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10.14.
1
2
2
)()()()(2
Boundary Plugon ConditionsCheck
0,2,: As
2
3
4
1
2
1
2
3
2
1
1
2
2][22
2cos
2
1
11
1
22
2
2
][2
0 with and hole on the forceresultant no
with(10.4.8) and (10.4.7) relations match with they since eappropriat are formsgiven The
3
2
1
1
2
)(
2
1
2
)(
)(
2
1
4
)(
2
4
)(
4
2
22
____________
4
4
2
23
2
4
4
2
23
2
222
2
2
2
2
2
4
4
2
23
4
2
3
2
2
22
−
−
−=+
===−=+−=+→
−
−
++
−=
−
−
+−
−=
=+−
+=
+
+=
+
+=
=+
===
−
−
+−=
+
−
−−=
−=
+=
−=
=
=
aez
aez
xyyxxyxyyx
xyxy
yx
xyyx
a
T
aT
a
T
zzzzivu
σTσTiσσTσσr
z
a
a
zz
za
T
z
a
a
z
T
z
Ta
z(z) + (z)"z i
r
a
T
zz
a
T
z
a
z
aT
(z) + (z)
σT,σ
z
a
a
z
T
z
z
a
a
z
z
T
z
z
Ta
z
z
aT
z
z
a
z
T
z
i
i
10.15.
d
z
h
i
azaz
e
zb(z)zaγ(z)
d
z
h
i
d
zi
d
zi
d
zi
e
C
zi
eezhezzzz
ezzzzeivuiuu
i
C
i
n
n
n
n
CCCC
i
ii
r
z
i
ii
r
)(
2
1
)0(2)(
2
relation boundary in ,:potentialscomplex for form series Using
)(
2
1)(
2
1)(
2
1)(
2
1
2
contour boundary thearound gintegratin and
)(2
1
by gMultiplyin
1
,with ,)()()()(
2
1
:Con condition Boundary
)()()(
2
1
)(
21
____________
1
____________
____________
−+
−
=−−−
==
−
=
−
−
−
−
−
−
=====
−
−
−
−
=+=+
−
−
−
=
=
−
−−
10.16.
identical are 10 and 8 Chapters from results the
1
)()(
2
2, that noting and stresses theseAdding
1
))((
)(
))((
)(2
1
)(
)(2
axis- thealong loadingfor (8.4.69) From
1
)()(
2
2
1
)(
1
)(
12
(10.5.18), From
22222
222
3
222
3
222
2
222
2
22222
22222
−
++
+
+
+−
−
−=+
==
−
++
+
+
−+
−
−=
−
+
+
−
−=
−
++
+
+
+−
−
−=
+
−
+
−
−=+=+
ayax
xa
yax
xaP
aDaR
DxRy
xR
xRy
xRP
yxR
yxRP
x
ayax
xa
yax
xaF
aazaz
Fa
yx
y
x
yxr
10.17.
−
−
−
+
=
−−
+
−−−
+
−
+
−=
+
−
+
+
=
+
−+
=−+
−=
+−
+
=−−
+
=
+−
+
=−
−
+
+−+−
+
=
++
+
+
−
+
=+−
+
−=
−
+
++
−=
−
+
+
+
−=+
2sin2cos
22
2sin3sin4
)1(2
sin
2cos3sin4
)1(2
sin
)1(
cos
2sin2cos
22
)1(2
)5(4
,)1(4
)3(4
)1(2
)3(
)1(2
)3(
)1(
)(
)1(
)3()3(
)1(
)(
1
)1(
)(
)1(
2
)1(
2
)()()1(
1
)1(
1
(10.6.2)solution general From
22
22
4
22
4
22
4
222
4
222
4
2
2323
4
2
2
r
P
r
P
r
P
r
rx
rx
ry
r
Py
rxy
r
Py
ryx
r
Px
iyx
r
P
yxyixyx
r
P
iyx
P
iyx
iyxP
i
r
Px
iyx
P
iyx
P
z
iYX
z
iYX
xy
yxyx
xy
yxyx
r
xy
yx
xy
xy
xyxy
yx
10.18.
(8.4.9)solution previous with matcheswhich ,0,1,1
stresses for the Solving
22)]()([22
2
22
2])()([2
)(,
2
)(
)4(0,0,0,,)2(0,0
(10.6.12) relations 6,-10 Examplein work previous From
)(,
2
)(
yields (10.4.7) formsolution general the,0, stresses field-farWith
2
2
2
2
2
2
2
2
2
2
2
1
2
32
2
11
11
=
+=
−=
==
=+−
=
+=
=+
−
==
====−===
=+=
===
=
−
=
−
rr
ii
rr
r
nn
n
n
n
n
n
n
xyyx
r
a
S
r
a
S
r
Sa
e
z
Sa
ez + z"z i
S
SS
z + z
z
Sa
zz
S
z
nbaabbSabnaa
zbzzaz
S
z
S
10.19.
( )
0 :infinityat Point
plane- inboundary elliptical1
)1()1(
that Note
sin)1( andcos)1( where,
]sin)1(cos)1[(
:plane- inboundary On
1
)(:Case Elliptical
0 :infinityat Point
elyappropriat match boundaries
:plane- inboundary On
)(:CaseCircular
22
2
22
2
→→
=
−
+
+
−=+=+=
−++=+=
=
+
==
→→
==
==
−
−
z
z
mR
y
mR
x
mRymRxiyx
mimRmeeRz
e
mRwz
z
eRze
R
wz
ii
i
ii
10.20.*
: PlotMATLAB
1
3
)1(
)1)(3(
12
32
)
2
(
12cos,
2
case For the
value)minimuma toescorrespond0 that (note,,
2
,002sin
1 withcase general gconsiderin,0)1)(1(2sin4
0)2cos212(2sin4)12cos2(2sin4
0set valuemaximum find To
12cos2
2cos212
)( :(10.7.14) equation From
22
2
2
22
2
2
+
−
−=
+
+−
−=
++
−+
=
−=
=
=
==
=−+
=−−+−+−
=
+−
−−+
=
m
m
S
m
mm
S
mm
mm
S
mmm
mmmmm
d
d
mm
mm
S
Angle, (degrees)
/S
10.21.
b
S
m
S
m
Sab
b
a
S
m
m
S
mm
m
Sba
mm
SSSS
mm
mm
mm
mm
-mmm
m
ba
ab
m
mm
mm
S
ba
ba
m
mm
mm
S
yx
x
x
2
)1(2
)1(2
2/at found is stressmax the,For
2
1
)1(2
12
)1(2
0at found is stressmax the,For
12cos2
)(, case For the
12cos2
2cos212
12cos2
2cos212
)(ion superposit Using
where;
1)2/(2cos2
)2/(2cos212
)(
where;
12cos2
2cos212
)(
2
2
max
2
2
2
21
12
2
2
2
2
22
)2(
1
1
2
1
2
11
)1(
=
−
=
−
==
=
−
+
=
+−
−
==
+−
===
+−
−++−
+−
−−+
==
−=
+
−
=
+−−
−−−+
=
+
−
=
+−
−−+
=
yy S=
+
(1) Horizontal Loading
(2) Vertical Loading
x
y
xx S=
a
b
x
y
a
b
10.22.*
: PlotMATLAB
12cos2
2sin4
0 boundary, hole On
12cos2
2sin2
2
)(
)(
2
2
)(
)(
)(
2
)(
)(
)(
,0 :stress field-far withproblem hole Elliptical
2
222222
22
22
_______
______
11
1
2
+−
−==
+−
−
=
++−
−
=
−
−
+
−
=
+
=+
=
−
=
===
mm
S
mm
S
mm
iS
m
iS
m
iS
w
wm
iS
dz
d
S
xyyx
/S (m=0)
/S (m=0.5)
/S (m=1)
10.23.
1
)(
)(
2
2
cos
2
2
2
cos
22
cos
2
cos,,2 :crack tipNear
1
2
cos
2
11
:8-10 ExampleFrom
2
2/)(3
2/3
2
)
2
()
2
(
2/)(2/)(
+
−
=+−
+
−
+
=
+
−
−
+
−=
−
+=
−+=+
+−
−
+
−−
+
−
+−
−
+
a
e
rr
eer
Sai
ar
Sa
arar
rr
r
S
ee
rr
r
ReS
err
er
err
er
ReS
i
ii
xyxy
yx
ii
i
i
i
i
yx
10.24.*
Using MATLAB y-contours of Equation (10.8.7) are shown
10.25.
()
()
−+
−
−
+−−=
−
−−
−
−
−−−=
−
−=+
:crack tip Near the
224
2
1
2
44
)()()()(2
22
2
22
22
2
22
22
____________
z
z
az
zzaS
zazz
S
az
a
z
S
az
zS
zzazz
S
zzzzivu
Isotropic Crack Tip y/K Contours
-1
-0.8
-0.6
-0.4
0.6
0.8
10.26.
0,
2
0
2
,0
2
0
2
2/ with,,
:fieldFar
checks,0
2
cos
)(
)0sin(
0
22
3
sin
)(
)0sin(
22
cos
:surfacescrack On
:conditionsboundary Check
)(3
22
)(3
sin
)(
sin
22
cos
)(Im)(Re
22
)(3
sin
)(
sin
22
cos
)(Im)(Re
,,
:figure) (see field stress explore toscoordinatepolar Use
2
,
2
)( : FunctionStress
2121
2/3
21
2/3
21
2
21
21
2
21
2/3
21
2
21
21
21
2/3
21
2
21
21
21
22
21
→=++−→=−−−→
===→
=
=
=+
+−
=
+
+
+
+−
+
−=
+
+=
−
+
−−
+
−=
−
−=
=+=−=
=−
−
=
xyyx
xy
y
y
x
ii
i
S
SS
S
SS
S
rrr
rr
r
S
rr
Sa
r
S
rr
Sr
Sa
S
rr
Sa
r
S
rr
Sr
AzZyzZ
S
rr
Sa
r
S
rr
Sr
AzZyzZ
erazerazrez
S
A
S
az
Sz
zZ
r
r2
a
a
r1
x
y
•
2
1
10.27.
( ) ( )
( )
( ) ( )
BzZyzZ
zZy
zZyzZ
zzZ
BzReyzIm(z) + (z)"z Im
zImyzzzzRe
zImyzRezzzzRe
zzziBz
iBzzzzzzzzRe
yx
xy
y
x
xy
y
x
y
y
−
−−=
=
−=
=
−
−
−=
=
=
+
+
=
−
=
−
−
=
−
−−=
−=
+
+
=
+
+
=
=
)(Re)(Im
)(Im
)(Im)(Re2
)(2)( function dWestergaar theDefining
)(2)(2][
)(2)]()()(2[
)(2)(4)]()()(2[
)()(2)(
)()()(20)]()()(2[
vanishes stress normal the,)0( axis- thealong problem, symmetric-skew For the
