# 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.
* An elastic disk spinning around its axis of symmetry

### 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
    ${\begin{aligned}{\text{(3)}}\qquad \sigma _{rr}&={\frac {1}{r}}{\frac {\partial \varphi }{\partial r}}+{\frac {1}{r^{2}}}{\frac {\partial ^{2}\varphi }{\partial \theta ^{2}}}+V\\{\text{(4)}}\qquad \sigma _{\theta \theta }&={\frac {\partial ^{2}\varphi }{\partial r^{2}}}+V\\{\text{(5)}}\qquad \sigma _{r\theta }&=-{\frac {\partial }{\partial r}}\left({\frac {1}{r}}{\frac {\partial \varphi }{\partial \theta }}\right)\end{aligned}}$
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,
    ${\begin{aligned}{\text{(7)}}\qquad \rho {\dot {\theta }}^{2}r&=-{\frac {\partial V}{\partial r}}\\{\text{(8)}}\qquad 0&=-{\frac {1}{r}}{\frac {\partial V}{\partial \theta }}\end{aligned}}$
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
    ${\begin{aligned}{\text{(13)}}\qquad \nabla ^{2}{}&={\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {1}{r}}{\frac {\partial}{\partial r}}+{\frac {1}{r^{2}}}{\frac {\partial ^{2}}{\partial \theta }}\\{\text{(14)}}\qquad \nabla ^{4}{}&=\nabla ^{2}{[\nabla ^{2}{}]}\end{aligned}}$
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
    ${\text{(18)}}\qquad \varphi =C_{1}r^{2}+C_{2}\ln(r)$
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
    ${\text{(19)}}\qquad \nabla ^{4}{\varphi }=64C_{0}=2\rho {\dot {\theta }}^{2}(1-\nu )$
or,
    ${\text{(20)}}\qquad C_{0}={\frac {\rho {\dot {\theta }}^{2}(1-\nu )}{32}}$
Therefore, the general solution is
    ${\text{(21)}}\qquad \varphi ={\frac {\rho {\dot {\theta }}^{2}(1-\nu )}{32}}r^{4}+C_{1}r^{2}+C_{2}\ln(r)$
The corresponding stresses are (from equations (3, 4, 5)),
    ${\begin{aligned}{\text{(22)}}\qquad \sigma _{rr}&=-{\frac {(3+\nu ){\dot {\theta }}^{2}\rho r^{2}}{8}}+2C_{1}+{\frac {C_{2}}{r^{2}}}\\{\text{(23)}}\qquad \sigma _{\theta \theta }&=-{\frac {(1+3\nu ){\dot {\theta }}^{2}\rho r^{2}}{8}}+2C_{1}-{\frac {C_{2}}{r^{2}}}\\{\text{(24)}}\qquad \sigma _{r\theta }&=0\end{aligned}}$
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
    ${\text{(26)}}\qquad \sigma _{rr}={\frac {\rho {\dot {\theta }}^{2}}{8}}(3+\nu )(a^{2}-r^{2})$
    ${\text{(27)}}\qquad \sigma _{\theta \theta }={\frac {\rho {\dot {\theta }}^{2}}{8}}\left[(3+\nu )a^{2}-(1+3\nu )r^{2}\right]$
    ${\text{(28)}}\qquad \sigma _{r\theta }=0$