A 2m long rectangular bar of 7.5 cm × 5 cm is subjected to an axial t

1.	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» Given: L = 2m; B = 7.5cm = 0.075m; D = 5cm = 0.05m P = 1000kN \(\delta\)L = 2mm = 0.002m \(\delta\)b = 10 × 10^-6 m. Longitudinal strain e_L = et = \(\delta\)L/L = 0.002/2 = 0.001 Lateral strain \(\delta\)b/b = 10 × 10^-6/0.075 = 0.000133 Tensile stress (along the length) \(\sigma\)_t= P/A = (1000 × 1000)/(0.075 × 0.05) = 0.267 × 10⁹ N/m² Modulus of elasticity, E = \(\sigma\)_t /e_t= 0.267 × 10⁹ /0.001 = 267 × 10⁹ N/m² Poisson’s ratio = Lateral strain/ Longitudinal strain = (\(\delta\)b/b)/(\(\delta\)L/L) = 0.000133/0.001

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»

Given:

L = 2m;

B = 7.5cm = 0.075m;

D = 5cm = 0.05m

P = 1000kN

\(\delta\)L = 2mm = 0.002m

\(\delta\)b = 10 × 10^-6 m.

Longitudinal strain

e_L = et = \(\delta\)L/L = 0.002/2 = 0.001

Lateral strain

\(\delta\)b/b = 10 × 10^-6/0.075 = 0.000133

Tensile stress (along the length)

\(\sigma\)_t= P/A = (1000 × 1000)/(0.075 × 0.05) = 0.267 × 10⁹ N/m²

Modulus of elasticity,

E = \(\sigma\)_t /e_t= 0.267 × 10⁹ /0.001 = 267 × 10⁹ N/m²

Poisson’s ratio = Lateral strain/ Longitudinal strain

= (\(\delta\)b/b)/(\(\delta\)L/L) = 0.000133/0.001