Determine the maximum uniformly distributed load:
A timber beam 150 mm wide & 300 mm deep is supported simply over a span of
4 m. determine the maximum uniformly distributed load that the beam may carry, if the stress is not to exceed 8 N/mm^{2}.
Solution
Breadth of beam, b = 150 mm
Depth of beam, d = 300 mm
Moment of inertia, I = (1/12) bd^{3}
= (1/12) × (150) × (300)^{3}
= 3.375 × 108 mm^{4}
Maximum bending stress, σ = 8 N/mm^{2}
Span of beam, l = 4 m
Extreme fibre distance, y = 150 mm
Bending stress, σ = ( M/ I) × y
or,
8 = (M / (3.375 × 10^{8})) × 150
∴ Maximum bending moment, M
= (8 × 3.375 × 10^{8}) /150
= 18 × 10^{6} N mm = 18kN m
But M = wl ^{2}/8
i.e. 18 = w × (4)^{2} / 8
∴ w = (18 × 8)/16
= 9 kN/m
∴ The maximum consistently distributed load the beam might carry = 9 kN/m.