# Introduction to Elasticity/Plate with hole in shear

## Circular hole in a shear field

 Elastic plate with circular hole under shear

Given:

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

The BCs are

at ${\displaystyle r=a}$
${\displaystyle {\text{(103)}}\qquad t_{r}=t_{\theta }=0~;~~{\widehat {\mathbf {n} }}=-{\widehat {\mathbf {e} }}~r\Rightarrow \sigma _{rr}=\sigma _{r\theta }=0}$
at ${\displaystyle r\rightarrow \infty }$
${\displaystyle {\text{(104)}}\qquad \sigma _{12}\rightarrow S~;~~\sigma _{11}\rightarrow 0~;~~\sigma _{22}\rightarrow 0}$

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 ${\displaystyle r^{-n}\,}$.

### Unperturbed Solution

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

Therefore,

${\displaystyle {\text{(106)}}\qquad \sigma _{12}=-\varphi _{,12}=S}$

Integrating,

${\displaystyle {\text{(107)}}\qquad \varphi _{,1}=-Sx_{2}+f(x_{1})\Rightarrow \varphi =-Sx_{1}x_{2}+\int f(x_{1})dx_{1}}$

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

${\displaystyle {\text{(108)}}\qquad \varphi =-Sx_{1}x_{2}=-S(r\cos \theta )(r\sin \theta )=-{\cfrac {Sr^{2}}{2}}\sin(2\theta )}$

or,

${\displaystyle {\text{(109)}}\qquad \varphi =-{\cfrac {Sr^{2}}{2}}\sin(2\theta )}$

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 ${\displaystyle \varphi }$ in such a way that

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

Recall,

${\displaystyle \varphi =\sum _{n=0}^{\infty }f_{n}(r)\cos(n\theta )+\sum _{n=0}^{\infty }g_{n}(r)\sin(n\theta )}$

where,

{\displaystyle {\begin{aligned}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{aligned}}}

So the appropriate stress function for the perturbation is

${\displaystyle {\text{(110)}}\qquad \varphi =g_{2}(r)\sin(2\theta )=\left(C_{2}r^{-2+2}+D_{2}r^{-2}\right)\sin(2\theta )}$

or,

${\displaystyle {\text{(111)}}\qquad \varphi =\left(C_{2}+D_{2}r^{-2}\right)\sin(2\theta )}$

Hence, the stress function appropriate for the superposed solution is

${\displaystyle {\text{(112)}}\varphi =-{\cfrac {Sr^{2}}{2}}\sin(2\theta )+\left(C_{2}+D_{2}r^{-2}\right)\sin(2\theta )}$

We determine ${\displaystyle C_{2}}$ and ${\displaystyle D_{2}}$ using the boundary conditions at ${\displaystyle r=a}$.

The stresses are

{\displaystyle {\begin{aligned}{\text{(113)}}\qquad \sigma _{rr}&={\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial r}}+{\cfrac {1}{r^{2}}}{\cfrac {\partial ^{2}\varphi }{\partial \theta ^{2}}}=\left(S-4C_{2}r^{-2}-6D_{2}r^{-4}\right)\sin(2\theta )\\{\text{(114)}}\qquad \sigma _{\theta \theta }&={\cfrac {\partial ^{2}\varphi }{\partial r^{2}}}=\left(-S+6D_{2}r^{-4}\right)\sin(2\theta )\\{\text{(115)}}\qquad \sigma _{r\theta }&=-{\cfrac {\partial }{\partial r}}\left({\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial \theta }}\right)=\left(S+6D_{2}r^{-4}\right)\cos(2\theta )\end{aligned}}}

Hence,

{\displaystyle {\begin{aligned}{\text{(116)}}\qquad \left.\sigma _{rr}\right|_{r=a}&=0=\left(S-4C_{2}a^{-2}-6D_{2}a^{-4}\right)\sin(2\theta )\\{\text{(117)}}\qquad \left.\sigma _{r\theta }\right|_{r=a}&=0=\left(S+2C_{2}a^{-2}+6D_{2}a^{-4}\right)\cos(2\theta )\end{aligned}}}

or,

{\displaystyle {\begin{aligned}{\text{(118)}}\qquad 4C_{2}a^{-2}+6D_{2}a^{-4}&=S\\{\text{(119)}}\qquad 2C_{2}a^{-2}+6D_{2}a^{-4}&=-S\end{aligned}}}

Solving,

${\displaystyle {\text{(120)}}\qquad C_{2}=Sa^{2}~;~~D_{2}=-{\cfrac {Sa^{4}}{2}}}$

Back substituting,

{\displaystyle {\begin{aligned}{\text{(121)}}\qquad \sigma _{rr}&=S\left(1-4{\cfrac {a^{2}}{r^{2}}}+3{\cfrac {a^{4}}{r^{4}}}\right)\sin(2\theta )\\{\text{(122)}}\qquad \sigma _{\theta \theta }&=S\left(-1-3{\cfrac {a^{4}}{r^{4}}}\right)\sin(2\theta )\\{\text{(123)}}\qquad \sigma _{r\theta }&=S\left(1+2{\cfrac {a^{2}}{r^{2}}}-3{\cfrac {a^{4}}{r^{4}}}\right)\cos(2\theta )\end{aligned}}}

## Example homework problem

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

 Elastic plate with a circular hole under pure shear

The stresses close to the hole are given by

{\displaystyle {\begin{aligned}{\text{(29)}}\qquad \sigma _{rr}&=S\left(1-4{\frac {a^{2}}{r^{2}}}+3{\frac {a^{4}}{r^{4}}}\right)\sin(2\theta )\\{\text{(30)}}\qquad \sigma _{\theta \theta }&=S\left(-1-3{\frac {a^{4}}{r^{4}}}\right)\sin(2\theta )\\{\text{(31)}}\qquad \sigma _{r\theta }&=S\left(1+2{\frac {a^{2}}{r^{2}}}-3{\frac {a^{4}}{r^{4}}}\right)\cos(2\theta )\end{aligned}}}
• 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., (${\displaystyle \tau _{\text{max}}/S}$) (shear) and (${\displaystyle \sigma _{\text{max}}/\sigma _{0}}$) (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, ${\displaystyle r=\infty }$. Therefore,

{\displaystyle {\begin{aligned}{\text{(32)}}\qquad \sigma _{rr}&=S\sin(2\theta )\\{\text{(33)}}\qquad \sigma _{\theta \theta }&=-S\sin(2\theta )\\{\text{(34)}}\qquad \sigma _{r\theta }&=S\cos(2\theta )\end{aligned}}}

To rotate the stresses back to the ${\displaystyle (x_{1},x_{2})}$ coordinate system, we use the tensor transformation rule

${\displaystyle {\text{(35)}}\qquad {\begin{bmatrix}\sigma _{11}&\sigma _{12}&\sigma _{13}\\\sigma _{21}&\sigma _{22}&\sigma _{23}\\\sigma _{31}&\sigma _{32}&\sigma _{33}\end{bmatrix}}={\begin{bmatrix}\cos \theta &-\sin \theta &0\\\sin \theta &\cos \theta &0\\0&0&1\end{bmatrix}}{\begin{bmatrix}\sigma _{rr}&\sigma _{r\theta }&\sigma _{rz}\\\sigma _{r\theta }&\sigma _{\theta \theta }&\sigma _{\theta z}\\\sigma _{rz}&\sigma _{\theta z}&\sigma _{zz}\end{bmatrix}}{\begin{bmatrix}\cos \theta &\sin \theta &0\\-\sin \theta &\cos \theta &0\\0&0&1\end{bmatrix}}}$

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

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

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

{\displaystyle {\begin{aligned}{\text{(39)}}\qquad \sigma _{11}&=-S\left[\sin(2\theta )\cos(2\theta )-\sin(2\theta )\cos(2\theta )\right]=0\\{\text{(40)}}\qquad \sigma _{22}&=S\left[\sin(2\theta )\cos(2\theta )-\sin(2\theta )\cos(2\theta )\right]=0\\{\text{(41)}}\qquad \sigma _{12}&=S\left[\sin(2\theta )\sin(2\theta )+\cos(2\theta )\cos(2\theta )\right]=S\end{aligned}}}

Hence, the far field stress BCs are satisfied.

The stresses at the hole (${\displaystyle r=a}$) are

{\displaystyle {\begin{aligned}{\text{(42)}}\qquad \sigma _{rr}&=S\left(1-4+3\right)\sin(2\theta )=0\\{\text{(43)}}\qquad \sigma _{\theta \theta }&=S\left(-1-3\right)\sin(2\theta )=-4S\sin(2\theta )\\{\text{(44)}}\qquad \sigma _{r\theta }&=S\left(1+2-3\right)\cos(2\theta )=0\end{aligned}}}

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

{\displaystyle {\begin{aligned}{\text{(45)}}\qquad {\text{at}}~\theta ={\frac {\pi }{4}}&&\sigma _{\theta \theta }=-4S\\{\text{(46)}}\qquad {\text{at}}~\theta ={\frac {3\pi }{4}}&&\sigma _{\theta \theta }=4S\end{aligned}}}

The maximum shear stress is given by

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

Therefore, the stress concentration factors are

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

The stress function used to derive the above results was

${\displaystyle {\text{(49)}}\qquad \varphi =-{\frac {S}{2}}r^{2}\sin(2\theta )+Sa^{2}\sin(2\theta )-{\frac {Sa^{4}}{2}}r^{-2}\sin(2\theta )}$

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

{\displaystyle {\begin{aligned}{\text{(50)}}\qquad 2\mu u_{r}&=-{\frac {S}{2}}\left[-2r\sin(2\theta )\right]+Sa^{2}\left[(\kappa +1)r^{-1}\sin(2\theta )\right]-{\frac {Sa^{4}}{2}}\left[2r^{-3}\sin(2\theta )\right]\\{\text{(51)}}\qquad 2\mu u_{\theta }&=-{\frac {S}{2}}\left[-2r\cos(2\theta )\right]+Sa^{2}\left[(\kappa -1)r^{-1}\cos(2\theta )\right]-{\frac {Sa^{4}}{2}}\left[-2r^{-3}\cos(2\theta )\right]\end{aligned}}}

or,

{\displaystyle {\begin{aligned}{\text{(52)}}\qquad u_{r}&={\frac {Sr\sin(2\theta )}{2\mu }}\left[1+(\kappa +1){\frac {a^{2}}{r^{2}}}-{\frac {a^{4}}{r^{4}}}\right]\\{\text{(53)}}\qquad u_{\theta }&={\frac {Sr\cos(2\theta )}{2\mu }}\left[1+(\kappa -1){\frac {a^{2}}{r^{2}}}+{\frac {a^{4}}{r^{4}}}\right]\end{aligned}}}

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

{\displaystyle {\begin{aligned}{\text{(54)}}\qquad u_{r}&={\frac {Sr\sin(2\theta )}{2\mu }}\left[1+\left({\frac {4}{1+\nu }}\right){\frac {a^{2}}{r^{2}}}-{\frac {a^{4}}{r^{4}}}\right]\\{\text{(55)}}\qquad u_{\theta }&={\frac {Sr\cos(2\theta )}{2\mu }}\left[1+2\left({\frac {1-\nu }{1+\nu }}\right){\frac {a^{2}}{r^{2}}}+{\frac {a^{4}}{r^{4}}}\right]\end{aligned}}}

At ${\displaystyle r=a}$,

{\displaystyle {\begin{aligned}{\text{(56)}}\qquad u_{r}&={\frac {Sa\sin(2\theta )}{\mu }}\left({\frac {2}{1+\nu }}\right)\\{\text{(57)}}\qquad u_{\theta }&={\frac {Sa\cos(2\theta )}{\mu }}\left({\frac {2}{1+\nu }}\right)\end{aligned}}}

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

{\displaystyle {\begin{aligned}{\text{(58)}}\qquad u_{r}&={\frac {4Sa\sin(2\theta )}{E}}\\{\text{(59)}}\qquad u_{\theta }&={\frac {4Sa\cos(2\theta )}{E}}\end{aligned}}}

The deformed shape is shown below

 Displacement field near a hole in plate under pure shear