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/mm². 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 × 10⁶ N mm

From the relationship, M / I = σ/ y , we get, I = (M × y )/ σ

i.e.

 (1/12) bd³   = ((46.875 ×10⁶)/8 )×      (d /12)

bd² = (46.875 × 10 ⁶× 12) / 16

= 35.15625 × 10⁶

Depth to width ratio, d/ b  = 1.5 or d = 1.5 b,

On substituting,   ∴ b (1.5b)² = (35.15625 × 10⁶ ) /2.25

b³ = 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