Pure shear:
Consider a square bar of unit width being subjected to uniform tension σx in x-direction and uniform compression σy in Y-direction which are equal in magnitude. Taking square element inside this bar at an angle of 45° to X-axis and Y-axis, we observe the effect of the applied tension and compression on this element.
Figure on the right shows the representation on Mohr’s circle for the circle of stresses representing the case of pure shear. OD represents the stress on the plane ab and cd perpendicular to x-y plane and inclined to 45° to x-axis. OD1 represent the stress acting on the plane ad and cb perpendicular to ab and cd.
It is further observed that normal stress on these planes is zero, and that shearing stress over the plane represented by the radius of the circle is equal to the tensile stress σx, so that.
If element abcd is isolated it will be in equilibrium under shearing stress alone as shown in the figure above, right side. Such a state is known as a state of pure shear. Pure stress can also be said to be a state of stress produced by tension in one direction and an equal compression in the perpendicular direction.
Comparing failure theories:
Considering a case of pure shear. Also assuming that the material has same yield point in tension and compression.
σx = – σz = τ ; σy=0
Maximum stress theory:
τmax = σx = σyp
Maximum shear stress theory:
τmax = (σx – (-σz))/2 = σyp/2
τmax = σx = σyp/2 = 0.5 σyp
Total strain energy criterion:
from which,
as τmax = σx
Taking µ = 0.5 for steel;
Distortion Energy Criterion:
From which;
Pressure Vessel guide 