Introduction to Elasticity/Polar coordinates

The Edge Dislocation Problem

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 $r_o\rightarrow\infty$ . 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

\varphi = f(r) \sin\theta

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

\begin{align} \sigma_{rr} & = \cfrac{1}{r}\cfrac{\partial\varphi}{\partial r} + \cfrac{1}{r^2}\cfrac{\partial^2\varphi}{\partial \theta^2} \\ \sigma_{\theta\theta} & = \cfrac{\partial^2\varphi}{\partial r^2} \\ \sigma_{r\theta} & = -\cfrac{\partial}{\partial r} \left(\cfrac{1}{r}\cfrac{\partial\varphi}{\partial \theta}\right) \end{align}

\nabla^4\varphi = \nabla^2{(\nabla^2{\varphi})} = \left(\cfrac{\partial^2}{\partial r^2}+\cfrac{1}{r}\cfrac{\partial}{\partial r}+ \cfrac{1}{r^2}\cfrac{\partial^2}{\partial \theta^2}\right) \left(\cfrac{\partial^2\varphi}{\partial r^2}+\cfrac{1}{r}\cfrac{\partial\varphi}{\partial r}+ \cfrac{1}{r^2}\cfrac{\partial^2\varphi}{\partial \theta^2}\right)

and

\begin{matrix} 2\mu u_r & = -\cfrac{\partial\varphi}{\partial r} + \alpha r \cfrac{\partial\psi}{\partial \theta} \\ 2\mu u_{\theta} & = -\cfrac{1}{r}\cfrac{\partial\varphi}{\partial \theta} + \alpha r^2 \cfrac{\partial\psi}{\partial r} \end{matrix}

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

f(r) = Ar^3 + \cfrac{B}{r} + Cr + Dr \ln r

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

\sigma_{rr} = \sigma_{\theta\theta} = \cfrac{D}{r} \sin\theta ~;~~ \sigma_{r\theta} = -\cfrac{D}{r} \cos\theta

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

\varphi = 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.

A cantilevered triangular beam with uniform normal traction

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

Solution:

Given:

\varphi = Cr^2(\alpha + \theta - \sin\theta\cos\theta - \cos^2\theta\tan\alpha)

Using a cylindrical co-ordinate system, the stresses are

\begin{align} \sigma_{rr} & = 2C\left(\alpha + \theta + \sin\theta\cos\theta - \tan\alpha + \cos^2\theta\tan\alpha\right) \\ \sigma_{r\theta} & = -2C + \cos^2\theta - \sin\theta\cos\theta\tan\alpha \\ \sigma_{\theta\theta} & = 2C\left(\alpha + \theta - \sin\theta\cos\theta - \cos^2\theta\tan\alpha\right) \end{align}

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

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

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

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

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

\varphi = -\frac{pr^2}{2(\alpha - \tan\alpha)} (\alpha + \theta - \sin\theta\cos\theta - \cos^2\theta\tan\alpha)

To satisfy compatibility, $\nabla^4{\phi} = 0$ . 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{align} F_1 = \int_{-\alpha}^0 (\sigma_{rr}\cos\theta - \sigma_{r\theta}\sin\theta) L d\theta = 0 \\ F_2 = \int_{-\alpha}^0 (\sigma_{rr}\sin\theta + \sigma_{r\theta}\cos\theta) L d\theta = -pL\\ M_3 = \int_{-\alpha}^0 L \sigma_{r\theta} L d\theta = \frac{pL^2}{2} \end{align}

You can verify these using Maple.

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

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