AdvancedBeamIndeterminateThree-momentContinuous

Two-span continuous beam

Add one support and the beam stops being solvable by statics alone — it becomes indeterminate. The classical way in is Clapeyron’s three-moment theorem, and this symmetric two-span case produces the results every engineer memorises: hogging over the middle support, and end reactions of only . Along the way we’ll see exactly why continuity makes the beam so much more efficient than two independent spans.

Figure 1.The problem: a two-span continuous beam on supports A, B and C, carrying a uniform load of 12 kN/m across both spans. We solve for the support moment and reactions below; the reactions and deflected shape come after.

Given

Configuration: 2 spans × 6 m
Supports: Pin · Roller · Roller
Span: 6 m
: 200 GPa
: 100 × 10⁶ mm⁴
: 12 kN/mentered −12

Step 1 — One support too many for statics

Three reactions, two equations — the missing one is about geometry, not force.

Add one support to a simply supported beam and it stops being solvable by statics alone. Count the unknowns: three supports mean three vertical reactions, but a vertically loaded beam offers only two useful equilibrium equations — = 0 and = 0. Three unknowns, two equations: the beam is statically indeterminate, with one redundant unknown, and no amount of clever moment-taking will crack it.

The missing equation is not about forces at all — it is about geometry. Whatever the reactions turn out to be, the deflected beam must remain one smooth, continuous curve passing over all three supports. Turning that geometric fact into algebra is exactly what the next step does.

Step 2 — Clapeyron’s three-moment theorem

One equation for the unknown support moment.

The classical way in is Clapeyron’s three-moment theorem. Take any three consecutive supports of a continuous beam and the theorem writes one equation linking the bending moments at those three points. The left side is pure geometry — the span lengths. The right side measures the loading: each span contributes a load term (for a full UDL it is ) built from the bending-moment diagram that span would have if it stood alone as a simply supported beam. Behind the algebra it is nothing more than Step 1’s compatibility condition worked out: the slope of the beam just left of B must match the slope just right of B.

Here the bookkeeping collapses. The outer ends are simple supports with nothing beyond them, so — only the interior moment is unknown, and the two equal spans collapse the theorem to a single line:

Substituting L = 6 m and w = 12 kN/m and solving:

(hogging)

The sign is the story: negative means hogging — over the middle support the curvature reverses and the top face of the beam goes into tension. One equation, one unknown, and the indeterminacy is gone.

Step 3 — The reactions

The middle support takes 62.5% of everything.

With known, the beam is unlocked — everything from here is plain statics. Treat span AB on its own: a simply supported beam carrying w, plus the end moment applied at B. The simple-span part sends = 36 kN to each end; the hogging end moment then redistributes 9 kN (= ) from the outer support to the middle. Then use symmetry:

(symmetry)

The middle support carries 62.5% of the total load — the reason continuous beams demand a strong interior column right where the hogging moment already peaks.

Step 4 — Shear and moment, span by span

The zero-shear point locates the sagging peak.

Walk span AB from the left. Shear starts at the +27 kN reaction and falls at the steady 12 kN/m: V(x) = 27 − 12x. It crosses zero at x = 2.25 m and reaches −45 kN just left of B, where the 90 kN reaction throws it back up; the second span then mirrors the first. The sagging moment peaks exactly where the shear vanishes:

Put the two peaks side by side: 30.375 kN·m of sagging inside each span against 54 kN·m of hogging over B. In a continuous beam the biggest moment usually lives at the support, not at midspan — the reverse of the simple-span habit.

Figure 2.Shear force diagram — the jump of 90 kN at the middle support.

Figure 3.Bending moment diagram — sagging 30.4 kN·m in each span (above the axis), hogging 54 kN·m over B (below the axis). We follow the US convention (sagging positive, plotted above the axis); the Beam Calculator has a Flip control for the opposite convention.

In practice — the default, and its fine print

Continuous beams are the default in real floors and bridges — walk under a concrete floor slab or a multi-span highway bridge and the members run unbroken over their supports. The reason is the redistribution you just computed: the hogging over the supports relieves the midspan, so a continuous member gets away with less depth and less steel than a chain of simple spans. But continuity buys sensitivity along with efficiency. Because the whole beam acts as one, a support that settles a few millimeters re-shuffles the moments along the entire length — the companion settlement example puts numbers on that price — and live load on one span alone can be worse for parts of the beam than load everywhere. A design office therefore never checks only the all-spans-loaded case: it also runs the pattern cases — alternate spans loaded for the worst sagging, adjacent spans loaded for the worst hogging at a support.

Step 5 — The deflected shape

Zero over every support, sagging between — compatibility made visible.

The deflected shape ties the whole solution together. Each span sags about 4.2 mm — the classical result at roughly 0.42L from an outer support — while directly over every support the deflection is zero by definition. Look closely near B: the curve flattens and its curvature reverses through the hogging region, then it passes over the middle support without a kink. That smooth, continuous curve over all three supports is the compatibility condition from Step 1 — no longer an abstract requirement but a shape you can see.

Figure 4.Deflection — about 4.2 mm within each span; zero (by definition) over every support.

Figure 5.The solved continuous beam — the reactions (purple; the big center reaction at B) and the deflected shape (dashed), sagging in each span.

✓

The proof

Hand calculation vs the solver.

Verified — hand calculation vs the solver, to round-off

Quantity	By hand	StructureCalcs
Hogging moment M_B	−54 kN·m (wL²/8)	−54 kN·m at 6 m
R_A = R_C	27 kN (3wL/8)	27 kN
R_B	90 kN (10wL/8)	90 kN
Max sagging moment	30.375 kN·m at 2.25 m (9wL²/128)	30.375 kN·m at 2.25 m
Max shear \|V\|	45 kN beside B	45 kN at 6 m
Max deflection	≈ 4.2 mm at ≈ 0.42L (wL⁴/185EI)	4.212 mm at 2.53 m

Every value was worked by hand with the classical method, then checked against this site’s solver — the same engine the Try it button opens. This agreement is re-run automatically on every build.

Now make it yours

Open this exact model in the calculator — then change a load, drag a support, and watch every diagram update in real time. The best way to build intuition is to break it and see what happens.

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