Bem Ultimo.pdf

APPLICATION OF BOUNDARY ELEMENT METHOD FOR SOME 3-D ELASTICITY PROBLEMS by KAPPYO HONG,

B.E., M.E.

A THESIS IN CIVIL ENGINEERING Submitted to the Graduate Faculty of Texas Tech University in Partial Fulfillment of the Requirements for the Degree of MASTER OF SCIENCE IN CIVIL ENGINEERING Approved

Accepted

Dean of the Graduate School

August, 1985

ACKNOWLEDGEMENTS I wish to express chairman

sincere appreciation

Dr. C. V. G. Vallabhan

guidance during research. Vann and

for his

to my committee

encouragement and

Also thanks are due to Dr. W. P.

Dr. W. K. Wray for their helpful suggestions.

I thank

my parents

for instilling

in me

education and for their financial support. my wife, Miroo,

and brother, Sunpyo,

support and encouragement.

IL

the value of

Also I thank to

for their

unwavering

TABLE OF CONTENTS page ACKNOWLEDGEMENTS

ii

LIST TABLES

v

LIST FIGURES

vi

CHAPTER I. INTRODUCTION

1

Purpose Of Study

2

Scope of Rearch

3

II. BOUNDARY ELEMENT METHOD

4

Introduction

4

Basic Equations in an Elastic Continuum

4

Equilibrium Equations

4

Strain-Displacement Equations

6

Stress-Strain Relations

7

Equations on The Boundary

8

Formulation Of Equations For The Boundary Element Method

9

Formulation of Boundary Integral Equations

9

Boundary integral Equation for an Interior Point Fundamental Solution

13 14

Boundary Integral Equation for a Boundary Point 111

17

page III. NUMERICAL ANALYSIS

21

Introduction

21

Discretization

22

Determination of Internal Displacement and Stresses

26

Numerical Integration

27

IV. EXAMPLE PROBLEMS

33

Introduction

33

Example Problem 1: Prescribed Traction on the Top Surface of Large Elastic Soil

34

Example Problem 2: Embedded Foundation With Prescribed Displacements on The Top Surface of Large Elastic Soil

45

Example Problem 3: a Solid Cube (Uniformly Loaded on one side only)

48

V. SUMMARY, CONCLUSIONS, AND RECOMMENDATIONS

59

Summary

59

Conclusions

59

Recommendations

60

REFERENCES

61

LIST OF TABLES Table

Page

1. Comparison of Vertical Stresses Between Analytical Solutions and Boundary Element Method due to Surface Loading

41

2. Comparison of Vertical Stresses due to Displacements

50

3. Comparison of Vertical Displacements at Internal Points due to Vertical and Horizontal Displacements

53

4. Results For Example 3 Subjected To Axial Tension...

57

y

LIST OF FIGURES

Figure

Page

1. Stress Tensor in Cartesian Coordinate System

6

2. Definition of General Problem

10

3. Definition of the Displacement and Traction

15

4. Kelvin's Problem

16

5. Augmented Boundary Surface

18

6. Definition of Spherical Coordinate System

18

7. Quadrilateral Boundary Element

23

8. Quadrilateral Element In Dimensionless Coordinate System

28

9. Idealized Semi-Infinite Region Problem

35

10. Discretization of Example 1

36

11. Discretization Top Surface

37

12. Boundary Conditions and Applied Loads

39

1 3. Applied Moment in Example 1 14. Comparison of Vertical Stresses due to Vertical Traction

40

42

15. Comparison of Vertical Stresse due to Horizontal Load

43

16. Comparison of Vertical Stresses due to Moment

44

17. Discretization of Top Surface of Example 2

45

18. Discretization of Finite Element Model 19. Boundary Conditions and Application of Displacements

47

20. Comparison of Vertical Stresses due to Vertical Displacements vi

49

51

Figure

Page

21 . Comparison Of Vertical Stresses Due To Horizontal Displacements

52

22. Comparison Of Vertical Displacements At Internal Points Due To Vertical Displacements

54

23. Comparison Of Vertical Displacements At Internal Points Due To Horizontal Displacements

55

24. Surface Elements Arrangements for Example 3

58

VLl

CHAPTER I INTRODUCTION For solving complicated equations representing physical problems, engineers and scientists have been using numerical techniques for several years. Most of them were using finite difference, solving

finite element

these

researchers

problems.

have

come

or

other numerical methods for

During

up

with

boundary element method [1,2,3]. ment

of

the classical

the a

last

decade,

technique

some

called

the

This method is an improve-

integral equation methods

used

in

mechanics and physics. The improvement in these methods lies primarily in the computerization of the method, so that method

can be

used

for

a wide

variety

of

problems

the in

continuum mechanics. Researchers will asic difference,

the

question:

while

the

finite

finite element and other numerical methods have

been successfully used in solving a wide variety of problems in mechanics and while their versatility has been established,

why develop another method ?

for the development of the

There are many

boundary element

reasons

method.

Most

numerical methods such as finite difference, finite element, etc.,

are

domain

methods,

and

discretize the whole domain. method,

the domain

they

But in the

equations are 1

require boundary

converted

that

one

element

into boundary

integral equations. If these integral equations are boundary integrals, be

then only the boundary of the continuum needs to

discretized.

The method

can be applied to solve a few

problems involving infinite continua easily The method in the

is gaining

literature

popularity

and

[3,5,11,13,15].

and accurately.

is ;videiy

reported

The advantages of

the

boundary integral methods over the domain methods are: 1. There

is a considerable reduction

unknowns to solve a problem; dimensional method

continuum

requires

surfaces of the domain.

the number of

for example, in a three

problem

elements

in

and

the boundary element nodes

only

on

the

This becomes very convenient

for input data preparation. 2. For many problems, one can get more accurate

results

using the boundary element method for the same number of unknowns and computer time. 3. For many problems in mechanics,

for example in fluid

mechanics, the input-output flow data are very nicely and accurately represented. 4. Modeling certain semi-infinite and infinite domains having linear properties is comparatively easy. Purpose of Study There

are

many problems

in solid mechanics where the

boundary element method can be laore advantageous

than other

numerical methods. relates

One such series of engineering problems

to those

in soil and rock mechanics

mass of solids is subjected on the surface. tions,

to forces

where a large

in a restricted area

Classical mathematical

closed form

solu-

such as Boussinesq, Kelvin, Mindlin, Cerutti, etc.,

[8,16,17] are

applicable

only to a limited geometry of the

continuum. If the loading surface is embedded into the semiinfinite continuum, structure, solve

the

continuum. to

like a footing or any other

one has to resort stresses The

and

to numerical

displacements

foundation

techniques

induced

boundary element method has

in

to the

been reported

be very successful in solving such problems. Scope of Research In this research,

the boundary

for solving a three dimensional linear,

elastic,

solid continuum,

homogeneous, and

program is developed

element method is

isotropic.

for this purpose.

which

used is

A computer

A detailed descrip-

tion of the theory of the integral equations in the boundary element deals

method with

the

is presented numerical

formulating the discrete Chapter IV

discusses

in Chapter II. integration

Chapter III

procedure

used in

boundary element system equations.

results of a few example

problems to

validate the overall methodology.

All conclusions

from this research are

in Chapter V

summerized

some recommendations for future research.

derived

along with

CHAPTER II BOUNDARY ELEMENT METHOD

Introduction It is found convenient

to present the

basic equations

in the theory of linear elasticity before the integral equations

for the

summary

boundary

of the basic

element

equations

method are showing the

presented.

relationships

between stresses, strains, displacements, and surface tions in a three dimensional Using these equations, derived.

continuum

the boundary

required in the

trac-

are presented here.

integral equations are

Finally, the fundamental solution

static problem

A

for an elasto-

formulation of the boundary

element method is also presented at the end of this chapter. Basic Equations in an Elastic Continuum The reference coordinates used here to represent a three dimensional

elastic

continuum

orthogonal cartesian axes denoted or simply by x. (1=1,2,3).

consist of

three

mutually

by x, y, z, by x^, x^,

Throughout this work,

x^

cartesian

tensor notation will be used for convenience with a range of 1 to 3 for the index notation. Equilibrium Equations Consider an infinitesimal parallelepiped body

referred

to a cartesian

coordinate system

x^,(i = 1,2,3).

The derivation

isotropic continuum is given

as shown in Figure 1 with of the

aquations

for

an

in standard text books [6,7,8]

and, hence, is not produced here.

The differential equation

of equilibrium at an interior point in the continuum

can be

written as: 3a 3T XX ^ xy 3a

3T

3x 3T

^

xz

"iT Here,

3T

3T

3y 3T

xz

^

^^'^'

^ ^ " °

3a zz

yz

* ^

3z

* -^T

* 2 =0

a / CT,,,, t^„„ ^re normal stresses XX yy zz

are shear stresses.

and

, T , T xy' yz' xz

T

X, Y, Z are body forces per unit volume

at that point. These equations can be symbolically rewritten using index notations as a. . . + b. = 0 13/3 1 where

a.

represents

the

stress

components

on a

face

perpendicular to the x. axis and in the direction of the x. 1 3 axis as shown in Figure 1. The positive directions of a,. 13 is also shown

in Figure 1.

volume in the i-direction.

b. is the body force

The comma after ij in the above

equation represents differentiation respect

to the corresponding axes

script following the comma.

per unit

of the stress a., represented

with

by the sub-

Figure 1. Stress Tensor in Cartesian Coordinate System Strain-Displacement Equations If u, V, w ment

or

functions in

displacements at

u. (i = 1,2,3) are continuous displacean elastic

continuum

representing

the

x., assuming that the body undergoes small

displacements, the components of strains at any point x. can be represented by derivatives of the displacement components

7 These relations

between

strain

and

^xy " 3x

3y

c- - IZ

^,

3v

S - 3y

V

displacement

are

as

follows [9,10]: X

3x

3w z ~ 3z Here, e , e , e X y z are

shear

3w

/O ox

= 3? " 3l

^^-^^

_ 3_u ^ ^zx ~ 3z "^ 3x are

strains.

normal

strains

and

Y / 'xy

Y / Y 'yz 'zx

These equations

can

be symbolically

rewritten as £.. .) 13 =Tr(u. 2 1,3. + u .3,i' where £..

is a strain tensor.

If i = j ,

£.. represent components of normal strain and

if i ^ j, they represent components of shear strain.

It is

to be noted that Y-• is the engineering

shear strain

which

are the corresponding

shear

is equal to

2£.. , where z. .

strain components. Stress-Strain Relations The next step in solving a problem of elasticity is

to

develop the stress-strain relations. Assuming a homogeneous, isotropic, linear, and elastic material, the general form of Hooke's law may be expressed as 'ij = ^ ^kk ^i: * 2G e,.

(2.3)

8 where A. and G are Lame's constants given by X = .....y^

...

,

(1+v)(1-v) '

where

E

is

the Young's

G =

2

^ ~ 2(1+2v)

modulus

of

elasticity and

Poisson's ratio of the material in the continuum.

v is

A general

form of the linear relationships of stresses and strains can be written as a. . = C. ., , e, , where 13 13 kl kl material property tensor.

C. ,, , is a fourth order 13 kl

Equations on The Boundary To solve an elasticity problem, we need equations on the boundary.

There

are

two

types of

boundary

conditions

prescribed: 1. Prescribed displacements on the boundary, T u.=u. i

l

,

o n f . .

i.e., (2.4)

l

2. Prescribed tractions on the boundary

f^.

These pre-

scribed surface tractions p. are related to the stress tansor as follows: Pi = ^11 ^1 * ^12 ^2 ^ ^13 ^3 ^2 = "^21 ""l "• ^22 "^2 " "^23 ^'^3

^'^

^2

^^'^^

P3 = '^31 ^1 * ''32 ^2 ^ ^33 ''^3 or symbolically p.=p.=a..n. on F-, '^i ^1 13 3 2 where n. are the direction cosines of the surface

F^.

Formulation of Sguations For The Boundary Element Method To derive

the boundary

integral equation,

a weighted

residual method is used here.

This is one of the classical

approximation

in numerical mathmatics

techniques used

and

can be applied to solve many types of equations. Explanation of this method [8].

along with some comparisons can

In the formulation of the boundary

be found in

integral equation,

an influence function is used as a weighting function.

The

details of the influence function will be discussed later. Formulation of Boundary Integral Equations For the general

three

body shown in Figure 2,

dimensional

isotropic

elastic

the field equations and correspond-

ing boundary conditions are: a. . . + b. = 0 ,

in the domain

Q

(2.6)

with boundary conditions u. 'i = u"i

on

r^ ' 1

(2.7)

p.. =- p. M.

on on

FF-, -,

(2.8)

2

1 ^ 1

where

F. is the boundary where displacements are prescribed

and F^

is the boundary where tractions are prescribed. The

entire boundary is F = F. '^ ^o If

u. is

assumed as the weighting functions,

weighted residual technique,

in

the

then the equilibrium equations

10

Figure 2: Definition of General Problem can be rewritten as + b. ) u. df2 = 0 (a iD/D 1 1 JQ In order to get the expressions the integral,

(2.9) for the above part

of

consider another integral of the type

(a.. u.) . d^ ID 1 /D M

(2.10)

Differentiating the integrand of equation 2.10, r (a., u*) . df2 =

f

(a.. . u*) dQ +

f

we have

a. . u. . dfi (2.11)

11 and (a. . u. ) . d^ = 13 1 ,3

JQ

(a u.) n. dF 13 1 3

• / ,

From equations 2.11

X

Q

{o

u.)

13r3

Substituting

p. u. dF r ^1 1

(2.12)

and 2.12:

dQ =

I

1

p.u! dF

a

u. . d^l (2.13)

J r ^1 1

equation 2.13

into the

equilibrium

equation

2.9: p. u. dF ^ ,F ^

/

a. . u. . ^^ ^'3

Q

d^

b. u. d!^ = 0 (2.14) Q ^ ^

Using the symmetry of the stress tensor, i.e. a.. = 13 1 ^ ^ I a . . u * . d ^ = I a. ^ (u, . + u. .) d^ 2 1,3 3/I a. . e. . d^ Substituting

equation 2.15

into

31

(2.15)

the

equilibrium equation

2.14: r

p. u* dF -

r

a.. e*. d^

b^ u^ dJ^ = 0

(2.16)

^ijkl

Using

^

For a linear elastic material, a.

13

%1

this identity,

f

a. . e * . dfi =

f

C.i. 3.,k l, £,k l, £.13. df^

Q

\l

\ l ^"^

(2.17)

12 Once again, using the Divergence Theorem;

I:„ '^ij '^i'-j ^" = j r °tj "i "j ^' 'L p* u. dF

(2.18)

and differentiating by parts: I

(a. . u. ) . dS^ =: f

f

a*. . u. di:^ +

a*. £. . d!;^ ^^

Q

(2.19)

^^

Substituting equations 2.17 and 2.18 into equation 2.19: a.. £.. do =

/ p. u. dr -

/

a.. . u. dQ

(2.20)

Now, equilibrium equation 2.16 becomes

p. u. dr +

/

p. u. dr

a. . . u. di^ +

^ r Q

b. u* d^ = 0 ^

(2.21 )

^

or a. . . u . di^+

r

p . u , dr

/

=

r p. u* dr +

jr ^ "

r

b. u* dn

(2.22)

Ja ^ ^

In order to remove the first domain integral in the above equations, we can assume a*. . + A. -, (s,q) = 0 13/3

11

in

^1

(2.23)

13 if s ?^ q,

A^^

if s = q,

A.^ is a Direc delta function

such that

where

= 0

A.T d^

= 1, if i = 1

A^-j^

= 0, if i j^ 1

s is the source point, q is the field point, 1 is the direction of force at s,

which means that the solution to this equation in the domain is the solution due to direction.

one

of

solution, rather

at node s in the

i-

For every 1-direction (1=1,2,3) in source point,

we have 3 sets of any

a load applied

equilibrium equations and the solution of

equilibrium *

*

p.. and

u..,

equations which

from

the

fundamental

are second order

tensors

than a vector. Boundary Integral Equation for an Interior Point

For any internal within the domain becomes

-u., due

Q,

point

's' where

the domain

to equation

a force is applied

integral in equation 2.22 2.23,

and in

case of

the

absence of body forces, u*(s) + r p*.(s,q) u.(q) dr(q)= [ u..(s,q) p.(q) dF(q) 1 J r ^^ ^ JT -• -• (2.24) where

i is j is

the direction of a unit force at s, the direction of displacements or tractions on

14 the boundary, p^.(s,q) and u^.(s,q) are the traction and displacement on the boundary in the j-th

direction due to a

unit load at s in the i-direction The equation 2.25 gives

the displacements

at any point

inside the domain fi in terms of the boundary values u.

and

Fundamental Solution The fundamental solution at the field

point 'q' given by

Hi

u..(s,q) represents displacement field

point, q,

source

point, s,

due to

in the j-direction

a unit force

as shown

at the

in i-direction

in figure 3.

at a

This notation is

applicable to tractions as well. For an which is

elastostatic problem, the fundamental solution,

written

as equation

2.25,

has been

derived by

Kelvin and his solution can be found in several books [5,6]. * 1 ,,-, .,.v . u. . = . ^ „ , ^ 1— {(3-4v) A. . + 13 167rG(1-v)r ^ 13 * 1 r9r , ,. ^ , Pii = 2 ^ ^ {(1-2v)A

3r i:— 3x. A +

3r , ;c } 3x. ' -, 3r 3r , 3 T^p T^p}

-{(1 - 2v) T ^ n. - ^ n. }] 3x. 3 3x. 1 1 3 (2.25) where r is the distance between the source point 's' and the field point 'q' (see Figure 4), and

15

Figure 3: Definition of the Displacement and Traction

16

components of u,, and p^.

Figure 4: Kelvin's Problem

IK

17 / J/2 2 2 2 , , r = (r^ r^) = r^ + r2 + r3 = |s - q| r^ = x^(q) - x^(s)

(2.26)

^ r. r = — i ^ — = -i ,i 3x^(q) r

Boundary Integral Equation for a Boundary Point In

the

preceeding

sections,

the

boundary

integral

equation for an interior point was derived. Considering that the unknown tractions and displacements on the boundary must be calculated

to obtain the solutions in the domain,

equa-

tion 2.24 must be modified in order to use on the boundary. Equation 2.24 is valid when the source point is the domain ^. r,

But when this point is moved to the boundary,

one of the integrals in equation 2.24 becomes

i.e.,

if the source point is equal to the field

such a case,

within

special techniques

must be

singular, point.

used to

In

evaluate

these singular integrals. Assuming Figure 5, as the

that

the body can be represented as shown in

the integral

sum of

on the surface

integral on the

These integrals are

evaluated

can

be considered

boundaries, at the

F-e

limit as

calculation of this integral e is faciliated spherical system of coordinates (Figure 6 ) .

and

e-)-0.

e. The

by employing a

i —

^

part of surface containing the point 's' Figure 5. Augmented Boundary Surface

^x

^2= ^ sin 9 sin ;^

Figure 6. Definition of Spherical Coordinate System

19 Consider

the

first

integral in

equation 2.24 as two

parts:

F-e and let

I =

lim

f

£ u.

p*.

dF ,

(2.28)

e *

Substituting for p.. from equation 2,25: I =

lim

u. ^ [- I ^ -[^ {(l-2v)A. . + 3 1 ^ | ^ } ^ f ^ {(1-2v)A. -» L^"^ •a^ I \ dx. ox. / I F^ 87r(1-v)r^ ^-'.j

- (1 - 2 v ) { ^ n. - ^ n.}dF] (2.29) 3x. 3 3x. 1 1 ^ 3 Now, for the particular case of the hemispherical region, 3x.

n. "1

r

=

e. '1

1

where e. are

the projection of the unit normal vector on i-

coordinate axes. r,.

The second term in equation 2.2 9 becomes, r,.

-

r,. r,.

And noting the fact that —^—

= 1,

=0 equation 2.30 becomes

I, = lim [ - r u.{(1-2v) A + 3r,. r,. } ^ ^ dF ^ €^0 -^ r ^ ^^ ^ ^ 87r(l-v)r^ (2.30) and

expanding

I = lim [ - I

£-0

equation 2.3 0

for the instance

when j = 1,

{u.^(1-2v) + 3u.^e.^e.^+ 3u2a.^ e2 +

JF^

. . sin8d8__d8 ,„, ^^3^1^3^ 87T(1-v) ^^^

,-, ^. . ^^'^^^

20 The integral is now independent of r and may be expressed in terms of

9 and (j) only.

with respect to The

same

9

result

and (J),

Integrating the equation 2.31

we get

applies for

I.= -0.5 u.

i = 2 and i = 3,

giving the

combined result as : I = -0.5 u.

(2.32)

1

Limitation of the right does

not affect

into equation

side integral

the integral.

2,24,

of equation 2.24

Substituting equation 2.32

the boundary integral equation

on the

boundary becomes C(s) u.(s) +r p*.(s,q) u.(q) dF(q)= ru*.(s,q) p,(q) dF(q) 1 Jr 13 3 Jr 13 3 (2.33) where C(s) = j

for a smooth boundary.

It should be noted that the smooth linear

boundary.

or higher

necessarily

lie

C(s) is equal to 0.5 only

And for the

order elements, on

a

smooth

general

case of

where the nodes do

boundary,

C(s)

for using not

cannot be

calculated easily in the case of three-dimensional problems. However,

calculation of

C(s) can be done

using a rigid body motion criterion.

alternatively by

CHAPTER III NUMERICAL ANALYSIS

Introduction In chapter II,

the boundary integral equation required

for the boundary element method

was derived.

The equation

2.34 is reproduced here for further development. In equation 3.1,

the

displacements at a

point s

on the

boundary are

related to displacements and tractions over the boundary and body forces over the volume: C(s) u^(s) + r p*j(s,q) Uj(q) dr(q)=

ru*j(s,q) p^(q)dr(q) (3.1 )

where

p^.(s,q)

and

and displacements,

u^.(s,q) are the fundamental tractions and

p.(q) and

and displacements on the boundary. tion

' j , ' either the

In each cartesian direc-

displacement

u.(q) are

precribed.

and

on the boundary,

0.5

u.(q) are the tractions

C(s) is equal

p.(q) or

the traction

to 1.0

in the domain

if the boundary

is smooth

as

mentioned before. Because the extremely

analytical

difficult,

integration of equation 3.1 is

a numerical approach

obtain a solution of the equation. in a three dimensional domain, 21

To solve

is necessary to this equation

the boundary which becomes a

22 surface

is discretized

using surface elements.

variations of tractions and displacements over can be obtained through The order of

kept

the

same

be the same in

each element

the use of interpolation functions.

the interpolation functions

and tractions can

Differeni:

this

for displacements

or different,

research

for

but they are

computational

and

programming efficiency. Using

Gaussian

integrations collection

are of

quardrature

performed

boundary

point in one element.

formulas,

sequentially

elements Then

the

for

over

tions on the

equation,

boundary

boundary

displacements at an internal point are necessary,

the entire

conditions

are

A X = B,

and

unknown displacements

are determined.

surface

a particular source

applied to form a system equation of the type by solving this

the

The

and trac-

stresses

calculated

and

wherever

using equations which are discussed later.

Discretization For a three dimensional

solid mechanics problem,

boundary is a surface and it can be divided into

the

a discrete

number of surface elements. In this resesarch, these surface elements are assumed to be plane quadrilateral elements with four nodes, which are further assumed to be on one plane. For

a typical

element

boundary values of u. and

as

shown

in

Figure

7,

the

p. defining the displacement

and

23 tractions

on the boundary are assumed

interpolation functions u. and

p. can

to vary according to

within the element.

be assumed

The values of

to vary depending on the

of degrees of freedom that the analyst

would like to imple-

ment in the model.

(x

number

31 ^32 ^33^

(^41 ^42 ^43^^

'^^U "^12 ^13^ (x 21 ""22 ""23^

Figure 7: Quadrilateral Boundary Element

24 Consider the case of boundary values of u. and discrete

o. on the '3

J

surface element given by an interpolation function

such that: u . =

3k

^j ^

k

jk Pk

(3.2)

where u^ and p, are the nodal values of displacements and tractions at the node n (n=l,2,3,4). Simply by choosing the appropriate interpolation function the

$,, or 4/., , 3k 3k'

surface

of

quadratic, etc.

the

u. and p. can have any 3 3 element

such

as

constant,

These functions are standard

functions similar

variation

on

linear,

interpolation

to those used in the conventional

finite

element formulation. Supposing

that for

the discretization

of the body,

M

elements on F are considered, and applying the equation 3.1 on the whole boundary with equation 3.2, we get: M ^ ^ C(s) u.(s) + Z { p..(s,m) ^.,(m) dF(m)} u,(m) "• m=1 J F "-^ ^^ ^ m M . (s,m) ^., (m) dF(m) } p!J(m) m=i 1=1

Ji J F_

'^

^^

"

m where LM

is the total number of boundary elements, and

F is the m-th boundary element, m

(3.3)

25 In

equation 3.3,

integrals

on the discrete surfaces.

are replaced by integrals

For the constant element in which

tractions and displacements are

constant over each element,

the equation 3.3 becomes

^

r

*

C(s) u.(s) + 2 {

^

p..(s,m) dF(m)}

m=1 JF M = 2 { m=l

m F

u.(m)

^^

^

u.,(s,m) dF(m)} ^^

p.(m) ^

(3.4)

For a domain with 'M' nodes, the equation 3.4, in matrix form,

becomes CU + fiU = GP.

If H = fi + C I, then the whole

equation becomes HU =: GP.

(3.5)

Equation 3.5 can be written where 3M comes

from the

as equation 3.6

number of

node multiplied by the number P are global vectors containing

in matrix form

degrees of freedom

of elements

M. • Here, U and

the displacements and trac-

tions on the boundary. ^ H H H

H

l.rl

H 1,2

H

2,1

H 2,2

H

3,1

3M,1

r

1,3

1 ,3M

u1

2,3

2,3M

u.

H 3,2

H 3,3

3,3M

u.

H

H

3M,2

3M,3

per

H

u 3M

3M,3iyi )

26

'1,1

'1,2

1,3

'1 ,3M

'2,1

'2,2

'2,3

'2,3M

'3,1

'3,2

'3,3

'3,3M (3.6)

'3M,1

^3M,2

Note.that tractions

^3M,3

M^ values of displacements and

(M^ + M^ = 3M) are

hence in the U and P vectors, may be

'3M,3M

gathered

into

known

3M M-, values of

on the boundary,

there are 3M unknowns,

a left-hand

side

vector X.

and which After

reordering the equations, we obtain, AX = B where A is

a square

vector X is tions.

matrix

formed by the

of 3M, unknown

The contributions of

included in vector B. yield

(3.7)

all remaining

the

fully

populated,

and

displacements and tracprescribed

Equation 3.7 may

values

now be

solved

are to

unknown displacements and tractions on

the boundary. Determination of Internal Displacement and Stresses To find the internal displacements discrete point equation.

within the domain,

and stresses at any

we can use the following

27 °VHn ki3 = -^a4 - ^ 1 - 2 ^ ) [A,.^3. r ,3. +A,k:3. r ,i. - A.13. r ,,K' ] + B r . r . r ,] ,1 ,3 ,k^

1 A r 4aiT(1-v)

(3.8) ^ '

. (1-2V) (6n^r^.r_. . n.A.^. n.A.^)

- n-4v)n^A..l

m=!l

,^J(,.,,

(3.9)

m=l

where D,ki3 ..

and

S,ki3 .. are the contributions of the k-th direction to a.. at internal point due to trac13

tions and displacements,

respectively.

a.. is the stress tensor at the discrate 13

point, and

u (m) is in i-axes at the .m-th boundary element, n r . 3r / 3x. 1

/I

ci= 1 , 3= 2, Y= 4

for two dimensional cases, and

a= 2, 3= 3, y= 5

for three dimensional cases.

Numerical Integration In assumed

this section, to

the shape of the surface boundary is

be a quadrilateral

on a plane

with four nodes

28 as shown

in Figure 8.

The quantities to be integrated in equation 3.4 are: p,, dF

and

/ .

4

/ s' dF JF.^1} 3

.3

«

0

-1 •1

•2

-1

Figure 8: Quadrilateral Element in a Dimensionless Coordinate System

29 *

where

u..(s,m) 13

discussed all

*

and

p..(5,m) are the fundamental ^^13

before.

In order

these functions must

coordinates ( r\ ,r\^)

is such

in Figure 8.

to perform

the integration,

transformed

to dimensionless

which are used to transform the quadri-

lateral into a square. ( n., , 12)

be

solutions

The set of dimensionless

that they vary from

Hera, the coordinate of

-1

coordinates

to

+1 as shown

the nodal points are

also interpolated such a s : X. =

N. X..

1

3

where

i = 1,2,3, r

j 1

f

3 =1,2,3,4

'

J

'

or x^ = N^ X^, . N , X^^ . N3 X j ^ - N4 X ,, X2 = N^ X,2 * "2 =^22 * ^

=^32 * ^4 ^ 42

'^•^'>

X3 = N, X,3 * N2 X23 * N3 X33 . N^ X 43 where N.1. are N^ = \

i'^ - n.^ ) (1 - n2)

N2 = ^ (1 + n^ ) (1 - 02) N3- = ^ (1 + n^ ) (1 N4 = ^ (1 - n^)(i

+02) .

n^)

and X.. is the cartesian coordinate of

the node ' j ' in the

31

'i' direction. Since terms of

the H,,

interpolation H^ coordinates,

the elements of area

dF

from

functions

are

expressed in

it is necessary to transform cartesian coordinates to the

30 n.^ ,

ri2

coordinates.

A

differential of area

dF

will

be

given by

dF =

dn•i-

1

->•

1^

3 3x.

3x.

3ni

3rii 1

3r|'1

3x.

3x-

3x.

3r|^

3n

3ri.

3x.

(3.12)

dn^ dri2

dn-

^

->

= g-li + 923 + g3k

dn^ dTi2 where 3x.

3x.

3x.

3x.

3rii

3n

3ri-i

3ri

3x.

3x.

3x.

3x.

3r|,

3r|

3n.

3n

3x.

3x.

3x-

3x,

3n-|

3ri

3n.

3n

2 ^

1 2 2

^

|G| is the Jacobian relating the elements of area in the two systems of coordinates. For the quadrilateral element. 3N

3x

1 3ri-

r-^(N.X. . ) 3 n -j 1 1T

3N

1

3n

=^11^

= r^(N.X,^) 3n-J i i 2

=

3n-,

1

=^21^ —

3N.

3N.

3x.

3^

X

12

3N^

3N

dT]^

1

=^ 31""

3N. X. 22

30^

sTT

X

41

3N X. "32

3rii

X '42

31 ^^3

3N2

9N^

3

3N2

3 ^ - 37T{^^i^i3^ = 3n7 '1 ^n"- J^ 3x.

3

9N

^22^

3N^

37i^ ^33'" T^

3N

3N

3N

3n:

3 X. 3r|2 31

21

^43

3ri2

X

41

3N. 3N. 3N ^^2 3 ^^1 -^'7 -'7 3 ^2 = 3^i^i^i2^ _ ^ ^3712 X 22^ _ + ^3n2 X _ ^ ^3n2 X"42 -M2 X x^ = a^r«n2 X **12^ ""32^

3x

3N.

3N

3N.

9n2 " 3n^^i^i3^ = 3TT^ ^13* 3n^ ^23* TT

3N

^33* ITT ^ 43

in matrix form we can write, 3N.

3N.

3N.

3N

3T?^ Sn"^ 3n"^

3n

3n

3n

3n

3x.

3N.

3N.

3N.

3N

3x

^2

3x_ 3x.^

3x_ 3x.^ ^2

3n

^2

3n.

3n

X1 1 X 21

X 22

X 23

X

X

X

31

3n

X 41

3N.

3N.

3N

3N^

X

3^ 3N.

3N.

3N.

3N^

3n"

3^

3^

3^ ;

1 2 X1 3

11

32

33

42

X 43

V

X1 2 X 1 3

X 21

X 22

'23

X 31

X 32

X 33

X

X

X ^'43

41

42

(3.13)

J

Substituting Jacobian

equation 3.13

| G | can be calculated.

^ C(s) u (s)+ I

i

m=l

{

m=l

= 2 {

JF

.m

r

into

equation 3.12,

the

Then from equation 3.4,

* p(s,m)

J Fm "-^

u.(s,m) 13

|G| dri.(m) dn..(m)} u.(m)

^

| G | dri,(m) dr|-,(m)} p. (m) ' ^ J

^ (3.14)

32 Replacing the integrals by summations again, M K ^ C(s) u (s) + Z { Z p..(s,m) |G|, (m) W, (m)} u.(m) i m=1 k=l ^^ K K 3 M K = 2 { Z m=1 k=1 where K

^ u..(s,m) |G|, (m) W, (m)} p,(m) ^^ K K 3

(3.15)

is the number of Gaussian integral points, and

W (m) is the weighting factor of m-th element, and | G L (m) is the Jacobian of m-th element. From the application of equation 3.15 to all M boundary elements, arises.

a

final system

of 3M

equations

(equation 3.7)

CHAPTER IV EXAiMPLE PROBLEMS Introduction The boundary integral equation was derived in and a numerical technique for

formulating

Chapter II

the equations in

discrete form has been described in Chapter III.

A boundary

element computer program using the Fortran language has been developed

to

solve

three

dimensional

problems where the material is and

homogeneous.

applied

to

linear, elastic,

The developed

solve the

following

continuum

a horizontal traction, embedded

foundation

subjected to

subjected

problems:

isotropic, will be

(1) an elastic

normal forces;

2) a

to a vertical traction,

and a moment on the surface;

in a semi-infinite

prescribed

continuum

computer program

solid in the form of cube subjected to semi-infinite

solid

displacements.

elastic

(3) an

continuum

The results

from

this program, are compared to those from analytical solutions or those from

the finite element method.

For each problem,

the entire surface of the continuum is disretized as surface boundary elements.

The required input data for the program

include: 1. Geometry A. Nodal numbers and B. Element numbers of surface elements 33

34 2. Material Properties A. Shear modulus and B. Poisson's ratio 3. Prescribed

boundary

values

in the form of surface

tractions or surface displacements for each element 4. Coordinates of internal points

where the

tractions

or displacements are desired. Example Problem 1; Prescribed Traction on the Top Surface of Large Elastic Soil The semi-infinite region to be solved by the discretization technique

requires

a finite boundary

at some

length

fairly far away from the region of applied tractions. classical solution, defining

the

established research, times

accurate

certain dimensions are established

finite boundary differently

for

a semi-infinite

the loaded

stresses and

for

and these lengths are to different

continuum

length as shown

results,

Using

be

problem.

In this

is dimensioned

as

in Figure 9.

5

For more

larger distances to the boundaries where

displacements are small

are necessary.

This

example deals with the prescribed tractions or moment on the boundary at the are

center of the top

surface and

the results

compared with known analytical solutions [11]. The discretizations of this example are shown in Figures

10

and 11.

elements

which

The continuum is consisting

discretized

of 57

elements

as 85

surface

at the top,

6

35

20

20

figure 9: Idealized Semi-Infinite ' Region Problem

36

20

20

iqure 10: Discretization of Example 1 Figur

37

^^'^^v^^ ^ • . : $ :

<J

^

^^i^ ^ ^

Figure 11: Discretization of Top Surface

38 elements for each side,

and 4 elements at the bottom.

total number of unknowns in this problem is 255 and,

The

hence,

the order of the G and H matrices is 255 X 255. Shear modulus used for this problem is 1000 psi and Poisson's ratio is 0.3. Loads are considered

to be applied at the center of the

top surface and are uniformly distributed over a square area of length 4 ft, as shown in Figure 11. Three cases of loading conditions for this problem are: 1. A uniformly distributed vertical load = 100 psf 2. A uniformly distributed horizontal load = 1OOpsf 3. A moment = 40 ft lbs applied by tractions on 4ftx4ft area The first two cases are easily incorporated in the analysis. For the third case, since the tractions are assumed constant over each element,

moment should be changed to two opposite

tractions as shown in Figure 12. Here, using three elements, a zero traction is applied

to the center element

and equal

and opposite tracions are applied to the outer elements. can

expect

that

the smaller

the distance

between

elements, the better simulation of the moment. varying stresses

are considered,

to get a better simulation, too. displacements

outer

If linearly

more elements are

needed

Normal components of

on the side surface and

bottom

We

the

surfaces are

39

'^t

fUR]]^

^2i:

^"'^Jifliiijj

imm b

mm

[ a 1^ b -mt

^ = a (a+b)

H =

Figure 12: assumed

q.

Applied Moment in Example 1

to be zero for the

boundary condition as shown

in

Figure 13. The vertical stresses beneath area due

to vertical

the corner of the

loaded

and horizontal tractions are given in

Table 1 and are plotted in Figures 14 and 15,

respectively,

where they are compared against Hell's solution [7]. Vertical stress at the Table 1.

same points

due to moment

The results are plotted

are shown

in Figure 16,

in

where the

result is compared with Giroud's solution [7]. From these results, it is shown that the boundary element solution for

a vertical traction is very good

for vertical

40 4 ' .• ,<

0

#

20

^

20' — (a) Application of Vertical Load

r—1 I

.

X

I

0

20

m

jw

Mr

m

k^

20 ' (b) Application of Horizontal Load Figure 13: Boundary Conditions and Applied Loads

41

Table 1 : Comparison of Vertical Stresses Between Analytical Solution and Boundary Element Method due to Surface Loading Vertical Stress

X3/a

Case 1 Vertical Load Hell's Solution B.E.M.

o^

(kips/ft^)

Case 2 Horizontal Load Holl's Solution B.E.M.

Case 3 Moment Giroud's B.E.M. Solution

0.2

.249

.312

.152

.140

.212

.246

0.4

.240

.261

.133

.115

.151

.189

0.5

.232

.242

.121

.105

.125

.159

0.6

.223

.226

.109

.096

.103

.132

0.8

.200

.198

.086

.077

.070

.090

1 .0

.175

.172

.067

.061

.047

.060

1 .2

.152

.149

.051

.048

.032

.041

1 .5

.121

.121

.035

.033

.023

.024

1 .6

.112

.113

.031

.031

.019

.020

1 .8

.097

.099

.024

.024

.016

.015

2.0

.084

.087

.019

.019

.012

.010

2.5

.060

.060

.011

.011

.008

.005 . ,,

—

_

42

Vertical Stress

0

0.1

0.2

o

(kips/ft ) 0.3

0.4-

0.8-

1.2-

1.6Depth (x3/a) 2.0-

2.4-

Roll's Solution B. E. iM.

2.8-

T^igure 14: Comparison of Vertical Stresses due to Vertical Traction

0

43

Vertical Stress

a

(kips/ft^

0.4 -

0.8-

1.2Depth (x3/a) 1.6-

2.0-

2.4-

2.8.-

Figure 15: Comparison of Vertical Stresse: due to Horizontal Load

44

Vertical Stress

a

(kips/ft^

0.4-

0.8-

1.2Depth (x^/a) 1 .6-

2.0-

2.^-

2.8-

Figure 16: Comparison of Vertical Stresses due to Moment

45 stresses at a distance of x-./a = 0.6 below side of the loaded area. have

about

point,

10 percent

For cases errors at

where 'a' is the

2 and 3,

the points

but as the number of elements

the results beyond

increases,

that

the error

decreases rapidly. Example Problem 2: Embedded Foundation With Prescribed Displacements on The Top Surface of Large Elastic Soil Here,

an embedded foundation such as a rigid footing on

the top surface is considered. problem are the same as

Dimensions of

in the example 1,

discretization of the top surface shown

in Figure 17.

elements,

this example

except that the

are slightly different as

The continuum is discretized

which consist of

with 87

59 elements at the top surface,

6 elements for each side, and 4 elements for the bottom side. Two cases of boundary conditions for this problem are: Case 1. A Vertical Displacement = 1 inch Case 2. A Horizontal Displacement = 1 inch Due to the unavailability of this problem, code

a finite element solution

(SAP V) is

discretization Figure 18.

an analytical solution for

employed

of the

to

finite

compare

using the the

element model

Fortran

results. is

The

shown

in

For the finite element discretization, 219 solid

block elements which have 8 nodes per element, and 320 nodes are

used.

The total

number of

unknowns

for the

finite

46

figure

17

D i s c r e t i z a t i o n of Top Surface of Example 2

47

Figure 13

: Discretization of Finite Element Model

48 element model is 960.

As the finite element method requires

compatibility of displacements at each node between elements, it is very

difficult to

use larger elements

away from the

loaded area to reduce the number of degrees of freedom. Since this is the surface,

a displacement boundary value problem at

special boundary

prescribe these displacements

elements

must be

used

in the SAP iv code.

boundary conditions on the sides and bottom in the previous example (see Figure 19). of top surface are shown there.

to

All the

are the same as The displacements

Results obtained from these

two methods are compared at: (1) x-/a = 0.625 (x ^= 2.5) for tractions and (2) at x^/a =1.25 (x

= 5.) for displacements,

as shown in Tables 3, 4 and Figures 20, 21, 22, and 23. From the results,

it is shown that the boundary element

solution agrees remarkably well

with SAP IV solution.

the displacement at internal points due displacement, good

the boundary element

agreement with those

The displacement loaded

area

calculated

using very

to applied vertical

solution

solution.

at any point away from the

the boundary e lement

accurately

shows fairly

of the finite element

and the stress

For

because

technique can be

of the nature

of the

solution using the fundamental solution. Example Problem 3: a Solid Cube (Uniformly Loaded on one Side Only) This simple problem

when solved

by the finite

element

49

23

7W

JmT

20 (a) Vertical Displacements

h0 ^

V

23

ifmjr

mm 20

(b) Horizontal Displacements

Figure 19: Boundary Conditions and Application of Displacements

50

-

•

t —

^-'

•

r^

AJ M-l

CO (0

a

Q)

•H

m m

^

Q)

s w

a m

u w

ro

<j\

^

ro ro

CN

,_

"^

vo

in

1

C •

s w•

M (d

o

CN

00

O

•«^

r**x>

CN 1

1

O 00

CN 00

00

in

ro

1

1

CN 1

o

CD r^ m fd (d - P N •H

CO

c\

"^

• CQ

0

u

CN

• ^

•H CN Q

U

4J

l^ CN

0 ffi

CN 00

cr> •«* in

00

o

o

T—

in 00

• ^

,—

r*a\

• ^

CN CN

CN

in

KO

o

o

o

vo

vo 1

CN 1

CN 1

^

o

'^

1

o 1

CN

fo

4J 4J

•H C V^ Q)

0) s

>

(U

u

M-i rd

-

•

0 r-i

T—

a c w 0 -H

>>i^

m Q

•H

M 0 (d -P

a B 0)

0 3 U 'O *. CM

0

s• w

•

r-

VO

r-

r-

r-

t—

00

r-

00

00

00

(N

V^

^

•^

CM

^

CN

TT CN

'
vo

CN

T—

CN

f^

m

m

00

in

T—

T—

vo

in

'^

r-


00 (N

r^

CN

in

CN

CN

r—

o o

O CN

o

T—

f

OQ

a m

•H

Q iH

Q) (d

to o

fd -P U -H M 0) >

•

in

CN

in

CN

s•

CN

in

T—

vo

H •

T—

CN

CN

.H

^
1

ro

><

in CN

o

in

o

o

in

r-»

in

CN

00 1

in 1

1

CN 1

1

o

in

o

in

r^

in

in CN

CN

ro

in

00

.

1

51

(U

en

fC

en

u1

d)

en en

tn 4J G) C 5-t (U 4J G w (U U •-( (d (T) i H

o

CU

•H en 4J •H

U C

a)

> ^ (d »+-i

( 0 —u 4J

U 0 0} >

c 0 -rH

M 0 fd -p

a fl> e 0 3 u ^ • •

o


0) i-

fr. CN <4-l

m

u -p CO

52

4J

00 CN

o

Q)

vi;

en fd

O 1

o

en o; en 4J

ress emen

t


c_->

•

CN

""•

X "" (_) •

o


o

u en •H

fd

0) j-i

-P

en

a

•H D 4J

u ^

CU ft > -^ iw

c

0

•H

o • o

en en

a

fd en

•

Q

en r e

mpar e to

1

CN

iH

r-j

1

CN

o

in fd

on o oriz

•

-P o

T—

0 3 U T!

o 00

•r-l

53

•H

^ •

T -

-'-'

s•

•

H • CQ

a

CO •H CN D

4J

(d en 4J Ti

r*

,_

T—

T—

p^

r—

^

vo vo o

r^

CT^ O

o

o

o

o

o

o

o

o

o

o

VO (Ts O

(N CN

00

^

^_

ro

1

O 1

O 1

o

1

CN

^ ^—

rsj 00

r^

«—

,—

o

o

o

1

fd -P

u 0c

B d) .H u (d

•

N •H J-I

•u H

S

•

w•

0 ffi

CUA->

cn ^1

•H

ro

d) 1—1

c c

iH

ro CN

cn fd

Q) fd

fd

T -

^

Q

(U > cn

iH

0

in

00

• ^

"^

^

T—

T—

o

o

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^



ro

T—

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00

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1

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C

-p Q) o Q) BQ) •H fd

-P 3

;

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54

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56 method yields exact solution.

But when the boundary element

method is employed to solve it, the results are exact,

even

though

this

the errors

problem,

in the

the boundary

solution.

results are

small.

element method will

And to demonstrate this,

For

not yield exact

this particular problem

is selected. A cube of dimensions of 2 inch on each side Figure

24,

is loaded

in a state

of uniaxial

applying 100 psi of normal traction on the top

as shown in tension by surface. The

cube is divided into 6, 12, or 2 4 surface elements, as shown in Figure 24.

If 6 elements are

used in a

problem,

there

will be a total number of 18 unknowns and the order of the G and H matrices will be 18x18. configuration is used,

When the 12 and

24

element

the order of the matrix will be 36 x

36 and 72 x 72 matrices, respectively. The results Table 4-

for this problem when v = 0.3

are given in

The internal stresses are calculated

at 2 points

namely, (.4, .4, .4) and (.6, .6, .6). Comparing reactions

the

results

with

the analytical solution,

at the bottom are found to be within 2 percentage

of the analytical

solution.

Maximum axial

which occur at the top surface, for 6 element

give

displacements,

17.6 percentage error

and 7.4 percentage error for 24 elements.

twelve element model gives

the most accurate

results.

A

57 Table 4: Results For Example 3 Subjected To Axial Tension Units : in, lb or psi N = Reaction at bottom ele. ( % error) Displacement at top ele. ( % error)

6

N = 12

N = 24

EXACT

100. ( 0.0)

102. ( 2.0)

100. ( 0.0)

100. ( 0.0)

.00122 (17.6)

.00130 (14.2)

.00137 ( 7.4)

.00148 ( 0.0)

94.60 ( 5.4)

96.47 ( 3.6)

94.29 ( 5.7)

100. ( 0.0)

94.55 ( 5.5)

96.18 ( 3.8)

94.29 ( 5.7)

100. ( 0.0)

Int. point 1 ( % error) Int. point 2 ( % error)

internal point-1:(.4, .4, .4) internal point-2:(.6, .6, .6)

53

(a)

1 1 1 1

1

1 1

1

— •^^ -

1

i

1

1

1

1 1

•

1

.....

(b) 6 elements

(c ) 12 e l e m e n ts

)

( d ) 2 4 e 1 e m e .T •:

Figure 24: Surface Elements Arrangements for Example 3

CHAPTER V SUMMARY, CONCLUSIONS, AND RECOMMENDATIONS Summary In this research, of boundary program

element

for

technique

solving

three-dimensional

basic principles of the development are

for stresses

isotropic,

reviewed.

and displacements

homogeneous,

due to Kelvin is

employed

displacements and tractions are quadrilateral element. tested

for three

in a

linear,

elastic, and solid continuum is developed. solution

A computer

and

The fundamental

in this program.

The

assumed constant within the

A constant quadrilateral element is

dimensional

elastic problems and results

are presented. A 6-point Gaussian quadrature integration formula is used for the formulation of the equations. Three example problems are solved and results are compared with known solutions. Conclusions Following conclusions are made in this research. 1 . The boundary

element method

can be

applied to three

dimensional elastostatic problems accurately. 2. For the problems

dealing

loaded

in a relatively small

method

offers

analysis.

a very

with the area,

convenient

large domains and

the boundary

and accurate

element

method for

The convenience is in the input data preparation. 59

60 The method is engineering

applicable to many problems

where

large

mass

of

soil

in foundation

medium

is

to be

considered. 3. Use

of 4-point

Gaussian

integration

inaccuracy in the equations and hence

leads to

some

6-point integraton is

found to be more accurate. 4. The method

yield poor results

for problems where the

surface volume ratio is large. Recommendations Following research are made for future research area. 1 . The method can be used

very

efficiently

many soil-structure interaction problems. that the finite

element method be

for solving

It is recommended

used for structures

and

where the surface volume ratio is large. 2. For numerical Gaussian

integration

computations.

integration, can be

both 4-point and 6-point

used to reduce

the number of

The 6-point formula can be used for integra-

tion of a equation near the area of stress concentration and 4-point integration stress concentration.

formula

for all

other area

away from

REFERENCES 1.

C. A. Brebbia. The Boundary Element Method For Engineers. London: Pentech Press, 1978.

2.

J. c. F. Telles. The Boundary Element Method Applied to Inelastic Problems. New York. Springer-Verlag, 1983.

3.

C. V. G. Vallabhan, J. Sivakumar, and N. R. Radhakrishnan, "Application of Boundary Elements for Soil-Structure Interaction Problems," 5th International Conference on Boundary Elements, University of Southampton, Southampton, UK. July, 1984

4.

C. A. Brebbia, J. C. F. Telles, and L. C. Wrobel. Boundary Element Techniques. New York,. SpringerVerlag, 1984.

5.

R. K. Nakaguma. "Three Dimensional Elastostatics Using The Boundary Element Method." Doctoral Dissertation. Dept. Of Civil Engineering. The University Of Southhampton. England. 1979.

6.

p. K. Banerjee and R. Butterfield. Boundary Element Methods in Engineering Science. New York, McGraw-Hill Co., 1980.

7.

H. G. Poulous and E. H. Davis. Elastic Solutions for Soil and Rock Mechnics. New York. John Wiley And Sons. 1974.

8.

A. I. Lur'e, D. B. McVean, and J. R. M. Radok. Three Dimensional Problems of The Theory of Elasticity. New York. Interscience Publishers. 1964.

9.

C. A. Brebbia and S. Walker. Boundary Element Techniques in Engineering. England. Pentech Press. 1978.

10.

Arthur P. Boresi. Advanced Mechanics of Materials. 3rd ed. New York. John Wiley and Sons. 1978

11.

D. J. Dansen. "Elasticity Problems With Body Forces." Boundary Element Techniques in Computer Aided Engineering. 5th International Conference on Boundary Elements, University of Southampton, Southampton, London Sep. 1983. 61

62 12.

C. A. Brebbia and J. Doningnez. "Boundary Element Methods Versus Finite Elements." Boundary Element Techniques in Computer Aided Engineering. 5th International Conference on Boundary Elements, University of Southampton, Southampton, London Sep. 1983.

13.

J. C. F. Telles and C. A. Brebbia. "Boundary Element Solutions For Half Plane Problems." Boundary Element Techniques in Computer Aided Engineering. 5th International Conference on Boundary Elements, University of Southampton, Southampton, London Sep. 1983.

14.

C. A. Brebbia. "Simulation of Engineering Problems Using Boundary Elements." Boundary Element Techniques in Computer Aided Engineering. 5th International Conference on Boundary Elements, University of Southampton, Southampton, London Sep. 1983.

15.

-M. Van Laethern and E. Back. "The Use of Boundary Elements to Represent The Far Field in Soil-Structure Interaction." Boundary Element Techniques in Computer Aided Engineering. 5th International Conference on Boundary Elements, University of Southampton, Southampton, London Sep. 1983.

16.

S. P. Timoshenko and J. N. Goodier, Theory of Elasticity. 3rd ed.New York, McGraw-Hill Co., 1982

17.

A. E. H. Love, A Treatise on the mathematical Theory of Elasticity. Dover Publication, N.Y. 1944