Introduction to Elasticity/Axially loaded wedge

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{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 ?

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