How to use it
- Choose simply supported (resting on two supports) or cantilever (fixed at one end).
- Choose a point load (in the middle, or at the tip of a cantilever) or a uniform load along the whole span.
- Enter the span in metres and the load in kN or kN/m.
- Enter E in GPa (steel about 200, aluminium about 70), I in cm⁴ and the section depth in mm for the stress.
Formulas
| Case | Max deflection | Max moment | Max slope | Reaction |
|---|---|---|---|---|
| Simply supported, P at mid-span | PL³ ÷ 48EI | PL ÷ 4 | PL² ÷ 16EI | P ÷ 2 |
| Simply supported, uniform w | 5wL⁴ ÷ 384EI | wL² ÷ 8 | wL³ ÷ 24EI | wL ÷ 2 |
| Cantilever, P at the tip | PL³ ÷ 3EI | PL | PL² ÷ 2EI | P |
| Cantilever, uniform w | wL⁴ ÷ 8EI | wL² ÷ 2 | wL³ ÷ 6EI | wL |
The bending stress at the outer fibre of a symmetric section is σ = M × (h ÷ 2) ÷ I. The formulas assume a straight elastic beam of constant section and small deflections; the beam’s own weight is not included unless you add it to w.
Worked example
A 3 m simply supported steel beam (E = 200 GPa, I = 2000 cm⁴ = 2 × 10⁻⁵ m⁴, depth 200 mm) carries 10 kN in the middle. EI = 4 × 10⁶ N·m².
- δ = 10,000 × 3³ ÷ (48 × 4 × 10⁶) = 1.406 mm, that is L/2,133
- M = 10 × 3 ÷ 4 = 7.5 kN·m
- σ = 7500 × 0.1 ÷ 2 × 10⁻⁵ = 37.5 MPa
- each support carries 5 kN; the ends rotate by 0.0806°
With 5 kN/m along 6 m instead, the deflection is 21.094 mm (L/284) and the moment 22.5 kN·m.
Tips
- Deflection grows with L³ or L⁴. Doubling the span of a uniformly loaded beam makes it sag 16 times more.
- Depth is the cheapest stiffness. For a rectangle I grows with h³, so a beam twice as deep is 8 times stiffer.
- This is not a design check. Real design also needs load factors, buckling, shear and connection checks.