Elasticity/Spinning disk

Problem 1:
A thin disk of radius $$a$$ is spinning about its axis with a constant angular velocity $$\dot{\theta}$$. Find the stress field in the disk using an Airy stress function and a body force potential.

Solution:
The acceleration of a point ($$r,\theta$$) on the disk is
 * $$ \text{(1)} \qquad

a_r = -\dot{\theta}^2 r ~; a_{\theta} = 0 $$ The body force field is
 * $$ \text{(2)} \qquad

f_r = \rho\dot{\theta}^2 r ~; f_{\theta} = 0 $$ Since there is no rotational acceleration, the body force can be derived from a potential $$V$$. The relations between the stresses, the Airy stress function and the body force potential are
 * $$\begin{align}

\text{(3)} \qquad \sigma_{rr} & = \frac{1}{r}\frac{\partial \varphi}{\partial r} + \frac{1}{r^2}\frac{\partial^2 \varphi}{\partial \theta^2} + V  \\ \text{(4)} \qquad \sigma_{\theta\theta} & = \frac{\partial^2 \varphi}{\partial r^2} + V \\ \text{(5)} \qquad \sigma_{r\theta} & = -\frac{\partial }{\partial r}     \left(\frac{1}{r}\frac{\partial \varphi }{\partial \theta} \right) \end{align}$$ where
 * $$ \text{(6)} \qquad

f_r = -\frac{\partial V }{\partial r} ~;~ f_{\theta} = -\frac{1}{r}\frac{\partial V}{\partial \theta} $$ From equations (2) and (6), we have,
 * $$\begin{align}

\text{(7)} \qquad \rho\dot{\theta}^2 r & = -\frac{\partial V}{\partial r} \\ \text{(8)} \qquad 0 & = -\frac{1}{r}\frac{\partial V}{\partial \theta} \end{align}$$ Integrating equation (7), we have
 * $$\text{(9)} \qquad

V = -\rho\dot{\theta}^2 \frac{r^2}{2} + h(\theta) $$ Substituting equation (9) into equation (8), we get
 * $$\text{(10)} \qquad

\frac{dh(\theta)}{d\theta} = 0 \Rightarrow h(\theta) = C $$ This constant can be set to zero without loss of generality. Therefore,
 * $$\text{(11)} \qquad

V = -\rho\dot{\theta}^2 \frac{r^2}{2} $$ The spinning disk problem is a plane stress problem. Hence the compatibility condition is
 * $$\text{(12)} \qquad

\nabla^4{\varphi} + \left(2-\frac{1}{\alpha}\right)\nabla^2{V} = 0 \Rightarrow \nabla^4{\varphi} + (1-\nu)\nabla^2{V} = 0 $$ where
 * $$\begin{align}

\text{(13)} \qquad \nabla^2{} & = \frac{\partial^2 }{\partial r^2} + \frac{1}{r}\frac{\partial  }{\partial r} + \frac{1}{r^2}\frac{\partial^2 }{\partial \theta} \\ \text{(14)} \qquad \nabla^4{} & = \nabla^2{[\nabla^2{}]} \end{align}$$ Now, from equations (11) and (13)
 * $$\text{(15)} \qquad

\nabla^2{V} = -\rho\dot{\theta}^2[1 + 1 + 0] = -2\rho\dot{\theta}^2 $$ Therefore, equation (12) becomes
 * $$\text{(16)} \qquad

\nabla^4{\varphi} = 2 \rho\dot{\theta}^2(1-\nu) $$ Since the problem is axisymmetric, there can be no shear stresses, i.e. $$\sigma_{r\theta} = 0$$ and no dependence on $$\theta$$. From Michell's solution, the appropriate terms of the Airy stress function are
 * $$\text{(17)} \qquad

r^2 ~; r^2\ln(r) ~; \ln(r) $$ Axisymmetry also requires that $$u_{\theta}$$, the displacement in the $$\theta$$ direction must be zero. However, if we look at Mitchell's solution, we see that $$u_{\theta}$$ is non-zero if the term $$r^2\ln(r)$$ is used in the Airy stress function. Hence, we reject this term and are left with
 * $$\text{(18)} \qquad

\varphi = C_1r^2 + C_2 \ln(r) $$ If we plug this stress function into equation (16) we see that $$\nabla^4{\varphi} = 0$$. Therefore, equation (18) represents a homogeneous solution of equation (16). The $$\varphi$$ that is a general solution of equation (16) is obtained by adding a particular solution of the equation.

One such particular solution is the stress function $$\varphi = C_0 r^4$$ since the biharmonic equation must evaluate to a constant. Plugging this into equation (16) we have
 * $$\text{(19)} \qquad

\nabla^4{\varphi} = 64 C_0 = 2 \rho\dot{\theta}^2(1-\nu) $$ or,
 * $$\text{(20)} \qquad

C_0 = \frac{\rho\dot{\theta}^2(1-\nu)}{32} $$ Therefore, the general solution is
 * $$\text{(21)} \qquad

\varphi = \frac{\rho\dot{\theta}^2(1-\nu)}{32}r^4 + C_1r^2 + C_2 \ln(r) $$ The corresponding stresses are (from equations (3, 4, 5)),
 * $$\begin{align}

\text{(22)} \qquad \sigma_{rr} & = -\frac{(3+\nu)\dot{\theta}^2\rho r^2}{8} + 2 C_1 + \frac{C_2}{r^2} \\ \text{(23)} \qquad \sigma_{\theta\theta} & = -\frac{(1+3\nu)\dot{\theta}^2\rho r^2}{8} + 2 C_1 - \frac{C_2}{r^2} \\ \text{(24)} \qquad \sigma_{r\theta} & = 0 \end{align}$$ At $$r = 0$$, the stresses must be finite. Hence, $$C_2 = 0$$. At $$r = a$$, $$\sigma_{rr} = \sigma_{r\theta} = 0$$. Evaluating $$\sigma_{rr}$$ at $$ r = a$$ we get
 * $$\text{(25)} \qquad

C_1 = \frac{(3+\nu)\rho\dot{\theta}^2 a^2}{16} $$ Substituting back into equations (22) and (23), we get
 * $$\text{(26)} \qquad

\sigma_{rr} = \frac{\rho\dot{\theta}^2}{8}(3+\nu)(a^2-r^2) $$
 * $$\text{(27)} \qquad

\sigma_{\theta\theta} = \frac{\rho\dot{\theta}^2}{8}\left[ (3+\nu)a^2 - (1+3\nu)r^2\right] $$
 * $$\text{(28)} \qquad

\sigma_{r\theta} = 0 $$