Problem 1:

A thin disk of radius ${\displaystyle a}$ is spinning about its axis with a constant angular velocity ${\displaystyle {\dot {\theta }}}$. 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 (${\displaystyle r,\theta }$) on the disk is

${\displaystyle {\text{(1)}}\qquad a_{r}=-{\dot {\theta }}^{2}r~;~~a_{\theta }=0}$

The body force field is

${\displaystyle {\text{(2)}}\qquad f_{r}=\rho {\dot {\theta }}^{2}r~;~~f_{\theta }=0}$

Since there is no rotational acceleration, the body force can be derived from a potential ${\displaystyle V}$. The relations between the stresses, the Airy stress function and the body force potential are

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

where

${\displaystyle {\text{(6)}}\qquad f_{r}=-{\frac {\partial V}{\partial r}}~;~f_{\theta }=-{\frac {1}{r}}{\frac {\partial V}{\partial \theta }}}$

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

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

Integrating equation (7), we have

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

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

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

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

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

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

${\displaystyle {\text{(12)}}\qquad \nabla ^{4}{\varphi }+\left(2-{\frac {1}{\alpha }}\right)\nabla ^{2}{V}=0~~\Rightarrow ~~\nabla ^{4}{\varphi }+(1-\nu )\nabla ^{2}{V}=0}$

where

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

Now, from equations (11) and (13)

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

Therefore, equation (12) becomes

${\displaystyle {\text{(16)}}\qquad \nabla ^{4}{\varphi }=2\rho {\dot {\theta }}^{2}(1-\nu )}$

Since the problem is axisymmetric, there can be no shear stresses, i.e. ${\displaystyle \sigma _{r\theta }=0}$ and no dependence on ${\displaystyle \theta }$. From Michell's solution, the appropriate terms of the Airy stress function are

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

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

${\displaystyle {\text{(18)}}\qquad \varphi =C_{1}r^{2}+C_{2}\ln(r)}$

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

${\displaystyle {\text{(19)}}\qquad \nabla ^{4}{\varphi }=64C_{0}=2\rho {\dot {\theta }}^{2}(1-\nu )}$

or,

${\displaystyle {\text{(20)}}\qquad C_{0}={\frac {\rho {\dot {\theta }}^{2}(1-\nu )}{32}}}$

Therefore, the general solution is

${\displaystyle {\text{(21)}}\qquad \varphi ={\frac {\rho {\dot {\theta }}^{2}(1-\nu )}{32}}r^{4}+C_{1}r^{2}+C_{2}\ln(r)}$

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

${\displaystyle {\begin{aligned}{\text{(22)}}\qquad \sigma _{rr}&=-{\frac {(3+\nu ){\dot {\theta }}^{2}\rho r^{2}}{8}}+2C_{1}+{\frac {C_{2}}{r^{2}}}\\{\text{(23)}}\qquad \sigma _{\theta \theta }&=-{\frac {(1+3\nu ){\dot {\theta }}^{2}\rho r^{2}}{8}}+2C_{1}-{\frac {C_{2}}{r^{2}}}\\{\text{(24)}}\qquad \sigma _{r\theta }&=0\end{aligned}}}$

At ${\displaystyle r=0}$, the stresses must be finite. Hence, ${\displaystyle C_{2}=0}$. At ${\displaystyle r=a}$, ${\displaystyle \sigma _{rr}=\sigma _{r\theta }=0}$. Evaluating ${\displaystyle \sigma _{rr}}$ at ${\displaystyle r=a}$ we get

${\displaystyle {\text{(25)}}\qquad C_{1}={\frac {(3+\nu )\rho {\dot {\theta }}^{2}a^{2}}{16}}}$

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

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

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

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