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13.15.
nintegratio of functionarbitrary thedropped have wewhere,2 gIntegratin
2
1
0
)1(2
1
21)21)(1(4
0)0,(
)1(421)1(421
2)(
12-13 ExerciseFrom
)1(4)1(4
43
)1(4)1(4)1(4
2
)1(4)1(4
2
0and0with,),(,),(,0
:s Potentialq Boussines- FormicAxisymmetr
)1(4
2 :ation Represent PapkovichGeneral
2
2
2
2
2
2
2
22
z
B
A
z
B
z
A
z
A
z
B
z
A
z
A
z
A
r
zA
B
zz
A
z
AzA
eeee
z
A
z
z
B
A
z
A
z
A
z
B
A
zA
B
z
Au
r
A
z
r
B
zA
B
r
u
ABzrBBzrAAAA
B
z
zzzz
z
zzzz
zzrz
z
z
zz
z
z
zz
zz
r
zzzr
=
=
=
−
−
−
+
−
+
−−
−=
−
+
−
+
−
+
−
−
−=
+++=
−
−
−
−
−
=
−
−
−
−
−=
−
+
−=
−
−
−=
−
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−=
======
−
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RA
Au
13.16.
ψ(z)(z)γzγ(z)ivu
y
B
i
x
B
ψ(z)
iAA
γ(z)
y
B
i
x
B
y
A
y
y
A
xi
x
A
y
x
A
xiAAivu
y
B
y
A
y
y
A
xAv
x
B
x
A
y
x
A
xA
x
A
yA
x
A
x
x
B
A
yAxA
B
x
Au
yxBByxAyxA
B
−
−=+
−=
−
=
−
+
=
+
−
+
+
+
−+−
−
=+
−
−
−−
−
=
−
−
−−
−
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++
−
−
−=
−
+
+
−=
=+=
−
+−=
)(2
43 with and
)1(4
Define
))(43(
)1(4
1
)(2
ntdisplacemecomplex theForming
)43(
)1(4
1
2
Likewise
)43(
)1(4
1
)1(4
1
)1(4
)(
2
),( and ),(),( where
)1(4
2 :ation Represent PapkovichlDimensiona-Two
21
2121
21
21
2
21
1
2
1
1
1
21
1
21 21 eeA
RA
Au
13.17.
Problem Kelvinfor the (13.3.10) relation in given that matches fieldnt displaceme This
02
1)21(2
)1(8
)1(4
)1(8
)1(82)1(82)1(4
)1(8)1(8)1(4
2
12-13 ExerciseFrom
2
,0 : Problemfor Kelvin Functions PapkovichTrial
1
3
2
3
2
2
3
=
++
−
−
=
−
−
−
=
−
−
−
−
=
−
−=
−
−=
==
u
R
z
RRν
P
R
r
Rν
P
R
rP
R
P
R
Pz
zR
P
zA
z
Rν
zrP
R
Pz
r
zA
r
u
R
P
AB
z
z
r
z
13.18.
+−+−
−
−=
−=
−+
−=+−=
→
−−
−−
→→
components stress nonzeroother for the Likewise
)(3)()21(
)1(8
),(),(
lim),(),(lim
0limit takeand ProblemsKelvin twoofion superposit use Problem, ForceDoublet For the
2/52232/322
2/52222/322
00
D
zrzrzrz
z
D
z
d
d
zrdzr
ddzrzr
d
D
rrr
d
rr
d
D
r
13.19.
directions allfor validbe thenshould results These
0
)1(2
)21(
1
1
)2(2
1
)1(8
)1(2
ionsuperposit Applying
0,
)1(8
)1(2
sexpression axis- previous thein
2
by replace axis,- thealong actinga Doublet For
cossin
)1(8
)1(2
,sin
1
)2(2
cos
)1(8
)1(2
sexpression previous thein
2
by replace axis,- thealong actinga Doublet For
cossin
)1(8
)1(2
,cos
1
)2(2
sin
)1(8
)1(2
as 18-13 Exercisein given wassolution theaxis,- thealong actinga Doublet For
principle. ionsuperposit apply the toshown as plane-, thein direction speciala consider we
state, stress symmetricy sphericalla produce willn DilatatioofCenter that theexpected isit Since
33
3
3
22
3
3
22
3
=++=
−
−
−=
−
+
−
+−
−
+
−=++=
=
−
+
=
−
+
=
+
−
+−
−
+
−=
−
−
+
−=
+
−
+−
−
+
−=
z
R
y
R
x
RR
z
R
y
R
x
RR
y
R
y
R
x
R
x
R
z
R
z
R
R
D
R
D
σ
R
D
zy
R
D
R
D
x
R
D
R
D
z
zx
x
13.20.
=
+
−
=
+==−=
++=+
+==
++=+
+=
+=
−==+−=−+=−+−
=
=−+=−
===
2
,
)1(2
21
that Note
,
2
:as written be tostresses thealloweswhich
)(222
2)2(22
relations stress thegives (A.9) into (A.2) relations Using
2,1 roots0)2)(1(02022)1(
equ. governing into substitue and :form theof solutionsfor Look
0
22
0
2
toreduces (A.12)relation and ,0),( :case symmetricy sphericall For the
2
3
2
1
3
2
1
2
2
1
2
22
2
2
2
1
K
CKC
R
K
K
R
K
K
dR
du
R
u
R
u
R
u
dR
du
R
u
dR
du
dR
du
R
u
dR
du
R
C
RCu
mmmmmmmm
ARu
u
R
dR
du
R
dR
ud
R
u
u
uuRuu
R
R
m
R
13.21.
ion.concentrat stress thereduceor relieve todimension additionalan
hasdomain ldimensiona- threea since expected be tois which So
2 result was ldimensiona- twoingcorrespond the8.4.2,section From
3
Factorion Concentrat Stress
3
)()(
2
2/)()()(
2
0
2
0)(
:) asen cavity tak of (raduis problemon ConditionsBoundary
,
2
20-13 Exercise From
32
2
max
3
3
3
21
3
2
1
3
2
1
3
2
1
3
2
1
DD
D
R
R
R
KK
K
KSaa
R
R
SaKSKS
a
K
K
a
K
Ka
a
R
K
K
R
K
K
−−
−
=
=====
=====
==−=
+==−=
13.22.
+
−
==
−
−
=
==
−
−
+
−
−
==
−
−
−
=
−
−
=
−
−
=
−=−=
+==−=
3
3
2
3
1
3
2
3
1
3
3
2
3
1
3
2
3
1
21
33
1
3
2
3
2
3
121
3
1
3
2
3
22
3
11
33
1
3
2
3
3
3
1
3
2
3
3
3
1
3
2
3
2
3
121
2
3
1
3
2
3
22
3
11
1
2211
3
2
1
3
2
1
2
1,1
0, :case special For the
1
)(2
)(
)(
become stresses theand
)(2
)(
,
)(,)(
:problem shell sphericalon ConditionsBoundary
,
2
20-13 Exercise From
R
R
RR
pR
R
R
RR
pR
ppp
RRR
RRpp
RR
RpRp
RRR
RR
RR
RRpp
K
RR
RpRp
K
pRpR
R
K
K
R
K
K
R
R
RR
R
13.23*.
:stressesboundary of Plots MATLAB
1
3
1
21
1,
1
)1(3
1
21
21
:inclusion theofboundary On the
:2/1 with case For the
1
21
1,
1
21
21
1
21
00)(
)1(2
21
)(
:ConditionsBoundary ProblemInclusion
2
,
)1(2
21
with
,
2
,
:20-13 Exercise From
3
3
3
3
3
2
3
12
2
2
1
11
2
211
3
2
1
3
2
1
2
2
1
SSSS
aR
S
R
a
S
R
a
S
SaKaCC
a
C
aCau
SCSKS
K
CKC
R
K
K
R
K
K
R
C
RCu
R
R
R
R
R
+
=
+
−
−==
+
−
=
+
−
+=
=
====
+
−
−==
+
−
+=
+
−
−=−==+=
+
−
===
=
+
−
=
+==−=+=
2.5
3
R/S
13.24.
21.-13 Exercise of results with matcheswhich
2
3
)57(2
1521
)57(2
315
)57(2
153
)57(2
1527
)2/,()0,()0,(
:directions and , along Tension (c)
loading field-far shear pure a toescorrespond alsowhich
57
15
)57(2
153
)57(2
1527
)0,()0,(
:directions n compressio and along Tension (b)
)57(2
3024
)57(2
153
)57(2
1527
)0,()0,(
:directions and along Tension (a)
)57(2
315
)2/,(
)57(2
1527
)0,(
)57(2
153
)0,(
:direction along ion with tensproblemcavity spherical for the relations stressGiven
max
max
max
SSSSS
aaza
zyxS
SSS
aza
xzS
SSS
aza
zxS
Sa
Sza
Sa
zS
z
z
z
z
=
−
−
=
−
−
+
−
+
−
−
−
=
=+=+==
−
=
−
+
+
−
−
=
=−==
−
−
=
−
+
−
−
−
=
=+==
−
−
==
−
−
==
−
+
−==
x
y
z
13.25*.
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
1
1.5
2
2.5
3
3.5
Poisson's Ratio
Stress Concentration Factor, max/S
(a)
(b)
(c)
13.26*.
-1 -0.5 0 0.5
1.75
1.8
1.85
1.9
1.95
2
2.05
2.1
2.15
2.2
2.25
Poisson's Ratio
Stress Concentration Factor
0.42
Case (b) has highest stress
13.27.
( )
( )
),(function stressAiry theof form usual by thegiven are stresses
0
2
1
0
2
1
00
2
1
2
1
2
1
2
1
2
1
2
1
)(2
2
2
),(,
2
,
2
1
,
2
1
ith function w stress Morera theUse
12,
2,
22,
1,
12,2332,1222,31
31
21,
1,
11,
2,
31,1221,1311,23
23
12,12,21,
23,
1,
13,
2,
23,1313,2333,12
12
2211
2
12,12
33
11,
31,
1,
31,31
22
22,
23,
2,
23,23
11
2
12,122,231,13
yx
zz + + -
zz + + -
-
z -
z -
yxzz
=
−++
−−=
=
=+
−+
−−=
=
−+
−+−=
+==
=
==
=
==
=
=
−=−=−=
