Question

A steel bar of rectangular cross section 20 mm x 10 mm is subjected to a pull of 20 kN in the direction of its length. Taking E = 0.2 x 10⁶ N/mm² and m = 10/3, find the length of the sides of the cross section and percentage decrease of area of cross section.

Answer 1

The strain in the direction of pull is

e₁ = \(\frac pm\) = \(\frac p{AE}\) = \(\frac{20000}{20\times10\times0.2\times10^6}\) = 5 x 10^-4

Lateral strain = - \(\frac{e_1}m\) = - \(\frac3{10}\) x 5 x 10^-4

Hence 20 mm side is decreased by 20 x 1.5 x 10^-4 = 0.0030 mm 

10 mm side is decreased by 10 x 1.5 x 10^-4 = 0.0015 mm 

New area of cross section = (20 - 0.003)(10 - 0.0015) = ( 200 - 0.06)

\(\therefore\) % decrease of area of cross section = \(\frac{0.06}{200}\) x 100 = 0.03