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 $$\begin{align} \sigma_{rr} & = \cfrac{1}{r}\cfrac{\partial\varphi}{\partial r} + \cfrac{1}{r^2}\cfrac{\partial^2\varphi}{\partial \theta^2} \\ \sigma_{\theta\theta} & = \cfrac{\partial^2\varphi}{\partial r^2} \\ \sigma_{r\theta} & = -\cfrac{\partial}{\partial r}                      \left(\cfrac{1}{r}\cfrac{\partial\varphi}{\partial \theta}\right) \end{align}$$ $$   \nabla^4\varphi = \nabla^2{(\nabla^2{\varphi})} = \left(\cfrac{\partial^2}{\partial r^2}+\cfrac{1}{r}\cfrac{\partial}{\partial r}+            \cfrac{1}{r^2}\cfrac{\partial^2}{\partial \theta^2}\right) \left(\cfrac{\partial^2\varphi}{\partial r^2}+\cfrac{1}{r}\cfrac{\partial\varphi}{\partial r}+            \cfrac{1}{r^2}\cfrac{\partial^2\varphi}{\partial \theta^2}\right) $$ 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.

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.
 * 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{align}

\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{align}$$ At $$\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{align}

0 & = 0 \\ -p & = 2C(\alpha - \tan\alpha) \end{align}$$ 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{align}

F_1 = \int_{-\alpha}^0 (\sigma_{rr}\cos\theta - \sigma_{r\theta}\sin\theta) L d\theta = 0 \\ F_2 = \int_{-\alpha}^0 (\sigma_{rr}\sin\theta + \sigma_{r\theta}\cos\theta) L d\theta = -pL\\ M_3 = \int_{-\alpha}^0 L \sigma_{r\theta} L d\theta = \frac{pL^2}{2} \end{align}$$ You can verify these using Maple.

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