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Chapter 7

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.

Step-by-Step

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.}

7.11
7.11a