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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: YSteel = 2 × 1011 N/m2)
Solution
Given: A = 1.2 mm2 = 1.2 × 10-6 m2, F = 120 N, Ysteel = 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
Notes
The answer given in the textbook is incorrect.
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