ME 2733 Chapter : Ch 8 HW Sol
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Equation 8. 1 is employed to solve this problem i. e. , Values for 0 140 mpa), 2a (3. 8 10 2 mm), and t (1. 9 10 4 mm) are provided in the problem statement. 8. 3 if the specific surface energy for aluminum oxide is 0. 90 j/m2, then using data in table 12. 5, compute the critical stress required for the propagation of an internal crack of length 0. 40 mm. We may determine the critical stress required for the propagation of an internal crack in aluminum oxide using equation 8. 3. Taking the value of 393 gpa (table 12. 5) as the modulus of elasticity, and realizing that values for s (0. 90 j/m2) and 2a (0. 40 mm) are given in the problem statement, leads to. = 33. 6 106 n/m2 = 33. 6 mpa. 8. 5 a specimen of a 4340 steel alloy with a plane strain fracture toughness of 54. 8 mpa m (50 ksi in. ) is exposed to a stress of 1030 mpa (150,000 psi).
