Introduction to Elasticity/Plate with hole in tension

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[edit] Plate with hole in a tensile field

An elastic plate with a circular hole under tension

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}

[edit] 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.

[edit] Perturbation

Similar to previous problem, but we simply choose terms from the Michell solution of the same form (i.e. containing cos(2θ)) 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)

[edit] 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{S}{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 in this case is 3 and is the same in both tension and shear.

[edit] Example homework problem

Consider the elastic plate with a hole subject to uniaxial tension.

Elastic plate with small circular hole under 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}
  • 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.

[edit] 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(srr1, r=a)),{cos});
srta := collect(simplify(eval(srt1, 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 θ = 0 or θ = π / 2.

At θ = 0, σθθ = − T.

At θ = π / 2, σθθ = 3T.

The maximum shear stress at r = a is τmax = 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 θ = π / 2 to the far field hoop stress

σθθ = T / 2(1 − cos2θ)

The ratio is

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

This ratio is 0.95 when r \approx 3.5 a, 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 μ with E / 2(1 + ν), and κ with 3 − ν / 1 + ν, we get

\begin{align} u_r & = \frac{Tr\cos(2\theta)}{2E}\left[ (1+\nu)+\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

Deformation of the hole under tension
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