Stresses in wedges

Wedge with Boundary Tractions

File:Wedge with surface traction.png Elastic wedge with normal and shear surface tractions

Suppose

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

Then

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

To find $f(\theta )$ plug into ${\mathrm{\nabla }}^4\phi =0$.

$\text{(2)}(\frac{d^2}{d{\theta }^2}+(n+2{)}^2)(\frac{d^2}{d{\theta }^2}+n^2)f(\theta )=0$

If $n\ne 0$ and $n\ne -2$,

$\text{(3)}\phi =r^{n+2}[a_1\cos \{(n+2)\theta \}+a_2\cos (n\theta )+a_3\sin \{(n+2)\theta \}+a_4\sin (n\theta )]$

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.

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

$\begin{array}{cc}{\sigma }_{rr}=r^{\lambda -1}[& -a_1\lambda (\lambda +1)\cos \{(\lambda +1)\theta \}-a_2\lambda (\lambda -3)\cos \{(\lambda -1)\theta \}\\ & -a_3\lambda (\lambda +1)\sin \{(\lambda +1)\theta \}-a_4\lambda (\lambda -3)\sin \{(\lambda -1)\theta \}]\text{(6)}\\ {\sigma }_{r\theta }=r^{\lambda -1}[& +a_1\lambda (\lambda +1)\sin \{(\lambda +1)\theta \}+a_2\lambda (\lambda -1)\sin \{(\lambda -1)\theta \}\\ & -a_3\lambda (\lambda +1)\cos \{(\lambda +1)\theta \}-a_4\lambda (\lambda -1)\cos \{(\lambda -1)\theta \}]\text{(7)}\\ {\sigma }_{\theta \theta }=r^{\lambda -1}[& +a_1\lambda (\lambda +1)\cos \{(\lambda +1)\theta \}+a_2\lambda (\lambda +1)\cos \{(\lambda -1)\theta \}\\ & +a_3\lambda (\lambda +1)\sin \{(\lambda +1)\theta \}+a_4\lambda (\lambda +1)\sin \{(\lambda -1)\theta \}]\text{(8)}\end{array}$

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 eiegnvalues for general wedge angles using graphical methods.

Special case : α = π = 180°

In this case, the wedge becomes a crack.In this case,

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

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

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

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.

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

Axially Loaded Wedge

File:Wedge with axial force.png Elastic wedge loaded by an axial force

The BCs at $\theta =\pm \beta$ are

$\text{(30)}{t}_r=t_{\theta }=0;\widehat{𝐧}=\pm \widehat{𝐞}\theta \Rightarrow {\sigma }_{r\theta }={\sigma }_{\theta \theta }=0$

What about the concentrated force BC?

• What is $\widehat{𝐧}$ 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

$\text{(31)}\widehat{𝐧}=\widehat{𝐞}r\Rightarrow {\sigma }_{r\theta }=t_r;{\sigma }_{\theta \theta }=t_{\theta }$

For equilibrium, $\sum F_1=\sum F_2=\sum M_3=0$. Therefore,

$\begin{array}{c}{P}_1+\underset{-\beta }{\overset{\beta }{\int }}[{\sigma }_{rr}(a,\theta )\cos \theta -{\sigma }_{r\theta }(a,\theta )\sin \theta ]ad\theta =0\text{(32)}\\ \underset{-\beta }{\overset{\beta }{\int }}[{\sigma }_{rr}(a,\theta )\sin \theta +{\sigma }_{r\theta }(a,\theta )\cos \theta ]ad\theta =0\text{(33)}\\ \underset{-\beta }{\overset{\beta }{\int }}[a{\sigma }_{r\theta }(a,\theta )]ad\theta =0\text{(34)}\end{array}$

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,

$\text{(35)}{\sigma }_{r\theta }=\frac{\partial }{\partial }r\left(\frac{1}{r}\frac{\partial }{\partial }\phi \theta \right)=0\Rightarrow \phi =r\eta (\theta )+\zeta (r)$

Hence,

$\text{(36)}{\sigma }_{\theta \theta }=\frac{{\partial }^2}{\partial \phi \partial r}={\zeta }^{{\text{'}}^{\prime }}(r)$

That means ${\sigma }_{\theta \theta }$ is independent of $\theta$. Therefore, in order to satisfy the BCs, ${\sigma }_{\theta \theta }=0$, i.e.,

$\text{(37)}\zeta (r)=C_{1}r+C_2\Rightarrow \phi =r\eta (\theta )+C_{1}r=r[\eta (\theta )+C_1]=r\xi (\theta )$

Checking for compatibility, ${\mathrm{\nabla }}^4\phi =0$, we get

$\text{(38)}{\xi }^{(IV)}(\theta )+2{\xi }^{{\text{'}}^{\prime }}(\theta )+\xi (\theta )=0$

The general solution is

$\text{(39)}\xi (\theta )=A\sin \theta +B\cos \theta +C\theta \sin \theta +D\theta \cos \theta$

Therefore,

$\text{(40)}\phi =r[A\sin \theta +B\cos \theta +C\theta \sin \theta +D\theta \cos \theta ]$

The only non-zero stress is ${\sigma }_{rr}$.

$\text{(41)}{\sigma }_{rr}=\frac{1}{r}[2C\cos \theta -2D\sin \theta ]$

Plugging into equation (33), we get

$\text{(42)}-D[2\beta -\sin (2\beta )]=0\Rightarrow D=0$

Hence,

$\text{(43)}{\sigma }_{rr}=\frac{2C}{r}\cos \theta$

Plugging into equation (32), we get

$\text{(44)}-P=C[2\beta +\sin (2\beta )]\Rightarrow C=\frac{-P}{2\beta +\sin (2\beta )}$

Therefore,

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

The stress state is

$\text{(46)}{\sigma }_{rr}=-\frac{2P\cos \theta }{r[2\beta +\sin (2\beta )]};{\sigma }_{r\theta }=0;{\sigma }_{\theta \theta }=0$

Special Case : β = π/2

A concentrated point load acting on a half plane.

$\text{(47)}{\sigma }_{rr}=-\frac{2P\cos \theta }{\pi r};{\sigma }_{r\theta }=0;{\sigma }_{\theta \theta }=0$

Displacements

$\begin{array}{cc}2\mu u_r& =-\frac{\partial }{\partial }\phi r+\alpha r\frac{\partial }{\partial }\psi \theta \\ 2\mu u_{\theta }& =-\frac{1}{r}\frac{\partial }{\partial }\phi \theta +\alpha r^2\frac{\partial }{\partial }\psi r\end{array}$

where

$\begin{array}{c}{\mathrm{\nabla }}^2\psi =0\\ \frac{\partial }{\partial }r\left(r\frac{\partial }{\partial }\psi \theta \right)={\mathrm{\nabla }}^2\phi \end{array}$

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

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

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

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

Hence,

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

Solving,

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

Therefore,

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

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,

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

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

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

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,

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

Therefore, from Tables,

$\begin{array}{cc}{\sigma }_{rr}& =C_1\left(\frac{2\cos \theta }{r}\right)+C_2\left(\frac{\cos \theta }{r}\right)+C_3\left(\frac{2\sin \theta }{r}\right)+C_4\left(\frac{\sin \theta }{r}\right)\\ {\sigma }_{r\theta }& =C_2\left(\frac{\sin \theta }{r}\right)+C_4\left(\frac{-\cos \theta }{r}\right)\\ {\sigma }_{\theta \theta }& =C_2\left(\frac{\cos \theta }{r}\right)+C_4\left(\frac{\sin \theta }{r}\right)\end{array}$

From traction BCs, $C_2=C_4=0$. From equilibrium,

$\begin{array}{cc}{F}_1+2\underset{\alpha }{\overset{\beta }{\int }}\left(\frac{C_1\cos \theta -C_3\sin \theta }{a}\right)a\cos \theta d\theta & =0\\ F_2+2\underset{\alpha }{\overset{\beta }{\int }}\left(\frac{C_1\cos \theta -C_3\sin \theta }{a}\right)a\sin \theta d\theta & =0\end{array}$

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

$\begin{array}{cc}{u}_1& =-\frac{F_1(\kappa +1)\ln |x_1|}{4\pi \mu }+\frac{F_2(\kappa +1)\text{sign}(x_1)}{8\mu }\\ u_2& =-\frac{F_2(\kappa +1)\ln |x_1|}{4\pi \mu }-\frac{F_1(\kappa +1)\text{sign}(x_1)}{8\mu }\end{array}$

where

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

and

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

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