Elasticity/Disk with hole

Disk with a central 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)} \qquad

\varphi = \sum^{\infty}_{n=0} f_n(r) \cos(n\theta) + \sum^{\infty}_{n=0} g_n(r) \sin(n\theta) $$ In the absence of body forces,
 * $$ \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) $$ Plug in $$\varphi$$.
 * $$\begin{align}

\nabla^2{\varphi} = & \sum^{\infty}_{n=0} \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^{\infty}_{n=0} \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{align}$$ or,
 * $$ \text{(86)} \qquad

\nabla^2{\varphi} = \sum^{\infty}_{n=0} F_n(r) \cos(n\theta) + \sum^{\infty}_{n=0} G_n(r) \sin(n\theta) $$ Therefore,
 * $$\begin{align}

\nabla^4{\varphi} = & \sum^{\infty}_{n=0} \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^{\infty}_{n=0} \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{align}$$ To satisfy the compatibility condition $$\nabla^4{\varphi} = 0$$, we need
 * $$\begin{align}

\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{align}$$ The general solution of these Euler-Cauchy type equations is
 * $$\begin{align}

\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{align}$$ We can use either to determine $$f_n(r)$$. Thus,
 * $$ \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} $$ or,
 * $$ \text{(93)} \qquad

r^2 f^{''}_n(r) + r f^{'}_n(r) - n^2 f_n(r) = A_1 r^{n+2} + B_1 r^{-n+2} $$ The homogeneous and particular solutions of this equation are
 * $$\begin{align}

\text{(94)} \qquad f^h_n(r) & = A_2 r^n + B_2 r^{-n}  \\ \text{(95)} \qquad f^p_n(r) & = A_1 r^{n+2} + B_1 r^{-n+2} \end{align}$$ Hence, the general solution is
 * $$ \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$$. If $$n = 0, 1$$, alternative forms are obtained. Thus,
 * $$\begin{align}

\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{align}$$ Terms in $$f_n$$ are chosen according to the specific problem of interest.

Traction BCs

 * at $$ r = a $$

\text{(100)} \qquad \sigma_{rr} = T_1(\theta) ~, \sigma_{r\theta} = T_2(\theta) $$
 * at $$ r = b $$

\text{(101)} \qquad \sigma_{rr} = T_3(\theta) ~, \sigma_{r\theta} = T_4(\theta) $$ Express $$T_i(\theta)$$ in Fourier series form.
 * $$ \text{(102)} \qquad

T_i(\theta) = \sum^{\infty}_{n=0} A_{ni}\cos(n\theta) + \sum^{\infty}_{n=0} B_{ni}\sin(n\theta) ~,i=1,2,3,4 $$ Terms in $$T_i$$ are chosen according to the specific problem of interest.