Elasticity/Spinning disk

Thin 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.

File:Elastic spinning disk.png An elastic disk spinning around its axis of symmetry

Solution:

The acceleration of a point ($r,\theta$) on the disk is

$\text{(1)}{a}_r=-{\dot{\theta }}^{2}r;a_{\theta }=0$

The body force field is

$\text{(2)}{f}_r=\rho {\dot{\theta }}^{2}r;f_{\theta }=0$

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

$\begin{array}{cc}\text{(3)}{\sigma }_{rr}& =\frac{1}{r}\frac{\partial \phi }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2\phi }{\partial {\theta }^2}+V\\ \text{(4)}{\sigma }_{\theta \theta }& =\frac{{\partial }^2\phi }{\partial r^2}+V\\ \text{(5)}{\sigma }_{r\theta }& =-\frac{\partial }{\partial r}\left(\frac{1}{r}\frac{\partial \phi }{\partial \theta }\right)\end{array}$

where

$\text{(6)}{f}_r=-\frac{\partial V}{\partial r};f_{\theta }=-\frac{1}{r}\frac{\partial V}{\partial \theta }$

From equations (2) and (6) , we have,

$\begin{array}{cc}\text{(7)}\rho {\dot{\theta }}^{2}r& =-\frac{\partial V}{\partial r}\\ \text{(8)}0& =-\frac{1}{r}\frac{\partial V}{\partial \theta }\end{array}$

Integrating equation (7), we have

$\text{(9)}V=-\rho {\dot{\theta }}^2\frac{r^2}{2}+h(\theta )$

Substituting equation (9) into equation (8), we get

$\text{(10)}\frac{dh(\theta )}{d\theta }=0\Rightarrow h(\theta )=C$

This constant can be set to zero without loss of generality. Therefore,

$\text{(11)}V=-\rho {\dot{\theta }}^2\frac{r^2}{2}$

The spinning disk problem is a plane stress problem. Hence the compatibility condition is

$\text{(12)}{\mathrm{\nabla }}^4\phi +(2-\frac{1}{\alpha }){\mathrm{\nabla }}^{2}V=0\Rightarrow {\mathrm{\nabla }}^4\phi +(1-\nu ){\mathrm{\nabla }}^{2}V=0$

where

$\begin{array}{cc}\text{(13)}{\mathrm{\nabla }}^2()& =\frac{{\partial }^2()}{\partial r^2}+\frac{1}{r}\frac{\partial ()}{\partial r}+\frac{1}{r^2}\frac{{\partial }^2()}{\partial \theta }\\ \text{(14)}{\mathrm{\nabla }}^4()& ={\mathrm{\nabla }}^2[{\mathrm{\nabla }}^2()]\end{array}$

Now, from equations (11) and (13)

$\text{(15)}{\mathrm{\nabla }}^{2}V=-\rho {\dot{\theta }}^2[1+1+0]=-2\rho {\dot{\theta }}^2$

Therefore, equation (12) becomes

$\text{(16)}{\mathrm{\nabla }}^4\phi =2\rho {\dot{\theta }}^2(1-\nu )$

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

$\text{(17)}{r}^2;r^2\ln (r);\ln (r)$

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

$\text{(18)}\phi =C_1{r}^2+C_2\ln (r)$

If we plug this stress function into equation (16) we see that ${\mathrm{\nabla }}^4\phi =0$. Therefore, equation (18) represents a homogeneous solution of equation (16). The $\phi$ 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 $\phi =C_0{r}^4$ since the biharmonic equation must evaluate to a constant. Plugging this into equation (16) we have

$\text{(19)}{\mathrm{\nabla }}^4\phi =64C_0=2\rho {\dot{\theta }}^2(1-\nu )$

or,

$\text{(20)}{C}_0=\frac{\rho {\dot{\theta }}^2(1-\nu )}{32}$

Therefore, the general solution is

$\text{(21)}\phi =\frac{\rho {\dot{\theta }}^2(1-\nu )}{32}{r}^4+C_1{r}^2+C_2\ln (r)$

The corresponding stresses are (from equations (3, 4, 5)),

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

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

$\text{(25)}{C}_1=\frac{(3+\nu )\rho {\dot{\theta }}^2{a}^2}{16}$

Substituting back into equations (22) and (23), we get

$\text{(26)}{\sigma }_{rr}=\frac{\rho {\dot{\theta }}^2}{8}(3+\nu )(a^2-r^2)$

$\text{(27)}{\sigma }_{\theta \theta }=\frac{\rho {\dot{\theta }}^2}{8}[(3+\nu )a^2-(1+3\nu )r^2]$

$\text{(28)}{\sigma }_{r\theta }=0$