Beam Theory – Bending Stress, Deflection & Composite Beams

Beam theory provides the fundamental relationships between applied bending moment, stress, strain, material elasticity, section geometry, curvature, slope and deflection. These relationships are widely used in the analysis of structural beams and mechanical components subjected primarily to transverse loading and bending.

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 elementary straight-beam bending theory, direct integration for beam slope and deflection, the moment-area method and the transformed section method for composite beams.

Introduction
Superposition Principle
Nomenclature
Simple Bending Theory
Deflection of Beams
Moment-Area Method
Composite Beams

Introduction

The elementary theory of beam bending relates stress, strain, material elasticity, section geometry and curvature under a defined set of assumptions.

The simple bending theory developed below initially considers pure bending: a beam subjected to bending moment without transverse shear force. The resulting bending relationships are also fundamental to the analysis of beams subjected to transverse loads, provided the assumptions of the theory are appropriate.

Superposition Principle

The principle of superposition is an important tool for solving beam-loading problems within the linear-elastic range.

For a beam subjected to several loads, the resulting reactions, shear force, bending moment, slope, deflection and stress may be obtained by adding the effects produced by each load acting separately, provided the response remains linear and the boundary conditions are unchanged.

Superposition requires linear behaviour. It should not be applied without qualification where material non-linearity, large deflection, changing contact conditions or other non-linear behaviour is significant.

Nomenclature

Symbol	Description	Typical SI Unit
A	Cross-sectional area	m²
E	Young's modulus	Pa (N/m²)
F	Applied force	N
I	Second moment of area about the neutral axis	m⁴
L	Beam length	m
M	Bending moment	N m
R	Radius of curvature of the neutral surface	m
V	Shear force	N
w	Distributed load per unit length	N/m
x	Distance along beam	m
y	Distance from neutral axis in bending-stress equations	m
v	Transverse beam deflection	m
Z	Elastic section modulus, I/y_max	m³
ε	Longitudinal strain	dimensionless
θ	Beam slope	rad
σ	Longitudinal normal stress	Pa (N/m²)

Engineering section-property tables commonly give I in mm⁴ or cm⁴ and Z in mm³ or cm³. Units must therefore be made consistent before using the equations below.

Simple Bending Theory

Consider a straight, prismatic beam of homogeneous material subjected to equal and opposite end moments, producing pure bending.

Straight beam subjected to pure bending showing the neutral axis and deformation into a circular arc

Assumptions of Elementary Bending Theory

The elementary Euler-Bernoulli bending theory is based on assumptions that include:

the beam is initially straight and of uniform cross-section;
the material is homogeneous and behaves linearly elastically;
Young's modulus is the same in tension and compression;
plane cross-sections remain plane and normal to the neutral axis after bending;
beam deflections and rotations are sufficiently small for the small-deflection approximation;
the beam is sufficiently slender for transverse shear deformation to be neglected in the elementary bending model.

Strain Distribution

Consider two initially parallel cross-sections separated by a small distance. When the beam bends, longitudinal fibres on one side of the neutral surface extend while fibres on the other side shorten.

The neutral surface contains fibres whose longitudinal length does not change. For a homogeneous beam in pure bending, its intersection with the cross-section is the neutral axis.

If the neutral surface bends to a radius R through a small angle θ, its length is:

L = R θ

A longitudinal fibre a distance y from the neutral axis has length:

L_{y} = (R + y) θ

The longitudinal strain in that fibre is therefore:

ε = \frac{L_{y} - L}{L} = \frac{y}{R}

Stress Distribution

For a linearly elastic material:

σ = E ε = \frac{E y}{R}

Hence:

\frac{σ}{E} = \frac{y}{R}

The normal bending stress therefore varies linearly with distance from the neutral axis.

Position of the Neutral Axis

For pure bending there is no resultant axial force across the section. Therefore:

\int σ dA = 0

Substituting σ = Ey/R gives:

\frac{E}{R} \int y dA = 0

For a homogeneous section this condition requires the neutral axis to pass through the centroid of the cross-section.

Beam cross-section showing elemental area and distance from the neutral axis

Moment Equilibrium

The bending moment produced by the longitudinal stress distribution must balance the externally applied bending moment:

M = \int σ y dA

Using σ = Ey/R:

M = \frac{E}{R} \int y^{2} dA

The second moment of area about the neutral axis is:

I = \int y^{2} dA

Therefore:

M = \frac{E I}{R}

Simple Bending Equation

Combining the strain, stress and moment relationships gives the fundamental simple bending equation:

\frac{M}{I} = \frac{E}{R} = \frac{σ}{y}

Maximum Bending Stress

The magnitude of the bending stress increases linearly with distance from the neutral axis. The maximum stress therefore occurs at the extreme fibre, where |y| is greatest:

σ_{\max} = \frac{M y_{\max}}{I}

The elastic section modulus is:

Z = \frac{I}{y_{\max}}

and hence:

σ_{\max} = \frac{M}{Z}

For a cross-section symmetrical about the neutral axis, the extreme tensile and compressive bending stresses have equal magnitudes. For an unsymmetrical section, the distances to the extreme fibres differ and the two stresses must be calculated separately.

Deflection of Beams

The deflected shape of the neutral axis is known as the elastic curve.

Small element of a deflected beam showing change in slope, arc length and deflection

For small rotations:

θ \approx \frac{d v}{d x}

The exact curvature of the elastic curve is:

\frac{1}{R} = \frac{\frac{d^{2} v}{d x^{2}}}{{[1 + {(\frac{d v}{d x})}^{2}]}^{\frac{3}{2}}}

For small slopes this reduces to:

\frac{1}{R} \approx \frac{d^{2} v}{d x^{2}}

Moment-Curvature Relationship

From the simple bending equation:

\frac{1}{R} = \frac{M}{E I}

Therefore, using the sign convention adopted for the beam:

E I \frac{d^{2} v}{d x^{2}} = M (x)

Sign convention: The signs of bending moment, slope and deflection depend on the convention selected. A consistent convention must be used throughout a calculation.

Slope and Deflection by Integration

For constant E and I, integration of the moment-curvature equation gives the beam slope:

\frac{d v}{d x} = \int \frac{M (x)}{E I} d x + C_{1}

A second integration gives the deflection:

v = \int \int \frac{M (x)}{E I} d x d x + C_{1} x + C_{2}

The constants of integration are obtained from the beam boundary and continuity conditions. For example, at a fixed end both deflection and slope are zero:

v = 0 \frac{d v}{d x} = 0

Load, Shear, Moment and Deflection Relationships

Using a commonly adopted sign convention:

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

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

Combining these with the beam equation gives, for constant EI:

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

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

For more complex loading arrangements see Singularity Functions and Shear Force and Bending Moment Diagrams.

Example – Cantilever Beam with End Load

Consider a uniform cantilever beam of length L carrying a concentrated downward load F at its free end.

Cantilever beam of length L carrying a concentrated force F at the free end

Taking x from the fixed support and using the sign convention shown, the bending moment is:

M (x) = - F (L - x)

Therefore:

E I \frac{d^{2} v}{d x^{2}} = - F (L - x)

Integrating:

E I \frac{d v}{d x} = - F (L x - \frac{x^{2}}{2}) + C_{1}

At the fixed end, x = 0 and dv/dx = 0, hence C₁ = 0. Integrating again:

E I v = - F (\frac{L x^{2}}{2} - \frac{x^{3}}{6}) + C_{2}

At x = 0, v = 0, hence C₂ = 0. At the free end, x = L:

Free-end slope

θ_{L} = - \frac{F L^{2}}{2 E I}

Free-end deflection

v_{L} = - \frac{F L^{3}}{3 E I}

Example – Simply Supported Beam with Central Load

Consider a simply supported uniform beam of span L carrying a central concentrated load F. Each support reaction is F/2 and the deflection is symmetrical about the centreline.

Simply supported beam of span L carrying a central concentrated load F

By symmetry, the slope at the centre is zero. The maximum deflection occurs at the centre:

v_{\max} = - \frac{F L^{3}}{48 E I}

The magnitude of the slope at either support is:

| θ_{support} | = \frac{F L^{2}}{16 E I}

Moment-Area Method

The moment-area method provides a graphical and analytical means of determining changes in beam slope and deflection from the bending-moment diagram.

Theorem 1 – Change in Slope

The change in slope between two points A and B is equal to the area under the M/EI diagram between those points:

θ_{B} - θ_{A} = \int_{A}^{B} \frac{M}{E I} d x

Theorem 2 – Tangential Deviation

The displacement of point B relative to the tangent drawn to the elastic curve at A is equal to the first moment of the M/EI diagram between A and B about B.

Example 1 – Cantilever

For the cantilever shown with an end load F:

Cantilever with end load and triangular bending moment diagram used for the moment-area method

The bending moment at the fixed end is:

M_{A} = - F L

The signed area of the bending-moment diagram is:

A_{M} = - \frac{F L^{2}}{2}

The centroid of the triangular bending-moment diagram is a distance 2L/3 from the free end B:

x_{c} = \frac{2 L}{3}

The free-end slope is therefore:

θ_{B} = \frac{A_{M}}{E I} = - \frac{F L^{2}}{2 E I}

and the free-end deflection is:

v_{B} = \frac{A_{M} x_{c}}{E I} = - \frac{F L^{3}}{3 E I}

Example 2 – Simply Supported Beam

Consider the symmetrical simply supported beam shown below. The beam carries two 10 kN loads positioned symmetrically about the centreline.

Simply supported beam with two 10 kN loads and associated bending moment diagram for moment-area calculation

For the example:

E = 210 GPa

I = 834 cm^{4} = 8.34 \times 10^{- 6} m^{4}

E I = 1.7514 \times 10^{7} N m^{2}

The bending-moment diagram areas between A and C are:

A_{1} = \frac{10 \times 1.8 \times 1.8}{2} = 16.2 kN m^{2}

A_{2} = 10 \times 1.8 \times 2 = 36 kN m^{2}

The centroid distances of these areas from A are:

x_{1} = \frac{2}{3} \times 1.8 = 1.2 m

x_{2} = 1.8 + 1 = 2.8 m

The slope at A is equal in magnitude to the area of the M/EI diagram between A and C. Therefore:

θ_{A} = \frac{(16.2 + 36) \times 10^{3}}{1.7514 \times 10^{7}} = 0.00298 rad

or:

θ_{A} \approx 0.171 °

By symmetry the slope at C is zero. The centre deflection is obtained from the first moment of the M/EI diagram between A and C:

δ_{C} = \frac{(16.2 \times 1.2 + 36 \times 2.8) \times 10^{3}}{1.7514 \times 10^{7}} \approx 0.00687 m

δ_{C} \approx 6.87 mm

Composite Beams

A composite beam contains two or more materials that act together during bending. Examples include bonded metal sections and reinforced or layered members.

When the materials remain linearly elastic and are adequately connected so that there is no significant slip between them, a composite beam may be analysed using the transformed-section method.

Important: Plane sections remaining plane means that compatible materials at the same distance from the neutral axis have the same longitudinal strain. The strain is not constant through the beam depth; under elementary bending theory it varies linearly with distance from the neutral axis.

Composite beam consisting of steel and aluminium sections

Strain Compatibility

For perfectly bonded materials at the same distance from the neutral axis:

ε_{S} = ε_{A}

Since ε = σ/E:

\frac{σ_{S}}{E_{S}} = \frac{σ_{A}}{E_{A}}

Therefore:

\frac{σ_{S}}{σ_{A}} = \frac{E_{S}}{E_{A}}

Modular Ratio

If aluminium is selected as the reference material, the modular ratio for steel relative to aluminium is:

n_{SA} = \frac{E_{S}}{E_{A}}

For force equivalence, the transformed area of steel is:

A_{eq} = n_{SA} A_{S}

If the layer depth is retained, its equivalent width is similarly multiplied by the modular ratio.

Transformation of a composite steel and aluminium beam into an equivalent single-material section

Composite Beam Example

Consider a 30 mm wide composite section comprising 20 mm deep layers of aluminium, brass and steel. Transform the section to an equivalent aluminium section and determine its neutral axis and second moment of area.

Composite aluminium brass and steel beam transformed into an equivalent aluminium section

Using:

E_{S} = 210 GPa

E_{B} = 105 GPa

E_{A} = 70 GPa

the modular ratios relative to aluminium are:

n_{SA} = \frac{210}{70} = 3

n_{BA} = \frac{105}{70} = 1.5

The equivalent aluminium widths are therefore 30 mm for the aluminium, 45 mm for the brass and 90 mm for the steel.

Taking the bottom of the section as the datum, the neutral-axis position is:

\bar{y} = \frac{30 \times 20 \times 50 + 45 \times 20 \times 30 + 90 \times 20 \times 10}{30 \times 20 + 45 \times 20 + 90 \times 20} = 22.7 mm

Using the parallel-axis theorem, the second moment of area of the transformed section about its neutral axis is approximately:

I_{x - x} = [\frac{30 20^{3}}{12} + 30 \times 20 {(50 - 22.7)}^{2}] + [\frac{45 20^{3}}{12} + 45 \times 20 {(30 - 22.7)}^{2}] + [\frac{90 20^{3}}{12} + 90 \times 20 {(22.7 - 10)}^{2}] \approx 89.5 \times 10^{4} mm^{4}

Stress in the Composite Section

For an applied bending moment of 1000 N m, the equivalent aluminium-section stress at the top surface is:

σ_{t} = \frac{M y}{I} = \frac{1000 \times 1000 \times (60 - 22.7)}{89.5 \times 10^{4}} \approx 41.7 N/mm^{2}

The equivalent aluminium-section stress at the bottom surface is:

σ_{b} = \frac{1000 \times 1000 \times 22.7}{89.5 \times 10^{4}} \approx 25.4 N/mm^{2}

Because the bottom layer is steel and the transformed section is based on aluminium, the actual steel stress is obtained by multiplying the equivalent aluminium stress by the steel-to-aluminium modular ratio:

σ_{S} = n_{SA} σ_{b} = 3 \times 25.4 \approx 76.2 N/mm^{2}

Composite-beam limitation: The transformed-section method assumes linear-elastic material behaviour and adequate bonding or connection to maintain strain compatibility. It is not appropriate where substantial interfacial slip, yielding or other non-linear behaviour occurs.

Related Beam References

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Beam Theory and the Fundamentals of Bending, Stress and Deflection

Contents

Introduction

Superposition Principle

Nomenclature

Simple Bending Theory

Assumptions of Elementary Bending Theory

Strain Distribution

Stress Distribution

Position of the Neutral Axis

Moment Equilibrium

Simple Bending Equation

Maximum Bending Stress

Deflection of Beams

Moment-Curvature Relationship

Slope and Deflection by Integration

Load, Shear, Moment and Deflection Relationships

Example – Cantilever Beam with End Load

Example – Simply Supported Beam with Central Load

Moment-Area Method

Theorem 1 – Change in Slope

Theorem 2 – Tangential Deviation

Example 1 – Cantilever

Example 2 – Simply Supported Beam

Composite Beams

Strain Compatibility

Modular Ratio

Composite Beam Example

Stress in the Composite Section

Related Beam References