Introduction to Elasticity/Polar coordinates

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[edit] The Edge Dislocation Problem

Stress due to an edge dislocation

Assume that stresses vanish at r = ri and that ri 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 σrr and σθθ are symmetric about the x2 − x3 plane. Similarly, it is probable that σrθ is symmetric about the x1 − x3 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 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 / r3, 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 ur = 0 at θ = 0 + and ur = b at θ = 2π − . 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 = ri and θ = 0 + . The final expressions for the displacements can then be obtained.

[edit] Sample homework problems

[edit] 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 α 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 α.

[edit] 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 θ = 0, tr = 0, tθ = − p, \widehat{\mathbf{n}}{} = \widehat{\mathbf{e}}{\theta}. Therefore, σθθ = − p and σrθ = 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 θ = − alpha, tr = 0, tθ = 0, \widehat{\mathbf{n}}{} = -\widehat{\mathbf{e}}{\theta}. Therefore, σθθ = 0 and σrθ = 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 tr = σrr and tθ = σrθ. 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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