Elasticity/Concentrated force on half plane

Concentrated force on a half-plane
From the Flamant Solution


 * $$\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}$$ and

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

If $$\alpha = -\pi\,$$ and$$\beta = 0\,$$, we obtain the special case of a concentrated force acting on a half-plane. Then,
 * $$\begin{align}

F_1 + 2\int_{-\pi}^{0} \left(C_1\cos^2\theta - \frac{C_3}{2}\sin(2\theta)\right) d\theta & = 0 \\ F_2 + 2\int_{-\pi}^{0} \left(\frac{C_1}{2}\sin(2\theta) - C_3\sin^2\theta\right) d\theta & = 0 \end{align}$$ or,
 * $$\begin{align}

F_1 + \pi C_1 & = 0 \\ F_2 - \pi C_3 & = 0 \end{align}$$ Therefore,

C_1 = - \frac{F_1}{\pi} ~; C_3 = \frac{F_2}{\pi} $$ The stresses are

\sigma_{rr} = -\frac{2F_1\cos\theta}{\pi r} - \frac{2F_2\sin\theta}{\pi r} ~; \sigma_{r\theta} = \sigma_{\theta\theta} = 0 $$ The stress $$\sigma_{rr}\,$$ is obviously the superposition of the stresses due to $$F_1\,$$ and $$F_2\,$$, applied separately to the half-plane.

Problem 1 : Stresses and displacements due to F2
The tensile force $$F_2\,$$ produces the stress field

\sigma_{rr} =- \frac{2F_2\sin\theta}{\pi r} ~; \sigma_{r\theta} = \sigma_{\theta\theta} = 0 $$

The stress function is

\varphi = \frac{F_2}{\pi} r\theta\cos\theta $$

Hence, the displacements from Michell's solution are
 * $$\begin{align}

2\mu u_r & = \frac{F_2}{2\pi}\left[(\kappa-1)\theta\cos\theta + \sin\theta - (\kappa+1)\ln(r)\sin\theta\right] \\ 2\mu u_{\theta} & = \frac{F_2}{2\pi}\left[-(\kappa-1)\theta\sin\theta - \cos\theta - (\kappa+1)\ln(r)\cos\theta\right] \end{align}$$ At $$\theta = 0$$, ($$x_1 > 0$$, $$x_2 = 0$$),
 * $$\begin{align}

2\mu u_r = 2\mu u_1 & = 0 \\ 2\mu u_{\theta} = 2\mu u_2 & = \frac{F_2}{2\pi}\left[-1 - (\kappa+1)\ln(r)\right] \end{align}$$ At $$\theta = -\pi$$, ($$x_1 < 0$$, $$x_2 = 0$$),
 * $$\begin{align}

2\mu u_r = -2\mu u_1 & =\frac{F_2}{2\pi}(\kappa-1)\\ 2\mu u_{\theta} = -2\mu u_2 & = \frac{F_2}{2\pi}\left[1 + (\kappa+1)\ln(r)\right] \end{align}$$ where
 * $$\begin{align}

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

Since we expect the solution to be symmetric about $$x = 0\,$$, we superpose a rigid body displacement
 * $$\begin{align}

2\mu u_1 & = \frac{F_2}{4\pi}(\kappa-1)\\ 2\mu u_2 & = \frac{F_2}{2\pi} \end{align}$$

The displacements are
 * $$\begin{align}

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

\text{sign}(x) = \begin{cases} +1 & x > 0 \\ -1 & x < 0 \end{cases} $$ and $$r = |x|\,$$ on $$y = 0\,$$.

Problem 2 : Stresses and displacements due to F1
The tensile force $$F_1\,$$ produces the stress field

\sigma_{rr} =- \frac{2F_2\cos\theta}{\pi r} ~; \sigma_{r\theta} = \sigma_{\theta\theta} = 0 $$ The displacements are
 * $$\begin{align}

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

Stresses and displacements due to F1 + F2
Superpose the two solutions. The stresses are

\sigma_{rr} = -\frac{2F_1\cos\theta}{\pi r} - \frac{2F_2\sin\theta}{\pi r} ~; \sigma_{r\theta} = \sigma_{\theta\theta} = 0 $$ The displacements are
 * $$\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}$$