Elasticity/Williams asymptotic solution

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+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{(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 eigenvalues 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_{r\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_{\theta\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_{r\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_{\theta\theta} & = \frac{K_{II}}{\sqrt{2\pi r}} \left[\frac{1}{4}\cos\left(\frac{\theta}{2}\right) + \frac{3}{4}\sin\left(\frac{3\theta}{2}\right)\right] \text{(29)} \qquad \end{align}$$