Elasticity/Plate with hole in tension

Plate with hole in a tensile field
The BCs are
 * $$\begin{align}

\text{(124)} & \qquad \text{at}~  r = a & &  t_r = t_{\theta} = 0 ~; \widehat{\mathbf{n}} = -\widehat{\mathbf{e}}~{r} \Rightarrow \sigma_{rr} = \sigma_{r\theta} = 0 \\ \text{(125)} & \qquad \text{at}~ r \rightarrow \infty & & \sigma_{11} \rightarrow T ~; \sigma_{12} \rightarrow 0 ~; \sigma_{22} \rightarrow 0 \end{align}$$

Unperturbed Solution
The unperturbed part of the Michell solution gives us

\varphi = \cfrac{Tx_2^2}{2} = \cfrac{T (r\sin\theta)^2}{2} = \cfrac{Tr^2}{4} - \cfrac{Tr^2\cos(2\theta)}{4} $$ or,
 * $$\text{(126)} \qquad

\varphi = \cfrac{Tr^2}{4} - \cfrac{Tr^2\cos(2\theta)}{4} $$ The first term is the axisymmetric term while the second term is the periodic term.

Perturbation
Similar to previous problem, but we simply choose terms from the Michell solution of the same form (i.e. containing $$\cos(2\theta)$$) and such that the stresses decay with increasing radius. The relevant terms from the table are:
 * $$ \text{(127)} \qquad

\ln(r) ~, \theta ~, r^{-2+2}\cos(2\theta) ~,r^{-2}\cos(2\theta) $$

Perturbed Solution
The perturbed solution is
 * $$ \text{(128)} \qquad

\varphi = \cfrac{Tr^2}{4} - \cfrac{Tr^2\cos(2\theta)}{4} + A\ln(r) + B\theta + C\cos(2\theta) + Dr^{-2}\cos(2\theta) $$ After applying the BCS, we get
 * $$ \begin{align}

\text{(129)}\qquad\sigma_{rr} & = \cfrac{T}{2}\left(1-\cfrac{a^2}{r^2}\right) + \cfrac{T\cos(2\theta)}{2} \left(\cfrac{3a^4}{r^4} - \cfrac{4a^2}{r^2} + 1\right)\\ \text{(130)}\qquad\sigma_{\theta\theta} & = \cfrac{T}{2}\left(1+\cfrac{a^2}{r^2}\right) - \cfrac{T\cos(2\theta)}{2} \left(\cfrac{3a^4}{r^4} + 1\right)\\ \text{(131)}\qquad\sigma_{r\theta} & = \cfrac{T\sin(2\theta)}{2} \left(\cfrac{3a^4}{r^4} - \cfrac{2a^2}{r^2} - 1\right) \end{align}$$

The stress concentration factor, often referred to as Kt, in this case is $$3$$ and is the same in both tension and shear.

Example homework problem
Consider the elastic plate with a hole subject to uniaxial tension.
 * Show that the stress function

\varphi = \frac{Tr^2}{4} - \frac{Tr^2\cos(2\theta)}{4} + A\ln(r) + B\theta               + C\cos(2\theta) + Dr^{-2}\cos(2\theta) $$ leads to the stresses

\begin{align} \sigma_{rr} & = \frac{T}{2}\left(1-\frac{a^2}{r^2}\right) +       \frac{T\cos(2\theta)}{2}        \left(\frac{3a^4}{r^4} - \frac{4a^2}{r^2} + 1\right) \\ \sigma_{\theta\theta} & = \frac{T}{2}\left(1+\frac{a^2}{r^2}\right) -       \frac{T\cos(2\theta)}{2}        \left(\frac{3a^4}{r^4} + 1\right)\\ \sigma_{r\theta} & = \frac{T\sin(2\theta)}{2}       \left(\frac{3a^4}{r^4} - \frac{2a^2}{r^2} - 1\right) \end{align} $$ or, in cartesian coordinates:

$$ \begin{align} \sigma_{xx}(r,\theta)&=T-T\frac{a^{2}}{r^{2}}(\frac{3}{2}\cos 2\theta+\cos 4\theta)+T\frac{3a^4}{2r^4}\cos 4\theta,\\ \sigma_{yy}(r,\theta)&=-T\frac{a^2}{r^2}(\frac{1}{2}\cos 2\theta-\cos 4\theta)-T\frac{3a^4}{2r^4}\cos 4\theta,\\ \tau_{xy}(r,\theta)&=-T\frac{a^2}{r^2}(\frac{1}{2}\sin 2\theta+\sin 4\theta)+T\frac{3a^4}{2r^4}\sin 4\theta \end{align} $$
 * Calculate the stress concentration factors at the hole, both in shear and in tension, and show that they are the same.  How far from the hole (in units of hole diameters) does the stress reach 95% of the far field (unperturbed) value?
 * Calculate the displacement field corresponding to this stress field (for plane stress). Plot the deformed shape of the hole.

Solution
We can use the following Maple code to show the above results. phi := T*r^2/4*(1 - cos(2*theta)) + A*ln(r) + B*theta + C*cos(2*theta) + D/r^2*cos(2*theta);

srr := 1/r*diff(phi,r) + 1/r^2*diff(phi,theta,theta); stt := diff(phi,r,r); srt := -diff((1/r*diff(phi,theta)),r);

srra := collect(simplify(eval(srr, r=a)),{cos}); srta := collect(simplify(eval(srt, r=a)),{cos});

eq1 := coeff(srra, cos(2*theta)); eq2 := coeff(srta, sin(2*theta)); eq3 := 1/2*(T*a^4+2*A*a^2)/a^4; eq4 := 1/a^2*B;

BB := solve({eq4=0},{B}); AA := solve({eq3=0},{A});

sol := solve({eq1=0,eq2=0},{C,D});

phi := subs(BB, phi); phi := subs(AA, phi); phi := subs(sol, phi);

srr2 := 1/r*diff(phi,r) + 1/r^2*diff(phi,theta,theta); stt2 := diff(phi,r,r); srt2 := -diff((1/r*diff(phi,theta)),r);

srr3 := collect(simplify(srr2),{cos}); stt3 := collect(simplify(stt2),{cos}); srt3 := collect(simplify(srt2),{cos});

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

\begin{align} \sigma_{rr} & = 0 \\ \sigma_{\theta\theta} & = T - 2T\cos(2\theta) \\ \sigma_{r\theta} & = 0 \end{align} $$ The maximum hoop stress is given at $$\theta = 0$$ or $$\theta = \pi/2$$.

At $$\theta = 0$$, $$\sigma_{\theta\theta} = -T$$.

At $$\theta = \pi/2$$, $$\sigma_{\theta\theta} = 3T$$.

The maximum shear stress at $$r = a$$ is $$\tau_{\text{max}} = 1.5T$$ while that at $$r = \infty$$ is $$0.5T$$.

Therefore, the stress concentration factor in tension is $$3T/T = 3$$, while that in shear is $$1.5T/0.5T = 3$$.

Both stress concentration factors are equal.

Let us look at the ratio of the hoop stress at $$\theta = \pi/2$$ to the far field hoop stress

\sigma_{\theta\theta} = T/2(1-\cos 2\theta) $$ The ratio is

\text{ratio} = 1 + \frac{3a^4}{2r^4} + \frac{a^2}{2r^2} $$ This ratio is 0.95 when $$r \approx 3.5a$$, i.e., at a distance of $$1.75$$ diameters from the center.

The given stress function is

\varphi = \frac{Tr^2}{4} - \frac{Tr^2\cos(2\theta)}{4} + A\ln(r) + B\theta + C\cos(2\theta) + Dr^{-2}\cos(2\theta) $$ Therefore, the displacement field from the Michell solution is
 * $$\begin{align}

2\mu u_r & =  \frac{T}{4}\left[(\kappa-1)r\right] -  \frac{T}{4}\left[-2r\cos(2\theta)\right] +  A\left[-\frac{1}{r}\right] +  C\left[(\kappa+1)r^{-1}\cos(2\theta)\right] +  D\left[2r^{-3}\cos(2\theta)\right]   \\ 2\mu u_{\theta} & = -  \frac{T}{4}\left[2r\sin(2\theta)\right] +  C\left[-(\kappa-1)r^{-1}\sin(2\theta)\right] +  D\left[2r^{-3}\sin(2\theta)\right] \end{align} $$ From the stress calculation step, we have

A = -\frac{Ta^2}{2} ~; B = 0 ~; C = \frac{Ta^2}{2} ~; D = -\frac{Ta^4}{4} $$ After substituting the constants and collecting terms,
 * $$\begin{align}

2\mu u_r & = \frac{Tr\cos(2\theta)}{2}\left[ 1 + (\kappa+1)\frac{a^2}{r^2} - \frac{a^4}{r^4}\right] + \frac{Tr}{4}\left[(\kappa-1) + 2\frac{a^2}{r^2}\right] \\ 2\mu u_{\theta} & =  -\frac{Tr\sin(2\theta)}{2}\left[ 1 + (\kappa-1)\frac{a^2}{r^2} + \frac{a^4}{r^4}\right] \end{align}$$ Replacing $$\mu$$ with $$\frac{E}{2(1+\nu)}$$, and $$\kappa$$ with $$\frac{3-\nu}{1+\nu}$$ (for plane stress conditions), we get
 * $$\begin{align}

u_r & = \frac{Tr\cos(2\theta)}{2E}\left[ (1+\nu)+4\frac{a^2}{r^2}-(1+\nu)\frac{a^4}{r^4}\right] + \frac{Tr}{2E}\left[(1-\nu)+(1+\nu)\frac{a^2}{r^2}\right] \\ u_{\theta} & =  -\frac{Tr\sin(2\theta)}{2E}\left[ (1+\nu) + 2(1-\nu)\frac{a^2}{r^2}+(1+\nu)\frac{a^4}{r^4}\right] \end{align}$$ At $$r = a$$,
 * $$\begin{align}

u_r & = \frac{Ta}{E}\left[1+2\cos(2\theta)\right] \\ u_{\theta} & = - \frac{2Ta}{E} \sin(2\theta) \end{align}$$ The deformed shape is shown below: In cartesian coordinates, the displacement field is given by
 * $$\begin{align}u_{x}(r,\theta)&=&\frac{Ta}{8\mu}\left[\frac{r}{a}(\kappa+1)\cos\theta + \frac{2a}{r}((1+\kappa)\cos\theta + \cos3\theta) - \frac{2a^3}{r^3}\cos3\theta\right],\\

u_{y}(r,\theta)&=&\frac{Ta}{8\mu}\left[\frac{r}{a}(\kappa-3)\sin\theta + \frac{2a}{r}((1-\kappa)\sin\theta + \sin3\theta) - \frac{2a^3}{r^3}\sin3\theta\right]\end{align}$$