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hmwk4b - ME541 Fatigue of Materials Homework 4 Solutions...

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ME541: Fatigue of Materials Homework 4 Solutions 3.1) Assuming that the edge crack is much smaller than the width of the plate allows use of the stress intensity factor, π a F σ K I = where F = 1.12. For the critical condition c Ic I π a ksi 40 1.12 π a σ in ksi 45 K K × = = = = . Solving for a c gives, a c = 0.321 in. 3.4) In this case, the stress, critical crack length, and fracture toughness are known ..... the question is, what is the required F(a/b)? For the lower strength material, = = = = π a F σ in ksi 105 K K Ic I in π 0.2 ksi (130/2) F × Solving for F (a/b) gives F=2.0379 ...... Solving for a/b in the equation for an edge cracked panel (F(a/b) = 1.12-0.231 (a/b) + 10.55 (a/b) 2 - 21.72 (a/b) 3 + 30.39 (a/b) 4 ) gives a/b=0.39 or b=0.512 in. Similarly for the higher strength material, F = 0.8412 which does not occur for the edge cracked panel. Thus for this combination of high stress, low fracture toughness and the specified critical crack size, the higher strength material cannot be used. The material change should not be approved with the maximum stress of 1/2 the yield strength.
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