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 diameter is 0.0045 mm. The Poisson's ratio will be

• A. $1/4$
• B. $1/3$
• C. $1/5$
• D. $1/6$

Answer 1

The correct option is B $1/3$

$Longitudinalstrain,{ϵ}_L=\frac{\Delta L}{L}=\frac{0.09}{200}$
$Laterialstrain,{ϵ}_L=\frac{\Delta D}{D}=-\frac{0.0045}{30}$
Poisson's ratio
$=\frac{{ϵ}_D}{{ϵ}_L}=\frac{0.0045}{30}\times \frac{200}{0.09}=\frac{1}{3}$