Question

A bar of uniform section is subjected to axial tensile loads such that the normal strain in the direction is 1.25 mm per m. If the Poisson's ratio of the material of the bar is 0.3, the volumetric strain would be

• A. $2\times {10}^{-4}$
• B. $3\times {10}^{-4}$
• C. $4\times {10}^{-4}$
• D. $5\times {10}^{-4}$

Answer 1

The correct option is D $5\times {10}^{-4}$

${\epsilon }_N=1.25mm$ per $m$
$\text{Lateral strain},{\epsilon }_L=-\text{Poisson's ratio}\times \text{longitudinal strain}$
$=-0.3\times 1.25mm$ per $m$
Volumetric strain.
${\epsilon }_V={\epsilon }_N+{\epsilon }_L+{\epsilon }_L$
$=1.25-0.3\times 1.25-0.3\times 1.25$
$=0.4\times 1.25=0.5mm$ per $m$
$=5\times {10}^{-4}$