Continuous Beam Calculator (Two Spans, Three-Moment Equation)

How it works

Three-moment equation with simply supported ends (MA = MC = 0): 2·MB·(L₁ + L₂) = −(w₁L₁³ + w₂L₂³)/4, so MB = −(w₁L₁³ + w₂L₂³) / [8(L₁ + L₂)] · RA = w₁L₁/2 + MB/L₁ · RC = w₂L₂/2 + MB/L₂ · RB = total load − RA − RC · maximum span moment = R²/(2w) at x = R/w from the end support.

A beam that runs continuously over three supports is statically indeterminate: equilibrium alone cannot give the reactions. Clapeyron’s three-moment equation adds the compatibility condition that the beam’s slope is continuous over the middle support, which gives the hogging moment MB directly; the reactions and span moments then follow from statics of each span. For two equal spans under the same uniform load it reproduces the classic results MB = −wL²/8, end reactions 0.375wL and centre reaction 1.25wL. Elastic analysis assumes a prismatic beam (constant EI) and unyielding supports.

Worked example

Spans of 5 m and 4 m carry 10 kN/m each. MB = −(10 × 125 + 10 × 64)/(8 × 9) = −1,890/72 = −26.25 kN·m. RA = 10 × 5/2 − 26.25/5 = 25 − 5.25 = 19.75 kN, RC = 10 × 4/2 − 26.25/4 = 20 − 6.56 = 13.44 kN and RB = 90 − 19.75 − 13.44 = 56.81 kN. The largest sagging moment in span 1 is 19.75²/(2 × 10) = 19.50 kN·m at 1.98 m from A, and in span 2 it is 9.03 kN·m at 1.34 m from C. The largest shear is 30.25 kN just left of support B.

Sources: Three-moment equation (Clapeyron’s theorem) · American Wood Council — Design Aid No. 6: Beam Design Formulas (case 29, two equal spans)

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