Question

A steel wire having a cross-sectional area of 1.2 mm² is stretched by a force of 120 N. If a lateral strain of 1.455 is produced in the wire, calculate the Poisson’s ratio.
(Given: Y_Steel = 2 × 10¹¹ N/m²)

Answer 1

Given: A = 1.2 mm² = 1.2 × 10^-6 m², F = 120 N, Y_steel = 2 × 10¹¹ N/m², Lateral strain = 1.455, F = 120 N

To find: Poisson’s ratio (σ)

`because "Y" = ("FL")/("A"l)`

`=> l/"L" = "F"/("YA")`

`=> l/"L" = 120/(2 xx 10^11 xx 1.2 xx 10^-6)`

`=> l/"L" = 50 xx 10^-5`

∴ Poisson’s ratio (σ) = `"Lateral Strain"/"Longitudinal strain"`

`= 1.455/(50 xx 10^-5)`

`= 1.455/(50 xx 10^-5) xx 2/2`

`= 2.910/(100 xx 10^-5)`

= 2.910 × 10³