e45_hw_1

# e45_hw_1 - E45 HW01 Due September 3 2010 Reference J.F...

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E45 HW01 Due September 3, 2010 Reference: J.F. Shackelford, Introduction to Materials Science for Engineers, 6 th Edition, Prentice Hall, New Jersey (2005) Problem 1 6.7 In normal motion, the load exerted on the hip joint is 2.5 times body weight. a. Calculate the corresponding stress (in MPa) on an artificial hip implant with a cross-sectional area of 5.64 cm 2 in a patient weighting 150 lb f . SOLUTION: 5.64 cm 2 = 0.000564 m 2 σ = P/A= (2.5)(150 lb f ) / (5.64*10 -4 m 2 ) = 6.65*10 5 Pa= 0.665 MPa b. Calculate the corresponding strain if the implant is made of Ti-6Al-4V, which has an elastic modulus of 124 GPa. SOLUTION: 124 GPa= 1.24*10 11 Pa ε= Δl/l 0 = σ/E= (6.65*10 5 Pa)/(1.24*10 11 Pa)= 5.36*10 -6 Problem 2 6.8 Repeat Problem 6.7 for the case of an athlete who undergoes a hip implant. The same alloy is used, but because the athlete weighs 200 lb f , a larger implant is required (with a cross-sectional area of 6.90 cm 2 ). Also, consider the situation in which the athlete expends his maximum effort exerting a load of five times

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## This note was uploaded on 11/19/2010 for the course E 45 taught by Professor Gronsky during the Fall '08 term at Berkeley.

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e45_hw_1 - E45 HW01 Due September 3 2010 Reference J.F...

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