Introduction to Elasticity/Flamant solution

The Flamant Solution

 Elastic wedge loaded by two forces at the tip
• This problem is also self-similar (no inherent length scale).
• All quantities can be expressed in the separated-variable form ${\displaystyle \sigma =f(r)g(\theta )}$.
• The stresses vary as ${\displaystyle (1/r)}$ (the area of action of the force decreases with increasing ${\displaystyle r}$). How about a conical wedge ?

From Michell's solution, pick terms containing ${\displaystyle 1/r}$ in the stresses. Then,

${\displaystyle \varphi =C_{1}r\theta \sin \theta +C_{2}r\ln r\cos \theta +C_{3}r\theta \cos \theta +C_{4}r\ln r\sin \theta }$

Therefore, from Tables,

{\displaystyle {\begin{aligned}\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{aligned}}}

From traction BCs, ${\displaystyle C_{2}=C_{4}=0}$. From equilibrium,

{\displaystyle {\begin{aligned}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{aligned}}}

After algebra,

${\displaystyle \sigma _{rr}={\frac {2C_{1}\cos \theta }{r}}+{\frac {2C_{3}\sin \theta }{r}}~;~~\sigma _{r\theta }=0~;~~\sigma {\theta \theta }=0}$

Special Case : ${\displaystyle \alpha =-\pi }$, ${\displaystyle \beta =0}$

${\displaystyle C_{1}=-{\frac {F_{1}}{\pi }}~;~~C_{2}={\frac {F_{2}}{\pi }}}$

The displacements are

{\displaystyle {\begin{aligned}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{aligned}}}

where

{\displaystyle {\begin{aligned}\kappa =3-4\nu &&{\text{plane strain}}\\\kappa ={\frac {3-\nu }{1+\nu }}&&{\text{plane stress}}\\\end{aligned}}}

and

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