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Beam Deflection

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Prerequisites

ArithmeticLogic
GeometryAlgebra
TrigonometryFunctions
VectorsLimits
Physics IDerivatives
StaticsGeneral ChemistryIntegrals
Mechanics of MaterialsMaterials Science

Beam deflection is the displacement of a beam's centroidal axis under load. A design can satisfy a strength limit and still fail in service if it bends too far, vibrates excessively, or loses alignment. This note develops the Euler-Bernoulli elastic-beam model and shows how to apply it with boundary conditions and superposition.

Assumptions and notation

The basic model assumes:

  • The beam is slender and has a prismatic or piecewise-prismatic cross-section.

  • Deflections and rotations are small.

  • The material is linear elastic, so $\sigma=E\varepsilon$.

  • Plane cross-sections remain plane and normal to the neutral axis.

  • Transverse shear deformation is negligible.

The last assumption is the main limitation of Euler-Bernoulli theory. Short, deep beams and beams made from low-shear-modulus materials may require Timoshenko beam theory.

Use the following symbols:

  • $x$: coordinate along the undeformed beam

  • $v(x)$: transverse deflection

  • $M(x)$: internal bending moment

  • $E$: Young's modulus

  • $I$: second moment of area about the bending axis

  • $w(x)$: distributed load, with a chosen sign convention

  • $P$: concentrated force

The flexural rigidity is $EI$. Larger $E$ or $I$ means a stiffer beam.

Elastic beam equation

For small slopes, beam curvature is approximately the second derivative of deflection:

$$ \kappa \approx \frac{d^2v}{dx^2}. $$

The moment-curvature relation is

$$ EI\frac{d^2v}{dx^2}=M(x), $$

where the sign may be negative under another convention. The important requirement is consistency between the moment diagram and the deflection direction.

For a constant $EI$, integrate twice:

$$ EI\frac{dv}{dx}=\int M(x)\,dx+C_1, $$
$$ EIv(x)=\int\!\!\int M(x)\,dx\,dx+C_1x+C_2. $$

The slope is $\theta(x)=dv/dx$. The integration constants are found from boundary conditions.

An equivalent fourth-order form follows from the load-moment relationships. For one common convention,

$$ \frac{dV}{dx}=-w(x),\qquad \frac{dM}{dx}=V(x), $$

so

$$ EI\frac{d^4v}{dx^4}=-w(x). $$

Use either the moment function or the load function, not both independently unless the signs and integration constants are checked carefully.

Boundary conditions

Boundary conditions describe what the supports permit or prevent.

Fixed end

At a built-in end, both displacement and rotation vanish:

$$ v=0,\qquad \frac{dv}{dx}=0. $$

Simple support

At a pin or roller, transverse displacement is zero and the support does not apply a couple:

$$ v=0,\qquad M=EI\frac{d^2v}{dx^2}=0. $$

Free end

At an unloaded free end, both internal moment and shear vanish:

$$ M=0,\qquad V=0. $$

If a force or couple is applied at the free end, the corresponding shear or moment condition is replaced by that applied end loading.

Continuity conditions

When a beam is divided into regions, deflection and slope are normally continuous at an ordinary point:

$$ v_1=v_2,\qquad \theta_1=\theta_2. $$

For a continuous beam with no concentrated couple, moment is also continuous. A concentrated force causes a jump in shear; a concentrated couple causes a jump in moment.

Standard loading cases

The following results apply to a prismatic beam with constant $E$ and $I$. Deflection directions are shown by magnitude; attach the sign required by your convention.

Cantilever with an end force

For a cantilever of length $L$ with a transverse force $P$ at the free end:

$$ \delta_{max}=\frac{PL^3}{3EI}, $$

at the free end, and the end rotation is

$$ \theta_{max}=\frac{PL^2}{2EI}. $$

Cantilever with a uniform load

For a constant load $w$ over the full length:

$$ \delta_{max}=\frac{wL^4}{8EI}, $$

at the free end, and

$$ \theta_{max}=\frac{wL^3}{6EI}. $$

Simply supported beam with a center force

For a centered point force $P$:

$$ \delta_{max}=\frac{PL^3}{48EI}, $$

at midspan.

Simply supported beam with a uniform load

For a uniform load $w$ over the full span:

$$ \delta_{max}=\frac{5wL^4}{384EI}, $$

at midspan.

The $L^3$ dependence for a point force and $L^4$ dependence for a distributed load make span length especially important in stiffness design.

Worked example: cantilever end load

A steel cantilever has length $L=1.20\ \mathrm{m}$, rectangular cross-section $b=30\ \mathrm{mm}$ and $h=60\ \mathrm{mm}$, and an end load $P=500\ \mathrm{N}$. Take $E=200\ \mathrm{GPa}$. Find the free-end deflection and rotation.

The section bends about the axis associated with $h$, so

$$ I=\frac{bh^3}{12} =\frac{(0.030)(0.060)^3}{12} =5.40\times10^{-7}\ \mathrm{m^4}. $$

Using the cantilever formulas,

$$ \delta=\frac{PL^3}{3EI} =\frac{(500)(1.20)^3}{3(200\times10^9)(5.40\times10^{-7})} =2.13\times10^{-3}\ \mathrm{m}. $$

Therefore, the free end deflects approximately $2.13\ \mathrm{mm}$ downward. The rotation is

$$ \theta=\frac{PL^2}{2EI} =3.33\times10^{-3}\ \mathrm{rad} \approx0.191^\circ. $$

The result is reasonable: doubling the length would increase the deflection by a factor of eight, while doubling the beam height would increase $I$ by a factor of eight and reduce the deflection by the same factor.

Superposition

Because the model is linear, responses from separate load cases can be added:

$$ v_{total}(x)=v_1(x)+v_2(x)+\cdots, $$

and likewise for slope, moment, and shear. Superposition is valid when:

  • Material behavior remains linear elastic.

  • Geometry changes are small.

  • Supports and contact conditions do not change.

For example, a cantilever carrying an end force and a uniform load has free-end deflection

$$ \delta=\frac{PL^3}{3EI}+\frac{wL^4}{8EI}, $$

with signs chosen according to the directions of the loads. Superposition does not mean that two separate load cases have to be solved with separate beams; it means their mathematically linear responses may be combined.

Choosing a solution method

Use direct integration when the load arrangement is simple and boundary conditions are clear. Use standard tables when the beam and loading match a known case. Use singularity functions or Macaulay brackets when loads change at several positions. Use the area-moment or virtual-work method when only one displacement or rotation is needed.

For a unit-load calculation, apply a unit force at the point and in the direction of the desired displacement. Then, for a linearly elastic beam,

$$ \delta=\int_0^L\frac{M(x)m(x)}{EI}\,dx, $$

where $M$ is from the real loading and $m$ is from the unit load. This is useful for beams with nonuniform loading or variable $EI$.

Common mistakes

  • Using $I=bh^3/12$ with the wrong dimension as the height. The dimension perpendicular to the bending axis is cubed.

  • Applying a simply supported formula to a cantilever or forgetting that a fixed end has zero slope as well as zero deflection.

  • Mixing millimetres and metres in $L$, $I$, or $E$.

  • Treating a distributed load's intensity $w$ as a total force without multiplying by its loaded length.

  • Assuming the maximum moment and maximum deflection occur at the same location in every beam.

  • Adding load effects when yielding, large deflection, changing contact, or other nonlinear behavior invalidates superposition.

  • Reporting a magnitude without stating the deflection direction or sign convention.

Practice problems

Problem 1: simply supported center load

A simply supported beam has $L=2.0\ \mathrm{m}$, $E=70\ \mathrm{GPa}$, $I=1.5\times10^{-6}\ \mathrm{m^4}$, and a centered load of $P=900\ \mathrm{N}$. Find the midspan deflection.

Solution.

$$ \delta_{max}=\frac{PL^3}{48EI} =\frac{(900)(2.0)^3}{48(70\times10^9)(1.5\times10^{-6})} =1.43\times10^{-3}\ \mathrm{m}. $$

The deflection is $1.43\ \mathrm{mm}$ downward.

Problem 2: cantilever uniform load

A cantilever has $L=0.8\ \mathrm{m}$, $E=100\ \mathrm{GPa}$, $I=4.0\times10^{-7}\ \mathrm{m^4}$, and a uniform load $w=300\ \mathrm{N/m}$. Find the free-end deflection.

Solution.

$$ \delta=\frac{wL^4}{8EI} =\frac{(300)(0.8)^4}{8(100\times10^9)(4.0\times10^{-7})} =3.84\times10^{-4}\ \mathrm{m}. $$

The free end deflects $0.384\ \mathrm{mm}$ in the load direction.

Problem 3: superposition and scaling

A cantilever has an end-load deflection of $1.2\ \mathrm{mm}$ and a uniform-load deflection of $0.8\ \mathrm{mm}$, both downward. What is the total elastic deflection? What happens to the total if $E$ is doubled while geometry and loads remain unchanged?

Solution.

By superposition,

$$ \delta_{total}=1.2+0.8=2.0\ \mathrm{mm}. $$

Every elastic deflection is proportional to $1/E$. Doubling $E$ halves the result, so the new deflection is $1.0\ \mathrm{mm}$.

Summary checklist

  1. Draw the beam, loads, supports, and coordinate system.

  2. Determine $M(x)$ or use a verified standard loading formula.

  3. Write $EIv''=M(x)$ with a consistent sign convention.

  4. Apply support, end, and continuity conditions.

  5. Add load cases only when linearity assumptions hold.

  6. Check units, direction, maximum location, strength, and serviceability.

Sources