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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.

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