A disk between vertebrae in the spine is subjected to a shea

A disk between vertebrae in the spine is subjected to a shearing force of 675 N. Find its shear deformation, taking it to have a shear modulus of 1.50 × 109 N / m2. The disk is equivalent to a solid cylinder 0.850 cm high and 3.00 cm in diameter.

Solution

Shear stress = Shear modulus * shear strain

F/A = Y ( dletaL / L )

(675) / ( pi (0.03/2)^2) = (1.50 x 10^9) ( deltaL / 0.850)

6.366 x 10^-4 = deltaL / 0.850

deltaL = 5.41 x 10^-4 cm .....Ans