Given that:
σx = 120KN/m2
σy = – 40 KN/m2
τxy = – 50KN/m2
(i) Calculation for Principle stress and their direction
Principal stresses is given by the equation:
σ1,2 = ½ [(120 – 40) ± {(120 + 40)2 + 4 x (–50)2 }1/2]
σ1,2 = 40 ± 94.34
σ1 = 134.34KN/m2
σ2 = – 54.34KN/m2
The Direction of the plane is given by the equation:
tan 2θ = 2τxy/ (σx − σy)
tan 2θ = (2 x (–50))/(120 + 40))
2θ = – 32º
θ = – 16º
Major principal plane = θ = – 16º
Miner principal plane = θ + 90º = 74º
(ii) Maximum shear stress is given by the equation:
τ = 1/ 2 (σ1 – σ2) sin 2θ
(Since Maximum Shear stress is at θ = 45º)
= ½ (134.34 + 54.34) sin 90º
= 94.34KN/m2
(iii) Resultant stress on a plane inclined at 40º with x-axis
Since
σn = ½ (120 – 40 ) + ½ (120 + 40) cos 80º – 50 sin 80º
= 40 + 13.89 – 49.24
= 4.65 KN/m2
τ = ((σx - σy)/2) sin 2θ - τxy cos 2θ
τt = ½ (120 + 40) sin 80º + 50 cos 80º
τt = 87.47KN/m2
Resultant stress is given by the equation
σr = √(σn2 + τt2)
σr = (4.652 + 87.472)1/2
σr = 87.6 KN/m2