Wedge loaded transversely by a concentrated load

Given:

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

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

${\displaystyle {\begin{aligned}\sigma _{rr}&={\frac {2}{r}}\left(C_{1}\cos \theta +C_{2}\sin \theta \right)\\\sigma _{r\theta }&=0\\\sigma _{\theta \theta }&=0\end{aligned}}}$

The constants ${\displaystyle C_{1}\,}$ and ${\displaystyle C_{2}\,}$ can be found by using the equilibrium conditions

${\displaystyle {\begin{aligned}2\int _{-\beta }^{\beta }\left(C_{1}\cos \theta -C_{2}\sin \theta \right)\cos \theta ~d\theta &=0\\P+2\int _{-\beta }^{\beta }\left(C_{1}\cos \theta -C_{2}\sin \theta \right)\sin \theta ~d\theta &=0\end{aligned}}}$

or,

${\displaystyle {\begin{aligned}C_{1}\left[2\beta +\sin(2\beta )\right]&=0\\P+C_{2}\left[\sin(2\beta )-2\beta \right]&=0\end{aligned}}}$

Therefore,

${\displaystyle C_{1}=0~;~~C_{2}={\frac {P}{2\beta -\sin(2\beta )}}}$

Hence, the stresses are

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