Elasticity/Plate with hole in shear

Circular hole in a shear field
Given:
 * Large plate in pure shear.
 * Stress state perturbed by a small hole.

The BCs are
 * at $$ r = a $$

\text{(103)} \qquad t_r = t_{\theta} = 0 ~; \widehat{\mathbf{n}} = -\widehat{\mathbf{e}}~r \Rightarrow \sigma_{rr} = \sigma_{r\theta} = 0 $$


 * at $$ r \rightarrow \infty $$

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

Unperturbed Solution

 * $$ \text{(105)} \qquad

\sigma_{11} = \sigma_{22} = 0 ~; \sigma_{12} = S $$ Therefore,
 * $$\text{(106)} \qquad

\sigma_{12} = -\varphi_{,12} = S $$ Integrating,
 * $$\text{(107)} \qquad

\varphi_{,1} = -Sx_2 + f(x_1) \Rightarrow \varphi = -Sx_1x_2 + \int f(x_1) dx_1 $$ Since $$\varphi$$ is a potential, we can neglect the integration constants (these do not affect the stresses - which are what we are interested in). Hence,
 * $$ \text{(108)} \qquad

\varphi = -Sx_1x_2 = -S (r\cos\theta) (r\sin\theta) = -\cfrac{Sr^2}{2}\sin(2\theta) $$ or,
 * $$ \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 $$\varphi$$ in such a way that


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

Recall,

\varphi = \sum^{\infty}_{n=0} f_n(r) \cos(n\theta) + \sum^{\infty}_{n=0} g_n(r) \sin(n\theta) $$ where,
 * $$\begin{align}

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{align} $$ So the appropriate stress function for the perturbation is
 * $$\text{(110)}\qquad

\varphi = g_2(r) \sin(2\theta) = \left(C_2 r^{-2+2} + D_2 r^{-2}\right) \sin(2\theta) $$ or,
 * $$\text{(111)} \qquad

\varphi = \left(C_2 + D_2 r^{-2}\right)\sin(2\theta) $$ Hence, the stress function appropriate for the superposed solution is
 * $$\text{(112)}

\varphi = -\cfrac{Sr^2}{2}\sin(2\theta) + \left(C_2 + D_2 r^{-2}\right)\sin(2\theta) $$ We determine $$C_2$$ and $$D_2$$ using the boundary conditions at $$r=a$$.

The stresses are
 * $$\begin{align}

\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_2r^{-2} - 6D_2r^{-4}\right)\sin(2\theta) \\ \text{(114)}\qquad\sigma_{\theta\theta} & = \cfrac{\partial^2\varphi}{\partial r^2} = \left(-S + 6D_2r^{-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_2r^{-4}\right)\cos(2\theta) \end{align}$$ Hence,
 * $$\begin{align}

\text{(116)}\qquad\left.\sigma_{rr}\right|_{r=a} & = 0 = \left(S - 4C_2a^{-2} - 6D_2a^{-4}\right)\sin(2\theta) \\ \text{(117)}\qquad\left.\sigma_{r\theta}\right|_{r=a} & = 0 = \left(S + 2C_2a^{-2} + 6D_2a^{-4}\right)\cos(2\theta) \end{align}$$ or,
 * $$\begin{align}

\text{(118)}\qquad 4C_2a^{-2} + 6D_2a^{-4} & = S \\ \text{(119)}\qquad 2C_2a^{-2} + 6D_2a^{-4} & = -S \end{align}$$ Solving,
 * $$\text{(120)}\qquad

C_2 = Sa^2 ~; D_2 = -\cfrac{Sa^4}{2} $$ Back substituting,
 * $$\begin{align}

\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{align}$$

Example homework problem
Consider the elastic plate with a hole subject to pure shear. The stresses close to the hole are given by
 * $$\begin{align}

\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{align}$$
 * Show that the normal and shear traction boundary conditions far from the hole are satisfied by these stresses.
 * Calculate the stress concentration factors at the hole, i.e., ($$\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, $$r = \infty$$. Therefore,
 * $$\begin{align}

\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{align}$$ To rotate the stresses back to the $$(x_1,x_2)$$ coordinate system, we use the tensor transformation rule
 * $$\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} $$ Setting $$\sigma_{rz} = 0$$ and $$\sigma_{\theta z} = 0$$, we get the simplified set of equations
 * $$\begin{align}

\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{align}$$ Plugging in equations (32-34) in the above, we have
 * $$\begin{align}

\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{align}$$ Hence, the far field stress BCs are satisfied.

The stresses at the hole ($$r = a$$) are
 * $$\begin{align}

\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{align}$$

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{align}

\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{align}$$ The maximum shear stress is given by
 * $$ \text{(47)} \qquad

\tau_{\text{max}} = \frac{1}{2}\left|\sigma_{rr}-\sigma_{\theta\theta}\right| = 2S $$ Therefore, the stress concentration factors are
 * $$\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
 * $$\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) $$ From Michell's solution, the displacements corresponding to the above stress function are given by
 * $$\begin{align}

\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{align}$$ or,
 * $$\begin{align}

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

\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{align}$$ At $$r = a$$,
 * $$\begin{align}

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

\text{(58)} \qquad u_r & = \frac{4Sa\sin(2\theta)}{E} \\ \text{(59)} \qquad u_{\theta} & = \frac{4Sa\cos(2\theta)}{E} \end{align}$$ The deformed shape is shown below