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in Physics by (24.5k points)

A bar of cross-section A is subjected to equal and opposite tensile force F at its ends. If there is a plane through the bar making an angle θ, with the plane at right angles to bar. (Fig. at the top of the page).

(a) Find the tensile stress at the plane in terms of F, A and θ?

(b) What is the shearing stress at the plane in terms of F, A and θ?

(c) For what value of θ, the shearing stress is maximum?

1 Answer

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Best answer

(a) Tensile stress = normal force/area

and area = A/cosθ

Normal force = F cosθ

So, tensile stress = (F cosθ)/(A/cosθ) = F cos2θ/A

(b) Shearing stress = (Tangential force)/area

Tangential force = F sinθ

Area = A/cosθ

So, Shearing stress = (F sinθ)/(A/cosθ) = F/A sinθ cosθ

= F/2A x 2sinθcosθ = F/2A x sin 2θ

(c) Tensile stress will be maximum, when

cos2θ = maximum = 1

or, cosθ = 1 = cos 0° θ

or, θ = 0°

(d) Shearing stress will be maximum, when

sin 2θ = maximum = 1 = sin 90°
or, 2θ = 90°
or, θ = 45°

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