Question

A steel rod of cross-sectional area $1{\text{m}}^2$ is acted upon by forces as shown in the figure. The total elongation of the bar is (Take $Y=2.0\times {10}^{11}{\text{Nm}}^{-2}$)

• A. $11\times {10}^{-7}\text{m}$
• B. $13\times {10}^{-7}\text{m}$
• C. $15\times {10}^{-7}\text{m}$
• D. $17\times {10}^{-7}\text{m}$

Answer 1

The correct option is B $13\times {10}^{-7}\text{m}$

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


We know that,

$\Delta l=\frac{Fl}{AY}$

So, we can calculate $\Delta l$ for all parts of the rod.

$\Delta l_{AB}=\frac{F_{AB}{l}_{AB}}{AY}=\frac{60\times {10}^3\times 1.5}{1\times 2\times {10}^{11}}$

$\Delta l_{AB}=4.5\times {10}^{-7}\text{m}$

$\Delta l_{BC}=\frac{F_{BC}{l}_{BC}}{AY}=\frac{70\times {10}^3\times 1}{1\times 2\times {10}^{11}}$

$\Delta l_{BC}=3.5\times {10}^{-7}\text{m}$

$\Delta l_{CD}=\frac{F_{CD}{l}_{CD}}{AY}=\frac{50\times {10}^3\times 2}{1\times 2\times {10}^{11}}$

$\Delta l_{CD}=5\times {10}^{-7}\text{m}$

$\Delta l=\Delta l_{AB}+\Delta l_{BC}+\Delta l_{CD}$

$\Delta l=(4.5+3.5+5)\times {10}^{-7}\text{m}$

$\Delta l=13\times {10}^{-7}\text{m}$