Question

A mild steel bar, $1.5m$ long, has a square section $40mm\times 40mm$. The bar is subjected to a two-dimensional stress, ${\sigma }_x=310N/mm\left(tensile\right)$ and ${\sigma }_y=300N/mm$ (compressive). $E=2\times {10}^{5}N/m{m}^2$, Poisson's ratio $\mu =0.3$. The elongation of the bar in the direction of ${\sigma }_x$ will be

• A. $1.25mm$
• B. $1.75mm$
• C. $2.25mm$
• D. $3mm$

Answer 1

The correct option is D $3mm$


${\epsilon }_x=\frac{{\sigma }_x}{E}-\frac{\mu {\sigma }_y}{E}$
$=\frac{1}{2\times {10}^5}[310-(0.3\left)\right(-300\left)\right]$
$\frac{\Delta L}{L}={\epsilon }_x$
$\Delta L={\epsilon }_x\left(L\right)={\epsilon }_x(1.5\times {10}^3)=3mm$