# Introduction to Elasticity/Polar coordinates

## The Edge Dislocation Problem

 Stress due to an edge dislocation

Assume that stresses vanish at ${\displaystyle r=r_{i}}$ and that ${\displaystyle r_{i}}$ is the radius of an undeformed cylindrical hole. Also stresses vanish at ${\displaystyle r_{o}\rightarrow \infty }$. Relative displacement ${\displaystyle b}$ 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 ${\displaystyle \sigma _{rr}}$ and ${\displaystyle \sigma _{\theta \theta }}$ are symmetric about the ${\displaystyle x_{2}-x_{3}}$ plane. Similarly, it is probable that ${\displaystyle \sigma _{r\theta }}$ is symmetric about the ${\displaystyle x_{1}-x_{3}}$ plane.

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

${\displaystyle \varphi =f(r)\sin \theta }$

In cylindrical co-ordinates, the gudir beta Airy stress function leads to

{\displaystyle {\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}}}
${\displaystyle \nabla ^{4}\varphi =\nabla ^{2}{(\nabla ^{2}{\varphi })}=\left({\cfrac {\partial ^{2}}{\partial r^{2}}}+{\cfrac {1}{r}}{\cfrac {\partial }{\partial r}}+{\cfrac {1}{r^{2}}}{\cfrac {\partial ^{2}}{\partial \theta ^{2}}}\right)\left({\cfrac {\partial ^{2}\varphi }{\partial r^{2}}}+{\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial r}}+{\cfrac {1}{r^{2}}}{\cfrac {\partial ^{2}\varphi }{\partial \theta ^{2}}}\right)}$

and

${\displaystyle {\begin{matrix}2\mu u_{r}&=-{\cfrac {\partial \varphi }{\partial r}}+\alpha r{\cfrac {\partial \psi }{\partial \theta }}\\2\mu u_{\theta }&=-{\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial \theta }}+\alpha r^{2}{\cfrac {\partial \psi }{\partial r}}\end{matrix}}}$

Proceeding as usual, after plugging the value of ${\displaystyle \varphi }$ in to the biharmonic equation, we get

${\displaystyle f(r)=Ar^{3}+{\cfrac {B}{r}}+Cr+Dr\ln r}$

Applying the stress boundary conditions and neglecting terms containing ${\displaystyle 1/r^{3}}$, we get

${\displaystyle \sigma _{rr}=\sigma _{\theta \theta }={\cfrac {D}{r}}\sin \theta ~;~~\sigma _{r\theta }=-{\cfrac {D}{r}}\cos \theta }$

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

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

## Sample homework problems

### Problem 1

Consider the Airy stress function

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

### Solution:

Given:

${\displaystyle \varphi =Cr^{2}(\alpha +\theta -\sin \theta \cos \theta -\cos ^{2}\theta \tan \alpha )}$

Using a cylindrical co-ordinate system, the stresses are

{\displaystyle {\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 ${\displaystyle \theta =0}$, ${\displaystyle t_{r}=0}$, ${\displaystyle t_{\theta }=-p}$, ${\displaystyle {\widehat {\mathbf {n} }}{}={\widehat {\mathbf {e} }}{\theta }}$. Therefore, ${\displaystyle \sigma _{\theta \theta }=-p}$ and ${\displaystyle \sigma _{r\theta }=0}$.

{\displaystyle {\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

${\displaystyle {C=-{\frac {p}{2(\alpha -\tan \alpha )}}}}$

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

${\displaystyle \varphi =-{\frac {pr^{2}}{2(\alpha -\tan \alpha )}}(\alpha +\theta -\sin \theta \cos \theta -\cos ^{2}\theta \tan \alpha )}$

To satisfy compatibility, ${\displaystyle \nabla ^{4}{\phi }=0}$. 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 ${\displaystyle r=L}$. The traction distribution on the surface ${\displaystyle r=L}$ are ${\displaystyle t_{r}=\sigma _{rr}}$ and ${\displaystyle t_{\theta }=\sigma _{r\theta }}$. The statically equivalent forces and moments are

{\displaystyle {\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).