The Three-Moment Equation, Worked Through

A beam continuous over its supports is statically indeterminate: more unknowns than equilibrium equations. The classical hand method is Clapeyron’s theorem of three moments, which relates the bending moments at three consecutive supports to the loading on the two adjacent spans.

The equation

For supports A, B, C with spans L₁ (A–B) and L₂ (B–C), constant EI:

M_A·L₁ + 2·M_B·(L₁ + L₂) + M_C·L₂ = −6·(A₁·ā₁/L₁ + A₂·b̄₂/L₂)

The right-hand side collects the simply-supported moment diagrams of each span. For common loads the right side has closed forms:

Load on span	Right-hand contribution
Uniform w over the span	−w·L³/4
Central point load P	−3·P·L²/16

Worked example: two equal spans, uniform load

Two spans of 5 m, uniform load 10 kN/m on both, pinned ends (M_A = M_C = 0).

One equation, at support B:

0·5 + 2·M_B·(5 + 5) + 0·5 = −(10·5³/4 + 10·5³/4)
20·M_B = −2·(1250/4) = −625
M_B = −31.25 kN·m

Negative — the beam hogs over the middle support, top fibre in tension. Exactly what you expect.

Getting reactions from the support moment

Cut the beam at B and take moments about B for the left span:

ΣM_B = 0:  R_A·5 − 10·5·2.5 + 31.25 = 0   →  R_A = 18.75 kN

By symmetry R_C = 18.75 kN, and vertical equilibrium gives R_B = 100 − 2·18.75 = 62.5 kN.

Check: ΣR = 18.75 + 62.5 + 18.75 = 100 kN = total load ✓. This equilibrium residual is exactly what the continuous beam calculator reports after every solve — a sound analysis drives it to zero.

More spans

With n supports you write one three-moment equation per interior support: a tridiagonal system, solved by hand for two or three spans, or by a solver for more. The calculator handles up to six spans and reports the full shear, moment, deflection and stress diagrams.