Determine the width and depth of the beam:
A timber beam of rectangular section is only supported over a span of 5 m. It carries an uniformly distributed load of 15 kN/m over the entire span. Determine the width and depth of the beam, if the bending stress is restricted to 8 N/mm^{2}. The depth to width ratio might be taken as 1.5.
Solution
Span of beam, l = 5 m
Uniformly distributed load, w = 15 kN/m
Maximum bending moment at centre, M = wl ^{2}/8
Thus, M =15 × (5) ^{2}/ 8
= 46.875 kN m
= 46.875 × 10^{6 }N mm
From the relationship, M / I = σ/ y , we get, I = (M × y )/ σ
i.e.
(1/12) bd^{3} = ((46.875 ×10^{6})/8 )× (d /12)
bd^{2} = (46.875 × 10^{ 6}× 12) / 16
= 35.15625 × 10^{6}
Depth to width ratio, d/ b = 1.5 or d = 1.5 b,
On substituting, ∴ b (1.5b)^{2 }= (35.15625 × 10^{6} ) /2.25
b^{3} = 35.15625 × 10
= 15.625 × 106
Therefore, we get b = 250 mm
d = 1.5 × 250 = 375 mm.
Thus, width of beam = 250 mm, and
Depth of beam = 375 mm