Question

A square beam and a circular beam of same material have the same length, same allowable stress and the same bending moment. The ratio of weights of the square beam to that of circular beam is
1. \(\frac{1}{2}\)
2. 1
3. \(\frac{1}{1.12}\)
4. \(\frac{1}{3}\)

Answer 1

Correct Answer - Option 3 : \(\frac{1}{1.12}\)

Concept:

The moment of resistance of a beam is given by:

M = σ × Z

where, M = Bending moment, σ = Bending stress,

Z = Section modulus = \(\frac{I}{A}\)

Calculation:

Given:

As all the two beams have the same allowable stress (σ), and same bending moment (M), therefore the section modulus (Z) of the two beams must be equal.

\({{\rm{Z}}_{\rm{s}}} = \frac{{{{\rm{a}}^3}}}{6}\)

where “a” is the side of a square beam

\({{\rm{Z}}_{\rm{c}}} = \frac{{\rm{\pi }}}{{32}}{{\rm{d}}^3}\)

Where “d” is the diameter of the circular beam.

Z_s =Z_c

\(\frac{{{a^3}}}{6} = \frac{{\rm{\pi }}}{{32}} \times {{\rm{d}}^3}\)

\({{\rm{a}}^3} = \frac{{3{\rm{\pi }}}}{{16}} \times {{\rm{d}}^3}\)

\({\rm{a}} = {\left( {\frac{{3{\rm{\pi }}}}{{16}}} \right)^{\frac{1}{3}}} \times {\rm{d}}\)

The weight of the beams is proportional to their cross-sectional area.

\(\frac{{{\rm{weight\;of\;the\;square\;beam}}}}{{{\rm{weight\;of\;circular\;beam}}}} = \frac{{{\rm{Area\;of\;square\;beam}}}}{{{\rm{Area\;of\;circular\;beam}}}}\)

\( \frac{{{\rm{weight\;of\;the\;square\;beam}}}}{{{\rm{weight\;of\;circular\;beam}}}} = \frac{{{{\rm{a}}^2}}}{{\frac{{\rm{\pi }}}{4}{{\rm{d}}^2}}} = \frac{4}{{\rm{\pi }}}{\left( {\frac{{\rm{a}}}{{\rm{d}}}} \right)^2} = \frac{4}{{\rm{\pi }}}{\left( {\frac{{3{\rm{\pi }}}}{{16}}} \right)^{\frac{2}{3}}} = \frac{1}{{1.12}}\)

The ratio of weights of the square beam to that of the circular beam is \(\frac{1}{{1.12}}\).