Stresses in wedges

Wedge with Boundary Tractions

Elastic wedge with normal and shear surface tractions

Suppose

• The tractions on the boundary vary as $r^{n}$.
• No body forces.

Then

${\text{(1)}}\qquad \varphi =r^{n+2}f(\theta )$

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 $n\neq 0$ and $n\neq -2$,

${\text{(3)}}\qquad \varphi =r^{n+2}\left[a_{1}\cos\{(n+2)\theta \}+a_{2}\cos(n\theta )+a_{3}\sin\{(n+2)\theta \}+a_{4}\sin(n\theta )\right]$

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.

The Williams' solution

• 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

${\begin{aligned}\sigma _{rr}=r^{\lambda -1}\left[\right.&-a_{1}\lambda (\lambda +1)\cos\{(\lambda +1)\theta \}-a_{2}\lambda (\lambda -3)\cos\{(\lambda -1)\theta \}\\&-a_{3}\lambda (\lambda +1)\sin\{(\lambda +1)\theta \}-a_{4}\lambda (\lambda -3)\sin\{(\lambda -1)\theta \}\left.\right]{\text{(6)}}\qquad \\\sigma _{r\theta }=r^{\lambda -1}\left[\right.&+a_{1}\lambda (\lambda +1)\sin\{(\lambda +1)\theta \}+a_{2}\lambda (\lambda -1)\sin\{(\lambda -1)\theta \}\\&-a_{3}\lambda (\lambda +1)\cos\{(\lambda +1)\theta \}-a_{4}\lambda (\lambda -1)\cos\{(\lambda -1)\theta \}\left.\right]{\text{(7)}}\qquad \\\sigma _{\theta \theta }=r^{\lambda -1}\left[\right.&+a_{1}\lambda (\lambda +1)\cos\{(\lambda +1)\theta \}+a_{2}\lambda (\lambda +1)\cos\{(\lambda -1)\theta \}\\&+a_{3}\lambda (\lambda +1)\sin\{(\lambda +1)\theta \}+a_{4}\lambda (\lambda +1)\sin\{(\lambda -1)\theta \}\left.\right]{\text{(8)}}\qquad \end{aligned}}$

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 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

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

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.

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

Axially Loaded Wedge

Elastic wedge loaded by an axial force

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,

${\begin{aligned}P_{1}+\int _{-\beta }^{\beta }\left[\sigma _{rr}(a,\theta )\cos \theta -\sigma _{r\theta }(a,\theta )\sin \theta \right]a~d\theta =0{\text{(32)}}\qquad \\\int _{-\beta }^{\beta }\left[\sigma _{rr}(a,\theta )\sin \theta +\sigma _{r\theta }(a,\theta )\cos \theta \right]a~d\theta =0{\text{(33)}}\qquad \\\int _{-\beta }^{\beta }\left[a\sigma _{r\theta }(a,\theta )\right]a~d\theta =0{\text{(34)}}\qquad \end{aligned}}$

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,

${\text{(45)}}\qquad {\varphi =Cr\theta \sin \theta ={\frac {-Pr\theta \sin \theta }{2\beta +\sin(2\beta )}}}$

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

${\begin{aligned}2\mu u_{r}&=-{\frac {\partial }{\partial }}{\varphi }{r}+\alpha r{\frac {\partial }{\partial }}{\psi }{\theta }\\2\mu u_{\theta }&=-{\frac {1}{r}}{\frac {\partial }{\partial }}{\varphi }{\theta }+\alpha r^{2}{\frac {\partial }{\partial }}{\psi }{r}\end{aligned}}$

where

${\begin{aligned}\nabla ^{2}{\psi }=0\\{\frac {\partial }{\partial }}{}{r}\left(r{\frac {\partial }{\partial }}{\psi }{\theta }\right)=\nabla ^{2}{\varphi }\end{aligned}}$

Plug in $\varphi =Cr\theta \sin \theta$,

${\begin{aligned}&{\frac {\partial }{\partial }}{}{r}\left(r{\frac {\partial }{\partial }}{\psi }{\theta }\right)=\nabla ^{2}{\varphi }\\\Rightarrow &{\frac {\partial }{\partial }}{}{r}\left(r{\frac {\partial }{\partial }}{\psi }{\theta }\right)={\frac {2C}{r}}\cos \theta \\\Rightarrow &r{\frac {\partial }{\partial }}{\psi }{\theta }=2C\ln r\cos \theta +A(\theta )\\\Rightarrow &{\frac {\partial }{\partial }}{\psi }{\theta }=2C{\frac {\ln r}{r}}\cos \theta +{\frac {A(\theta )}{r}}\\\Rightarrow &\psi =2C{\frac {\ln r}{r}}\sin \theta +{\frac {\eta (\theta )}{r}}+\xi {r}\end{aligned}}$

Plug $\psi$ into $\nabla ^{2}{\psi }=0$,

${\begin{aligned}&{\frac {1}{r^{3}}}\eta ^{''}(\theta )+{\frac {1}{r^{3}}}\eta (\theta )+\xi ^{''}(r)+{\frac {1}{r}}\xi ^{'}(r)-{\frac {4C}{r^{3}}}\sin \theta =0\\\Rightarrow &\eta ^{''}(\theta )+\eta (\theta )+r^{3}\xi ^{''}(r)+r^{2}\xi ^{'}(r)-4C\sin \theta =0\end{aligned}}$

Hence,

${\begin{aligned}\eta ^{''}(\theta )+\eta (\theta )-4C\sin \theta &=b\\r^{3}\xi ^{''}(r)+r^{2}\xi ^{'}(r)&=-{\frac {b}{r^{3}}}\end{aligned}}$

Solving,

${\begin{aligned}\eta (\theta )&=-2C\theta \cos \theta +d\cos \theta +e\sin \theta +b\\\xi ^{'}(r)&={\frac {f}{r}}+{\frac {b}{r^{2}}}\end{aligned}}$

Therefore,

${\begin{aligned}2\mu u_{r}&=2\alpha C\ln r\cos \theta +(2\alpha -1)C\theta \sin \theta +\alpha (e-2C)\cos \theta -\alpha d\sin \theta \\2\mu u_{\theta }&=-2\alpha C\ln r\sin \theta +(2\alpha -1)C\sin \theta +(2\alpha -1)C\theta \cos \theta -\alpha d\cos \theta -\alpha e\sin \theta +\alpha fr\end{aligned}}$

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,

${\begin{aligned}u_{r}&={\frac {\alpha C}{\mu }}\ln \left({\frac {r}{L}}\right)\cos \theta +{\frac {(2\alpha -1)C}{2\mu }}\theta \sin \theta \\u_{\theta }&=-{\frac {\alpha C}{\mu }}\ln \left({\frac {r}{L}}\right)\sin \theta +{\frac {(2\alpha -1)C}{2\mu }}\theta \cos \theta -{\frac {C}{2\mu }}\sin \theta \end{aligned}}$

The displacements are singular at $r=0$ and $r=\infty$. At $\theta =0$,

${\begin{aligned}u_{r}&={\frac {\alpha C}{\mu }}\ln \left({\frac {r}{L}}\right)\\u_{\theta }&=0\end{aligned}}$

Is the small strain assumption satisfied ?

The Flamant Solution

Elastic wedge loaded by two forces at the tip

• 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 ?

From Michell's solution, pick terms containing $1/r$ in the stresses. Then,

$\varphi =C_{1}r\theta \sin \theta +C_{2}r\ln r\cos \theta +C_{3}r\theta \cos \theta +C_{4}r\ln r\sin \theta$

Therefore, from Tables,

${\begin{aligned}\sigma _{rr}&=C_{1}\left({\frac {2\cos \theta }{r}}\right)+C_{2}\left({\frac {\cos \theta }{r}}\right)+C_{3}\left({\frac {2\sin \theta }{r}}\right)+C_{4}\left({\frac {\sin \theta }{r}}\right)\\\sigma _{r\theta }&=C_{2}\left({\frac {\sin \theta }{r}}\right)+C_{4}\left({\frac {-\cos \theta }{r}}\right)\\\sigma _{\theta \theta }&=C_{2}\left({\frac {\cos \theta }{r}}\right)+C_{4}\left({\frac {\sin \theta }{r}}\right)\end{aligned}}$

From traction BCs, $C_{2}=C_{4}=0$. From equilibrium,

${\begin{aligned}F_{1}+2\int _{\alpha }^{\beta }\left({\frac {C_{1}\cos \theta -C_{3}\sin \theta }{a}}\right)a\cos \theta d\theta &=0\\F_{2}+2\int _{\alpha }^{\beta }\left({\frac {C_{1}\cos \theta -C_{3}\sin \theta }{a}}\right)a\sin \theta d\theta &=0\end{aligned}}$

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

${\begin{aligned}u_{1}&=-{\frac {F_{1}(\kappa +1)\ln |x_{1}|}{4\pi \mu }}+{\frac {F_{2}(\kappa +1){\text{sign}}(x_{1})}{8\mu }}\\u_{2}&=-{\frac {F_{2}(\kappa +1)\ln |x_{1}|}{4\pi \mu }}-{\frac {F_{1}(\kappa +1){\text{sign}}(x_{1})}{8\mu }}\end{aligned}}$

where

${\begin{aligned}\kappa =3-4\nu &&{\text{plane strain}}\\\kappa ={\frac {3-\nu }{1+\nu }}&&{\text{plane stress}}\\\end{aligned}}$

and

${\text{sign}}(x)={\begin{cases}+1&x>0\\-1&x<0\end{cases}}$

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