Circular hole in a shear field
edit
Elastic plate with circular hole under shear
Given:
Large plate in pure shear.
Stress state perturbed by a small hole. The BCs are
at
r
=
a
{\displaystyle r=a}
(103)
t
r
=
t
θ
=
0
;
n
^
=
−
e
^
r
⇒
σ
r
r
=
σ
r
θ
=
0
{\displaystyle {\text{(103)}}\qquad t_{r}=t_{\theta }=0~;~~{\widehat {\mathbf {n} }}=-{\widehat {\mathbf {e} }}~r\Rightarrow \sigma _{rr}=\sigma _{r\theta }=0}
at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikiversity.org/v1/":): {\displaystyle r \rightarrow \infty }
(104)
σ
12
→
S
;
σ
11
→
0
;
σ
22
→
0
{\displaystyle {\text{(104)}}\qquad \sigma _{12}\rightarrow S~;~~\sigma _{11}\rightarrow 0~;~~\sigma _{22}\rightarrow 0}
We will solve this problem by superposing a perturbation due to the hole on the unperturbed solution. The effect of the perturbation will decrease with increasing distance from the hole, i.e. the effect will be proportional to
r
−
n
{\displaystyle r^{-n}\,}
Unperturbed Solution
edit
(105)
σ
11
=
σ
22
=
0
;
σ
12
=
S
{\displaystyle {\text{(105)}}\qquad \sigma _{11}=\sigma _{22}=0~;~~\sigma _{12}=S}
Therefore,
(106)
σ
12
=
−
φ
,
12
=
S
{\displaystyle {\text{(106)}}\qquad \sigma _{12}=-\varphi _{,12}=S}
Integrating,
(107)
φ
,
1
=
−
S
x
2
+
f
(
x
1
)
⇒
φ
=
−
S
x
1
x
2
+
∫
f
(
x
1
)
d
x
1
{\displaystyle {\text{(107)}}\qquad \varphi _{,1}=-Sx_{2}+f(x_{1})\Rightarrow \varphi =-Sx_{1}x_{2}+\int f(x_{1})dx_{1}}
Since
φ
{\displaystyle \varphi }
(108)
φ
=
−
S
x
1
x
2
=
−
S
(
r
cos
θ
)
(
r
sin
θ
)
=
−
S
r
2
2
sin
(
2
θ
)
{\displaystyle {\text{(108)}}\qquad \varphi =-Sx_{1}x_{2}=-S(r\cos \theta )(r\sin \theta )=-{\cfrac {Sr^{2}}{2}}\sin(2\theta )}
or,
(109)
φ
=
−
S
r
2
2
sin
(
2
θ
)
{\displaystyle {\text{(109)}}\qquad \varphi =-{\cfrac {Sr^{2}}{2}}\sin(2\theta )}
Note that we have arranged the expression so that it has a form similar to the Fourier series of the previous section.
For this we have to add terms to
φ
{\displaystyle \varphi }
The unperturbed solution continues to be true as
r
→
∞
{\displaystyle r\rightarrow \infty \,}
The terms have the same form as the unperturbed solution,i.e.,
s
i
n
(
2
θ
)
{\displaystyle sin(2\theta )\,}
The new
φ
{\displaystyle \varphi }
r
−
n
{\displaystyle r^{-n}\,}
Recall,
φ
=
∑
n
=
0
∞
f
n
(
r
)
cos
(
n
θ
)
+
∑
n
=
0
∞
g
n
(
r
)
sin
(
n
θ
)
{\displaystyle \varphi =\sum _{n=0}^{\infty }f_{n}(r)\cos(n\theta )+\sum _{n=0}^{\infty }g_{n}(r)\sin(n\theta )}
where,
f
0
(
r
)
=
A
0
r
2
+
B
0
r
2
ln
r
+
C
0
+
D
0
ln
r
f
1
(
r
)
=
A
1
r
3
+
B
1
r
+
C
1
r
ln
r
+
D
1
r
−
1
f
n
(
r
)
=
A
n
r
n
+
2
+
B
n
r
n
+
C
n
r
−
n
+
2
+
D
n
r
−
n
,
n
>
1
{\displaystyle {\begin{aligned}f_{0}(r)&=A_{0}r^{2}+B_{0}r^{2}\ln r+C_{0}+D_{0}\ln r\\f_{1}(r)&=A_{1}r^{3}+B_{1}r+C_{1}r\ln r+D_{1}r^{-1}\\f_{n}(r)&=A_{n}r^{n+2}+B_{n}r^{n}+C_{n}r^{-n+2}+D_{n}r^{-n}~,~~n>1\end{aligned}}}
So the appropriate stress function for the perturbation is
(110)
φ
=
g
2
(
r
)
sin
(
2
θ
)
=
(
C
2
r
−
2
+
2
+
D
2
r
−
2
)
sin
(
2
θ
)
{\displaystyle {\text{(110)}}\qquad \varphi =g_{2}(r)\sin(2\theta )=\left(C_{2}r^{-2+2}+D_{2}r^{-2}\right)\sin(2\theta )}
or,
(111)
φ
=
(
C
2
+
D
2
r
−
2
)
sin
(
2
θ
)
{\displaystyle {\text{(111)}}\qquad \varphi =\left(C_{2}+D_{2}r^{-2}\right)\sin(2\theta )}
Hence, the stress function appropriate for the superposed solution is
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikiversity.org/v1/":): {\displaystyle \text{(112)} \varphi = -\cfrac{Sr^2}{2}\sin(2\theta) + \left(C_2 + D_2 r^{-2}\right)\sin(2\theta) }
We determine
C
2
{\displaystyle C_{2}}
D
2
{\displaystyle D_{2}}
r
=
a
{\displaystyle r=a}
The stresses are
(113)
σ
r
r
=
1
r
∂
φ
∂
r
+
1
r
2
∂
2
φ
∂
θ
2
=
(
S
−
4
C
2
r
−
2
−
6
D
2
r
−
4
)
sin
(
2
θ
)
(114)
σ
θ
θ
=
∂
2
φ
∂
r
2
=
(
−
S
+
6
D
2
r
−
4
)
sin
(
2
θ
)
(115)
σ
r
θ
=
−
∂
∂
r
(
1
r
∂
φ
∂
θ
)
=
(
S
+
6
D
2
r
−
4
)
cos
(
2
θ
)
{\displaystyle {\begin{aligned}{\text{(113)}}\qquad \sigma _{rr}&={\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial r}}+{\cfrac {1}{r^{2}}}{\cfrac {\partial ^{2}\varphi }{\partial \theta ^{2}}}=\left(S-4C_{2}r^{-2}-6D_{2}r^{-4}\right)\sin(2\theta )\\{\text{(114)}}\qquad \sigma _{\theta \theta }&={\cfrac {\partial ^{2}\varphi }{\partial r^{2}}}=\left(-S+6D_{2}r^{-4}\right)\sin(2\theta )\\{\text{(115)}}\qquad \sigma _{r\theta }&=-{\cfrac {\partial }{\partial r}}\left({\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial \theta }}\right)=\left(S+6D_{2}r^{-4}\right)\cos(2\theta )\end{aligned}}}
Hence,
(116)
σ
r
r
|
r
=
a
=
0
=
(
S
−
4
C
2
a
−
2
−
6
D
2
a
−
4
)
sin
(
2
θ
)
(117)
σ
r
θ
|
r
=
a
=
0
=
(
S
+
2
C
2
a
−
2
+
6
D
2
a
−
4
)
cos
(
2
θ
)
{\displaystyle {\begin{aligned}{\text{(116)}}\qquad \left.\sigma _{rr}\right|_{r=a}&=0=\left(S-4C_{2}a^{-2}-6D_{2}a^{-4}\right)\sin(2\theta )\\{\text{(117)}}\qquad \left.\sigma _{r\theta }\right|_{r=a}&=0=\left(S+2C_{2}a^{-2}+6D_{2}a^{-4}\right)\cos(2\theta )\end{aligned}}}
or,
(118)
4
C
2
a
−
2
+
6
D
2
a
−
4
=
S
(119)
2
C
2
a
−
2
+
6
D
2
a
−
4
=
−
S
{\displaystyle {\begin{aligned}{\text{(118)}}\qquad 4C_{2}a^{-2}+6D_{2}a^{-4}&=S\\{\text{(119)}}\qquad 2C_{2}a^{-2}+6D_{2}a^{-4}&=-S\end{aligned}}}
Solving,
(120)
C
2
=
S
a
2
;
D
2
=
−
S
a
4
2
{\displaystyle {\text{(120)}}\qquad C_{2}=Sa^{2}~;~~D_{2}=-{\cfrac {Sa^{4}}{2}}}
Back substituting,
(121)
σ
r
r
=
S
(
1
−
4
a
2
r
2
+
3
a
4
r
4
)
sin
(
2
θ
)
(122)
σ
θ
θ
=
S
(
−
1
−
3
a
4
r
4
)
sin
(
2
θ
)
(123)
σ
r
θ
=
S
(
1
+
2
a
2
r
2
−
3
a
4
r
4
)
cos
(
2
θ
)
{\displaystyle {\begin{aligned}{\text{(121)}}\qquad \sigma _{rr}&=S\left(1-4{\cfrac {a^{2}}{r^{2}}}+3{\cfrac {a^{4}}{r^{4}}}\right)\sin(2\theta )\\{\text{(122)}}\qquad \sigma _{\theta \theta }&=S\left(-1-3{\cfrac {a^{4}}{r^{4}}}\right)\sin(2\theta )\\{\text{(123)}}\qquad \sigma _{r\theta }&=S\left(1+2{\cfrac {a^{2}}{r^{2}}}-3{\cfrac {a^{4}}{r^{4}}}\right)\cos(2\theta )\end{aligned}}}
Example homework problem
edit
Consider the elastic plate with a hole subject to pure shear.
Elastic plate with a circular hole under pure shear
The stresses close to the hole are given by
(29)
σ
r
r
=
S
(
1
−
4
a
2
r
2
+
3
a
4
r
4
)
sin
(
2
θ
)
(30)
σ
θ
θ
=
S
(
−
1
−
3
a
4
r
4
)
sin
(
2
θ
)
(31)
σ
r
θ
=
S
(
1
+
2
a
2
r
2
−
3
a
4
r
4
)
cos
(
2
θ
)
{\displaystyle {\begin{aligned}{\text{(29)}}\qquad \sigma _{rr}&=S\left(1-4{\frac {a^{2}}{r^{2}}}+3{\frac {a^{4}}{r^{4}}}\right)\sin(2\theta )\\{\text{(30)}}\qquad \sigma _{\theta \theta }&=S\left(-1-3{\frac {a^{4}}{r^{4}}}\right)\sin(2\theta )\\{\text{(31)}}\qquad \sigma _{r\theta }&=S\left(1+2{\frac {a^{2}}{r^{2}}}-3{\frac {a^{4}}{r^{4}}}\right)\cos(2\theta )\end{aligned}}}
Show that the normal and shear traction boundary conditions far from the hole are satisfied by these stresses.
Calculate the stress concentration factors at the hole, i.e., (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikiversity.org/v1/":): {\displaystyle \tau_{\text{max}}/S}
) (shear) and (
σ
max
/
σ
0
{\displaystyle \sigma _{\text{max}}/\sigma _{0}}
Calculate the displacement field corresponding to this stress field (for plane stress). Plot the deformed shape of the hole.
Far from the hole,
r
=
∞
{\displaystyle r=\infty }
(32)
σ
r
r
=
S
sin
(
2
θ
)
(33)
σ
θ
θ
=
−
S
sin
(
2
θ
)
(34)
σ
r
θ
=
S
cos
(
2
θ
)
{\displaystyle {\begin{aligned}{\text{(32)}}\qquad \sigma _{rr}&=S\sin(2\theta )\\{\text{(33)}}\qquad \sigma _{\theta \theta }&=-S\sin(2\theta )\\{\text{(34)}}\qquad \sigma _{r\theta }&=S\cos(2\theta )\end{aligned}}}
To rotate the stresses back to the
(
x
1
,
x
2
)
{\displaystyle (x_{1},x_{2})}
(35)
[
σ
11
σ
12
σ
13
σ
21
σ
22
σ
23
σ
31
σ
32
σ
33
]
=
[
cos
θ
−
sin
θ
0
sin
θ
cos
θ
0
0
0
1
]
[
σ
r
r
σ
r
θ
σ
r
z
σ
r
θ
σ
θ
θ
σ
θ
z
σ
r
z
σ
θ
z
σ
z
z
]
[
cos
θ
sin
θ
0
−
sin
θ
cos
θ
0
0
0
1
]
{\displaystyle {\text{(35)}}\qquad {\begin{bmatrix}\sigma _{11}&\sigma _{12}&\sigma _{13}\\\sigma _{21}&\sigma _{22}&\sigma _{23}\\\sigma _{31}&\sigma _{32}&\sigma _{33}\end{bmatrix}}={\begin{bmatrix}\cos \theta &-\sin \theta &0\\\sin \theta &\cos \theta &0\\0&0&1\end{bmatrix}}{\begin{bmatrix}\sigma _{rr}&\sigma _{r\theta }&\sigma _{rz}\\\sigma _{r\theta }&\sigma _{\theta \theta }&\sigma _{\theta z}\\\sigma _{rz}&\sigma _{\theta z}&\sigma _{zz}\end{bmatrix}}{\begin{bmatrix}\cos \theta &\sin \theta &0\\-\sin \theta &\cos \theta &0\\0&0&1\end{bmatrix}}}
Setting
σ
r
z
=
0
{\displaystyle \sigma _{rz}=0}
σ
θ
z
=
0
{\displaystyle \sigma _{\theta z}=0}
(36)
σ
11
=
σ
r
r
cos
2
θ
+
σ
θ
θ
sin
2
θ
−
σ
r
θ
sin
(
2
θ
)
(37)
σ
22
=
σ
r
r
sin
2
θ
+
σ
θ
θ
cos
2
θ
+
σ
r
θ
sin
(
2
θ
)
(38)
σ
12
=
σ
r
r
−
σ
θ
θ
2
sin
(
2
θ
)
+
σ
r
θ
cos
(
2
θ
)
{\displaystyle {\begin{aligned}{\text{(36)}}\qquad \sigma _{11}&=\sigma _{rr}\cos ^{2}\theta +\sigma _{\theta \theta }\sin ^{2}\theta -\sigma _{r\theta }\sin(2\theta )\\{\text{(37)}}\qquad \sigma _{22}&=\sigma _{rr}\sin ^{2}\theta +\sigma _{\theta \theta }\cos ^{2}\theta +\sigma _{r\theta }\sin(2\theta )\\{\text{(38)}}\qquad \sigma _{12}&={\frac {\sigma _{rr}-\sigma _{\theta \theta }}{2}}\sin(2\theta )+\sigma _{r\theta }\cos(2\theta )\end{aligned}}}
Plugging in equations (32-34) in the above, we have
(39)
σ
11
=
−
S
[
sin
(
2
θ
)
cos
(
2
θ
)
−
sin
(
2
θ
)
cos
(
2
θ
)
]
=
0
(40)
σ
22
=
S
[
sin
(
2
θ
)
cos
(
2
θ
)
−
sin
(
2
θ
)
cos
(
2
θ
)
]
=
0
(41)
σ
12
=
S
[
sin
(
2
θ
)
sin
(
2
θ
)
+
cos
(
2
θ
)
cos
(
2
θ
)
]
=
S
{\displaystyle {\begin{aligned}{\text{(39)}}\qquad \sigma _{11}&=-S\left[\sin(2\theta )\cos(2\theta )-\sin(2\theta )\cos(2\theta )\right]=0\\{\text{(40)}}\qquad \sigma _{22}&=S\left[\sin(2\theta )\cos(2\theta )-\sin(2\theta )\cos(2\theta )\right]=0\\{\text{(41)}}\qquad \sigma _{12}&=S\left[\sin(2\theta )\sin(2\theta )+\cos(2\theta )\cos(2\theta )\right]=S\end{aligned}}}
Hence, the far field stress BCs are satisfied.
r
=
a
{\displaystyle r=a}
(42)
σ
r
r
=
S
(
1
−
4
+
3
)
sin
(
2
θ
)
=
0
(43)
σ
θ
θ
=
S
(
−
1
−
3
)
sin
(
2
θ
)
=
−
4
S
sin
(
2
θ
)
(44)
σ
r
θ
=
S
(
1
+
2
−
3
)
cos
(
2
θ
)
=
0
{\displaystyle {\begin{aligned}{\text{(42)}}\qquad \sigma _{rr}&=S\left(1-4+3\right)\sin(2\theta )=0\\{\text{(43)}}\qquad \sigma _{\theta \theta }&=S\left(-1-3\right)\sin(2\theta )=-4S\sin(2\theta )\\{\text{(44)}}\qquad \sigma _{r\theta }&=S\left(1+2-3\right)\cos(2\theta )=0\end{aligned}}}
The maximum (or minimum) hoop stress at the hole is at the locations
where
d
σ
θ
θ
/
d
θ
=
−
8
S
cos
(
2
θ
)
=
0
{\displaystyle d\sigma _{\theta \theta }/d\theta =-8S\cos(2\theta )=0}
θ
=
π
/
4
{\displaystyle \theta =\pi /4}
θ
=
3
π
/
4
{\displaystyle \theta =3\pi /4}
(45)
at
θ
=
π
4
σ
θ
θ
=
−
4
S
(46)
at
θ
=
3
π
4
σ
θ
θ
=
4
S
{\displaystyle {\begin{aligned}{\text{(45)}}\qquad {\text{at}}~\theta ={\frac {\pi }{4}}&&\sigma _{\theta \theta }=-4S\\{\text{(46)}}\qquad {\text{at}}~\theta ={\frac {3\pi }{4}}&&\sigma _{\theta \theta }=4S\end{aligned}}}
The maximum shear stress is given by
(47)
τ
max
=
1
2
|
σ
r
r
−
σ
θ
θ
|
=
2
S
{\displaystyle {\text{(47)}}\qquad \tau _{\text{max}}={\frac {1}{2}}\left|\sigma _{rr}-\sigma _{\theta \theta }\right|=2S}
Therefore, the stress concentration factors are
(48)
σ
max
S
=
4
;
τ
max
S
=
2
{\displaystyle {\text{(48)}}\qquad {\frac {\sigma _{\text{max}}}{S}}=4~;~~{\frac {\tau _{\text{max}}}{S}}=2}
The stress function used to derive the above results was
(49)
φ
=
−
S
2
r
2
sin
(
2
θ
)
+
S
a
2
sin
(
2
θ
)
−
S
a
4
2
r
−
2
sin
(
2
θ
)
{\displaystyle {\text{(49)}}\qquad \varphi =-{\frac {S}{2}}r^{2}\sin(2\theta )+Sa^{2}\sin(2\theta )-{\frac {Sa^{4}}{2}}r^{-2}\sin(2\theta )}
From Michell's solution , the displacements corresponding
to the above stress function are given by
(50)
2
μ
u
r
=
−
S
2
[
−
2
r
sin
(
2
θ
)
]
+
S
a
2
[
(
κ
+
1
)
r
−
1
sin
(
2
θ
)
]
−
S
a
4
2
[
2
r
−
3
sin
(
2
θ
)
]
(51)
2
μ
u
θ
=
−
S
2
[
−
2
r
cos
(
2
θ
)
]
+
S
a
2
[
(
κ
−
1
)
r
−
1
cos
(
2
θ
)
]
−
S
a
4
2
[
−
2
r
−
3
cos
(
2
θ
)
]
{\displaystyle {\begin{aligned}{\text{(50)}}\qquad 2\mu u_{r}&=-{\frac {S}{2}}\left[-2r\sin(2\theta )\right]+Sa^{2}\left[(\kappa +1)r^{-1}\sin(2\theta )\right]-{\frac {Sa^{4}}{2}}\left[2r^{-3}\sin(2\theta )\right]\\{\text{(51)}}\qquad 2\mu u_{\theta }&=-{\frac {S}{2}}\left[-2r\cos(2\theta )\right]+Sa^{2}\left[(\kappa -1)r^{-1}\cos(2\theta )\right]-{\frac {Sa^{4}}{2}}\left[-2r^{-3}\cos(2\theta )\right]\end{aligned}}}
or,
(52)
u
r
=
S
r
sin
(
2
θ
)
2
μ
[
1
+
(
κ
+
1
)
a
2
r
2
−
a
4
r
4
]
(53)
u
θ
=
S
r
cos
(
2
θ
)
2
μ
[
1
+
(
κ
−
1
)
a
2
r
2
+
a
4
r
4
]
{\displaystyle {\begin{aligned}{\text{(52)}}\qquad u_{r}&={\frac {Sr\sin(2\theta )}{2\mu }}\left[1+(\kappa +1){\frac {a^{2}}{r^{2}}}-{\frac {a^{4}}{r^{4}}}\right]\\{\text{(53)}}\qquad u_{\theta }&={\frac {Sr\cos(2\theta )}{2\mu }}\left[1+(\kappa -1){\frac {a^{2}}{r^{2}}}+{\frac {a^{4}}{r^{4}}}\right]\end{aligned}}}
For plane stress, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikiversity.org/v1/":): {\displaystyle \kappa = (3-\nu)/(1+\nu)}
. Hence,
(54)
u
r
=
S
r
sin
(
2
θ
)
2
μ
[
1
+
(
4
1
+
ν
)
a
2
r
2
−
a
4
r
4
]
(55)
u
θ
=
S
r
cos
(
2
θ
)
2
μ
[
1
+
2
(
1
−
ν
1
+
ν
)
a
2
r
2
+
a
4
r
4
]
{\displaystyle {\begin{aligned}{\text{(54)}}\qquad u_{r}&={\frac {Sr\sin(2\theta )}{2\mu }}\left[1+\left({\frac {4}{1+\nu }}\right){\frac {a^{2}}{r^{2}}}-{\frac {a^{4}}{r^{4}}}\right]\\{\text{(55)}}\qquad u_{\theta }&={\frac {Sr\cos(2\theta )}{2\mu }}\left[1+2\left({\frac {1-\nu }{1+\nu }}\right){\frac {a^{2}}{r^{2}}}+{\frac {a^{4}}{r^{4}}}\right]\end{aligned}}}
At
r
=
a
{\displaystyle r=a}
(56)
u
r
=
S
a
sin
(
2
θ
)
μ
(
2
1
+
ν
)
(57)
u
θ
=
S
a
cos
(
2
θ
)
μ
(
2
1
+
ν
)
{\displaystyle {\begin{aligned}{\text{(56)}}\qquad u_{r}&={\frac {Sa\sin(2\theta )}{\mu }}\left({\frac {2}{1+\nu }}\right)\\{\text{(57)}}\qquad u_{\theta }&={\frac {Sa\cos(2\theta )}{\mu }}\left({\frac {2}{1+\nu }}\right)\end{aligned}}}
Now
μ
=
E
/
2
(
1
+
ν
)
{\displaystyle \mu =E/2(1+\nu )}
(58)
u
r
=
4
S
a
sin
(
2
θ
)
E
(59)
u
θ
=
4
S
a
cos
(
2
θ
)
E
{\displaystyle {\begin{aligned}{\text{(58)}}\qquad u_{r}&={\frac {4Sa\sin(2\theta )}{E}}\\{\text{(59)}}\qquad u_{\theta }&={\frac {4Sa\cos(2\theta )}{E}}\end{aligned}}}
The deformed shape is shown below
Displacement field near a hole in plate under pure shear
![{\displaystyle {\begin{aligned}{\text{(39)}}\qquad \sigma _{11}&=-S\left[\sin(2\theta )\cos(2\theta )-\sin(2\theta )\cos(2\theta )\right]=0\\{\text{(40)}}\qquad \sigma _{22}&=S\left[\sin(2\theta )\cos(2\theta )-\sin(2\theta )\cos(2\theta )\right]=0\\{\text{(41)}}\qquad \sigma _{12}&=S\left[\sin(2\theta )\sin(2\theta )+\cos(2\theta )\cos(2\theta )\right]=S\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/62e8c34f2f2ae2dfc85d130284af3d5cf1d8a000)
![{\displaystyle {\begin{aligned}{\text{(50)}}\qquad 2\mu u_{r}&=-{\frac {S}{2}}\left[-2r\sin(2\theta )\right]+Sa^{2}\left[(\kappa +1)r^{-1}\sin(2\theta )\right]-{\frac {Sa^{4}}{2}}\left[2r^{-3}\sin(2\theta )\right]\\{\text{(51)}}\qquad 2\mu u_{\theta }&=-{\frac {S}{2}}\left[-2r\cos(2\theta )\right]+Sa^{2}\left[(\kappa -1)r^{-1}\cos(2\theta )\right]-{\frac {Sa^{4}}{2}}\left[-2r^{-3}\cos(2\theta )\right]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/78bd8689956acb88a1cc4c0859e77ed7a49c9f59)
![{\displaystyle {\begin{aligned}{\text{(52)}}\qquad u_{r}&={\frac {Sr\sin(2\theta )}{2\mu }}\left[1+(\kappa +1){\frac {a^{2}}{r^{2}}}-{\frac {a^{4}}{r^{4}}}\right]\\{\text{(53)}}\qquad u_{\theta }&={\frac {Sr\cos(2\theta )}{2\mu }}\left[1+(\kappa -1){\frac {a^{2}}{r^{2}}}+{\frac {a^{4}}{r^{4}}}\right]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7656b5cf9571fa0b62ca1dfd977270f3b852fff6)
![{\displaystyle {\begin{aligned}{\text{(54)}}\qquad u_{r}&={\frac {Sr\sin(2\theta )}{2\mu }}\left[1+\left({\frac {4}{1+\nu }}\right){\frac {a^{2}}{r^{2}}}-{\frac {a^{4}}{r^{4}}}\right]\\{\text{(55)}}\qquad u_{\theta }&={\frac {Sr\cos(2\theta )}{2\mu }}\left[1+2\left({\frac {1-\nu }{1+\nu }}\right){\frac {a^{2}}{r^{2}}}+{\frac {a^{4}}{r^{4}}}\right]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce9c8f83816b2e58e1d51f8080ba30af484c4ffa)
