weight) per unit volume,
S
area,
= part
x
M — number
maximum
A
value of
vt
level,
M=
Y = body
=
Pa =
applied
displacement corresponding to
the joint forces
P a is now appropriate. Pu can be treated as
Pih
Since a
applied to the total structure,
is
force
2.
contribution of the joint force
we
volume,
of points at which joint forces are applied; here the
comment concerning
joint force
=
surface loading or traction per unit surface
of the surface on which surface loading acts,
nodal (joint) forces at local
and
V
volume,
51
Pig
applied at point
i
a local
in the global sense. Later
more convenient to add contributions of concentrated when we consider potential energy of the entire body; becomes clearer when the total structure is considered.
shall see that
it is
joint or nodal forces their relevance
The terms
Eq. (3-21 a) are essentially similar to those in Eq.
in
now we
except that
assign
volume
(3- 19a),
to the element instead of treating
it
as a
spring.
For the present, we assume that the cross-sectional area A of the element is
constant; then Eq. (3-2 la) reduces to
- A\
Tyvdy -
n,
=
^
Ty
is
the (surface) loading per unit length along the centerline of the
Here
idealized line
o y e y dy
column
Yvdy-
£ Pav
t
(3-21b)
.
(Fig. 3-7).
Equation (3-2 lb) can
now
be expressed
in
terms of the local coordinate
system by using the transformation of Eq. (3-2) as
dy
=
(3-22)
-L-dL.
Therefore, 1
11,
=
^
Next we
ay€ y
dL-^C
YvdL-l-j TyvdL-tP»Vi'
(«lc)
J"'
and a y from Eqs.
substitute for v, €y9
(3-8), (3-13),
and
(3-14),
respectively, in Eq. (3-21c) to obtain, in matrix notation,
n* = T
j\ _1
F
(1
x
1)(1
-yj "'
^WL "t£
[C]
{€'
x
1)(1
[N] (1
x
x
2) (2
x
(1
1)
dL-^
T,
{q} 1)(1
x
[N]
1)
x 2)(2 x
Pu (1
x
f dL
{q} 1)(1
x
1)
(3-21d)
v,
l)(l
x
1)
One- Dimensional Stress Deformation
52
Chapter 3
or
=^
n
{qf
[W
x
x
(1
Al
2) (2
x
-yj
[B]
x 1)(1 x
1)(1
[N] (1
[C]
Y
{q}
x
2) (2
[N]
x
1)(1
x
2) (2
1)
M r
x
x
2) (2
x
1)(1
(3-2 le)
T
1
(1
1)
dL
fy dL-
{q}
dL
{q}
(1
1)
x
1)(1
x
1)
where Y and Ty are assumed to be uniform. Equation (3-21) represents a quadratic function expressed in terms of v and v 2 In matrix notation, transposing in Eq. (3-2 Id) is necessary to make .
x
the matrix multiplication in {e } r [C]{€ >,} consistent so as to yield the scalar 3;
a y e y = Eel m Eq. (3-2 lc). The need for transposing will become when we expand the terms in Eq. (3-2 lc). The last term denotes sum-
(energy) term clear
mation,
P uv
l
+P
M=
2 iV 2 , if
2.
EXPANSION OF TERMS
We now them
consider the
first
three terms in Eq. (3-2 le) one by one
and expand
as follows:
First term
"-*!>
1
—
{v\
2v v 2 l
+
..{:)-
v\)dL.
(3-23a)
Second term:
AIY
W Here
Y
is
assumed
fy
for uniform
[y(l
to be
-
L)v,
+ i-(l +
L)v 2 ~\^dL.
(3-23b)
uniform gravity load per unit volume. Similarly,
the third term yields
W
pi
Finally, the
'
C
¥
sum of
Eq. (3-2 le) gives
J',
[t (1 "
L)Vl
+
T
(1
+ L >*\ dL
-
(3 " 23c >
Eqs. (3-23a), (3-23b), and (3-23c) and the last term in
One- Dimensional Stress Deformation
Chapter 3
¥l',[i Notice that
2) to find its
+
invoke the principle of
minimum
+ L>
2
|
-
C P,^.
if J
minimum
.
potential energy (see Chapter x
\\
2
=
i-(l
value by differentiating E^, with respect to v and v 2 as
Tyl
gig
L) Vl
a quadratic or second-order function in Vj and v 2
II p is
Now we
-
(1
53
T
(1
<" 2 ^
" +
L)dL
"
2v ^ dL
Pl/
=
~ ^f\\
t
(3 " 25a)
°'
T
(1
+
L)dL
X
~
^2
T
\
Although we used
(1
+
L)flfL
~
P2/
=
(3 " 25b)
°-
and partial differentiations to find advanced applications to multidimensional problems, it is more concise and convenient to use calculus of variations; then both Eq. (3-25a) and Eq. (3-25b) are written together as
the
minimum of lip,
differential calculus
for
SU P = where 6
is
the variation notation.
We
(3-26)
0,
shall briefly discuss variational princi-
ples at various stages, but the majority of derivations in this
book
will
be
obtained by using differential calculus.
INTEGRATION For the one-dimensional problem, the integrations in Eq. (3-25) are simple; in fact, the first term in these equations is independent of the coordinate L
;
hence integrations involve only constant terms. Thus,
aeV 41
L
1
(2»,
-
2v 1 )dL
= ^(2t>, -
-1
=
4\
2v 2 )L
4) AIY 2
One- Dimensional Stress Deformation
54
Chapter 3
Integrations of other terms in Eq. (3-25) lead to
^0>, ^fi-Vi
v2)
+v
2
-Mf-TA- P u =
)-^-^- P
2l
0,
(3-27a)
= 0.
(3-27b)
In matrix notation,
AE
1
-1
I
1
1
or
= {Q}.
M{q} Here it is
[k]
=
stiffness
(3-28b)
matrix of the element, and with the linear approximation
identical to the matrix of stiffness influence coefficients in matrix struc-
tural analysis;
{Q}
=
it is composed of body and the joint loads. It is
element nodal load vector, and
force (due to gravity), surface traction or loading,
interesting that the finite element derivations with the linear result in
approximation
AIY
lumping of the applied loads (equally) at the two nodes:
= weight of the element, where Al = volume of the element, and fy = total l
surface load.
We note that if higher-order approximations were used, the load come out
vector would not necessarily
as
lumped loading; we
shall consider
such cases at later stages.
Equation (3-28) provides a general expression that can be used repeatedly to find stiffness relations for
In the foregoing,
we
all
elements in the assemblage.
derived element equations by differentiating the
expanded function for Tl p This is possible because there are only two variables, v and v 2 Advanced problems, however, involve a large number of variables, and we commonly write the results directly in terms of matrix equations. For instance, differentiation of II P in Eq. (3-21e), .
.
x
dn£
_Q
*n*-o. dv 2 leads to
*L
[BY (2
x
[B]
[C] 1)(1
x
1)(1
x
dL{q}
=
Al 2
(2
2)
Y dL
[Nf _1
x
1)(1
x
+
1)
dL
f,
(2x
1)(1
x
1)
+
{P//} (2
/=
x
1)
1,2,
or [k]{q]
= {Q},
(3-28b)
Chapter 3
One- Dimensional Stress Deformation
55
where ai f
1
h=%
[B]
T
E[B]dL
and (Q}
The terms
in
=
y J'
[NTF^L
+ -J-J"
[Nff/L -
{P,}.
Eq. (3-28) are the same as in Eq. (3-25), except that here they The transpose in [N] r arises because after
are arranged in matrix notation.
differentiations [Eq. (3-25)] the force terms yield
order of 2 in the final results. After the process
?.
load vector that has an
on n^, the a set of (linear) simultaneous equations [Eq. (3-28b)]. Often, the stiffness matrix [k] is called an operator, which means that when operated on (nodal) displacements, the results are the (nodal) forces.
end
result
o. differentiation
is
COMMENT The primary
task in finite element formulation can be considered to
involve derivation of element equations. involve assembly, patching
The subsequent steps essentially up or combination of elements, and use of linear
algebra for solution of the resulting simultaneous equations. Step
5.
Assemble Element Equations
to
Obtain Global Equations
POTENTIAL ENERGY APPROACH Although, for simple understanding, we considered the equilibrium of a single element in the foregoing step, it is necessary to emphasize that it is the equilibrium of the entire structure in which we are interested. Consequently, we look at the total potential energy of the assemblage and find its stationary (minimum) value. The procedure of assembling element equations can be interpreted through minimization of total potential energy.
As an example, we
consider the column (Fig. 3-1) divided into three
elements with four nodes (Fig.
3-8).
Here we have numbered nodes starting from the top of the column and have measured the global coordinate y as positive downward. This is just for convenience; for instance, since the loads act downward, they are positive, and displacements are positive. It is, however, possible to measure y as positive upward or downward from any convenient point and to number nodes in other consistent fashions.
By using Eq. (3-24) we can write the potential energies H p for each element in the assemblage and add them together to obtain total potential energy IP, as
One- Dimensional Stress Deformation
56
Global
Local
Nodes
Displacement
Nodes
Properties
Displacement
S ©
©
A
Chapter 3
" y
©
© © © 1 J
v2
A
An to ,
©
©
«3
© ©
1
J
/
»2
v?
A © ©
v4
Figure 3-8 Discretization of column and numbering.
2
+
»/
—
1
^fj"'
W - 2v v 3
+ vl)dL
4
^f
[y
-
i>.
+
yO
+ L ^] rfL
y^ J'
[yd -
L) Vl
+
-1(1
+ L)v^dL
y^
[yd " L> + yd + L)»4 ]rfi
fir
3
3
J'
~
¥
-
%^ J'
/',
- P!,«, -
[t°
~~
L)v '
+ T (I + LKl rfL
[yd " iK + yd +
(PL
+ P*u)v - (P|, + 2
£)»,]<«.
P\,)v 3 (3-29)
One- Dimensional Stress Deformation
Chapter 3
57
Here we assume that A,E, /, Y, and fy are different for different elements and an element number, e denotes an element, the super= total number of elements. Note script on P denotes an element, and that the local joint or point loads at common nodes are added together and 2 yield the global joint or point loads at these points. For instance, PJ, + P U gives global point load at global node point 2. For global equilibrium we minimize 11^ with respect to all four nodal displacement unknowns, v u v 2 v 3 and v 4 Thus, their subscript denotes
M
,
dU
=
dv f
^f -
^
<2
^ -
.
,
2v ^ dL
-
^r f y 1
~
({
L)dL
1
[yO -
J
- Ph =
L)\dL
(3-30)
0,
and so on. In combined variational notation we write
5n p
(3-31a)
which denotes cttPp
dn
dU p [
dv 3 dll'p
dv 4 After the required integrations and arranging the four equations in matrix notation,
j
we
AiEi
h
\-AiEi ll
obtain
-AiEi h
A\E h
X
v\
A2E2
-A2E2
h -A2E2 h
h
V2
j
!
!
A2E2
A 3 E3
-A E
h
h
h
3
3
V
3
i
-A E 3
A 3 E3 h
3
h A\\\Y\
_j_
Ty\ll
2
2
Axhfx
fylh
+
P\,
A2I2Y2
fylh 2
^3/3^3 fylh A2I2Y2 fylh -^— + -2T + -l~ + -2~ A Y — + 2 ^p ,
,
3 l3
~2
3
,
7^3/3
1
3
2/
Pii+Ph (3-32a)
Ph + Pl
+ One- Dimensional Stress Deformation
58
Chapter 3
or
=
[K]{r]
(3-32b)
{R},
where [K] = assemblage stiffness matrix, {r} = [v v 2 v 3 v 4 ] = assemblage nodal displacement vector, and {R} = assemblage nodal load vector. r
l
METHOD
DIRECT STIFFNESS
The foregoing approach can be explained and understood alternatively through the direct stiffness method [6]. We note, however, that the basic idea of assembly evolves essentially as a result of the minimization of total potenenergy.
tial
A
close look at Eq. (3-32a)
added together;
this
shows that the
common node
corresponding to the
coefficients
indicated by enclosing
is
common node
stiffness
for elements
them
in
dashed
or influence
and 2 are
1
lines.
Simi-
added together. This interpretation can lead to the familiar concept of obtaining the assemblage matrix by adding
larly,
loads at the
are also
individual element matrices of contributing elements through the direct
stiff-
ness approach.
Let us express the matrix equations for the three elements by labeling the terms with subscripts as
—* —>
Global Local 1
»
A E X
2
2
"
1
2
1
2
1
-r
Local Global \v\
>vA
i_ [vl
>v 2 \
X
_-l
h
\
—> —>
Global Local 2
l
3
2
2 1
"
1 A 2 E2 h _-l
2
-r
3
4
>
1
2
3
l
"
1 A E h _-l 3
4
2
3
+
r 1/ (3-33a)
Local Global \v\
i_ [vl
—>
Local
¥
3
>v 2
>v
3
_A Global
*
-r
\
\ 2 l2
Y2
^}
f
f
i
+
(3-33b)
Local Global
h\ 3
i_ [v 2
—>v — vj 3)
>
_A
3 l3
Y n\ 3
, ,
Ty3 l
3
K;:l-
(3-33c)
Chapter 3
One- Dimensional Stress Deformation
Here the superscript indicates the element; for instance, node 2 for element 1, and so on (Fig. 3-8).
59
v\
is
the displacement
at
To assemble order of
these three relations,
we note
that the total global degrees
and hence the assemblage matrix and load vector are of the Consider Eq. (3-34) in which we assign
of freedom are 4.
4,
A,E
X
h
h
-A E X
AxE
X
A 2 E2
x
-
-A2E2
!
v\
h
h
-A2E2
A2E2
h
h
\
fAthfi 2
—A3E3 h
A3E3
+
h
\
v2
— A3E3
A3E3
h
h
.
=v
v\
v\
3
v4
+ 7^, +p}/
AxhY\
Tylh
2
2
A2I2Y2
Tylil
+ p\l +
A212Y2 2
+
Tylh 2
==
(3-34)
<
blocks of 4
x 4
2
2
A 3 hY 3
Ty3h
+
A3I3Y3
+n
+ p32i
for the assemblage stiffness matrix
blage load vector.
fy 3 l 3
and 4 X
1
for the assem-
Now we insert the coefficients of the matrices for the three
elements [Eq. (3-33)] into the proper locations in Eq. (3-34). For instance, the coefficient in Eq. (3-33b) for element 2 corresponding to the local indices
added to the global location
(2, 2) is
(3, 3)
and so on. Similarly the nodal
loads corresponding to the local index 2 are added to the global location 3
and so on.
We
notice that Eq. (3-34)
direct stiffness
approach
the displacements at the that
is,
is
is
the
same
as Eq. (3-32).
We
also see that the
based essentially on the physical requirement that
common
nodes between elements are continuous;
there exists interelement compatibility of displacements at the
common
nodes.
BOUNDARY CONDITIONS we
now we have
concentrated only on the properties of the column. Next consider the physical conditions that support the column in space, because
Until
the foregoing stiffness properties are called into action only is
when
the
column
supported. This leads us to the concept of boundary conditions or con-
straints.
As
the
name
displacement or
implies, a its
boundary condition denotes a prescribed value of on a part of the boundary of the structure or
gradient(s)
One- Dimensional Stress Deformation
60
body.
A
boundary condition
we
space. After
how
us
tells
body (column)
the
introduce these conditions,
Chapter 3
we have
is
supported in
a structure that
is
ready
to withstand applied forces; in lieu of these conditions, Steps 1-6 resulting in
tell us only about the capability of the column to withstand and not how they are withstood. In other words, without boundary
Eq. (3-32)
forces
conditions, the stiffness matrix [K] ishes,
and there can be an
infinite
singular, that
is
number of
is,
its
determinant van-
possible solutions. Hence, the
equations in (3-32) cannot be solved until [K]
is
modified to
reflect
the
boundary conditions. TYPES OF BOUNDARY CONDITIONS As discussed in Chapter 2, we encounter three kinds of boundary conditions. They are prescribed (1) displacements (or other relevant unknowns), (2) slopes or gradients of unknowns, or (3) both. They can also be called first or Dirichlet, second or Neumann, and third or mixed boundary conditions, respectively.
For
instance,
at the base
is
if
gradient or slope fully fixed,
it
the
specified, is
column
is
the
it is
specified,
constitutes the
supported such that the axial displacement
it
first is
condition;
if
the base
the second condition;
is
fixed such that
and
if
the base
is
mixed condition.
The boundary conditions
in
terms of the given displacement are often
called geometric or forced, while those in terms of gradients are often called
natural: the latter can often relate to generalized forces prescribed
on the
boundary.
HOMOGENEOUS OR ZERO-VALUED BOUNDARY CONDITION As an that
is,
at
illustration, let us
nodal 4
assume that the displacement at the column base, one can specify a nonzero displace-
(r 4 ) (Fig. 3-8), is zero;
ment also if the base experiences a given settlement. The boundary condition can be simulated in the finite element equations by properly modifying the assemblage equations [Eq. (3-32)]. To understand this modification, we shall consider one of the available procedures by writing Eq. (3-32) in symbolic form as 1
•R.
\Vt
v2 >
^32
We
delete the fourth
^34
K43
^44 _
row and
=
(3-35)
<
v3
"-33
JV
i«J
the fourth co 1 umn corresponding to v
which leads to the modified equations "
23
33_
>r <
A
as
'-':
^'3,
Jtl >
= R
1
>
<
2
,*3,
(3-36a)
One- Dimensional Stress Deformation
Chapter 3
61
or
= {R}.
[K]{r]
(3-36b)
The overbar denotes modified matrices. The assemblage
matrix [Eq. (3-32)]
stiffness
symmetric and banded or
is
Symmetry implies that k u = kJt and handedness implies nonzero values of ktJ occur only on the main diagonal and a few off-
sparsely populated. that
,
diagonals, whereas other coefficients are zero. These properties, which occur in
many
make
engineering problems,
economical. This
solution of the equations easier and
achieved by storing in the computer only the nonzero
is
we
elements within the banded zone. In Appendix 2
discuss the solution of
systems of such simultaneous equations.
NONZERO BOUNDARY CONDITIONS nonzero value, then the procedure is which can include the homogeneous condition as a special case. For instance, assume that v 4 = 5. Here Eq. (3-35) is modified by setting the coefficient k Ait equal to 1, all other coefficients in row 4 as 0, and R A = S. Therefore If the specified displacement has
somewhat
different,
ku
fc,2
k 21
k 21
V
"
k 23
R>
v2 <
k 32
k 33
These equations can
now be
v3
£34 _
1
= *
2
(3-37a)
,
1
v4
«3 c5 .
solved for v u v 2 v 3 and v 4 Often, ,
.
,
when we take advantage of
when
dealing
symmetry of a matrix for reducing storage requirements, it becomes advisable and necessary to restore the symmetry of the matrix in Eq. (3-37a) that is broken by the modification. It can be done by substracting from R l9 R 2 and R 3 the quantities k l4 X 3, k 24r x 3, and k 34r x (5, respectively. Thus, the equations with large matrices (Appendix 2)
the
,
reduce to
k 21
V
0"
*I1
k 23
R,
v2
1
>
kzi
k 33
Here kt4 x 8
(i
=
1, 2,
3) is
(3-37b)
R
v3 Ij
-0
<
k 34r x 6
v*.
transposed to the right-hand
side.
For instance,
for the third row,
0xt),+
k 32 v 2
+
k
k,*
x S
=R
or
X
v
l
+ k 32 v + 2
k 33 v 3
+
=R 3
k 34 S.
(3-37c)
The foregoing boundary conditions expressed in terms of the unknown displacement are the forced or geometric constraints. The natural boundary
;
:
One- Dimensional Stress Deformation
62
Chapter 3
conditions usually do not need the special consideration required for the
geometric boundary condition. The natural boundary condition, as a zero value for a derivative (slope) of the cally in
the
an integrated sense
sum of
the
in the finite
unknown,
is
if specified
satisfied
automati-
element formulation. In other words,
computed values of the
derivative at the
boundary vanishes
unknowns and
different categories
approximately. Different problems involve different
we
of boundary conditions, which Step
Solve for Primary
6.
Equation (3-32) set for the
is
shall discuss.
Unknowns Nodal Displacements :
a set of linear algebraic simultaneous equations.
column problem
is
linear
because the coefficients
K
tJ ,
The
which are
composed of material properties (E) and geometric properties (/, A), are constants and do not depend on the magnitude or conditions of loading and deformations. For instance, we have assumed linear Hooke's law and small strains and deformations in the formulation of the foregoing equations. We note here that if we assume material behavior to be nonlinear and large strains and deformations, then Eq. (3-32) will be nonlinear. Although we shall briefly discuss nonlinear
topic
is
behavior at later stages, detailed discussion of
beyond the scope of this
The unknowns assemblage vector
this
text.
in Eq. (3-37) are the nodal displacements given by the r [v x v 2 v 3 v 4 ] in which v 4 is already known. The
{r}
=
equations can be solved by using direct, iterative, or other methods; see
Gaussian elimination and a number of its modifications are one sets of direct methods used for solution of the finite element equations. The direct procedure involves two steps: elimination and back
Appendix of the
2.
common
substitution
To
Example
We
[5, 7].
illustrate the solution
procedure,
we now
consider two examples.
3-1
assume the following data for the column problem
(Fig. 3-8):
Cross-sectional area,
E = 1000 kg/cm 2 = 10 cm; A — 1.0 cm 2
Surface traction,
fy =
Body
Y=
Modulus of elasticity, Element length,
weight,
Boundary condition,
;
/
;
v4
=
1
kg/cm
.0
0.5
kg/cm 3
;
0.0.
We also assume that the column has uniform cross-section and material properties. Substitution of these values into Eq. (3-28) leads to the following element and
assemblage equations
One- Dimensional Stress Deformation
Chapter 3
1000 x
nw
1
10
x 10 x
1
lJl v2
63
a*it\) *H\) +
or
-100"
100
3c:KM;i
100^
100
v
and so on for the other two elements. Assembly of the three elements [Eq.
(3-32)]
yields "
100
100
100
200
100
100
200
Vi
[2.5]
v2
5.0
100
100_
100
100
100
200
100
100
200
5.0
,Vt.
.2.5]
5|
10
+
(
^3
=
Introduction of the boundary condition v A
=
>
i
-100
r
'
10 5
leads to the modified equations
"
In the expanded form,
-
100v 2
-100^ +
200v 2
-
100v 2
lOOvj
For solving these equations,
first
+0 +
= =
100v 3 200^3
7.5,
(3-38a)
15.0,
(3-38b)
15.0.
(3-38c)
we
follow the elimination procedure. Equation
=
7.5
(3-38a) gives
100^! Substitution for
100^
in
+
100v 2
.
Eq. (3-38b) yields
-7.5
-
100v 2
+
200^2
-
100^3
=
15
or
=
100v 2
Now, Eq.
22.5
+
100v 3
.
(3-38c) gives
-22.5
-
I00v 3
+
200v 3
=
15
or 37.5 v3
100
cm.
Back substitution leads to v2
100
cm
and 67.5 V\
100
cm.
Example 3-2 Instead of uniform
Y
and
fyy
equal to 10 kg at the top, that
example we apply only a concentrated load node 1. We can look upon this load as a global
in this is,
at
,
:
One- Dimensional Stress Deformation
64
joint load
P lg
.
Chapter 3
In this case, the final equations are
-
lOOvj
-100^!
100v 2
+ -
200v 2 100v 2
=10,
-f
- 100^3 = + 200v = 3
(3-39)
0, 0.
Note that the equations on the left-hand side are the same as before only the rightis changed. This gives us an idea that for linear problems, where Kij do not change, for a given structure, the elimination process is required only once since we can store the elimination steps in the computer. For different loadings only the back substitution needs to be performed. Thus, we obtain ;
hand-side load vector
Step
7.
vi
=
v3
=^cm
cm
-3$q
v4
=
(specified)
Solve for Secondary Unknowns; Strains and Stresses
For the displacement formulation based on potential energy, nodal disWe call them primary because they are the main unknowns involved in the formulation and in Eq. (3-32). The secondary unknowns are those derived from the primary unknowns evaluated in Step 6. For stress analysis problems, these are strains, stresses, moments, shear forces, etc. These designations will change depending on the type of formulation procedure used (see Chapter 1 1). placements are the primary unknowns.
STRAINS AND STRESSES By
substituting the
computed displacements from Step computed as follows
strains in the elements are
Example For element
3-1 (continued)
1
ey (D
=
1
iQ
[-l
1
-7.5 1000"
and
Similarly, the strains for elements 2
€y
^=
3 are
—22.5 1000
and
—37 tyK
}
(67.5)
l]^o(
5
1000
60
(
6 into Eq. (3-13),
One- Dimensional Stress Deformation
Chapter 3
Axial stresses in the three elements can
The negative
strains
-
OyV)
=
^§
^
=
:
(3)
100Q
now
x 1000
-
X 1000
mi x
100 °
be derived by using Eq. (3-14): -
-7.5 kg/cm 2
=
-22.5,
=
-37.5.
sign denotes compressive stresses.
Example 3-2
The
a y {\)
-7.5
65
and
(continued)
stresses are
-10 €y(l)
1000'
-10 ey {2)
1000'
€y (3)
1000'
-10 and
(D
<7,(2)
OyO) Step
= ~10.0 kg/cm 2 = - 10.0, = - 10.0.
,
Interpretation and Display of Results
8.
Figure 3-9 shows plots of displacements and stresses for the two previous examples. For Example 3-1, the distribution of displacements within each
element
is
linear,
need not be
whereas for the entire column, the displacement distribution The computed stresses are constant within each element,
linear.
whereas the actual distribution over the entire column can be linear. Existence of the constant stress states is dictated by the assumed displacement model, which is linear. In
Example
the exact value
3-2, the computed displacements and stresses are the same from closed form solutions, because for the exact solution,
oy
=
_4A =
^=
as
-10 kg/cm 2
1
and Vl
_ PQl) _
~ AE ~
10 x 30 1000 x 1
~
30 100
An important observation can be made at this stage. The accuracy of the assumed approximation will depend on the type of loading, geometry, and material properties. In Example 3-1, the loading varies from zero at the top to the highest value at the base for the gravity load Y and is uniformly
/
One- Dimensional Stress Deformation
66
= 0.675
v,
©
cm
Chapter 3
a = -7.500 kg/cm 2
22.50
©
—Computed
= 0.600
v2
j
Actual
v3
©
37.50
= 0.375
/
/
/ / /
v4 =
© w,
0.000
Displacements
Stresses
(a)
0.30
W
-10.0 kg/cm 2
cm
t-
/ I
Computed and exact
/
— v2
©
/
1
©
1 1 1 i
©3
v 4 = 0.00
Disf )lacements
Stresses
(b)
Figure 3-9 Distribution of computed displacements and stresses, (a) Results for
Example
3-1. (b) Results for
distributed for the surface loading
Example
Ty The computed .
of the actual, and the accuracy of the
3-2.
stresses are
an average
element computations
finite
is
consid-
ered satisfactory.
For Example
3-2, the
numerical solution
is
identical with the exact solu-
A number of other factors can influence the accuracy of numerical predictions and can require addi-
tion,
because there
is
only one concentrated load.
tional considerations for
improvements to account for the
factors.
One- Dimensional Stress Deformation
Chapter 3
67
The accuracy of the predictions can be improved by using two methods: mesh and/or (2) higher-order approximation models. The decision as to which approach should be used depends on the characteristics of the problem, and trade-offs exist with respect to accuracy, reliability, and com(1) finer
puter cost.
becomes essential to use higher-order models. If the column geometry and loading, it may be useful to consider higher-order models. For instance, we can use a quadratic approximation Very often
it
(Fig. 3-8) has irregular
for v as
v
= \L{L = [Ntfq},
X)v x
+ \UL +
\)v 2
+
(1
-L
2
K
(3-40)
where [N]
=
-
1)
{qf
=
[±L(L
±L(L+1) (1-L 2 )]
and [«,
The element here has
three nodes (Fig. 3-10), with the third
middle of the element.
It is
In
fact,
we can choose
interpolation functions
node
in the
not necessary to have the third node in the middle.
the node anywhere within the element; then the
TV,
will
be different for such choices. L =
©
l = +1
®
©
Figure 3-10 Element with quadratic or second-order approximation
model.
The
steps of the finite element formulations can be repeated to derive the
required equations for the higher-order element; this
an exercise; see Prob.
is left
to the reader as
3-8.
FORMULATION BY GALERKINS METHOD In the case of the variational approach,
we applied
the procedure to a single
generic element and used the results recursively for deriving properties of all elements. Then the individual element results were combined to obtain the also emphasized assemblage equations for the entire discretized body.
We
that in reality the variational procedure it
was only
for convenience that
to the entire
domain, and
to analyze element
by element.
was applied
we chose
assemblage equations [Eq. (3-32)] were derived by considering the total potential energy of the discretized body.
To
illustrate this concept, the
first
:
One- Dimensional Stress Deformation
68
As dure
explained in Chapter 2 and in Appendix
is
1,
Chapter 3
Galerkin's residual proce-
also applied so as to minimize the residual over the entire domain.
Another important aspect of this procedure is that the approximation function and the functions q> [Eq. (2-12)] are defined over the total domain. While applying Galerkin's method for finite element analysis, we use the interpolation functions N as the weighting functions [Eq. (2-14)], which are defined over the entire domain. In view of the foregoing characteristics of Galerkin's method and because for the variational procedure we have defined Ni relevant simply for an element, it is necessary to clarify a number of aspects and explain the relevance of TV,- over the total domain. {
t
Explanation and Relevance of Interpolation Functions
Figure 3-1 1(a) shows an approximation v to the exact solution v* defined for the total
domain of a one-dimensional body. The approximate or trial domain can be expressed in
function v for the displacement over the entire the sense of Eq. (2-12) as
M
v
m
=V£
(3-41)
JVjwJ,
k=\ j=\
M
where the superscript k denotes an element, is the total number of elements, and m is the number of interpolation functions per element. For the column with three elements (Fig. 3-8), Eq. (3-41) can be expressed as follows
Sum for
v
Sum
=
elements, k
=2 ;=i
1,2,3:
N)v)
- £ ;=i
Njvj
oxer interpolation functions, j
=
+
£
N)v).
(3-42a)
j=\
1, 2,
,
m;
= N\v\ + N\v\ + N\v\ + + Nlvl + N\v\ + N\vl + N\v\ + + N mvm + N]v] + N\v\ + N\v\ + + N m vl linear approximation [Eq. (3-6)] m = 2; therefore
v
• •
2
•
•
•
•
•
•
2
3
For the
v
=
N\v\ 4-
-f
N\v\
N\v\
+
+
N\v\
+
(3-42b)
N\vl
N\v\.
(3-42c)
Here N\ denotes interpolation function for node 2 of element 1, and so on, where the superscript denotes an element. Figure 3-1 1(b) shows the plots of the linear interpolation functions over the three elements. Since v\
=v u
One- Dimensional Stress Deformation
Chapter 3
3
-K
:
;
N?
\
2
2 2 N k vk
v =
N?
69
N^
N?
N3
Vt •
/
Vt
;<
/^
(b)
NJT 1
+ Ni
subscript => local node
f
subscript => element
i-1
i
+
1
(0 Figure 3-11 Approximation and interpolation functions in Galerkin's method, (a) Approximation for v over total domain D. (b) Interpolation functions over
node
v\=v\ = v
v 2 vl ,
D.
Interpolation function for
(c)
i.
= v\
v 3 and v\ ,
=v
4
,
we have
= N\v + (N + N\)v + (Nl + N\)v + = £ A^- = #i*i + #2*2 + N v + N v l
x
2
2
3
3
2
A
N\v A (3-43)
A,
N denotes interpolation functions around node general N => N\ + N\. For the linear case [Fig. 3-1 1(c)], where
1;
t
2
for instance, in
One- Dimensional Stress Deformation
70
y yt
N (y)
- yt-x = NV,
(3-44a)
-y = N'U
t
y i+i
y
-yt
bi+i
y t -\
-yt-\
Chapter 3
S t -u
t
5,-1
i+ v.
—y
y
t
-\
.
/i-i
_
Si,
where s
the
1,
and
/,
denotes the length of
The definition of N in Eq. (3-44a) is valid only for points within domain D. For each end node one of the values will be relevant. Thus interpolation function around node i has nonzero values only in the two
element the
the local coordinates
is
(3-44b)
-y
?*+!
i.
t
adjoining elements
A
i
—
1
and
i
and zero values
in all other elements.
= cross-sectional area
<*Z&
"Y
X*# Figure 3-12 Equilibrium of column segment.
With the foregoing explanation, we are ready for formulating finite element equations for the
to use Galerkin's
column problem.
derive the governing differential equation for the column, Fig. 3-12. sidering the equilibrium of forces for an elemental
method
We
first
By con-
volume of the column,
we have
Acy
- Aay - A^dy + f(y)dy =
or
dGy
_
~dy'~
f(y).
(3-45a)
oidajdy from Eq. (3-14a) after insertion of e y from Eq. (3-12a) dropping the subscript to E for convenience)
Substitution gives (after
AE,d
2
v*
dy
f(y).
(3-45b)
One- Dimensional Stress Deformation
Chapter 3
Here v*
is
the exact solution,
and we have considered /(j) as surface
f(y) per unit length per unit surface area.
Pa
body force Y(y) and joint load form as
traction
possible to consider f(y) as the latter can be defined in a function
;
PuW = Ptiyj) where Puiyj)
71
x S(y
It is
- yj
(3-46a)
),
= magnitude of joint load at pointy and S(y — yj)
is
the Dirac
delta function,
&(y-yj) The
=
y
1
l
'
Z
yj '
(3
(3-6))
is
R(v)
=
EA^ -f.
(3-47)
2
For convenience we wrote / instead off(y). Minimization of
N
t
46b )
residual for the differential equation (3-45b) (and the linear approxi-
mation function
to
"
R with
respect
[Eq. (3-6)] leads to
VSEA w- f N dy=o '
(3 48)
'
)
where h is the total length of the column equal to y 4 the coordinate of the last node (Fig. 3-8). Integration by parts for the first term gives ,
£' BA&NA = EA%l,N [ - EA t
and hence Eq.
(3-48)
[ % f *.
(3-49a)
•
is
<%->=l>^-*4«> We
inserted
the
y
/,,
(3-49b)
direction cosine of angle between the axis of the
axis, as unity, since cos
9y
= cos =
column and
1.
Now if v from Eq. (3-43) is substituted into Eq. (3-49b), we obtain
/= The index notation
l,2,3,4,y=l,
implies set
i
i
i
i
= =2 =3 =4 1
and and and and
= varyy = vary j = vary 7 = vary j
1, 2, 3, 4, 1, 2, 3,
1
,
4,
2, 3, 4,
1, 2, 3,
4,
2, 3, 4.
(3-50)
One- Dimensional Stress Deformation
72
which leads
to
~dN,
dN,
dN
dy
dy
dy
dN2 dN
dN,
EA Jyi
Chapter 3
2 •
2 "
dy
dy
dy
dN,
dN,
dNz
dy
dy
dN
dy
dy
dN2
dN,
dy
dy
dN,
dN,
dy
dN,
dN,
dy
dy
'
dN,' •
\
f
V
dy
dN2 dN, dN2 '
dy
'
dy
dy
dN,
dN,
dN,
dy
dy
dy
dN,
dN,
dN,
dN,
dy
dy
dy
{
dy
dy
"
dy
dN4 dN2 dN4
dN,
.dy
3
dy '
dy
dN,'
dy
<*>N, dy
fN,
fN2 \dy
r
i\
dy.
+ EA\
^N dy
2
(3-5 la)
fN, fN,
or
R
ffi It is
R
important to note that the terms
total length
and that the
f in
(3-51b)
fR,
Eq. (3-51) are integrated over the
boundary term (dv/dy)Ni
limits in the
fj
are defined
at the ends.
Terms such
r
as
dN,
dN,
dy
dy
dN
!'
dy
+ vanish because to
N = 3
in region
y 4 Thus only those terms .
y
dy
dy
Jy s
y3
3
dy
x
to
dN,
dN,
dy
dy
i"
dy
y 2 and
N =
in regions
t
dN^
dN2
dy
dy
t
dy
dN,
EA
dy
r Jy
t
dN2 dN2 dy
dN± m
dy
1
y 3 and two ele-
to
side reduces to
)
dy
dN^
dN
dy
dy
dN,
dN
dy
dy
m
dy
dN2
y2
that have nonzero values in one or
ments contribute to the integration. Hence, the left-hand
dN,
dy
dN
3 '
dN
3
2
dy
dN
'
3
dy or [K]{rj
'
dy
3
dN4
dy
dy
dy
dN<
dN4
dNt
dy
dy
%
.
dN,
m
dy _
(3-52a)
+
i
i
!
;
!
|
& *> 5>|
*
i
*>
<5>j
t
5
J
$
#
^ §>j
!
•
!
!
j!
•
i
I> §>! 5
f
t
j
|
73
One- Dimensional Stress Deformation
74
Note
Chapter 3
that the terms enclosed in the dashed lines represent [B] r [B] for the
three elements. This indicates that the contributions of the element stiffness properties are added such that compatibility at
common
nodes
is
assured.
Integrations for the relevant limits will lead to the assemblage stiffness matrix [K].
For
instance,
Assuming elements of equal length
= =
si
*2
Now
ds
x
= (\/l)dy
dN\
dy
ds
ds±
_L
u
'
dy
x
we have
(y
-ydlh
(y 3
- y)li-
= (—l/l)dy.
and ds 2
dN\
/,
and
I
(3-53b)
By noting
that
dN = dN
*
*
dy
ds
>-
dy
ds 2
we have
= 7T
#22
and
f(lXl)
(3-53c)
Jo
finally 1
EA
[K]
-1 2
-1
-1
2
1
I
(3-54a)
-1 which is the same as [K] in Eq. (3-32) for constant element length and form A and E. The first term on the right-hand side of Eq. (3-5 la) denotes
uni-
>
1
ffNXdy ryi
y 1
\
1
'fN{dy
+
{R;
fN\dy
I
2
n \ Jy 2
+
fNldy
n \
(3-54b)
;
2
2
fN\dy\
\"fN\dy\ which
is
the
1
k
v
same
/
as the contribution to the assemblage load vector {R}
by
T, in Eq. (3-32).
The second term on
the right-hand side in Eq. (3-5 la) denotes essentially
the end boundary terms, because
points
1
and
4.
For
instance,
N
t
have nonzero values only
at the
end
Chapter 3
One- Dimensional Stress Deformation
dv Ndy since
N-
N\
=
at
dy
nodes
1
\,x
and
dy
Hence
4.
15
it
reduces to
(-» dv dy
R
AEi
(3-54c)
m> dv
1
at
nodes
1
and
4,
we have
(-<)) R
(3-54d)
K)
dy/*
J
Equation (3-54d) represents natural boundary conditions defined through prescribed values of (normal) derivative of the displacement. The term AE(dv/dy)
= AEe y = Ao y
denotes (internal) force, and hence
{R.}
=
(3-54e)
where F, and FA are forces at the boundary nodes. Comments. If an external point load is applied
at node 1, F, equals the no external load there, F = 0. When the column base has a prescribed boundary condition, say v4 = 0, then the term
prescribed external load. If there
F4
will
not appear in the
(3-51)] are
is
{
final results
when
the assemblage equations [Eq.
modified for the given boundary condition. In other words, the
natural boundary condition at a point will not have influence on the final results if that point has a prescribed geometric
boundary condition.
important to understand the foregoing idea that Galerkin's residual procedure is applied to the entire domain and that its application yields It is
boundary conditions which should be properly use of Galerkin's method
we may
interpreted. In the subsequent
by element, with a clear understanding of the foregoing implications. When Galerkin's procedure is applied to an element, we can look upon consider
it
to be applied element
the interpolation functions for the element to have nonzero values within
.
One- Dimensional Stress Deformation
76
Chapter 3
themselves and zero elsewhere. For instance, in Fig. 3-11, N\ and N\ have nonzero values within element 2 but zero values at other locations in the domain.
The foregoing finite element equations from the Galerkin method are same as those from the variational approach. This happens
essentially the
usually in the case of problems mathematically classified as self-adjoint
a rather elementary
manner
of the symmetric operator matrix linear or the
problem
is
[8].
In
self-adjointness can be attributed to the existence [K]. If the
governing equations are non-
non-self-adjoint, the
two
results
can be
different.
An example of a non-self-adjoint problem is given in Chapter 8.
COMPUTER IMPLEMENTATION Although we could perform hand calculations for the one-dimensional column problem, almost all problems solved by using the finite element method involve large matrices, and recourse has to be made to electronic computation.
To understand computer
codes in a progressive manner,
duce a rather simple code (program)
DFT/C-1DFE
in
we plan
Chapter
6.
to intro-
This code
can handle the problems of axial deformation, one-dimensional flow, and one-dimensional temperature distribution or consolidation.
The reader should study problems
in
Examples
3-1
the code in Chapter 6 and use it for solving the and 3-2 and some of the subsequent problems.
OTHER PROCEDURES FOR FORMULATION In this section,
we
shall introduce alternative
methods of formulation by
using variational principles such as the principle of stationary complementary energy
and the mixed principle [5,9-11]. Although hybrid principles
are found to be successful for a in this
number of situations, they
will
not be covered
elementary text except for one example for Torsion in Chapter
1 1
Comment. For an undergraduate course, the instructor may wish postpone coverage of the following material for a
later stage or for
to
a graduate
course.
COMPLEMENTARY ENERGY APPROACH IT C is defined as the sum of the complementary and the complementary potential of the external loads
The complementary energy strain energy
W
U
c
•
U = U + Wpc C
C
.
(3-55a)
Chapter 3
One- Dimensional Stress Deformation
Graphical interpretation of Uc
shown
is
in Fig. 3-4.
11
According to the principle
of stationary (minimum for an elastic body in equilibrium) complementary energy,
we have
8U = 8U + 5Wpc = C
C
or
= 5U - SW = 0,
<5n c
where
W
c
is
the complementary
because the potential strains,
C
is
lost into
(3-55b)
C
work of external
loads.
sign occurs
and displacements,
n = <
W MW dv ~ If PFWdS, T
* JIf V
where [D]
is
S2
is
=
the part of the
For the column problem, Eq.
[C]"
1 ,
{u} is the vector
boundary on which
of prescribed
{u} are prescribed.
(3-55c) can specialize to
A p o A Cydy - p T vdy.
n =
y
c
express
them
(3-55d)
y
In the case of the complementary energy approach,
unknowns and
(3-55c)
st
the strain-stress matrix
displacements, and
as
The minus
work. In terms of the components of stresses,
in
terms of nodal
we assume the
stresses, or joint
stresses
or nodal
Hence, the approach is also called the stress or equilibrium method. For matrix structural analysis, this procedure can specialize to the well-known force method. Advanced study and applications of this approach are beyond the scope of this text however, we shall illustrate the method by using the rather elementary problem of the axial column. We express stresses (
;
to where [NJ
is
=
[N.]{Q n },
the matrix of interpolation functions and {Q„}
nodal or joint forces related to a the element [5,6,9, 12,
The
(3-56a)
statically
is
the vector of
determinate support system for
13].
surface tractions {T} can be expressed as {T}
where [NJ
is
= [NJ{Q
(3-56b)
n },
a relevant matrix of interpolation functions.
Substitution of Eqs. (3-56a) and (3-56b) into II C [Eq. (3-55d)] gives
U = c
T i J]J {Q n } [KY[D]\N a ]{Q n }dV V
-
J|
{QJTNJW&
(3-57)
St
Differentiation of II C with respect to {Q„}
[f]{QJ
=
(q)
and equating to zero leads to (3
-
58a )
One- Dimensional Stress Deformation
78
Chapter 3
where [f ]
=
JJJ
[N.r[D][NJ
(3-58b)
V
and {q}
[f] is
=
r JJtNJ {u}«.
called the element flexibility matrix
and
(3-58c)
{q} is the vector
of prescribed
displacements.
The element flexibility equations can be assembled into global equations and then solved for the unknown forces (stresses). This procedure can be difficult and cumbersome. It is often convenient to transform the element matrix into the element
flexibility
stiffness matrix.
Example 3-3 Consider the column problem (Fig.
and
stress
approaches are shown
3-7).
The
force systems used in the displacement
in Fig. 3-13. In the case of the latter, the flexibility
is formed by eliminating the degree of freedom corresponding to the rigidbody motion; it is done by constraining one of the ends while the other is loaded, which yields a stable element.
matrix
Fo = F
©
'///////A v?
=0
©
777777777} v\
u
=0
fi=,-F
(a)
(b)
Figure 3-13 Force systems, (a) Displacement approach, (b) Stress
approach.
For
F acting at node
1,
v2
= 0, and for F acting at node 2, Vt =
0.
Under
these
circumstances the axial stress can be expressed as C7 y
Then use of Eq.
= ^[lW
1
}
=
[S a ]{Q n }.
(3-58b) gives the flexibility matrix as
(3-59)
One- Dimensional Stress Deformation
Chapter 3
79
P±.J_± dy
[f]
A E A
J_
(3-60)
AE The equation
relating the reactions
is
given by (3-61)
From
matrix structural analysis, the
can be transformed to the
flexibility relation
stiffness relation as [9, 12, 13]
[k]
[f]-
=
1
[G][fr
AE
which
is
same matrix
the
[fFtGF
!
[owr -AEl
:
I
I
-AE
AE
I
I
i
m. (3-62)
_
Eq. (3-28).
[k] as in
Comment. The complementary energy approach can
also be used in con-
junction with the concept of stress functions. This approach can be easier to
implement and understand ever,
it is
we have
somewhat
in
comparison to the foregoing procedure. Howformulate with one-dimensional problems;
difficult to
illustrated its application for the
two-dimensional torsion (Chapter
11).
MIXED APPROACH In the mixed approach, both the element displacements and stresses are
assumed to be unknowns. The formulation results in coupled sets of equawhich nodal displacements and stresses appear as unknowns.
tions in
Variational
Method
In the mixed method, the variational functional, often Hellinger-Reissner principle,
nR =
JJJ
({«f{e}
is
expressed as
known
as the
[5, 10]
- *WTD]M - mx})dv
V
JJ
iuYmdSi
-
Si
Here (w
—
{(u
JJ
- u)f{T}
{P„) r {q,}.
u) denotes the difference between the actual
placements along the boundary
(3-63)
St
S2
,
and {PJ and
{q,}
and prescribed
dis-
are the vectors of applied
nodal or joint loads and displacements, respectively.
One- Dimensional Stress Deformation
80
Chapter 3
The stationary value of IT* yields a set of equilibrium and stress-displacement equations. In these equations both the displacements and the stresses simultaneously appear as primary unknowns. Residual Methods
These methods can be used to formulate finite element equations based on governing differential equations. For example, Galerkin's method can be applied to the equilibrium and stress-displacement (stress-strain) equations, which results in coupled equations in displacements and stresses. Example 3-4
We illustrate
column problem assumed to be linear:
the mixed approach for the
the vertical displacement
is
= #1 - L)v +
v
= Here [NJ
The
is
the
same
axial stress
where
=
[a
x
polation functions.
Variational
As in Eq.
+ L)v 2
(3-8d)
(3-8e)
[NJ{q}.
as [N] in Eq. (3-8d).
can also be assumed linear as
ay
r {
i(l
x
(Fig. 3-7).
a2
]
= =
i(l
- L)a, +
i(l
+
L)o 2
[NJ{aJ,
the vector of nodal stresses
is
Note
(3-64)
that here
and [NJ
we have assumed [NJ
=
is
the matrix of inter-
[NJ
=
[N].
Method
Substitution of
v,
a y and ,
-*[
ey into the special form of Eq. (3-63) leads to
{
T [NY~[N){
-
(TV,
+ P2 v 2
(3-65)
).
Here for convenience we have not included the body force, surface traction, and the term related to difference in displacements. These terms can, however, be included without
difficulty.
Now we differentiate 11* with
respect to
a u a2
({tf„})
and v u v 2
({q})
and equate
the results to zero:
|^ = => y
!
f
[W\B)dL
{q}
_
^
f
[NFrNtfLto}
=
0,
(3-66a)
One- Dimensional Stress Deformation
Chapter 3
^
=
0=>y['
[BRN]d!L{
+
{0}
=
81
(3-66b)
[P,,},
or
+ [k TU ]{q] = 0, IKuYiQn) + {0} = {Q}, [k TT ]{
(3-66c)
or {0}|
=
(3-67a)
LfCrTTo] "Jl{q}i
1
[QV
where
3E
AT 6E
A[
A[
6E
3£J
A]_
[Nf[N]JL
[K
[k T J
=
y
[
[N] r [B]^L
(3-67b)
2
2
-/I
^4
(3-67c)
=
2J
2
and
« in
(3-67d)
13
For further illustration, consider the following properties of the column divided two elements (Fig. 3-14):
A = Pi /
= =
E= Figure 3-14
cm 2
,
10 kg, 10 cm,
1000 kg/cm 2
Column problem
for
.
mixed procedure.
a<
f 10
1
1)
v?»
cm (D o%\ r
®
al
10
,
2,
(D W \i
()=* element
2
,vl
2
»
cm
V7b77>
®
2) 2) o 2 ,v 2 {
v3 = °
One- Dimensional Stress Deformation
82
Chapter 3
Substitution of these properties into Eqs. (3-67) gives the element equations as
Element
-10
-10
1
3000
6000
2
-10
-10
6000 ____
3000 ____
2
1"
1
1
2
<*2 >
1
1
2
=
—
<
or
«
--
(3-68)
10
Vl
2
2
f
Ox
2
2
Element 2
1
,
f
v2
The element equations can be assembled now. For convenience, in the following we have rearranged the nodal unknowns as [c v a 2 v 2 a 3 v 3 ]; hence the assem!
x
bled equations are
10
10
3000
6000
-
_J_ 2
/
10 - 6000
1
a,
2
1
4
__1_
10
10
1
6000
3000
2 1
=
=
(3-69)
1
v2
2
Here the boundary condition v 3
\
10
Vx
2
20 3000
10
6000
/
\
Ox
<7
_
3
[V 3 )
i
0/
has been introduced as described previously.
Solution of Eq. (3-69) leads to
= —10 kg/cm 2 a 2 = -10, <7 = -10, Ox
3
,
vx
cm,
V 2'-
cm,
v3
=
10
(prescribed).
These results are similar to those in Example 3-2. We may note that this is rather an elementary problem used simply to illustrate the procedure. Also, the results would have been different if we had used other orders of approximation and different loading
and geometry.
Comment. The matrix in Eq. (3-69) contains zeros on the main diagonal. For the standard Gaussian elimination, this can cause computational difficulties since the
Then
it
is
zeros can appear as "pivots" in the denominator (Appendix
necessary to use the idea of partial and complete pivoting
2). [7],
which involves exchanges of rows and columns during the elimination procedure. Mathematically, the matrix [K] in Eq. (3-69)
is
positive semidefinite in
constrast to the positive-definite character of the stiffness matrix in the dis-
placement approach.
Galerkin's
With
Method
we used
the displacement formulation,
[Eq. (3-45b)]. It
was derived on the
the governing equation
basis of the equilibrium equation [Eq.
and the stress-strain or stress-displacement equation [Eq. (3-14a)]. use both the equilibrium and the stress-strain displacement equation for formulating the mixed approach with Galerkin's method. The residuals according to the two equations are (3-45a)]
We can
* = §* -
fy
(3-70a)
and
**
=
-% + *'
Assuming the approximation models (3-64),
we
for v
(3 - ?0b)
and a y
as in Eqs. (3-8)
and
have, according to Galerkin's method,
^ 4'( dy
T^N dy =
(3-71a)
t
and
g + £)#,, =
<).
(3-71 b)
After proper integration by parts, Eqs. (3-7 la) and (3-7 lb) will lead to the
same
results as in Eq. (3-67), obtained
by using the variational procedure.
COMMENT we have presented formulaby variational and residual methods. In the variational methods, the attention has been given to the displacement, complementary, and mixed In this chapter and in Chapters 7 and 11
tions
procedures, while Galerkin's procedure has been used mainly in the residual
methods.
A
number of
useful hybrid procedures [11] are available in the
variational methods; Gallagher
As
procedures.
[9]
has presented examples of some of these
described in Appendix
1,
other residual methods are also
possible.
In this elementary text,
it
will
be
difficult to
cover the total range of the
methods of formulation. Hence, we have chosen only a few of them, particularly those which can be illustrated easily with simple problems.
BOUNDS discussed rather qualitatively that different approaches yield different bounds to the exact solution. In the case of the variational procedures, if the physical and mathematical requirements are fulfilled, the poten-
In Chapter
1
,
we
83
One- Dimensional Stress Deformation
84
tial
Chapter 3
(displacement) and complementary (stress) energy approaches yield,
respectively, lower
and upper bounds to the exact solution for displacement.
The approximate
(algebraic) value of potential energy with a given approximation function is higher than the exact or minimum value (Fig. 3-6). That is, the approximate stiffness is higher than its exact value. Consequently, the approximate displacements u, are lower than the exact displacements. This is depicted in Fig. 3-1 5(a), which shows the convergence behavior
mesh is refined. The physical explanation is that the displacement approach yields a stiffer or stronger structure than what it really is, because the assumed displacement
as the
functions, although continuous within elements, provide an approximate (global) distribution of displacements. Indirectly, this introduces additional
supports or constraints in the structure and makes
it stiffer.
In the case of the complementary energy approach, the approximate value
of complementary energy flexibility
is
higher than the exact or
minimum
value.
The
value from a numerical approximation has higher value than
the corresponding exact value
;
that
is,
the stiffness has lower value than the
Consequently, the approximate displacements are higher
exact stiffness.
than the exact displacements, Fig. 3-1 5(b).
The
approach yields a discontinuous distribution of displacement. may be construed to introduce gaps and overlaps in the structure. This makes the structure weaker or less stiff than what it really is. stress
Physically, this
Figure 3-15 Symbolic representation of bounds in analysis, (a) stress
Bounds
in
finite
element
displacement approach, (b) Bounds in
approach.
[K]
Number
Number
of nodes
of nodes
(a)
u => displacements
[f]
Number
Number
of nodes
(b)
of nodes
One- Dimensional Stress Deformation
Chapter 3
85
Thus the two approaches bound the exact solution from below and from [Figs. 3-1 5(a) and (b)]. This property is useful, and often we can bound
above
the exact solution by using both approaches.
Comment. Although the mixed and hybrid approaches can
yield satis-
factory and often improved solutions, they do not possess bounds.
ADVANTAGES OF THE FINITE ELEMENT METHOD At this stage, it may be ment method.
useful to
some of the advantages of the
list
finite ele-
Since the properties of each element are evaluated separately, an obvious
advantage element. is
no
is
that
we can
incorporate different material properties for each
Thus almost any degree of nonhomogeneity can be
medium hence
restriction as to the shape of the
geometries cause no
difficulty.
Any
;
included. There
arbitrary
and
irregular
type of boundary conditions can also be
accommodated
easily; this can be an important factor. have not yet discussed other factors such as nonlinear behavior and dynamic effects. Subsequently, and during study beyond this book, the reader will find that the method can easily handle factors such as nonlinearities,
We
and time dependence. method that makes it so appealing and powerful for solution of problems in a wide range of disciplines in engineering and mathematical physics. This generality allows application of formulations and codes developed for one class of problems to other classes of problems with no or minor modifications. For instance, one-dimensional stress deformation (Chapter 3), flow (Chapter 4), and time-dependent flow of heat or fluids (Chapter 5) can be solved by using the same code (Chapter 6) with minor modifications. Similarly, formulation and code for such field problems as torsion, heat flow, fluid flow, and electrical and magnetic potentials (Chapters 11 and 12) have almost identical bases of formulations and aribitrary loading conditions, Finally,
it is
the generality of the
codes.
PROBLEMS 3-1.
N
L and interpolation functions N^ and 2 if the local coordinates were measured from the quarter point. See Fig. 3-16.
Derive local coordinate
Figure 3-16
© ® -*~
y
•
1-
r*7/4-H
©
One- Dimensional Stress Deformation
86
(a) For L = (v - y )/(l/4),L(l) = -v N L), 2 = i(l + L). (b) For L = (y i(3 1, and Ni = |(1 - L), N2 = ^(1 + 3L).
Solution:
3-2.
-1, L(2)
3
Derive element equations
if
Chapter 3
the value of the elastic
=
L(l)
3 )/(3//4),
modulus
N =
and t -$, L(2)
3,
=
=
E and the area A
vary linearly as
E=
[{{\
-L)
JO +L)}<
and
A=[\{\ -L) where
A and A 2 and
£j and
x
£2
^(1
+^)]{^}'
are the values at nodes
1
and
2, respectively.
Answer: [k]
where the scalar X
=
A
1
_-l
"!]
given by
is
~0
6/
Mi
E
x
A,
E2
]
-i 3-3.
Consider the column /
of each element
=
in Fig. 3-17
J" 'A,
1
10^0 0^01 1
0_
E, {
A2
E2
with linearly varying E, A, and
Ty
.
Length
10 cm. Derive element equations, assemble them, intro-
duce boundary condition v A = 0, and solve for displacements, of displacements and stresses.
strains,
stresses. Plot variation
Figure 3-17 2 E, = 1000 kg/cm 2 _Aj =0.5 cm T y1 = 0.5 kg/cm
©
A ©
A ©
E 3 = 1500 Ao = 0.5
2000
A ©
and
Chapter 3
One- Dimensional Stress Deformation
87
Partial solution:
Equations for element 2:
soor
J
p
-1 Assemblage equations:
0~
-7 15
-8
-8
16
(V
0.20
h
0.55
<
-8
-8 i'i
3-4.
0.29732 cm, v 2
8_
0.26875 cm, v 3
0.65
*>3
.V 4
0.40,
,
0.17500 cm, and v 4
=
0.
fy to vary linearly from zero at and compute displacement and stresses.
In Example 3-1, consider surface loading the top to 2
kg/cm 2
at the base,
Hint: Consider
fy = Then the load vector
-L)
H(l
|(1
+L)]{J")-
is
af + fy «-*K."+ *3 /
3-5.
For Example and stresses.
3-1,
3-6.
Use the code and 3-2.
in
3-7.
For Example
3-1,
stresses as the
number of nodes
Consider
yi
assume that the base
settles
by
0.1
cm. Find displacements
Chapter 6 and obtain computer solutions for Examples
4, 8, 16,
3-1
study convergence of solutions for displacements and are increased using the code in Chapter
and 32 nodes; plot
results,
6.
and comment on improvement
in accuracy, if any.
Partial solution:
See Fig. 3-18. Note that in view of the linear model and plane is not significantly influenced by the
axial problem, the displacement solution
refinement of mesh. 3-8.
Sketch the variations of
TV,-
in
Eq. (3-40) for the element in Fig. 3-10, and A and E to be constant.
derive the element stiffness matrix assuming Solution:
2
-16
2
14
-16
16
-16
32.
14 [k]
3-9.
Derive element
= AE
stiffness for
"ST
quadratic displacement (Prob. 3-8)
varies linearly as 'A,
A=[#l -L)
i(l
+^){ //}
if
the area
One- Dimensional Stress Deformation
88
3-10.
By
Chapter 3
using the data of Example 3-1, solve for the displacements and stresses in
Prob. 3-8. 3-11.
Derive interpolation functions
3-12.
=
TV,-, i
model when the intermediate node
is
1, 2, 3,
for a quadratic approximation
situated at one-third the distance
or right-hand node.
the
left-
(a)
Derive load vector {Q} due to surface loading T,
and quadratic v
in
=
i(l
- L)f + yi
i(l
fy
given as
+ L)fyt
Eq. (3-40).
Figure 3-18 Results with
mesh refinement and convergence,
(a)
Distribution of displacement v for eight elements, (b) Distribution
of stress ay for eight elements, (d)
Convergence of
(c)
Convergence of displacements,
stress at midsection.
from
One- Dimensional Stress Deformation
Chapter 3
,
89
i
At top
«^^
u.y
0.6
At midsection^ 0.5
0.4
-
0.3
-
0.2
-
0.1
i
i
4
l
i
8
Number
i
i
l
12
16
12
16
of elements
(0
_
40.0
CM
E
1
30.0
°
20.0
10.0
1— 8
4
Number
of elements
(d)
Figure 3-18 (Continued)
t
90
One- Dimensional Stress Deformation
Chapter 3
Solution:
iL{ {Q}
\
J "'
+
i(l
+ L)T„]dL
the load vector due to gravity given by
Y= Assume
L)T„
+ 2Ty J
127V,
Compute
-
l)ki(l
(1-£*)J
I
(b)
Dl
= -j\r UUL + >
-L)F, +i(l +L)f2
i(l
.
linear variation for v.
Answer:
= Ali2Y + Y2] l
[Q] (c)
Find the load vector TV
if
= iL(L -
!)?„
+
^L(L
+
1)7V 2
+
(1
- L*)f„
Answer:
Tv + 2TV
4TV
I
3-13.
27V,
,
+ 2fw + 16fJ
Idealize as closely as possible the irregularly shaped
solve for displacements
and
stresses
column
in Fig. 3-19;
by using DFT/C-1DFE.
L E
=2000 kg/cm 2
Y
=
Ty
= 2.0 kg/cm = 10 kg
P
1
.0
kg/cm 3
y//////////////y
Figure 3-19
3-14.
Consider the coordinate system in Fig. 3-2(c) and derive the quadratic
inter-
polation function model [Eq. (3-40)] from the following generalized coordi-
nate model for v
:
v
=
a
t
+
a2x
+
a3 v2
=
[
]{a}.
One- Dimensional Stress Deformation
Chapter 3
91
Solution:
{a}
[A]-»{qJ,
—y\y*
yjyz
"
(y 2
=
[A]" 1
=
- y\)(yi - y\) -O3 + y 2 - y\)(ys - yi)
(y 2
)
(y 2
(y 2
[N]
=
3
)
(y*
3
)
(y 3
y\)(yz
- y\)
(y 2
'
- y 2 )(y - y\) ~(y 2 + yi) - y 2 )(y - yi) 3
3
-1
1
Sy 2 -
y\y 2
- yi)(y - y 2 yz + y\ - y\){y - y 2
1
- y\){yz - y 2 )
(^3
- ^2) (^3 - ^i)J
[][A]-»
-y 2 )(y - y -(y - y\)(y - ys) (y - yi)(y - y 2 — y\){yz — y\) (y 2 - yi){y - y 2 (y - y 2 )(y - Ji)J] Because y 3 - y = 7/2, y - y 2 = -7/2, y2 - y = and (y - y )Kl/2) = (y {y 2
3)
3
t
L,
we
)
3
finally
)
3
x
7,
3
3
have
= [iL(L-l) iUL +
[N]
1)
1
-£*].
REFERENCES [1]
Zienkiewicz, O. F.
C,
C,
Irons, B. M., Ergatoudis,
Dimensional Analysis," I,
[2]
[3]
and
J.,
Ahmad, S., and Scott, Two- and Three-
"Isoparametric and Associated Element Families for Bell, K., eds.),
in Finite
Element Methods in Stress Analysis (Holland,
Tech. Univ. of Norway, Trondhein, 1969.
Bazeley, G. P., Cheung, Y. K., Irons, B. ML, and Zienkiewicz, O. C, "Triangular Elements in Plate Bending-Conforming and Nonconforming Solutions," in Proc. Conf. on Matrix Methods of Struct. Mechanics, Wright Patterson Air Force Base, Dayton, Ohio, 1965.
Strang, G. and Fix, G.J. An Analysis of the Finite Element Method, PrenticeEnglewood Cliffs, N.J., 1973. ,
Hall, [4]
Bathe, K. sis,
[5]
J.,
and Wilson, E. L., Numerical Methods Englewood Cliffs, N.J., 1976.
Desai, C.
S.,
and Abel,
Nostrand Reinhold, [6]
[7]
[9]
Element Analy-
J. F.,
New
Introduction to the Finite Element Method,
Van
York, 1972.
Turner, M. J., Clough, R. W., Martin, H. C, and Topp, L. C, "Stiffness and Deflection Analysis of Complex Structures," /. Aeronaut. Sci., Vol. 23, No. 9, Sept. 1956. Fox, Press,
[8]
in Finite
Prentice-Hall,
L.,
An
New
Introduction to Numerical Linear Algebra, Oxford University
York, 1965.
Prenter, P. M., Splines and Variational Methods, Wiley,
New
Gallagher, R. H., Finite Element Analysis Fundamentals, Englewood Cliffs, N.J., 1975.
York, 1975. Prentice-Hall,
One- Dimensional Stress Deformation
92
[10]
Washizu, K., Variational Methods
in Elasticity
and
Chapter 3
Plasticity,
Pergamon
Press, Elmsford, N.Y., 1968. [11]
Pian, T. H. H., and Tong, P., "Finite Element Methods in
Mechanics,"
in
Advances
in
Continuum
Applied Mechanics, Vol. 12, Academic Press,
New
York, 1972. [12]
Przemieniecki,
New [13]
J. S.,
Theory of Matrix Structural Analysis, McGraw-Hill,
York, 1968.
Rowan, W. H., Hoadley, P. G., and Hackett, R. M., Computer Methods of Structural Analysis, Prentice-Hall, Englewood Cliffs,
Beaufait, F. W., N.J., 1960.
;
ONE-DIMENSIONAL FLOW
Flow of heat or
fluid
through solids
is
a problem that
is
frequently
encountered in engineering. In general, such a flow occurs in three spatial
we can assume the flow to occur one dimension. Examples of these problems are heat flow through uniform bars, fluid flow through pipes of uniform cross section, vertical flow in a medium of large extent, and flow in and out of vertical banks and cuts and toward vertical retaining structures. As in Chapter 3, the flow domain (Fig. 4-1) can be idealized as a onedimensions. For some problems, however, essentially in
dimensional line with the material properties attached to the
line.
The
governing differential equation for the one-dimensional steady-state flow can
be expressed as
/(*)
dx-
=
#x),
(4-1)
where k x is the material property, coefficient of permeability for flow through porous media {LIT) or thermal conductivity for heat flow;
two problems.
important and useful to note that Eq. (4-1) is similar to Eq. (3-45b) that governs one-dimensional deformation in a column. This indicates that It is
the
phenomenon of deformations
follow similar natural behavior.
in a
We
column and the one-dimensional flow
can even consider an interpretation that 93
Chapter 4
One- Dimensional Flow
94
(a)
(b)
A
A ©
A ®
©
®
(c)
Figure 4-1 Idealization for flow in pipe, (a) Flow through pipe, (b)
One-dimensional idealization,
(c)
Discretization in three ele-
ments.
flow or rate of change of deformation due the (axial) load and the flow or rate of change of potential or temperature due to applied fluid head or temperature are analogous.
Equation (4-1) governs steady or time-independent flow of heat and fluid through one-dimensional media; in fact there are a number of other phenomena such as one-dimensional moisture migration that are also governed by similar equations. The finite element formulation for Eq. (4-1) can therefore be used for 5,
we
all
these problems with only
minor modifications. In Chapter by using the problems of heat
shall derive equations for transient flow
flow and consolidation, while in this chapter
we
use steady fluid flow through
media as the example problem. We may note, however, that the finite element formulation for all these problems will essentially be the same except for different relevance for material properties and meanings of the unknowns such as temperature and fluid head. For the fluid flow problem, then, (p is the fluid head or potential = ply + z, where p = pressure, y = density of fluid, z = elevation, and k x = coefficient of permeability in the x direction. The forcing function f(x) can take the form of applied fluid flux intensity q(x). Now, we follow the steps rigid
(Chapter 2) required in the
finite
element formulation.
Step
As
Chapter
in
Step
We
Choose Element Configuration
1.
we
3,
use the one-dimensional line element [Fig. 3-2(c)].
Choose Approximation Function
2.
choose a linear approximation model for
=
(p
fltj
+
cc 2
fluid potential as
x
(4-2a)
or
where
=N
N = ^(1 — L), N =
^-(1
2
x
+N
l(Pl
2
=
+ L),
and
element nodal fluid heads. Equation (4-2) here
q>
(4-2b)
[N]{
is
r
{^„}
=
[
,
p2
]
is
the vector of
similar to Eq. (3-7) except that
replaces v.
The requirements for choosing the approximation function for
required only for the nodal values of (p. In other words, since
is
the highest order of derivation in the energy function, given subsequently in
Eq.
only up to order The completeness requirement is allows for rigid body motion and
(4-6), is 1, the interelement compatibility is required
equal to
1
—
= 0, that
1
is,
for nodal heads.
satisfied since the linear function,
Eq.
for constant state of gradient
Eq. (4-3a).
Step
,
Define Gradient-Potential Relation
3.
and Constitutive
The
gx
(4-2),
relation
Law
between the gradient of
(p
and
analogous to the definition of strain e y in Eq.
(p
can be defined in a manner
(3- 12a);
hence
«.-£ where gx
The
is
the gradient of
(«a)
with respect to x.
(p
relevant constitutive law describes the flow behavior through porous
media. The simplest constitutive law that
we can
use
is
Darcy's law (Fig.
4-2),
given by vx
where v x
= velocity
= -k x d£ = -k xgx
in the
x
direction
(4-3b)
,
and gx
= hydraulic
gradient.
The
negative sign occurs because the velocity decreases with gradient (Fig. 4-2).
The value of
the gradient can be obtained by differentiating
q>
in
Eq. (4-2)
with respect to x as
3?
= fx
[
? ({
" L)
L)
'
(4_4a)
95
;
One- Dimensional Flow
96
9x
=
Chapter 4
3y?
3^
Figure 4-2 Relation between velocity and gradient, Darcy's law.
Use of the chain
rule of differentiation leads to
= 4-1-1 £ dx I
As
before [Eq. (3-13)] [B]
i]
* \ fl
=
[B]{
is
the gradient-potential transformation matrix.
(4-4b)
Substitution of Eq. (4-4b) into Eq. (4-3) gives
-[RPKfJ, where [R] case
it is
= matrix
of coefficients of permeability
simply the scalar k x
.
It is
(4-5)
;
for the one-dimensional
interesting to note the similarity
between
Eqs. (4-4b) and (3-13) and between Eqs. (4-5) and (3-15); the gradient and velocity terms here are, respectively, analogous to the strain in the stress-deformation
and
stress
terms
problem.
Derive Element Equations
Step
4.
The
finite
element equations can be derived by using either a variational For problems with certain mathematical
principle or a residual procedure.
properties such as self-adjointness, valid variational principles are available.
For certain other problems,
it
may
not be possible to establish a mathemat-
ically consistent variational principle.
Residual methods such as Galerkin's
procedure can be used to derive
element equations for such problems
finite
methods can be used as general procedures. For the problem governed by Eq. (4-1) we shall first use a variational
in fact, the residual
principle. Later in this chapter,
method.
we
shall illustrate the use
of Galerkin's
One- Dimensional Flow
Chapter 4
The
variational principle corresponding to Eq. (4-1) can be written as
r^k
Q=A where
97
A
is
x
(^fdx - pfydx,
the cross-sectional area, assumed to be uniform,
(4-6)
and q
is
the
prescribed fluid flux; here units of q are (L 2 /T).
In Eq. (4-6)
we used
the symbol
Q
to denote a
system through which the flow occurs; this term
is
measure of energy in the analogous to the energy
measure IT^ for the stress-deformation problem. It is only to distinguish this measure for different problems that we have used different symbols in fact, we can use any convenient symbol. Note the analogy between the strain energy density term \o y e y in Eq. (3-21 a) and the term ^v x gx in Eq. (4-6) and between the potential of the external load fyv in Eq. (3-2 la) and q
(4-6).
Substitution of d
H=A
into Eq. (4-6) leads to
q>
[" #
Differentiation of
O
with respect to f
<5n
= o=
-
I""
and
l
^=
m]{
(4-7)
yields
0,
(4-8)
1^ = 0, or
A
f" [BYk x [B]dx{
=
P q[Wdx,
(4-9a)
or
M{
where
[k]
{Q}.
(4-9b)
= element property (permeability) matrix, [k]
=A
(" •>
[WkJWx
XI
i-'J'
and {Q}
=
{WkAWL,
(4-9c)
= element nodal forcing parameter vector {Q}= J{"qlWdx Xl
h;
q[N] T dL.
(4-9d)
:
Evaluation of
[k]
}
and (Q
Substitution for [B] and [N] in Eqs. (4-9c) and (4-9d) and integrations lead to
Ak.
[BY[B]dL
[k]
2
Ak
r[-
w-D
r
qi
{Q}
2 J_,
Notice that Eq. (4-10)
is
(4-10)
dL
=
li(l+L)J~
ql
(4-11)
2
[i
similar to Eq. (3-28) except for the fact that
has replaced E. The lumping of half the applied flux at each node
is
kx
the result
of the linear approximation model, and for higher-order models, the forcing vector can be different.
The element equations can thus be expressed Ak. I
Step
As
as
ql/2
1
(4-12)
-1
ql/2
Assemble
5.
explained in Chapter
3,
we can
differentiate the total of
elements and obtain the assemblage equations. The procedure similar to the direct stiffness
ment
that the fluid heads at a
is
Q
f
for all
essentially
approach and is based on the physical requirecommon node between two elements are equal.
is analogous to the interelement compatibility for displacements in Chapter 3. Then for a domain discretized in three elements (Fig. 4-1) with uniform area of cross section, we have
This
~
Ak x
1
-1
I
-1
'
1
[
2
-1
-1
2
-1
\
-1 1
>=
2
ql
(4- 13a)
2'
2
93
W
1 L
,
or
[K]{r}={R],
where
[K]
=
(4-13b)
assemblage property (permeability) matrix,
vector of nodal heads, and {R}
= assemblage
{r}
= assemblage
vector of nodal forcing
parameters. Step
6.
Solve for Potentials
Solution of Eq. (4-13) after introduction of the prescribed boundary conditions allows computations of nodal heads. the following properties
98
To
illustrate this step,
assume
:
Example
,
4-1
= = — =
A kx q /
Boundary
cm 2
1.0
,
cm/sec,
1
cm 2 /sec,
0.0
10 cm.
conditions:
2.0 cm, 1.0
We
cm.
use Eq. (4-12) to find the element equations. For instance, for element
number
1
we have 1
x
1
icm:
10
and so on for the other elements. The assembled equations are 0"
-1
1
2
-1
-1
2
1
9i cp 2
=
>
-1 The boundary conditions set
k
1 1
k 44
= =
<
3
1
L
are introduced as follows 1
and
ku
=0,
j
=
2, 3, 4,
1
and
k Ai
=0,
/
=
1, 2, 3,
r =2,
=
R<
1
1.
Therefore
0~
1
2
_2_
(Pi
10 <
.
>
TO
the
first
equation
>
<
10
1_
From
=
1
we have
q>
=
t
2.
From 9>3
10
"^
1
the second equation, _
10" =
"
10
or -
but
(p
x
=
+2
- "?3
-
=
2; therefore 2
From (p A
(p 3
2.
(i)
the third equation
-tp 2 but
- = 0,
=
1
.
+
2(p 2 - -4
=
=
Therefore,
-
+
2(Pz
=
1.
(ii)
99
:
One- Dimensional Flow
100
Equations
(i)
and
(ii)
Chapter 4
lead to
+
-7(p 2
=
4^3 3(p
2
:
Hence
and 2
-f =
2.
Hence
=f-
Step
7.
Secondary Quantities
The secondary quantities for the fluid flow problem are the velocities and The knowledge of
quantity of flow. these quantities. Velocities.
Element
1
:
= Similarly for elements 2
and
0.0333 cm/sec.
3
"- =
10
-' "III
\3 cm/sec. 0.0333
^,
= -^[-1 =
Quantity of flow
I?
is
0.0333 cm/sec.
:
Qf = v x A = where
i]{*
the velocity
0.0333
x
1
= 0.0333 cc/sec,
normal to the cross-sectional area A, and the subscript 1 and so on.
in v xl denotes element
Step
8.
Interpret and Plot Results
The computed values of cp and
v x are plotted in Fig. 4-3.
We can
see that
the finite element computations give exact solutions both for fluid heads and velocities.
This
is
because the cross section and material properties are
One- Dimensional Flow
Chapter 4
2
101
cm
1 0.0333 cm/sec
(b)
Figure 4-3 Plots of
finite
element computations,
(a)
Fluid heads,
(b) Velocity.
uniform, and constitutive (Darcy's) law
is
linear.
nonuniform cross section and nonlinearity can for (p and v.
FORMULATION BY GALERKIN'S METHOD The residual corresponding to Eq. (4-1) Rif)
=
k x %£
As
indicated in Chapter
3,
yield nonlinear distributions
is
- q.
(4-14)
The details of derivation are almost identical to those for the column problem in Chapter 3. The final results for the three-element system (Fig. 4-1) will be >
f
1
-1
1
2
1
-1
2
1
dx). \
-1
|(
Lid 2
2 >
-1
1 \
[K]{r]
+
k
= {R r + }
{RJ
(4-1 5a)
<
2
4
= {R}.
(*%1
1
(4-15b)
The boundary terms {R £ } include flux, k x (d
PROBLEMS 4-1.
Solve Example 4-1 by considering a nonzero value of q.
4-2.
Use code DFT/C-1DFE and
4-3.
Use code DFT/C-1DFE and compute the 30
cm
verify the results in
Example
4-1.
steady-state temperature in a bar
long with the following properties,
= 1.0, = 0,
kx <7
/= and temperatures
10 cm,
at the ends,
T(0)
7(30 cm)
= =
100 degrees, degrees.
BIBLIOGRAPHY Desai, C.
S.,
"An Approximate Solution ASCE, Vol. 99, No. IR1,
Drain Div., Desai, C.
S.,
ments
wicz, O.
Desai, C.
"Finite Element
in Fluids,
S.,
C,
Vol.
eds.),
"Seepage
I
in
for
Flow
in
(Gallagher, R. H., Oden,
Wiley,
Engineering (Desai, C.
Methods
for Unconfined Seepage," /. Irrigation
1973.
New
Porous Media," J.
York, 1975, Chap.
T., Taylor,
in Finite Ele-
C, and
Zienkie-
8.
Porous Media," in Numerical Methods in Geotechnical S., and Christian, J. T., eds.), McGraw-Hill, New York,
1977, Chap. 14.
Pinder, G.
F.,
and Gray, W. G.,
Finite
face Hydrology, Academic Press,
Remson,
I.,
Hornberger, G. M., and Molz,
face Hydrology, Wiley-Interscience,
102
Element Simulation York, 1977.
in
Surface and Subsur-
New
New
F. J., Numerical Methods York, 1971.
in
Subsur-
5 ONE-DIMENSIONAL TIME-DEPENDENT FLOW Introduction to Uncoupled
and Coupled Problems
In this chapter
we
treat
the case is
when
in which temperature or fluid pressures These effects can occur in two ways. For
problems
act in addition to external loading.
the magnitudes of temperature or fluid pressures are
relatively easier to include
them
in the finite
known,
it
element formulations because
Some examples known temperature distribution in a structure and known fluid pressure in a porous body. The general case occurs when temperature is unknown just like the displacement. Then we need to consider interaction or the effects can be superimposed o r considered as uncoupled.
of these are
coupling between deformation and thermal
effects.
UNCOUPLED CASE As an
illustration, let us
perature Tis to cause a
assume that the
known
€,.
where
a'
,
f*
The
stress-strain relation
ay
is
known change
column
in tem-
[Fig. 5- 1(a)],
= *'T,
(5-1)
= coefficient of thermal expansion.
e y the effective elastic strain e yn
of a
effect
strain e yo in the
If the total strain
is
denoted by
given by
= *, -
€y
(5-2)
-
can then be written as
= Ee = E(e, yn
e yo ).
(5-3)
103
One- Dimensional Time- Dependent Flow
104
Chapter 5
EA
(a)
1000 kg
10cm Area = 1.0 cm 2
®a' v,
= coefficient of thermal expansion = 0.0000065 cm/cm-0°F
=0
10cm E =
J
10
2x 10 6 kg/cm 2
©
cm
© (b)
Figure 5-1 Thermal effects in a bar idealized as one-dimensional, (a)
and
Bar and one-dimensional properties.
idealization, (b) Finite element
mesh
One- Dimensional Time- Dependent Flow
Chapter 5
For deriving
Tl p
= A-
element equations,
finite
energy function [Eq. f"
a,€„dy
-
r Yvdy - P T vdy - £ Puv
A
y
t
related to the external forces are the
same
as before.
term related to the strain energy can be written by using Eq.
U'
(5- 4)
.
Jyi
Jyi
Note that the terms
modify the potential
to
(3-21)], as
Jyi
first
we need
105
The
(5-3) as
= ±\" E(e y -(J(e y -eJdy Jyi
A C y2
=4
C
=U +U 1
can drop the
last
Ee y €yo dy
A C
+
f
Jyi
Jyi
We
yz
Ee)-A\
y2
Ectfy. Jy
t
(5-5a)
2
term of Eq.
contribute to element equations
(5-5a), since,
when
Up
is
being a constant,
differentiated.
the problem in Chapter 3 [Eq. (3-21)], the only
it
will
not
Thus, compared to
new term
that enters
is
the
second term in Eq. (5-5a), which after substitution for e y from Eq. (3-13) can
be expanded as
U2 =
Ap[Vl
v2
]^-^Ee
yo
dy
=
A{qf
J"
T [B] [C]{eyo }dy
Jyi
2
-\L)\ aU-Vi
+
(v 2 L)
= AEe (-v + v U2 is the part of the strain energy relevant to e yo
1
Here same as loads.
(5-5b)
2 ).
U in Eq. (3-21) and corresponds to the Now differentiation of 11^ with respect to v
yo
;
the other part
strain x
U
x
is
the
due to the external
and v 2 gives
Mi =
AEey X-l),
(5-6a)
^=
AEeJY),
(5-6b)
or in matrix notation,
=AOy II
\
!*
One- Dimensional Time- Dependent Flow
106
Chapter 5
In general terms,
{Q Finally, if
=A
}
f" [BY[C]{e yo ] dy.
(5-7b)
j yi
we add {Q
}
in Eq. (5-7) to Eq. (3-28), the modified element
equations are [k]{q}
=
{Q}
+
{Q
(5-8)
},
where all terms except {Q } have the same meanings as before. We call {Q } an additional, extra, correction, "initial," or residual load vector. Thus the known strains due to temperature changes become an additional load which is superimposed on the external load {Q}. It appears on the right-hand side of Eq. (5-8) and does not contribute to the left-hand side, which contains the unknown displacements. Initial Stress
As can be can act as
seen from Eq. (5-7), the
"initial" stress or pressure.
known temperature or fluid pressure instance, we may know from the
For
solutions of the Laplace equation (Chapter 12) the distribution of seepage
dam; such
pressures below a
pressures can be converted as additional load
vectors by using Eq. (5-7).
Residual Stresses
The concept of "initial" term "residual,"
is
strain or stress,
which can be
called
by a
common
important and useful for incorporation of residual
conditions caused by temperature, fluid pressure, creep, lack of
phenomena. This concept has
also been used widely in a
fit,
and other
number of techniques
for nonlinear analysis.
Example
5-1
We now illustrate the concept tion.
The
details are
shown
tions to include the effect of a
solve for displacements
and
of
load vector by using the foregoing formula-
initial
in Fig. 5-1 (b). It
is
required to derive the element equa-
in temperature of 1 00°F and then to combined (uncoupled) superimposed thermal
change (increase)
stresses for
and external load = 1000 kg at the top of the bar. According to Eq. (5-1), the initial strain is
effects
= aT =
€yo
Hence, the resulting
0.0000065 x 100
= 0.00065 cm/cm.
stress [Eq. (5-3)] is
E€yo
=
2 x 10 6 x 0.00065
=
1300 kg/cm 2
,
1
One- Dimensional Time- Dependent Flow
Chapter 5
and the
initial
load [Eq. (5-7)]
is
{Q The assembled
initial
}
=
x 1300
1
{~ J}-
load vector {R} for the three elements
-1 1
=
{Rol
13( 1
-
is
-1300 1
=
»
<
1300,
1
By
107
following the procedure in Chapter 3 and by adding {R
vector,
}
to the external load
we obtain 1
2 x 10« x
-1 2
-1
-1
2
1
1
0"
\v 2 >
<
10
-1
-1
-1300
o
\v\
=
i
»
+
<
^3
.-1000,
1_ [v* .
1300.
-1300
300
.
Note that here (Fig. 5-1) we have measured y positive in the upward direction. The boundary condition v = is now introduced, and the solution of the resultx
ing equations yield v\
=
v2
=0.0015,
v3 V4
The
total strains
due to the
0.000
cm
(specified),
= 0.0030, = 0.0045. effects
of both the external loads and thermal effects
are found as
[B]{q}=y[-1
€,
€yi €y2 €yt
= = =
»C)
+0.00015, cm/cm, +0.00015, +0.00015.
These strains are equal to the sum of the strain due to the external load, ey -0.00050, and that due to the thermal effect, €yo = 0.00065.
=
COMMENT This example
concept of
initial
is
included mainly as an elementary illustration of the
load vector.
It
does not necessarily represent a practical
One- Dimensional Time- Dependent Flow
108
Chapter 5
Moreover, unless the ends are constrained, the temperature not influence the stresses in the bar.
situation.
may
change
TIME-DEPENDENT PROBLEMS As
illustrations we shall consider two problems governed essentially by the same equation. They are time-dependent temperature effects [1] and timedependent deformations with expulsion of fluid from the pores (of a homogeneous medium); in geomechanics, the latter is called consolidation [2]. Both problems can be represented essentially by the following second-
order partial differential equation
called a parabolic equation:
[1],
dT*
d 2 T*
where T* = temperature (or fluid pressure) and a = thermal diffusivity = k/pc, where k = thermal conductivity (Btu/hr-ft-0°F or W/m K), p = density (lbm/ft 3 or g/cm 3 ), c = specific heat (Btu/lbm-°F or J/kg.K), x = space coordinate, and t = time (Fig. 5-2). It is again interesting and useful to note that Eq. (5-9a) is similar to Eqs. (3-45b) and (4-1) that govern deformations in a column and flow in a medium, respectively. The difference is that now we have the extra term on the right-hand side to account for the time dependence of the phenomenon. It is often more convenient and necessary, particularly for problems involving nonhomogeneities and layering, to express Eq. (5-9a) as •
/9 2
k for the thermal
problem
and
[1]
At*
T*
-6^ =
ec
(5 " 9b >
-W
as
*L*g = m,*! dx 2
yw for the consolidation
k
problem
[2, 3, 4];
v
(5-9c) v
dt
'
here/?*
= (excess)
pore water pres-
the coefficient of permeability (ft/sec or cm/sec), y w is the unit coefficient of volume comweight of water (lb/ft 3 or g/cm 3 ), and v sure,
is
m =
pressibility (ft 2 /lb or
cm
2
/g).
The boundary conditions
associated with Eq. (5-9) can be expressed as
initial conditions, (i)
and boundary
(iii)
We
is
= f *(*),
< x < h, < 0, t
(5-10)
conditions, (ii)
where h
T*(x, 0)
T*(0,t) T*(h,t)
= f*(t), = f (h), h
t>0,
(5-1 la)
/>0,
(5-1 lb)
the total length of the domain.
note that for the time-dependent problem extra boundary conditions
occur as
initial
conditions [Eq. (5-10)] to define the
initial
or starting condi-
One- Dimensional Time- Dependent Flow
Chapter 5
Element
109
i
(a)
t-At
t
+ At
(b)
Figure 5-2 Solution in time domain, (a)
Time
variation of tem-
perature in a typical element, (b) Finite difference approximation for
first
derivative.
tions of the body. In the case of static or time-independent problems in
Chapters 3 and 4 such
We
initial
conditions are not required.
shall first follow the steps for the
thermal problem and then
illustrate
the case of consolidation. Step
1.
Discretize and Choose Element Configuration
The one-dimensional medium Step
2.
is
divided into line elements, as in Fig. 3-2.
Choose Approximation Model
We choose a linear model to express temperature within an element as r = i(i-L)r (0 + i(i-f-x)r2 (0 1
= [N(x)]{T «}, n
(5-12)
=
:
1
One- Dimensional Time- Dependent Flow
10
where
{T„}
= [T T
T
Here
the vector of unknown temperatures at nodes
2 ],
x
respectively.
T
x
Chapter 5
and
T2
are time dependent,
and [N]
1
and 2,
dependent on the
is
spatial coordinate x. As stated in Chapter 4, this function satisfies the requirements for approximation function for T for the one-dimensional flow.
Step
Define Gradient-Temperature and Constitutive Relation
3.
The following heat flow and
is
rate equation can be
q'
where
q'
= rate
2
or
m
Step
(5-13)
W/m-K), and A
= area normal
to
= thermal x
direction
2
).
4.
First
^
= -kA 6
of heat flow in the x direction (Btu/hr or W), k
conductivity (Btu/hr-ft-°F or (ft
assumed to describe the mechanism of
used as the constitutive relation:
Derive Element Equations
we
consider the transient heat flow problem. Use the following
variational functional at a given time level for obtaining the element equations
[5, 6]
°
A
['iH£y + % T i dx - \j Tdx
(5 - M) -
where q = applied heat or fluid flux and CI denotes the variational functional. We have here used the symbol Q to distinguish from the n p used for potential energy in the stress-deformation problems.
By
using Eq. (5-12),
we have
g = ^[i(i-i-)r =
1
+i(i+L)r2
]
(5-15a)
[B]{T„}
and
£J= t
= £[#! -
L)T,
+ ^(1 + L)T2
]
dT, dt
[N]
\dT2
"fcl
,
I
dt
[N]{f„} ;
(5-15b)
One- Dimensional Time- Dependent Flow
Chapter 5
111
where {T„} r = \t f 2 ] and the overdot denotes derivative with respect to time. Note that [N] is function of space coordinates and hence constant for the time derivative. Substitutions from Eqs. (5-15a) and (5-15b) into Eq. (5-14) yield x
A_l_ 2 2
F„Y[BY
j
+ ^^{t} T lWmTn}dL (5-16)
Expansion of Eq. (5-16) leads to (assuming a to be constant)
(5-17a)
1
-1
-1
1
J
V-4 2.
#1#2
#i# 2
--f J"
(JV.7,
+N T )dL. 2
(5- 17b)
2
Further expansion of the matrix terms leads to
Q
2T T2 x
(N\t,T,
(N T X
We now
X
+
T\)dL
+ N,N
+N
2
2
t,T2
+ N N t T + N\t T )dL x
2
2
x
2
2
T2 )dL.
(5-18)
Q
take differentiation (variation) of
important difference between
this variation
with respect to 7\ and
and the variation
in
T2 An .
Chapters 3
and 4 may be noted. Here, the functional involves time derivatives, f and f 2 During the variations with respect to T and T2 we make an assumption that t t and t 2 remain constant. This assumption makes the variational x
.
x
,
:
One- Dimensional Time- Dependent Flow
112
approach mathematically dQ.
Ace
Wt-TxlF'j: + 4[j dQ
_ ~ 2l
At*.
dT,
(
differentiation yields
2T >~ 2TJ dL
+ NyNjjdL - St
(Nit,
N
t
dL
= 0,
(5-19a)
J"
(-2T
2
The
less rigorous.
Chapter 5
l
+ 2T2 )dL
+ 4[^(N N2 T +Nlt 1
l
2
)dL
= 0.
,dL
-#J>
(5-19b)
Therefore,
dQ dT
~
x
Aa
f,
2/ ,
- T )dL
(7\
2
1
+ dQ dT2
^J
^a
+ N N t - lf.(l) =
(tfff ,
ti>
r,
+r
2 )
+
^f*
(5- 19c)
0,
2)
2
x
QTtNjt + l
ATif 2 )
(5-19d) Integration of the terms gives
^= A - T g = ^(-r, + -5j-(7\
+
2)
%(2t,
+1 2)
r2 ) + f(t, + 2f 2 ) -
Rearranging
^a /
in
" 1
|-
= 0,
f=
(5-20a)
o.
(5-20b)
matrix notation leads to
-r
_-l
i_
"2
1"
kr-6- .1
2_
f^il
,
^/
ft
-* r
>l
7-
_W
2
_?' 2
(5-2 la) 1
or
LKKTJ
+
[kjffj
= {Q(r)}.
(5-2 lb)
Layered Media In the case of layered media, the formulation based on Eq. (5-9b) or (5-9c) will lead to the following element equations
:
One- Dimensional Time- Dependent Flow
Chapter 5
i
-i
i
i
r2
r
a, 1 2
1
2
a-fo
113
(5-21c)
where A, = ^4/:// and A 2 = Apcl/6 or ^j = Ak/yJ and A 2 = AmJ/6, corresponding to Eq. (5-9b) for the thermal problem and Eq. (5-9c) for the consolidation problem. In the case of the latter, temperature is replaced by pore water pressure. Although the same symbol q is used for prescribed flux in Eqs. (5-21a) and (5-21c), their units will be different and should be defined carefully. Here [k a ] = element thermal diffusivity matrix, [kj = element matrix related to time dependence, and {Q(0} = element nodal vector of forcing (flux) parameters, which can be time dependent. The matrices in Eq. (5-21) can be expressed as [K]
=
A P[Bfa[B]
(5-22a)
J XI
[k,]
{Q(t)}
=A =
\" [W[N]dx,
(5-22b)
I" [Wdx.
(5-22c)
J Xl
Solution in
Time
Additional derivations are required in the time-dependent
problems
because of the appearance of the second term on the left-hand side of Eq. (5-21).
Up
using the
we have discretized the domain only in the spatial need to obtain solutions in time. This can be done by
to Eq. (5-21)
direction, x.
Now we
finite difference [7]
respect to time.
The
first
type of discretization for the derivatives with
derivative dT/dt can be written approximately as
[Fig. 5-2(b)]
where At = time increment. Equation (5-23) essentially gives the slope of the chord joining points A and B as an approximation to the continuous derivative dT/dt. By using Eq. (5-23), we can now write approximations to the time derivative at the two nodes of an element [Fig. 5-2(b)]
^r.(' + ^)-r.(O dT\
The time
We
_T + 2 (t
At)
-T
integration scheme [Eq. (5-23)]
2 (t)
is
t
t
(5.24a)
(524b)
the Euler-type procedure
[7].
use this simple scheme only for the sake of introduction. In fact, a
number of improved and mathematically superior schemes such as the Crank-Nicholson procedure are commonly used in finite element applications.
One- Dimensional Time- Dependent Flow
114
Chapter 5
Substitution of Eq. (5-24) into Eq. (5-21) leads to
or
(w + iw)
;i;:3**^w
<->
or
>#:s-«* where
= ft.] +
[k]
=
{0}
{Q(r
+
^ft] A/)}
+ [k,]{™
At any time
/, the terms on the right-hand side in Eq. (5-26) are usually known, {Q(/ + At)} from the specified forcing functions and the second term from the known values of T at the previous time level. Because the initial conditions [Eq. (5-10)] are given, we know the initial values of Tat any point
at time
/
= 0.
Hence, we can solve Eq. (5-26) for Tat
the previous time level are
The
+
At, since the
Tat
known.
solution process for the time-dependent problem thus involves two
and then (2) propagation or marching in time The foregoing problem belongs to the class of problems problems, since we start from an initially known state.
steps: (1) discretization in space
at various time levels.
called initial value
Step
4.
We now
Derivation by Galerkin's
Method
consider formulation of
finite
element equations [Eq. (5-21)] by
using Galerkin's residual method. Equation (5-9a) can be written as
tF-t?-*
(5 - 27a >
or
a (
fe-|) r * = °-
(5 - 27b)
or
LT*
= 0,
(5-27c)
where L is the differential operator. Denoting the approximate solution by we have the residual R, *<*>
= a 0-w-
T,
(5 " 28)
One- Dimensional Time- Dependent Flow
Chapter 5
On
the basis of the explanation in Chapter
Assuming
for the Galerkin formulation.
3,
linear
we
115
consider a generic element
approximation for an element
we have
as in Eq. (5-12),
r = i(i-L)r (0 I
+ i(i+L)r (0 2
= t NT
(5-29)
t.
t
Here the interpolation functions
T
peratures
are functions of x,
TV,
We
are the functions of time.
t
and the nodal tem-
note here that for a given time
t,
Eq. (5-29) gives variation of the temperature along the length of the element.
At
time dependence
this stage, the
only, 7^(0,
Now
T2 (t)
is
included in the temperatures at the nodes
[Fig. 5-2(a)].
according to Galerkin's method, (5-30a)
£ Integration by parts of the
—
—
3— dx a3 ** &<
L
first
-
term leads to
NA
a 3-
dx
'L
-f
J„
-3dt
N dx =
(5-30b)
t
or x
(
\dTdN
[
dT M N dx
W
,
„dT
= *
"dx-lti^
+L
T=
into Eq. (5-30c) leads to
I Substitution of
X2 ,
{
,
£ NT t
t
<
(5-30c)
<
i
=
l,2,y=l,2. (5-3 la)
The index notation has an implication In matrix notation, we have dNj dx J Xl
u
dN1 dN2
dl^^dNi
.
dx
dx
dx
dN2 dN2
dx
dx
3.
dT,
t
dx
similar to that explained in Chapter
Ty
dx
+
r
dx
dt
N,N2
>dx
dT2
N {
dt
J
\..dT
dx (5-3 lb)
dT dx-
The two terms on the left-hand [k,]
as in Eq. (5-21).
side yield the
same element equations [kj and
:
One- Dimensional Time- Dependent Flow
116
Because
xu
N = x
at
x 2 and
N
at x,
2
and A^
=
1
Chapter 5
and
at x,
7V2
=
1
at
the left-hand side yields
— 0i
i-(©,i
(5-32)
which denotes (joint) fluid fluxes at nodes 1 and 2, respectively. As explained in Chapter 3, when we assemble the element equations, the terms in Eq. (5-32) can vanish, except those at the ends, and they can constitute conditions specified at the ends.
Step
5.
As an
Assembly for Global Equations
one-dimensional bar of metal (Fig. 5-la)
illustration, consider a
through which heat flow occurs. The bar (Fig. 5-3) with four
divided into three elements
is
nodes; the elements are assumed to be of uniform length
A
A
A
©
©
L = -1
L = +1
©
©
Figure 5-3 Discretization of bar.
and cross
Use of Eq. (5-21) allows generation of element equations By observing the fact that temperatures at adjacent continuous, we can add the element equations to obtain global
section.
for the three elements.
nodes are
equations as follows
"1-1
Ace I
~2
TV T2
0"
2-1 0-1 2-1 T 0-1 J
-1
1
^
>
6At
3
1_
4
"2
R*
=
*2
1
^
>
<
R
6At
3
A
_0
t+At
t;
4
1
T2 T
1
2_
J
0"
T
1
1
_0
t+At
0~
1
4
1
4
1
1
4
1
1
2_
3
t
.
t
r1 T
(5-33a)
3
J
t
,
or in matrix notation,
^[K *]{r} f+Af + ^[K nw a
t
+ At
= Wt+
»+£k%
Here [K*] and [K*] are the assemblage matrices and
{r}
r}„
(5-33b)
and {R} are the
:
Chapter 5
One- Dimensional Time- Dependent Flow
assemblage nodal
unknown
(temperature) and
and the superscript
vectors, respectively,
117
known
forcing function
asterisk denotes assemblage matrices
Note that we can take out a// and l/6At as common if the assumed to be homogeneous; then Eq. (5-33b) can be written as
in Eq, (5-33a).
material
is
Ar[K*]{r}, +A
where AT = to be unity)
ccAt/l
2
+ |[K*]{r} r+Ar = ^{R}, +A + -g-[K*]{r}„
(5-33c)
,
,
is
a nondimensional time increment (with area
Btu pc
ft
3
lbm-°F Btu
hr-ft-°Flbm
A assumed
ft^
hr*
Therefore,
AT = 5!
*L '
hr
ft
2
= nondimensional.
Boundary Conditions
At
t
= 0,
we have
scribe values of
+
at
At
l9
at
t
<
We
=f
as uniform
.
choose to pre-
Hence Eq.(5-33)
is
7-[KJ]{r} 0+Af
The
the initial condition [Eq. (5-10)].
T T2i T3i and r4
initial
+
conditions are applied at
Now we
=
^-[K*]{r} 0+Af
/
{R} 0+/
6At
[KJ{r}
.
(5-34)
= 0.
t
can introduce end (geometric) boundary conditions r,
=r,(Q,
r =r
4 (/z,
4
*)
= *i.
'>o,
o
=S
t>
4>
(5-35) o.
A procedure
similar to the one described in Chapter 3 can be used. Here S and (5 4 are the known values applied at nodes 1 and 4, respectively, and can be assumed to remain constant with time. For clarity of illustration let us assume that Ar = 1, T(x, 0) = T and {R} = {0} then Eq. (5-34) can be expressed as x
;
0"
"4
I sym.
T
1
i
o
o-
T2
i
4
o
-i
T,
sym.
$
£
4
r4
§
1
-! f
7-1
r,
4-
Af
3
6
+
+ 3
\
2
6
"*"
6
T, (5-36)
£_0
6
^
_i
6
^-*
3
+
_j_ _f_o "*"
6
To
li 3
\2
One- Dimensional Time- Dependent Flow
118
It is
important to note that the right-hand side essentially yields an equivalent
or residual load vector {R f
Chapter 5
known temperatures
corresponding to the
}
at
= 0. now be
Equation (5-36) can [Eq. (5-35)].
modified to include boundary conditions
The modification can be performed by
setting
KAA — J = 2, 4, 2, 3, J =
*n =1,
1,
Ku=0, K4J = 0,
(5-37)
1,
R =5 U t
=S
R*
A9
which leads to 0"
'
1
f
-i
i
!
I I.
Note
that Eq. (5-38) lost
its
'Ti
s,
T2 T Tt
T
\
(5-38)
r.'
3
A,
A/
symmetric nature. Often,
it is
economical and
convenient from the computational viewpoint to restore the symmetry of the matrix.
It
can be achieved, as described
Chapter
in
3,
by modifying further
the other equations as
j
=
7
= 1,2,3,
Kn = 0, Kj, = 0, and the right-hand term
0"
|-|o
0-1
f 1
Step
6.
(5-39a)
as
"1
o
2, 3, 4,
Solve for Primary
[TV
8.
\T2
To
\t3
r.
k.
At
.
+ +
§<5,
1
(5-39b)
i*
4
<5 4
J
Unknowns
Example 5-2
To
illustrate solution
of Eq. (5-39b),
let
us adopt quantitative values for
T 5u and ,
S 4 as follows:
f = Si
S4
= =
0.0,
< x < h, < 0,
10degrees|
20 degrees/
t
f>Q
(5-40)
1
One- Dimensional Time- Dependent Flow
Chapter 5
Then Eq.
(5-39b) for the
first
time increment
=
(t
0"
+
119
At) becomes
10
(
5_0
f
6
(5-41 a)
> 1
"I
oo I
[T4t
1
2oJo
Therefore,
!^2
S'3
6
'
7Y
00 3
6
'
or
- 5T = -5T + \6T = \6T2 2
3
50,
3
100,
or by multiplication by 5 and 16, respectively, of the two equations,
8or2 - 25 T = -8or2 + 2567 231 T = 3
250
3
1600
3
1850 =
=>TT = : 3
we have
8.01
and
16T2
-
16T2
-
-
x 1850
=
50,
=
50,
231
40.04
T2 =
5.63.
SECOND TIME INCREMENT For
the next time increment
1
-1
-1
0"
Tl \
5
I
-i
!
_5
T2 T
+
At
fi
l
o
i
i
i
(/
]
4
[T4
At
on
=
+
\
Lo
2 At
Introduction of boundary conditions,
i o
5.63
1 I
8.01
<5i
=
tJ
we have
[25.63] ]
o
|
2A/),
10 f
40.53
=h 57.67
>
o
3
1
+
20
10 and S 4
Af
=
(5-41 b)
48.0lJ
20, yields
10
01 (7V
1
1
-I <
-1
8
1
T2 T
40.53 >
3
=
x 10
6 (
48.01
x 20
1_ IT*] 2Af
20 10 90.43/6 (5-41 c)
148.01/6
20
One- Dimensional Time- Dependent Flow
120
Chapter 5
Therefore 8
-
It
+
5
T -
|r
3
=
90.43
6
'
148.01
6
or
- 5r = -5T2 + 16T = 16T2
3
90.43,
3
148.01,
or
80r2 80r2
+
25T3 256r3
= =
452.15,
2368.16,
or
231T3 =2820.31,
T3 =
12.209,
and
- 5 x 12.209 = 16r2 = 151.4756, T2 = 9.467, 16jT2
90.43,
and so on for other time increments. Step
7.
Compute
the Derived or Secondary Quantities
The derived quantities can be rate of flow and quantity of flow. For example, from Eq. (5-13) rate of heat flow q' for an element can be written as
=
-kA[B]{T„}
= =^[-i Since
we know T and T2 x
(5-42a)
!]{£}•
for each element, computation of q'
forward. For instance, for element 2 at time At,
T = x
10 and
is
straight-
T2 =
5.63.
Therefore g'(2)
= =^(-10 =
+5.63)
4.37^ Btu/unit
time.
(5-42b)
ONE-DIMENSIONAL CONSOLIDATION Consolidation
the
is
[3, 8, 9]
phenomenon which describes time-dependent deformamedium such as soil under applied (external)
tion in a saturated porous
The material deforms with time while the liquid or water in the phenomenon involves coupling or interaction between deformation and pressure in the fluid. Under a number of assumptions [2], it is possible to approximate the phenomenon to occur only in one (vertical) direction; then the stressdeformation behavior of the skeleton of the medium and the behavior of fluid can be treated separately. The stress-strain behavior is expressed through an loading.
pores gradually squeezes or diffuses out. This
effective stress
concept given by
a=o'+p, (5-43) = total applied stress (Fig. 5-4), a' = effective stress carried by the skeleton, and p = pore fluid (water) pressure. The stress-strain or
where a soil
constitutive relation for the deformation behavior of the skeleton can be
expressed as (see Ref.
[2]
or other undergraduate texts on geomechanics)
Aa'
= -—Ae,
(5-44)
av
where a v
= coefficient compressibility; e = void ratio,
proportional to axial
and A denotes change due to a load increment. Under the above assumptions, the governing equation for one-dimensional consolidation is given by Eq. (5-9c); for homogeneous media, it can
or vertical strain
;
be expressed as
*&-£ where
cv
<5
-
45 >
= coefficient of consolidation = k{\ + e )/y w a k = coefficient of y w = unit weight of water, y = vertical coordinate, and v,
permeability, eQ
= initial void ratio.
The
constitutive law for the flow behavior
is
Darcy's
law:
v=-*- dP. Equation (5-45)
is
(5-46)
of the same form as Eq. (5-9) for heat flow, and
all
steps of
the finite element formulations are essentially the same.
We
shall therefore detail
an
illustrative
problem for consolidation by
using the results derived previously.
Consider a consolidating mass (Fig. 5-4) divided into three elements [Fig. 5-4(d)].
Assume
the following properties:
Length of element,
/
Coefficient of consolidation,
cv
Applied vertical load,
a
= 10 cm; = cm /sec; = g/cm 1
1
2
2
121
One- Dimensional Time- Dependent Flow
122
Chapter 5
TTTTT
ty
(a)
(b)
p(0,t) =
p(y, 0) = p
y
10
cm
10
cm
A A
= a
2H Saturated
—
30 cm
(
n
10
cm
^^^%m% v&z p(2H,t) =
^(2H,t)
(d)
=
3y
(c)
Figure 5-4 One-dimensional consolidation, (a) Consolidating mass,
One-dimensional idealization,
(b)
(c)
Boundary conditions,
(d)
Discretization.
By using Eq. " 1
10
(5-2 la),
we
obtain, for each element (with
1
-r
[2
11
-1
i_
{;]+? _1
2_
A
=
QM
1),
(5-47)
or [k.]{p„}
The simple forward
(™ +
+
[k,]{p„}
= {Q(0}.
difference Euler-type integration process will lead to
**€L =m))- + ^ti
(5 - 48)
One- Dimensional Time- Dependent Flow
Chapter 5
123
Assembly of the three element equations gives 1
-1
-1
2
0"
Pi
-1
10
6
4
p3 t
1
10
1
1
4
1
Pi
1
2
Pa
0)
P(0,
t)
>(2H,
t)
= 1.0g/cm = 0.0, = 0.0,
f+Al
2 ,
Pi (5-49)
assumed homogeneous medium and A7" = 1. initial and boundary conditions (Fig. t
Pi 1
Assume
p(y
\P*\
4
R>
Pa
Pi
'2
Ri
JU.
1
12
lj .Pa. + Al
Ri
We
Pi
10
1
)
-1
o
L
01 Pi
>1
0
5-4) as h
,t<0,
t>0,
(5-50a)
t>0.
Here we assume that the top and bottom of the consolidating mass are pervious, and hence the pore water pressure there times; this constitutes the
first
possible to incorporate impervious boundary, that
type condition. For instance,
dp (2H, By
is
is,
the second or
we can have impervious r)
essentially zero at all
or Dirichlet-type boundary condition.
=
0,
t
>
0.
base,
It is
Neumann-
and then (5-50b)
it is not necessary to make any boundary condition. The condition of no flow across the base implied in Eq. (5-50b) is achieved simply by leaving the boundary node as it is and treating it just like other nodes where p is assumed as the unknown and obtained through the finite element computations. The procedure for introduction of the boundary conditions [Eq. (5-50a)] is essentially the same as for the temperature problem. An important observation can be made at this stage. The problem of consolidation is a coupled problem involving interaction between the deformation of the skeleton of the medium and the pressure in the pores of the medium. However, it is only because of one-dimensional idealization and the accompanying assumptions [2, 8, 9] that we are able to consider separately the deformation [Eq. (5-43)] and pore pressures [Eq. (5-45)]. In reality and in a mathematical sense, the effects of the two phenomena, deformation and fluid pressure, are coupled.
In the
finite
element formulation herein,
special modification for this
COMPUTER CODE The code DFT/C-1DFE mentioned
in Chapters 3 and 4 and described in Chapter 6 can be used to solve the problems of transient heat flow and
consolidation by specifying the required option.
At
this stage, the reader
may
study the portion of the code relevant to these problems. In the following are described
some
from application of this code.
results
Example 5-3 Figure 5-5 shows a one-dimensional idealization of a homogeneous into 10 elements
and
Initial
The following
11 nodes.
conditions:
T(x, 0) or p(x, 0)
Boundary conditions:
divided
=
100 units,
0=0, or p(2H,t) = 0;
7(0, /) or p(0,
T(h,t)
a or
medium
properties are given:
cv /
/
= = =
1
unit,
1
unit,
0.1 unit.
Temperature Problem. In the case of heat flow, this problem represents timedependent cooling of a bar initially at a temperature of 100° whose ends are cooled to 0° and kept at that temperature for all subsequent time levels.
A
pictorial distribution of
the bar
is
shown
computed temperatures
in Fig. 5-6(a). Distribution
at various time steps
along the bar at various times
is
along
shown
in Fig. 5-6(b).
Thus, sional
it
is
possible to
compute
distribution of temperatures in a one-dimen-
medium by using the finite element
procedure. The temperatures and quantity
of heat flow can be found at any point in the bar and at any time
level,
giving the
under cooling or heating due to a given change in temperature. It is possible to include different geometrical and material properties for each element; hence, nonhomogeneous media can be easily accommodated. It is also possible to include natural boundary conditions, that is, boundaries that are insulated against heat. For complete insulation, it can be done just by leaving the node at that boundary free, that is, by allowing the finite element procedure to compute temperature at the node. This is possible because the formulation procedure includes the natural or gradient boundary condition automatically, in an integrated sense. Consolidation. In this case, the problem represents transient deformations of a porous saturated medium such as a soil foundation subjected to a load of 100 units. Because the medium is saturated, at / 0, the fluid carries all the applied load (pressure) and the initial conditions are p(y, 0) = 100 and a'(y, 0) = 0. As time elapses, the pressure dissipates and the load is gradually transferred to the soil skeleton and a' increases; at / = oo, o'iy, oo) = 100 units. If the top and bottom entire thermal history
<
boundaries are pervious,
The time-dependent temperature (Fig. 124
we can assume
that p(0, t)
=p(2H,
t)
=
0.0 at
all
times.
distribution of the p(y, t) are similar to the distribution of
5-6). Often, in
geomechanics,
it is
convenient to define degree of
:
One- Dimensional Time- Dependent Flow
Chapter 5
125
© A" © A
A
©
Figure 5-5 Finite element mesh.
consolidation
U as
[2]
U=
1
\7
pdy
hM* e
Pody where p
M=
is
the initial pressure,
(7T/2)(2/
+
Tv =
(5-51)
=
nondimensional time factor
cv t/H 2
,
and
1).
Figure 5-7(a) shows the distribution of U versus time factor Tv Figure 5-7(b) shows values of excess pore water pressure along the depth at various values of U and Tv The foregoing results indicate that the finite element procedure can be used for consolidation or settlement computations of foundations idealized as one-dimensional and subjected to (vertical) structural loads. The procedure can yield history of settlements and pore water pressures with time and the final settlements under .
.
a given load increment.
Once
the pore pressure at any time under a given load increment
Eq. (5-43) can be used to compute change in
From
a\
since a, the applied load,
the knowledge of the change in a' and the material property a m
possible to find the change in strain void ratio (Ae)
from Eq.
(5-44).
The
is is
found,
known.
it is
then
settlement
can be found by multiplying the strain by the total length of the medium Av(t)
where Av(t)
is
=
Ae{t)2H,
the vertical settlement at time
t.
(5-52)
(d
126
)
1
'ajruejadwaj.
o
X)
Q
B
* 2
B E R u ') jo
p
c
C
c a C= x 09
8!
c
±«
43 <_
C B ^jjj
-,
c
X V >
',)
c X u
.2
g c uu sO
o ed
>
! u o u s3 c cti
i-
s
5
E
127
1
'ajniejaduj9j_
09
K o F
'CI
y ed >
0.10
0.08
0.06
j!
0.04
/ 0.02
/ 0.20
/
/
0.40
/
/
0.80
0.60
1.0
Degree of consolidation, U (a)
Excess pore water pressure, p/p
0.25
0.50
0.75
1.0
(b)
Figure 5-7 Finite element results for consolidation problem, (a)
Time
factor vs. degree of consolidation, (b) Pore water pressure vs. depth.
128
One- Dimensional Time- Dependent Flow
Chapter 5
It is
129
also possible to include layered systems with horizontal stratification.
An
example of such a system follows.
Example
5-4.
Media
Consolidation of Layered Clay
Figure 5-8(a) shows a layered system of soil with four layers
shown
[10].
The material prop-
any convenient units can be assigned to the dimensions and to the properties. Program DFT/C-1DFE was used to solve the problem. Figure 5-8(b) shows distribution of excess pore water pressures at various time levels during consolidation. The problem of temperature distribution in layered media can be solved almost identically. It can be seen that the distribution of pore water pressure at the interfaces between the layers is not continuous. If the magnitudes of material properties between two layers differ widely, the discontinuity can cause difficulties. Then it may be necessary to derive criteria for restricting spatial and timewise meshes [8, 9, 11]
erties are
in the figure [10]
Figure
5-8
;
Consolidation
layered system
[10]. (b)
in
system,
layered
(a)
Details
of
Dissipation of pore water pressure with
time.
I
Pervious
k
.
E
Layer
o CN
1
cv =
10
m 2 /day, m v
k = 0.1
= 0.01
m 2 /t
m/day
'
>
i
i
b CM
Layer 2
c
=5. m„ = 0.001, k = 0.005
Layer 3
c
=20,
i
i
v
E
o CN
-
'
i
k
c v = 40,
E Layer 4
00 CM
rrv,
m
k = 0.04
r
i
Pervious
(a)
v
= 0.01. k = 0.20
0.001
One- Dimensional Time- Dependent Flow
130
Chapter 5
Excess pore pressure, p/p 0.4
0.6
0.8
1.0
(b)
Figure 5-8 Contd.
or to use alternative formulation in terms of velocities or stream functions as
unknowns.
As
explained before, once the pore water pressures are known,
it is
easy to find
the deformations at any time level.
PROBLEMS 5-1.
Derive an expression for the "initial load vector" for a line element with linear approximation for/? and due to given fluid pressure p For p = 10 kg/cm 2 compute {Qo}; solve the problem in Fig. 5- 1(b) with an external load of .
,
One- Dimensional Time- Dependent Flow
Chapter 5
1000 kg at the top by including the [Qo} 5-2.
effect
of p
Hint:
.
yi
= -A
{BYPo dy.
\
Formulate the finite element equations for the temperature (or consolidation) problem with linear variation of areas and material properties as
A =i(l -L)A +^(1 + L)A 2 X
a=i(l -L)a, +i(l + 5-3.
Solve Example
5-2 for
L)a 2
,
.
two time steps with the properties as before except the
conditions, which vary as
initial
7*(0, 0)
T(h/2, 0)
T(h, 0) 5-4.
131
= = =
10,
20, 10.
Derive equations for the three elements in Fig. 5-3 with the following properties
:
Element
1
:
k, p, c;
Element
2:
2k, p, c;
Element
3
3k, p,
:
c.
Assemble the equations and solve for the conditions that
= 0.0, t) = 10, t) = 20,
T(x, 0) 7X0, T(h,
Assume
that
ment and 5-5.
A u &i and A 2
,
ct 2
t> t>
h,
t
< 0,
0, 0.
are tne values at the
two end nodes of an
ele-
that they vary linearly.
Consider gradually refined mesh for both space and time, and by using DFT/ C-1DFE, study the behavior of the numerical solution for one-dimensional heat flow (or consolidation); assume the boundary conditions in Example 5-3.
Use
2, 4, 8, 16,
1.0, 2.0, 10.0.
and 32 elements and At
Make various combinations
=
0.05, 0.1, 0.2, 0.25,
and
0.5
of these values and obtain numer-
Plot error vs. AT = QcAt/l 2 Define error as the difference between the numerical solution (p) and the exact solution T*(p*)\ ical solutions.
.
error
=
T*
— T
or
p*
— p,
where P*
= £ J*
-
cos
nn){sm^)e-^^^
whereto is the uniform initial pressure, for the consolidation problem. Hint: Note that with increasing value of At, say beyond 2.0, the solution will become less and less accurate [8, 9]. 5-6.
Solve Example 5-2 with forcing fluxes given as q
5-7.
By using DFT/C-1DFE,
solve
Example
5-3
=0.1
unit/unit length.
by considering only half the
One- Dimensional Time- Dependent Flow
132
Chapter 5
depth with the boundary condition at midsection as {dpldy){H, t) = 0.0. Hint Since the problem is symmetrical about the midsection, dp/dy = 0. :
5-8.
By
using
DFT/C-1DFE,
dary, that
solve
Example
5-3 for
an impervious bottom boun-
is,
Ty^H,t)=0. 5-9.
Formulate the consolidation problem if the soil were deposited gradually with 1 unit of depth to 10 units of depth in 10 years [12]. Assume linear
time from
Assume required parameters.
variation of the deposition.
REFERENCES [1]
Carslaw, H.
S.,
and Jaeger,
J.
C,
Conduction of Heat
in Solids,
Clarendon
Press, Oxford, 1959. [2]
[3]
Terzaghi, K., and Peck, R. New York, 1955. Desai, C.
S.,
B., Soil
Mechanics
in
Engineering Practice, Wiley,
and Johnson, L. D., "Some Numerical Procedures
for Analysis
of One-Dimensional Consolidation," in Proc. Symp. on Appl. of Finite Element Methods in Geotech. Eng., C. S. Desai (ed.), Waterways Expt. Station,
Vicksburg, 1972. [4]
Schiffman, R.
L., private
[5]
Courant,
and Hilbert, D., Methods of Mathematical Physics, WileyYork, 1965.
R.,
Interscience, [6]
communication.
New
Desai, C. S., and Abel, J. F., Introduction Nostrand Reinhold, New York, 1972.
[7]
Crandall,
[8]
Desai, C.
S. H.,
to the Finite
Engineering Analysis, McGraw-Hill,
Element Method, Van
New
York, 1956.
and Johnson, L. D., "Evaluation of Two Finite Element FormuOne-Dimensional Consolidation," Comput. Struct., Vol. 2, 1972,
S.,
lations for
pp. 469-486. [9]
Desai, C.
S.,
and Johnson,
for Consolidation," Int. J. [10]
L. D., "Evaluation of
Num. Methods
Some Numerical Schemes
Eng., Vol.
7,
1973, pp. 243-254.
Schiffman, R. L., Stein, J. R., and Jones, R. A., "PROGRS-I, A Computer Program To Calculate the Progress of Ground Settlement Reference Manual," Report No. 71-4, Computing Center, University of Colorado,
—
Boulder, 1971. [11]
Desai, C.
S.,
and Saxena,
K., "Consolidation Analysis of Layered Aniso-
S.
Num.
tropic Foundations," Int. J.
Anal. Methods Geomech., Vol.
1,
No.
1,
Jan.-Mar. 1977, pp. 5-23. [12]
Koutsoftas, D. C, and Desai, C. Finite Elements: Solution of 76-17, Dept. of Civil Engg.,
Some
VPI
&
S.,
"One-Dimensional Consolidation by
Practical Problems," Report No. VPI-E-
SU, Blacksburg, Va,
May
1976.
COMBINED COMPUTER CODE FOR ONE-DIMENSIONAL DEFORMATION, FLOW, AND TEMPERATURE/ CONSOLIDATION By now
the reader
element method. 3, 4,
and
must have
realized the general nature of the finite
A comparison of the derivations and the results in Chapters
5 indicates that there
is
significant similarity in the theory
and
formulation of the three categories of problems. The physical phenomena of
deformation [Eq. (5-9)]
(3-45)], steady
flow [Eq.
forcing function, either a load or a flux,
medium these
(4-1)],
and transient flow [Eq.
are governed by analogous differential equations.
in similar
is
The
effect
of a
transmitted or diffused through a
manners, thus indicating the close relationship between
phenomena.
All but very trivial problems that are solved by using the finite element method require use of the computer. The amount of information to be digested and processed is so great for most problems that it is not possible to perform the calculations manually. Hence, a knowledge of and exposure to the development and use of computer programs or codes become necessary.
To introduce this topic gradually, we present in this chapter details of a code for the three one-dimensional problems covered in Chapters 3, 4, and 5. It is shown here that the generality of the method permits use of essentially the
same program
Flow,
The program is called Deformation One-Dimensional Finite Element with sample input/output and documentation is
for all three problems.
Temperature/Consolidation,
DFT/C-1DFE.
A
listing
=
provided at the end of the chapter. In the following, details of various stages of the code are described. 133
;
PHILOSOPHY OF CODES
A
computer code should be concise, but
same time,
at the
should be
it
properly documented so that the user can understand and employ
spending an undue amount of time. This
is
it
without
particularly true of a code
To achieve this purpose, and to make we have provided sufficient commentary
written for a beginner, as in this book.
the code relevant to academic aims, in the
code and
in the following.
STAGES The explanation of
number of
the code can be divided into a
include input, computer implementation, and output.
stages,
which
These stages are
described in the subsequent listing of the code. Stage
Input Quantities
1.
This step deals with the input quantities such as the
common
title,
parameters, material properties, and nodal and element data.
The first card gives details of the number of the problem and a number of problems can be executed in the same run. If the problem number NPROB is set equal to zero (a blank card), the program will automatically exit from the computer. The second card includes information on number of nodes (NNP), number of materials (NMAT), number of surface or boundary traction cards (NSLC), option for whether body force is applied or not (NBODY), and option for choosing the category Input Set
the
1.
of the problem
title
;
1
of problem stress deformation, flow, or temperature/consolidation :
semibandwidth (IBAND), which in the case of the for the code is 2; and number of selected time desired
(NTIME). The
third card or cards (as
give the properties relevant to the category
;
linear
levels at
many
that
is,
as
(NOPT)
approximation used
which output
number of
is
materials)
deformation, steady flow,
or time-dependent flow; Table 6-1 shows the meaning of the terms for the
weight).
body weight DENS denotes material density (unit For consolidation of nonhomogeneous media y wi = unit weight of
water
input in place of
categories. In the case of
is
RO.
Input Set 2. This set gives the data on the nodal points. Here
M denotes
number and KODE designates the boundary condition at the denotes the y coordinate (or any one-dimensional coordinate) of the
the node point
node.
Y
node
point.
^or 134
VLY
gives the specified value of the
an explanation of symbols, see pp. 138-139.
boundary condition
;
it is
Combined Computer Code for One- Dimensional
Chapter 6
TABLE
135
Material Properties for Various Problems
6-1
Terms
AMV
RO
DENS
Remarks
E
1.0
1.0
7
Eq. (3-14)
E*
1.0
1.0
Ji
1.0
1.0
y
1.0
1.0
yi
PROP Stress deformation
Homogeneous Nonhomogeneous Steady flow
kora
Homogeneous Nonhomogeneous
ki
or
a,-
Eq. (4-3b)
Time-dependent Temperature
Homogeneous Nonhomogeneous
a
1.0
1.0
y
Eq. (5-9a)
ki
Ci
Pi
yi
Eq. (5-9b)
y
Eq. (5-9a)
yi
Eq. (5-9c)
Consolidation
Homogeneous Nonhomogeneous *i
=
1,2,...,//
=
Cy
1.0
1.0
k
m
ywi
(
number of
tied in with the value of
vi
different materials.
KODE.
If
KODE = 0,
the node
is
"free,"
and we
VLY to be a concentrated load or forcing function; if KODE = 1, VLY = specified value of displacement, fluid head, temperature, or
can input then
pore water pressure. Note that if
gradient or slope
zero,
is
should be provided for
Within input nodal data. This
set 2, is
all
if
we
there
input
is
a natural boundary condition,
KODE = 0.
i.e.,
In general, these data
nodes, except in the case of the following.
we have made
applicable only
generated nodes are "free," that
is,
if
provision for automatic generation of the nodes are equidistant,
and
if
the
KODE = 0, then nodal data are provided
first and last node in the region where these conditions are Nodal data should be provided whenever there is a change in the length and properties of the element and in the boundary condition. We also compute the length / (ALL) of the element and store it for
only for the satisfied.
subsequent computations. Input Set 3.
An
The
total
number of elements
NEL
is
set
equal to
NNP—
1.
by the two end node (global) numbers and a number corresponding to the type of material of that element. Hence, for each element, four quantities are input: element number M, node numbers IE(M, 1) and IE(M, 2), and material type number IE(M, 3). The value of the latter varies from 1 to NMAT. element
As
is
identified
explained in the user's guide
matically the element properties
if
(p. 140), it is
possible to generate auto-
certain conditions are fulfilled.
Combined Computer Code for One- Dimensional
136
Chapter 6
Three different variations of (element) areas can be assigned. In the case all elements, IAREA = 1 and only one value of AREA(l)
of uniform area for is
input. If
,
IAREA =
first and the last element and the computer assigns intermediate area by variation. If all elements have different areas, IAREA = 3,
2,
areas are input for the
referred to their midsections,
assuming a linear and areas for all elements (at their midsections) are input. Input Set 4. Here NSLC denotes the number of surface traction loading (fy) cards, and TY denotes the value of surface loading. For instance, if in Fig. 6-2, Example 6-2, Ty is applied on two of the elements, NSLC = 2. The computer converts this loading to lumped loads at the two surrounding nodes according to Eq. (3-28). KEL denotes the element on which TY is applied. If no TY is specified, set NSLC = 0, then no data are required for surface tractions.
Input Set
5.
This set contains three subsets of data for the time-dependent
temperature or consolidation problems. The
increment (DT), total time tions,
(TOTIM)
and option (INOPT)
first
subset contains the time
desired for the time-dependent solu-
specifying uniform,
for
linear,
or arbitrary
variation for the initial conditions.
In the second subset, input the specific values of time levels TIM(I) at
which output
The
is
desired; the
number of these values is equal to NTIME. on initial conditions T(x, 0) [Eq. (5-10)].
third subset contains data
Here, according
INOPT, data
are input as uniform, linearly varying, or
arbitrary values of initial temperature at the nodes. If
INOPT =
1,
input
simply one value UINIT(l) of initial temperature of pore pressure. If INOPT
= 2,
input the values at the
first
and the
last
nodes, and the computer will
provide intermediate values by linear interpolation. If
INOPT =
3,
input
temperature or pore pressure for each node.
Note: The information in input set 5 is not required for stress-deformation and steady flow problems, that is, when NOPT = 1 or 2. Stage
2.
Initialize
Various quantities are set;
NCT
initialized here.
TIME =
0.0
and
NCT =
are
denotes the number of time steps and can be printed out at each
time step. H(I)
is
set
equal to UINIT(I), the
initial
conditions. Matrices
] in Eqs. (5-22a) and and (5-22c)] and R(I) are the element and assemblage forcing
QK(I,J) and QP(I,J) are the element matrices [kj and [k
f
(5-22b), respectively. Vectors Q(I) [Eqs. (3-28), (4-11),
[Eqs. (3-32), (4-13),
and
(5-33b)]
vectors, respectively. Matrix A(I,J) stores the assemblage matrices.
AK(I,J)
is
used repeatedly for the time solution and
during the Gaussian elimination process;
its
value
is
Matrix
is
modified at each step
set
equal to A(I,J) before
Combined Computer Code for One- Dimensional
Chapter 6
starting a
new time
step.
For the
stress
137
and steady flow problems, A(I,J) and
AK(I,J) are computed once, and no such resetting
is
required. AP(I,J)
the
is
assemblage matrix corresponding to [K *] [Eq. (5-33b)]. f
Intermediate quantities required for computation of the time factor in the consolidation problem are evaluated here. Stage
3.
Compute Element Matrices
Element matrix
computed
for
all
gravity load are
[k]
and load vector {Q} due to surface traction or
flux are
elements one by one. The contributions at nodes due to the
computed according
to Eq. (3-28)
and are added to the
element load vector {Q}. Stage
At
4.
Assemble
this stage,
we add
the contributions of the element matrix and load
vector to the appropriate locations in the global matrix [K] and load vector is done by identifying the global nodes corresponding and 2 for each element.
{R} [Eq. (3-32)]. This the local nodes
Stage
5.
1
to
Concentrated Forces
The concentrated
forces
VLY(I) are added
to the corresponding locations
in vector {R}.
Stage
6.
Boundary Conditions
The boundary conditions
are introduced here to modify the assemblage
matrices and load vectors [Eqs. (3-32), (4-13), and (5-33)]. In view of the fact that
we have taken advantage of bandedness and symmetry, make this modification.
special logic
is
used to
Stage
7.
Time Integration
This is operational only for the time-dependent problem (NOPT = 3 and and pertains to Eq. (5-33). Here the equivalent load vector [K,]{r} is computed and added to the assemblage vector {R}.
4)
r
Stage
The
8.
final
Solve Equations
equations are solved by performing the Gauss-Doolittle elimina-
tion procedure.
For
NOPT =
1
or
2,
this is
performed only once. For
:
:
Combined Computer Code for One- Dimensional
138
NOPT = 3 and 4, steps = TOTIM/DT.
it
is
Chapter 6
performed as many number of times as the time
The space of vector R(I)
used here to store the values of computed
is
displacements, fluid heads, temperatures, or pore water pressures. Stage
This
Set {R}
9.
is
f
=
{H}
=
relevant only for
computed are stored
{R} f+Af
NOPT =
3
and
4.
Here the values of R(I) just
in H(I) as the value {R} f at the previous time stage.
Output Quantities
Stage 10.
Depending on the value of NOPT, computed
results are printed
out as
follows
NOPT =
1
=2 =3 =4
results for stress-deformation
results for flow
results for
problem,
problem,
time-dependent temperature problem,
results for consolidation
problem.
In the case of consolidation, additional quantities such as time factor and
degree of consolidation are also printed out.
Explanation of Major Symbols and Arrays
A =
AK =
Assemblage matrix which is set equal to AK; only once for stress deformation and steady flow but at every time step for transient flow. Assemblage matrix which is computed only once. This matrix corresponds to [K][Eqs. (3-32), (4-13) or (5-33)].
ALL = AP = AREAEL =
DENS = DT = GWT = H=
Length of each element. Assemblage matrix corresponds to [K«] [Eq. (5-33)]. Area of elements. If IAREA = 1, input one value of uniform area, first and last values, = 3 input all values.
= = =
2 input
Density (unit weight) for different materials.
Time increment
[Eqs. (5-23)
and
(5-33)].
Contribution of gravity or body force at an element node.
Vector that stores Rt+At
IAREA =
=
computed
at
initial /
+
values of temperature (or pore pressure) and also
A/ time
level for use as
R
t
for the subsequent time level.
Option for specifying areas of elements for uniform area over all elements. 1 :
2: for linearly varying areas; input values for only the first 3: for arbitrary variation, that
is,
and
input area for each element.
last
elements.
:
Combined Computer Code for One- Dimensional
Chapter 6
IBAND =
139
Semibandwidth. Set 2 for the one-dimensional problem with linear approximation.
= Node 1 of element M. = Node 2 of element M. — Material type of element M. INOPT = Option for initial conditions = uniform temperature or pressure at all nodes. = 2 linearly varying temperature or pressure input values only for the first and
IE(M,1) IE(M,2) IE(M,3)
1
:
;
:
the last nodes.
= 3: arbitrary; input values for all nodes. KEL = Element(s) on which surface tractions (TY) are applied. KODE = Code for boundary conditions: = specified displacement, potential, temperature, or pore = 0: free node where concentrated force can be applied. NBODY = for no body force, = if body force specified. NCT = Step of time increment can be printed out if desired. NMAT * Number of materials. NNP = Number of nodes. NOPT = stress deformation, = 2 steady flow, = 3 transient 1
:
pressure.
1
;
1
temperature,
=
4
consolidation.
NPROB = Problem number if = 0, program exits from the computer. NSLC = Number of surface traction cards. NTIME = Number of time levels at which output is desired. PROP = Material property: = E, elastic modulus for stress deformation. = k, coefficient of permeability for flow. = a, thermal diffusivity for temperature (for homogeneous medium). = c coefficient of consolidation (for homogeneous medium). For ;
v
,
layered
media, see Table 6-1.
QK = QP =
R=
Element matrix [k] in Eqs. (3-28) and (4-9) and [k a ] in Eq. Element matrix [k ] in Eq. (5-21). Assemblage forcing vector {R} in Eqs. (3-32), (4-13), and
(5-21).
r
(5-33).
Computed
nodal displacements, potential, temperature, or pressure are stored in
this
vector after Gaussian elimination.
RO = TF = TIM = TIME = TITLE = TUINIT =
TY = UAV =
Mass Time
density of material. factor.
Selected time levels at which output
is
desired.
Elapsed time. Title
and description of problem. initial values = (sum of initial pore pressures) x (number of nodes).
Total of
Applied surface traction load. Average degree of consolidation
at a given time.
UINIT =
Values of
USUM =
Average dissipation of pore pressure at a given time. Value of applied boundary condition. If KODE = 1,
VLY =
initial
temperature or pore pressure at time
placement, potential, temperature, or pressure;
Y=
it
0.
implies specified dis-
KODE =
0,
it
implies free
node where forcing parameters (concentrated load) can be specified. y coordinate for stress deformation (of vertical column and strut) and for consolidation.
=
if
<
x coordinate
for flow
and temperature.
c Si
Q
o
L.
Ql
*-
co
o
-o
'I
£ Si
=3
_
o I
E
Q -I
o
CJ LUI
•
H
s
"5
x
(D i
CO in
.Q
— ^
140
oc
O
irf
C
.—
15 C E *
CN in
00
s-
E
^
a O .2
u.
s
co »-
<
IS
is
—
_J
-D
3 UJ
S w
<
«
—c
>
4->
Q.
©3^
CD 09 ro
r
CO CD
C*
1ro
03
>
4->
C
•V
C
CO
CD
+_,
r 0) n C
a> i_ CD
to oj
= (0 >
D C a. — ~c —>:
*-
i 2£
03
fc
<*>
it
c
c
g>
—
4-
M-
o "-T D »_ il_ o o O "+CN CO
1
II
2
fe
E
OH
II
II
CO
c >
to < a S = s O "-
_2>
<
o5
IO "-H
h-
<
oi
E
C
S
s
Ou-
3 3
o U_
LLI
141
'
o 2 C Q.
"8
CO
—
2!
o
il z
£
o o c E .c
o o
c o
c/>
3
.2
§ k"
-
o M—
'5
o
CN
co
15
ro
'-^
II
II
n
*-
!tr
!»=
° °
'-o
CO
— O)
>
a>
CO
E
o
+_<
liJ
+-
CO
1
3
a 3 o
h-
< nr
u
142
O u_
— ~
-
T
^ 1-
LU
c O
co —
'6
<
§1 = O
C o
-a 2?
"5
r-
°C6
C° g tr
Combined Computer Code for One- Dimensional
Chapter 6
Note
1: (a) If the
143
nodal points are equidistant, data are required only
for the
and the
first
last
node. The data for intermediate
nodes are generated by the computer. If a boundary condition or force
is
applied at a node, data must be input for such
nodes, (b)
KODE = 0:
Node
is
free,
and concentrated force can be
applied.
KODE =
1
Boundary condition
:
potential,
in terms of displacement,
temperature,
or pore pressure
is
specified.
KODE is set equal to zero for the nodes that are generated by the computer.
Note
2: If the elements
RO, DENS, data
have same material properties,
PROP, AMV,
and the last element; data for intermediate elements are generated by the computer, and their PROP, AMV, RO, and DENS are set equal to those for the previous element. If there is a change in material property, data must be separately provided. For instance, for 1 1 elements, if there is a change in material property at element 6, then data for elements 1 and 5 and 6 and 1 1 should be provided.
Example
See Fig. 6-1.
are required only for the
first
6-1
By using DFT/C-1DFE, compute displacements and axial
stresses. Plot
the results.
10 kg
©
E u
o
(3) E= 1000 kg/cm / = 10 cm (for each = 1.0 cm 2
E~
©A =0
u
o
element)
Vl
.it
\/h/////,
© Figure 6-1
Example 6-2 See Fig. 6-2. stresses.
By using DFT/C-1DFE, compute and
plot
displacements and
"
Combined Computer Code for One- Dimensional
144
© (3) ^S
Chapter 6
E= 1000 kg/cm 2 /=10cm _A= 1.0 cm
©T
= 1.0 kg/cm 3 y = 0.5 kg/cm
y
mW0 Figure 6-2
Example 6-3
The pipe in Fig. 6-3 is subjected to potentials and 4, respectively. Use DFT/C-1DFE, and tials and velocities.
(p
10cm +-« 10cm ,
^
2
'
© A
= 100.0
cm 2
=
find
©
2 and 1 at the two end points 1 and plot distributions of poten-
.
+-«
10cm »
©
©
k x = 1.0 cm/sec,
,
i
i
1
q = 0.0
Figure 6-3
Example 6-4
The bar in Fig. 6-4 is initially The boundary conditions are 7(0,
The coefficient of diffusivity CC and adopt A/ = 1.0 unit.
at a
/)
=
= 1
temperature of 100°F; that
T(10, .
/)
=
0,
t
is
100°F.
is, 1 1
nodes,
> 0.
Divide the bar into 10 elements, that
10 units
=
T(x, 0)
i
I
Figure 6-4
Use DFT/C-1DFE, and compute the
distribution of temperature at various
time levels as the bar cools.
PROBLEMS See the problems in Chapters 3-5.
LISTING
N
)
)
AND SAMPLES OF INPUT/OUTPUT:
DFT/C-1DFE C
C C C £
Q C £ C C
20
q
MAIN FINITE ELEMENT C3DE FOh ONE-DIMENSIONAL DEFORMATION. ^LOW, MA I TEMPERATURE AND CONSOLIDATION MAIN OFT/C-iDFC PROGRAM NAME MAIN DEVELOPED BY C.S.OESAI **t^*^»**Wt»**A ********** ******** ************»*--****** *t***»**W«****yAIN DIMENSION A(41,3), AK(4l,3), AP(41,3>, H(*l), >(4l), UK(2,2), 0P(2main 1,2), 0(2), LP(2), TIM|20I« Y(4l), IE(40,3), Vt.Y(4l), &ROPIIO), ARFMAIN 2AEL12J), DENSC101, K0DE(41), ALLIED, TY(2DJt KELC201, ^IICI, AMVMAIN MAIN 3(10), UINIT(41), TITLE(ld) ****** »**W** **? * *********** **V** :**.**** *****: ***.******** ft*******y^J\J MAIN *»* STAGE 1***- INPUT QUANTITIES ***A ****** **************»*********************»> *^ FPR EXPLANATION OF VARIOUS STAGES CHAPTEF 6 MAIN SEE *** INPUT SET I MAIN REAO 920, NPRUB. TITLE MAIN IF INPRQ3.LE.0I GO TO 910 MAIN WRITE (6,930) NPROB, TITLE «AIN WC ITE (6,940) MAIN WRITE (6,950) MAIN *** PROBLEM PARAMETERS MAIN REAO (5,960) NNP , NMA T, NS LC , NBODY ,NCPT , I 3 ANO, NT ME MAIN WRITE (6,9/0) NNP.NMATtNSLCtNBODY.NOPT, IbANOtNTIME MAIN MAIN *** MATERIAL PROPERTIES WRITE (6,990) MAIN DO 3 1=1, NM AT MAIN READ (5,980) PROPC I ) , AMV (I , PC ( 1 ) , DENS MAIN CONTINUE MAIN WRITE (6,1000) I, PROP (I ),AMV( I),PC( I) ,UENS(I) ,I=1,NMAT) MAIN *»* INPUT SET 2 MAIN *** NODAL POINT OATA MAIN WR ITF (6,1010) MAIN N=l MAIN READ (5,1020) M, KOOE M , Y( M , VLY( M MAIN IF (M-N) 50,80,60 MAIN CONTINUE MAIN WRITE (6,1030) M MAIN GO TO 910 MAIN AUTOMATIC GENERATION OF NCDAL POINT OATA * * MAIN DF=M+1-N MAIN RY=(Y(M)-Y(N-1) )/DF MAIN KGDE(N) = MMN Y(N)=Y(N-l)+RY MAIN VLY(N)=0.0 MAIN IF (N.EQ.l) GO TO 100 MAIN **** ** COMPUTE MAIN ELEMENT LENGTH ALL(N-l)=Y(N)-Y(N-l) ^AIN MAIN CONTINUE WRITE (6,1040) N,K0DE(N) ,Y(N),VLY(N) MAIN N=N+1 MAIN IF (M-N) 110,90,70 MAIN IF (N.LE.NNP) GO TO 40 MAIN ******«** a* **************** ********^*******w****************it*5»**MA IN *** MAIN INPUT SET 3 *** ELEMENT OATA MAIN jp
Jf
X.
<
^
I
C
(
)
33
I
(
C C
43 53
C
60 70
80 C
93 100
110 C C C C
123 130 140
(
)
)
******
(
I
)
)
145
35
40 50 50 7")
80 90 103 110 J20 133 ]^Q 150
160 170 130 190 200 210 220 233 240 250 260 270 283 293 300 313 320 330 340 350 363 373 380 390 400 410 423 430
440 450 460 473 480 490 533 510 520 533 540 55) 560 57Q 583 590 600 613 62
3
630 640 650
1 E
)
)
C
190
E
(
N,2
210
)
=
I
E
(
N- 1 , 2
MAIN MAIN MAIN MAIN MAIN MAIN «AIN MAIN MAIN MAIN MAIN MAIN MAIN MAIN MAIN MAIN MAIN MAIN
1
)
IE(N,3)=IE(N-l,3) (M-N) 173,170,130 (NEL-N) 180,180,120 CONTINUE INPUT ELEMENT AREAS READ (5,960) IAREA GO TO (190,210,230), IAREA READ (5,980) AREAEL(l) IF IF
DO 200
200
)
IE(N,i)=IE(N-l,l)+i I
160 170 180
))
and Samples (cont.)
Listing 150
))
1
= 1, NEL
AREAEK I)=AREAEL(l
)
GO TO 240 READ (5,980) AR AEL( 1 ,AR?AE L NEL AL=Y(NNP)-Y( I) -(Al_L( 1 +A LL NEL /2 SLUPE=(AREA5L( NEL -ARE AEL 1 / AL NEL1=NEL-1 (
)
)
(
)
)
)
(
)
.
)
DISTY=0.0 DO 220
I
=2, NEL
AREAEL
(I
)
=
AREAEL(
1
)
/2 + S L OP E*D STY (
I
)
)
.0
I
CONTINUE GO TO 240
230 240 250 Q C c Q
260 270 280 C C C C
290
300 310
320
330 340 350
(
(
I
,
I
(
)
,
fc (
>
,
I
(
,
(
)
9
30
*** MAIN 940 INPUT SET 4 *** sukface traction cards MAIN 950 ******************** ** *************-*********** **** ****«•*** * »*****«v«AIN 960 MAIN 973 IF (NSLC.EU.O) GO TO 280 WRITE (6,1080) MAIN 980 MAIN 990 DO 263 I=1,NSLC READ (5,1090) KEL(I),TY(I) MAIN1000 MAIN1010 DO 270 I=1,NSLC WRITE (6,1100) I ,KEL I , TY I MAIN1020 (
(
)
)
CONTINUE MAIN1030 **********************************»:******************************* na sjio40 *** INPUT SHT 5 MAIN1050 *** DATA FOR TIME DEPENDENT PRPRLFMS MAIN1060 ****** ** ************ ******** ******** *********** ******************* m,a mio 70 IF (N0PT.LT.3) GO TO 370 MAIN1080 WRITE (6,1110) MAIN1090 READ (5,1120) DT,T3TIM, INOPT MAIN1100 WRITE (6,1130) DT,TOTIM, INOPT MAIN1110 WRITE (6,1140) MAIN1120 1= 1 ,NT I ME READ (5,980) MAIN1130 T IM( MAIN1140 DO 290 I=1,NTIME WRITE (6,1150) I,TIM(I) MAIN1150 MAIN1160 INOPT =1 UNIFORM INITIAL COND I TI CNS,=2 LINEAR, =3 ARBITRARY MAIN1170 WRITE (6,1160) GO TO (300,320,340), INOPT MAIN1180 READ (5,980) UINIT(l) MAIN1190 MAIN1200 DO 310 1=1, NNP MAIN1210 UINIT( I)=UINIT(1) GO TO 350 MAIN1220 MAIN1230 READ (5,980) UIN IT 1 ,UINIT NNP MAIN1240 NNPl=NNP-l MAIN1250 DO 330 I=2,NNPi MAIN1260 AL=Y(NNP)-Y(l) MA IN 1270 SL0PE=(UINIT(NNP)-UINIT( 1) )/AL MAIN1280 UINITt I)=UINIT( I +SLOPE*Y I MAIN1290 GO TO 350 I =1 MAIN1330 NNP READ (5,980) Ul NI T( MAIN1310 CONTINUE = 1, NNP MAIN1320 DO 360 MAI.M1330 WRITE (6,1170) I,UINIT(I) i
i
I
)
,
(
(
)
(
1
360
773 780 793 800 810 923 R30 MA I 840 MAIN 850 MAIN 860 MAIN 37 3 MAIN 883 MAIN 890 MAIN 9)3 MAIN
READ (5,980) AREAEL ), 1=1 NEL CONTINUE DO 2 50 M=1,NEL A° E a:- L w M, 2 WRITE (6,1070) M, E M, 1 I E m, 3) ******************** ******** ************ ********* V ******** < **r****MAlN
(
C
50 76 3 7
\!
0ISTY=DISTY *(ALL( I-D +ALL 220
660 670 680 690 700 713 720 730 740
146
(
I )
,
,
Listing
and Samples (cont.)
370
*t*****r*****a ****************** **************-p******t*»**********MAlN13'V0 *** STAGE 2*** INITIALIZE MAIN1350 **»********************+**********************»•************** MA IN1360 CONTINUE MAIN1370 NCT =
MA IN 1380
380
TIME=0.0 00 380 I=l,NNP IF (N0PT.LT.3) UINIT(I)=0.0 H(I)=UINIT(I) R(I)=0.0 00 383 J= 1,1 BAND A(I,J)=0. AK(I,J)=0. AP(I,J)=0.
MAIN1390 MAIN1400 MAIN1410 MAIN1420 MAIN1430 MAIN1440 MAIN1450 MAIN1460 MAIN1470
DO 520 M=l ? NEL
MAINU80
Q C C
II1=IE(M,1) MAIN1490 II2=IE(M,2) MAIN1500 ALEN=ABS(Y( I II )-Y( 112) MAIN1510 MT=IE(M,3) MAIN1520 DO 390 1=1,2 MAIN1530 Q(I)=0.0 MAIN1540 DO 390 J=i,2 MAIN1550 QK(I,J)=0. MAIN1560 QP(I,J)=0. HAIN1570 IF (N0PT.LT.4) GO TO 410 MAIN1530 FOR CONSOLIDATION (OF LAYERED MEDIA) FIND TIME FACTOR ON THF PAS I S MA I M1590 OF AVERAGE CV, THIS IS AN APPROXIMATION , ALTERNATIVELY THIS MAIN1600 CAN BE DONE ON THE BASIS OF ONE OF THE LAYERS MAIN1610 MAIN1620 ANEL=NEL MAIN1630 CV=0.0 MAIN1640 DO 400 MM=1,NEL MAIN1653 MTT=IE(MM,3) MAIN1660 CV=CV+PROP (MTT)/(RC(MTT)*AMV(MTT)) MAIN1670 CVA=CV/ANEL HH=( Y(NNP)-Yd) )/2.0 MAIN1680 MAIN1690 TFF=CVA/(HH*HH) MAIN1700 CONTINUE ****************** *******-****»****<-*********v»9**"e******>"» ****#***VAIN1710 *** STAGE 3*** COMPUTE ELEMENT MATRICES MAIN1720 ****** ft*************************************************** ********MAIN1730 MAIN1740 IF (N0PT.LT.3) GO TO 450 IF (N0PT.EU.4) GO TO 420 MAIN1750 TEMPERATURE PROBLEM MAIN1760 0Rl=(AREAEL(M)*PR3P(MT) )/ALEN MA IN 17 70 MAIN 1780 DR2=(AMV(MT)*R0(MT)*ALEN)/DT MAIN1790 GO TO 440 MAIN1800 CONSOLIDATION PROBLEM MAIN1310 DRl=PROP(MT)/(RO{MT)*ALEN) 0R2=(AMV(MT)*ALEN)/DT MAIN1820 MAIN1830 TUINIT=0.0 MAIN1840 DO 430 1=1, NNP MAIN1850 TUINIT=TUINIT+UINIT( I) MAIN1860 CONTINUE MAIN1870 QK(1,1)=DR1 MAIN1830 QK(2,2)=QK(l,l ) MAIN1890 QM1,2)=-QK<1,1) MAIN1900 QK(2,1)=QK(1,2) QP(1,1)=0R2*(1. 0/3.0) MAIN1910 MAIN1920 QP(l,2)=QPll,l)/2. MAIN1930 QP(2,2)=QP(1,1) MAIN1940 QP(2,1)=QP<1,2) MAIN1950 GO TO 500 MAIN1960 CF=(AREAEL(M)*PROP(MT) l/ALLCMI MAIN1970 QK(1,1)=CF MAIN1980 QK(2,2)=CF MAIN1990 QK(1,2)=-CF MAIN2000 QK(2,1)=-CF )
390 C C C
400
410 C
C q
C
C
420
430 440
450
147
Listing C C C
C
460
470 480 C
490 500
510 520 530
C C
C C C C
540 C C
C C
550 C C
q C
560
and Samples (cont.)
**************************************************** ************** ma J nj20 10 *** STAGE 4*** ASSEM8LE MAIN2020 ****************************************************************** ma I N2030 IF (NSLC.EQ.O) GO TO 480 MAIN2040 COMPUTE ELEMENT FORCE VECTOR Q DUE TC TRACTION OR FLUX MAIN2050 00 460 IM=1,NSLC MAIN2060 MK=KEL(IM) MAIN2070 IF (MK.EQ.M) GO TO 470 MAIN2080 CONTINUE MAIN2090 GO TO 480 MAIN2100 Q( 1)=Q(1)+(TY( IM)*ALL(MKI)/2.0 MAIN2U0 MAIN2120 Q(2)=Q(2)*U(1) CONTINUE MAIN2130 ADD FORCING VECTOR DUE TO BODY FORCE OR FLUX TO ELEMENT LOAD VECTOMA I N2140 IF (NBODY.NE.l) GO TO 490 MAIN2150 GWT=(AREAEL(M)*ALL=AK(I,J)+QK(LL,MM) MAIN2290 IF (N0PT.LT.3) GO TO 510 MAIN2300 AP( I,J)=AP( I,J)*QP(LL,MM) MAIN2310 MMN2320 CONTINUE MAIN2330 CONTINUE VAIN2340 NCT=NCT+1 IF (N0PT.GE.3) TIME=TIME+DT MAIN2350 MAIN2360 IF (N0PT.EQ.4) TF=TFF*TIME HAIN2370 DO 540 1=1, NNP MAIN2380 DO 540 J=l,IBAND **** IMPORTANT NOTE ***** IF NO (TIME DEPENDENT) FORCING MAIN2390 ELEMENT PARAMETERS SUCH AS FLUX ARE APPLIED, THEN INITIALIZE VECTOMA IN2400 R AT THE START OF EACH TIME STEP** MAIN2410 IF TIME DEPENDENT FORCING MAIN2420 PARAMETERS ARE APPLIED THEN STAGES 3 AND 4 SHOULD 8E PERFORMED MAIN2430 AT EACH TIME LEVEL AND THE ELEMENT CONTRIBUTIONS ADDED TO P MAIN2440 CONCENTRATED TIME DEPENDENT FOPCES CAN HOWEVFP 3E APPLIED MAIN2450 IF (N0PT.GE.3) R(I)=0.0 MAIN2460 A( I,J)=AK( I, J) ****************************************************************** MA I N2470 MAIN2480 *** STAGE 5***C0NCENTPATED FPRCFS MAIN2490 ADD CONCENTRATED FORCES TO ASSEMBLAGE LOAD VECTOR R ****************************************************************** ma N2500 MAIN2510 DO 550 1=1, NNP IF (KODE(I).NE.O) GO TO 550 MAIN2520 R(I )=R(I)+VLY{ I) MAIN2530 MAIN2540 CONTINUE ****************************************************************** MA I N2 550 6*** *** STAGE BOUNDARY CONDITIONS MAIN2560 *************************************************** ******** *******m A I N2570 MAIN2590 ADD K(ALPHA) *K(T) MAIN2600 DO 560 1=1, NNP MAIN2610 DO 560 J=1,IBAND MAIN2620 A( I,J)=A( I,J)*AP(I,J) MAIN2630 CONTINUE MAIN2640 DO 600 N=1,NNP IF (KODE(N).EQ.O) GO TC 590 MAIN265D MAIN2660 BOUND=VLY(N) MAIN2670 DO 580 M=2,IBAN0 MAIN2680 K=N-M*i MAIN2690 IF (K.LE.O) GO TO 570 I
148
Listing
570
580
590 600 C C
C
R
and Samples (cont.) MMN2700 MAIN2710 MAIN2720 MAIN2730 MAIN2740 MAIN2750 MAIN2760 HAIN2770 MAIN2780 MAIN2790 MAIN2800 MAIN2810 MAIN2820 MAIN2830
*** STAGE 7*** TIME MAIN2850 INTEGRATION ************ **«********* ¥ *****r***«»x*********:»,* T:«**********±r»**«-«*AT N2860 IF (N0PT.LT.3) GO TO 650 MAIN287D 00 640 1=1, NNP MAIN2880 MAIN2890 BB=0. IF (I.EQ.i) GO TO 620 MAIN2900 MAIN2910 IP=I-1 »1AIN2920 KK=1,IP 00 610 MAIN2930 JJ = I«-l-KK MAIN2940 COMPUTE RIGHT HANO SIDE 'EQUIVALENT LOAD* K(T)* R
610 620
630
640 650 C C q
660 670
680
8*** SOLVE EQUATIONS MAIN3080 *** STAGE MAIN3090 EQUATION SOLVER - GAUSS- DOOL ITTLE ELIMINATION PROCEDURE ******************************************************************** MA I N3100 MAIN3110 NRS=NNP-1 NR=NNP HAIN3123 *MN3130 DO 670 N=l,NRS MAIN3140 M=N-1 MAIN3150 MR=MIN0(IBAN0,NR-M) MAIN3160 PIV0T=A(N,1) MAIN3170 DO 670 L=2,MR MAIN3180 C = A(N,L)/PWOT = MAIN3190 1 M+L MAIN3200 J=0 MAIN3210 DO 660 K=L,MR MAIN3220 J=J+1 A( I, J)=A(I , J)-C*A(M, K) MAIN3230 A(N,L)=C MAIN3240 DO 680 N=1,NRS MAIN3250 M=N-1 MAIN3260 MR = MIN0( IBAND,NR-M) MAIN3270 C=R(N) MAIN3230 R(N)=C/A(N,1) MAIN3290 DO 680 L=2,HR MAIN3300 I=M+L MAIN3310 R( I )=R( I)-A(N,L)*C MAIN3320 R(NR)=R(NR)/A(NP,1) MAIN333} DO 690 1=1 MRS «AIN33^0 N=NR-I MMN3350 M=N-1 MAIN3360 MR=MIN0( IBANU»NR-M) MAIN3370 ,
149
Listing
and Samples (cont.)
00 690 K=2,MR
MAIN3380 MAIN3390 R(N)=R(N)-A(N,K)*R(L) MAIN3400 ******************** ** r ******* ******** ************************** **MAIN3413 *** STAGE 9*** SE T R(T)=H( )= RCT+OT) MAIN 3420 ************************************** ************************** **MAIN3430 00 700 1=1, NNP MAIN3440 H(I|=R
690 C C C
700 C
c c
710 C
720
730 C
)
740 C C
750 760 C
770 C c
780
790
800
810
820 C
830
840
150
(
I
Listing
850 860
870
880
890 900 910
C C
920 930 940 950 960 970
,
E
and Samples (cont.) MAIN4060 MAIN4070 MAIN4080 MAIN4090 MAIN4100 MAIN4110 MAIN4120 MAIN4130
00 850 I=1,NTIME DIF=ABS(TIME-TIM(I)I IF (OIF.LT.TOLERI GO TO 860 CONTINUE GO TO 530 CONTINUE USUM=0.0
DO 870 1=1, NNP UZ=R(I)/TUINIT USUM=USUM*UZ UAV=1.-USUM WRITE (6,1310) TIME,TF,UAV PRINT OUT NOOAL PORE PRESSURES WRITE (6,1320) 00 880 1=1, NNP WRITE (6,1330) I,R(I) IF (UAV.GE.0.98) GO TO 900
MAI N4 140
MAIN4150 MAIN4160 MAIN4170 MAIN4180 MAIN4190 MAIN4200 MAIN4210 MAIN4220 MAIN4230 GO TO 530 MAIN4240 CONTINUE MAIN4250 CONTINUE MAIN4260 GO TO 20 MAIN4270 CONTINUE MAIN4280 WRITE (6,1340) MAIN4290 STOP **#*****************»#********#**** *****«***!»««**************.****maIN^300 MAIN4310 MAIN4320 FORMAT (I5,3X,18A4) MAIN4330 FORMAT (/1H1,10X,8HPR08LEM=, 15, 3H.. ,18A4////) FORMAT MAIN4340 1QX,16HINPUT QUANTITIES////) PARAMETERS//) MAIN4350 FORMAT (10X,38HINPUT TABLE 1A .. PROBLEM FORMAT (1615) MAIN436D FORMAT (5X.39HNUMBER OF NODE POINTS . . .=, I 5/5X,39HNU"MA I N4370 . . .= , 15/ 5X ,39HNUMBEP 1BER OF MATERIALS OF TRACTION CMAIN4380 ..MAIN4390 ...=,I5/5X,39HCPTI0N FOR BODY FORCE =0 OR 1 2ARDS I5/5X, 39HSEM I-MAIN4400 3.=,I5/5X,39H0PTI0N FOR PROBLEM TYPE . . .= OF OUTPUT TIME MAIN4410 4BAND WIDTH . . .= I 5/ 5X , 39HNUMBER 15) MAIN4420 5LEVELS MAIN4430 FORMAT (8E10.3) FORMAT (////10X,36HINPUT TABLE IB. . . MATE R I AL PROPER T I S//5X ,55H MMAIN4440 1AT K RO/OEN OF WAT MATDENS//) MAIN4450 C OR MV FORMAT 5X,I5,2X,E10.3,2X,E10.3,2X,E13.3,2X,E10.3) MAIN4460 FORMAT (////10X,33HINPUT TABLE 2 NODAL POINT DATA//5X,40H NODE MAIN4470 DISP/FORCE/) MAIN4480 KODE Y-COCRD FORMAT (2I5,2E10.3) MAIN4490 MAIN4500 FORMAT (10X,19HERR0R IN NODE CAR0=,I5) MAIN4510 FORMAT (5X,I5,2X,I5,2X,E10.3,2X,E12.3) FORMAT (////10X,30HINPUT TABLE 3 .. ELEMENT DATA//5X, 40HEL NO N0MAIN4520 AREA/) MAIN4530 NOOE J MTYPE IDE I MAIN4540 FORMAT (5X,21HERR0R IN ELEMENT CARD, 15) MAIN4550 FORMAT (5X,I5,2I8,2X,I6,2X,E10.3) FORMAT (////10X,35HINPUT TABLE SURFACE TRACTI0NS//5X,26HNUMBMAIN4560 TRACTION/ MAIN4570 1ER ELEM FORMAT (I5,E10.3) MAIN4580 FORMAT (5X,I6,2X,I5,2X,E10.3) MAIN4590 FORMAT (////10X,51HINPUT TABLE 5A DATA FOR TIME DEPENDENT PR0BMAIN4600 MAIN4610 1LEMS//) FORMAT (2E10.3,I5) MAIN4620 FORMAT (5X,15HTIME I NCREMENT=, ElO. 3,2X ,20HT0TAL SOLUTION T IME= , E10MAIN4630 1.3,5X,7H0PTI0N=,I5) MAIN4640 FORMAT ////10X.44HI NPUT TABLE 5B, DATA FOR OUTPUT TIME L EVELS//5MAI N4650 IX,26HNUMBER OUTPUT TIME/) MAIN4660 FORMAT (5X,I6,10X,E12.3) MAIN4670 FORMAT (////10X.37HINPUT TABLE 5C. . INITIALS CONDITIONS //5X, 25HNOMAIN4680 10E TEMP/PRES/) MAIN4690 FORMAT (5X,I4,10X,E10.3) MAIN4700 FORMAT (//1H1,10X,18H0UTPUT QUANTITIES) MAIN4710 FORMAT (////10X,46H0UTPUT TABLE I .. STRESS-DEFORMATION PROBLEM /MAIN4720 MAIN4730 1/) (
,
980 990 1C00 1010
1020 1030 1040 1050
1060 1070 1080 1090 1100 1110
1120 1130 1140 1150 1160
1170 1180 1190
(
(
151
J
Listing 1200 1210 1220 1230 1240 1250 1260 1270 1280 1290 1300
1310 1320 1330 1340
and Samples (cont.)
FORMAT /5X,27H NODE DISP LACEMEN T /) MAIN4740 FORMAT /5X,24HELEM S TRESS) MAIN4750 FORMAT /5X,27HELEM VELOCITY/) MAIN4760 FORMAT 5X, I4,10X,E10.3) MAIN4770 FORMAT ////10X,32HCUTPUT TABLE 1 FLOW PROBLEM/) MAIN4780 FORMAT 5X,I5,10XtE12.3I MAIN4790 FORMAT //10X, 14HELAPSED TIME =,E1 0.3//) MAIN4800 FORMAT /5X.24HN0DE POTE NTIAL) MAIN4810 FORMAT ////10X,38H0UTPUT TABLE 1 . . TEMPERATURE PROBLEM//) MAIN4820 FORMAT /5X»28H NODE TE MPERATURE///) MAIN4830 ////10X,52H0UTPUT TABLE 1 .. RESULTS FOR CONSOLIOATICN PRMAIN4840 FORMAT MAIN4850 10BLEM/ FORMAT 5X,13HELAPSED TIME=tE10.3, 2X.12HTIME FACTOR E10.3,2X. 24HDMAIN4860 MAIN487D 1EGREE OF CONSOLIDATIONS 10. 3/) FORMAT /10X,28H NODE POR E PRESSURE//) MAIN4880 MAIN4890 FORMAT 10X, I5,10X,E13.4) /////17H ** JOB FORMAT MAIN4900 END ** MAIN4910 END (
(
(
<
(
(
( (
(
(
(
(
(
)
(
PROBLEM
1.
EXAMPLE 6-1/1-D STRESS DEFOR/POINT LOAD AT TOP/ CONSTANT AREA INPUT QUANTITIES
INPUT TABLE
PROBLEM
1A
PARAMETERS
NUMBER OF NODE POINTS NUMBER OF MATERIALS NUMBER OF TRACTION CARDS OPTION FOR BODY FORCE =0 OR 1 CPTION FOR PROBLEM TYPE SEMI-BAND WIDTH NUMBER OF OUTPUT TIME LEVELS
INPUT TABLE
1
MAT
K
I
0.100E 04
INPUT TABLE
NODE
KOOE 1
J
B.
. .
MATERI AL PROPERTIES OR MV
RO/DEN OF WAT
0.100E 01
0.100E 01
C
2
..
NODAL POINT DATA
Y-CCORD O.OOOE 00 O.IOOE 02 0.200E 02 0.3 ODE 02
DISP/FORC!
O.OOOE 00 O.OOOE 00 O.OOOE 00
O.IOOE 02
MATDENS O.OOOE 00
.
Problem INPUT TABLE EL NO
NODE
3
NODE
I
1.
(cont.)
..
ELEMENT DATA
J
MTYPE
AREA
O.IOOE 01 O.IOOE 01 O.IOOE 01
1 1 1
OUTPUT
QUANTITIES
OUTPUT TABLE
DISPLACEMENT
NODE
-O.OOOE -O.IOOE -0.200E -0.300E
1
2 3
4
1
2 3
2.
00 00 00 00
STRESS -O.IOOE 02 -O.IOOE 02 -O.IOOE 02
EL EM
PROBLEM
STRE SS-OEFO* MATI ON PRG5LEM
..
1
EXAMPLE
6-2/1-D
STRESS DEFOR/SURFACE
AND BODY FORCES/CONSTANT AREA INPUT QUANTITIES
INPUT TABLE
PROBLEM
1A
PARAMETER S
NUMBER OF NODE POINTS NUMBER OF MATERIALS NUMBER OF TRACTION CARDS OPTION FOR BODY FORCE =0 OR 1 OPTION FOR PROBLEM TYPE SEMI-BAND WIDTH NUMBER OF OUTPUT TIME LEVELS VELS
INPUT TABLE
NODE
KODE 1
.=
1
3
• • .
=
1
. . .
=
2
. .
.=
RO/DEN OF WAT
O.IOOE 01
O.IOOE 01
C
2
i
. .
OR MV
K
O.IOOE 04
*
. .
= = = .
.MATERI AL PROPERTIES
INPUT TABLE IB. MAT
. . .
• • •
..
MATDENS
-0.500E 00
NODAL POINT DATA
Y-COOPD O.OOOE 00 O.IOOE 02 0.200E 02 0.300E 02
DISP/FORCE O.OOOE O.OOOE O.OOOE O.OOOE
00 00 00 00
153
Problem INPUT TABLE EL NO
NODE
3
NODE
I
(cont.)
2.
..
ELEMENT DATA
J
MTYPE
1
2
2
2
3
1
3
3
4
1
INPUT TABLE
NUMBER
ELEM
1
1
2
2
3
3
OUTPUT
4
1
TRACTION
-O.IOOE 01 -O.IOOE 01 -O.IOOE 01
QUANTITIES
NODE
STRESS-DEFORMATION PROBLEM
DISPLACEMENT -O.JOOE -0.375E -0.600E -0.675E
ELEM
00 00 00 00
STRESS
0.375E 0.225E 0.750E
1
2 3
3.
O.IOOE 01 O.IOOE 01 O.IOOE 01
SURFACE TRACTIONS
..
OUTPUT TABLE
PROBLEM
AREA
1
EXAMPLE
02 02 01
6-2/1-D
STRESS DEFOR/SURFACE
AND BODY FORCES/LINEAR AREA INPUT QUANTITIES
INPUT TABLE
1A
PROBLEM
NUMBER OF NODE POINTS NUMBER OF MATERIALS NUMBER OF TRACTION CARDS CPTION FOR BODY FCRCE =0 OR 1 CPTICN FOR PROBLEM TYPE SEMI-BAND WIDTH NUMBER OF OUTPUT TIME LEVELS 154
PARAMETERS
Problem INPUT TABLE IB. MAT
C OR MV
RC/OEN OF WAT
0.1O0E 04
0.100E 01
0.100E 01
Y-COORD O.OOOE 0.100E 0.2 00E 0.300E
INPUT TABLE
NOOE
NO
I
I
1
2 3
2 3
3
NUDE
INPUT TABLE
NUMBER
2 3
OUTPUT
1
2 3 4
ELEM
DISP/FORCE O.OOOE O.OOOE O.OOOE O.OOOE
00 02 02 02
..
ELEMENT DATA
J
MTYPF 1
4
I
1
4 ..
00 00 00 00
AREA
2 3
0.150E 01 O.IOOE 01 0.500E 00
SURFACE TRACTIONS
-O.IOOE 01 -O.IOOE 01 -O.IOOE 01
QUANTITIES
OUTPUT TABLE
NODE
-0.500E 00
TRACTION
ELEM 1
MATDENS
NODAL POINT DATA
..
2
KODE 1
EL
.MATERIAL PROPERTIES
K
INPUT TABLE
NODE
.
(cont.)
3.
STRESS-D6FCRMATICN PROdLFM
DISPLACEMENT -O.OOOE -0.242E -0.442E -0.567E
30
00 00 00
STRESS
1
-0.242 6
2 3
0.200E
02 02
-0.125E 02
155
PROBLEM
EXAMPLE
4.
STRESS DEFOR/SURFACE
6-2/1-D
AND BODY FORCES/ARBITRARY AREA INPUT QUANTITIES
INPUT TABLE
1A
PROBLEM
..
NUMBER OF NOOE POINTS NUMBER OF MATERIALS NUMBER OF TRACTION CARDS CPTION FOR BODY FORCE =0 OR CPTICN FOP PROBLEM TYPE SEMI-BAND WIDTH NUMBER OF OUTPUT TIME LEVELS
INPUT TABLE
1
MAT
K
1
O.IOOE 04
INPUT TABLE
NODE
B. ..
OR MV
RO/DEN OF WAT
0.100E 01
0.100E 01
..
2
.
MATERIAL PROPERTIES C
NOCAL POINT DATA
Y-COORD
KODE
DISP/FCRCE
O.OOOE 00 O.IOOE 02 J.20DE 02
1
O.OOOE O.OOOE O.OOOE O.OOOE
D.3 00E 02
INPUT TABLE
NODE
EL NO
I
1
1
2 3
2
NODE
3
NUMBER
ELEM
..
ELEMENT DATA
J
-ITYPtf
2 3 4
INPUT TABLE
4
..
00 00 00 00
AREA 1 1
I
0.150E 01 O.IOOE 01 0.500E 00
SURFACE TRACTIONS
TRACTION
2
-O.IOOE 01 -O.IOOE 01
3
-0,1 ODE 01
1
156
3
PARAMETERS
MATDENS
-0.500E 00
Problem
QUANTITIES
OUTPUT
OUTPUT TABLE
STRESS-DEFORMATION PROBLEM
..
L
DISPLACEMENT
NODE
-O.OOOE -0.242F -0.442E -0.567E
1
2 3
4
00 00 00 00
STRESS
EL EM
-0 .242E 02 -0 .200E 02 -0 L25E 02
I
2 3
PROBLEM
(cont.)
4.
.
5.
EXAMPLE AREA
6-3/1-D
STEADY FLOW/CONSTANT
INPUT QUANTITIES
INPUT TABLE
PROBLEM
1A
PARAMETERS
NUMBER OF NODE POINTS NUMBER OF MATERIALS NUMBER OF TRACTION CARDS OPTION FOR BODY FORCE =0 OR 1 OPTION FOR PROBLEM TYPE SEMI-BAND WIDTH NUMBER OF OUTPUT TIME LEVELS
INPUT TABLE IB. ..MATERIAL PROPERTIES MAT
K
1
0.1J0E 01
INPUT TA3LE
NODE
KODE 1
1
C
OP M V
RO/OEN OF wAT
0.100E
0.100E 01
2
..
01
MATOENS J.OD.)E 00
NODAL POINT OATA
Y-COQRD O.OOOE 00 O.iOOE 02 0.200E 02 0.3 00E 02
OISP/FORCE 0.200E O.OOOE O.OOOE O.IOOE
01
00 00 01
157
Problem INPUT TABLE EL NO
NODE
1
1
2
2 3
3
OUTPUT
(cont.)
..
ELEMENT DATA
J
MTYPE
3
NODE
I
5.
2 3
1
4
i
QUANTITIES
FLOW PROBLEM
NODE
POTENTIAL 0.200E 0.167E 0.133E 0.100E
01 01 01 01
VELOCITY
ELE*
0.333F-01 0. 3336-01 0.333E-01
1
2 3
6.
0.100E 03 0.100E 03 0.100E 03
i
OUTPUT TABLE
PROBLEM
AREA
EXAMPLE5-2/CHAPTER5/HANDCALCULATION INPUT QUANTITIES
INPUT TABLE
PROBLEM
1A
PARAMETERS
NUMBER OF NODE POINTS NUMBER OF MATERIALS NUMBER OP TRACTION CARDS OPTION FOR BODY FORCE =0 OR I OPTION FOR PROBLEM TYPE SEMI-BAND WIDTH NUMBER OF OUTPUT TIME LF VELS
INPUT TABLE MAT
1
INPUT TABLE
KODE 1
1
158
. .
2
.
4 1
..=
...= ...=
3 2
=
3
. . .
MATERI A L PROPERTIES OR ^V
RO/DEN OF WAT
0.100E 01
0.100E 01
C
K
3.10JE 01
NODE
B.
...= ...= ...=
..
NODAL POINT DATA
Y-COORD O.OOOE 00 0.100E 01 0.200E 01 0.30DE 01
DISP/FCRCF 0.100E O.OOOF O.OOOF 0.200E
02 00 00 02
XATDENS D.OOOE 00
Problem INPUT TA3LE _
VCOc
',;
3
\EOE
I
(cont.)
..
ELEMENT DATA
J
"TYPE
APE/
O.IOOE 31
1
1
2
1
2
2 3
3
1
:.
<*
1
O.IOOE 01
3
INPUT T^bLr time
6.
INCREMENT^ O.IOOE
^:tal
31
INPUT TABLE '-.'•-E
5A
T
PI-
I-E DEPENDENT
SOLUTION TINE= 0.3OOE
53..
DATA
OUTPUT
3
Lt^i.
..
iooe ci
«=C^
11
==:~LEMS DPTION-
OUTPUT TIME LEVELS
TINE
O.IOOE 31 D.200E 01 0.300E 01
1
2 3
5c.
input ta3le
initials conditions
te«f/?pes
s^:e
O.OOOE 00 0.0 00= 00 I.OOOE 33 C.C3GE JO
1
2 3
4
OUTPUT
QUANTITIES
OUTPUT TAoLc
EL^SE.^ IOC
E
1
2
3 H
1
..
TIM.E
TEMPERATURE P^C?LE W
=
3.13GE
TEMPERAT
31
: .
E
9. IOOE 02 3.5b3E 31 3. 331' 31 2.213? 32
159
1
Problem
(cont.)
6.
ELAPSED TIME NODE
=
J.
2006 01
TEMPERATURE
.
0.100E 0.968E 0.129E 0.200E
ELAPSED TIME
=
02 01 02 02
0.300E 01
TEMPERATURE
NODE
0.
IOOE
02
0.117E 02 D.150E 02 0.2O0E 02
PROBLEM
EXAMPLE 6-4/TRANSIENT HEAT FLOW/CON STANT AREA AND INITIAL TEMPERATURE
7.
INPUT QUANTITIES
INPUT TABLE
PROBLEM
1A
PARAMETERS
NUMBER OF NODE POINTS NUMBER OF MATERIALS NUMBER OF TRACTION CARDS CPTION FOR BODY FORCE =0 OR 1 OPTION FOP PROBLEM TYPE SEMI-BAND WIDTH NUMBER OF OUTPUT TIME LEVELS
INPUT TABLE IB. MAT
K
1
O.IOOE 01
INPUT TABLE
NODE
KODE
1
1
2 3
4 5
6 7 8 9
10 11
160
1
. .
I
3 2
12
MATERI AL PROPERTIES OR MV
RO/DEN OF WAT
O.iOOE 01
O.IOOE 01
C
2
11
..
NODAL POINT DATA
Y-COORO O.OOOE O.IOOE 0.200E 0.300E 0.400E 0.500E 0.600E 0.700E 0.800E 0.900E O.IOOE
00 01 01 01 01 01 01 01
01 01 02
DISP/FORCE O.OOOE O.OOOE O.OOOE O.OOOE O.OOOE O.OOOE O.COOF O.OOOF O.OOOE O.OOCE O.OOOE
00 00 00 0) 00 00 00 00 00 00 00
NATDENS
O.OOOF 00
Problem IN°UT TABLE
NODE
NO
EL
3
NODE
I
7.
(cont/
..
ELEMENT TATA
J
MTYPE
AP5A
1
l
2
2
2 3
3
3
4
4
4
5
5
5
&
6
6
7
]
3
]
7 8
7
a 9
L
9
10
INPJT TA3LE
THE INCREMENT^
.1 ]
]
10 11
9
5A
1
1
..
TOTAL
O.IOOE 3C
J.100E 01 O.IOOE 01
]
]
E
>1
0.100E 01 O.IOOE 01 0.100E 01 0. 100E
01
O.IOOE 3 1 O.IOOE 01 O.IOOE 01
DATA FC^ TIME DEPENDENT PROBLEMS
SOLUTION TI*E =
^
.
30 OE
03
OPTION:
INPUT TA5LE 53.. DATA FO^ OUTPUT TINE LEVELS
NUMBER
OUTPUT
TIME .
3
3
j 3
,
100E On 50DE 00 ICOE
01
200E 01 300E 01 .400E CI 600E 01 OOE 30E
01 02
.150E
02 02
3
3 1
200E
0.300E 02
INITIALS CONDITIONS
INPUT TA?1 NCOS
TEMP/?" =S
1
3.10JE
2
0.1 DOE
13
3
O.IOOE
03 33
5
10
O.IOOE 03 O.IOOE 13 O.IOOE 03 O.ICDE 03 O.IOOE 13 O.IOOE B3
11
0.1 DOE
0. 133E
6 7 8
9
03
03
161
Problem
7.
OUTPUT
QUANTITIES
(cont.)
1
..
ELAPSED TIME
=
O.IOOE 00
TEMPERATURE
NODE
1
-0.030E 0.884E 0.101E 0.999F
00
O.IOOE O.IOOE 0.999E 0.101E 0.8846 O.OOOF
33 03 02 03 02 00
02 03 02 0. 100E 03
2 3
4 5
6 7
8 9
10 11
ELAPSEO TIME NODE
1
2 3
4 5
6 7
8 9
10 11
ELAPSEO TIME NODE
1
2 3
4 5
0.500E 00
=
T
EMP=RATURE
-0.030E 0.633E 0.935E 0.99RE O.IOOE
03
3.1005 3.133E 0.993E 0.935E 3.633E 0.0 DOE
03 03 02 02 02 33
=
03 OZ 02 32
3.103E 31
TEMPERATURE
-3. 33JE
J 3
0.49OE 32 0.822E j J.963E 32 0.995E 02 P.
100E
6
0.
7 8 9
0.99 5E 3.960E 0.822E 0.*96E O.OOOE
13 11
162
TEMPERATURE PROBLEM
OUTPUT TA3L5
33 02 32 32 J?.
:o
Problem
7.
EL-^SEO TIME
=
(cont.)
0.200E 31
TEMPERATURE
MODE
0.000E 3.3725 0.669E 0.3575 3.946E 3.974E 3.948E 0.857E 3.669E 0.3725 O.OOOE
ELADSEO TIME
=
5
6 7
8 9 ]
11
33E
0.300E OL
3.576E 0.767E 0.877E 0.913E 0.877E 0.767E 0.576E 0.311E O.OOOE
3
*
N
02 32 02 00
-O.OOOE 30 0.311E 02
I
2
ELAPSEO TIME
32 32
TEMPERATURE
N00 E
I
3J 02 02 02 02
=
32 02 32
02 02
02 32
02
00
O.OOOE 01
TEMPERATURE
-0.333E 00 J.271E 32 0.539E 32 3.691E 02 0.3025 32 0.
9 10
11
»*0E
0.302E 0.6 9 IE 3.5395 3.271E O.OOOE
02 32
02 02 32
33
163
Problem ELAPSEO TIM E
7.
=
NODE
-0.000E 0.217E 0.411E 0.564F 0.662E 0.695C 0.662E 0.564E 0.411E 0.217E 0.000E
I
* 5
6 7
8
9
10 11
=
02
02 02 02
02 02 02
02 00
0.800E 01
-O.OOOE 00
1
0. 177E 0.336E 0.463E 0.544E 0.572E 0.544E 0.463E 'J. 336c 0.177E O.OOOE
2 3
4 5
6 7
3
9
10 11
=
02 02 02 02 02 02 02 02 02 00
O.iOOE 02
NODE
TEMPE^ATUR
1
-O.OOOE 00
2 3
4 5
6 7
8 9
10 11
164
00 02
TEMPERATURE
NODE
ELAPSEO TIME
0.600E 01
TEMPERATURE
2 3
ELAPSED TIME
(cont),
3.145E 0.2 76E 0. 330E J.446E 0.469E 0.446E 0.380E 0.276E 0.145F O.OOOE
02 02
02 02 02 02 02 02 02
00
Problem
(cont.)
7.
ELAPSED TIME
3.150E 02
=
TEMPERATURE
NODE
O.OOOE 00 0.383E 01
1
2 3 4
0. 168£
0.231E 0.272E 0.286E 0.272E 0.2315 0.163E 0.883E O.OOOE
5
6 7 8 9
10 11
PROBLEM
oz 02 02 02 01
01 02 01 JO
EXAMPLE 5-4/CONSOLIDATION OF LAYERED MEDIA
8.
INPUT QUANTITIES
NPUT TABLE
PROBLEM
1A
NUMBER OF NODE POINTS NUMBER OF MATERIALS NUMBER OF TRACTION CARDS CPTION FOR BODY FORCE =0 OR OPTION FOR PROBLEM TYPE SEMI-BAND WIDTH NUMBER OF OUTPUT TIME LEVELS
K
1
0.400E-01 0.200E 00 0.50QE-02 0.100E 00
2 3
4
INPUT TABLE
NODE
KODEE
1
L
2
3 4 5
6 7 8
9
10
21 4 1
4 2 8
MATERIAL PROPERTIES
INPUT TABLE V MAT
PARAMETERS
OR MV
C
0.100E-02 0.100E-01 0.10GE-02 0.100E-0L
kO/DEN
OF
0.100E 0.100E 0.100F 0.100E
WAT
MAT DENS
01 01
O.OOOE on O.OOOE 00
01 01
O.OOOE 00
0.000»=
00
NODAL POINT DATA
2
Y-CCORD O.OOOE 0.400E 0.800E 0.120E 0.160E
00 01 31
02 02
0.2 00E 02 0.2 40E 02
0.280E 02 0.320E 02 0.360E 02
DISP/FORCE O.OOOE O.OOOE O.OOOE O.OOOE O.OOOE O.OOOE O.OOOE O.OOOE O.OOOE O.OOOE
00 00 00 00 00 00 00 00 00 00
165
Problem
0.400E 0.440E 0.480E 0.520E 0.560E 0.600E 0.640E 0.680E 0.720E 0.760E 0.800E
11 12
13 14 15 16 17 18 19
20 21
1
INPUT TABLE
NODE
NO
I
3
NODE
1
2 3
3 4
4
5
6
7 8
7 8
9
9 10 11 12
10 11 12 13 14
AREA
2 2 2
11 12
2
13 14 15
2
15 16 17 18 19
13 14 15 16 17 18 19
20
4
20
20
21
<
TIME INCREMENT= 0.500E 00
1
2
3
4 5
6 7
3 3
16 17 18 19
01
3
4 4 4
..
O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE 0.100P O.IOOE O.IOOE
01 01 01 01
01 01 01 01
01 01 01 01 01 01 01
DATA FOR TIKE DEPENDENT PROBLEMS
TOTAL SOLUTION TIME= 0.200E 03
INPUT TABLE 5B.
NUMBER
01 01
01 0. 100E 01
10
5A
00 00 00 00 00 00 00 00 00 00 00
O.IOOE 0.100E O.IOOE O.IOOE
5 6 7 8 9
INPUT TABLE
166
ELEMENT DATA MTYPE
2
6
32 02 02 02 02 02 02 02
..
1
5
0.000E 0.000E O.OOOE O.OOOE O.OOOE O.OOOE O.OOOE O.OOOE O.OOOE O.OOOE O.OOOE
02 02 02
J
2 3
4
(cont.)
8.
OUTPUT
.
DATA
«=0R
TIME
0.5 00E 0.650E 0.280E 0.600E 0.985E O.llOE 0.150E 0.200E
00 01 02 02 02 03 03 03
OPTION=
OUTPUT TIME LEVEL!
1
Problem [NPUT TABLE
5C.
8.
INITIALS CONDITIONS
TEMP/PRES
DE
2 3 4
0.100E 01 O.IOOE 01 0.100E 01 O.IOOE 01
5
0.1 JOE 01
6
O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE O.IOOE
1
7 3
9
10 11
12 13 14 15
16 17 18 19
20 21
OUTPUT
NODE 1
2 3
4 5
6 7
8 9
10 11 12
13 14 15 16 17 18 19
20 21
ELAPSED TIME= 0.650E 01
01 01
01 01 01 01 01 01 01 01 01 01 01 01
01 01
QUANTITIES
OUTPUT TABLE ELAPSED TIME= 0.500E 00
(cont.)
I
..
RESULTS FOR CONSOL OAT ON PROBLEM I
TIME FACTOR= 0.695E-02
DEGR.FE
OF
1
CONSOLIDATION 0.1*4E 00
PORE PRESSURE
-O.OOOOE 0.5435E 0.8194E 0.9286E 0.9718E 0.9890E 0.99616 0.9996E 0.9999E O.iOOOE 0.1000E O.IOOOE O.IOOOE O.IOOOE O.IOOOE 0.9999E 0.9996E 0.9968E 0.9720E 0.7549E O.OOOOE
00 00 00 00 00 00 00 00 00 01 01 01
01 01 01 00 00
00 00 00 00
TIME FACTQR= 0.904E-01
DECREE OF CONSOLIDATION 0.335E 00 167
Problem NODE 1
2 3
4 5
6 7 8 9
10 11 12
13 14 15 16 17 18 19
20 21
ELAPSED TIM£= 0.280E 02
NODE
1
2 3
4 5
6 7 8
9 10 11
12 13 14 15 16 17 18 19
20 21
ELAPSEO TIME= 0.600E 02
NODE 1
2
168
8.
(cont.)
PORE PRESSURE
-O.00O0E 00 0.1533E GO 0.3 033E 00 00 00 00 00
0.4474E 0.5836E 0.7116E 0.8320E 0.9465E 0.9655E 0.9734E 0.9865E 0.9909E 0.9924F 0.9857E 0.9503E 3.8463E 0.8040^ 0.6923E 0.5118? 0.273 OF 0.0000E
00 00 00 00 00 00 00 00 00 00 00 00 00
TIMF F<\CTOk FACTORS= 0.3895 00
DECREE OF CONSOLIDATION
0. 539E
00
PORE PRESSURE
-J.OOOOF •jO 0.1182E UO 0.2360F 0.3532E JJ 0.4693E 00 0.5840E no 3.6970E 31 0.8080E 00 0.8275E 0.8421E 00 0.8516E 00 0.8560E 0.8550E 00 0.6973E 00 0.5071E 00 0.2768E 00 0.2538E 00 0.2108E 00 0.1509E 00 0.7867E- -01 O.OOOOE 00 TIME FACTOR= 0.834E 00
PORE PRESSURE
-O.OOOOE 00 0.9214E-01
DEGREE OF CONSCL IDAT ION= 0.672E 00
Problem
8.
(cont.)
0.1840E 00 0.2752E 00 0.3656E 00 0.4548E 00 0.5425E 00 0.6284E 00 0.6434E 00 0.6542E 00 0.6607E 00 0.6629E 00 0.6608E 00 0.4828E 00 0.2902E 00 0.8632E-01 0.7420E-01 0.5837^-01 0.4087E-01 0.2094E-01 O.OOOOE 00
3 4 5
6 7 8
9
10 11 12 13
14 15 16 17 18 19 20 21
ELAPSFD TTME= 0.985E 02
NODE 1
2 3
4 5
6 7 e
9 10 11 12 13 14 lb 16 17
18 19 20 21
ELAPSFD TI^E= 0.110E 03
NODE 1
2 3 4 5
6 7
FflCTQK= 0.137E 01 TIME FflCTCK=
DFGRFE n F fONSFt. IDAT ICN= 0.765E 00
PORE PRESSURE
O.OOOOE 00 0.6731E- -01 CO 00
0.1344E 0.2011E 0.2670F 0.332 2F 0.3962E 0.4539E 0.4698F 0.4776E 0.4822F 0.433 7F 0.4820E 0.3446F 0.1979E
oc >0
00 oc
00 00
00
JO 0.4575E- -01 0.37A>8E- -01 0.2ti91E- -01 0.1959E- -01 0.9 894E- -02
O.OCOOF TIME FACTQR= 0.153^ 01
DEGREE OF CC*SOL IOAT
IC1N=
0. 78hE
PORE PRESSURE
-O.OOOOE 00 0.6125E-01 0.1223E 00 0.1629E 00 0.2430E 00 0.3022E 00 0.3605E 00 169
00
Problem
9 10 11 12 13 14 15 16 17 18 19
20 21
ELAPS C D TIMS= 0.150E
33
1
2 3
4 5
6 7 8
9
10 11 12
13 14 15 16 17 18 19
20 21
0.4175E 0.4274F 0.4345E 0.4308E 0.4401E 0.4385E 0.3130F 0.1793E
00 00 00 00
00 00 00
0.4060E- -01 0.3330E- -01 0.2547E- -01 3.1722F- -01 0.8684E- -02 0.0000E 00
1
2 3
4
-O.OOOOE 00 0.4439E-01 0.8803E-01 3.1317E 00 0.1749F 00 0.2175E 00 0.2595E 00 0.3005E 00 0.3077E 00 0.3128F 00 0.3158E 00 0.3167E 00 0.3156F 00 0.2250F UO 0.1284E 00 0.2847F-01 0.2326E-01 0.1772E-01 J.1195F-01 0.6016E-02 O.OOOOE 00
-O.OOOOE 00 3.2922E-01 0.5 83 5E-01 0.8729F-01
5
0.U59E 0.1442F u,1720 c 0.1992E J.2J39E 0.2073E 0.2093E 0.210GE
7 8
10 11 12
DEGREE OF
PORE PRESSURE
6
9
DEGREE OP CONSOLIDATION* 0.846E 00
PORE PRESSURE
TIME FACTOR= 0.278E 01
NODE
170
(cont.)
TIME F4CT0R= 0.2J9F 31
NUDE
ELAPSEO TIME= 0.200E 03
8.
00 00 00 00 00 00 00 JO
C
CNSCL IOAT ION= 0.698E 00
Problem 8 (cont.) 13
14 15 16 17 18 19
20 21
** JOB
0.2092E 00 0.1491E 00 0.3507E-01 0.1882E-01 0.1536E-01 0.1170E-01 0.7887E-02 0.3969E-02 O.OOOOE 00
END **
171
BEAM BENDING AND BEAM-COLUMN
INTRODUCTION Problems of beam bending and beam-column analysis using one-dimensional shows a beamcolumn subjected to the transverse load p(x) and axial load P. We first treat the case of bending only, without the load P. Under the usual assumptions of beam bending theory [1], the governing differential equation can be written as idealization are considered in this chapter. Figure 7- 1(a)
dx 1
(fW^)=
/J
(4
(7-la)
= transverse displacement; F(x) — EI(x), flexural rigidity; and = coordinate along the centroidal axis of the beam. If F(x) assumed
where w*
x
is
uniform along the beam, Eq. (7-la) specializes to (7-lb)
p(x). dx'
Here we have used z
as the vertical coordinate [Fig. 7- 1(a)]
and w
as the
corresponding transverse displacement.
We now Step
1.
follow the various steps of
element formulation.
Discretize and Choose Element Configuration
With the one-dimensional [Fig. 7- 1(b)],
172
finite
with the rigidity
idealization the
Flumped
beam can be
at the line.
The
replaced by a line
idealized
beam
is
now
— Beam Bending and Beam-Column
Chapter 7
Cross
173
pT
section
(a)
fTTp( x
p
>
-*s (b)
(c)
hv A ©
©
^
^.M, u,.P
2
w i,Qi
,M 2
w 2 ,Q 9 (d)
Beam bending and beam-column, and axial loads, (b) One-dimensional cretized beam, (d) Generic element.
Beam
with trans-
Figure 7-1
(a)
verse
idealization, (c) Dis-
discretized into one-dimensional line elements [Fig. 7-l(c)]. is
shown Step
A generic element
in Fig. 7- 1(d).
2.
Choose Approximation Model
In the case of (one-dimensional) column deformation (Chapter dealt only with plane deformations.
continuity of such structures,
it
To
3),
satisfy the physical condition
was necessary
to satisfy interelement
we of
com-
;
Beam Bending and Beam-Column
174
fulfill
problem by using only
[Fig. 7-2(a)]. It was and mathematical requirements of the
nodal displacement
patibility at least with respect to
therefore possible to
Chapter 7
the physical
first-order (linear)
approximation. In contrast to the
plane deformations, for realistic approximation of the physical conditions in the case of bending,
it is
necessary to satisfy interelement compatibility with
respect to both the displacements
As
of displacement [Fig. 7-2(b)].
and
slopes, that
a consequence,
is, first it
derivative (gradient)
becomes necessary
to use
higher-order approximation for the displacement in the case of bending. Since
it is
necessary to provide for interelement compatibility for slopes also,
we can add
slope at the
node
as
an additional unknown. This leads to two vv and slope 6 = dw/dx, at each node
primary unknowns, displacement
hence, for an element there are a total of four degrees of freedom:
node
1
and
vv 2
,
2
at
node 2
wu
X
at
[Fig. 7-l(d)].
O
0A00A0
A
Local nodes Elements
Subscript^ SuDerscript
node element
local =»
A
A (a)
(b)
Figure 7-2 Requirements of interelement compatibility, (a) Inter-
element compatibility for axial deformation (Chapter
element compatibility for
beam
3). (b)
Inter-
bending.
The commonly used
interpolation approximation model for displacement any point s x/l (x *i)//, where s = local coordinate, x = global coordinate of any point, x = global coordinate of point 1, and / = length of the element, is given by
w
at
=
—
=
x
w (x)
=Nw +N6 +Nw +
w(x)
=
l
l
2
t
3
2
N,0 2
(7-2a)
or [N]{q}.
(7-2b)
Chapter 7
Here{q} r =[w! 6
X
polation functions
Beam Bending and Beam-Column
175
= [N N N N
the matrix of inter-
w2
N
t,
9 2 ] and [N]
=
i
2
t
3
4 ] is
1, 2, 3, 4,
N = - 3s + 2s N = /s(l -2s + s N = s (3 - 2s), 2
1
t
3 ,
2 ),
2
(7-3)
2
3
Nt In Fig. 7-3,
=
we have shown
N
literature the functions
t
2
ls (s
-
1).
plots of
N
=
i
i9
1, 2, 3, 4.
are called Hermitian functions
^
In mathematical
[2, 3].
N
:
©
© -^_N
5^*
'3
N4
_^-
Figure 7-3 Plots of
It is
2
worthwhile here to
N
it
i
=
1, 2, 3, 4,
illustrate that the
[Eq. (7-3)].
matrix [N] in Eq. (7-2) can be
derived by following the procedure outlined in Chapter
with polynomial functions. Here w(x)
where
a,
= generalized
matrix notation as
= «j
-f
we can cc 2
x
+
3,
where we started
use the following cubic polynomial:
x2
+
a 4 x-
(7-4)
coordinates. Equation (7-4) can be expressed in
Beam Bending and Beam-Column
176
=
w (x )
X2
x
[1
Differentiation of w(x) with respect to
^=Q =a
2
+
Chapter 7
X2]
x
(7-5)
leads to
2a 3 x
= [0 2x = [*']{«}.
+
3a 4 x 2
3x 2 ]{a}
1
(7-6)
Here the prime denotes derivative with respect to x. To express w(x) in term of nodal values of w and 0, we use Eqs. (7-5) and (7-6) to first evaluate their values at nodes 1 and 2 as '
Wl (x
"10
= oy
*i(*=0) w 2 (x = I)
[q]
0"
«1
1
(7-7)
<
1
M* = 0.
_0
2
/
I
1
2/
/
3
a3
3/ 2 _
.«4
or
=
{q}
[
(7-8a)
4]{a}.
Therefore,
{a}=[A]-'{q}.
(7-8b)
Finally, substitution of {a} into Eq. (7-5) leads to
w(X)
= [*][A]-'{q},
(7-9)
where
[A]
-3// 2
L If
we perform
2//
the multiplication
Comment on Requirements
for
3
-2// l//
3// 2
_1 [][A]
,
2//
2
-1//
3
l//
we obtain
2
[N] in Eq. (7-2).
Approximation Function
The approximation function [Eq. (7-2)] is conformable since it provides up to n — 1 = 2 — 1 = 1 derivative of w, that is, for both w and its first derivative [Fig. 7-2(b)], where n = 2 is the
for interelement compatibility
highest order of derivative in the potential energy function in Eq. (7- 14a)
below.
It
also satisfies the completeness criterion since
it
allows for rigid
Beam Bending and Beam-Column
Chapter 7
body motion because of (slope), Eq. (7-6).
n
=
3
is
a, in Eq. (7-4)
177
and for constant
states of strain
Notice that the complete expansion [Eq. (3-3d)] up to
retained in Eq. (7-4).
PHYSICAL MODELS It
could be very instructive to the student to explain some of the principles in
the finite element formulation by using simple models.
For
instance, the
concept of interelement compatibility provided by approximation functions of different orders can be illustrated by preparing models of a beam. Details
of construction and use of such models are given in Appendix Step
3.
3.
Define Strain-Displacement
and Stress-Strain Relationships
From beam bending theory
[1],
the relevant strain-displacement relation
d w du _ ~ au _ z a*\v ~ zw = —„= -izi = dx dx 2
*
v
€ (*> z )
where u
=
axial displacement
derivative. In
fn " im
r,
(7
.
:
is
1() )
and the superscript (two) primes denote second
bending problems the constitutive law
commonly
is
expressed
through the moment-curvature relationship
M[x)
= F(x}w"(x).
(7-11)
Differentiation of w(x) in Eq. (7-2) twice with respect to x, through use of the chain rule of differentiation,
d d = -T7Tr ^ dx ds 1
A and
I
d1
d2
1
v (1 " 10 12
^2=72-^2' dx 2 ds I
2
(7
:
)
leads to
"
where
[B]
„/ v (x)
d2W
= S? = =
= transformation
differentiation.
1
d2W
1
d1
nvT-ir
/-,
-,
F SP = Frf? [N]{q}
-j
v
(7-13b)
[B]{q},
matrix;
I
(7 " 13a)
'
its
coefficients are obtained
by proper
For example,
= -j-[-6s -
6s 2
1(1
-4s+
3s 2 )
6s
-
s
2
l(3s
z
-
9, 2s)]
(7- 13c)
Beam Bending and Beam-Column
178
Chapter 7
and
d2w dx 2
_}_d_(dw\ I
=
ds\ds)
JL[_6
+
-4/
\2s
+
-
6
6/j
12s
61s
-
21]
•
(7- 13d)
1*2
Step
4.
Derive Element Equations
ENERGY APPROACH
We now derive
use the principle of stationary (minimum) potential energy to
element equations. The potential energy for the
assuming only surface traction loading p(x),
Up = Substitution of w and II,
where
F is
[*'
w" from
= \Fl JoV =
Differentiating
IT^,
pw
\
element,
(7- 14a)
dx.
Eqs. (7-2) and (7-13d) into Eq. (7-14a) leads to
{q}W[B]{q}
-/f
'
(7-14b)
\N\p{q}ds,
^0
assumed uniform, and from
the transformation dx
beam [4, 5]
Xi
-
%F(w") 2 dx
expressed as
is
s
=
(x
—
x^/l,
we have
substituted
Ids.
w l9
with respect to
l9
u> 2 ,
and
2
term to zero, we obtain four element equations expressed
and equating each in
matrix notation
as 1
1
Fl\ [BY[B]ds{q}=l\ [Wpds
(7-15a)
Jo
Jo
or
M{q}
=
(7-1 5b)
{Q},
where \k]
= Fl C
[BY[B]ds
Ja
and {Q}
which
[k]
in the
= F/f Jo
expanded form 6(25 2(3* 6(2* 2(3* -
=
/
'
f Jo
[NT/»&,
is
\)ll*\ 2)// l)// 2 1)//
After the integrations,
p^JJ
2(^-2)
-6(2,-!)
ffi^iyj^
(? . 16a)
Beam Bending and Beam-Column
Chapter 7
ri2
F
[k] = ~~
/
3
6/
-12
6/
4/ 2
-6/
2/
12
-6/
sym.
179
2
(7- 16b)
4/ 2
and, assuming the surface traction
p where
/7j
=(1 -
p
varies linearly as
^s/? 2
5)/?!
(7-1 6c)
are values of loading at nodes
and p 2
-3s - 2s ls(l -2s + s 2
(I
and
1
2, respectively,
3 )
2
{Q}
= /f
2
)
1-5]
[S
- 2s) (s - 1)
(7-16d)
s (3 ls
2
J>2
or 7/?,
+
3/? 2 \
^-(3/7,
-f-
2/7 2 )
(7-16e)
«»=2i<
+
-^-(2^
3/> 2 )|
DERIVATION USING GALERKIN'S METHOD On the basis of the explanation in Chapter element.
The
residual
is
we consider one
3,
generic
given by (7-17)
Therefore, use of the Galerkin
F
I
(
S"
method
p Ntdx
Integration by parts of the
=
gives
°
9
/=1
'
2 3 4 '
(7-18)
-
'
)
first
term leads to
d*w dN, dx 2 dx
w from
Substitution of
0.
(7- 19a)
Eq. (7-2) leads to
<£»§**-£**+'£*
,d
2
w dN
t
dx 2 dx
= 0,
i=l,2,3,4, y=l,2,3,4
(7-19b)
or
*
I,
2
2
"**«
Jxi
^'^
'
f
L
dxl dx
(7- 19c)
Beam Bending and Beam-Column
180
Chapter 7
Expansion of the term on the left-hand side leads to ~N'[ 2
N'i J X\
Nim N'm
N'iN'( 1
N'iNT
o
N'iN'l
aw
N?
sym.
w,
N'i
t
(7-20a)
>dx.
•
w2
1
_
e2
Here the double prime indicates second derivative with respect to proper transformations for differentiations and integrations, will yield the
same
x. After
this expression
results in Eq. (7-1 6b).
term on the right-hand-side will lead to the load vector in Eq. (7-1 6e). The remaining two terms on the right-hand side yield (internal) joint shear and moment forces. Let us consider
The
first
r
(7-20b)
iVj
dx'
Noting the properties of
N
we
(Fig. 7-3),
{
find that this reduces to
/
(f)\
(7-20c)
(4s) which gives "joint" shear forces
two nodes.
at the 2
rd w
dNA
Similarly,
X2
dx 2 dx leads to
K (
dx 2 dx)\ xi
Fd
2
wdN \\ x 2
\
dx 2 dx)\ x
(F
d 2 wdN 3 \\ x dx 2 dx)\ x
V /
\
2 f d w dNA
dx 2 dx
Xi \
)\ xu
(7-20d)
Beam Bending and Beam-Column
Chapter 7
181
or
m. X
dx>) 2
(7-20e)
\
because
~
dx
dN2 dx
~
and x 2)
at x,
dx
dN4 _,
at
dx
x and x 2 x
,
and at x>
dx
dN,_ dx which
yield "joint"
there were (7-20c)
and
no
moment
Q
Sit
xu
two nodes.
forces at the
when
externally applied joint forces,
(7-20e)] are
assembled for
all
It
may
be noted that
if
joint load vectors [Eqs.
elements in the discretized body,
only the terms related to the two ends of the entire
beam
will remain,
and minus
the other terms will be zero due to alternating plus
whereas
signs.
The
nonzero end terms will denote the boundary conditions; for instance, for a simply supported beam the end term in Eq. (7-20e), which denotes moment, will vanish. Note that the element equations from both the energy and residual procedures will be essentially the same. Futher details of applications
of the Galerkin and other residual methods are given in Appendix
1.
Steps 5 to 8
We
shall illustrate the steps
of assembly and computation of primary and
secondary quantities by using an example. Figure 7-4 shows a length
L
—
20
cm and cross-sectional
area
= 2 cm
2
cm deep x
(2
beam with 1
cm wide)
kg/cm 2 It is subjected to a uniform surface traction p(x) = 100 kg/cm. The beam is divided into two elements of length / = 10 cm each. Use of Eqs. (7- 16b) and (7-16e) leads to the following (local) stiffness
and
E=
10 6
.
relationship for both elements (with / 3
16
x
W
15
15-3 100
-15
|
cm 4 ):
w
15"
50
'
1
0, <
-3
-15
3
15
50
-15
-15
w2
100_
e2
=
3'
5
1000
(7-21)
6
"
3
,-5,
Beam Bending and Beam-Column
182
p(x) = 100
Chapter 7
kg/cm
M Mi M tn L
Ut*
n
t
L = 20 1
e E . 10 kg/cm:
cm
cm
A
© ©
Global
Local
A
© ©©
10cm
O A
© ©
10cm
Nodes Elements Figure 7-4 Example for
By following
beam
bending.
the assembly procedure as described in Chapter
3,
we
obtain
the following global stiffness relation for the two elements: 3
15
15
100
-3
-15
15
50
-
3
15
-15
50
-
3
15
The beam
(Fig. 7-4)
r
3]
0i
5
6
3
15
w2
-15
50
2
-15
3
50
-15
(7-22a)
-15 I* Wy
3
kJ
100_
-5j
supported at two unyielding ends; therefore, the
is
boundary conditions are w
(hV
200
6
16
-
0"
t
= w = 0. Introduction of these constraints into 3
Eq. (7-22a) leads to modified global equations as
0"
1
(w>r f
15
100
-3
-15
15
50
-15
50
-
6
16
3
15
-15
50
w2 1
200
15
50
=
6
(7-22b)
<
r W>3
1
-15
°1 5
0i
100_
kJ
,-5J
Beam Bending and Beam-Column
Chapter 7
183
Solution of the four simultaneous equations gives
w, 0j
= 0.0000 (given), = 0.0500 radian,
= =
w2 2
0.3125 cm,
w3
0.0000 rad.,
0,
CLOSED FORM SOLUTIONS From beam bending theory w(x)
dw_
~
dx which
yield the
same
computations. That
method
yields the
[1]
= 0.0000 (given), = -0.0500 rad.
the closed form or "exact" solutions are
P x (T3 {L 24F
2Lx 2
p (fZ 24F K
6x 2
+
+
x 3 ),
(7-23)
4x 3 ),
(7-24)
results as those in Eq. (7-22c)
from the
with the cubic approximation model, the
is,
same
(7-22c)
results as the closed
finite
element
finite
element
form solutions insofar as the
displacements and slopes are concerned.
SECONDARY QUANTITIES Let us now consider the secondary quantities: moments (M) and shear forces (V). To find moment we substitute relevant nodal displacements
M
into
d2w dx 2
M where d 2 w/dx 2 is defined For element 7,
(7-25)
in Eq. (7- 13d).
0.0000
M(at$
=
=
0)
0.0500
-£[-6
4/
-21],
6
>
0.3125 .0.0000
-°' 25F
=
i
\
=
0.08334
x
\®l
=
833.4 kg-cm
lilarly,
M(at
M For element
s
-
(at s
0.5)
-
1)
- -0/250F -
15F
~°f 2
-
-3333.3 kg-cm,
-5833.4 kg-cm.
2,
0.3125
M(ats
=
0)
= -^[-6
-4/
0.0000 6
-21}.
*
0.0000
-0.0500,
=
-5833.4 kg-cm.
Beam Bending I and Beam-Column
184
Chapter 7
Similarly,
M(2LtS
=
-0.05F
0.5)
I
=
M(aXs
0A25F 1) I
The closed form
M= which
at
x
=
//4
solution for
=
2
moment
=
—3333.3 kg-cm,
-833.4 kg-cm.
is [1]
\2p (-Lx 24
-Fw'Xx)
and x
=
2
+
x 2 ),
(7-26)
1/2 gives
m(x =
-j\
=
M (x = -L\ = Figure 7-5(a) shows the bending
-3750.0 kg-cm,
-5000 kg-cm.
moment diagrams from
the finite element
computations and from the closed form solution. Figure 7-5 Comparisons for bending
two elements,
(a)
moments and shear
Bending moment diagrams,
(b)
forces:
Shear force dia-
grams.
~j^ 833.4
Finite element solution (a)
Closed form solution
500 kg
Finite element solution
(b)
Beam Bending and Beam-Column
Chapter 7
Next we compute shear force
is
given in closed form as (7-27)
yields
V (x
=
^=
0)
= y) = V (x = L) = (*
From
which
d3w dx 3
V which
V,
185
-1000.0
kg,
0.0kg, 1000 kg.
the finite element computations,
dx
^-[12
Therefore, for element
ds
I
6/
-12
(7-28)
6/]
/,
[0.00001
K=-^[12
6/
-12
0.0500 500.25 kg,
6/]
0.3125
0.0000
and for element
2,
V
= 0J5F = + 500.25 kg.
Figure 7-5(b) shows plots of shear force from closed form and
finite
element
computations.
Comments. As noted previously, both methods
yield the
displacements and slopes. However, for moments, the
finite
same values of element method
whereas the closed form solution shows continuous distribution. Also, introduction of only Wj in wthe finite element procedure does not yield zero moments at the ends, which is required for the assumption of simple support in the closed form solution. Overall, the magnitude of the moments from the finite element computations
yields a bilinear distribution [Fig. 7-5(a)],
is
satisfactory.
In the case of the shear forces, however, the
finite
element computations
show a wide disparity as compared with the closed form results [Fig. 7-5(b)]. At the ends the difference between the two results is high. As discussed in Chapter 3, there are two possible methods by which we can improve computations of bending moments and shear forces: (1) refine mesh and/or
(2)
choose different (higher-order) approximation models.
:
MESH REFINEMENT The beam is now
divided into four elements of
Fig. 7-6. Substitution of E,
/,
/,
/
=
5
cm, as shown in
and/? into Eqs. (7-16b) and (7-16e) leads to
the following general element equation
30
12
10
6
125
100
X 2 x 3
-12 -30
30"
50
100_
20
5
cm
5
Figure 7-6
1
0i
w2
250
kJ
-2500/12,
cm
cm
Mesh
25 ° f
2500/12
-30
12
ym
M
5
cm
refinement for
5
beam
cm
bending.
Assembly of the four element equations and introduction of the boundary conditions w,
=
w
5
=
lead to the followi ng modifie< J ass embk igee qui itions:
"10
0~
30
100
-30
-12
-30
24
30
50
50
-12
30
200
-30
50
-12
-30
24
30
50
-
-12
30
-
-30
50
0.00\
208.33
w2
500.00
02
0.00
W3
500.00
333.33
l
200
-12
-30
30
50
200
-
-12
30
-
-30
50
50
-
-30
w4
500.00
04
0.00
W5
1
30
i
0.00
24
_
=
100_
,051
0.00 ,
-208.33y
Solution of these equations gives
Wi vv
5
l
e,
= 0.0000, = 0.0000, = 0.0500, = 0.0500.
w2
=
0.2227,
H> 3
=
0.3125,
H> 4
=
0.2227,
2
=
0.0344,
03
=
0.0000,
4
=
-0.0344,
Figure 7-7 shows comparisons between bending
moments and shear
forces
from the finite element analysis and those from the closed form solution. It
186
Beam Bending and Beam-Column
Chapter 7
5
cm
cm
5
187
cm
5
5
cm
,
208-^
Closed form Finite element a
243
(b)
Figure 7-7 Comparisons for bending
moment
four elements, (a) Bending
moments and
shear forces:
diagrams, (b) Shear force dia-
grams.
can be seen that the computed values are than the comparisons
now
closer to the exact solutions
two-element approximation.
in Fig. 7-5 for the
HIGHER-ORDER APPROXIMATION As the next step from the cubic approximation, we can adopt
a fifth-order
approximation as follows: w(x)
= =
ol
:
[1
—
a2x
X
X2
—
a3 A'
.\-
:
—
J cc X B 4x
2
— -
—
a 5 x4
a 6x5
3
(7-29)
where {a} 7
The
first
and second
$
"
=
[a,
%2
a
3
%4
a.
a.]-
derivatives are obtained as
= [0 =£ ax
x
3x 2
4x 3
4
5.x ]{ai
(7-30)
Beam Bending and Beam-Column
Chapter 7
and
d 2w dx 2
w where
vv"
at the
two nodes leads to
\2x 2
6x
2
[0
20x 3 ]{a},
(7-31)
the second derivation or curvature. Evaluation of w, 6,
is
"10
'Wt
o
"
10
1
w"
{q}=<
and w'
>
2
=
IIP
w2
2/
1
/5
/4
2
4/
6/
12/ 2
3/
2
_0
(7-32a)
P
3
5/
4
20/ 3 _
or
=
{q}
(7-32b)
[A]{a}.
Therefore
{a}=[A]-{q},
(7-33)
where "
P
~ 9
I
[A]"
1
=±
l
-10/ 6
-6/ 7
15/
5
8/
6
-6/
4
-3/
5
9
/2
-3/ 8 /2
10/ 6
-15/
7
3/ /2
-/
6
6/
/2
5
4
- -4/ 7 7/
6
- -3/ 5
P/2
-P -P/2_
Hence,
uir)=[
As i
before, multiplication of [
=
1, 2,
.
.
.
,
6,
-1
(7-34)
leads to the interpolation functions
N
ti
given by
N = t
10s 3
1
N = l(s -
6s 3
2
Ni
= N = N, 5
P (s
10j
3
-
2
-
/(-4s
3
+ +
- 6s - 3s
155 4 8s 4
+
3s 3
5 ,
5
),
-
3s 4
- 6s - 7s - 3s
s 5 ),
(7-35)
5
\5s*
,
4
5
),
-V
Use of patibility
these
comand curvature w". This would provide
interpolation functions will enforce interelement
of displacement w, slope
6,
Beam Bending and Beam-Column
Chapter 7
189
an improvement in the approximation to the continuity of the deformed beam. The shape functions are plotted in Fig. 7-8. Problem 7-9 refers to derivation of equations for the beam bending problem with the fifth-order approximation. We have given partial results and comments with Prob. 7-9.
BEAM-COLUMN If in addition to the lateral
load
P
[Fig. 7-l(a)].
effect is called a
To
then the
load the
beam
beam
is
subjected to a constant axial
also acts as a
column, and the combined
beam-column.
simplify the problem,
we assume
that the axial load
is
small in corn-
Figure 7-8 Interpolation or shape functions for fifth-order model.
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Beam Bending and Beam-Column
190
Chapter 7
Figure 7-8 (cont.)
parison with the lateral load and that the problem
assumptions, Step
As
we can superimpose
the bending
and
is
linear.
Under
these
axial load effects.
1
before, the
beam
is
divided into one-dimensional line elements.
Step 2
Now
with the cubic model
we have
of the element: w, 9 for bending [Eq.
three degrees of freedom at each end
(7-2)]
and u
for the axial deformation
Beam Bending and Beam-Column
Chapter 7
[Fig. 7- 1(a)].
191
Thus the total number of degrees of freedom for the element (w„ $ u u x h< 2 ,0 2 ,w 2 ). As in Chapter 3, we choose linear
[Fig. 7-l(d)] is six
\
approximation for axial deformation (denoted here as u instead of as v in Chapter 3); accordingly, in terms of local coordinate s,
=
u(x)
[\
—
s
(7-36)
[NJ{qJ.
s]
For the bending part, we use the cubic function [Eq. model can be now written as
The combined
(7-2)].
interpolation
["Ml
[NJ
'
x
(1
"
[0]
2)
(1
x
fe.J
4)
(7-37)
w(x)
x
(1
where [NJ and
[NJ
[0]
[q b ] are the
same
2)
(1
x
and
as [N]
4).
{q} in
Eq. (7-2); here
we have used
the subscript b to denote bending.
Steps 3 to 5
The strain-displacement and bending problem are given
stress-strain relations for the axial
in Eqs. (3-1 2a)
and
(3-15)
and
and the and
in Eqs. (7-10)
(7-11), respectively.
The
potential energy
bending [Eq.
II,
=/
j
and
(7-14)]
is
i-F(w") 2
Here we assume
P
given as essentially the
axial
-
deformation [Eq.
-I
V
energies due to
Hence
wpds
-
Pu.
a concentrated uniform axial load; the term
is
the potential of load P.
Substitution for
E@ffds
Al
sum of the
(3-21)].
w
[Eq. (7-2)],
W
[Eq. (7-13)], u [Eq. (7-36)],
(7-38)
Pu denotes and du/dx
[Eq. (3-12)] leads to
II,
= § £ {q
T b}
[B b Y[B b ]ds{q b }
-
AEl
£
{q a F[BJ
r
[BJ^{q
fl
}
-/['[NJfqJ^-PINJfq.}. Here the subscripts a and b denote
Now we
axial
(7-39)
and bending nodes, respectively. w l ,0 l ,u i ,w2 ,02 , and w 2 and
differentiate TL P with respect to
equate the results to zero to obtain a
set
of
six linear algebraic equations.
After proper integrations and rearrangement of terms, and assuming
uniform
in
p
to be
Eq. (7-1 6e), the resulting element equations can be expressed as
Beam Bending and Beam-Column
192
Y1E1
6EI
\2E1
/3
/2
P
™\ o »jo
\
i
rv
12
Chapter 7
f j
Pl
\
2
!
4EI
6EI
I
I
2
6EI
12EI /
1
0,
ft
3
W
2
12
W2
pi 2
e2
pi* 12
1
«j. J
sym.
I
1
EA
EA
/
I
-
EA
1
(7-40)
P "2
1
P
_
I 1
We
)
|
have arranged the element relations such that the superimposition of
The top relation is the same and the bottom relation as in Eq. (3-28). The remaining steps of assembly and introduction of boundary conditions are essentially the same as before. The reader should at this stage the bending and axial effects can be seen clearly. as in Eq. (7-15)
undertake solution of Prob.
7-8.
COMMENT What happens
if
the load
P
between the bending and axial
is
not small and
effects
if
the coupling or interaction
needs to be considered ? For instance,
we ignored the bending effects that may be caused by P. Under these circumstances, the problem can no longer be considered as linear,
in the foregoing,
and the principle of superimposition usually does not hold. One of the is buckling. This type of nonlinearity falls under the category of geometrical nonlinearity and is beyond
manifestations of such a nonlinear problem the scope of this text.
OTHER PROCEDURES OF FORMULATION we shall discuss the use of the complementary energy and mixed principles for the derivation of finite element equations. Only a simple introduction to these procedures is included for advanced study the reader may consult texts and publications referenced elsewhere in the text, e.g., in Chapters 3 and 15. In this section
;
Complementary Energy Approach
For generating a
stable structural element for
eliminate rigid-body degrees of freedom. For the
beam bending, we need to beam element [Fig. 7- 1(d)],
such degrees of freedom are those relevant to the conditions of displacement
Beam Bending and Beam-Column
Chapter 7
and rotations
beam
at
one of the nodes. This leads to an element Then the complementary energy
or cantilever (Fig. 7-9).
193
like a built-in is
expressed as
a special case of Eq. (3-55c) as i
f*
2
fw,l
1
(7-41)
n\N^
l Q-|
,
Figure 7-9 Stable element for bending.
The
variation of
M along x can be expressed as M = xQ -M
moment
x
l
(7-42)
-1]
[x
Note
that the
moment replaces
M
-\}dx
(7-43a)
-3/
2 2/ 21
(7-43b)
-3/ 3/1-3/ relation between the
flexibility
is
[x
The
primary unknown. The
stress as the
matrix according to Eq. (3-58a)
IT
x
6
end nodal forces
is
given by (7-44a)
M
2
=IQ X
Mu
(7-44b)
or
-1
ia
/
2,
-1
[G]
(7-44c)
(7-44d)
M,
Now
the inverse of
[f ] is
LJ /
By using Eq.
(7-45)
3
b/
2/2
(3-62), we obtain the element stiffness matrix [k], which will be same as that in Eq. (7- 16b). If the variation of moment were different from linear, we would obtain a different stiffness matrix.
the
Beam Bending and Beam-Column
194
Chapter 7
As stated earlier (Chapter 3), use of the stress function approach can be more straightforward and easier. It is illustrated in Chapter 11. Mixed Approach
Here the formulation involves two primary unknowns: displacement w and moment (stress) M. We assume the same cubic function [Eq. (7-2)] for w as and linear variation for
M M=
- s)M + sM
(1
l
= [N m ]{M
(7-46)
2
n ],
M
M
T = where is the moment at any point, [M n } [M 2 ] is the vector of nodal and matrix of interpolation functions. moments, [NJ is the A special form of Eq. (3-63) for beam bending can be written as t
n,
= - i- j" M±Mdx +
J"
M^dx - {Qf{q},
(7-47)
where
M
{Qf=[e,
t
Q
M
2
2]
and r {q)
Substitution of
=K
,
e1i
>v 2
M and w" from Eqs. (7-46) and
(7- 13d), respectively, into
IIr leads to
{M.w AJ
lf[
+ {M„r
I
r [NjqN'Trfxfa}
- {QF{q}.
(7-48)
For a stationary value, we take derivations of and (w l5 6 l9 w 2 9 2 ). Thus
II* with respect to (Af,,
M
2)
,
fe = -A£ [Njr[N-k/x{M + ^g|
P [N J
T
[N"]rfx{q}
=
-1
(7-49a)
0,
= o + j" [NJ[N"]V.x{M„} -
{Qj
=
0,
(7-49b)
Beam Bending and Beam-Column
Chapter 7
195
or in matrix notation,
M
'
t
M
2
~~[6fJ 1
—
(7-49c)
<
M,
0,
a
w2 e ll
Mi,
where [k Tr ]
= -^J"[N m F[NJ^ 2
_/_
fTd-^) Wl - J)
ds
IF Jo
'2
J_ 6F
fcJ
o2
1"
(7-49d)
2
1
= rtNJTN"]^ —s
-Hi' i//
[-6 + 2^-
-4Z + 6/5
6-125
6ls-2l]ds
s
-1
1//
0'
(7-49e) i//
1//
1
Hence the element equations are _/_
IF
6F
J_
6F
J_ 3F
_J_
j_
/
M
\
/
2
I
!
-1 _1_
\
/
l_
=
e, (7-50)
<
M
o.
x
1
™2
Qi
/
4-
As an elements.
illustration,
o
we
The assemblage
consider the
r beam
relation for the
M
2
2
*
1
in Fig. 7-4, divided in
two elements
is
two
obtained by
|
Beam Bending and Beam-Column
196
Chapter 7
observing compatibility of nodal moments, displacements, and rotations; this gives
-1
/
1
3F
/
/
1
6F
/
1
1
/
/
(f
h
° (
W\
-1 1
11
2
/
1
6F
/
3F
/
6F
/
1
1
/
/
Mi
M
H'2
QT-Qr M - .\/f
/
1
/
/
3F
1
1
/
/
I
(7-5 la)
l)
02
6F
>
M
0i
/
l
Q\
2
M,
1
1 /
QT
H'3
0_
1
03'
M
\
(
2)
2
or
R
[K]{r]
Here the superscripts
By
(1)
and
(2)
(7-51b)
denote element numbers.
example of
substitution of the numerical values as in the
Fig. 7-4,
we
have, for uniform loading, 5
1
106
10
-1
2.5
1
106
10
;
o
w
1
1
"10
Hi
10
-1
01
2.5
1
10 6
10
10 106
2
2.5
10
10 6
2
1
10
/
M
fo°
2
1
H'2
~I6
|= 1000
•
02
\
2.5
1
5
10 6
10
106
1
1
10
10
fo
'
Af3 \_
H'3
1
We
2
o
o_;
>
can see that at the midsection the slope 9 2
boundary conditions
w
(7-51c)
1
10
t
=
w
3
=
= 0.
10 1
12
Introduction of the
into Eq. (7-5 lb) leads to
\r
Beam Bending and Beam-Column
Chapter 7
5
w
2.5
-1
1
1
10«
10
-^:
10
1
10
-l
01
12 2.5
10 10*
1
10*
10
2.5
~I6
1
~I6«
1
1
10
10
Mz
10
-
:.
)= lOOOj
)
r-5id>
1
0: 2.5
1
1
~w
10
Vi
1
10 1
Note Chapter
12/
that the assemblage matrix in Eq. (7-51) has zeros
such a system
If
10
h
1
3.
we
is
shall
on the diagonal.
solved by using Gaussian elimination, as discussed in
need to resort to partial or complete pivoting.
Solution of Eq. (7-5 Id) leads to
Af
:
*, d.
= -833.33 kg-cm. = 0.00 cm (given). = 0.04999 rad.
M_ = -5S33.33 mg-cm. w z = 0.312495 cm. =0.0000
0-
-V 3
-833.33 kg-cm, 0.000 (given).
-0.05100
rad.
rad.
method. The unknowns, whereas in the displacement approach they were derived from computed displacements and slopes. The results are the same because we assume linear variation for moments. This is a rather elementary problem included simply to illustrate the mixed procedure. These
difference
results are the is
same
that here the
as in the case of the displacement
moments
are primary
PROBLEMS 7-1.
(a)
Derive
stiffness
matrix
[k] for the
beam bending problem
if
EI
= F varies
linearly within the element as
f = y-.F - \ F = s. s = (x — x :
:
where (b)
V = :
Derive
1
—s
[k] if
and
\" ;
)
:
the area of the
beam
A = A \ :
7-2.
:
.
I.
varies linearly within the element as :
- A \z z
Derive the load vector due to uniform body force with the approximation function for w as in Eq.
.
Z acting in "-2 . I
Solution:
Q
r
4 i2'±
1 1
—r
the z direction,
:
Beam Bending and Beam-Column
198
7-3.
Derive if
by a
matrix
stiffness
beam
the
is
the
[k] for
supported on an
Chapter 7
beam bending problem
elastic
(with uniform F)
foundation which can be represented
of linear elastic springs with uniform spring constant k f (F/L)
series
(Fig. 7-10).
k f = foundation spring constant
Figure 7-10
Part/a/ solution:
The
tion spring support
Beam on
elastic foundation.
additional term in 11^ [Eq. (7-14a)] due to the founda-
given by
is
(k f w)wds
Jo
folTNTMNHq}*, which
will lead to the
stiffness
following contribution by the spring support to the
matrix
-
+ 2s ls(l - 2s + s s (3 - 2s) Is (s - 1) + 2s [1 - 3s 1
3s 2
3
2
1
f
i*-*/ J. Jo'
) >
X
2
2
2
=
13
11/
9
35
210
70
/a
13/
105
420
,
3
ls(l
-
- 2s
+
s 2)
s 2 (3
--
ls 2 (s - -
2s)
Ws
13/"
420 I
2
140
k sym.
13
11/
35
210 /2 1( >5_
7-4.
_1 in Eq. (7-9) and show that the result yields Multiply [0][A] 4, in
7-5.
Eq.
Perform
N
it i
=
1, 2, 3,
(7-3). all
(7-1 6a) to
steps of multiplications
Eq.
(7-1 6b)
and from Eq.
and integrations (7-1 6c) to (7-1 6e).
for going
from Eq.
Beam Bending and Beam-Column
Chapter 7
7-6.
Derive {Q}
if
199
the load varies as
p{s)
=
+N
NiPi
2
p2
+N
p
,
where
= 2s 2 - 35 N = s(2s -1) No = 45(1 -s) Ni
-
-
2
-
denotes the node at midpoint of the element. 7-7.
:
end fixed against movement and rotation and the other end settles by 0.05 cm. 7-8.
One One end
Solve the example problem in Fig. 7-4 with the following conditions (a)
By using two elements
and with constant
(Fig. 7-4)
free, (b)
axial load
P=
50 kg,
perform the entire process of formulation, assembly, introduction of boundary conditions, and solution of equations. Assume boundary conditions Wj 7-9.
=
VV 3
Compute
=
0.
and load vectors due
the stiffness matrix
face traction
p with
body force
to
Z and
sur-
the fifth-order approximations [Eq. (7-34)].
Solution:
K=Fl Jof[BF[B]
1805 2
4-
3
3
)
where
[BF
/
= /
2
3 )
3
3
)
3
/
)
Partial results:
= 17.41 F// k l2 - 175.24 F// 2 ^ 4 = 8.57 F// 2 k 33 = 0.088 F/l body force Z and surface load p, the load vector 3
kit
,
5
For constant
,
is
'
/
{Q}
/
= AZ
Compute tion
(7-34)]
/
/120
3
>
+
/
2
/10
/120
Pi 1/2
-/ 2 /10
-/ 2 /10
the load vector for the
model [Eq.
/10
1/2
,/
7-10.
3
2
H2
'
111
'
3
/120.
/
3
/120,
beam bending element with
and surface loading varying p(s)
=
(1
-
s)pi
+
sp 2
.
the interpola-
linearly as
Beam Bending and Beam-Column
200
Chapter 7
Solution:
Pi
Wt +
-Pi
*lft+A<*-*>] Jpj_
Pi
280 >)
*120^ V12C
{Q}={
{% + Y42-Pi)]
[3 1
(Pl.
V120
+ Pi 210 _|_
Consider two elements (Fig. 7-4) with different values of E:
7-11.
Element
Element
1:
E=
10 6
2:
£=
2 x 10 6 kg/cm 2
kg/cm 2
,
.
Assemble and solve for displacements, slopes, moments, and shear forces for the other properties as for the example of Fig. 7-4. Comment on the distribution of these quantities around the junction of the two elements. 7-12.
Assume
the area of the
beam element
A
to vary linearly as, see Prob. 7- 1(b)
- N A + N2 A 2 X
X
and derive element equations for uniform loading. Assuming the area to vary shown in Fig. 7-11, compute displacements, slopes, movement, and shear forces for the loading and properties of the example in Fig. 7-4.
linearly as
Note: Since area varies, corresponding variation
in
moment
of inertia should
also be considered.
10
—
2
cm
cm 2
10
—
-*
4
cm
cm 2
3 cm'
Figure 7-11
REFERENCES [1]
Timoshenko,
S.,
Strength of Materials,
Van Nostrand Reinhold, New York,
1956. [2]
[3]
Strang, G., and Fix, G. J., An Analysis of the Hall, Englewood Cliffs, N.J., 1973.
Finite
Element Method, Prentice-
Prenter, P. M., Splines and Variational Methods, Wiley,
New
York, 1975.
Beam Bending and Beam-Column
Chapter 7
[4]
[5]
Desai, C. S., and Abel, J. F., Introduction Nostrand Reinhold, New York, 1972.
Crandall,
S.
H., Engineering Analysis,
to the Finite
McGraw-Hill,
201
Element Method, Van
New
York, 1956.
ONE-DIMENSIONAL MASS TRANSPORT
INTRODUCTION
A
number of problems in various disciplines of engineering involve the phenomenon of mass transport. This can include transport through diffusion and convection of chemicals, pollutants, contaminants, and dissolved salts we shall treat mass transport for simple one-di-
in water. In this chapter,
mensional idealizations.
The
differential
be stated as
where
Dx
equation governing one-dimensional mass transport can
[1]
is
the dispersion coefficient, c*
(concentration of pollutant or dissolved
W
is
salt),
the
vx
unknown is
or state variable
the velocity or convection
x and t are the space and same as the first term in the flow equation [Eq. (5-9)] and represents the phenomenon of diffusion. The second term denotes the process of transport by convection. parameter,
is
the applied source or sink, and
time coordinates. The
FINITE
first
term
is
essentially the
ELEMENT FORMULATION
In this and the following chapters,
main
steps, while
labels.
202
some of
the
we
common
shall often detail
and
label only the
steps will be discussed without explicit
One-Dimensional Mass Transport
Chapter 8
—
203
L
H T \
\ \
\ \ \ \
'a)
(b)
A © A © A © k ©
©
L=+1
L = -1
(c)
Figure 8-1 One-dimensionai mass transport, (a) diffusion
and convection,
(b)
Mass transport by
One-dimensional idealization,
(c)
Dis-
cretization.
The domain of mass transport (Fig. 8-1) is idealized as one-dimensional. Use of the line element and the linear approximation gives the approximation for c as
=
c
i(l
-
L)c
x
+
i(l
+ L)c
(8-2a)
2
or c
where
L
r
{q}
=
[cj
=
c 2 ],
[N]{q]
and
c
is
= E Vet,
4.
=
U2,
(8-2b)
and medium.
the approximation to the concentration c*
denotes local coordinate; in Fig. 8-1, Step
i
it
also denotes length of the
Derivation of Element Equations
For derivation by Galerkin's method, the
residual
R
is
(8-3)
One- Dimensional Mass Transport
204
R
Weighing
with respect to
TV,-
yields
|r(S As explained ever,
Equation
in
Chapter
3,
we
(8-4)
is
NjCj) N dx = 0. t
shall treat the elements
understood that the procedure
it is
Chapter 8
one by one. How-
applied for the entire medium.
is
expanded as
^
£[ilD>i&™
Nrt- wJSEg£* dx
W - jf(LNJ>)Ntdx = 0. Integrating by parts the stant,
(8-4)
first
term
Eq. (8-5) and assuming
in
Dx
(8-5)
to be con-
we have
JT^-icEW }N dx = D
x N,$i:
t
dn
x
*/
1
X\
The first term on the right-hand side in Eq. (8-6) denotes Neumann-type boundary conditions and can be specified as a known value of the normal derivative dc/dn. The term D x (dc/dn) £j is the (known) flux on the end boundaries. Substitution of Eq. (8-6) into Eq. (8-5) leads to element equations in
matrix form as [E]{q}
+
[E,]{q}
=
(8-7)
{Q},
where {q} = [dcjdt dc 2 /dt], [E] and [E ] are the element property matrices, and {Q} is the element forcing parameter vector. These matrices are defined r
f
as follows: [E]
= r([BYD x [B]dx +
[BYvJS\dx)dx
(8-8a)
9
[E,]= r'[NF[N]
(8-8b)
and {Q}
=
p [WWdx - D N^ x
(8-8c)
n
Here [B] is the usual transformation matrix [Eqs. (3-13) and (4-4)] obtained by taking the proper derivative of c [Eq. (8-2)]. In writing Eq. (8-8a) we assumed, only for simplicity, that v x second term in Eq. (8-5)
is
is
constant with respect to x; hence the
not included.
We
note that
it is
not necessary to
make this assumption because that term can be easily included. The evaluation of [E], [EJ, and {Q} is as follows:
One- Dimensional Mass Transport
Chapter 8
[-1
[E]
- /[_] -
=
We
ij
205
\}dL
+ 2[
i_
i
DJl - vJ2 -DJl - vjT DJl + vJ2_\ -DJl - vJ2
(8-9)
note here the important characteristic that the element matrix [E]
nonsymmetric ;
mass transport. [E,]
this
property
is
is
contributed by the convection part of the
Now
=
^
2 "2
1
L
2
[1
-L
+
1
L]dL
11 (8-10)
2
1
This matrix
[Eq
is
from the time-dependent term
similar to the matrix arising
(5-21)].
Wl\
(8-11)
1]
(4
2
The first term in Eq. (8-11) indicates that applied sink or source quantity is lumped equally at the two nodes. As discussed in previous chapters, only the terms at the end nodes remain when the second part of {Q} is considered. Step
5.
Assembly
Consider the three-element mesh (Fig.
Combination of element
8-1).
matrices for the three elements leads to the assemblage equations as
D x /l -Dx ll +
v x /2 v x l2
- D x /l -
2D x /l +
-D x + v x /2 /l
1
v x /2
Q
f
L
-D x \l-v x \2 2D x jl +
(c
-D x /l-v x /2
-D x + v x /2
D x /l +
/l
10 0" 14 10 14
"2
1
v x /2j
)c 3
U
dd/dt
1
daldt
2
<
1
dajdt
2_
dctldt,
Wl = E±{ 2
.
2 ,
1
,
(8-12a)
One- Dimensional Mass Transport
206
The contribution of
the boundary terms
Chapter 8
not included because
is
we assume
that
|(A0 =
(8-13a)
and
c(0,t)=c.
(8-13b)
In matrix notation, Eq. (8- 12a) becomes
+
[K]{r]
where
and
[K]
[K,] are the
KSjt)
=
{R},
(8-12b)
assemblage property matrices, {R}
blage forcing parameter vector, and
{r} is
is
the assem-
the vector of assemblage nodal
unknowns. Solution in
A
Time
number of
integration schemes can be used to solve the matrix equa-
For
tions in the time domain.
form as
difference
instance, Eq. (8-1 2b) can be expressed in finite
[2]
- 0)W,) +
RK»{rU +
+ o-^W-»^{*L We
where 6 is a scalar. ing on the value of 6.
= ^,
we
(8 - 14)
obtain various implicit and explicit methods depend-
If
6
=
1,
the fully implicit
method
is
obtained, and
if
obtain the Crank-Nicholson scheme. These schemes possess dif-
ferent mathematical properties such as convergence
and
stability
which can
influence the quality of the numerical solution. Detailed discussion of these
aspects If
is
we
beyond the scope of this
text.
use the following scheme to approximate the
first
derivatives in
- ^[K ]){r} r
(8-16)
Eq. (8-14),
then Eq. (8-14) becomes (for 6 ([K]
+
^[K,]){r} f+A
,
£)
~ 2{R} f+A -
Since the values of {R} f+A and ,
=
,
{r},
([K]
f
known, we can compute {r} r+Ar at the The procedure starts from the first step
are
next time step by solving Eq. (8-16).
when the values of {r} at t = are prescribed as initial conditions. The boundary conditions are prescribed as known values of {r} nodes. Equation (8-16)
is
at given
modified for these conditions before the solutions
are obtained at various time levels.
CONVECTION PARAMETER v x For the solution of Eq. (8-16), the value of the convection velocity v x should be known. It can often be obtained by using available data or formulas.
One- Dimensional Mass Transport
Chapter 8
One
of the ways to compute
it is
207
to first solve the flow or seepage equations
which allow computation of v x Comments. In most engineering problems the assemblage matrices are symmetric and banded. In the case of mass transport, the matrix [K] in Eq. (8-16) is banded but not symmetric. Consequently, as discussed in Appendix 2. we have to store in the computer core the coefficients on the entire band (2B-1). This aspect will involve relatively more computer time in solving the equations as compared to the time required for symmetric and banded systems. The nonsymmetric property of [K] can be considered to be a characteristic of the non-self-adjoint nature of the problem. The behavior of the numerical solution can be influenced significantly due to the existence of the convection term. If the magnitude of the convection term is relatively large, the system [Eq. (8-1)] is predominantly convective. (Chapter
4),
.
This can render the solutions more susceptible to numerical instability. the other hand,
On
the diffusion term predominates, the numerical solution
if
can be well behaved.
Example
To
8-1
illustrate a
few time steps
in the solution of
Eq. ($-16). we adopt the following
properties:
D,= *= 1=
(i-
1
(L
1
[L\
W= At Here L denotes illustrate the
length.
may
Tk
{MTU)
=
•Tk
1
M mass and T
procedure, and
n.
1
time.
These properties are chosen simply to
not necessarily represent a
field situation.
Boundary- conditions.
aO.n =
(ML
1.0.
3 )
at
node
1.
t
>
0. (Fig. 8-1).
Initial conditions: ci.v.
At time i
-i
t
—
=
-1
At,
from Eq. "2
0"
2
-i
-4
2
_
-4
_3
0)
=
(.8-16).
0.0.
we have 0~\ 'ci
1
14 10 14
J_ j
1
l_
_0 1
f
= -
1
i -i -4-
!_
c:
c3
2_|/ [C4
o
Ar
i
o"
2-1
-i \
r<0.
2
0-4
2 1
-f 4_
.
x x
1
~2
1
1
4
1
1
4
1
1
:_
6
_0
°"\
\C\
c2 Cl ie4
One- Dimensional Mass Transport
208
Because c
x
=
c2
=
c3
C4
=
0at
=
/
o,
0"
7
7
Cj
t
_7
20
1
Cl
3
6
=r
>
— 37
20
-i
Chapter 8
6 1
13
i
6
-
1
,c A
=
Introduction of the boundary condition c x
<
>
c3 Ar
gives
1
Cl
1
f
20
i
7
6
J 1
6
L
1
—3
20
1
13
-i
6
C2
=
<
|c 3
[oJ
- lc 4
Solution by Gaussian elimination leads to c,
c2 c3 c4
= = = =
1.00 (given),
509 x 10" 4
,
260 x 10" 5
,
200 x 10~ 6
.
The next stage, t = / + A/ = 2Af, can now be performed by using the values of c computed at the end of t = At as initial conditions and so on for other time steps. Example 8-2
A solution to Eq. (8-1) was obtained by Guymon [3] using a variational procedure. We present here his results, which compare numerical predictions with a closed form solution for a problem in
idealized as one-dimensional.
terms of the nondimensional quantities x/L, c/c
the total length of the one-dimensional
The
initial
The results are presented and v x l/D x where L is
v x t/L,
,
medium.
conditions are
=
c(x, 0)
and the boundary condition
0,
is
c(0,f)
That
,
=
c
.
is,
Co
(0,0
=
1
Table 8-1 shows a comparison between the numerical solutions and the closed form solution. The former are obtained for two conditions: v x l/D x = 0.5 and v x l/D x = 0.25 at a time level v x t/L = 0.5. The closed form solution is obtained from [3]
where
erfc denotes error function.
One- Dimensional Mass Transport
Chapter 8
TABLE
209
Comparison Between Finite Element Predictions and Closed Form Solution at v x t L = 0.5 [3]
8-1
Closed
Finite
Element Solution
Form
Dx =
Dx =
x/L
Solution
1.000
1.000
1.000
0.1
0.976
0.975
0.972
0.2
0.928
0.927
0.923
0.3
0.851
0.850
0.845
0.4
0.-45
0."43
0.738
0.5
0.616
0.614
0.606
0.6
0.478
0.475
0.463
0.7
0.345
0.341
0.324
0.8
0.230
0.228
0.209
0.9
0.142
0.145
0.132
1.0
0.080
0.060
0.080
Comment. The
vx l
0.25
vx l
0.50
magnitude of the convective less accurate. This term has significant influe: s on numerical solutions of the one-, two-, and threedimensional problems involving diffusion-convection phenomena. term vj
results indicate that as the
D., increases, the
numerical solutions are
REFERENCES [1]
Bear,
J..
Dynamics of Fluids
in
Porous Media. American Elsevier,
New
York,
1972. [2]
Richtmeyer. R. D.. and Morton. K. W.. Difference Methods for Initial-Value New York. 1957.
Problems. Wiley-Imerscience. [3]
Guymon. G.
L..
A
Finite
Element Solution of the One-Dimensional Diffu6. No. 1. Feb. 1970,
sion-Convection Equations." Water Resour. Res.. Vol. pp. 204-210.
BIBLIOGRAPHY Amend. fate
cal
Contractor. D. N.. and Desai. C. S.. "Oxygen Depletion and Sulin Strip Mine Spoil Dams." in Proc. 2nd Intl Conf. on NumeriMethods in Geomechanics. Blacksburg. Ya.. C. S. Desai fed.). June 1976. J.
H..
Production
ASCE. New York,
1976.
Cheng. R. T.. '"On the Study of Convective Dispersion Equations." in Proc. Intl Symp. on Finite Element Methods in Flow Problems. Swansea. U.K., Jan. 1974, University of Alabama Press, Huntsville, 1974.
One- Dimensional Mass Transport
210
Chapter 8
and Contractor, D. N., "Finite Element Analysis of Flow, Diffusion Water Intrusion in Porous Media," in Proc. U.S. -Germany Symposium on Theory and Algorithms in Finite Element Analysis, Bathe, K. J., Oden, J. T., and Wunderlich, W. (eds), M.I.T., Cambridge, Mass., Sept. 1976.
Desai, C.
and
S.,
Salt
L., Scott, V. H., and Hermann, C. R., "A General Numerical SoluTwo-Dimensional Diffusion-Convection Equation by the Finite Element Method," Water Resour. Res., Vol. 6, No. 6, Dec. 1970, pp. 1611-1617.
Guymon, G. tion of
Proceedings, First International Conference on Finite Elements in Water Resources,
Gray,
W.
G., and Pinder, G. F. (eds.), Pentech Press, London, 1977.
Segol, G., Pinder, G. F., and Gray,
W.
G.,
"A
Galerkin Finite Element Technique
for Calculating the Transient Position of the Salt Res., Vol. 11,
Smith,
I.
No.
2,
Water Front," Water Resour.
April 1975, pp. 343-347.
M., Farraday, R. V., and O'Connor, B. A., "Rayleigh-Ritz and Galerkin
Finite Elements for Diffusion-Convection Problems," Water Resour. Res., Vol. 9,
Wu,
No.
3,
1973, pp. 593-606.
S., and Contractor, D. N., "Finite Element Procedure for Water Intrusion in Coastal Aquifers," Report No. VPI-E-76-23, Department of Civil Engineering, Virginia Polytechnic Institute and State University,
T. H., Desai, C.
Salt
Blacksburg, Va., July 1976.
:
ONE-DIMENSIONAL
OVERLAND FLOW
INTRODUCTION Overland flow can include problems
in hydraulics
and hydrology such as
runoff due to rainfall, flow in shallow open channels and in rivers in flood plains,
and flood routing. The problem
is
three-dimensional, but
be approximated by one-dimensional idealization, and for treatment,
we
shall consider the
this
it
can often
introductory
one-dimensional case.
The equations governing overland flow
and
consist of the continuity
momentum equations developed by Saint- Venant. The equations for gradually varying flow in open channels can be assumed as follows [1-5] Continuity equation:
(i)
dQf.dA + dt dx (ii)
Momentum
(9-1)
0.
equation:
dQ
*+&5) -**-*>-*£
<«>
dx~
Here
Qf
is
the discharge in overland flow (Q f0 ) or channel flow (Q fc ) r is excess rainfall in overland ;
the lateral inflow per unit length of flow plane
flow
bed
(r )
and net inflow
slope,
S
in the
for overland
—
channel caused by overland flow
and Sc
for channel
the area of overland flow or channel flow
;
x
;
Sf is
is
(r c )
;
S
the friction slope
is ;
the
A
is
the coordinate in the flow
211
212
One- Dimensional Overland Flow
Chapter 9
Ac Section C-C
0-0
Section
(a)
Lateral overland flow
I
I
I
I
I
I
due to
I
I
I
I
I
I
I
I from overland flow,
rainfall, r„
(b)
Channel inflow I
y
I
I
I
I
1
rr
i
(c)
Figure 9-1
Representations
direction
;
y
is
overland
for
Overland and channel flow, rainfall, (c) Channel flow. (a)
(b)
and channel
flows,
Overland flow due to (excess)
the coordinate in the depth of flow direction
;
and
/
is
the
time [Fig. 9-1 (a)].
Various assumptions lead to the so-called kinematic wave approximation [6],
according to which the
momentum equation reduces to S
=
Sf
.
(9-2b)
Equation (9-2b) can be approximated by using Manning or Chezy equations. use Manning's equation, given by
We
:
One- Dimensional Overland Flow
Chapter 9
K=!^/y n
3
213
S 1/2
(9-3a)
or
Q f = h^.RpS 1/2 A V
where
is
the velocity of flow,
Rh
is
(9-3b)
i
=
the hydraulic radius
area/wetted
and n is the Manning roughness coefficient. As a consequence of the use of Manning's formula, the finite element procedure essentially perimeter,
involves solution of Eq. (9-1).
APPROXIMATION FOR OVERLAND AND CHANNEL FLOWS Figure 9- 1(a) shows an arbitrary domain with a channel. The rainfall excess
Q f0 which discharges into the channel, resulting in the Q fc A number of investigators have considered finite element
causes overland flow
channel flow
.
solution for general [7-11].
As
In the
first
and one-dimensional
a simple approximation, stage, the overland flow
we can
idealizations for overland flow
solve the
due to excess
problem
rainfall, r
in
two
stages.
[Fig. 9- 1(b)], is
computed by solving Eqs. (9-1) and (9-3) for the overland flow. The results of flow from the first stage are considered as the input flow r c in the channel [Fig. 9-l(c)]. Then Eqs. (9-1) and (9-3) are solved for computing the channel flow. The same finite element formulation can be used for both stages.
ELEMENT FORMULATION
FINITE Step
Discretize and
1.
The
Choose Element Configuration
domain [Fig. 9-2(a)] is idealized as shown in Fig. domain is divided in two parts: overland part [Fig. 9-2(c)] and channel part [Fig. 9-2(d)]. The overland part is divided into a suitable number of zones depending on the physical characteristics such as the bed slope and topography. Each zone of the overland domain and the 9-2(b).
arbitrary flow
The
channel
is
Step
2.
idealized
replaced by one-dimensional line elements [Fig. 9-2(e)].
Choose Approximation Functions
The area A and
the discharge
Qf
can be assumed as unknowns and
represented by using linear approximation
A(x9
t)
= (l-s)A (t) + sA 1 l
=
(t)
[N]{A„(r)}
= jlN Afi) t
(9-4)
214
One- Dimensional Overland Flow
Chapter 9
Divisions idealized on the basis
of topographic and other properties
(a)
—
1
+
1
r-
~~
Overland
^
0)
1
.c
1
--*
1
— — — - <>,--<
-
1
^--
1
-*
>1
L_
(b)
(d)
(c)
1
s
= x//
(e)
Figure 9-2
Finite
(b) Idealization, (c)
element
discretization,
Overland flow,
(d)
(a)
Actual
Channel flow,
(e)
region,
Generic
element.
and
Qf (x,
t)
= (1 - s)Q n {f) + sQn {t) = mQfjm = t N,Qfl {t),
is
= [A
the vector of nodal areas and [Q fn } T the vector of nodal discharges.
where {A„} T
l
A 2]
is
(9-5)
= [Qfl Qf2
]
.
One- Dimensional Overland Flow
Chapter 9
215
In view of the use of Manning's equation, the values of
we
various time levels, and as
area
A
as the time-dependent
Qf
are
known
at
shall see later, the solution involves only the
unknown.
Derivation of Element Equations
Step
4.
We
use Galerkin's residual method to derive the individual element
equations. Minimization of the residual of Eq. (9-1) leads to
(9-6)
i Use of Eqs.
(9-4)
and
(9-5) gives
=
(9-7a)
[BHQ,.}
and
^= = When
-
[1
s]{A„]
5
[N]{A„}
=
[N](^>
(9-7b)
we obtain
Eqs. (9-7a) and (9-7b) are inserted into Eq. (9-6), X
n[NHN]rfjr{A„}+ J x\
JfX\
'[N] r [B]{Q /n }rfx= ["
[Wrdx
(9-8a)
J X\
or
[1—5
r
1
s]ds {k " ]
-
s
-r[
(-2/1 Ir
After performing integrations,
V2
11
•
'
'
1
Qn - Q ;-
1
Ir ;
—1—
\
dA |_1
2j [dt
(9-8b)
rrrh-
we have element equations
[dAA dt
- Qf i)ds
'
<
(9-9a)
>
2
7 J
,
1
.
1
or [k]{A„}
Step
5.
=
(9-9b)
{Q}.
Assembly
Consider a mesh made of three elements of lengths 9-3(a)].
The element equations can be assembled
as
/,,
/2
,
and
/3
[Fig.
One- Dimensional Overland Flow
216
~2h
Chapter 9
0" dAA
/,
dt (21,
+
2/ 2 )
dA 2
/,
dt
(ll 2
l2
+
dA 21,)
)
3
dt
Symm
dA 4
h
Qh Qh [Qh ffi>.
dt\
2/ 3 _
Qh Qh + Qh - Qh Qh + Qh - Qh Qh
(Vi
+4
(9-10a)
<
hr 2
+
hr z
IVs
J
or in matrix notation [K]{A„}
=
{R g }
+
{R,}
=
{R}
(9-10b)
where the superscript on Q f denotes an element and {AN } = [A A 2 A 3 A A ]. Here we have used {A^} for the asemblage vector in place of [r] to avoid T
x
confusion with
r that is
used for applied inflow.
Figure 9-3 Discretization and timewise solution, (a) Three-element discretization, (b) Finite difference
approximation for
first
deriva-
tive.
(?) Globa node I
— 2
©A © A
t
(b)
1
•-
©
(a)
+ At
1
A®
Local node
/l\ Element
One- Dimensional Overland Flow
Chapter 9
We
use the most simple time integration, often called Euler formula to
approximate the
derivative in Eq. (9-10), [Fig. 9-3(b)].
first
jA^jf
iAy]r+Ar lAtfJ
If
we
(9-11)
At
substitute Eq. (9-11) into Eq. (9-10), the result
^[K]{A^J f+Af
=
[R Q }
t
+
now
Equation (9-12) can
The
+ JptKKA*},.
(9-12)
be solved by using the "marching" process in
solution starts at time
t
= 0, when we can assume that
A (x,0)=0] i
where
{R r }, +Af
is
CONDITIONS
INITIAL
time.
217
N
the upper
Q fi (x,0)=0\ = number of nodes. We boundary
(or
i=
further
an end node)
N,
1,2,
(9-1 3a)
assume that discharge
at all times
;
that
2/(0,0=0. The values of excess
rainfall for
is
zero at
is,
(9-13b)
overland flow and resulting inflow due to the
excess rainfall for the channel flow are prescribed as
= F(x,
r(x, t)
(9-13c)
t),
where the overbar denotes known quantity. Solution in
With
the conditions in Eq.
J.
f
0"
"2/, /,
Time
1,
2/,
6At
+
h
2/ 2
h
+
2/ 2
.0
/,
h
2/ 3
2/J
Mil A2
I
=
+
Si
+ hh
At
is
(9-14)
Ay
UJ
I3P3
t = At. These values by using Manning's equation in which the bed slope S and coefficient n are prescribed, and the knowledge of areas and prescribed values of average widths permit computation of R h At the next time step t = At + At = 2At, the vectors {R Qit* and (\IAt)\\L]{K N }to are now available from the solutions at t = At, while {FN } is known at all time levels hence
Solution of Eq. (9-14) gives values of areas of flow at
are
now used
to find nodal quantities of flow
.
;
[K]{A„j
The procedure
is
2Af
,}*
+
{R,h*
+ ^[kha*}*.
repeated until the desired time level
is
reached.
(9-15)
Step
Solution for {\ N }
6.
Example
As a simple
9-1
procedure
illustration of the solution
in time,
we adopt
the following
properties:
/= r = At =
Element length,
Flow due
to excess rainfall,
Time increment, Here and
in the
following
1.0m; 10
may
simple properties in order to illustrate
not necessarily represent a
Substitution in Eq. (9-14) yields equations at "2 1
1
6
0"
1
x 0.166
t
=
At
1
A2
1
1
4
1
A3
1
2_
A4
field situation.
as
At
4
_0
assumed constant
0.166 hr.
we have chosen
the procedure; these properties
m 2 /hr,
-1
1.0
x 10"
1
2 ,
2 2 1
At
or
0~
~2
1
1
4
1
1
4
1
A3
1
2_
A4
0.5
'Ai
A*
1
(9-16)
'
_0
Comment.
'
.0.5. At
At
We note here that the matrix differential equations [Eq.
sent an initial value problem.
The
the stiffness matrices (Chapter
3),
resulting matrix [K]
is
and can be solved as expressed
solution by Gaussian elimination gives the areas of flow at
A = A2 = A3 = A4 = l
0.166
(9-9b)] repre-
not singular, in contrast to
m
=
t
in
Eq. (9-16). The
At:
2 .
Now we use the Manning formula [Eq. (9-3)] to find nodal flows Q f by assuming the following data:
w
= 0.3,
5=
0.1;
Average width of flow Hence,
at
nodes
:
at global
x
= 0.5 m,
= 0.6 m,
w3
= 0.75 m, u> 4 =
-0.166/0.5
Q n =^(o.33)°- 667 (0.1)
218
w2
= 0.330 m, = Rh2 =0.166/0.6 0.277 m, Rh3 =0.166/0.75 = 0.220 m, RM =0.166/1.3 = 0.128 m, nodes at time = At as
R hl
which gives flows
w
/
-
5
x 0.166
0.125
m
3
/hr,
1 .3
m.
One- Dimensional Overland Flow
Chapter 9
1.49
Qf:
(0.277)°- 667 (0.1)
-
5
x 0.166
=
0.110,
-
5
x 0.166
=
0.095,
0.3
Q f3 =
1.49.
^y(0.220)°- 667 (0.1) 1.49,
(0.128)
Qf
To
219
-
667
x
(O.l) 05
x 0.166
=
0.066.
0.3
evaluate the
first
part of the forcing vector in Eq. (9- 10a),
necessary to
it is
convert the above flows at global nodes into flows at local nodes, Fig. 9-3(a). If
we
assume that the flow at a global node is divided equally among the common local nodes of two adjoining elements, and that only half of the flow at an end node is that node, then Q} = 0.0625, Q} 2 = 0.055; Q fl = 0.055, = 0.0475, and Q} 2 = 0.033. Q} time = A/ + A/, we now have, using Eqs. (9-10a) and (9-15),
effective
At
2
at
0.0475;
x
=
v
t
2
1
1
4
0~ 1
'Ai
0.0075
A2
0.0150
1
0.5 1.0
+
>
6
Q} 2
x 0.166
1
4
1
A3
1
2_
A,
0.0220
1.0
0.0145
2 At
'
0.5
Af
(0.1667
0.1667
+
6
(9-17)
x 0.166
0.1667 10.1667
.0
Solution of Eq. (9-17) leads to nodal areas at for the required
/
=
2A/.
The procedure
is
repeated
number of steps.
Example 9-2 field problem solved by Ross [10] using a finite element procean area of 136 square miles around a portion of the South River in Virginia. Field observations of overland and river flow were available from a gaging station at the base of the area considered. The region was divided into three subregions upper South River, Back Creek, and lower South River containing
Figure 9-4 shows a
dure.
It
consists of
—
—
three channels.
The
finite
element formulation
is
applied independently to three
overland flows and three channel flows and the results are added to obtain computed flow at the gaging station. Special provision was
modate
Figure 9-4 also shows one of the [10];
it
made
in the
formulation to accom-
existing flood retention structures [10]. finite
element discretizations used by Ross
contains three overland flood plains (OFPs) that are idealized as 24 one-
dimensional elements. The elements are labeled as IA1, meaning idealized strip(s) in flow plane,
flow planes,
strip,
and element
and elements are shown
first
flow plane,
in the strip. Detailed properties of the in
Table
9-1.
The
three channels were
discretized similarly; the properties are given in Table 9-2. Values of excess rainfall
220
One- Dimensional Overland Flow
Chapter 9
Legend Overland flow plane boundary Surface element boundary
Channel element Primary channel
Secondary channel Flood detention structure and reservoir
IIB1
Figure 9-4 Finite element
watershed
mesh
(24 elements) for upper South river
[10].
based on the rainfall during Hurricane Camille, the night of August
which
total rainfall in the region
ranged from about
3 to
1 1
in.,
19, 1969, in
were used as the
in-
put for overland flow.
Figure 9-5 shows a typical comparison between the computed and observed
(hydrograph) values of discharges during the period of the hurricane. For these results a value of time
Comments.
A
We
increment At
=
900 seconds was used.
have presented simple problems only as an introduction. factors such as multidimensional effects and physical
number of other
One- Dimensional Overland Flow
Chapter 9
TABLE 9-1
221
Surface Geometry and Properties of 24-Element Mesh
[10]
Average
OFP
Strip
Element
No.
No.
No.
A B 2
C
D 2
E F 2
G H A
II
2
B 2
C 2
D 2
E F 2
A
III
B
TABLE
9-2
Length (m)
Width
Slope
Manning
(m)
(m/m)
n
2334 4023 4023 2334 4023 4023 2334 4023 4023 2334 4023 2736 2736 2855 2655 2736 2736 2655 2655 2736 2655 2655 2655 2414
9012 2374 7121 5177 7215 6196
0.028
0.297
0.016
0.343
0.010
0.336
0.020
0.323
0.065
0.316
0.025
0.309
5981
0.024
0.244
2982 5825 5338 2843 4855 5552
0.103
0.292
Element
No.
No.
Length (m)
properties of the flow tions. Also,
0.340
0.006
0.338
0.041
0.286
0.024
0.266
1475
0.108
0.300
4426 3018
0.043
0.300
0.111
0.280
5968
0.071
0.291
4493
0.111
0.300
5432
0.047
0.286
2950 4734 5512
0.011
0.231
0.066
0.308
0.019
0.294
1931
0.003
0.264
3058
0.057
0.121
[10]
Slope
Manning
(m/m)
n
1
9495
0.013
0.100
2
5686
0.004
0.085
3
5686 5686 5901
0.001
0.085
4
III
0.310
0.017
Geometry and Properties of Channel
Channel
II
0.027
1
0.001
0.105
0.039
0.120
2
5901
0.010
0.115
3
5901
0.005
0.115
1
3862
0.002
0.130
domain may require additional advanced considera-
numerical characteristics such as convergence and
require special attention for
more complex problems. Hence,
stability
may
the derivations
One- Dimensional Overland Flow
222
Chapter 9
600.0
500.0
\
1
400.0
o *
/
f
Calculated
*
Recorded
\\
/
/
/
/
J
;
/
/
-o
\\
/
»
/
300.0
\
J
1
n*--
/
- "*<^
200.0 /
/
/
/
'
100.0
/
//
J/ 0.0 SdfeHBE 0.0
1
i
10.0
5.0
20.0
15.0
Time
25.0
1
1
1
30.0
35.0
40.0
(hr)
Figure 9-5 Comparison of observed and computed discharges of
gaging station, At
and
=
results presented herein
900 sec
[10].
should be treated simply as illustrations for the
one-dimensional idealization. For practical applications and multidimensional problems, the user should consider other relevant characteristics.
REFERENCES [1]
"Numerical Solution of Flood Prediction and River Regulation I Derivation of Basic Theory and Formulation of NumerMethods Attack," Report No. IMM-200, Institute of Mathematical
Stoker,
J. J.,
Problems, Report ical
Science, [2]
[3]
[4]
New York
Abbott, M.
:
University,
New
York, 1953.
and Ionescu, F., "On the Numerical Computation of Nearly Horizontal Flows," /. Hyd. Research, Vol. 5, No. 2, 1967, pp. 96-117. B.,
Brutsaert, W., "De Saint-Venant Equations Experimentally Verified," Hydraulics Div. ASCE, Vol. 97, No. HY9, Sept. 1971, pp. 1387-1461.
/.
J. A., and Woblhiser, D. A., "Difference Solution of ShallowWater Equations," /. Eng. Mech. Div. ASCE, Vol. 93, No. EM2, April 1967,
Liggett,
pp. 39-71. [5]
Prince, R. K., "Comparison of Four Numerical Methods for Flood Routing," /. Hydraulics Div. ASCE, Vol. 100, No. HY7, July 1974, pp. 879-899.
One- Dimensional Overland Flow
Chapter 9
[6]
Lighthill, M.
Movement
in
and Whitham, G.
J.,
Long
B.,
223
"On Kinematic Waves,
Rivers," Proc. R. Soc. London Ser. A, Vol. 229,
I:
Flood
May
1955,
pp. 281-316. [7]
[8]
Judah, O. M., "Simulation of Runoff Hydrographs from Natural Watersheds by Finite Element Method," Ph.D. thesis, Virginia Polytechnic Institute and State University, Blacksburg, Va., Aug. 1972.
Al-Mashidani, G and Taylor, C. "Finite Element Solutions of the Shallow Water Equations Surface Runoff," in Proc. Intl Symp. on Finite Element Methods in Flow Problems, University of Wales, Swansea, U.K., Jan. 1974, Oden, J. T., Zienkiewicz, O. C., Gallagher, R. H., and Taylor, C. (eds.), Uni.
,
,
—
versity of [9]
[10]
[1 1]
Alabama
Press, Huntsville, 1974, pp. 385-398.
Taylor, C., Al-Mashidani, G., and Davis, J. M., "A Finite Element Approach to Watershed Runoff," /. Hydrol, Vol. 21, No. 3, March 1974. Ross, B. B., "A Finite Element Model To Determine the Effect of Land-Use Changes on Flood Hydrographs," M.S. thesis, Virginia Polytechnic Institute and State University, Blacksburg, Va., Nov. 1975.
Taylor, C, "A Computer Simulation of Direct Run-off," in Proc. Intl Conf. on Finite Elements in Water Resources, Princeton University, Princeton, N.J., July 1976, Gray, W. G., and Pinder, G. F., (eds.), Pentech Press, London, 1977, pp. 4.149-4.163.
ONE-DIMENSIONAL STRESS WAVE PROPAGATION
INTRODUCTION 5, 8, and 9, time-dependent problems of heat and fluid flow, mass and overland flow were considered. Although these problems were represented by different mathematical equations, the finite element solutions
In Chapters transport,
in time involved first-order derivatives with respect to time.
Now we consider
a different kind of time-dependent problem which involves second-order time derivatives in the finite element equations.
When a time-dependent force caused by factors such as an impact, blast, and earthquake loading impinges on a medium, it is transmitted through the medium as a (stress) wave. Generally such waves propagate in all the three spatial directions. Under certain circumstances and assumptions, it is possible to idealize the
medium
as one-dimensional.
Consider a homogeneous bar of uniform cross section (Fig. 10-1). A timedependent force Px (t) acting on the bar causes vibrations in the bar, and a (stress)
wave propagates
to
and
fro in the bar.
equation for the one-dimensional case, often is
given by
where a x
is
[1,
The governing
known
as the
differential
wave equation,
2]
the axial stress in the
material of the bar, and u
is
x
direction,
p
is
the mass density of the
the axial displacement. Equation (10- la)
is
a
statement of the dynamic equilibrium at an instant of time and can be 224
One- Dimensional Stress Wave Propagation
Chapter 10
225
(a)
s
=
s
©
= +1
© (0
(b)
Figure 10-1 One-dimensional wave propagation, (a) Bar subjected to time-dependent load, (b) Idealization
and
discretization,
(c)
Generic element.
derived by using Newton's second law. the
force;
acceleration
term p(d 2 u/dt 2 )
and
Px (i)
is
= pii
derivatives,
E
is
the elastic
is
is
the
assumed to be
is
=E^. L dx
(10-2a)
we obtain fo x dx
where
law
Ee L€x
ox first
left-hand side denotes internal
the external force. If the material
linearly elastic, the stress-strain
Taking
The
denotes inertia force, where u
=E dx
modulus and e x
(10-2b)
2
is
the gradient of u or axial strain.
Substitution of Eq. (10-2b) into Eq. (10-la) gives dh4.
dx 2
dt 2
+ PM.
(10-lb)
The quantity */E/p = c x is called the velocity of elastic wave propagation. A number of other physical phenomena such as propagation of waves of sound and vibrations of strings are also governed by Eq. (10-1), which in mathematical terms
FINITE
classified as a hyperbolic
is
equation
[3, 4].
ELEMENT FORMULATION
For the body idealized as one-dimensional, we use and linear approximation as
= N u, + N u = [N]{q], N = s, and s = (x — x u
where N,
1
—
5,
2
x
2
2
t )/l.
line
elements (Fig. 10-1)
(10-3)
:
Step
Derive Element Equations
4.
This step can be achieved by using the variational (Hamilton's) principle and the principle of virtual work [4, 5]; these two are essentially statements of the same phenomenon. Here we shall illustrate the use of the virtual work principle [4], which is somewhat easier to understand for the introductory treatment of the dynamics problem. Equilibrium of the dynamical system at an instant of time requires satisfaction of the virtual
{«x f8{*x
JJJ
work equation
W=
V
[4, 5]
+
{F x Yd{u}dV
JJJ
JJJ
{Px Yd{n}dV,
where S denotes a small virtual change or perturbation and Fx lent body force per unit volume caused by the inertial effect,
Fx The
(10-4a)
V
V
=-P dt
is
the equiva-
(10-4b)
2
quantities required in Eq. (10-4a) can be evaluated as
6x
~
dx
~
(10-5a)
1]
L
I
or {€}
=
[B]{q]
duy dt
du 2 \dt) (10-5b)
[N]{q],
and {d
2
uA
dt
2
2
d u2 [dt 2
= since the
Af-
(10-5c)
[N]{q};
are functions of space coordinates only, the time derivatives
apply only to the nodal displacements. Substitution of
W
r
(JJJ
{e}, {u},
and
{u} into
[BF[C][BJ^){q}
=
-W +
226
Eq. (10-4a) leads to r
(JJJ
{<5qf
JJJ
P[W[^W)W [N]{P^K.
(10-6)
One- Dimensional Stress Wave Propagation
Chapter 10
Since {dq} represents arbitrary virtual changes,
JJJ
[BF[C][B]^K{q}
=
-JJJ
p[W[K]dV{q]
we have
+
{X\ T
JJJ
F*W
(10-7a)
V
V
V
227
or
+
M{q}
We
[m]{q]
= {Q«}.
(10-7b)
we could have obtained these results by using Hamilton's and by properly differentiating the associated variational function with respect to the components of {q} and equating the results to zero. The terms [k] and {Q} have the same meaning as before [Eq. (3-28)]. The additional matrix [m] is called the element mass matrix. For the line element with uniform area A and constant P r at nodes, the matrices are note that
principle
[5]
-1
M = AE
(10-8a) 1
AIP X (Q
{Q(01
(10-8b)
2
and
w-^f'f/ki
s]ds
pAl
(10-8c)
6
The matrix
in Eq. (10-8c)
is
called consistent because
it is
consistent (variational) principle; in other words, the mass
may
Often,
it
Apl
divided equally
among
Because the lumped matrix
On
the
two nodes:
Apl
"1
0"
(10-8(1)
2 is
1
diagonal,
it
can offer computational advan-
the other hand, the consistent matrix can be
mathematical analyses. Detailed consideration of scope of this Step
5.
more accurate is beyond
this aspect
for
the
text.
Assemble Element Equations
The element equations can now be added such
that interelement con-
tinuity of displacements (and accelerations) are ensured at
Then
distributed to
first
m
tages.
is
term on the right-hand side of Eq. (10-7a). be convenient to approximate it as lumped, where the total mass
the nodes consistent with the
is
derived from the
for the three-element discretization (Fig. 10-2) [K] (4
X
+
{r}
4) (4
x
1)
[M] X 4)
(4
=
(f)
(4
x
1)
{R(0}, x 1)
(4
common
nodes.
we have (10-9)
One- Dimensional Stress Wave Propagation
228
L = 30
Chapter 10
cm j|1
kg t
A ® A ® A ®
©
Figure 10-2
Mesh
for one-dimensional
medium.
where [M] is the assemblage mass matrix and the other terms have the same meanings as in Eq. (3-32). Equation (10-9) represents a set of (matrix) partial differential equations and is the result of the discretization of physical space in the first phase. The time dependence is contained in {?} = d 2 {r}/dt 2 and the next phase involves ,
discretization in time in order to find solution in time.
As
in the case
of Eqs. (5-21),
(8-7),
and
(9-9),
Eq. (10-9) represents a time-
dependent phenomenon. The difference is that here we need to consider second derivatives instead of the first derivative in the other three equations.
Time Integration
we assume that the acceleration {r} varies between a time step At from time level / to t + At [Fig. 10-3(a)]. As a result, the velocities will be approximated as quadratic and the displacements as cubic [Figs. 10-3(b) and (c)]. As presented in Refs. [5] and [6], this assumpAs
a simple approximation,
linearly
l
L near approximation
Quadratic
,
/-r
y
n t
1t
+ At
(b)
(a)
t (0 Figure 10-3 Approximations for time integration, (a) Acceleration, (b) Velocity, (c) Displacement.
One- Dimensional Stress Wave Propagation
Chapter 10
229
tion leads to finite difference discretization of Eq. (10-9) as
M'U*
=
(10-10a)
{«},-*,
where [K]
{R},.^
= [K]-^L[M].
=
(10-10b)
- ^[M]({r} -
{R},.*
-
Ar{f} r
r
-^(Ar) 2 {r}
r
(10-10c)
).
starts from time t = 0, {r} {r} and {r) boundary (displacement) and initial (velocity and acceleration) conditions/ Thus the vector {R}a.. for the first time step t = — At = At is known since {R} is known at all time levels. Thus we can solve Eq. (10-10a) at t = At and compute displacements at At as {r}^. Once the {r} Af are known, velocities and accelerations at t = At are found from [5, 6]
Because the time integration
are
known from
l*U
,
= 5
m« =
Wo)
-
is
-
2{f}
^{f}
(10-1 la)
,
^(M* - Wo) - 5 Wo - m*
The procedure can now be continued nAt, where nAt
,
the given
(io-i lb)
for subsequent time steps 2At, 3 At,
the total time period for which the solution
is
.
.
.
,
desired.
and t — At may not be the best procedure for the time integration. We have used it mainly for a simple illustration. Alternative and mathematically superior schemes are available and used for practical problems ([4] and the bibliography).
The assumption of
Example
linear acceleration
t
10-1
Consider the propagation of an (Fig. 10-2).
between
Assume
wave
elastic
in a bar discretized in three
elements
the following properties:
Element length.
10cm;
/ ==
cm :
A
==
E
==
Density,
P
= = 10"
Wave
Cx
Element area. Elastic
modulus.
velocity,
1
;
lOOOkg'cm 2 11
= J—f
These simple properties are chosen only to
;
cm 4
kg'sec :
=
109
cm
illustrate the
;
sec
procedure; they
may
not
necessarily represent field situations.
An
approximate
size
of the time step can be found as
At
=
— =^= 10 c 7
10-6 sec
(10-12)
.
x
This size
is
often called the characteristic time step. Substitution of these
into Eqs. (10-8a)
and (10-8b) and an assemblage
[Eqs. (10-9)
and
numbers
(10-10)] leads to
:
One- Dimensional Stress Wave Propagation
230
~
10 _
x 10~ n
1
[M]
2-1 0-1 2-1 0-1 "2 10 0" x 10 14 10 14
-1
x 100C
1
[K]
0"
-1
1
Chapter 10
>
1.
>
6
1
_0
1
2_
and hence 1
-1
-1
2
-1
-1
2
"
[K]
=
10 2
=
1
-1 2
-1
-1
2
INITIAL
is
zero at
all
The
initial /
=
+
1
4
1
1
2_
10 0" 14 10 14
10 2
(10-13)
1
_0
1_
is
1
2_
that the displacement at the fixed end of
times; that Wl (0,
zero at
4
1
CONDITIONS
The boundary condition (Fig. 10-2)
0"
1
"2
-1
-1
1
_0
0"
_
BOUNDARY AND
+
-1 1_
-1 10 2
6 x 10" x 10 10" 6 x 10~ 6 x 6 11
-1
_
~
"2
0"
node
1
is,
t)
=
(10-14a)
0.
conditions are that displacements, velocities, and acceleration are
0: v(x, 0)
u{x, 0) ii(x,
0)
= = =
0,
(10-14b)
0, 0.
An initial condition defines the state of the body at the start or initiation ing in terms of the displacement and/or
An external
force
is
its
of load-
derivatives.
assumed to be applied
at the free end,
node
4, as
a constant
force
F4 (t) = With
1
(10-14c)
kg.
these conditions, the load vector {R}
is
~2
(R}r + Af
—
[R}f+Af
+
6 x lO" 11 x 10 10" 6 x 10" 6 x 6
1
_0
+
10- 6 {f}, +-i-x
10- 12 {r},).
0"
1
4
1
1
4
1
1
2_
(M,
(10-15)
J
One- Dimensional Stress Wave Propagation
Chapter 10
Using Eqs. (10-13) to (10-15), we can write Eq. (10-10)
10
:
:
i
+
At
=
At as
(10-16)
<
ir 3
12.
.0
/
lw 4
.
Ar
1
1
introduce the boundary condition
=
u { (At) in
=
10 u«r 14 10 w? = 14 1
Now we
t
0"
2 10
at
231
Eq. (10-16), which leads to
T600 600
(10-17) 300.
Solution of Eq. (10-17) by Gaussian elimination (with u
x
prescribed) gives
u' 1
f
\
=
I
(10-18a)
(
U3
uj
1
Ar
Substitution of Eq. (10-1 8a) and the
and
(10-1 lb) allows
[
300 }
initial
computation of
conditions [Eq. (10-14b)] in Eqs. (10-
velocities
and accelerations
at
/
=
a)
At as
\
|o\
dt
(°)
f°l
du 2 dt
du 3
>
=
3
-{
>
-
>
1Q- 6
2<
I
(10-18b)
2
dt
du 4
1
\300
dt
\
I
At
1
1
ol
/ o
/
k
and d 2u
x
\
dt* \
d 2 u2 dt 2
d 2 u.
~
6 )
10" 6 x 10" 6
J=-A 10" 6
)
f
>
-2<
>-(10-18c)
dt 2
d 2 uA
i
300
dt 2
J
Ar
Equation (10-10a) can
now
\
be solved for time level
/
=
/o
2At, 3At, and so on.
We note here that the foregoing numerical calculations are meant only as an illustration of the procedure
and not necessarily
For the latter, one and temporal meshes.
as acceptable solutions.
needs to program the procedure and select
optimum
spatial
;
Example 10-2
Now we
present results from a problem in one-dimensional
solved by
Yamada and Nagai
The
[7].
wave propagation
properties of the problem (Fig. 10-4) are as
follows:
L or
Total length of bar,
Number
of nodes
Number
of elements
500
mm
/
Area of cross section assumed
arbitrarily
1
elasticity,
;
;
Density,
Wave
=
= 51; = 50; = 10mm; = mm 2 £ = 20,000 kg/mm 2 p = 0.008 kg msec 2 /mm 4 c x = 5000 mm/msec; At = ljc x = 0.002 ms.
Length of each element,
Modulus of
h
velocity,
Time increment.
h
=:
500
mm
|
h
i
*-
—
=
1
mm/ms
f
Figure 10-4 One-dimensional wave propagation
[7].
BOUNDARY CONDITIONS t)
=
(A, t)
=
u(h,t)
=
u(0,
du
0.
1
mm/ms.
or
^-(h,t)
t.
Ox
The 10/5000
results in
=
is,
after
analytical solutions [Figs. 10-5(a)
mations, respectively.
mass
sistent
at time / = 0.08 ms (for At = I/c x = 40 time steps, are compared with closed form
terms of particle velocity
0.002 ms), that
It
and
(b)] for consistent
and lumped mass approxi-
can be seen that for the characteristic time step, the con-
yields exact solution,
whereas the lumped mass formulation
is
not that
accurate.
The
size
of the time step can have significant influence on the numerical solu-
shown in Figs. 10-5(c) and (d) for both approximations for At = Again the lumped mass formulation is not close to the analytical solution. On the other hand, the consistent mass formulation shows oscillations and inaccuracies, particularly in the vicinity of the wave front. The aspect of numerical stability and comparison of lumped and consistent masses is wide in scope, and the reader may consult the references and the bibliography. tions.
0.5l/c x
232
This
.
is
— 2.0
Analytical solution
"""Numerical solution
t
ms
= 0.08
1.0
HMXXWJSOOO.
21
11
31
41
51
41
51
41
51
41
51
Node number (a)
Analytical solution
000 Numerical solution
2.0
- 0.08 ms
t
:•• o
1.0
......
••
oo
°
o a
11
21
31
Node number (b)
Analytical solution
000 Numerical solution
2.0
t= 0.08 ms 1.0 <
.a
!
,E
oo
°
1
11
31
21
Node number (c)
Analytical solution ~E
000 Numerical solution
2.0
t
>
J
= 0.08
ms
1.0
>
o
I °-
o 11
1
21
31
Node number (d)
Figure 10-5 Results for wave propagation in bar
mass: At At = 0.5
= I
l'c x
cx
.
.
(d)
[7]. (a)
Lumped mass: At = I c x (c) Lumped mass: At = 0.5 Hc x
(b)
.
Consistent
Consistent mass:
.
233
Damping
Most natural systems possess damping. It can be easily included dynamic equations; the resulting element equations will then be
-
[k]{qj
where
[c]
material ([4]:
damping
the
is
is
+
[c]{q}
[m]{q}
=
(10-7c)
[Q(t)\.
matrix. Determination of
and
a very important topic
discussed
is
in the
damping properties of various publications
in
the bibliography).
PROBLEMS 10-1.
Compute
mass matrix for the quadratic approxima-
the element consistent
tions [Eq. (3-40)]:
u
where
=
\L(L -
1
-
)«,
( 1
- L2
)
WL
-
uQ
-
1
)w 2 ,
denotes the middle node.
Solution:
Apir\
[m]
4
UL
J.,
-
L(\
Apl
10-2.
Compute
1)1 2 )
\[L(L
-
(1-L
1)
L{L-\)]dL
2 )
1)J
8-4"
16 8
64
L-4
8
120
-
-L
(1
8 16.
the element consistent
mass matrix for cubic (Hermitian function)
u:
u
=
(1
-
3s 2
-5 2 (3
+
2s 2 )u
l
+
-
ls(s
l) 2
^ OX
-2s)u 2 +ls 2(s
- 1)^-
Solution:
[m]
10-3. 10-4.
22
54
22
4
13
-3
-22
Apl 420
54
13
156
-13
-3
-22
Perform computations for three time steps
in
4.
Example
10-1.
Consider a single-element discretization which can be replaced by a spring-
mass system.
It is
subjected to a forcing function
Specialize Eq. (10-10a) for this single element
time steps.
234
-13"
156
Assume
properties as in
Example
Px
as
shown
in Fig. 10-6.
and obtain solutions for two
10-1
.
Compare numerical
results
One- Dimensional Stress Wave Propagation
Chapter 10
i
235
P.tlii m
k
Figure 10-6
with the closed form solution from U{t)
where p 1
=
k/m, where k and
respectively. 10-5.
The
initial
=p y(l -
m
cos pt),
are the stiffness
conditions are: u(x, 0)
=
and mass of the
it(x,
0)
=
spring,
0.
Prepare a computer problem based on Eq. (10-10) and solve Probs. 10-3 and 10-4.
REFERENCES [1]
A
Love, A. E. H.,
Treatise on Mathematical Theory of Elasticity, Dover,
New
York, 1944. [2]
[3]
Timoshenko, York, 1970.
Carnahan, Wiley,
[4]
New
Desai, C.
S.,
B.,
and Goodier,
Luther, H.
[6]
N., Theory of Elasticity, McGraw-Hill,
and Wilkes,
J.
New
O., Applied Numerical Methods,
York, 1969.
and Christian,
S.,
Engineering, McGraw-Hill, [5]
A.,
J.
Numerical Methods York, 1977.
J. T., (eds.),
New
Desai, C. S., and Abel, J. F., Introduction Nostrand Reinhold, New York, 1972.
to the Finite
in
Geotechnical
Element Method, Van
Wilson, E. L., "A Computer Program for the Dynamic Stress Analysis of Underground Structures," Report No. 68-1, University of California, Berkeley, Calif., 1968.
[7]
Yamada,
Y.,
and Nagai,
Y., "Analysis of
One-Dimensional Stress Wave by
the Finite Element Method," Seisan Kenkyu, Journal of Industrial Science,
Univ. of Tokyo, Japan. Vol. 23, No.
5,
May
1971, pp. 186-189.
BIBLIOGRAPHY Archer, Div.
J. S.,
"Consistent Mass Matrix for Distributed Mass Systems,"
ASCE,
Vol. 39,
No. ST4, Aug.
Clough, R. W., "Analysis of
Structural Vibrations
and Dynamic Response," in Yamada, Y., and
Proc. U.S. Japan Seminar, Tokyo, 1969 (Gallagher, R. H.,
Oden,
J.
T. eds.), University of
/. Struct.
1963.
Alabama
Press, Huntsville, 1971.
One- Dimensional Stress Wave Propagation
236
Chapter 10
Desai, C. S. (ed.), Proc. Symp. on Appl. of Finite Element Methods Waterways Expt. Station, Vicksburg, Miss., 1972.
Desai, C. S.
(ed.),
Proc. Second Intl Conf. on
burg, Va., 1976, Vols.
Desai, C.
S.,
and Lytton, R.
Parabolic Equation," Idriss,
I.
I, II, III,
M., Seed, H.
B.,
Num. Methods
Num. Methods
Geomech., Blacks-
of Finite Element Schemes for
Eng., Vol. 9, 1975, pp. 721-726.
Seriff, N., "Seismic
Finite Elements," /. Geotech. Eng. Div.
in
Geotech. Eng.,
1976.
L., "Stability Criteria
Int. J.
and
ASCE,
in
Response by Variable Damping Vol. 100, No. GT1, 1974, pp.
ASCE,
1-13.
Kreig, R. D., and Key, S. W., "Comparison of Finite Element and Finite Difference Methods," Proc. ONR Symp. Num. Methods Struct. Mech., University of
Illinois, Sept. 1971, (Fenres, S.J.,
brich,
W.C.
eds.)
Academic
Press,
Perrone, N., Robinson, A.R., and Schno-
New
York, 1973.
Newmark, N. M., "A Method of Computation ASCE, Vol. 85, No. EM3, 1959, pp. 67-94.
of Structural Dynamics," Proc.
Nickell, R. E., "Direct Integration Methods in Structural Dynamics," Mech. Div. ASCE, Vol. 99, No. EM2, April 1973.
/.
Eng.
TORSION
INTRODUCTION Until
now we have
considered problems that can be idealized as one-dimen-
They involved only
sional.
unknown
only one
at
elements and, except for
line
beam
bending, had
each node. Next, we advance to problems that can be
idealized as two-dimensional.
To
We
start with, let
choose torsion
unknown
us consider a problem in stress-deformation analysis.
first
problems (Chapter the torsion problem.
field
We
because
at each point (node). 12), are
shall illustrate here a
it
A
involves only one degree of freedom or
number of other problems,
often called
governed by equations similar to those for
number of formulation procedures
:
displace-
and hybrid and mixed, which are based on the energy principles. For all these formulations, we approximate the behavior of a bar [Fig. 11 -1(a)] subjected to torsion by considering essentially the ment,
stress or equilibrium,
behavior of the cross section of the bar [Fig. 11 -1(b)]. In the semi-inverse
method of Saint- Venant,
it is
assumed that the twisting of the bar
is
composed
of the rotations of the cross sections of the bar as in the case of a circular bar
and of warping of the cross sections
[1, 2].
sections; the latter
As a consequence, no normal
is
constant for
stress exists
all
cross
between the longi-
is no distortion of the planes of cross and hence the strain components e x €yi and y xy vanish, and only pure shear components y x2 and y y2 will remain.
tudinal fibers of the bar. Also there section,
,
237
238
Torsion
Chapter 11
y, v
(b)
(a)
Warping
(c)
Figure 11-1 Torsion of bar. (a) Bar subjected to torsion, (b) Cross section of bar. (c) Warping.
For the two-dimensional approximation, we
shall consider discretization
involving triangular elements (Fig. 11-2); quadrilateral elements are treated in
Chapter
12.
For the triangular
known can be adopted,
case, a linear
approximation for the un-
while for the quadrilateral isoparametric elements,
a bilinear approximation can be chosen (Chapter first
two
12).
This constitutes the
steps in the finite element formulation.
TRIANGULAR FINITE ELEMENT Before considering subsequent steps, ties
we
shall first detail
some of the proper-
of the triangular element.
As
discussed in Chapter
coordinate systems for
3, it is
finite
advantageous to use the concept of local
element formulations. For the triangular
ele-
ment, the local or area coordinates are often defined in terms of component
Chapter 11
239
Torsion
(x 3 .y3
(0,0,
P(x, y)
©
)
(L,,L 2 .L3
1)
u=o
U =0
Figure 11-2 Triangular element.
areas,
A l9 A 2
,
and A 3
local coordinates
(Fig.
L l9 L
2
Then
11-2).
L
t
-\-
L2
-\-
L3 =
nondimensional form, the
3
i
Since
in the
and L are defined as
,
1,
=
(11-1)
1,2,3.
there are only
two independent
corresponding to the two global coordinates x and
between the global and the local coordinates
y.
local coordinates
The
relationship
given by
is
(11 -2a)
y
=L
3 x
y
x
+L
2
y
2
+L
3
y3
= 2= Ajn i
* where [N]
is
= m*&
(ll-2b)
the matrix of interpolation functions
the local coordinates
L
{
;
{x n } T is
}
1
N
it
which are the same as
these are plotted in Fig. 11-3.
=
[x M
x2
x3
y
%
y2
y3
]
the vector of nodal coordinates. Equation (11-2) represents linear variations
for coordinates at
any point P(x,
y).
Figure 11-3 Distribution of interpolation functions
N
t
,
i
=
1, 2, 3.
:
240
Chapter 11
Torsion
The
inverse relation corresponding to Eq. (11-2)
is
N =L =
2^(^23
+
M+
a iy)>
N =
2 2^( ^3i
+
b2x
+
a 2 y),
J_(2A l2
+
b3x
+
a 3 y),
t
x
2
=
L2
AT 3
2A
(ll-3a)
or ~
AY
"2^23
6,
a
2^3
62
flj
'
1
x
1
ATa
A ,. 7
where 2A
=a
3
b2
of the triangle;
and the
origin
—
A 23
_2^ 12
= ^63 —
a2b3 is
1
X
•
A3
«3.
#3^
=a
'
(ll-3b)
.*. 2
the area of the triangle
b
l
—
a b2 x
whose
and so on; and a and b are given {
t
;
A
is
the total area
vertices are
nodes
2, 3
as differences between
various nodal coordinates
a
x
a2 a3
=x - x =x -x =x - x u 2
3
3
x
is
b3
2
u
x
b2
,
The unknown, represented by a
where [N]
b
,
=y - y =y -y =y - y 2
3
3
ly
t
2
general symbol
= A^w, + N = [N]{q},
2
u2
+N
3
,
(H-3c)
.
w,
can be expressed as
u3 (ll-4a)
the matrix of interpolation functions and yields linear variation
of u over the triangle. Since [N] in Eqs. (11-2) and (11-4) call this
element an isoparametric triangular element.
FINITE
ELEMENT FORMULATION
is
the same,
we can
Displacement Approach First
torsion
we
consider the displacement approach. According to Saint- Venant
[1, 2], in
body forces, the differential (Laplace) equahomogeneous and isotropic bar can be expressed
the absence of
tion governing torsion in a
as
dx 1
+
dy 2
~
(11-5) '
where x and y are the global coordinates in the plane of cross section of the bar (Fig. 11-1). The warping function, y/, is related to the displacement w in the z direction, and is assumed to be constant along the length of the bar
,
Chapter 11
241
Torsion
and function only of x and y
(Fig. 11-1):
w=6y/(x,y) where
is
moment,
M
(ll-6a)
>
the twist of the bar per unit length under the applied twisting [Fig.
t
1
l-l(a)].
The warping function denotes a measure of warpThe other two displacement
ing of the cross sections of the bar [Fig. 11 -1(c)].
components, u and
v,
are given by
= —6zy, = Qzx.
u v
The boundary condition along the
(11 -6b)
surface of the bar
is
given by
(£->)£-(£ + *)£=° where
s
is
measured along the surface or boundary. Equation (11-7) indicates
that the shear stress normal to the boundary, r„ z Step
(11-7)
,
is 0.
Gradient-Unknown Relation and Constitutive Law
3.
By using Eq.
the gradient or strain-warping function relation
(11-6),
is [2]
'dw
dx [«J
where
{e}
=
[y xz
,
du
dw y
yz ]
+ is
(% - >)
y«
dz >
,dy r
+
—
<
dv dz
>
—
(11-8)
<
»©
y„
+ *)l
the vector of (shear) strain components.
In the general three-dimensional problem, there are six nonzero
compo-
However, for the foregoing two-dimensional idealization, there are only two nonzero shear-stress components, x xz = r zx and r yz = tzy corresponding to the two shear-strain components. The constitutive rela-
nents of stress
[2].
,
tion can be expressed as
r xi
Gy X2
G
Gy yz
.0
(11-9)
where
{
r
=
[t xz
r yz ]
is
GJ|
the vector of (shear) stress
components and
G
is
the shear modulus. Step
4.
Derive Element Equations
For the displacement approach, the potential energy function correspondis given by [3, 4]
ing to Eq. (11-5)
where h
is
the length of the bar, and
A
denotes area.
242
Chapter 11
Torsion
According to the general expression [Eq. y/,
(11 -4a)] for linear variation
of
we have y/(x, y)
= N y/ + N = [N r ]{q r 1
1
2 y/ 2
+N
3
y/ 3
(ll-4b)
},
where
y/{x, y) is
the warping function at any point (x, y) in the triangular
element and {q^} r
=
[y/ i
y/ 2
y/ 3 ] is
the vector of nodal values of the warp-
ing function.
Requirements for the Approximation Function
As
discussed in Chapter
the approximation function in Eq. (ll-4b)
3,
should satisfy continuity within an element, interelement compatibility, and completeness. Equation (11 -4b)
form of
linear
y/
which
=
+
a,
the transformed version of the polynomial y/
as
a2x
+
a 3 >>,
(ll-4c)
continuous within the element.
is
The
is
approximation for
highest order of derivative in the energy function [Eq. (11-10)]
is 1;
hence the approximation function should provide interelement compatibility
— 1=0, that
warping functions. The linear approximawarping functions across interelement boundaries. This is illustrated in Fig. 1 1-4. Since y/\ — y/\ and y/\ yt\ and since only one straight line can pass through two points, the variations of warping function along the edge AA of element 1 and BB of element 2 must coincide. That is, at the common boundary, compatibility of y/ is
up
to order
1
is,
for
tion indeed provides for compatibility of
=
Figure 11-4 Interelement compatibility.
^2 = warping
at
node 2 of element
1,
and so on
Chapter 11
Torsion
Note
fulfilled.
243
that for linear approximation this does not necessarily imply
interelement compatibility of higher orders, such as the
first
derivatives.
body motion (term a ) and constant state of strains or gradients of y/ (terms a : .v and a : v\ it is complete. Notice that in this problem, there exist two such states: dy/ dx and dy/ dy. Furthermore, it contains all terms up to the order Since the approximation function in Eq. (ll-4c) provides for rigid :
n
=
polynomial expansion for the two-dimensional problem. This
in the
1
idea can be explained by using the polynomial expansion represented (by Pascal's triansle) as follows:
order or degree
(ri)
of polynomial
n=0 x
X
2
To
y
x zy z
=
1
3
?
*r
approximation
see that for the linear
and including n
to
1
xy 2
x 2y
x~
Wc can
y
xzy
x3
up
XV
constant
1
Linear
2
Quadratic
3
Cubic
4
Quartic
in Eq. (1 l-4c), all the three
terms
are provided.
find the values of the derivatives in Eq. (11-8),
we use
the following
rule of differentiation:
_ ^ sbL — " dx± dx dx dX
§2Ll
du_
dy
_
~
dX-
dy/
dy
dX
_ dX
:
dy
:
§3L
dX
dx
:
_
$2i±
dy/
_ dX
dX
dy
X
For example, by differentiation of and (ll-4b). we obtain
z
(/
=
3
1, 2,
m
^L dX
dx
2
iiai Ui-na;
3
dy/
(1 1-1
dN
lb)
3
3)
and
y/
with respect to x
in Eqs. (11-3)
£ = Jx.~T4 i2A: -
>
~
{2a ^ jx ta d_
dx J2(2A ::
= Since
h*>
A Z3
,
b X --
~ " V)
^*^ ~ N** ~ dk (Ni¥i ~
~
b*x
_
-
b,x
- a.yi^iX-U: - X
azy)1
+&> + &>
etc.,
y/ t
(i
=
:
y/ :
_
NiWi)
-X
3
y/ 3 )
(1M2a)
are constants, they
differentiation. Since the
Nz¥z
*****
do not contribute to the results after 3) are values at the nodes and not
1. 2.
244
Chapter 11
Torsion
functions of
N
t
and
they also do not contribute to the differentiation.
(x, y),
Similarly, dy/
Y1 -ar.+a* 2A Y2 fy~2A '
Equations (11-1 2a) and
(1 1-1
E
¥
2b) are combined in matrix notation as -
=
>
dy/
(ll-12b)
2A YZ
'
l
2A
~b,
b2
b3
J*\
a2
a3_
¥i
Vs.
dy,
(ll-12c)
Here
[B]
is
the strain-warping function transformation matrix.
Hence Eq.
(11-8)
becomes
-By
OX
to =
(ll-13a) 1
Ox
dy\
By (ll-13b)
Ox
Now n,
Eq. (11-10) can be expanded as
Gh6 2
pas' -*+']+ [©•+*+*]* (ll-14a) dy/
GhO 1 [[ 2
+b
JJ
dx
[dy/
dyf]
Idy
dy] dy
+
2
dy/
JSx
dy/
dy
_
w (ll-14b)
\dxdy.
x]
Substitution from Eqs. (11-12) and (11-13) into Eq.
n,
Ghd
r {q,} [B]^[B]{q,}
(ff
+
[y m
xm ]
)dxdy,
+
(1
l-14b) gives t
-y>
2{qwY\Bf
(11-15)
:
Chapter 11
= (*j +
where x m
mean
Torsion
values. It
+
x2
= {y + y + ^
x 3 )/3 and y m
2
x
make
not necessary to
is
245
this
3 )/3
assumed
are the
assumption, however.
We
can
x and y from Eq. (11-2) and pursue the derivations, which will be somewhat more involved. Now we invoke the principle of stationary (minimum) potential energy hence substitute for
;
dU p
=
*n,
dn p
o
(11-16)
dy/ 2
dUp dy/ 3
which leads to
GhO 2
[B]
Note that the
=
[B]dxdy[%}
term
last
Equation
results.
T
JJ
is
JJ
[Bfi
in IT^, being constant,
(11 -17a)
_M
dxdy.
(1
l-17a)
does not contribute to the
can be written in matrix form as
= (
fcjfar}
where [kj
GhO 1
(ll-17b)
the element stiffness matrix:
[kj
=
Ghd 1
[B]
JJ
T
[B]dxdy
(ll-17c)
A
and {Q^}
is
the equivalent nodal load vector
m= They can be evaluated
Ghe>\\{w\_
t
6,-
(ll-17d)
\dxdy.
as follows ~b t
a,
o2
a2
6, L^3
Since a and
ym
a "3.
b
t
LM
1
b,
dxdy.
are constants, the integral equals the area [k r ]
=
(ll-18a)
;
hence
Gh0 2 A[B] T [B]
or
G/z0
fc
:
6|
4^ sym.
+
a\
b2b 3 b\
+ +
a 2 «3 a\
(ll-18b)
:
:
246
Torsion
Chapter 11
and b
>
0\
b2
a2
\
GhO-
[Q,
jy
k
y,
dxdy
— x,
(ll-19a)
a3. a
Gh0 2 A 2A
y,
a
b2
x,
b3
b,y
Gh6
Tl
:
b 2 y* [b 3 y r
-
a xm l
a 2 xm
a,x m
,
We note that the term GhO
2
is
(ll-19b)
common on both
lem
governed by the Laplace equation [Eq.
is
Step
and can and the prob-
sides of Eq. (1 l-17a)
be deleted, because of the assumption that the bar
is
isotropic
(11-5)].
Assembly
5.
Example
Torsion of Square Bar:
11-1.
Warping Function Approach Equations
(1
l-18b) and
(1
l-19b) can
now
be used to generate stiffness matrices and
For instance, Fig. 11 -5(a) shows a square bar, 2 cm x 2 cm, discretized in four elements. Figure 11 -5(b) shows the four elements with their local and global node numbers. For this example assume the following properties load vectors for
all
the elements in a discretized body.
G = h
6
The area of each illustrate the
The
A =
triangle
1
= =
cm
1
N/cm 2
1
cm,
1
rad/cm.
2 .
,
These properties are chosen
in
order to
procedures; they do not necessarily refer to practical problems.
quantities a t
and b for the four elements are as follows {
Element 1:
— x2 = - 1 = -1 2 = -1, b = y2 - y 3 = = -1, a2 62=^3-^1=1-0 = 1, xi — x = - = 0, = 2, b = vi - y 2 = 03 .= x 2 — Xi =2 x m = (0 + 2 + l)/3 :l,^=(0+0 + l)/3=|. a
x
= =
x3
1
t
1
3
(ll-20a)
3
and
Element 2:
= a2
1
a3
and x m
=
(1
+
+
0)/3
h =0 b2 = 1 - 2 1
1,
= -1, = 0,
= i, ym =
^,3=2 (2
+
+
— l)/3
(11 -20b)
Chapter 11
Torsion
0~(O,
247
(2,0)
0)
Global node
(a)
1
_®
®.
A
Local node
dement
(b)
Figure 11-5 Torsion of square bar. (a) Square bar and mesh, (b)
Elements with local and global nodes.
Element 3: a,
a2 tf 3
and * m
=
(2
+
2
= 1 -2 = -1, = 2- 1 =1, =2-2= + l)/3 = f,
m
=
b2
= 1 -0 = 1,
2
1
=
1,
(ll-20c)
= - 2 = -2, + 2 + l)/3 = 1. 63
0, >>
-
bi
=
(0
Element 4: fll
a2 ^3
and x m
=
(2
+
= i -0= =2 - = 1
1,
6j
1,
62
=2- = = -2= -1, 1
1,
1
= - 2 = -2, 6 = 2 - 2 = 0, + l)/3 = 1, ym = (2 + 2 + l)/3 = f. 3
(ll-20d)
248
Torsion
Chapter 11
Substitution of these values into Eqs. (ll-18b) and (ll-19b) gives the following:
Element
1:
Global
(ll-21a)
Element 2: Global Local 1
I
3
1
^3
1
^1
2
y\ = vi
2
1
2
5
-2
L-2
yi-
(ll-21b)
Element 3:
^2
Global Local i
I
2
1
^1 2
4
(ll-21c)
L-2
5
Element 4: Global Local i 1
(ll-21d) 3
-2
-2
we have shown the relations between local and global node numbers. For the particular node numbering chosen in this example the coefficients of the element matrices are the same. In fact, even if a different numbering is used, if the elements are of equal dimensions, [k ¥ ] can be generated once it is obtained for one element. The assembly procedure is carried out by observing the fact that the values of y/ at common nodes are compatible. Thus, by adding the local coefficients in
In Eq. (11-21) the superscript denotes an element, and
appropriate locations in the global relation,
we
obtain
-
,
249
7br^ K7/2
Chapter 11
>\
Global
2
4
3
5
1
"
1
(2
+
(-2 -
(-2
2)
(2
2
+
2)
1
3
+
(2
4
(2
_(-2-
2)
(
-2 -
+
- 2) + 44
(-2
2)
:-2-2) (-2-2)
2)
^2
2)
(-2-2)
2)
4 5
2)
(4
+
4
^4 4)J
|-i=o
'
f
+ f=4
1
(ll-22a)
»
-f + f = + 2-2 +
,-2
2
=
or '
-4" Vii
4
-4 -4 -4
4 4 4
_-4
-4
-4
-4
1
16
V2 1
<^3
-16
(11 -22b)
>
^4
16_ Vs,
0,
or :K]{r]
where
[K]
is
functions,
=
{R },
(11 -22c)
the assemblage stiffness matrix, {r}
and [R]
is
is
the vector of global nodal warping
the assemblage nodal forcing parameter vector.
Since the values of the warping function are relative, the boundary conditions
can be introduced by assigning datum value of y/ to one of the nodes. Then we obtain relative values of nodal warping functions. For example, assume y/^
Then
the modified equations are obtained by deleting the
Eq.(ll-22b) (see Chapter "
-4"
>
4
-4
-14
=
= cm
Step
6.
/rad (given),
y/ 2
-8 (ll-22d)
<
-4 16.
y/si
Solution of these equations yields the primary y/ l
0.
in
8'
>2
-4
4
2
=
row and column
3):
4
_-4
first
shall
=2,
y/ 3
=
unknowns -2,
as
y/ 4 =
o,
y/s
Secondary Quantities
The secondary quantities can be the shear stresses and twisting moment. Equation (11-9) can be used to find the shear stresses in the four elements:
250
Torsion
Element
Chapter 11
1: dy/
*xz >
-y m
dx
=
G9<
>
-
G0<
dy/
T„
*m
idyl
X
1
&1V1
+
b 2 y/ 2
a,y/,
+
a2^
+
6 3 ^:
1
X
1
(11-23)
1
^V:
:
Substitution from Eq. (ll-20a) then gives \? X2
0+1X2 -1x0-1x2 + 2
if-lX
\
x
-i
x
HJH1 Element 2:
Element 3: Txi T...
Element 4:
The shear
stresses are plotted in Fig. 11-6.
TWISTING
MOMENT
The expression
Mt
=
for twisting
G6
moment
y JI i~ fc
~
x
A
Often
M /G6 t
is
for the twisting
t
w ^ x2Jr
called the torsional constant
moment
bar
in the
is
given by
y2 dxdy
and
'
[2, 3, 4]
(11_24a)
)
is
used to express solution
in the bar.
we assume x m and y m as average constant values associated with the two terms in the integrand and since dy/jdx and dy/jdy are constant for
If first
M
Chapter 11
Torsion
251
the linear approximation, the integral in Eq. (ll-24a) can be approximated as
§ - |, *. - |, [-y.A$l + XmA(gj
m
+ *. + ,„)}
(,1.24b)
where [Eq. (ll-34a)]
xm
=
-7-(x\
+
?»
=
-g-C^?
+ y\ + ^3 + 7l72 + ^2^3 + ^1^
*1
+
x\
+
XjX-2
+
*2* 3
+
Xi* 3 )
and
M
is
the total
number of
elements,
the contribution to twisting
m
3 ).
denotes an element, and
M
tm
denotes
moment by element m.
V'V Stresses
shown
at
centroids of elements
2
cm
Figure 11-6 Plots of computed shear stresses: warping function
approach.
The twisting
total twisting
moment can now be found
moments. Substitute
for dy//dx
as the
and dy//dy from Eq.
sum of element (11-12),
coordinates of the nodes of each element lead to the following:
Element 1: dy/
dx dy/
i(-l
x0+lx2 + 0x0)=l,
^(-1 x
0-
^=-4x1 +
1
1
x
2
+
x (-1)
2
x 0)- -1
+ |= -f
and the
252
Chapter 11
Torsion
Element 2:
^L
=
-1x0-2x0] =
J[_l x (-2)
^ = }[lx (-2) M
tl
= -\
X
1
-
x
x
0]
=
1,
-1,
l+Jx(-l) + |=~t + f.
Element 3:
^ = J(lx2+lxO-2xO)=l, = -1, x + x ^ = J(-i x 2 4-
0)
1
6
Element 4:
*£ = jS!
Afr«
iKl
JI1
-1
X
X (-2)
= i[lxO-Mx = — 47 _
1
x
_
24 6
(-2)
= —1
+ X 0] =
1,
-
2
-1,
-1-
24.
3
x
0]
=
6
Therefore
M
t
= M + M„ + A/„ + Mu = _^(4 + 4 + 8 + 8) -f i(8 + = _^ j_ M = _g + 10.6666 = 2.6666 N-cm. (l
8
+
24
+
24)
COMPARISONS OF NUMERICAL PREDICTIONS AND CLOSED FORM SOLUTIONS Twisting
Moment
Based on the stress function approach presented subsequently, a closed form solution for the twisting moment for a square bar is given by [2]
M, ~0.1 406(70(2 a) 4
(11-25)
:
>
253
Torsion
Chapter 11
or
Mi
=
where a
~ 0.1406(2*)*,
is
^-0.1406(2 x
~ 0.1406 x ~ 2.250 cm The value computed from the error
form value of the
the half width of the bar. Hence, the closed
torsional constant
l)
4
16 4 .
the finite element procedure
2.666; hence
is
is
em)r
inn = 2.2500-2.6666 vXl0 ° 12500
~
—18.5%.
Shear Stresses
From
the stress function approach (see next section), the closed
form
values of the components of shear stresses, Eq. (11 -28a) below, can be
obtained from the following expression for closed form solution for stress function in a rectangular bar
32G0c'-
9*
3
7l
rt
d(p
- dy'
t
=1,3,5,..
\6G6a n1
[2]
^lL
1
j
_
n3
±
„=]1.37s,...
cos h(nny>-2a) -\
CQS
cos hinnb 2d)\
sin^r2.] ^(-l)^ n h{nnb 2d) 1
0S
cos
(nnx\ \ 2a /
W \2a
(11 -26a)
)
(11 -26b)
_dg> _ \6G0a
r*
(7,X
7T^
1
!Lzl1
j
n=l,3,5,...
_
cos
h{nnyad)
~\
cos h(nnb;2a)\
«
/nnx\ ^ \2a
)
(ll-26c)
where
a:
and y are measured from the center point of the
bar,
and a and b
denote half the dimensions of the cross section of the bar, respectively. In the finite element analysis, we assumed linear approximation for the warping function; hence, the distribution of shear stresses is constant at every
point in the element. This element can, therefore, be called the constant strain or stress triangle (CST). If
we had used
a higher-order approximation,
shear stresses would vary from point to point in the element. In view of the
constant values for the shear stresses, 11-6, to attach
them
it is
often customary, as
at the centroids of the elements.
shown
in Fig.
For the purpose of com-
254
Chapter 11
Torsion
we compute
parison and error analysis below,
form or exact must be understood that this is only an approximation, since the computed shear stresses are constant over an element as a part of the formulation, and are attached at the closed
values of shear stresses at the centroid of an element.
It
the centroid only for convenience.
The
is important and wide in scope. Our purpose by presenting comparisons between computed and closed form solutions for some of the problems in this chapter. In making these comparisons we have considered only typical elements and nodes. For Example 11-1, let us choose element 3. The centroidal coordinates of is
subject of error analysis
to introduce the topic
element are
this
x
= 0.667
Then use of Eqs. (ll-26b) and
and
y
= 0.000.
(ll-26c) gives closed
form values of shear
stresses as tJ,
- 0.000 TV/cm
2 ;
= 0.775 N/cm
r*
2 .
These and the subsequent closed form solutions are obtained by using n in Eq. (11-26). The errors in the two components are then given by
=
25
error in xxx
and error
in z yz
= 0.000 - 0.000 = 0.00%, Tyz — r Tyz - r* x
„
_
0.775
-
10
0.667
0.775
=
14.00%.
Comment: The foregoing comparisons show
that despite linear approxi-
mation, and the rather crude mesh, the computed results yield
realistic solu-
same order of magnitude. Here we used a coarse mesh mainly to illustrate hand calculations. In general, however, one should use a finer mesh, and the solution will improve significantly. For instance, with a finer mesh, say with 200 elements, the computed value of the torsional constant for the square bar would involve an error of less than 1 .0 %. tions of the
STRESS APPROACH The
torsion problem can be expressed by using the concept of stress function,
approach represents the same phenomenon as represented by the warping function, but it is an alternative or dual representation for
[1-7]; this
Chapter 11
Torsion
The governing
torsion.
geneous medium,
in
255
equation for a linear, isotropic, and homo-
differential
terms of (p, according to Prandtl
[1, 2], is
V^-W— -^
Q
where
is
26
(11-27)
the equivalent forcing parameter, which can be a prescribed
known
function of
(x, y).
The two nonzero shear
terms of
stresses in
=
>
Ty <
.
between the
(ll-28a)
>
dcp
TL yz relation
are
d(p
1 X2
The
cp
5iJ
stress function
and warping function
is
(ll-28b)
*-«&+«) The energy function associated with Eq.
(11-27)
is
the complementary
energy, expressed as (see Chapter 3)
w =
n =u + c
c
u -
c
c
w,
or 2
n
«=JJi[® + (g)>^-JJe^'
*"•*
or
dx
n = 2b
dxdy
'
Here tial
£/c is
the complementary strain energy,
of external loads, and
The unknown T
[q s ]
=
[cp
cp
x
W
is
the
work of
for a triangular element
= Nl9l
(p
where
-
IT (20)prfjcrf>\
(11 -29b)
dcp
cp 2
cp 3 ].
-N The
-N
2 cp 2
e
is
is
the complementary poten-
=
now
expressed as
[N s ]{q s ],
gradient-^? relation
(ll-4d)
is
dcp'
gj {g}
2
W
external loads.
>
—
dx [B]{q,!
<
(11-30)
dcp
g,
Substitution of Eqs.
n = c
(1
l-4d)
idyl
and
(1 1-30)
£ JJ {q s y[BY[D][B]{q s }dxdy
-
into Eq. (1 1-29) yields
jj (20)[NJ{q,}^,
(ll-31a)
:
256
Torsion
Chapter 11
where 1
=
P>]
1
is
the strain-stress matrix.
By invoking
the principle of stationary
(minimum)
complementary energy, we have f'
sn c =
n «
o
(ll-31b)
*S,-0. »3
or
JJ
=
[BY[D}[B]dxdy{q,}
(20)[N]Wy,
JJ
(ll-31c)
or
M{qJ =
(H-31d)
{Q,},
where [kj is the element property or flexibility matrix and [QJ is the ment nodal forcing parameter vector. They are evaluated as follows
=
[kj
=
^jjlB7lB]dxdy
+
~b\
x
b\
4GA
^[BY[B]
+ +
b b2
a\
1
ele-
+a a b + a a bl + al
a a2
b b3
x
3
a\
b2
2
3
l
x
sym.
3
(11-32)
and IN,
«U-2»JJ N The
integrations in Eq.
formula
(1
1-33)
2 }
dxdy
(11-33)
.
can be evaluated
in closed
form by using the
[4]
s
a!jg!y!
N\N{N\dxdy (a
+£+y+
2A,
(11 -34a)
2)!
A
where
!
denotes factorial and
|[ N.dxdy
=
IT
!
=
1
.
For example,
N\N\N\ dxdy 1!0!0!
(1+0 +
+
2A 2)!
= 3
2A X 2~
A_ 3
(ll-34b)
Chapter 11
257
Torsion
Hence
{
26A
(ll-34c)
3
which implies that the applied forcing function
is
distributed equally at the
three nodes.
Example
11-2.
Torsion of Square Bar: Stress Function
We now The
Approach
consider an example of torsion of a square bar 4
properties
G and 9
are as before.
Our
cm x
4
cm
(Fig. 11-7).
idea in choosing this and subsequent bars
of different cross sections for different approaches
is
to illustrate different discretiza-
by special characteristics such as symmetry of the cross sections. In view of symmetry in the distribution of the stress function, we can consider and discretize only one-quarter of the bar. In fact, as is done in the next example, only an eighth of the bar can be discretized. The finite element mesh for the quarter of the bar is shown in Fig. 11-7, and it is identical to the mesh used in Example 11-1. Because the coordinates and the geometry of this mesh are the same as before, the tions governed
1
©
(0,2)
(2,2)
\a/
©
a\©a
/ A \ ©
L_
(0, 0)
L
(2, 0)
©
_l
4< ;m
Figure 11-7 Torsion of square bar: stress function approach.
'"
258
Chapter 11
Torsion
element
stiffness
matrices are essentially similar.
The element load vector
is,
how-
by Eq. (ll-34c). Assembly of the four element matrices and
ever, different, given
load vectors lead to the global equations as '2'
-41 9i -4 r
4
-4
-4
2
2
-4
16_
2
2
.
.4,
(ll-35a)
or
=
[K]{r]
[R]
BOUNDARY CONDITIONS According to Prandtl torsion, the boundary condition
=
(p
If
we choose
(p
=
we have
0,
is
or constant along the boundary.
(ll-36a)
(Fig. 11-7).
=
=
<^4
=
(ll-36b)
0.
Equation (ll-35a) is now modified by deleting the rows and columns corresponding to
4
4
L-4
\(p
16Jl^ 5
(ll-35b)
U
3
'
Solution by Gaussian elimination gives
?l
=| =
=
*
=
=
2.66667 TV/cm, 1.33334,
and ?2 Step
6.
*
0.
Secondary Quantities
SHEAR STRESSES The [B]
stresses
can
now
be computed using the available values of
(p.
Use of the
matrix in Eq. (ll-12c) in Eq. (11-30) gives
dp dx dcp_
J_P>i 2A i_fli
b2
b{
a2
a2
(11-37) .
[dy Substitution of at and b t from Eqs. (11-20) leads to stresses in the four elements:
Element
1: dip
dx dip
dy.
r4 |J"-1
l
o-
h
- M
(H N/cm
2 .
259
Torsion
Chapter 11
we have
Therefore, from Eq. (ll-28a),
dcp
dy >
N cm
2 .
d(p_
Txi
Element
dx
2.
N 'cm
:
Element 3:
N cm
i:i Element
: .
4.
eh>In view of the fact that the stress function [Eq.
and
(
1
l-4dj]
is linear.,
element are constant. Hence, as before, we can
stresses in the
the shear strains
element
call this
a constant-strain or -stress triangle (CST) element.
MOMENT
TWISTING
According to the
M
t
approach, the twisting
stress
=
l
||
tpdxdy
= f
2
(p
|
moment M,
m dxdy,
is
given by
[2]
(ll-38a)
|
'
A
where
m=
1, 2,
. . . ,
Af,
"a
M = number of
elements. Equation (ll-38a) represents
twice the volume under the stress function distribution over the cross section of the
bar (Fig. 11-7). For a generic element
M
(p
tu
1
1
we have
m dxdy
[Ntf -
N2f2 -
Nsfftbcdy
2
-y«P?
Summing
(ll-38b)
(P\
over the four elements, we have
M
2x1/8
t
2 x 3
2 x
„
,
f-° 0+3
4 3
1
1
0-0
1
0-0
3
2 x 3
2 (32
7.1111 N-cm.
(ll-38c)
260
Chapter 11
Torsion
Therefore, total
moment
for the bar
M, = 4 x Step
1
l-8(a)
and
for the entire bar (4 linear
=
28.4444 N-cm.
(ll-38d)
show plots of computed stress
functions and shear stresses
7.1111
Interpretation and Plots
8.
Figure
is
approximation
and are attached
(b)
cm x is
at the
4 cm) and the quarter bar, respectively. Note that because
used, the shear stresses are constant within each element
element centroids.
Figure 11-8 Results for torsion of square bar:
approach,
(a)
stress
function
Distribution of computed stress functions in entire
bar. (b) Shear stresses in quarter bar.
2.360
Exact distribution of
\p
2.666
--?<: Section
A-A .2.666
3fc-. 1.333 Section B-B
(a)
y^yz ?
\ N
^4° t 3
jo
,
V /\ A
i
shown
at centroids
/ L
*
/
Stresses
3
° \
\
/
(b)
\
*- x. r„.
COMPARISONS The exact value of (p
Stress Function.
node
at
(x
1
=
0,
y
=
0)
from Eq.
(1
l-26a)
is
f Therefore, the error
=
2.360
is
err0r
=
2.360-2.666 inn X 10 ° 1360
=
-13.00%.
Shear Stresses. Use of Eqs. (11 -26b) and ical element number 1 {x 1/3) as 1, y
=
(11 -26c) gives the shear stresses in typ-
—
T„ = -0.246 AVcm 2 T„
=
1.060.
Therefore, the errors are error in x xz
error in x yz
= -0.246 — = -25.00%
0.000
=
1.060-1.333
=
-26.00%.
=-gg
inn x 100
Torsional Constant. According to Eq. (11-25), the torsional constant
^~ ~ From
0.1406(2 x 2)*
36.0000
cm 4
.
computed
the foregoing, the error in the error
is
= 36.0000-28.4444 =-7
Jo
torsional constant
is
inft x 100
^21.00%.
We have used a coarse mesh so that hand calculations can be performed. the errors are rather high.
a finer
mesh
is
The computed
Hence,
solutions can be improved significantly
if
used.
BOUNDS Figure 11-9 shows the computed values of torsional constants from the warping (displacement) and stress function approaches in comparison with the closed form
M
'Gd for the 2 cm x 2 cm bar of Ex. by 16 as per Eq. (11-25). The displacement approach yields an upper bound to the true value of torsional constant, whereas the stress function approach gives a lower bound solution. value.
For
this
comparison, the values of
t
11-1 are multiplied
Example
11-3.
Torsion of Square Bar with Finer Mesh: Stress Function Approach
Figure ll-10(a) shows a refined mesh for an eighth of the square bar in Fig. 11-7. In view of the symmetry,
it is
possible to reduce the problem to analysis of only an
261
:
:
262
Torsion
Note: The
number
Chapter 11
of nodes plotted refer to the specific problem only.
44.000 Warping function (Example 11-1)
42.000
40.000
38.000
36.000 8
6
10
Number
12
14
16
18
of nodes
34.000 Stress function
(Example 11-4)
32.000
Stress function
30.000
(Example
1
1-3)
Stress function
(Example 11-2)
28.000
26.000
Figure 11-9 Bounds in torsion problem.
eighth of the bar. Thus,
now we have
effectively eight elements for a quarter of the
bar instead of four as in Example 11-2. Figure ll-10(b) shows local node numbers for the elements.
In the following
we
give brief details of computations of element equations for
the four elements (with values of
ax
Element
1
=
= — 1,
0,
a2
a3
=
1
b
;
G
and 6 as before)
— — 1,
x
1
1
1
2 1
b2
=
1,
b3
=
b2
=
\,
b3
=0;
0;
Xm
"J»
sin
=i
ym
"J
j
0"
1.
Element 2: ci\
=
—1, a 2
=
0,
a2
=
1
~
1,
2
-1
-1
1
_-l
= i;
263
Torsion
Chapter 11
©
©(1,0)
©(0,0)
(a)
Y' T vz
A
(b)
Figure 11-10 Torsion of square bar: stress function approach, finer
mesh,
(a)
Mesh
for eighth bar of Fig.
1
1-7. (b)
Local node numbers.
Element 3:
ax
= — 1,
a2
= 1,
«3
= 0;
bx
= 0,
b2
=
1,
b3
= — 1;
*m
=f
,
ym
=\\
|,
7m
=
0"
1
-1
1
2
-1
-1 1_
Element 4:
=
L
1
-1
-1
2
-1
-1, 62
= 1,
63
=0;
0'
-1 !_\{
tr
•
_
5
f;
:
:
264
Chapter 11
Torsion
Assembly of the element equations leads to the following global equations
1-1 -1
0"
4-2-1
0-2 0-1
0-2
4
2-10 0-2-1 4
(pi
2
03
4
C
1
-1
1
[*
-f
(ll-39a)
2 2
1_
\
,
1
,
Introduction of the boundary conditions g>4
=
=
=
0.0
leads to modified assemblage equations as
=h +4p -2
-
-
2
(ll-39b)
3
solution of which gives (p
x
(p*
(p 5
p6
The shear
stresses
= 2.333 TV/cm, — 1.666, = 1.500, = 0.000' = 0.000 prescribed. = 0.000 >
according to Eqs. (11-37) and (ll-28a) are
follows
Element
1: dq>
0.166
dy
N/cm
dy dx
0.666
Element 2: 0.166
N/cm f
:
1.666
Element 3: 0.000
) I
N/cm
:
1.5000/
Element 4: >
N/cm 2
.
;
now found
as
Chapter 11
Torsion
265
.1.666 2.333-
©
©
© Along nodes 1-2-4
-1.500
2.333
©
©
©
Along nodes 1-3-6 (a)
y-^
Figure 11-11 Results for torsion of square bar: stress function Distribution of computed stress func-
approach, refined mesh,
(a)
tions in quarter bar. (b)
Computed
shear stresses in eighth bar.
Figure 11-11 shows plots of computed values of
(p
and shear
stresses for the
refined mesh.
COMPARISONS Torsional Constant. (11-38)
M
torsional constant for the eighth bar according to Eq.
i
r/7
2 [( L\ 3
t
= Then
The
is
'
3
4J5 3 )
'
V3
3/"
1
'
3.8888 N-cm.
the total value of torsional constant
XM, = 8x
3.8888
is
=
31.1 100
N-cm.
3
"*"
3 J
266
Chapter 11
Torsion
Therefore, the error in the numerical solution
err0F
=
36.0000
-
is
31.1100
3^000
inn X 10 °
~ 14.00%. The computed value of the torsional constant from the finer mesh is shown and yields an improved lower bound to the true value of MJG9. Stress Function. From Eq. (ll-26a), the closed form value of (p at node
in
Fig. 11-9
(x
=
v
0,
=
0)
=
hence, the error
1
found to be
is
is
e rr or
Shear Stresses. For element
2.360,
^
=
2.360-2.333
=
1.00%.
=
(x
3,
5/3,
y
x lnft 100
=
2/3),
use of Eqs. (ll-26b) and
(11 -26c) yields errors as
error int*,
error in x yz
= -0.194 = -19.00%
0.000
=
1.900-1.500
=
21.00%
For the foregoing comparisons,
j-^
inn x 100
can be seen that the numerical solutions
it
with the finer mesh are closer to the closed form values. The error in the torsional constant reduced from 21.00% to 14.00%, and that in the shear stresses from 25.00 to 19.00 and from 26.00 to 21.00% in r xz and T yz respectively. Note that the ,
comparisons of errors
in the shear stresses are
not rigorous, since they are not
same point; however, the general trend shows reduction
necessarily at the
in the
error.
The numerical
solution can be improved significantly with progressively refined
meshes. Such refinement should follow certain criteria for consistent comparisons; these are briefly discussed in Chapter 13
Example
To examine
11-4.
Computer Solution
(Example
13-5)
for Torsion of
and
in Ref. 4.
Square Bar
the influence of a different (higher-) order element
on the numerical
we now consider a square bar divided into four (square) quadrilateral elements [Fig. 11 -12(a)]. Note that the size of the bar here is 2 cm x 2 cm, while in solution,
the previous examples with the stress function approach the size of the bar
4
cm x
was
4 cm.
The computer code FIELD-2DFE used to solve
this
rilateral element,
problem. This code
which
is
covered
tribution of the stress function as
in
and Appendix 4 was based on a four-node isoparametric quadChapter 12. This element has a bilinear disdetailed in Chapter 12
is
compared
to the linear distribution within the
Chapter 11
267
Torsion
y
©
©
®
(1.D
(0, 1)
A
©
©
A @o,o
A ©
G=
1
e =
1
A d,o)
(5)
@
x
(a)
(0.00)
(0.00)
(0.00)
±0.1
^ 0.87
(0.48)
(0.00)
(0.38)
^J0.24 0.24
(0.62)
^J0.39 0.39
(0.48)
J0,7 0.1
(0.00)
[b]
[c]
Figure 11-12 Torsion of square bar with quadrilateral element, (a) Finite
element mesh for quarter bar. (b) Computed nodal stress
functions in parentheses,
(c)
Computed
shear stresses.
is an improvement in the assumed approximation within the element. Figure ll-12(b) shows the computed values of stress functions at the node points. Computed values of the shear stresses i xz and t yz are plotted in Fig. 11 -12(c). The computed value of the torsional constant is 2.046.
triangular element used in this chapter. Thus, there
COMPARISONS Torsional Constant.
constant
The
error in the
computed and exact values of the
is
-
error
2.250 2.046 x 100 2.250
^9.00'
torsional
268
Torsion
The value of
M /G9 = 2.046
=
x 16
t
32.74
Chapter 11
is
shown
(p
(Eq. 11 -26a) at node
in Fig. 11-9
and
yields a
lower bound to the exact solution.
=
0,
Shear Stresses. The shear stresses are compared for two typical elements,
1
The
Stress Function.
y
=
0)
is
closed
form value of
1
(x
found to be
=
hence, the error
0.590,
is
error
=
0.590-0.620
^-^
inn x 100
= -5.00% and
2:
Element 1
=
(x
0.25,
y
=
0.25)
-
errorinr,^
error in t yz
Element 2
=
(x
0.75,
y
=
-
^- ^ !^
238
-
=
1.00%,
=
0.238-0.240
-
-1.00%.
fyrvi
24
1AA x 10 °
0.25)
- (-0.100) ^-^
-0.104
=
error in i X2
'-
=
-4.00%,
=
0.853-0.870
=
-2.00%.
error in t yz
nfcTa
Use of the 4-node isoparametric in the solution for the values of
inn x 100
x inn ^^0
quadrilateral provides an overall
improvement
shear stresses, and torsional constant.
For better accuracy one should use finer meshes with a computer code. With about 100 elements one can expect results of acceptable accuracy for all practical purposes. The cost of such computations with a computer code is
not high.
REVIEW AND COMMENTS In the displacement approach,
we assume approximate displacement
or warp-
ing functions. These functions are chosen such that they satisfy physical
continuity of the
body or continuum up
mation. The potential energy functions,
and
its
Up
is
to a certain degree or level of approxi-
then expressed in terms of the assumed
stationary value yields approximate equilibrium equations.
In other words, for the assumed compatible function, equilibrium of forces is
satisfied
only in an approximate sense.
Chapter 11
Torsion
269
In the case of the stress approach, the assumed stress functions that
complementary energy
satisfy equilibrium are substituted into the
II C
.
Its
stationary value leads to approximate compatibility equations.
both cases, for the assumed unknowns, either the equilibrium or is fulfilled only approximately. For example, in the case of the displacement formulation for torsion, the stress boundary conditions over
Thus
in
compatibility
the surface are not necessarily satisfied. That
is, as required by Saint- Venant's normal to the boundary computed from the displacement procedure may not be zero. Figure 11 -13(b) shows the variation of t„ along the boundary AB of the discretized region of a triangle [Fig. 11-1 3(a)] subjected to torsion [8-10] computed from a displacement and a hybrid formulation. It can be seen that t„ is not zero as required by the theo-
torsion, the shearing stress t„
retical
We
assumptions.
can overcome some of these deficiencies by using hybrid and mixed
procedures.
HYBRID APPROACH
A wide variety of procedures is available for formulating the hybrid approach One can assume displacement
and stresses on the assume stresses inside the element and displacement along the boundaries. Here we shall consider [4, 8-11].
boundaries of the element. Conversely, a special case of the former
Since
it is
inside the element
it is
possible to
[9].
found that the major error
in the results
from the displacement
we
shall choose the elements on the boundary, shown shaded in Fig. 11-14, for the hybrid formulation. For the elements in the interior the displacement formulation will be used. Since the displacement approach has already been formulated, we give now details of the hybrid approach for the elements on the boundary. Consider a generic boundary triangular element (Fig. 11-14). The stress function is assumed within the element as
formulation occur in the stresses at the boundaries,
=N =
1
q>
l
+N
2
+N
2 q> 2
(ll-4d)
[NJ{q,},
where the subscript s denotes stress function. The general expression for the warping function in the element is defined as y/
= N.y/, + N = tN„]{q r
2 y/ 2
+Ny 3
3
(1Mb)
}.
If
we denote
the warping associated with the three sides of the element as
Vi3> Vsi* an d ^i2> respectively, specialization of
y/
in Eq. (ll-4b) along the
rad/mm
'a)
Hybrid
••v^ -
4-4
-1
[,
I
/
/
A. i
!/
A
A
\/\/]/\/\/V
-2 L Displacement
(b)
Figure 11-13 Torsion of triangular bar with hybrid approach (a) Finite
edge AB.
270
[9].
element mesh, (b) Distribution of normal shear stress on
A
111
Torsion
Chapter 11 Y
Boundary element
731
^Bou
W 7 CD—
~^W Inner element
=-
"•,2
and boundary elements
Figure 11-14 Discretization
in
hybrid
approach.
sides of the element leads to
y23 ¥si
^12
Here we have used
^ +N
relation
2
y + (1 - N )y/ + N y/ = N l¥l + y2 + (1 - tf )^ = (1 - ^ Vi + ^2^2 + y/
=
.
t
3
•
3
.
2
definitions of
+N = 3
and the angle of
Nu N
2
,
and
N
3
3
,
3
,
3
.
(11-40)
as in Eq. (11-3)
and the
1.
As suggested by Yamada
M
2
3
x
et al. [9],
we
shall also
add the twisting moment and follow
twist per unit length 9 to the formulation
their procedure.
Step
4.
Element Equations
In the case of the hybrid stress approach,
we
define a modified complemen-
tary energy expression, TI ch per unit thickness of the bar as follows
[9, 11]
,
u = ue + wph = u ch
c
:
w
h,
"-AJ[©'+(K)>*-£»*«iA
= iHere
W
h
jj
{*}
T
\P]{"}dxdy
-
j {TYMdS
denotes the work of boundary forces, and
(11-41)
we have used
the subscript
h to denote the hybrid formulation. In very simple words, the procedure in the case of stress
H
is
combines complementary strain energy Uc as approach and potential of external loads ph as in the
called hybrid, because in
ch it
W
case of the displacement approach.
In Eq. (11-41), {T} r
= [M
t
r 23 t 31 t 12 ]
and normal components of shear
stresses
is
the vector of twisting
on the boundary,
{\\f b
T }
moment
=
[6 y/ 23
272
Chapter 11
Torsion
Wn
is
W\i\
the vector of corresponding angle of twist
with the sides of the element, {a}
defined in Eq.
is
and warping associated and
(1 l-28a),
0"
1
[D] 1
which
the inverse of stress-strain matrix [C].
is
We
assume
linear elastic
behavior.
W
Uc and ph in IT cA we need to derive the following results boundary element. For the triangular element, the stress vector {
evaluate
,
for the is
Uxz
w=
—
r
dy
=
cp 2
(ll-42a)
lies
b,
dx
.
Because the element
a3
d(p '~2A
[*»
boundary,
a2
01
,
i
l
on the boundary such that the side 2-3 is along the which we assume as zero; then Eq. (ll-42a)
= constant,
specializes to
f*„l
1
*
f "
(ll-42b)
<*>.
or •}
=
(ll-42c)
[PHK,
where [P]
and
{P}
= {
x
is
2A\ b
just a scalar. Substitution of Eq. (ll-42c) into
U
c
[Eq. (11-41)] leads to
u
c
=m jj^ffow\dxd m T
(ll-43a)
y
wmm 2
(ll-43b)
-
X 2A X 2A //« ,
'a
_Ari (a] + SA 2 G where
l\ 3
side 2-3
= (a + 2
b\)
3U:H
b}) _lh(t>\
(ll-43d)
&GA
= (x — 3
x2 ) 2
(11 -43c)
-f (y 2
—y
2 3)
denotes the length of the
and [H]
= _
JJ
\PYtPmdxdy
*23
4G^
(11-44)
Chapter 11
Torsion
The second pan tkms T over
in
273
Eq. (11-41) denotes the potential of the surface trac-
and can be
the entire surface of the element per unit thickness
expressed as IV. ,
=
M.(
T
7r.Vr.dS-
'iiViidS r
y,
x
'-^- dS
(ll-45a)
|
(ll-45b)
-;5.
Here r23 and so on are constants because of the linear interpolation function. The components of {T} r = [M. t/ ;: u-. ¥-.i\ are evaluated as follows: .
:
{—ymx„
M.
(-y-a
f-4
-
xmx^dxdy
:
1
(11-46)
and by
referring to Fig. 11-14. r ::
r
~.
~_
—(a.b; — b.a-.)p. 2Al ::
x: sin
r,,
cos a
'
'
'
_ — (a-.b: — —
a b•
|
imponant
a
'
— It is
—
=
b a :
(11-47) l
b-a-
to note that the stress r23
)(p
_
l
0.
:
normal
to the
boundary
zero, as required by the torsion problem. In matrix notation,
\f
1
b;
identically
-xjt
y-.-i
:
is
we have
Ul
«1 f»3
(11-4&0
r:: *12-
k
hi
{f}
=
«3 /::.
::
The
surface integrations relevant to
performed Eq.
11-45
easily.
For
{\f/ b }
of Eq. (11 -45b) in
instance, substitution for
u :: from .
W+
can be
Eq. (11-40) into
gives
:
because JS
di-4Sb)
[R;p
=
dN
l23
2
l-NJvt + N&JKhs
along side
2-3. Finally, integration
(ll-49a)
of the expression
274
in
Torsion
Chapter 11
Eq. (ll-49a) leads to
J2
The other two
y/ 23
= k^{y/i +
dS
(ll-49b)
Vi\
3
integrals are evaluated similarly.
Then Eq.
(11-45)
becomes
W = {tY\{y }dS ph
(ll-50a)
b
e
(P}Wy
(ll-50b)
+
\hiiWi
2A g>1
^~ ai
b^
¥i)j
ymA
(b 2
-x m ^
(a 2
+ +
b 3 )/2
0,
a 3 )/2
(fl 3
+ +
&,)/2
(6 1
«i)/2
(fl,
+6 )/2 + a )/2 2
2
2
1^3 (ll-50c)
=
m [G]{%] = {P}W[L]{q,}. T
(ll-50d)
Here
[G]
=
(ll-50e)
PT[L],
and
-2y m A
{L}
b
Here we used the
Now, we
relations a
substitute for
x
U
e
+
a2
and
b2
b3
a2
a3
x
(ll-50f)
2x m A
a,
+
W
a3
—
_
and
^ +
Z> 2
U = i{pf[H]{p} - {pf[G]{q,}. differentiating U with respect to {f} = p, and
By
Z>
3
= 0. (1 1-51)
ch
to zero,
+
ph in Eq. (11-41) to obtain
ch
equating the results
we obtain
dUeh
[HP] - [GHqJ
=
0,
(ll-52a)
or [H]{P)
=
(ll-52b)
[G]{q r ],
or
»}=[H]-»[G]{q r }, which expresses the relation between nodal ing functions for the boundary elements.
(11-52C) stress functions
and nodal warp-
:
Element
Matrix
Stiffness
We now
derive element stiffness matrix [kj relevant to the hybrid
approach. For
we
this,
express the strain energy as
P = tf*FhK*}where
T
{q w }
= [6
y/ x
y/ 2
the vector of generalized nodal displace-
is
y/ 3 ]
(H-53a)
ments. The complementary strain energy expression
U
e,
after substitution
of Eq. (ll-52c), leads to
m =
u = c
T
\n\m
Mq,}W([H]-y[H][H]-[G]{q,} {qjqGF[H]^[G]{q 2"14^
Here, since [H] defined by Eq.
(1
1-44)
is
(ll-53b)
v
usually symmetric,
we have
([H]
_1 r )
= [H]-'. Equating the above two expressions for stiffness
U and U we obtain the element c,
matrix \k h ] as
[k*]=[GF[H]-i[G] [Gf
AGA
(ll-54a)
[G]
Hz A 2 (ym en + xm b\) 2
Al 223
A(y m ai -A{y m a (ym a\
23
y m ai
L-(y m ai
x
+
x m bi)y-^—
+x m b
+
—A{y m a\
-2 A
1
){—^-)
x m b\) 2
+ xm bi + xm bi)
ym a\
+
x
+
+
a2
~2~ )
—A(ym ai+x m bi)(2A)
1
j(-2A)(2A)
-j(2A)(-2A)
\(2A)(2A)
— (ym ai +
x m bi
x m bi)~ (ll-54b)
-1
0-1 a2
2A\
xm b\)[
^(-2A)(-2A)
1
Here we used again the following a
+
1
relations
= 0,
b
x
-f b 2
+
63
= 0,
and a 3 b2
—
a b2
a2b 3
x
2A.
Inner Elements
We choose to use the displacement approach to define the stiffness matrix for the elements inside the bar.
By following
the procedure outlined previ-
ously in this chapter, the stiffness matrix with warping functions can be 275
276
Torsion
Chapter 11
derived as
4A 2 (xl
+
w-&
yl)
-
2A{a\x m
+
b\
biy m )
2A(a 2 x m
a\
b\b 2
sym.
b\
-
+ +
2A(a 3 x m
b 2 ym)
a\ci2
b\bz
a\
b2b 3
b\
Both
[k A ]
and [kj express
relations
-
b 3 ym)
+ 01^3 + a 2a + a\
(11-55)
3
between generalized forces and
dis
placements {q w } and have identical forms as
and
M{qJ =
{Ql
(11 -56a)
WW =
(Ql
(11 -56b)
where
Ay m
~^*m
&i
T
flj
2
{Q}
'M,'
^2 2
«2
*3
*3
2
2
2
Ft
ft-
(ll-56c)
F2
F3 1
F ,F2 and F3 represent one-half of the resultant shearing forces on the sides 2-3, 3-1, and 1-2, respectively.
and
l
,
Computation of Boundary Shear Stresses
For an element with and (ll-52c),
side 2-3 along the
boundary, we have, from Eqs.
(11-42)
2A
_:;h
-
AL;;Hw» (0
—a l\i\_
l
{a
^- b x x m ) l ym
b {a y m x
l
-b
x
xm)
—a
x
0i
(11-57) 6,
¥3 Assembly
For elements on the boundary and inside the bar, we use element equaand (ll-56b), respectively. Since both are expressed in terms of nodal displacements, the assembly follows the same rule that the displacements at common nodes are compatible. In view of the fact that the hybrid approach satisfies the stress boundary conditions, the results from the proposed procedure can yield improved
tions (ll-56a)
:
Chapter 11
277
Torsion
accuracy in the stresses at and around the boundary. In the following,
we
present an example using
puter solutions by
Example
We
Yamada
hand
calculations
and then
results
first
from com-
et al. [9].
Torsion of Square Bar: Hybrid Approach
11-5.
consider torsion of the square bar 2
cm x
cm shown
2
in Fig. 11-5. In this
example the local node numbering is changed as shown in Fig. 11-15 to comply with the numbering for the boundary element in the foregoing hybrid formulation.
©
Global node Local node
1
/j\ Element
©
©
(0, 0)
(2, 0)
x
Figure 11-15 Torsion of square bar: hybrid approach.
All the four elements
lie
on the boundary and hence are treated as boundary
elements in the sense of the hybrid approach.
boundary elements; it is
necessary to use Eqs.
(1
incidental that all elements are
It is
both boundary and inner elements
in general,
l-54b) and
will occur.
Hence
for deriving the element
(1 1-55), respectively,
matrices for the boundary and inner elements. In the following are given salient details of the
Element
element and assemblage equations
1:
Q\
a2 3
Here
A =
1
2,
b\
—1,
b2
=y - ys = o, =yi -y\ = -h
= X2 ~ X = — 1,
b3
=y\ -yi =
= *3 — =x — x
cm 2 and we ,
*i
x3
= =
t
xm
=
1,
>23
=
2,
2
y m =i,
GA
have assumed that
G=
1
1
N/cm 2
.
i>
278
Chapter 11
Torsion
Use of Eq. (ll-54b) leads
to element equations:
I
5
1
2<
I
i
2
3<-
4
r
Mi = ~
I
Local
(0)
Mf, f
l
I
J
1
I
I
I
}V1 2
1
¥2
1
-1
L-f is
2-\
2
T
7
We assume that
Global
lj ly 3
J
uniform over the bar, and hence a special number
both local and global numbering the twisting moment for element
I
is
assigned to this degree of freedom.
1
and so on.
common to
M
Element 2: a,
=0,
br
a2
=
b2
1,
=2,
= =
b3
-1, -I.
=h GA
1
/23
'23 I
5
1«-
I
1
3«-
4
2
5
1
_
id-
ri
Mtl
=
=
>
2
T
1
2
-1
1
-
u Wz\
1
I
I
1
5
2
3
3
1
<
¥2
1
|
1
\
\
[k„] 2
Local
j
1
Element 3: a\
*2 03
*
= = = =
=
-2,
0,
6,
-1.
62
1,
63
= 1, = 1,
.Vm
=
5
1,
GA 23
4<-
Global
3^-
W-
if
-¥
-*£
1
^
-1
ion 3
Local
(6
M>
¥1
-1
¥2
1_ .Wsl
I
J
i
1
I
I
tl
denotes
Chapter 11
Element
Torsion
279
4:
xm
= = = =
/'23 23
=
a\
ai #3
-2,
6,
=0,
*3
=
1, 1,
"I,
y m =4,
1,
2,
'| 3
3*--
Global
3^-
^ [k;»]4
—
10
Local
mu
0—1
1
3
3
1
1
I
I
¥1 "T
^ ^
-1
1
-1
¥2
1_
Vs.
,
Combination of the foregoing four element equations by observing compatibility
of nodal
y/'s
leads to assemblage equations:
208
12
9
3
12 3
12 3
2
-1
-1
2
12 T
-1
1
W1
2
1
¥2
-1
2
¥4 0_
These equations can for
6
=
1
the solution
now
f
S
m=l
M
tm )
¥1 1
-1
-
e
.¥s.
=
,
, ,
be modified for the boundary condition
y/^
=
0.
Then
1
1-1
with
is
= 1, ¥1
=
® (prescribed),
¥2 =2, ¥s
¥* ¥s These
= = =
-2, 0, 0.
results for y/'s are essentially the
the warping function approach.
We
same
as in the case of
have considered
Example
this rather
simple example
hand calculations. For acceptable accuracy and in order to improved boundary stresses in the hybrid procedure, one should use a finer mesh. In Example 1 1-6 we shall consider a computer solution for the mesh in Fig. 11-13 where both boundary and inside elements occur. The results in Example 11-6 will illustrate the advantage of the hybrid approach.
mainly to
illustrate
exploit the advantage of
:
SHEAR STRESSES For the four elements
boundary, we use Eq. (11-57) to evaluate the shear
at the
stresses:
Element
1:
2(2 x
O-i
2"
0.
(0.666)
=
-2
J)
N/cm 2
.
lo.oooJ
Element 2: 1
0"
' 1
4 _2 x §
WvJ
-2J -2
2
0.000)
f
N/cm 2
.
1—0.666/
Element
3. 1
"
Zxz
0"
1
" 4
\
lyz)
_-^ 000]
-2
N/cm 2
2J
2
.
lo. 666)
Element
4. 1
_20
2
-2
ri--[
0_
-2 f
—0.6661
N/cm I
:
O.OOOJ
These results are the same as those in Example 11-1 (Fig. 11-6); they show that boundary is zero. It was only incidental that with the four-element mesh chosen the warping function approach also satisfied the zero stress boundary condition. With a different and arbitrary mesh, the warping function approach, in general, would not satisfy this condition. the shear stress normal to the
TWISTING MOMENT The torsional constant can be found by 1//
into the
280
first
substituting the
of the above assemblage equations
computed
results for
281
Torsion
Chapter 11
M^ [fxl = This value
is
same
the
_ 4x2 + 4x( _ 2)]
1.7777 N-cm.
as that
from the
approach and has an error
stress function
of about 21.00%.
Example
Torsion of Triangular Bar:
11-6.
Hybrid Approach, Computer Solution Figure 11-1 3(a) shows a bar in the shape of an equilateral triangle. Analysis for torsion for this bar using the hybrid
method
symmetry, only one-sixth of the triangle
is
is
presented by
discretized.
Yamada
The
et al. [9].
Due
to
properties of the system
are
Poisson's ratio,
E= =
Shear modulus,
G=
Young's modulus,
2.1
x 10 4 kg/mm 2
,
0.3,
=
=0.81 x
10 4
kg/mm 2
Table 11-1 compares the values of warping along the side
AB
[Fig. ll-13(a)]
Angle of
6
twist,
>
2(]
=
1.194
'
.. fl
,
x 10~ 4 rad/mm.
obtained from the closed form solution, the displacement method, and the hybrid
TABLE
Warping Along the Contour
11-1
AB
[9]»
Displace-
Node
Exact
Hybrid
ment
Solution
Method
Method
B
11
0.370
0.368
0.375
T
13
0.957
0.942
0.975
21
1.924
1.884
1.921
29
2.772
2.721
2.766
i
A
36
3.471
3.419
3.467
43
3.887
3.821
3.868
49
3.973
3.913
3.960
55
3.962
3.632
3.659
60
2.978
2.918
2.960
63
2.398
2.352
2.385
66
1.865
1.834
1.854
69
1.354
1.330
1.347
73
0.898
0.885
0.896
76 79
0.530
0.524
0.529
0.270
0.268
0.270
*h>
x
10 3
mm.
282
Chapter 11
Torsion
method. The two numerical results are close to the exact solution, without significant difference between them. As noted before, Fig. ll-13(b) shows comparisons between computed values of the shear stress normal to the boundary r n from the displacement and the hybrid methods. In contrast to the solution by the displacement method, the hybrid approach yields zero values of T„, as required by the theory. Figures 11 -16(a) and (b) show comparisons between shear stresses r xz and t yz along
AB
as
computed from exact solution and the displacement and hybrid
methods. The
latter
boundary but
in the interior also; this aspect
improves computations of the shear stresses not only along the is indicated in Table 11-2, which lists the shearing stresses at the centroids of typical elements adjacent to side AB (Fig. 11-13). These results also show the improvement in accuracy provided by the hybrid
approach as compared to the
TABLE
11-2
results
from the displacement approach.
Comparison of the Shearing Stresses AT THE CENTROID OF TRIANGULAR ELEMENTS ADJACENT to AB [9]*
Element
2
48
81
99
122
-7.136
-6.354
-4.351
-2.259
-0.128
-7.168
-6.264
-4.117
-2.251
-0.125
-7.145
-6.884
-4.759
-2.563
-0.225
12.354
10.803
7.209
3.348
0.168
12.416
10.853
7.131
3.900
0.216
12.541
10.511
6.653
3.628
0.071
-0.003
-0.101
-0.164
-0.282
-0.027
0.000
0.000
0.000
0.000
0.000
0.083
-0.706
-0.795
-0.406
-0.159
Ixz
Exact solution
Hybrid
method Displacement
method
tyz
Exact solution
Hybrid
method Displacement
method
Exact solution
Hybrid
method Displacement
method
*Unit:
kg/mm 2
.
Exact solution
-2
Hybrid method Displacement method
-4 -
t x2
(kg/mm 2
)
(a)
r yz
(kg/mm 2
)
14
+.* 12
10
—
Exact solution
Hybrid method Displacement method
(b)
Figure 11-16 Comparisons for shear stresses x xz along contour
AB.
(b)
Shear
stress xyz
[9].
(a)
Shear
stress
along contour AB.
283
MIXED APPROACH In the mixed formulation, both the displacements (warping) and stresses (stress function) within (including the
to be
unknown. Hence,
boundaries) the element are assumed
for a triangular element,
V
=
¥
=
[N ,]{q r }
[N
N
x
N N N 2
t
N
2
3
MJfo.
N N N t
[NJ {q
=
r
where
{
x y2 ]
[t xz
is
The
* XZi
XZl
* XZ3
2
(ll-58b)
(q c 3
.
the vector of shear stresses at any point, [N^] and
are the interpolation functions, {q^}
= [i
(ll-58a)
3]
lyzt
Tyz*
*yz 3 ]
strain displacement relation
tutive relation for linearly elastic
is
is
is
the vector of nodal warping,
given in Eq. (11-8), and the consti-
and homogeneous material 1
and
the vector of nodal shear stresses.
is
0"
(11-59) 1
Step
We
Derive Element Equations
4.
[12, 13].
expressed as
n* =
A
T
Noor and mixed approach can be
procedure presented by
t
dU
=
the
variational function for the
w w - *ww - JJJ^. -
c
is
[f xz
If
V
vm^ds,
(i
l-eo)
Si
the complementary energy density [Eq. (11-29)], given by
dU
{a}
essentially
[4, 12]
JIf V
where
follow
shall
Anderson
fy2 ]
is
c
=
^
X2
TJ^J ^
*XM
dV
the shear stresses at the boundaries; {n}
is
the vector of
outward normal to the boundary; V is the volume of the element; [x} T = [x y] and S 2 is the part of boundary on which tractions are prescribed. Substitution for [a] in Eq. (ll-58a), {e} in Eq. (11-8), and dUc in n* leads ;
to
284
285
Torsion
Chapter 11
n*
=
{q.f[N.F[B]{q^K
|j{ V
-
{q.F[N,F[D][NJ{q„}rfK
JJJ V
-6
{q.F[N„F[N,]{x„}
JJJ V
-
r
[[ {q,} 5
Here
{x„}
=
x2 x 3 y y2 y 3 ]
[x x
[N r F{n}[NJ{q„Jrf5.
x
the vector of nodal coordinates, given by
is
x\VN N N 2
x
z
~~
yJ and
[B]
is
Nt
L
(11-62)
N Nj 2
3
defined in Eq. (ll-12c).
Application of the variational principle to TIr leads to
The
(11-61)
2
its
stationary value.
variations or differentiations are performed independently
and simul-
taneously with respect to nodal stresses and nodal warping functions; thus
dUM <5ru
=
Wq.
o
(11-63)
-{q
This leads to two sets of equations,
fc]
ftJ'
LfcrF
{Q.:
{q.
J {q
[0]
(ll-64a)
{0}
where, for the triangular element (Fig. 11-2),
fcj (6
x
=
— JJJ [NJ (6
6)
x
PU = jjj [NJ r (6x3)
(6
x
r
x
[B]
2) (2
x
^^z,
[NJ
[D]
2) (2
2) (2
x
(ll-65a)
6)
(ll-65b)
dtofy
3)
and
{Q.}=0jJ{N.f (6
x
1)
(6
x
2) (2
^{4
[N]
x
6)
(6
x
(ll-65c)
1)
The last term in Eq. (11-61) represents boundary conditions. Here we have assumed that the boundary conditions occur as prescribed shear
286
Chapter 11
Torsion
stresses
and that their components normal to the boundary vanish. The components do not contribute when the variation, Eq. (11-63), is
tangential
performed, hence the
last
term in Eq. (11-61) does not appear in the calcula-
tion of forcing parameter vector {Q,,}.
Evaluation of Element Matrices and Load Vector
We now
computations of the element properties for the
illustrate
angular element. For unit length, AT,
N
2
«--*jj
N,
0" 3
Lo
N,
N Ns
N
^2
fN,
N
AT,
2
//l U.
N_ 3
2
m
m
,
N,N N,N
Mt/ft
"
3
Nl
}
Nl
hB
dxdy.
m
sym.
N,N2
If1*3
Nl
NN 2
3
Nl The
integrations can be performed in closed
N\dA
=
2A
\\
form for ;
A
2!0!0! (2
+
2)!
and
jy
NN x
2
dA
=
2A
1!1!0!
A
(4)!
12
Therefore,
0"
"2
1
[kj
A
UG
1
2
1
1
1
2
1
1
1
2_
instance,
tri-
Chapter 11
Torsion
287
Now ~JV
0"
i
tf z
N
w-]J
-
b3
b2
bi
1
3
dA
2A
N
2
_0
^3_ r
N bi
Nb Nb N b, N b 2
4
t
N
a
_N
a 3
2
1
6
and
t
x
2
2
3
2
t
N,a 2
N,a 3
N N
N N
3
-Air N a
N b,Nb Nb
N,b 2
t
2
x
l
a2
3 a2
2
3
3
3
2
a3
3
a 3_
dA
~
"*,
b2
b3
*,
b2
b3
*,
b2
b3
*1
a2
a3
a2
a3
a2
a3_
0"
~N,
'*>'
N
Xl
2
w-fj
N W _0
r
AXi
=
6A
N
pv,
A^3
-\-
[_o
t
2
N
0"
3
AT,
N
2
N
3
x3
_
2
>'2
JV 3 _
^3>
X 2 ~ -x 3
x,
+
*1
+ x2 + 2x
2x 2 - - x 3
T2'
3
>•
-y* yi
+
2y 2
- ->' 3 2>'3,
xample
11-7.
To rsion of Squ are
The node numbers and details for tion) in Fig. 11-4 are shown in Fig.
the
I lar
:Mixed Approach
problem of the square bar
(2x2
cross sec-
11-17. In the following are given essential details
i
288
Chapter 11
Torsion
of the computations of element and assemblage matrices based on Eqs. (11-64) and (11-65).
Equation (ll-64a) can be written
in the
expanded form as
-2
-1
2b
2b 2
2b{
-1
-2
2b i
2b 2
2b 3
?XZ2
-1
-1
2b 3
Txz 3
A \2G
2b
2b 2
-2
-1
-1
2a
2a 2
2a 3
-1
-2
-1
2a i
2a 2
2a 3
*yz2
-1
-1
-2
2a\
2a 2
2a 3
•yz%
2b\
2bi
2b\
2a i
2a \
2a\
2b 2
2b 2
2bi
2a 2
2a 2
2a 2
_2b 3
2b 3
2b 3
2a 3
2a 3
2a 3
+ x2 + x + 2x 2 + X x\ + x 2 + 2x 2yi + y 2 + y \= QAhi + 2y 2 + y 121 y\ + yi + 2y (2x\ X\
3\ 3
3
3
3
)•
3
W\
\V3l (ll-64b)
Element 1: «i
a2
a3
x 13
1
4
-2
-1
-1
-1
-2
-1 1
14
=2,
= -1 = -1 =
-2 2
-1
-2 -1
4
4
-2 -2
-2 -2
= -1, =h =h Global
3
-1
-1
b2
ym
2
-1
=0,
b3
1
-2
bi
1
4
-2 -2 -2 -2 -2 -2
2"
1
It XZ\ \ *>
2
2
-2 -2 -2
*
3
XZi
We
13 1
Txz,
5
4
tyz\
2
14
y \ =( yz
)
Tyzt
Vi
0_
/4\
1
1
2 5
15
¥2
3
Wsl
6
have assigned global numbers by rearranging the unknowns as shown in Fig. 11-17. That is, in the subsequent assemblage equations, the global numbers are assigned such that the three unknowns at a given node appear consecutively. This is
done simply for convenience. Element 2: ax
=0,
bi
=2,
"2
= =
1,
b2
-1,
b3
= = =
a3
Xm =i,
ym
-1,
-1, 1,
)
(7, 8, 9) (
289
Torsion
Chapter 11
® (10, 11, 12)
T xz3< r vz3< ^3
(
r xz4' r yz4' ^4>
(2,0)
(4,5,6) (t xz ,,t
<
©
r xz2' r vz2'
iM
Global nodes
/l\ Element Local nodes
1
(1, 2, 3)
=
(r xz1
,
T yz1
,
\p y
)
Global degrees of freedom
Figure 11-17 Torsion of square bar: mixed approach.
13
7
-2
-1
-1 -1
14
-2 -2 -2
-2 -2
2
-1
-1
-2
-2
Global 1
-2" /t
-2 -2 -2
\
f2\
-2
13
*«l
1
7
***,
1
1
*jr*i
4
14
5
8
{*» -1
4
-2 -2
3<-
9
Tr*i
^ r
3
2
^1
15
^2
9
W
3 y
U)
3
Chapter 11
Torsion
290
Element
3:
=0,
«i
= —h = 1, =
#2
a3 v 13
10
-2
-1
-1
-1
-1
-2
14
b
5
11
5
x
Z>3
ym
2
2
-2
-2
2
2
Global
12*
1
2
2~
T„\
6\
2
2
^*Z2
7
4
2
2
*«,
7
10
2
Ty*i
4
Tr*«
3
Ty*s
5
-2 -2 -2
-1
2
= 1, =h
-4 -4 -4 -1
-4
-2,
6
15
-2
-4
=
2 i
2
0_
13
14 >•
5 11
¥1
15
¥2
6
W3/
\0l
12
Element 4:
= = = =
ax
a2 a3
x 13
10
14
7
11
-2,
0,
l,
63= -1
1, 1
.Vm
8
15
=4, Global
12
-2
-1
-1
2"
-1
-2
-1
-1
-1
-2 -2
-1
-1
-1
-2
-1
-1
-1
-4
-4
-2 -4
2
2
2
2
2
2
-2
-2
-2
2
2
2
-4 -4 -4
1
r X2 \
'4\
13
2
~XZ2
5
10
2
r X23
3
7
2
ty*\
6
14
2
lyzt
7
11
2
Tyzi
>
=
<
7
¥l
12
¥l 0_ \¥3
8
15
9
The foregoing four element equations can now be assembled by observing
inter-
element compatibility. It is difficult
hand calculations the (modified) assemblage equations and it becomes necessary to use an (available) equation of equations (see Appendix 2). In this elementary treatment,
to solve by
(involving 14 unknowns), solver for large sets
it
291
Torsion
Chapter 11
will suffice
only to illustrate the mixed approach and the foregoing steps, because
considered beyond our scope. Solutions for this
detailed consideration of the topic
is
and the other relevant
Problems are
exercises in
left
to the inquisitive
and advanced
reader.
The foregoing completes our treatment of the torsion problem. Before we proceed to the next chapter in which a 4-node quadrilateral element is described, it is appropriate to present a brief description of the topic of (static) condensation which
is
often used in finite element applications.
STATIC CONDENSATION Often
it
may be
particularly single
convenient and necessary to use nodes within an element,
when
element
is
it
is
required to use higher-order elements. Sometimes a
divided into subcomponent elements by using the inner
nodes. For instance, Fig. 11-18 shows a line and a quadrilateral element
each with one inner node. The
element has two end or external or
line
primary nodes and one inner node, and the quadrilateral has four corners or primary nodes and one inside node.
Figure 11-18 Static condensation,
(a)
Line element, (b) Quadri-
lateral.
Inner node
©
© 'a'
©
292
Chapter 11
Torsion
The
finite
element equations
[k]{q}={Q] are derived
on the
(ll-66a)
basis of unknown degrees of
freedom
at all the nodes.
Thus
the formulation includes improved (higher-order) approximation yielded
by the use of the inner node. In the case of the quadrilateral [Fig. 11-1 8(b)], the element equations are often obtained by adding individual element equations of the four
unknowns
component triangles. Thus the total number of element and 10 for the line and the quadrilateral, respectively.
in {q} are 3
These element equations can now be assembled to obtain global equations, in which the degrees of freedom at the inner node(s) will appear as additional unknowns, and to that extent the number of equations to be solved will be increased. It is
unknowns
possible to temporarily eliminate the
by creating an equivalent element relation procedure of doing
this
in place
called static condensation. It
is
at the inner
node
of Eq. (ll-66a). The is
possible to
do
this
because the unknowns at the inner node do not participate in the interele-
ment compatibility at the element sides; that is, the unknowns at the inner node are not needed for the direct stiffness assembly procedure. The procedure of static condensation involves solution of the unknowns at the inner node in terms of those at the primary nodes. To understand it, we write Eq. (ll-66a) in a partitioned form as [q,}|
IK]
= ""
fCQJl
(ll-66b)
ItQJJ
{q,]J
{Q p } and {q,}, {Q,} denote vectors of nodal unknowns and loads primary and inner nodes, respectively. The second equation in Eq. (ll-66b) can be used to solve for {q,} as where
{q p },
at the
{q,}
= [k,,]-'({Q,) -
Substitution of {q,} into the
dk,J
-
first
[k,,F{q,}).
equation of Eq.
frJfcJ-fcjFXq,}
= {Q,} -
(1
(1
l-67a)
(1
l-67b)
l-66b) leads to
fcJfcJ-'tQ,!
or
where
[k]
and {Q} are the condensed element
(stiffness)
and nodal load
vectors,
unknown {qj now contains the degrees of freedom only at the external primary nodes thus when [k] and {Q} are used to assemble element equations, we have reduced from the total equations to be solved equations corresponding to the unknowns at the inner node. respectively. Notice that the vector of
;
After the assembled equations are solved,
if
desired,
it is
possible to retrieve
Chapter 11
the
unknowns
static
293
Torsion
at the inner
node by using Eq.
condensation can be found in
many
(ll-67a). Further details of
publications, including Ref.
[4].
PROBLEMS Invert the matrix [A] in Eq. (11-2) given by
11-1.
1
and obtain the matrix
in
r
i
x2
x3
yi
yy
EHB
Eq. (ll-3b).
mesh in Fig. 11-10 and treat the problem as torsion of a triangular By employing the warping function approach and setting boundary condition as y/ 2 = 0, compute nodal warping functions, shear stresses, and Use
11-2.
the
bar.
twisting
With the mesh
11-3. 11-4.
moments.
1
in
With the mesh
Prob. 11-2, solve for torsion using the hybrid approach.
in
Prob. 11-2, derive element equations using the mixed
approach. 11-5.
Solve for torsion of the square bar in Fig.
1
1-19 by using the warping func-
tion approach.
1
©
®
®„
\A A\ a\ © © A \A\A® \ A\
(0, *)
'
i
\A
©,
\*-,
„ *i
1
-*
©
(0,0)
\
G=
1
6 =
1
,®
(2,0)
Figure 11-19 11-6.
Solve Prob. 11-5 by using the stress function approach.
11-7.
Solve Prob. 8 as
1
1 1-5 by using the hybrid approach. Consider elements boundary elements and the remaining as inner elements.
4, 6,
and
This and some of the other problems will require use of an available equation solver once
the global equations are obtained.
294
11-8. 11-9.
Torsion
Chapter 11
Solve Prob. 11-5 by using the mixed approach.
Consider the mesh in Fig. 11-20 and solve for torsion by using the stress it may be necessary to use a computer. The code FIELD-2DFE can be used for solving the entire problem. Alternatively, the
function approach. Here,
assemblage matrix can be computed by hand calculations, and its solution can be obtained by using an available code for solution of simultaneous equations.
©
©
©
®
(2,0)
(0,0)
Figure 11-20
11-10.
For the mesh
in Fig.
1
1-21, use the
mixed procedure and derive the element
equations [Eq. (ll-64a)].
Figure 11-21
.
Chapter 11
11-11.
For the torsion problem in Fig. 11-7, treat the bar as having a cross section cm x 2 cm. With boundary conditions (p = (p 2 =
1
discuss
its
Answer: 11-12.
295
Torsion
accuracy.
=
0.6667;
M /G9 = t
1.777, error
=
21.00%.
For the circular bar of 2-cm diameter, compute nodal stress functions for mesh shown in Fig. 11-22. Find shear stresses and the twisting moment = 0.005 deg/cm. by using a computer code. Assume G = 10 7 N/cm 2 and the
Figure 11-22
11-13.
For the
bar a
elliptic
= 2 cm
stress functions, shear stresses,
code.
Assume G
=
1
and 6
=
and b = 1 cm (Fig. 11-23), compute nodal and twisting moment by using a computer 1.
1
cm
Figure 11-23
11-14.
By
using the warping function approach, find shear stresses and torsional
constant for the triangular bar divided into two elements (Fig. 11-24).
Assume G
=
1
and
=
1
)
296
Torsion
Chapter 11
^© (0,0)
Figure 11-24
11-15.
Use
the stress function approach to solve for torsion of the triangular bar
inProb. 11-14. 11-16.
By
using the code
FIELD-2DFE
mesh and obtain
or other available code, refine the
progressively (Example 11-4), say 1,4, 16,64, etc., elements,
results for stress functions
M
t
vs.
the numerical solution. 11-17.
Higher-order approximation. Use the quadratic interpolation function for q>
with a triangular element (6 nodes) and derive the element matrix and
load vector. See Fig. 11-25. Partial results:
Assume
W Note
9
=
tyl
that this function includes
=
all
[N]{qJ,
9*
terms up to n
9s\
=
2 in the polynomial
expansion discussed under "Requirements for the Approximation Function."
Figure 11-25
Local coordinates, L
Node
(1,0,0) (0,
1,0)
(0, 0,
(2
\
,
1
-. 0) v;, 2
,
(O.I.I) (1.0.1)
Chapter 11
297
Torsion
-1)1
[1.1(21,,
-
I 2 (2I 2 L 3 (2L [N]
3
1) 1)
]
4I 2 I 3 4I 3 Ii
[BJ
H
(2x
[0]-,
=
,
[qj .
.
[Blq,
&>]-
[0]
1)
where
t] = [(4Z.,
-1)6,
(4I 2 -1)^2
(4L3-1)63
-l)a 2
(4I 3 -l)fl3
4{IiZ)2-I:6i)
4(L z b 3
-L
3
b2)
4(Z. 3 6,
+£,63)]
and fc]
= [(4L,-l)a,
(4I :
4(I.,
+£2*1)
4d*« 3 +X*«2) 4(L3fli +JL,a3)J.
Computation of {Q,} will require integrations over element area and vant boundary surfaces.
rele-
REFERENCES [1]
Love, A. E. H., The Mathematical Theory of
Elasticity,
Dover,
New
York,
1944. [2]
[3]
Timoshenko, York, 1951.
Herrmann.
S.,
and Goodier,
N., Theory of Elasticity. McGraw-Hill,
New
L. R., "Elastic Torsional Analysis of Irregular Shapes," /. Eng.
Mech. Div. ASCE, Vol. [4]
J.
91,
No. EM6, Dec.
Desai. C. S.. and Abel, J. F., Introduction Nostrand Reinhold, New York. 1972.
[5]
Bach, C, and Baumann.
[6]
Murphy.
R., Elastizitat
1965, pp. 11-19.
to the Finite
Element Method. Van
and Festigheit. Springer.
Advanced Mechanics of Materials. McGraw-Hill.
G.,
Berlin. 1924.
New
York,
1946. [7]
Yalliappan.
S..
Anisotropic Bars."
and Pltmano. /.
Struct. Div.
V. A.,
ASCE.
'Torsion Vol. 100,
of
Nonhomogeneous
No. ST1,
Jan. 1974, pp.
286-295. [8]
Yamada,
Kawai,
T.. Yoshimura, N., and Sakurai, T., "Analysis of the Problems by Matrix Displacement Method," Proc. 2nd Conf. Matrix Methods in Structural Mechanics, Wright Patterson Air Force Base, Dayton, Ohio, 1968, pp. 1271-1289.
Y..
Elastic-Plastic
[9]
Yamada.
Y.,
Nakagiri.
S..
and Takatsuka. K.. "Analysis of Saint-Yenant
Torsion Problem by a Hybrid Stress Model/' in Proc. Japan-U.S. Seminar on Matrix Methods of Structural Analysis and Design, Tokyo, Aug. 1969.
298
[10]
Chapter 11
Torsion
Yamada,
Nakagiri,
Y.,
S.,
and Takatsuka, K.,
"Elastic-Plastic Analysis of
Saint-Venant Problem by a Hybrid Stress Model," Engineering, Vol. 5, 1972, pp. 193-207. [11]
Int. J.
Num. Methods
Pian, T. H. H., and Tong, P., "Finite Element Methods in
Mechanics,"
in
Advances
in
in
Continuum
Applied Mechanics, Vol. 12, Academic Press,
New
York, 1972. [12]
Noor, A. K., and Andersen,
CM.,
"Mixed Isoparametric Elements
Saint-Venant Torsion," Comput. Methods Appl. Mech. Eng., Vol. pp. 195-218. [13]
Noor, A. K.,
(private communication).
6,
for
1975,
;
OTHER FIELD PROBLEMS POTENTIAL, THERMAL, AND FLUID FLOW
INTRODUCTION The problem of torsion (Chapter 11) and a number of other problems that we shall consider in this chapter are often known as field problems. They are governed essentially by similar differential equations, which are special cases of the following general equation [1]:
U £) + U k
The
k
+
U %) + °=°%-
(12 - la)
k
associated boundary conditions are q>
kx
>%)
d
£l + kjjgt, + x
= p(t)
k^Ll,
+
<*(
1
on S
-
cpo)
(12-2a)
!
+
iff)
=
on S 2 and S 3
0,
.
(12-2b)
Here
cp
is
the
unknown
(warping, stress function, velocity potential, stream
head or potential); k x ,ky and k z are material properties in the x, y, and z directions, respectively; Q is the applied (heat, fluid, and so on) flux; c is specific heat or effective porosity, and so on S is the part of the boundary on which (p is prescribed and S 2 is the part of the boundary on which the intensity of flux q is prescribed. In the case of heat flow a(^ —
function, temperature, electrical potential, fluid
;
,
{
,
,
299
Potential, Thermal,
300
and Fluid Flow
Chapter 12
the direction cosines of the outward normal to the boundary, t denotes time, and the overbar denotes a prescribed quantity. In this book, we shall consider only two-dimensional steady-state problems; that is, the problem is independent of time and the right-hand side of Eq. (12-la) vanishes. Also, for simplicity, only homogeneous materials are
considered; then Eq. (12-la) reduces to
kx %g
+ k*j* + Q =
and the boundary conditions
to
=f
q>
(12-lb)
on 5,
(12-2c)
and kx
We note that Eqs.
d
/x
lx
+ k,fy, + q =
(1 1-5)
and
(1 1-27)
on S 2
(12-2d)
.
have the same form as Eq. (12-lb).
POTENTIAL FLOW The
potential flow of fluids
is
governed by a special form of Eq. (12-lb),
dx 1
+
dy 2
or
V> =
(12-3)
0.
which is called the Laplace equation. Here V 2 is a differential operator. The assumptions commonly made are that the flow is irrotational that is, fluid particles do not experience net rotation during flow, the friction between the fluid and surfaces is ignored, and the fluid is incompressible [2]. Some practical problems where this kind of flow can be assumed are flow over weirs and through pipes (with obstructions). Equation (12-3) is based on the basic requirement that the flow is con;
tinuous; that
is,
where v x and v y are components of velocity in the x and y direction, respectively. The flow problem can be represented in terms of either the velocity potential
q>
or the stream function
that for the stress-deformation
the
same manner
respectively.
as
the
y/.
This dual representation
problem
in the sense that
(p
and
is
y/
similar to
are used in
displacement (warping) and stress functions,
Chapter 12
The
Potential, Thermal,
relations
and Fluid Flow
301
between the velocity components and
q>
and
y/
are given by
[2]
x
~
dx (12-5a)
and
57 (12-5b) tfy
"' =
Substitution of i\ and
p, into
?
Eq. (12-4) leads to the Laplace equation as in
Eq. (12-3).
Boundary Conditions
For flow through a cylindrical pipe (Fig. 12-1) a boundary condition is and the wall have the same normal velocity. Hence, if the wall stationary, we have
that the fluid is
V.w =V..n =0, where n
is
V
and
the unit
Vw
(12-6)
are the velocities of the fluid
normal
and the
wall, respectively,
and
vector. Substitution of Eq. (12-5a) into Eq. (12-6) yields d(P n
-
-0
d(P n
Sv
(12-7a)
,
n
S,-^'
Figure 12-1
From
Flow
in pipe.
Fig. 12-1.
dx »,
dn dv
_ ~
dy
(12-8a)
ds
dx
(12-8b)
Potential, Thermal,
302
Hence Eq.
dx dx dn
is
Chapter 12
(12-7a) transforms to dtp
which
and Fluid Flow
,dydy
+
similar to Eq. (12-2b). This
condition (Chapters 2 and
The
3).
the flow or
is
(12-7b)
0,
dn
dy dn
Neumann-type boundary boundary condition
potential or Dirichlet
is
onS
=
x
(12-8c)
.
Often both the flow and potential boundary conditions occur together, which is
called the
FINITE
mixed condition.
ELEMENT FORMULATION
For two-dimensional
idealization,
we can
use a triangular or quadrilateral
element. Formulation with a triangular element will be essentially identical
one described for torsion in Chapter 11. Hence, we present a formulaby using a four-node quadrilateral element (Fig. 12-2). It is possible to
to the
tion
use either
(p
or
y/
(or both) for the finite element formulation.
sider a formulation with
We
first
con-
(p.
t
(x 4
,
y4
)
M.+1)
/§T
@7( +1
/
(-1,-1)
<
+1
^^-^_~®/^ (x 2
,
y2
>
s
)
(+1,-1) »~
Figure 12-2 Quadrilateral isoparametric element.
The velocity potential is a scalar and has one value at any point. For the four-node quadrilateral, there are thus four nodal degrees of freedom, and a bilinear
model
for
q>
at
any point can be written as
(p
= g, +
9
= mm*
CC 2
X
3
y
+
&*xy
(12-9a)
or (12-9b)
where
=
[]
x y xy] and {a} r = [<%! a 2 a 3 a 4 ] is the vector of generalized Note that the term xy yields a bilinear distribution compared
coordinates.
Chapter 11;
by the triangular element discussed
in a given direction the distribution is linear,
pt (Pi
^3 ^4
in
however. Evalua-
nodes yields
at the four
303
[1
to the strictly linear distribution given
tion of
and Fluid Flow
Potential, Thermal,
Chapter 12
= a, + a
2
x
+ y + 83^2 +a + a 3^4
x
= + <*2*2 =a +ax = ai + a *4 <*1
2
,
1
4 j> 3
3
2
+ aA x y u + a ^2> +ax + xy x
;
4
4
o^ 4
2
J
3 .y 3
,
4
(12-10a)
4i
or {q,}
where
[A]
(12-10b)
[A]{a],
the square matrix of coordinates of the nodes
is
^4]-
[
=
We
r {q,,}
=
can solve for {a} as
{*}=[Ar{q,} and
and
(12-lOc)
substitute the results into Eq. (12-9b) to yield
9
The product
mwr i%i = IMW = £ N&-
=
l
!
[
(12-11)
gives the matrix of interpolation functions [N],
and
in terms of the local coordinates s
N N N
(Fig. 12-2) are given
/
t
=i(l-s)(l-t),
2
=#l+s)(l-t),
which
by
(12-12)
=i(l+s)(l+t% N<=±(l-s)(l+t). 3
Here,
s, t
forty,
The
i
are nondimensionalized local coordinates. Figure 12-3 shows plots
=
1,2,3,4.
global coordinates x,
y
at
any point
in the
element can also be
N
expressed by using the same interpolation functions
i9
S Nx t
t
(12-13a)
y
= S N,y„ i=l
or
x \- ~N, N2 N N y\N, N r T where {x„} = [x, x x x and {y„} = [y, y 3
0~ {{*„
4
#3
2
2
3
4]
2
are both expressed
4
(12-13b)
n
y 3 y4].
This approach in which the geometry, that
unknown
N _ \{y
is, x, y coordinates and the by using the same interpolation functions
:
.
Potential, Thermal,
304
and Fluid Flow
©
©
Figure 12-3 Distributions of interpolation functions
N
t
is
called the isoparametric concept.
cept offers a
Chapter 12
number of advantages
As
in
N
(
,
i
=
1, 2, 3, 4.
stated previously, use of this con-
terms of easier differentiations and
integrations.
The approximation function
in
Eqs. (12-9) and (12-11)
satisfies
the
requirements of continuity, comformability and completeness for the flow
problem governed by Eqs. (12-1) and (12-14). It does not, however, include all terms in the polynomial expansion, see Chapter 1 1 Step
4.
Use of
Derive Element Equations either a variational or a residual procedure for the
problem gov-
erned by Eq. (12-3) will yield essentially the same results. We consider the following variational function for the two-dimensional idealization
™-JR[@' +(£)>* We
note here that the terms
d Jt d \_ JP. 2 dx dx
and
i-^z 2 dy dy
(12-14)
Chapter 12
Potential,
Thermal and Fluid Flow
305
are similar to those in Eqs. (3-21) and (4-6) and can be considered to represent
a measure of energy.
The of
tp
by taking partial derivatives
derivatives in Eq. (12-14) can be obtained
x and y
[Eq. (12-11)] with respect to
_ ~
gx
dtp
dx
_ ~
dtp ds ds dx
_ ~
dtp
:
dtp dt
+
~di
dx (12-15)
dtp g>
Since the
N
dy
ds
dtp^dt_
ds dy
dt
dy
are expressed in terms of local coordinates
t
s, t,
we
use the
following mathematical results based on the chain rule of differentiation in
order to find gx and gy
:
dN = dN
dx dx ds
tr
ds
t
dNt== dN
dx
dx
dt
t
dN
dy dy ds {
,
6N
.
t
7 '
dy }
dt
which as a general
rule, in
dy dt'
matrix notation, can be expressed as
d)
id] dx
[*1 Ts
dx
dy
ds
ds
d
dx
[dt,
Tt
dy d Tt_ Idyl
>
The matrix
[J] is
=
dx (12-16b)
[J]«
d Idyl
often referred to as the Jacobian matrix. Equation (12- 16b)
represents a set of simultaneous equations in which d/dx
unknowns. Solution by Cramer's
Id] dx >
=
[J]-i<
d
and d/dy are the
rule gives
>
d
M
dt
dy
dyl (d\
dt
ds
ds
dx
dx
d
(12-17) dt
where the determinant
\J\ is called the 7
,
Jl
The terms
in
=
dx dy
~dJdI
{dt,
dx dy
(12-18)
dt ds'
Eq. (12-18) can be evaluated by using expressions for x and y
Eq. (12-13)] and
57
_
ds_
Jacobian,
N
t
[Eq. (12-12)].
For instance,
" t)Xl + *° + s)(l " t)Xl + i(l -f s)(l + t)x + 1(1 + 0*J = -1(1 - 0*i + i(l - t)x + 1(1 + 0*3 {X l fs *
"
5)(1
W
3
2
i(l
+
0*4
(12-19a)
306
Thermal and Fluid Flow
Potential,
Chapter 12
and so on, and
and so on. Then
-
-i(l
ds
(12-19b)
evaluated as
|/| is
ii-sitewf")-®**")] (12-20a)
By
setting
i
=
4 and j
1, 2, 3,
=
1, 2, 3,
4 for each
the
i,
summation
in Eq.
(12-20a) leads to |/| in matrix notation as
\-t -s + -l+t 1+j s-t -\-s 1-5 -1 —t S +
-l+s~
t
X "gL-M
Xt
i
X,
X-i
-s-t
yi
l+t
(12-20b)
t
Expansion of /
1
|
in Eq. (12-20b) gives
= H(*i - x )(y - y - (x - x )(y - y + s[(x - x ){y, - y ) - (x, - x )(y - y + t[(x - x Xy - y - (x, - x )(y - y )]} = i[^13^24 - *24.Fl3) + S(x 34 y l2 ~ *12j>34) + ^23^14 -^14^23)], where x u = x — Xj and y = y — yj. 4)
2
3
4
3
3
2
2
4)
s
3)
l
2
4 )]
3
4
i}
t
Use of Eq.
A
2
2
(12-20c)
3
t
(12-17) allows computations of ds/dx, dt/dx, ds/dy,
and dt/dy
in Eq. (12-15) as o
ds
_
1
dydk_
(iyis
dy
1
_
1
($*N
t
\
ai . 2 u\
Similarly,
dy' dt
?i
from Eq.
(12-11),
)
f±dN y \ = ~\T[\k~~SF 1
dy- \J\\M Now
(12-21b)
Xi
\J]\&i dt t
(12-21c)
T
(12-21d)
Xi )'
ds
we have
dN % = -£» dtp
{
*
m and
dq>
dNim
f^-fff*
(12-22)
x
an4 Fluid Flow
Potential, Thermal,
Chapter 12
and (12-22)
Substitution of Eqs. (12-21)
307
into Eq. (12-15) finally leads to
d
dx (12-23a) dtp
& The
indicial notation
instance, the setting
i
=
first
and/
1
dN m V(dN x
+ ,
When
=
h h i/'W ~df ~ ~5T ~J7)
(dN dN
1, 2, 3,
dN
6N,
v
(dN dN2
dN,
x
,
3
is
obtained by
dN
2\
dN dNA ~W) y \
(ON, dN,
,
W x
added to the other three terms obtained by 4; i 3, j — 1, 2, 3, 4; and i = 4, j = 1, 2, 3,
this is
1, 2, 3,
row
"
3\
\W ~dT ~ ~6T ~di) yz ^\d7Tt x
first
4 as
dN, dN,\
1
\\T\\
Eq. (12-23) indicates double summation. For
in
term in the summation for the
=
Xi
=
setting 4,
(12-23b)
i
=2
we have
and j
d
Similar evaluation gives dtp/dy.
After relevant substitutions and rearrangements, 9\
#12
#13
#14
B
-"23
B
B"24
=
M{q,},
#11 B-,21 A
rr -"22
(12-24a)
or {g}
(12-24b)
where
#u
8|/|
0>24
-^34^ -y23*)>
1
#12
#13
8|J|
=
(-^13 +^34^+^140.
1
8T7t( l
#14 8|/|
_
;
->
24
+y^s
-yut),
Ol3 -^12^+^230. (12-24c)
1
#21
8|/|
8|/|
—
\X 13
X 34 S
1
8|7|
X 34 S
x \2 S
8~L7T 1
#24
+
1
B 22 B 23
(~ *24
(—
13
+
+
*2 3 0,
^140»
"T"
x l2 s
*140>
-
x 23 t).
Potential Thermal and Fluid Flow
308
Now, we
substitute {g} into
np =
T
Qp
[Eq. (12-14)] to yield
(12-25)
l\lBY[H[B]dxdy{q,l
{q,}
Chapter 12
A
where [I]
=
0"
1
[C] 1
Taking derivatives of Q^, with respect to
en, that
=
h
we
obtain (12-26a)
0,
is,
d
dQ,
= )==> sq p
=
(12-26b)
o,
d
dQ,
=
which leads to r {H} JJ[BnB]rf^{q,}
=
(12-27a)
A
or
\\lW[R]dxdy{%}
=
(12-27b)
or
Mi,} = where [kj
is
(12-27c)
("I.
the element property matrix:
[kj
=
jj[BY[B]dxdy.
(12-28)
A
Numerical Integration
The
coefficients
k9ll
of [kj need integration. They look like
= ffah + Blt)dx4y JJl(8|J
(j 2 4
-y 4S-y 23 3
t)
:
A
+
/
8
)2
|
|
(-^24
+
*34*
+
X 23
2
]^^
(12-29)
y
Potential, Thermal,
Chapter 12
and so on.
It
can be
difficult to
and
and Fluid Flow
309
evaluate these integrals in closed form, and
it
perform the integration numerically. The idea of numerical integration in finite element analysis is similar to integration by using well-known formulas such as the trapezoidal and Simpson rules. In finite element applications, the Gauss-Legendre [3] formula is
often convenient
is
often used. In general terms, numerical integration can be expressed as
efficient to
P F(x)dx = 2 F(
)W
Xi
(12-30)
t,
1=1
Jxi
where i denotes an integration point, m is the number of such points at which the function is evaluated and summed [Fig. 12-4(a)], and the are the weighting function. That is, an approximate value of the integral in Eq. (1230) is obtained by finding the summation of the values of the function at a number of points multiplied by a weighting function at each point. For the stiffness matrix in the two-dimensional space of the quadrilateral, we have
W
t
=
[K]
f_JjBY[B]\J\dsdt m
_n
^S E[B(*„f j=l j=l
y )F[B(*„f,)]
|/(*,,/,)
|
W,»0,
(12-31a)
m
and n are the number of integration points in the two coordinate and and Wj are corresponding weights. Note that we transformed the integral to local coordinates by using \J\. As an example, we can use m =2 and n — 2 as shown in Fig. 12-4. Then total integration points are 4, and we can write Eq. (12-3 la) as
where
W
directions [Fig. 12-4(b)]
W-t
t
[Bfe ORBfe,
W
The magnitudes of (s t t) and which gives them in ready-made {,
Example It is
12-1.
t
t t)]\J(s t , t t)
|
W
(12-31b)
t.
can be obtained from available literature
tables
[3].
Numerical Integration
useful to understand the subject of numerical integration because
many
finite
element applications.
it is
used in
We illustrate here some of the steps for numerical
integration over a (square) quadrilateral [Fig. 12-4(c)]. In view of the fact that
hand
can be lengthy and cumbersome, details of only one term will be attempted here further details on this element are also given in Chapter 13. First we compute the common terms required for integration of term
calculations of
all
terms of matrix
[k,,] ;
ku
inEq.
From
(12-29).
Eq. (12-20c):
x3
=0- = 1
-1,
= 1 - = 1, x4 = l -0 =
x4 *34
—
x3
1,
^24
= y 2 - y* =
yi3
=y\ -y$
yi2=yi -
2
o
-
l
= -i
=0- = l
-l
=o -o = o,
F(x)
F(x)
1
III
1
1
1
II.
Si
s s
2 3
sa
= ~a, t, = + a, t :
=
+a U #
- -a, t A
+a +a
(1. 1)
(0, 1)
©
,
=
0, y,
=
2=1,72=0
(-1, 1)
(1,
+
+
4
3
'(3)
D
3
= 1.Y3 =
1
4
=
1
0,
v4 =
-f- Integration points
0,0
©
Point
1
2
+
+
(-1,-1)
1
2 3 4
(1,-D *
t
-0.577 -0.577 +0.577 +0.577
»-
1
(0.0)
s
-0.577 +0.577 +0.577 -0.577
i
1,0 (c)
Figure 12-4 Numerical integration, (a) Schematic representation of numerical integration of a function, (b) Numerical integration
over quadrilateral,
310
(c)
Example of numerical
integration.
:
= x — x2 = — 1 = — 1, = x — x3 = 1 — 1 =0, = A _ x4 = 1 - 1 = 0,
A'12
x23 Xl4
and Fluid Flow
Potential, Thermal,
Chapter 12
x
>'
34
2
j>
14
= = =
y 23
-j
>'
3
vj
y2
311
- }'a = - y4 = -y =
-
1
3
1
1
1
= = =
0.
-1,
-1,
and \J\
=
-
-§-{[(-D(-i)
U
U
8
where A
is
only the
first
-
(D(-i)]
8
=
unit 2
1
term contributes to |/| and is
(-1X0)]
-
r[(0X-l)
- (OX-DD
4
8
the area of the square
square, which
-
sfdXO)
constant for
all
.
its
Note
that for a (square) quadrilateral
value
one-fourth of the area of the
is
points in the square.
Now, numerical
integration in
Eq. (12-29) can be expressed by using Eq. (12-3 lb) as
*,n
=S r=
»?ife, U)
-
Bi^s;. fJM J(*fc
tt)\
W
i
{,
=
1, 2, 3, 4.
i
From
the available numerical integration tables, such as in Ref.
[3],
we can
choose an appropriate number of points of integration and weighting function
W
t.
For a 2 x 2 or four-point integration with Gauss-Legendre quadrature, the points of integrations are as shown in Fig. 12-4(c), and the weighting functions are W\ = = W+ = 1. Here we have used the local coordinates of the integration 1V2 = 3 points only up to third decimal; for computer implementation, values with higher decimal points are used. The computations for B :1 and B 2l are shown below in
W
tabular forms
B\\
=
B2i =
g
I
y
I
(?24
—
8|7|(- A'24
J34-S"
—
>'23^)»
- x34S -
x 23 t).
Points of Integration
(-0.577, -0.577)
B 2i
(0.577,
-0.577)
(0.577,0.577)
1.577
1.577
0.423
2
2
2
2
1.577
0.423
0.423
1.577
Therefore,
-
(-0.577,0.577)
[Bh(s2i
t2 )
W,(*3, h) [flf,(J4,
U)
- Bl {s 2 - Bl(s + Bl (s A x
3
x
,
,
,
t 2 )]\J(s 2
,
t 3 )]\J(s 3
,
t 2 )\
t3
W
2
)\W3
U)]\J(s A U)\H>\ ,
0.423
© Potential, Thermal,
312
=
=
X
+
(-0.423) 2
l
X
^(9.9477
Other terms of Step
W
T
[k 9 ]
+
+ (-
[(_1 577)2 -
+
+
(-0.423) 2
0.7157)
and the load
and Fluid Flow
1
-
577 ) 2
+
(-0.423) 2
= ^|^ =
(1-577)2
+
Chapter 12
+
(-0.423)2
(-1.577) 2 ]
0.667.
vector(s) can be evaluated in a similar
manner.
Assembly
5.
Equations (12-27) are assembled such that the potentials at common nodes
The
are compatible.
assemblage equations are
final
= {0}.
tKJ{r,}
Under (12-32)
(12-32)
boundary conditions, Eq. and then solved for nodal potentials.
the application of only Dirichlet-type is
modified for given
(p t
Applied Fluxes
In addition to the flow caused by difference in velocity potentials, the
system can be subjected to a number of physical forcing parameters.
A fluid flux Q can be applied as a source or a sink, concentrated at node(s), within the flow domain. Also, a fluid flow q can be applied at the boundary.
Then
the
boundary condition [Eq. dtp
2? dn If
we provide °-
-
for
Q
and
q,
M©
+
q
(12-7b)]
=
(12-33)
0.
the variational function in Eq. (12-14)
2
becomes
2
-
+
2
A
Use of the
becomes
fl^ - J>*
<
12 - 34 >
stationary principle then leads to
WW =
C2-35a)
{Q3.
where {Q}
=
JJ
[NY{Q}dxdy
+
J
[Wl$ds
(12-35b)
A
= is
{Qi}
+
{Q 2 }
the nodal forcing parameter vector. Evaluation of [Q]
This vector can be evaluated by using numerical integration. The expanded forms of the two parts of {Q} are given below. The first part is
Chapter 12
Potential, Thermal,
{Qi}
=
\\lW{Q}dxdy
=
i
Ni(st
,
N (s h N (s h N4(s Here we assumed
Q
The second part
ii
[W{Q}\J\
j
ti
dsdt
)\w
i
t t)}
2
t t)
3
tt)
i9
313
l
£
= i:ms ,t )Y{Q}\j(s i
and Fluid Flow
Q\J(s
t
,t t
)\w
t
(12-36)
.
t t)
to be a uniform flux.
relevant to the
is
boundary
only.
As an
illustration,
consider side 1-2 of an element (Fig. 12-5) subjected to components q x and qy Then the second part for q x specializes to
.
N N
t
{Q 2 }
2
=
f
A
(12-37a)
q x dS.
r
3
Figure 12-5 Boundary fluxes.
We
—
1,
i(l
note that along side
and
s varies
- s\ N = 2
{(W
=
£(1
from
1-2,
3
w-
*)
+
*)
coordinate
t
has a constant value equal to
+1. Therefore, the N specialize 0, and NA = 0. Hence, we have
to
+ s), N = 10
f
—1
qx dS =
{
K 1
£
o
N = {
*)
fcl'IrfS.
.
to
(12-37b)
and Fluid Flow
Potential, Thermal,
314
Since s
= 2S/l
l9
we have dS
where
is
lx
Chapter 12
=
lj-ds
=
\J\ds,
(12-38)
the length of side 1-2. Therefore, fl
=
{Q 2 }
i
f,
-s]
+
*
(12-37c)
>q x ds,
o
which reduces to
=
(Q 2 }
^
(12-37d)
This indicates that the total flux on side 1-2 in the x direction equally between nodes the
x and y
1
and
2.
is
divided
Contributions of other applied fluxes in both
directions can be evaluated in an identical manner.
STREAM FUNCTION FORMULATION Formulation by using stream function with velocity potential
(p.
The
differential
g+$=
as an
y/
is
similar to that
vy = o,
or
and the corresponding variational functional
«*»
unknown
equation for flow with
y/ is
(12-39)
is
-am ©>* +
(12-40)
A
Stream function
y/
can
now be
V=[N„]{q„}
By following
expressed as
=
£N„,.^.
(12-41)
a procedure similar to the one above, the element equations
are (12-42)
[Mq,}={0], where [kj
is
the element property matrix
and {%}
vector of nodal stream functions.
The boundary conditions
are expressed as y/
where S
{
is
the part
on which
=
y/ is
y/
onS b
prescribed.
T
= WiWi
¥s V*]
ls
the
Secondary Quantities
The
velocities
can be computed by using Eq.
(12-5).
For the velocity
potential approach, Pi
^11
#12
^13
#14
.-#21
-#22
^#23
#24j
fi
(12-43)
(Pa)
and
for the stream function approach.
V: -#21
-#22
-#23
_ L -#11
""-#12
_ ^13
Vi
^24 ~~
(12-44)
-#14.
Wa)
The quantity of flow Q f across
now found
can be
a section
A-A
in
an element
(Fig. 12-6)
as
QfM =AV
(12-45)
n,
where A is the cross-sectional area at section A-A and Vn is the velocity normal to the section. Vn can be computed from the two components v x and v y for an element as vXt sin 9
where
i
denotes an element and 9
the horizontal.
The
is
—
Vy,
cos 9,
the angle between the section
total flow across a section
fj
(12-46a)
2,A V t
where j denotes the cross section and
(12-46b)
H
M
A-A and
can be found as
is
the total
number of elements
across the cross section. Figure 12-6 Computation of quantity of flow.
315
Example
Potential
12-2.
Flow Around Cylinder
shows the problem of uniform around a cylinder of unit radius. The flow domain is 8 x 8 units. Due to symmetry, only one-half of the domain is discretized, as shown in Fig. 12-7(b). (a)
Velocity Potential Solution. Figure 12-7(a)
fluid flow
Figure 12-7 Analysis for potential flow around cylinder, (a) Flow
around
cylinder, (b) Finite element
(0, 0) etc.,
mesh
for half flow domain.
denote coordinates.
I
i
1
„ *
1
- ±J
4 y
t^ 4^
m
—D +->
'c
00
-
1
'
i
L = 8 units (a)
(8,4)
(8,0) 2§) ^"^
Orinin for fnr Origin finite
element
analysis
316
(b)
(4J
"- Origin for closed
(46)
(gl)
form solutions
Potential, Thermal,
Chapter 12
The boundary conditions (p
(p
— =
1
and Fluid Flow
317
for the velocity potential are
unit along the
upstream boundary,
nodes 1-5,
unit along the
downstream boundary,
nodes 51-55.
(12-47)
The computer code FIELD-2DFE
(for further details, see
Appendix
4)
was used
to obtain numerical solutions. Solutions for velocity potential are obtained by setting
NTYPE =
2 in the code.
Figure 12-8 shows computed distribution values of velocity potentials at nodes in half of the flow
tions
from
domain. Table 12-1 gives comparisons between closed form solu[2] and numerical results for cp at the nodes along the line y = 0.
Streeter
The formula
for closed
form solution
where [
=
is
the uniform undisturbed
0)
-
8 [Fig. 12-7(a)],
= L)]/L = (1 r = (x 2 — y 2
1
)
measured from x
[2]
+ y) cos 6 -
u(r
U
=
given by
is
0)/8
2
,
a
velocity
=
is
0.125,
(12-48)
0.50,
in
L
=
the negative
x direction domain
length of the flow
the radius of cylinder,
and 6
is
= =
the angle
axis.
TABLE
Comparison of Computed and Exact
12-1
Solutions for
Node
Computed
Exact
26
0.5000
0.5000
31
0.3743
0.3780
36
0.2500
0.2765
41
0.1875
0.2132
51
-0.03125
0.0000*
Prescribed.
Stream Function Solution. Specification of
(b)
2DFE
NTYPE =
3 in the
code FIELD-
permits solution for the stream function approach. For this solution, the fol-
lowing boundary conditions were used:
W = y/
=
0.5 along the top
nodes 5-55, (12-49)
0.0 along the
bottom nodes, 1-16-26-36-51.
Figure 12-9 shows the distribution of computed values of y/. Table 12-2 gives comparisons between closed form solutions and numerical predictions for y/ along the line of nodes 26-30.
For the (0.5
-
line considered,
0.0)/4
=
0.125.
The closed form
6
solution
y/
= u(r -y)sin#.
=
90 deg, and
r
=
(x 2
is
given by
[2]
(12-50)
— y2
1
)
2
=
y.
The value of
U=
'"
'
'
—"
—*
'
'
paquosajj
8
o lo o—
*
1 1
1
//
/
i J
/
o LO o—
1
/o
/o /
/
^ °
LO
r^-
LO
o CO CO
o —
*
CO r>.
o —
1^
I
1 1
^ d
\
\
•**
\o
V \
w \ • \ o V*"""
\
\ \^
\
»fr
CO
\ o \
CO r^
r^ r^
"~"
o
o
0) W)
o —
1
00 00
o —
\\ d *
00 r-
o —
Oi 00
o o
o o
o o
,
,_
—
— paquosajj
318
o
T
—
0)
Q.
^M
QX OJ
c
\\
vv
_
'
o o "—
T3
,
•2
"
O D
\o
00 00
o
o a
C £ £ -c
d
00 00
CD
j-
-C T3
c
s CO I
o a
.2
\
>^
C
c
o
«N *°
\
o
CTl
0)
**>
1
00 LO
^
o
o o
/
O
*
1
M
JO
,
3 a F3
o
+3
U
o Z a> w o O
00 <4
>
CO
a.
aj
re
(U
c c
o c
o o
CD
o
D
h 3 00 to
—
'
1 i
.
1
,
05 °° 1
n o N
D n n
J
a> 1
ol
a
/ / /^
/
r^ ^
/
°-
/
—
\
Id
\
d —
^ o —
\
"
\
V
\
<©
^ \°
\
\
^ *\ o
\
V
_-.
/ /
/in
s* I
\
M
-
\
-
\ CN \ d \—
j
o
as
£
-a
s
— s^ CO CO
o d
\.
=
-—
0)
-a
2:5
.
g
\^
\.
\.
O _ J^ c CO O C C o o o o c c 3 3 >4- Hb b ra
_
*-•
a
—
•«->
CO CO i- CN
co CN
d
"-"
"
? CN
CN
p^
m d —
f^ CO
d —
d — 1
o c
•a
o 3
C £
o
U
9s ri
V
3
M
«— «—
CO CO
d
r
a s
c C
d \ ° \
53 a>
° °
'o
\
o o
d —
d
^ CN
CN
d
d
319
Potential, Thermal,
320
TABLE
and Fluid Flow
Chapter 12
Comparisons Between Computed and Exact Solution for y/
12-2
Node
Exact
26
0.0000
0.0000*
27
0.1473
0.1529
28
0.2625
0.2746
29
0.3678
0.3892
30
0.4688
0.5000*
Computed
Prescribed.
Figure 12-10 shows comparisons between nondimensionalized values of the x component of velocity v x along the section of nodes 26-30. The closed form value is obtained from [2]
or 1
The numerical
+~
(12-51)
results plotted in Fig. 12-10 are the values of v x in element(s) adjoin-
ing the section.
The comparisons
in Tables 12-1
and 12-2 indicate
that the finite element predic-
The plots for v x /U in Fig. and stream function approach give potential approach yields the lower bound
tions yield satisfactory solutions for the flow problem.
12-10
show
that both the velocity potential
satisfactory comparisons.
The
velocity
Figure 12-10 Comparisons of velocities at midsection.
2.0
V*—-Closed form Stream function
i_ 1.0
f Velocity potential
i.i. i.i.i y =
26
28
27
Nodes
29
30
Potential, Thermal,
Chapter 12
and Fluid Flow
321
and the stream function approach the upper bound of the exact velocity. numerical solutions can be improved by using finer meshes.
If neces-
sary, the
THERMAL OR HEAT FLOW PROBLEM In the case of the heat flow problem, the general governing equation essentially the
are different.
is
same as Eq. (12- la) except that the meanings of various terms The unknown can now be the temperature J at a point; k x ,
ky and k z are thermal conductivites; Q is the (internal) applied heat flux; and q is the (surface) intensity of heat input. In addition, there is the possibility of heat transfer due to the difference between the temperature of the medium, T, and the surroundings, TQ The differential equation for twodimensional steady-state heat flow then becomes ,
.
T k** + k* dx dy y
1
z
2 =
(12-52)
0,
and the boundary conditions are
T= 1
on S
(12-53a)
and kx
tx
k
d~x
where a
The
is
>o\
C>
ol(T
-T )-q =
onS
Q
2
and 5
3
,
(12-53b)
the heat transfer coefficient.
finite
element formulation with the quadrilateral element
will result
in equations similar to Eq. (12-35) with the addition of the following
term in
CI, [Eq. (12-34)],
i
which
- T Y-ciS.
(12-54)
[kj and {Q 3 } to the element
will lead to additional contributions
matrix and the forcing parameter vector, respectively:
For evaluation of
[kj
=
{<2 3 }
=
[k a ]
f
f
a[N] r [N]
(12-55a)
*\Nr{T
(12-55b)
we need
]dS.
to integrate along the sides (boundaries)
of the element. As an illustration, consider side 1-2 (Fig. 12-5) as before:
= ^^lW[K]dS c
[K]
-*£
i(l
-
*)
>[£(l-s)
i(l-s)
0]dS
.
Potential, Thermal,
322
i(l
-
i(l
*£
2
* )
i(l
and Fluid Flow
+
Chapter 12
2
s ) s)
2
dS 0.
"W olI x
i
o
o-
i
(12-56)
2
Computation of {Q 2 }
similar to that for q x in Eq. (12-37a).
is
Finally the element equations for heat flow will be
(M + where
{q r
f - [^ T T
Example
As a
IW){qr} 2
12-3.
3
= {Q} = {6,} +
{Qi}
+
{Qil
(12-57)
rj.
Two-Dimensional Heat Flow
simple illustration, consider steady-state heat flow in a rectangular block of
The heat flow domain is divided The boundary conditions are assumed as follows:
unit thickness (Fig. 12-11).
12 nodes.
= 7X3, y) =
7X0, y)
into 9 elements with
100 deg, (12-58)
25 deg.
conductivities are assumed to be equal as k x = ky = 1 Table 12-3 shows the computed values of temperatures at various nodes ob-
The thermal
tained by using the code
FIELD-2DFE.
TABLE
12-3
Node
Numerical Results for Temperatures Temperature (deg)
1
100.0000*
2
100.0000*
3
100.0000*
4
74.9998
5
74.9998
6
74.9998
7
49.9999
8
49.9999
9
49.9999 25.0000*
10 11
25.0000*
12
25.0000*
'Prescribed.
Potential, Thermal,
Chapter 12
and Fluid Flow
©
323
©.
A
a
—-
©
©
A ©
©^
3 units
Figure 12-11 Heat flow in two-dimensional body.
SEEPAGE Seepage
is
defined as flow of fluid, usually water, through porous (soil) media.
The governing equation
is
similar to Eq. (12- la) except the
meanings of
various terms are
Unknown,
——z
(p
k x ky k 2 ,
the total fluid head or potential,
is
coefficients of permeability in the x, y, z
,
(12-59)
directions, respectively,
Q = q
=
applied (internal) fluid flux, applied (surface) intensity of fluid,
where p is the fluid pressure, y = density of water, and z is the elevation head [Fig. 12-12(a)]. For two-dimensional steady-state confined flow, the element equations are identical to Eq. (12-35) except for the meaning of the terms. Often seepage is
defined as confined or unconfined
occurs through a saturated ditions
medium
and does not involve the
[4].
so-called free or phreatic surface. In the case
of unconfined seepage, on the other hand, surface.
In the confined category, seepage
subjected to prescribed boundary con-
Both categories can involve
we encounter
the free or phreatic
either steady, or unsteady or transient
conditions.
When 12-12(b)],
and surface of seepage mixed boundary conditions on these surfaces:
the seepage occurs with a free surface
we have
the
q>
=
z
[Fig.
(12-60a)
and
** 911
= o.
(12-60b)
Potential, Thermal,
324
and Fluid Flow
Chapter 12
U (a)
Free surface
Surface seepage
©
0,0,0
(2.0)
(0.0) :
:
M: -:0 -
(2.0)
(0.0)
,a5,
(c)
Figure 12-12 Seepage through porous media, (a) Three-dimensional seepage, (b) Free surface
and mixed boundary conditions,
(c)
Steady
seepage in porous medium.
At the free surface, the pressure is atmospheric; hence, p = and cp = z. The second condition implies that the velocity of flow normal to the free surface is zero. The two conditions [Eq. (12-60)] must be satisfied simultaneously. The mixed conditions render the problem nonlinear. Consequently, in contrast to the steady confined flow, which requires only one finite
element solution,
it is
often necessary to perform iterative analysis in the case
of free surface flow. This requires a number of topic
is
beyond the scope of
finite
element solutions. This
this text; the interested reader
can consult
Refs. [4]-[6].
Example
12-4.
Steady Confined Seepage
Figure 12-1 2(c) shows a rather simple problem of seepage through porous (soil) media with coefficients of permeability k x = ky = 0.005(L/r) and thickness =
Chapter 12
1
Potential.
The boundary conditions
unit.
q O
= =
Thermal and Fluid Flow
are
= =
atx
2
x
at
The computed values of
at
525
0.
(12-61)
2 units.
(12-61)
various nodes are shown in Fig. 12-12
parentheses.
The computed values of
velocities are
shown
in
Table 12-4. These values com-
pare closely with the velocities computed from Dairy's law:
TABLE
Computed Seepage Velocities
12-4
Element
I
1
0.005
-0.45
10"*
2
0.005
-0.1S
10-*
=
.
As expected, the
--:
= S dx
0.005
=
0.005.
d
\2
components of velocity are very small, almost equal 12-46b was found to be equal to 0.01 unit.
;.
quantity of flow [Eq.
Example
V,
'.,
to zero.
Steady Confined Seepage Through Foundation
12-5.
The foregoing problem is rather simple and can permit hand calculations. consider a problem that is more difficult and needs use of the computer.
A
problem of steady confined flow through the foundation
(dam. sheet
pile)
shown
is
with different coefficients is.
kx
=
k:
\
this
included in the
The
.'
1
we
of a structure
The foundation consists of two layers of permeability. The soil is assumed to be isotropic, that in Fig. 12-13' a).
assumption finite
soil
Now
is
not necessary because anisotropic properties can be
element procedure. The structure
itself is
assumed
to be
impervious. The two-la>ered foundation rests on a material which has a very low permeability; hence, implies a zero flow Stead;, rtuid tively,
For
this
i
at
the
depth
natural or
heads o, and
ot"
'.'.
m.
impervious base
Neumann) boundary
:"-
act :n the
is
assumed. This
condition.
upstream and downstream
sides, respec-
problem, we have assumed
a = g, =
1
m. (12-62)
0.
which constitutes the geometric (or forced or Diriehlet) boundary condition. Because the problem is linear, results for this boundary condition can be extended to any multiple of fm and fd with proportional difference in the head. For instance, the = 100 m and (p d = 50 m subsequent results can be used to derive solutions for by multiplying the computed nodal heads by 50. The foundation medium extends toward "infinity" in the lateral direction. Because we can include only a finite pan of this extent in the analysis, an approximation must be made to fix the discretized end boundaries. For instance, in
^
20
m
Impervious base (natural boundary condition) (a)
=
1
m
<£
/
3<£
Discretized end boundary
=
^
m
@
©
(b)
^z
Discretized end boundary
1
7^
V//////////////A
" i
1
|
V
\ \ \ i
\ 1
\
1
\
l
i
l
50
40 30
20
10 Percent
(c)
Figure 12-13 Steady confined seepage through layered foundation, (a) Details of foundation, (b) Finite element mesh, (c) Solution for nodal potentials: equipotential lines.
326
)
52"
and Fluid Flo*
Potential. Thermal,
Chapter 12
m (equal to the
en the end boundaries at a distance of 20
Fig. 12-
width of the stru::_:e Groin the edges of the structure; that at that distar.ee the Bold r
ten! ia
:
!
•••ill
be app:
on the upstream and downstream
potentials
this involves
an assumption :he applied
sides. Often,
it
may
be necessary to
perform parametric studies to find the distances at which such conditions can be this topic appears in Chapter 13.
assumed. Further discussion on Figure 12-13 is
needed
b
>r
s
:
at the interface
mesh m ith -5 modes and 72 elements. A nodal line between the two layers. The geometric boundary condi-
a finite
upstream and downstream nodes are p. = 1 m and p d = m. resre:have dq> dz = 0: being the natural boundary condition, it is satisfied automatically in an integrated sense. Consequently, in the tions at the
me bonom boundary we
A:
fin te
element analysis,
problem of the
is
we computed
fluid
heads at nodes
symmetric, we could have solved
medium
is,
al
mis bounda:;-
Since the
by considering mesh only for one-half
: Then at the centerline we can assume a natural boundf/dx = 0. This is valid since at the centerline the rate of
[Fig. 12-13
dary condition; that
it
1
change of g van: sr.es Figure 12-13 ; thowa the finite element computations for nodal heads ob:a ined by using the code FIELD-2DFE. The equipotential lines are shown as a percentage of the total head difference fB — fd .
From on the
the results in Fig. 12-15
structure.
Moreover, the
:
.
we :an
finite
velocities in all or selected elements
find the seepage :~::.es
easing
uplift
element procedure allows computations of
and the quantity of flow
at selected secticr.s
AM
this
information can be used for analysis and design of structures founded on porous
soil
foundations through which seer-ge occurs.
ELECTROMAGNETIC PROBLEMS In the case of steady-state electromagnetic problems the governing differential equation, in the
absence of
in
Eq. (12-la). reduces to the Laplace
equation
or
if
kx
=
A.\.
V-r = where
V
is
(12-
0.
the electric or magnetic potential.
V:
is
the Laplacian operator.
and k x and k y are the material properties; in the case electrical flow they are the permittivities in the x and y directions, respectn ely. The finite element formulations and details will follow essentially the same procedures as in the :^se of the field problems covered in this chapter and Chapter 11. For further details on this topic the reader can consult various :"
:
publications
Thus,
[7, 8].
we once again observe
the similarities of various
phenomena
in
:
328
Potential,
Thermal and Fluid Flow
Chapter 12
and physics and how the method provides a common ground for their solutions. different disciplines in engineering
COMPUTER CODE
finite
element
FIELD-2DFE
problems and permits solution of torsion, and heat flow. The user needs to supply appropriate properties relevant to the specific problem and specify codes for the type of problem as follows
This
is
a general code for
field
potential flow, steady-state seepage,
Problem
Code,
Torsion
NTYPE 1
Potential flow
2
Velocity potential
Stream function Seepage Heat flow
In
fact,
5
the code can be used easily for other steady-state problems such as
and magnetic
electrostatic
Appendix
3
4
flow. Further details of the
code are given in
4.
PROBLEMS 12-1.
Derive in detail (see
12-2.
Chapter
[k]
and {Q}
for potential flow
by using a triangular element
11).
Find {Q 2 } [Eq. (12-37)]
if
q y and q x were applied on sides 1-2 and 2-3,
respectively.
12-3.
By using numerical
integration,
compute the
coefficients
k9tt and k 9li
for
the square element in Fig. 12-4(c). 12-4.
By hand
calculation, find the element matrix [k] for steady-state heat flow
in a square plate of unit thickness (Fig. 12-14) discretized in four equal
square elements. The boundary conditions are
T(x,y)
=
100 deg,
7X2,>0=Odeg, and ky
=
ky
=
1
unit.
Assemble the equations for the four elements and 5, and 6. Hint: You may use the results
solve for temperatures at nodes 4,
and Fluid Flow
Potential, Thermal,
Chapter 12
v
©
©
329
t(9)
(0,2) (2,2)
A
A,
©
©
A© A
©
©
(0, 0)
(2, 0)
Figure 12-14
indicated in Eq. (12-3 la) and the numerical integration 12-1. Solution:
12-5.
Tat nodes
4, 5,
=
and 6
shown
in
Example
50 deg.
=
In Prob. 12-4, consider heat flux q 0.1 unit per unit length along the side 3-6-9 and compute temperatures at nodes 4, 5, and 6. Hint: You may use the results in Eq. (12-37) for finding the forcing function vector {Q}.
12-6.
In Prob. 12-4, consider a source (or sink) the temperatures at nodes, 4,
5,
and
6.
Q = 0.1
Hint:
units at
You may
node
6
and find
use Eq. (12-36) to
evaluate {Q}, which will need numerical integration similar to that used in
Example 12-7.
12-1.
Use the same domain such as a
soil
as in Prob. 12-4 but
kx
and applied
fluid potentials
=
ky
=
(p{2, v)
(a) fluid potentials at
in all the elements;
12-8.
12-9.
(c)
= =
nodes
10 cm, 5
cm.
4, 5,
Consider Prob. 12-7 but include 4, 5,
and
and
6; (b) velocities v x
and
vy
the quantity of flow across a cross section.
Consider Prob. 12-7 but include q = 0.1 and find fluid heads at nodes 4, 5, and 6.
heads at nodes 12-10.
and
cm/sec
0.1
equal to (p{0,y)
Compute
composed of a porous medium
with
cm 3 /sec-cm
Q = 0.1 cm
3
at
along the side 3-6-9
node 6 and compute
fluid
6.
Consider Example 12-5. Prepare three meshes with discretized boundaries
and 40 m from the edges of the structure. By using a computer compare the results from the three analyses and offer comments on the effect of the distance of the end boundaries on the numerical predicat 10, 20,
code,
tions of the heads
under the structure.
330
Potential, Thermal,
and Fluid Flow
By using FIELD-2DFE or another
12-11.
Chapter 12
available code, solve for steady tem-
perature distribution in a composite material for the conditions
shown
in
Fig. 12-15.
Insulated
T=
T=
100 deg
0.0 deg
8 units Figure 12-15
12-12.
By
using
FIELD-2DFE
or another available code, solve for seepage in the
foundation of the sheet pile wall shown in Fig. 12-16.
Figure 12-16
12-13.
Derive the
finite
element formulation for
node isoparametric quadrilateral
field
(Fig. 12-17).
problems by using an
eight-
and Fluid Flow
Potential, Thermal,
Chapter 12
331
(-1,1) (1, 1)
©
/©
®\
(-1,0)
(1,0) ,
J
®
.
(-1,1)
•D
(0
Figure 12-17
Partial results:
Assume
=
u /
[A7"=<
where
[iV]{q},
- 5)(1 - /)(1 ~S ~t)\ -id s){\ - t){\ -s-t) -1(1 - s){\ - r)(l - 5 - f) -id - s){\ - r)(l -s-t) id - **xi - t) -i(l
i(l
-5)(1 -/*)
id -
{qf /1(1
&"-<
Usee)f
u3
u4
u5
u6
t)(2s
JJ
-5)(2/-5)\
1(1
-5)(2r-5)
t)
i(l
-5)(2r-5)
i(l
-
§pr
=<
)
t) y
I
-/(1-J)
then leads to
[brc][B]^
and {Q}
=
5) )
)
2
etc.,
-
-id - * "/(I ~ *) 2 id - *
t)
)
S )(2t
2
2
/
U B ],
/i(l
r
Eqs. (12 -16), (12- 17),
=
u-
t)
-t)(2s-t)
s(\ id -s{\ -id -
[k]
-
f)(25
?
\
u2
[«i
-t)(2s-t))
#1 1(1
1(1
-
**xi
id -5)(1-^)
,
=
>>
\\[XY{Q}dxdy
-
j
^
[NF[»«-
i
REFERENCES [1]
[2]
[3]
Desai, C. S., and Abel, J. F., Introduction Nostrand Reinhold, New York, 1972.
Streeter, V.
L.,
to the Finite
Fluid Dynamics, McGraw-Hill,
New
Element Method, Van
York, 1948.
Abramowitz, M., and Stegun, I. A., (eds.), Handbook Mathematical Functions with Formulas, Graphs, and Mathematical Tables, Applied Math. Series 55, National Bureau of Standards, Washington, D.C., 1964.
[4]
Desai, C.
and Christian,
S.,
[5]
Desai, C.
J.
T. (eds.), Numerical
New
Engineering, McGraw-Hill,
Methods
Desai, C.
S.,
[8]
98,
Silvester, P., and Chari,
M.
Magnetic Field Problems,"
IEEE
in
Geotech. Eng.,
Fields by Finite Elements,"
/.
Soil
No. SMI, 1972.
V. K., "Finite Element Solution of Saturable
Chari, M. V. K., and Silvester,
PAS
FEM
"Seepage Analysis of Earth Banks Under Drawdown,"
Mech. Found. Eng., ASCE, Vol. [7]
Geotechnical
"Finite Element Procedures for Seepage Analysis Using an
S.,
Isoparametric Element," in Proc. Symp. on Appl. of Waterways Expt. Station, Vicksburg, Miss., 1972. [6]
in
York, 1977.
PAS
Trans., Vol. P.,
IEEE
89, 1970, pp. 1642-1651.
"Analysis of Turbo-Alternator Magnetic
Trans.
Power Apparatus and Systems, Vol.
90, 1971, pp. 454-464.
BIBLIOGRAPHY Desai, C.
S.,
ments
"Finite Element
in Fluids,
kiewicz, O.
Doherty, W.
C,
P.,
Vol.
(eds.),
1
Methods
for
Flow
in
Porous Media,"
(Gallagher, R. H., Oden,
Wiley,
Wilson, E.
L.,
New
J.
T., Taylor,
York, 1975, Chap.
and Taylor, R.
in Finite Ele-
C, and
Zien-
8.
L., "Stress
Analysis of Axisym-
metric Solids Utilizing Higher Order Quadrilateral Finite Elements," Report 69-3, Struct. Engg. Lab., Univ. of Calif., Berkeley, Jan. 1969.
Ergatoudis,
I.,
Irons, B. M., and Zienkiewicz, O.
C, "Curved,
Quadrilateral Elements for Finite Element Analysis," 4,
No.
1,
Int. J.
Isoparametric
Solids Struct., Vol.
1968.
Zienkiewicz, O. C, Mayer, P., and Cheung, Y. K., "Solution of Anisotropic Seepage by Finite Elements," /. Eng. Mech. Div. ASCE, Vol. 92, No. EMI, 1966.
332
TWO-DIMENSIONAL STRESS-DEFORMATION ANALYSIS
INTRODUCTION number of one-dimensional problems and two-dimensional problems with only one unknown or degree of freedom at a point, we are now ready to consider a different class of two-dimensional problems. This class involves analysis of stress and deformations with more than one degree of freedom at a point. After studying a field
Most real problems are three-dimensional. Under certain assumptions, which can depend on the geometrical and loading characteristics, it is possible to approximate many of them as two-dimensional. Such two-dimensional approximations generally involve two categories: plane deformations and bending deformations. In the case of plane deformation, we encounter subcategories such as plane stress, plane strain, and axisymmetric; in the case of bending, we deal with problems such as bending of plates, slabs, and pavements.
PLANE DEFORMATIONS Plane Stress Idealization
Figure 13-1 shows a (thin)
beam and
applied in the plane of the structure, that small
compared
a plate subjected to loads that are is,
in the x-y plane.
to the x-y dimensions of the body.
The
thickness
is
Such loadings are often 333
Two- Dimensional Stress- Deformation Analysis
334
L
^
y,v
Chapter 13
y,v
-p-J
(a)
(b)
Figure 13.1 Plane stress approximation, (a) Beam, (b) Plate.
membrane
referred to as in-plane or
(stretching).
Under
these conditions,
the assumption that the variation of stresses with respect to the body,
is
constant,
it is
z,
that
is,
and
across
reasonable to assume that out of the six com-
ponents of stresses in a three-dimensional body [1-3] three of them, a 2 Tgx and r yz can be ignored in comparison to the remaining three, a xi a y and ,
,
T xy This idealization .
nonzero
,
,
stress
is
called plane stress
and involves only the following
components,
w
(13-1)
which are functions of the coordinates
x,
y
only.
The corresponding com-
ponents of strains are €*
=
W
(13-2) [y,
a and € denote components of normal stress and strain and t and y denote components of shear stress and strain, respectively, and {a} T = [o x a y T xy] and {e} r = [e x e y y xy] are the vectors of stress and strain components. In view of the plane stress assumption, we need to consider only two components of displacements at a point, u and v, in the x and y In Eqs. (13-1) and (13-2),
directions, respectively. If
we
and isotropic materials, the
restrict ourselves to linear, elastic,
material behavior can be expressed by using the generalized Hooke's law for the three
components of
Gv
and
stress
=
E vE
—V +
vE
,
€x
E 2(1
strain [1, 2].
v)'
+
Thus
„
,2'->'>
(13-3a)
Two- Dimensional Stress- Deformation Analysis
Chapter 13
335
or in matrix notation,
w=
1
V
V
1
1
rate] 1
€],
(13-3b)
E
and v are Young's
-v 2
J
where [C] is stress-strain or constitutive matrix and modulus and Poisson's ratio, respectively. Plane Strain Idealization
is large compared to the x-y dimensions and where the loads are acting only in the plane of the structure, that is, the x-y plane, it can be assumed that the displacement component w in the z direction is negligible and that the in-plane displacements w, v are independent of z. This approximation is called plane strain, in which case the nonzero stress components are given by
In cases where the thickness
(Fig. 13-2)
w= and a z
= v(a x +
a y). The
(13-4)
stress-strain relationship for this idealization
is
expressed as
1-v
W=
v 1
[C]{€] (1
+
v)(l
-
-v
€
2v)
1
sym.
.
(13-5)
-2v
y, v
z,
y~^
w
x u '
UL
(b)
(a)
Figure 13-2 Plane strain approximation, (a) Strip load, (b)
Long
underground tunnel.
Axisymmetric Idealization
Figure 13-3 shows a body symmetrical about
its
centerline axis
and
subjected to a load symmetrical about the axis. In view of the symmetry, the
Two- Dimensional Stress- Deformation Analysis
336
j
H
'
\
Chapter 13
'
t
,
>z,
w
—
Figure 13-3 Axisymmetric approximation.
components of stress are independent of the circumferential coordinate 6. As a consequence, we have the following nonzero stress and strain components
[2]:
'Or
^r
€e
e
\>
\
{€}=rT €
(13-6)
g
[yj Dr this
case
V
— [C]{€] (1
+
v)(l
-
is
V V
V 1
2v) sy m.
-v
[el 1
(13-7)
-2v
Strain-Displacement Relations
From we can
the theory of elasticity, assuming small strains and deformations, define the following strain-displacement relations for the three
idealizations [2]:
:
Plane Stress and Strain.
du
(
dx dv dy
=
t>
du y xy
(13-8a)
dv
+
\dy
dx
Axisymmetric:
du
\
dr
u
=
<
r
(13-8b)
<
dw dz
du
Initial Stress initial
dw
.
dz^
'»,
dr)
and Strain: As explained
in
Chapter
5, it is
possible to include
or residual strains or stresses existing in the structure before the load
is
applied.
The
initial strain (or stress) state
temperature, (fluid) pressure, creep in the case
of temperature,
=
e
where dT
caused by factors such as known and geostatic stresses. For instance,
may be
effects,
= T— T
is
T \
*dT,
(13-9)
the change in temperature
and a
is
the coefficient of
thermal expansion.
We
define total strain e as the
sum of
the effective elastic strain, e % and
the initial strain {€}
where
{e
} is
strain case
[C]"
1
=
= [q({e}-{e which
[D], in
{€„},
(13-10)
and
the vector of initial strains
to The matrix
+
=[C]" '{«}
= [q{€«}
})
(13-11)
[D], the strain-stress matrix, for the
plane
is
—v 1
[D]
1
-v
-
V
7
1
-
1
(13-12)
v
337
ELEMENT FORMULATION
FINITE
As shown
in Fig.
element discretization will involve two-
13-4, the finite
dimensional elements such as triangles and quadrilaterials (squares, rectangles, trapezoids) in the
x-y plane. The third dimension
by specifying unit thickness for plane plane
(z) is
generally included
strain or a thickness h in the case of
stress.
y.v P(x,y) (L 1( L 2
,L 3
)
P(x, y) (s, t)
Figure 13-4 Discretization with triangular and quadrilateral ele-
ments.
Detailed properties of the triangular and quadrilateral elements have
been covered
in
Chapters
the quadrilateral element
and 12. Here we shall discuss in detail the use of and then state briefly the use of the triangular
11
element.
As
stated previously, there are
point P(x, y) [Fig. 13-4]. point in the element as u ( x ,y)
v(x,y)
We
two unknown displacements
can write approximation models for
= «i + & = fi -F
2
x
fi 2
t
x
+ + Prf +
-r
cc 3
y
cc 4
u, u,
v at a v at a
xy, (13-13a)
fi A xy,
or
=
{u}
where 338
ju}
r
=
[u
v],
[a}
T
= [a
/1
(13-13b)
[
cc 2
a 3 a4
^
f} 2
fi 3
fi 4 ],
and [O]
is
the
Chapter 13
~i::.\
::'
Di'j
».o-D:
1
coordinates. Evaluation
=
-
2
—
:
—
2;X
—
Pz x
P:
— —
:
2:V.
Pis
.
:::
':Yr
o-ces gives
c:
— a, — PS-
2.5.4.
or
a
•vnere
=>
q-
ments and \V Eo.
1
15- 14b
:<
:<
:
:
:«*'..
.
;
A Tne rrccuct ere t~e
[^[[Aj-
V
"
a^
>:
q
bo.ut.ion ::
c,
;
-
v*
r
\
i
-
5
--e
-
h i
.
that
is.
-,.
= * " 5X1 - 0, = ifl -^Xl+0, 1
(15-16)
= K1 - *Xi + 0-
are local coordinates (Figs. 12-2
The geometr;.
15-15
q
ons hV. "*-'"*
dehned n Eg
V;
;
tne vec
:s
,.
matrix of nocal
"A"
V;
5.
i
:
i
:
v =
Here
i
:
S
15-14:
for "a" gives
:
\v v
i
the square iS
:s
i
the
::.
;.
re expressed b> using :he sam;
and 15-4
coordinates
i:
any point
erpolation functions
r=VYx
t
ine
e
emeni. car
V
1=17-4 il3-l"ai
V
.
V-
1
•
"
^
'
::
il5-l-b>
(2x8)
(2x1)
wnereW = [x, This leads
xa
:o :ne
jc3
xJandfoF
dehninon
o:
(8x1)
=
|>, y2 yl :he elemen: is :
dement. Requirements for Approximation Function
within the dement. This
polynomial form as
is
in Eq.
sutisrieu since >
15-15
we have
i
~
z r zy\i: r ::
Two- Dimensional
340
Stress- Deformation Analysis
For plane problems, the approximation models must
Chapter 13
satisfy the inter-
element compatibility at least up to derivative of order zero; that
placement between adjoining elements must be compatible. This
is
is,
the dis-
tied in with
the highest order of derivative n in the energy function, Eq. (13-21) below.
Since n
=
bility is
equal to n
1
in Eq. (13-21), the
—
1
minimum
order for interelement compati-
=0.
The approximation of Eqs. distributions of u
(13-13) and (13-15) yields continuous bilinear and v within an element and linear distributions along the
element boundaries.
common
It
can be seen (Fig. 13-5) that the displacements along
two elements afe compatible since only the straight line can pass through two nodal displacements common to both elements. For the sides of
two-dimensional plane deformation problems, the approximation function provides for rigid body displacements and constant states of strains: e x
,
However, the function does not include all terms in the polynomial expansion represented by the Pascal's triangle, Chapter 11. e y y xy ,
.
Equal
nes a 1 b, and a 2 b 2 coincide
Equal Figure 13-5 Interelement compatibility.
Plane Stress Idealization First in
we
Chapter
€x
By following the procedure components can be evaluated as
consider the plane stress idealization. 12, the strain
_ du dx
~
_ ~
duds ds
\J\
,
du dt
dx^Ttfa
&ML'\ds
dt
dt
dsF
J
(13-18a)
j
Similarly,
dv dy
_ ~~
dvd±
dvdt^
ds dy
dt dy
(13-18b)
Two-Dimensional Stress-Deformation Analysis
Chapter 13
341
and 7xy
~
du ^ dv + dx dy .
_ ~
dt\ (duds (du d± j_dz dy §£\
(dv $1 (§1 ds \ds dx
.
\ds dy
dt
'
i,
dy)
'
dv §i\ dr dx)
_ ,
OH.
(13-18c)
dt
'
we have
Finally,
B 12
~B n
B 21
B 13
B 22
B 14 B lx B xx
B 24
B 23
B 23 B l2
B 22 B 12
B 24
{q}
(13-19)
B 14
or (13-20)
[B]{q},
where the Step
We is
B
are defined in Eq. (12-24c)
is
defined in Eq. (13-14b).
[3]
t If wmm* * 6
where {X} r
Ty
{q}
use the principle of stationary potential energy; the potential energy
-
h
JJ
A
[Tx
and
Derive Element Equations
4.
given by
n' =
tj
MT*kfc* - h \ {
3 - 21 )
A
=
[X Y]
components of body
the vector of
is
r forces; {T}
the x and y
=
components of surface tractions in and h is the (uniform) thickness of the element. Substitution for {u} and {e} from Eqs. (13-15) and (13-20) in Eq. (13-21) ]
is
the vector of
direc-
tions, respectively;
yields
H,
=4
r fa}
|1
(1x8)"
-
h
{qf
(1
x
8)
(1
By taking
x
8)
[C]
x
3) (3
[W x
(8
J5l
r
x
(8
\\
h {qf
[B]
(8
3) (3
x
{q} (8
8)
x
1)
{X} dxdy
2) (2
x
1)
{f } dS.
[Nf x
dxdy
[B]
2) (2
x
(13-22)
1)
partial derivatives of 11^ with respect to
u t9 v lt
etc.,
and equating
to zero,
6U,
(13-23)
0. d{<\]
which leads to the eight element equilibrium equations as
M (8
x
=
(q) 8) (8
x
1)
(Q) x 1)
(8
=
{Qil x 1)
(8
+
{Q 2 } (8
x
1)
(13-24)
j
:
Two- Dimensional Stress- Deformation Analysis
342
where
[k] is
Chapter 13
the element stiffness matrix, [k]
= h ft [Bf[C][B]dxdy,
(13-25)
A
and {Q}
is
the element nodal load vector,
=
{Q}
+
{Qi}
Evaluation of
[k]
{Q,}
=
h
JJ
\NY{X}dxdy
+
h
js
\NY{T}dS
and [Q
The coefficients of [k] are functions of local coordinates s and more convenient to perform numerical integration as follows
M - h S M*, tdriQMst, where
{st ,
t t)
The
is
first
t,)]\J(s l9 tt)
I
x
and
t,
w
(N
2 or four-point
= 4)
i
and so on. integration
used.
part
{QJ
of the load vector can be computed as follows:
{Qi}=AStN(*„0] r {X}»V If
we assume a uniform body
13-6(a)],
it is
(13-27)
t.
denote the local coordinates of the integration point
Often, for quadrilateral elements, 2 (Fig. 12-4)
(13-26)
X = 0,
force intensity
the expanded form of {Q,}
Y
(per unit volume) [Fig.
is
0~
~N,
N
(13-28a)
2
N,
N<
N N N_
)\W
J(st9
N,
tt
(13-28b)
t.
i;
t }
_0 The
subscript (s (i
instance, the fifth
t t)
t
(Sl,tt)
denotes that the matrix
component of 1(5)
1
evaluated at points
ti
)\J(s h
ti
)\W Y,
/,•).
For
(13-28c)
i
y direction at node point The second part {Q 2 } arises due to surface
gives nodal force in the
(s h
{Q,},
hT N (s h l
is
1,
and so on.
on the boundary of an element. Often it is possible to evaluate this_part by using closed form integration. For instance, consider fxl and fyl as applied tractions applied
Two- Dimensional Stress- Deformation Analysis
Chapter 13
343
(a)
tttttttttt
®
(b)
Figure 13-6 Loading on quadrilateral element,
tractions
(b)
Surface tractions.
on
side 1-2 [Fig. 13-6(b)].
(a)
Body
force,
Then
0"
r^
N N
{Q 2 }
=
1
2
2
Txi
N< h
dS.
(_'
Nt
N N _0
2
3
N,_
(13-29a)
Two- Dimensional Stress- Deformation Analysis
344
Now, along
side
1
=—
s
,
1
N = i(l A^ = i(l x
2
A^ 3 A^4
= i(l = iO
to
and
1
- s)(l + s)(l + *X1 - *X1
= —1
t
-t)= i(l -t)= j(l
Chapter 13
therefore,
;
- 5), + ^), (13-29b)
+0=0, +0=0,
substitution of which leads to a line integral as (1 (1
{Q 2 }
We
hU
+
"
s)/2 s)j2
dS.
l
(1
-s)j2
(i
+
(13-29c)
y\
m
used here the transformation relation in Eq. (12-38).
integrations,
Upon
required
we have IT T.
o
{Q 2 } 1
.
where
lx
= length of side
1-2.
(13-29d)
y
y\
w
This implies that the applied load
equally at the two nodes pertaining to the side 1-2. This the fact that the interpolation function [Eq. (13-15)]
of the quadrilateral. If results
may
we
is
is
is
distributed
a consequence of
linear along the sides
use a different (higher-order) approximation, the
not be similar. Furthermore, in the case of the higher order
it may be easier to perform numerical integration, as was done for {Q,}. The assembly of element equations can be achieved by following the
approximation,
principle that the displacements at the
procedure
is
essentially the
same
common
nodes are compatible. The
as illustrated in Chapters 3, 4, 5, 7, 11,
and
12 and as illustrated subsequently in Examples 13-1 and 13-2.
The assemblage equations
are modified for the
boundary conditions
in
terms of prescribed values of u and v on part(s) of the boundary. Solution of
Two- Dimensional
Chapter 13
Stress- Deformation Analysis
Then
the resulting equations gives nodal displacements.
are
computed by using Eqs.
(13-20);
and
(13-3), (13-5),
345
strains
and
and
stresses
(13-7), respec-
tively.
Triangular Element
After the derivations in Chapter
1 1
for the torsion problem,
it is
relatively
straightforward to derive element equations for the plane stress idealization
using the triangular element (Fig. 11-2). Use of the linear function, Eq. (ll-4a) for both u
and v
various requirements (for two-dimensional
satisfies
plane deformations).
Various terms for the triangular element are stated below: [u
[N]
~N
=
(13-30a)
v],
N
N
2
,
3
_0
W=
"2
["l
V
«3
b2
~b t
a2
#!
N
t
N
2
V2
t
(13-30b) 3
V 3 ],
(13-30c)
0"
b3
M=A where the
N
N,
a3
fl t
fl 2
a3
b
b2
b3_
are defined in Eq. (11 -3b)
x
(13-30d)
*
and the b and a are given t
t
in
Eq.
(ll-3c).
The general form of
the element equations
is
identical to that for the
quadrilateral element [Eq. (13-24)]; only the orders of various matrices are different.
The
stiffness
matrix has the order of 6
[k]
x
and
6
is
given by
= h[BY[C][B] j\ dA hA[BY[C][B].
The load vector has the order
6x1.
(13-31)
For a uniform body weight
Y (X
= 0)
[Fig. 13-7(a)],
[Q The uniform
T l
}
= ^[0
surface traction
is
7
divided equally
belonging to the side on which traction acting
on
side
1
[Fig. 13-7(b)]
{Q 2 y
=
7
is
(13-32a)
7].
among
applied.
For
the two nodes
instance, for
Txl
we have
^ZkZi[0
110
0],
(13-32b)
Two- Dimensional
346
Chapter 13
Stress- Deformation Analysis
(b)
Figure 13-7 Loading on triangular element, (a)
Body
force, (b) Sur-
face traction.
where /, is the length of side 1. Contributions of tractions and on other sides can be similarly evaluated. Example
in the
y
directions
Details of Quadrilateral Element
13-1.
Figure 13-8 shows a square isoparametric element (see also Example 12-1, Fig. 12-4) subjected to a surface traction are
E=
10,000 kg/cm 2 and v
First
we show we have
(12-20c),
fx
equal to
brief details for the
jc
13
*24
X34 12
*23
x 14
1
1
-
kg/cm on
side 2-4. Material properties
computation of matrix
=o - = -1, = 1 -0-1, = l -0-1, = 0-1 - -1, = =
1
= 0.30.
1
= 0, = 0,
[B].
= -1, 7,3=0-1 = -1, - = 0, y 12 = y„ = - = 0, 7 14 =0-1 = -1, ^ 2 3=0-l = -1. y 1Af
=
-
1
1
1
Referring to Eq.
'
Two- Dimensional Stress- Deformation Analysis
Chapter 13
347
y i
n
(0.1)
1
"©
kg/cm
® +
+
(-0.577,
(0.577,
0.577)
0.577)
(-0.577,
(0.577,
-0.577)
-0.577)
+
+
s
,©
©,
(0,0)
>
(1,0)
+
Integration points
Figure 13-8 Integration over square element.
Then |/|
= i[(-D(-D +
/[(0)(-i)
= id + = 0.25, which
is
-
s[\
-
x
(-1X0)]
(0)(-i)]
1)
one-fourth of the area of the square. Therefore, use of Eq. (12-24c) gives
T_! _
B 8
Bi 2
xs-{-\)t]=\{-\ +
x 0.25
= y[ + l +
B 13 =±[\
x
s
+0XS +
B 14 = y[~l +1 X
S
B 22 — \[-\ x ~2~L
°
+
(-1)/]
(1)/]
+
= y(l -
= y(l + X
t)
= y(-l + S\
= -lxs-0xt] — \{-\ ~ 2 x
x/
xn
,j
° X
'}
=
B 2 * = y[l + (-«» -
x
/]
= y(l -
=
y
[1
"
/),
/),
+
*23
Now
(-D(-l) +
(_1) *
the product [B] r [q[B] in [k] [Eq. (13-27)]
T
is
(1
+ S)
>
s).
given by
s),
/),
I
•*
•*•*
•*•*
Tt^t
cq
T*
f
I
'"t
cq
cq
co"
cq
cq
oq
cq
cq
cq
cq
cq
cq
on o n u+u u+u u+u u+u
r»
o +u « oq
— cq
cq
—
m
c*
u+u
u+
cq
—
•*
c*
oqoq
oq
cqoq
cqoq
o o
r*
N
—
N
f>
o
U+O U+U — Oq
~ N N
,
oqoq N
N
oqcq
on M
cq
cq
oq
cq
^^h Mr)
oqoq
oqoq
oqoq
p^«
r<
N
r* •*
— «
oqcq o
N ~
cq
cq
<>
r*
~ N n
^
„
N N
^ N
cqoq cqoq o
N
oq
tori
u +
£
cq
cq
cq
N«
^
«Q
cq
Z
oq
cq
cq
oq
oq
cq
^
„Z
cq
cq
-
-
^
N:
.,
_1_ r
,
^_ oq
—
***
t)
""I
Cj
^Lj -}"^
u+ £
cq
"
cq
cq
oq ^<
cq
£
^
«
«
CQ
I
g CO
cq
cq*
fin
-
^
tt\ <*>
^* m
r>
cq
,
^„
CJ -"
cq
oq^^
N «
u +G
^
flq
_
«^
cq
oq
cq
~*
**>
oq
n
f»vrl
°5
r
,
oq
^
cq
^ £ on ofnV^'NJ u+u
cq
«nr» .
oq
^
„
oqoq
cq
oq^^
Z
M
oqoq
cqoq
U+
oqcq _
oqoq
cqoq
cq
oq
cq~
r»ri
oqcq
Z
oq
com
oq
-*
oq*
J +
-i
oq
oqoq
cq
cqoq
cq
t»>
N cq
oqoq
+U
r*
oq
m —
348
*•*
•*-*
cq
r
»
_L
r
•»
Two- Dimensional Stress-Deformation Analysis
Chapter 13
Here we have arranged the vector
=
[qf
As
{q} as
u2
vl
[u l
u3
v2
we need
indicated in Eq. (13-27),
349
to
w4
v3
v 4 ].
compute the terms one by one
at
each
of the four integration points (Fig. 13-8) and add them together to obtain the matrix [k].
As an
Buisi,
Bn(s2 Bu(s 3
t2)
,
t3 )
The
"49.45
[k]
=
C
are obtained by using
tj
final result for the [k]
at the four points are
found as
= i(-l 0.577) = -0.789, = ^(1 - 0.577) = 0.212, = i(l + 0.577) = 0.789, = i(-l*+ 0.577) = -0.212,
Bui**, U)
and so on. The values of the
tt )
-
tt)
,
Bn(s h
of
illustration, the values
matrix
E and
v.
is
17.86
-30.22
-13.74
-24.73
-17.86
5.49
1.38"
49.45
1.38
5.49
-17.86
-24.73
-1.38
-30.2
49.45
-17.86
5.49
-1.38
-24.73
17.86
49.45
1.38
-30.22
17.86
-24.73
49.45
17.86
-30.22
-1.38
49.45
1.38
5.49
49.45
-17.86
102
(13.34a)
sym.
49.45.
Use of Eq. (13-29d) (Q}r
=
leads to the load vector {Q} as
0.50
0.0
[o.O
0.0
0.50
0.0
0.0
(13-34b)
0.0].
Introduction of the boundary conditions ui leads to
=
=
v\
=
v2
v3
=
=
W4
=
v4
two equations as 4945w 2
4-
549w 3
549«!
+
4945w 3
= =
0.50, 0.50.
Solution by Gaussian elimination yields
u2
Use of Equation
{
=
=
u3
=
= =
1.000
kg/cm 2
,
0.300, 0.000,
stresses
Ox
are
x 10" 4 cm.
[C][B]{q] leads to element stresses as
ax =
and the principal
0.91
=
1.000 kg/cm 2
,
0.300
computed from
^ i V(<7* - O Y + y
—
(13-34c)
Xly.
—
Thus we obtain the quantities displacements, stresses, and strains required and design of a structure idealized as a two-dimensional plane problem.
for analysis
:
Example
13-2.
Evaluate element
Triangular Element stiffness
two triangular elements stresses
matrices for the problem in Example 3-1 subdivided into (Fig.
13-9).
Assemble and solve for displacement and
with the following data.
The surface traction fy on Boundary conditions
«(0, 0)
=
t
=
11(0, 1)
v(l,
=
side 2-4 of element 2
i)
= =
v(0, 0)
=
1
kg/cm per
unit thickness.
0.0,
=0.0, v(0, 1)
= =
(13-35) 0.0, o.O.
Figure 13-9 Discretization with triangular element.
Thickness h =
1
cm
Element 2 1
Local node
(T) Global node
—Degree 350
of freedom
1
Two- Dimensional Stress-Deformation Analysis
Chapter 13
We
1
351
1 (Fig. 13-9). Here we have numbered local degrees and 6, corresponding to local displacements u u 2 u 2 Vi, v 2 and 3 respectively. The global degrees of freedom are numbered by assigning consecutively two indices to each node corresponding to displacement components u, v. respectively. Thus in Fig. 13-9 there are a total of 8 degrees of freedom. The terms required for finding the matrix [B] are evaluated as follows:
first
consider element
of freedom as 1.2, i-
,
3, 4, 5,
,
x
,
,
.
a
:
az a2
= = =
.r
— Xj = —
3
- x3 = - x, =
x,
xz
bt
=j
0.
b2
=
1.
bi
=y -y
1.
-->-3
= "I -y = 1.
:
v3
:
:
=0,
;
and
2A
=
—
a2 b2
az b 3
=
1
1
-
=
>
1
cm :
Eq. (13-30d>.
"-I
0"
1
y
B =
-1 -1
For plane [k]
=
we use Eq.
stress conditions,
-1
1
1
•
1
C
13-36;
1
13-3(b)J
and
1
13-31) to yield
MBFicm — 1
o
V
---l
o
ooo-
1
-1
1
= A^
1 1
i
•l
1
7
J !
2
l
10
v
-1
1
13-3
"a.)
1
and hence f
Element
Global—
1 I
Local
1
-1
2
1
-d -
V)
-<1
- - V)
-
v
-(
- - V)
I
V 1
3
—V
-
V
H
3
-v-j
—V
2
-; 1
2.1
V
i
3
W=
-
1
j
13-3-r.
-(1
" - VI
3
1
—
V
svm.
5
1
_
6
Local
Element 2
4—431obal
:
Two- Dimensional Stress- Deformation Analysis
352
The
matrix for element
stiffness
2,
Chapter 13
which has the same area and dimensions as
can be deduced from that of element 1 simply by properly exchanging the node numbers. For instance, in Fig. 13-9 we have marked local numbers for
element
1
element
2,
,
and the corresponding global numbers are shown
at the
bottom of
Eq. (13-37b).
With each node having two degrees of freedom, we assign global and local numbers to them. Thus the global numbers are (1, 2), (3, 4), (5, 6), and (7, 8) for the four nodes, 1, 2, 3, and 4, respectively. The local numbers for the nodes are (1, 4), (2, 5), and (3, 6) for the three local nodes 1, 2, and 3, respectively. Since there is no load on any of the sides of element 1, there is no contribution to the load vector. In the case of element 2, the surface traction of 1 kg/cm on side 2-4 yields
Element 1 Global
Element 2
Local 1
1
7
2
3
5
|h 1
[Q]
X 2
1
1
Element
assign
E=
[
Global
^
(
Local
we
3
5
3
4
2
8
5
4
6
6
6
4
(13-37c) '
.0.
If
Global
10,000 and v
=
0.3,
Eq. (13-37b) reduces to
1
1 i
i
M = 10,000 1.82
"
1.35
-1.00
-0.35
2
-1.00
1.00
3
-0.35
4
0.65
5
-0.35
6
_-0.30
1
-0.30"
0.65
-0.35
0.00
-0.30
0.00
0.00
0.35
-0.35
0.35
0.00
-0.30
-0.35
1.35
-0.35
-1.00
0.00
0.35
-0.35
0.35
0.00
0.300
0.00
-1.00
0.00
1.00_
0.30 (13-38)
Local]
Element 2 Globall
Assembly of the follows
stiffness
matrices and load vectors yield global relations as
oooooooo
o
~|« <* :=
as
a
(S
sT
a
&
7 •n © o m o en o en © © © © o O o © 7 + o en d in m © © en en en m 8 o © o O o © © + + 1
1
1
1
1
1
1
en o o
in en
in en
O
©
© © o © O © in
>n en
in en «-"
^ 3
5*
in m © en en en © © © •n © en © ~ © © © © © 7 in in in m m © © en en en © en O © © © en © © © © © © © 7 © in in in •n © © © en en rn en © © © © © © ^ © 8 © © in m © m © en en ©
•n •n o 8 8 en O en en © o O O © ©
in
as
1
•n en "^
'o
1
1
1
1
,-H
1
1
1
1
1
1
© © 8 © © © © © © © © in •n © © en m in en © en en © © © © © © © m © © en en © © © © © m + en in en
in en
1
in en
i
1
1
|
o o
in en
in 8 en 7 in o en o o © o © o o + 7
o o
© © ©
1
1
m
en
© 1
1
1
r © © in in in m © en © en 8 en © 8 en en — o © © © © © © +
o o — m o en o
7 1
(A
in en
«n en
1
s
1
1
1
1
+
1
1
© © DO © rN ©"
1
© en © in >n •n © © en en en © 8 © © © © © © 1
1
1
1
353
:
Two- Dimensional
354
Stress- Deformation Analysis
Chapter 13
Introduction of the four boundary conditions in Eq. (13-35)
is achieved by rows and columns corresponding to u u v it v2 , u 3 , t> 3 , and v 4 that is, global degrees of freedom 1, 2, 4, 5, 6, and 8, respectively. This leaves only two
deleting the
,
equations 1.35« 2
=
-0.35« 4
2^§o =
0.000091
and
-0.35m 2
+
=
1.35w 4
0.000091.
Solution by Gaussian elimination gives
=
u2
These
=
u4
results agree with the solution
x 10" 4 cm.
0.91
by using the quadrilateral element.
Comment on Convergence In the case of the displacement formulation, algebraic value of the potential energy in the system
As
is
higher than the exact energy at equilibrium.
is improved by using higher-order approximation and/or by using refined mesh, the energy converges to the exact value. Hence, an element which has lesser value of approximated stiffness can be
the quality of the formulation
considered to be superior (see Fig. 3-15). Often, the trace of the stiffness matrix
The
trace
is
defined as
2K
ih
that
trace for the quadrilateral element
is
used as a measure of the
stiffness.
sum of the diagonal elements. The 4945 x 8 = 39,560, whereas the trace the
is,
is
= 59,344. This shows that use of the which contains an extra term (xy) in the approximation model can yield better solutions. Such an inprovement may not be evident with the use of only one or two elements. However, for larger problems the improvement can be significant. For general mathematical analysis we need to use the concept of the eigen system of the matrix. This topic is discussed in advanced texts.
for the triangular element
is
7418 x 8
quadrilateral element
COMPUTER CODE The computer code PLANE-2DFE (Appendix
4)
permits linear elastic
analysis of bodies idealized as plane stress, plane strain, or axisymmetric.
The
code incorporates the four-node isoparametric element [Fig. 13-4]. The quadrilateral can be degenerated to be a triangle by repeating the last node. In the following are presented typical problems solved by using this code. Example
13-3.
Analysis of Shear Wall
Figure 13-10(a) shows a shear wall, which
The
wall has an opening 2
m wide
x
5
m
may
constitute part of a building frame. deep in the lower floor level. The prop-
Chapter 13
erties
Two- Dimensional Stress- Deformation Analysis
355
of the wall are
= 2.1 x 10 9 kg/m 2 v (columns, beams, and wall) = 0.3, Thickness of columns and beams = 0.3 m, Thickness of wall = 0.15 m, £ (columns,
Loading
is
beams, and wall)
shown
as
,
in Fig. 13-10(a).
By using the code PLANE-2DFE or other available codes, analyze the loaddeformation behavior of the wall assuming plane stress conditions. Obtain two sets of results with two values of E for the shear wall with lateral loads only. 1.
E^m =0.21 x
10 9
2.
£ wall =
x 10 9 kg/m 2
,
0.0021
kg/m 2
,
.
Plot the deflected shape of the wall the
first
value of £"wall
=
and the distribution of vertical stress a y for x 10 9 kg/m 2 at selected sections. Compare the results, expected from the conventional beam bending theory
0.21
qualitatively, with the results
Beam 10,000 kg 1
I— -4—4
n
!-! 1
m
I
-4
—
i
i
1
m
4--I-I
JBeam'
k
m- —
1
20,000 kg
m
h •Column Wall
Opening
Wall
n Column
I
I
'///////////////////A
y////////////////////y
8m (a)
Figure 13-10 Analysis of shear wall, (a) Elevation of shear wall.
Two- Dimensional Stress- Deformation Analysis
356
Chapter 13
y ,
10,000 kg
^
20,000 kg
i
A A
©A
A ®
A
Origin-^ W/a
A w/AW/A
© ©
wA
w//,W/A
X
©
(b)
Figure 13-10 (Contd.) (b) Finite element mesh.
and
offer
comments concerning
the influence of the stiffness of the shear wall
on the
load-deformation behavior. Figure 13-10(b) shows a
finite element mesh for the shear wall; relatively finer used near the two horizontal beams. Figure 13-1 1(a) shows the computed deflected shape of the left end vertical side of the structure for the two values of
mesh
is
stiffnesses for the shear wall.
Figure 13-1 1(b) shows the deflected shapes of the
entire shear wall for the
two conditions.
Figure 13-12 shows A-A, B-B, and C-C [Fig.
13-10(a)].
vertical (bending) stress
along three horizontal sections
We have assigned the stresses at the center of the
element for sketching the variation shown in Fig. 13-12.
Comments. Conventionally, shear walls are often designed by assuming them beams idealized as one-dimensional. Here the usual assumptions of beam bending are considered to be valid (see Chapter 7). In view of the irregular geometry and the large area of the structure, such an assumption may not be valid and can give results that are different from reality. to be thick
Two- Dimensional Stress-Deformation Analysis
Chapter 13
357
A two-dimensional finite element analysis permits inclusion of irregular geometry
and the multidimensional aspects of the problem.
displacements, stresses, and bending
moments
It
allows computations of
in all elements; thus
one can compute
concentrations of stresses at critical locations and perform improved analysis and design as compared with the conventional procedures. For instance, Fig. 13-12 shows that the distributions of bending stress o y at various sections are not linear as assumed in the conventional beam bending theory. The bending stresses show irregular and nonlinear behavior depending on the geometry, material properties, and loading conditions. Moreover, it is possible to perform parametric studies to include the influence of the stiffness of the wall on the behavior of the frame consisting of beams and columns. With a finer mesh, it is possible to identify the zones of stress concentrations, particularly
Example
13-4.
Analysis of
Figure 13-13 shows a concrete the bottom layer
is
near the corners.
Dam
on Layered Foundation
dam resting on
a foundation consisting of two layers;
underlain by a rock mass extending to considerable depth. There
exists a crack at the crest of the gallery.
By using PLANE-2DFE and assuming plane
— -n
Scale:
Horizontal: '
'
1
cm-
3 1.5 x 10"
m
Vertical:
'
1
cm
= 0.788
m
/ /
E^-O^xlOHg/m 2
E^n-0.21 x10 7 kg/m
:
(a)
Figure 13-11
shapes for
Computed
deflections for shear wall, (a) Deflected
left vertical side.
Two- Dimensional
358
Scale:
Horizontal:
1
Vertical:
1
Stress- Deformation Analysis
cm = 7.5 x 10 cm = 0.788 m
E^ = 0.21 E vwaii =
4
Chapter 13
m
x10 9 kg/m 2
°- 21 *
10? k g/ m2
r I
i
I
i I
-T
7
///
// '/
!
/ '
/
i/i/
//
/
/
/
i (b)
Figure 13-11 (Contd.) (b) Deflected shapes of shear wall.
strain conditions, solve for displacements
dam and
foundation.
Concrete
in
The
stresses
under the self-weight of the
dam:
E= v = y = Layer
and
properties are given as follows:
432 x 10 6 psf
(=
2.1
x 10 9 kg/m 2 ),
0.3,
150
pcf(= 2400 kg/m 3 ).
1:
E = 144 x 10 4 psf (= 7.0 x v = 0.4, y = 100pcf(= 1600 kg/m 3
).
10 6
kg/m 2 ),
T*o- Dimensional Stress- Deformation Analysis
Chapter 13
Scale
cm
1
for a
= 1.57
m
104 kg
m
359
:
+ Tension - Compression
A— 0' B— 0-
C— 0-
,
Figure 13-12
Computed
[Fig. 13-l(Xa)] for
Layer
distributions
£w ,u =
0.21
x 10»
of c >
kgm 2
at
A- A, B-B,
C-C
.
2.
E=
144 x 10 5 psf
(=
7.00 x 10 8 kg
v
=0.3,
7
= HOpcf ('= 1760kgm
m
2 ),
3
j.
Partial Results
Figure 13-14 shows a finite element discretization for the dam. In this figure, the nodes and elements are numbered in the x direction. It may be computationally economical to number them in the y direction, because according to Eq. ( 13-39) below, this numbering will reduce the bandwidth B. At the same time, it will need a greater number of data cards. The mode of numbering will depend on the given problem and capabilities of the code and the computer. In general, however, selection of a
numbering system
that minimizes
B may
be a better strategy.
360
Www
iZ
994
a 'o
361
Two- Dimensional
362
Stress- Deformation Analysis
Chapter 13
Discretization of "infinite" masses such as the foundation of the special treatment. Since only a "finite"
zone can be included
dam
requires
element mesh, we must include adequate extents in the mesh in the vertical and lateral directions so that approximate boundary conditions can be defined. For instance, if in in the finite
mesh we include sufficiently large distances in the two directions, it can be assumed that the displacements at such large distances are negligible. Thus in Fig. 13-14 we have included a lateral distance of about twice the width of the dam from the edge of the dam and have assumed that the horizontal displacement u at that distance is approximately zero. For the problem in Fig. 13-13, a rigid rock boundary is available at a distance of 50 ft (15.25 m) below the base of the dam; hence, we have assumed both displacements to be zero. If such a rigid boundary is not available, we can go to a sufficiently large depth, say about three to five times the width of the dam, and assume either v = and u free or both u = v = at such a discretized boundary. For new problems, the analyst may need to perform a parametric study the
to decide the extents of discretized boundaries.
The crack at the crest of the gallery is introduced by providing node numbers on both sides of the crack. In Fig. 13-15 is shown the distribution of vertical displacements along the ground
level for gravity loads
and for both gravity and hydrostatic
loads.
Detailed analyses of stress concentrations around the crack and of such other aspects as the effect of hydrostatic forces are
Comments.
on
On
left
to the reader.
the basis of the results in Figs. 13-14 and 13-15 and the results
stress distributions
which can be plotted from the same computer
results,
possible to perform (preliminary) design analysis for the behavior of
it is
dams on
multilayered foundation systems.
The computed concentrations of stresses near (shear) stress in the concrete
whether the compressive crete. Similar soils.
check
show us if the tensile we can check
is
within allowable compressive strength of the con-
conclusions can be drawn regarding the stresses in the foundation
Moreover, if
stress
the crack can
within allowable limits. Similarly,
is
if
the design limits allowable deformations,
it
may
be possible to
the loading causes deformations within allowable limits.
In this example, we have considered essentially only loading caused by gravity and have assumed linear material behavior. Hence, the results can be treated as preliminary. For a final design one needs to consider other loadings such as hydrostatic load due to water in the reservior, earthquake loading, uplift due to seepage through foundation, and loading caused by variations in temperature difference between the dam and the surrounding atmosphere. Nonlinear behavior of the soil and rock foundations will influence the behavior of the dam. Finite element procedures that can allow inclusion of these factors are available and will be stated in advanced study of the method and its applications, Chapter 15. For the present, we note that it is possible to perform a detailed stress-deformation analysis of dams on nonlinear foundations; the latter can include both soils and jointed rock masses. In addition to computations of stresses and deformations for various loading conditions, the
metric studies. For instance, tion
it is
method can
possible to find
also be used for performing para-
optimum locations for instrumenta-
and for (underground) structures such as caverns and
galleries.
3 Gfl
-a
c a
I
363
Example
As discussed results that
Beam Bending
13-5.
in the earlier chapters, a numerical
procedure should generally yield
converge or approach the exact solution as the mesh
is
refined con-
To study convergence for the two-dimensional problems, let us consider a (deep) beam (Fig. 13-16) subjected to uniformly distributed load of 1000 kg/m. The beam is divided into three progressively refined meshes as shown in Fig. 13-16. sistently [3].
Note
mesh
that a refined
includes the previous coarse mesh; this
is
required for
mathematical convergence. This and other requirements for mesh refinement are discussed in Ref.
[3].
Table 13-1 shows computed values of vertical and horizontal displacements at
The number of nodes N,
points A, B, and C(Fig. 13-16).
TABLE
the
maximum
difference
Results for the Uniform Loading
13-1
Displacement x 10" 3 (m) Point
N D B
==
(/>
9
4
10
25
6
14
81
10
22
+
1)/
NB 2
A
u
Point
u
V
B
Point
V
//
C V
900 0.0478 -0.1948 -0.0448 -0.1924 -0.00045 -0.0766 -0.0645 -0.2796 -0.00073 -0.1119 0.00079 -0.1278 39,204 0.0761 -0.3198 -0.0726 -0.3173 4,900 0.0678 -0.2821
between any two node numbers D, the semibandwidth B, and the quantity NB 2 shown in the table. The semibandwidth B is computed from the formula
are also
# = (/> +
l)/
(13-39)
where /is the number of degrees of freedom at a node; here/= 2. Figure 13-1 7(a) shows convergence behavior of computed displacements u, v at points A, B, and C with the number of nodes, and Fig. 13-17(b) shows their relationship with NB 2 In both cases the computed displacements tend to converge as N and NB 2 increase. .
The quantity
NB 2 is proportional
to the time required for solution of the global
equations, which constitutes a major portion of the time for the finite element solu-
Figure 13-1 7(b) shows that refinement of mesh results in a faster increase
tion.
NB 2
in
computational effort or cost. In other words, for a given desired accuracy, if the number of nodes are increased, the quantity NB 2 increases at a higher rate than the gain in accuracy. Thus there exists a trade-off between accuracy and ,
that
computer
is,
effort or cost of a finite
Figures 13-1 8(a), three meshes.
(b),
and
(c)
element solution
show
[4].
distributions of the bending stress
We can see that the distribution of bending stress
o x for
the
improves with mesh
refinement.
Comments. The
results of this
example
a finite element solution by progressive refined, the
number of elements
for data preparation
As noted 364
illustrate that, in general,
mesh
we can improve
refinement. Obviously, as the rnesh
is
increase with corresponding increases in the effort
and computer
time.
in previous chapters, a finite
element solution can also be improved by
1000 kg/m
/
II!.., A ® i
@
A
E = 'c
A
A
©
2x10 9 kg/m 2
t>=0.3 Thickness = 0.2
5
-*
m
m
]
(a)
1®
©
A
A ©
t
C i
A
1©
© 5
m
(b)
©
® A
64A
5
'c
;
=
©
A
© (c)
Figure 13-16
mesh,
(c)
Beam
bending,
(a)
Four-element mesh,
(b) 16-element
81 -element mesh.
365
CO
O I
v,
X
A denotes
v
displacement at point
A
and so on
0.40 v,
A
0.30
0.20
0.10
30
40
Number
50
60
of nodes,
80
90
1
00
N
(a)
0.30
0.20 v,
C
u,
A
/. 0.10
/
jL \u, B
/U, C
/ 10,000
20,000
30,000
I
40,000
NB 2 (b)
Figure 13-17 Behavior of numerical solutions for displacements, (a)
Displacement
vs.
number of nodes:
13-16). (b) Displacements vs.
366
NB 2
:
points
points A, B,
A,B, and C (Fig. and C(Fig. 13-16).
—
'
—
;
15,810 kg cnv
*^H
—
»•
—
«*
2,670 kg cm? (a)
45,000 kg cm 2
cm
1,660 kg
:
a :
u
-
j-
23,220 kg cm? (b)
^^
A
/24,000
70 0O0
—*—^-<
ss * R
v
^r ££
'
•>
8
«.
1
J
5
-
$
]
'<, I
—
tf >
[
•.
:
J 1
j| 423 [c
Figure 13-18 Distributions of bending
stress,
ax
at typical sections.
367
Two- Dimensional Stress- Deformation Analysis
368
Chapter 13
using higher-order elements. For instance, for the plane problems,
we can use
quadratic or second-order approximation for displacement within the element. This
nodes
will require additional
and
for the triangle there will be six nodes,
;
for the
quadrilateral there will be eight nodes. There will be a corresponding increase in the
of the element matrices and the number of equations to be solved. Which of the two approaches, refinement of mesh with a lower-order approximation or higher-order approximation, should be used will depend on the type of size
problem. There are trade-offs
wide
The
in
one approach over the other. This subject
in using
is
scope and usually needs parametric studies for a specific problem on hand.
analyst
may need
to use both approaches
and then derive
criteria for their
use
for the given problem.
PROBLEMS 13-1.
Compute
the
13-2.
Compute
N
t
[A]
load vector
initial
triangular element -1
in
number
1
{Q
}
for a given initial strain {e
for the
}
in Fig. 13-9.
Eq. (13- 14c) and then find the interpolation functions
in matrix [N] [Eq. (13-15)].
the details of the derivation of € x €y and y xy in Eq. (13-18).
13-3.
Fill in
13-4.
Obtain the load vector {Q 2 } for a traction load
,
,
fy
applied on side 4-1
[Fig. 13-6(b)].
13-5.
Evaluate the
stiffness
matrix for the triangular element in Example 13-2 by
rearranging the load vector as [q}
13-6.
Find the element
T
=
[u l
u2
vt
v2
u3
v 3 ].
matrix for the triangular element in Fig. 13-9 for
stiffness
plane strain conditions. Hint: Use [C] from Eq. (13-5). 13-7.
Evaluate the element
stiffness
matrix for the triangular element in Fig. 13-9
for axisymmetric idealization. 13-8.
By using PLANE-2DFE or another problem
(Fig. 13-10) with
available program, solve the shear wall
an additional
and compare with those without the 13-9.
kg at the and stresses
vertical load of 10,000
centerline of the wall. Plot the results in terms of displacements vertical load.
In Example 13-4, consider hydrostatic force caused by a water level of
m) above
ground surface,
due to the and obtain distributions of displacement along the ground surface and plot a x a y and r xy within the dam. Compare results with those from the analyses without the hydrostatic force. Figure 13-15 shows the deflected shape of the ground surface with hydro-
40
ft
(12
the
weight of the dam. Use
in addition to the load
PLANE-2DFE
,
,
static loading.
13-10.
Vary the
relative values of
E in the two foundation
layers in Fig. 13-13
and
study their influence on displacements and stresses. 13-11.
For the beam in Fig. 13-16, consider a concentrated vertical load of 5000 kg at point A, and study convergence of displacements vs. TV and NB 2
13-12.
Figure 13-19 show a
.
finite
element mesh for the
beam
in
Example
3-5,
with
<3
<8
° x
E cr>
cn
II
II
©
MOO II
LU
^
-C
41
<$
©
O
© ^3 369
Two- Dimensional
370
Stress- Deformation Analysis
Chapter 13
a hole around the centerline. For the concentrated load shown, obtain the finite
element solution and study the concentration of stresses around (the
corners of) the hole. 13-13.
Figure 13-20 shows a is
medium
subjected to external loading.
The medium
=
of "infinite" extent. Assuming discretized boundaries (u
v
and
distances of four times the width of the loading in the lateral
=
0)
at
vertical
on a suitable mesh. Choose your own value of B, material medium, and the loading. By using PLANE-2DFE or another code, evaluate stresses and deformations in the medium assuming (a) concentrated load at the center point with plane strain and axisymmetric approximations, (b) strip load with plane strain approximation, and (c) cir-
directions, decide
properties of the
cular load with axisymmetric approximation.
Change
the boundaries to
6B, 8B, and 10B in both directions and examine the displacements near the discretized
end boundaries and near the loading. ^Concentrated load
Q_
x
=4,6,8,10,
Strip or circular load.
_:
T3.
Discretize
X 2 =4,6,8, 10,
Discretized end boundaries
y
Figure 13-20 Load-deformation analysis for "infinite" medium.
13-14.
Figure 13-21 shows an underground tunnel (3-m diameter) with a structural lining of thickness equal to 30
cm. The tunnel
is
excavated in a (homo-
geneous) rock mass of large lateral and vertical extent.
It is
required to
compute the changes in the states of stress and deformations due to the tunnel excavation. The material properties are shown in the figure. By using the mesh shown in the figure (or any other suitable mesh) and a computer code, compute stresses and deformations (a) before the tunnel is excavated, that is, under the gravity loading. Here use the entire mesh including that in the zone of the tunnel and lining, (b) Consider the elements in the tunnel zone to be removed, and include the elements for the lining. It may be necessary to renumber the nodes and elements. Compare the two analyses and draw conclusions regarding the influence of tunnel excavation on the states of stresses and deformations, particularly in the vicinity of the tunnel.
Chapter 13
Two- Dimensional Stress- Deformation Analysis
20
///&/£>
371
m
//W/AZ
Rock E =
6x10 8
kg/m 2
^=0.3
u =
3 7 = 2400 kg/m
v free
Lining
E = 2x 10 9 kg/m 2 v = 0.25
u = v free
V77777777777777777777777777777777777777777777777,
U = O, V = o
Figure 13-21 Analysis of underground tunnel.
REFERENCES [1]
Popov, E.
P., Introduction to
Cliffs, N.J.,
Mechanics of
and Goodier,
Timoshenko, York, 1951.
[3]
Desai, C. S., and Abel, J. F., Introduction Nostrand Reinhold, New York, 1972.
Abel,
J.
F.,
S.,
and Desai, C.
Bending," Proc. 2148.
Englewood
N., Theory of Elasticity, McGraw-Hill,
[2]
[4]
Solids, Prentice-Hall,
1968.
ASCE,
S.,
J.
to the Finite
New
Element Method, Van
"Comparison of Finite Elements for Plate Vol. 98, No. ST9, Sept. 1972, pp. 2143-
J. Struct. Div.,
MULTICOMPONENT SYSTEMS BUILDING FRAME AND FOUNDATION
INTRODUCTION Very often, in practice, configuration of a system or structure is such that its approximate simulation may require the use of or it may be beneficial to use more than one type of idealization. For instance, if we need to idealize a three-dimensional building frame and its foundation (Fig. 14-1), it is convenient and economical to treat the building frame as composed of onedimensional beams and columns and two-dimensional slabs and plates. As a rather crude approximation, the foundation can be included by assuming the structure resting on a bed of (individual) springs representing the foundation (Fig. 14-1); this approach is often referred to as a Winkler foundation. Thus, the system contains three components beam-columns, slabs or plates, and foundation. Many other situations in stress-deformation analyses and field problems require such multicomponent idealizations. We shall illustrate here the problem of a building frame (Fig. 14-1). For simplicity we shall consider an orthogonal frame with only horizontal and vertical members. Extension to the general case of inclined members can be achieved by appropriate trans:
formations
372
[1-3].
Chapter 14
Building
Frame and Foundation
373
zation of supports
Idealized
%£>
e
column
Idealized fou ndation
Figure 14-1 Building frame: multicomponent system.
VARIOUS COMPONENTS As
stated earlier, we need to consider structural components that act as beam-columns, slabs and plates, and foundation (springs).
Beam-Column In Chapter 7 we have already derived the element equations for a beamcolumn. However, now we need to consider the possibility of loads causing
bending in both x and y directions. Figure 14-2 shows the beam-column idealized as one-dimensional.
The degrees of freedom for the element consist of axial displacement w, displacements u and v, and rotations 9 X yi and 6 Z about the three
lateral
,
Here the subscript denotes derivative; for instance, 6 X = dw/dx. Hence, the beam-column element has 6 degrees of freedom at each node and a total of 12 degrees of freedom. For the time being, if we do not consider rotation or twist around the z axis, we have a total of 10 degrees of
axes, respectively.
:
374
Building
Frame and Foundation
Chapter 14
w2
A,E,I X ,T,I 2
z
/I
Bending
Axial load
z
Bending
in
x-direction
in
y-direction
Figure 14-2 General beam-column element.
freedom. The nodal displacement vector
W= where 0.^
As
in
[«i
0*1
= (dw/dx) Chapter
7,
lf
"2 yl
0*2
0,i
»i
^2
0,2
w, i
w 2 ],
(14-1)
= (dw/dy) u and so on.
we assume
the following approximation functions
= a, + a z + a v(z) = a + a z + a w(z) =a + a u(z)
then given by
is
2
5
6
9
10 z.
3
z2
7z
2
+a +a
4z
3
z
3
8
,
(14-2a)
,
(14-2b) (14-2c)
:
Chapter 14
Here
u(z)
Frame and Foundation
Building
and
v(z)
375
correspond to bending displacements in the x and y and w(z) denotes shortening or extension due to axial
directions, respectively,
(end) loads.
As
discussed in Chapter
Eq. (14-2) can be transformed to
7,
express w(z), v(z), and w(z) in terms of interpolation functions and their nodal values as [Eq. (7-37)]
= Nxl u + Nx2 = N v + Ny2
u(z) v(z)
yl
i
+N w 2
l
2
yl
x
w(z)=N w
+ Nx3 u + Nx4 x2 + N v + Ny4 6 y2
xl
t
2
2
y3
(14-3a)
,
(14-3b)
,
(14-3c)
,
where
Nxl
N
-
+ 2s ls(\ -2s + s 5 (3 - 25), Is (s - 1), 1
>!
yl
Nx3 Nx = N,
3* 2
3 ,
2 ),
(14-3d)
2
>'3
2
,
and AT,
N = 9
where
s
is
By
(14-3e) s,
the local coordinate given by S
where f
-s,
1
=z—
z
x
and z
x
=-r
(14-4)
the coordinate of node
is
1.
following the procedure outlined in Chapter
7,
we can
derive the
following element equations
mii)
'{Q»]
=
(14-5a)
{Q} {Q.3
where a,[kj
[0]
[0]
[0]
ajkj
[0]
M .
[0]
12
[kj
=
[kj
[0]
6/
4/
=
2
(14-5b)
fcJJ
-12
6/"
-6/
2/ 2
12
-6/
(14-5c)
sym.
4/2
/
P
3
(14-5d)
1
[kw] /
-1
(14-5e)
376
Building
Frame and Foundation
{QF = A |W{X}
and the subscripts a and b denote
+
Chapter 14
fW{f}
(14-6)
Jo
axial
and bending modes,
respectively.
Plate or Slab
The general loading conditions on the frame can cause plane deformaand twisting; for convenience we consider only the first two. Figure 14-3 shows the two effects, which can be superimposed if we assume
tions, bending,
small strains and deformations.
Membrane
Effects
For the in-plane or membrane loading we can assume the plane stress and use the element equations derived in Chapter 13, Eqs. (13-24) and (13-31) for the quadrilateral and triangular elements, respectively. idealization
Figure 14-3 Plate
Membrane patibility
elements
and interelement compatibility,
or in-plane behavior, (b) Bending behavior,
between plate and beam-column.
L (b)
(a)
e-
u
(c)
(a)
Com-
Bending
For isotropic
flat plates,
Dn /(9
4
vv*
{-d^
D=
where v
is
Eh
3
/I2(l
—
the Poisson's ratio,
v
the governing differential equation 4
w*
2d w* d w*\ d^W + -d^) =p
+ ,
),
is
written as: , tA
,
E is
2
is
4
_
(14_7)
the elastic modulus, h
the plate thickness,
is
the transverse displacement
and p
is
the applied
surface traction.
Our aim
herein
is
to
show
essentially the treatment of a
multi-component
system for a building frame in which plates or slabs occur as one of the main
components. problem.
bending
It is
not our intention to go into
It will suffice
is
essentially similar to the other
considered. Here
we
much detail
of the plate bending
to say that the finite element procedure for plate
problems that we have previously
shall describe only the salient features of the
problem.
For convenience, we consider only orthogonal frames; hence, we have only orthogonal (horizontal or vertical) plates. For a general formulation, plates with other configurations can be handled without
much
difficulty.
For the rectangular element in Fig. 14-3(b) we consider three degrees of freedom at each node, namely the transverse displacement w; the rotation about y axis, 9 X = dw/dx, and the rotation about x axis, 6 y = dw/dy. Together with the two in-plane displacements u and v [Fig. 14-3(a)], there are a total of five degrees of freedom at each node. As shown in Fig. 14-3(c), we have an equal number of degrees of freedom for the beam-column and the plate element. Consequently, we can observe the requirement of interelement compatibility between adjacent plate and beam-column elements. The transverse displacement w can be approximated by using hermite interpolation functions as in the case of the beam bending problem [Chapter 7 and Eq. (14-3d)]. In fact, the interpolation functions for the two-dimensional problem can be generated by proper multiplication of the hermitian functions defined for the x and y directions. For example, we can adopt an approximation for
w
as
[3, 5]
= N^N^w, + Nx2 N 6 xl + Nxl Ny2 + Nx3 Nyt w + Nx ,N x2 + Nx3 Ny2 6 y2 + Nx3 Ny3 w + Nx4 Ny3 x3 + Nx3 NyA 6 y3 + Nxl Ny3 w + Nx2 Ny3 9 xA + Nxl N ,6 = [N, N N Nt .-. N l2 ]{q (14-8a) = [NJfo}, the matrix of interpolation functions, N = Nxl NyU N =
w(x,y)
yl
yl
yl
2
3
2
where [NJ
Nx3 N
yl
,
is
etc.,
y,
y
4
3
b]
{
2
where 377
378
Building
Frame and Foundation
Nxl =l- 3s + ls\. Nx2 = as(s - l) Nx3 = s (3 - 25), tf, = as (s -
+ 2V l) btit Ny3 = (3 20, W = bt\t - 1).
2
f
2
x
{a
and
1,
1),
{q^F is
= j>/6, < < r
=
[Wl
0*1
1,
y4
[Fig. 14-3(b)], s
= x/a,
and
^2
^yl
(14-8b)
2
of the element
6) denotes the size t
2
3t
,
2
<
1
2
,
Here
-
N.
2
4
-
Chapter 14
0*2
H> 3
0,2
0*3
™4
0,3
0*4
yi]
the vector of nodal unknowns.
The
N N
interpolation functions
N = Nxl N = (1 -
the definition of inter-
etc., satisfy
,
+
3s 2
yl
t
2
l9
we have
polation functions. For instance,
-
25 3 )(1
+
2t
3
= = 0, = 0, = 0, =h =h - 3t +
2/
3
3r
2
(14-9a)
)
for
5=0, '=0, 5
= =
5
=0,1
o
_ — ^, _
5
5
1,1
1,1
1
#,
f=0, f
f
AT,
=
1,
#,
=
l,
tf,
,>
i
AT,
l
s=i,i'=0,
Now N = Nx2 Nyl =
and so on. tive
of
N
2
2
with respect to x
^2 = ox
Then
±a (3S
tf,
-
2
2
1) (1
_
45
+
1)(1
-
-3/ 2
for
N2
the case of
N
2
,
W
2 ldx
f
=
1,/
=0,
0.0
0.0
*
=
l,f
=
1,
0.0
0.0
5
=0,
=
1,
0.0
0.0
J
= i, = 0, t
9a/64
16
5
= £, = 0, /
Sa/64
—?
s
= |, r = o,
3a/64
5
N
2r
5
5
Plots of #,,
(
+
= 0,
s == o,
2
,
t
= o, = £, — ^, — %, and
0.0
1.0
r
0.0
^
4a/64
(?A^2
).
The
first
deriva-
gr /en by
is
2
05(5
=
1,
/^
are
shown
3
(14-9b)
).
(rad)
5_
16
i -iin Figs. 14-4(a), (b),
and
(c).
In
the slope correspc nding to the degree of freedom 6 xl at node
1
:
Chapter 14
Building
Frame and Foundation
379
(a)
Figure 14-4 Plots
of
bending, (a) Ny. (b)
is
unity. It
interpolation
typical
N2
(c)
.
functions
for
plate
dN2 jdx.
can thus be shown that
other functions in Eq. (14-8a) satisfy
all
the definition of interpolation functions.
The
strain (gradient)-displacement relation for plate 2
d w]
z<
d2w dy 2
=
d2 w
w yy
<
z[Bb ]{ qi
The strain-displacement transformation matrix required derivatives of w from Eq. (14-8). The constitutive law is expressed through moments and second derivatives or curvatures
Mxx
M
"1
V
V
1
is
the relation between the
"
Eh" ~ 12(1
-v
w w
2
)
1
-v 2
[C]{€] is
obtained by finding the
\
yy
Here h
(14-10)
2w xy
dxdv>
[
[6]
w xx
dx 2 (€]
bending is given by
•
the plate thickness.
w J (14-11)
380
Frame and Foundation
Building
The
Up
potential energy for the plate bending
= "T
If
problem
Chapter 14
is
& T[C^ dxdy - h JJ {^YM dxdy - j
expressed as
(tfMd&
(14-12)
A
The components of
can be found by taking appropriate and by using the transformations
{e}
derivatives of w [Eq. (14-8)] -=-
dx
= — ->-
and
= [BJ{q
Substitution of {e}
= —--=-.
-=-
dy
a ds
(14-13) v 7
b dt
and w = [NJ{q 6 } in IT^ and taking variacomponents of {q b } lead to its stationary
A}
tion of Il p with respect to the (minimum) value as
= 0=>|&L =
(?n,
partial
(14-14a)
0,
which leads to the element equations as
KK*} =
(14-I4b)
{Q b },
where \kb ]
=
h
{Q,}
=
h
\\
[B b Y[C][Bb ]dxdy
\\
[Wmdxdy
and
+ The
coefficients
f
[N] T {t}ds.
of [k b ] can be found in closed form by integration. Their
values are tabulated in Ref.
[4].
The load vector {Q b } can
also be
found
in
closed form. For instance, for uniform transverse surface load/? on the plate,
F—
fO iv»J
:
:
[ P ab |_
4
P a2 b
pab 1
pab
pa 2 b
24
24
4
24
2
pa b
pab
"24
24
= P^[6
2
a
b
6
pab
pa b
"24
4 a
2
-b
6
pab 2
pab
"24 pab
4
2
24 ]
-a
-b
6
-a
b].
(14-15)
Here we have considered only one of the many possible and available fact, the function in Eq. (14-8) can be improved by adding the additional degree of freedom Q xy = d 2 w/d x d y [4, 5]. approximations for plate bending. In
Assembly
The assembly of the element equations for beam-columns and plates can be achieved by using the direct stiffness approach, which assures the interelement compatibility of nodal displacements and rotations.
Chapter 14
As
Building
381
numbers of degrees of freedom and beam-columns, the assembly procedure at
stated earlier, since there are equal
for the junctions of plates
such junctions gives no
difficulty.
The element equations the
Frame and Foundation
membrane behavior
for the
beam-column
are given in Eq. (14-5a). for
of the plate in Eq. (13-24). and for the bending
behavior of the plate in Eq. (14-14b).
The
assemblage equations can be expressed as
final
[K]{r}={R}.
(14-16)
These are modified for the boundary conditions and then solved for the nodal
unknowns. Then the
strains,
stresses,
bending moments, and shear forces
can be evaluated as secondary quantities. Representation of Foundation
Figure 14-5 shows an approximate representation for a foundation by using a series of (independent) springs. Structural supports provided by adjacent structures (Fig. 14-1) can also be simulated by using this concept. In the case of foundations, the constants for the springs. k f can be tests on soil samples and or from field ,
evaluated on the basis of laboratory
experiments. They can be evaluated also from the concept of subgrade reaction
[7].
:
5
Figure 14-5 Idealization
column,
of
foundation
by
sp:;-zs.
[a]
Beam-
(b) Plate or slab.
Since the springs are assumed to be independent, their stiffness coefficients
can be added directly to the diagonal coefficients of the global matrix [K] [Eq. (14-16)]. For instance., if kft denotes the stiffness coefficient of spring support at node i in the x direction, then it is added to the diagonal element. Kjj, where j denotes the corresponding global degree of freedom. We note that the spring supports can be specified in the direction of translations (x, y, z)
and rotations
(9 X
,
6 y ).
COMPUTER CODE
A computer program STFN-FE that foundations
is
allows analysis of building frames and
described in Appendix
Some examples
4.
solved by using this
code are described below.
Example
Plate with Fixed Edges and Central
14-1.
Load
Figure 14-6 shows a square plate 25.4 cm x 25.4 cm divided into four equal elements [3]. The thickness of the plate is 1.27 cm, with E = 2.1 x 10 5 kg/cm 2 and v =0.3. The concentrated load P = 181.2 kg. According to the closed form approach [6], the maximum central deflection is given by 0.0056 x
P x D
(25.4) 2
(14-17)
where
D=
Eh3 12(1
~~ - v2 =
21 X
1Q5(L27)3
12(1
)
-
=
0.09)
3 94 x 3jm
10* 1U
*
Therefore 0.0056 x 181.2 x 25.4 2 3>94
The value of
central deflection
x 1Q4
,, = nm 0.0166 cm.
computed from the
finite
element analysis
is
0.01568 cm.
25.40
cm
Figure 14-6 Square plate with central load.
Example
14-2.
Plate (Beam) with
Two
Two
Fixed Edges,
Free Edges, and Central Load
Figure 14-7(a) shows a plate fixed at two ends and free at the other two plate can also be approximately treated as a beam.
0.453 kg
is
applied at the center point.
A
The value of
[3, 8].
This
concentrated load equal to
E=
7.0
x 10 5 kg/cm 2 was
assumed. finite element mesh for the plate. In Fig. 14-8 are shown computed values of transverse displacements at center section C-C of the plate
Figure 14-7(b) shows a the
[Fig. 14-7(b)].
We can compute the displacement by using the results from strength of materials [9] if
the plate
is
and (b) show the moments and conjugate beam, respectively. Displacements approach are shown in Fig. 14-8 in comparison with the finite
treated approximately as a beam. Figures 14-9(a)
distribution of bending
computed from
this
element solutions. For instance, the
382
maximum
deflection at the central section
is
Chapter 14
Building
Frame and Foundation
383
0.453 kg
cm
1.905
0.3175
jL
€ 26.67
cm
T
cm
(a)
®r£
&
M
A
©*A
(b)
Figure 14-7 Analysis for (thin)
beam bending
[8].
(a)
Beam
with
fixed ends loaded centrally, (b) Finite element mesh.
———
——
Finite element solution
Strength of materials solution
Figure 14-8 Comparisons for displacements
from
:
computed and closed
solutions.
PL — \92EI ~ 3
W max m ax
~~
=
453 x 26.67 3 192 x 7.0 x 10 5 x 0.00508
(14-18)
0.0126 cm,
where l
Example
A
14-3.
~
1.905
12
x 0.3175 3
=
0.00508.
12
Frame with Lateral Wind Loads
one-dimensional idealization of a part of the building frame [Fig. 14- 10(a)] is shown in Fig. 14-10(b) [10]. The loading and other properties are shown in Fig. 14-10(b).
384
Building
Frame and Foundation
Chapter 14
1.510 kg-cm
1.510
1.510
(a)
1.510
1.510 EI
(b)
Figure 14-9 Procedure for closed form solution for displacements, (a)
Bending moment diagram,
(b)
Conjugate beam.
732 cm
732 cm
(a)
Figure 14-10 Details of building frame
[10]. (a)
Plane layout.
Figure 14-11 shows the computed deflections and the deflected shape of the frame. Table 14-1 shows the computed values of bending bers in comparison with those
moments
in various
mem-
from the conventional portal method of frame analy-
E = 2.1 I
x
I
Area
x
106 kg/cm 2
= 27346 cm 4 = 1544 cm 4
y
A
= 94.84
cm 2
© 1268
A 2536
®
kg.
A
A
kg-
A
A
A
A
A
732 cm
732 cm 77777
-777777,
777777,
®
©
®
(b)
Figure 14-10 (Contd.) (b) Single idealized frame.
Scale: Horizontal (displacement):
1
cm
= 0.0394
cm
(0.4298, 0.0022 cm)
~~7 /
/ /
/
/
/
/
/
/
/ /
/
-J
/
_J
/
/ /
/
/
z
/ /
/ / / 1
/
/ / /
/
»-
Figure 14-11 Deflected shape of frame.
385
386
Building
Frame and Foundation
Chapter 14
based on the assumption of a point of inflection at the midsection of can be seen that the values from the two methods differ widely at various locations. The finite element approach shows the redistribution of moments sis;
the latter
a member.
is
It
in the frame.
TABLE
14-1
Member 1
2 3
Comparison of Bending Moments for the Frame [Fig. 14- 10(b)]
End
6 7 8
9 10
Example
14-4.
Method
x (kg-cm)
Finite Element
1
203,103
327,224
2
203,103
191,750
1
67,701
21,474
2
67,701
95,712
1
406,205
355,139
2
406,205
258,680 120,860
4 5
Portal
M
1
135,402
2
135,402
17,260
1
203,103
312,616
2
203,103
179,523
1
67,701
30,914
2
67,701
99,495
1
270,803
213,223
2
270,803
190,466
1
270,803
189,069
2
270,803
210,436
1
67,701
95,714
2
67,701
85,694
1
67,701
87,565
2
67,701
99,495
Building
Frame with Floor Slabs
Figure 14- 12(a) shows the layout of a building frame idealized by using a one-
dimensional beam-column, two-dimensional plate membrane, and bending elements; Fig. 14-1 2(b) shows the loading at the two floor levels. Two analyses were performed one without floor plates and the other with floor plates. The properties ;
of the frame material are
E = 2.1 x 10* kg/cm 2 Ix = 27,346 cm 4 ly = 27,346 cm 4 A = 94.84 cm 2 v = 0.3.
,
,
,
Area of cross
section,
,
Table 14-2 shows a comparison between moments (Mx ) at typical nodes for the It can be seen that without floor slabs,
analyses with and without floor slabs.
the middle section of the frame will have higher loads (moments) that those at the
Chapter 14
Building
TABLE
14-2
Frame and Foundation
Comparison of Moments
(Mx )
387 at Typical
Nodes Plates
Node
Not Included (kg-cm)
Plates Included
(kg-cm)
3
32,597
31,289
6
93,747
104,875
12
49,077
22,741
15
158,825
116,723
21
32,576
32,386
24
93,712
105,206
A
©
First
floor level
Figure 14-12 Analysis of "three-dimensional" building frame frame.
[10]. (a)
Details of
Building
388
Frame and Foundation
5817 kg
634
Chapter 14
5817 kg
11,633 kg
— 23,266
11,633
11,633
1268—
5817 634-
5817
11,633
j.
y
i Second floor
level
10,274
20,548
1268-*-1First floor level (b)
Figure 14-12 (Contd.)(b) Loadings at
first
and second
floor levels.
edges. Moreover, the displacements at the central section are significantly reduced if
floor plates are included.
is
0.24393
cm and
Example
To examine
14-5.
For
instance, the
that without plates
is
y displacement
at
node 12 with
plates
0.40225 cm.
Effect of Springs at Supports
on the behavior of the frame shown in Fig. k 6x (kg-cm/rad) is introduced at the three supports (Fig. 14-13). The magnitudes of k 0x were varied as shown in Table 14-3. The results indicate that a very low value of k 6x essentially models a simple support, whereas a the effect of restraint
14-10(b), a rotational spring
Chapter 14
Building
Frame and Foundation
389
A A A
A
732 cm
A
732 cm
At, k.
Figure 14-13
Frame
with
rotational
restraint
represented
by
spring.
very high value produces results close to those with total restraint (see Table 14-1).
Near the value of k 6x between 10 8 and 10 7 the point of inflection can be near the midsection, as assumed in many conventional analyses for frames.
TABLE kex (kg-cm/rad) lOis
lOio 10 9 108
107
10 5 103
14-3
Moments
M
x (kg-cm)
End
Member
1
for Typical Members
Member
1
327,011
312,412
2
191,627
179,406 312,151
1
326,690
2
191,927
179,727
1
323,835
309,812
2
194,620
182,581
1
297,794
287,448
2
219,196
208,307
1
164,984
162,128
2
344,492
335,943
1
3,297
3,260
2
497,105
489,437
1
4
3
2
500,217
48,422
5
Member
8
-188,945 -210,299 -189,132 -210,524 -190,816 -212,537 -206,185 -230,761 -284,898 -322,438 -380,988 -433,436 -382,943 -435,691
TRANSFORMATION OF COORDINATES As we
discussed earlier, it is convenient and useful to adopt a local coordinate system for an element different from the global coordinate system that is
:
390
Building
used to define the entire body.
Frame and Foundation
We
Chapter 14
note, however, that our final
develop element and assemblage relations in the global or
aim
common
is
to
coordi-
nate system.
Although the
were used for applications in
local coordinate systems
Chapters 3-13, the formulation process involved transformations that yielded
For
the element equations in the global system.
instance, in the case of the
one-dimensional line element (Chapters 3-10), the direction of the local system was the same as that of the global system, and the transformations occurred essentially in the derivatives and integrations. The isoparametric
formulation involved use of global displacements in the development of
element equations. Hence, both these instances did not involve transformation of coordinates.
For certain
beam-columns, nonorthogonal and curved structures (e.g., shells), it is often systems whose directions are different from the global
situations such as inclined
slabs in the building frame,
necessary to use local systems.
Then
it is
required to transform the element relations evaluated in
the local system to those in the global system. Such transformation
is
achieved
by using a transformation matrix consisting of direction cosines of angles between the local and the global coordinates. Figure 14- 1 4(a) shows a beam-column element in an (orthogonal) global reference system, x, y,
z,
with a local (orthogonal) system,
x',
y\
z'
attached
with the element. For this case, the transformation matrix can be expressed as
[t]
where l xi
mx
,
etc.
ly
my
ni
I,
m,
n,
(14-19a)
are the direction cosines of angles between the local
and
global axes; for instance l x represents the direction cosine of the angle
between
As
x'
and x axes and so on.
let us consider the beam-column element in the two-dimensional x-y plane, Fig. 14- 14(b); the local coordinate x' is inclined
at i
s
a simple illustration,
an angle of a with the global axes
x.
Here the transformation matrix
[t]
given by
m,
cos a
sm a
m
sin
a
cos a
(14-19b)
[t]
J, The transformation matrix
x
[t] is
orthogonal, that
is, its
inverse equals
its
transpose [t]"
We {Q ( }
can
now
1
=[t] r
(14-19c)
and and load vector evaluated
write transformations for various quantities. If [kj
respectively denote element stiffness matrix
Chapter 14
Frame and Foundation
Building
x, y, z
391
Global Local
x', y', z
(a)
x
y
XcL
©
X
lb)
Figure 14-14 Transformation sional, (b)
of coordinates,
(a)
Three-dimen-
Two-dimensional.
with respect to the local system x\ /,
W=
z',
then
T [T] [ki][T]
(14-20a)
and
{
= imft)
(14-20b)
where [kj and {Q g } are the corresponding matrix and vector referred to the global system, x, y,
z.
In the foregoing, the transformation matrix [T] (14-19a); for instance, [T] in Eq. (14-20)
is
is
composed of [t]
in Eq.
given by
[/]
(14-20c)
[71 [t]
.
[/].
The transformation between displacements can be
m = mw
written as
(14-20
392
Building
where
[u,]
and
[u g ]
Frame and Foundation
Chapter 14
are the vectors of displacements at a point for local
and
global systems respectively.
The element equations
at the global level are lk.]{q,}
where
{q g } is the vector
=
now
expressed as (14-22)
{Q,}
of element nodal unknowns obtained by using Eq.
(14-21).
Once the global
relations are obtained, they can be assembled
by using com-
the direct stiffness assembly procedure by fulfilling the inter-element
unknowns. The foregoing concept can be used and extended for other one-dimensional, and two- and three-dimensional elements. The process is usually
patibility of the
straightforward. For further details, the reader can consult various references
including those in Chapter 15.
PROBLEMS 14-1.
Derive the load vector {Q} in Eq. (14-6) for uniform body force form traction f causing bending. Hint See Chapter 7.
X and uni-
:
14-2.
Evaluate the matrix [B 6 ] in Eq. (14-10). Hint: Find second derivatives of
w
in Eq. (14-8a) as indicated in Eq. (14-10). 14-3.
Derive the part of the
stiffness
matrix for the beam-column element cor-
responding to the twist about the z
axis.
Solution:
[k,]
where
G is
the shear
= GJ
1
-1
!]
modulus and / is the polar moment of inertia. The corzX and z2 at nodes 1 and 2, respectively.
responding degrees of freedom are
REFERENCES [1]
[2]
Martin, H. C, Introduction to Matrix Methods of Structural McGraw-Hill, New York, 1966. Tezcan,
S. S.,
Struct. Div.
"Computer Analysis of Plane and Space Vol. 92, No. ST2, April 1966.
Analysis,
Structures," /.
ASCE,
S., and Patil, U. K., "Finite Element Analysis of Building Frames and Foundations," report, Department of Civil Engineering, Virginia Polytechnic Institute and State University, Blacksburg, Va., 1976-1977.
[3]
Desai, C.
[4]
Bogner, F. K., Fox, R. element-Compatible
L.,
and Schmidt, L. A., "The Generation of Interand Mass Matrices by the Use of Interpolation
Stiffness
Chapter 14
Building
Formulas,"
in Proc.
Frame and Foundation
Second Conf. on Matrix Methods
393
in Struct.
Mech., Wright
Patterson Air Force Base, Ohio, Oct. 1965. [5]
[6]
[7]
[8]
S., and Abel, J. F., Introduction Nostrand Reinhold, New York, 1972.
Desai, C.
Timoshenko, S., and Kieger, Hill, New York, 1959.
S.
to the Finite
Element Method, Van
W., Theory of Plates and
Terzaghi, K., and Peck, R. New York, 1967.
B., Soil
Durant, D., "Analysis of a
Plate Bending
Mechanics
in
Shells,
McGraw-
Engineering Practice, Wiley,
Problem Using Code STFN-FE,"
course project report, Department of Civil Engineering, Virginia Polytechnic Institute [9]
and State University, Blacksburg, Va., 1977.
Timoshenko,
S.,
Strength of Materials,
Van Nostrand Reinhold, New York,
1930. [10]
Malek-Karam, project report, tute
A., "Design and Analysis of a Two-Story Building," M.E. Department of Civil Engineering, Virginia Polytechnic Insti-
and State University, Blacksburg, Va., 1976.
PRELUDE TO ADVANCED STUDY
AND APPLICATIONS
In compliance with our aim to present an elementary treatment,
covered relatively simple problems in Chapters 1-14.
It
is
emphasize, however, that one of the basic advantages of the
method
lies in its
we have
necessary to finite
element
capacity to permit solutions of complex problems for which
conventional solutions are not available or are
difficult.
The reader interested in the detailed knowledge of the method and its wideranging applications would now need to pursue advanced aspects of the method.
The complexity of problems rendered tractable by the finite element method arises due to many factors. These can include irregular and arbitrary shapes of structures and continua, complex boundary and initial conditions and loading, nonhomogeneous materials, nonlinear constitutive or stressstrain
behavior, existence of liquids in the media, interaction between
dissimilar media,
and time dependence.
Higher-order approximations and mesh refinement, improved element formulations, relative merits and trade-off analyses, mathematical properties
—convergence,
stability,
and consistency
number of other aspects method are important. Following are some of
—of various
related to the theory
formulations, and a and mathematics of the
these topics and their significance. Many of the advanced topics can be quite involved and can constitute subjects of significant depth by themselves. It will indeed be very difficult to describe them in detail. Hence, we have essentially listed topics and subtopics relevant to
394
;
.
Prelude to Advanced Study and Applications
Chapter 15
engineering.
The bibliography
at the
395
end of the chapter contains publications
that can be consulted for the study of these topics.
THEORETICAL ASPECTS 1
Formulation Procedures a.
Variational potential, complementary, mixed, and hybrid approaches :
Ritz and Raleigh-Ritz methods. b.
Weighted residuals:
collocation,
least
squares,
subdomain,
and
Galerkin. 2.
Transformations of Coordinate Systems
3.
Isoparametric Formulations
4.
Higher-Order Approximations
5.
a.
Polynomial, Lagrangian, Hermitian, and spline interpolation.
b.
Requirements for approximation functions.
c.
Mesh
d.
Higher-order approximation versus mesh refinement.
refinement.
Mathematical Aspects a.
Basis of formulation
b.
Accuracy, convergence,
c.
Initial
linear
and nonlinear operators, normed
stability, consistency, error
spaces.
bounds.
and boundary value problems.
d. Integration
6.
:
of equations.
e.
Time
f.
Singularities.
g.
Eigenvalue problems,
h.
Solution of equations.
integration
and
properties.
Factors a.
Arbitrary geometries.
b.
Nonhomogeneities.
c.
Composite materials. Boundary conditions, nonlinear boundary conditions,
d.
free-surface
problems. e.
Loading.
dynamic, and repetitive, Monotonic and path dependent, iii. Time dependent. Interaction effects: between two dissimilar media such and soil and structure and water. i.
Static,
ii.
f.
7.
Constitutive (or Stress-Strain) Behavior a.
Linear
elastic.
b. Piecewise linear elastic. c.
Higher-order
elastic.
as structure
Prelude to Advanced Study and Applications
396
Chapter 15
d. Hypoelastic. e.
Elastic-plastic. i.
ii.
Perfectly plastic: nonfrictional. Perfectly plastic: frictional.
iii.
Strain softening,
iv.
Cap models and
f.
Viscoelastic.
g.
Creep.
critical state
concepts.
h. Thermoviscoelastic. i.
8.
Endochronic.
Nonlinear Analysis a.
Material and geometric nonlinearities.
b. Incremental, iterative, c.
9.
Initial stress, strain
and mixed procedures.
procedures.
Coupled Problems a.
Thermoelastic.
b.
Consolidation.
c.
Liquefaction.
10. Fluid
Flow
Multidimensional seepage and consolidation. 11.
Environmental Problems:
Mass
Transport, Diffusion, and Convection
Mechanics and Hydrodynamics Thermodynamics Eigenvalue Problems Computer Implementation a. Bandwidth and wavefront techniques.
12. Fluid 13.
14. 15.
16.
Evaluation and Comparisons of Various Schemes
BIBLIOGRAPHY Brebbia, C. A., and Connor,
J. J.,
Fundamentals of Finite Element Techniques,
Butterworth's, London, 1973.
Cook, R. D., Concepts and Applications of
Finite
Element Analysis, Wiley,
New
York, 1974. S., and Abel, J. F., Introduction Nostrand Reinhold, New York, 1972.
Desai, C.
Desai, C.
S.,
neering,
to the Finite
Element Method, Van
and Christian, J. T. (eds.), Numerical Methods McGraw-Hill, New York, 1977.
in
Geotechnical Engi-
Finlayson, B. A., The Method of Weighted Residuals and Variational Principles, Academic Press, New York, 1972.
Gallagher, R. 1975.
H,
Finite
Element Analysis, Prentice-Hall, Englewood
Cliffs, N.J.,
Prelude to Advanced Study and Applications
Chapter 15
Huebner, H., The
Finite
Element Method for Engineers, Wiley,
Martin, H. C, and Carey, G. Hill, New York, 1973.
F., Introduction to Finite
Oden,
J.
T., Finite
J. T.,
New
—Fundamentals and
Elements of Nonlinear Continua, McGraw-Hill,
and Reddy,
J.
N.,
Variational
Methods
York, 1975.
Element Analysis, McGraw-
Norrie, D. H., and de Vries, G., The Finite Element Method Applications, Academic Press, New York, 1973.
Oden,
397
in
New York,
1972.
Theoretical Mechanics,
Springer, Berlin, 1976.
Pinder, G.
demic
F.,
and Gray, W. G., Finite Elements New York, 1977.
in
Subsurface Hydrology, Aca-
Press,
Prenter, P. M., Splines and Variational Methods, Wiley,
New
York, 1975.
Strang, G., and Fix, G. J., An Analysis of the Finite Element Method, Prentice-Hall, Englewood Cliffs, N.J., 1973.
C, The Finite Element Method York, 1971.
Zienkiewicz, O. Hill,
New
in
Engineering Science,
McGraw-
APPENDIX
VARIOUS
NUMERICAL PROCEDURES SOLUTION OF BEAM BENDING PROBLEM
INTRODUCTION To
illustrate the
use of energy procedures and of the methods of weighted
an example of beam bending by and finite difference methods the former is based on the minimization concept and is often considered a forerunner of the finite element method. The coverage of different methods herein is intended only as an introduction to various procedures and to give the reader an idea of the available schemes detailed study of these procedures is beyond the scope of this elementary treatment. Before the problem of beam bending is considered, we shall give further details of the methods of weighted residuals (MWRs) introduced in Chapter 2. A good introduction and applications of are given by Crandall [1]. As stated in Eq. (2-12), the trial or approximation function in the residuals, in this
appendix we
shall solve
using these procedures. Moreover,
we
also consider the Ritz
;
;
MWR
MWR
is
expressed as
u
=
±
(2-12)
«,?,.
The undetermined parameters a, are chosen such that the the domain D vanishes. This is usually done in an average R(x) (Fig. 2-7) with respect to weighting functions
f
Jd
398
R(x)W (x)dx i
= 0,
i
=
1, 2,
.
.
.
W ,
t
(x).
«.
residual R(x) over
sense by weighting
Thus, (Al-1)
Appendix
Solution of Beam Bending Problem
1
D is
For a one-dimensional problem, the domain
399
simply the linear extent of
the body.
VARIOUS RESIDUAL PROCEDURES
Wu and depending
There are a number of ways to choose
W
we
ti
on the choice of
obtain different procedures.
In the case of the collocation method,
W = 3(x - x
(Al-2a)
t).
t
Then RixdSix
-
= 0,
x) t
=
1, 2,
1,
x
=x
[0,
x
^x
i
.
.
.
,
(Al-2b)
n,
where
=
3 is
the Dirac delta function. This
selected
number of
A 1-1 (a),
the residual
The
i
means
if
t9
that the residual
is
equated to zero at a
points in the domain. For instance, as is
=
equated to zero at n
shown
in Fig.
5 points.
domain can be divided into a number of subdomains [Fig. and the residual is integrated and equated to zero over each subdomain. This yields the subdomain method. Here the weighting functions are
A
1
total
-1(b)],
-
,+1
x 'J < x and x
0,
t
'
>
(Al-3a)
xt+l
.
Then
p
1
R(x)dx
= 0,
=
i
In the case of the least-squares
1, 2,
...,«-
method
1.
[Fig. Al-l(c)], the
(Al-3b) weighting
functions are chosen to be
*-g. and
(Al-4)
minimization of the integrated square residual as
this leads to
f Jd
R 2 (x)a dx = t {
i
(Al-5)
0.
In Galerkin's method [Fig. Al-l(d)], the weighting functions are chosen as the coordinate functions from Eq. (2-12), and hence f
Jd
*(*)?,(*)=
0,
/
=
1,2, ...,«.
This expression implies that the functions
q> t
are
made orthogonal
(Al-6) to the
residual R{x). In finite element applications, the approximation or trial
Solution of Beam Bending Problem
400
Domain D
Appendix 1
,
£ R(3)
12
!
R
I 4
1
3
5
(x=)
=
= 1,2
i
5
(a)
R(x)dx =
-1
+
0,
i
= 1,2,3, 4
f:
1
(b)
2
JR
(x)a, = 0,
= 1,2
5
n
Ry?,dx = 0,
Figure Al-1 Methods of weighted residuals,
Subdomain.
functions are functions
N
t ;
(c)
=
1, 2, 3,
4
(a) Collocation, (b)
Least squares, (d) Galerkin.
commonly then the
i
expressed in terms of shape, interpolation, or basis
N
(
are usually chosen as the weighting functions.
BEAM BENDING BY VARIOUS PROCEDURES The beam and
details are
shown
in Fig.
A 1-2. Assuming flexural rigidity to be
uniform, the governing differential equation residual
is
given by Eq. (7- lb), and the
is
R(x)
=
F^ -
P(x),
(Al-7)
Appendix
Solution of Beam Bending Problem
1
401
Collocation Finite difference
5
Subdomain 3L/4
L/2
Galerkin, Least squares,
(c)
Ritz
0-L
A
(d)
A
©
Finite element
©
Figure Al-2
Beam bending with
finite difference, (b)
© methods,
different
Subdomain.
(c)
(a) Collocation,
Galerkin, least squares, Ritz.
(d) Finite element.
where F = £7 is the p(x)
is
w is the assumed transverse displacement,
flexural rigidity,
the forcing function,
and x
the coordinate.
is
The boundary conditions
associated with Eq. (7- lb) can be expressed as
W (X
= 0) =
2
d w, dx 2
The
first set
tions
w(x
=L)=0,
Ax
0)7
dx
L)
=
(Al-8a)
(Al-8b)
0.
:
[Eq. (Al-8a)] represents the essential or forced
and the second
set [Eq.
boundary condi-
(Al-8b)] represents the natural boundary
conditions.
We now choose unknown w*
the following trial or approximation function for the
:
w
=
a, sin
^+a
2
sin
^+
a3
sin
^+
a4
sin
^
= t *0M, where the a are the undetermined parameters and the t
(Al-9) (p t
are the
known
:
Solution of Beam Bending Problem
402
1
Here we have chosen w* in terms of the trigonometric functions, Chapter 7 the approximation function was chosen in terms of
functions.
whereas
Appendix
in
interpolation functions
N
t.
Note that the function w in Eq. (A 1-9) satisfies the boundary conditions in Eq. (A 1-8) at the two ends of the beam. In Chapter 7, when we used Galerkin's method for each element, then only the geometric boundary conditions [Eq. (Al-8a)] at the ends were used to modify the assemblage
equations; as explained in Chapter
boundary conditions
the natural
3,
[Eq. (Al-8b)] are satisfied automatically in an integrated sense.
To
express R(x) in terms of
w and
its
derivative,
we
differentiate the
expression in Eq. (A 1-9) four times as
=
-T-;
1*cl x sin -j-
+ where X
256A 4 a 4
+
16/ 4 a 2
sin
-j-
+ 8U
4
a3
sin
-j—
sin^,
(Al-10)
= n/L.
Now we shall consider the solution of the beam bending problem by using a number of different procedures. For this illustration, the following prop-
assumed
erties are
E=
X
10
10 6 psi,
= 10 in., A = in. x in. = in. PA = 500 and PB = 1000 lb/in. L
1
2
1
1
,
Collocation
As shown 4L/5,
i
=
A 1-2, we chose four points at x = L/5,
in Fig.
1, 2, 3, 4,
at the supports.
{
from the end A. Note
instance, for Xj
X4a
l
+
sin
4.77
x
103
7.72
x
7.72
x 10 3 ai
4.77
x
103
103
2L/5, 3L/5, and identically zero
j-
= x )=0,
1
t
= 1,2,3,4.
(Al-lla)
= L/5, we have
~ + 16A a 4
256A 4 a 4
and so on. The
is
Then we have R(x
For
that the residual
sin
2
sin
^^ -
^ y + 8U a
Pb
4
~
Pa
3
sin
^~
±--Pa =
(A-l lb)
resulting equations are
+ + -
1.23
x 10^2 10 4 a
7.63
x
7.63
x 10 4 a 2
1.23
x 10*a 2
2
+ 6.25 - 3.86 - 3.86 + 6.25
x 10 4 a 3 x 105a 3 x 10 5 a 3
x
105
+ + -
= 10^4 = 10 6 a 4 = 10 6 a 4 =
x 10«a 4
600,
1.98
x
700,
1.98
x
1.22
x
1.22
(Al-12) 800, 900.
:
Appendix
Solution of Beam Bending Problem
1
403
Solution of these equations gives values of a, as a,
a3
= 0.11374399, = 0.00033150,
a2 a3
= =
-0.00105974, (Al-13)
-0.00001564.
Hence, the approximate solution according to the collocation method
w
0.11374 sin
^ - 0.00106
-0.00001564
sin
^+
0.0003315
sin
is
^*
4nx
(Al-14)
sin
Subdomain Method Here,
we have
A
[Fig.
1
-2(b)] rL/4
R(x)dx
o,
R(x)dx
o,
R(x)dx
0,
R(x)dx
0.
Jo r 2L/4
J L/4
(Al-15) r
3
LI 4
J 1L 4
L [
J
3Z.
4
After integrations, the resulting four equations are 7.57
x lCPai
1.83
x
10 4
1.83
x 10 4 ai
7.57
x 10 3 ai
ai
+ + -
1-19
x 10*a 3
4.93
x 105a 3
x 10 5 a 2
+ -
4.93
x 10 5 a 3
x
+
1.19
x 10 6 a 3
2.07
x
2.07
x 105a 2
2.07 2.07
105
10 5 a 2
+ -
3.31
x 10^4
3.31
x 10*a 4
+ -
3.31
x 10 6 a 4
3.31
x 10 6 a 4
= = = =
1410,
1720,
(Al-16) 2030, 2340.
Solution of these equations leads to the following approximation
w
=
0.12388 sin
^-
0.00004724
0.00151 18 sin
^+
0.00078719
4nx
sin
^ (Al-17)
sin
Least-Squares Method
In this procedure, the weighting functions are
&«*»•
(Al-18)
Therefore,
W — sin-p, x
W
3nx 3
sin
W,
2nx sin
L
W = sin Anx —fjjr
.
A
>
Solution of Beam Bending Problem
404
According to the least-squares method
[Fig.
A
1
Appendix
1
-2(c)],
*(*)*§& =
1 or
=
0,
sin
—j- dx =
0,
sin
—j- dx =
0,
!*(*) sin
^
0.
R(x)
I
R(x)
sin -j-
dx
(Al-19)
R(x)
The
final
4.06
four equations and the resulting approximate solution are
x 10 4 ai x
ai
x
ai
Oxai w
=
5
0.11760 sin
-
+ 0xa + 0xa + 3.29 x 10 6 a + 0xa
+ 0xa 2 + 6.49 x 10 a 2 x a2 + + 0xa 2
3
3
3
^ - 0.001226
0.00003830
sin
^
sin
3
+ + + +
0xa 4 0xa 4 x 1.04
^+
x
= 4774.65 = -795.77, a4 = 1591.55, 10 7 a 4 = -397.89,
0.0004837
sin
(Al-20)
^ (Al-21)
•
Galerkins Method It
is
method
incidental that the weighting functions
are the
same
in this specific case,
same
as the functions
(p t
W
t
in the least-squares
used for the Galerkin method. Hence,
both the Galerkin and least-squares methods yield the
solutions.
Ritz
Method
In the Ritz method [1,2] the potential energy in the body (beam) trial functions, and the resulting expression
expressed in terms of the
minimized with respect to a For instance, f
a,.
is is
This leads to a set of simultaneous equations in
.
^ = \[ F &) dx -\j wdx
<
A1 - 22 >
with
p{x)=Pa
+ ^(Pb-Pa)-
(Al-23)
Solution of Beam Bending Problem
Appendix 1
405
Minimization of IT, with respect to a gives t
dll p
= 0, ~
da 2
U
'
(Al-24)
dUp
0,
da 3 dll p
= 0.
(?a 4
For
problem, the equations are the same as in the Galerkin and
this specific
least-squares methods,
and the approximate solution
is
the
same
as in
Eq. (Al-21).
is
Comment: It may be noted that in the Ritz procedure the potential energy minimized for the entire beam in other words, the limit of the integral is ;
from
to L.
The concept is thus
similar to the finite element
method (Chapters
method, the minimization of achieved for the domain composed of a patchwork of elements.
3-5), except that in the case of the finite element
11,
is
Finite
Element Method
method can be achieved in a manner Chapter 7, with the subdivision consisting of two elements [Fig. A 1 -2(d)]. We can use Eq. (7-15) for generating the twoelement equations and then performing the assembly with boundary conditions, and the solutions of the resulting assemblage equations are [3]
The
solution by the finite element
identical to that covered in
"
12
30
30
100
-12 -30
F -12 -30
24
125
30
30
(
50
50
-12
30
200
-12 -30
-30
12
-30
50-30 Solution after introduction of w = w = w = 0.1171875, w = 0.000000, 6 = 0.0007295, 0, = 0.036665, x
rv 2
2
Finite Difference
'
3
w2 >
50
rw
3
100_ [e 3l
1437.5'
1250.0
e,
30
t
W;
=
3750.0
(Al-25)
<
417.0
2312.5
-1875.0,
gives
w
3
= 0.0000000,
e
(A 1-26) -0.0383335.
Method
Before the era of the
finite
element method, the
was the commonly used technique
method and mathe-
finite difference
for problems in engineering
Solution of Beam Bending Problem
406
matical physics. Details of this
However, we
method, the reader can
The
beyond the scope of
are
this
book.
beam problem using this method mainly to of commonly used numerical methods. For study of
shall solve the
complete the discussion this
method
Appendix 1
refer to various textbooks
method
finite difference
[1].
based on the concept of replacing the
is
continuous derivative in the governing differential equation by approximate
For
finite differences.
Fig.
instance, various derivatives are
approximated as
(see
Al-3)
Second
w i+1 — w Ax
dw
First derivative:
dx ~
derivative:
d2 w dx 2
w,_,
—
"
t
Third derivative:
Fourth derivative:
d4 w dx 4
w i-2 -i
Substitution of
/
+
t
i
_
1
2Ax
w i+1
2
2w,_
—
t
2w i+l
+
(Al-27)
w i+2
2(Ax) 3 yv t
—
4w
f
_,
+
6vv,-
—
+
4w /+1
w
t
Ax 4
to replace the fourth-order derivative in Eq. (7- lb) leads to
—
4w -_ t
1
+
6w
t
—
4w i+l
+
w i+2
= 0,
1, 2, 3, 4,
A
1
5 for six points in the
-2(a)] gives six
(A 1-28)
Pf
Ax'
into five segments [Fig.
tion of the
Ax
2w
— w,-_ +
p Wj-2
f
t
Ax 2
d3 w _ dx 3
Use of Eq. (Al-27)
^ w — w _i „w M—w ~~
beam domain
boundary conditions [Eq. (Al-8)]
finally gives
four simultaneous
Figure Al-3 Finite difference approximation.
2
i-1
i
divided
simultaneous equations. Introduc-
i
+
1
i
+
2
i
+3
x
:
Appendix
Solution of Beam Bending Problem
1
wu w2
equations in
vv 3
,
,
5w — 4h>! + w — + i
\
and
u> 4
:
+ w — 4w + 6w - 4w
4w 2 6h> 2
4w 2
w2
407
2
3
3
3
+ + — +
=0.0115,
w4 4w 4 5w 4
= 0.0134, = 0.0154, = 0.0173.
(A 1-29)
Solution.
with
w
==
=
w5
w
t
w
3
= 0.070656, = 0.115968,
w 2 =0.114432, w4
(A 1-30)
= 0.073344,
0.
COMPARISONS OF THE METHODS To compare
from various methods, we first state the closed form on the strength of materials theory [4]
results
solution for the displacement based
w
=
^(L - 2Lx + x 2
3
3
) -|-
)X
(/7
f80ir^l
(3^
4
-
10L 2 * 2
+
Results for displacements at typical locations on the
7L 4 ).
(Al-31)
beam by
using
various methods and the closed form solution [Eq. Al-31] are compared in
Table
AM. TABLE
Al-1
Comparisons for Displacements Deflection,
Location from
Method
2
in.
4
in.
End A
5 in.
6
in.
8 in.
Collocation
0.067
0.110
0.113
0.109
0.068
Subdomain
0.072
0.117
0.123
0.118
0.075
Galerkin
0.068
0.111
0.117
0.112
0.071
Least squares
0.068
0.111
0.117
0.112
0.071 0.071
Ritz
0.068
0.111
0.117
0.112
Finite element
0.067
0.110
0.117
0.069
Finite difference
0.071
0.114
—
0.116
0.116
0.073
Strength of
0.068
0.111
0.117
0.112
0.071
materials
The
results for the first five
methods and the
last
procedure are obtained
and by substituting various values (Al-31). Since we did not have a point at x = 5, there is no direct result at that point for the finite difference method. In the case of the finite element
of*
into Eqs. (Al-14), (Al-17), (Al-21),
Solution of Beam Bending Problem
408
Appendix
1
method, once the nodal displacements and slopes are obtained, values at other points can be obtained by substitution of (local) coordinates into Eq. (7-2). We can also compute moments and shear forces by using the results with second and third derivatives, as we did in Chapter 7. For examples of solutions of problems similar to the beam bending and other problems solved by using different procedures presented herein, the reader can refer to various publications such as Crandall [1]. For the foregoing beam bending problem, the results for displacements from various procedures are close to each other and to the result from the closed form solution. These comparisons are presented only for the sake of introducing the reader to some of the available numerical procedures. The merits of the finite element
method become evident as we solve problems with and geometric properties and
greater complexities in factors such as material
loading characteristics.
REFERENCES [1]
Crandall,
[2]
Abel,
[3]
Desai, C.
J.
S.
H., Engineering Analysis, McGraw-Hill,
F., (private S.,
[4]
Timoshenko, 1956.
S.,
York, 1956.
communication).
and Abel,
Nostrand Reinhold,
New
J. F.,
New
Introduction to the Finite Element Method,
Van
York, 1972.
Strength of Materials,
Van Nostrand Reinhold, New York,
APPENDIX
SOLUTION OF SIMULTANEOUS EQUATIONS
INTRODUCTION Most problems
solved by using numerical methods result in a set of algebraic
simultaneous equations of the form [Eqs. (2-21) and (2-22)] [K]{r}
= {R}.
(A2-1)
These equations can be linear or nonlinear;
we have
in this text
essentially
dealt with linear problems.
In general, Eq. (A2-1)
is
often expressed in matrix notation as [A]{x}
= {b)
(A2-2a)
or simply as
Ax where
[A]
is
= b,
(A2-2b)
the matrix of (known) coefficients such as
the vector of
parameters such as {R}.
*n*i 021*1
a Hl x
x
K
(J
[Eq. (2-22)], {x}
is
and {b} is vector of (known) forcing In expanded form Eq. (A2-2) can be written as
unknowns such
as
+ 012*2 + + 022*2 + +
an2 x 2
+
{r},
•
• •
•
•
•
•
•
•
+ +
a ln xn
+
a nn xn
a lH xn
=b =b
l9
29
(A2-3)
=b
n,
409
410
Appendix 2
Solution of Simultaneous Equations
where n
number of unknowns and denotes
the
is
number of
the total
equations.
METHODS OF SOLUTION The two common methods
for solution of Eq. (A2-3) are the direct and Gaussian elimination and a number of its modifications
iterative procedures.
are examples of direct methods, and Jacobi, Gauss-Seidel, successive over-
and symmetric successive overrelaxation (SSOR) are We shall briefly illustrate some of
relaxation (SOR),
examples of the
iterative techniques [1-3].
these techniques.
Gaussian Elimination
This
is
equations.
perhaps the simplest and a It is
common method
for solving linear
based on the idea of creating a sequence of equivalent systems
of equations by using a number of steps of elimination, and then solutions for the
unknowns x
lent systems
are obtained by a process of back substitution.
The equiva-
have the same solutions, but each successive sequence
than the previous one. Let us consider the system of equations
and denote the
sequence by superscript
initial
+ +
«iV*i fllV*i
)
tfi
tfixt «ffl*a
*i+.«ff*a
+ + +
•
•
•
•
•
•
+ +
.-.
flfite
is
simpler
Eq. (A2-3)
(1) as
= M» = W\
a[»xn
in
+ a£x =b n
n
(A2-4a)
>\
or
=b m
A (1) x
(A2-4b)
.
step of elimination, we use the first equation in Eq. (A2-4a) and by appropriate multipliers so as to annihilate the first terms of all the subsequent equations. For instance, if the first equation is multiplied by k 2l — — (tfiV/fliV) an d then added to the second equation, we have
In the
multiply
first
it
(a« +Xi*iV)*i
+
(flffl
+
l
*2ia 2 l)x 2
+
•
•
•
+
+
(affi
Aaiaffitei
=
W
}
+ hiW)
(A2-5)
or fl£>X„ In
= b?\
where a 222 == dil + A 21 « 2 and so on and the superscript (2) denotes the second sequence. Similarly, we can multiply the first equation by ( 1 )
i
A1 31
_ —
fl 3
1
-TIT'
flu
A1 41
—
"4 1 (i)>
flu
1 •
•
•
»
*nl
—
u n\ (1)
flu
1
Appendix 2
Solution of Simultaneous Equations
and add to the second, the end of
first
third, etc., equations to obtain a modified
1
v
I'M
+ a^ixi + + a%x + + a™x + 2
2
+ a$x + 2
In the next step,
we
•
•
•
•
•
•
•
+ aiite + a%xn — + flff*. —- o
/>< 2 >
/)
•
'
•
+
•
•
aff x.
=
(2 > 3
(A2-6a)
,
.(2) b«
multiply the second equation in Eq. (A2-6a) by
/32
42
wr
~
~wr
•
x *»
•
- — sgf
results of the third, fourth, etc., equations,
«iV*i
sequence at
step of elimination as (l)
and add the
41
!)
+ + + +
«iy*a „(2) Y
"22-M
1
+ fl,Vxj + a&Xt + «JV* + a®x S
3
+
+ + + +
which leads to
+ affjc. = 6 + t&x. = 6i2 + a»>*„ = W, + a%x = Ai», (1)
•
•
'i
•
>
>,
•
•
(A2-6b)
n
+ fl& *3 + Finally, at the end of n — elimination steps, we shall have a.'M" + ai¥*2 + <*,V*s + «JV*4 + + ai'.'x, = M 2 + «Sx2 + ai 3>x + a$JC + + tfi»*. = b{ \ + + a$x + aflx, + + a%x = 6< + + + a^xt + + ai?x = b»\ }
•
1
1
•
•
.',
•
2
4
3
•
•
•
•
•
•
3
n
3
>,
n
(A2-6c)
fl
We
(»-l)
r
I
fl
(»-l)
y
_
A(n-l)
note that Eqs. (A2-6a), (A2-6b), and (A2-6c) are equivalent to the
original Eq. (A2-4) in the sense that they
(A2-6c)
is
x n can be obtained Back
The
all
have the same solution. Equation
a triangular system of equations, and the solution for the directly
from the
unknown
last equation.
Substitution
first
(A2-6c) as
step here
is
the solution for
x n from
the last equation of Eq.
:
412
Solution of Simultaneous Equations
*»
Appendix 2
= £fcu
(A2-7a)
n.n
Now
the solution for
xn _ can be found from the x
*.->
=
^
—
(n
l)th equation as
(A2-7b)
^fe^*-
Since x„ is known from Eq. (A2-7a), xa _, can be found easily. The process can be repeated until the solution for x, is obtained. The foregoing is the basic Gaussian elimination technique. A number of
modifications and alternatives such as Crout, Jordan, Aitken, and GaussDoolittle can be used depending
on the
characteristics of the system equa-
tions [1-3].
Banded and Symmetric Systems In finite element applications, very often
we encounter systems of equa-
banded and symmetric. In banded matrices, nonzero coeffioccur only on the main and adjacent diagonals, and other locations
tions that are cients
have zero a tj
=a
jt .
coefficients.
For
Moreover, the system
is
very often symmetric; that
instance, the following represents a
banded
is,
(tridiagonal)
symmetric system of equations
+
flii*i+tfi2*2+ #12*1
+ + + +
=b u
-f-
=b + + #23*3 + =b 23 *2 + #33*3 + #34*4 + + #34*3 + #44*4 + #45*5 = +#35X4 + 055X5=65. +
#22*2 tf
(A2-8)
2i
3
,
^>4>
Bandedness and symmetry of the equations allow
significant simplifica-
tions in the foregoing general elimination procedure. It
is
necessary to store
only the nonzero elements in the computer and only the coefficients on the
main diagonal and the upper or lower diagonals
[4].
SOLUTION PROCEDURE Almost
all
systems of equations resulting from the
are relatively large
portion of computational effort in a solution of these equations. It
is
not
the subsequent iterative procedures
For sake of introduction, we elimination process.
finite
element analysis
and require the use of the computer. In finite
element solution
fact,
is
a major
spent in the
difficult to program the elimination and on the computer.
shall
now
illustrate the use
of the foregoing
:
Example A2-1.
Solution by Gaussian Elimination
Consider the system of three equations, Eq. (3-38). In Chapter
we
equations by a form of Gaussian elimination. Here
3,
we
solved these
follow the foregoing general
procedure 100 (1) jc!
-
+ +
100 (1) * 2
-lOO^*! + 200 (1) * 2 (1)
^!
-
100 (1) x 2
(1)
*3
100 (1) *3
200 (1) *3
= = =
7.5 (1)
,
15.0 (1 \ 15.0 (1)
(A2-9)
.
Details:
+ 10Q(i) _ Therefore
we have
+
the second equation as
(200
-
100) (2) x 2
-
100 (1) * 3
=
15.0 (1)
+
100 (2) * 2
-
100 (2) * 3
=
22.5 (2)
+
7.5 (1)
or
Since the third equation already has zero in the
lOO^*!
- 100 + 100 - 100
(1)
(2)
x2
(2)
Now
A 32
=
+100/100
+
=
*2
+ - 100 + 200
term, the modified equations are 7.5 (1 \
(2)
*3
(1)
*3
= =
22.5 (2) , 15.0 (2)
.
Therefore, the third equation becomes
1.
+
x2
first
.
(200
-
100) (3) X3
=
15.0 (2)
=
37.5 f3)
+
22.5 (2)
or
+ Hence, the
final
+
100 (3) x 3
modified equations are
IOO^jcj
- 100 + 100 +
(1)
(2)
The back
.
jc 2
*2
+ - 100 + 100
7.5 (1) (2)
(3)
;t3
=
jc 3
=37.5 (3)
22.5
,
(2) ,
.
substitution gives
*3
_37.5
~
100 22.5
100
+
100 100
37.5
(A2-10)
100
60 100 7.5
100
+
100 100
60 100
67.5 100'
413
—
——
—
1
Iterative Procedures
In the iterative procedure, an estimate
made
is
for the
successively corrected in a series of iterations or trials. is
continued until convergence, which
is
The
unknowns x and iterative
is
procedure
often defined by selecting a small
number by which the final solution differs from the solution in the previous iteration. The simplest iterative procedure is the Jacobi scheme [1-3]. acceptable
To it
illustrate this
procedure, consider the set of equations (A2-3) and express
as
a lt xP (
021*
O)
1
o) 3
i*i
o)
0*i*i
+ + +
a 12 x?>
+
* 30)
+
0)
+
*23*3
a 23 x 20)
+ + +
a 33 x 3 0)
+
•
a n2 x 20)
+
aB2 x 3"
+
- - -
0)
tf22*2
tf
13
where the superscript (0) denotes compute the values of x as Y (l) Xi
0n
Y 2(l) _ — X
Y 3(l)
— —°\2 xY
_ —
— — 21 XY fl
a 22
_ — 031 —Y —
x
—
Now we
—— X 0_nl
l
— 032 —Y
2
— XY
2
~7,
0_n2.
x
a nn
compare \x\
' ' '
•
= bu 2nX™ = b =b a 3n x
+ + +
a ln x™
+
a nn x^
2t
{
x\ l)
x)
\3
0u
xy (0) 3
_
032 Y (0) — X3
-
x
0)
{
3
=b
\n
«u
_
.
.
a 22
_
.
.
_ 03« —Y
.
~7,
033
(O)
——
0_n3_
a nn 0)
(0)
n
(0)
"
b |
we
x
-\
j_ -f-
I
'
—#2 a 22
—
b3
(A2-llb)
'T,
033
j_ ^3
...
Xv (0) 3
first iteration,
1
_ —a 2n XY
.
(0) n
(A2-lla)
,
m9
In the
_ —a— xY
...
"
(0)
with x\ {
0)
n
033
Y (0)
a nn
•
a 22
(0)
7,
•
initial estimates.
— —G —
(0)
033
Y n(l) x
(0)
2
•
-f-
033
by finding the difference
\
i
=
1, 2,
.
.
.
(A2-12a)
,n,
where e is a small number. If this condition is satisfied, we accept x} l) as the approximate solution; otherwise, we proceed to the next iteration. In general at the rath iteration we have xv (m) l
_ —
X2
—
x (m>
— ai\
an
= _q
3
\
fl33
x nm) (
414
=
an
y (m-l) 2
_
'
l
_
Cl\n
.
# 23 v (m•* 3
1)
ain v (m-l)
.
-
«32 **2 v (m-l) 033
Qnl -*2 v (mQnn
1)
_
Q/»3
a nn
v (m-l)
an
3
an
l
x (m-l)
Ojjym-i)
an
_
x (m-l)
Qn\ v (m-l) x
a nn
fll2
x (m-l) 3
_
.
,
a22
an
a 3n v (m-l) x" 7. #33
as3
—
•
an
(A2-llc)
•
a nn
j
:
Appendix 2
Solution of Simultaneous Equations
and the convergence condition \xi
m ~ l) x\
-
m)
|
415
is
<
€,
i
=
1, 2,
.
.
,
.
(A2-12b)
n.
The foregoing procedure can be slow to converge; that is, number of iterations before an acceptable solution is
a large
it
may
require
obtained.
The
improved by using the Gauss-Siedel procedure in which the solution for an unknown during an iteration is used in the comrate of convergence can be
unknowns
putation of subsequent
—
v (m)
•*2
x3
—
A21 Y (m)
—
A3 1 urn) X A33
flln
1)
_
fl/»2
xv (m) 2
a nn
Example A2-2.
first
=
_
A«3
iteration
|
*<>
=
^(0.6125)
€
=
|
Second
x
(i)
( o.575)
2)
_ A^^fm-l) an n
_|_
+
.
bi_
aii
_ £1^-1) +
^_,
(A2-13)
A33
**. fl/m
x
We
x 2°>
=
guess a solution set as
=
x 3°>
0.50,
=
+
1^0(0.5)
+
^=
+ i|
a575
|
(o) |
= =
0.50.
0.1125
0.04375
> > >
'
0.6125,
=
(A2-14a)
0.38.25.
0.005, then the convergence check
is
0.005
0.005>
not acceptable.
(A2-15a)
0.005J
iteration:
^.=^(0.6125)
4
.
+T55
100
_ _
.
we have
=
x a)
_
x (m) 3
a nn
x (i) -jc<°>| =0.075
|
.
b\
All
A22
0.50,
4o
we choose
.
..
xv (m) 2
*i"=Jo>- 5>
Tf
i
Solution by Gauss-Siedel Procedure
x<°>
for the
_
A2 3 ^(wi-l) 3 A22
Consider the equations in Eq. (A2-9).
Then
v (m-\)
All
A3 2 „(m) -*2 A33
j
a„ n
3
flu
A22
— flwl
y(m)
Al2 „(mx2 flu
—
=
x? =
+1±
±^(0.6875)
+
^(0.38125)
^(0.609375)
+
45s
=0.6875,
+
^= =
0.609375,
0-3796875.
(A2-14b)
416
Solution of Simultaneous Equations
Appendix 2
Convergence check. jc(2)
-
*< 2 >
= = x[ x^\ = x \n\ l)
\
> < <
0.1125 0.003 1 25
0.0015625
0.005^
not acceptable.
0.005 }
(A2-15b)
0.005,
Third iteration.
=
•(3)
100 77^(0.609375) 100
+H^ 100
100 0.684375) 200
+
^(0.3796875)
^(0.6070312)
+
^
x?
=
V
7.5
15
200
=
0.684375,
=
0.6070312,
=
0.3785156.
(A2-14c)
Convergence check: \
X U) - x (2)| =
0.003125
xy = _x = |
0.002344
|
0.001172
.12)
|
Hence the
x (3)
j2)
final solution at the
< 0.005] < 0.005 > < 0.005J
end of three Xl
x2 x2
= = =
(A2-15c)
acceptable.
iterations
is
0.6844,
0.6070, 0.3785.
We can improve on this solution by setting a more severe condition on convergence, say € = 0.001. Then additional iterations are needed for an acceptable solution. For
large systems of equations generated in the finite element applica-
becomes necessary to use improved iterative schemes. One such improved scheme is successive overrelaxation (SOR) [2]. To show the motivation of this method, we first write the Jacobi and Gauss-Siedel schemes
tions,
it
in matrix notation as follows [2]
:
Jacobi:
x (m
>
=(L+
U)x lm
~
l
+
>
(A2-16a)
fi.
Gauss-Siedel.
Lx (m)
L and U are
the lower
+
Ux (m
~
l
>
+
and upper triangular parts of the matrix For instance,
the constant term at a given stage of iteration.
A2-2, at
m=
1,
100
Too ioo 200
(A2-16b)
fi.
u
[A] in
and fi is Example
Appendix 2
Solution of Simultaneous Equations
100 100
417
1
100 200
U 7.5
fi-m
15 15
We
where
can
now
x(m>
= X (m-D
write _j_
^[(^(m)
+
ux (m
~
+
l)
P)
-
x {m
~
X)
(A2-16c)
l
the overrelaxation factor and the second term
co is
on the right-hand Rearrangement of terms in Eq. (A2-16c)
side denotes the correction term.
gives
x (m) which
=(/-
is
coL)-
x
[{\
a statement of
For
significantly.
-
(o)I
SOR
+
coU]x (m
SOR
[2].
details the reader
may
~
l)
+
co(I
- coL)-
1
/},
(A2-17)
accelerates the solution procedure
consult Refs.
[1, 2].
COMMENTS In the foregoing, we have presented only a very elementary introduction to some of the solution procedures. There are also available a number of other schemes and subschemes. Moreover, there are a number of aspects related to the numerical characteristics of the set of equations that can influence the
accuracy and
reliability
of a procedure. For instance, the magnitudes of the
diagonal elements ati which become "pivots" in the elimination procedure
can influence the computational characteristics. The initial guess or estimate of the unknowns and the value of o become important in the iterative procedures.
of the available procedures and the beyond the scope of this book. The reader
Detailed descriptions
characteristics of the equations are
can refer to many publications [1-4] for further study.
REFERENCES [1]
[2]
L.,
New
York, 1965.
2,
Numerical Linear Algebra, Oxford University Press,
T., A Survey of Numerical Mathematics, Addison-Wesley, Reading, Mass., 1972.
Bathe, K.
J.,
Prentice-Hall, [4]
to
Young, D. M., and Gregory, R. Vol.
[3]
An Introduction
Fox,
and Wilson, E. L., Numerical Methods in Englewood Cliffs, N.J., 1976.
Desai, C. S., and Abel, J. F., Introduction Nostrand Reinhold, New York, 1972.
Finite
to the Finite
Element Analysis,
Element Method, Van
APPENDIX
PHYSICAL MODELS
INTRODUCTION may
often be instructive and useful to construct physical models to illussome of the principles of the finite element theory. There can be a number of possibilities for constructing such models. Here we consider details of some of the simple models that the author has tried.
It
trate
MODELS FOR CONVERGENCE As
stated in Chapter
plastic.
As shown
1,
one can construct models made of cardboard or can illustrate the idea of convergence as
in Fig. 1-8, they
polygons of an increasing number of sides are drawn inside or outside of a circle.
MODELS FOR BEAM BENDING One of
the
cretization
(Chapter 7)
main ideas here can be the illustration of the principle of disand interelement compatibility with respect to the order of
approximation functions.
The models can be made of a described subsequently were
RTV
Products Department, General Electric, Watford, 418
The models Rubber (Silicone
suitable flexible material.
made from
Silicon
New
York). This material
Physical Models
Appendix 3
419
can be cast into molds and is available in different colors. Its properties are suitable for illustrating deformations with small loads and hand pressure. Figure A3-1 shows a plastic mold used to cast the models. Figure A3-2(a) shows a model of a continuous beam. The model in Fig. A3-2(b) was cast
Figure A3-1
Mold made
Figure A3-2 Physical model for
beam.
beam
of plastic.
bending, (a) Continuous
Physical Models
Mesh with "rigid" elements: linear apMesh with deformable elements: higher-order
Figure A3-2 (Contd.) (b)
proximation,
(c)
approximation.
A flexible string was passed through holes in each The holes are provided by (copper) pipe, which makes each element considerably stiff. Under a load, the composite will deform such that gaps into four pieces (elements).
element.
may
develop at the junctions, thus violating interelement compatibility.
With
sufficient tightness in the string, the
model may not show gaps and can
indicate displacement compatibility at the junctions, but slopes at the junc-
Appendix 3
Physical Models
421
may be considered to illustrate a linear approximation function for the transverse displacement w. Figure A3-2(c) shows a model made in four elements. The elements are
tions can be seen to be different. This
connected at the junctions by using hooks at the top and at the bottom of the beam. This arrangement permits deformation of each element and improved connection at the junction as compared to the arrangement in Fig. A3-2(b).
Under
load,
the
model deforms without gaps and
indicates
improved
interelement compatibility. This can be considered to illustrate higher-order
approximation for w.
MODEL FOR PLATE BENDING (Chapter 14) Figure A3-3 shows a model for plate bending sheet.
The four elements
[Fig. A3-3(a)].
made out of
(thin) plastic
are riveted (or glued) at the edges of the
A screw arrangement [Fig.
A3-3(b)]
is
model
provided at the middle
of the base plate such that the top of the screw touches the bottom (at the center) of the four elements.
Figure A3-3 Physical model for plate bending, (a) Four-element mesh, (b) Arrangement for applying "load."
Physical Models
422
By moving sively)
the screw
up and down, the
Appendix 3
plate assembly can be (progres-
deformed. The interelement compatibility in terms of displacement
and d 2 w/dxdy) can be qualitatively illustrated by observing the movements at the junctions of the plates. The effect can be accentuated by attaching colored tapes on either side of a
(w) and
its
three derivatives (dw/dx, dw/dy,
junction.
COMMENTS The foregoing
are examples only of simple models. Other similar
and
sophisticated models can be constructed to facilitate introduction of various
concepts in
finite
element analysis.
APPENDIX
COMPUTER CODES
INTRODUCTION Descriptions of a
number of computer codes
Most of the codes
described herein have been prepared such that the beginner
are included in this appendix.
can understand and use them with relative ease. They are relevant to many topics covered in this book, and hence can be used by the teacher and the student for solving specific problems. Brief statements of other codes relevant to a given topic are also given
problems for
;
these codes
class illustrations or for
may be used
for solving
advanced
term projects.
The codes described here can be made available to the reader and the and teaching purposes at the cost of reproduction, mailing and other required costs. For details such as descriptions of teacher essentially for personal
background theory,
guide,
user's
release of codes, the reader
may
sample problems, conditions for the
contact the author. In addition to the codes
described here, other advanced codes can also be
CHAPTER
made
available to the user.
5
CONS-1DFE: One-Dimensional
Consolidation
with Nonlinear Properties
This
is
dependent
a modification of settlement
DFT/C-1DFE
analysis
described in Chapter 6 for time
of foundations
with
nonlinear
material
properties.
423
Other Codes on Consolidation
Codes for time dependent settlement analysis of foundations
idealized as
plane strain and axisymmetric consolidation of problems such as layered
foundations and earth banks are available. Both linear and nonlinear (plasticity)
models are included for material behavior. Capabilities for time
dependent loading and simulation of construction sequences such as embank-
ment and excavation
CHAPTER
also exist.
7
BMCOL-1DFE:
Analysis of Axially and Laterally Loaded
Beam-Columns such as
Piles
and Retaining Walls;
Linear and Nonlinear Analysis
This code can solve problems such as beams, beam-columns (Chapter
beams on deformable foundations,
axially
and
laterally
retaining walls idealized as one-dimensional. Linear strain behavior for
deformable supports such as
loaded
piles,
and nonlinear
soil
7),
and
stress-
foundations can be
included.
The output
in the
is
form of nodal displacements and
rotations,
and
bending moments.
CHAPTER
8
MAST-1DFE: This code
is
One-Dimensional Diffusion-Convection
based on the formulation described
in
Chapter
8.
Other Codes on Diffusion-Convection
This
is
a two-dimensional code for solution of situations such as
salt
water intrusion and other concentration problems.
CHAPTER
10
WAVE-1DFE:
One-Dimensional
Wave
Propagation
This code is based on the formulation described in Chapter 10. Wave propagation for problems idealized as one-dimensional such as in bars and soil
424
media can be
solved.
CHAPTERS
AND
11
FIELD-2DFE: Field Problems
12
Analysis of Two-Dimensional Steady State :
Torsion, Potential Flow, Seepage, Heat Flow
This is a common code for solution of a number of steady state field problems governed by similar differential equations (Chapters 11 and 12).
The following
table gives the details of each
problem and
its
corresponding
option code.
Option
Code
NTYPE
Problem Torsion
Relevant Problems
1
(Chapter 11) Potential
Output Quantities
Torsion of bars of regular or
Nodal
irregular cross sections
stresses, twisting
Flow through
Nodal
stress functions, shear
moment
Flow
Velocity
2
pipes and open
channels
Potential
Stream Function Seepage (flow through porous media) Heat Flow
Flow through
3
4
velocity potentials,
velocities, quantity of flow
pipes and open
Nodal stream
functions,
channels
quantity of flow
Steady state confined seepage through pipes, foundations
Nodal heads,
of
dams and
sheet piles,
velocities,*
quantity of flow
and
earth banks and wells
Steady state heat flow in bars,
5
plates, slabs,
and other plane
Nodal temperatures, quantity of heat
bodies
The formulation
is
based on four-node isoparametric quadrilateral
element with linear constitutive laws; for instance, for seepage, Darcy's law
assumed to be
is
valid.
Other Codes on Flow Problems
Codes
for two-dimensional transient free (phreatic) surface seepage
through plane bodies such as earth banks, dams, and wells. Code for three-dimensional steady and transient free surface flow through arbitrary three-dimensional porous bodies such as earth banks, dams, and junctions between various structures, e.g., abutment
and dam.
Code plates
for two-dimensional time dependent heat flow in bodies such as
and
bars.
425
.
CHAPTER
13
PLANE-2DFE: Two-Dimensional
Plane Stress-Deformation:
Linear Analysis
This code
is
capable of analysis for problems idealized as plane
plane strain, and axisymmetric (Chapter
13). It
can handle linear
stress,
elastic analy-
of engineering problems such as (approximate) bending analysis of beams and inplane or membrane analysis of flat plates or slabs, shear walls, earth and concrete dams and slopes, foundations, underground pipes and tunnels, and sis
retaining walls.
Surface and nodal point loads as well as self weight load can be applied.
A
version of the code with graphic option which can plot zones of equal
and displacement intensities is also available. The finite element formulation is based on a four-node isoparametric quadrilateral element (see Chapter 13) and on linear elastic stress-strain law. The output is in the form of nodal displacements and element stresses. stress
Other Codes for Plane Bodies
1
A
code for two-dimensional nonlinear analysis of problems idealized and axisymmetric is available. The stress-strain behavior can be nonlinear elastic, elastic-plastic (critical state model), with linear elastic behavior as a special case. The code has an additional feature of handling soil-structure interaction problems this is done by providing a special nonlinear interface or joint element between the structure and soil. The formulation is based on a four-node isoparametric quadrilateral element (see Chapter 13). Some examples of engineering problems are: retaining structures, axially loaded structures such as pile foundations, cylindrical tanks, dams and slopes, and underground as plane strain
;
structures.
In the case of pile foundations, the code also has the facility of outputting design quantities such as bearing capacity, wall shear
and point loads,
friction
in addition to
nodal displacements and element
stresses. 2.
This
is
a code for three-dimensional analysis with linear and nonlinear and plasticity) constitutive models. It has a provision for
(elasticity
interface elements.
426
CHAPTER STFN-FE:
14 Analysis of
Frame
Structures and Foundations:
Linear Analysis
In a general building frame type of problem the beams and columns are approximated as one-dimensional beam-column elements, the slabs as plates subjected to inplane and bending (Chapter 14), and the foundation is replaced by equivalent spring elements. The spring elements can also be used to simulate supports provided by members such as adjoining structures. Surface and point loads can be applied in all the three coordinate directions. is in the form of nodal displacements and and bending moment and shear forces.
The output stresses,
A
rotations, element
modified version of this code includes nonlinear characteristics for
foundation media.
427
INDEX
Advantages of finite element method, 85 Approximation functions, 39, 43, 44, 95 176 (see also Approximation models) complete, 44 conformable, 44 requirements for selection, 43
Approximation models, 19, 43, 109, 173 338 (see also Approximation functions) 338 beam bending, 173 requirements, 173, 174 for time-dependent problems, 109 generalized coordinate, 39, 175 linear, 39-43, 95, 109 quadratic, 67 requirements for, 43 beam bending, 176 field problems, 95,304 column, 43 plane deformaions, 43, 339 bilinear,
Assemblage property matrix, 31 Assemblage vector of nodal forcing parameters, 31
Assembly, 55, 98, 116, 181, 182, 184 direct stiffness method, 58 of element equations, 55 (potential) energy approach, 55 Axisymmetric idealization, 335, 370 for footing, 370
for
torsion, 242
Area coordinate (see Local coordinates) Assemblage equations, 30-33 building-foundation, 381 column, 58-62 field problems, 312 flow, 98 mass transport, 205 modified, 33,60, 61 overland flow, 216 plane stress, 349, 353 stiffness matrix, 58 torsion, 248, 258, 264, 279, 288 wave propagation, 227 Assemblage nodal vector, of unknowns, 31
Banded matrices, 412 Band width methods,
61-63, 396 Basis functions (see Interpolation functions) Beam bending, 15, 172, 173, 364, 369, 398,
400,418 approximation models, 173 physical models for, 418-421 Beam-column, 172, 173, 189, 373
Boundary conditions,
27, 28, 31, 32, 59, 60-62, 99, 108, 117, 123, 207, 217, 231, 249-
279,301,317,325,395 categories of, 27
concept, 59 Dirichlet, 60 essential, 27, 31
explanation, 59-62 first,
60
forced, 27, 31, 60 geometric, 27, 31, 60 heat flow, 117 homogeneous, 27, 60 in consolidation, 123 in Galerkin's method, 74, 75, 180 in mass transport, 207, 208 in overland flow, 217-219
429
430
Index
Boundary conditions
(cont.):
in potential flow, 301, 302,
317
in seepage, 325
stream function approach, 317 in temperature problem, 1 17 in
258,264,279 wave propagation, 230-232
in torsion, 249, in
mixed, 60 modification for, 32, 33, 60-62 natural, 31,60
Neumann, 60 nonzero, 61 types of, 60 zero-valued, 60
Bounds,
7,
11,83-85,261,320,321
lower, 11,84 symbolic representation, 84 upper, 1 1 84 Building-foundation systems, 372, 392 spring supports, 372, 388 ,
Calculus of variations, 24 Collocation, method of, 26, 27, 47, 398-408 Comparisons: numerical predictions, and closed form solutions, 65-67, 183-185, 252, 261, 267, 282, 283, 316-325 in column problem, 65-67 in beam bending, 183-185 in potential flow, 316-321 in seepage, 325 Compatibility, 7, 30, 43 interelement, 30, 43 for approximation functions, 43 Complementary energy, 24, 47, 76-79, 192,
271,295,296 for beam bending, 192 for one-dimensional problem, 76-79 in torsion, 269 modified expression for hybrid approach,
271 principle of, 47 stationary, 24
Completeness of approximation functions, 44,45,243 Pascal's triangle, 243 applications:
Computer codes,
Constant-strain-triangle (CST), 259 Constant-stress-triangle, {see Constant strain triangle)
Constitutive laws, 23, 95, 110, 395 {see also Stress-strain laws) Constraints {see Boundary conditions) Continuity, 4, 8, 30 conditions, 30 for approximation functions, 43
Convection, 202 Convergence, 7,
8, 10, 44, 87, 131, 186,
354,
364,418
comments on, 354 in beam bending, 364 in consolidation, 131 in heat flow, 131
mesh refinement, 87 monotonic, 44 one-dimensional column, 87, 88 physical models, 419 Coordinate systems: element, 36 global, 36 local,
36
Coupled problems,
103, 121, 396
Damping, 234 Darcy'slaw, 15,24,45,95 Degree of consolidation, 124, 125 Degree of freedom, definition of, 21 Derived quantities, {see Secondary quantities) Diffusion-convection, 202-210 Direct methods, for solution of equations, 410,413 Direct stiffness method, 58 Discretization,
1, 5, 13, 18,
22, 36, 173, 203,
213,225,246 Discretizations of "infinite" boundaries, 329 Displacement approach, 39, 240, 281, 282,
293,295,338,375-381 for torsion, 240 Displacement vector:
assemblage, 58 element, 54 nodal, 54 Dynamic problems, 224-236
beam-column, 424 building-foundation, 382, 427 field problems, 266, 328, 330, 425 flow problems, 102, 124,425 heat flow, 158-164
mass transport, 424 plane deformations, 354-368, 426 stress-deformation, 76 torsion, 266, 293-296 wave propagation, 424 philosophy of, 134 stages in, 134 Condensation {see Static condensation)
Conformable functions, 44 Consistent mass matrix {see Mass matrix) Consolidation, 108, 113, 121-132 one-dimensional, 108, 121-130 layered media, 112, 113, 129 Constant-strain-line element, 46
Electromagnetic problems, 327, 332 Element coordinate system, 36, {see also Local coordinate system) Element equations, 24, 29, 47, 53-55, 96-98, 110-113, 178, 227, 245, 256, 275, 286,
308,314,342-345,380,391 assembly of, 55
beam bending,
178, 179
beam-column, 192, 375 column, 47, 54 consolidation, 121-123 flow, 96-98 heat flow, 110-113 mass transport, 204, 205 methods of formulation, 24 overland flow, 215 permeability matrix, 97 plate,
380
Index
431
Element equations (com.):
Foundation, representation
torsion, 241, 245, 256, 271, 284-290, 308,
314 two-dimensional stress-deformation, 341 wave propagation, 227 Element property matrix, 29 Energy procedures, 24 Equations of equilibrium, 50 Errors, 13,22,28, 131,254,281 in method of weighted residuals, 28 in numerical solutions: torsion, 254, 261,265-267,281 Essential
boundary conditions,
31
(see also
Forced boundary conditions) External loads, 47-51 potential of, 47, 48 work done by, 48
Field problems, 299, 330 Finite difference method, 113, 400 Euler-type integration, 113
Crank-Nicholson procedure, 113 Finite elements: definition, 18, 19, 36 nodal line, 19 nodal point, 19
one-dimensional, 19 two-dimensional, 19 three-dimensional, 19
Element Formulation, 17-34
beam bending,
172
beam-column, 189, 374 column, 50, 6", "6, "9 diffusion-convection, 202 electromagnetic, 327 flow: consolidation, 121 fluid, steady, 95, 101, 302, 314, 325 heat, steady, 321 heat, transient, 109 overland flow, 213
plane
(stress, strain,
plate,
377
axisymmetric), 338
seepage, 325 spring, 48 torsion, 240, 254, 269, 284
wave propagation, 225 Element Method:
Finite
advantages
of, 85
basic concept,
spring supports, 372, 381
Galerkin's method, 26-28, 47, 67, "1, 83, 101. 114,
179,203,213,399-408
explanation, 68 for beam bending, 1"9 forces, representation in. "1 for diffusion-convection, 203, 209 for flow problem, 101, 102 for heat flow, 114 for one-dimensional heat flow, 1 14 for one-dimensional stress-deformation, 6" for overland flow, 215 interpolation functions, relevance to, 68 interpretation ox. "5 mixed approach, 80-83 Gaussian elimination, 63. 99. 119. 183. 19". 208, 218, 231, 249. 264. 279, 349, 410. 413 elimination, 410 back substitution. 411 Generalized coordinates, 39-41, 175 Geometric boundary conditions, 31 {see also Forced and Essential boundary conditions)
Geometric
invariance, for approximation models, 45 Global coordinate systems, 36-39
explanation, 36
types of, 19,20 Finite
of, 372, 381
1
finite element, definition of, 18, 19 interpretation of results, 33 steps in, 17, 18-33 nodal line, 19 nodal point, 19 Flow problems, 93, 299, 332, 396 similarity with deformation problem, 93
Flow, quantitvof, 315 Fluxes, 312-314, 328, 329 Formulation procedures, 47, 395 variational, 47 residual, 47 Forced boundary conditions, 31 {see also Geometric and Essential boundary conditions)
Forcing function, 28
for one-dimensional problem, 38 Global equations, (see Assemblage equations) Gradient, of fluid head. 23 Gradient-potential relation, 95 Gradient-temperature relation, 1 10 Gradient-unknown relation, 22
Heat flow, 108-120, 299, 321, 328, 300. 331 finite element formulation, 108-114. 321 one-dimensional, 108 two-dimensional, 322 uncoupled analysis, 103-10". 337 change in temperature, 103, 337 Hermite interpolation, 175, 377 Higher order approximation, 6", 185-199, 296, 330, 395 (see also Higher order elements) beam bending, 185-189 line element. 6" quadratic, 6", 8"-89
quadrilateral, 330, 331 triangular element, 296. 29" Higher order elements, 6". 8". 368 (see also
Hieher order approximation) Hooke'slaw, 15,23,43,62 Hvbrid approach, 24. 4". 269, 281, 282, 293, 29", 298 in torsion,
269
"Infinite"' media, discretization of, 362. 370 conditions, 108, 11". 123. 20". 21".
Initial
230, 232 in consolidation, 123 in heat tlow, 108, 11"
mass transport, 207, 208 overland flow, 217-219 in wave propagation, 230-232 Initial load vector, 106, 130, 337, 368 in in
1
1
432
Index
Initial strain,
due to temperature, 103-107,
337
337 value problem,
Initial stress, 106,
Initial
1
14
Integration, 53 (see also Numerical integration)
Interpolation functions, 39, 41-43, 69, 174, 175, 239, 339, 377, 379
42 Hermitian, 175 in Galerkin's method, relevance, 69 in plate bending, 377 Interpretation of results, 33, 65, 66, 100-101 definition,
Interelement compatibility, 22, 30, 43, 59, 242, 360, 376 for approximation functions, 43 in assembly,
59
for different problems, 43 in plate problem, 376 in
two-dimensional problems, 242
Isoparametric elements, 19, 42, 302, 332, 339, 395 concept of, 19 explanation, simple, 42 quadrilateral element, 302, 330-339 eight node, 330, 331 four node, 302, 339 Isotropy, for approximation models, 45 Iterative methods, solution of equations, 410,
414 Gauss-Siedel, 415
Jacobi,414,416
SOR, 414, 416, 417 SSOR,410,417 Jacobian, 305 determinant, 41, 305, 306, 311, 360 Joint forces, explanation of, 51 global level, 51 local level, 5
Laplace equation,
15,
300
Least squares, method of, 26, 27, 47, 398-408 Load vector, 54-58, 72-76, 106, 179, 192, 204,
215,342-345,380,391 additional, 106
assemblage, 58 correction, 106 for beam bending, 178-179 for beam-column, 192, 376 for column, 54 for flow, 98, 312 for heat flow, 110-113,321 for mass transport, 204, 205 for overland flow, 215 for plate, 380 for torsion, 246, 257, 276, 284-290, 308,
312-314 two-dimensional stress-deformation, 341-349 for wave propagation, 227 in Galerkin's method, 72-76 initial, 106 nodal, 54 residual, 106 Local and global coordinates, 36-43 relation between, 42
Local coordinate systems, 36-39, 85, 174 alternatives for, 38 explanation, 36 for triangular element, 239 in beam bending, 174 in one-dimensional problems, 38 in two-dimensional problems, 238, 239 Local coordinates, {see Local coordinate systems)
Mass matrix, 227,
230, 234-236
consistent, 227
lumped, 227, 234 Mass transport, one-dimensional, by diffusion-convection, 202 saltwater intrusion, 210
Mesh refinement, 67, 185-187, in beam bending, 185-189 in
202, 209, 210
261
one-dimensional problems, 67, 87
in torsion, 261
Methods of formulation, 24 energy procedures, 24 residual methods, 24 Method of weighted residuals, 26, 27, 80, 398-408 (see also Weighted residuals,
methods
of)
Minimization, of potential energy, 48 example, 48-5 manual, 48-50 mathematical, 50, 51 Mixed approach, 47, 79, 194, 284-298 Hellinger-Reissner principle, 79 in beam bending, 194 in torsion, 284 residual method, 80 Galerkin, 80 variational method, 79 Mixed procedure (see Mixed approach) Multicomponent systems, 372 Natural boundary conditions, 31
Nodal Nodal
line, definition,
19
point, definition, 19, 36 Nonlinear analysis, 106, 396 Numerical integration, 308-312, 332, 342, 349 for quadrilateral element, 309-312
Operator: differential, 27 stiffness matrix, 55
Order of approximation: approximation) Overland flow, 211 approximation for, 213
(see
Higher order
finite element formulation, 213 kinematic wave approximation, 212 surface runoff, 223
for
Pascal's triangle, 45, 243
Patch
test,
45
Permeability, co-efficient of, 15 Physical models, 11, 177,418-420 for beam bending, 177, 419, 420 for plate bending, 421 , 422 Plane deformation, 30, 333 (see also
dimensional problems)
Two-
Index
433
Plane strain idealization, 335, 357 for dam and foundation, 357 for footing, 370 for tunnel, 370 Plane stress idealization, 333, 340 for beam, 334, 364 for plate, 334 for shear wall, 354 Plates:
bending, 371, 376, 377, 382, 393, 421 physical models, 421 plane stress idealization, 334 membrane effects, 334, 376 Polynomials: expansion, 44 fifth order, 187
approximation models, 39
quadratic, 67 Potential energy, 15-26, 47-53, 55, 268 approach for assembly, 55 definition, 25 minimum of, 25 mimimum principle of, 22, 47 minimization of, 47, 48
manual, 48 mathematical, 50 of elastic bodies in equilibrium, 26 in torsion, 268 potential of external loads in, 25, 47-51 stationary value of, 24, 47 strain energy in, 25 Potential flow, 299, 300, 328 around cylinder, 316 stream function, 314 velocity potential, 302 Potential, of external loads, 25,47-51
Primary unknowns, 33, 62, 98-100, 118-120, 183
nodal displacements, 62 nodal temperatures, 118-120 solution for, 33, 62 Primary quantities {see Primary unknowns) Principal strains, 15 Principal stresses, 15, 349
Quadratic approximation, 67, 87-89 Quadrilateral element, 238, 302-314, 338, 346
Requirements, for approximation functions,
43-45,110,176,242,339 compatibility, 44
completeness, 44 continuity, 43 for beam bending, 176 for field problems, 95,304 for time-dependent problems, 110 for torsion, 242 for two-dimensional
stress-deformation,
339 Residual, 7, 27, 399, 401 in
method of weighted
minimum
Initial
load
vector)
Residual methods, {see
Method of weighted
residuals)
Residual stress, 106 {see also
Initial stress)
Ritz method, 404
Secondary quantities, 33, 64, 100, 120, 183, 249, 258, 315 {see also Secondary unknowns) in column problem, 64 in
beam bending,
183
moments, 183 shear forces, 185
m heat flow problem,
cubic, 175
for
Residual load vector, 106 {see also
residuals, 27
of, 7
minimization of, 27 schematic representation of, 28
120 quantity of flow, 120 quantity of flow, 100 velocities, 100
Secondary unknowns,
{see
Secondary quanti-
ties)
Seepage, 323-332, {see also Fluid flow) categories, 323, 324, 327 confined, 324 through formulation, 325 Shape functions {see Interpolation functions) Simultaneous equations, 33, 62, 410-417 Gaussian elimination, 410 linear, 62 nonlinear, 62 methods of solution: direct, 410, 413 iterative, 410, 414 Solution in time, 113, 1 18-120, 206, 217, 228, {see also
Time
integration)
approach, 113, 206 Solution of equations, {see Simultaneous finite difference
equations) Spring supports, 372, 388 Static condensation, 291 Stationary value, 7, 24, 47 Steady-state flow, 15 Steps in the finite element method, 17, 18-33 Stiffness matrix, 54-61, 178, 179 banded, 61 assemblage, 58 element, 54 as operator, 55 sparsley populated, 61
symmetric, 61, 412 Strain-displacement relation, 22, 23, 45, 177 Strain displacement transformation matrix, 46 Strain energy, 25, 47-51 internal, 25 Stream function formulation, for potential flow, 314 Stress approach, 254 in torsion, 254 Stress concentration, 368 Stress-deformation, 35, 45, 71, 333, 372 finite element formulation, 338 one-dimensional, 35-91 governing equation, 71 two-dimensional, 333 plane stress, 333 plane strain, 335 axisymmetric, 335
434
Index
Stress-strain matrix, 46 Stress-strain relation, 22, 23, 45, 46, 177, 395 {see also Constitutive laws)
Stress-wave propagation, 224 one-dimensional, 224 finite element formulation, 225
Subdomain method,
Transformation matrix, 46, 96, 177, 390 Transformation, of coordinates, 389, 395 Two-dimensional problems: torsion, 237 field problems, 299 stress-deformation, 333
26, 27, 399, 403
Uncoupled problems, 103-108, 337 Temperature problem, 103-127 Thermal conductivity, coefficient of Time-dependent problems: approximation model, 109
15, 108
coupled, 103 one-dimensional, 103 heat flow, 108 consolidation, 108 solution in time, 113 uncoupled, 103
Thermal flow, 299
Time
{see also Heat flow) integration, 113, 206, 217, 228, 229 {see also Solution in time)
Wavefront methods, 396 Weighting functions, 28, 399-409 in method of weighted residuals, 28 Weighted residuals, methods of, 26,
Prandtl, 255 Saint- Venant, 237, 269
approach, 254 two-dimensional approximation, 237, 238 Trade-off, accuracy vs computer time, 368 stress
residuals,
finite
80, 399-409 collocation, 26, 399-408
Galerkin, 26, 399-408 illustration of, 398-408 least squares, 26,
399-408
subdomain, 26, 399-408
26,27 Triangular
Variational methods, 24, 30, 80-82, 96-98 Variation notation, 48 Variation of properties, 43, 131 within element, 43 linear, 43, 86 Vector of nodal unknowns, 29 of nodal displacements, 29 Virtual work, principle of, 226
225 hyperbolic equation, 225
mixed approach, 284
method of weighted
initial stress,
Wave equation,
Torsion: displacement approach, 240 finite element formulation, 240 hybrid approach, 269
Trial solution, in
load vector, 106 106 residual stress, 106 initial
,
element, 238-240, 325-350
trial
solution in, 26, 398
27, 29,
HH
ELEMENTARY FINITE ELEMENT HI Chandrakant
The
basic
Desai
S.
aim of
this text
is
to provide a simple elementary treatment of
the finite element method that can be understood without advanced prerequisites required
by other available
texts,
and at the same time, present
a wide range of problems that can lead toward exploitation of the method for solution of
This
is
complex problems
in
engineering and mathematical physics.
the only book available that presents both the theory and the
potential for applications of the finite element
method
most elemenmethod and
at the
tary level possible, and yet, brings out the generality of the
covers a wide range of topics in engineering. Because of the simple and general approach, the text is ideally suited to the undergraduate and the beginner. The comprehensive treatment of a number of topics will also
be useful to the graduate student and the advanced reader. To keep the details at as simple a level as possible, considerable attention is given to
problems idealized as one-dimensional. This approach facilitates an easier understanding and renders itself to hand calculations of a large number of example and practice problems. The one-dimensional topics include: one dimensional stress deformation, one-dimensional steady state flow, onedimensional time dependent flow (heat flow and consolidation), beam
bending and beam-column, one-dimensional mass transport, one-dimensional wave propagation and one-dimensional overland flow. To illustrate the generality of the method, details and applications of a computer code are included.
To understand the strengths and
potential of the method, the reader is then leadlo two-dimensional problems. Here the approach has been to advance gradually from simple two-dimensional to more difficult two-dimensional problems
chapters have been obtained by using different and residual procedures. In the case of the former,
in various
pies ie
main
aiieii
n
is
given to the principle of stationary potential energy, is primarily used for the residual schemes.
hod
>yer illustrative applications of various residual schemes
ocedures, solutions and simultaneous equations,
nd details of relevant computer codes.
PBEtfflOE HAfii; IN £
0-13-256636-2