Question

The beams, one having square cross-section and another circular cross- section, are subjected to the same amount of bending moment. If the cross sectional area as well as the material of both the beams are the same then

• A. maximum bending stress developed in both the beams is the same
• B. the circular beam experience more bending stress than the square one
• C. the square beam experience more bending stress than the circular one
• D. as the material is same both beams will experience same deformation

Answer 1

The correct option is B the circular beam experience more bending stress than the square one

Let d= diameter of circular cross-section
a=side of square cross-section
Since cross-sectional area of square and circular
cross-section are equal
$\therefore \frac{\pi }{4}{d}^2=a^2$

For circular cross-section,
${\sigma }_c-\frac{My}{I}=\frac{M\frac{d}{2}}{\frac{\pi d^4}{64}}=\frac{32M}{\pi d^3}$

Square cross-section.

${\sigma }_s=\frac{My}{I}=\frac{M\frac{a}{2}}{\frac{a^4}{12}}=\frac{6M}{a^3}$

$Now,\frac{{\sigma }_c}{{\sigma }_s}=\frac{\frac{32M}{\pi d^3}}{\frac{6M}{a^3}}=\frac{32a^3}{6\pi d^3}$

$=\frac{32}{6\pi }{\left(\frac{a}{d}\right)}^2\left(\frac{a}{d}\right)$

$=\frac{32}{6\pi }\times \left(\frac{\pi }{4}\right)\times \left(\sqrt{\left(\frac{\pi }{4}\right)}\right)$

$=\frac{32}{24}\sqrt{\frac{\pi }{4}}=\frac{32\sqrt{\pi }}{48}>1$

$\therefore {\sigma }_c>{\sigma }_s$