An aluminum alloy rod with 10 mm diameter was subjected to a 5-kN tensile load. After the load was applied, the diameter...

Given:

Original diameter of rod, d = 10 mm

Tensile load, P = 5 kN

Reduced diameter of rod, d’ = 9.997 mm

Yield strength = 139 MPa

The formula for lateral strain is given by,

Where,

∆d = change in diameter

d = original diameter

 

The formula for Poisson’s ratio is given by,

 

The formula for Young’s modulus is given by,

 

The formula for stress is given by,

Where,

P = Load

A = Cross-sectional area

 

Also,

Yield strength = 0.2% of Young’s modulus (E)

The lateral strain is calculated as,

The stress is calculated as,

The Young’s modulus will be calculated as,

The longitudinal strain is calculated as,

So, the Poisson’s ratio is calculated as,

Answer:

Poisson’s ratio of the material, µ = 0.3275