Elasticity/Plate with hole in tension

The perturbed solution is

${\displaystyle {\text{(128)}}\qquad \varphi ={\cfrac {Tr^{2}}{4}}-{\cfrac {Tr^{2}\cos(2\theta )}{4}}+A\ln(r)+B\theta +C\cos(2\theta )+Dr^{-2}\cos(2\theta )}$

After applying the BCS, we get

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

The stress concentration factor, often referred to as K_t, in this case is ${\displaystyle 3}$ and is the same in both tension and shear.

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 (${\displaystyle r=a}$) are

${\displaystyle {\begin{aligned}\sigma _{rr}&=0\\\sigma _{\theta \theta }&=T-2T\cos(2\theta )\\\sigma _{r\theta }&=0\end{aligned}}}$

The maximum hoop stress is given at ${\displaystyle \theta =0}$ or ${\displaystyle \theta =\pi /2}$.

At ${\displaystyle \theta =0}$, ${\displaystyle \sigma _{\theta \theta }=-T}$.

At ${\displaystyle \theta =\pi /2}$, ${\displaystyle \sigma _{\theta \theta }=3T}$.

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

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

Both stress concentration factors are equal.

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

${\displaystyle \sigma _{\theta \theta }=T/2(1-\cos 2\theta )}$

The ratio is

${\displaystyle {\text{ratio}}=1+{\frac {3a^{4}}{2r^{4}}}+{\frac {a^{2}}{2r^{2}}}}$

This ratio is 0.95 when ${\displaystyle r\approx 3.5a}$, i.e., at a distance of ${\displaystyle 1.75}$ diameters from the center.

The given stress function is

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

Therefore, the displacement field from the Michell solution is

${\displaystyle {\begin{aligned}2\mu u_{r}&={\frac {T}{4}}\left[(\kappa -1)r\right]-{\frac {T}{4}}\left[-2r\cos(2\theta )\right]+A\left[-{\frac {1}{r}}\right]+C\left[(\kappa +1)r^{-1}\cos(2\theta )\right]+D\left[2r^{-3}\cos(2\theta )\right]\\2\mu u_{\theta }&=-{\frac {T}{4}}\left[2r\sin(2\theta )\right]+C\left[-(\kappa -1)r^{-1}\sin(2\theta )\right]+D\left[2r^{-3}\sin(2\theta )\right]\end{aligned}}}$

From the stress calculation step, we have

${\displaystyle A=-{\frac {Ta^{2}}{2}}~;~~B=0~;~~C={\frac {Ta^{2}}{2}}~;~~D=-{\frac {Ta^{4}}{4}}}$

After substituting the constants and collecting terms,

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

Replacing ${\displaystyle \mu }$ with ${\displaystyle {\frac {E}{2(1+\nu )}}}$, and ${\displaystyle \kappa }$ with ${\displaystyle {\frac {3-\nu }{1+\nu }}}$ (for plane stress conditions), we get

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

At ${\displaystyle r=a}$,

${\displaystyle {\begin{aligned}u_{r}&={\frac {Ta}{E}}\left[1+2\cos(2\theta )\right]\\u_{\theta }&=-{\frac {2Ta}{E}}\sin(2\theta )\end{aligned}}}$

The deformed shape is shown below:

In cartesian coordinates, the displacement field is given by

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