# Introduction to Elasticity/Polar coordinates

## The Edge Dislocation Problem

Assume that stresses vanish at $r=r_{i}$ and that $r_{i}$ is the radius of an undeformed cylindrical hole. Also stresses vanish at $r_{o}\rightarrow \infty$ . Relative displacement $b$ is prescribed on each face of the cut.

The edge dislocation problem is a plane strain problem. However, it is not axisymmetric.

It is probable that $\sigma _{rr}$ and $\sigma _{\theta \theta }$ are symmetric about the $x_{2}-x_{3}$ plane. Similarly, it is probable that $\sigma _{r\theta }$ is symmetric about the $x_{1}-x_{3}$ plane.

These probable symmetries suggest that we can use a stress function of the form

$\varphi =f(r)\sin \theta$ In cylindrical co-ordinates, the gudir beta Airy stress function leads to

{\begin{aligned}\sigma _{rr}&={\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial r}}+{\cfrac {1}{r^{2}}}{\cfrac {\partial ^{2}\varphi }{\partial \theta ^{2}}}\\\sigma _{\theta \theta }&={\cfrac {\partial ^{2}\varphi }{\partial r^{2}}}\\\sigma _{r\theta }&=-{\cfrac {\partial }{\partial r}}\left({\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial \theta }}\right)\end{aligned}} $\nabla ^{4}\varphi =\nabla ^{2}{(\nabla ^{2}{\varphi })}=\left({\cfrac {\partial ^{2}}{\partial r^{2}}}+{\cfrac {1}{r}}{\cfrac {\partial }{\partial r}}+{\cfrac {1}{r^{2}}}{\cfrac {\partial ^{2}}{\partial \theta ^{2}}}\right)\left({\cfrac {\partial ^{2}\varphi }{\partial r^{2}}}+{\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial r}}+{\cfrac {1}{r^{2}}}{\cfrac {\partial ^{2}\varphi }{\partial \theta ^{2}}}\right)$ and

${\begin{matrix}2\mu u_{r}&=-{\cfrac {\partial \varphi }{\partial r}}+\alpha r{\cfrac {\partial \psi }{\partial \theta }}\\2\mu u_{\theta }&=-{\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial \theta }}+\alpha r^{2}{\cfrac {\partial \psi }{\partial r}}\end{matrix}}$ Proceeding as usual, after plugging the value of $\varphi$ in to the biharmonic equation, we get

$f(r)=Ar^{3}+{\cfrac {B}{r}}+Cr+Dr\ln r$ Applying the stress boundary conditions and neglecting terms containing $1/r^{3}$ , we get

$\sigma _{rr}=\sigma _{\theta \theta }={\cfrac {D}{r}}\sin \theta ~;~~\sigma _{r\theta }=-{\cfrac {D}{r}}\cos \theta$ Next we compute the displacements, in a manner similar to that shown for the cantilever beam problem. The displacement BCs are $u_{r}=0$ at $\theta =0+$ and $u_{r}=b$ at $\theta =2\pi -$ . We can use these to determine $D$ and hence the stresses.

Rigid body motions are eliminated next by enforcing zero displacements and rotations at $r=r_{i}$ and $\theta =0+$ . The final expressions for the displacements can then be obtained.

## Sample homework problems

### Problem 1

Consider the Airy stress function

$\varphi =C~r^{2}~(\alpha +\theta -\sin \theta \cos \theta -\cos ^{2}\theta \tan \alpha )$ • Show that this stress function provides an approximate solution for a cantilevered triangular beam with a uniform traction $p$ applied to the upper surface. The angle $\alpha$ is the angle subtended by the free edges of the triangle. A cantilevered triangular beam with uniform normal traction
• Find the value of the constant $C$ in terms of $p$ and $\alpha$ .

### Solution:

Given:

$\varphi =Cr^{2}(\alpha +\theta -\sin \theta \cos \theta -\cos ^{2}\theta \tan \alpha )$ Using a cylindrical co-ordinate system, the stresses are

{\begin{aligned}\sigma _{rr}&=2C\left(\alpha +\theta +\sin \theta \cos \theta -\tan \alpha +\cos ^{2}\theta \tan \alpha \right)\\\sigma _{r\theta }&=-2C+\cos ^{2}\theta -\sin \theta \cos \theta \tan \alpha \\\sigma _{\theta \theta }&=2C\left(\alpha +\theta -\sin \theta \cos \theta -\cos ^{2}\theta \tan \alpha \right)\end{aligned}} At $\theta =0$ , $t_{r}=0$ , $t_{\theta }=-p$ , ${\widehat {\mathbf {n} }}{}={\widehat {\mathbf {e} }}{\theta }$ . Therefore, $\sigma _{\theta \theta }=-p$ and $\sigma _{r\theta }=0$ .

{\begin{aligned}0&=0\\-p&=2C(\alpha -\tan \alpha )\end{aligned}} Hence, the shear traction BC is satisfied and the normal traction BC is satisfied if

${C=-{\frac {p}{2(\alpha -\tan \alpha )}}}$ At $\theta =-alpha$ , $t_{r}=0$ , $t_{\theta }=0$ , ${\widehat {\mathbf {n} }}{}=-{\widehat {\mathbf {e} }}{\theta }$ . Therefore, $\sigma _{\theta \theta }=0$ and $\sigma _{r\theta }=0$ . Both these BCs are identically satisfied by the stresses (after substituting for $C$ ). Hence, equilibrium is satisfied.

$\varphi =-{\frac {pr^{2}}{2(\alpha -\tan \alpha )}}(\alpha +\theta -\sin \theta \cos \theta -\cos ^{2}\theta \tan \alpha )$ To satisfy compatibility, $\nabla ^{4}{\phi }=0$ . Use Maple to verify that this is indeed true.

The remaining BC is the fixed displacement BC at the wall. We replace this BC with weak BCs at $r=L$ . The traction distribution on the surface $r=L$ are $t_{r}=\sigma _{rr}$ and $t_{\theta }=\sigma _{r\theta }$ . The statically equivalent forces and moments are

{\begin{aligned}F_{1}=\int _{-\alpha }^{0}(\sigma _{rr}\cos \theta -\sigma _{r\theta }\sin \theta )Ld\theta =0\\F_{2}=\int _{-\alpha }^{0}(\sigma _{rr}\sin \theta +\sigma _{r\theta }\cos \theta )Ld\theta =-pL\\M_{3}=\int _{-\alpha }^{0}L\sigma _{r\theta }Ld\theta ={\frac {pL^{2}}{2}}\end{aligned}} You can verify these using Maple.

Hence, the given stress function provides an approximate solution for the cantilevered beam (in the St. Venant sense).