Elasticity/Williams asymptotic solution

Williams' Asymptotic Solution

Ref: M.L. Williams, ASME J. Appl. Mech., v. 19 (1952), 526-528.

File:Williams solution mode I.png The Williams' solution

• Stress concentration at the notch.
• Singularity at the sharp corner, i.e, ${\sigma }_{ij}\to \mathrm{\infty }$.
• 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

$\text{(4)}U=\frac{1}{2}\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{r}{\int }}{\sigma }_{ij}{\epsilon }_{ij}rdrd\theta =C\underset{0}{\overset{r}{\int }}{r}^{2a+1}dr$

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 $n=\lambda -1$.

$\begin{array}{cc}\text{(5)}\phi =r^{\lambda +1}[& a_1\cos \{(\lambda +1)\theta \}+a_2\cos (\lambda -1)\theta +\\ & a_3\sin \{(\lambda +1)\theta \}+a_4\sin (\lambda -1)\theta ]\end{array}$

The stresses are

$\begin{array}{cc}{\sigma }_{rr}=r^{\lambda -1}[& -a_1\lambda (\lambda +1)\cos \{(\lambda +1)\theta \}-a_2\lambda (\lambda +1)\cos \{(\lambda -1)\theta \}\\ & -a_3\lambda (\lambda +1)\sin \{(\lambda +1)\theta \}-a_4\lambda (\lambda +1)\sin \{(\lambda -1)\theta \}]\text{(6)}\\ {\sigma }_{r\theta }=r^{\lambda -1}[& +a_1\lambda (\lambda +1)\sin \{(\lambda +1)\theta \}+a_2\lambda (\lambda -1)\sin \{(\lambda -1)\theta \}\\ & -a_3\lambda (\lambda +1)\cos \{(\lambda +1)\theta \}-a_4\lambda (\lambda -1)\cos \{(\lambda -1)\theta \}]\text{(7)}\\ {\sigma }_{\theta \theta }=r^{\lambda -1}[& +a_1\lambda (\lambda +1)\cos \{(\lambda +1)\theta \}+a_2\lambda (\lambda +1)\cos \{(\lambda -1)\theta \}\\ & +a_3\lambda (\lambda +1)\sin \{(\lambda +1)\theta \}+a_4\lambda (\lambda +1)\sin \{(\lambda -1)\theta \}]\text{(8)}\end{array}$

The BCs are ${\sigma }_{r\theta }={\sigma }_{\theta \theta }=0$ at $\theta =\alpha$.Hence,

$\begin{array}{cc}0=r^{\lambda -1}\lambda [& +a_1(\lambda +1)\sin \{(\lambda +1)\alpha \}+a_2(\lambda -1)\sin \{(\lambda -1)\alpha \}\\ & -a_3(\lambda +1)\cos \{(\lambda +1)\alpha \}-a_4(\lambda -1)\cos \{(\lambda -1)\alpha \}]\text{(9)}\\ 0=r^{\lambda -1}\lambda [& -a_1(\lambda +1)\sin \{(\lambda +1)\alpha \}-a_2(\lambda -1)\sin \{(\lambda -1)\alpha \}\\ & -a_3(\lambda +1)\cos \{(\lambda +1)\alpha \}-a_4(\lambda -1)\cos \{(\lambda -1)\alpha \}]\text{(10)}\end{array}$

The BCs are ${\sigma }_{r\theta }={\sigma }_{\theta \theta }=0$ at $\theta =-\alpha$.Hence,

$\begin{array}{cc}0=r^{\lambda -1}\lambda [& +a_1(\lambda +1)\cos \{(\lambda +1)\alpha \}+a_2(\lambda +1)\cos \{(\lambda -1)\alpha \}\\ & +a_3(\lambda +1)\sin \{(\lambda +1)\alpha \}+a_4(\lambda +1)\sin \{(\lambda -1)\alpha \}]\text{(11)}\\ 0=r^{\lambda -1}\lambda [& +a_1(\lambda +1)\cos \{(\lambda +1)\alpha \}+a_2(\lambda +1)\cos \{(\lambda -1)\alpha \}\\ & -a_3(\lambda +1)\sin \{(\lambda +1)\alpha \}-a_4(\lambda +1)\sin \{(\lambda -1)\alpha \}]\text{(12)}\end{array}$

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),

$\text{(13)}{a}_3(\lambda +1)\cos \{(\lambda +1)\alpha \}+a_4(\lambda -1)\cos \{(\lambda -1)\alpha \}=0$

Subtracting equation (10) from (9),

$\text{(14)}{a}_1(\lambda +1)\sin \{(\lambda +1)\alpha \}+a_2(\lambda -1)\sin \{(\lambda -1)\alpha \}=0$

Adding equations (11) and (12),

$\text{(15)}{a}_1(\lambda +1)\cos \{(\lambda +1)\alpha \}+a_2(\lambda +1)\cos \{(\lambda -1)\alpha \}=0$

Subtracting equation (12) from (11),

$\text{(16)}{a}_3(\lambda +1)\sin \{(\lambda +1)\alpha \}+a_4(\lambda +1)\sin \{(\lambda -1)\alpha \}=0$

Therefore, the two independent sets of equations are

$\text{(17)}\left[\begin{array}{cc}(\lambda +1)\sin \{(\lambda +1)\alpha \}& (\lambda -1)\sin \{(\lambda -1)\alpha \}\\ (\lambda +1)\cos \{(\lambda +1)\alpha \}& (\lambda +1)\cos \{(\lambda -1)\alpha \}\end{array}\right]\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

and

$\text{(18)}\left[\begin{array}{cc}(\lambda +1)\cos \{(\lambda +1)\alpha \}& (\lambda -1)\cos \{(\lambda -1)\alpha \}\\ (\lambda +1)\sin \{(\lambda +1)\alpha \}& (\lambda +1)\sin \{(\lambda -1)\alpha \}\end{array}\right]\left[\begin{array}{c}{a}_3\\ a_4\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

Equations (17) have a non-trivial solution only if

$\text{(19)}\lambda \sin (2\alpha )+\sin (2\lambda \alpha )=0$

Equations (18) have a non-trivial solution only if

$\text{(20)}\lambda \sin (2\alpha )-\sin (2\lambda \alpha )=0$

• 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 $\phi =a_4\sin (0)$. 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,

$\text{(21)}\lambda =\frac{1}{2},1,\frac{3}{2},$

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

$\text{(22)}{a}_1=\frac{A}{2}\sin \left(\frac{\alpha }{2}\right);a_2=-\frac{3A}{2}\sin \left(\frac{3\alpha }{2}\right)$

where $A$ is a constant. The singular stress field at the crack tip is then

$\begin{array}{cc}{\sigma }_{rr}& =\frac{K_I}{\sqrt{2\pi r}}[\frac{5}{4}\cos \left(\frac{\theta }{2}\right)-\frac{1}{4}\cos \left(\frac{3\theta }{2}\right)]\text{(23)}\\ {\sigma }_{r\theta }& =\frac{K_I}{\sqrt{2\pi r}}[\frac{3}{4}\cos \left(\frac{\theta }{2}\right)+\frac{1}{4}\cos \left(\frac{3\theta }{2}\right)]\text{(24)}\\ {\sigma }_{\theta \theta }& =\frac{K_I}{\sqrt{2\pi r}}[\frac{1}{4}\sin \left(\frac{\theta }{2}\right)+\frac{1}{4}\sin \left(\frac{3\theta }{2}\right)]\text{(25)}\end{array}$

where, $K_I$ is the { Mode I Stress Intensity Factor.}

$\text{(26)}{K}_I=3A\sqrt{\frac{\pi }{2}}$

If we use equations (18) we can get the stresses due to a mode II loading.

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