Question

A 2m long rectangular bar of 7.5 cm × 5 cm is subjected to an axial tensile load of 1000kN. Bar gets elongated by 2mm in length and decreases in width by 10 × 10 ^{– 6} m. Determine the modulus of elasticity E and Poisson's ratio of the material of bar.

Answer 1

Given: 

L = 2m; 

B = 7.5cm = 0.075m; 

D = 5cm = 0.05m 

P = 1000kN 

δL = 2mm = 0.002m 

δb = 10 × 10 ^{– 6}m. 

Longitudinal strain e_L = e_t = δL/L = 0.002/2 = 0.001 

Lateral strain = δb/b = 10 × 10 ^{– 6}/0.075 = 0.000133 

Tensile stress (along the length) σ_t = P/A = (1000 × 1000)/(0.075 × 0.05) = 0.267 × 10⁹ N/m² 

Modulus of elasticity, E = σ_t /e_t = 0.267 × 10⁹ /0.001 = 267 × 109 N/m²

Poisson’s ratio = Lateral strain/ Longitudinal strain = (δb/b)/(δL/L) = 0.000133/0.001