# Introduction to Elasticity/Axially loaded wedge Elastic wedge loaded by an axial force

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

${\text{(30)}}\qquad t_{r}=t_{\theta }=0~;~~{\widehat {\mathbf {n} }}{}=\pm {\widehat {\mathbf {e} }}{\theta }\Rightarrow \sigma _{r\theta }=\sigma _{\theta \theta }=0$ ### What about the concentrated force BC?

• What is ${\widehat {\mathbf {n} }}{}$ 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

${\text{(31)}}\qquad {\widehat {\mathbf {n} }}{}={\widehat {\mathbf {e} }}{r}\Rightarrow \sigma _{r\theta }=t_{r}~;~~\sigma _{\theta \theta }=t_{\theta }$ For equilibrium, $\sum F_{1}=\sum F_{2}=\sum M_{3}=0$ . Therefore,

{\begin{aligned}P_{1}+\int _{-\beta }^{\beta }\left[\sigma _{rr}(a,\theta )\cos \theta -\sigma _{r\theta }(a,\theta )\sin \theta \right]a~d\theta =0{\text{(32)}}\qquad \\\int _{-\beta }^{\beta }\left[\sigma _{rr}(a,\theta )\sin \theta +\sigma _{r\theta }(a,\theta )\cos \theta \right]a~d\theta =0{\text{(33)}}\qquad \\\int _{-\beta }^{\beta }\left[a\sigma _{r\theta }(a,\theta )\right]a~d\theta =0{\text{(34)}}\qquad \end{aligned}} 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,

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

${\text{(36)}}\qquad \sigma _{\theta \theta }={\frac {\partial ^{2}}{\partial \varphi \partial r}}=\zeta ^{''}(r)$ That means $\sigma _{\theta \theta }$ is independent of $\theta$ . Therefore, in order to satisfy the BCs, $\sigma _{\theta \theta }=0$ , i.e.,

${\text{(37)}}\qquad \zeta (r)=C_{1}r+C_{2}\Rightarrow \varphi =r\eta (\theta )+C_{1}r=r[\eta (\theta )+C_{1}]=r\xi (\theta )$ Checking for compatibility, $\nabla ^{4}{\varphi }=0$ , we get

${\text{(38)}}\qquad \xi ^{(IV)}(\theta )+2\xi ^{''}(\theta )+\xi (\theta )=0$ The general solution is

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

${\text{(40)}}\qquad {\varphi =r\left[A\sin \theta +B\cos \theta +C\theta \sin \theta +D\theta \cos \theta \right]}$ The only non-zero stress is $\sigma _{rr}$ .

${\text{(41)}}\qquad \sigma _{rr}={\frac {1}{r}}\left[2C\cos \theta -2D\sin \theta \right]$ Plugging into equation (33), we get

${\text{(42)}}\qquad -D\left[2\beta -\sin(2\beta )\right]=0\Rightarrow D=0$ Hence,

${\text{(43)}}\qquad \sigma _{rr}={\frac {2C}{r}}\cos \theta$ Plugging into equation (32), we get

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

${\text{(45)}}\qquad {\varphi =Cr\theta \sin \theta ={\frac {-Pr\theta \sin \theta }{2\beta +\sin(2\beta )}}}$ The stress state is

${\text{(46)}}\qquad {\sigma _{rr}=-{\frac {2P\cos \theta }{r[2\beta +\sin(2\beta )]}}~;~~\sigma _{r\theta }=0~;~~\sigma _{\theta \theta }=0}$ ### Special Case : $\beta =\pi /2$ A concentrated point load acting on a half plane.

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

{\begin{aligned}2\mu u_{r}&=-{\frac {\partial }{\partial }}{\varphi }{r}+\alpha r{\frac {\partial }{\partial }}{\psi }{\theta }\\2\mu u_{\theta }&=-{\frac {1}{r}}{\frac {\partial }{\partial }}{\varphi }{\theta }+\alpha r^{2}{\frac {\partial }{\partial }}{\psi }{r}\end{aligned}} where

{\begin{aligned}\nabla ^{2}{\psi }=0\\{\frac {\partial }{\partial }}{}{r}\left(r{\frac {\partial }{\partial }}{\psi }{\theta }\right)=\nabla ^{2}{\varphi }\end{aligned}} Plug in $\varphi =Cr\theta \sin \theta$ ,

{\begin{aligned}&{\frac {\partial }{\partial }}{}{r}\left(r{\frac {\partial }{\partial }}{\psi }{\theta }\right)=\nabla ^{2}{\varphi }\\\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{aligned}} Plug $\psi$ into $\nabla ^{2}{\psi }=0$ ,

{\begin{aligned}&{\frac {1}{r^{3}}}\eta ^{''}(\theta )+{\frac {1}{r^{3}}}\eta (\theta )+\xi ^{''}(r)+{\frac {1}{r}}\xi ^{'}(r)-{\frac {4C}{r^{3}}}\sin \theta =0\\\Rightarrow &\eta ^{''}(\theta )+\eta (\theta )+r^{3}\xi ^{''}(r)+r^{2}\xi ^{'}(r)-4C\sin \theta =0\end{aligned}} Hence,

{\begin{aligned}\eta ^{''}(\theta )+\eta (\theta )-4C\sin \theta &=b\\r^{3}\xi ^{''}(r)+r^{2}\xi ^{'}(r)&=-{\frac {b}{r^{3}}}\end{aligned}} Solving,

{\begin{aligned}\eta (\theta )&=-2C\theta \cos \theta +d\cos \theta +e\sin \theta +b\\\xi ^{'}(r)&={\frac {f}{r}}+{\frac {b}{r^{2}}}\end{aligned}} Therefore,

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

{\begin{aligned}u_{r}&={\frac {\alpha C}{\mu }}\ln \left({\frac {r}{L}}\right)\\u_{\theta }&=0\end{aligned}} Is the small strain assumption satisfied ?