A steel bar 2 m long, 40mm wide and 20mm thick is subjected to an axial pull of 160 KN in the direction of its length. Find the change in length, width and thickness of bar. Take E= 200Gpa and Poisson’s ratio= 0.3

Change in length $(\delta l)=\frac{P*L}{AE}$

Given, $P=130*10^3 N, L=4200mm$

$W=35mm, t=25mm$

$E=200GPa, E=200*10^3 N/mm^2$

$\mu=0.3$

$\delta L=\frac{130*10^3*4200}{35*25*4200)10^3}$

$\delta L=1.48mm$

$Poisson's Ratio=\frac{Lateral strain}{Longitudinal strain}$

Calculations of changes in width : -

$Lateral strain=\frac{\delta w}{w} or \frac{\delta t}{t}$

$Longitudinal strain=\frac{\delta L}{L}$

$\mu=\frac{\frac{\delta w}{w}}{\frac{\delta L}{L}}$

$0.3=\frac{\frac{\delta w}{35}}{\frac{2}{4200}}$

$\delta W=0.022mm$

Calculation of change in thickness:

$\mu=\frac{\frac{\delta t}{t}}{\frac{\delta L}{L}}$

$0.3=\frac{\frac{\delta t}{25}}{\frac{2}{4200}}$

$\delta t=0.015mm$