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