— David Wagner 2009/10/18 12:39

√ME 5463-001 Fracture Mechanics, Homework #13

1. Determine the failure stress of a semi-infinite body in tension if the stress is perpendicular to a semi-elliptical surface crack with a major axis length of 10mm and a minor axis length of 2.0mm [Kc = 20 MPa(m)1/2].

2. Determine Keff of a plate with a single-edged crack and subjected to a tensile stress (σ = 60MPa; a = 5mm; σy = 60MPa.

Problem 1.

  • 2c = 10 mm ⇒ c = 0.005 m.
  • 2a = 2 mm ⇒ a = 0.001 m.

Phi approx {3 pi}/8 + {pi a^2}/{8 c^2} = Phi approx {3 pi}/8 + {1*10^{-6}pi}/{2 * 10^{-4}} approx 1.1938.

K_c approx {1.12 sigma_c sqrt{pi a}}/{Phi } ⇒ sigma_c approx {K_c Phi}/{1.12 sqrt{pi a}} = sigma_c approx {20*10^6 *1.1938 }/{1.12 * sqrt{pi*0.001}} = 3.80 x 108 = 380,000 ksi.

Problem 2.

  • Assume flow stress is equal to yield stress: sigma_0 = sigma_y.
  • K = sigma sqrt{pi a}

r_y = {1/{2 pi}}(K/{sigma_0})^2 = {1/{2 pi}}({sigma sqrt{pi a}}/{sigma_0})^2 = {1/{2 pi}}({60*10^6 *sqrt{pi*0.005}}/{60*10^6})^2 = {1/2} 0.005 = 0.0025.

a_{eff} = a + r_y = a+ Delta p = a+{1/{2 pi}}({K_{eff}}/{sigma_0})^2 ⇒

K_{eff} = sigma_0 sqrt{2 pi r_y} = 60*10^6 sqrt{2 pi * 0.0025} = 7.52 x 106.


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