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Lecture Notes in Engineering Edited by C. A. Brebbia and S. A. Orszag
35
WTang
Transforming Domain into Boundary Integrals in BEM A Generalized Approach
Spri nger-Verlag Berlin Heidelberg New York London Paris Tokyo
Series Editors C. A. Brebbia . S. A. Orszag
Consulting Editors J. Argyris . K.-J. Bathe' A. S. Cakmak . J. Connor' R. McCrory C. S. Desai' K.-P. Holz . F. A. Leckie' G. Pinder' A. R. S. Pont J. H. Seinfeld . P. Silvester' P. Spanos· W. Wunderlich' S. Yip
Author Prof. Weifeng Tang East China University of Chemical Technology 130 Mei-Iong Road Shanghai 200237 PRChina
ISBN-13:978-3-540-19217-6 e-ISBN-13:978-3-642-83465-3
001: 10.1007/978-3-642-83465-3
This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microfilms or in other ways, and storage in data banks. Duplication of this publication or parts thereof is only permitted under the provisions of the German Copyright Law of September 9, 1965, in its version of June 24, 1985, and a copyright fee must always be paid. Violations fall under the prosecution act of the German Copyright Law.
© Springer-Verlag Berlin, Heidelberg 1988
The use of registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use.
216113020-543210
ABSTRACT
In this work a new and general approach to transform
domain integral terms into boundary integral terms in BEM
formulations is presented. The technique can be used
for both potential and elasticity problems. The method
is based on expanding the integrands in domain integrals
into Fourier series, which ensures convergence of results.
The corresponding Fourier coefficients can be calculated
analytically or numerically. The mathematical implementa
tion and corresponding programming are described in this
thesis. Numerical applications using the present approach
validate and illustrate how the method can be used in
engineering practice, including the application for
elasto-plastic analysis. The present approach is a
general and reliable transformation technique.
CONTENTS
CHAPTER 1 GENERAL INTRODUCTION
1-1 Numerical Methods
1-2 Domain Methods 2
1-3 Boundary Element Method 3
1-4 The Main Procedures and Features of BEM 5
1-5 The Subject of this Work 7
1-6 Contents of the Present Work 9
1-7 The Cartesian Tensor Notation 11
CHAPTER 2 POTENTIAL PROBLEMS
2-1 Introduction 12
2-2 The Boundary Integral Formulation 13 for Potential Problems
2-3 The Boundary Element Method for 22 Potential Problems
2-4 Motivation and General Ideas 27
2-5 Fourier Analysis 30
2-6 Basic Formulations for Transforming 35 the Domain Integrals into the Boundary for 2-D Problems
2-7 Numerical Approaches 42
2-8 Numerical Accuracy of the Transformation 46 Formula
2-9 Some Further Discussions 68
2-10 Examples 73
2-11 The Transformation Formula for 3-D 90 Poisson's Equation
2-12 Applications in Time-dependent Problems 95
2-13 Application in Non-linear Problems 99
CHAPTER
V
Page
3 LINEAR ELASTOSTATICS
3-1 Introduction 101
3-2 Basic Relationships of Elasticity 102
3-3 Fundamental Solution for Elastostatics 105
3-4 Somigliana Identity 108
3-5 The Boundary Integral Equations of 112 Elastostatics
3-6 The Boundary Element Method in Elasticity 116
3-7 Basic Formulations for Transforming 2-D 117 Elasticity Domain Integrals to the Boundary
3-8 Numerical Implementation 132
3-9 Results of Numerical Experiments 140
CHAPTER 4 APPLICATIONS IN ELASTICITY AND ELASTOPLASTICITY
4-1 Introduction 158
4-2 An Example of Gravitational Loading 160
4-3 An Example with a More General Type 166 of Distributed Loading
4-4 Relationship between Plastic Stresses 169 and Plastic Strains
4-5 The Governing Equations for Elasto- 177 Plasticity
4-6 Numerical Analysis using Finite 181 Fourier Series
4-7 Application to Elasto-plastic Problems 185
CHAPTER 5 PROGRAMMING
5-1 Potential Problems
5-2 Elasticity Problems
5-3 Elasto-Plasticity Problems
CHAPTER 6 GENERAL DISCUSSION AND CONCLUSIONS
REFERENCES
194
196
198
201
204
CHAPTER 1 GENERAL INTRODUCTION
1-1 NUMERICAL METHODS
For the last two or three decades, scientists and
engineers have used numerical methods as an important
tool in many different areas. This significant fact
has its inexorable historical trend and it is the
inevitable outcome of the recent developments in science,
technology and industry.
Analytical methods have been developed for a long
period and have produced a great amount of successful
results, but they failed to solve most practical engineering
problems with complicated boundary conditions or irregular
geometry. It is also very difficult to solve non-linear
or time-dependent problems using analytical approaches,
even if they are very simple. On the other hand, research
on analytical methods has provided a solid foundation for
different types of numerical methods.
Because of the rapid developments of science and
technology it is now necessary to solve complicated
problems using more efficient and accurate approaches
than before. Not only problems with complicated boundary
conditions or irregular configurations require solutions
but also non-linear or time-dependent problems must be
solved.
Computer hardware and software have developed at an
unexpected high speed. During the last thirty years, ithaz
become possible for scientists and engineers to use
numerical methods with computers easily. This has
2
stimulated scientists and engineers to improve some
classical numerical methods (such as finite difference
method) and to establish new numerical methods (such as
the finite element method and boundary element method).
For all these reasons, numerical methods have
rapidly developed in the areas of mechanics and engineering.
Furthermore, a new discipline, computational mechanics
has already emerged and become a very important and
active branch of mechanics.
The development of computational mechanics provided
excellent tools for different engineering areas. The
numerical solutions can now be obtained accurately and
efficiently even for very complicated boundary conditions,
non-linear problems and time-dependent problems. Moreover
engineers can nowadays use many kinds of software packages,
which are easily available in the software-markets and
can solve problems in their own specialities.
1-2 DOMAIN METHODS
Currently, there are some important domain methods
in the area of computational mechanics, such as finite
difference method (FDM), finite element method (FEM)
and weighted residual method (WRM). All these methods
have already proved their efficiency to solve differential
equations corresponding to complex engineering problems.
For instance, during the last two or three decades, the FEM
has been developed to a high degree of sophistication
both in terms of its fundamental theory and of software
packages.
3
In contrast with analytical methods, numerical methods
give no one formula type expression of the results, but
some discretized and approximate results. The domain
of the problem is discretized into some subdomains,
using grids (in FDM) or elements (in FEM), then the
results are calculated at certain points, which are chosen
in advance, such as intersections of subdomains or Gauss
integration points. The governing differential or
integral equation is reduced into a finite set of
linear algebraic equations, which is suitable for com
puting. In FDM, for example, every order of difference
of functions is required instead of the derivatives.
Based on some variational principles or weighted residuals
in FEM or WRM, for instance, a linear combination of trial
functions replaces the unknown function. FEM has more
flexibility than other methods because the approximate
expression of the trial functions is only valid in one
of the subdomains. Moreover, very complicated geometrical
domains with arbitrary boundary and initial conditions,
as well as non-linear and time-dependent problems can be
treated by means of iterative and incremental procedures.
1-3 BOUNDARY ELEMENT METHOD
Since the publication of the first book called
"Boundary Elements" in 1978 [ 1), BEM has been developed
rapidly and many new applications in engineering have now
been produced. The technique is an important alternative
method in the area of computational mechanics.
4
BEM is based on well established theoretical
foundations, such as boundary integral equations,
fundamental solutions of partial differential equations
and weighted residual methods. These are combined with
numerical techniques, such as discrete method, numerical
integrations, increment methods, iterative technique and
others.
Early in this century, Fredholm established the
theory of integral equations, then the integral equations
and the boundary integral methods were applied in the
area of mechanics. Some Russian scientists made significant
contributions, particularly Mikhlin, Muskhelishvili and
Kupradze [2,3,41. At that time, integral equations were
considered to be a different and powerful type of
analytical method. Hence, because of the difficulty to
obtain the solutions analytically, it was hard to apply
them in engineering. More recently computer techniques
and numerical methods have provided excellent foundations
for using integral equations in the area of mechanics.
In 1963, Symm and Jaswon presented a numerical method to
solve boundary integral equations for potential problems
using Dirichlet, Neumann or Cauchy boundary conditions
[6,71. For elasticity problems, Cruse and Rizzo presented
the direct integral formulation, in which all variables
are original physical quantities [8-101. It is important
to point out that the direct BEM is more suitable for
solving engineering problems. This direct method is now
more widely used in the integral equation approach.
5
Many books [1,11,1~] and proceedings [7,10,13-21) have been
published and international conferences are held
periodically. Scientists of mechanics and engineers
have recently worked on BEM in the areas of time-dependent
and non-linear problems.
1-4 THE MAIN PROCEDURES AND FEATURES OF BEM
Generally speaking the main procedures of BEM are
the following:
A. Transform the differential equation with boundary
and initial conditions into the corresponding
boundary integral equation:
Using weighted residual method and integration by
parts (or using other equivalent methods, such as
Green's second identity or Somigliana identity), the
governing differential equations can then be made
equivalent to integral equations. After considering
the fundamental solutions, which satisfy the same
operators as the governing equations and are solved
by applying a Dirac delta-function, hence the domain
integrals are replaced by boundary ones. However,
if there are some other functions on the right-hand
side of the governing equation, there remains another
kind of domain integral term in the integral equation.
B. Solve the boundary integral equation numerically:
After expressing the variables by means of inter
polation functions and discretizing the intagral
equations, one can calculate all the boundary integrals
as summations of the values on all boundary elements.
6
Hence the final formulations can be reduced to a
set of linear algebraic equations. The set of
equations can then be solved to obtain all boundary
values.
C. Find the results at internal points of the domain
Afterwards, if values at internal points are required,
they can be calculated from the boundary solutions.
The developments and applications of BEM are recent,
but the method has shown some remarkable advantages over
domain methods.
These advantages can be summarised as follows:
A. BEM requires only the discretization of the boundary
of the domain. So that the mesh generation and the
preparation of initial input-data are considerably
simplified. For the same reason, the total degrees
of freedom of the problem are reduced considerably.
For these reasons BEM systems are easier to use and
to interface with other CAD systems.
B. The application of fundamental solutions in BEM not
only improves the accuracy of the results, but also
permits to represent boundary conditions at infinity.
Hence, BEM avoids taking a large mesh to present
infinite domain problems, while FEM requires a very
large number of elements to solve the same problems.
C. Both the unknown functions (displacement in elasticity
or value of field in the potential problems) and their
derivatives (traction in elasticity or flux in the
potential problems) are present in the BEM formulationj
hence, it is a mixed type formulation which gives the
7
same degree of accuracy for the derivatives as for
their functions using BEM. This makes the BEM more
suitable for cases of stress concentration and high
temperature gradient.
There are some other advantages in BEM, such as the
possibility of using discontinuous elements and simple
mesh refinements.
It is important to realise, however, that the BEM
also has some disadvantages over the FEM. For inGtance,
BEM formulations are usually more difficult than FEM ones
and the BEM systems of equations are fully populated, while
the FEM are not.
These disadvantages can be solved by using available
software systems developed by experts (such as BEASY [22]),
efficient numerical methods and large computers. Regarding
the last point, it is worth mentioning that the formulations
of BEM are especially suitable for the type of supercomputers,
which are rapidly becoming available for solving engineering
problems.
1-5 THE SUBJECT OF THIS WORK
As already mentioned, one of the obvious advantages
of using BEM is that it requires discretization only on
the boundary. Unfortunately this feature is generally
lost when source terms are present in the governing
differential equations. For example, when internal
distributed body forces (in elasticity) or internal
distributed sources (in potential problems) exist in
8
the domain, also in time-dependent and in non-linear
problems, domain inte-gral terms appear in the formulations
of the boundary integrals. Usually there are two
approaches to compute the domain integral terms in these
cases, first, in some special cases the body force (or
source) term can be transformed into boundary integral
terms. For instance, in the case of potential problems,
if the source function is constant or a harmonic: in
elasticity problems, if the body force is a gravitational,
centrifugal or therm& load [25-28]. The second approach
consists in dividing the domain into a certain number of
cells and then computing the domain integral numerically
as a summation of values over the cells. For this purpose,
quadrature techniques using Monte Carlo method and an
adaptive quadrature algorithm were presented in referencest24.
29-31]. These algorithms, however, consume a great
amount of CPU time and sometimes fail to get the results
within the specified accuracy.
The disadvantage of the second approach is that it is
necessary to define internal mesh points and cells over
the domain, so that a part of the main advantage of BEM
is lost [23]. In addition, the approach can not be used
for the case of infinite or semi-infinite domain, as it is
impossible to divide these domains into a finite number of
cells.
In order to transform the domain into boundary
integrals, a method called dual reciprocity (DRM) has
been developed by Brebbia and Nardini [32-35]. This method
can be used both in potential and elastic problems and some
9
accurate results have been obtained. Another advantage of
this method is that it introduces no new influence matrices
in the formulations, so it is computationally very efficient.
Its main drawback is that the body force functions are
expressed using an incomplete set of functions, and the
expressions are not sufficiently accurate. This can
cause considerable errors in some results.
In this thesis, a generalized method of transforming
the domain integrals into boundary ones is presented both
for potential and elasticity problems. Although some
expressions are similar to those presented in the
DRM , the idea and the derivations of formulations in
this thesis are independent work and their numerical
implementation is different. In addition, the approximate
functions used here are a complete set which is a.:comp1ete1y
different type from the ones used in DRM.
1-6 CONTENTS OF THE PRESENT WORK
In Chapter 2 a short review of BEM formulations in
potential problems is presented. Then, the method of
transforming the domain into boundary integrals is presented
and the formulations are derived for 2-D and 3-D potential
problems. Afterwards, the numerical approach for computing
the source integrals is introduced. In order to examine
the numerical accuracy of these formulations, tests have
been carried out, comparing results with those obtained
using the original domain integral. It is possible to
conclude from these results that the present formulation
is accurate. Some Poisson's equation examples are presented
10
and discussed. Finally it is shown how to generalize this
method for time-dependent and non-linear problems.
Chapter 3 is concerned with elasticity problems. All
the contents are almost parallel to potential problems.
Although the basic formulation is similar to the one for
potential cases, the problem is now more complex due to
the relative complexity of the elasticity equations. The
numerical approach is also more complex. Some problems,
which do not occur in potential problems, are discussed
in numerical implementation.
Applications in elasticity are presented in Chapter 4.
As has been mentioned above, body force integral terms
can be due to non-linearities. This is the case in
plasticity, even in the absence of body forces due to the
existence of the plastic strains. Using incremental methods,
the plastic strains can be obtained from the previous step
and a domain integral formulated. Then this integral can
be transformed to the boundary using the approach previously
described.
Chapter 5 describes the programs developed throughout
this work, with special emphasis to the programming of the
transformation formulae.
Chapter 6 presents a general critical discussion of
the topics investigated in the previous chapters and gives
the conclusions and suggestions for further work.
11
1-7 THE CARTESIAN TENSOR NOTATION
Throughout this work the Cartesian tensor notation
is used. This notation not only permits expressions to
be written in compact forms, but also is useful in deriva-
tion and proof of theorems. Such notation makes use of
subscript indices (1,2,3) to represent the spatial
coordinates (Xl, X2, X3).
These are some rules:
A. A repeated index in a term implies a summation with
respect to this index over its range.
B. The Kronecker delta symbol 0ij and the permutation
symbol e ijk are used throughout.
The definitions of 0ij and e ijk are as follows:
0"" = jl 1.] 0
when i j
when i ;. j
when any two indices are the same
when i,j,k are an even permutation of 1,2,3
otherwise
C. The spatial derivatives are indicated by a comma
and the index corresponding to that derivative, for
example, u" stands for au/ax1." • ,1.
CHAPTER 2 POTENTIAL PROBLEMS
2-1 INTRODUCTION
Many phenomena of mechanics and physics can be
reduced to the solutions of potential problems, such as
heat conduction, diffusion, flow of ideal fluid, flow
in porous media, torsion, electrostatics and others.
The Poisson's equation is discussed, first, in this
chapter. It is shown how a problem governed by a partial
differential equation and with prescribed boundary con
ditions can be recast into a boundary integral equation
form. Because there is a source term on the right-hand
side of the Poisson's equation, the boundary integral
equation involves a domain integral term too. Even
though the integrand is a known function in order to
compute the numerical value of this integral the domain
of the problem under consideration can be divided into
some cells {subdomainsl and these cells must be numbered
for numerical procedures. However in order to avoid
dividing the domain into subdomains, the source function
in the Poisson's equation can be expanded as a Fourier
series. Since each function in this series has some
particular features, the domain integral can be transformed
into a summation of boundary integrals. The error of the
integral values depends on how accurately the source function
can be expressed by this series. After deriving the
formulations, the numerical approach and the results for
examining are presented. Some examples are then presented
13
to describe the validity of the approach and finally
applications in time-dependent and non-linear problems
are described.
2-2 THE BOUNDARY INTEGRAL FORMULATION FOR POTENTIAL PROBLEMS
For steady potential problems the governing equation
can be expressed as Poisson's equation, assuming that
there exist sources b(x) inside the domain Q. These
sources can be due, for instance, to internal heat genera-
tion for heat conduction problems. The potential function
u is governed by Poisson's equation:
b(x)
and the following boundary conditions
A. Dirichlet type boundary condition:
u(x) u(x)
B. Neumann type boundary condition:
q(x) = q(x)
where V2 is the Laplace operator:
V is the Nabla operator:
and x is a spatial point with its coordinates xi
e; is unit vector along the axis x • i'
au q an
(2-2-1)
(2-2-2)
(2-2-3)
14
n is the unit outward normal to the boundary r,
u and q are prescribed values of the function u and
its normal derivative over the boundary. The total
boundary is given by r = r1 + r2 (Fig. 2-2-1).
Multiplying Eq.(2-2-1) by a function u* and
integrating it over all the boundary, one can express
Eq. (2-2-1) as an integral equation as follows:
J u*(E,;,x) [V 2 u(x) - b(x)]dn(x) = 0
n (2-2-4)
where u* is the fundamental solution of Laplace's equation,
i.e. u* satisfies the following equation
V2 U *(E,;,X) -t,(E,;,x) (2-2-5)
where E,; is an arbitrary point in the space. ~(E,;,x) is
the Dirac delta-function which has the following
properties:
and
~(E,;,x) {: for E,; " x
for E,; = x
J u(x)~(E,;,x)dn(x) u(E,;)
n
u*
u*
1 21T
1 41ir
1 In(r) for 2-D
for 3-D
(2-2-6)
15
Figure 2-2-1 Notation
16
Considering Green's second identity
J (u av - v aU)df an an (2-2-7)
f
and letting v = u*
one obtains the following equation from Eq. (2-2-4)
J [ * au au* )df - J u an - u an-r n
Using the notations
q au an and q* au*
an-
u* b(x)dn
and considering the integral feature of the Dirac delta-
function, the previous equation becomes
J u*q df - J q*u dr -
f f
Taking into consideration
(2-2-3), one obtains
J u*q dr J u*q df +
r fl
and
J q*u dr J q*u df + r f1
J u* b(x)dn
n u (t;) (2-2-8)
the boundary conditions Eqs (2-2-2)and
J u*q dr
f2
J q*u df
f2
Eq. (2-2-8) is valid for any point ~ inside the domain n.
But in order for this equation to be valid for any value
17
of ~ including those on the boundary r, the point ~ needs
to be taken to the boundary r. After accounting for the
jump in the second integral on the left -hand side of
(2-2-8)
J = J q*u dQ Q
the boundary integral equation (2-2-8) becomes
c(~)u(O J u*q dr - J q*u dr - J u*b(x)dQ (2-2-9) Q r r
Eq. (2-2-9) is the basic integral equation for BEM
potential problems.
The value of c(~) depends on the position of point
lj as follows:
c = 1 for an internal point ~
c = 0 for an external point ~ (2-2-10)
c = 1 + lim J q* dr for ~ on the boundary r E+O r E
In order to c"a1cu1ate the value of c (~), assume that the
domain under consideration can be augmented by a small
region r E ' which is a part of a sphere (3-D case) or
a circle (2-D case) centred at point ~ with radius E
(Fig.2-2-2). So the integral J can be separated into
two parts
J J q*u r-r
E
(2-2-11)
18
{}
Figure 2-2-2 Boundary f augmented by f£
19
The local polar (r,e) coordinate is introduced for
the 2-D case (Fig. 2-2-3). Considering the direction of
n along the radius, r = E and dr = Ede, one can write
q* 1 - 2nE
substituting this relationship into J 2 of Eq. (2-2-11), one
obtains
J q*u dr
r E
For the 3-D case, one can use spheroidal coordinates
(r,e,$) as shown in Fig.2-2-4. Therefore, relationships
are:
q* 1 - 4nE3
substituting them into J 2 of Eq. (2-2-11), one has
q*u dr
21T n/2
1 E2
- u!~) J I cos$ d~ de
o ~1
_ u(~) (1-sin~ ) 2 1
The geometric meaning of I r E
i dr (for 2-D) and J r E
Figure 2-2-3
Figure 2-2-4
20
----. ---
Evaluating c(~) for two-dimensional
Evaluating c(~) for three-dimensional
21
(for 3-D) is the plane and the solid angle respectively
which is spanned on r£ with centre ~ (Fig.2-2-3). As £ tends to zero, the values of these angles are equal to
the external plane angle (for 2-D) and external solid
angle (for 3-D) at point ~ on r. The values c(~) equal the
internal plane angle divided by 2n(for 2-D) and internal
solid angle divided by 4n(for 3-D) at point on ~ •
The fact of c(~) = I when ~ is located inside the
domain Q and c (~) = 0 when ~ is located outside ·the domain
Q is clear by looking at the geometric meaning of J 2 •
After taking into account the jump on the boundary
and expressing c as Eq. (2-2-10), the integral J is
considered as a Cauchy principal value integral because
when £ tends to zero in Eq.(2-2-11) J 1 is a Cauchy
principal integral. The existence of the integral J 2
can be proved if u(x) satisfies a Holder condition [45] at
point ~ as follows
lu(x) - u(~) I ~ B rcx
where Band cx are positive constants.
There is also a domain integral term on the right-hand
side of Eq.(2-2-9) as follows
I(~) = I u*b(x)dQ Q
(2-2-12)
For some particular cases, such as b(x) equals a con-
nt8.nt: or when b (x) is a harmonic function in Q, this
domain integral can be transformed into equivalent boundary
integrals, but for an arbitrary function b(x) numerical
22
quadrature is usually applied on each cell.
Besides the Poisson's equation, domain integral
terms also exist for the diffusion equation, time-dependent
problems and non-linear problems.
2-3 THE BOUNDARY ELEMENT METHOD FOR POTENTIAL PROBLEMS
This section presents the numerical implementation
of Eq. (2-2-9) for 2-D problems using only linear continuous
elements. In addition, the source function b(x) can be
assumed identically zero for simplicity as later on the
study will be extended to include the case of a more
general b(x) function.
In order to solve Eq. (2-2-9) numerically, one needs
to compute the integrals in Eq. (2-2-9) numerically
therefore, two important concepts need to be introduced.
A. BOUNDARY ELEMENT DISCRETIZATION
The boundary integrals, f, are computed as a r
summation of a certain number of numerical integrals, each
of which is carried out on a segment of the boundary.
This kind of segment is regarded as a boundary element
(Fig.2-3-1). Thus the integrals in Eq.(2-2-9) can be
expressed as
f NE
f u*q dr L u*q dr]
r j=1 rj
(2-3-1)
f NE
f q*u dr L q*u dr]
r j=l r. J
where NE is the total number of elements.
23
Figure 2-3-1 Discretization of boundary elements
24
B. INTERPOLATION FUNCTION
Using interpolation functions, only some values of
variables u and q are needed. The points, at which these
values are considered are, called nodes. The variables
elsewhere can be expressed in terms of these values at the
nodes. Using interpolation functions for continuous linear
elements, for instance, both ends of each element are
taken as nodes and the value of variables at every point
between the nodes can be expressed as a linear combination
of the values at the nodes (Fig.2-3-2).
(2-3-2)
where u jl and u j2 represent the potential values of the
nodes. qjl and qj2 are fluxes at the nodes. ~1 and ~2
are interpolation functions, i.e.
~1 J, (1-n) J, (1+n) (2-3-3)
where n is the local coordinate of the boundary element.
Notice that the length of every element in the local
coordinate is 2, Le. n e [-1,+1].
as:
After that, one can express the integral over element j
I u*q dr
rj
where
J q*u dr J q*[4>l 4>21
r. r. J J
gj1 J u* 4>1 dr
r. J
gj2 J u* 4>2 dr
r. J
hj1 J q* 4>1 dr
r j
hj2 J q* 4>2 dr r.
J
26
t'l f u j 11 J rr=[h j1 hj21
u j2 u. 2 J l J
(2-3-4)
(2-3-5)
Note, because all u*, q* and 4>1' 4>2 are known functions,
the results of hj1' hj2' gj1 and gj2 can be evaluated
by Eqs(2-3-5). The integrals in (2-3-5) are usually
computed by means of numerical quadrature, which is
described in Section 2-6. However, when the source point
~ is located on the element j, Eqs(2-3-5) can be calculated
analytically [1,121.
Substitution of Eqs(2-3-1) and (2-3-4) into Eq. (2-2-9)
leads to the following matrix equation:
H U G Q (2-3-6)
27
where matrix H is assembled with the elements h j1 , hj2 and c
and matrix G is assembled by gj1' gj2 (j = 1,NE). U and
Q are vector, the components of which are potentials and
fluxes at the nodes, i.e.
(2-3-7)
Q
After replacement of certain components in U and Q
by their prescribed values, given by the boundary conditions,
and rearrangement, Eq.(2-3-6) becomes a set of linear
algebraic equations, from which all unknown values of
potential and flux at the nodes can be solved. Furthermore,
the potential at the internal points can be computed from
Eq. (2-2-9) numerically.
2-4 MOTIVATION AND GENERAL IDEAS
When the function b(x) of Eq. (2-2-9) happens to be
a particular case, such as constant or sine function, the
domain integral on the right-hand side of Eq. (2-2-9)
1(0 J u*(~,x)b(x)dn(x) n
can be transformed into a boundary integral.
Let us suppose the function b(x) satisfies
IPb (x) = 6b (x)
(2-4-1)
(2-4-2)
28
where B is a non-zero constant.
After substituting Eq. (2-4-2) into (2-4-1), one
obtains
1(1;;) i J u* V' 2 b(x)Ml
n
and then considering Green's second identity (2-2-7)
and letting u
1(1;;) 1 B
u*, v = b, one has
J u* ab J an dr - q* b(x)dr + r r
J b(x)V'2 u *drl J rl
Considering u* is a fundamental function of Laplace
equations (Eq. (2-2-5», the last integral of previous
equation can be rewritten as
- J b(x) V' 2 u* (I;;,x)drl = b(l;;)
rl
Hence the previous equation now becomes
1(1;;) 1 B J u*
r ab J an dr - q*b dr - b (0 ]
r (2-4-3)
When the point I;; is on the boundary r, one calculates
the second integral on the right-hand side of Eq. (2-4-3)
as a Cauchy principal value integral. Then one can
rewrite Eq. (2-4-3) as follows:
I ( 1;;) 1 B J u*
r
ab dr an J q*b dr - c(l;;)b(I;;)]
r (2-4-4)
29
where c is as given in Eq. (2-2-10).
Hence, if the function b(x) in the domain integral
(2-4-1) satisfies Eq. (2-4-2), then this domain integral
(2-4-1) can be transformed into a boundary integral as
shown in Eq. (2-4-4).
An arbitrary function b(x) generally does not satisfy
Eq.(2-4-2) but based on the above idea, one can expect
that, if
A. Here choose a set of functions {bi (x) I i=l, ••. , oo} ,
and each function among this set satisfies Eq. (2-4-2),
B. Every arbitrary function b(x) can be expanded as a
c.
series of this set of functions, i.e.
b(x) = L kibi(x) i
(2-4-5)
When the set of functions {b.} is a complete set,then a 1.
finite part of series (2-4-5) instead of the infinite
series can express the original function b(x) with
sufficient accuracy. Thus in all the following
sections, the index i of the summation in the series
be only taken from 1 to a certain number N. In fact,
in this case, the meaning of 'equal to' in the
formulae will be 'approximately equal to'.
After substituting Eq. (2-4-5) into Eq. (2-4-1), it
becomes
I(O u*b. drt 1.
(2-4-6)
Then, using Eq. (2-4-4), the domain integral of Eq. (2-4-6)
can be expressed as follows
30
I(O J obi J u* az:l dr -r r
(2-4-7)
In order to satisfy the above conditions, one can use
trigr>nometric function as a set and expand b (x) as a
Fourier series. Because trigonometric functions
constitute a complete set, when one takes enough
terms in Eq. (2-4-5), this finite part series can
express the funcion b(x) accurately enough. This idea
allows us to transform any general domain integral into
a boundary one in potential problems.
2-5 FOURIER ANALYSIS
In this section a short review of Fourier analysis
is presented [36,67].
First, one defines a one dimensional function fIx)
as periodic if, and only if, there exists a positive
number 2a, such that for every x in the domain of
function f
fIx + 2a) = f(x) (2-5-1)
The number 2a is called a period of fIx).
Consider a function fIx) which is periodic with
period 2n, and suppose that it can be represented in the
interval [-n, n] by the following infinite trigonomp.tric
series
fIx) ko +I (k~ cos nx + k~ sin nx) n
(2-5-2)
31
where ko ' kl and k 2 (for n=l, ... ) are constants. In n n order to determine these unknown constants, one can use
the orthogonality of harmonics, i.e.
11
f sin(nx)dx 0
-11 11
J cos(nx)dx 0
-11 11
J sin(nx) sin(rnx)dx= 11 a (2-5-3) nm -11 11
J cos (nx) cos (mx>dx= 11 a nm -11
11
J sin(nx) cos (mx)dK= 0
-11
To find ko,one can integrate series (2-5-2) term by term
over the interval [-11, 111. Considering the properties
(2-5-3), one obtains
1 211
11
f -11
f(x)dx (2-5-4)
since f(x) is a known function which is assumed to be
integrable.
In a similar way to the above, to find kl or k 2 (for n n
n = 1,2, ... ) one multiplies each side of series (2-5-2)
by con(nx) or sin(nx) and then integrates from -11 to 11,
assuming again that term-by-term integration is justified.
using the orthogonality of sines and cosines and reducing,
32
one obtains
1T kl I
J fIx) sin(nx)dx n 1T -1T
1T
k 2 I
J fIx) cos (nx)dx (2-5-5) -n 1T -1T
Formulae (2-5-4) and (2-5-5) are regarded as Euler
formulae and the constants ko' k~ and k~ defined above
are called the Fourier coefficients of function fIx).
In this case, the series (2-5-2) is called the Fourier
series representation of function fIx).
Next, if the periodic function fIx) has a period
of 2a in the interval [-a, a), one can transform it into
a periodic function of y with period 21T, i.e.
y 1T a x (2-5-6)
Therefore, the function f(ay/1T) has a Fourier series
representation as given by Eqs(2-5-2) and (2-5-3).
substituting (2-5-6) in these equations, one obtains the
Fourier series expressed in terms of the variable x with
period 2a, Le.
fIx) . (n1TX)] s~n --a
(2-5-7)
where
33
a
ko 1 I f(x)dx 2a
-a a
kl 1 I fIx) (n 1TX) (2-5-8) n ""a cos -a dx -a
a
k 2 1 I fIx) sin (n;x) dx -n a
-a
Up to now, the function fIx) with period 2a is expanded as
a series (2-5-7) only formally. A theorem which gives
the conditions under which a Fourier series converges
to this function is called a Fourier theorem. One of such
theorems is referred to as Dirichlet theorem, i.e.
If fIx) is a bounded periodic function which in
any period has at most a finite number of discontinuous
points then the Fourier series of function fIx) converges
to fIx) at all points where fIx) is continuous and
converges to the average of the right- and left-hand limits
of fIx) at each point where fIx) is discontinuous.
The conditions of the above theorem usually are called
the Dirichlet conditions. Obviously,the Dirichlet conditions
are Valid for piecewise continuous functions. Although the
Dirichlet theorem is a sufficient condition for conver-
gence, almost all functions of practical engineering can
satisfy these conditions. Therefore one supposes ~ll the
functions used throughout this thesis satisfy the Dirichlet
conditions.
The following theorem gives the asymptotic behaviour
of the Fourier coefficients of a periodic function
fIx) [67].
34
THEOREM: As n becomes infinite, the coefficients
k~ and k~ in the Fourier expansion of a period function
satisfying the Dirichlet conditions always approach zero
at least as rapidly as c/n where c is a constant independent
on n. If the function has one or more points of discon
tinuity, then either k~ or k~, and in general both, can
decrease no faster than this. In general, if a function
f(x) and its first m-l derivatives satisfy the Dirichlet
conditions and are everywhere continuous, then as n becomes
infinite, the coefficients k~ and k~ in the Fourier series
f f () d 1 . dl / m+l o x ten to zero at east as rap1 y as c n • If
in addition, the mth derivative of f(x) is not everywhere
continuous, then either k~ or k~ , and in general both,
m+l can tend to zero no faster than c/n •
From the above theorem, generally speaking, the
smoother the function, the faster its Fourier series
expansion converges.
Finally, when f(x) possesses certain symmetry prop-
erties, the coefficients in its Fourier expansion become
especially simple. Suppose that f(x) is an even function,
i.e. f(-x) = f(x). The coefficients in the Fourier series
of f(x) (2-5-8) become
a
ko 1 I f(x)dx a
0
a kl 2 I (mrx) (2-5-9) n a f(x) cos a dx
0
k 2 0 n
35
Conversely, if fIx) is an odd periodic function, i.e.
f(-x)= -fIx), the coefficients in the Fourier series of
fIx) (2-5-8) become
k kl 0 0 n
(2-5-10) a
k~ 2
f fIx) . (mrx) dx s~n --n a a 0
The properties of Fourier analyses of two or three
dimensional functions are similar to these for one
dimension~ function and these will be used in this thesis
directly.
2-6 BASIC FORMULATIONS FOR TRANSFORMING THE DOMAIN INTEGRALS INTO THE BOUNDARY FOR 2-D PROBLEMS
In this section, one derives the basic formulae,
which are expressed using only boundary integrals. For
simplicity, the domains are assumed to be two dimensional,
so the vector x in all the equations has two components
x 1 and x 2' and ~ = (~ 1 ' ~i . Let us suppose the function b(x) = b(x1 ,x2 ) in Eq. (2-2-1)
is an arbitrary function, which satisfies the Dirichlet
conditions mentioned in the previous section in the domain
n of the problem under consideration. One extends b(x)
from the original domain n to the domain n f , which involves
domain nand is[2a1 X 2a 2 ] (shown in Fig.(2-6-1)). Then one
carries out the periodic extension of b(x) with the period
2a1and 2~ with respect to xl and x 2 directions respectively.
36
-----'----------2a,
Figure 2-6-1 Notation
37
From the above hypothesis, one can expand the function
b(x1,x2) as a convergent Fourier series. Therefore one
takes the finite part summation of Fourier series to
express the function b(x) approximately as follows:
b(x)
where
f 0
fl n
f2 n
f3 n
f4 n
fl nm
f2 nm
f3 nm
f4 nm
1
k f o 0 +
nrrx 1 cos (--) a 1
nrrx 1 sin (--) a 1
nrrx2 cos (--) a 2
nrrx 2 sin (--) a 2
fl. f3 n m
fl. f 4 n m
f2.f3 n m
f2.f4 n m
4 N 4 N M L L L L L
R,=1 n=1 R,=1 n=1 m=1
(2-6-1)
(2-6-2)
38
k 1
J J b(x)drl 0 4a1a 2
rl f
k£ 1 J
( f£(x)drl
2a 1a 2 J b(x) (2-6-3)
n n rlf
k£ = 1 J J b(x)f;m(X)drl nm a 1a 2
rl f
After substituting Eq. (2-6-1) into the domain
integral term (2-4-1), one obtains
1(0
+
J u*b drl
rl
u*f o
4 N M J L L L k£ u*f£ £=1 n=l m=l nm nm
rl
drl (2-6-4)
Consider that for each of the functions f o ' f~(x) or
f~m(x) there exists a corresponding function uo ' u~(x) or
u£ (x), which satisfies one of the corresponding equations nm
as follows
1,2,3,4 n,m 1, 2, ...
(2-6-5)
these functions u can be expressed as
39
1 (x 2-X 20 )2] Uo "4 [(x l -x lO )2 +
! -[:~l' fR. R. 1,2 n R. (2-6-6) u n
_ [:~) 2 fR- R- 3,4 n
where x lO and x 20 are arbitrary constants.
Using Eqs. (2-6-5)one by one and the second Green's
identity and considering the source point ~ can be
located anywhere, one has
- f q*uo dr - cUo(~)
r
f u*f~ f u* V2 UR. f dUR-
dQ dQ u* n dr n an Q Q r
- f R. dr - cu~(~) (2-6-7) q*u n
r
f u*fR. f u* V2 UR. f dUR.
dn dQ u* nm dr nm nm ----an Q Q r
- f R. dr - R-q*u cUnm (~) nm r
40
where c are the same as in Eq. (2-2-10).
Substituting Eqs. (2-6-7) into (2-6-4), one can transform
each term of the domain integral on the right-hand side of
Eq. (2-6-4) into boundary integrals. After defining
a go an Uo
R. Cl fR. (2-6-8) gn an n
R. a fR. gnm= an nm
the Eq. (2-6-4) can be rewritten as a transformation formula
for the domain integral 1(~):
1(0 = I u*b dSl So [ I u*g dr- I q*u dr - c U o (~)] + 0 0
Sl r r
4 N sR. I i I q*f i f~(~)] + + L L u*gn dr - dr - c
.11.=1 n=1 n n r r
4 N M i I i I *f i fi (~)] + L L L Bnm u*gnm dr - q nm dr - c R.=1 n=1 m=1 nm
r r
(2-6-9)
where
1,2
3,4
41
(2-6-10)
The first term of Eq. (2-6-4) can also be transformed
as follows. Let us introduce a function v* defined as
r 2 1 v* = an [In(r) + 1] (2-6-11)
and that function v* satisfies
u* (2-6-12)
The first integral on the right-hand side of Eq. (2-6-4)
can now be expressed as
f u* dn n
= f av* dr an r
(2-6-13)
Now, having substituted Eqs. (2-6-7) and (2-6-13) into
Eq.(2-6-4), we obtain
I (E,;) f u*b(x)dn ao f av* dr + an
n r
4 N a2. f
2-
f *f2. f2. (E,;)] + + L L u*9n dr - dr - c 2.=1 n=l n q n n
r r
4 N M 2. I 2. f *f2. - c f~m(E,;)] + L L L anm [ u*qnm dr - dr
2.=1 n=1 m=1 q nm
r r (2-6-14)
The above Eq. (2-6-9) or (2-6-14) is the final formula
which transforms the domain integral of the body source the
term into boundary integrals. This is calledAtransformation
42
formula for simplicity throughout this thesis. For two
dimensional problems, one has
u*
q*
v*
<lv* <In
1 1 2n In (i)
1 <lr - 2nr an
1 <lr - an [2r(ln r-1) + r1 <In
2-7 NUMERICAL APPROACHES
..f-[2 I (1) 11 <lr 871 n i + an
The function b(x) on the right-hand side of Poisson's
equation (2-2-1) is known, so the coefficients of its
Fourier series can be determined by Eqs. (2-6-3). After
that, all the functions and constants are known in Eq.
(2-6-9) or (2-6-14). In this case, the source integral
term can be computed by means of numerical quadratures
with formula (2-6-9) or (2-6-14). Using numerical quad-
ratures directly instead of using interpolation functions
one can obtain more accurate results. To this purpose,
one discretizes all the boundary r into NE elements and
calculates every boundary integral in Eq.(2-6-9) as a
summation of the values over all these elements:
J f(x)dr
r
NE L
j=l J f(x)dr
rj
When computing the values of the integral over every
element, using standard Gauss quadrature as follows [38,461
43
J f(x)dr r.
(2-7-1)
J
where NP is the number of Gauss integral points chosen.
Sp is the local coordinate of the Gauss integral point p
and wp is the corresponding weight value Both sp
and w can be found in the tables, which are included in p
references [38,12]. Because the values are listed in
those tables in terms of normal local coordinates in the
interval [-1,+1], it is necessary to transform the length
of element, rj,into the length 2. In Eq. (2-7-1) IJI is
the values of the Jacobian of this transformation which
gives
IJ I = r. /2 J
It can be proved, if the integrand is continuous
on the interval [A, B] that the results of Gauss quadrature
converge to the exact value of the integral as the number
of Gaussian points increases. For regular integrals
Gauss quadrature gives good results with few integration
points.
But this standard Gauss quadrature (2-7-1) is not
suitable for the case when the source point s is on the
element, because of singularities of functions u*, q*
and ~*/an. In this case, a special coordinate transform-
ation can improve the accuracy of results significantly
[39] •
Let us consider the singular integral I as follows
I
+1
J f(x)dx
-1
44
(2-7-2)
in which fIx) is singular at a point x in the interval
[-1,11.
Now, one chooses a non-linear transformation as
follows
x (z) az 3 + bz 2 + cz + d (2-7-3)
with the following conditions on the boundary and at the
singular points:
x(l) = 1
x(-l) =-1
dxl = 0 dz -z
~~~ 1_ 0 z
(2-7-4)
where the image point z is the point, which corresponds
to the original singular point x. The third equation
in Eqs. (2-7-4) means that the Jacobian value of this
transformation equals zero at the singular point z and
the fourth equation in Eqs (2-7-4) means the Jacobian is
an extreme value, which can be chosen as extremely small.
So, after using the transformation (2-7-3) the singularity
of integrand can be smoothed by the Jacobian, and Eq. (2-7-2)
becomes
I
+1
J f(x(z))dx/dz dz
-1
45
(2-7-5)
From the constrained conditions (2-7-4) of this
transformation one obtains all these coefficients in
Eq. (2-7-3) as follows
a = 1/Q
b - 3z/Q
d -b
Where z is simply the value of z which satisfies x(z) x,
this value can be evaluated by
z = (xx* + Ix*I)1/3 + (xx* - Ix*I)1/3 + x
and
x* = x:Z - 1
Thus, after transformation the integral (2-7-2) becomes
I (2-7-6)
Using this transformation, the order of accuracy in
numerical integration can be improved up to one to two
orders in comparison with the standard Gauss quadrature.
The value c in Eq. (2-6- 9) was expressed as Eq. (2-2-10)
In regard to the constants x 10 and x 20 in the first
of Eqs (2-6-6), the values of (x 10 ' x 20 ) can be taken as
coordinates of a certain reference point. This reference
point can be chosen at an arbitrary position, such
46
as the origin of coordinates or the middle point of domain
Q. It must be avoided that taking the magnitude of slO or
x 20 is significantly different from that of xl or x 2 '
otherwise it causes some numerical error.
2-8 NUMERICAL ACCURACY OF THE TRANSFORMATION FORMULA
In this section the numerical accuracy of formula
(2-6-9) or (2-6-14) is examined. For this purpose, let
the source function b(x) equal each term in Eq.(2-6-1)
one by one and then compute the value of the integral using
formula (2-6-9). Notice that all the integrals are
implemented only on the boundary. In order to compare the
results the same domain integrals are computed also by
means of the original domain integral method, the procedures
of which are as follows:
A. Divide the domain Q into some subdomains.
B. Compute the values of this integral over these subdomains.
C. Add all the values of the subdomain integrals to obtain
the original domain integral.
When the above step B is carried out, the standard Gauss
quadrature is applied over the subdomain. If the source
point is located outside this subdomain, one has
f NP1 NP2
Is(~) f(x)dQ L L f(~ , ~ )w iJi Q P1=1 P2=1 PI P2 P1P l
s (2-8-1)
where iJi is the Jacobian of transformation for Q s into
the standard 2-D Gauss quadrature region [2 x 21, i.e.
47
IJI = A /4 s (2-8-2)
where As represents the area of the subdomain Os.
When the source point ~ is within the subdomain Qs'
the same transformation as shown in Eq. (2-7-3) can be used
in both the xl and x 2 directions.
In order to compare the results for the two different
methods, one selects the same subdivision of the domain and
boundary (shown in Fig. 2-8-1). The number of boundary
elements is NE = 2(NX + NY), where NX and NY are the numbers
of divisions in xl and x2 directions. The numbers of
Gaussian points NP, NP1 and NP2 for standard Gauss quadrature
are 4 or 6, which depend on the distances between the source
point and the middle point of the boundary element or the
subdomain. If the distance is less than 1.5 times the
length of the boundary element or the side of the subdomain,
NP (or NP1, NP2) equals 6 otherwise it equals 4. For the
singular boundary element or subdomain, in which the trans-
formation (2-7-3) is carried out, the number of integral
points is 8. Furthermore, without losing generality, the
area of Fouri~r expansion, Qf iS2a1 X2a2 and it involves
the integral domain O. The source point ~, when located
on the boundary r, is notated ~l or ~2. If the source
point is located outside or within the domain, it is notated
~3 or ~4· The locations of these source points are shown
in Fig. '2-8-1. The results of domain integrals, which are
defined as
I(~) J u*b(x) dQ(x)
°
48
20
15
• nesh points 10
5
2 7 10
Figure 2-8-1 Geometry and mesh
49
are computed with Eq. (2-6-9) and the original domain
integral method respectively, the results of which are
noted as II and 12 separately in Tables 2-8-1 - 2-8-10.
In these tables, the wave numbers of harmonics (2-6-2) are
nand m in the xl and x 2 directions respectively.
From these tables, some interesting facts can be drawn:
A. If the wave number n or m is less than 6, II and 12
coincide with each other exactly in the numerical sense.
It means, in this case, the Eq.(2-6-9) is correct both
theoretically and numerically.
B. While the wave number n or m is increasing, the errors
of these results increase as well, so the differences
between II and 12 become more pronounced. This kind of
numerical error is also reasonable, because in every
boundary element or every side of a subdomain there are
many wave numbers, when compared with the number of
integral points NP, NPI or NP2. For example: when n = 24
in every boundary element along xl direction there are 2
waves of harmonics, it means that the signs of these
harmonics alternate four times along one element, but the
number of integral points, NP, is only four per regular
element. Obviously it causes unacceptable errors. The
same case happens for the results of the subdomain integrals
1 2 •
C. Even if there are as many as four waves of harmonics
in three boundary elements (that is nor m equals sixteen),
the results still have four or five digit accuracy.
50
Table 2-8-1 Comparison of boundary inte,gral II
with domain integral I2
I = I u*.1 d!1
!1
NX I1 I2 E:. NY
3 -0. 8272959X102 -0.8272960X102
6 -0.8272958X102 -0.8272957X102
E:.! 9 -0.S272958X102 -0.8272958X102
12 -0. 8272959X1 02 -0. 8272965X102
15 ":'0. 8272959X102 -0.8272972X102
3 -0.7534999X102 -0. 753500OX102
6 -0. 7535000X102 -0. 7535000X102
E:.2 9 -0. 7535000X102 -0. 7535007X102
12 -0. 7534999X102 -0.7535001X102
15 -0. 7534999X102 -0.7535009Xl02
51
Table 2-8-2 Comparison of boundary integral 11
with domain integral 12
~ n
1
2
3
4
5
~l 6
7
10
15
25
1
2
3
4
~2 5
6
7
10
15
25
I J u* cos(n;;) dQ Q
11
-0.7622810 X101
0.5229649 X102
0.5056158 X101
-0.3344568 X101
0.1191101 X102
-0.1082302 X102
-0.8292571 X101
0.2712600 X100
-0.4247177 X101
0.2479330 X101
-0.8491453 X101
0.4558578 X102
0.5711345 X101
0.1174665 X100
0.1297357 X102
-0.9624903 X101
-0.9006425 X101
0.1432590 X100
-0.3697073 X101
0.2233753 X101
12
-0.7622812 X101
0.5229648 X102
0.5056158 X101
-0.3344566 X101
0.1191102 X102
-0.1082301 X102
-0.8292586 XlI)1
0.2712435 X100
-0.4247263 X101
0.2420987 X101
-0.8491419 X101
0.4558580 X102
0.5711358 X101
0.1174508 X100
0.1297358 X102
-0.9624891 X101
-0.9006439 X101
0.1432513 X100
-0.3697100 X101
0.2183748 X101
52
Table 2-8-3 Comparison of boundary integral II
with domain integral 12
I J * 'n(~) dn u S1 10 n
E; n 11 12
1 -0.7424272 X102 -0.7424271 X102
2 -0.9916166 x101 -0.9916165 X101
3 0.2555545 X10 2 0.2555545 Xl02
4 -0.4099361 X101 -0.4099360 X101
5 0.8766171 X10 1 0.8766171 X101
6 0.1361520 X102 0.1361522 X102
E; 1 7 -0.7063324 X101 -0.7063317 X101
10 -0.1041555 X102 -0.1041555 X102
15 0.2715321 X101 0.2715315 X101
25 0.1675875 X101 0.1604687 X101
1 -0.6699394 X102 -0.6699390 X102
2 -0.1106592 X102 -0.1106590 X102
3 0.2007587 X102 0.2007588 X102
4 -0.4384451 X101 -0.4384450 X10
5 0.9765131 X101 0.9765128 X101 E;2
0.1482711 X102 0.1482712 X102 6
7 -0.4656162 X101 -0.4656166 X101
10 -0.1085651 X102 -0.1085651 X102
15 0.3093648 X101 0.3093693 X10'
25 0.1968400 X101 0.1919084 X101
Table 2-8-4
I; n
1
2
3
4
5
I; 1 6
7
10
15
25
1
2
3
4
1;2 5
6
7
10
15
25
53
Comparison of boundary integral II
with domain integral I2
I f u* cos (n27Tl) dS"l S"l
I1
0.1025681 X102
0.5387757 X102
-0.1768966 X102
-0.2668609 X101
0.5235638 X101
-0.1596658 X102
0.6657169 X101
o 9456897 X101
0.2848718 X1~1
-0.1464234 X101
0.1295744 X102
o 5021597 X102
-0.2207117 X102
-0 4883914 X101
0.5886055 X101
-0.1325916 X102
0.9007068 Xl01
0.7555257 X101
0.3923257 X101
-0.1937091 X101
I2
0.1025684 X102
0.5387759 X102
-0.1768966 X102
-0.2668607 X101
0.5235623 X101
-0.1596659 X102
o 6657169 Xf01
o 9456928 X101
0.2849329 X101
-0.1491409 X101
0.1295761 X102
o 5021592 X102
-0.2207116 X102
-0.4883922 X101
0.5886029 X101
-0.1325917 X102
0.9007059 X101
0.7555306 X101
0.3922994 x101
-0.1962734 X101
54
Table 2-8-5 Comparison of boundary integral II
with domain integral 12
£;
£; I
£;2
I f u* sin(n;;) dn
n
n 11
1 -0.7482952 X10 2
2 0.1685810 X102
3 0.2697564 X10 2
4 -0.1301431 X102
5 0.1242537 X102
6 -0.2266829 X10 1
7 -0.1027152 X102
10 0.3230910 X100
15 -0.4582665 X101
25 -0.2598Q64 X101
1 -0.6848425 X10..!
2 0.2121402 X102
3 0.2658022 X102
4 -0.1591316 X102
5 0.9124641 X101
6 -0.3630992 X101
7 -0.9268256 X101
10 0.1419310 X101
15 -0.3825251 X101
25 -0.1944320 X101
12
-0.7482947 X102
0.1685812 X102
0.2697564 X102
-0.1301432 X102
0.1242537 X102
-0.2266842 X102
-0.1027155 Xl02
0.8232314 X100
-0.4582321 X101
-0.2510424 X101
-0.6848415 X102
0.2121401 X102
0.2658024 X102
-0.1591315 X102
0.9124645 X101
-0.3630999 X101
-0.9268279 X101
0.1419359 X101
-0.3825188 X101
-0.1874890 X101
55
Table 2-8-6 Comparison of boundary integral II
with domain integral 12
~ n m
1 1
1 2
2 1
1 3
2 2
3 1
1 4
~l 2 3
3 2
4 1
1 1
1 2
2 1
1 3
~2 2 2
3 1
1 4
2 3
3 2
4 1
I J u* cos(n;;) cos(m2~Y) dn n
11 12
0.2918378 ~101 0.2918366 X101
0.5592705 X101 0.5592709 X101
-0.5243825 X101 -0.5243832 X101
-0.4876962 X101 -0.4876965 X101
-0.3356133 X102 -0.3356137 X102
-0.3528005 X101 -0.3528009 X101
-0.1588836 X101 -0.1588832 XH)1
0.9215223 X101 0.9215222 X101
-0.4111199 X10' -0.4111189 X101
-0.1495466 X101 -0.1495464 X101
0.2733293 X101 0.2733322 X101
0.6142702 X101 0.6142701 X101
-0.7863301 X101 -0.7863300 X101
-0.4531018 X101 -0.4531018 X101
-0.3039432 X102 -0.3039432 X102
-0.3782635 X101 -0.3782633 X101
-0.1569167 X101 -0.1569179 X101
0.1338504 X102 0.1338503 X102
-0.4924170 X101 -0.4924174 X101
-0.1025702 X10-1 -0.1025700 X10-1
56
Table 2-8-7 Comparison of boundary integral 11
with domain integral 12
E;, n m
1 1
1 2
2 1
1 3
2 2
E;, I 3 1
1 4
2 3
3 2
4 1
1 1
1 2
2 1
1 3
2 2 E;,2
3 1
1 4
2 3
3 2
4 1
I =J u* cos(nnx) s1'n(mny ) d n 10 20" Q
11 12
-0.7075934 X101 -0.7075955 X101
0.4749635 X101 0.4749632 X101
0.4715874 X102 0.4715874 X102
0.3588468 X101 0.3588460 X101
-0.8670514 X101 -0.8670522 X101
0.4808176 X101 0.4808191 XlO 1
-0.3405131 X101 -0.3405133 X101
-0.1618861 X102 -0.1618862 X102
-0.5755103 X101 -0.5755108 X101
-0.2804108 X101 -0.2804120 X101
-0.7857435 X101 -0.7857417 X101
0.4437354 X101 0.4437364 X101
0.4143626 X102 0.4143625 X102
0.3840045 X101 0.3840088 X101
-0.1287099 X102 -0.1287098 X102
0.5514474 X101 0.5514489 X101
-0.3119807 X101 -0.3119800 X101
-0.1610427 X102 -0.1610427 X102
-0.6095337 X101 -0.6095333 XlO 1
0.1185471 X100 0.1185437 X100
57
Table 2-8-8 Comparison of boundary integral 11
with domain integral 12
I I * . (n 1TX) (m1TY) dSl u I:!l.n 10 cos 20 Sl
~ n m 11 12
1 1 0.8825028 X10 1 0.8825037 X101
1 2 0.4819830 X102 0.4819831 X102
2 1 0.4297675 X101 0.4297669 X101
1 3 -9.1527316 X102 -0.1527316 X102
2 2 0.7400675 X101 0.7400671 X101
3 1 -0.1230353 X101 -0.1230359 X101 ~l
-0.2078539 X101 -0.2078537 X1.0 1 1 4
2 3 -0.7196420 X101 -0.7196427 X101
3 2 -0.1586730 X102 -0.1586731 X102
4 1 -0.1256455 X101 -0.1256454 X101
1 1 0.1152898 X102 0.1152899 X102
1 2 0.4465325 X102 0.4465328 X102
2 1 0.4179062 X101 O. 4~;?9073 X101
. 1 3 -0.1963381 X102 -0.1963380 X102
~2 2 2 0.8256349 X101 0.8256357 X101
·3 1 -0.3480577 X101 -0.3480571 X101
1 4 -0.4353125 X101 -0.4353124 X101
2 3 -0.6880232 X101 -0.6880241 X101
3 2 -0.1338276 X102 -0.1338276 X102
4 1 -0.1998341 X101 -0.1998334 X101
58
Table 2-8-9 Comparison of boundary integral II
with domain integral 12
~ n m
1 1
1 2
2 1
1 3
2 2
3 1
~ 1 1 4
2 3
3 2
4 1
1 1
1 2
2 1
1 3
2 2 ~2
3 1
1 4
2 3
3 2
4 1
I J u* sin(~~X) Sin(~cr) dn
n
11 12
-0.6710900 X102 -0.6710900 X102
0.1452079 X102 0.1452079 X102
-0.9240564 X101 -0.9240568 X101
0.2394382 X10 2 0.2394383 X102 ,
0.6998923 X101 0.6998926 X101
0.2289640 X10 2 0.2289640 X102
-0.1129755 X102 -0.1129755 X102
0.4891803 X101 0.4891800 X10 1
-0.2101770 X101 -0.2101773 X101
-0.3598833 X101 -0.3598852 X101
-0.6089127 X102 -0.6089122 X10 2
0.1887393 X10 2 0.1887393 X102
-0.1031246 X102 -0.1031246 X102
0.2364258 X102 0.2364259 X102
0.6770080 X101 0.6770090 X101
0.1824759 X102 0.1824760 X102
-0.1415146 X102 -0.1415146 X102
0.5440886 X101 0.5440889 X101
-0.5698493 X101 -0.5698484 X101
-0.3653636 X101 -0.3653646 X101
59
Table 2-8-10 Comparison of boundary integral Il
with domain integral I2
~ n m
1 1
1 2
2 1
1 3
2 2
~3 3 1
1 4
2 3
3 2
4 1
1 1
1 2
2 1
1 3
2 2 ~4
3 1
1 4
2 3
3 2
4 1
I Jf u* . (nnx) . (mny) dn s~n --ro s~n 20
n
Il I2
-0.4055582 Xl02 -0.4055582 Xl02
0.1226002 Xl02 0.1226003 Xl02
-0.2847070 Xl01 -0.2847061 Xl01
0.6456060 Xl01 0.6456055 Xl01
0.4988304 Xl01 0.4988313 X101
0.1210875 Xl02 0.1210877 Xl02
-0. 133441 0 Xl02 -0.1334412 Xl02
-0.2021419 Xl01 -0.2021427 Xl01
-0.3698058 Xl01 -0.3698051 Xl01
-0.7289629 Xl01 -0.7289616 Xl01
-0.9541643 Xl02 -0.9541643 Xl02
0.7483069 Xl01 0.7483064 Xl01
-0.1921332 Xl02 -0.1921332 Xl02
0.3226102 Xl02 0.3226102 Xl02
0.2354544 X101 0.2354541 X10 1
0.2946567 X102 0.2946566 Xl02
-0.5857213 Xl01 -0.5857200 XlO 1
0.6655581 Xl01 0.6655580 Xl01
-0.1936955 X101 -0.1936950 Xl01
-0.1324903 Xl01 -0.1324900 Xl01
60
To decrease the numerical errors in B, one needs
only to take more integral points or divide the boundary
using finer meshes.
Table 2-8-11 presents the results obtained using the
finer meshes. It can be found that when NE equals 24
(that means there is one wave along every boundary element),
the relative error of the result decreases to only 4 x 10- 6
with Eq. (2-6-9). By contrast, for the same division as
in the original subdomain integral method, the results do
not improve as much as those for Eq. (2-6-9). This can be
explained by noticing that the domain integral has one
more dimension and here it causes more errors than the
boundary integrals.
In addition, formula (2-6-14) has the same accuracy
as formula (2-6-9).
All these mean that formula (2-6-9) or (2-6-14) gives
reliable results for the values of the domain integrals.
After testing the transformation formulae term by
term in series (2-6-1), some further examples are examined.
The exact results of following examples are using the
refined mesh, i.e. they are in the numerical sense.
EXAMPLE 2-8-1:
Let us assume that the source function b(x) is as
follows
b(x) (2-8-3)
The Fourier [:;erien e::presnio;'l of thin function is
Table 2-8-11
I
NX
~ NY
3
6
9
~2 12
15
18
21
exact solution:
61
Comparison of boundary integral II
with domain integral 12
J u* sin(nl~x)dn n
11
0.1968400 X10 1
0.1969215 X101
0.1969207 X101
0.1969207 X101
0.1969207 X101
0.1969207 X101
0.1969207 X101
.0.1969207 X101
n = 25
12
0.1919084 X101
0.2019204 X101
0.1968971 X101
0.1969194 X101
0.1969230 X101
0.1969200 X101
0.1969203 XHl1
62
Table 2-8-12
The results of Example 2-8-1
2 x 5
Relative errors of (%)
N I I function
2 -0.6785131 x 10 2 1.11 2.72
~1 5 -0.6713483 x 10 2 0.039 0.75
10 -0.6710841 x 10 2 -0.00082 0.19
exact. -0.6710896 x 10 2
2 -0.6153646 x 10 2 -1. 06 2.79
~2 5 -0.6090678 x 10 2 -0.026 0.76
10 -0.6088849 x 10 2 0.0046 0.19
exact. -0.6089122 x 10 2
63
(2-8-4)
The geometry and meshes are shown in Fig. 2-8-2
and the results are listed in Table 2-8-12. In this table
the relative errors of Fourier series (2-6-1) and the
results of the integrals are listed as well. From these
relative errors, it is clear that the convergence of ~e
results is faster than that of Fourier series. For example,
when N increases from 2 to 5, the relative errors of the the
Fourier expression of A function are improved from 2.72%
to 0.19%, but the relative errors of integral results are
improved rapidly from 1.11% to -0.00082% at point ~1.
This is because from relationships (2-6-10) one can find
that the coefficients of B tend to zero faster than the
coefficients of Fourier series as n is increasing.
EXAMPLE 2-8-2:
Let us suppose the source function is in the form of
This is a periodic function with a period 4a1 . Obviously,
if one takes the expansion domain nf as 4a l in the xl
direction, one can expect the solutions to have the same
accuracy as in Table 2-8-2 with n = 1. Let us see what
happens if one expands it as a periodic function with a
64
a 2 ------,
15
10
5
o ~~~ ___ ~ __ _L __ ~
2
Figure 2-8-2
7
Geometry and meshes for Examples 2-8-1, 2-8-2 and 2-8-3
period 2a1 , i.e.
b(x) 1 + 1 'IT 'IT
65
(2-8-5)
The results are listed in the Table 2-8-13 with the
same meshes in Fig. 2-8-2 as in example 2-8-1. From
Table 2-8-13 one can find that the relative errors of
integrals are much smaller than those of Fourier series
ofb(x).
EXAMPLE 2-8-3:
In this example one supposes function b(x) is equal
to unity. In this case, the expression of Fourier series
only has the constant term ko = 1. It means that one
expands b(x) as an even function. After periodic extension
of b(x), this function equals unity everywhere. The results
for the same mesh shown in Fig. 2-8-1 are listed in Table
2-8-1. The results have six or seven digit accuracy. Here
also to test the discontinuous function, one carried out
the period extension for b(x) as an odd function. Then
one can express it as a Fourier series in the form of
b(x) 1 4 'IT
y __ 1 __ sin (2n-1) nX1 n=l 2n-1 a 1
(2-8-6)
Using the Fourier series (2-8-6), one computes the
integral (2-6-14). The results are listed in Table 2-8-14,
from which one can also find the convergence of integrals
is faster than that of Fourier series. Obviously, the
accuracies of results in Table 2-8-14 are worse than those
shown in Table 2-8-1. Because the periodic extension is as
66
Table 2-8-13
The results of Example 2-8-2
2 x 5
Relative error of (%)
N I I function
5 -0.5993685 x 10 2 -0.027 0.91
~l 10 -0.5994077 x 10 2 -0.031 0.17
20 -0.5991896 x 10 2 0.0052 0.0498
exact. -0.5992106 x 10 2
5 -0.5507037 x 10 2 -0.016 0.98
~2 10 -0.5507957 x 10 2 -0.032 0.18
20 -0.5505929 x 10 2 0.0038 0.051
exact. -0.5506117 x 10 2
67
Table 2-8-14
The results of EXqmple 2-8-3
2 x 5
Relative errors of
~ N I I ( %) function(%)
2 -0.836827 x 10:1 -1.15 13.59
5 -0.827162 x 10:1 0.016 7.1
~1 10 -0.827076 x 10:1 0.027 3.69
Eg. (2-8-6) 15 -0.827288 x .• l0:l 0.00097 2.40
2 -0.767789 x 10:1 -1. 90 13.01
5 -0.752942 x 10:1 , 0.074 6.46 ~2
10 -0.753207 x 10:1 +0.039 3.29
15 -0.753461 x 10:1 0.0052 2.19
2 -0.847423 x loa -2.43 21. 85
5. -0.835158 x loa -0.95 7.14
~1 10 -0.827229 x loa 0.008 4.62
Eg. 20 -0.827246 x loa 0.006 2.44 (2-8-7)
2 -0.773523 x loa -2.66 21.15
5 -0.759431 x loa -0.79 6.47 ~2
10 -0.753249 loa 0.033 4.48 x
20 -0.753457 x loa 0.0057 2.19
~ = ~1 -0.827296 x loa exact.
~ = f,;2 -0.753500 x 10:3
68
an odd function for b(x), the values of b(x) have some
discontinuities on the lines as xl = 0, aI' 2a l , •••.
In contrast, b(x) is constant as an even function and has
continuous derivatives. According to the theorem in section
2-5, the above results are reasonable. If one considers
the period extension for b(x) as an odd function both in
Xl and x 2 directions, the function can be represented as
b(x) 1 16 1f2
N M 1 L L ...,-:::--~...,...,,----~
n=l m=l (2n-l) (2m-I)
. ( 2 n -1) 1TX 1 s l.n.------=-
. ( 2m-I) 1TX 2 s l.n.------=- (2-8-7)
Integrating with the same mesh, one lists the results also
in Table 2-8-14. It is clear the accuracy of the results
is much worse than before, because now the discontinuity
exists not only in the xl ' but also in the x 2 direction.
2-9 SOME FURTHER DISCUSSIONS
Based on the results discussed in the previous section,
if the source function can be expressed as a convergent
Fourier series, then sufficiently accurate solutions of
the formula (2-6-9) or (2-6-14) can be expected with enough
terms in the Fourier series. Usually only very
few terms are sufficient for the continuous function.
Let us consider that the distributed source function
b(x) is defined over all regions of the problem under
consideration, that is, the domain nb equals the domain n.
69
In this case, the boundary rb coincides with r. However,
if the source function b(x) is defined only on a part of
the domain n of the problem, i.e. ~ is a subdomain of nand
one can now apply two different approaches.
One of them is that the source function b(x) can be
extended to all the domain n, for example, a new function
b(x) is defined over the whole domain n as follows
J b(x) b(x) = 1 0 (2-9-1)
Then, one can compute the values of the integral (2-4-1)
with the transformation formula (2-6-9) and integrate along
the boundary r of the domain n.
The other approach is to compute the value along the
boundary rb of thesubdomain nb , noticing that
I (~) J u*b(x) dn
n (2-9-2)
Usually, the second approach is more efficient and
accurate than the first one, i.e. computing the integral
along the boundary r b • In addition, sometimes it simplifies
the expansion in Fourier series. For example, in Fig.
(2-9-1), b(x) only exists over the subdomain nb , which
is involved by n and the function b(x) is
b(x) 1: (2-9-3) otherwise
70
Figure 2-9-1 Notation
71
The extension of the function b(x) can be carried out
as follows
b(x) = 1 everywhere (2-9-4)
After making the expansion domain nf coincide with nb ,
in this case, only a constant term exists in the Fourier
series of b(x). The value of domain integral is computed
by the integral on the right-hand side in Eg. (2-9-2).
Obviously, it is an efficient and accurate approach.
However, how to choose the integral boundary Pb and
the expansion domain nf strongly depends on the features
of the function b(x). The following section shows some
examples with numerical results.
When function b(x) is a piecewise function, the
Fourier theorem ensures that the Fourier series converges
to the average of the right-and left-side limits. Con
sider, for instance, that the function b(x) as shown in
Fig.2-9-2 (a) is expanded as a Fourier series for periodic
functions of period a l • Since b(O) ~ b(a l ),
the Fourier series converges to the value [b(O) + b(a l )]/2
= b(a l )/2 at the points xl 0 and a l (Fig. 2-9-2 (b».
This kind of discontinuity in Fourier series does not
produce any problems when using Eg. (2-6-9) or (2-9-14),
even in the case that the integral boundary Pb coincides
with the discontinuous lines of Fourier series. However,
in this case, more terms of Fourier series must be taken
for obtaining a sufficiently accurate expansion. For
this reason, it is better to avoid discontinuity when
choosing the expansion domain nf and the integrating
boundary Pb . In this example, the same function b(x)
72
(a) (b)
Figure 2-9-2 Periodic extensions of function b(x)
73
can be extended as a periodic function of period 2a l
on nf and computed on the boundary of fb with interval
[0, all (shown in Fig.2-9-2 (c». This way is more
efficient and accurate than the approach shown in Fig.
2-9-2 (b).
2-10 EXAMPLES
After testing the transformation formulae (2-6-9)
and (2-6-14) in the previous section, one presents here
some examples applying this transformation technique to
Poisson's equation. In these examples, only continuous
linear elements are used and all the integrals in the
form of (2-4-1) are computed with formula (2-6-9).
Besides the results and relative errors listed
later, some results are given related to the analyses
of errors, how to choose the extension of function b(x) ,
how to choose the expansion domain nf etc.
Comparing with the original subdomain integral,
one finds that the transformation formula is an efficient
and accurate approach to compute the integral (2-4-1).
EXAMPLE 2-10-1:
In this example, one assumes that the source
function b(x) in Poisson's equation (2-2-1) is the form
of
b(x) x " n (2-10-1)
74
and can easily find out the theoretical solution as
follows
l:iafxf 1 x~ u = - 12 1
(2-10-2)
and
dU x~
a1xf 1
dn - 3a1 (2-10-3)
One considers that the geometry of domain ~ (Fig.2-10-1) is a
circle centred at the origin of coordinates with radius
a 1 = 10. The prescribed boundary conditions are
Dirichlet type over all the boundary of the problem.
The given values for boundary conditions at all the mesh
points are obtained from expression (2-10-2).
The source function then can be expressed by
Fourier series as
b(x)
Results at boundary points and internal points
are listed in Table 2-10-1. From this table, one can
find:
A. The errors, which correspond to mesh A and N=5, can
be reduced using the finer mesh B with the same
number of terms N=5 in the Fourier series expression.
B. With the same mesh B, the results are not apparently
different between N=5 and N=10.
C. Some results of integral (2-4-1), which are computed
with formula (2-6-9), are listed in Table 2-10-2.
From them, it is clear then when the mesh is
refined from A to B (for there N=5), the results
change about 3%,but using the same mesh B
75
• internal points
x Itesh points
4·( 4
Figure 2-10-1 Geometry and meshes in Example 2-10-1
76
Table 2-10-1 The Results of Example 2-10-1
Fluxes at the boundary points
Mesh N 1 2 3 4
A 5 675.41 571.33 222.45 -22.29 1.31% 1. 57% -2.94% -
B 5 668.85 564.83 226.40 -8.30 0.33% 0.41% -1.21% -
B 10 668.64 564.50 226.67 -8.52 0.30% 0.36% -1. 09% -
Exact 667.67 562.50 229.17 0
Potential values at the internal points
Mesh N 5 6 7 8 9 10
A 5 -224.93 918.42 1974.17 103.80 351. 31 746.11 5.44% 1.13% 0.67% 33.20% 13.60% 8.66%
B 5 240.48 914.18 1964.85 87.71 321.13 701.33 3.41% 0.66% 0.19% 12.56% 3.84% 2.14%
B 10 240.24 914.16 1965.01 87.49 321.10 701. 55 3.31% 0.65% 0.20% 12.28% 3.80% 2.17%
Exact 232.54 908.20 1861. 06 77.92 309.25 686.65
77
Table 2-10-2 The Results of Integral (2-4-1) in
Example 2-10-1
At boundary points
Mesh N 1
A 5 -53684.75
B 5 -55072.12
B 10 -55098.05
* At internal points
Mesh N 5
2
-53100.58
-54435.98
-54461.98
6 7
A 5 -0.398901 -0.429945 -0.476575
B 5 - 0.411500 -0.442669 -0.489507
B 10 -0.411777 -0.442952 -0.489796
3 4
-51936.06 -51354.02
-53162.21 -52526.11
-53194.01 -52560.28
8 9 10
-0.396652 -0.433191 -0.463712
-0.409169 -0.434585 -0.475912
-0.409450 -0.434878 -0.476236
* All the figures have to be multiplied by 10 5
78
and increasing the number of terms from N=5 to
N=10, the results are improved only about 0.05%.
D. From above A, Band C, one can see that the errors
are mainly due to use of too coarse meshes.
EXAMPLE 2-10-2:
In order to compare results, let us set the potential
function u(x) in the form of
2 2 lIx1 u(x) = (2.) cos(-)
11 2a x c: n (2-10-5)
This function satisfies Poisson's equation with the
source function b(x) as
(2-10-6)
so that, the flux function on the boundary is
x <:: f (2-10-7)
where n 1 is the first component of direction cosine
on the boundary f, i.e. n 1 = cos(n,x1 ) and n is outward
unit normal of f •
. The representation in Fourier series of function
(2-10-6) is
b (x) 2 11
1 N (_l)n nTIx 1 11 L 1 cos (-a-)
n=l '4 - n 2
(2-10-8)
The configuration of domain n of the problem under
consideration, mesh points and internal points are shown
79
in Fig.2-l0-2. The expansion domain for the Fourier
series, nf is [-a l ~ xl ~ aI' -a2 ~ x 2 ~ a21 and a l = 10,
a 2 = 20.
The boundary conditions at all mesh points are
prescribed as potential values from Eq. (2-10-5). The
results of potential at internal points are listed in
Table 2-10-3. One can see accuracy of results is
sufficiently good even for only N=2. They improve
rapidly from N=2 to N=5 especially at poi~ 3~
The values of integral (2-4-1), which are computed
by formula (2-6-9), are listed in Table 2-10-3 as well.
Comparing the accuracies of potential and integrals
in Table 2-10-3, one can find that the errors in potentials
at internal points are caused by the coarse meshes. The
relative errors of the values of integral (2-4-1) are
improved by more than thirty times from N=2 to N=5.
But the relative erors of the results of potential at the
internal points are only improved two or three times
from N=2 to N=5 at internal points I' and 2'.
EXAMPLE 2-10-3:
This example is for the case for which the source
function b(x) exists only partly in the domain n of the
problem under consideration. One supposes that:
b(x) = 1: (2-10-9) elsewhere
and the boundary conditions are homogeneous, i.e.
u o on r
80
15
3~ • internal points
10 l' 2'
x zresh points . .
5 2 3
0~~2~------~7----------
Figure 2-10-2 Geometry and mesh in Example 2-10-2
81
Table 2-10-3 Results of Example 2-10-2
The Values of Potential at the Internal Points
Internal Point N l' 2' 3'
2 34.020 27.185 30.858 0.088% 0.245% 0.130%
5 33.9BO 27.15B 30.B16 -0.029% +0.146% -0.006%
Exact 33.990 27.119 30.B1B
The Values of Integral (2-4-1) at the Mesh Points
Mesh Point
N 1 2 3
2 60.3416 55.4421 57.0107 0.7018% 0.6929% 0.9480%
5 59.9369 55.0704 56.5014 0.0265% +0.0162% 0.0462%
Exact 59.9211 55.0612 56.4753
82
The geometry of the domain and mesh points are
shown in Fig. (2-10-3).
As mentioned in section 2-9, one can extend the
function b(x) as an even function, so that the expression
of Fourier series of b(x) can be represented as follows
b(x) = 1 everywhere
Hence, in this case, the integral (2-4-1) is only
integrated over the domain Qb as shown on the right
hand side in Eq. (2-9-2) and the boundary of integration
in formula (2-6-9) is r b • Dividing rb into boundary
elements, one can obtain the results of (2-6-9) for
certain source points. In the numerical practice one
finds that only four boundary elements, each of which
is one side of the rectangular domain Qb' are sufficient
to obtain the exact results in the numerical sense. The
values of integral (2-4-1) for various mesh points are
listed in column 'exact'in Table 2-10-4. The values of
potential, which are approximated numerically, are listed
in the same column as 'exact' too. But the errors of
potential values are due to computing the first and
second integrals on the right-hand side of Eq. (2-2-9)
numerically, and are not due to computing the third
integral in Eq.(2-2-9) with the transformation formula
(2-6-9) •
One can also expand function b(x) as a Fourier
series in the domain Q, i.e. the expansion domain Qf
coincides with Q.
83
I I I I I I : I I
14 __ ~ __ fL-
o r. I b b I
--- /; ---f' 11--I r I I I I I I 0 I
-- _fL_f~-F-I I I I I I
• internal points
x rresh points
I I I
Figure 2-10-3 Geometry and mesh in Example 2-10-3
84
Table 2-10-4 Results of Example 2-10-3
The results of integral (2-4-1) at the boundary points
Number N = 2 N = 5 Exact of points
1,4 -6.8559508 -7.1768198 -7.0897632
-3.30% 1. 23%
2,3 -6.6487422 -6.9796948 -6.8836842
-3.41% 1. 39%
5 -7.1871481 -7.5382462 -7.4431710
-3.44% 1. 28%
6,9 -6.0212121 -6.3417339 -6.2561951
-3.76% 1. 37%
7,8 -4.9700890 -5.2666259 -5.1861911
-4.17% 1. 55%
10,13,16 -5.4122548 -5.5643382 -5.5176592
-1.91% 0.85%
11,12 -4.9976969 -4.9998822 -4.9908090
0.14% 0.18%
14,15 -3.5861070 -3.6087379 -3.5955679
-0.26% 0.37%
The results of potential at the internal points
Number N = 2 N = 5 Exact of points
17 -0.0823950 -0.1218647 -0.1121903
-26.56% 8.62%
18 -0.1125917 -0.1588545 -0.1474590
-23.65% 7.73%
19 -0.0697340 -0.1012834 -0.0936967
-25.57% 8.10%
20 -0.2401351 -0.2758389 -0.2670178
-10.07% 3.30%
21 -0.2331309 -0.2518989 -0.2470396
-5.63% 1.97%
85
In this case, one obtains more complex Fourier
coefficients as follows
ko
kl n
k 2 n
k 3 n
k~ n
1 d 3 ) -- (d - d ) (d -a 1a 2 2 1 4
(d4-d3) ~ nTTd 2 sin (--) nTTd 1 ]
- sin(--)
-
-
2nTTa 2
(d4-d3 )
2nTTa2
(d2-d1 )
2nTTa1
(d2-d1 )
2nna 1
1 n2nm
1 - n2nm
1
a 1 a 1 .
G nTTd 2 cos (--) a 1
nTTd 1 ] - cos(--) a 1
sin ( __ 4) -~ nTTd a 2
nTTd 3 ] sin (--) a 2
~ nnd4 nnd 3 J cos (--) - cos (--) a 2 a 2
mnd 3 ] - sin(--) a 2
mnd 3 ] - cos(--) a 2
mTTd 3 J - sinC·-) a 2
= -- ~ nnd2 cos (--) -a 1
nTTd 1 l r mTTd 4 cos (-a-) ICOS (-a-) TTnm 1 .. ' 2
mTTd 3 ] - cos(--) a 2 .
(2-10-10)
where a 1 6, a 2 = 10
d 1 x(2), d 2 = x(3)
d 3 y(15), d 4 = y(14)
86
The results obtained integrating along the boundary
r with formula (2-6-9), are shown in Table 2-10-4.
By comparison, the first approach, which is integration
along rb , is much more efficient and accurate than the
latter one, obviously.
EXAMPLE 2-10-4:
This example analyses the torsion of a shaft with
keys as shown in Fig. 2-10-4. The governing equation of
this problem is Poisson's equation. By using a dimension-
less torque function, u, the value of b(x) can be set
to two. The shear stresses in a section are expressed
by
au T ~ zy
(2-10-11)
au T -ay zx
Because of symmetry, only a quarter of the shaft
is calculated.
Obviously, the simple&way is to extend b(x) as
an even function and then to expand b(x) with only ko
equal to two in the Fourier coefficients, the others
are all zero. The results are shown as case A in Table
2-10-5.
.I
I I
I I
D
/ I/,'"
I /'".'" / /'/
I ~/ I~
MESH2
B A
\ I \ I
\ 1 'I
l
MESH 3
\ I \1 \I \I
87
14
mash points
Figure 2-10-4 Geometry and meshes in Example 2-10-4
88
Table 2-10-5 The Shear Stresses in Example 2-10-4
Item Mesh N A B B' C D
BEASY I a zx 110.0 301. 0 180.1 78.4 95.2
a 0 0 -310.4 -111. 8 -55.1 zy
II a zx 112.8 253.6 195.3 75.2 88.3
a 0 0 -338.2 -107.4 -51. 0 A zy
III a 104.8 zx
204.8 156.7 63.6 91.8
a 0 0 zy -271. 4 -90.8 -53.0
III 5 a 102.1 zx 198.9 152.3 61.8 87.6
a 0 0 -263.7 -88.2 -50.6 zy
B III 10 a 103.6 200.3 153.3 62.2 89.9 zx a 0 0 -265.5 -88.8 -51. 9 zy
III 20 a 104.5 202.4 154.9 62.7 90.8 zx a 0 0 -268.3 -89.6 -52.4 zy
III 5 a 98.9 zx 174.2 113.2 53.4 80.8
a 0 0 zy -196.1 -76.3 -46.3
c III 10 a zx 99.4 191. 6 146.6 59.5 87.4
a 0 0 zy -254.0 -85.0 -50.5
III 20 a 102.0 zx 197.0 151. 3 61.3 89.5
a 0 0 zy -262.1 -87.6 -51.7
89
By comparison, b(x) is extended as an odd function
as well. There are two extensions for function b(x)
as follows:
One of them is
x > 0 b(x) (2-10-12)
x < 0
Then after expanding the Fourier coefficients are
k 0 0
kR. 0 n
j 8 k 2 nlT n
0
kR. 0 nm
Another is
b (x) -j' - 2
R. 1,3,4
n 1,3,5, ••• , (2N+1)
n 2,4, •••
x.y > 0
x.y < 0 (2-10-13)
and the corresponding Fourier coefficients are
k 0 0
kR. 0 n kR. 0 R. nm 1,2,3
j" k4 ~mlT2 nm
n, m = 1, 3, ••• , (2N+ 1)
n or m = 2,4, •••
90
In the first one, b(x) is discontinuous on the
axis y in Eq. (2-10-12), but in Eq. (2-10-13) b(x) is
discontinuous on both axes x and y. The corresponding
results are listed as cases Band C in Table 2-10-5.
The results in Table 2-10-5 are compared with
solutions found using the BEASY code [22] with quadratic
boundary elements. Obviously, from the shear stresses
at the singular points Band B', one can find the results
of case A are excellent. In contrast, in cases Band C
one can not expect very accurate solutions. They also
require much more CPU times than that in case A because
of the discontinuity of b(x) in cases Band C.
2.11 THE TRANSFORMATION FORMULA FOR 3-D POISSON'S
EQUATION
In order to derive the formulations, which trans-
form the domain integral into boundary ones in 3-D,
a similar procedure as in section 2-6 can be carried
out.
Suppose there the function b(x) in Eq. (2-4-1) is
a piecewise continuous function in a 3-D region
After periodic
extension of the function b (x) , this can be expanded
as a convergent Fourier series:
b(x) b(xl'x2,x3)
6 kR.fR. +
12 kR. fR. kofo + L L L L L +
R.=l n n R.=l nm nm n n m
8 + L L L L kR. fR.
i=ln m p nmp nmp (2-11-1)
91
where f 1 a
f I nlfx 1 nlfx 1
cos (-a-) f;! sin (--) n 1 n a 1
nlfx 2 f4
nlfx 2 f3 cos (-a-) sin (--) n 2 n a 2
fS nlfx 3
f6 nlfx 3 cos (--) sin (--) n a 3 n a 3
fl flf3 f;! flf4 nm n m nm n m
f3 f2f3 f4 f;!f4 nm n m nm n m
fS f3fs f6 f3f6 nm n m nm n m
e t4f s f8 f4f6 (2-11-2) nm n m nm n m
f9 fSfl flo f s t2 nm n m nm n m
f I I f6fl f 12 f6f2 nm n m nm n m
fl e f3fS nmp n m p
f;! flf3f6 nmp n m p
f3 flf4fS nmp n m p
e flf4f6 nmp n m p
fS f;!f3fS nmp n m p
f6 f;!f3f6 nmp n m p
e f;!ef s nmp n m p
f6 f;!f4f6 nmp n m p
92
and the Fourier coefficients are as follows:
ko 1 III b(x)dll 8a 1a 2a 3
II
kR. 1 III b (x) f~(X)dll n 4a 1a 2a 3 II
kR. 1 fJI b(x) f~m(X)dll (2-11-3) nm 2a1a 2a 3 II
kR. 1 III b(x) f~mp(X)dll nmp a 1a 2a 3 II
Now we define some functions in the following way:
R. u nmp
a fR. _(-1.)2
nn n a
fR. _(~) 2 nn n
a fR. _ (-2) 2
nn n
fR. nm
n2 Ull.) 2 + a 1
fR. nm
n2 Ull.) 2 + a 2
fR. nm
n2 [(ll.) 2 + a 3
R. 1,2
R. 3,4
R. 5,6
R. 1,2,3,4 (2-11-4) (~) 2J a 2
R. 5,6,7,8 (~) 21 a 3
R. 9,10,11,12 (~)2J a 1
where x10 ' x 20 ' x 30 are constants, which are the coordinates
of reference point xo'
93
f . iii The above unct~omuo,un' unm and u nmp satisfy
the following equations respectively:
\]lU f 0 0
\]lU~ fi n
\]lU i fi = nm nm
(2-11-5)
\]lU i fi nmp nmp
Then, defining
a go an u
0
i a fi gn an n
i a fi gnm = an nm
(2-11-6)
i a fi gnmp = an nmp
and
1,2
3,4
5,6
(2-11-7)
94
k~ nm
~ 1,2,3,4 1T2 [(..!!..) 2 + (~) 21
a 1 a 2 _
~ k~
nm ~ 5,6,7,8 Snm [ n 2 (~) 2] 1T2 (-) +
a 2 a 3
k~ nm
~ 9,10,11,12 [ n 2 (~) 2J 1T2 (-) +
a 3 a 1
k~ nmp
and using the second Green's identity, we obtain the
transformation formula for 3-~ as follows:
I (E,;) fU*bdr.l
r.l
6 + L L
~=1 n
12
df - f q* Uo df - c Uo 1 +
f
f ~
df - f q* f~ dr u*gn - c n f f
f~ n
+ L L I f ~
u*gnm df - f q* f~ df - c ~=1 n m f
nm f
8 ~
f * ~ f f~ + I I I L Snmp u gnmp df - q* df ~=1 n m p nmp
f f
- c f~ nmp
1 +
f~ 1 + nm
(2-11-8)
where c
u*
95
c(~) is shown in Eq.(2-2-10) and
1 411r q* 1 ar
- 411r2 an
The coefficient c(~)depends on the position of the
source point ~.
The numerical techniques for computing formula
(2-11-8) are similar to those mentioned in section 2-7.
2-12 APPLICATIONS IN TIME-DEPENDENT PROBLEMS
Not only does the source function term produce domain
integrals in potential problems, but also time-dependent
or non-linear terms can cause this type of domain integral
in boundary'e1ement formulations. A similar method, which
transforms domain into boundary integrals for the source
term, can be applied to these problems. In this section
time-dependent problems are discussed. After recasting
the partial differential equation into a boundary integral
form the similarities between the previous and the present
problems are pointed out.
Diffusion problems are governed by the following
equation:
V2 u(x,t) a at u(x, t) x (: g
with boundary conditions of the two types:
Dirichlet B.C. u(x,t)
Neumann B.C. q(x,t)
= u(x,t)
a an u(x,t)
(2-12-1)
q(x,t)
(2-12-2)
96
and initial conditions
u(x,O) = uo(x) (2-12-3)
where u, q and Uo are known functions.
To recast Eq. (2-12-1) with boundary and initial
conditions (2-12-2) and (2-12-3) into a boundary integral
equation, it is usual to combine with finite differences.
The derivative of function u with respect to time t,
d at u , can be expressed in finite difference form as
follows
d at u(x,t) 1 ~t [u(x,t + ~t) - u(x,t)] (2-12-4)
where ~t is a sufficiently small time step. Eq. (2-12-1)
can then be rewritten as
1 V2 U (X,t + ~t) - ~t u(x,t + ~t) 1
- ~t u(x,t)
(2-12-5)
Now the fundamental solution of the above Helmholz
equation, u*, is introduced [40,41]:
- ~(cx) (2-12-6)
u* for 2-D
(2-12-7)
where Ki is the modified Bessel function of the second
kind of order i.
97
Applying the procedures described in section 2-2,
the differential equation (2-12-5) can be rewritten as
a boundary integral equation, i.e.
c(t;) u(t;) J u*q dr -
r J q*u dr + It J u*u dn r n
(2-12-8)
where the unknowns u and q are at time step t+~t, and u
is a known function of u(x,t) over all the domain n from
previous time step t.
Using the finite difference method with the time-
independent fundamental solution (2-12-7), Eq. (2-12-8)
can be solved in series of time steps. To obtain suffic-
iently accurate results, small 6t time steps must be used
and after every step the solutions at a sufficient number
of internal points must be computed to find u which can
then be expressed as a Fourier series. For these
reasons this method usually requires large CPU times.
Instead of using the time-independent fundamental
solution, one can apply the time-dependent fundamental
solution u* in the form of [42,43]
u* 1 d/2 exp [- 41r (~~l)l H (t-l) 411 (t-l) ]
(2-12-9)
where d is the number of spatial dimensions of the
problem. H is Heaviside function. The function u*
possesses the following properties:
98
a 1J 2 u*(E;,x,t-T) - at u*(E;,x,t-T) - ll(E;,x)ll(t,T)
(2-12-10)
After applying the fundamental solution (2-12-9)
one can rewrite Eq. (2-12-1) as a boundary.integral
equation, i.e.
c(t,)u(E;)
t
J J u*q dfdT -
o f
t
J J q*u dfdT + o f
(2-12-11)
Note that in the domain integral on the right-hand side
of the above equation there is only the known function
Uo from the initial condition. Comparing with Eq. (2-2-9)
and setting the notation b(x) = -uo(x) , one finds that
Eq. (2-12-11) is of the same type as Eq. (2-2-9) excepting
the different fundamental solutions u*. The Fourier
expansion (2-6-1) can be used now after setting the
fundation b(x) = -uo(x). Because of the different
fundamental solutions, Eqs. (2-6-5) must be changed as
follows
1J 2 u a f -at u 0 0 0
(2-12-12)
1J 2 u a fll. -at u n n n
1J 2 u lI. a lI. fll. nm -at u nm nm
Consequently, the transformation formula (2-6-9)
or (2-6-14) can be used in the same form for the domain
integral in Eq. (2-12-11) .
99
The same procedure can be applied for hyperbolic
time-dependent problems.
2-13 APPLICATION IN NON-LINEAR PROBLEMS
Usually non-linear equations are solved numerically
by applying an iterative procedure: substituting the
results of a previous step into the non-linear term, which
can then be considered as a known function. Thus, for
each step only a linear problem is solved.
Suppose a non-linear equation is given by
L(u) + N(u, .•. ) R(x) (2-13-1)
where L is a linear operator, function N is a non-linear
function of unknown function u and its derivatives, R
is a function of the spatial coordinates. Using
iterative procedures, Eq. (2-13-1) can be expressed as
follows for the s step.
(2-13-2)
where Us and u s _ 1 are the solutions for the sand s-l
steps, the former are unknown and the latter are known
from the previous s-l step. In this case it is possible
to solve Eq.(2-13-2) using boundary element methods
with the fundamental solution u*, which satisfies the
linear operator:
L(u*) + lI(Cx) 0
After having defined b(x) = R-N, one can transform
the differential Eq. (2-13-1) with its boundary conditions
100
into a boundary integral equation with a domain integral
term as
1(0 J u*b(x)dl1 11
J u*[R(x) - Njdl1 11
(2-13-3)
This integral can be transformed into boundary integrals
as explained before.
CHAPTER 3 LINEAR ELASTOSTATICS
3-1 INTRODUCTION
In this chapter the bodies under consideration are
assumed to be homogeneous and isotropic. It is also
assumed that their behaviour can be analysed using linear
theory of elasticity which implies that:
A. Within certain limits the behaviour of the material
can be represented by a linear relationship between
stresses and strains.
B. The relationship between displacements and strains
is also linear, that is second or higher order
derivatives of displacements with respect to spatial
coordinates can be neglected.
A short review of the basic linear theory of elasto
statics is presented first, including definitions of
stresses and strains, relationship between stresses and
strains, stress equilibrium equations, compatibility
equations and boundary conditions. In order to derive
the basic boundary integral equation of elastostatics,
the Somigliana identity and the fundamental solutions of
elasticity are also presented. Then the fundamental
concepts of BEM in elasticity are described, using the
2-D case for simplicity.
The main concepts are introduced regarding computation
of body forces integrals, following the same idea as
described in the previous chapter, i.e. using the trans
formation formula, which evaluates the body force domain
integrals in terms of only boundary integrals. The
102
numerical implementation of this technique is discussed
and numerical results presented. From these solutions
it is clear that the transformation formula is reliable
for computing the body force integrals.
3-2 BASIC RELATIONSHIPS OF ELASTICITY
Let us define the state of stress at a point as a
stress tensor
o = (0 .. ) ~J
(3-2-1)
In the absence of body moments, the following relationships
apply for the tensor components
o .. ~J
o .. J~
(3-2-2)
Introducing the distributed internal body force
vector b = (b i ), the following equilibrium equations over
the body can be obtained
0 ... + b). ~J,~
o (3-2-3)
When the traction vector ~ acting at a point on the
boundary of the body is represented by ~ = (Pi)' the
following relationship holds over the boundary
p.=o .. n. ~ J~ )
(3-2-4)
where (n j ) represents the direction cosines of the outward
normal vector on the boundary.
A body is displaced from its original place or
configuration under the action of forces. Let (Xi) denote
the initial position of a point Q, (x.+u.) its deformed ~ ~
position i then (ui ) represents the displacement
103
vector of point Q (Fig.3-2-1). According to the linear
theory, the strains can be represented by the following
Cauchy infinitesimal strain tensor:
Eo 0 = ~(uo 0 + Uo 0)
1J 1,J J,1 (3-2-5)
To solve for displacements from Eq. (3-2-5) uniquely,
the compatibility equations of strains must be satisfied:
Eo 0 k. + Ek • 0 0 - Eo k o. - Eo. 0 k 1J, '" "',1] 1 ,]'" ]",,1
o (3-2-6)
All the above relationships are independent of
material properties. If the material is linear and
isotropic, the linear Hooke's law relating stress and
strain tensors can be stated in the form of
(3-2-7)
or inversely
Eo 0
1] (3-2-8)
where v is Poisson's ratio and G is the shear modulus.
~ and ~ are Lame's constants. Notation E is Young's
modulus.
For an isotropic material only two of these constants
are independent and can be related to one another by the
following relationships
Ev (l+v) (l-2v)
G E
2 (1+v)
2Gv 1-2v
(3-2-9)
104
Figure 3-2-1 Displacement
105
Hence in linear elasticity, the variables are (ui ),
(a .. ) and (E: •. ) and the equations are Eqs. (3-2-3), (3-2-5) ~J ~J
and (3-2-7) (or (3-2-8)). These equations can be further
manipulated and the equilibrium equation can be written
as the Navier equation, in which the variables are the
displacement components, i.e.
G + _G_ b uj,kk 1-2v Uk,kj + j o (3-2-10)
Using Eqs. (3-2-4), (3-2-5) and (3-2-7), one can also
obtain the traction boundary condition on the boundary
as follows
2Gv 1- 2 " uk, k n; + G (u. . + u. .) n. = p.
v ~ ~,J J,~ J ~ (3-2-11)
For the case of elastostatics, the boundary conditions
can be expressed as the fdlowing two types
The displacement B.C. u j on r 1 (3-2-12)
and the traction B.C.
(3-2-13)
where u i and Pi are prescribed displacements and tractions
and the total boundary is r = r 1 + r 2 (Fig.3-2-2).
3-3 FUNDAMENTAL SOLUTION FOR ELASTOSTATICS
The fundamental solutioffi for displacement u~. are ~J
the solutions of the Navier Eqs. (3-2-10) for an unbounded
domain when applying particular concentrated body forces
as follows
106
Figure 3-2-2 Boundary conditions
107
(3-3-1)
In this case, the fundamental solutions Uij satisfy
o (3-3-2)
The body force components b! in Eq. (3-3-1) correspond to
a positive unit load applied at a point ~ in the direction
of the unit vector e i • Here u~.(~,X) represent the ~J ::. .
displacements in j direction at the point x corresponding
to a unit force bi applied at point ~ in the i direction.
After substituting the fundamental solutions of
displacements u*ij into Eq.(3-2-11), the fundamental
traction components prj can be found. Then from the
definition of strain tensor in Eq. (3-2-5) and the
relationship (3-2-7) the corresponding fundamental
solutions of strain (E'!j) and stress (Gij) tensor components
are obtained. The fundamental solut~ons due to Kelvin
can be represented as follows
1 16n(1-v)Gr [(3-4v)oij + r,ir,j]
for 3-D (3-3-3)
1 8n(1-v)G [(3-4v)ln r 0ij - r,ir,j]
1
for 2-D plane strain
(3-3-4)
C3r { [(l-2v) 0 .. + B r . r . h-- -4an(1-v)ra ~J,1 ,J ~n
- (1-2v)(r .n. - r .n.)]} ,1 J ,J 1
(3-3-5)
108
and
[(1-2\1)(r k O" + r .o'k) 8aH(1-\l)Gra , 1J ,J 1
1
-r 'O'k + B r .r .r k1 ,1 J ,1 ,J , (3-3-6)
* O'k'(Cx) J 1 [(1-2\1) (r k O .. + r .ok'
4aH(1-\l)r a , 1J ,J 1 1
- r . 0J'k) + B r .r .r k1 ,1 ,1 ,J ,
(3-3-7)
where r
for 3-D and 2-D plane strain respectively, B a+1.
The notation £jki and 0jki represent the components of
strain £jk and stress 0jk at point x due to a unit point
load applied at s in i direction:
£jk £jki e. 1
(3-3-8)
°jk °jki e. 1
Note, the plane strain expressions are valid for plane
stress only if \I is replaced by \1/(1+\1).
3-4 SOMIGLIANA IDENTITY
In order to derive the boundary integral equation
of linear elasticity, the weighted residual method can
be used [1,401, although one can use the Somigliana
identity as will be shown here.
Consider a body defined by ~ with its boundary r
(Fig. 3-4-1), which is in equilibrium under some given
loads and displacements. This state is represented by
the variables 0 .. , 1J
109
Figure 3-4-1 Elastic body with region n and boundary r
110
In addition, suppose there is another domain Q*
with boundary r*, which contains the body Q+r of the
problem under consideration (Fig. 3-4-2) and with the
same properties as Q. As before, this domain is also
in equilibrium with variables a~" E~, , u~, p~ and b~. ~) ~) 1. ~ 1.
Now, the well known reciprocity principle, which can
be inferred by simple symmetry of the constitutive
equation of stresses and strains, is introduced, i.e.
(3-4-1)
After integrating by parts both sides of Eq. (3-4-1)
and considering the equilibrium equations under the Q
integrals, one obtains
f bk u~ dQ + f Pk u~ dr n r
(3-4-2)
which corresponds to Betti's second reciprocal work
theorem.
Eq. (3-4-2) can be further modified by assuming uj'
a!)" E~" p~ are fundamental solutions which are caused ... .1.) )
by bI = 6(~,x)ei' where ~ and x belong to Q*. Therefore,
if ~ c Q too, the first integral in Eq. (3-4-2) becomes
(3-4-3)
Furthermore, considering the point loads bk are
applied separately in each direction, the corresponding
displacements and tractions can be rewritten in the form
of
111
Figure 3-4-2 General region Q* + r* containing the body Q + r
112
(3-4-4)
p~ = p~. (Cx)e. J 1.J 1.
where Uij(~'x) and pij(~'x) represent the results in
j direction at the point x corresponding to a unit point
force applied at point ~ in the i direction. After that,
substituting Eqs. (3-4-3) and (3-4-4) into Eq. (3-4-2) ,
Eq. (3-4-2) can be rewritten in function of components
J ulj (ex) Pj (x) dr (x)
r - J Pij(~,x)Uj(x)dr(x) +
r
+ J u.*. (~,x)b. (x)d~(x) 1.J J
(3-4-5)
~
The above Eq. (3-4-5) is called Somigliana identity.
3-5 THE BOUNDARY INTEGRAL EQUATIONS OF ELASTOSTATICS
This section considers the derivation of the boundary
integral equation applying the Kelvin fundamental solutions
and Somigliana identity presented in the previous two
sections.
As in the case of potential problems, Eq. (3-4-5) must
be modified for when the point ~ is on the boundary r.
For this purpose, the boundary r of domain ~ can be
augmented by a small region r£ (Fig.3-5-1). Therefore,
the second integral on the right hand side of Eq. (3-4-5) can
be written as
and
113
r
Figure 3-5-1 Augmented boundary
J Pij(~,X)Uj(x)dr r
+ lim J £-+0 r£
lim £-+0
(3-5-1)
p~.(~,x)dr ~J
After that, the first integral on the right-hand side
of Eq.(3-5-1) is evaluated in the Cauchy principal
value sense [44], the existence of which can be proved
if u(x) satisfies a Holder condition [45] at the point
~ in the form
114
where B and a are positive constants.
The first integral on the right-hand side of Eq.
(3-4-5) has no contribution from f£ since
(3-5-2)
Therefore, after defining Cij(~) in the form
(3-5-3)
one obtains the boundary integral equation of linear
elastici ty, i. e.
c .. (Ou.(O 1.) ) J u~. (~,x)p. (x)df(x) -
1.) ) f
- J p~. (I;,x)u. (x)df(x) 1.) )
+ J uij(I;,X)bj(x)dn(x) n f
(3-5-4)
It relates the surface displacements and surface tractions
With the boundary conditions, the unknown displacements
and tractions on the boundary can be solved numerically
by means of BEM.
Once the displacements and tractions on the boundary
f are solved, Eq. (3-5-4) can also be used to find out the
displacements at any internal points I; inside the domain n.
o
115
If the stress state at internal points is required,
they can be obtained by:
A. Taking derivatives of Eq. (3-5-4) with respect to
the coordinates of ~ to obtain the strain tensor.
B. Substituting the strains into Hooke's law (3-2-7).
The expression of stresses at internal points can be
reduced to
J Uijk(~,x)pk(x)dr(x) -
r
- J pijk(~,x)uk(x)dr(x) + J Uijk(~,x)bk(x)dn(x) r n
(3-5-5)
The tensor terms Uijk and pijk corresponding to
Kelvin fundamental solutions are in the form of
- 0* ijk (3-5-6)
G ar --"'----::-B {B an [( 1-2\1)0 .. r k 2a~(1-\l)r ~J ,
+ \I(o;k r . + 0J'k r .) - y r .r .r kl + L,J ,~ ,~ ,J ,
+ B\I(n.r .r k + n.r .r k) + (1-2\1) (Bnkr .r . + ~ ,J, J ,~ , ,~ ,J
where a = 2 and a = 1 for 3-D and 2-D plane strain
respectively. B = a+1 and y = a+3. For 2-D plane
stress problems, as before, it is necessary to use
116
-v v/(1+v) instead of v in all above equations.
3-6 THE BOUNDARY ELEMENT METHOD IN ELASTICITY
The boundary integral equation with boundary con-
ditions can be solved numerically in potential problems
as shown for Eq. (2-2-9). The same approach can be used
in elastostatics.
For the discretization of Eq.(3-5-4), the boundary
r is approximated by a certain number of elements. For
2-D linear elements, the edges of each element are taken
as nodes, at which the values of variables are considered.
In this case, the displacement 0' traction and body force
can be expressed with their components:
u = {::) {::) b
{::) (3-6-1)
and the fundamental solutions can be written as matrices:
p* u*
For linear elements, linear interpolation functions are
used
~1 ~(1-n) (3-6-3)
~2 ~(1+n)
Then, the variables ~ and p can be expressed with the
values at nodes and interpolation functions ~1 ' ~2 :
<P 1u j1 + <P 2Uj2
<P 1Pj1 + <P 2Pj2
117
(3-6-4)
where u j1 ' u j2 and Pj1, Pj2 are displacements and
tractions at nodes 1 and 2 of the element j. After that,
in a similar way to that in Section 2-3 - by considering
the prescribed boundary conditions one can obtain a set
of linear algebraic equations, which contains all the
unknown variables at the nodes. These unknown variables
can be solved from this set of linear algebraic equations.
Then, if the displacements and stresses at the internal
points are required, they can be computed from Eqs. (3-5-4)
and (3-5-5) numerically.
During the above numerical procedures, domain
integrals, which exist in the last terms in Eqs. (3-5-4) and
(3-5-5), are usually computed by subdividing the original
domain into subdomains or cells and using numerical
quadrature formulae over subdomains. The approach however
is expensive as it requires domain integrations, but the
following section presents instead a transformation
technique which can transform domain into boundary integrals.
3-7 BASIC FORMULATIONS FOR TRANSFORMING 2-D ELASTICITY
DOMAIN INTEGRALS TO THE BOUNDARY
In Eqs.(3-5-4) and (3-5-5) there exist domain
integral terms of the forms
J u~. (I;,x)b. (x)d!1(x) ~J J
(3-7-1)
!1 and
118
J u!jk(~,x)bk(x)dQ(x) Q
(3-7-2)
Using Somigliana identity, one derives in this
section the formulations to transform domain integrals
such as (3-7-1) and (3-7-2) into boundary integrals. The
Somigliana identity describes the relationship between
displacements u j ' tractions Pj and body forces b j
distributed on the boundary and in the body respectively.
Note, that in the Somigliana identity, the displacements
and tractions are present only in the boundary integrals
and the body forces are in the domain integrals. There-
fore if components of displacements and tractions, which
correspond to the body forces in Eq.(3-7-1), can be
found, the domain integral term in Eqs. (3-7-1) can be
transformed into boundary integrals. The general idea
is similar to that represented for potential problems,
for it is difficult to find the solutions u j and Pj
corresponding to arbitrary body force functions b .• J
The series expansion method will again be applied. Only
2-D problems are studied here for simplicity.
In order to expand the components of body forces
as series in integral (3-7-1), let us choose the 2~D
complete set of functions:
{ f f R. fR. I 0 = 1 4 o'n' nm'" , ••• "n,m 1, ••• , oo} (3-7-3)
where f o ' f~ and f~m are shown in Eqs.(2-6-2).
This is not only a complete set, which means that the body
force function can be expressed in a sufficiently precise
119
way, but also the actual displacements and tractions
corresponding to the functions (3-7-3) can be found easily.
Suppose the body force functions b1 (x) and b 2 (x) are
bounded and periodic functions of periods 1a1 and 2a2 in
xl and x 2 directions respectively. The components of
the body force can be expanded as Fourier series in the
region nf :2a1 x2a2 as follows
b. J
4 N M L L L
1=1 n=l m=l k~ fR,
Jnm nm
j = 1,2 (3-7-4)
The coefficients of the Fourier series are
kjO 1
J b. (x)dn 4a1a 2 J nf
k~ 1 J
R,
In 2a1a 2 bj(x)fn(x)dn
nf
(3-7-5)
k~ 1 J b. (x) fR, (x) dn
Jnm a 1a 2 J nm nf
In order to find out the solutions of displacements
and tractions corresponding to each term of the expansion
in Eq. (3-7-4), all the terms in Eq. (3-7-4) are classified
in three groups:
(A) Constants
(3-7-6)
120
(B) The function of sine or cosine
4 N kR, fR, b l I I
R,=I n=1 In n
(3-7-7) 4 N
kR, fR, b 2 I I R,=I n=1 2n n
(e) The multiple harmonics
4 N M L I I
R,=l n=l m=l
(3-7-8)
4 N M I I I
R,=I n=1 m=1
First, one takes body forces in Group (A) and, for
simplicity omit the subscript 0 from k lO and k 20 in the
following derivations. Notice that now kl and k2 are
known constants. Let us assume that the stress components
can be written in the form of
(3-7-9)
where c l ' c 2 ' c 3 and c 4 are unknown constants and x IO '
x 20 are arbitrary reference coordinates.
Substituting stresses (3-7-9) and body forces (3-7-6)
into the equilibrium equation (3-2-3),one obtains
121
(3-7-10)
Considering the relationships (3-2-8), the strains
can be rewritten as follows
l+v [(1-v)c 1 (x 1-x10 ) - VC 2 (X 2-X 20 )] £11 ~
l+v [-vc 1 (x 1-x 10 ) + (I-v) c 2 (x 2-x20 ) 1 (3-7-11) £22 ~
l+v [c 3 (x1-x10 ) + c 4 (x 2-x20 )] £12 £21 ~
It can be found that the above strains satisfy the 2-D
compatibility equation (3-2-6) as follows:
2 £12,12 (3-7-12)
Integrating the first two Eqs. (3-7-11), one finds
the following displacements
(3-7-13)
Fom Eqs. (3-7-13), one can obtain the shear stress
(3-7-14)
122
Comparing Eq. (3-7-14) with the last equation of Eqs.
(3-7-11), one has
(3-7-15)
From Eqs. (3-7-15) and (3-7-10), one can easily find all
the constants c 1 ,c 2 , c 3 and c 4 ' which are involved in the
expressions of stress (3-7-9), strain (3-7-11) and dis-
placement (3-7-13) components, i.e.
2 k1 c 1 - 2-v
2 k2 c 2 - 2-v
(3-7-16) v
k2 c 3 2-v
v k1 c 4 2-v
Furthermore, one can express the tractions as
(3-7-17)
123
where n 1 and n 2 are components of direction cosine of
the unit outward normal vector on the boundary r.
After obtaining displacements (3-7-13) and tractions
(3-7-17) corresponding to the constant body forces (3-7-6)
and considering Somigliana identity (3-4-5), one can
represent the constant domain integrals as boundary
integrals, i.e.
f u!j b j drI = - f u!j Pj df + f p!j u j df + CijU j n r r
(3-7-18)
where u j ' Pj are functions known from (3-7-13) and
(3-7-17). c .. are given in Eqs. (3-5-3). ~J
Let us consider the body forces from group (B),
because in this there are several terms. To find actual
displacements and tractions corresponding to these forces,
one rearranges the functions in Group (B) in the following
four kinds of combinations:
l. b 1 kl e In n
b 2 k~ f~ 2n n
2. b 1 k~ f.:l In n
b 2 k 1 fl 2n n
3. b 1 k 3 f3 In n (3-7-19)
b 2 k4 f4
2n n
4. b 1 k4 f4 In n
b 2 k 3 f3 2n n
124
Let us consider the body forces b l and b 2 from
(3-7-19 1) as an example and find out the corresponding
solutions for displacements and tractions:
(3-7-20)
where n' = nn/a 1 .
Here one uses notations kl and k2 instead of k~n and k~n
for simplicity.
In order to satisfy the equilibrium equation (3-2-3), one
can expect the stresses to be
(3-7-21)
Substituting the expressions for the body forces and
stresses - (3-7-20) and (3-7-21) - into the equilibrium
Eq. (3-2-3), one finds
o (3-7-22)
From Eq. (3-2-8), one can write the strains in the form of
where
d f2 2 n (3-7-23)
125
d l 1 [(H21.1)c l - Ac2l 41.1 (HI.I)
d 2 1
[-Ac 1 + (I..+21.1) c 2l (3-7-24) 41.1 (1..+1.1)
d 3 1
21.1 c 3
or
1:;) ['-' -v 0
] {:;l l+v I-v 0 (3-7-25) ~ -v
0 0 1
They must satisfy the compatibility equation (3-7-12),
hence one has
n '2 d 2 0
therefore
substituting the above relationship into the second
equation in (3-7-24), one obtains the following
relationship
(3-7-26)
Taking into consideration the relationships between
constants C l ' c 2 and c 3 Eqs. (3-7-22) and (3-7-26), one
can solve for these unknown constants provided the
determinant is not equal to zero, i.e.
126
n' 0 0
0 0 -n' + 0
-A 1.+21.1 0
that is n' 2 (;\+21.1) + 0
Hence n' nn,'a l is non-zero, then
A + 21.1 + 0
or
v + ~ and v + I (3-7-27)
The conditions (3-7-27) can be satisfied by every
compressible material. One now obtains the following
results
kl c i - i1'
v kl ( 3-7-28) c 2 - I-v i1'
k2 c 3 i1'
substituting these results into Eq. (3-7-25), one finds
the coefficients in Eqs. (3-7-23) as
(l+v) (1-2v) c i (l-v) E i1'
o (3-7-29)
so that the displacements can be written as
(1+\1) (1-2\1) k1 f I
(1+\1) E fiT 2 n
and tractions
1 fiT
1 -fiT
127
(3-7-30)
(3-7-31)
For the other combinations of body forces in Eqs.
(3-7-19 2. 3. and 4.), one just lists all the solutions
as shown in Table 3-7-1.
Consequently, from the combinations of body forces
b j in the forms of (3-7-7', one can find out the
corresponding displacements u j and tractions Pj' Then the
domain integrals (3-7-1) can be expressed in the form of
Eq. (3-7-18), in which u j and Pj must be found from the
corresponding columns in Table 3-7-1
Similarly from Group (C) in Eq. (3-7-8), one can
obtain stresses, strains, displacements and tractions
corresponding to certai~ combinations of body forces and
these solutions are listed in Table 3-7-2. The existence
condition of all these solutions is the same as for
(3-7-27).
The only reason for choosing the combinations of body
forces as given in Tables 3-7-1 and 3-7-2 is that it is
then easy to find out the solutions of displacements and
tractions corresponding to each combination of body forces.
Tab
le
3-7
-1
b.
b1
~
b2
all
02
2
01
2
o ..
~J
c1
c2
c3
-------
Th
e S
olu
tio
ns
co
rresp
on
din
g to
B
od
y
Fo
rces
in
Eq
s.
(3-7
-19
)
1 2
3 4
k fl
1
n k
f2
1 n
kf3
1
n k
fit
1 n
k f2
2
n k
fl
2 n
k e
2 n
k f3
2
n
c f2
1
n c
fl
1 n
C f
3
1 n
C f
it
1 n
c f2
2
n c
fl
2 n
C f
3
2 n
C f
it
2 n
c fl
3
n c
f2
3 n
C f
it
3 n
C f
3
3 n
-kIln
' k2
/n'
.-..Y..
... kin
"
1-\1
2
-_
\1-
kin
"
1-\1
2
-.-..
Y.....
kin
' 1-
\1
1 1~\I k
2/n
' k2
/n"
-k
2/n
"
k2
/n'
-k2/n
' -k
Iln
"
kin
"
1 ~ -
~
----
I
I'J
(XI
£ ..
~J
Ui
Pj
Tab
le
3-7
-1
(Co
nti
nu
ed
)
1
£1
1
d f2
1
n
£2
2
d f2
2
n
£1
2
d fl
3 n
(l+
v)
(1-2
v)k
1 d
l -
(l-v
)E n
'
d2
0
d3
(l+
v)
k2
E n
'
d u
l _
--.!.
f I
n'
n
2d
3 U
2 --
f2
n'
n
I P
I 11
' [-k
1 f~
nl +k2f~n21
-l~
v kl
f~n2
+k2f
~nl
P2
n
'
wh
ere
n
' n1
T a
1 n
" n1
T a
2
2
d fl
1
n
d fl
2
n
d f2
3
n
(l+
v)(
1-2
v)
kl
(l-v
)E n
'
0
(l+
v)k
2 -
E n
'
d --.!.
f2
n'
n
2d
3 -ri' f
~
kIf~nl-k2f~n2
n'
l~v
klf~n2-~2f~nl
n'
3
d f3
I
n
d f3
2
n
d f'
3 n
0
(l+
v)
(l-2
v)
k2
(l-v
)E
n"
(I+
V)k
l -
E
nil
2d
3 -~ f~
d 2f"
n"
I~v
k2f~
nl-k
If~n
2 n
"
k2f~
n2-k
lf~n
l n
"
4
d f'
I n
d f'
2 n
d f3
3
n
0
(l+
v)
(l-2
v)
k2
-
(l-v
) E
n"
(l+
v)
kl
E
nit
2d
3 fi"
"" f~
d
2 -
fiTt f~
-I~V
~2f~
nl+k
If~n
2 n
" ..
-k2f~n2+klf~nl
n"
,
N c.o
Tab
le
3-7
-2
Th
e S
olu
tio
ns
Co
rresp
on
din
g to
B
od
y
Fo
rces
in
Eg
s. (
3-7
-8)
bi
I
o ..
>
J
E ..
>J
u. J
Pj
wh
ere
n' b
1
b2
all
02
2
01
2
c1
c2
c3 Ell
E2
2
E12
d1
d2
d3
u1
u2
PI
P2
-
n~
a1
1
kl~
k f3
2
nm
clf~
m
c f'
2 nm
c f'
3 nm
n* [-
k1
+m
'c3
]
;.
[-k
2+
n'c
3]
m'e
1k
1+
n'e
2k2
(I-v
) (m
' 2
+n
,:2
) 2
d f'
1
nm
d f'
2 nm
d f'
3 nm
d1
-fiT
f~
d2
-ffiT
f~
m
c1
f:un
nl +c3
f~mn
2
c3fA
mnl+
c2f~
mn2
m'
mn
a2
e1
nr2
\)
-m
l2 (
I-v
) e
2
2 I
3 4
k f3
1
run
k f'
1 nm
k
f'
1 ru
n
k f>
2
nm
k f'
2
nm
k f'
2 nm
clf~
m c
f3
1 nm
c
f'
1 nm
c2f~
m c
f3
2 nm
c
f'
2n
m
c f'
3 nm
c
f'
3 nm
c
f'
3n
m
;,
[k1
+m
'c3
' n1
, [-
k1-m
'c3
' n1
, [k
1-m
'c3
'
ml,
[k2
+n
'c3
J m
l, [k
2-n
'c)J
,;
, [-
k2-
n'c
3,
m'p
. 1k
1+
n'e
2k
2 m
'elk
l-n
'e2
k2
-m
'e1
k1+
n'e
2k
2 (1
-\1
) (m
' 2
+n
ti)
a
(I-v
) (m,2+n'~).2
(1-v
) (m
' ~+n' i
) 2
I d
f'
1 ru
n d
f3
1 nm
d
f'
1 nm
I
d f'
2 nm
d
£3
2
nm
d f'
2 nm
d f'
3 nm
d
f'
3 nm
d
f3
3 nm
(l~vl
[(l-
vlc
1 _
c2
'
(l~Vl
[-c
1 +
(l
-vl
c2
,
l+v
~c3
~ f
3
n'
nm
_ ~ f
' n
' nm
d
1 f'
n'
nm
~ f
' m
' nm
~ f
' m
' nm
_
d2
£'
m'
nm
clfAmnl+c3f~n2
c1 f~mnl +c3f~n2
clf~mnl+c3f3n2
c3f~
mn2+
c2f~
mn2
c3f~mnl+c2f~mn2
c3f~
mnl+
c2f~
rnn2
.-
m,2
v -
n,2
{l-
v)
w
o
131
One can now express the domain integral (3-7-1) as
a summation of boundary ones:
1. (0 = f u~.(t,;,x)b.(x)dn(x) ~ ~] ]
n =-f Uij(Pj)o dr + f pij(uj)o dr + cij(uj)o
r r
£11 nIl [ [ Uij(Pj)~ dr - [ Pij(Uj)~ dr - Cij(Uj)~l
4 N M I I I
£=1 n=l m=l
(3-7-32)
where (u.) , (p.) correspond to Eqs.(3-7-13) and ] 0 ] 0
(3-7-17), (u.)l and (p.)l can be found in Table 3-7-1 ] n ] n
I £ and (uj)nm and (Pj)nm in Table 3-7-2. The superscript
£ is the type of combination of body forces. Note, that
only boundary integrals exist on the right-hand side of
formula (3-7-32). All the results presented above are
valid for plane stresses by transforming v into v = v/(l+v).
As regards the integrals (3-7-2), one can take
derivatives of Eq.(3-7-32) and consider the matrix c .. is ~]
a unit matrix for every internal point t,;. Finally the
transformation formula for integral (3-7-2) can also be
obtained simply by replacing Uij and pij by Uijk and pijk:
132
J Uijk(~,x)bk(x)dn(X) n
=-J Uijk (Pk) 0 df+ J p ~ . k (uk) d f + c (u . ) ~J 0 ~ 0
r f
4 N
J Uijk(Pk)~ J Pijk(Uk)~ R.
- L L [ df - df -c(ui)n1 R.=1 n=l f r
4 N M
J Uijk(Pk)~m J Pijk(uk)~m - L L L df - df R.=1 n=l m=l f f
(3-7-33)
where Uijk and pijk are as shown in Eqs. (3-5-6) and
(3-5-7) . c is equal to 0ij'
3-8 NUMERICAL IMPLEMENTATION
In order to compute the results of Eqs. (3-7-32) and
(3-7-33), one needs to discretize the problem using boundary
elements and applying numerical integration techniques.
As mentioned in potential problems, the integral along
the boundary is divided into several integrals over the
boundary elements:
J F(x)df f
NE L
j=l J F(x)dr 1
fj
(3-8-1)
Because the displacement u j and tractions Pj in the
integrands of integrals (3-7-32) and (3-7-33) are continuous
functions over every element j, the singularity is only due
133
to the fundamental solution components Uij and pij .
There are some kinds of singularities for 2-D problems,
such as In(r) and l/r, in Eqs. (3-3-4) and (3-3-5), there-
fore for the purpose of numerical implementation one can
classify the integrals into three types, i.e.
A. Those that can be evaluated using fue standard Gauss
quadrature for regular functions.
B. Those that require a transformation (2-7-3) for the
functions involving the factor of In(r) in the integrand.
c. Those that can be computed using Kutt quadrature
(finite-part integration) applicable for functions of
the type l/r.
Types A, and B. have already been described in section
2-7, therefore, one needs to discuss here only the case C.
For this case, the finite-part integral technique is used
later. The finite-part integral technique and the corres-
ponding tables can be found in some references [47-49].
In order to use the tables directly, one must first
scale the interval of the integral from [a,b] to [0,1]:
I
b
J fIr) dr r-a
a
1
J f[(b-~)t + a] dt (3-8-2)
o
Then one can find the local coordinates of the integral
points t. and the weights w. in the finite-part integral 1 1
tables and express the integral as
I N L f[(b-a)t i + a]wi
i=l (3-8-3)
134
Although the form of Eq. (3-8-3) is similar to that of
Gauss quadrature, their deduction and tables are completely
different.
Moreover, when one evaluates the kind of integral
(3-8-1), in which p~. exists, they are computed in the 1)
meaning of Cauchy principal value as mentioned in section
3-5. For this reason, if the interval [a,b] involves the
singular point s, the integral I becomes
b
I f 1 f(r)dr r-s
a
s-£ b
lim f f(r) dr + f
f(r) dr ] (3-8-4 ) £~o
r-s r-s a s+£
where f(r) is supposed to satisfy the Holder condition at
point sand f(s) + 0, so that the singularity is actually
of order one. Note that the same value of £ must be taken
in the first and second integrals on the right-hand side
of Eq. (3-8-4) in Cauchy principal integral. Hence, after
transforming Eq. (3-8-2) there is an additional term in the
formulation. Using Eq. (3-8-2), Eq. (3-8-4) can be rewritten
as
I
b
f ~ f(r)dr a
lim £~o
£/Is-al
[ f 1
lim £~o
s-£ b
f f ~) dr + f a s+£
1
f (r) dr ] r
f[(a-s)t+s] dt + t f f[(b-s)t+s] l
t dtJ £/ Ib-s I
135
£/ls-al f[(b-s)t+sl dt +
t lim £-+0
[J 1
1 f[(a-~)t+sl dt + J
£/ls-al
= lim £'-+0
£/ls-al + J f[(b~S)t+sl d~
£/lb-sl
£ '
[ f 1
1
f[(a-~)t+sl dt + f £ '
Ib-sl + f(s) ln -Is-al (3-8-5)
The above integral (3-8-5) can be evaluated numerically
by the finite-part integral as follows
NP NP I -~ f[(a-s)x +slw + p p ~ f [ (b- s ) x + s 1 w
p p p=l p=l
(3-8-6)
Consequently, when the finite-part integrals are used to
compute the terms in Eq.(3-7-32) in the form of
the addition term u(s) ln [I~=:I) must be considered as
shown in Eqs.(3-8-5) and (3-8-6).
The matrix c .. in Eq.(3-7-32) can be considered now. ~)
From Eqs. (3-5-3) and (3-3-5), one has
and
136
J pij (t;,x) dr(x)
r£
( 3-8-7)
1 dr 4n(l-v)r {[(l-2v)oij + 2r,ir,jl an -
- (l-2v) (r .n. + r .n.)} (3-8-8) ,1) ,) 1
when the source point t; is on the boundary r. One supposes
the boundary is augmented by a small boundary rc which is
a part of a circle centred at point; with radius £ (Fig.
3-8-1). If one uses the polar coordinates and considers
r = £ on the boundary r£, one has
£ cos e
X2 = £ sin e
r , 1 cos e
r ,2 sin e
dr 1 'Yii
grad r x !! o (3-8-9)
137
Figure 3-8-1 Boundary r + re:
138
substituting Egs. (3-8-9) into Eg. (3-8-8), one obtains
I .. (i;) = ~J
1
f p~. (i;,x)dr ~J
r £ 92
41f (l-v) f 1. [ (1- 2 v) <5.. + r . r . 1 rd 9 r ~J, ~ , )
(3-8-10)
The components of matrix I .. can be integrated as follows: ~J
(3-8-11 )
Then the matrix c ij can be expressed by Egs. (3-8-11)
as follows
<5 •• + I .. (0 ~J ~J
(3-8-12)
There are constants x lO and x 20 in Egs. (3-7-13)
and (3-7-17), which are the coordinates of a reference
point as mentioned in section 3-7. When one calculates
formula (3-7-32) or (3-7-33), the values of (x 10 ' x 20 )
can be taken as the coordinates of the point near the
middle of domain n or, for simplicity, the origin of
coordinates. Only do not let mag'nitudes xl and x 10 or
x 2 and x 20 be significantly different, which causes some
rounding errors for numerical reasons.
139
Finally, one wants to mention here that in Eqs. (3-7-13)
and Tables 3-7-1 and 3-7-2 the displacements can involve
some other terms, which correspond to the arbitrary rigid
body motion (translated and/or rotated) and pure shear
state. These rigid body motion and pure shear state terms
have no influence on the results of integrals (3-7-1) and
(3-7-2). For testing the rigid body motion, one can assume
the following displacements
(3-8-13)
where w is the rotation angular velocity, (x IO ' x 20 )
are the coordinates of the centre of rotation. u 10 and
u 20 are components of the translated motion. In this case
the tractions corresponding to displacements (3-8-13)
vanish.
For testing the pure shear state, one takes all the
variables as follows
u 1 k 1 y
u 2 k 2x
0 0 11
0 0 22 0 0 E E
12 21 l+V 12 (3-8-14)
E 0 11
E 0 22 E E l.j(k1 + k 2 ) 12 21
PI 0 21 n 2
P2 0 n 1 12
140
Substituting u. and p. in Eqs .• (3-8-13) and (3-8-14) J J
into formulae (3-7-32) and (3-7-33), one obtains results
which are all numerically zero, so that, one has the
following conclusion: in Eqs. (3-7-32) and (3-7-33) the
displacements u. and tractions p. involve arbitrary rigid J J
body motion and pure shear state without any influence
to the results of integrals (3-7-1) and (3-7-2). So
in the Eqs. (3-7-13) and (3-7-17) and the expressions of
u. and p. in Tables 3-7-1 and 3-7-2, one can add some J J
terms, which correspond to the rigid body motion and pure
shear state, without changing the values of integrals
(3-7-32) and (3-7-33).
3-9 RESULTS OF NUMERICAL EXPERIMENTS
This section discusses some results, which are
obtained for examining Eqs. (3-7-32) and (3-7-33) numeric-
ally. In order to test the accuracy of Eqs. (3-7-32) and
(3-7-33) thoroughly, one studies the body forces b 1 (x)
and b 2 (x) term by term for each combination in Eqs. (3-7-6),
(3-7-7) and (3-7-8). Besides using formulae (3-7-32) and
(3-7-33), integrals (3-7-1) and (3-7-2) are computed for
comparison also using the original domain integrals as
well. The domain is subdivided into several cells in
advance and the numerical quadrature formulae and trans-
formations used to compute the integrals (3-7-1) and
(3-7-2) are the same as those mentioned in section 2-8.
Because of this they will not be repeated here.
The results for the two methods mentioned above are
listed in Tables 3-9-1 to 3-9-16. The integral domain n
141
expansion domain nf and discretizations of boundary
elements and subdomains (cells) are the same as shown
in Figure 2-8-1, which is for testing the transformation
formula in potential problems in section 2-8. From these
results, one can also obtain a set of conclusions as
those in section 2-8. Therefore, formulae (3-7-32) and
(3-7-33) can be used for transforming the domain integrals
(3-7-1) and (3-7-2) into boundary integrals provided one
chooses suitable element lengths.
It is important to specify how to choose the boundary
rb for computing the boundary integrals in transformation
formulae and the domain of the Fourier expansion, Qf .
In principle, those are chosen according to the features
of body force functions bi(x). When the periodic extensions
of bi(x) are carried out, the less discontinuous are
the functions and their derivatives the faster will be
the convergence of the solution. Here it is also emphasized
that the convergence for formulae (3-7-32) and (3-7-33)
is faster than for their Fourier series as mentioned in
section 2-8.
~ n
r1
r1
r2
rL
. 1
2 1
2 8
1 -0
.109
710X
10-5
0.
2411
490X
10-7
-0
.;09
810X
10-5
0.
2411
439X
10-7
III
t:r
I-
' (l
)
2 0.
6110
476X
10-5
-0
.814
7151
X10
-6
0.61
1047
3X10
-5
-0.8
1471
59X
10-6
H
.....
w
I
3 0.
8866
362X
10-6
0.
1809
437X
l0-7
O
. 886
6364
X1·
0-6
0.18
0940
7X10
-7
J"'""
I \0
I .....
4 -0
. 222
0403
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-6
0.17
7581
9X10
-6
-0.2
2204
11X
l0-6
0.
1775
824X
10-6
~l
5 0.
1368
458X
l0-5
-0
.121
5229
X1
0 ~
o.13
6845
8X10
-5
-0. 1
2152
20X
l0-6
:8'_
__
__
~
(")
c .....
0
..... *
r1"
;3
w.
::r
'0
6 -0
. 127
1625
Xl0
-5
0.15
4515
8Xl0
-6
-0.1
2716
23X
10-5
0.
1545
163X
10-6
III
J"
'""I
0..
11
0 .....
7 -0
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6687
Xl·
0-5
0.77
4362
1X10
-7
-0.1
0466
87X
10
5 0.
7743
695X
10-7
X
;3
ti
l III
0
t:r
.....
::l
w.
::l
8 0.
1800
425X
l0 6
-0
.105
5635
Xl0
-6
0.18
0043
2X10
-6
-0.1
0556
20X
l0-b
0
X
.....
I"tl
::l
.j>
.
1 -0
. 116
8440
X10
-5
-0.2
9297
84X
l0-6
-0
.116
8423
X10
-5
-0. 2
9297
42X
1 0-
6 0.
. r1
" t:r
N
:8
(l
) 0
\Q
C
2 0.
5638
448X
l0-5
-0
. 505
2277
Xl 0
-6
0.56
3845
1X10
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-0.5
0522
66X
l0-6
X
11
::l
III
0.
. I-
' III
11
3 0.9637214Xl0-~
0.46
3545
7X10
-6
O. 9
6372
17X
1 0-
6 0.
4635
461X
10-6
H
'<
t:r
.....
.. .....
.....
~2
4 0.
1109
685X
l0-7
0.
4148
128X
l0-6
0;
1109
358X
l0-7
O.
4148
129X
10~6
::l
r1
" (l
)
5 0.
1408
068X
l0-5
-0
. 115
5626
Xl0
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0.11
4080
67X
10-5
0.
1155
627X
10-6
0
\Q
0 11
ti
l III
6 -0
. 118
5283
Xl0
-5
-0.4
7115
70X
l0-8
-0
. 118
5823
X1
0-5
. -0
.471
1408
Xl0
-8
f I-
'
o ~
H
X
..... -
~
7 -0
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7829
Xl0
-5
-0.1
0581
38X
10-6
-0
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7831
X10
-5
-O.1
0581
47X
l0-6
t:r
8 0.
2229
765X
l0-7
-0
.204
2218
Xl0
-6
0.2;
2·97
90X
10-7
-0
. 204
2223
X1
0-6
IV
0
E; n
rl
rl
r2
r~
'1
'')
'1
>-3
1 -0
. 880
172
5Xl 0
-5
0.10
6278
2Xl0
-5
-0.8
80
1725
Xl 0
-5
O.1
0627
82X
l0-5
III
0
-H
I-
' ....
CD
2 -0
. 147
7261
Xl0
-5
0.20
9800
8Xl0
-7
-0. 1
4772
63X
l0-5
0.
2093
UO
OX
10-7
.....
W
I
3 0.
2863
316X
l0-5
-0
.488
7780
Xl0
-6
0.28
6331
7Xio
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8877
88X
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1.
0 I N
4 -0
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3961
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0.76
4693
5Xl0
-7.
-0.3
0839
76X
l0-6
0.
7646
866X
l0-7
E;
1
:0--
---.
c: .... *
~
()
0.11
4116
9Xl0
-5
-0.4
6651
66X
l0-7
0.
1141
168X
l0-5
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5163
Xl0
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i 5
u.
.... 0
rt
S .....
::T
'0
6 0.
1660
195X
l0-5
-0
. 124
3190
Xl0
-6
0.16
6019
6Xl0
-5
-0. 1
2431
88X
l0-6
III
X
0
-11
-
0 ....
7 -0
.736
0188
X10
-6
o • 1
6042
65X
l 0-6
-0
.736
0183
Xl0
-6
0.16
0426
9Xl0
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0-
S en
u
. III
0
.... ::l
X
::l
8
· 0.
5887
573X
l0-7
0.
1284
062X
l0-9
0.
5887
579X
l0-7
0.
1315
370X
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0
0-
H
HI
.... :0
.... '
" .::
.
1 -0
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4160
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0.45
7190
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W
X
0 c: ::l
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6635
84X
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9130
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0
. III
11
3 0.
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9226
Xl 0
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0.24
8534
6Xl0
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0U92
2UX
10-6
0
-'<
I-
' ....
4 -0
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5762
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0.31
5750
0Xl0
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-0.2
9157
52X
l0-6
0.
3157
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E;
2 0.
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843X
l0-5
0.
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5
0.12
0384
3Xl0
6 O
. 172
0595
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0.40
9925
5Xl0
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0.17
2059
6Xl0
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0.40
9921
4Xl0
-7
II ::l
rt
en
CD
....
\Q
::l
11
-f III
I-
'
~~
H
....
I-'
7 -0
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4408
Xl0
-6
O. 1
5859
37X
l 0-6
-0
. 570
4404
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0.15
8594
6Xl0
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0-
N
8 0.
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l0-7
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9325
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·0.5
4542
87X
l0-7
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. 930
9359
Xl0
-7
0
~ r'
r'
r~
r~ .
I n
1 2
1 2
>-3
0.43
0663
0X10
-7
0.10
0946
6X10
-5
0.43
0665
7X10
-7
0.10
0946
2X10
-5
,
1 i
III
tr
......
(1)
2 -0
.787
9702
X10
-6
0.54
4470
6X10
-5
-0. 7
8797
02X
1 0
-6
0.54
4470
6X10
-5
H ...
IN
I
3 :"
0.10
5397
6X10
-6
-0. 1
7584
42X
1 0
-5
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66X
1 0
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10-5
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"t
\0
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4 0.
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10-b
-0
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6893
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2368
75X
10-o
:0
----
-..
~ 1
5 0.
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O
. 562
1677
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0.
1019
903X
10-6
0.
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10-6
~
n s::
... 0
"'>
1-rt
" a
u.
::T
'0
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2104
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10-6
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7127
X1o
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0406
8X10
-6
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7171
26X
10-5
III
J'
"t
0.
t1
0 ...
>< a
III
7 -0
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i846
X;0
-7
0.61
4214
9X10
-6
-0.3
2511
55X
1o-7
0.
6l42
143X
10-6
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0
tr
... ::
l u
. ::
l
8 -0
. 656
5576
X1
0-7
0.
2366
767X
10-7
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5486
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6709
3X10
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1 0.
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607X
10-7
0.
1438
303X
10-5
0.
1904
696X
10-7
.0
. 14 3
8248
X1
0-5
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x ::s
-0
Il-
.... H
I
2 3
-0. 1
2367
31X
10-6
-0
. 953
9099
X1
0-6
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6738
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5391
Q9X
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. 1
1 0.
2878
616X
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O
.128
0342
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0.28
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::J
::s rT
tr
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t:
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11
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0 III
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-.....
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6919
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0.
4136
916X
10-5
tr
11
.....
H
"<
... ...
2
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7806
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61X
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· 0.
5620
047X
10-6
....
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0
rT
CD
1 3
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6285
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. 220
9712
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-0
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9703
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0-5
<Q
tr
11
N
III
2 2
0.31
2555
7X10
-6
0.54
6049
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-6
0.31
2547
4X10
-6
0.54
6050
1X10
-6
~2
.....
H
til
.... -
3 1
-0. 7
4123
12X
10-7
-0
. 386
2464
X1
0-6
-0
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224O
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.... ::s
1 4
0.42
2505
6X10
-7·
-0. 3
4778
28X
1 0
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O. 4
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45X
1 0
-7
-0. 3
4777
29X
1 0
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t 0=
1
X
2 3
0.86
9222
4X1
0-7
-0
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0.
8691
403X
10-7
-o
.944
7700
X10
-6
'-'
n 0 _
.
til ~
0=
1
'-'
~ ~3
Tab
le
3-9
-9
Co
mp
ari
son
o
f b
ou
nd
ary
in
teg
ral
I!.
wit
h
do
mai
n in
teg
ral
I~.
~J
~J
Iij
(1
;)
. u~.J~,x)bk(x)dQ(x)
J ~J]{
(nn
x)
b1
co
s 1
0
b2
o Q
1 1
1 2
n II
I 11
2 12
2 II
I
1 -0
.621
161
Xl0
U
0.16
0007
Xl0
-1
-0.1
3020
4 X
l0u
-0.6
2114
7 X
l0u
2 -0
.107
825
Xl0
l -0
.124
332
Xl0
0 -0
.228
044
Xl0
0 -0
.107
823
Xl0
l
3 0.
8845
02 X
l00
-0.4
8842
3 X
l0-3
0.
1560
88 X
l00
0.88
4489
Xl0
0
4 0.
1209
16 X
101
-0.1
6519
1 X
l0-3
0.
3498
36 X
l00
0.12
0910
Xl0
1
5 0.
5081
36 X
l0-l
-0
.463
455
Xl0
-l
0.97
1343
Xl0
-1
0.50
7743
Xl0
-1
6 -0
.346
407
Xl0
0 0.
2760
42 X
l0-1
-0
.170
180
Xl0
0 -0
.346
354
Xl0
0
7 -0
.583
137
Xl0
0 0.
2633
87 X
l0-1
-0
.223
335
Xl0
0 -0
.583
100X
l00
8 -:
0.40
9894
Xl0
0 -0
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010
Xl0
-2
-0.7
7358
2 X
l0
1-0.
4098
67 X
l00
9 0.
3064
59 X
l00
0.26
4594
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-l
0.11
4655
' Xl0
0 0.
3063
45 X
l00
10
0.28
2352
Xl0
0 0.
9370
20 X
l0-3
0.
1264
38 X
l00
0.28
2219
Xl0
0
2 112
0.16
0005
Xl0
-1
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2433
2 X
l00
-0.4
8788
0 X
l0-3
-0.1
6505
6 X
l0-3
-0.4
6346
1 X
l0-l
0.27
6037
Xl0
-l
0.26
3389
Xl0
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3794
6 X
l0-2
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6654
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0.13
0187
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\)
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2801
4 X
l00
0.15
6070
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0
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9758
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0
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0877
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-l
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1 X
l00
1-0.
2232
06 X
l00
1-0.
7732
73 X
l0-l
0.11
4518
Xl0
0
0.12
6273
Xl0
0 L
..
-
U1 o
~ n 1 2 3 4 5 6
~3
7 8 9 10 T
ab
le
3-9
-10
C
om
par
iso
n o
f b
ou
nd
ary
in
teg
ral
Iij
do
mai
n
inte
gra
l Iij
Iij
(F,
;) J Ui
jk(~
,x)b
k(x)
dQ(x
) Q
b1
(niT
x)
sin
10
b
2
1 III
1 112
1 12
2 2 III
0.83
0808
Xl0
0 0.
1824
02 X
l00
0.13
3178
Xl0
0 0.
8308
03 X
l00
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60
14
5 x
l00
0.17
5922
Xl0
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-0.1
9267
5 X
l00
0.96
0125
X10
0
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2648
3 X
l0l
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5003
9 X
l0-l
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659
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0 0.
1264
79 X
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0.47
1857
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0 0.
2629
65 X
10-1
0.
4047
19 X
10-1
0.
4718
64 X
100
0.86
4016
Xl0
0 -0
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495
Xl0
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0.29
3438
X10
0 0.
8640
00 X
l00
0.44
44<
S X
l00
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6708
5 X
10-1
0.
1960
57 X
l00
0.44
4352
Xl0
0
0.13
6033
Xl0
0 0.
1499
62 X
l0-l
-0
.299
035
Xl0
-1
0.13
6054
X10
0
-0.4
9501
6 X
l00
1-0.
3446
91
X10
-2
-0.1
8516
2 X
100
0.49
4893
X10
0
-0.4
2933
6 X
l00
0.52
7043
Xl0
-3
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2699
2-X
l00
0.42
9254
X10
0
0.14
1107
X10
0 0.
3175
40 X
l0-l
0.
4722
64 X
l0-1
0.
1410
28 X
100
---
o
2 112
2 122
0.18
2402
Xl0
0 0.
1331
68 X
l00
0.17
5917
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0.19
2650
Xl0
0 (J
1
-0.5
5003
8 X
l0-l
0.
3196
03 X
100
0.26
2971
X
10-1
0.
4047
89 X
l0 -1
-0.2
7049
2 X
l0-l
0.
2933
55 X
l00
-0.4
6708
8 X
l0-l
0.
1959
66 X
l00
0.14
9956
Xl0
-l
0.29
8746
Xl0
-l
-0.3
4470
3 X
l0-2
0.
1850
14 X
l00
0.52
7593
Xl0
-3
0.12
6892
Xl0
0
0.31
7545
Xl0
-l
0.47
1328
Xl0
-l
--
~ n 1 2 3 4 5 6
I--
~3
7 8 9 10
Tab
le
3-9
-11
C
om
par
iso
n o
f b
ou
nd
ary
in
teg
ral
I!.
1]
wit
h
do
mai
n in
teg
ral
I~.
1]
Iij (
E,;)
J Ui
jk(~
,x)b
k(x)
dn(x
) n
bl
0 b
2 co
s (n
7TY)
20
11
1 1
2 2
11
112
122
III
112
0.1.
2722
7 X
10-1
0.
3105
02 X
10-
1 -0
.980
660
X10
0 -1
0.
3104
82 X
10 -
0.12
7383
X10
0.82
5340
X10
-2 -
0.27
8873
X10
0 -0
.992
261
X10
0 0.
8270
67 X
10-2
-0
.278
870
X10
0
-0.1
0573
1 X
l0-1
-0.
4627
72 X
l0-1
0.
1828
53 X
l01
-0.1
0608
7 X
10-1
-0
.462
749
X10
-1
0.13
3679
X10
0 0.
2891
90 X
10-1
0.
1300
69 X
101
0.13
3620
X10
0 0.
2891
71
X10
-1
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0967
2 X
l0-1
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8842
59 X
10-2
-0
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360
X10
0 -0
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455
X10
-1
-0.8
8465
9 X
l0-2
-0.2
2300
9 X
100
0.63
9096
X10
-1
-0.1
2169
7 X
101
-0.2
2290
6 X
l00
0.63
9096
X10
-1
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6656
6 X
101
0.57
1043
X10
-1
-0.5
0850
0 X
100
0.26
6254
X10
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0.57
1107
X10
-1
0.20
0599
Xl0
0 0.
1170
43 X
10-1
0.
7378
66 X
100
0.20
0479
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0 0.
1170
85 X
10-1
~.135203 X
100
-0.4
1000
8 X
10-1
0.
7857
93' X
100
0.13
5092
X10
0 -0
.410
082
Xl0
-1
-0.1
0391
0 X
100
-0.6
7088
3 X
10-1
-0
.258
859
X10
0 ~0
.103
822
Xl0
0 -0
.671
001
X10
-1
---------
2 12
2
0.98
0692
Xl0
0
0.99
2294
X10
0
0.18
2861
X
101
0.13
0081
X
101
0.82
7404
Xl0
0
-0.1
2171
8 X
l01
-0.5
0856
3 X
100
0.73
8105
X10
0
0.78
6014
Xl0
0
~.259033 X
l00
U1
r-:l
~ n 1 2 3 4 5 6 7
~3
8 9 10
Tab
le
3-9
-12
C
ompa
r:is
on o
f b
ou
nd
ary
in
teg
ral
I i j
wit
h
do
mai
n in
teg
ral
I~.
~J
I ..
(~)
=
! ul.~'k(~,x)bk(x)dn(x)
~J
J J
n b
1 ::!
: 0
1 111
1 112
1 12
2
-0.5
9030
8 X
10-1
0.
3806
12 X
100
0.84
2310
X10
0
0.17
2246
X10
-1
0.48
9891
X10
-1
-0.1
6537
6 X
101
-0.6
1725
3 X
10-1
-0.
1478
77 X
100
-0.1
1705
8 X1
01
0.24
2442
X10
-2 -
0.24
0081
X10
-1
0.14
9773
Xl0
1
0.19
1204
X10
0 -0
.456
698
X10
-1
0.13
2311
X
101
-0.3
4218
1 X
10-2
-0.
3954
00 X
10-1
-0
.773
660
X10
-1
-0.2
2512
7 X
100
0.36
2702
X10
-1
-0.1
0045
3 X
101
-0.7
6664
9 X
10-1
0.
5669
71.X
10-1
-0
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083
X10
0
0.15
7129
X10
0 0.
5218
41
X10
-1
0.47
5333
X10
0
0.18
6373
X10
0 -0
.182
212
X10
-1
0.60
2097
X10
0
b2
sin
(nn
y)
20
2 111
2 112
-0.5
9035
6 X
10-1
0.
3806
05 X
100
0.17
2533
X10
-1
0.48
9873
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-1
-0.6
1689
0 X
10-1
-0
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874
X10
0
o 23
9076
Xl 0
-2
-0.2
4004
8 X
10-1
0.19
1122
X10
0 -0
.456
708
X10
-1
-0.3
4221
4 X
10-2
-0
.395
458
X10
-1
-0.2
2501
1 X
100
0.36
2685
X1U
-1
-0.7
6595
5 X
10-1
0.
5670
42 X
10-1
0.15
7019
X10
0 0.
5219
16 X
10-1
0.18
6220
X10
0 -0
.182
281
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-1
2 122
I 0.
8423
01
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0
-0.1
6538
2 X
101
-0.1
1706
5 X
101
, I
0.14
9780
X10
1 :
0.13
2327
X10
1
-0.7
7366
4 X
10-1
-0.1
0047
6 X
10'
-0.7
9422
1 X
100
0.47
5541
X
100
0.60
2399
X10
0
U1
W
~.
n m
1 1
1 2
2 1
1 3
2 2
3 1
~3
1 4
2 3
3 2
4 1
~
Tab
le
3-9
-13
C
om
par
iso
n o
f b
ou
nd
ary
in
teg
ral
I:.
.
~J
wit
h
do
mai
n in
teg
ral
I~.
~J
Iij (
0
J Uijk(~,x)bk(x)dQ(x)
Q
b1
1 I" 1 112
1 122
-0.1
40
19
3 X
100
-0.5
37
83
8 X
10-1
-0.
2806
63 X
10-2
0.49
5116
X10
0 -0
.546
981
X10
-1
0.78
3757
X10
-1
-0.2
4515
2 X
100
0.22
6282
X10
0 -0
.453
927
X10
-2
0.33
5846
X10
0 0.
1067
66 X
100
-0.9
3308
1 X
10-1
0.86
0420
X10
0 0.
2329
24 X
100
. 0.
1363
81
X10
0
0.19
2607
X10
0 0
•. 481
649
X10
-1 -
0.78
0953
X10
-2
-0.2
1959
4 X
100
0.13
4118
X10
0 0.
8689
72 X
10-2
0.58
7276
Xl0
0 -0
.432
963
X10
0 -0
.188
041
X10
-1
-0.6
9424
3 X
100
0.48
1790
X10
-1 -
0.86
8414
X10
-1
0.29
6994
X10
0 -0
.424
330
X10
-2
0.43
3999
X10
-1
(n1T
X)
(m1T
Y)
co
s 1
0
co
s 20
b
2
2 I"
2 112
-0.1
4018
9 X
100
-0.5
3781
3 X
10-1
0.49
5105
X10
0 -0
.546
951
X10
-1
-0.2
4514
7 X
100
0.22
6276
X10
0
0.33
5837
X10
0 0.
1067
60 X
100
0.86
0418
X10
0 0.
2329
17 X
100
0.19
2603
X10
0 0.
4815
65 X
10-1
-0.2
1958
8 X
l00
0.13
4107
Xl0
0
0.58
7260
X10
0 -0
.432
948
X10
0
-0.6
9423
2 X
lOO
0.
4816
96 X
10-1
0.29
6978
X10
0 -0
.424
386
X10
-2 o
2 12
2
-0.2
8020
7 X
10-2
0.78
3605
X10
-1
-0.4
5314
9 X
10-2
-0.9
3433
5 X
10-1
0.13
6355
X10
0
-0.7
8142
6 X
10-2
0.86
8969
X10
-2
-0.1
8825
4 X
10-1
-0.8
6825
5 X
10-1
0.43
3799
X10
-1
Ul
.j:>.
~ n 1 1 2 1
£;3
2 3 1 2 3 4
Tab
le
3-9
-14
C
om
pari
son
o
f b
ou
nd
ary
in
teg
ral
I~.
1J
wit
h
do
mai
n in
teg
ral
I~.
1J
Iij (
0
J Ui
jk(~
,x)b
k(x)
dQ(x
) Q
b1
sin
(n
;;)
co
s (m
;l)
b2
o
m
1 11
1 1 112
1 12
2 2 I"
2 112
1 0.
1786
17 X
100
-0.3
3271
5 X
100
-0.1
3552
5 X
10-1
0.
1786
16 X
100
-0.3
32
70
5 X
100
2 -0
.647
370
X10
0 -0
.342
414
X10
0 -0
.689
566
X10
-1
-0.6
47
36
7 X
100
-0.3
4240
4 X
100
1 -0
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583
X10
0
-0.7
3927
2 X
10-1
-0.
9345
49 X
10-3
-0
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578
X10
0 -0
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282
X10
-1
3 -0
.41
06
05
X10
0 0.
6367
12 X
100
0.50
6452
X10
-1
-0.4
1060
2 X
100
0.63
6693
X10
0
2 0.
7621
41
X10
0 -0
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292
X10
-1
0.11
3999
X10
0 0.
7621
23 X
100
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4921
9 X
10-1
1 -0
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99
65
X10
0 0.
1009
15 X
100
1-0.
2641
74 X
10';'1
-0
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99
55
X10
0 0.
1009
11
X10
0
4 0.
2625
66 X
100
0.62
8186
X10
0 0.
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76 X
10-1
0.
2625
64 X
100
0.62
8153
X10
0
3 0.
5105
57 X
100
0.14
8305
X10
0 0.
2141
04 X
10-1
0.
5105
43 X
100
0.14
8291
X
100
2 0.
1028
35 X
101
0.10
3941
X
100
0.20
4724
X10
0 0.
1028
31
X10
1 0.
1039
38 X
100
1 0.
9309
65 X
10-1
-0
.70
41
80
X10
-2
0.21
9268
X10
-1
0.93
0984
X10
-1
-0.7
05
05
6 X
10-2
2 122
-0.1
35
55
2 X
10-1
-0.6
8947
3 X
10-1
-0.9
2793
9 X
10-3
0.50
6530
X10
-1
0.11
3977
X10
0
-0.2
6402
9 X
10-1
-0.3
35
53
5 X
10-1
-0.2
14
28
9 X
10-1
0.20
4675
X10
0
-0.2
19
25
0 X
10-1
U1
U1
E;, n 1 1 2 1
E;,s
2 3 1 2 3 4
Tab
le
3-9
-15
C
om
par
iso
n o
f b
ou
nd
ary
in
teg
ral
I~. ~J
m
1 2 1 3- 2 1 4 3 2 1
wit
h
do
mai
n in
teg
ral
I~.
~J
Iij (
E;)
J
Uij
k(T
I,x
lbk
(xld
Q(x
l Q
bI
0
1 I11
1 I12
1 I22
0.35
4809
X10
-1 -
0.60
0064
X10
-1
0.25
8646
X10
0
0.36
8907
X10
-1
0.30
7772
X10
0 -0
.265
865
X10
0
-0.8
8294
2 X
10
-2 -
0.10
4416
X10
0 0.
5966
09 X
100
-0.6
3763
2 X
10-1
0.
1212
75 X
100
0.49
0090
X10
0
-0.6
0969
1 X
10-2
0.
5351
02 X
100
0.60
3650
X10
0 -
-0.8
2513
0 X
10-1
0.
6610
94 X
10
0.40
9206
X10
0
-0.3
0770
5 X
10-1
-0.
7567
19 X
10-1
0.
4429
97 X
100
0.88
2097
X10
-2
0.20
8585
X10
0 o
1113
28 X
1 01
-0.8
6429
3 X
10-1
-0.
3992
75 X
100
0.42
2950
X10
0
-0.5
4096
5 X
10-2
0.
1795
57 X
100
0.83
2501
X
10-1
--
------
b2
cos(~~)cos(m;l)
2 I11
2 I12
0.35
4871
X
10-1
-0
.600
000
X10
-i
0.36
8973
X10
-1
0.30
7747
X10
0
-0.8
8406
9 X
10-2
-0
.104
403
X10
0
-0.6
3777
0 X
10-1
0.
1212
54 X
100
-0.6
1087
7 X
10-2
0.
5350
57 X
100
-0.8
2527
5 X
10-1
0.
6610
13 X
10-1
-0.3
0793
3 X
10-1
-0
.756
570
X10
-1
0.88
4556
X10
-2
0.20
8548
X10
0
-0.8
6444
9 X
10-1
-0
.399
249
X10
0
-0.5
4107
6 X
10-2
0.
1795
23 X
'-OO
--
----
---
2 I22
-0.2
5865
8 X
100
-0.2
6587
8 X
100
~
0.59
6631
X
100
0.49
0119
X10
0
0.60
3673
X10
0
0.40
9237
X10
0
0.44
3044
X10
0
-0.1
1133
3 X1
01
0.42
2981
X
100
0.83
2798
X10
-1
E; n
m
1 1
1 2
2 1
1 3
·
E;3
2 2
3 1
1 4
2 3
3 2
4 1
Tab
le
3-9
-16
C
om
pari
son
o
f b
ou
nd
ary
in
teg
ral
1~, ~J
wit
h
do
main
in
teg
ral
1~,
~J
1ij
(t;)
J u!
jk(~
,x)b
k(x)
dn(x
) n
bl
1 111
1 112
1 122
0.11
8509
X10
-1
0.52
4739
X10
-1
0.87
3266
X10
0
0.78
7263
X10
-2 -
0.3
55
48
8 X
100
1-0.
8835
90 X
100
0.64
8160
X10
-1 -
0.8
71
41
7 X
10-1
t-0
.411
675
X10
0
-0.1
05
34
0 X
10-1
-0
.93
33
88
X10
-1
0.16
2865
X1
01
0.67
5941
X
10-1
0.
4644
55 X
100
0.42
3902
X10
0
-0.2
98
94
3 X
10-2
-0.
1569
71
X10
0 0.
2627
49 X
100
0.11
6968
X10
0 0.
5805
95 X
10-1
0.
1161
08 X
101
-0.1
17
74
4 X
100
0.17
4248
X10
0 0.
7828
64 X
100
-0.1
71
69
2 X
10-2
v.
697'
:!54
X10
0 0.
2658
64 X
100
-0.8
48
86
2 X
10-1
0.
6541
03 X
10-2
0.
2794
99 X
100
o b
2 ,
(n 1T
X)
(m1T
Y)
s~n 1
0 c
os
20
2 111
2 112
0.11
8662
X10
-1
0.52
4693
X10
-1
0.78
8871
X
10-2
-0
.35
54
73
X10
0
0.64
8275
X10
-1
-0.8
71
30
4 X
10-1
-0.1
0566
9 X
10-1
-0
.93
32
67
X10
-1
0.67
4416
X10
-1
0.46
4416
X10
0
-0.2
9946
5 X
10-2
-0
.15
69
47
X10
0
0.11
6913
X10
0 0.
5805
08 X
10-1
-0.1
17
76
8 'X
100
0.17
4217
X10
0
-0.1
7226
0 X
10-2
0.
6978
72 X
100
-0.8
4901
9 X
10-1
0.
6544
35 X
10-2
2 122
-0.8
7329
6 X
100
-0.8
8362
1 X
100
-0.4
11
69
8 X
100
0.16
2872
X10
1
-0.4
23
92
6 X
100
0.26
2760
X10
0
0.11
6119
X10
1
0.78
2916
X10
0
0.26
5874
X10
0
0.27
9530
X10
0 ,
U1
-....I
CHAPTER 4 APPLICATIONS IN ELASTICITY AND ELASTO-PLASTICITY
4-1 INTRODUCTION
The previous chapter discussed how formulae (3-7-32)
and (3-7-33) can be used to transform the domain integrals
(3-7-1) and (3-7-2) into boundary integrals and examined
the accuracy of these formulae. The present chapter
shows some applications of these concepts to solve elasto-
statics and elasto-plasticity problems.
It is not difficult to apply the previously deduced
formulae to elastostatics provided that the body forces
are known. After evaluating the Fourier coefficients for
the body force functions analytically, the transformation
formulae (3-7-32) and (3-7-33) can be used directly. The
accuracy of the solutions depends on the accuracy of the
Fourier series representations and usually the relative
errors of the solutions are better than those of the
Fourier series. This is because the order of the coeff-
~ ~ ~ ~ icients of (Pj)n' (Pj)nm or (uj)n' (uj)nm shown in
Tables 3-7-1 and 3-7-2, is one or two orders higher than
the order of Fourier coefficients of b j , when n (or m)
becomes infinite. One can see that they approach zero
faster than the Fourier coefficients as n (or m) becomes
infinite. For the continuous body force functions, only
a few terms of Fourier expansions need to be considered,
hence the CPU time required when using transformation
formulae is less than the CPU required when using the
original domain integrals, because with these formulae
159
the integrations are only performed on the boundary,
i.e. they are one dimension less than domain integrals.
In addition, one does not need to divide the domain into
cells. In the cases of discontinuous body force functions,
however, more terms of Fourier series must be taken to
obtain sufficiently accurate solutions.
A short review of plasticity will now be presented
and then the boundary integral equations for elasto
plasticity will be deduced.
Applications of boundary integral equations to
plasticity have been developed only since 1971 starting
with a paper by Swedlow and Cruse [53] and then followed
by a few other references [54-55]. Recent works
[56-59] especially have shown how the boundary element
formulations for three-, two-dimensional and semi
infinite plasticity problems can be successfully developed
and those formulations are used in the present work.
In the area of plasticity problems, there are three
kinds of formulations in boundary element methods:
initial strain, initial stress and fictitious traction
and body force approaches. All of them require some
types of domain integrals to be computed. In principle,
each of these domain integrals can be transformed into
boundary ones with a similar technique as previously
shown, but here one will consider only the fictitious
traction formulation.
The domain integral terms in plasticity are due to
not only the actual body forces, but also tQe plastic
strains or stresses. These plastic strains also constitute
160
fictitious traction and body force components. As one
solves the plastic problems by means of an incremental
procedure, the prescribed loads or displacements applied
to the body or on the boundary are divided into increments,
which are applied step by step. For each step an
iterative technique is used until convergent solutions
are obtained. The plastic strains and stresses are
obtained only at a certain number of points at the end of
each iteration. Because no analytical expressions generally
exist for them, the Fourier coefficients must be calculated
from these finite values numerically. The numerical
method applied here is also described in this chapter.
Unfortunately, it is time consuming to compute the Fourier
coefficients. The main reason is that the Fourier
coefficients must be recalculated at the end of each
step during the iterative procedures. The advantage of
this method is however that it requires only the dis
cretization of the boundary and there is no need to
subdivide the domain into subdomains (cells) which need
to be numbered in advance.
4-2 AN EXAMPLE OF GRAVITATIONAL LOADING
In this example a simple body force, i.e. gravitational
load, is applied to an elastic body of rectangular cross
section. The problem is assumed to be in plane strain.
The boundary conditions are free surfaces on three
sides and the other is constrained in the x direction.
161
In order to avoid the rigid body motion in y direction,
the middle point of that side is fixed (shown in Fig.
4-2-1). The properties of material are as follows:
E gm, g 10, m 250
For this kind of constant body force function, the
Galerkin tensor can be used, which is related to the
fundamental solution Uij by the following expression
[60-62) :
There
1 Gij,kk - 2(1-v) G*ik,kj (4-2-1)
G~. consists of two (for 2-D) or three (for 3-D) ~J
vectors each of them corresponding to the infinite space
in which the unit load is applied in the direction i.
For this infinite space the Galerkin tensor can be
represented as follows
Gij 1 ro .. 8nG ~J
(for 3-D) (4-2-2)
Gij 1 1
8nG rZln (-) o .. r ~J
(for plane strain) (4-2-3)
In order to satisfy Eq.(4-2-1) for 2-D plane strain,
a constant term must be added to the fundamental solution
for displacements (3-3-4), i.e.
u~. ~J
1 8n(1-v)G [(3-4v)ln rOij -
o
(7-8v) - r .r . + --2-- o .. J (for plane strain) (4-2-4) ,~ ,J ~J
162
///L_LLL / / / /' /'//// ~ J
! -y
I I I I
I j j j 2. 0'
5
j j j I 2 L
Figure 4-2-1 Geometry, body force and boundary conditions
163
substituting the Galerkin tensor into domain
integrals:
1. f * b j drl 1 Uij
Q
f (G~. kk -1
G\ k') b. dQ (4-2-5) 1J, 2(1-\1) 1 , J J
Q
one can transform the above domain integral into boundary
integrals for the case of a constant body force b i • The
complete formulation can be found as follows [12]:
where
and
f Uij(s,x)bj(x)dQ(x)
Q
(for 3-D)
(for plane strain)
These domain integrals can be computed with formulae
(3-7-32) and (3-7-33) also integrating only on the
boundary. The meshes and internal points are shown in
Figure 4-2-2 and the results are presented in Table
4-2-1. For the constant body force, the Fourier expansion
results present the same accuracy as those shown in
Tables 3-9-1. In addition, comparing the results with
the formulation using the Galerkin tensor, they are
exactly the same in the numerical sense.
164
• INl'ERNAL POINTS
x MESH POINTS
A • C A • C -A • C
B B I· MESH I MESH II MESH III
Figure 4-2-2 Meshes and internal points
Tab
le
4-2
-1
mes
h p
oin
t
A
I B
C
A
II
B
C
A
III
B
I C
I
Th
e D
isp
lacem
en
ts
and
S
tresses at
Inte
rnal
Po
ints
u u
a a
x y
xx
xy
0.23
473
X10
-2
-0.4
1677
X10
-7
0.38
350
X10
4 -0
.772
48 X
10-3
0.30
738
X10
-2
-0.6
0943
X10
-7
0.25
279
X10
4 0.
3474
2 X
10-2
0.23
008
X10
-2
-0.3
8354
X10
-3
0.37
826
X10
4 0.
4859
1 X
102
0.23
393
X10
-2
-0.5
8844
X10
-8
0.38
273
X10
4 -0
.475
88 X
10-3
0.30
655
X10
;;2
-0.8
3701
X10
-8
0.25
266
X10
4 0.
2641
7 X
10-3
0.22
939
X10
-2
-0.3
8152
X10
-3
0.37
651
X10
4 0.
5086
9 X
102
0.23
375
X10
-2
-0.1
98
72
X10
-7
0.38
227
X10
4 -0
.567
44 X
10-4
I 0.3
0627
X10
-2
-0.2
7570
X10
-7
0.25
238
X10
4 -0
.484
94 X
10-3
i 0.
2291
9 X
10-2
-0
.38
09
6 X
10-3
0.
3760
1 X
104
0.51
754
X10
2
a yy
-0.5
0304
X10
2
-0.1
5619
X10
2
-0.5
1258
X10
2
-0.5
2201
X
102
-0.2
8905
X10
2
-0.3
3713
X10
2
-0.5
2203
X10
2
-0.2
9514
X10
2
-0.3
2237
X10
2
ozz
0.11
354
X10
4
0.75
368
X10
3
0.11
194
X10
4
0.11
325
X10
4
0.74
931
X10
3
0.11
194
X10
4
0.11
311
X10
4
0.74
829
X10
3
0.11
183
X10
4
I I
0)
C11
166
4-3 AN EXAMPLE WITH A MORE GENERAL TYPE OF DISTRIBUTED
* LOADING
This section presents an example with body force
functions in the form of
b 1f. (1fX) 1 2a Sl.n 2a
(4-3-1)
It is not difficult in this case to find out the
analytical solutions for stresses, strains and displacements
which satisfy equilibrium equation (3-2-3), the relation-
ship between stresses and strains (3-2-8) and the
definition of strains (3-2-5), are given by
°11 cos G:)
°22 v cos G:) (4-3-2)
°12 °21 sin G:)
1-v 2 cos G:) £11 -E-
£22 0 (4-3-3)
1 sin G:) £12 £21 2G
u l 1-v 2 2a . (1fX) -- - Slon-
E 1f 2a (4-3-4)
2a cos (~X) u 2 - TIG .. a
* In this section the coordinate Xl is replaced by x for simplicity
167
The domain of the problem under consideration is
defined as n: [0 ~ x ~ 5, 0 < y ~ 4), a = 10 and
the mesh points, internal points are shown in Fig. 4-3-1.
The boundary conditions are prescribed as
on r 1 , r3
and (4-3-5)
on r 2 , r 4
the values of ui and Pi at nodes on the boundary are
evaluated by Eqs. (4-3-4) and (4-3-2) together with the
following equation:
p. = o .. n. ~ J~ J
where nj (j = 1,2) are the components of the outward
normal on the boundary r.
In this case, body forces can be expressed as
Fourier series with only one term, if one chooses the
domain for the harmonic expansion nf : [-2a < x ~ 2a).
Using formulae (3-7-32) and (3-7-33),one can expect to
obtain with n = 1 results as accurate as those shown
in Tables 3-9-2 and 3-9-3. Later on one will adopt the
notation I 2a for the solutions obtained with this
expansion.
One can also select the domain of harmonic expansion
nf : [-a ~ x ~ a) to see what influence this has on the
results. The Fourier series for the body forces b i are
now in the form of
168
• intemal points .37
xmesh points -36 ,
• 29 - 30 '31 -32 -33 4 r2
-:35 .~
f- ~4
r1
5
Figure 4-3-1 Geometry, mesh and internal points
169
1 N (_l)n . (nn ) a I -1---- sJ.n a x
n=l '4 - n 2
1 1 a + 2a I (-l)~n cos(:n x )
n=l n 2 - '4
(4-3-6)
After setting E = 2 x 10 6 and v = 0.3, one can
divide the boundary into elements as shown in Figure
4-3-1. The solutions are listed in Tables 4-3-1 and
4-3-2, in which displacements and stresses at the internal
points are presented. The relative errors of the solutions
are also shown in these tables. One can see that the
solutions, even for N = 1, are quite accurate by
comparison with solutions I 2a -and solutions I 2a are very
accurate comparing with the analytical solutions.
4-4 RELATIONSHIP BETWEEN PLASTIC STRESSES AND PLASTIC
STRAINS
Up to now, the materials under consideration have
been assumed as elastic, which means:
A. After unloading, the configuration of the body for
the original unstrained configuration is recovered.
B. The deformations of the body depend only on the final
stresses, not on the previous load history or strain
path.
In plasticity, these two conditions are not satisfied,
residual strains are expected to occur when the loads are
removed and the final deformations depend not only on the
final stresses, but also on the path stress history from
the beginning of yielding.
Tab
le 4
-3-1
D
isp
lace
men
ts a
t In
tern
al
Po
ints
in
Exa
mpl
e 4-
3-1
All
th
e nu
rrbe
rs o
f d
isp
laca
nen
ts m
ust
be
mu
l tip
ied
by
10
-6 i
n t
his
tab
le.
I2a
N=1
N=2
u1
u2
u1
u2
u1
u2
u1
29
0.37
388
-8.1
85
3
0.46
578
-8.1
81
5
0.34
858
-8.1
89
3
0.37
206
-1. 1
1 %
-0
.24
%
23.1
9 %
-0
.29
%
-7.8
0 %
-0
.19
%
-1.5
9 %
30
0.74
288
-7.9
76
0
0.82
419
-7.9
73
5
0.72
148
-7.9
78
7
0.74
046
-0.9
1
%
-0.2
3 %
9
.94
%
-0
.26
%
-3.7
6 %
-0
.19
%
-1.2
3 %
31
1.09
90
-7.6
28
6
1.11
766
-7.6
29
5
·1.0
78
8
-7.6
29
5
1.09
66
-0.8
6 %
-0
.23
%
6.1
4
%
-0.2
2 %
-2
.68
% -
0.2
2 %
-1
.07
%
32
1.43
66
-7.1
50
2
1.51
80
-7.1
55
0
1.41
53
-7.1
49
8
1.43
43
-0.8
1
%
-0.2
4
%
4.81
%
-0
.17
%
-2.2
8 %
-0
.24
%
-0.9
7 %
33
1. 7
506
-6.5
47
7
1.84
37
-6.5
55
0
1.72
65
-6.5
47
1
1.74
75
-0.7
2 %
-0
.28
%
4.5
6
%
-0.1
7
%
-2.0
9 %
-0
.29
%
-0.9
0 %
34
1.10
29
-7.6
33
5
1.16
19
-7.6
36
2
1.08
82
-7.6
36
2
1.10
08
-0.5
0 %
-0
.16
%
4.81
%
-0
.13
%
-1.8
3 %
-0
.13
%
-0.6
9 %
35
1.10
00
-7.6
29
8
1.17
28
-7.6
31
4
1. 0
812
-7.6
31
4
1.09
77
-0.7
7 %
-0
.21
%
5
.80
%
-0.1
9 %
-2
.46
%
-0.1
9 %
-0
.97
%
36
1.09
99
-7.6
29
5
1.17
30
-7.6
29
7
1.08
09
-7.6
29
7
1.09
76
-0.7
8 %
-0
.22
%
5.8
2 %
-0
.21
%
-2
.48
%
-0.2
1
%
-0.9
8 %
37
1.10
27
-7.6
33
2
1.16
35
-7.6
321
1.08
72
-7.6
32
1
1. 1
007
-0.5
2 %
-0
.17
%
4
.97
%
-0
.18
%
-1.9
2 %
-0
.18
%
-0.7
1
%
-
N=3
u2
-8.1
85
0
-0.2
5 %
-7.9
76
2
-0.2
2 %
-7.6
291
-0.2
2 %
-7.1
50
8
-0.2
3 %
-6.5
48
0
-0.2
7 %
-7.6
31
5
-0.1
9 %
-7.6
29
3
-0.2
2 %
-7.6
31
2
-0.1
9%
-7.6
36
7
-0.1
2 %
An
aly
tica
l
solu
tio
ns
u1
u2
0.37
808
-8.2
05
3
0.74
970
-7.9
941
1.10
85
-7.6
461
1.44
83
-7.1
67
3
1.76
34
-6.5
65
8
1.10
85
-7.6
461
1.10
85
-7.6
46
1
. 1.1
085
-7.6
461
1.10
85
-7.6
461
I I ,
-....J o
Tab
le 4
-3-2
S
tress
es
at
Inte
rnal
Po
ints
in
Exa
mpl
e 4-
3-1
An
aly
tica
l I2
a N
=1
N=2
N
=3
solu
tio
ns
011
012
0"22
<1
11
012
CT22
CJ
'i1
0'l2
0"
22
CTi1
012
OL2
0i
1 01
2 0"
"22
29
0.99
650
0.13
143
0.30
713
1.00
620
0.13
244
0.30
894
0.99
593
0.13
076
0.30
670
0.99
263
0.13
200
0.30
669
0.99
144
0.13
053
0.29
743
0.51
%
0
.69
%
3
.26
%
1
.49
%
1
.47
%
3.8
7 %
0
.45
%
0.1
8 %
3.
11
%
0.1
2 %
1
.13
%
3.11
%
30
0.96
925
0.26
022
0.29
836
0.98
502
0.26
185
0.30
432
0.97
428
0.25
910
0.29
806
0.95
547
0.26
096
0.29
573
0.96
593
0.25
882
0.28
978
0.3
4 %
0
.54
%
2
.96
%
1.9
8 %
1
.17
%
5.0
2 %
0
.86
%
0.11
%
2
.86
%
-1.0
8 %
0.8
3 %
2
.05
%
31
0.92
472
0.38
215
0.28
303
0.93
913
0.38
417
0.28
612
0.93
913
0.38
108
0.28
612
0.90
262
0.38
239
0.27
792
0.92
388
0.38
268
0.27
717
0.0
9 %
-0
.14
%
2.1
2
%
1.6
5 %
0
.39
%
3
.23
%
1.6
5 %
-0
.42
% 3
.23
%
-2.3
0 %
-0
.08
% 0
.27
%
32
0.86
374
0.49
771
0.26
117
0.87
338
0.49
989
0.26
051
p.88
412
0.49
715
0.26
676
0.84
218
0.49
721
0.25
627
0.86
603
0.50
000
0.25
981
-0.2
6 %
-0
.46
% 0
.53
%
0
.85
%
-0.0
2 %
0.2
7 %
2
.09
%
-0.5
7 %
2.6
8 %
-2
.75
% -
0.5
6 %
-1
.36
%
-...J
33
0.78
769
0.60
734
0.23
243
0.79
366
0.60
908
0.23
453
0.80
390
0.60
740
0.23
677
0.77
577
0.60
658
0.22
951
0.79
335
0.60
876
0.23
801
-0.7
1
% -
0.2
3 %
-2
.34
%
0
.04
%
0
.05
%
-1.4
6 %
1
.33
%
-0.2
2 %
-0
.52
%
-2.2
2 %
-0
.36
% -
3.5
7 %
34
0.91
487
0.3
77
82
0.2
8132
0.
9242
3 0.
4171
7 0.
2839
1 0.
9242
3 0.
3652
9 0.
2839
1 0.
8994
6 0.
3778
4 0.
2756
1 0.
9238
8 0.
3826
8 0.
2771
7
-0.9
7 %
-1
.27
% 1
.50
%
0.0
4
%
9.01
%
2
.43
%
0.0
4 %
-4
.54
% 2
.43
%
-2.6
4 %
-1
.26
% -
0.5
6 %
35
0.92
129
0.37
990
0.28
137
0.93
449
0.40
108
0.28
376
0.93
449
0.37
330
0.28
376
0.90
063
0.37
985
0.27
652
0.92
388
0.38
268
0.27
717
-0.2
8 %
-0
.73
% 1
.52
%
1.1
5 %
4.
81
%
2.3
8 %
1
.15
%
-2.4
5 %
2.3
8 %
-2
.52
% -
0.7
4 %
-0
.23
%
36
0.92
884
0.38
440
0.28
426
0.94
207
0.38
778
0.28
384
0.94
207
0.40
152
0.28
384
0.90
715
0.39
899
0.27
632
0.92
388
0.38
268
0.27
717
0.5
4
%
0.4
5 %
2
.56
%
1
.97
%
1.3
3 %
2.
41
% 1
.97
%
4.9
2
%
<:.4
1 %
-1
.81
%
4.2
6
%
-1.1
.31
%
37
0.93
729
0.38
658
0.28
293
0.94
937
0.37
648
0.28
389
0.94
938
0.41
139
0.28
309
0.91
985
0.40
393
0.27
524
0.92
388
0.38
268
0.27
717
1.4
5 %
1
.02
%
2.0
8 %
2
.76
%
-1
.62
% 2
.42
%
2.7
6
%
7.5
0 %
2
.42
%
-0
.44
%
5.5
5 %
-U
.7U
%
172
The problem of formulating physical relations
describing the actual behaviour of a material during
plastic flow is a very complex one. In order to simplify
it, one considers the behaviour of a specimen stressed
in simple tension shown in Figure 4-4-1. If one expresses
the total strain as composed of elastic strain £e and
plastic strain £P, one has
£
e ° where £ = E .
(4-4-1)
When the stress has not reached the yield point y:
a - y < 0
only the pure elastic behaviour is obtained for the
loading.
Once a exceeds Y, however, one has the following
relationship from Figure 4-4-1:
After rearranging the previous equation, one obtains
Y + (4-4-2)
1
If the stress a is less than 00' the behaviour is
elastic. For general stress state, the yield condition
at a point is assumed to satisfy some criteria. One of
them commonly used is the von Mises yield criterion
which can be written as
a
ao-_
y
173
/1 / I
I I / I
I I / I
~~----------~--------~---------- &
Figure 4-4-1 Uniaxial stress-strain diagram showing elastic and plastic strains
174
F(o .. ,k) = 13J2 - a ~J 0
o (4-4-3)
where J 2 is the second invariant of the stress deviator
tensor:
s .. ~J
(4-4-4)
In Eq. (4-4-3), one can define I.3J"2 as an equivalent
s~ress ~, so that the von Mises yield criterion can be
expressed as
a e
a o (4-4-5)
which is used in the examples shown in this chapter.
In order to consider strain hardening, one applies
the load step by step in small increments after yielding.
Each increment loading step creates increments of strain
and stress tensor components, i.e. U, E .. } and {t1O .. } • . ~J ~J
Next one considers the relationship between the
plastic strain increments and stresses. Assume that a
loading path is found to reach a certain level and to
have then Elj plastic strains. Then, the load is increased
by a small increment, the additional plastic strains
produced are ~Elj and the total strain Eij are expressed
as follows:
(4-4-6)
where
175
£7. are total elastic strains including the ~J
current load increment.
The definition of modified total strain is
£! . ~J
£ •. - £~. ~J ~J
(4-4-7)
Considering £7. satisfy the Hooke's law (3-2-8), ~J
one can rewrite Eq. (4-4-7) as
£! . ~J
The equivalent strain is defined by
(4-4-8)
(4-4-9)
where e!. is the modified strain deviator tensor: ~J
(4-4-10)
Therefore, plastic strain increments are computed
by the following expression derived from the well-known
Prandtl-Reuss equations [63,64],
(4-4-11)
where ~£p is the equivalent plastic strain increment: e
I (4-4-12)
176
which can be calculated through the uniaxial stress-
strain curve. Thus one has
3G e:et - 0e 3G + H' (4-4-13)
where H' is the slope of uniaxial tensile curve given
by
H' do e
de: P
one can find H' from (4-4-2) after considering (4-4-5)
as
H' (4-4-14)
One has now all the relationships needed. For a
step and during the iterative procedure, the convergence
of ~e:P can be checked at the nodes. The recent procedures e
can be described as follows:
A. After the stresses are obtained from the boundary
element formulation, e:ij can be computed from (4-4-8).
B. The equivalent strain e:et is calculated with Eq.(4-4-9).
C. Next, from Eq. (4-4-13) the equivalent plastic strain
increment ~e:P can be obtained. It must be greater e
than zero, otherwise set it equal to zero.
D. Finally, the plastic strain increment ~e:lj can be
found from Eq. (4-4-11).
Repeat the above procedures until ~e:P are convergent e at every node.
177
4.5 THE GOVERNING EQUATIONS FOR ELASTO-PLASTICITY
In this section the basic differential equations and
boundary integral equations for elasto-plasticity are
presented. Here only some formulations are given without
detailed derivations, which can be found in reference
[12] •
First one define the strain increments as:
~(l1u .. + l1u .. ) ~,J J,~
(4-5-1)
where l1£ij , l1£~j and l1£lj are total, elastic and plastic
strain increments respectively.
The equilibrium equation in incremental form can
be rewritten as
l10.. . + l1 b J. ~J,~ o
and on the boundary one has
in n (4-5-2)
on r (4-5-3)
The Hooke's law (3-2-7) is still valid for the
elastic part:
l10ij = 2~(l1£ij - l1£lj) + A(l1£kk - l1£kk)&ij
(4-5-4)
The above Eq.(4-5-4) can be rewritten as:
l10 .. ~J
(4-5-5)
where l101 j are the components of the initial stresses,
which satisfy relationship:
t:.of? • 1.)
178
(4-5-6)
In the above equations t:.of? and t:.£f? can be 1.) 1.)
considered as "initial stress" and "initial strain"
terms.
Eq.(4-5-5) is valid for 3-D and plane strain problems,
but for plane stre$V must be replaced by v = vft1+v)in the
expression of A in (3-2-9).
The t:.£~k in Eq. (4-5-4) is as follows [12]
(4-5-7)
(for plane stress)
substituting Eq. (4-5-5) into Eqs. (4-5-2) and (4-5-3)
and considering Eq. (4-5-1), one obtains
G • + __ G__ • uUj,kk 1-2v UUk,kj 2G t:.£f? . - b.
1.),1. )
and (4-5-8)
+ G(t:.u .. + t:.u .. In. 1.,) ),1.)
(4-5-9)
Eq. (4-5-8) is the incremental form of the Navier
equation and Eq. (4-5-9) is incremental form of the traction
boundary condition.
179
The above expressions can alternatively be written
in the following form:
G A + _G_ 6 Lluj,kk 1-2\1 Uk,kj - 6'0.
J (4-5-10)
and
6Pi (4-5-11)
where 6b j and 6Pi are fictitious body forces and fictitious
tractions defined by
and
lib J. - 6oi? . 1J,1
(4-5-12)
(4-5-13)
After the concepts of fictitious body forces and
fictitious tractions are introduced in expressions (4-5-12)
and (4-5-13), one can find that there are no substantial
differences between the Navier equations (4-5-10) and
(3-2-10) and traction conditions (4-5-11) and (3-2-11)
except for the following:
A. Now in the incremental form.
B. The fictitious tractions and body forces instead
of previous actual ones.
All the boundary integral equations, fundamental
solutions and boundary implementations mentioned in
Chapter 3 are still valid for plasticity in the incremental
180
form with the concept of "fictitious":
c .. L'lU J. 1J I pij L'lU j df + I Uij L'lb j dQ f Q
(4-5-14)
Using the concepts of initial strains or initial
stresses instead of using fictitious force, one produces
the following boundary integral equations:
c .. llu. I u~. llpj df - I pij L'lu. df + 1J J 1J J
f f
+ I u~. lib. d~ + I ajki llE~k dQ 1J J
(4-5-15)
Q ~
c .. llu. I u~. lip. df- I pij llu. df + 1J J 1J J J
f f
+ I u* L'lb. d~ + I E'\. L'la~k dQ J J 1
(4-5-16)
Q ~
where Eijk and aijk are the same as Eqs. (3-3-6) and
(3-3-7) .
By comparing with the elasticity formulation, one
can see that there is another term in the domain integral,
which is due to the plasticity phenomenon. This term is
usually computed using subdomain integrations. However,
using the transformation technique previously discussed,
one can transform these terms into boundary integrals.
As one can not have the analytical expressions of plastic
stress or strain increments , during the iterative
181
procedures, one obtains the values of 601j and 6Elj
at a certain number of points only. The locations of
these points are chosen in advance. Hence, the Fourier
coefficients must be evaluated numerically and this will
be discussed in the next section.
The boundary element methods to solve the boundary
integral equation (4-5-14), (4-5-15) or (4-5-16) are
completely similar to those mentioned before, so one does
not need to repeat them in this chapter.
4-6 NUMERICAL ANALYSIS USING FINITE FOURIER SERIES
Let us suppose again that b(x) is of the period 2n
as in Section 2-5, but that its values are known only at
a discrete set of equally spaced points in this period
interval [65,66), i.e. at the 2N+l points
in/N i -N, -N+l, .•. , O,l, ••• ,N
According to the Fourier theorem, Fourier series converges
to the average of the right- and left-hand limits of b(x)
at each point where b(x) is discontinous, so one sets
b(-n) bIn) ~[lim b(x) + lim b(x)) (4-6-1)
After that, one has 2N independent points, which may be
expected to determine the coefficients of 2N terms in
the finite Fourier series:
b(x) kl cos nx + n
N-l I
n=l k 2 sin nx n (4-6-2)
182
Let us denote the ith coordinate as
in/N i -N+l, -N+2, ... ,D,1, ... ,N
so that the 2N independent values b i
The finite set of functions
(4-6-3)
b(x i ) are prescribed.
{I, cOS(nx), sin (mx) I n=l, ... ,N m=l, ... ,N-l}
is an orthogonal set under summation over the set of
points (4-6-3):
N t: n = m 1, ... , N-l L sin nx. sin mx.
i= -N+l 1 ]-# n m
N L sin nX i cos mx. D
i= -N+l 1
N
{,: n = m 1, ... ,N-l
L cos nx. cos mx. n = m D,N i -N+l 1 ~
n ,m
(4-6-4)
Now an approximation is assumed in the form of
b(x) =
N ko + L
n=l kl cos nx +
n
N-l L sin nx
n=l
(4-6-5)
and the least-square criterion is adopted as follows:
183
N N L [b(x i )- ko L kl cos nXi
i= -N+l n=l n
N-l L k 2 sin nXi )2 min
n=l n (4-6-6)
This leads to the following conditions:
N N L [b(xi ) - k L kl cos nx.
i= -N+l a n=l n ~
N-l L k 2 sin nxil 0
n=l n
(4-6-7)
N N L cos pxi[b(xi ) - k L kl cos nXi
i= -N+l a n=l n
N-l L k 2 sin nxil 0
n=l n
P l, ••• ,N
N N L sin pxi[b(xi ) - k L kl cos nx.
i= -N+l a n=l n ~
N-l L k 2 sin nxil 0
n=l n p l, ••• ,N-l
Considering Eqs. (4-6-4) and (4-6-7) , one obtains
1 N
k 2N L b(xi ) a i= -N+l
1 N
kl N L b (xi) cos nx. n = l, ••• , N-l n i= -N+l ~
(4-6-8)
1 N
kl 2N L b (xi) cos Nx. N i= -N+l ~
184
1 N -N L b(xl.') sin nxl.'
n= -N+1 n = 1, ..• ,N-1
A derivation for two-dimensional functions is
completely analogous to the one shown above for a
one-dimensional function and one has
where
4 N ko + L L
JI.=l n=l
k~ Nm
1 N
k ~ = 0 nM
4NM L n= -N+1 m
4 N M L L L
JI.=l n=l m=l
(4-6-9)
n=l, ... ,N m 1, ••• ,M
n = 1, ••• ,N-1
JI. = 1,3
1 N M NM L L b(x , x ) fJl. (x , xm)
n= -N+1 m= -m+1 n m nm n
1 N M 2NM L L
n= -N+1 m= -M+1 b(x n
JI. 1,3
1 N M 2NM L L b(xn
n= -N+1 m= -M+1 xm) fJl. (x , xm) Nm n
JI. 1,2
1 N M 4NM L L b(xn n= -N+1 m= -M+1
(4-6-10)
1
185
Using formulations (4-6-10), one can calculate the
coefficients of finite Fourier series from several
discrete values of the function and express the function
approximately as a Fourier series. Hence, in elasto-
plasticity, after every step in the iteration, one can
obtain the plastic strains and plastic stresses at certain
points inside the domain and on the boundary, so that
essentially, one can compute the domain integral terms
in Eqs. (4-5-14), (4-5-15) and (4-5-16) with the trans-
formation technique mentioned before.
If the period of the functions bi(x) is not equal to
2n, the transformation can be carried out as shown in
Eq. (2-5-6) of Section 2-5. Similarly the corresponding
finite Fourier expansion and the formulation of Fourier
coefficients can be obtained.
4-7 APPLICATION TO ELASTO-PLASTIC PROBLEMS
As mentioned in Section 4-5, the boundary integral
equation for elastoplastic problems can be equivalent
using (4-5-14), (4-5-15) or (4-5-16). There are two
kinds of domain integrals in each of those formulations.
One kind is due to the body forces b i and can be trans
formed into boundary integrals with the formula (3-7-32),
which has already been discussed before. Another is due
to the plastic stress increments 6a~. or the plastic 1)
strain increments 6£lj. They can also be computed
with the same technique. For example, by means of initial
strain one can compute the last domain integral in Eq.
(4-5-15) with formula (3-7-33) because aijk
186
However when one calculates the stresses at internal
points, there is a domain integral in the initial strain
formulation, i.e. [12]
(4-7-1)
In order to compute it using boundary integrals, a
derivation analogous to the one shown in Eqs. (3-7-32) to
Cl·/-3l j.s u:1~d. The space derivatives of (3-7-33) with
respect to the spatial coordinate, yields a new trans-
formation formula.
Thu:'> in principle, using any of the formulations,
(i.e. initial strain, initial stress or fictitious body
forces and tractions), the domain integrals involved in
the boundary integral equations can be taken to the boundary.
In the present section, only the fictitious body force
and traction formulation (4-5-14) is discussed mainly
because of its simplicity, i.e.
A. After replacing b i and Pi by bi and Pi' the
formulation is the same as that of elasticity.
B. In the expression of fictitious body forces (4-5-12)
there are derivatives of the plastic strains ~£~ •• , ~],~
which are difficult to obtain numerically, but when
the finite Fourier series of the plastic strains
in the form of (4-6-9) is used, it is easy to evaluate
analytically these space derivatives.
After solving Eq. (4-5-14) , one obtains the displace-
ments and tractions at all nodes on the boundary. Then
the internal stresses can be computed by using Eqs.(4-5-1)
and (4-5-4) as follows [12]:
187
60ij J Uijk 6Pk df - J pijk 6uk df +
f f (4-7-2)
+ J Uijk 6bk dn - Cijk~ a
6Ek~ n
EXAMPLE 4-7-1
The simplesiexample is a rectangular specimen under
tensile force. The geometry of the specimen, mesh
points and internal points are shown in Fig. 4-7-1.
The material properties are
A. The body is composed of an elastic perfectly plastic
material (shown in Fig. 4-7-2 A), i.e. El = O.
B. The material has hardening properties (shown in
Fig. 4-7-2 B), i.e. EI
and, in addition, E = l~, v = 0.33. Yield point 00
1.3*10~.
The final displacement is 6 = 0.015.
The solutions are shown in Fig. 4-7-3.
EXAMPLE 4-7-2
In this example, a solid plane strain specimen is
indented by a rigid flat punch (shown in Fig.4-7-4).
The properties of material are the same as those in
Example 4-7-1 (Fig. 4-7-2 B). The final displacement of
the rigid punch is 6 = 0.02. The internal points, at
which the plastic strains are computed and the Fourier
coefficients are evaluate, can be seen in Fig. 4-7-4. The
dimensions are as follows:
188
• intemal p:>ints
x Iresh points
.- . I 1 I I I
..----T--'- --
I I 2 I I
~---t- - - -+ - ___ l-
I I I I 1 1 1/
3
ct. Figure 4-7-1 Geometry of the specimen in Example 4-7-1
a
y
189
a
y
~L-____________________ ,-£ ____________________
Figure 4-7-2 Uniaxial stress-strain diagram in Examples4-7-1 and 4-7-2.
6 .... ., 0
B 0 0 M r-~ ~ -" a g " ., I"il
Z9L
M M
,,; " "
cr,
~1 I I I I
\~
190
..... I
rI ... Cl) ..... 0.
~ ><
I"il
'" o s:: o .... .... ::I ..... o ., Cl) .r: E-o
M I ..... I ...
o
00
o
.... o
N .,
o
B 13.5
h 8.5
b 6
191
Accurate results are shown in Fig. 4-7-5. By
comparison with the results presented in reference [121,
the tractions and plastic strains at the point A (in Fig.
4-7-4), at which the solution is discontinuous, are
less than 5% different.
192
x mesh points
• internal points
. I 1
I
H-++~ H--tt-H-i I~ I ._-LLj ____ . ~~: _____ b __ ~ ___ ·_-_____ B ____________ ~
Figure 4-7-4 Plane strain punch problem
h
193
>I,Q
I
\ N I
I"-I .... .... _0
0
\ Q) .... 0. e ., I< iii
..... 0
~
\ 0 .r! .oJ ::J .... 0 M
VI -0
\ Q) 0
.<::
\ E-t
III
\ I
l"-I ....
\ . \ Q) J.< ::J
'", 0>
.r! r.. N
0
0
.", .
'" "" ."" '" _0 . 0
'" ~9L
"';1 .,; o
CHAPTER 5 PROGRAMMING
In this chapter, a general description of the programm
ing is presented. These programs were developed for
implementating the transformation formulae derived in
chapters two and three. They can be applied to solve
potential, elasticity and elasto-plasticity problems.
All these programs are coded in FORTRAN 77 and have
data structures suitable for most computers and micro
computers. For simplicity of preparation of the input
data file a special pre-processor is used [51].
5.1 POTENTIAL PROBLEMS
The program BEPOLI, for solving the potential problems
allows for general distributed source functions using a
linear continuous element. It consists of the following
subroutines.
1. Subroutine INPUT
The function of this subroutine is to read initial data
from input-data file.
The input data can be divided into the following
types:
A. General control numbers, such as the number of
boundary elements, the number of nodes on the boundary f of
the problem under consideration, the number of
internal points.
B. Basic numbers needed for the transformation formula,
such as number of boundary elements on f b , along
which the transformation formula is integrated, the
195
periods of source function in xl and x 2 directions,
the number of terms in finite Fourier expression Nand
M.
C. Geometry description, such as the coordinates of
nodes on boundary rand r b , the coordinates of
internal points, and the connecting node numbers
for every boundary element.
D. The coefficients of the Fourier series of the source
function.
E. Prescribed boundary conditions.
2. Subroutine MATC
The values of C are calculated for each node according
to the formulation (2-2-10).
3. Subroutine BFPO
The integrals due to the source function are computed
at every node required using transformation formula.
In this subroutine, a suitable number of integration
points is determined for each boundary element according
to the distance between the middle point of this element
and the source point. A judgement is made: if the source
point is located on the boundary element over which the
integrals are carried out, the transformation (2-7-3)
is used automatically, otherwise the original Gauss quad
rature is used.
4. Subroutine MATHG
According to the formulation (2-3-5), one computes
the elements of matrices Hand G through all the boundary
196
and assembles them into the global matrices Hand G. Then,
after substituting the prescribed boundary conditions and
adding the integrals due to the source function, a set
of linear algebraic equations is created by rearranging
the original matrices.
5. Subroutine SOLV
The function of this subroutine is to solve the Eet
of linear equations with pivoting.
6. Subroutine INTER
This subroutine computes the potentials at the
internal points. The difference between this program and
that for Laplace equation is only the last integral term
of Eq. (2-2-9), the values of which have been computed
after calling subroutine BFPO.
7. Subroutine OUTPUT
This subroutine arranges the output data including
the values of potential and flux at the boundary nodes
and the values of potential at the internal points.
A main program controls the above subroutines.
5-2 ELASTICITY PROBLEMS
The program BEELLI uses linear continuous element to
solve two-dimensional elasticity problems with general
body forces. This program consists of the following
subroutines.
197
1. Subroutine INPUT
Besides reading the initial data the same as that
in potential problems, this subroutine reads some basic
data from input-data file, such as properties of material
(Young's modulus, Poisson's ratio) and the type of problem,
i.e. plane strain or plane stress.
2. Subroutine MATe
In this subroutine, the matrices c(s), the components
of which are expressed in Eqs. (3-8-11), are computed at wbere
the pointsAintegrals (3-7-1) or (3-7-2) are required
during the computations.
3. Subroutine BFEL
In this subroutine, the integrals (3-7-1) are computed
at all the mesh points on the boundary r and the internal
points in the domain with the transformation formula
(3-7-32). The integrals (3-7-2) are computed at all the
internal points with the formula (3-7-33).
If the source point s is located on the element over
which the integrals are carried out, the transformation
(2-7-3) and the finite-part integral (3-8-6) are used.
4. Subroutine MATHG
This subroutine computes the matrices Hand G element
by element and assembles them into the global matrices H
and G.
Then, after substituting the prescribed boundary
conditions into the matrix equation and rearranging the
equation, one obtains a set of linear algebraic equations.
198
The values of integral (3-7-1) are added to the
corresponding locations on the right-hand side of the
linear equations.
5. Subroutine SOLV
This subroutine is the same as that used in program
BEPOLI for solving the set of equations.
6. Subroutine INTER
The functions of this subroutine are to calculate
the displacements and stresses with formulations (3-5-4)
and (3-5-5). The statements are almost the same as for
the case of the program without body forces except that
one now adds the last integrals of Eqs. (3-5-4) and
(3-5-5) to the solutions. The values of these integrals
have already been computed when calling the subroutine
BFEL.
7. Subroutine OUTPUT
The function of this subroutine is arranging the
ouput-data file, which produces the displacements and
tractions at the boundary nodes and the displacements,
stresses and displacements at the internal points.
A main program calls the subroutines listed above.
5-3 ELASTO-PLASTICITY PROBLEMS
The program BEEPLI uses the fictitious tractions
and fic-titious body forces applied to study elasto
plastic problems.
The program BEEPLI is based on the program BEELLI and
is extended to cover iterative and incremen~techniques.
199
In addition, a new subroutine is necessary in order to
obtain the Fourier coefficients numerically. The
differences between this program and the program BEELLI
are as follows.
1. Subroutine FOCO
The functions of this subroutine are:
A. After obtaining the plastic strains the Fourier
coefficients of plastic strain increments are
calculated with Eqs.(4-6-10).
B. Then, the Fourier coefficients of fictitious traction
increments ~p. and fictitious body force increments ~
~b. are combined with Eqs. (4-5-12) and (4-5-13). ~
2. Subroutine ITERA
From step A to step D, described in the last part
of section 4-4, is carried out for certain increment of
boundary conditions.
After executing subroutine ITERA, there are two
possible options.
A. If the equivalent plastic strain increment ~E~ converges,
(i.e. the differences of Eet between two adjacent
iterative steps are less than the prescribed error
at every node), the program proceeds to the next
incremental step.
B. If the solutions do not converge, the values of ~EP e
and ~Elj obtained during the last iterative step are
recorded and a new iterative procedure is started.
200
3. In addition to the above, the actual tractions are
calculated with Eq. (4-5-13) after each iterative step.
4. In the main program, the increment loops are added.
After obtaining convergent solutions, one can then add
the increment to the prescribed boundary conditions and
continue with the iterative procedures.
CHAPTER 6 GENERAL DISCUSSION AND CONCLUSIONS
The present work describes a way to transform domain
integrals into boundary integrals in boundary element
formulations. The transformation is presented for both
potential and elasticity problems with detailed treatments
of how they can be implemented numerically. The results
tested term by term demonstrate the accuracy of this
method. The transformation formulae are proved to be
reliable using the numerical methods and the programs
developed in this work. For the case of arbitrary source
or Body force functions, the solutions obtained by the
transformation technique are sufficiently accurate
provided that enough number of terms are taken in the
Fourier series. The examples presented demonstrate
that this type of transformation technique is feasible
for applications to Poisson's or elasticity type problems
including cases with arbitrary source or body force
functions. The numerical solutions are accurate for
engineering problems when a relatively low number of terms
of Fourier series is taken. Therefore the transforma
tion technique is also efficient in computer time.
The obvious advantage of the approch is that only
the discretizatio~ of the boundary is necessary and
hence its efficiency is mainly due to the equations
having one degree less of dimensionality.
The approach can be extended to the non-linear or
time-dependent problems. In these cases the Fourier
coefficients need to be calculated numerically which
202
presents the following problems:
A. In order to obtain accurate Fourier expansion
numerically, a sufficiently large number of integral
points must be taken and both operations, i.e.
computing the solutions at the internal points and
finding the Fourier coefficients require a consider
able amount of CPU time.
B. After each step of iteration, the Fourier coeffic
ients must be recalculated. This additional
computation makes the technique less efficient than
the original domain integral method.
In spite of the above, the transformation technique
presents a general and alternative way to solve non
linear problems without having to discretize the domain
cells.
Throughout this work, the boundary element integrals
can be computed numerically following two different
approaches: a. using direct numerical integrations to
compute the boundary integrals presented in the trans
formation formulae, which contain the harmonic functions
for Fourier series. b. using polynomial interpolation
functions one can express the harmonic functions over
each element in terms of nodal values, with which the
boundary integrals can be computed using a series of
influence coefficients and nodal values of the harmonic
functions.
The former approach is used throughout this work.
The advantage of it is great accuracy even using coarse
boundary elements. The latter approach is more efficient
than the former; particularly in non-linear problems
203
the integrals have to be computed at each itera~ion.
In this case, the computations for the transformation
formulae will consume almost the same time as taking
only one term in the Fourier expansion. It can be done
as further work including the error analysis of results.
Throughout this work, harmonic functions are chosen
to expand the source or body force functions. One of
the advantages of this set of functions is that it is
easy to find out the solutions which satisfy the original
differential operators of the problems. Another is that
arbitrary functions can be expanded as a convergent
Fourier series in most practical engineering problems.
Despite these obvious advantages, there exist
certain disadvantages for this technique. One of them is
that convergence may not be fast enough in some particular
cases, such as for discontinuous functions. Another
disadvantage is that it is necessary to find out the
Fourier coefficients required in the transformation.
One way of increasing the efficiency of the method is
to compute the Fourier coefficients analytically, which
can only be done in the case of having analytical
expresions for the source or body force functions. Other
wise, a numerical approach must be used. This usually
leads to loss of accuracy and requires considerably
more CPU time
A suggestion for further work will be to investigate
the numerical behaviour and computer efficiency of complete
functions other than Fourier series. This work may intro
duce new difficulties but it is worth pursuing.
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Lecture Notes in Engineering
Edited by CA Brebbia and SA Orszag
Vol. 1: J. C. F. Telles, The Boundary Element Method Applied to Inelastic Problems IX, 243 pages. 1983.
Vol. 2: Bernard Amadei, Rock Anisotropy and the Theory of Stress Measurements XVIII, 479 pages. 1983.
Vol. 3: Computational Aspects of Penetration Mechanics Proceedings of the Army Research Office Workshop on Computational Aspects of Penetration Mechanics held at the Ballistic Research Laboratory at Aberdeen Proving Ground, Maryland, 27-29 April, 1982 Edited by J. Chandra and J. E. Flaherty VII, 221 pages. 1983.
Vol. 4: W.S. Venturini Boundary Element Method in Geomechanics VIII, 246 pages. 1983.
Vol. 5: Madassar Manzoor Heat Flow Through Extended Surface Heat Exchangers VII, 286 pages. 1984.
Vol. 6: Myron B. Allen III Collocation Techniques for Modeling Compositional Flows in Oil Reservoirs VI, 210 pages. 1984.
Vol. 7: Derek B.lngham, Mark A. Kelmanson Boundary Integral Equation Analyses of Singular, Potential, and Biharmonic Problems IV, 173 pages. 1984.
Vol. 8: Linda M. Abriola Multiphase Migration of Organic Compounds in a Porous Medium A Mathematical Model VIII, 232 pages. 1984.
Vol. 9: Theodore V. Hromadka II The Complex Variable Boundary Element Method XI, 243 pages. 1984.
Vol. 10: C. A. Brebbia, H. Tottenham, G. B. Warburton, J. M. Wilson, R. R. Wilson Vibrations of Engineering Structures VI, 300 pages. 1985.
Vol. 11: M. B. Beck Water Quality Management: A Review of the Development and Application of Mathematical Models VIII, 108 pages. 1985.
Vol. 12: G. Walker, J. R. Senft Free Piston Stirling Engines XIV, 286 pages. 1985.
Vol. 13: Nonlinear Dynamics of Transcritical Flows Proceedings of a DFVLR International Colloquium, Bonn, Germany, March 26, 1984 VI, 203 pages. 1985.
Vol. 14: A. A. Bakr The Boundary Integral Equation Method in Axisymmetric Stress Analysis Problems XI, 213 pages. 1986.
Vol. 15: I. Kinnmark The Shallow Water Wave Equation: Formulation, Analysis and Application XXIII, 187 pages, 1986.
Vol. 16: G. J. Creus Viscoelasticity - Basic Theory and Applications to Concrete Structures VII, 161 pages. 1986.
Vol. 17: S. M. Baxter C. L. Morfey Angular Distribution Analysis in Acoustics VII, 202 pages. 1986.
Vol. 18: N. C. Markatos, D. G. Tatchell, M. Cross, N. Rhodes Numerical Simulation of Fluid Flow and Heat/Mass Tranfer Processes VIII, 482 pages. 1986.
Vol. 19: Finite Rotations in Structural Mechanics Proceedings of the Euromech Colloquium 197, Jablonna 1985 VII, 385 pages. 1986.
Vol. 20: S. M. Niku Finite Element Analysis of Hyperbolic Cooling Towers VIII, 216 pages. 1986.
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