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.

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{align}

\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{align}$$ The traction boundary conditions in terms of components of the stress tensor are
 * $$\begin{align}

\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{align}$$

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{align}

\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{align}$$

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{align}

0 & = \mu \left(A\sin\alpha + B\cos\alpha\right) \\ S & = \mu \left(-A\sin\alpha + B\cos\alpha\right) \end{align}$$ 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{align}

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{align}$$ 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) $$