A 30mm dia. Solid circular aluminum rod 3m long, is subject to an axial pull of 100 KN. Taking E=70 GN/m^2 and Poisson’s Ratio =1/3, determine the elongation, change in the diameter and change in volume of the rod. Also, find the bulk modulus.

Given data:

diameter d=30mm

Length L=3mm=3000mm

Force F=100KN

Modulus of elasticity $(E)=70GN/m^2=70*10^3 N/mm^2$

Poisson's ratio $\mu=\frac{1}{3}$

Now, $\sigma stress=\frac{force or axial pull}{area of cross section}=$ $\frac{100*10^3}{\frac{\pi}{30^2}} N/mm^2$

Stress=$141.47 N/mm^2$

Now, modulus of elasticity=$\frac{stress (\sigma)}{strain (\epsilon)}$

$70*10^3=\frac{141.47}{\epsilon}$

$\epsilon=2.021*10^{-3}$

$\epsilon=\frac{change in length}{original length}$

$2.021*10^{-3}=\frac{\Delta}{3000}$

$\Delta L=6.063$

Now, Poisson's ratio $\sigma= \frac{lateral strain}{longitudinal strain}$

$\frac{1}{3}=\frac{\frac{\delta d}{d}}{\frac{\Delta}{L}}=$ $\frac{\frac{\delta d}{30}}{\frac{6.063}{3000}}$

$\Delta d=0.02021 mm$

$Now, E=3K(1-2 \mu) or E=2G(1+\mu)$

Using first equation as bulk modulus is asked, not modulus of rigidity

$70*10^3=3K(1-2\frac{1}{3}$

$70*10^3=3K(\frac{3-2}{3}$

$K=70*10^3 N/mm^2 or 70 KN/mm^2$

Now, $bulk modulus=\frac{stress}{volumetric strain}$

$70*10^3=\frac{141.47}{\epsilon_v}$

$\epsilon_v=2.021*10^{-3}$

$\epsilon_v=\frac{change in volume}{original volume}$

$2.021*10^{-3}=\frac{\Delta v}{\frac{\pi}{4}*d^2*L}$

$2.021*10^{-3}=\frac{\Delta v}{\frac{\pi}{4}*30^2*3000}$

$Change in volume (\Delta V)=4285.68 mm^3$