# Elasticity/Polar coordinates
## The Edge Dislocation Problem

* Stress due to an edge dislocation
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).