Question

A vertebra is subjected to a shearing force of 686 N. Find the shear deformation, taking the vertebra to be a cylinder 3.29 cm high and 7.69 cm in diameter

Answer 1

The shear deformation is

$$\delta_s=\frac{FH}{AG}=\frac{4FH}{\pi d^2G}=\\\space\\ \delta_s=\frac{4·686·0.0329}{\pi·0.0769^2·8·10^{10}}=6.07·10^{-8}\text{ m}.$$