Stresses in wedges

Wedge with Boundary Tractions
Suppose Then
 * The tractions on the boundary vary as $$r^n$$.
 * No body forces.
 * $$\text{(1)} \qquad

\varphi = r^{n+2} f(\theta) $$ To find $$f(\theta)$$ plug into $$\nabla^4{\varphi} = 0$$.
 * $$\text{(2)} \qquad

\left(\frac{d^2}{d\theta^2} + (n+2)^2\right) \left(\frac{d^2}{d\theta^2} + n^2\right) f(\theta) = 0 $$ If $$n \ne 0$$ and $$n \ne -2$$,
 * $$\text{(3)} \qquad

\varphi = r^{n+2} \left[ a_1 \cos\{(n+2)\theta\} + a_2 \cos(n\theta) + a_3 \sin\{(n+2)\theta\} + a_4 \sin(n\theta)\right] $$ The corresponding stresses and displacements can be found from Michell's solution. We have to take special care for the case where $$n = 0$$, i.e., the traction on the surface is constant.

Williams' Asymptotic Solution
''Ref: M.L. Williams, ASME J. Appl. Mech., v. 19 (1952), 526-528.''
 * Stress concentration at the notch.
 * Singularity at the sharp corner, i.e, $$\sigma_{ij}\rightarrow\infty$$.
 * William's solution involves defining the origin at the corner and expanding the stress field as an asymptotic series in powers of r.
 * If the stresses (and strains) vary with $$r^{\alpha}$$ as we approach the point $$r = 0$$, the strain energy is given by
 * $$\text{(4)} \qquad

U = \frac{1}{2}\int_0^{2\pi}\int_0^r\sigma_{ij}\varepsilon_{ij}~r~dr~d\theta = C\int_0^r r^{2a+1} dr  $$ This integral is bounded only if $$a > -1$$. Hence, singular stress fields are acceptable only if the exponent on the stress components  exceeds $$-1$$.

Stresses near the notch corner

 * Use a separated-variable series as in equation (3).
 * Each of the terms satisfies the traction-free BCs on the  surface of the notch.
 * Relax the requirement that $$n$$ in equation (3) is an integer. Let $$ n = \lambda-1$$.
 * $$\begin{align}

\text{(5)} \qquad \varphi = r^{\lambda+1} \left[\right. & a_1 \cos\{(\lambda+1)\theta\} + a_2 \cos{(\lambda-1)\theta} +\\ & a_3 \sin\{(\lambda+1)\theta\} + a_4 \sin{(\lambda-1)\theta}\left.\right] \end{align}$$ The stresses are
 * $$\begin{align}

\sigma_{rr} = r^{\lambda-1} \left[\right. & -a_1\lambda(\lambda+1)\cos\{(\lambda+1)\theta\} -a_2\lambda(\lambda-3)\cos\{(\lambda-1)\theta\} \\ & -a_3\lambda(\lambda+1)\sin\{(\lambda+1)\theta\} -a_4\lambda(\lambda-3)\sin\{(\lambda-1)\theta\}\left.\right] \text{(6)} \qquad \\ \sigma_{r\theta}= r^{\lambda-1} \left[\right. & +a_1\lambda(\lambda+1)\sin\{(\lambda+1)\theta\} +a_2\lambda(\lambda-1)\sin\{(\lambda-1)\theta\} \\ & -a_3\lambda(\lambda+1)\cos\{(\lambda+1)\theta\} -a_4\lambda(\lambda-1)\cos\{(\lambda-1)\theta\}\left.\right] \text{(7)} \qquad \\ \sigma_{\theta\theta}= r^{\lambda-1} \left[\right. & +a_1\lambda(\lambda+1)\cos\{(\lambda+1)\theta\} +a_2\lambda(\lambda+1)\cos\{(\lambda-1)\theta\} \\ & +a_3\lambda(\lambda+1)\sin\{(\lambda+1)\theta\} +a_4\lambda(\lambda+1)\sin\{(\lambda-1)\theta\}\left.\right]\text{(8)} \qquad \end{align}$$ The BCs are $$\sigma_{r\theta} = \sigma_{\theta\theta} = 0$$ at $$\theta = \alpha$$.Hence,
 * $$\begin{align}

0= r^{\lambda-1} \lambda \left[\right. & +a_1(\lambda+1)\sin\{(\lambda+1)\alpha\} +a_2(\lambda-1)\sin\{(\lambda-1)\alpha\} \\ & -a_3(\lambda+1)\cos\{(\lambda+1)\alpha\} -a_4(\lambda-1)\cos\{(\lambda-1)\alpha\}\left.\right] \text{(9)} \qquad \\ 0= r^{\lambda-1} \lambda \left[\right. & -a_1(\lambda+1)\sin\{(\lambda+1)\alpha\} -a_2(\lambda-1)\sin\{(\lambda-1)\alpha\} \\ & -a_3(\lambda+1)\cos\{(\lambda+1)\alpha\} -a_4(\lambda-1)\cos\{(\lambda-1)\alpha\}\left.\right] \text{(10)} \qquad \end{align}$$ The BCs are $$\sigma_{r\theta} = \sigma_{\theta\theta} = 0$$ at $$\theta = -\alpha$$.Hence,
 * $$\begin{align}

0= r^{\lambda-1} \lambda \left[\right. & +a_1(\lambda+1)\cos\{(\lambda+1)\alpha\} +a_2(\lambda+1)\cos\{(\lambda-1)\alpha\} \\ & +a_3(\lambda+1)\sin\{(\lambda+1)\alpha\} +a_4(\lambda+1)\sin\{(\lambda-1)\alpha\}\left.\right] \text{(11)} \qquad \\ 0= r^{\lambda-1} \lambda \left[\right. & +a_1(\lambda+1)\cos\{(\lambda+1)\alpha\} +a_2(\lambda+1)\cos\{(\lambda-1)\alpha\} \\ & -a_3(\lambda+1)\sin\{(\lambda+1)\alpha\} -a_4(\lambda+1)\sin\{(\lambda-1)\alpha\}\left.\right]\text{(12)} \qquad \end{align}$$ The above equations will have non-trivial solutions only for certain eigenvalues of $$\lambda$$, one of which is $$\lambda = 0$$. Using the symmetries of the equations, we can partition the coefficient matrix.

Eigenvalues of λ
Adding equations (9) and (10),
 * $$\text{(13)} \qquad

a_3(\lambda+1)\cos\{(\lambda+1)\alpha\} + a_4(\lambda-1)\cos\{(\lambda-1)\alpha\} = 0 $$ Subtracting equation (10) from (9),
 * $$\text{(14)} \qquad

a_1(\lambda+1)\sin\{(\lambda+1)\alpha\} + a_2(\lambda-1)\sin\{(\lambda-1)\alpha\} = 0 $$ Adding equations (11) and (12),
 * $$\text{(15)} \qquad

a_1(\lambda+1)\cos\{(\lambda+1)\alpha\} + a_2(\lambda+1)\cos\{(\lambda-1)\alpha\} = 0 $$ Subtracting equation (12) from (11),
 * $$\text{(16)} \qquad

a_3(\lambda+1)\sin\{(\lambda+1)\alpha\} + a_4(\lambda+1)\sin\{(\lambda-1)\alpha\} = 0 $$ Therefore, the two independent sets of equations are
 * $$\text{(17)} \qquad

\begin{bmatrix} (\lambda+1)\sin\{(\lambda+1)\alpha\} & (\lambda-1)\sin\{(\lambda-1)\alpha\} \\ (\lambda+1)\cos\{(\lambda+1)\alpha\} & (\lambda+1)\cos\{(\lambda-1)\alpha\} \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ and
 * $$\text{(18)} \qquad

\begin{bmatrix} (\lambda+1)\cos\{(\lambda+1)\alpha\} & (\lambda-1)\cos\{(\lambda-1)\alpha\} \\ (\lambda+1)\sin\{(\lambda+1)\alpha\} & (\lambda+1)\sin\{(\lambda-1)\alpha\} \end{bmatrix} \begin{bmatrix} a_3 \\ a_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ Equations (17) have a non-trivial solution only if
 * $$\text{(19)} \qquad

\lambda\sin(2\alpha) + \sin(2\lambda\alpha) = 0 $$ Equations (18) have a non-trivial solution only if
 * $$\text{(20)} \qquad

\lambda\sin(2\alpha) - \sin(2\lambda\alpha) = 0 $$
 * From equation (4), acceptable singular stress fields must have $$\lambda > 0$$.Hence, $$\lambda = 0$$ is not acceptable.
 * The term with the smallest eigenvalue of $$\lambda$$ dominates the  solution.  Hence, this eigenvalue is what we seek.
 * $$\lambda = 1$$ leads to $$\varphi = a_4 \sin(0)$$. Unacceptable.
 * We can find the eiegnvalues for general wedge angles using  graphical methods.

Special case : α = π = 180°
In this case, the wedge becomes a crack.In this case,
 * $$\text{(21)} \qquad

\lambda = \frac{1}{2}, 1, \frac{3}{2}, $$ The lowest eigenvalue is $$1/2$$. If we use, this value in equation (17), then the two equations will not be linearly independent and we can express them as one equation with the substitutions
 * $$\text{(22)} \qquad

a_1 = \frac{A}{2}\sin\left(\frac{\alpha}{2}\right) ~; a_2 = -\frac{3A}{2}\sin\left(\frac{3\alpha}{2}\right) $$ where $$A$$ is a constant. The singular stress field at the crack tip is then
 * $$\begin{align}

\sigma_{rr} & = \frac{K_I}{\sqrt{2\pi r}} \left[\frac{5}{4}\cos\left(\frac{\theta}{2}\right) - \frac{1}{4}\cos\left(\frac{3\theta}{2}\right)\right] \text{(23)} \qquad \\ \sigma_{\theta\theta} & = \frac{K_I}{\sqrt{2\pi r}} \left[\frac{3}{4}\cos\left(\frac{\theta}{2}\right) + \frac{1}{4}\cos\left(\frac{3\theta}{2}\right)\right] \text{(24)} \qquad \\ \sigma_{r\theta} & = \frac{K_I}{\sqrt{2\pi r}} \left[\frac{1}{4}\sin\left(\frac{\theta}{2}\right) + \frac{1}{4}\sin\left(\frac{3\theta}{2}\right)\right] \text{(25)} \qquad \end{align}$$ where, $$K_I$$ is the { Mode I Stress Intensity Factor.}
 * $$\text{(26)} \qquad

K_I = 3A\sqrt{\frac{\pi}{2}} $$ If we use equations (18) we can get the stresses due to a mode II loading.
 * $$\begin{align}

\sigma_{rr} & = \frac{K_{II}}{\sqrt{2\pi r}} \left[-\frac{5}{4}\sin\left(\frac{\theta}{2}\right) + \frac{3}{4}\sin\left(\frac{3\theta}{2}\right)\right] \text{(27)} \qquad \\ \sigma_{\theta\theta} & = \frac{K_{II}}{\sqrt{2\pi r}} \left[-\frac{3}{4}\sin\left(\frac{\theta}{2}\right) - \frac{3}{4}\sin\left(\frac{3\theta}{2}\right)\right] \text{(28)} \qquad \\ \sigma_{r\theta} & = \frac{K_{II}}{\sqrt{2\pi r}} \left[\frac{1}{4}\cos\left(\frac{\theta}{2}\right) + \frac{3}{4}\cos\left(\frac{3\theta}{2}\right)\right] \text{(29)} \qquad \end{align}$$

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 : β = π/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 ?

The Flamant Solution
From Michell's solution, pick terms containing $$1/r$$ in the stresses. Then,
 * This problem is also self-similar (no inherent length scale).
 * All quantities can be expressed in the separated-variable form  $$\sigma = f(r)g(\theta)$$.
 * The stresses vary as $$(1/r)$$ (the area of action of the force  decreases with increasing $$r$$).  How about a conical wedge ?

\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}$$ From traction BCs, $$C_2 = C_4 = 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 $$

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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