Ref: M.L. Williams, ASME J. Appl. Mech., v. 19 (1952), 526-528.
The Williams' solution
Stress concentration at the notch.
Singularity at the sharp corner, i.e,
σ
i
j
→
∞
{\displaystyle \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
α
{\displaystyle r^{\alpha }}
r
=
0
{\displaystyle r=0}
(4)
U
=
1
2
∫
0
2
π
∫
0
r
σ
i
j
ε
i
j
r
d
r
d
θ
=
C
∫
0
r
r
2
a
+
1
d
r
{\displaystyle {\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
{\displaystyle a>-1}
−
1
{\displaystyle -1}
Stresses near the notch corner
edit
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
{\displaystyle n}
n
=
λ
−
1
{\displaystyle n=\lambda -1}
(5)
φ
=
r
λ
+
1
[
a
1
cos
{
(
λ
+
1
)
θ
}
+
a
2
cos
(
λ
−
1
)
θ
+
a
3
sin
{
(
λ
+
1
)
θ
}
+
a
4
sin
(
λ
−
1
)
θ
]
{\displaystyle {\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
σ
r
r
=
r
λ
−
1
[
−
a
1
λ
(
λ
+
1
)
cos
{
(
λ
+
1
)
θ
}
−
a
2
λ
(
λ
+
1
)
cos
{
(
λ
−
1
)
θ
}
−
a
3
λ
(
λ
+
1
)
sin
{
(
λ
+
1
)
θ
}
−
a
4
λ
(
λ
+
1
)
sin
{
(
λ
−
1
)
θ
}
]
(6)
σ
r
θ
=
r
λ
−
1
[
+
a
1
λ
(
λ
+
1
)
sin
{
(
λ
+
1
)
θ
}
+
a
2
λ
(
λ
−
1
)
sin
{
(
λ
−
1
)
θ
}
−
a
3
λ
(
λ
+
1
)
cos
{
(
λ
+
1
)
θ
}
−
a
4
λ
(
λ
−
1
)
cos
{
(
λ
−
1
)
θ
}
]
(7)
σ
θ
θ
=
r
λ
−
1
[
+
a
1
λ
(
λ
+
1
)
cos
{
(
λ
+
1
)
θ
}
+
a
2
λ
(
λ
+
1
)
cos
{
(
λ
−
1
)
θ
}
+
a
3
λ
(
λ
+
1
)
sin
{
(
λ
+
1
)
θ
}
+
a
4
λ
(
λ
+
1
)
sin
{
(
λ
−
1
)
θ
}
]
(8)
{\displaystyle {\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
σ
r
θ
=
σ
θ
θ
=
0
{\displaystyle \sigma _{r\theta }=\sigma _{\theta \theta }=0}
θ
=
α
{\displaystyle \theta =\alpha }
0
=
r
λ
−
1
λ
[
+
a
1
(
λ
+
1
)
sin
{
(
λ
+
1
)
α
}
+
a
2
(
λ
−
1
)
sin
{
(
λ
−
1
)
α
}
−
a
3
(
λ
+
1
)
cos
{
(
λ
+
1
)
α
}
−
a
4
(
λ
−
1
)
cos
{
(
λ
−
1
)
α
}
]
(9)
0
=
r
λ
−
1
λ
[
−
a
1
(
λ
+
1
)
sin
{
(
λ
+
1
)
α
}
−
a
2
(
λ
−
1
)
sin
{
(
λ
−
1
)
α
}
−
a
3
(
λ
+
1
)
cos
{
(
λ
+
1
)
α
}
−
a
4
(
λ
−
1
)
cos
{
(
λ
−
1
)
α
}
]
(10)
{\displaystyle {\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
σ
r
θ
=
σ
θ
θ
=
0
{\displaystyle \sigma _{r\theta }=\sigma _{\theta \theta }=0}
θ
=
−
α
{\displaystyle \theta =-\alpha }
0
=
r
λ
−
1
λ
[
+
a
1
(
λ
+
1
)
cos
{
(
λ
+
1
)
α
}
+
a
2
(
λ
+
1
)
cos
{
(
λ
−
1
)
α
}
+
a
3
(
λ
+
1
)
sin
{
(
λ
+
1
)
α
}
+
a
4
(
λ
+
1
)
sin
{
(
λ
−
1
)
α
}
]
(11)
0
=
r
λ
−
1
λ
[
+
a
1
(
λ
+
1
)
cos
{
(
λ
+
1
)
α
}
+
a
2
(
λ
+
1
)
cos
{
(
λ
−
1
)
α
}
−
a
3
(
λ
+
1
)
sin
{
(
λ
+
1
)
α
}
−
a
4
(
λ
+
1
)
sin
{
(
λ
−
1
)
α
}
]
(12)
{\displaystyle {\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
λ
{\displaystyle \lambda }
λ
=
0
{\displaystyle \lambda =0}
Adding equations (9) and (10),
(13)
a
3
(
λ
+
1
)
cos
{
(
λ
+
1
)
α
}
+
a
4
(
λ
−
1
)
cos
{
(
λ
−
1
)
α
}
=
0
{\displaystyle {\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),
(14)
a
1
(
λ
+
1
)
sin
{
(
λ
+
1
)
α
}
+
a
2
(
λ
−
1
)
sin
{
(
λ
−
1
)
α
}
=
0
{\displaystyle {\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),
(15)
a
1
(
λ
+
1
)
cos
{
(
λ
+
1
)
α
}
+
a
2
(
λ
+
1
)
cos
{
(
λ
−
1
)
α
}
=
0
{\displaystyle {\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),
(16)
a
3
(
λ
+
1
)
sin
{
(
λ
+
1
)
α
}
+
a
4
(
λ
+
1
)
sin
{
(
λ
−
1
)
α
}
=
0
{\displaystyle {\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
(17)
[
(
λ
+
1
)
sin
{
(
λ
+
1
)
α
}
(
λ
−
1
)
sin
{
(
λ
−
1
)
α
}
(
λ
+
1
)
cos
{
(
λ
+
1
)
α
}
(
λ
+
1
)
cos
{
(
λ
−
1
)
α
}
]
[
a
1
a
2
]
=
[
0
0
]
{\displaystyle {\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
(18)
[
(
λ
+
1
)
cos
{
(
λ
+
1
)
α
}
(
λ
−
1
)
cos
{
(
λ
−
1
)
α
}
(
λ
+
1
)
sin
{
(
λ
+
1
)
α
}
(
λ
+
1
)
sin
{
(
λ
−
1
)
α
}
]
[
a
3
a
4
]
=
[
0
0
]
{\displaystyle {\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
(19)
λ
sin
(
2
α
)
+
sin
(
2
λ
α
)
=
0
{\displaystyle {\text{(19)}}\qquad \lambda \sin(2\alpha )+\sin(2\lambda \alpha )=0}
Equations (18) have a non-trivial solution only if
(20)
λ
sin
(
2
α
)
−
sin
(
2
λ
α
)
=
0
{\displaystyle {\text{(20)}}\qquad \lambda \sin(2\alpha )-\sin(2\lambda \alpha )=0}
From equation (4), acceptable singular stress fields must have
λ
>
0
{\displaystyle \lambda >0}
λ
=
0
{\displaystyle \lambda =0}
The term with the smallest eigenvalue of
λ
{\displaystyle \lambda }
λ
=
1
{\displaystyle \lambda =1}
φ
=
a
4
sin
(
0
)
{\displaystyle \varphi =a_{4}\sin(0)}
We can find the eigenvalues for general wedge angles using graphical methods.
edit
In this case, the wedge becomes a crack.In this case,
(21)
λ
=
1
2
,
1
,
3
2
,
{\displaystyle {\text{(21)}}\qquad \lambda ={\frac {1}{2}},1,{\frac {3}{2}},}
The lowest eigenvalue is
1
/
2
{\displaystyle 1/2}
(22)
a
1
=
A
2
sin
(
α
2
)
;
a
2
=
−
3
A
2
sin
(
3
α
2
)
{\displaystyle {\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
{\displaystyle A}
σ
r
r
=
K
I
2
π
r
[
5
4
cos
(
θ
2
)
−
1
4
cos
(
3
θ
2
)
]
(23)
σ
r
θ
=
K
I
2
π
r
[
3
4
cos
(
θ
2
)
+
1
4
cos
(
3
θ
2
)
]
(24)
σ
θ
θ
=
K
I
2
π
r
[
1
4
sin
(
θ
2
)
+
1
4
sin
(
3
θ
2
)
]
(25)
{\displaystyle {\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
{\displaystyle K_{I}}
(26)
K
I
=
3
A
π
2
{\displaystyle {\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.
σ
r
r
=
K
I
I
2
π
r
[
−
5
4
sin
(
θ
2
)
+
3
4
sin
(
3
θ
2
)
]
(27)
σ
r
θ
=
K
I
I
2
π
r
[
−
3
4
sin
(
θ
2
)
−
3
4
sin
(
3
θ
2
)
]
(28)
σ
θ
θ
=
K
I
I
2
π
r
[
1
4
cos
(
θ
2
)
+
3
4
sin
(
3
θ
2
)
]
(29)
{\displaystyle {\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}}}
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48e4ac5d36ed4f6c8521e84f86639d8690232400)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7816cb15964da4747197e64d0fc7f56ac394e630)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e395881836e45678c860806fae499a77e232b9af)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d39884212b7c5b8f1bd298475b13f8bddd09b9ee)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8cc8addd2234aa7669e18edf4a1730fbd3eb90cf)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6ef16db3ac7822ec165d5f0a3f4323d91ea1e06)