Flamant Solution - Forces Acting On A Half-plane

Forces Acting On A Half-plane

For the special case where, the wedge is converted into a half-plane with a normal force and a tangential force. In that case

C_1 = - \frac{F_1}{\pi} ~;~~ C_3 = -\frac{F_2}{\pi}

Therefore the stresses are

\begin{align} \sigma_{rr} & = -\frac{2}{\pi~r} (F_1~\cos\theta + F_2~\sin\theta) \\ \sigma_{r\theta} & = 0 \\ \sigma_{\theta\theta} & = 0 \end{align}

and the displacements are (using Michell's solution)

\begin{align} u_r & = -\cfrac{1}{4\pi\mu}\left[F_1\{(\kappa-1)\theta\sin\theta - \cos\theta + (\kappa+1)\ln r\cos\theta\} + \right. \\ & \qquad \qquad \left. F_2\{(\kappa-1)\theta\cos\theta + \sin\theta - (\kappa+1)\ln r\sin\theta\}\right]\\ u_\theta & = -\cfrac{1}{4\pi\mu}\left[F_1\{(\kappa-1)\theta\cos\theta - \sin\theta - (\kappa+1)\ln r\sin\theta\} - \right. \\ & \qquad \qquad \left. F_2\{(\kappa-1)\theta\sin\theta + \cos\theta + (\kappa+1)\ln r\cos\theta\}\right] \end{align}

The dependence of the displacements implies that the displacement grows the further one moves from the point of application of the force (and is unbounded at infinity). This feature of the Flamant solution is confusing and appears unphysical. For a discussion of the issue see http://imechanica.org/node/319.

Read more about this topic:  Flamant Solution

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