Question

A steel bar 20mm wide 10mm thick and 2m long subjected to axial pull 20kn along its length find changes in dimension and its volume e=2×10^5 and 1\m = 0.3

Answer 1

Answer:

Step-by-step explanation:

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)