Question

A mild steel bar of square cross-section 40 mm x 40 mm is 400 mm long. It is subjected to a longitudinal tensile stress of 440 N/mm${}^2$ and lateral compressive stress of 200 N/mm${}^2$ in perpendicular direction. E = 2 x ${10}^5$ N/mm${}^2$, μ = 0.3. What is the approximate elongation of the bar in the longitudinal direction?

• A. 0.44 mm
• B. 0.88 mm
• C. 0.22 mm
• D. 1 mm

Answer 1

The correct option is D 1 mm

The lateral compressive stress is applied on per
perpendicular direction (not directions). So considering 200 N/$m{m}^2$lateral stress in one direction. Therefore, strain in longitudinal direction

${ϵ}_i=\frac{1}{E}({\sigma }_1-\mu {\sigma }_2)$

$=\frac{1}{2\times {10}^5}\times (440+0.3\times 200)$
$=\frac{1}{400}$
$\therefore$ Elongation of bar in longitudinal direction
$={ϵ}_i\times 400$
$=1Mm$