Question

A cylindrical rod with length (1.7 m) and radius (2 cm) is fixed from one end and a mass of (20 kg) is attached to the other free end. The elongation in the rod is (0.5 mm). If this cylinder is exposed to an excess pressure of (180 atm.) above the atmospheric pressure its volume decreases by (1.2 %). Find Poisson’s ratio for the rod material.

Answer 1

Answer:

0.44

Explanation:

Poisson's ratio is:

ν = (3K − E) / 6K

where K is the bulk modulus and E is Young's modulus.

Young's modulus is:

E = FL / (AΔL)

where F is the force, L is the initial length, A is the cross sectional area, and ΔL is the change in length.

E = (20 kg × 9.8 m/s²) (1.7 m) / (π (0.02 m)² × 0.0005 m)

E = 0.530×10⁹ Pa

Bulk modulus is:

K = -ΔP / (ΔV/V)

where ΔP is the change in pressure, ΔV is the change in volume, and V is the initial volume.

K = -(180 atm × 101325 Pa/atm) / (-0.012)

K = 1.52×10⁹ Pa

Therefore, the Poisson's ratio is:

ν = (3(1.52×10⁹ Pa) − 0.530×10⁹ Pa) / 6(1.52×10⁹ Pa)

ν = (3(1.52) − 0.530) / 6(1.52)

ν = 0.442

Rounded to 2 significant figures, the Poisson's ratio is 0.44.