Elasticity/Flamant solution

The Flamant Solution
From Michell's solution, pick terms containing $$1/r$$ in the stresses. Then,
 * This problem is also self-similar (no inherent length scale).
 * All quantities can be expressed in the separated-variable form  $$\sigma = f(r)g(\theta)$$.
 * The stresses vary as $$(1/r)$$ (the area of action of the force  decreases with increasing $$r$$).  How about a conical wedge ?

\varphi = C_1 r \theta\sin\theta + C_2 r\ln r \cos\theta + C_3 r \theta\cos\theta + C_4 r\ln r \sin\theta $$ Therefore, from Tables,
 * $$\begin{align}

\sigma_{rr} & = C_1\left(\frac{2\cos\theta}{r}\right) + C_2\left(\frac{\cos\theta}{r}\right) + C_3\left(\frac{2\sin\theta}{r}\right) + C_4\left(\frac{\sin\theta}{r}\right) \\ \sigma_{r\theta} & = C_2\left(\frac{\sin\theta}{r}\right) + C_4\left(\frac{-\cos\theta}{r}\right) \\ \sigma_{\theta\theta} & = C_2\left(\frac{\cos\theta}{r}\right) + C_4\left(\frac{\sin\theta}{r}\right) \end{align}$$ From traction BCs, $$C_2 = C_4 = 0$$. From equilibrium,
 * $$\begin{align}

F_1 + 2\int_{\alpha}^{\beta} \left(\frac{C_1\cos\theta - C_3\sin\theta}{a}\right)a\cos\theta d\theta & = 0 \\ F_2 + 2\int_{\alpha}^{\beta} \left(\frac{C_1\cos\theta - C_3\sin\theta}{a}\right)a\sin\theta d\theta & = 0 \end{align}$$ After algebra,

\sigma_{rr} = \frac{2C_1\cos\theta}{r} + \frac{2C_3\sin\theta}{r} ~; \sigma_{r\theta} = 0 ~; \sigma{\theta\theta} = 0 $$

Special Case : α = -π, β = 0


C_1 = - \frac{F_1}{\pi} ~; C_2 = \frac{F_2}{\pi} $$ The displacements are
 * $$\begin{align}

u_1 & = - \frac{F_1(\kappa+1)\ln|x_1|}{4\pi\mu} + \frac{F_2(\kappa+1)\text{sign}(x_1)}{8\mu} \\ u_2 & = - \frac{F_2(\kappa+1)\ln|x_1|}{4\pi\mu} - \frac{F_1(\kappa+1)\text{sign}(x_1)}{8\mu} \end{align}$$ where
 * $$\begin{align}

\kappa = 3 - 4\nu & & \text{plane strain} \\ \kappa = \frac{3 - \nu}{1+\nu} & & \text{plane stress} \\ \end{align}$$ and

\text{sign}(x) = \begin{cases} +1 & x > 0 \\ -1 & x < 0 \end{cases} $$