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 $\sigma =f(r)g(\theta )$.
• The stresses vary as $(1/r)$ (the area of action of the force decreases with increasing $r$). How about a conical wedge ?

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

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

Therefore, from Tables,

$\begin{array}{cc}{\sigma }_{rr}& =C_1\left(\frac{2\cos \theta }{r}\right)+C_2\left(\frac{\cos \theta }{r}\right)+C_3\left(\frac{2\sin \theta }{r}\right)+C_4\left(\frac{\sin \theta }{r}\right)\\ {\sigma }_{r\theta }& =C_2\left(\frac{\sin \theta }{r}\right)+C_4\left(\frac{-\cos \theta }{r}\right)\\ {\sigma }_{\theta \theta }& =C_2\left(\frac{\cos \theta }{r}\right)+C_4\left(\frac{\sin \theta }{r}\right)\end{array}$

From traction BCs, $C_2=C_4=0$. From equilibrium,

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

After algebra,

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

Special Case : α = -π, β = 0

$C_1=-\frac{F_1}{\pi };C_2=\frac{F_2}{\pi }$

The displacements are

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

where

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

and

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