Question

A piece of steel 200 mm long and 20 mm x 20 mm cross section is subjected to a tensile force of 40 kN in the direction of its length. Calculate the change in volume. Take 1/m = 0.3. E = 2.05 x 10⁵ N/mm² .

Answer 1

e₁ = \(\frac{40000}{(20\times20)(2.05\times10^5)}\) = 4.88 x 10^-4

e₂ = e₃ = - \(\frac{e_1}m\) = - 4.88 x 0.3 x 10^-4 = -1.464 x 10^-4

δV/V = e₁ + e₂ + e₃ = [4.88 - (1.464 x 2)]10^-4 = 1.952 x 10^-4

V = 200 x 20 x 20 = 80000 mm³ 

\(\therefore\) δV = 1.952 x 80000 x 10^-4 = 15.62 mm³