An aluminum alloy rod with 10 mm diameter was subjected to a 5-kN tensile load. After the load was applied, the diameter was measured and found to be 9.997 mm.If the yield strength is 139 MPa, calculate the Poisson’s ratio of the material.
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