Elasticity/Concentrated force on half plane

Concentrated force on a half-plane

File:Concentrated force on half plane.png Concentrated force on a half plane

From the Flamant Solution

$\begin{array}{cc}{F}_1+2\underset{\alpha }{\overset{\beta }{\int }}\left(\frac{C_1\cos \theta -C_3\sin \theta }{a}\right)a\cos \theta d\theta & =0\\ F_2+2\underset{\alpha }{\overset{\beta }{\int }}\left(\frac{C_1\cos \theta -C_3\sin \theta }{a}\right)a\sin \theta d\theta & =0\end{array}$

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,

$\begin{array}{cc}{F}_1+2\underset{-\pi }{\overset{0}{\int }}(C_1{\cos }^2\theta -\frac{C_3}{2}\sin (2\theta ))d\theta & =0\\ F_2+2\underset{-\pi }{\overset{0}{\int }}(\frac{C_1}{2}\sin (2\theta )-C_3{\sin }^2\theta )d\theta & =0\end{array}$

or,

$\begin{array}{cc}{F}_1+\pi C_1& =0\\ F_2-\pi C_3& =0\end{array}$

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

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

File:Stress conc force half plane.png Stress due to concentrated force $F_2$ on a half plane

The stress function is

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

Hence, the displacements from Michell's solution are

$\begin{array}{cc}2\mu u_r& =\frac{F_2}{2\pi }[(\kappa -1)\theta \cos \theta +\sin \theta -(\kappa +1)\ln (r)\sin \theta ]\\ 2\mu u_{\theta }& =\frac{F_2}{2\pi }[-(\kappa -1)\theta \sin \theta -\cos \theta -(\kappa +1)\ln (r)\cos \theta ]\end{array}$

At $\theta =0$, ($x_1>0$, $x_2=0$),

$\begin{array}{cc}2\mu u_r=2\mu u_1& =0\\ 2\mu u_{\theta }=2\mu u_2& =\frac{F_2}{2\pi }[-1-(\kappa +1)\ln (r)]\end{array}$

At $\theta =-\pi$, ($x_1<0$, $x_2=0$),

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

where

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

Since we expect the solution to be symmetric about $x=0$, we superpose a rigid body displacement

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

The displacements are

$\begin{array}{cc}{u}_1& =\frac{F_2(\kappa -1)\text{sign}(x_1)}{8\mu }\\ u_2& =-\frac{F_2(\kappa +1)\ln |x_1|}{4\pi \mu }\end{array}$

where

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

and $r=|x|$ on $y=0$.

Problem 2 : Stresses and displacements due to F₁

The tensile force $F_1$ produces the stress field

${\sigma }_{rr}=-\frac{2F_2\cos \theta }{\pi r};{\sigma }_{r\theta }={\sigma }_{\theta \theta }=0$

File:Stress1 conc force half plane.png Stress due to concentrated force $F_1$ on a half plane

The displacements are

$\begin{array}{cc}{u}_1& =-\frac{F_1(\kappa +1)\ln |x_1|}{4\pi \mu }\\ u_2& =-\frac{F_1(\kappa -1)\text{sign}(x_1)}{8\mu }\end{array}$

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

$\begin{array}{cc}{u}_1& =-\frac{F_1(\kappa +1)\ln |x_1|}{4\pi \mu }+\frac{F_2(\kappa -1)\text{sign}(x_1)}{8\mu }\\ u_2& =-\frac{F_2(\kappa +1)\ln |x_1|}{4\pi \mu }-\frac{F_1(\kappa -1)\text{sign}(x_1)}{8\mu }\end{array}$

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