Question

A steel rod of cross-sectional area $1 \,m ^{2}$ 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

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 $
$\therefore l =1.3 \,\mu m$