#### Question

A steel wire having cross-sectional area 2 mm2 is stretched by ION. Find the lateral strain produced in the wire. (Given : Y for steel = 2 x 10^{11} N / m2, Poisson's ratio δ = 0.29)

#### Solution

Given: A = 2 mm^{2} = 2 x 10^{–6} m^{2}

F = 10 N

Y_{steel} = 2 x 10^{11} N/m^{2}

δ = 0.29

To find: Lateral strain

Formula:

1) Y = longitudinalstress/longitudinalstrain

2) δ = lateralstrain/longitudinalstrain

Calculation: Longitudinal stress = `F/A`

`= 10/(2xx10^(-6))`

= 5 x 10^{6} N/m^{2}

Using formula i.,

longitudinal strain = longitudinalstress/Y

`= (5 xx 10^6)/(2xx10^11)`

= 2.5 x 10^{–5}

Using formula ii,

lateral strain = δ x longitudinal strain

= 0.29 x 2.5 x 10^{–5}

lateral strain = 7.25 x 10^{–6}

Lateral strain produced in the wire is 7.25 x 10^{–6}.

Is there an error in this question or solution?

#### APPEARS IN

Solution A Steel Wire Having Cross-sectional Area 2 Mm2 is Stretched by Ion. Find the Lateral Strain Produced in the Wire. (Given : Y for Steel = 2 X 1011 N / M2, Poisson'S Ratio δ = 0.29) Concept: Definition of Stress and Strain.