Question

19.
A bar of cross section A is subjected to two equal
and opposite tensile forces as shown. Consider a
cross-section BB' as shown in figure. The
shearing stress at this plane is:-
​

Answer 1

FN = Normal force = F cos θ

AN = Area = A / cos θ

Tensile stress = Normal force / Area = F cos θ / (A / cos θ) = F cos2θ / A

Shearing stress = Tangential force / Area = F sin θ / A cos θ = F sin 2θ / 2A

The shearing stress is maximum if sin 2θ = 1 or 2θ = 900 or θ = 450

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