Continuous Beams

Continuous beams extend over more than one span and are supported at more than two points. Fixed or built-in beams have rotational restraint at their supports. These arrangements are generally statically indeterminate, so the support reactions and bending moments cannot normally be obtained from static equilibrium alone.

Need help choosing a beam calculation? See the Beam Design and Analysis Guide for guidance on beam types, loading, bending, shear, deflection and selecting the appropriate RoyMech calculation method.

This page covers fixed-fixed beams and the analysis of continuous beams using Clapeyron's three-moment theorem, including worked examples for two-span and three-span beams.

Nomenclature
Fixed or Built-In Beams
Fixed Beam with Uniformly Distributed Load
Continuous Beams
Clapeyron's Three-Moment Theorem
Moment-Diagram Areas and Centroids
Example 1 – Two-Span Continuous Beam
Example 2 – Three-Span Continuous Beam

Nomenclature

Symbol	Description	SI Unit
A	Area of free bending-moment diagram used in the three-moment theorem	N m²
E	Young's modulus	Pa (N/m²)
F	Concentrated force	N
I	Second moment of area	m⁴
L, l	Beam or span length	m
M	Bending moment	N m
R	Support reaction	N
V	Shear force	N
w	Distributed load intensity	N/m
W	Total applied force	N
x	Distance along beam	m
y	Beam deflection	m
θ	Beam slope	rad

The signs of bending moment, shear force and deflection depend upon the adopted sign convention. A consistent convention must be maintained throughout each calculation.

Fixed or Built-In Beams

A fixed-fixed beam, sometimes called a built-in or encastre beam, is restrained against translation and rotation at both ends. In the usual idealised horizontal arrangement, both the deflection and slope are zero at the supports.

Unlike a simply supported beam, a fixed beam develops support moments. The restraint changes the bending-moment distribution and generally reduces the maximum span deflection compared with an equivalent simply supported beam.

Depending upon the loading, the bending moment may change sign between the supports, producing points of contraflexure. A typical fixed beam is shown below.

Fixed beam showing loading, support moments and points of contraflexure

The support reactions and end moments cannot generally be determined from the equations of static equilibrium alone. Compatibility conditions based on beam slope and deflection are also required.

Beam Differential Relationships

The relationships between distributed load w, shear force V, bending moment M and beam deflection y are:

V = \frac{d M}{d x}

\frac{d V}{d x} = - w

\frac{d^{2} M}{d x^{2}} = - w

For a beam of constant flexural rigidity EI, using the sign convention adopted here:

V = \frac{d M}{d x} = E I \frac{d^{3} y}{d x^{3}}

- w = E I \frac{d^{4} y}{d x^{4}}

For further development of these relationships see Beam Theory and Shear Force and Bending Moment Diagrams.

Fixed Beam with Uniformly Distributed Load

Consider a fixed-fixed beam of span l carrying a uniformly distributed load w over the complete span.

Derivation for a fixed-fixed beam carrying a uniformly distributed load

Differential Equation

E I \frac{d^{4} y}{d x^{4}} = - w

First Integration – Shear Force

E I \frac{d^{3} y}{d x^{3}} = - w x + C_{1}

Second Integration – Bending Moment

E I \frac{d^{2} y}{d x^{2}} = - \frac{w x^{2}}{2} + C_{1} x + C_{2}

Third Integration – Slope

E I \frac{d y}{d x} = - \frac{w x^{3}}{6} + \frac{C_{1} x^{2}}{2} + C_{2} x

At the left fixed support:

\frac{d y}{d x} = 0 at x = 0

At the right fixed support:

\frac{d y}{d x} = 0 at x = l

Applying the second condition gives:

- \frac{w l^{3}}{6} + \frac{C_{1} l^{2}}{2} + C_{2} l = 0

Fourth Integration – Deflection

E I y = - \frac{w x^{4}}{24} + \frac{C_{1} x^{3}}{6} + \frac{C_{2} x^{2}}{2}

Since y = 0 at x = 0 and y = 0 at x = l, the fixed-end boundary conditions give:

C_{1} = \frac{w l}{2} C_{2} = - \frac{w l^{2}}{12}

Bending Moment

Substitution of the integration constants gives:

M = E I \frac{d^{2} y}{d x^{2}} = \frac{w}{12} (- 6 x^{2} + 6 l x - l^{2})

At either fixed end:

M_{end} = - \frac{w l^{2}}{12}

At the centre of the span:

M_{centre} = \frac{w l^{2}}{24}

Shear Force

V = E I \frac{d^{3} y}{d x^{3}} = w (\frac{l}{2} - x)

Thus:

V = \frac{w l}{2} at x = 0

V = - \frac{w l}{2} at x = l

A continuous beam extends over two or more spans and normally has more than two supports. The intermediate supports introduce additional reactions, making the beam statically indeterminate.

A classical method of analysing such beams is Clapeyron's three-moment theorem.

Continuous beam and free bending moment diagrams used in Clapeyron's three-moment theorem

The three-moment theorem relates the bending moments at three consecutive supports. It is obtained by considering compatibility of beam slope at the common support between two adjacent spans.

Clapeyron's Three-Moment Theorem

Constant Section and Supports at the Same Level

For two adjacent spans of constant section and flexural rigidity, with the supports at the same level:

M_{A} L_{1} + 2 M_{B} (L_{1} + L_{2}) + M_{C} L_{2} = - 6 (\frac{A_{1} x_{1}}{L_{1}} + \frac{A_{2} x_{2}}{L_{2}})

Here A₁ and A₂ are the areas of the free bending-moment diagrams for the two spans, and x₁ and x₂ locate their centroids as shown in the diagram.

Different Flexural Rigidity in Adjacent Spans

If the adjacent spans have different second moments of area, the more general expression is:

\frac{M_{A} L_{1}}{I_{1}} + 2 M_{B} (\frac{L_{1}}{I_{1}} + \frac{L_{2}}{I_{2}}) + \frac{M_{C} L_{2}}{I_{2}} = - 6 (\frac{A_{1} x_{1}}{L_{1} I_{1}} + \frac{A_{2} x_{2}}{L_{2} I_{2}})

This form assumes the same Young's modulus in the adjacent spans. If E also differs between spans, the corresponding flexural rigidities EI must be retained explicitly.

Moment-Diagram Areas and Centroids

Application of the three-moment theorem requires the area and centroid of the free bending-moment diagram for each loaded span.

Uniformly Distributed Load

Free bending moment diagram area and centroid for a simply supported span carrying a uniformly distributed load

For a simply supported span of length L carrying a uniformly distributed load w:

M = \frac{w L x}{2} - \frac{w x^{2}}{2}

A_{1} = \int_{0}^{L} (\frac{w L x}{2} - \frac{w x^{2}}{2}) d x

A_{1} = \frac{w L^{3}}{12}

x_{1} = \frac{L}{2}

Concentrated Load

Free bending moment diagram area and centroid for a simply supported span carrying a concentrated load

For the concentrated-load case shown in the original diagram:

A_{1} = \frac{F x (L - x)}{2}

x_{1} = \frac{L + x}{3}

Example 1 – Two-Span Continuous Beam

This example considers a two-span continuous beam with simple supports at the two outer ends. The bending moments at supports A and C are therefore zero.

1. Bending Moments

Two-span continuous beam with distributed and concentrated loading

Three-moment theorem calculation for the two-span continuous beam

For span AB:

L_{1} = 2 m

A_{1} = \frac{10 2^{3}}{12} = \frac{20}{3} kN m

x_{1} = 1 m

For span BC:

L_{2} = 3 m

A_{2} = \frac{400 \times 1}{2} = 200 kN m

x_{2} = \frac{3 + 1}{3} = \frac{4}{3} m from C

Applying the three-moment theorem:

M_{A} 2 + 2 M_{B} (2 + 3) + M_{C} 3 = - 1086.67

Since the outer supports are simple:

M_{A} = M_{C} = 0

M_{B} = - 108.67 kN m

2. Shear Forces and Reactions

Shear force and support reaction calculations for the two-span continuous beam

The support reactions are then obtained by considering the equilibrium of each span, including the calculated support moment.

3. Shear Force and Bending Moment Diagrams

Shear force and bending moment diagrams for the two-span continuous beam

Example 2 – Three-Span Continuous Beam

This example considers a three-span continuous beam with simple supports at the two outer ends A and D.

Three-span continuous beam with concentrated loads

Therefore:

M_{A} = M_{D} = 0

Spans AB and BC

Three-moment calculation for spans AB and BC

From the free bending-moment diagrams:

A_{1} = 0.525 x_{1} = 0.567 A_{1} x_{1} = 0.2975

A_{2} = 1.12 x_{2} = 0.667 A_{2} x_{2} = 0.7467

Applying the three-moment theorem to spans AB and BC:

4.4 M_{B} + 1.2 M_{C} = - 5.518

Hence:

M_{B} = \frac{- 5.518 - 1.2 M_{C}}{4.4}

Spans BC and CD

Three-moment calculation for spans BC and CD

A_{1} = 1.12 x_{1} = 0.533 A_{1} x_{1} = 0.5973

A_{2} = 0.36 x_{2} = 0.467 A_{2} x_{2} = 0.168

Applying the three-moment theorem to spans BC and CD:

1.2 M_{B} + 4 M_{C} = - 4.247

Substituting the expression for M_B and solving the two simultaneous equations gives:

M_{B} = - 1.051 kN m M_{C} = - 0.746 kN m

Resulting Bending-Moment Diagram

Resulting bending moment diagram for the three-span continuous beam

The final support moments are therefore:

M_{A} = 0 M_{B} = - 1.051 kN m M_{C} = - 0.746 kN m M_{D} = 0

Continuous Beam Analysis Considerations

The three-moment theorem provides a direct classical method for analysing many continuous beams. It combines static equilibrium with compatibility of beam deformation and is particularly convenient for a sequence of prismatic spans.

Where span stiffness varies, the appropriate values of flexural rigidity must be included. Support settlement, elastic supports and other imposed displacements also require the appropriate additional terms in the compatibility equations.

For more complicated beam systems, alternative methods include moment distribution, energy methods, matrix stiffness methods and finite-element analysis.

Continuous and Fixed Beam Analysis

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