Introduction to Elasticity/Transversely loaded wedge

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[edit] Sample homework problems

[edit] Wedge loaded transversely by a concentrated load

Given:

A wedge of infinite length with a concentrated load \mathbf{P} = P~\widehat{\mathbf{e}}_{2} per unit wedge thickness at the vertex. Plane stress/strain.

Wedge loaded transversely by a concentrated load

Find:

The stress field in the wedge.

[edit] Solution

From the Flamant solution, we know that the stress field in the wedge is

\begin{align} \sigma_{rr} & = \frac{2}{r}\left(C_1\cos\theta+C_2\sin\theta\right) \\ \sigma_{r\theta} & = 0 \\ \sigma_{\theta\theta} & = 0 \end{align}

The constants C_1\, and C_2\, can be found by using the equilibrium conditions

\begin{align} 2\int_{-\beta}^{\beta} \left(C_1\cos\theta - C_2\sin\theta\right)\cos\theta ~d\theta & = 0 \\ P + 2\int_{-\beta}^{\beta} \left(C_1\cos\theta - C_2\sin\theta\right) \sin\theta ~d\theta & = 0 \end{align}

or,

\begin{align} C_1 \left[2\beta + \sin(2\beta)\right] & = 0 \\ P + C_2\left[\sin(2\beta) - 2 \beta\right] & = 0 \end{align}

Therefore,

C_1 = 0 ~;~~ C_2 = \frac{P}{2\beta - \sin(2\beta)}

Hence, the stresses are

\begin{align} \sigma_{rr} & = \frac{2P\sin\theta}{r\left[2\beta-\sin(2\beta)\right]}\\ \sigma_{r\theta} & = 0 \\ \sigma_{\theta\theta} & = 0 \end{align}
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