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Question

A bar of cross section A is subjected to equal and opposite tensile forces F at its ends. Consider the plane through the bar making an angle θ with a plane at right angles to the bar shown in the figure.
Shearing stress at this plane, in terms of F, A and θ
1247325_db552ac4b6b7451ca85c32321309ab80.png

A
F sin θ
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B
F sin2 θA
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C
F sin θcos θA
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D
F cos2 θA
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Solution

The correct option is D F cos2 θA
The resolved part of F along the normal is the tensile force on this plane and the resolved part parallel to the plane is the shearing force on the plane

Area of plane section = Asecθ

Tensile stress = ForceArea = FcosθAsecθ

FAcos2θ

Shearing stress applied on the top face

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