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— David Wagner 2009/10/25 22:49

√ME 5463-001 Fracture Mechanics, Homework #15

6.7 A 1/4 inch-thick wide aluminum (sigma_ys = 40 ksi) tension panel has an edge crack 0.80 inch long. Based on fracture tests with a similar material for the same thickness, the fracture toughness has been measured as 45 ksi sqrt(in) for slowly applied loads.

(a) Compute the diameter of the plastic zone for both plane-stress and plane-strain conditions.

(b) Based on your engineering judgment and the information available to you, which of the two plane conditions prevails? Why?

© What is the magnitude of the largest remotely applied steady-state stress that the panel can support without failure?

(d) What would be the consequence of an impact load on your answer to part ©?

Part (a)

Plane Stress: r_b = {1/{2 pi}} (K_I/{sigma_{ys}})^2 =1/{2pi} (45000/40000)^2 = 0.20 in.

Plane Strain: r_y = 0.067 in.

Part (b)

For plane strain: 0.25 = B > 2.5(K_c/sigma_{ys})^2 = 3.16

This is not so, so plane stress conditions prevail.

Part (c)

sigma_c = K_c/{1.12 sqrt{pi a}} = sigma_c = 45000/{1.12 sqrt{0.8*pi}} = 2534 psi

Part (d)

The specimen could withstand a higher impact load.


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