Question

A metal wire is fixed vertically from one end. If a mass (0.33 Kg) is hanged to the other end, it will elongate by (1 mm) and if this end is twisted by a couple of (1.5 × 10^-5 N m), its angle of twist is (60°). Find Poisson’s ratio knowing that wire radius is (0.16 mm). Calculate also the bulk modulus of wire material if its original length equal to (2 m).

Answer 1

Answer:

ν = 0.45

K = 2.5×10¹¹ Pa

Explanation:

Poisson's ratio is:

ν = E/(2G) − 1

where E is Young's modulus and G is the shear 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 = (0.33 kg × 9.8 m/s²) (2 m) / (π (0.00016 m)² × 0.001 m)

E = 8.04×10¹⁰ Pa

Shear modulus is:

G = τL / (Jθ)

where τ is the torque, L is the length, J is the second moment of inertia, and θ is the angular deflection.

G = (1.5×10⁻⁵ Nm) (2 m) / ((π/2 (0.00016 m)⁴) (60° × π / 180°))

G = 2.78×10¹⁰ Pa

The Poisson's ratio is therefore:

ν = (8.04×10¹⁰ Pa) / (2 × 2.78×10¹⁰ Pa) − 1

ν = 0.446

And the bulk modulus is:

K = E / (3 − 6ν)

K = (8.04×10¹⁰ Pa) / (3 − 6 × 0.446)

K = 2.48×10¹¹ Pa

Rounded to 2 significant figures, the Poisson's ratio is 0.45 and the bulk modulus is 2.5×10¹¹ Pa.