Elasticity/Transversely loaded wedge

Sample homework problems

Wedge loaded transversely by a concentrated load

Given:

A wedge of infinite length with a concentrated load $𝐏=P{\widehat{𝐞}}_2$ per unit wedge thickness at the vertex. Plane stress/strain.

File:Wedge with vertical force.png Wedge loaded transversely by a concentrated load

Find:

The stress field in the wedge.

Solution

From the Flamant solution, we know that the stress field in the wedge is

$\begin{array}{cc}{\sigma }_{rr}& =\frac{2}{r}(C_1\cos \theta +C_2\sin \theta )\\ {\sigma }_{r\theta }& =0\\ {\sigma }_{\theta \theta }& =0\end{array}$

The constants $C_1$ and $C_2$ can be found by using the equilibrium conditions

$\begin{array}{cc}2\underset{-\beta }{\overset{\beta }{\int }}(C_1\cos \theta -C_2\sin \theta )\cos \theta d\theta & =0\\ P+2\underset{-\beta }{\overset{\beta }{\int }}(C_1\cos \theta -C_2\sin \theta )\sin \theta d\theta & =0\end{array}$

or,

$\begin{array}{cc}{C}_1[2\beta +\sin (2\beta )]& =0\\ P+C_2[\sin (2\beta )-2\beta ]& =0\end{array}$

Therefore,

$C_1=0;C_2=\frac{P}{2\beta -\sin (2\beta )}$

Hence, the stresses are

$\begin{array}{cc}{\sigma }_{rr}& =\frac{2P\sin \theta }{r[2\beta -\sin (2\beta )]}\\ {\sigma }_{r\theta }& =0\\ {\sigma }_{\theta \theta }& =0\end{array}$