Question

A cylindrical specimen of a metal alloy is stressed elastically in tension. The original diameter was 8 mm and a force of 15,700 N produces a reduction in diameter of 5x10-3 mm (magnitude). Compute Poisson's ratio for this alloy if its modulus of elasticity is 140 GPa.

Answer 1

The value of poisson's ratio of the material = 0.28

Step-by-step explanation:

Given data

Force F = 15700 N

Diameter D = 8 mm

Change in diameter ΔD = 0.005 mm

Modulus of elasticity E = 140 ×
`10^(3)` M Pa.

Stress induced in the shaft due to applied force

`\sigma = (F)/(A)`

Area A =
`(\pi)/(4) d^(2)`

⇒ A =
`(\pi)/(4) 8^(2)` = 50.24
`mm^(2)`

Now stress
`\sigma = (15700)/(50.24)`

⇒ Stress
`\sigma` = 312.5 M pa ------ (1)

Now strain in the element is given by

∈
`= (\sigma)/(E)`

⇒ ∈ =
`(312.5)/(140000)`

⇒ Strain ∈ = 2.232 ×
`10^(-3)`

We know that poisson's ratio is given by

⇒
`\mu =`
`((change in diameter)/(original diameter) )/(strain)`

Change in diameter ΔD = 0.005 mm

Diameter D = 8 mm

Strain ∈ = 2.232 ×
`10^(-3)`

⇒
`\mu = ((0.005)/(8) )/(0.002232)`

⇒
`\mu =` 0.28

This is the value of poisson's ratio of the material.