Cantilever Deflection — Uniform Load

Spread a load along a cantilever instead of hanging it at the tip and the coefficient becomes $wL^4/8EI$. A 2 m steel cantilever with $I = 10^7$ mm⁴ carrying 5 kN/m droops $5000 \times 16/(8 \times 2.0\times10^{11} \times 10^{-5}) = 5$ mm. Compare it against the same total load — 10 kN — concentrated at the free end, which gives $PL^3/3EI = 13.3$ mm. The distributed case is exactly three-eighths of that, because most of the load sits closer to the fixed end where it has less leverage.

The fourth power on length is the part that governs design. Add a third to the projection of a balcony, from 1.5 m to 2 m, and the deflection multiplies by $(2/1.5)^4 = 3.16$ for the same slab and the same load per square metre. Cantilevers are the members where serviceability nearly always beats strength: the usual limit is $L/180$ on the projection, which for a 2 m balcony means 11 mm, and there is no visual reference point out there to hide the sag against, so people notice a droop on a cantilever that they would never see mid-span.

Everything here also assumes the fixed end is genuinely fixed. It rarely is. A bracket bolted to a flexible column, a joist pocketed into masonry, or a slab cantilevering off a beam that itself twists will all rotate at the support, and that rotation adds $\theta L$ directly to the tip deflection — often more than the flexural term this formula computes. Also remember to superimpose: a balcony carries its own uniform weight and a rail load at the tip, and the two deflections simply add.