Introduction to Elasticity/Flamant solution

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[edit] The Flamant Solution

Elastic wedge loaded by two forces at the tip
  • This problem is also self-similar (no inherent length scale).
  • All quantities can be expressed in the separated-variable form σ = f(r)g(θ).
  • The stresses vary as (1 / r) (the area of action of the force decreases with increasing r). How about a conical wedge ?

From Michell's solution, pick terms containing 1 / r in the stresses. Then,

\varphi = C_1 r \theta\sin\theta + C_2 r\ln r \cos\theta + C_3 r \theta\cos\theta + C_4 r\ln r \sin\theta

Therefore, from Tables,

\begin{align} \sigma_{rr} & = C_1\left(\frac{2\cos\theta}{r}\right) + C_2\left(\frac{\cos\theta}{r}\right) + C_3\left(\frac{2\sin\theta}{r}\right) + C_4\left(\frac{\sin\theta}{r}\right) \\ \sigma_{r\theta} & = C_2\left(\frac{\sin\theta}{r}\right) + C_4\left(\frac{-\cos\theta}{r}\right) \\ \sigma_{\theta\theta} & = C_2\left(\frac{\cos\theta}{r}\right) + C_4\left(\frac{\sin\theta}{r}\right) \end{align}

From traction BCs, C2 = C4 = 0. From equilibrium,

\begin{align} F_1 + 2\int_{\alpha}^{\beta} \left(\frac{C_1\cos\theta - C_3\sin\theta}{a}\right)a\cos\theta d\theta & = 0 \\ F_2 + 2\int_{\alpha}^{\beta} \left(\frac{C_1\cos\theta - C_3\sin\theta}{a}\right)a\sin\theta d\theta & = 0 \end{align}

After algebra,

\sigma_{rr} = \frac{2C_1\cos\theta}{r} + \frac{2C_3\sin\theta}{r} ~;~~ \sigma_{r\theta} = 0 ~;~~ \sigma{\theta\theta} = 0

[edit] Special Case : α = − π, β = 0

C_1 = - \frac{F_1}{\pi} ~;~~ C_2 = \frac{F_2}{\pi}

The displacements are

\begin{align} u_1 & = - \frac{F_1(\kappa+1)\ln|x_1|}{4\pi\mu} + \frac{F_2(\kappa+1)\text{sign}(x_1)}{8\mu} \\ u_2 & = - \frac{F_2(\kappa+1)\ln|x_1|}{4\pi\mu} - \frac{F_1(\kappa+1)\text{sign}(x_1)}{8\mu} \end{align}

where

\begin{align} \kappa = 3 - 4\nu & & \text{plane strain} \\ \kappa = \frac{3 - \nu}{1+\nu} & & \text{plane stress} \\ \end{align}

and

\text{sign}(x) = \begin{cases} +1 & x > 0 \\ -1 & x < 0 \end{cases}
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