Question

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 Q, with the plane at right
angles to the bar
a) Find the tensile stress at this plane in terms of F, A and Q
b) What is the shearing stress at the plane in terms of F, and Q.
c) For what value of Q is tensile stress a maximum.

Answer 1

Answer:

Consider Angle thita in place of Q

Answer of 3

Tensile Strength maximum when Q =0°

Answer 2

Here is your answer:

Explanation:

1) Tensile stress = Normal force/Area

Normal force = F cosθ

Tensile Stress = F cosθ / (A/ cosθ) = F cos^2θ /A

2)

Shearing stress = Tengential Stress / Area

tengentia force  = F cosθ

       Area = A /  cosθ

Shearing Stress = F cosθ / (A/ cosθ)

                           = F/ A Sinθ Cosθ

                           = F2 Sinθ Cosθ/ 2A (divide and multiply by 2)

                           = F/ Sin^2 θ /2A

3) Tensile Stress

                            = F cos ^2θ /A

Tensile Stress ∝ cos2θ

                            = cos2θ = Maximum = 1

                            = cosθ   = 1

                         θ = Cos ^-1 (1)