Elasticity/Axially loaded wedge

Axially Loaded Wedge
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{align}

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{align}$$ 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{-P r\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{align}

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{align}$$ where
 * $$\begin{align}

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

& \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{align}$$ Plug $$\psi$$ into $$\nabla^2{\psi} = 0$$,
 * $$\begin{align}

& \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{align}$$ Hence,
 * $$\begin{align}

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

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

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

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

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