Elasticity/Plate with hole in tension

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Plate with hole in a tensile field[edit | edit source]

An elastic plate with a circular hole under tension

The BCs are

Unperturbed Solution[edit | edit source]

The unperturbed part of the Michell solution gives us


The first term is the axisymmetric term while the second term is the periodic term.

Perturbation[edit | edit source]

Similar to previous problem, but we simply choose terms from the Michell solution of the same form (i.e. containing ) and such that the stresses decay with increasing radius. The relevant terms from the table are:

Perturbed Solution[edit | edit source]

The perturbed solution is

After applying the BCS, we get

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

Example homework problem[edit | edit source]

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

leads to the stresses

or, in cartesian coordinates:

  • 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[edit | edit source]

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) +   

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 () are

The maximum hoop stress is given at or .

At , .

At , .

The maximum shear stress at is while that at is .

Therefore, the stress concentration factor in tension is , while that in shear is .

Both stress concentration factors are equal.

Let us look at the ratio of the hoop stress at to the far field hoop stress

The ratio is

This ratio is 0.95 when , i.e., at a distance of diameters from the center.

The given stress function is

Therefore, the displacement field from the Michell solution is

From the stress calculation step, we have

After substituting the constants and collecting terms,

Replacing with , and with (for plane stress conditions), we get

At ,

The deformed shape is shown below:

Deformation of the hole under tension

In cartesian coordinates, the displacement field is given by