Elasticity/Antiplane shear example 1

Example 1

Given:

The body $-\alpha <\theta <\alpha$, $0\le r<a$ is supported at $r=a$ and loaded only by a uniform antiplane shear traction ${\sigma }_{\theta z}=S$ on the surface $\theta =\alpha$, the other surface being traction-free.

File:Antiplane shear ex1.png A body loaded in antiplane shear

Find:

Find the complete stress field in the body, using strong boundary conditions on $\theta =\pm \alpha$ and weak conditions on $r=a$.

[Hint: Since the traction ${\sigma }_{\theta z}$ is uniform on the surface $\theta =\alpha$, from the expression for antiplane stress we can see that the displacement varies with $r^1=r$. The most general solution for the equilibrium equation for this behavior is $u(r,\theta )=Ar\cos \theta +Br\sin \theta$]

Solution

Step 1: Identify boundary conditions

$\begin{array}{cc}\text{at}r& =0;u_r=0,u_{\theta }=0\\ \text{at}r& =a;u_r=0,u_{\theta }=0,u_z=0\\ \text{at}\theta & =-\alpha ;t_{\theta }=0,t_r=0,t_z=0\\ \text{at}\theta & =\alpha ;t_{\theta }=0,t_r=0,t_z=S\end{array}$

The traction boundary conditions in terms of components of the stress tensor are

$\begin{array}{cc}\text{at}\theta & =-\alpha ;{\sigma }_{\theta r}=0,{\sigma }_{\theta \theta }=0,{\sigma }_{\theta z}=0\\ \text{at}\theta & =\alpha ;{\sigma }_{\theta r}=0,{\sigma }_{\theta \theta }=0,{\sigma }_{\theta z}=S\end{array}$

Step 2: Assume solution

Assume that the problem satisfies the conditions required for antiplane shear. If ${\sigma }_{\theta z}$ is to be uniform along $\theta =\alpha$, then

${\sigma }_{\theta z}=\frac{\mu }{r}\frac{\partial u_z}{\partial \theta }=C$

or,

$\frac{\partial u_z}{\partial \theta }=\frac{Cr}{\mu }$

The general form of $u_z$ that satisfies the above requirement is

$u_z(r,\theta )=Ar\cos \theta +Br\sin \theta +C$

where $A$, $B$, $C$ are constants.

Step 3: Compute stresses

The stresses are

$\begin{array}{cc}{\sigma }_{\theta z}& =\frac{\mu }{r}\frac{\partial u_z}{\partial \theta }=\mu (-A\sin \theta +B\cos \theta )\\ {\sigma }_{rz}& =\mu \frac{\partial u_z}{\partial r}=\mu (A\cos \theta +B\sin \theta )\end{array}$

Step 4: Check if traction BCs are satisfied

The antiplane strain assumption leads to the ${\sigma }_{\theta \theta }$ and ${\sigma }_{r\theta }$ BCs being satisfied. From the boundary conditions on ${\sigma }_{\theta z}$, we have

$\begin{array}{cc}0& =\mu (A\sin \alpha +B\cos \alpha )\\ S& =\mu (-A\sin \alpha +B\cos \alpha )\end{array}$

Solving,

$A=-\frac{S}{2\mu \sin \alpha };B=\frac{S}{2\mu \cos \alpha }$

This gives us the stress field

${\sigma }_{\theta z}=\frac{S}{2}(\frac{\sin \theta }{\sin \alpha }+\frac{\cos \theta }{\cos \alpha });{\sigma }_{rz}=\frac{S}{2}(-\frac{\cos \theta }{\sin \alpha }+\frac{\sin \theta }{\cos \alpha })$

Step 5: Compute displacements

The displacement field is

$u_z(r,\theta )=\frac{Sr}{2\mu }(-\frac{\cos \theta }{\sin \alpha }+\frac{\sin \theta }{\cos \alpha })+C$

where the constant $C$ corresponds to a superposed rigid body displacement.

Step 6: Check if displacement BCs are satisfied

The displacement BCs on $u_r$ and $u_{\theta }$ are automatically satisfied by the antiplane strain assumption. We will try to satisfy the boundary conditions on $u_z$ in a weak sense, i.e, at $r=a$,

$\underset{-\alpha }{\overset{\alpha }{\int }}{u}_z(a,\theta )d\theta =0.$

This weak condition does not affect the stress field. Plugging in $u_z$,

$\begin{array}{cc}0& =\underset{-\alpha }{\overset{\alpha }{\int }}{u}_z(a,\theta )d\theta \\ & =\frac{Sa}{2\mu }\underset{-\alpha }{\overset{\alpha }{\int }}(-\frac{\cos \theta }{\sin \alpha }+\frac{\sin \theta }{\cos \alpha }+C\frac{2\mu }{Sa})d\theta \\ & =\frac{Sa}{2\mu }\underset{-\alpha }{\overset{\alpha }{\int }}(-\frac{\cos \theta }{\sin \alpha }+\frac{\sin \theta }{\cos \alpha }+C\frac{2\mu }{Sa})d\theta \\ & =\frac{Sa}{2\mu }{\left[(-\frac{\sin \theta }{\sin \alpha }-\frac{\cos \theta }{\cos \alpha }+C\theta \frac{2\mu }{Sa})\right]}_{-\alpha }^{\alpha }\\ & =\frac{Sa}{2\mu }(-2\frac{\sin \alpha }{\sin \alpha }+2C\alpha \frac{2\mu }{Sa})\\ & =-\frac{Sa}{\mu }+C\alpha \end{array}$

Therefore,

$C=\frac{Sa}{2\mu \alpha }$

The approximate displacement field is

$u_z(r,\theta )=\frac{S}{2\mu }(-r\frac{\cos \theta }{\sin \alpha }+r\frac{\sin \theta }{\cos \alpha }+a\frac{1}{\alpha })$