Question

Suppose a vertebra is subjected to a shearing force of 420 N. Find the shear deformation in m, taking the vertebra to be a cylinder 2.6 cm high and 3.6 cm in diameter. The shear modulus for bone is 80 • 109 N/m2.

Answer 1

Answer:

$1.34103\times 10^{-7}\ m$

Explanation:

F = Shearing force = 420 N

$L_0$ = Initial length = 2.6 cm

d = Diameter = 3.6 cm

r = Radius  = $\frac{d}{2}=\frac{3.6}{2}=1.8\ cm$

E = Young's modulus = $80\times 10^9\ N/m^2$

A = Area = $\pi r^2$

Change in length is given by

$\Delta L=\frac{FL_0}{AE}\\\Rightarrow \\\Rightarrow \Delta L=\frac{420\times 0.026}{\pi 0.018^2\times 80\times 10^9}\\\Rightarrow \Delta L=1.34103\times 10^{-7}\ m$

The shear deformation of the vertebra is $1.34103\times 10^{-7}\ m$