Elasticity/Polar coordinates

The Edge Dislocation Problem

File:Elastic edge dislocation.png 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 $r_o\to \mathrm{\infty }$. 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

 Template:Center top$\phi =f(r)\sin \theta$Template:Center bottom

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

 Template:Center top$\begin{array}{cc}{\sigma }_{rr}& =\frac{1}{r}\frac{\partial \phi }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2\phi }{\partial {\theta }^2}\\ {\sigma }_{\theta \theta }& =\frac{{\partial }^2\phi }{\partial r^2}\\ {\sigma }_{r\theta }& =-\frac{\partial }{\partial r}\left(\frac{1}{r}\frac{\partial \phi }{\partial \theta }\right)\end{array}$Template:Center bottom

Template:Center top${\mathrm{\nabla }}^4\phi ={\mathrm{\nabla }}^2({\mathrm{\nabla }}^2\phi )=(\frac{{\partial }^2}{\partial r^2}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2})(\frac{{\partial }^2\phi }{\partial r^2}+\frac{1}{r}\frac{\partial \phi }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2\phi }{\partial {\theta }^2})$Template:Center bottom and

 Template:Center top$\begin{array}{cc}2\mu u_r& =-\frac{\partial \phi }{\partial r}+\alpha r\frac{\partial \psi }{\partial \theta }\\ 2\mu u_{\theta }& =-\frac{1}{r}\frac{\partial \phi }{\partial \theta }+\alpha r^2\frac{\partial \psi }{\partial r}\end{array}$Template:Center bottom

Proceeding as usual, after plugging the value of $\phi$ in to the biharmonic equation, we get

 Template:Center top$f(r)=A{r}^3+\frac{B}{r}+Cr+Dr\ln r$Template:Center bottom

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

 Template:Center top${\sigma }_{rr}={\sigma }_{\theta \theta }=\frac{D}{r}\sin \theta ;{\sigma }_{r\theta }=-\frac{D}{r}\cos \theta$Template:Center bottom

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

$\phi =C{r}^2(\alpha +\theta -\sin \theta \cos \theta -{\cos }^2\theta \tan \alpha )$

• 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.

File:Triangle cantilever.png A cantilevered triangular beam with uniform normal traction

• Find the value of the constant $C$ in terms of $p$ and $\alpha$.

Solution:

Given:

$\phi =C{r}^2(\alpha +\theta -\sin \theta \cos \theta -{\cos }^2\theta \tan \alpha )$

Using a cylindrical co-ordinate system, the stresses are

$\begin{array}{cc}{\sigma }_{rr}& =2C(\alpha +\theta +\sin \theta \cos \theta -\tan \alpha +{\cos }^2\theta \tan \alpha )\\ {\sigma }_{r\theta }& =-2C+{\cos }^2\theta -\sin \theta \cos \theta \tan \alpha \\ {\sigma }_{\theta \theta }& =2C(\alpha +\theta -\sin \theta \cos \theta -{\cos }^2\theta \tan \alpha )\end{array}$

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

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

Hence, the shear traction BC is satisfied and the normal traction BC is satisfied if

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

At $\theta =-alpha$, $t_r=0$, $t_{\theta }=0$, $\widehat{𝐧}=-\widehat{𝐞}\theta$. 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.

$\phi =-\frac{p{r}^2}{2(\alpha -\tan \alpha )}(\alpha +\theta -\sin \theta \cos \theta -{\cos }^2\theta \tan \alpha )$

To satisfy compatibility, ${\mathrm{\nabla }}^4\varphi =0$. 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

$\begin{array}{c}{F}_1=\underset{-\alpha }{\overset{0}{\int }}({\sigma }_{rr}\cos \theta -{\sigma }_{r\theta }\sin \theta )Ld\theta =0\\ F_2=\underset{-\alpha }{\overset{0}{\int }}({\sigma }_{rr}\sin \theta +{\sigma }_{r\theta }\cos \theta )Ld\theta =-pL\\ M_3=\underset{-\alpha }{\overset{0}{\int }}L{\sigma }_{r\theta }Ld\theta =\frac{p{L}^2}{2}\end{array}$

You can verify these using Maple.

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