Question

01. A short cast iron column carries a load of50 T. If the original diameter is 8 cm, E = X106 kg/sq.cm and Poisson's ratio = 0.25, theincrease in diameter of column in 'cm' wouldbe(a) 0.00318(b) 0.00256(c) 0.002(d) 0.00280

Answer 1

The increase in diameter after stretching is (c) 0.002

Given:

Poisson's ratio = 0.25 = μ

Area density = 1 × 10¹⁰ kg/m² = A

Force = 50 Tons = 5 × 10⁴ kg = F

Diameter = 8 × 10⁻² m = d

Radius = 4 × 10⁻² m = r

Explanation:

The Poisson's ratio is given by the formula:

μ = (Transverse strain)/(Longitudinal strain)

μ = (F/A)/(Strain)

0.25 = ((5 × 10⁴)/(1 × 10¹⁰))/(Strain)

0.25 = (5 × 10⁻⁶)/(Strain)

Strain = (5 × 10⁻⁶)/0.25

Strain = 0.00002 m = 0.002 cm

∴ Increase in diameter = 0.002