Question

A steel wire having cross sectional area 1.5 mm² when stretched by a load produces a lateral strain 1.5 x 10^-5. Calculate the mass attached to the wire.

(Y_steel = 2 x 10¹¹ N/m², Poisson’s ratio σ = 0.291,g = 9.8 m/s²)

Answer 1

Given that A = 1.5 mm², lateral strain = 1.5 *10⁻⁵,

Y_steel = 2 * 10¹¹ N/m², σ = 0.291 and g = 9.8 m/s²

Poisson's ratio, σ =`"Lateral strain"/"Longitudinal strain"`

∴Longitudinal strain = `(1.5*10^-5)/0.291=5.14*10^-5`

Longitudinal stress = Y * Longitudinal strain

                                =2*10¹¹*5.14*10^-5

∴Longitudinal stress =10.28*10⁶

But longitudinal stress =`(mg)/A`

→10.28*10⁶ = `(M*9.8)/(1.5*10^-6)`

∴M =`(10.28*10^6*1.5*10^-6)/9.8=15.42/9.8`

∴M = 1.58kg