Physics:Michell solution

Short description: Elasticity equation

In continuum mechanics, the Michell solution is a general solution to the elasticity equations in polar coordinates ($r,\theta$) developed by John Henry Michell in 1899. The solution is such that the stress components are in the form of a Fourier series in $\theta$.

Michell showed that the general solution can be expressed in terms of an Airy stress function of the form $$\begin{array}{rl}\phi (r,\theta )& =A_0{r}^2+B_0{r}^2\ln (r)+C_0\ln (r)\\ & +(I_0{r}^2+I_1{r}^2\ln (r)+I_2\ln (r)+I_3)\theta \\ & +(A_{1}r+B_1{r}^{-1}+{B_1}^{\prime }r\theta +C_1{r}^3+D_{1}r\ln (r))\cos \theta \\ & +(E_{1}r+F_1{r}^{-1}+{F_1}^{\prime }r\theta +G_1{r}^3+H_{1}r\ln (r))\sin \theta \\ & +\sum _{n=2}^{\mathrm{\infty }}(A_n{r}^n+B_n{r}^{-n}+C_n{r}^{n+2}+D_n{r}^{-n+2})\cos (n\theta )\\ & +\sum _{n=2}^{\mathrm{\infty }}(E_n{r}^n+F_n{r}^{-n}+G_n{r}^{n+2}+H_n{r}^{-n+2})\sin (n\theta )\end{array}$$ The terms $A_{1}r\cos \theta$ and $E_{1}r\sin \theta$ define a trivial null state of stress and are ignored.

Stress components

The stress components can be obtained by substituting the Michell solution into the equations for stress in terms of the Airy stress function (in cylindrical coordinates). A table of stress components is shown below.

$\phi$	${\sigma }_{rr}$	${\sigma }_{r\theta }$	${\sigma }_{\theta \theta }$
$r^2$	$2$	$0$	$2$
$r^2\ln r$	$2\ln r+1$	$0$	$2\ln r+3$
$\ln r$	$r^{-2}$	$0$	$-r^{-2}$
$\theta$	$0$	$r^{-2}$	$0$
$r^3\cos \theta$	$2r\cos \theta$	$2r\sin \theta$	$6r\cos \theta$
$r\theta \cos \theta$	$-2r^{-1}\sin \theta$	$0$	$0$
$r\ln r\cos \theta$	$r^{-1}\cos \theta$	$r^{-1}\sin \theta$	$r^{-1}\cos \theta$
$r^{-1}\cos \theta$	$-2r^{-3}\cos \theta$	$-2r^{-3}\sin \theta$	$2r^{-3}\cos \theta$
$r^3\sin \theta$	$2r\sin \theta$	$-2r\cos \theta$	$6r\sin \theta$
$r\theta \sin \theta$	$2r^{-1}\cos \theta$	$0$	$0$
$r\ln r\sin \theta$	$r^{-1}\sin \theta$	$-r^{-1}\cos \theta$	$r^{-1}\sin \theta$
$r^{-1}\sin \theta$	$-2r^{-3}\sin \theta$	$2r^{-3}\cos \theta$	$2r^{-3}\sin \theta$
$r^{n+2}\cos (n\theta )$	$-(n+1)(n-2)r^n\cos (n\theta )$	$n(n+1)r^n\sin (n\theta )$	$(n+1)(n+2)r^n\cos (n\theta )$
$r^{-n+2}\cos (n\theta )$	$-(n+2)(n-1)r^{-n}\cos (n\theta )$	$-n(n-1)r^{-n}\sin (n\theta )$	$(n-1)(n-2)r^{-n}\cos (n\theta )$
$r^n\cos (n\theta )$	$-n(n-1)r^{n-2}\cos (n\theta )$	$n(n-1)r^{n-2}\sin (n\theta )$	$n(n-1)r^{n-2}\cos (n\theta )$
$r^{-n}\cos (n\theta )$	$-n(n+1)r^{-n-2}\cos (n\theta )$	$-n(n+1)r^{-n-2}\sin (n\theta )$	$n(n+1)r^{-n-2}\cos (n\theta )$
$r^{n+2}\sin (n\theta )$	$-(n+1)(n-2)r^n\sin (n\theta )$	$-n(n+1)r^n\cos (n\theta )$	$(n+1)(n+2)r^n\sin (n\theta )$
$r^{-n+2}\sin (n\theta )$	$-(n+2)(n-1)r^{-n}\sin (n\theta )$	$n(n-1)r^{-n}\cos (n\theta )$	$(n-1)(n-2)r^{-n}\sin (n\theta )$
$r^n\sin (n\theta )$	$-n(n-1)r^{n-2}\sin (n\theta )$	$-n(n-1)r^{n-2}\cos (n\theta )$	$n(n-1)r^{n-2}\sin (n\theta )$
$r^{-n}\sin (n\theta )$	$-n(n+1)r^{-n-2}\sin (n\theta )$	$n(n+1)r^{-n-2}\cos (n\theta )$	$n(n+1)r^{-n-2}\sin (n\theta )$

Displacement components

Displacements $(u_r,u_{\theta })$ can be obtained from the Michell solution by using the stress-strain and strain-displacement relations. A table of displacement components corresponding the terms in the Airy stress function for the Michell solution is given below. In this table

$\kappa =\{\begin{array}{ll}3-4\nu & \mathrm{f}\mathrm{o}\mathrm{r}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{i}\mathrm{n}\\ \frac{3-\nu }{1+\nu }& \mathrm{f}\mathrm{o}\mathrm{r}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\\ \end{array}$

where $\nu$ is the Poisson's ratio, and $\mu$ is the shear modulus.

$\phi$	$2\mu u_r$	$2\mu u_{\theta }$
$r^2$	$(\kappa -1)r$	$0$
$r^2\ln r$	$(\kappa -1)r\ln r-r$	$(\kappa +1)r\theta$
$\ln r$	$-r^{-1}$	$0$
$\theta$	$0$	$-r^{-1}$
$r^3\cos \theta$	$(\kappa -2)r^2\cos \theta$	$(\kappa +2)r^2\sin \theta$
$r\theta \cos \theta$	$\frac{1}{2}[(\kappa -1)\theta \cos \theta +\{1-(\kappa +1)\ln r\}\sin \theta ]$	$-\frac{1}{2}[(\kappa -1)\theta \sin \theta +\{1+(\kappa +1)\ln r\}\cos \theta ]$
$r\ln r\cos \theta$	$\frac{1}{2}[(\kappa +1)\theta \sin \theta -\{1-(\kappa -1)\ln r\}\cos \theta ]$	$\frac{1}{2}[(\kappa +1)\theta \cos \theta -\{1+(\kappa -1)\ln r\}\sin \theta ]$
$r^{-1}\cos \theta$	$r^{-2}\cos \theta$	$r^{-2}\sin \theta$
$r^3\sin \theta$	$(\kappa -2)r^2\sin \theta$	$-(\kappa +2)r^2\cos \theta$
$r\theta \sin \theta$	$\frac{1}{2}[(\kappa -1)\theta \sin \theta -\{1-(\kappa +1)\ln r\}\cos \theta ]$	$\frac{1}{2}[(\kappa -1)\theta \cos \theta -\{1+(\kappa +1)\ln r\}\sin \theta ]$
$r\ln r\sin \theta$	$-\frac{1}{2}[(\kappa +1)\theta \cos \theta +\{1-(\kappa -1)\ln r\}\sin \theta ]$	$\frac{1}{2}[(\kappa +1)\theta \sin \theta +\{1+(\kappa -1)\ln r\}\cos \theta ]$
$r^{-1}\sin \theta$	$r^{-2}\sin \theta$	$-r^{-2}\cos \theta$
$r^{n+2}\cos (n\theta )$	$(\kappa -n-1)r^{n+1}\cos (n\theta )$	$(\kappa +n+1)r^{n+1}\sin (n\theta )$
$r^{-n+2}\cos (n\theta )$	$(\kappa +n-1)r^{-n+1}\cos (n\theta )$	$-(\kappa -n+1)r^{-n+1}\sin (n\theta )$
$r^n\cos (n\theta )$	$-n{r}^{n-1}\cos (n\theta )$	$n{r}^{n-1}\sin (n\theta )$
$r^{-n}\cos (n\theta )$	$n{r}^{-n-1}\cos (n\theta )$	$n{r}^{-n-1}\sin (n\theta )$
$r^{n+2}\sin (n\theta )$	$(\kappa -n-1)r^{n+1}\sin (n\theta )$	$-(\kappa +n+1)r^{n+1}\cos (n\theta )$
$r^{-n+2}\sin (n\theta )$	$(\kappa +n-1)r^{-n+1}\sin (n\theta )$	$(\kappa -n+1)r^{-n+1}\cos (n\theta )$
$r^n\sin (n\theta )$	$-n{r}^{n-1}\sin (n\theta )$	$-n{r}^{n-1}\cos (n\theta )$
$r^{-n}\sin (n\theta )$	$n{r}^{-n-1}\sin (n\theta )$	$-n{r}^{-n-1}\cos (n\theta )$

Note that a rigid body displacement can be superposed on the Michell solution of the form

$\begin{array}{rl}{u}_r& =A\cos \theta +B\sin \theta \\ u_{\theta }& =-A\sin \theta +B\cos \theta +Cr\\ \end{array}$

to obtain an admissible displacement field.

See also

• Linear elasticity
• Flamant solution
• John Henry Michell