Question

A steel bar 2 m long, 20 mm wide and 10 mm thick is subjected to a pull of 20KN in the direction of its length. Find the changes in length, breadth and thickness. Take E = 2 × 10⁵ N/mm² and poisons ratio 0.30.

Answer 1

Longitudinal strain = δL/L = stress/ modulus of elasticity 

= (P/A)/E = P/AE = 20 × 10³ /{(20 × 10) × (2 × 10⁵)} = 0.5 × 10 ^{– 3} 

Change in length δL = longitudinal strain x original length 

= (0.5 × 10 ^{– 3}) × (2 × 10³) =1.0 mm (increase) 

Lateral strain = Poisson's ratio × longitudinal strain = 0.3 × (0.5 × 10 ^{– 3}) = 0.15 × 10^-3 

The lateral strain equals δb/b and δt/t 

Change in breadth δb = b × lateral strain = 20 × (0.15 × 10 ^{– 3}) 

= 3 × 10 ^{– 3} mm (decrease) 

Change in thickness δt = t × lateral strain = 10 × (0.15 × 10 ^{– 3}) 

= 1.5 × 10 ^{– 3 mm} (decrease)