Elasticity/Disk with hole

Disk with a central hole

File:Elastic disk with hole.png 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 $\theta$, e.g., ${\sigma }_{11}(r,\theta )={\sigma }_{11}(r,\theta +2m\pi )$.

An Airy stress function appropriate from this situation is

$\text{(83)}\phi =\sum _{n=0}^{\mathrm{\infty }}{f}_n(r)\cos (n\theta )+\sum _{n=0}^{\mathrm{\infty }}{g}_n(r)\sin (n\theta )$

In the absence of body forces,

$\text{(84)}{\mathrm{\nabla }}^4\phi ={\mathrm{\nabla }}^2({\mathrm{\nabla }}^2\phi )=0;{\mathrm{\nabla }}^2=(\frac{{\partial }^2}{\partial r^2}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2})$

Plug in $\phi$.

$\begin{array}{cc}{\mathrm{\nabla }}^2\phi =& \sum _{n=0}^{\mathrm{\infty }}[f_n^{{\text{'}}^{\prime }}(r)\cos (n\theta )+\frac{1}{r}{f}_n^{\text{'}}(r)\cos (n\theta )-\frac{n^2}{r^2}{f}_n(r)\cos (n\theta )]+\\ & \sum _{n=0}^{\mathrm{\infty }}[g_n^{{\text{'}}^{\prime }}(r)\sin (n\theta )+\frac{1}{r}{g}_n^{\text{'}}(r)\sin (n\theta )-\frac{n^2}{r^2}{g}_n(r)\sin (n\theta )]\text{(85)}\end{array}$

or,

$\text{(86)}{\mathrm{\nabla }}^2\phi =\sum _{n=0}^{\mathrm{\infty }}{F}_n(r)\cos (n\theta )+\sum _{n=0}^{\mathrm{\infty }}{G}_n(r)\sin (n\theta )$

Therefore,

$\begin{array}{cc}{\mathrm{\nabla }}^4\phi =& \sum _{n=0}^{\mathrm{\infty }}[F_n^{{\text{'}}^{\prime }}(r)\cos (n\theta )+\frac{1}{r}{F}_n^{\text{'}}(r)\cos (n\theta )-\frac{n^2}{r^2}{F}_n(r)\cos (n\theta )]+\\ & \sum _{n=0}^{\mathrm{\infty }}[G_n^{{\text{'}}^{\prime }}(r)\sin (n\theta )+\frac{1}{r}{G}_n^{\text{'}}(r)\sin (n\theta )-\frac{n^2}{r^2}{G}_n(r)\sin (n\theta )]\text{(87)}\end{array}$

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

$\begin{array}{cc}\text{(88)}{F}_n^{{\text{'}}^{\prime }}(r)+\frac{1}{r}{F}_n^{\text{'}}(r)-\frac{n^2}{r^2}{F}_n(r)& =0\\ \text{(89)}{G}_n^{{\text{'}}^{\prime }}(r)+\frac{1}{r}{G}_n^{\text{'}}(r)-\frac{n^2}{r^2}{G}_n(r)& =0\end{array}$

The general solution of these Euler-Cauchy type equations is

$\begin{array}{cc}\text{(90)}{F}_n(r)& =A_1{r}^n+B_1{r}^{-n}\\ \text{(91)}{G}_n(r)& =C_1{r}^n+D_1{r}^{-n}\end{array}$

We can use either to determine $f_n(r)$. Thus,

$\text{(92)}{f}_n^{{\text{'}}^{\prime }}(r)+\frac{1}{r}{f}_n^{\text{'}}(r)-\frac{n^2}{r^2}{f}_n(r)=A_1{r}^n+B_1{r}^{-n}$

or,

$\text{(93)}{r}^2{f}_n^{{\text{'}}^{\prime }}(r)+r{f}_n^{\text{'}}(r)-n^2{f}_n(r)=A_1{r}^{n+2}+B_1{r}^{-n+2}$

The homogeneous and particular solutions of this equation are

$\begin{array}{cc}\text{(94)}{f}_n^h(r)& =A_2{r}^n+B_2{r}^{-n}\\ \text{(95)}{f}_n^p(r)& =A_1{r}^{n+2}+B_1{r}^{-n+2}\end{array}$

Hence, the general solution is

$\text{(96)}{f}_n(r)=A_1{r}^{n+2}+B_1{r}^{-n+2}+A_2{r}^n+B_2{r}^{-n}$

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

$\begin{array}{cc}\text{(97)}{f}_0(r)& =A_O{r}^2+B_0{r}^2\ln r+C_0+D_0\ln r\\ \text{(98)}{f}_1(r)& =A_1{r}^3+B_{1}r+C_{1}r\ln r+D_1{r}^{-1}\\ \text{(99)}{f}_n(r)& =A_n{r}^{n+2}+B_n{r}^n+C_n{r}^{-n+2}+D_n{r}^{-n},n>1\end{array}$

Terms in $f_n$ are chosen according to the specific problem of interest.

Traction BCs

at $r=a$

$\text{(100)}{\sigma }_{rr}=T_1(\theta ),{\sigma }_{r\theta }=T_2(\theta )$

at $r=b$

$\text{(101)}{\sigma }_{rr}=T_3(\theta ),{\sigma }_{r\theta }=T_4(\theta )$

Express $T_i(\theta )$ in Fourier series form.

$\text{(102)}{T}_i(\theta )=\sum _{n=0}^{\mathrm{\infty }}{A}_{ni}\cos (n\theta )+\sum _{n=0}^{\mathrm{\infty }}{B}_{ni}\sin (n\theta ),i=1,2,3,4$

Terms in $T_i$ are chosen according to the specific problem of interest.