Elasticity/Transversely loaded wedge

Wedge loaded transversely by a concentrated load
Given:

A wedge of infinite length with a concentrated load $$\mathbf{P} = P~\widehat{\mathbf{e}}_{2}$$ per unit wedge thickness at the vertex. Plane stress/strain.

Find:

The stress field in the wedge.

Solution
From the Flamant solution, we know that the stress field in the wedge is
 * $$\begin{align}

\sigma_{rr} & = \frac{2}{r}\left(C_1\cos\theta+C_2\sin\theta\right) \\ \sigma_{r\theta} & = 0 \\ \sigma_{\theta\theta} & = 0 \end{align}$$ The constants $$C_1\,$$ and $$C_2\,$$ can be found by using the equilibrium conditions
 * $$\begin{align}

2\int_{-\beta}^{\beta} \left(C_1\cos\theta - C_2\sin\theta\right)\cos\theta ~d\theta & = 0 \\ P + 2\int_{-\beta}^{\beta} \left(C_1\cos\theta - C_2\sin\theta\right) \sin\theta ~d\theta & = 0 \end{align}$$ or,
 * $$\begin{align}

C_1 \left[2\beta + \sin(2\beta)\right] & = 0 \\ P + C_2\left[\sin(2\beta) - 2 \beta\right] & = 0 \end{align}$$ Therefore,

C_1 = 0 ~; C_2 = \frac{P}{2\beta - \sin(2\beta)} $$ Hence, the stresses are
 * $$\begin{align}

\sigma_{rr} & = \frac{2P\sin\theta}{r\left[2\beta-\sin(2\beta)\right]}\\ \sigma_{r\theta} & = 0 \\ \sigma_{\theta\theta} & = 0 \end{align}$$