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This is to certify that the
thesis entitled
ON THE PROBLEM OF A PLANE, FINITE, LINEARELASTIC
REGION CONTAINING A HOLE OF ARBITRARY SHAPE:
A BOUNDRY INTEGRAL APPROACH
presented by
Ali Reza Mir Mohamad Sadegh
has been accepted towards fulﬁllment
of the requirements for
Ph.D. . Metallur Mechanic
degree in gy ’ S
and Material Science
Date §A%&8
07 639
gJ , “ .
ifs partqu In , . _. “.wntﬁ ._
n
. . l. Easlmk~FlASTlC '
i  "1" ‘\:._ 9.1!.3 OE.
Metallurgv, nr .AHigs JL; ﬁtterinl miiance  . _
19'8
Lc:?,'
3;“9N THE PROBLEM OF A PLANE, FINITE, LINEAR ELASTIC
REGION CONTAINING A HOLE OF ARBITRARY SHAPE:
A BOUNDARY INTEGRAL APPROACH
‘1‘.”
BY
Mir Mohamad Sadegh
Ali Reza
4': , '.
‘P‘ ' .A
. . 'A ‘ < A.“
13“" V
, 'v‘f‘lv ..
{7. M '.“ I
I
v
A DISSERTATION
' ‘ Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of H ,
DOCTOR OF PHILOSOPHY
1‘ i
1978
G
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.\_‘ .
ABSTRACT
ON THE PROBLEM OF A PLANE, FINITE, LINEAR—ELASTIC
REGION CONTAINING A HOLE OF ARBITRARY SHAPE:
A BOUNDARY INTEGRAL APPROACH
By
Ali Reza Mir Mohamad Sadegh
Previous boundary integral equation methods have been
developed for problems of twodimensional elastostatics
which yield excellent results everywhere except near the
boundary. This presents a major disadvantage for problems
in which a hole, slot or sharp crack is present, since
such an opening must be considered as boundary. Thus,
results in the vicinity of the hole are not reliable. In
this dissertation a new formulation of the boundary inte
gral method is presented which eliminates this inaccuracy
on and near the opening. This is done by replacing the
kernel of the integrand (the influence function) with one
which includes the effect of the opening. This influence
function is determined in terms of the complex potential
functions for an infinite elastic plane containing the
opening and subjected to a concentrated line load at an
arbitrary point. This is accomplished using the
Muskhelishvili method of plane elasticity. Potential
functions are found for the cases of a circular hole, an
g
Ali Reza Mir Mohamad Sadegh
ptical hole and a sharp crack. The determination of
ii! functions for other opening shapes is also discussed.
z'bOundary integral equation method is then applied to
finite regions containing either a circular hole, an
‘iptical hole, or a sharp crack. The results are pre
ted and compared with exact solutions and experimental
.‘ults where available.
To my brother, my mother
and my father
.. ' . k ‘ , M '21. an if}
ACKNOWLEDGEMENTS .
It is my pleasure to take this opportunity to express
my deepest appreciation and gratitude to my major advisor,
Professor Nicholas J. Altiero, for his contributions,
guidance, and encouragement during the course of this
investigation, and also for his friendship and painstaking
review of this manuscript.
Grateful appreciation is expressed to Professor
flatacommittee, is appreciated. Thanks are extended to Professor
iﬁrilf Financial support for this research was provided by
‘ V
i;;§{¢he National Aeronautics and Space Administration under
i ,
“;grant number NSC3101. Partial financial aid was provided
Special thanks are due to my family, especially my
V'her, for their encouragement and moral support, and for
5genuine interest they have always shown in my work.
TABLE OF CONTENTS
,r Page
;"ACKNOWLEDGEMENTS . . . . . . . . . . . . . . . . . . iii
... LIST OF TABLES . . . . . . . . . . . . . . . . . . . vi
=:,; LIST OF FIGURES. . . . . . . . . . . . . . . . . . . viii
; LIST OF APPENDICES . . . . . . . . . . . . . . . . . xi
<.¢a LIST OE SYMBOLS. . . ._. . . . . . . . . . . . . . . xii
‘ INTRODUCTION . . . . . . . . . . . . . . . . . . . . 1
“a
’CHAPTER I  BACKGROUND AND PRELIMINARIES . . . . . 5
1.1 AN INTEGRAL EQUATION METHOD . . s
1.2 THE MUSKHELISHVILI METHOD: A
COMPLEx VARIABLE METHOD IN
ELASTICITY. . . . . . . . 25
1.3 CAUCHY INTEGRALS AND RELATED
. THEOREMS. . . . . . . . 34
:“CHAPTER II  GENERAL SOLUTION AND A MAPPING _
.**. TECHNIQUE. . . . . . . . . . . . . . . 38
11.1 INTRODUCTORY REMARKS. . . . . . 38
11.2 A MAPPING TECHNIQUE . . . . . . 38
:3 11.3 GENERAL SOLUTION. . . . . . . . 43
' GHAPTER III  CIRCULAR HOLE IN A FINITE TWO
,. DIMENSIONAL REGION . . . . . . . . . . 52
111.1 INTRODUCTION. . . . . . . . . . 52
III.2 DERIVATION OF THE INFLUENCE
FUNCTIONS USING KNOWN POTEN
TIAL FUNCTIONS. . . . . . . . . 53
111.3 DERIVATION OF THE INFLUENCE
FUNCTIONS USING A MAPPING
TECHNIQUE . . . . . . . . . . . 58
iv
H O “'7'?
Cmt ll.
P'v
.v.
.
‘4
h,
L.“
.'
.‘L _
.CHAPTER IV
CHAPTER V
CHAPTER VI
III.4 THE BOUNDARY INTEGRAL EQUATION
METHOD APPLIED TO PLANE FINITE
REGIONS WEAKENED BY A CIRCULAR
HOLE. . . . .
ELLIPTICAL HOLE OR SHARP CRACK IN A
FINITE TWO DIMENSIONAL REGION. . . .
IV.1 INTRODUCTION. . . . . . . . . .
IV.Z DERIVATION OF THE INFLUENCE
FUNCTION USING THE MAPPING
TECHNIQUE: THE ELLIPTICAL
HOLE PROBLEM. . . . . . . .
IV.3 DERIVATION OF THE INFLUENCE
FUNCTION: THE SHARP CRACK
PROBLEM . . . . . . . . . . . .
ON THE PROBLEM OF AN ARBITRARILY
SHAPED HOLE IN A TWODIMENSIONAL
REGION . . . . . . . . . . . . . . . .
v.1 INTRODUCTION. . . .
V.2 THE CONTOUR OF AN ARBITRARILY
SHAPED HOLE AND THE MAPPING ’
FUNCTION. . . . . . . . . . . .
v.3 ON THE INFLUENCE FUNCTION OF
A PARTICULAR CLASS OF OPENING,
CASE 1. . . . . . . . . ... . .
V.4 ON THE INFLUENCE FUNCTION OF
A PARTICULAR CLASS OF OPENING,
CASE 2. . . . . . . . . . .
v.5 ON THE INFLUENCE FUNCTION OF
A MORE GENERAL CLASS OF
OPENING . . . . . . . . . . .
~ CLOSURE. . . . . . . . . . . . . . .,.
Page
73
92
92
93
116
156
156
157
167
181
202
206
210
253
Table
3.1
LIST OF TABLES
Stresses and displacements in a rectangular
region containing a circular hole at the
origin, Case 1 . . . . . . .
Rectangular region containing a circular
hole at the origin, Case 2 . .
Rectangular region containing a circular
hole at the origin, Case 3 .
Stresses and displacements in a rectangular
region containing a nonsymmetrically
located circular hole.
Stress and displacement in a circular plane
containing a circular hole at the origin,
Case 1 . . . . . . . .
Circular plane containing a circular hole
at the origin, Case . . . . . . .
Circular plane containing a circular hole
at the origin, Case 3 . . . . . . .
Stress and displacement of a circular plane
containing a nonsymmetrically located
circular hole. . . . . . . .
Stress and displacement of a rectangular
plane containing an elliptical hole at the
origin . . . . . . . . . . . . . . . .
Rectangular plane containing an elliptical
hole (inclined major axis) at the origin,
Case 1 . . . . . . . . . . . . . . .
Rectangular plane containing an elliptical
hole (inclined major axis) at the origin,
Case 2 . . . . . . . . . . . . . .
Rectangular plane containing a nonsymmet
rically located elliptical hole (inclined
major axis). . .
vi
Page
81
83
84
85
88
89
90
91
136
137
138
139
Page
Stress and displacement of a circular plane
containing an elliptical hole at the origin.
Circular plane containing an elliptical hole
(inclined major axis) at the origin, Case 1. 144
Circular plane containing an elliptical hole
(inclined major axis) at the origin, Case 2. 145
Circular plane containing an elliptical hole
(inclined major axis 30°) nonsymmetrically
located. . . . . . . . . . . . . . . . . . . 146
Stress and displacement of a rectangular
plane containing a sharp crack at the
origin . . . . . . . . . . . . . . . . . . . 149
Rectangular plane containing an inclined
sharp crack at the origin, Case 1. . . . . . 151
Rectangular plane containing an inclined
sharp crack at the origin, Case 2. . . . . . 152
Rectangular plane containing an inclined
nonsymmetrically located sharp crack . . . . 154
Sharp crack (notch) on the side of a
rectangular plane. . . . . . . . . . . . . . 155
LIST OF FIGURES
Figure Page
1.1 Finite twodimensional region with pre
scribed traction all around the boundary . . 9
Half plane subjected to a concentrated
force on the boundary. . . . . . . . . . . . 9
Region of Figure 1.1 embedded successively
in half planes . . . . . . . . . . . . . . 10
Boundary value problem in plane elasticity . 10
Region R embedded in an infinite plane . . . 12
Concentrated line load . . . . . . . . . . . 39
The problem of interest expressed as the
superposition of two problems. . . . . . . . 41
Mapping the auxiliary problem to the disc. . 42
Fundamental problem expressed as super
position of two problems . . . . . . . . . . 60
Mapping the auxiliary problem to a unit
disc . . . . . . . . . . . . . . . . . . . . 60
A unit circular hole in a plane finite
region with prescribed traction on the
boundary 0 C I O I I O I I O O O I C I I O O 7 4
Region R embedded in an infinite plane con
taining a circular hole at the origin. . . . 74
Circular hole symmetrically placed in a
finite rectangular plate under uniaxial
tension. . . . . . . . . . . . . . . . . . 80
Circular plane, containing a circular hole,
subjected to radially uniform tension over
a portion of the boundary. . . . . . . . . . 87
viii
§
A
Figure Page
4.1 Superposition of the problem of ellip
tical hole . . . . . . . . . 95
4.2 Mapping the auxiliary problem to a unit
disc . . . . . . . . . . . . . . . . 96
4.3 A sharp crack in an infinite plane . . . . . 117
4.4 An elliptical hole in a plane finite
region with prescribed traction on the
boundary, Be . . . . . . . . . . . . . . . . 123
4.5 A horizontal slit in a plane finite
region with prescribed traction on the
boundary, BS . . . . . . . . . . . . . . . 123
4.6 The finite region Re and RS, with sub
divided boundary and c0ncentrated line
loads. . . . . . . . . . . . . . . . . . 124
4.7 Region Re, embedded in an infinite plane
containing an elliptical hole at the
origin . . . . . . . . . . . . . . . 125
4.8 Region RS, embedded in an infinite plane
containing a horizontal slit at the origin . 126
4.9 Rectangular plane weakened by an ellip
tical hole at the origin . . . . . . . . . . 134
4.10 Circular plane weakened by an elliptical
hole at the origin . . . . . . . . . . . . 141
.11 Rectangular plane weakened by a sharp
crack at the origin. . . . . . . . . . . . . 148
Different contours for 2 and c = 1 . . . 159
2 and o
. The first application of the methods of poten
49. tial theory to classical elasticity theory was introduced
ti.
—~ by E. Betti [7] in 1872. Later this work was expanded by
:é; Somigliana [8], Lauricella [9] and others. In particular,
, Bettifs contribution, i.e., the general method of inte
ITS; grating the equations of elasticity, was simply a develop
;Ii:ment of the potential methods of Green and Poisson. Thus,
gsome fundamental results from potential theory should
first be discussed. Let a function 6 be the solution of
#Laplace 5 equation throughout a region R.
V2¢ = o in R (11)
6 = f on GR;
%%= g on 8R2 (1.2)
1" 3R; + 8R2 is the boundary of R. Note that
’17Fere r is the distance between two points in R, is
a,“
L
A”)
.
L1
A
.
_ I»
if;' found.
‘ ¢
6
. I
‘,;a,singular solutIOn to Laplace's equation. Combining 1/r
11;”with the solution ¢ in the classical Green's theorem of
integral calculus [10] results in the identity,
4(2)
1
Where 2 is
1 l 3 1
If j:R[g(z°)FTZ:TFT  f(zo) 55 (TIITIITJ ]' d5(zo)
(1.3)
any point in R and Z0 any point on 8R; Since
f and g are both needed everywhere on 3R, only one of
To
" them is known at each point, then the other has to be
accomplish this [ll13], consider taking the
limit of equation (1.3) as Z approaches a boundary point
'T: Z}, on 3R.
Ami)  24,—,
’nfthe lim
"eted in
zf.
g g is
The result is:
[[gczo) mi—m  H20) 5392—; (mf—Jw] d5(zo)=0
BR
(1.4)
it [11]. Thus, this integral is to be inter
the Cauchy principal value sense. Equation
Solving the integral equation (1.4) for either
given, or g, if f is giVen, leads to the
car
lea
Een'
6t .‘
.1“.
n\nv
for:
and

i
7
In spite of this classical foundation of the boun
, dary integral equation method, the literature contains at
__£5 least two seemingly distinct formulations for the treat
ment of elasticity problems. One of these, due to Rizzo
et a1. [1416] and Cruse [l7], follows directly from
Somigliana's identity of elasticity [18]. The other
_formu1ation due to Massonnet [19] and extended by Altiero
and Sikarskie [20] attacks the problem by embedding the
region in an infinite plane and distributing a layer of
body force on the proposed boundary in such a way that the
desired solution is produced within the region of interest.
Both approaches will be discussed in this chapter and the
latter will be employed in the subsequent analysis.
The formulation of the boundary integral equation
method due to Rizzo [14] is based on Somigliana's identity:
\
. 'Uq(Z) = [B Ii;q(Z,Zo) ti(Zo) dS(Zo)
 LHij;q(Z,zo) Ui(zo) nJ(Zo) dSCZo) (15)
+
where U is the displacement vector, Ii_q(Z,Zo) is the
l ' 9
[,ith component of the displacement at 2 produced by a
’gnit force applied in the q direction at 20 in an infinite
‘Qedium, and ti and Hij°q(z’z°) nj(Zo) are the components
‘1 i'q’ respectively. In three dimensions,
( . ,
AJEEBESI
7",.
Q
6
f
!
prete
tract
9
'12“.
31.2“
8
Taking the limit as Z approaches a point 2; on B
from the inside leads to
% Uq(zi) + /; Hij;q(zr.zo) Ui(zo) nj(zo) d5(zo)
= L11;q(21120) ti(zo) (15(20) (106)
where the integral on the lefthand side is to be inter
preted in the Cauchy principal value sense. If the
traction is prescribed everywhere on B, then the right
hand side of equation (1.6) is known and a system of
singular integral equations can be solved for the boundary
displacement U. The interior displacements can then be
foundfrom equation (1.5). If the displacements are
prescribed everywhere on B, then the lefthand side of
equation (1.6) is known but the resulting set of integral
equations are not singular. For the mixed boundary condi
tions, some of the equations are singular and some are
not .
The second formulation of the boundary integral
method, i.e., that of Altiero and Sikarskie [20], is an
extension of the work of Massonnet [19]. Massonnet intro
duced a method for solution of traction boundary value
problems in which the real body is embedded in a series
of "fictitious" half planes which are sequentially
tangent to the real boundary. To demonstrate the idea,
consider a finite two dimensional region with a prescribed
traction all around the boundary, Figure 1.1. Choose the
«LA
Figure 1.1. Finite twodimensional region with pre
in ha}
10
'~<
Figure 1.3 Region of Figure 1.1 embedded successively
Thalf planes.
T 19';
l
f”
I .‘ :‘T‘l “ m
,3... . I
. , i X
'7 ' a,
1" i
a"
Er 1.4 Boundary value problem in plane elasticity.
11
simple radial stress distribution, i.e., a half plane
subjected to a concentrated line load on the boundary, as
a fundamental singular stress field, Figure 1.2. This
solution is well known [21]. Then, draw the tangent to a
point, Z0, of the real boundary and consider the half plane
extending indefinitely below this tangent, Figure 1.3.
In other words, the body has been embedded in a succession
of half planes. An unknown "fictitious" line load is
introduced at each point of tangency. A vector boundary
integral equation for the unknown fictitious tractions
results when one forces satisfaction of the traction
boundary conditions of the original problem.
An approach somewhat similar to Massonnet's has been
developed [22] for anisotropic regions subject to traction
boundary conditions. This approach, later, was extended
by Altiero and Sikarskie [20] to mixed boundary value
problems. Consider a two dimensional, linear elastic
region R with boundary B as shown in Figure 1.4. For
prescribed boundary conditions, i.e., tractions and/or
displacements on B, the stress field and displacement
field in the region R are to be determined. The region
R will be embedded in an infinite (fictitious) plane of
the same material and thickness as R, Figure 1.5. The
influence function which satisfies the equations of
elasticity, i.e., H (Z,Zo) and Ii,q(Z,Zo), are known
ij;q
[23], where H. (Z,Zo) is the ijth stress component at
UN
a field point 2 due to a unit line load in the q direction
. —.
w
an infinite plane.
acti
sclu
.\~
DC
a
\ll
r
y 13
at a source point 20 and Ii;q(Z,Zo) is the displacement in
the i direction at Z due to the unit line load at Z0. Con—
sider now a fictitious layer of body force P* (unknown)
acting along the contour B, see Figure 1.5. Since the
problem is linear, then the superposition of fundamental
solutions leads to the determination of stresses and dis
placements at a point 2 as follows:
0.. Z = H.. Z,Z P* Z d Z
13() [B 13;q( o) q( o) $( 0)
Ui(Z) = [B Ii;q(Z,Zo) Pane) d5(zo) (1'7)
where Z0 is now on the bOundary B and ds(Zo) is an element
of length along B at 20. Then all equations of linear
elasticity are satisfied by equations (1.7) since they
represent the superposition of fundamental solutions.
In order to solve the boundary value problem of interest,
the boundary conditions on B are yet to be satisfied.
These conditions are:
U. = U: on B (1.8)
where nj is the j—component of the outward unit normal
to a point on B and Pi, U: are the specified traction and
displacement components, respectively. Note that one may
also specify one traction component and one displacement
compc
are l
‘0
Ant
14
component at a particular boundary point, provided they
are mutually orthogonal. Let the interior point Z approach
a boundary point 21, on B, Figure 1.5. Then the stresses
and displacements, equations (1.7), must satisfy the boun
dary conditions of equations (1.8). Thus, substitution of
equations (1.7) into equations (1.8) leads to
I P§(Z1) + S¢i Hij;q(zl’z°) Pa(ZO) njCZi) ds(Zo)
= Pﬁtzi)
Z1 on B1: (1.9)
9531,;qczi,zo)1>;(zo)dsczo) = Uiczl)
Z1 on Bu (1.10)
Note that the subscript 1 refers to a coordinate direc
tion at a boundary point 21. Equations (1.9) and (1.10)
represent coupled integral equations in the unknown fic—
titious traction P*. Note that the singularity has been
extracted from equation (1.9) and the integral of this
equation is to be interpreted in the Cauchy principal
value sense. Equations (1.9) and (1.10) contain several
types of problem. For the first fundamental problem of
IJlane elasticity, i.e., traction boundary conditions only,
12he vector equation (1.9) is to be used. For the mixed
boundary value problem, both equations appear but not in
15
the same direction at the same point, i.e., if a traction
is specified in the i direction at Z1, then equation (1.9)
holds; and if a displacement is specified, then equation
(1.10) holds.
Like the Rizzo formulation, equations (1.9), i.e.,
the traction boundary value problem, are singular. However,
for displacement boundary value problems, equations (1.10)
are used and these are not singular. For mixed boundary
value problems, some of the equations will be singular
and some not.
It is felt that the formulation of Altiero and
Sikarskie is preferable for the following reason. In the
method of Rizzo, one must first perform integration around
the boundary before the required integral equations are
defined. This is clearly not necessary in the Altiero
and Sikarskie formulation, where one merely needs to
specify the tractions and displacements themselves and
the righthand sides of the required integral equation
are immediately known. Therefore, the Altiero and
Sikarskie formulation will be used here.
Note the fact that the Altiero and Sikarskie formu
lation is not restricted to embedment in an infinite
plane. Massonnet, as discussed earlier, used embedment
in a succession of half planes. However, to obtain
singular equations for the traction problem, and there
fore more numerically efficient equations, this approach
requires tangency of the half plane to the embedded body
successively around the boundary.
16
The Massonnet approach is therefore somewhat cumber
some and particularly inconvenient, particularly for the
solution of anisotropic elasticity problems [22]. Also,
it is very difficult to apply this method to multiply
connected regions. Whichever formulation is used, the
fundamental solutions for the fictitious region should
be simple. This is best satisfied by the infinite plane.
Once P* has been determined, the stresses and dis~
placements at any field point can be determined by sub
stituting P* into equations (1.7). These stresses and
displacements represent the solution to the boundary
value problem of interest within R. The influence
functions, i.e., the stress and displacement fields in
component form, due to a concentrated line load P*ds in
an infinite plane are given by Love [23]. These are:
* = _ 1 * 2 2 2_ 2
Hxx;q Pq 4rr“ [erx(a1rx + azry) + P;ry(a3rX azry)
1 2 2 2 2
*= * _ *
Hyy;qpq 4nru [erx(a3ry azrx) + Pyry(a1ry+a2rx)]
1 2 2 2 2
*=— * *
xy;qpq 4nr“ [ery(a1rx + azry) + Pyrx(a2rx+a1ry)]
(1.11)
P* =  1 [P*(a.rzlog r + asrz) — P*a5r r
x;qq 4,,2 x y y xy
1
a =  _ * * 2 2
y;qPq 4rr2 [ ansrxry + Py(aur log r + asrx)]
(1.12)
where rx and ry are the x,y components of the radius
vector from Z to 20 and the constants a1 through as for the
problem of plane strain are
a1 = (32v)/(1v)
a2 = (lZv)/(1V)
a3 = (l+2v)/(lv)
(34v)(1+v)/(1v)
9:
t
l
(1+v)/(1v) (1.13)
as
and, for plane stress, v is replaced by v* in all the
coefficients of equation (1.13) where:
\J
* =
V 1——+v
The influence functions found in equations (1.11) and
(1.12) can be obtained using the complex potential func—
tions associated with the concentrated line load in an
infinite plane. This will be discussed further in the
next section.
The solution to any boundary value problem of plane
elasticity is contained in equations (1.9), (1.10), and
(1.7). For tractions specified everywhere on B, equations
(1.9) are to be solved for P*. These values of P* are
then substituted into equations (1.7) to find the stresses
and displacements at any field point. Equations (1.7)
may give the displacement field to within a rigid body
displacement. The rigid body displacement, however, can
A
rq/A—J
Or«I
'V
18
be eliminated by suitably prescribing sufficient boundary
displacement information. For displacements specified
everywhere on B, equations (1.10) are solved for P* and
equations (1.7) again used to find the stress and dis
placement fields. For mixed conditions at a point either
the x component equation (1.9) and the y component of
equations (1.10) or the converse must be satisfied.
To obtain a numerical solution to equations (1.9)
and (1.10), the boundary is first replaced by an Nsided
polygon with sides of arbitrary length ASi. The resultant
boundary data over the interval ASi.is now defined at
the midpoint of each interval as follows:
P*. = / P*d , P*. = f P*d
X1 AS. x S Y1 AS. Y s
1 1
P . = f Psds , P . = f Psds
*1 AS. X Y1 AS. Y
1 1
Uxi = f Uids , U i = / Usds (1.14)
ASi V Asi Y
Note that the superscript * implies the fictitious com
ponent. Multiplying equations (1.9) and (1.10) by ds(Z1)
leads‘to
%P€(ZI) dS(ZI) + ﬁHij;q(zl’Z°) Pa(Zo) (15(20) . nj(21)dS(Zi)
= Pfcz.) ds(Z.)
19
9% Ii;q(zi,zo) Pa(zo) d5(zo) ' dS(Zi)
= U§(Z1) ds(21) (1.15)
Integrating equations (1.15) over boundary interval ASi
and assuming that the influence functions are independent
of 20 over a particular interval yields:
1
P*PZ +¢Hn Z,Z P*Z .2 d2
1‘ 1( I) B 1];q( 1 a) q( a) n3( 1) S( I)
g Pi(zl)
YE Ii;q(Zi,Zo) Pa(Zo) dSCZI) = Ui(Zi) (1~16)
71§Eppint of an interval by the interval length. This is
2sufficient for all intervals except for the second equa
r7
‘ w‘.‘
.. c..
' ,J—_‘
) '
. “4.1z".
. ..‘ a ~—e————“
.. u.
where
over
bound
repre
nunbe
(cap—1
’73
l 20
Si P§(Z1) + 1 Ii.q(zl!ZO) Pafzo) AS(ZI)
o= ’
20192,
= Ui(z.) (1.17)
where g. = _/ 1., (21,20) ds(Z1) (1.18)
1 As. 1’q
1
over the interval which includes 2; = 20. Note that the
boundary points Z0 and Z; in the discrete terms will now
represent the centerpoint location of the intervals,
numbered counterclockwise. Separating x and y components,
one obtains
l
2' P;(Zl) + :1[H H’XX q(Zi,Zo) P*(Zo) nx(Z1)
2,921
+ ny qCZl,Zo) P*(Zo ) ny (Z )] AS(ZI) = PXCZI)
]2'P;(Zl) + :1[H H'X)’ q(zi,zo) P*(Zo) 11x (21)
0:
Z0#Z1
(21.20) Pa(Zo) ny(21)] AS(Z;) = PYCZI) (1.19)
M
EXP;(ZI) + ;§;1 [Ix;q(21’z°) Pa(zo)] A5(Z1) = Ux(zl)
H
YYBQ
gyp;(zl) + ::;1 [Iy;q(zl,zo) PaCZOI] AS(ZI) ' Uy(zl)(120)
where the influence functions can now be written as
Um
I.
liq
s” '
D'J
3.4x
; l
3'4
21
(21,20) P;(Zo)
q(zi,z o) P*(Zo) = Hij' 'x
y(21,20) P;(Zo) ’ '
Ii q(21,Zo) Pa(Zo) = Ii;X(Z1,Zo) P;(Zo)
y(zi,zo) P;(Zo)
i,j = x,y (1.21)
.th
.2IASubstituting equations (1.21) into (1.19) and (1.20) and
,¥,$‘ rearranging the equations leads to
.‘.,.: N
. LIZ!) + :E; { [H “xx x(21,20) nx (2,)
a:
Zo¢Zx
+ MXY .x(zi,zo) n y(zi)] P;(Zo) + [Hxx;y(zl’z°) DXCZI)
+ H (21,20) ny(Zl)] P;(Zo) } AS(Zl) = Px(ZI)
XYEY
A) f _1 g [H H,,. x(21.2 o) nxtzil
2 22.2.
(21,20) HXCZII
x(zi,zo) ny(zr)] P;(Zo) + [ny;y
Hyy;y(21.zo) ny(zi) ] P;(Zo) 2 ASCZI) = Px(ZI) (122)
*,
gx'ka
(JO
‘~<
I
b—rJ
‘ < :0
xi
vi
LL)
ii
‘33..
u
'3‘ 22
';;5
‘n;”.P;(Z1) + g { Ix;x(zl’z°) P;(Zo)
20’21
k.
I:(u + Ix;y(zl,zo) P;(Zo) }AS(Z1) = Ux(zl)
éiz *
.'r.'?v"’§‘z” + XX W‘) PM
" 20‘21
+ 1y;y(21,zo) P;(Zo) } 5(21) = UYCZI) (123)
Qiyrﬁquations (1.22) and (1.23) can be written in the compact
. 9}" "7: form:
i=1,...N (1.24)
: , my(21.20) nxczz) + ny;y(zl.zo) nycz,)]Ascz.)
ﬂ
4 v
 “‘AfD
C..
1]
0..
1]
*4
C)
23
Cij = [ny;x(zl,zo) nx(zl) + H
yy;x(21.zo) ny(Zx)]AS(z1)
Dij = [ny;y(zl,zo) nX(Z1) + Hyy;y(zl,zo) ny(zl):]AS(Z1)
= = 1
Kxi Kyi 2
Bvxi = Px(Zl) , vai = Py(Zl) (1.25)
for a traction condition specified and
Aij = IX;X(Z1,ZO) AS(Z1) Bij = Ix;y(Zl,Zo) AS(Z1)
Cij = Iy;x(Zl,Zo) AS(Z1) Dij = Iy;y(ZI,Zo) AS(Z1)
Kxi = gx Kyi = gy
Bin = UX(Z1) BVyi = Uy(Z1) (1.26)
or a specified displacement condition. There are several
methods for solving equation (1.24) for the fictitious
traction. The first and simplest method is iteration.
Iteration works particularly well for traction boundary
value problems, equations (1.22). An initial choice of
fictitious fractions equal to the actual tractions pro
duces fairly rapid convergence. For the mixed problem,
equation (1.23), the iteration, in general, does not con
verge. A second method is matrix inversion or elimina
tion. Equation (1.24) can be written in the matrix form
r1
XX
x}.
X‘.‘
24
[[Kx]  [AH [B] {p;} €=%{Bvx}
where all submatrixes can be found using equations (1.25)
and/or (1.26). Equation (1.27) is written more compactly
W =33)
Once the fictitious tractions are determined, the
as:
stresses and displacements are found from the numerical
approximation of equations (1.7):
N .
= * ‘ *
0xx 5%; [H xx; X(F’ i) Px xi + Hxx;y(F’1) Pyi]
i5
0 = F P*i + H F ' P*.
U i=1 [Hm KC ’1) yy;y( ’1) >71]
'Mz
_ *i ' *
XY [Hm X(F’ i) PX + ny;y(F’1) Pyi]
1=1
N
=  *  *
Ux 1:21 [Ix;x(F’l) Pxi + Ix;y(F’1) Pyi]
N
= I ' P*. + I F,' P*.] 1.29
Uy E [y;x(F’1) x1 y;),( 1) Yl ( )
where F is a field point and P*xi’ P;i are the components
of the known fictitious traction at the interval 1.
25
It is clear from equations (1.29) that the stresses
and displacements can be found at small expense anywhere
in the field by simple summation.
It is important to note that the embedment in an
infinite plane can also be used for multiply connected
domains, such as a region containing a hole. However,
the hole would need to be treated as boundary. Discreti
zation of the boundary would therefore cause inaccuracy
of the solution near the edge of the hole where the solu
tion is most important. The goal of this dissertation
is to eliminate the contour of the hole from the boundary
and to find a new influence function for the problem,
which contains the effect of the hole, thus improving the
accuracy of the solution along and near the hole. To
accomplish this goal, the Muskhelishvili method will be
employed.
I.2 THE MUSKHELISHVILI METHOD: A COMPLEX VARIABLE
METHOD IN ELASTICITY
After the formulation of the linear theory of
elasticity had been largely completed (by the middle of
the nineteenth century), functions of a complex variable
were introduced into plane elasticity problems in 1909
by Kolossoff [24] who, together with Muskhelishvili [25],
developed the theory. However, nearly forty years elapsed
before the theory, based on Kolossoff's idea, was brought
to a successful conclusion. This was accomplished, in the
main, by a group of Russian mathematicians inspired by
the work of Muskhelishvili. The development has been
26
described by Muskhelishvili in two works [26,27]. The
general solution of the fundamental biharmonic boundary
value problem can be made by means of two analytic func
tions of a complex variable. Consider the biharmonic
boundaryvalue problem
vzv2 U(x,y) = o in R
U 6 = fe(s) on GR (1.30)
’
and let
V2U = X(x,y) (1.31)
Then, clearly, the function X is harmonic in R. Note that
a harmonic function is a singlevalued function of class
C2 which satisfies Laplace's equation in R, i.e., V2X=0.
For every harmonic function there is a conjugate harmonic
function which satisfies V2Y=o where the function X + iY
is an analytic function. Every analytic function is a
C°° function because it has a series expansion. Also, an
analytic function satisfies the CauchyRiemann equations
and the Cauchy integral formulae. Thus, every analytic
function is a harmonic function [28].
The complex conjugate of function X, i.e., Y(x,y),
can be easily found by the CauchyRiemann equations to
within an arbitrary constant. Thus, an analytic function
of a complex variable 2 = X + iy can be constructed
27
F(Z)
X + iY.
Let
¢(2) =21ff1=(2)dz
= X° + iY° (1.32)
where X° and Y° are the integrated functions of X and Y.
Then, ¢(Z) is analytic and its derivative is
ax° . ay°
+
¢'(Z) = M 1 ﬁ— = %(X+iY).
From the CauchyRiemann equations, it is clear that
o = o = 1
X,X Y,y IX
0 = _ o = l
x,y Y,x 4Y
Let
H(x,y) = U  X9x  Y9y (1.33)
Then it is easy to verify that H(x,y) is a harmonic
function, because
v2 (U  x‘ZxY‘iy) = VZU  v.V(X‘3x)  vV(Y‘2y)
and the fact that X°, Y° are harmonic leads to
Solri
The d,
4“
28
x  2vx° 2VY°
l 1 1 1
X  7X + 2'Y  7X  '2Y
Thus, H(x,y) is a harmonic function, the complex conjugate
of which can be easily found. Calling this conjugate
K(x,y), function can be constructed so that:
X(Z) = H(x,y) + iKCX,y) (1 34)
Solving equation (1.33) for U leads to:
U = XQX + YQV + H(x,y)
and substituting the analytic functions of equations
(1.34) and (1.32) into the above equation, the biharmonic
function is obtained in terms of the two analytic functions,
¢(Z) and x(Z):
U = Re[7¢(2) + x(2ﬂ (1.35)
Since ¢(Z) and x(Z) are analytic functions, it follows
that U(x,y) is of class Cm in R. Denoting the complex
conjugate values by bars, the equation can also be
written as
2U = 7¢CZ1 + Z¢(21 + XCZ) + Xizj (136)
The determination of stresses and displacements in terms
of the two analytic functions will now be discussed. The
29
stresses can be written in terms of the biharmonic
function:
0 = U
XX ,YY
Oyy = U,xx
oxy = U,xy (1.37)
This leads to:
Oxx + ioxy = 1(U, + 1U, ),y
 ' = U + 'U 1.38
xx loxy ( .x 1 ,y).x ( )
Let
W(Z) = x'(Z)
Then, from equation (1.36), the expression U x + iU y can
’ ,
be written
U + iU y = ¢(Z) + Z¢'121 + WZZ) (1.39)
,X
Calculating the derivatives of equation (1.39) with respect
to x and y and substituting into equations (1.38) leads to:
 ¢'(Z) + ¢'(Zi  Z¢"IZ)  @TTZT'
Q
+
H0
Q
I
¢'(Z) + ¢'(21 + Z¢"ZZi + W‘ZZ)
Q
I
Ho
Q
II
Stresses in terms of two analytic functions, ¢(Z) and
U(Z), can now be written as:
30
OXX + Oyy = 4Re [¢'(Z)]
ny  Oxx + ZioXy = 2[7¢"(Z)+W'(Z)] (1.40)
Finally, displacements in terms of the two analytic func
tions in the compact formula will be
2u(UX + in) = a¢(Z)  z¢'izi  W225 (1.41)
where
a = 34v for planestrain problem
or
_ 3v _
a  l+v for plane stress problem
Stresses and displacements can be found individually, using
equations (1.40) and (1.41), and are:
on = Re [2¢'(Z)  Z¢"(Z)  mm]
or”, = Re [2¢'(2) + Z¢"(Z) + ‘1“(Z)]
ox), = Im [7W (2) + \v'm]
u = Im [mm  2W  m)”.
U = Im [a¢(Z)  ZCD'iZS  (NJ/2p (1.42)
31
Now that stresses and displacements have been formulated
in terms of the two analytic functions, ¢(Z) and 9(2),
the structure and arbitrariness in the definition of the
two functions is an issue to be discussed. If the state
of stress in the region R is specified, from equations
(1.40), one can prove that the singlevalued analytic
functions ¢(Z) and U(Z) could be determined to within a
linear function Ci+y and a constant B, respectively [29].
In addition, if the displacements are prescribed, follow—
ing equation (1.41), one can find that
and
aY  B = 0
Hence, when the stresses are given, the three constants
c, y, B will be chosen in such a way that
NC) = 0
Im¢'(0) = O
1(0) = 0 (1.43)
and when the displacements are given, a suitable choice
of y will be assured by the condition
Mo) = 0 (1.44)
Thus, using the conditions (1.43) and (1.44), the func
tions ¢(Z) and U(Z) will be determined uniquely [27].
32
The structure of the two analytic functions for a
finite and infinite simply connected regions has been
discussed in [27].
Since the state of stress and the displacements can
be expressed by means of the two complex functions ¢(Z)
and f(Z), the fundamental boundaryvalue problems of
plane elasticity lead to the determination of these func
tions from prescribed values of certain combinations of
these functions on the boundary of the region.
Beginning with the first boundaryvalue problem in
which tractions are prescribed on the boundary, the
biharmonic function in terms of applied tractions, f(s),
can be written as
U,x + iU,y = f(s) on SR
The equation (1.39) leads to:
¢(Z) + Z¢'(Zj + W125 = f(s) on ER (1.45)
The corresponding boundary conditions of the second
boundaryvalue problem follow from equation (1.41):
a¢(Z)  Z¢'125  [25 = g(s) on 3R (1.46)
where g(s) is a prescribed displacement function on the
boundary. From either equations (1.45) or (1.46) one can
obtain the two complex functions. However, mapping the
region R into the inside or outside of a unit circle makes
the determination of the two functions much simpler.
Suppose the mapping function
33
Z = w(2;) (1.47)
maps point in the region R, Z plane, into a unit circle
Iclsl. The mapping function for a finite region where the
origin is taken in the interior can be represented as a
power series
Z=chn Iclsl
n=l n
whereas for an infinite region, where the origin is an
exterior point, the function is given by:
n
Z=%+Zn=o kn: lclsl
The boundary conditions, equations (1.45) and (1.46), can
then be written as
¢1(c)+ w“) “m, c + m. r. = Fm
aqua)  W?) m; c  ‘77. c = cm (1.48)
w' 2;
where
¢[w(c)] = ¢1(C) and w[w(;)] = vim
Equations (1.48) can now be solved for the two functions
¢1(C) and W1(;) by a power series expansion method or
integrodifferential equations using Cauchy integral formulae
[27]. Since the solution of the integrodifferential
34
equation reduces to the solution of the standard Fredholm
integral equation, then the existence of a solution of
equations (1.48) would follow, almost directly, from the
Fredholm theory [27].
1.3 CAUCHY INTEGRALS AND RELATED THEOREMS
Since the integrodifferential equations method will
be used to determine the two complex functions, it is
important to discuss Cauchy integrals and related theorems
briefly. The proof of the following theorems has been
presented in [30] and in [27].
Suppose R+ is a finite open simply connected region
enclosed by the contour B described in a counterclockwise
sense. Denote the region exterior to (R++B) by R and
the points on the boundary B by t. Let f(Z) be a complex
function analytic (holomorphic) in R+ and continuous on C.
Then
2%]; 131,1". = £(2) for 2512* (1.49)
and
"2'31?ij ﬁfgﬁ dt = fwm for 2512*
2%i./; %L%l dt = 0 for ZER (1.50)
Equation (1.50) is a necessary and sufficient condition that
the continuous function f(t) defined on B can be the boundary
value of a function analytic in R+. Let f(Z) be a complex
fu
an
may
Khol
the
35
function analytic in R including the point at infinity
and continuous on B. Then
2%1'1; «f—E? dt = f(°°) for ZcR+ (1.51)
2%1']; i—(él dt = f(w)  f(Z) for ZeR' (1.52)
The condition (1.51) that the Cauchy integral have a con
stant value in R+ is both necessary and sufficient for the
continuous function f(t), defined on B, to be the boundary
value of a function analytic in R.
Let ¢(t) be a complex function which satisfies the
Holder condition on an arc L. Then the Cauchy integral
<1>1(Z)= 7%1/14 $4? dt (1.53)
may be shown to be a sectionally analytic function in the
whole plane cut along the arc L. Further, the limiting
values ¢+(t), o(t) may be shown to exist on L and satisfy
the relations
¢1+(to)  ¢1'(to)= we)
¢1+(to) + ¢1(to) =‘#T./: isig dt (154)
where to is a point on L and the integral in equation
(1.54) is represented as a principal value. The assumption
that ¢(t) satisfies the Holder condition is sufficient for
the existence of the principal value. These results are
referre
[26].
Si
determi
followi
used fr
and let
for the
This is
U5, the
36
referred to as the Plemelj formulae. They are derived in
[26].
Since the unit circular region will be used in the
determination of principal value of some integrals in the
following chapters, the following integral form will be
used frequently. Let a and b be constants where b/a°[w(c)] = ¢>°(Z)
119(5) = W°[w(c)] = “(2)
and the derivatives are
44
*
¢ '(Z) ° w'(C)
*
¢1'(C)
19m 1W2)  w'(c)
* * *
¢1"(c) = d2 "(2)  w'ZU.) + cb '(Z) ~ w"(c) (2.2)
To find the influence function, the derivatives of equa
tions (2.1) are needed, so
¢°'(2) + ¢*'(Z)
¢'(Z) =
‘1“(2) = w°'(z) + v*'(2)
¢"(2) = ¢°"(Z) + ¢*"(2) (2.3)
i:
where ¢ '(Z), ¢*"(Z) and W*’(Z) can be easily found from
equations (2.2):
Q)
*
*1 = ¢1'(C)
<1 (2) C
¢*vc(z) = ¢I"(C) _ ¢I'(C)'w"(C)
w'ZICI 45T3(C)
*
w*'(2) = 54% (2.4)
Substituting equations (2.4) into equation (2.3) and recon
sidering equations (2.1), the requirements for the influ
ence function become:
*
¢(Z) = ¢°(Z) + mm
1(2) = W2) + vim
¢*'(c)
I _ 01 + 1
4) (Z) r 49 (Z) W
I — i w "(C
1(Z)‘P°(Z)+—ﬁyrﬁjl
* *
¢1v(z) = ¢0!1(Z)+ W .. (bl US$2.62) (C) (2.5)
The complex potential functions for an infinite plane with
a concentrated force, P, at a point 20, ¢°(Z) and W°(Z),
are known (Maskhelishvili [27], Sokolnikoff [29], Green
and Zerna [31]).
(130(2) = ' 2% ln(ZZo)
W0(Z) = C1 mim ln(ZZo) + 7&1? ' 2?.20 (2.6)
To find the complex potential functions for the problem
of Figure 2.2(A), it is necessary to find the complex
potential function for the problem of Figure 2.3(D), i.e.,
¢I(c) and 11(c).
Since ¢:(c) and WT(§) have to be analytic in the
domain, then as mentioned in Chapter I, just one boundary
condition is necessary to find the complex potential
functions, i.e., either of equations (1.45) or (1.46).
Note that in the original problem, Figure 2.2(A), the
boundary of the hole is tractionfree. Recall equation
(1.45) for the traction boundary condition [29]:
46
f(s) = £1 + if2 + const. = 0
Then equation (1.45) becomes:
¢(t) + t o'it) + Wit) = 0 on C (2.7)
where t represents the values of Z on the contour C. Sub
stituting equations (2.1) into (2.7) leads to
¢*(t) + t ¢*‘(t) + P*(t) =  ¢°(t) + t ¢0'(t) + w°(t)
on C (2.8)
Clearly, the lefthand side of equation (2.8) is in
the form of the traction boundary condition for the problem
of Figure 2.2(C), since ¢*(t) and W*(t) are the boundary
values of 6*(2) and P*(Z). Also, the righthand side of
equation (2.8) is known, since ¢°(t) and W°(t) are the
values of ¢°(Z) and W°(Z) on the fictitious contour in the
problem of Figure 2.2(B).
Since the boundary condition for the problem of
Figure 2.3(C) is known (equation 2.8), then the boundary
condition for the problem of Figure 2.3(D) can be obtained
by transforming (2.8) to the c plane using the transforma
tion functions
Z = w(§)
so that the boundary transforms by:
t = w(0)
where 0 represents the values of c on the circumference of
47
the disc. Hence, equation (2.8) becomes:
¢T(o) + 9491 ¢’1"(o) + We) = (We) + ”(0) <12 0
“m o m
+ ngETZ) (2.9)
* i: 9:
where ¢1(o) and W1(o) are the boundary values of ¢1(;)
and WT(;), respectively. Also, ¢?(o) and W9(o) are the
boundary values of ¢9(c) and W?(c), respectively.
The righthand side of equation (2.9) is known, so let
F(0) = [p9(0) + “(0) ¢?'(0) + 19(0)] (2.10)
(0'10)
then, equation (2.9) becomes:
¢1'(0) + 11(0) = F(0) (2.11)
¢I(0) + “(0)
:3
This is the mixed boundary condition for the problem of
Figure 2.3(D) from which the two analytic functions
¢:(c) and PT(;) will be found. It is necessary to point
out some characteristics of ¢t(c) and W?(;) before
proceeding.
As mentioned in section 1.2, ¢:(;) and Wt(c) must be
analytic (holomorphic) inside y, the unit circle. Also,
without loss of generality, it can be assumed that
¢I(0) = 0. Thus, ¢:(c) and WT(;) may be developed for
Icl<1 in power series of the form
00
¢I(5) = g;; akck , vI(c) = ;E% bkck (2.13)
48
where, in the first series, the constant term is absent
*
because of the condition ¢1(0) = 0. Furthermore,
_ —k * —k
¢TZC1 = Ea; akC , V1(C) = 2E; HR: (2.14)
Let g approach the boundary 7, i.e., c+o. Note that since
the radius of the disc is equal to one, then:
00 = l (2.15)
Equations (2.13) and (2.14) are valid for the boundary
9: i:
values ¢1(0) and 11(0). Substituting equation (2.15)
into equations (2.14) along with equations (2.13) for the
boundary values, these become:
¢I(o) =2; akok , 111(0) = 1:0 bkok (2.16.a)
¢§105 = E:; akak , Wfloi = 25% Skak (2.16.b)
Equations (2.16.a) show that ¢:(o) and Wf(o) have poles at
infinity, so they are analytic functions inside the unit
circle. Also, equations (2.16.b) show that 57(3) and TTTET
have poles at the origin, so they are analytic outside of
the unit circle.
Using this analysis and employing the Cauchy integral
formulas, the complex potential functions for the problem
of Figure 2.3(D), ¢:(c) and W?(;), can be computed. To find
¢:(§), let both sides of equation (2.11) be multiplied by
49
l . do
Zni oc
where C is a point inside 7, the unit circle.
Integrating both sides of the equation counterclockwise
around the unit circle leads to:
Li ¢1(0)d0+7L_¢w(01¢1“105d0
2N1 n1
+2111¢W1T§Td=2:1;;552
it 3!
Since ¢1(o) is analytic inside y and 91(0) is analytic
(2.16)
outside y, then due to Cauchy integral formulas (section
1.3), equation (2.16) can be written as:
* l w(o) ojioi —;T—7_ l dg F(0)
¢1(C) + TIT—{é m —O'C d0 + WI 0  W Y OC do
(2.17)
Note that the third integral of equation (2.16) becomes:
Y 0:...
where I???) is a constant.
Equation (2.17) is an intergodifferential equation
for ¢T(§). It contains an unknown constant WTTET, which
can be determined by letting c = O and imposing the condi
*
tion ¢1(0)= . Thus, if the value of W1(0) in equation
50
(2.17) is chosen arbitrarily and the corresponding solution
*
for ¢1(;) is found, then the actual value of WfIUI can
i:
be computed from the condition ¢1(0) = 0. This is due to
the fact that if ¢f*(;) is any solution of (2.17) for a
given wf(0), and if ¢f*(0) = auto, then ¢f*(;)ao is a solu
tion of (2.17) with T¥TUT replaced by TTTUT+ao. Thus,
WT(0) can be tentatively fixed, say VfTUT = 0. Also, as
mentioned in section 1.2, in order to have a unique solu
tion for ¢T(§) and WT(;), the following conditions must
be satisfied:
3! it
¢1(0) = 0 11(0) = 0 (2.18)
* .
To find W1(c), take the conjugate of equation (2.11) and
multiply both sides of the equation by
11 do
2“ C
where g is a point inside y.
Integrating both sides of the equation counterclock
wise around the unit circle leads to:
a
1 f dbin 5 £517 ¢ '(01
Zﬂl Y o C do + 21w'(o) 3 C do
1 P*(o) _ l o
mi 'L‘er, «5me 55;?“ (2'19)
An argument similar to that presented for reducing
equation (2.16) to equation (2.17) can also be presented
here to obtain:
51
it
1 m7 ¢1'(0) *  1 f 13:"
W*m mwdw‘mm'my e.; d“
(2.20)
Substituting condition (2.18) into equations (2.17) and
(2.20) and rearranging leads to
* = 1 f F(o) _ 1 $1.1m W o
¢1(C) "ZN—f Y 0"; (10 W Y w 0 0; C10 (2.21)
i:
* _ l Fioi l 5‘. (oi o '(o)
*1“) ‘ mi 0; do ' 273 Y 31(0) ér, do “2“
It is easy to reduce the solution of the integrodifferen
tial equation (2.21) to the solution of the standard
Fredholm integral equation. The existence of a solution
of equation (2.21) would then follow, almost directly,
from the Fredholm theory.
The second integrals of the righthand sides of
equations (2.21) and (2.22) are left in general form since
function w(C) has not yet been Specified.
CHAPTER III
CIRCULAR HOLE IN A FINITE TWODIMENSIONAL REGION
111.1. INTRODUCTION
The effect of a circular hole on the stress distri
bution in an elastic region has attracted considerable
attention for the past seventy years. The effect of a
circular hole on an infinite plate subjected to uniaxial
tension was first solved by Kirsch [32]. This work was
extended to other load conditions by Bickley [33].
Howland [34] solved the problem of a long strip weakened
by a circular hole subjected to uniaxial tension. Other
load conditions were considered by Savin [35]. The effect
of a circular hole on the stress distribution in a finite
elastic region has been treated numerically and experi
mentally using several methods.
In this chapter, the solution of the problem of a
finite plane elastic region containing a circular hole and
subjected to traction boundary conditions is presented.
This is the first implementation of the mapping technique
and boundary integral equation method. In section 2 of
this chapter, some known complex potential functions [36]
are used to find the influence function for an infinite
domain weakened by a circular hole. In section 3, the
52
53
Muskhishvili method is used and the mapping technique is
employed to determine the influence function directly.
It is shown that the two results are identical. In the
last section some example problems are considered and the
results are compared to some known solutions, where
available. The computer program for the computation is
included in Appendix B.
111.2 DERIVATION OF THE INFLUENCE FUNCTIONS
USING KNOWN POTENTIAL FUNCTIONS
Consider an infinite elastic plane with a circular
cavity of radius a centered at the origin. Let a force
P act at a point 20 where IZoI>a. Bhargava and Kapoor
[36] have constructed the potential functions, ¢(Z) and
W(Z), for this problem. They assume, following Green and
Zerna [31], that the complex potentials are of the form:
¢(2) = P {1n(zz )  a 1n (ziz—M A (zﬁ)’1
2nio+li ° 7? Z?
+ a 1n 2} (3.1)
(2) = F a 1n (ZZ ) + 2 2 (22 )1 + In (22:)
“7—HT a+ O 0 p 0 7;
+ B (zf‘—:—)'1 + c (ZiaY2  ln 2 + 112'1 + E z‘z}(3.2)
20 o
This choice clearly gives prOper singularities at the point
of action of the concentrated force 20. Also, it satis
fies the condition of zero stresses at infinity. The
54
unknown constants can be found from the condition that
and 0
normal and tangential stresses, orr re’
are zero at
the boundary of the hole, i.e.,
(0
rr  10re)2=o 
The boundary condition can be written in terms of ¢(Z) and
W(Z) following Muskhelishvili [27] as follows:
0,,  iOre = ¢'(Z) + 67177  eZietf ¢"(2) + 1'(Z)1 (3.3)
Substituting equations (3.1) and (3.2) into equation (3.3)
leads to:
2
A=§B 120
(1286
2
B=3a20B 120
F 82
2
c=B 12%
B“
2
n=8 12. I:(Ol'L)Zo
82 P 82
E =  E a a2
P
where 52  Z°Z° Th f ' 1 d ° f p
 7;T' US, or an 150 ate p01nt OTCC
acting at the point 20, the complex potentials at the point
Z are:
SS
2
¢(2) — 2“ 5+1 3  ln(ZZo)  a 1n(2%:) + a ln 2
0
F 20201 a2 1
+ (z—)
ZTTTOH'I) ; 322: n
P' 0.2 20201 0.2 '1
1(2) = {a ln(ZZo) + ln(2——)  1n 2  ——————(Z——)
ZTTiOH'1) 0 70 To
z Z 1 o2 2 20201 . 1 P — _ 1
+ 02?: (22%) + ‘20 Z) + 71W“ 3 2°“ 2°)
2  __ _ _ _
+ a 20(Z%E) l  a 202 1  i%— 2 1  a 322 2:
Without loss of generality, assume a = 1. Then 6(2) and
U(Z) become:
P 22 1
MD = —(——T) 3  ln(ZZo)  a 1n( ° 1;
2” 0+ 270
+ P 20201 . 1
Z"(“+1) 23 2201
P 22 l 2 2 l
W(2) ='_T—__I 3a ln(ZZo) + 1n(——£——)  —£—£——
211 CHI 220 Zial
+ ZoZofl + Zofo’l$+ P To + 02%
(2201)2 220 2"E“+15 z'z° 2701
_ aZogz'tl _ 292'; (3.3)
56
Let these potential functions be written as:
¢(z) = 7FTETTj{¢I(Z)} + 2FTSTTT {¢II(Z)}
“1) = 113+ {‘1’1(Z)}* man) {111(2)}
where 61, ¢II’ WI and W11 can be found by comparison to
equations (3.3).
Since ¢'(2), ¢"(Z) and w'(Z) will be needed, they
will be listed here:
4"”) = minim} * Wis—WWI}
1v .. P 11 P ..
¢ (2) _ n a+ {¢I } + n a+ {¢II}
w'(z) = fingTT {wi(z)} + 21(g117 {111(2)} (3.4)
l a
Z(ZZo‘1)
1 222 1
¢"(2)=+m_ + °
I ° 22(2201)2
¢' (2) =  
II 7% (2201):
57
_ 2
¢ii(z) = 2020 1 . 2.2.0
7% (2201)3
W'(Z) = _ 2b  __EZ§___ + o202o+1 + 2o
I (Z'Z°) (2201)2 7°22 7?
, _ l (Zo2ol)Zo 2(Zo2o—l)2o
w (23n‘3‘ +_____+ 
11 ° 2(2201) (2201)2 (2201)3
 2335;}. (3.5)
2022
Hence, the influence function can be easily found as
described in section 1.2. They are
Hxx;qP3 = Re [2¢'(Z)  Z¢"(2)  w'(2)]
*= 1 I! v
Hyy;qPq Re 12¢ (Z) + 2¢ (2) + W (2)]
*= 1! v
ny;qpq 1m [2o (2) + W (2)]
l
* = _ 
Ix:qpq ZﬁRe [o¢(Z) Z¢'125 W125]
*=_l_  ' 
Iyquq 2m 1m [a¢(2) 2¢ 125 W125] (3.6)
Substituting (3.4) into (3.6) leads to
Hxx;q(Z,Zo)Pa(Zo) = 2115:17'512¢i(z)  Zei'(2)  wi(2)] p*
+ 12¢],(2)  7111(2)  wi1(2) 1 P3]
58
R g  H + 1
+ 12¢i1(z) + Z¢]i(Z) + wil(2) 5 Pt]
Im
“xy;q‘z’z°)P3(Z°) = 2113:1111 7¢i'(z) + 11(2) 1 P*
Wham + 111(2) 5 W1
Ix;q(z,z )P;(zo) — 153%5117[1a¢1(2)  26:177‘ 1125 1 P*
+
18811”) + z11157:j ’ 115Zj 15“]
1y;q(z,zo)1>a(zo)  ﬁmy [{e¢1(2)  z—(‘Tei 2 — T71 z 113*
+1e¢n(z) + 2‘1—(‘74511 z  mu 2 [15*] (3.7)
where ¢I(Z), 11(2). ¢i'(2), ¢II(21. ¢i,(2), ¢ii(z)» w,(2).
Wi(2), WII(Z) and WiI(Z) are defined by equations (3.4).
These influence functions will be used to solve a simple
problem by the boundaryintegral method in section 4 of
this chapter.
111.3 DERIVATION OF THE INFLUENCE FUNCTIONS
USING A MAPPING TECHNIQUE
The mapping technique which is presented in Chapter
II is now employed to obtain the influence functions. Con
sider the problem of an infinite plane having a circular
59
hole of radius a at the origin and a concentrated point
force P acting in the plane at some points 20 where IZol>a.
This problem can be expressed as the superposition of two
problems, Figure 3.1.
In the problem of Figure 3.1(B), the concentrated
point force P acting at the point 20 in an infinite plane
is considered. In the problem of Figure 3.1(C), the
infinite plane contains a circular hole with a prescribed
traction acting on its circumference. The traction on the
circular hole is equal in magnitude and opposite in direc
tion to the generated traction on a circular contour in
the problem of Figure 3.1(B). By adding the solutions
to the problems of Figure 3.1(B) and 3.1(C), the zero
traction on the hole of the problem of Figure 3.1(A) is
obtained.
The solution of the problem of Figure 3.1(B) is well
known (Muskhelishvili [27]) so that the required traction
can be found. Also, the problem of Figure 3.1(C) may be
handled by mapping into a unit circle (disc), Figure 3.2.
Clearly, the mapping function for this problem is
2=w(c)=%
which is conformal and one to one [37]. Without loss of
generality, let a = 1. Let the complex potential function .
for the problems of Figure 3.1(A), 3.1(B), and 3.1(C) be
¢(Z), W(Z); ¢°(Z), W°(Z); and ¢*(Z) and W*(Z), respectively.
60
(C)
Figure 3.1 Fundamental problem expressed as super
position of two problems.
Figure 3.2 Mapping the auxiliary problem to a unit
disc.
61
Then the complex potential functions for the problem of
Figure 3.1(A) are
M2) ¢°(Z) + ¢*(Z)
NZ) ‘1’°(Z) + W*(Z)
where the derivatives are given by equations (2.2) through
(2.5) in section 11.3. To find ¢*(Z) and W*(Z), the
transformed complex potential functions which are given
by equations (2.21) and (2.22) will be used. It is first
necessary to calculate the integrals of equations (2.21)
and (2.22).
I; = 7.13% :37: ‘bgéoj do (3.8)
:2 mi 273%.. (>313) .. .3...
Taking the derivative of equation (2.16) leads to
M'CO) = Z: kakok (3.10)
k=1
and the complex conjugate is:
¢’1"(o§ = k2; kEkEkl
Since 03 = 1, then
62
W o = 2:1 szko’k"1 (3.11)
The mapping function and its derivative, evaluated on the
boundary, are
= l ' =  __
MO) 0 , w (o) 02
so that
“(0) =  3— (3.12)
w'ioi o3
Multiplying equations (3.11) by (3.12) gives:
(0(0) . W; z: kako'k'z (3.13)
w'io) k=1
From equation (3.13), it is clear that the righthand side
is an analytic function outside of y, the unit circle.
The value of the righthand side at infinity is zero.
Thus, due to the Cauchy integral formulas, the principal
value of the integral of equation (3.8) leads to:
_ 1 (0(0) W 
I1" mi 00 O 0; d0  0 (3.14)
To calculate the integral of equation (3.9), consider the
complex conjugate of equation (3.13), which is
w 0 =  03 (3.15)
Multiplying equations (3.10) by (3.15) leads to:
63
oo
¢f'(o) = kzl  kakok+3 (3.16)
+
Clearly, the righthand side of equation (3.16) is analytic
inside the unit circle. Hence, following the Cauchy
integral formulas, the principal value of the integral
of (3.9) leads to:
12=m5€ 3‘30) ‘1’??? do =  4: ewe) (3.17)
Substituting integrals (3.14) and (3.17) into the general
formulation for ¢f(g) and W§(;) (equations [2.21] and
[2.22]), the complex potential functions for a circular
disc with a specified boundary value, F(o), are obtained:
M02)  7.13% if) do (3.18)
‘P’fm = 7,13% SC: do + c3¢*'(c) (3.19)
Y ,
For the case considered, F(o) and Fig) will now be calcu
lated. Rewriting equation (2.10) and taking the conjugate
leads to:
13(0) =  («33(0) + %W1 o + W o] (3.20)
F105 =  [¢‘,’(0) + % ¢‘{'(o) + 13(0)] (3.21)
Substituting equation (3.7) into (2.6), the transformed
complex potential functions ¢2(c) and Wg(§) are:
12
F l
.773/
64
¢2(c) =  71%ETT) 1n (ligii) (3.22)
_ P' lZC P 2c
12(c)  a ZFTETTjln ( o ) + 2ﬁ(a+1) 1320; (3.23)
Taking the derivative of equation (3.22) and substituting
c = 0 into equations (3.22) and (3.23) leads to:
P
(133(0) = " m 1H (13500)
. _ P 1
¢3 (O) ' ’ 2nCa+l) (1200)
0 _ ﬁ 1 ZOO P _ 700
WI(O)  a Znia+I$ 1n ( ) + n a+ 1Zoo
Taking the complex conjugates of equations (3.24) along
with equations (3.12) and (3.15) will provide all the
terms on the righthand side of equations (3.20) and
(3.12). Then F(o) and Fioi will be
= l'ZoO _ _ + — l‘ZoC (3.25)
No) Q{1n ( O ) a In (a 70)} Q{_o(o70)}
FTE) = Q‘{g§%%§%l} + o'{1n (020)  a 1n.l;§19} (3.26)
_ P
where Q  7ET6:T73
Substituting equation (3.25) into (3.18) leads to
73795 ”(12°C 9 1( J)
l. _ n 0 0
$1 0 c do Zni Y C do
U+§$I.¢§ (1‘Z°°) do (3.27)
Y 0(0'70) (O'C)
65
(l;%12) is an
Recalling the discussion in section 1.3, In
analytic function outside of y, the unit circle. Because
Zo>l, the function has two essential singular points at
on and o = 1/20. The function is defined at infinity as:
l'ZoC
[1n ( O
[It—J
= In (’20)
0V
then
1 1 200
1 9;. C o ) _
2?: Y O_C do  1n (~20) (3.28)
Clearly, ln(oZo) is an analytic function inside y, the
unit circle. Hence, the Cauchy integral formulas lead to:
mf l——:——n(0; 20) d0 = ln(C' “2.0) (3.29)
Also
{ l'ZoO
1 (1Z°O) do = Residu
7F? 36 ocozo)(oc)
Y 0(0'Zo)(0'5)
O=C
+ Residu I { 1'Z°O
0:0 0(O'ZO)(O'C)
or
1 (lZoq) = 120; 1
'TFI _______.+ ___. (3.30)
Y 0(oZo)(oc) c(cfo) Zoc
66
Substituting the integrals of equations (3.29) and (3.30)
into equation (3.27), the complex potential function can
be obtained:
cptcc) = Q{1n(zo)  a ln(cm} + M 1'Z°7° } (3.31)
70(C‘Zo)
To obtain Wf(c) it is necessary to find the integral of
equation (3.19), i.e.,
= 1 .96 Flo)
I3 2N1 Y 0—; do
Then from equation (3.26):
_ 6(07 )do — 1 (02 )
I3 ’ 733$ (1zoo§(o;) 1 2%? Y no; 0 do
lZoo
. ln( )
 3?%'5§ 0: do (3.32)
where
—l+ C(O'Z°) do = Residu 0(OZ°)
ZNI SE (1 2003(0C) Ulc{ Zo(o;;)(oC) }
+ Residu C(Oiz°)
1 Z°(OZF)(OC)
O:
7?
After some simplification:
1 0(07 ) _ g 1Z 2
my: (Izooﬁoc) ‘10 ‘ ' 2‘; ‘ _°—'°‘
(3.33)
67
Substituting equations (3.33), (3.29) and (3.28) into
equation (3.32) leads to:
2%"? it?“ = Q{‘ 2;: $23} + U{1n(c°Zo)
 a ln(20)} (3.34)
Taking the derivative of equation (3.31) gives:
¢T1(C)= Q{ '01 } + Q’{1_'Z_£E£ . _'1_____2.} (3.35)
C'Zb Zo (C‘Zo)
Substituting equations (3.35), (3.34) into equation (3.19)
and following some simplification, the other complex
potential function is:
me) = Q { 7;“;  lit71  953—}+Q{1n(cZo)  a ln(z.)
_ 1z.Zo_ . ____C3 } (3.36)
To (C’Zo)2
As is discussed in section 1.2, the two complex potential
functions ¢f(§) and ?f(c), expressed by equations (3.31)
and (3.36), are not a unique set of functions. Since the
origin of the coordinates is within 7, then, following
section 1.2, the uniqueness conditions for ¢f(c) and Wf(c)
are:
68
¢1(0) = 0 , WT(0) = 0 (3.37)
Conditions (3.37) lead to the unique complex potential
functions:
We) Q{a ln(7.)  a ln(a70)}+ q{___1'zozo
702
1 12020 . 1 } (3.38)
70 620
_ _ c _ 6:3 + — _  
wt)  Q{ 73— th} Q{1n(c 2,) ln( To)
_ l;ZQZh.. __£3___.} (3.39)
2. (cZo)2
These can be rewritten as:
ma) Q {mm} + 6{¢f1(c)}
8 {W0} + 8 {18102)}
81%;)
where ¢f(c), 6f1(c), W?(;) and W¥I(c) can be obtained by
comparison to equations (3.38) and (3.39) and are given
in Appendix A.
Then
¢§'(c) = Q {¢f‘(c)} + Q {¢§i(c)}
¢f"(c) = Q {¢f"(c)} + — {¢fi'(c)}
Q
wt'cc) = Q {wg'm } + Q {mm} (3.40)
69
where
¢’f'(c:) =  °‘_
C‘Zo
me.) =  —_1'§°7° . _1_
Z0 (C'Zo)2
¢*H(C) = a
I (C‘70)2
cb’fi'hz) = ———1'Z°Z°  ____2
79 (C‘zh)3
2 _ _
wg'cc) =  .21.  8.; L24 3:0)
o (C'Zo)
W’I‘i(z;) = __1_  1ZoZo , r.’(c3Zo)
5'20 Z. (C'Zo)3 (3.41)
Complex potential functions of equation (2.6) can be
rewritten as:
¢°(Z) = Q.ln(ZZo)
11°(z) = Q 7° + (5.111(220) (3.42)
ZZo
and the derivatives are:
¢°'(Z)='Q°2T0
¢°"(2) =+Q—————1
(ZZo)2
‘1’“(2) = Q 7° +Q 0‘ (3.43)
(ZZb)2 2’78
70
Substituting equations (3.40), (3.42) and (3.43) into
equations (2.5) leads to:
4(2) = Q { 1n(2zo) + ¢I(2) }+ 6{¢II(z.)}
8(2) = Q {272 + WI(2)}+ Ma 1n(2 2.) + 8* II(:)}
 0
¢*'(c) (2)
, _ 1 + I —
‘1’ (Z) ' Q{zzo W2.)— (25%}
' _ Z wf'(C) a + w¥i(C)
w (Z)  Q{m+m}+q{zzo W}
4"(2) _ Q{ 1 ¢I"(2) ¢I'(2;)w"(2)}
(ZZo)2 (u'2(c) w'3CE)
¢*"(c) ¢*'(c)w"(c)
+ 6{—L——— II } (3.44)
w'2(C) w'3(c)
where ¢I'(c), ¢I"(c). 4* II(2). ¢II'(2), WI'CC) and VI I(c)
are defined by equations (3.41).
Substituting equations (3.44) into equations (3.6)
along with the mapping function, w(§) = 1/C, and its
derivatives, leads to the influence functions for a
circular opening:
Hxx;q(z’z°)Pa(Z°) = Re: Q*<;q(z’7‘°)1’q(zo)nx1 + ny;q(Z,Zo)Pq(Zo)ny1 A 1
j#i
= Pxi (1:1,..N)
N
p. . Z{
1 9. , + H z z P* z . A8.
_7%_ + j=1 ny;q(Z’Z°)Pq(Z°)nX1 yy;q( ’ 0) q( °)ny1 1
j¢i
= Pyi (1:1,..N) (3047)
where the influence functions Hij,q(Z,Zo)Pa(Zo) given by
equation (3.44) and the resultant fictitious traction on
a given interval is represented by:
* = * . * .=
Pqi PXi + 1Pyi (1 1,..N)
In equation (3.47) nxi and nyi are the components of
the unit normal to the interval i. Also, Pxi and Pyi are
the x and y component of the real resultant traction
applied to the mesh i.
Considering the influence functions, equation(3.44)
and splitting each equation into two components of P*,
i.e., P; and P;, leads to:
II
I:
'U
a.
+
I:
’U
a.
* .
Hxx;q(z’z°)Pq(Z°) xx;x X XX;y
II
I!
’U
a.
4.
IE
’U
3&
H y;q(Z.Zo)1’c"I(Zo)
y yy;x ° x YY3Y °
II
V
3}
.4.
”U
a
“ y;q(Z,Zo)Pa(Zo)
x ny;x ° x ny;x
76
t = . * . *
Ix;q(Z,Zo)Pq(Zo) Ix;x PX + Ix;y Py
I Z,Z P* Z — I ° P* + I  P* 3.48
ysqc °) q( °) y;x x yzy y ( )
where Hij;q and Ii;q can be easily found by comparing
equations (3.48) with (3.44), see Appendix A.
Substituting equations (3.48) into equations (3.47)
leads to:
N
P*.
X1 * * *
+ :E:: 01 ,  P + H , P )  n . + (H ,  P
2 j= xx,x x xx,y y x1 xy,x x
j#i
. it o = °=
+ ny;y Py) nyi :ASi PXi (1 1,..N)
N
12*. Z:
_§£ . * . * . . k
+ i=1 (nyzx PX + ny;y P y) ”xi + (Hyy;x PX
j¥i
. a . = .=
+ Hyy;y Py) nyi }ASi PXi (1 1,..N) (3.49)
Rearranging equations (3.49):
N
P*.
XI + [H  11 . + H ° n ] ° P* + HI ' n
7 .=1 xx;x x1 xy;x yi xi xx;y xi
jfi
* =
+ HXY.Y nyi] Pyi :AS xi
' N
+” Z}
1 . , . * .
+ j=1 [ny;x nxi + Hyy;x nyi] Pxi + [ny;y nxi
i=1
. . * = . .
+ Hyy;y nyi] Pyi :ASi Py1 (3 50)
N
1 * . it . * = °=
7 PXi + ;§;(Aij pXi + Bij Pyi) BVXi (1 1,2..N)
j¥i
N
% P;i + :E:(C..  P*. + D.. ' P*.)= BV . (i=1,2..N)(3.51)
.= 13 x1 13 y1 y1
where:
ij xx;x xi xy;x yi
B.. = H , ° n . + H , ° n .
1] XX:Y X1 xy,y Yl
C.. = H ,  n . + H ,  n .
1] XV,X X1 yy,x Yl
D.. = H  . + H  n .
1) xy;y nx1 yy;y y1
(i,j=l,...N)
Euqations (3.51) represent a set of 2N equations with 2N
unknowns, i.e., P*
t ' =
xi and Pyi for 1 1,...N. Methods for
obtaining the solution have been discussed in section 1.1.
78
Writing equation (3.51) in matrix form:
.. B.. P*. BV .
1) 13 x1 x1
C.. D.. P*. BV .
_ 1] 1] d y1 y1
(3.52)
Note that the diagonals of submatrices [Aij] and [Dij] are
1/2 and the diagonals of submatrices [Bij] and [Cij] are
zero
Equation (3.52) can be solved by matrix inversion,
iteration, or elimination (Faddeeva [38]). Once the
fictitious tractions are found, then the stress and dis
placement at the point F can be easily found following
section 1.2. These are:
. * , *
(HXX;X(F,20) P . + Hxx;y(F.2o) PyI)
xx . x1
1
N
o = :E: [H ;X(F,Zo)  P§i  H (F,z,)  p*.]
YY i=1 yy yy;y Yl
N
OxY = g;; [ny;x(F’Z°) ' P§i + HXY;Y(F’Z°) . p;i]
N
Ux = i=1 [IX;X(F,ZO)  p;i + Ix;y(F,Zo)  P;i]
N
Uy = 22; [Iy;x(F,Z )  P;i + 1y;y(F,z,)  P;i] (3.53)
79
EXAMPLE III.l
A Rectangular Plane Weakened by a Circular Hole
Consider the rectangular region (10cm x 20cm) of
unit thickness (h = 1cm) which is weakened by a circular
hole of radius r = 1cm at the origin, see Figure 3.5. A
uniformly distributed traction (w = 1.0 MPa) is applied
to the t0p and the bottom of the rectangular region as
shown. The boundary has been subdivided into sixty
equallyspaced meshes, each of length 1.0cm, i.e., 10
meshes are defined on each of the top and bottom edges
and 20 meshes on each vertical edge.
The field points, the points where the stress and
displacement are calculated, are chosen along the x,y axis
and include points on the edge of the hole. These are
also shown in Figure 3.6.
The data, i.e., the coordinates of the nodal points,
X(I) and Y(I), the resultant of the traction on each sub
division (calculated by the trapezoidal rule), BVX(I)
and BVY(I), and the coordinates of the field points, XF(I)
and YF(I), are read into the program (Appendix B). The
results are presented in Table 3.1.
The results are compared to the theoretical solution
of a long strip weakened by a circular hole subjected to
uniaxial tension (Howland [34] and Savin [35]). The
program required 35 seconds of CPU time on a CDC 6500
computer.
u b 6 5 7 a 5 4 3' 2 
42 001
H3 *
n4 ‘
Ins 4
FIG "
I7
.35 ..
no . ‘
“K
#20 [J ‘
20 cm
I
3' ‘:_\ “w“: 5 ¥ X
IHHIHH
q
Figure 3.5 Circular hole symmetrically placed in a
finite rectangular plate under uniaxial tension.
81
Table 3.1. Stresses and displacements in a rectangular
region containing a circular hole at the
origin, Case 1
Geometry: rectangular plane (10 x 20cm2) (1 cm thickness)
Load: w = 1.0 MPa
Eccentricity: X0 = 0.0 Y0 = 0.0
' E = 70000 MPa, u = 26315.79 MPa, v = 0.33
Field Coordinates U U
Point X Y Oxx Oyy xy x y
N0. cm cm (MPa) (MPa) (MPa) microns microns
1 1.0 0.0 70.0 3.13128 0.0 o.oo14 0.0
2 1.2 0.0 0.32793 2.1486 0.0 o.oo152 0.0
3 l 4 0.0 0.38043 1.70146 0.0 0.00154 0.0
4 1.8 0.0 0.31226 1.33672 0.0 0.00155 0.0
5 0.0 1 0 1.ll98 0.0 0.0 0.0 0.00281
6 0.0 1 2 0.44908 0.0235 0.0 0.0 0.00291
7 0.0 l 4 0.18639 0.10502 0.0 0.0 0.00298
8 0.0 1.8 0.02031 0.36858 0.0 0.0 0.00314
9 1.0 0.0 0.0 3.13128 0.0 0.0014 0.0
10 0.0 l.0 1.1198 0.0 0.0 0.0 0.00275
AY
Available Solution: 1 f f I I t_f ‘
Field
Point
No. References
1 0 =3.14 MP 34 35
S oxx= “1'11 MPa [341.[351 \J V
9 =3.14 MP 34 35
a”, a ( 1.( 1
10 0xx= l.11 MPa [341,[35]
1 H I I I 4
\w
82
To see the effect of the size of the plane on the
stress and displacement solutions, smaller rectangular
planes, 8cm x 16cm and 6cm x 12cm, weakened by the circu
lar hole of radius r = 1cm at the origin were considered.
The results are presented in Tables 3.2 and 3.4.
The program has been written in such a way that, if
different dimensions of the rectangular plane are needed,
only one character, WR, is to be changed. Note that the
proportionality of the long side to the small side remains
constant and equal to 2.0. Also, for different locations
of the hole, the new coordinates of the center of the hole
XO,YO must be read into the program. Finally, the example
of the problem of a rectangular plane (9cm x 19cm) weakened
by an unsymmetrically located circular hole is solved and
the results are presented in Table 3.4. Again, the CPU
time was 35 seconds for each run on a CDC 6500 computer.
EXAMPLE 111.2
A Circular Plane Weakened by a Circular Hole
Let a circular plane of radius R = 6cm and unit
thickness (h = 1cm), which is weakened by a circular hole
of radius r = lcm at the origin, be considered, see Figure
3.6. A radially uniform distributed load (w = 1.0 MPa) is
partially applied to the tOp and the bottom of the outer
circumference, as shown. The boundary has been subdivided
into sixty equally spaced meshes each of which covers 6
degrees of angle (0.6283cm) numbered from the top and
83
Table 3.2 Rectangular region containing a circular hole
at the origin, Case 2
eometry: rectangular plane (8 x 16cm2) (1 cm thickness)
oad: w = 1.0 MPa
ccentricity: X0 = 0 Y0 = 0
= 70000 MPa, u = 26315.79 MPa, v = 0.33
ield Coord1nates
oint X Y 0xx oﬁy XY UX UY
NO cm cm (MP3) (M a) (MP3) [microns hicrons
1 1.0 0.0 ‘ 0.0 3.2212 0.0 0.218 0.0
2 1 z 0.0 0.3339 2.2013 0.0 0.229 0.0
3 1.4 0.0 0.382 1.738 0.0 0.232 0.0
4 1.8 0.0 0.3040 1.3619 0.0 0.234 0.0
5 0.0 1.0 —1.1821 0.0 0.0 0.0 0.404
6 0.0 1.2 0.4851 0.0294 0.0 0.0 0.418
7 0.0 1. 0.2090 0.0992 0.0 0.0 0.427
8 0.0 1.8 0.0293 0.3662 0.0 0.0 0.450
9 1.0 0.0 0.0 3.2212 0.0 0.218 0.0
10 0.0 l.0 1.1821 0.0 0.0 0.0 0.392
0V
4111111“
([[[[[10
K.
84
Table 3.3 Rectangular region containing a circular hole
at the origin, Case 3
Geometry: rectangular plane (6 x 12cm2) (1 cm thickness)
Load: w = 1.0 MPa
. Eccentricity: X0 = 0.0 Y0 = 0.0
E = 70000 MPa, u = 26315.79 MPa, 0 = 0.33
Field Coordinates U U
Point X Y Oxx O y Oxy X y
No. cm cm (MPa) (Mga) (MP8) microns microns
1 1.0 0.0 ' 0.0 3.4341 0.0 0.244 0.0
2 1.2 0.0 0.3456 2.3206 0.0 0.257 0.0
3 1.4 0.0 0.3827 1.8198 0.0 0.260 0.0
4 1.8 0.0 0.2724 1.4061 0.0 0.265 0.0
5 0.0 1.0 1.306 0.0 .0 0.0 0.406
6 0.0 1.2 0.555 0.0425 0.0 0.0 0.429
7 0.0 1.4 0.2486 0.0852 0.0 0.0 0.439
8 0.0 1.8 0.0361 0.3580 0.0 0.0 0.460
9 1.0 0 0 0.0 3.4341 0.0 0.244 0.0
10 0.0 l.0 1.306 0.0 0.0 0.0 0.390
IY
( no
‘111111‘1
([[[[[v
85
Table 3.4 Stresses and displacements in a rectangular
region containing a nonsymmetrically located
circular hole
Geometry: rectangular plane (9 x l8cm2) (1 cm thickness)
Load: w = 1.0 MPa
Eccentricity: X0 = 0.5 cm Y0 = 1.5 cm
 E = 70000 MPa, u = 26315.79 MPa, 0 = 0.33
Field Coordinates o o 0 U U
Point X Y xx y xy x y
No. cm cm (MPa) (MPa) (MPa) microns microns
1 1.0 0.0 0.0 3.16445 0.0 0.3267 0.6244
2 1.2 0.0 0.3313 2.169040.0010>0.3376 0.6248
3 1.4 0.0 0.3841 1.715260.0013 0.340 0.6251
4 1.8 0.0 0.31466 1.343610.00150.3415 0.6253
5 0.0 1.0 ~1.1624 0.0 0.0 0.1193 1.033
6 0.0 1 2 r0.4736 0.0269 0.0038 0.1180 1.044
7 0.0 1.4 ~0.2027 0.102510.0047’—0.1184 1.053
8 0.0 l 8 ~0.0294 0.3697 0.0046 0.1192 1.0763
9 l.0 0 0 0.0 3.19990 0.0 0.0989 0.626
10 0.0 1.0 [1.15149 0.0 0.0 0.11812 0.2181
AY’
I Y 4
1H t I
d :;
/ h
\J
1 1 [4 [I 1)
86
counterclockwise. The field points are chosen along the
x,y axis and include points on the edge of the hole.
These are also shown in Figure 3.6.
The data, i.e., the coordinates of the nodal points,
X(I) and Y(I), the resultant of the traction on each sub—
division (calculated by the trapezoidal rule), BVX(I)
and BVY(I), and the coordinates of the field points, XF(I)
and YF(I), are read into the program (Appendix B). The
results are presented in Table 3.5. The program required
36 seconds of CPU time on a CDC 6500 computer.
The effect of the radius on the stress and displace
ment solution has also been considered by solving the
problem for R = 4.8cm and 3.6cm and r = 1cm. The results
are presented in Tables 3.6 and 3.7.
To obtain the solution for different radii of the
plane, one has to change the character WR which is the
ratio of the desired radius to the R = 6cm. Also, for
a different location of the hole, the new coordinates of
the center of the hole, XO,YO must be read into the pro
gram. To see the effect of eccentric placement of the
circular hole on the stress and displacement field, the
example of a circular plane (R = 5.4cm) weakened by an
unsymmetrically located circular hole of radius r = lcm
is solved and the results are presented in Table 3.8.
Again, the CPU time was 36 seconds for each run on a CDC
6500 computer.
87
X 1‘ ‘I
1 I ” “
‘ 3 ~\ J/f
\b 5 I I
." \ 16"
4 \
_ a, \\
{3 \
.0 \\
\
ll \ '
\
‘0 \u/”££“\y
o \ I
‘I! b \ I
\ I
.M \\ I
\ I
5‘5 \ I
0‘
1'6 \ \\ ’ﬁ 4" —X>
. , J
. / \
6‘ I
6 / \
v / \
\ I, \ I...
o I R'
I l/
b I {/1
I, /l
V I //
I
v I /
/ /
I‘ I
._ ‘ 1’
~. 36 \3
// / ‘1 “In...” \,
Figure 3.6 Circular plane, containing a circular hole,
subjected to radially uniform tension over a portion of the
boundary.
88
Table 3.5 Stress and displacement in a circular plane
containing a circular hole at the origin,
Case 1
A
‘—
Geometry: circular plane R = 6 cm (1 cm thickness)
Load: w = 1.0 MPa
Eccentricity: X0 = 0.0 Y0 = 0.0
E = 70000 MPa, u = 26315.79 MPa, v.= 0.33
Field Coordinates 0 U U
Point X Y Oxx yy Oxy x y
No. cm cm (MPa) (MPa) (MP8) microns microns
1 1.0 0.0 0.0 2.948 0.0 0.272 0.0
2 1.2 0.0 0.2975 1.9438 0.0 0.284 0.0
3 1.4 0.0 0.32771 1.4742 0.0 0.287 0.0
4 1.8 0.0 0.2345 1.06002 0.0 0.290' 0.0
5 0.0 1.0 r1.70816 0.0 0.0 0.0 0.323
6 0.0 1.2 ~0.8751 0.0846 0.0 0.0 0.380
7 0.0 1.4 I0.5090 0.0332 0.0 0.0 0.388
8 0.0 1.8 0.2547 0.31029 0.0 0.0 0.409
9 1.0 0.0 0.0 2.948 0.0 0.267 0.0
10 0.0 1.0 (1°7081 0.0 0.0 0.0 0.318
=‘X.
89
Table 3.6 Circular plane containing a circular hole at
the origin, Case 2
Geometry: circular plane R = 4.8 cm (1 cm thickness)
Load: w = 1.0 MPa .
Eccentricity: X0 = 0.0 Y0 = 0.0
E = 70000 MPa, u = 26315.79 MPa, v.= 0.33
Field Coord1nates o 0 U U
Point X Y xx gy xy x y
No. cm cm (MPa) (M a) (MP3) microns microns
1 1.0 0.0 0.0 3.0268 0.0 0.288 0.0
2 1.2 0.0 0.3077 1.9808 0.0 0.3011 0.0
3 1.4 0.0 0.3414 1.4806 0.0 0.304 0.0
4 1.8 0.0 0.2502 1.0181 0.0 0.306' 0.0
5 0.0 1. 1.8281 ‘0.0 0.0 0.0 0.347
6 0.0 1.2 0.9085 0.0886 0.0 0.0 0.362
7 0.0 1.4 0.528 0.0397 0.0 0.0 0.412
8 0.0 1.8 0.232 0.339 0.0 0.0 0.434
9 1.0 0. 0.0 3.0268 0.0 0.282 0.0
10 0.0 l.0 51.8281 0.0 0.0 0.0 0.340
=‘X
90
Table 3.7 Circular plane containing a circular hole
at the origin, Case 3
Geometry: circular plane R = 3.6 cm (1 cm thickness)
Load: w = 1.0 MPa
Eccentricity: X0 = 0.0 Y0 = 0.0
E = 70000 MPa, 0 = 26315.79 MPa, v= 0.33
Field Coordinates o 0 U U
Point X Y xx . yy xy x y
No. cm cm (MPa) (MPa) (MPa) microns microns
l 1.0 0.0 0.0 3.2127 0.0 0.327 0.0
2 1.2 0.0 0.3298 2.0625 0.0 0.341 0.0
3 0.0 0.3686 1.4878 0.0 0.344 0.0
4 1.6 0.0 0.2748 0.9105 0.0 0.344 0.0
S 0.0 1.0 2.093 0.0 0.0 0.046 0.352
6 0.0 1.2 1.0129 0.0980 0.0 0.0 0.407
7 0.0 1.4 0.550 0.0526 0.0 0.016 0.417
8 0.0 1.8 ~0.136 0.3982 0.0 0.0 0.485
9 1.0 0.0 0.0 3.2127 0.0 0.319 0.0
10 0.0 1.0 (20937 0.0 0.0 0.042 0.343
IIY‘
—.—X
91
Table 3.8 Stress and displacement of a circular plane
containing a nonsymmetrically located
circular hole
Geometry: circular plane R = 5.4 cm (1 cm thickness)
Load: m = 1.0 MPa
Eccentricity: X0 = 1.0 Y0 = 2.0
E = 70000 MPa, u = 26315.79 MPa, v= 0.33
Field Coordinates o 0 U U
Point X Y xx yy xy x y
No. cm cm (MPa) (MPa) (MPa) microns microns
1 1.0 0.0 0.0 3.239 0.0 0.0 0.919
2 1.2 0.0 0.325 2.161 0.019 0.0116 0.9163
3 1.4 0.0 0.358 1.667 F0.0346 0.0166 0.9118’
4 1.6 0.0 0.259 1.239 0.0629 0.0232 0.898
5 0.0 1.0 1.532 0.0 0.0 0.358 1.226
6 0.0 1.2 r0.6160 0.0417 0.05507 0.344 1.239
7 0.0 1.4 80.2224 0.11601 0.0629 0.332 1.247
8 0.0 1.8 0.1352 0.4443 0.04002 0.2523 1.309
9  0.0 0.0 2.779 0.0 0.509 0.782
10 l.0 1.5716 0.0 0.0 0.1903 0.5704
[1,,
CHAPTER IV
ELLIPTICAL HOLE OR SHARP CRACK IN A FINITE
TWODIMENSIONAL REGION
IV.1 INTRODUCTION
Problems associated with stress concentration around
holes in structures have motivated the effort to solve
problems of plane elastic regions weakened by elliptical
holes or sharp cracks. The solution for the stress near
an elliptical hole in an infinite plane subjected to a
uniform load was first obtained by Inglis [39] using
complex potentials. Later this problem was examined
experimentally by Durelli and Murray [40]. A method for
the determination of stresses and displacements near the
tip of a sharp crack in an infinite plane subjected to in
plane load was developed in an infinite series form by
Westergaard [41]. The effect of holes of more general
shape on infinite planes has received considerable atten
tion, most notably by Muskhelishvili [27]. The problem
of an elliptical hole or a sharp crack in a long strip
subjected to uniform tension and compression has been
treated experimentally and numerically using several
methods and techniques. Yet, no solution for an arbitrary
plane region weakened by an ellipse or crack is available.
92
93
In this chapter the solution for the problem of an
arbitrary, finite, two dimensional elastic region, weakened
by an arbitrarily located and oriented ellipse or sharp
crack, is presented. This is the extension of the imple
mentation of the mapping technique and boundary integral
equation method. In section 2 of this chapter, the
Muskhelishvili method is used and the mapping technique
is employed to determine the influence functions for an
elliptical hole. In section 3 this influence function is
extended to a sharp crack. Finally in the last section,
some example problems are solved for an elliptical hole
and a sharp crack at different orientations. These
solutions are compared to some available experimental
[42,43] and analytic [44] results. The computer programs
are included in Appendices C and D.
IV.2 DERIVATION OF THE INFLUENCE FUNCTION USING
THE MAPPING TECHNIQUE: THE ELLIPTICAL HOLE
PROBLEM
In this section, the influence function for an
infinite plane region containing an elliptical hole is
derived. Consider an infinite plane containing an ellip
tical hole at the origin and a concentrated point force P
acting in the plane at some point 20, where 20 lies on or
outside of the ellipse, i.e.,
x
o
"<
o
(4.1)
”I
N
+
°'
N
\V
._n
94
where Z0 = Xo + iY0 and a,b are the semimajor and semi
minor axes of the elliptic hole. The problem can be
expressed as the superposition of two problems, see
Figure 4.1. The problem of Figure 4.1(B) is simply that
of a concentrated point force P applied at 20 in an
infinite plane and the problem of Figure 4.1(C) is that
of prescribed traction acting on an elliptic hole in an
infinite region.
This applied traction on the elliptic hole is equal
in magnitude and opposite in direction to the traction
generated on an elliptic contour in the problem of Figure
4.1(B), by the concentrated point force P.
Adding the solutions of Figures 4.1(B) and 4.1(C),
the zero traction condition on the hole of the problem
of Figure 4.1(A) is obtained. The solution to the problem
of Figure 4.1(B) is known (Muskhelishvili [27]). Thus,
the required traction can be found.
To solve the problem of Figure 4.1(C), it is neces
sary to map this problem into a unit circle (disc), see
Figure 4.2. It is easy to verify that the mapping function
2 = R(%+ m)
for R > 0 and 0 < m s l (4.2)
transforms the region exterior to the ellipse into a unit
circle Iclgl (Churchill [37]), provided R and M are taken
as:
95
.oHo; Hmowpawaao mo EoHn0ha on“ mo :oﬁuﬁmompoasw
R: «mC
H.v ohsmﬂa
3c
6261 N N
\J
[L
t
x
96
.omﬁv was: m ow EoHLOHQ xpmwﬁﬂxsm ecu mcwmmwz N.v owsmﬂm
«m: «E
ocmi N
\/ T
«u
97
a+b
R = T and m  3—4'5 (4.3)
where a and b are the semi—major and semiminor axes of
the ellipse, respectively, and are equal to:
a = R(1+m) , b = R(1m)
The mapping function is conformal for, if w'CC) is con
sidered, i.e.,
w'(c) = ¥L+n1 o<
II
R(1+m) Cos 0
*<
ll
R(lm) Sin 0 (4.4)
If m = o, the ellipse becomes a circle and the transforma
tion function equation (4.2) becomes w(c) = R/c. How
ever, it will be seen that several expressions derived in
98
this chapter will be singular when m = o and therefore
the analysis is invalid for the case of the circular hole.
Since the case of the circular hole has already been
treated, the following restrictions will be placed on m:
o
ll
mozZoo+1 (4.26)
U7
ll
oZZoo+m (4.27)
Solving equation (4.5[a]), the mapping function, for c,
yields:
mcz — Z; + 1 = O (4.28)
As discussed earlier, the mapping function, equation
(4.5[a]), represents a conformal mapping, i.e., for every
point Z exterior to the ellipse, there exists only one
104
corresponding point in the a plane interior to the circle.
Since equation (4.28) is of the quadratic form and has
two complex roots, then one root has to fall inside ; and
the other root has to fall outside y, the unit circle.
The two quardratics, equations (4.26) and (4.28), have
the same coefficients. Thus, equation (4.26) has two
roots, one inside and one outside y. Denote the root
inside Y by ri and the root outside y by r0. Then equa
tion (4.26) can be written as
A = mozZoo+l = m(ori)(oro) (4.29)
where
For examination of equation (4.27), consider the following
mapping function:
2 = 4(2) = 2 +13 (4.30)
This function maps points in the plane exterior to the
ellipse onto points in the plane exterior to the unit
circle. This mapping function is also conformal. For,
if w'(§) is considered
w'(;) = 1  IL of"(c:) = Q (¢1§"(c))+ Q (¢’f'1'(c) )
9’1"“) = Q (W§'(c))+ Q(‘¥’ﬁ(c))
where ¢f‘(c), ¢f"(c), ¢fi(c), ¢ff(c), Wf'(c) and WfiCC) are
given in Appendix C.
The complex potential functions of Figure 4.1(B) and
their derivatives are given by equations (3.42) and (3.43).
Applying superposition and adding the two sets of poten
tial functions, expressed by equations (3.42) and (4.49),
leads to the potential functions of the problem of Figure
4.1(A):
113
M2) (23 ln(Z'zo) + ”(03+ 53¢f1u):
Y(Z)
Q g z' + Wf(c) £+ Q';a ln(ZTo) + Wf1(c)£ (4 50)
Zo
Since ¢'(Z), ¢"(Z) and W'(Z) are also needed for the
influence functions, they are written here:
1 ¢*'(C) ¢¥i(§)
¢'(Z) = Q3* ZZo +'—TG:f—+ W3
and)  Q3 ——°——Z+w71—WC)+Q:M)°‘ +51%“)
(ZZo)2 “"
¢*"(c) ¢*(c)w"(c) ¢*"(c)
¢"(Z) = 3 1 + I _ I £+ Q: I
(ZZo)2 w'ZCC) w'3(C) w'2(c)
(4.51)
¢*i(c)w"(c) :
w'3(c)
Finally, substituting equations (4.50) and (4.51) along
with the mapping function and its derivatives, equations
(4.5), into equation (3.6) leads to the influence functions
for an infinite region weakened by an elliptic hole:
Hxx;q(Z,Zo)Pa(Zo)
H z,z P*
yy;q( o) q(Zo)
+
4.
+
+
c“ n
———————— ¢f (c) 
114
Re Q*<:ZE%F.+
2:2
mcz—l
¢f'(c)  7 [———l——
(2'20)2
2:3
(mCZl)3
W§'(c))
3'”
WW .1 on]
To
(zZo)2
C2
mgzl
1 n
z [EETEI§? + ¢§ (c)
¢§'(c)] 
ff(c)
(mczl)“
To
2:3
(Z'Zo)2
(mCZl)3
6* “2
mczl
Z .___£:___ ¢
(mczl)“
. :2
mczl
wg'm)
*n C) _‘ 2§3
(mczl)3
II ¢ii(g)]*
115
2
a + C who)
2'70 mczl
Im Q* Z'[ 1 + ———£:—— ¢*"(C)
(. (Z’Zo)2 (chl)“ I
*
ny;q(Z,Zo)Pq(Zo)
2; *1 20
‘ ——— (b (C) 
(m<:,21)3 I ] (ZZo)2
3 2
 ———3£——— ¢¥i(C3 + a + C W§'(c)
(mczl)3 ZTo mczl
I 2,2 P* z
x;q( o) q( 0)
Re %( 0L ln(Z'Zo) + 0‘ 43?“)
_ —2
z i + §e—— ¢g'(c)  2°
ZZo m; 1 {20
3! :2
‘Y*IZC5)+ %(a ¢§I(c)  ZLEzl d) CO]
d ln(Z—Zo)  V¥ETE{)
H»
H"
116
I (Z Z )P*(Z ) = Im 91  a ln(ZZ ) + a ¢*(c)
y;q 9 0 q 0 2U 0 I
_ '1 :2 ! Z0
Z + _ ¢I C ' ' TIE)—
Z'z—o 111C, ’1 Z
+
it  '52
9E a ¢¥I(C) Z [:m32_1 ¢II:C:]
a ln(ZZo)  WTIlCi (4.52)
Hence, the influence functions for an infinite plane with
an elliptical cavity at the origin are found.
IV.3 DERIVATION OF THE INFLUENCE FUNCTION:
THE SHARP CRACK PROBLEM
Considering m = l, a sharp crack along the xaxis
between x = 2 and x = 2 is obtained, Figure 4.3. The
transformation function
_ _ 1
Z  m(C)  E'+ C (4.53)
transforms the whole region exterior to the crack into a
unit circle c<1. Substituting m = 1 into equations
(4.52) leads to the influence functions for the crack
problem:
’K
(2,0) (2,0)
V><
Z .plane
Figure 4.3 A sharp crack in an infinite plane.
*
HXX;q(Z’Z°)Pq(ZO)
ll
5U
(D
O
8'
I I
N
o
+
N
W
N
'9
H}
r\
n
\J
2 2
 cil wf‘(c))+ 5* (cf: ¢ﬁ(c)
 7' __£:L___ ¢*n(;)  _£¥::___ ¢*'(;):]
[(cZl)“ II (c21)3 II
2
 0L  “9—— $1102)
zzo c2—1
Re Q* 7:5— + 2C2 ¢*'(c) + 7
'20 C2 I
___1__
(Z'Zo)2
+ c“ n 2:3 .
(CZ1)“ ¢f (c)  (;2_1)3 ¢f (c)]
20 + C2 (11* ( 2C2
 ——————— ' c) + 6* ¢*'(c)
(27.)2 Czl I c21 II
H Z,Z P* Z
yy;q( o) q( o) 1
4.
H
xy;q(z,zo)Pg(zo)
Ix;q(Z,Zo)Pa(Zo)
119
= Im Q" '2' ————l——— + C“
*H
(ZZo)2 (C‘°'1)"CbI (C)
 2:3 (3,...” 7 '
(CZ1 3 I C  0 + Cl '
) (Z‘Zo)2 Cz'l W? (C)
+‘Q* 2’ —£:L——— ¢*"(c) —3¥::——
(2:21)“ II  (c21)3 Hm
+ a C2
ZZO + 2 WT'CC))
II
50
(D
 Z[ 1 + E2 ¢g Z
zZo 521 I C  Z 2  I C
 0
7:
+ %(0 ¢fI(C) ' 2 ° _Ez ¢IIEC3
1
:2
a ln(Zzo)  W§}TETJ)‘
120
Iy;q(Z’Z°)Pa(Z°)  a ln(ZZo) + a ¢f(c)
ll
H
5
r0
_ —2
ZZo Cz'l 7’20
9—5.. _Lij.
Zu a ¢II(C) Z 321 ¢Ii C
4.
a ln(2'20) " 15511;; g (4.54)
where ¢f(c). ¢f1(c), ¢f'(c). ¢fi(c), ¢f”(c). ¢ff(c). Wf(c).
Wf1(c), Wf'(;) and Wfi(c) are given in Appendix D.
Hence, the influence functions for an infinite plane
with a horizontal sharp crack lying on the xaxis at the
origin are found.
One must exercise some care when using these influence
functions in that a singularity will occur when Z = 70.
This case will now be considered. First solve the trans
formation function, equation (4.S3), for C.
c = g 1 //Z2/4  1T (4.54)
Then write the two roots inside and outside 7, equations
(4.29) and (4.34), for the case (m = l):
ri,o = %} : //Z%/4  1 (4.55)
t. = Zo/Z i J/Zﬁ/4  1 (4.56)
121
If the field point, Z, is equal to the complex conjugate
of the load point, 20, then equations (4.54) and (4.56)
will be identical (; = ti), since lclsl and Itil<1. Note
that this makes the righthand side of integral III,
equation (4.41), infinite. Thus, integral III has to be
reevaluated. Substituting ti = C into the integral leads
t0:
M(Cr.)(0r )
111 = 7%; I. 1 0 do
‘Y (Oto)(oz;)2
Thus,
m(or.)(or ) m(ot.)(o_r )
III = Residuel 1 0 = é% oft o 
o=C (OtoMO'C)2 o 0:;
or
III — 71—. ¢m(o'ri)(°‘ro) do = (ZmC'Zo)(C'to)'(mC2ZOC+1)
ﬂl Y (U'to)(OC)2 (C'to)2
This change modifies the complex potential function ¢§(C),
equation (4.47), to:
r0 0
r ‘C t C
¢’;(<:)=Q{1n(° )a1n(‘§ )}
_.:(ZmC'Zo)(C‘to)'(mC2'ZoC+1)
+ Q
(cto)2
Zot 1
 ——9——} (4.57)
2
t0
122
Clearly, this change affects the other complex potential
function and all the derivatives. The modified functions,
¢f(c). ¢fI(c). ¢f'(c). fi(c). ¢f”(c), ¢ff(c). Wch).
Wf1(c), Wf'(c) and T?i(c) for this special case are given
in Appendix D.
Now that the influence functions have been obtained
it is important to notice that at the tips of the crack
where Z = :2 the stress influence functions will become
infinite as expected. These two points are the only
singular points in the plane.
IV.4 THE BOUNDARY INTEGRAL EQUATION METHOD APPLIED
TO A PLANE FINITE REGION WEAKENED BY AN ELLIP
TICAL HOLE OR A CRACK
In this section, two classes of problems will be con
sidered. These are: (1) a plane finite region subjected
to traction boundary condition t and weakened by an ellip
tical hole, Figure 4.4; and (2) a plane finite region sub
jected to traction boundary condition t and weakened by a
crack, Figure 4.5.
Solutions will be obtained by embedding the regions
Re (region with elliptical hole) and Rs (region with a
sharp crack) in infinite (fictitious) planes of the same
material as Re and Rs’ containing an elliptical hole,
Figure 4.7, or a crack, Figure 4.8, respectively.
In the treatment of either of these problems, the
boundary is divided into a finite number of divisions, N,
of equal or unequal length. A concentrated line load,
which is the resultant of the traction on each division,
is then applied at the center of the division, i.e.,
123
““L
\ v
\m/
Re
t8,
Figure 4.4 An elliptical hole in a plane finite
region with prescribed traction on the boundary, B
.
Y
A
r
Rs
k/ KB.
Figure 4.5 A horizontal slit in a plane finite
region with prescribed traction on the boundary, Bs'
124
45%
v'
I
.J/'—d——“\
__—/’
\ ¢x
\7RLX
f2»
(«~44
=~<
Figure 4.6 The finite regionsRe and Rs, with sub
divided boundary and concentrated line loads.
125
Figure 4.7 Region Re, embedded in an infinite
plane containing an elliptical hole at the origin.
126
Figure 4.8 Region Rs, embedded in an infinite
plane containing a horizontal slit at the origin.
127
P . = p ds
x1 A51 x
P . = ds
y1 {5, py
1
and for the fictitious tractions
* = *
PXi AS pX ds
i
P*. = p ds
yl Asi Y
where A81 is the ith interval and i = 1,...N, see Figure
(4.6). The trapezoidal rule is used to approximate these
integrals. Following section 1.1, the fictitious traction
P* around the fictitious boundary can be found from:
N
P*. 2
x1 . . o
T .j_1(Hxx;q(z,zo)pa “xi + “xy;q(z’z°)Pa nyi)Asl
jv‘i
= P _
X].
N
P*. ' Z(
1 * * . .
+ + 1 ny;q(Z,Zo)Pq nxi + Hyy;q(z’z°)Pq ny1)ASi
i
gnu.
1L II
= Pyi for 1 = 1,...N (4.58)
128
where the resultant fictitious traction on a given interval
is represented by
13*
. = P*. + i P*
q1 x1
yi (i = 1,...N) (4.59)
and the influence functions Hij'q(z’z°) are given by equa
tions (4.54), for a crack.
Note that, in equatlon (4.58), nxi and nyi are the
components of the unit normal to the division i and Pxi
and Pyi are the x and y component of the real resultant
traction applied to the division i, i.e.,
P . = P . + i P . (i = 1,...N)
Substituting the components of the resultant fictitious
traction, equation (4.59), into the influence functions
for an elliptical hole or a slit, equations (4.52) or
(4.54), and rewriting them leads to:
Hxx;q(Z,Zo)Pa(Zo) = Hxx;x  p; + Hxx;y  p;
Hyyzq(z’z°)P3(z°) = Hyy;x p; + Hyy;y ' P;
ny;q(z’z°)Pa(Z°) = ny;x P; + nysy ° P;
Ix;q(Z,Zo)Pa(Zo) = Ix;x ° P; + 1*;y P;
1y;q(z,zo)Pa(zo) = 1y;x P; + 1y;y P; (4.60)
where Hijq(Z’Z°)’ which represents the ijth stress com
ponent at a point 2 due to a unit load in the q direction
129
at a source point 20, and Ii;q(Z,Zo), which represents the
ith displacement component at the point 2 due to the unit
load in the q direction at the source point 20, can be
easily found by comparing equations (4.60) with either
equations (4.52) for the elliptical hole (see Appendix C)
or equations (4.54) for the crack (See Appendix D).
Substituting equations (4.60) into equations (4.58)
and rearranging leads to:
N
2:
o o . *
2 + j=1 [Hxx;x nxi + ny;x nyi] Pxi
jti
* =
+ [Hxx,y nx1 + ny;y nyi] Pyi) AS1 Pxi
N
Pic
;1 *
+ 32:; ([HXY,X x1 + “max ny1] PM
j¢i
. . . * =
+ [ny;y nxi + Hyy;y nyi] Pyi)ASi Pyi (4.61)
or writing equation (4.61) in the form of equations (1.24)
leads to:
‘1 * * ' * =
szi * ;(Aij Pxi + Bij Pyi) 3in
N
1 * t t 
fpyi " Zj=1(cij Pxi ” Dij Pyi)‘ vai
j#i
for i=l,2,...N (4.62)
where
ij = Hxx;x nxi + ny;x nyi
ii = Hxx;y “xi + ny;y nyi
C.. = H n . + H n
13 xy;x y1 yy;x xi
= H
ii xysy “xi + H
n .
YY3Y Yl
Equations (4.62) are a set of 2N linear algebraic equations
with 2N unknowns, i.e., P;i and P;i for i = 1,...N. The
methods for obtaining the solution have been discussed in
section 1.1.
Clearly, from equation (4.61),'one can conclude that
1 1
AijZ Dij‘z
B.. 0.0 C..
13 13
0.0
for i j. In matrix form, equations (4.62) become:
131
P o o B o q P*. BV U
13 13 x1 x1
*
This system of equations can be solved for Pxi and P*. by
matrix inversion as follows:
1
*
Pxi ’ Aij Bij ‘ Bin
P*. _ c.. D.. . BV .
y1 13 13 y1
(4.64)
or by other methods for solving systems of linear equa
tions (Faddeeva [38]).
Let F be a field point at which the stresses and
displacements are to be found. Then, using Appendix C or
Appendix D, the stresses and displacements at the field
point due to a unit load at a boundary point such as 20,
for the plane finite region containing either the ellipti
cal hole or the slit, i.e., (F,Z°) and Ii,q(F,Zo),
H..
13:4
can be found. The known fictitious tractions, i.e., equa
tion (4.64), will now be applied to find the real stresses
and displacements at the field point. These stresses and
displacements are
n1
132
N
<3 == 2: [H
* *
xx . l xx;x(F’Z°)Pxi + Hxx;y(F’z°)Pyi]
*
;X(F,ZO)P;1 + H y;y(F,ZO)Pyi]
O
YY i=1 YY Y
. * *
(F,Zo) Pxi + H y”(lazo) Pyi]
Oxy i=1 xy;x x
. * . *
[IX;X(F,ZO) Pxi + Ix;y(F,Zo) P ]
N
U = .23 [I
* *
i=1 Y:X(F’Z°) Pxi + 1y;y(F’z°) Pyi]
Some example problems will now be considered. The plane
stress or plane strain problem can be considered by choosing
the apprOpriate value for a in the complex potential
function. For generalized plane stress:
3v
a = +V
pa
and for generalized plane strain:
a = 34v
133
EXAMPLE IV.1
A Rectangular Plane Weakened by an Elliptical Hole
Consider the rectangular region (10cm x 20cm) of
unit thickness (h = 1cm) which is weakened by an elliptical
hole described by
x = (1+m) Cos 0
y = (lm) Sin 6
om 1 X
r L 2.3 1
1. a
1 an L4 , _36 ._ _ , 1 3;
1 1 1 1 1 1 1 1 1 1
L IOCM :1
Figure 4.9 Rectangular plane weakened by an ellip
tical hole at the origin.
135
coordinates of the field points, XF(I) and YF(I), are
read into the program as the data input (Appendix E).
The results are presented in Table 4.1.
The results are compared to the theoretical solution
of an infinite plane weakened by an elliptical hole sub
jected to a uniaxial tension [39] and solution of a long
strip weakened by an elliptical hole subjected to uniform
tension [42]. The program required 41 seconds of CPU time
on a CDC 6500 computer.
Two angles of inclination of the ellipse, 6 = 30°, 60°,
are also considered and these results are presented in
Tables 4.2 and 4.3. Note that the rectangular boundary
has been embedded in the infinite domain at inclination 0
to the "horizontal" ellipse.
The program has been written in such a way that if
different angles of inclination are desired, only one
character, THETA, is to be changed. Also, different sizes
of the ellipse, i.e., different a and b, can be obtained
in each case by changing the character, M, in the program.
For different locations of the center of the hole, the
new coordinates of the center of the hole, Xo,Yo, must be
read into the program. The problem of a rectangular plane
subjected to uniform load and weakened by an elliptical
hole with major axis 2a = 3.6cm and minor axis 2b = 0.4cm
centered at X0 = 1.5cm, Y0 = 2.0cm and inclined at an angle
of 6 = 30° is solved. The results are presented in Table
4.4. Again, the CPU time was 42 seconds for each run on
a CDC 6500 computer.
136
Table 4.1 Stress and displacement of a rectangular plane
containing an elliptical hole at the origin
Geometry: rectangle 10 x 20 cm2 (1 cm thickness)
Load: w = 1.0 MPa ,
Eccentricity: X0 = 0 Y0 = 0, Angle: e = 0.0
E = 70000 MPa, u = 26315.79 MPa, v = 0.33, m = 0.5
‘ Coordinates
Field
Point X Y 0xx 0 y Oxy Ux Uy
No. cm cm (MPa) MPa (MPa) Juicrons microns
1 1.5 0.0 0.0 7.4638 0.0 0.359 0.0
2 1.7 0.0 1.0963 2.50561 0.0 o.291 0.0
3 2.3 0.0 0.4496 1.4263 0.0 0.251 0.0
2. 0.25124 1.2630 .0 0.253 0.0
5 0. 0.5 l.107l 0.0 .0 0.0 0.545
6 0.0 0.8 0.5836 0.0139 0.0 0.0 0.547
7 .0 1.7 0.0108 0.30356 0.0 0.0 0.564
8 0.0 2.4 0.0566 0.5381 0.0 0.0 0.605
9 l.5 0.0 0.0 7.4638 0.0 0.359 0.0
10 0.0 0.5 l.1071 0.0 0.0 0.0 0.545
1V
Available Solution: 4 t f f f f f H
Field
Point
No. Reference
1 =7.4
oyy MPa [42)
1 0 =7.0 MPa [39] <::::> a;
yy (infinite plane)
5 Oxx= l.lSMPa [42]
v 1 11 1 1 1 v
137
Table 4.2 Rectangular plane contaihing an elliptical hole
(inclined major axis) at the origin, Case 1
Geometry: rectangular plane (10cm x 20cm x 1cm)
Load: w = 1.0 MPa
 Eccentricity: X0 0 Y0 = 0, Angle: 6 = 30°
E = 70000 MPa, u = 26315.79 MPa, v = 0.33, m = 0.5
Field Coordinates 0 U U
Point X Y xx yy xy x y
No. cm cm (MPa) (MPa) (MPa) microns microns
l 1.5 0.0 0.0 5.34611 0.0. 0.1923 0.2319
2 1.7 0.0 0.8876 1.8630 0.9115 0.1345 0.2241
3 1.9 0.0 0.7232 1.3653 0.8148 0.1120 0.2313
4 2.8 _0.0 0.4109 0.9319 0.6010 0.075 0.295
5 0.0 0.5 0.4309 0.0 0.0 0.2514 0.3946
6 0.0 0.8 0.0934 0.0103 0.2562 0.214 0.3905
7 0.0 1.1 0.1047 0.0784 0.4023 0.1983 0.3878
8 0.0 2.4 0.2976 0.4357 0.5113 0.2319 0.4190
9 1.5 0.0 0.0 5.3461 0.0 0.1923 0.2319
10 0.0 0.5 0.4309 0.0 0.0 0.2514 0.3905
W”
.{ 1
\111111“
f
//
*w
1111111
138
Table 4.3 Rectangular plane containing an elliptical hole
(inclined major axis) at the origin, Case 2
Geometry: rectangular plane (10cm x 90cm x 1cm)
Load: w = 1.0 MPa
 Eccentricity: X0 = 0 Y0 = 0, Angle: 0 = 60°
E = 70000 MPa, u = 26315.79 MPa, v = 0.33, m = 0.5
Field Coordinates 0 o 0 U U
Point X Y xx yy xy x y
No. cm cm (MPa) (MPa) (MPa) microns microns
1 1.5 0.0“ 0.0 1.0337 0.0 0.1539 0.2344
2 1.7 0.0 0.453 0.5588 0.9191 0.1903 0.2263
3 1.9 0.0 0.5660 0.4398 0.8190 0.2086 0.2334
4 2.8 '0.0 0.69189 0.30451 0.5902 0.2853 0.2951
5 0.0 0.5 0.97538 0.0 0.0 0.2564 0.0818
6 0.0 0.8 0.9229 0.0593 0.2649 0.2191 0.0653
7 0.0 1.1 0.8810 0.1063 0.4156 0.2031 0.05381
8 0.0 2.4 0.7922 0.2095 0.5315 0.2385 0.0321
9 1.5 0 0 0.0 1.0337 .0 0.1539 0.2344
10 0.0 0. 0.9753 0.0 0.0 0.2564 0.0818
AY’ .y
‘111111‘
.1
“O
w111111
139
Table 4.4 Rectangular plane containing a nonsymmetri
cally located elliptical hole (inclined major
axis)
Geometry: rectangular plane (9cm x 18cm x 1cm)
Load: w = 1.0 MPa
. Eccentricity: X0 = 1.5 Y0 = 2.0, Angle: = 30°
E = 70000 MPa, u = 26315.79 MPa, v = 0.33, m = 0.8
Field Coordinates 0 U U
Point X Y xx yy xy x y
No. cm cm (MPa) (MPa) (MPa) microns microns
l 1.8 0.0 1 0.0 17.1905 0.0 0.2434 0.3892
2 1.9 0.0 2.0736 2.9680 1.6635 0.6380 0.5744
3 2.1 0.0 1.1269 1.7636 1.1255 0.7245 0.6193
4 3.2 0.0 0.4972 0.9748 0.6480 0.8071 0.7640
5 .8 0.0 0.2913 1.2050 0.4385 0.8116 0.8020
0.0 0.2 0.5871 0.0 0.0 1.1892 0.7932
0.0 0.25 0.5355 0.0015 0.0332 1.1808 0.7930
8 0.0 0.3 0.4849 0.0027 0.0656 1.1728 0.7927
9 0.0 0.4 0.3873 0.0030 0.1277 1.1585 0.7920
10 0.0 0.7 0.1350 0.0151 0.2847 1.1261 0.7898
AY’
§\‘1HH‘1
1"” A
1 1111\11
140
EXAMPLE IV.2
A Circular Plane Weakened by an Elliptical Hole
Let a circular plane of radius R = 6cm and of unit
thickness (h = 1cm) be weakened by an elliptical hole at
the origin described by:
X = (1+m) Cos 0
Y = (lm) Sin 0
o2 in equation (5.5) or n>2 in equation (5.2), it is very
cumbersome to find the inverse Of the transformation func
tion even though some numerical technique could be employed,
but the following two methods appear to be the most power
ful methods for determining the inverse transformation
function of equation (5.2).
Method 1: Power Series Expansions for the
Inverse Transformation Function
The transformation function Of equation (5.2) is
written in the following form:
N
_ R(1+c) . 1 lc 2 26 Z _. n+1
Z — __2———' C { 1 + 1+c C + 1+c n=1 (dn lhn)C } (5'6)
and letting
W R(1+c)
Z 01ka } (5.7)
164
where
(11:0
_ 28 . lc
a2  1+c (dllhl) + 1+c
a = 26 (d ih ) for k=1 3 4 N
k 1+c kl kl ’ ’ ""
ak = 0 for k>N
Equation (5.7) is a special case of:
where
173—2126 (kV+Q)/P)1al(<
ll
R(Cos 0 + e Cos 20)
..<
I!
R(c Sin 0  6 Sin 20) (5.10)
where Oﬁwpmpuﬁnpm cm wcﬁcwmucoo ocmHm muﬂcwmcﬁ mo EoHnowm one
«my
a.m owsmﬁa
«S
.6ch N
x?
184
.omﬁw pmﬁsopwo was: m oucﬁ
coauowwu wowﬁamm ow pouoonn3m oHo: may mo anpoum may Meagan: m.m owzmﬂm
«m: «S
6:30. a ..U mcmBTN
U
185
let 9 = (Mm/£)/§. Then equation (5.48) can be written
as follows:
4:17? 9 93
or
w = 9(1 + (2'2 + m'“) (5.49)
where
and
>a
II
(In  r)/£2
Finally, equation (5.49) can be written in the following
form:
w = (2% + Z ako‘Zk} (5.50)
k=1
where a; = 1, a2 = A and ak = 0 for k>3.
The inverse power series of equation (5.50) can now
be obtained following equation (5.8) where, in this case,
P = q = l and V = 2. Thus,
(5.51)
:0
II
2
MIN
H
4.
M 8
m
w
2I
N
w
W
where
186
and the aﬁn)'s can be formulated as
n
aén) = ”In k “E n (kn)
Since ok = 0 for k>3 and a = 1, then
(n) = n kn
Gk (kn) A
so that
k
e = 1 2 (2H) (“MM1
k lZE n=l 11 kn (5.52)
which leads to:
81 = '1
82 = '1  A
83=‘2'4}\
s. = s  15x  3A2
14  56A  28).2
85
It is clear that the Bk's are not converging very rapidly,
a deficiency of this method. However, the Bk's are all
functions of powers of A, which indicates that there would
be some general closed form for equation (5.51):
9
W = 1  p,  m  92x2  93x3 ........ (5.53)
187
The coefficient pi's will now be found. To find Do, let
A = 0 in equation (5.53), then
% = 1  p, (5.54)
Also, when A = 0, the transformation function, equation
(5.49), becomes
92  WQ + 1 = 0
or
QiQ+i=0
w2 W w2
. Q
S01v1ng for W leads to:
Q = l i Vl4/W2 (5.55)
W 2
The minus sign is not valid since the limit of the solu
tion, equation (5.55), must approach unity to satisfy
conformality of the transformation function, i.e., infinity
is transformed to infinity. Equating the righthand sides
of equations (5.55) and (5.54) leads to:
l  Vl4/W2
00 = T (5'56)
Now let n = k in equation (5.52). Then
1 2k1 k 0 (ZkZ)!
Bk ‘ T77? ( 1c ) (o) A ' ' ETTFTITT‘
188
or
2p 1
BP+1 = ' (P) PTI
Substituting into the inverse transformation function,
equation (5.51) leads to:
2 2P wzPz
Comparing (5.57) and (5.54), it is found that
z 2P wzpz
Co = (p) W— (558)
Now let k  n = l in equation (5.52). Then
1 Zkl kl 1 Zkl k
Bk=—2‘k‘1_(k1)(1)>\+—2E1_ (k)(o)x°
Substituting Bk into the inverse transformation function,
equation (5.51), yields:
_2 w 2P'2 E 2k2 2k
(2 P)  A ( ) w
pO= P I I k=1 kZ
Comparing this equation with equation (5.53) leads to:
=1) = (I?) ' PTT'(I))
Substituting k = P + l and the binomial coefficient above
into equation (5.59) leads to:
2P 2P2 z ZP _ _
01=Z (p)w———+ (P)W2P2 (5.60)
Taking the derivative of po, equation (5.58), and con
structing % paw
2P
1 _ E 2P2
Substituting equations (5.61) and (5.58) into the right
hand side of equation (5.60) leads to the expression:
_ 1 .
01  ' Do ‘ 7 DOW (562)
Taking the derivative of pa in its closed form, equation
(5.56), and calculating the righthand side of equation
(5.62) leads to:
190
2
_ 0°
pl — 1_Zpo (5.63.61)
Similarly, one can find oz in terms of be, which will lead
to better convergence of equation (5.53). Thus, the
inverse transformation function becomes
2
n = w (1  p0  19390 A— ...) (5.63)
where
_ 1  ./14/w2
00" 2
This was an example of using the power series expansion to
find the inverse transformation function, even though the
method did not prove efficient. The second method, con
tinued fractions, will also give the coefficient of 12,
i.e., oz. The second method will now be applied. Rewrite
equation (5.49) in the following form:
n“  w n3 + oz + A = 0
Dividing the equation by W“ leads to
n . n 3 n 2 , 1 A =
(w) ‘ (w) + (w) F+T 0
or
_L
g
($3 c%1)+ (€342 53+ w = o (5.64)
In order to check the accuracy and efficiency of this
method with the previous method, let the following
till I11]!
191
assumption be made:
r = 1  1%, (5.65)
and
_ 1  Vl4/W2
o0  7 (5.66)
Constructing (po  pi) using equation (5.66) leads to:
Substituting these results into equation (5.64) leads to:
f(T) = I‘(1"‘1)3 + (Tl)2 (Do0%) + Mmofﬂz = 0 (567)
Equation (5.67) is a polynomial in F of the form of equa
tion (5.9), which can now be solved by the continued
fractions method.
Let the first estimate to the required root F of the
equation f(F) = 0 be 60. Then, after some simplification,
the function and the derivatives of the function at the
estimated root are:
f(Oo) = (pap312
f'(oo) = (co1)2 (2001)
f"(po) = 2(5003) (00'1)
f'"(Do) = 6(4 po3)
f""(po) = 24
192
f(n)(po) = 0 for n>4
Thus
2
n = _ f(Do) = 400
E'Zpoj ‘
f"(oo) _ Spa3
5 "' 77—72 ' Do ' (1po)(2po1)
.= f'" 00 = 400'3
T 77—13 " Do (Spa31Tpo1)
6 = fun(po) = 1
ail (Do) 2{Do3
Then d1 = +n, d2 = ﬁg and d3 = U(E'Y)
= _t(T6)
d“ U(g €_.Y
Substitution into the continued fractions leads to:
 d1
T  00 + + 2
01'
O
“M'W
InlYEi
For simplicity, consider the first numerator. Then expand
ing as a binomial series and just choosing the first two
terms, the equation leads to:
193
I‘=oonn2€+ .....
Substituting equation (5.65), n and a into the above equa
tion leads to
02 01(50 3)
$2 = W(l  p  ° 1  ° °' ~12 ..... ) (5.68)
° 1'2“ (1'Zoo)3(1'oo)
where
_ 1  Vl4/W2
pO‘ 2
The inverse transformation function has now been obtained
by the two methods. Comparison of the two equations (5.63)
and (5.68) shows that the first three terms of both equa
tions are exactly the same except that the second method,
continued fractions, provides one more term. It is also
clear that the second method was more efficient. The first
method, however, can be more useful in some special cases.
Returning to the solution of the problem of Figure
5.7(A) and equation (5.46), the complex potential functions
will now be obtained following the general procedure pre
sented in section 11.3.
Let ¢°(Z) and W°(Z) be the complex potential functions
for the problem of Figure 5.7(B) and ¢*(Z) and 1*(2) be
the complex potential functions for the problem of Figure
5.7(C). Hence, the potential functions for othe problem
of Figure 5.7(A) are
194
¢°(Z) + 4*(2)
¢(Z)
W(Z) W°(Z) + P*(Z) (5.69)
where ¢°(Z) and W°(Z) are known [27]. The transformed
complex potential functions of ¢*(Z) and P*(Z) can be
obtained in a manner similar to that presented in section
11.3. These are
1 5'1 1
£2) (10 ' mfi‘ 20_(£2_ (b1 0 (10 (5.70)
_ 1 F
¢T(C)  7;? Y o w'(0) OC
_ 1 FIG) 1 1 wio) ¢*'(o)
WT(C)  W Y 0'; (10 mi (”'(0) 0'; (10 (5.71)
where F(o) and ET?) are given by equations (3.20) and
(3.21). Equation (5.70) is an integral equation in which
the second integral can be evaluated. Let the transforma
tion function of equation (5.48) and its complex conjugate
be evaluated at c = 0:
(11(0) =§+ 26 + r03
1'
(1)10) mo+%+—
0.3
Then the derivative and conjugate derivatives are
w'(o) =  1L + 1 + 3ro2
2
o
m=m62+2+§£
0,2
g(o) = o(m+2oz+ro“)
w'ioi 3r+Rozmo“
2.3111211. ZakoZk
m m2 02 k=2
=  % O  Z 616'21‘” (5.72)
k=1
where the ak's are the coefficient of the expansion.
Similarly,
mo“+202+r
(3ro“+£ozm)
wloi =
(L) 10')
= X Bk6‘2k'1 (5.73)
k=0
where the Bk's are the coefficient of the expansion.
Note that ¢f(o) and of'(o) are analytic inside Y and
$¥TTET is analytic outside 7. Thus, following section
111.3:
196
¢*'(0) = z kakok'1 (5.74)
1 k=1
M'io) = kzl kit—kol<+1 (5,75)
Multiplying equation (5.72) by (5.75) leads to:
(0(0) r z — k+2 2 — Ok+1 Z on oZk+1
ETTTET =  a  kako  kak  k
w'io) k=1 k=1 k=1
=  % 510  %§ 52  2;; end.“
where en is a coefficient of a power series expressed in
00
n
terms of 5k and ak. Since the summation 2 e o is an
n
e=l
analytic function outside y, and the value of the summation
at infinity is zero, then, following section 1.3, the value
of the second integral in equation (5.70) is
_ 2 _
—  %alz;  71]?— a2 (5.76)
Note that the 51‘s are the complex conjugates of the coef
ficients of the expansion series of ¢f(c) which are unknown.
Since (2r/m)az is a constant term and does not have any
effect in obtaining ¢f(§), it will be omitted. Substituting
equation (5.76) into (5.70) leads to:
l F(o)
ZFT Y oc
r — 2r —
“—1.81; ' F32 (5.77)
do +
¢f(c) =
197
In order to obtain a closed form for ¢f(c), 51 has to be
evaluated. This can be easily determined. Since
n
1 1 1 C C
—=——=_+__+ + +
0.: 0(1%) G 02 omI
then
_L 1% ”0) do = Z c k (5.78)
where the ck's can be found by substituting the expansion
of l/(oc) into (5.78):
_ 1
CR  Zni
F o)
Sé. géero
The Ck's could also be obtained by taking the kth deriva
tive of equation (5.78) and setting C = 0. The function
¢§(;), however, is analytic inside 7 and has the series
form given by equation (2.13). Hence, substitution of
equations (5.78) and (2.13) into (5.77) leads to:
z 1.:
k=1
00
where
198
Finally, equating the real parts and the imaginary parts
leads to evaluation of 51:
(5.79)
51 = m{Re(c1) , . Im(c;)
mr m+r }
Substituting equation (5.79) into (5.77) leads to the
closed form expression for the transformed complex poten
tial function:
¢§(c) = %f :52) do + r{B;—E§f1—)—  i .Ilmn10;_1)} c (5.80)
In order to find Wf(c), the second integral of equation
(5.71) must first be calculated. Multiplying equation
(5.73) by equation (5.75)
w'ioi ¢*'(o) = mo"+SLoz+r . 2 ka Okl
w (o) 1 (3ro“+£ozm) k=1 k
It is clear that the numerator of the equation above is
analytic inside 7 and the denominator has five roots, one
of which is zero and four of which can be found by solving
the biquadratic
1,
3ro + £02  m = 0
Some of these roots are inside 7 and some outside. Thus,
the following integral can be evaluated by the Cauchy
integral theorem presented in section 1.3:
199
*1 lb 2
..L, ¢ 31%?) ¢0~ go) do = "“1 ”C +1" WE'VE) * 89?)
Y 4(3rc“+8czm)
where g(c) represents the sum of the residues of
(m0“+202+r) ¢f'(0)
0(3r0“+£02m)(0c)
at the roots which are inside y. Thus, the other trans
formed complex potential function, i.e., equation (5.71),
becomes:
E is 2
WT(C) = 2%T.9£.8 do  (m; +£§ +r)  ¢f'(c)  g(C)
C(3rc“+£C2m)
(5.81)
Recall that F(o) and ET?) were given by equations (5.25)
and (5.26). The transformed complex potential functions
02(0) and 92(0) can be easily found by substituting the
transformation function, equation (5.48), into equations
(2.6). The results are
_ 2 4
02(0) = _ Q 1n m ZOU+§U +r0 (5.82)
 _ 2 1+
1112(0) = Q ZOO + Q 1 a 1n "1 2004'th +1.0 (5.83)
m200+£02+r0“ O
For simplicity, let
200
A(0) m — 200 + £02 + r0“ (5.84)
r + £02  2003 + m0“ (5.85)
B(o)
Taking the derivative of equation (5.82), obtaining the
complex conjugate of equations (5.82) and (5.83), and sub
stituting into equations (5.25) and (5.26) leads to the
expressions for F(o) and FIE). The calculation is omitted
and the results are
F(O) = Q {1n Ago)  a 1n Biz) }+ Q{62 A(.§)} (5.86)
16 {322141133  . 1,, Am
where A(o) and B(o) are given by equations (5.84) and
(5.85).
The two remaining integrals in the potential func
tions of equations (5.80) and (5.81) can now be calculated
since F(o) and ET?) are known.
To evaluate these two integrals, note that the con
formality of the mapping function m(c) implies that the
equation A(o) = 0 has one root inside and three roots
outside Y (since the polynomial of Zw(§) = 0 and A(0) = 0
are identical).
Denote the inside root by ri and the three outside
roots by r01, r02, and r03. Root ri has a significant
role in the calculation of the two integrals whereas ro ,
1
r02, and r03 need not be calculated. Root ri can be
201
calculated by the inverse transformation function using
either of the two methods presented at the beginning of
this section.
Now A(o) may be rewritten as:
A(o) = r(0ri)(0r01)(0r02)(0r03)
Similarly, equation B(o) = 0 is identical to Zw(l/c) = 0,
one root of which is outside Y and three of which are
inside. Denote the outside root by to and the inside
roots by til’ tiz and tia' Thus B(0) may be rewritten as:
B(o) = m(otil)(0tiz)(0tia)(0to)
Using these forms of A(o) and B(o), the two integrations
may be computed quite simply in a manner similar to that
presented in the previous section. The results are:
7%; §§%l do Q{111r(cr01)(cr02)(cr03)aIh1m(cto))1
'Y
3
(0t. )t? A(t. )
+ c2A(c) 1k 1k 1k
8 {—BTEI— + k=1 B(tik)(tik;) } (5'88)
1 F10) _ d. B(o) + B C
2N1 0c d0  Q‘130'A(0)(0;) 2C )
7 0:0 6 A(c)
B(ri)(0ri)
r;A(r1)(r1;)
+
}+ o{1nm(cto)
a 1nr(crol)(cr02)(cr03)} (5.89)
202
Substituting equations (5.88) and (5.89) into equations
(5.80) and (5.81) leads to the complex transformation
functions ¢f(;) and wf(;). The procedure for obtaining
the influence functions from ¢f(§) and Wf(§) is very
straightforward. This is done by substitution of these
functions into equation (5.69) and subsequent substitution
of this result into equation (3.6).
v.5 ON THE INFLUENCE FUNCTION OF A MORE
GENERAL CLASS OF OPENING
In the previous two sections, two cases of opening
were considered and the influence functions were found.
In this section a general form of this special kind of
opening is discussed. .
Consider an infinite p1ane bounded by the contour
given by equations
><
II
R( Cos 0 + e Cos N 0)
..<
ll
R(c Sin 0  8 Sin N 0)
where o(0 0 \D\XPO ~90
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END
221
D. COMPUTER PROGRAM FOR A CIRCULAR PLANE CONTAINING
A CIRCULAR HOLE (follows)
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END
APPENDIX C
THE POTENTIAL FUNCTIONS AND THE INFLUENCE
FUNCTIONS FOR AN INFINITE PLANE REGION
CONTAINING AN ELLIPTICAL HOLE
The following complex functions were used in the
influence functions for an elliptical hole, equations
(4.52).
* roC tog
¢Ic:)1n(ro) aln(to)
2
¢* ( ) m;2z; +1 (mti‘z°ti*1)c 1
c = _ _ . _
*, = 1 _ a
2 z +1
¢*'(c) = PD  mti Oti
II
(c270;+m)2 (totimtiv
¢f"(c) = '1 + °‘
(«zr0)2 (cto)2
226
227
[2m(c2foc+m)2(mc2Zoc+1)](CZZoc+m)2(2cZo)[PD]
¢*"(c) =
II (CZZoé+m)3
2
+ 2 (mti'ZOti+1)
(totiMcti)3
2
_ (r.Z r.+m)C 2
wf(c) = 52 7°C*m + 1 ° 1 Elé—iﬂl ¢f‘(c)  m
+
mczZoc+1 rim(ro'ri)(‘3'rij 1mr,2
2
VEICC) ‘ 1n ( 2 )  a 1n ( 0 ) + C(C +m) ¢*1(C)
O o 1mC2
rITorwm 2
V*'(C) = 'PD  1 1 + g(c +m) . ¢*"(c)
I (mCZ‘Z°C+1)2 m(roriMcri)2 1m;2 I
+ (352+m)(1mc2)+2mc2(cz+m) ¢f'(c)
(lmcz)2
' " 1 __(1 C(CZ'HII n
W* (C)  __  _ + _______1 ¢* (C)
II C to C re 1'mC2 II
+ L3gz+m)(1mc2)+2mc2(c2+m)
¢*'(c)
(1mc2)2 II
PD = (ZmC'Zo)(C2'ZoC+m) ‘ (ZC‘Zo)(mC2‘ZoC+1)
Substituting the components of the resultant fictitious
traction, equations (4.59), into the influence functions
for an elliptical hole, equations (4.52),
228
xxx = Re 2E2— + lg: ¢§'(C) ' 7 —1—'— + Cu ¢*"(C)
’ ° mc21 (Z'Zo)2 (mczl)“ ‘
 253 ¢I'(C)] + £9—  0‘
(mC2'1)3 (2'7032 Z'70
2
 (m:21) WI'(C) /2"(“+1)
xx.y = Re i zlé— + 2C2 (¢f'(c)  ¢*'(c))  7' ———l———
, mC2_l II (Z'ZO)2
2:3
+ (mcij1)“ (¢f”(c)  ¢fgcc)) m;21)3 (¢f‘(c)
 ¢fi(c)) ] + EE%§:3—+ EIEZ  EEEETIS (v? (C)
 W¥i(q))‘) /2w(a+1)
Hyy;x = Re 73%; + m:§f1 ¢§'(c) + 2'[ng%:33+ (m::1)“ ¢§"(c)
_ (m::j1)3 ¢§'(c)]  13%;??? + 2p;o
+ C 2 v '
m;21 w; (c) l/(2w(a+1))
229
HWY = Re 1(2320 + 2‘32 (¢’f’(c)  ¢ﬁ(c))+ 4 1
’ mCZl (ZZo)2
+ (mcff1)“ (¢f"(c)  ¢fg(c)) (m:fj1)3 (¢f‘(c)
_ 11(0)]  (2302  2:170 + ME; (wf'm
 mun) /2n(a+1)
ny;x = Im 2‘ [(zlzo)2 + (mg1)“ ¢’I‘"(c)  (m::jl)3 ¢’f'(c)]
 (2;?)2 + zjzo + "IS1 wi'cc) /(2n(a+1))
nyw = Im 1(7[(Z:o)2 + (mail.1)“ ( I /2n(a+1)
X§Y
YSX
YSY
235
_ —2
R€<{‘<11n(z‘zo) + a ¢f’(c)  Z [ 1' + _C ¢§’(c5]
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Zo
ZZo
’ a ln(Z‘Zo) ‘ W¥icj‘}//¢ynp(a+u
Re{i ( on ln(ZZo) + a 013?“) ‘ ¢fi(§)) ' Z [i;‘
ZZo
32 (W  ¢ﬁ(c))1  2° + a INT2°)  ”1*“?
czl
WIIECJ) ) } /4Uﬂ(a+l)
Im{ a ln(ZZo) + a ¢§'(c)  z ['—1—+ 32 m :1
270 321
_ZZ—z—  a 1n(7zo)  Wilt) 5/4"P‘4*"
 0
Im{i(a ln(ZZo) + a (cb’f'UE)  mm)  2 [Eli
+ a 1n(7Zo)
32 (¢*'(c)  ¢*'(c))]  2°
32.1 I II 720
(wflzgi  *IIICI))}/47TUCG+1)
APPENDIX E
COMPUTER PROGRAM FOR PLANE, FINITE REGION
CONTAINING AN ELLIPTICAL HOLE
OR A SHARP CRACK
A computer program was employed for the numerical
computation of the stresses and the displacements at the
field points of a twodimensional region containing an
elliptical hole or a sharp crack. A listing of that
program for the rectangular region subjected to uniaxial
tension (w = 1.0 MPa) and containing an elliptical hole
or a sharp crack and for the circular p1ane subjected to
a uniformly radially tension (w s 1.0 MPa) and containing
an elliptical hole are presented in this appendix.
A. INPUT DATA
The following information must be provided as input
(this is the order of appearance in the program).
PR  Poisson's ratio of the material
BMUD  modulus of elasticity of the material
NML  total number of subdivisions on the boundary
NFP  the number of field points at which the stresses
and displacements are to be computed
M  parameter used to describe semimajor axis
(a
elliptical hole
1+M) and semi—minor axis (b = lM) of
236
THETA
XO’YO "
WR 
X(I).Y(I)
I=1,NML
IPLANE
BVX(I): '
BV (I+NML)
I= ,NML
XF(I)9 '
YF(I)
I=1,NFP
237
angle of inclination of the elliptical hole or
the sharp crack with respect to the xaxis
location of the center of the rectangular or
circular plane with respect to the center of
the ellipse or crack
"width ratio" for the rectangular p1ane
(WR width/10 cm) and for the circular plane
(WR radius/6.0 cm)
coordinates of the outer boundary points,
specified counterclockwise (for circular p1ane
case, this input has been included in the
program)
this character specifies the type of the
plane problem, i.e., for plane stress, IPLANE=1
and for plane strain, IPLANE=2
x and y components of boundary tractions speci
fied at each subdivision
coordinates of the field points at which the
stresses and displacements are to be computed
B. OUTPUT DATA
The following information is obtained as output (this
is in the order of the appearance in the output).
PR,BMUD 
SHMUD 
NML,NFP 
Poisson's ratio and modulus of elasticity
shear modulus of elasticity
total number of subdivisions on the boundary
and number of field points at which the stresses
and displacements are computed
THETA
Xo,Yo
WR
X(I),Y(I)
I=1,NML
BV (I) 
BVx(I+NML)
I=I,NML
PSX(I) 
PSY(I)
I=1,NML
XF(I)2 '
YF(I)
I=1,NFP
SIGMAXX, 
SIGMAYY,
SIGMAXY,
UX,UY
238
angle of inclination of the elliptical hole
or the sharp crack with respect to the xaxis
location of the center of the rectangular or
circular p1ane
magnification of the size of the plane for the
rectangular plane (WR = width/10 cm) and for
the circular plane (WR = radius/6.0 cm)
coordinates of the outer boundary points, listed
counterclockwise
x and y components of boundary tractions speci
fied at each subdivision
components of the fictitious traction on the
boundary, represented by the concentrated loads
Pii’ P;i at the center of each one of the sub
divisions. These are computed by solving a
system of linear equations (4.61) (LEQTlF,
computer library).
location of the field point at which the
stresses and displacements are computed
components of stress and displacement at each
of the field points. These are computed using
equation (4.65).
C. THE COMPUTER PROGRAM FOR A RECTANGULAR PLANE
CONTAINING AN ELLIPTICAL HOLE OR A SHARP
CRACK (follows)

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