First slide

Elastic Modulie

Question

A cylindrical rod is subjected to axial tensile force as shown in figure. If Poisson's ratio of the material of the rod is 0.25 and longitudinal strain produced in the rod is 0.05, then percentage change in the volume of the rod is

Moderate

A

2.5%
B

7.5%
C

5%
D

4%

Solution

$V = {πr}^{2} l \Rightarrow \frac{ΔV}{V} = \frac{Δl}{l} + 2 . \frac{Δr}{r}$
But $\frac{Δr}{r} = - μ . \frac{Δl}{l}$
$\begin{array}{l} ∴ \frac{ΔV}{V} = \frac{Δl}{l} (1 - 2 μ) = 0 .05 (1 - 2 \times 0 .25) = 0 .025 \\ \Rightarrow \frac{ΔV}{V} \times 100 % = 0 .025 \times 100 % = 2 .5 % \end{array}$

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