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/mm².

Solution

Breadth of beam, b = 150 mm

Depth of beam, d = 300 mm

Moment of inertia, I = (1/12) bd³

= (1/12) × (150) × (300)³

= 3.375 × 108 mm⁴

Maximum bending stress, σ = 8 N/mm²

Span of beam, l = 4 m

Extreme fibre distance, y = 150 mm

Bending stress, σ = ( M/ I) × y

or,

8 =       (M / (3.375 × 10⁸)) × 150

 ∴ Maximum bending moment, M

= (8 × 3.375 × 10⁸) /150

= 18 × 10⁶ 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.