Question

A metal bar 5 cm x 5 cm section is subjected to an axial compressive load of 500 kN. The contraction on a 20 cm gauge length is found to be 0.5 mm and the increase in thickness 0.045 mm. Find the value of Young’s modulus and Poisson’s ratio.

Answer 1

Given P = 500 kN, l = 20 cm, Δl = 0.05 cm, Δt = 0.0045 cm. 

Area of cross section A = 5 x 5 = 25 cm² 

Longitudinal strain 

ε = Δl/l = 0.05/20 = 0.0025 ( compressive) 

stress σ = P/A = 500 x 10³ /25 x 10^-4 = 200 MPa (compressive) 

Young’s Modulus 

E = σ/ε = 200 x 106 /0.0025 = 80 GPa 

Lateral strain 

Δt/t = 0.0045/5 = 0.0009 (tensile) 

Poisson’s Ratio 

v = Lateral strain/Longitudinal = 0.0009/0.0025 = 0.36