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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 crosssectional 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 Ti6Al4V, 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 crosssectional 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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