Question 7.18: Three beams have the same length, same allowable bending str...

Given :
Fig. 7.24 shows a square, a rectangular and a circular section.

Let     x = Side of a square beam
b = Width of rectangular beam
∴       2b = Depth of the rectangular beam
d = Diameter of a circular section.
The moment of resistance of a beam is given by,

M = σ × Z

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

Section modulus of a square beam

=\frac{I}{y}=\frac{\frac{bd^3}{12}}{(\frac{d}{2})}=\frac{x\cdot x^3}{12}\times \frac{2}{x} \quad (∵  b = d = x) \\ \space \\ =\frac{x^2}{6}

Section modulus of a rectangular beam

=\frac{\frac{db^3}{12}}{\frac{d}{2}}=\frac{\frac{b\times (2b)^3}{12}}{(\frac{2b}{2})} \quad (∵  d = 2b) \\ \space \\ =\frac{b\times 8b^3}{12}\times \frac{2}{2b}=\frac{2}{3}b^3

Section modulus of a circular beam

=\frac{\frac{\pi d^4}{64}}{\frac{d}{2}}=\frac{\pi d^4}{64}\times \frac{2}{d}=\frac{\pi d^3}{32}

Equating the section modulus of a square beam with that of a rectangular beam, we get

\frac{x^3}{6}=\frac{2}{3}b^3

∴         b^3=\frac{3x^3}{6\times 2}=\frac{x^3}{4}=0.25x^3

or        b=(0.25)^{1/3}x=0.63x                  …(i)

Equating the section modulus of a square beam with that of a circular beam, we get

\frac{x^3}{6}=\frac{\pi d^3}{32}

∴        d^3=\frac{32x^3}{6\pi} \text{ or } d=\Big(\frac{32}{6\pi}\Big)^{1/3} \cdot x=1.1927x

The weights of the beams are proportional to their cross-sectional areas. Hence

\frac{\text{Weight of rectangular beam}}{\text{Weight of square beam}} = \frac{\text{Area of rectangular beam }}{\text{Area of square beam}} \\ \space \\ =\frac{b \times 2d}{x \times x}=\frac{0.63x \times 2 \times 0.63x}{x \times x}\\ \space \\ = \pmb{0.7938.}

and       \frac{\text{Weight of circular beam}}{\text{Weight of square beam}}=\frac{\text{Area of circular beam}}{\text{Area of square beam}} \\ \space \\ \quad \quad =\frac{\frac{\pi d^2}{4}}{x^2}=\frac{\pi d^2}{4x^2}=\frac{x \times (1.1927x)^2}{4x^2} \quad (∵  d = 1.1927x) \\ \space \\ \quad \quad = \pmb{1.1172.}