A steel wire having cross sectional area 1.5 mm^{2} 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 10^{11} N/m^{2}, Poisson’s ratio σ = 0.291,g = 9.8 m/s^{2})

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#### Solution

Given that A = 1.5 mm^{2}, lateral strain = 1.5 *10^{−5},

Y_{steel} = 2 * 10^{11} N/m^{2}, σ = 0.291 and g = 9.8 m/s^{2}

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*10^{11}*5.14*10^{-5}

∴Longitudinal stress =10.28*10^{6}

But longitudinal stress =`(mg)/A`

→10.28*10^{6} = `(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

Concept: Definition of Stress and Strain

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