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[edit] Thin spinning disk

[edit] 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.

An elastic disk spinning around its axis of symmetry

[edit] Solution:

The acceleration of a point ( $r,θ$ ) 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. $σ r θ = 0$ and no dependence on $θ$ . 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 $u θ$ , the displacement in the $θ$ direction must be zero. However, if we look at Mitchell's solution, we see that $u θ$ 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)),

$\begin{align} \text{(22)} \qquad \sigma_{rr} & = -\frac{(3+\nu)\dot{\theta}^2\rho r^2}{8} + 2 C_1 + \frac{C_2}{r^2} \\ \text{(23)} \qquad \sigma_{\theta\theta} & = -\frac{(1+3\nu)\dot{\theta}^2\rho r^2}{8} + 2 C_1 - \frac{C_2}{r^2} \\ \text{(24)} \qquad \sigma_{r\theta} & = 0 \end{align}$