Question

A mild steel bar of square cross-section $25mm\times 25mm$ is $1m$ long. It is subjected to bi-axial stress ${\sigma }_x=480N/m{m}^2$ (Tension) and ${\sigma }_y=400N/m{m}^2\left(Compression\right)$. $E=2\times {10}^{5}N/m{m}^2,\mu =0.3$, what is the elongation of the bar in mm in $x$ direction?

• A. $1.0$
• B. $1.5$
• C. $2.0$
• D. $3.0$

Answer 1

The correct option is D $3.0$

The elongation of the bar $x-$direction will be
$\frac{\Delta L}{L}=\frac{{\sigma }_x}{E}-\mu \frac{{\sigma }_y}{E}$
$\Rightarrow \Delta L=\frac{L}{E}[{\sigma }_x-\mu {\sigma }_y]$
$\Rightarrow \Delta L=\frac{1000}{2\times {10}^5}[480-0.3\times (-400\left)\right]$
$\Rightarrow \Delta L=3mm$