Question

A steel rod of cross-sectional area $1m^2$ is acted upon by forces shown in the figure. Determine the total elongation of the bar in $\mu m$.
[Take $Y=2.0\times 10^{11}N/m^2$ ]

Answer 1

Answer: 1.3

The action of forces on each part of rod is shown below


As, $l=\frac{FL}{AY}$
$\therefore l_{PQ}=\frac{\left(60\times 10^3\right)\times 1.5}{1\times 2\times 10^{11}}$
$=4.5\times 10^{-7}m$
$l_{QR}=\frac{\left(70\times 10^3\right)\times 1}{1\times 2\times 10^{11}}$
$=3.5\times 10^{-7}m$
and $l_{RS}=\frac{\left(50\times 10^3\right)\times 2}{1\times 2\times 10^{11}}$
$=5.0\times 10^{-7}m$
The total extension
$l=4.5\times 10^{-7}+3.5\times 10^{-7}+5.0\times 10^{-7}$
$=13\times 10^{-7}m$
$=1.3\times 10^{-6}m$
$\therefore l=1.3\mu m$