# Elasticity/Antiplane shear example 1
## Example 1

Given:
The body $-\alpha <\theta <\alpha$, $0\leq r<a$ 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.
* 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
    ${\begin{aligned}{\text{at}}~r&=0~;~~u_{r}=0,u_{\theta }=0\\{\text{at}}~r&=a~;~~u_{r}=0,u_{\theta }=0,u_{z}=0\\{\text{at}}~\theta &=-\alpha ~;~~t_{\theta }=0,t_{r}=0,t_{z}=0\\{\text{at}}~\theta &=\alpha ~;~~t_{\theta }=0,t_{r}=0,t_{z}=S\end{aligned}}$
The traction boundary conditions in terms of components of the stress tensor are
    ${\begin{aligned}{\text{at}}~\theta &=-\alpha ~;~~\sigma _{\theta r}=0,\sigma _{\theta \theta }=0,\sigma _{\theta z}=0\\{\text{at}}~\theta &=\alpha ~;~~\sigma _{\theta r}=0,\sigma _{\theta \theta }=0,\sigma _{\theta z}=S\end{aligned}}$
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
    ${\begin{aligned}\sigma _{\theta z}&={\frac {\mu }{r}}{\frac {\partial u_{z}}{\partial \theta }}=\mu \left(-A\sin \theta +B\cos \theta \right)\\\sigma _{rz}&=\mu {\frac {\partial u_{z}}{\partial r}}=\mu \left(A\cos \theta +B\sin \theta \right)\end{aligned}}$
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
    ${\begin{aligned}0&=\mu \left(A\sin \alpha +B\cos \alpha \right)\\S&=\mu \left(-A\sin \alpha +B\cos \alpha \right)\end{aligned}}$
Solving,
    $A=-{\frac {S}{2\mu \sin \alpha }}~;~~B={\frac {S}{2\mu \cos \alpha }}$
This gives us the stress field
    $\sigma _{\theta z}={\frac {S}{2}}\left({\frac {\sin \theta }{\sin \alpha }}+{\frac {\cos \theta }{\cos \alpha }}\right)~;~~\sigma _{rz}={\frac {S}{2}}\left(-{\frac {\cos \theta }{\sin \alpha }}+{\frac {\sin \theta }{\cos \alpha }}\right)$
Step 5: Compute displacements
The displacement field is
    $u_{z}(r,\theta )={\frac {Sr}{2\mu }}\left(-{\frac {\cos \theta }{\sin \alpha }}+{\frac {\sin \theta }{\cos \alpha }}\right)+C$
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 $r=a$,
    $\int _{-\alpha }^{\alpha }u_{z}(a,\theta )d\theta =0~.$
This weak condition does not affect the stress field. Plugging in $u_{z}$,
    ${\begin{aligned}0&=\int _{-\alpha }^{\alpha }u_{z}(a,\theta )d\theta \\&={\frac {Sa}{2\mu }}\int _{-\alpha }^{\alpha }\left(-{\frac {\cos \theta }{\sin \alpha }}+{\frac {\sin \theta }{\cos \alpha }}+C{\frac {2\mu }{Sa}}\right)d\theta \\&={\frac {Sa}{2\mu }}\int _{-\alpha }^{\alpha }\left(-{\frac {\cos \theta }{\sin \alpha }}+{\frac {\sin \theta }{\cos \alpha }}+C{\frac {2\mu }{Sa}}\right)d\theta \\&={\frac {Sa}{2\mu }}\left[\left(-{\frac {\sin \theta }{\sin \alpha }}-{\frac {\cos \theta }{\cos \alpha }}+C\theta {\frac {2\mu }{Sa}}\right)\right]_{-\alpha }^{\alpha }\\&={\frac {Sa}{2\mu }}\left(-2{\frac {\sin \alpha }{\sin \alpha }}+2C\alpha {\frac {2\mu }{Sa}}\right)\\&=-{\frac {Sa}{\mu }}+C\alpha \end{aligned}}$
Therefore,
    $C={\frac {Sa}{2\mu \alpha }}$
The approximate displacement field is
    $u_{z}(r,\theta )={\frac {S}{2\mu }}\left(-r{\frac {\cos \theta }{\sin \alpha }}+r{\frac {\sin \theta }{\cos \alpha }}+a{\frac {1}{\alpha }}\right)$