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
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B
r
u
ABzrBBzrAAAA
B
z
zzzz
z
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z
z
zz
z
z
zz
zz
r
zzzr
=
=
=
−
−
−
+
−
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−=
−
+
−
+
−
+
−
−
−=
+++=
−
−
−
−
−
=
−
−
−
−
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−
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−
−
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−
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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
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y
y
A
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x
B
x
A
y
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A
xA
x
A
yA
x
A
x
x
B
A
yAxA
B
x
Au
yxBByxAyxA
B
−
−=+
−=
−
=
−
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−
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+
+
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−
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−
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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
=
++
−
−
=
−
−
−
=
−
−
−
−
=
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−
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==
u
R
z
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P
R
r
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P
R
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R
P
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z
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R
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r
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r
u
R
P
AB
z
z
r
z
13.18.
 
 
+−+−
−
−=
−=
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→
−−
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→→
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
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D
rrr
d
rr
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13.19.
directions allfor validbe thenshould results These
0
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1
1
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1
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ionsuperposit Applying
0,
)1(8
)1(2
sexpression axis– previous thein
2
by replace axis,– thealong actinga Doublet For
cossin
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1
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cos
)1(8
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sexpression previous thein
2
by replace axis,– thealong actinga Doublet For
cossin
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1
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sin
)1(8
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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
=++=
−
−
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−
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−
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−
+
−=++=
=
−
+
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=
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−
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−
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z
R
y
R
x
RR
z
R
y
R
x
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y
R
y
R
x
R
x
R
z
R
z
R
R
D
R
D
σ
R
D
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D
R
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x
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D
R
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13.20.
=
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=
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===
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
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R
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u
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u
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u
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u
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u
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C
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mmmmmmmm
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u
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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
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2
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2
0
2
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:) 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
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a
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13.22.
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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
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ppp
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K
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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
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R
R
R
R
R
+
=
+
−
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+
−
=
+
−
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=
====
+
−
−==
+
−
+=
+
−
−=−==+=
+
−
===
=
+
−
=
+==−=+=
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
=
−++
−−=
=
=+
−+
−−=
=
−+
−+−=
+==
=
==
=
==
=
=
−=−=−=