Beam Deflection — Simply Supported, Uniform Load

The odd-looking 5/384 comes straight from integrating the parabolic moment diagram of a uniformly loaded simply supported beam twice; it works out to 0.013021. A 10 kN/m load on a 5 m span with E = 200 GPa and I = 10⁻⁴ m⁴ sags δ = 5 × 10 000 × 5⁴ ÷ (384 × 200 × 10⁹ × 10⁻⁴) = 0.00407 m, about 4 mm, or L/1230 — a comfortably stiff floor.

Now the fourth power. Going from a 4 m to a 5 m span with everything else unchanged multiplies the deflection by (5/4)⁴ = 2.44, and it is this exponent, not stress, that governs most residential floor framing: the joists are almost always chosen for bounce, not for strength. A useful shortcut is that the uniform-load deflection is 5/8 of the deflection a single mid-span point load of the same total magnitude would cause, so wL total spread out sags less than the same weight concentrated at the middle. Enter w in force per unit length (N/m, kN/m, lbf/ft) and I as a plain number in m⁴; forgetting the tributary width when turning a floor pressure into w is the classic mistake here.