Question

A wire has Poisson`s ratio of 0.5. If it stretched by an external force to produce longitudinal strain of 2x10-3 and its original diameter was 2mm, find out the final diameter after stretching.

Answer 1

Given:

A wire has poisson's ratio of 0.5

longitudinal strain produced by external force = 2 ×$10^{-3}$

Original diameter of wire = 3 mm .

To Find:

Final diameter of wire

Solution:

Poisson's ratio is defined as the ratio of transverse strain and longitudinal strain.

It is dimensionless number.

Poisson's ratio is denoted by letter $\mu$ .

$\mu = \frac{-transverse strain}{longitudinal strain}$

$\mu = \frac{-E_{t} }{E_{l} }$

longitudinal strain = 0.002

transverse strain = - .002(0.5)= -0.001

Original diameter = 2 mm.

The diameter of wire will decrease.

transverse strain = $\frac{change in dimension}{original dimension}$

change in dimension = -2 (.001)= -0.002

Final diameter after stretching = original diameter + change in diameter

Final diameter = 2 + (-0.002)

Final diameter = 1.998 mm

The final diameter of wire after stretching is 1.998 mm .

Answer 2

The final diameter after stretching is 4 mm.

Explanation:

The Poisson's ratio is given by the formula:

μ = Et/El

Where,

Et = Transverse strain

El = Longitudinal strain = 2 × 10⁻³ (Given)

On substituting the values, we get,

0.5 = Et/(2 × 10⁻³)

Et = 2 × 10⁻³ × 0.5 = 0.002 × 0.5

∴ Et = 0.001 = 1

Now, the transverse strain is given by the formula:

Et = Δl/l

Where,

Δl = Change in diameter

l = Original diameter = 2 mm

On substituting the values, we get,

1 = Δl/2

∴ Δl = 2 mm

The final diameter is given by the formula:

Final diameter = Original diameter + Change in diameter

Final diameter = 2 + 2

∴ Final diameter = 4 mm