First slide

Elastic Modulie

Question

A rad AB of uniform cross sectional area is under the action of an axial tensile force as shown in figure. Poisson's ratio of the material of the rod is 0.4 and percentage increase in the volume of the rod is 2%. Then longitudinal strain produced in the rod is

Moderate

A

0.05
B

0.2
C

0.1
D

0.01

Solution

Volume $V = {πr}^{2} l \Rightarrow \frac{ΔV}{V} = (\frac{Δl}{l}) + 2 . (\frac{Δr}{r})$
Now $\frac{Δr}{r} = - μ (\frac{Δl}{l}) where μ =Poisson's ratio$
$∴ \frac{ΔV}{V} = \frac{Δl}{l} - 2 μ \frac{Δl}{l} \Rightarrow \frac{ΔV}{V} = (1 - 2 μ) \in where \in =longitudinal strain$
$∴ \in = \frac{(ΔV / V)}{(1 - 2 μ)} = \frac{0 .02}{(1 - 2 \times 0 .4)} = 0 .1$

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