# Introduction to 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{aligned}{\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{aligned}} The stresses are

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

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

{\begin{aligned}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{aligned}} 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 $\lambda$ ${\text{(13)}}\qquad a_{3}(\lambda +1)\cos\{(\lambda +1)\alpha \}+a_{4}(\lambda -1)\cos\{(\lambda -1)\alpha \}=0$ Subtracting equation (10) from (9),

${\text{(14)}}\qquad a_{1}(\lambda +1)\sin\{(\lambda +1)\alpha \}+a_{2}(\lambda -1)\sin\{(\lambda -1)\alpha \}=0$ ${\text{(15)}}\qquad a_{1}(\lambda +1)\cos\{(\lambda +1)\alpha \}+a_{2}(\lambda +1)\cos\{(\lambda -1)\alpha \}=0$ 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 : $\alpha =\pi =180^{o}\,$ 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{aligned}\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{aligned}} 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{aligned}\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{aligned}} 