Example:

A simply supported beam contains a width of 100 mm & a depth of 150 mm. This is loaded along uniformly distributed load over the whole span of 3 m. If the permissible shear stress is 3 N/mm², discover the value of the uniformly distributed load on the beam.

Solution

Assume w be the uniformly distributed load on the beam.

After that, maximum shear force, F at the support = (wl/2) = w × (3/2) = (3/2) w

Maximum shear stress at the neutral axis = (3/2) ×  (F/bd)

This value equals 3 N/mm².

Thus,   ∴ (3/2) ×  (F/bd)  = 3

(3/2) × (3/2) w × 1/ (100 × 150) = 3

∴  w = 20000 N/m

∴  Value of consistently distributed load on the beam = 20 kN/m.