Question 7.19: A beam is of square section of the side ‘a’. If the permissi...

Given :
Bending stress = σ

1st Case
Fig. 7.25 (a) shows the square beam section when two sides are horizontal.

Let M_1 = Moment of resistance of the square beam when two sides are horizontal.
Moment of resistance is given by,

M = σ × Z

∴        M_1 = σ × Z_1              …(i)

where Z_1 = Section modulus

= \frac{I}{y_{\max}}=\frac{\frac{a\times a^3}{12}}{a/2}=\frac{a^4}{12}\times \frac{2}{a}=\frac{a^3}{6}

∴       M_1=\sigma\times \pmb{\frac{a^3}{6}.}

2nd Case

Fig. 7.25 (b) shows the square beam section when one diagonal is vertical.
Let M_2 = Moment of resistance of the beam in this position
∴       M_2 = σ × Z_2

where Z_2= Section modulus for the section shown in Fig. 7.25 (b).

=\frac{I_2}{y_{\max}}=\frac{2\times \frac{bh^3}{12}}{(\frac{a}{\sqrt{2}})} \\ \space \\ \Big(∵  \text{M.O.I. of a triangle about its base } = \frac{bh^3}{12}. \text{ There are two triangles} \Big) \\ \space \\ = \frac{\frac{2}{12} \times \sqrt{2}a(\frac{a}{\sqrt{2}})^3}{(\frac{a}{\sqrt{2}})} \quad \Big(∵ \text{ Here base } = b = \sqrt{2}a \text{ and } h=\frac{a}{\sqrt{2}}\Big) \\ \space \\ =\frac{a^3}{6\times \sqrt{2}}

∴               M_2=\sigma\times \pmb{\frac{a^3}{6\times \sqrt{2}}.}

Ratio of moment of resistance of the section in two positions

=\frac{M_1}{M_2}=\frac{(\frac{\sigma \times a^3}{6})}{(\sigma \times \frac{a^3}{6 \times \sqrt{2}})}=\sqrt{2}=\pmb{1.414.}