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

Solution :

Given :

cross sectional area = 5.6 cm square

The corresponding stress

$\sigma = \frac{F}{A}$

  $=\frac{2.5 \times 175 \ lb\times (1 \ N/ 0.2248 \ lb)}{5.6 \ cm^2 (1 \ N/100 \ cm)^2}$

  $=\frac{1946.17}{0.056 \times 10^{-2}}$

  $= 3.475 \times 10^6 \ N/m^2$

∴ $\sigma = 3.475 \ MPa$

And the strain is

$\epsilon = \frac{\sigma}{E}$

   $=\frac{3.475}{124 \times 10^3}$

   $= 2.80 \times 10^{-5}$