Question

A bar of diameter 30 mm is subjected to a tensile load such that the measured extension on a gauge length of 200 mm is 0.09 mm and the change in the diameter is 0.0045 mm. The Poisson’s ratio will be ...........

• A. 0.15
• B. 0.25
• C. 0.33
• D. 0.45

Hint:

Poisson's ratio relates lateral and axial strains. In general, a higher Poisson's ratio indicates a greater tendency for a material to deform laterally when stretched or compressed.

Answer 1

The Correct Option is C

The Poisson's ratio \( \nu \) is defined as the negative ratio of lateral strain to axial strain. It is given by the formula: \[ \nu = -\frac{\text{Lateral strain}}{\text{Axial strain}} = -\frac{\Delta d / d}{\Delta L / L} \] where:
- \( \Delta d \) is the change in diameter (0.0045 mm),
- \( d \) is the original diameter (30 mm),
- \( \Delta L \) is the extension (0.09 mm),
- \( L \) is the original length (200 mm).
We can calculate the strains: \[ \text{Axial strain} = \frac{\Delta L}{L} = \frac{0.09}{200} = 0.00045 \] \[ \text{Lateral strain} = \frac{\Delta d}{d} = \frac{0.0045}{30} = 0.00015 \] Now, substituting into the Poisson's ratio formula: \[ \nu = -\frac{0.00015}{0.00045} = 0.33 \] Thus, the Poisson’s ratio is \( 0.33 \).