Elasticity/Axially loaded wedge

Axially Loaded Wedge

Elastic wedge loaded by an axial force

The BCs at $\theta =\pm \beta$ are

$\text{(30)}{t}_r=t_{\theta }=0;\widehat{𝐧}=\pm \widehat{𝐞}\theta \Rightarrow {\sigma }_{r\theta }={\sigma }_{\theta \theta }=0$

What about the concentrated force BC?

• What is $\widehat{𝐧}$ at the vertex ?
• The traction is infinite since the force is applied on zero area. Consider equilibrium of a portion of the wedge.

At $r=a$, the BCs are

$\text{(31)}\widehat{𝐧}=\widehat{𝐞}r\Rightarrow {\sigma }_{r\theta }=t_r;{\sigma }_{\theta \theta }=t_{\theta }$

For equilibrium, $\sum F_1=\sum F_2=\sum M_3=0$. Therefore,

$\begin{array}{c}{P}_1+{\int }_{-\beta }^{\beta }[{\sigma }_{rr}(a,\theta )\cos \theta -{\sigma }_{r\theta }(a,\theta )\sin \theta ]ad\theta =0\text{(32)}\\ {\int }_{-\beta }^{\beta }[{\sigma }_{rr}(a,\theta )\sin \theta +{\sigma }_{r\theta }(a,\theta )\cos \theta ]ad\theta =0\text{(33)}\\ {\int }_{-\beta }^{\beta }[a{\sigma }_{r\theta }(a,\theta )]ad\theta =0\text{(34)}\end{array}$

These constraint conditions are equivalent to the concentrated force BC.

Solution Procedure

Assume that ${\sigma }_{r\theta }(r,\theta )=0$. This satisfies the traction BCs on $\theta =\pm \beta$ and equation (34). Therefore,

$\text{(35)}{\sigma }_{r\theta }=\frac{\partial }{\partial }r\left(\frac{1}{r}\frac{\partial }{\partial }\phi \theta \right)=0\Rightarrow \phi =r\eta (\theta )+\zeta (r)$

Hence,

$\text{(36)}{\sigma }_{\theta \theta }=\frac{{\partial }^2}{\partial \phi \partial r}={\zeta }^{{\text{'}}^{\prime }}(r)$

That means ${\sigma }_{\theta \theta }$ is independent of $\theta$. Therefore, in order to satisfy the BCs, ${\sigma }_{\theta \theta }=0$, i.e.,

$\text{(37)}\zeta (r)=C_{1}r+C_2\Rightarrow \phi =r\eta (\theta )+C_{1}r=r[\eta (\theta )+C_1]=r\xi (\theta )$

Checking for compatibility, ${\mathrm{\nabla }}^4\phi =0$, we get

$\text{(38)}{\xi }^{(IV)}(\theta )+2{\xi }^{{\text{'}}^{\prime }}(\theta )+\xi (\theta )=0$

The general solution is

$\text{(39)}\xi (\theta )=A\sin \theta +B\cos \theta +C\theta \sin \theta +D\theta \cos \theta$

Therefore,

$\text{(40)}\phi =r[A\sin \theta +B\cos \theta +C\theta \sin \theta +D\theta \cos \theta ]$

The only non-zero stress is ${\sigma }_{rr}$.

$\text{(41)}{\sigma }_{rr}=\frac{1}{r}[2C\cos \theta -2D\sin \theta ]$

Plugging into equation (33), we get

$\text{(42)}-D[2\beta -\sin (2\beta )]=0\Rightarrow D=0$

Hence,

$\text{(43)}{\sigma }_{rr}=\frac{2C}{r}\cos \theta$

Plugging into equation (32), we get

$\text{(44)}-P=C[2\beta +\sin (2\beta )]\Rightarrow C=\frac{-P}{2\beta +\sin (2\beta )}$

Therefore,

$\text{(45)}\phi =Cr\theta \sin \theta =\frac{-Pr\theta \sin \theta }{2\beta +\sin (2\beta )}$

The stress state is

$\text{(46)}{\sigma }_{rr}=-\frac{2P\cos \theta }{r[2\beta +\sin (2\beta )]};{\sigma }_{r\theta }=0;{\sigma }_{\theta \theta }=0$

Special Case

$\beta =\pi /2$

A concentrated point load acting on a half plane.

$\text{(47)}{\sigma }_{rr}=-\frac{2P\cos \theta }{\pi r};{\sigma }_{r\theta }=0;{\sigma }_{\theta \theta }=0$

Displacements

$\begin{array}{cc}2\mu u_r& =-\frac{\partial }{\partial }\phi r+\alpha r\frac{\partial }{\partial }\psi \theta \\ 2\mu u_{\theta }& =-\frac{1}{r}\frac{\partial }{\partial }\phi \theta +\alpha r^2\frac{\partial }{\partial }\psi r\end{array}$

where

$\begin{array}{c}{\mathrm{\nabla }}^2\psi =0\\ \frac{\partial }{\partial }r\left(r\frac{\partial }{\partial }\psi \theta \right)={\mathrm{\nabla }}^2\phi \end{array}$

Plug in $\phi =Cr\theta \sin \theta$,

$\begin{array}{cc}& \frac{\partial }{\partial }r\left(r\frac{\partial }{\partial }\psi \theta \right)={\mathrm{\nabla }}^2\phi \\ \Rightarrow & \frac{\partial }{\partial }r\left(r\frac{\partial }{\partial }\psi \theta \right)=\frac{2C}{r}\cos \theta \\ \Rightarrow & r\frac{\partial }{\partial }\psi \theta =2C\ln r\cos \theta +A(\theta )\\ \Rightarrow & \frac{\partial }{\partial }\psi \theta =2C\frac{\ln r}{r}\cos \theta +\frac{A(\theta )}{r}\\ \Rightarrow & \psi =2C\frac{\ln r}{r}\sin \theta +\frac{\eta (\theta )}{r}+\xi r\end{array}$

Plug $\psi$ into ${\mathrm{\nabla }}^2\psi =0$,

$\begin{array}{cc}& \frac{1}{r^3}{\eta }^{{\text{'}}^{\prime }}(\theta )+\frac{1}{r^3}\eta (\theta )+{\xi }^{{\text{'}}^{\prime }}(r)+\frac{1}{r}{\xi }^{\text{'}}(r)-\frac{4C}{r^3}\sin \theta =0\\ \Rightarrow & {\eta }^{{\text{'}}^{\prime }}(\theta )+\eta (\theta )+r^3{\xi }^{{\text{'}}^{\prime }}(r)+r^2{\xi }^{\text{'}}(r)-4C\sin \theta =0\end{array}$

Hence,

$\begin{array}{cc}{\eta }^{{\text{'}}^{\prime }}(\theta )+\eta (\theta )-4C\sin \theta & =b\\ r^3{\xi }^{{\text{'}}^{\prime }}(r)+r^2{\xi }^{\text{'}}(r)& =-\frac{b}{r^3}\end{array}$

Solving,

$\begin{array}{cc}\eta (\theta )& =-2C\theta \cos \theta +d\cos \theta +e\sin \theta +b\\ {\xi }^{\text{'}}(r)& =\frac{f}{r}+\frac{b}{r^2}\end{array}$

Therefore,

$\begin{array}{cc}2\mu u_r& =2\alpha C\ln r\cos \theta +(2\alpha -1)C\theta \sin \theta +\alpha (e-2C)\cos \theta -\alpha d\sin \theta \\ 2\mu u_{\theta }& =-2\alpha C\ln r\sin \theta +(2\alpha -1)C\sin \theta +(2\alpha -1)C\theta \cos \theta -\alpha d\cos \theta -\alpha e\sin \theta +\alpha fr\end{array}$

To fix the rigid body motion, we set $u_{\theta }=0$ when $\theta =0$, and set $u_r=0$ when $\theta =0$ and $r=L$.Then,

$\begin{array}{cc}{u}_r& =\frac{\alpha C}{\mu }\ln \left(\frac{r}{L}\right)\cos \theta +\frac{(2\alpha -1)C}{2\mu }\theta \sin \theta \\ u_{\theta }& =-\frac{\alpha C}{\mu }\ln \left(\frac{r}{L}\right)\sin \theta +\frac{(2\alpha -1)C}{2\mu }\theta \cos \theta -\frac{C}{2\mu }\sin \theta \end{array}$

The displacements are singular at $r=0$ and $r=\mathrm{\infty }$. At $\theta =0$,

$\begin{array}{cc}{u}_r& =\frac{\alpha C}{\mu }\ln \left(\frac{r}{L}\right)\\ u_{\theta }& =0\end{array}$

Is the small strain assumption satisfied ?

Elasticity