Disk with a central hole
edit
An elastic disk with a central circular hole
Under general loading, for the stresses and displacements to be single-valued and continuous, they must be periodic in
θ
{\displaystyle \theta }
, e.g.,
σ
11
(
r
,
θ
)
=
σ
11
(
r
,
θ
+
2
m
π
)
{\displaystyle \sigma _{11}(r,\theta )=\sigma _{11}(r,\theta +2m\pi )}
.
An Airy stress function appropriate from this situation is
(83)
φ
=
∑
n
=
0
∞
f
n
(
r
)
cos
(
n
θ
)
+
∑
n
=
0
∞
g
n
(
r
)
sin
(
n
θ
)
{\displaystyle {\text{(83)}}\qquad \varphi =\sum _{n=0}^{\infty }f_{n}(r)\cos(n\theta )+\sum _{n=0}^{\infty }g_{n}(r)\sin(n\theta )}
In the absence of body forces,
(84)
∇
4
φ
=
∇
2
(
∇
2
φ
)
=
0
;
∇
2
=
(
∂
2
∂
r
2
+
1
r
∂
∂
r
+
1
r
2
∂
2
∂
θ
2
)
{\displaystyle {\text{(84)}}\qquad \nabla ^{4}{\varphi }=\nabla ^{2}{(\nabla ^{2}{\varphi })}=0~;~~\nabla ^{2}{}=\left({\cfrac {\partial ^{2}}{\partial r^{2}}}+{\cfrac {1}{r}}{\cfrac {\partial }{\partial r}}+{\cfrac {1}{r^{2}}}{\cfrac {\partial ^{2}}{\partial \theta ^{2}}}\right)}
Plug in
φ
{\displaystyle \varphi }
.
∇
2
φ
=
∑
n
=
0
∞
[
f
n
″
(
r
)
cos
(
n
θ
)
+
1
r
f
n
′
(
r
)
cos
(
n
θ
)
−
n
2
r
2
f
n
(
r
)
cos
(
n
θ
)
]
+
∑
n
=
0
∞
[
g
n
″
(
r
)
sin
(
n
θ
)
+
1
r
g
n
′
(
r
)
sin
(
n
θ
)
−
n
2
r
2
g
n
(
r
)
sin
(
n
θ
)
]
(85)
{\displaystyle {\begin{aligned}\nabla ^{2}{\varphi }=&\sum _{n=0}^{\infty }\left[f_{n}^{''}(r)\cos(n\theta )+{\cfrac {1}{r}}f_{n}^{'}(r)\cos(n\theta )-{\cfrac {n^{2}}{r^{2}}}f_{n}(r)\cos(n\theta )\right]+\\&\sum _{n=0}^{\infty }\left[g_{n}^{''}(r)\sin(n\theta )+{\cfrac {1}{r}}g_{n}^{'}(r)\sin(n\theta )-{\cfrac {n^{2}}{r^{2}}}g_{n}(r)\sin(n\theta )\right]\qquad {\text{(85)}}\end{aligned}}}
or,
(86)
∇
2
φ
=
∑
n
=
0
∞
F
n
(
r
)
cos
(
n
θ
)
+
∑
n
=
0
∞
G
n
(
r
)
sin
(
n
θ
)
{\displaystyle {\text{(86)}}\qquad \nabla ^{2}{\varphi }=\sum _{n=0}^{\infty }F_{n}(r)\cos(n\theta )+\sum _{n=0}^{\infty }G_{n}(r)\sin(n\theta )}
Therefore,
∇
4
φ
=
∑
n
=
0
∞
[
F
n
″
(
r
)
cos
(
n
θ
)
+
1
r
F
n
′
(
r
)
cos
(
n
θ
)
−
n
2
r
2
F
n
(
r
)
cos
(
n
θ
)
]
+
∑
n
=
0
∞
[
G
n
″
(
r
)
sin
(
n
θ
)
+
1
r
G
n
′
(
r
)
sin
(
n
θ
)
−
n
2
r
2
G
n
(
r
)
sin
(
n
θ
)
]
(87)
{\displaystyle {\begin{aligned}\nabla ^{4}{\varphi }=&\sum _{n=0}^{\infty }\left[F_{n}^{''}(r)\cos(n\theta )+{\cfrac {1}{r}}F_{n}^{'}(r)\cos(n\theta )-{\cfrac {n^{2}}{r^{2}}}F_{n}(r)\cos(n\theta )\right]+\\&\sum _{n=0}^{\infty }\left[G_{n}^{''}(r)\sin(n\theta )+{\cfrac {1}{r}}G_{n}^{'}(r)\sin(n\theta )-{\cfrac {n^{2}}{r^{2}}}G_{n}(r)\sin(n\theta )\right]\qquad {\text{(87)}}\end{aligned}}}
To satisfy the compatibility condition
∇
4
φ
=
0
{\displaystyle \nabla ^{4}{\varphi }=0}
, we need
(88)
F
n
″
(
r
)
+
1
r
F
n
′
(
r
)
−
n
2
r
2
F
n
(
r
)
=
0
(89)
G
n
″
(
r
)
+
1
r
G
n
′
(
r
)
−
n
2
r
2
G
n
(
r
)
=
0
{\displaystyle {\begin{aligned}{\text{(88)}}\qquad F_{n}^{''}(r)+{\cfrac {1}{r}}F_{n}^{'}(r)-{\cfrac {n^{2}}{r^{2}}}F_{n}(r)&=0\\{\text{(89)}}\qquad G_{n}^{''}(r)+{\cfrac {1}{r}}G_{n}^{'}(r)-{\cfrac {n^{2}}{r^{2}}}G_{n}(r)&=0\end{aligned}}}
The general solution of these Euler-Cauchy type equations is
(90)
F
n
(
r
)
=
A
1
r
n
+
B
1
r
−
n
(91)
G
n
(
r
)
=
C
1
r
n
+
D
1
r
−
n
{\displaystyle {\begin{aligned}{\text{(90)}}\qquad F_{n}(r)&=A_{1}r^{n}+B_{1}r^{-n}\\{\text{(91)}}\qquad G_{n}(r)&=C_{1}r^{n}+D_{1}r^{-n}\end{aligned}}}
We can use either to determine
f
n
(
r
)
{\displaystyle f_{n}(r)}
. Thus,
(92)
f
n
″
(
r
)
+
1
r
f
n
′
(
r
)
−
n
2
r
2
f
n
(
r
)
=
A
1
r
n
+
B
1
r
−
n
{\displaystyle {\text{(92)}}\qquad f_{n}^{''}(r)+{\cfrac {1}{r}}f_{n}^{'}(r)-{\cfrac {n^{2}}{r^{2}}}f_{n}(r)=A_{1}r^{n}+B_{1}r^{-n}}
or,
(93)
r
2
f
n
″
(
r
)
+
r
f
n
′
(
r
)
−
n
2
f
n
(
r
)
=
A
1
r
n
+
2
+
B
1
r
−
n
+
2
{\displaystyle {\text{(93)}}\qquad r^{2}f_{n}^{''}(r)+rf_{n}^{'}(r)-n^{2}f_{n}(r)=A_{1}r^{n+2}+B_{1}r^{-n+2}}
The homogeneous and particular solutions of this equation are
(94)
f
n
h
(
r
)
=
A
2
r
n
+
B
2
r
−
n
(95)
f
n
p
(
r
)
=
A
1
r
n
+
2
+
B
1
r
−
n
+
2
{\displaystyle {\begin{aligned}{\text{(94)}}\qquad f_{n}^{h}(r)&=A_{2}r^{n}+B_{2}r^{-n}\\{\text{(95)}}\qquad f_{n}^{p}(r)&=A_{1}r^{n+2}+B_{1}r^{-n+2}\end{aligned}}}
Hence, the general solution is
(96)
f
n
(
r
)
=
A
1
r
n
+
2
+
B
1
r
−
n
+
2
+
A
2
r
n
+
B
2
r
−
n
{\displaystyle {\text{(96)}}\qquad f_{n}(r)=A_{1}r^{n+2}+B_{1}r^{-n+2}+A_{2}r^{n}+B_{2}r^{-n}}
This form is valid for
n
>
1
{\displaystyle n>1}
. If 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 n = 0, 1}
, alternative forms are obtained. Thus,
(97)
f
0
(
r
)
=
A
O
r
2
+
B
0
r
2
ln
r
+
C
0
+
D
0
ln
r
(98)
f
1
(
r
)
=
A
1
r
3
+
B
1
r
+
C
1
r
ln
r
+
D
1
r
−
1
(99)
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}{\text{(97)}}\qquad f_{0}(r)&=A_{O}r^{2}+B_{0}r^{2}\ln r+C_{0}+D_{0}\ln r\\{\text{(98)}}\qquad f_{1}(r)&=A_{1}r^{3}+B_{1}r+C_{1}r\ln r+D_{1}r^{-1}\\{\text{(99)}}\qquad 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}}}
Terms in
f
n
{\displaystyle f_{n}}
are chosen according to the specific problem of interest.
Traction BCs
edit
at
r
=
a
{\displaystyle r=a}
(100)
σ
r
r
=
T
1
(
θ
)
,
σ
r
θ
=
T
2
(
θ
)
{\displaystyle {\text{(100)}}\qquad \sigma _{rr}=T_{1}(\theta )~,~~\sigma _{r\theta }=T_{2}(\theta )}
at
r
=
b
{\displaystyle r=b}
(101)
σ
r
r
=
T
3
(
θ
)
,
σ
r
θ
=
T
4
(
θ
)
{\displaystyle {\text{(101)}}\qquad \sigma _{rr}=T_{3}(\theta )~,~~\sigma _{r\theta }=T_{4}(\theta )}
Express
T
i
(
θ
)
{\displaystyle T_{i}(\theta )}
in Fourier series form.
(102)
T
i
(
θ
)
=
∑
n
=
0
∞
A
n
i
cos
(
n
θ
)
+
∑
n
=
0
∞
B
n
i
sin
(
n
θ
)
,
i
=
1
,
2
,
3
,
4
{\displaystyle {\text{(102)}}\qquad T_{i}(\theta )=\sum _{n=0}^{\infty }A_{ni}\cos(n\theta )+\sum _{n=0}^{\infty }B_{ni}\sin(n\theta )~,~~i=1,2,3,4}
Terms in
T
i
{\displaystyle T_{i}}
are chosen according to the specific problem of interest.