Let x and y directions be selected as shown in the figure. Then
px = – 100 N/mm2 , py = – 75 N/mm2 , q = – 50 N/mm2

= – 87.5 – 51.54
i.e., p2 = – 139.04 N/mm2

Let principal plane of p1 make angle θ with x-axis. Then

The planes of maximum shear stresses are at 45° to the principal planes. These planes are shown in Fig. (b).
