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Stress Concentration

Example: A flat plate with a cross-section of h =3 in by w = 0.5 in is subjected to a centric

load P as shown in the figure below. The plate has two holes drilled into it at locations A and B.

The diameter of the hole at A is d_A = 0.5 in. The diameter of the hole at B is d_B = 1.5 in.

Suppose the allowable stress is σ_all = 16 ksi . Find the maximum value for P.

Strategy: Calculate the nominal stress in the plate then find the stress concentration factors

at the locations of each hole. The stress concentration factor, K, depends on the ratio of the

diameter of the hole to the width of the plate. Then apply K = σmax / σave where σave is

the nominal stress over the narrowest section of the plate of area td.

In this example σmax = 16 ksi.

Note: (1/2)d + d_hole + (1/2)d = h So at A, d + d_A = h, d = h ˗ d_A = 3 ˗ 0.5 = 2.5 in

Likewise at B: d = h ˗ d_B = 3 ˗ 1.5 = 1/5 in

The nominal tensile stress in the plate is P/A = P/(3/2) = (2/3)P ksi

From the chart for plates with holes drilled in them:

At A 2r/D = (1/2)/3 = 1/6 At B 2r/D = (3/2)/3 = ½

So at A K = K_A = 2.6 and at B K = K_B = 2.18

Also from the area of the narrowest section at A is td = (1/2 in)(5/2in) = 5/4 in²

So at A 2.6 = 16/[(4/5)P] the result is P = 7.7 ksi

Also from the area of the narrowest section at B is td = (1/2 in)(3/2in) = 3/4 in²

So at B 2.18 = 16/[(4/3)P] the result is P = 5.5 ksi

The allowable loading then must be the lesser of the two values for P which is 5.5 ksi.

Note: The stress concentration at the smaller hole is higher.

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