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## Q. 7.4

A rectangular beam 200 mm deep and 300 mm wide is simply supported over a span of 8 m. What uniformly distributed load per metre the beam may carry, if the bending stress is not to exceed 120 N/mm².

## Verified Solution

Given :
Depth of beam,    d = 200 mm
Width of beam,    b = 300 mm
Length of beam,  L = 8 m
Max. bending stress, $σ_{\max} = 120 N/mm^2$

Let w = Uniformly distributed load per metre length over the beam
(Fig. 7.11 (a) shows the section of the beam.)
Section modulus for a rectangular section is given by equation (7.7).
(7.7):          $Z=\frac{I}{y_{\max }}=\frac{b d^3}{12 \times\left(\frac{d}{2}\right)}=\frac{b d^3}{12} \times \frac{2}{d}=\frac{b d^2}{6}$

∴              $Z=\frac{b d^2}{6}=\frac{300 \times 200^2}{6}=2000000 mm ^3$

Max. B.M. for a simply supported beam carrying uniformly distributed load as shown in Fig. 7.11 is at the centre of the beam. It is given by

$M=\frac{w \times L^2}{8}=\frac{w \times 8^2}{8} \quad (∵ L = 8 m) \\ \space \\ = 8w Nm = 8w × 1000 Nmm \\ \space \\ = 8000w Nmm \quad (∵ 1 m = 1000 mm)$

Now using equation (7.6), we get
(7.6):  $M = σ_{\max} . Z$

$M = σ_{\max .} . Z$

or        8000w = 120 × 2000000

∴              $w=\frac{120 \times 2000000}{8000}=30 \times 1000 N / m = \pmb{3 0 k N / m.}$  