Introduction to Elasticity/Antiplane shear example 1

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[edit] Example 1

Given:

The body − α < θ < α, 0 \le r < a is supported at r = a and loaded only by a uniform antiplane shear traction σθz = S on the surface θ = α, 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 σθz is uniform on the surface θ = α, from the expression for antiplane stress we can see that the displacement varies with r1 = r. The most general solution for the equilibrium equation for this behavior is u(r,θ) = Arcosθ + Brsinθ]

[edit] 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 σθz is to be uniform along θ = α, 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 uz that satisfies the above requirement is

uz(r,θ) = Arcosθ + Brsinθ + 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 σθθ and σrθ BCs being satisfied. From the boundary conditions on σθz, 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 ur and uθ are automatically satisfied by the antiplane strain assumption. We will try to satisfy the boundary conditions on uz 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 uz,

\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)
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