A cylindrical specimen of a metal alloy 10 mm (0.4 in.) in diameter is stressed elastically in tension. A force of 15,000 N (3370 lbf) produces a reduction in specimen diameter of 7 * 10-3 mm (2.8 * 10-4 in.). Compute Poisson’s ratio for this material if its elastic modulus is 100 GPa (14.5 * 106 psi).

The Poisson's ratio is 0.36

Explanation:

Given-

Diameter = 10 mm = 10 X 10⁻³m

Force, F = 15000 N

Diameter reduction = 7 X 10⁻3 mm = 7 X 10⁻⁶ mm

The equation for Poisson's ratio:

ν =

and

εₓ = Δd / d₀

εz = σ / E

    = F / AE

We know,

Area of circle = π (d₀/2)²

= F / π (d₀/2)² E

εz = 4F / πd₀²E

Therefore, Poisson's ratio is

ν = - Δd ÷ d / 4F ÷ πd₀²E

ν =  -d₀ΔDπE / 4F

ν = (10 X 10⁻³) (-7 X 10⁻⁶) (3.14) (100 X 10⁻⁹) / 4 (15000)

ν = 0.36

Therefore, the Poisson's ratio is 0.36

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