shear strain - 10 points first correct procedure?

Question

Between each pair of vertebrae of the spine is a disc of cartridge of thickness 0.5 cm. Assume each disc has a radius of 0.04 m. The shear modulus of cartridge is 107 N/m2. A shear force of 20 N is applied to one end of the disc while the other end is held fixed. What is the resulting shear strain?

Answer 1

The thickness of 0.5 cm can be rewritten as 0.005 m, but it isn't used for this problem. The shear strain is γ = F/(AG), where F is shear force, A is cross-sectional area, and G is the shear modulus. This is single shear, so the shear force is equal to the applied force. γ = F/(AG) = F/(π(r^2)G) = (20 N)/(π(0.04 m)^2(107 N/m^2)) = 37.19. That's far too huge to be the correct shear strain. It means that the non-fixed face moves nearly 40 times the thickness of the disc! I suspect that the shear modulus is actually 107 GPa or at least 107 MPa, instead of just 107 Pa as you provided. With 107 GPa, the strain is 3.72 x 10^-8, and using 107 MPa, the strain is 3.72 x 10^-5. Either result would be more reasonable. We expect that rigid objects will deflect only a tiny fraction of their own size under normal loading.

For reference, Pa stands for paschals, the SI measure of pressure and stress. 1 Pa = 1 N/m^2, 1 MPa = 1 x 10^6 Pa, 1 GPa = 1 x 10^9 Pa