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

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.

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 $\alpha$.

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.

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