Question

A disk between vertebrae in the spine is subjected to a shearing force of 425 N. Find its shear deformation, taking it to have a shear modulus of 1.70×10^9 N/m^2. The disk is equivalent to a solid cylinder 0.700 cm high and 6.50 cm in diameter. a) 5.27 x 10^-7 m

b) 1.54 x 10^-6 m
c) 3.08 x 10^-6 m
d) 6.16 x 10^-6 m

Answer 1

option A

Step-by-step explanation:

given,

shear force = 425 N

Shear modulus = 1.7 × 10⁹ N/m²

disk equivalent to solid cylinder

height = 0.7 cm and diameter = 6.50 cm

`\delta = (VL)/(AG)\\\\\delta = (425* 0.007)/(0.25 * \pi d^2* 1.7* 10^9)\\\\\delta = (425* 0.007)/(0.25 * \pi* 0.065^2* 1.7* 10^9)\\\\\delta  = 5.27 * 10^(-7) m`

hence, the correct answer is option A