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+1 vote
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in Physics by (57.5k points)

A plane element is subjected to following stresses σx = 120KN/m2 (tensile), σy = 40 KN/m2 (Compressive) and τxy = 50KN/m2 (counter clockwise on the plane perpendicular to x-axis) 

find 

(1) Principle stress and their direction 

(2) Maximum shearing stress and its directions. 

(3) Also, find the resultant stress on a plane inclined 40º with the x-axis. 

1 Answer

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Best answer

Given that: 

σx = 120KN/m

σy = – 40 KN/m

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/m

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/m

τ = ((σ- 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 = √(σn+ τt2

σr = (4.652 + 87.472)1/2 

σr = 87.6 KN/m

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