Question

A steel specimen 1.5cm² in cross section stretches 0.05mm over 5 cm gauge length under an axial load of 30 KN. Calculate the strain energy stored in the specimen at this point. If the load at the elastic limit for specimen is 50KN, Calculate the elongation at the elastic limit.

Answer 1

Cross sectional area of specimen A = 1.5 cm² = 1.5 × 10^–4 m²

Increase in length over 5cm gauge length δL = 0.05mm = 0.05 × 10^–3 m

Axial load W = 30KN

Load at elastic limit = 50KN

Strain energy stored in the specimen

 = σ²A.L/2E =1/2. W. δL = 1/2× (30 × 1000) × 50 × 10^–3

U = 0.75 Nm or J

Also E = W.L/A.δL

= {(30 × 1000) × (5/100)}/{(1.5 × 10^–4) × (0.05 × 10^–3)}

= 200 × 10⁹ = 200 GN/m²

Elongation at elastic limit, δL

δL = W.L/A.E = {(50 × 1000) × (5/100)}/{(1.5 × 10^–4) × (200 × 10⁹)}

δL = 0.0000833 m = 0.0833 mm