Continuous Beam Analysis: Methods, Formulas & Workflow

📅 Updated: Sep 2026 👤 By MultiTooSite Team ⏱️ 12 min read

Continuous beam analysis deals with beams that span over more than two supports (three or more). Unlike a simple beam (statically determinate), a continuous beam is statically indeterminate — support reactions and moments cannot be found from equilibrium equations alone.

📋 What's Inside

Why Continuous Beams Are Indeterminate
Main Analysis Methods
Qualitative Behaviour vs Simple Spans
Key Assumptions
Practical Workflow
Relation to Simple Beam Calculator

1. Why Continuous Beams Are Indeterminate

For a continuous beam with \( n \) spans ( \( n+1 \) supports):

Unknowns	Count
Vertical reactions	\( n + 1 \)
Support moments (if continuous over supports)	\( n - 1 \) (or more if fixed ends)
Equilibrium equations available	Only \( 2 \) global equations (\( \sum F_y = 0 \), \( \sum M = 0 \)) plus local joint equilibrium

Degree of static indeterminacy for a continuous beam on simple supports is typically:

\[ i = r - 3 \]

(for a plane beam), or more precisely for \( n \) spans with simple end supports: \( i = n - 1 \).

Extra equations come from compatibility of displacements (deflection and slope continuity at intermediate supports).

2. Main Analysis Methods

A. Clapeyron's Theorem of Three Moments (Classic Hand Method)

For any two adjacent spans \( L_i \) and \( L_{i+1} \) with flexural rigidities \( EI_i \), \( EI_{i+1} \):

\[ M_{i-1} L_i + 2M_i (L_i + L_{i+1}) + M_{i+1} L_{i+1} = -6\left( \frac{A_i \bar{x}_i}{L_i} + \frac{A_{i+1} \bar{x}_{i+1}}{L_{i+1}} \right) \]

(when \( EI \) is constant the form simplifies further).

\( M_{i-1}, M_i, M_{i+1} \) = support moments
\( A_i \bar{x}_i \) = moment of the free (simple-span) bending-moment diagram about the left/right support

This produces a system of linear equations for all intermediate support moments.

B. Slope-Deflection Method

Member end moments are expressed in terms of joint rotations \( \theta \) and fixed-end moments (FEM):

\[ M_{AB} = M_{AB}^F + \frac{2EI}{L}(2\theta_A + \theta_B - 3\psi) \] \[ M_{BA} = M_{BA}^F + \frac{2EI}{L}(2\theta_B + \theta_A - 3\psi) \]

where \( \psi \) is the chord rotation (usually zero for continuous beams without settlement). Joint equilibrium \( \sum M = 0 \) at each support yields the rotations, then the moments.

C. Moment Distribution (Hardy Cross)

Iterative hand method still useful for quick checks:

Compute fixed-end moments for each span.
Calculate distribution factors at each joint:
\[ DF_{ij} = \frac{K_{ij}}{\sum K} \]
where stiffness \( K = 4EI/L \) (far end fixed) or \( 3EI/L \) (far end hinged).
Distribute unbalanced moments, carry over half to the far end, repeat until convergence.

D. Stiffness / Matrix Method (Modern)

The beam is assembled from element stiffness matrices. For a prismatic beam element:

\[ \begin{Bmatrix} M_i \\ M_j \end{Bmatrix} = \frac{EI}{L} \begin{bmatrix} 4 & 2 \\ 2 & 4 \end{bmatrix} \begin{Bmatrix} \theta_i \\ \theta_j \end{Bmatrix} + \begin{Bmatrix} M_i^F \\ M_j^F \end{Bmatrix} \]

Global system \( \mathbf{K}\boldsymbol{\theta} = \mathbf{F} \) is solved for rotations, then end moments and reactions are recovered. This is the basis of almost all structural analysis software.

3. Qualitative Behaviour vs Simple Spans

Quantity	Simple Beam (same total load)	Continuous Beam
Positive midspan moment	Higher (\( wL^2/8 \))	Lower
Negative moment at supports	Zero	Significant (often \( \approx wL^2/8 \) to \( wL^2/12 \))
Max deflection	Larger	Smaller
Redundancy	None	High (failure of one span does not collapse the whole beam)
Material economy	Less efficient	Usually more efficient for the same loads

Typical approximate coefficients for equal continuous spans under uniform load (service):

End span positive moment \( \approx wL^2/11 \)
Interior positive moment \( \approx wL^2/16 \)
Interior support moment \( \approx -wL^2/11 \) to \( -wL^2/12 \)
End support reaction \( \approx 0.4wL \), interior \( \approx 1.1wL \)

(Exact values depend on the number of spans and relative stiffnesses.)

4. Key Assumptions

Linear-elastic material
Prismatic members (or piecewise prismatic)
Small deflections (Euler-Bernoulli theory)
No axial deformation coupling (pure bending)
Supports do not settle (or settlement is known and included)
Loads are static

5. Practical Workflow for Hand / Spreadsheet Analysis

Idealise the continuous beam (spans, \( EI \), support conditions).
Compute fixed-end moments for each span under the real loads.
Choose a method (Three Moments, Slope-Deflection or Moment Distribution).
Solve for support moments.
Find reactions from equilibrium of each span.
Draw shear and moment diagrams (including the negative support moments).
Check deflection if required (integration or virtual work).
Design for the envelope of positive and negative moments.

6. Relation to the Simple Beam Calculator

The free beam load calculator you already have treats single-span simply supported and cantilever cases (statically determinate). Continuous beams require the extra compatibility equations above, so a full continuous-beam tool needs either:

Three-moment / slope-deflection solver for a user-defined number of spans, or
A stiffness-matrix engine.

Want to go deeper? A continuous-beam calculator can be built for 2–4 equal or unequal spans using UDL and point loads, based on either the Three-Moment theorem or the stiffness method. Alternatively, individual methods can be explored with worked numerical examples and moment distribution steps.

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Engineering Disclaimer: This guide is for educational and preliminary analysis purposes only. Continuous beam analysis for real structures must account for actual support conditions, material properties, load combinations, code requirements, and serviceability limits. Final structural designs must be verified by a licensed structural engineer.