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.

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Accepted Answer

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

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): [latex] Z=\frac{I}{y_{\max }}=\frac{b d^3}{12 imes\left(\frac{d}{2} ight)}=\frac{b d^3}{12} imes \frac{2}{d}=\frac{b d^2}{6} [/latex]

∴ [latex] Z=\frac{b d^2}{6}=\frac{300 imes 200^2}{6}=2000000 mm ^3 [/latex]

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

[latex] M=\frac{w imes L^2}{8}=\frac{w imes 8^2}{8} \quad (∵ L = 8 m) \ \space \ = 8w Nm = 8w × 1000 Nmm \ \space \ = 8000w Nmm \quad (∵ 1 m = 1000 mm) [/latex]

Now using equation (7.6), we get
(7.6): [latex]M = σ_{\max} . Z[/latex]

[latex]M = σ_{\max .} . Z[/latex]

or 8000w = 120 × 2000000

∴ [latex] w=\frac{120 imes 2000000}{8000}=30 imes 1000 N / m = \pmb{3 0 k N / m.}[/latex]

Q. 7.4

Chapter 7

Q. 7.4

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