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 ...........
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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.
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 \( 0.33 \).
