Elasticity/Plate with hole in shear

Circular hole in a shear field

File:Elastic plate with hole shear.png Elastic plate with circular hole under shear

Given:

• Large plate in pure shear.
• Stress state perturbed by a small hole.

The BCs are

at $r=a$

$\text{(103)}{t}_r=t_{\theta }=0;\widehat{𝐧}=-\widehat{𝐞}r\Rightarrow {\sigma }_{rr}={\sigma }_{r\theta }=0$

at $r\to \mathrm{\infty }$

$\text{(104)}{\sigma }_{12}\to S;{\sigma }_{11}\to 0;{\sigma }_{22}\to 0$

We will solve this problem by superposing a perturbation due to the hole on the unperturbed solution. The effect of the perturbation will decrease with increasing distance from the hole, i.e. the effect will be proportional to $r^{-n}$.

Unperturbed Solution

$\text{(105)}{\sigma }_{11}={\sigma }_{22}=0;{\sigma }_{12}=S$

Therefore,

$\text{(106)}{\sigma }_{12}=-{\phi }_{,12}=S$

Integrating,

$\text{(107)}{\phi }_{,1}=-S{x}_2+f(x_1)\Rightarrow \phi =-S{x}_1{x}_2+\int f(x_1)d{x}_1$

Since $\phi$ is a potential, we can neglect the integration constants (these do not affect the stresses - which are what we are interested in). Hence,

$\text{(108)}\phi =-S{x}_1{x}_2=-S(r\cos \theta )(r\sin \theta )=-\frac{S{r}^2}{2}\sin (2\theta )$

or,

$\text{(109)}\phi =-\frac{S{r}^2}{2}\sin (2\theta )$

Note that we have arranged the expression so that it has a form similar to the Fourier series of the previous section.

Perturbed Solution

For this we have to add terms to $\phi$ in such a way that

• The unperturbed solution continues to be true as $r\to \mathrm{\infty }$.
• The terms have the same form as the unperturbed solution,i.e., $sin(2\theta )$ terms.
• The new $\phi$ leads to stresses that are proportional to $r^{-n}$.

Recall,

$\phi =\sum _{n=0}^{\mathrm{\infty }}{f}_n(r)\cos (n\theta )+\sum _{n=0}^{\mathrm{\infty }}{g}_n(r)\sin (n\theta )$

where,

$\begin{array}{cc}{f}_0(r)& =A_0{r}^2+B_0{r}^2\ln r+C_0+D_0\ln r\\ f_1(r)& =A_1{r}^3+B_{1}r+C_{1}r\ln r+D_1{r}^{-1}\\ f_n(r)& =A_n{r}^{n+2}+B_n{r}^n+C_n{r}^{-n+2}+D_n{r}^{-n},n>1\end{array}$

So the appropriate stress function for the perturbation is

$\text{(110)}\phi =g_2(r)\sin (2\theta )=(C_2{r}^{-2+2}+D_2{r}^{-2})\sin (2\theta )$

or,

$\text{(111)}\phi =(C_2+D_2{r}^{-2})\sin (2\theta )$

Hence, the stress function appropriate for the superposed solution is

$\text{(112)}\phi =-\frac{S{r}^2}{2}\sin (2\theta )+(C_2+D_2{r}^{-2})\sin (2\theta )$

We determine $C_2$ and $D_2$ using the boundary conditions at $r=a$.

The stresses are

$\begin{array}{cc}\text{(113)}{\sigma }_{rr}& =\frac{1}{r}\frac{\partial \phi }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2\phi }{\partial {\theta }^2}=(S-4C_2{r}^{-2}-6D_2{r}^{-4})\sin (2\theta )\\ \text{(114)}{\sigma }_{\theta \theta }& =\frac{{\partial }^2\phi }{\partial r^2}=(-S+6D_2{r}^{-4})\sin (2\theta )\\ \text{(115)}{\sigma }_{r\theta }& =-\frac{\partial }{\partial r}\left(\frac{1}{r}\frac{\partial \phi }{\partial \theta }\right)=(S+6D_2{r}^{-4})\cos (2\theta )\end{array}$

Hence,

$\begin{array}{cc}\text{(116)}{{\sigma }_{rr}|}_{r=a}& =0=(S-4C_2{a}^{-2}-6D_2{a}^{-4})\sin (2\theta )\\ \text{(117)}{{\sigma }_{r\theta }|}_{r=a}& =0=(S+2C_2{a}^{-2}+6D_2{a}^{-4})\cos (2\theta )\end{array}$

or,

$\begin{array}{cc}\text{(118)}4C_2{a}^{-2}+6D_2{a}^{-4}& =S\\ \text{(119)}2C_2{a}^{-2}+6D_2{a}^{-4}& =-S\end{array}$

Solving,

$\text{(120)}{C}_2=S{a}^2;D_2=-\frac{S{a}^4}{2}$

Back substituting,

$\begin{array}{cc}\text{(121)}{\sigma }_{rr}& =S(1-4\frac{a^2}{r^2}+3\frac{a^4}{r^4})\sin (2\theta )\\ \text{(122)}{\sigma }_{\theta \theta }& =S(-1-3\frac{a^4}{r^4})\sin (2\theta )\\ \text{(123)}{\sigma }_{r\theta }& =S(1+2\frac{a^2}{r^2}-3\frac{a^4}{r^4})\cos (2\theta )\end{array}$

Example homework problem

Consider the elastic plate with a hole subject to pure shear.

File:Elastic plate with hole shear.png Elastic plate with a circular hole under pure shear

The stresses close to the hole are given by

$\begin{array}{cc}\text{(29)}{\sigma }_{rr}& =S(1-4\frac{a^2}{r^2}+3\frac{a^4}{r^4})\sin (2\theta )\\ \text{(30)}{\sigma }_{\theta \theta }& =S(-1-3\frac{a^4}{r^4})\sin (2\theta )\\ \text{(31)}{\sigma }_{r\theta }& =S(1+2\frac{a^2}{r^2}-3\frac{a^4}{r^4})\cos (2\theta )\end{array}$

• Show that the normal and shear traction boundary conditions far from the hole are satisfied by these stresses.
• Calculate the stress concentration factors at the hole, i.e., (${\tau }_{\text{max}}/S$) (shear) and (${\sigma }_{\text{max}}/{\sigma }_0$) (normal).
• Calculate the displacement field corresponding to this stress field (for plane stress). Plot the deformed shape of the hole.

Solution

Far from the hole, $r=\mathrm{\infty }$. Therefore,

$\begin{array}{cc}\text{(32)}{\sigma }_{rr}& =S\sin (2\theta )\\ \text{(33)}{\sigma }_{\theta \theta }& =-S\sin (2\theta )\\ \text{(34)}{\sigma }_{r\theta }& =S\cos (2\theta )\end{array}$

To rotate the stresses back to the $(x_1,x_2)$ coordinate system, we use the tensor transformation rule

$\text{(35)}\left[\begin{array}{ccc}{\sigma }_{11}& {\sigma }_{12}& {\sigma }_{13}\\ {\sigma }_{21}& {\sigma }_{22}& {\sigma }_{23}\\ {\sigma }_{31}& {\sigma }_{32}& {\sigma }_{33}\end{array}\right]=\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}{\sigma }_{rr}& {\sigma }_{r\theta }& {\sigma }_{rz}\\ {\sigma }_{r\theta }& {\sigma }_{\theta \theta }& {\sigma }_{\theta z}\\ {\sigma }_{rz}& {\sigma }_{\theta z}& {\sigma }_{zz}\end{array}\right]\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0\\ -\sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]$

Setting ${\sigma }_{rz}=0$ and ${\sigma }_{\theta z}=0$, we get the simplified set of equations

$\begin{array}{cc}\text{(36)}{\sigma }_{11}& ={\sigma }_{rr}{\cos }^2\theta +{\sigma }_{\theta \theta }{\sin }^2\theta -{\sigma }_{r\theta }\sin (2\theta )\\ \text{(37)}{\sigma }_{22}& ={\sigma }_{rr}{\sin }^2\theta +{\sigma }_{\theta \theta }{\cos }^2\theta +{\sigma }_{r\theta }\sin (2\theta )\\ \text{(38)}{\sigma }_{12}& =\frac{{\sigma }_{rr}-{\sigma }_{\theta \theta }}{2}\sin (2\theta )+{\sigma }_{r\theta }\cos (2\theta )\end{array}$

Plugging in equations (32-34) in the above, we have

$\begin{array}{cc}\text{(39)}{\sigma }_{11}& =-S[\sin (2\theta )\cos (2\theta )-\sin (2\theta )\cos (2\theta )]=0\\ \text{(40)}{\sigma }_{22}& =S[\sin (2\theta )\cos (2\theta )-\sin (2\theta )\cos (2\theta )]=0\\ \text{(41)}{\sigma }_{12}& =S[\sin (2\theta )\sin (2\theta )+\cos (2\theta )\cos (2\theta )]=S\end{array}$

Hence, the far field stress BCs are satisfied.

The stresses at the hole ($r=a$) are

$\begin{array}{cc}\text{(42)}{\sigma }_{rr}& =S(1-4+3)\sin (2\theta )=0\\ \text{(43)}{\sigma }_{\theta \theta }& =S(-1-3)\sin (2\theta )=-4S\sin (2\theta )\\ \text{(44)}{\sigma }_{r\theta }& =S(1+2-3)\cos (2\theta )=0\end{array}$

The maximum (or minimum) hoop stress at the hole is at the locations where $d{\sigma }_{\theta \theta }/d\theta =-8S\cos (2\theta )=0$. These locations are $\theta =\pi /4$ and $\theta =3\pi /4$. The value of the hoop stress is

$\begin{array}{ccc}\text{(45)}\text{at}\theta =\frac{\pi }{4}& & {\sigma }_{\theta \theta }=-4S\\ \text{(46)}\text{at}\theta =\frac{3\pi }{4}& & {\sigma }_{\theta \theta }=4S\end{array}$

The maximum shear stress is given by

$\text{(47)}{\tau }_{\text{max}}=\frac{1}{2}|{\sigma }_{rr}-{\sigma }_{\theta \theta }|=2S$

Therefore, the stress concentration factors are

$\text{(48)}\frac{{\sigma }_{\text{max}}}{S}=4;\frac{{\tau }_{\text{max}}}{S}=2$

The stress function used to derive the above results was

$\text{(49)}\phi =-\frac{S}{2}{r}^2\sin (2\theta )+S{a}^2\sin (2\theta )-\frac{S{a}^4}{2}{r}^{-2}\sin (2\theta )$

From Michell's solution, the displacements corresponding to the above stress function are given by

$\begin{array}{cc}\text{(50)}2\mu u_r& =-\frac{S}{2}[-2r\sin (2\theta )]+S{a}^2[(\kappa +1)r^{-1}\sin (2\theta )]-\frac{S{a}^4}{2}[2r^{-3}\sin (2\theta )]\\ \text{(51)}2\mu u_{\theta }& =-\frac{S}{2}[-2r\cos (2\theta )]+S{a}^2[(\kappa -1)r^{-1}\cos (2\theta )]-\frac{S{a}^4}{2}[-2r^{-3}\cos (2\theta )]\end{array}$

or,

$\begin{array}{cc}\text{(52)}{u}_r& =\frac{Sr\sin (2\theta )}{2\mu }[1+(\kappa +1)\frac{a^2}{r^2}-\frac{a^4}{r^4}]\\ \text{(53)}{u}_{\theta }& =\frac{Sr\cos (2\theta )}{2\mu }[1+(\kappa -1)\frac{a^2}{r^2}+\frac{a^4}{r^4}]\end{array}$

For plane stress, $\kappa =(3-\nu )/(1+\nu )$. Hence,

$\begin{array}{cc}\text{(54)}{u}_r& =\frac{Sr\sin (2\theta )}{2\mu }[1+\left(\frac{4}{1+\nu }\right)\frac{a^2}{r^2}-\frac{a^4}{r^4}]\\ \text{(55)}{u}_{\theta }& =\frac{Sr\cos (2\theta )}{2\mu }[1+2\left(\frac{1-\nu }{1+\nu }\right)\frac{a^2}{r^2}+\frac{a^4}{r^4}]\end{array}$

At $r=a$,

$\begin{array}{cc}\text{(56)}{u}_r& =\frac{Sa\sin (2\theta )}{\mu }\left(\frac{2}{1+\nu }\right)\\ \text{(57)}{u}_{\theta }& =\frac{Sa\cos (2\theta )}{\mu }\left(\frac{2}{1+\nu }\right)\end{array}$

Now $\mu =E/2(1+\nu )$. Hence, we have

$\begin{array}{cc}\text{(58)}{u}_r& =\frac{4Sa\sin (2\theta )}{E}\\ \text{(59)}{u}_{\theta }& =\frac{4Sa\cos (2\theta )}{E}\end{array}$

The deformed shape is shown below

File:Shear hole deform.png Displacement field near a hole in plate under pure shear

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