Continuous Beam Moment⚠ unverified
Mechanical / Beams · Compute the negative bending moment over an interior support of a continuous beam
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| w | w | N/m | 1.0 | Uniformly distributed load per unit length |
| L | L | m | 1.0 | Span length |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | M | N*m | Magnitude of the support (negative) bending moment, in newton-metres (N*m) |
The science & history
Understanding the Parameters
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Distributed load $w$ — the uniform load per unit length. The moment is linear in $w$: it is simply the load intensity scaled by the span geometry.
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Span length $L$ — the distance between supports. The moment grows with the square of the span, the same $L^2$ scaling as every distributed-load moment: doubling the span quadruples the bending moment. This is why long continuous spans need disproportionately deeper sections.
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The coefficient 1/12 — the signature of the fixed-fixed, uniform-load case. It is smaller than the simply-supported $1/8$ because the end fixity carries part of the load through negative support moments, relieving the mid-span. Different support/load combinations give different coefficients (cantilever $1/2$, propped cantilever $1/8$ at the fixed end, etc.).
Derivation (Approaching a Proof)
Because the beam is indeterminate, equilibrium alone is not enough — a compatibility condition (zero rotation at the fixed ends) is needed. Using the slope-deflection (or unit-load / moment-area) method for a prismatic span of stiffness $EI$ under uniform load $w$:
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Treat the fixed-end moments $M_{AB}$, $M_{BA}$ as unknowns. The end rotations from the applied load plus the end moments must both vanish (built-in supports): $$\theta_A = 0, \qquad \theta_B = 0.$$
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The rotation at end $A$ of a simply-supported span under uniform load is $\theta_{A,0} = \dfrac{wL^3}{24EI}$, and the rotation produced by end moments $M_{AB}, M_{BA}$ is $\theta_A = \dfrac{L}{6EI}(2M_{AB} - M_{BA})$ (from slope-deflection).
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Imposing $\theta_A = 0$ and, by symmetry, $M_{AB} = M_{BA} = M$, the load and moment rotations must cancel: $$\frac{wL^3}{24EI} = \frac{L}{6EI}\,(2M - M) = \frac{ML}{6EI}.$$
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Solving: $M = \dfrac{wL^2}{12}$.
The $EI$ cancels — the fixed-end moment depends only on load and span, not on stiffness. Symmetry of the loading and boundary conditions is what makes the two end moments equal. The mid-span moment then follows from equilibrium: $M_{\text{mid}} = \dfrac{wL^2}{8} - \dfrac{wL^2}{12} = \dfrac{wL^2}{24}$.
Dimensional check. $[M] = (\text{N/m})\,\text{m}^2 = \text{N}\cdot\text{m}$. ✓
History and Development
Continuous-beam analysis was a central problem of 19th-century structural engineering. Clapeyron's theorem of three moments (1857) gave the first systematic method for continuous beams over many supports; Hardy Cross's moment distribution method (1930) made indeterminate frames tractable by hand and popularised fixed-end-moment coefficients like $wL^2/12$, which are tabulated in every structural handbook. These coefficients survive today as the sanity-check against finite-element frame models.
Related Concepts: Beam Moment At X, Beam Bending Stress, Section Modulus, Rectangular Moment of Inertia, Moment Of Inertia I Beam
Notes: This is the fixed-end moment for a single uniformly loaded span with both ends fully fixed. Real continuous beams over multiple supports need the three-moment or moment-distribution method for the actual support moments; $wL^2/12$ is the fixed-end starting value those methods distribute.