Question

If you stand on one leg, the load exerted on the hip joint is 2.4 times your body weight. Assuming a simple cylindrical model for a hip implant, with a cross-sectional area of 5.6 cm2, estimate the following:

a) The corresponding stress on the hip implant in a 175-lb individual: ………………. [MPa] (0.2)
b) If the hip implant is made of Ti-6Al-4V (120 GPa elastic modulus), what is the strain for the current loading conditions? ………………. (0.1)
c) Complete solution (please show your work on a scrap paper, scan it, and insert the image below) (0.1)

Answer 1

Given :

cross sectional area = 5.6 cm square

The corresponding stress

`$\sigma = (F)/(A)$`

`$=(2.5 * 175 \ lb* (1 \ N/ 0.2248 \ lb))/(5.6 \ cm^2 (1 \ N/100 \ cm)^2)$`

`$=(1946.17)/(0.056 * 10^(-2))$`

`$= 3.475 * 10^6 \ N/m^2$`

∴
`$\sigma = 3.475 \ MPa$`

And the strain is

`$\epsilon = (\sigma)/(E)$`

`$=(3.475)/(124 * 10^3)$`

`$= 2.80 * 10^(-5)$`