Disk with a central hole

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 ${\displaystyle \theta }$, e.g., ${\displaystyle \sigma _{11}(r,\theta )=\sigma _{11}(r,\theta +2m\pi )}$.

An Airy stress function appropriate from this situation is

${\displaystyle {\text{(83)}}\qquad \varphi =\sum _{n=0}^{\infty }f_{n}(r)\cos(n\theta )+\sum _{n=0}^{\infty }g_{n}(r)\sin(n\theta )}$

In the absence of body forces,

${\displaystyle {\text{(84)}}\qquad \nabla ^{4}{\varphi }=\nabla ^{2}{(\nabla ^{2}{\varphi })}=0~;~~\nabla ^{2}{}=\left({\cfrac {\partial ^{2}}{\partial r^{2}}}+{\cfrac {1}{r}}{\cfrac {\partial }{\partial r}}+{\cfrac {1}{r^{2}}}{\cfrac {\partial ^{2}}{\partial \theta ^{2}}}\right)}$

Plug in ${\displaystyle \varphi }$.

${\displaystyle {\begin{aligned}\nabla ^{2}{\varphi }=&\sum _{n=0}^{\infty }\left[f_{n}^{''}(r)\cos(n\theta )+{\cfrac {1}{r}}f_{n}^{'}(r)\cos(n\theta )-{\cfrac {n^{2}}{r^{2}}}f_{n}(r)\cos(n\theta )\right]+\\&\sum _{n=0}^{\infty }\left[g_{n}^{''}(r)\sin(n\theta )+{\cfrac {1}{r}}g_{n}^{'}(r)\sin(n\theta )-{\cfrac {n^{2}}{r^{2}}}g_{n}(r)\sin(n\theta )\right]\qquad {\text{(85)}}\end{aligned}}}$

or,

${\displaystyle {\text{(86)}}\qquad \nabla ^{2}{\varphi }=\sum _{n=0}^{\infty }F_{n}(r)\cos(n\theta )+\sum _{n=0}^{\infty }G_{n}(r)\sin(n\theta )}$

Therefore,

${\displaystyle {\begin{aligned}\nabla ^{4}{\varphi }=&\sum _{n=0}^{\infty }\left[F_{n}^{''}(r)\cos(n\theta )+{\cfrac {1}{r}}F_{n}^{'}(r)\cos(n\theta )-{\cfrac {n^{2}}{r^{2}}}F_{n}(r)\cos(n\theta )\right]+\\&\sum _{n=0}^{\infty }\left[G_{n}^{''}(r)\sin(n\theta )+{\cfrac {1}{r}}G_{n}^{'}(r)\sin(n\theta )-{\cfrac {n^{2}}{r^{2}}}G_{n}(r)\sin(n\theta )\right]\qquad {\text{(87)}}\end{aligned}}}$

To satisfy the compatibility condition ${\displaystyle \nabla ^{4}{\varphi }=0}$, we need

${\displaystyle {\begin{aligned}{\text{(88)}}\qquad F_{n}^{''}(r)+{\cfrac {1}{r}}F_{n}^{'}(r)-{\cfrac {n^{2}}{r^{2}}}F_{n}(r)&=0\\{\text{(89)}}\qquad G_{n}^{''}(r)+{\cfrac {1}{r}}G_{n}^{'}(r)-{\cfrac {n^{2}}{r^{2}}}G_{n}(r)&=0\end{aligned}}}$

The general solution of these Euler-Cauchy type equations is

${\displaystyle {\begin{aligned}{\text{(90)}}\qquad F_{n}(r)&=A_{1}r^{n}+B_{1}r^{-n}\\{\text{(91)}}\qquad G_{n}(r)&=C_{1}r^{n}+D_{1}r^{-n}\end{aligned}}}$

We can use either to determine ${\displaystyle f_{n}(r)}$. Thus,

${\displaystyle {\text{(92)}}\qquad f_{n}^{''}(r)+{\cfrac {1}{r}}f_{n}^{'}(r)-{\cfrac {n^{2}}{r^{2}}}f_{n}(r)=A_{1}r^{n}+B_{1}r^{-n}}$

or,

${\displaystyle {\text{(93)}}\qquad r^{2}f_{n}^{''}(r)+rf_{n}^{'}(r)-n^{2}f_{n}(r)=A_{1}r^{n+2}+B_{1}r^{-n+2}}$

The homogeneous and particular solutions of this equation are

${\displaystyle {\begin{aligned}{\text{(94)}}\qquad f_{n}^{h}(r)&=A_{2}r^{n}+B_{2}r^{-n}\\{\text{(95)}}\qquad f_{n}^{p}(r)&=A_{1}r^{n+2}+B_{1}r^{-n+2}\end{aligned}}}$

Hence, the general solution is

${\displaystyle {\text{(96)}}\qquad f_{n}(r)=A_{1}r^{n+2}+B_{1}r^{-n+2}+A_{2}r^{n}+B_{2}r^{-n}}$

This form is valid for ${\displaystyle n>1}$. If ${\displaystyle n=0,1}$, alternative forms are obtained. Thus,

${\displaystyle {\begin{aligned}{\text{(97)}}\qquad f_{0}(r)&=A_{O}r^{2}+B_{0}r^{2}\ln r+C_{0}+D_{0}\ln r\\{\text{(98)}}\qquad f_{1}(r)&=A_{1}r^{3}+B_{1}r+C_{1}r\ln r+D_{1}r^{-1}\\{\text{(99)}}\qquad f_{n}(r)&=A_{n}r^{n+2}+B_{n}r^{n}+C_{n}r^{-n+2}+D_{n}r^{-n}~,~~n>1\end{aligned}}}$

Terms in ${\displaystyle f_{n}}$ are chosen according to the specific problem of interest.