Two parallel and opposite forces, each of 4000 N, are applied
tangentially to the upper and lower faces of a cubical steel block 25
cm on a side. Calculate:

a.the angle of shear; and


b.the displacement of the upper surface relative to the lower surface.
Take the shear modulus of steel as 8 ´ 1010 Nm-2.

a) the shear stress is force/area = 4000/.25^2 N/m^2 = 64000 N/m^2.
the tangent of the angle is the shear stress divided by the shear
modulus, or 8 * 10^-7.  So the angle is 8*10^-7 radians or 4.6 * 10^-5
deg.

b) the displacement is 8*10^-7 * 25cm or .0002 mm.