Beam Deflections

Learning Objectives

  • Understand the concept of the elastic curve and deflection in beams.
  • Apply the Double Integration Method to determine the equations of the elastic curve.
  • Utilize boundary conditions to solve for integration constants.
  • Calculate maximum deflections for standard loading conditions.
  • Apply Alternative Methods such as Superposition, Area-Moment, and the Conjugate Beam Method.
  • Formulate continuous moment equations using Macaulay's Method (Singularity Functions).
  • Apply the Virtual Work (Unit Load) method to calculate deflections.

Elastic Curve

The curved shape a beam assumes when it deforms away from its original longitudinal axis due to transverse forces.

Deflection and SLS Design

The perpendicular distance from the original neutral axis to the deformed neutral axis is the deflection. Calculating this vertical displacement is essential for Serviceability Limit State (SLS) design to ensure structures remain functional and aesthetically acceptable under everyday loads.

Double Integration Method

Differential Equation of the Elastic Curve

The deflection of a beam is related to the bending moment by the differential equation of the elastic curve:

Differential Equation of the Elastic Curve

Relates bending moment to deflection.

EId2ydx2=M(x)EI \frac{d^2y}{dx^2} = M(x)

Variables

SymbolDescriptionUnit
EEModulus of Elasticity (Material Stiffness).-
IIMoment of Inertia (Geometric Stiffness).-
M(x)M(x)Bending Moment as a function of xx.-
yyvertical deflection.-

Double Integration Method Procedure

  1. Formulate Moment Equation: Determine M(x)M(x) for the beam segments.
  2. First Integration (Slope): Integrate M(x)M(x) to get the slope equation EIθ(x)=∫M(x)dx+C1EI \theta(x) = \int M(x) dx + C_1.
  3. Second Integration (Deflection): Integrate again to get the deflection equation EIy(x)=∬M(x)dxdx+C1x+C2EI y(x) = \iint M(x) dx dx + C_1 x + C_2.
  4. Boundary Conditions: Use support conditions to solve for constants C1C_1 and C2C_2.

Boundary Conditions and Maximum Deflection

Boundary conditions are known physical constraints at the supports that allow us to solve for the integration constants.

  • Pinned or Roller Support: Deflection is zero (y=0y = 0). Slope (θ\theta) is generally non-zero.
  • Fixed Support: Deflection is zero (y=0y = 0) AND Slope is zero (θ=0\theta = 0 or y′=0y' = 0).
  • Free End: Deflection and slope are unconstrained. Moment and shear are zero only when no end couple or end force is applied; otherwise the boundary values follow those applied actions.

Locating Maximum Deflection

An interior deflection extremum occurs where the slope of the elastic curve is zero (y′=θ=0y' = \theta = 0). Also check interval endpoints: a cantilever's maximum deflection can occur at its free end with nonzero slope.

  • For a symmetrically loaded simply supported beam, this is always at the midspan (x=L/2x = L/2).
  • For an asymmetrically loaded beam (e.g., a point load off-center), the maximum deflection is not at the load, nor is it at the midspan. You must set the slope equation to zero (y′=0y' = 0) and solve for xx to find the location, then plug that xx into the deflection equation.

Alternative Methods

Method of Superposition

For linear elastic beams, the deflection due to multiple loads is the sum of the deflections caused by each load individually. This allows us to use standard formulas for simple cases (e.g., point load, uniform load) and add them up. This method is incredibly powerful because it turns complex loading problems into the addition of simple, standard cases, circumventing the need for complex calculus.

Common Formulas (Max Deflection)

  • Simply Supported (Uniform Load ww): δmax=5wL4384EI\delta_{max} = \frac{5wL^4}{384EI} (at center).
  • Simply Supported (Point Load PP at center): δmax=PL348EI\delta_{max} = \frac{PL^3}{48EI}.
  • Cantilever (Point Load PP at end): δmax=PL33EI\delta_{max} = \frac{PL^3}{3EI}.
  • Cantilever (Uniform Load ww): δmax=wL48EI\delta_{max} = \frac{wL^4}{8EI}.

Area-Moment Method

The Area-Moment method is a semi-graphical technique that relates the slope and deflection of a beam directly to the properties of the M/EIM/EI diagram.

  • First Area-Moment Theorem: The change in slope between any two points on the elastic curve equals the area of the M/EIM/EI diagram between those two points. θB/A=AreaM/EI\theta_{B/A} = \text{Area}_{M/EI}.
  • Second Area-Moment Theorem: The vertical deviation (vertical distance) of point B from the tangent line drawn at point A is equal to the "moment of area" of the M/EIM/EI diagram between A and B, taken about point B. tB/A=(AreaM/EI)⋅xˉBt_{B/A} = (\text{Area}_{M/EI}) \cdot \bar{x}_B.

Conjugate Beam Method

Conjugate Beam Method

The conjugate beam method is an ingenious technique that maps the problem of finding slopes and deflections (a geometric problem) entirely into a problem of finding shear and moments (a statics problem) on a fictitious "conjugate" beam.

The Fundamental Analogy:

  1. The "Load" (ww) on the conjugate beam is defined exactly as the M/EIM/EI diagram of the real beam.
  2. The internal "Shear" (VV) in the conjugate beam at any point equals the Slope (θ\theta) of the real beam at that point.
  3. The internal "Bending Moment" (MM) in the conjugate beam at any point equals the Deflection (yy) of the real beam at that point.

Because deflection and slope correspond directly to moment and shear, the boundary conditions of the real beam must be mathematically transformed into new boundary conditions for the conjugate beam.

Conjugate Support Mapping Rules

  1. Real: Simple End Support (Pinned/Roller) →\rightarrow Conjugate: Simple End Support (Pinned/Roller) (Reason: Real beam has y=0,θ≠0y=0, \theta \neq 0. Conjugate must have M=0,V≠0M=0, V \neq 0. A simple support satisfies this).
  2. Real: Fixed End Support →\rightarrow Conjugate: Free End (Reason: Real beam has y=0,θ=0y=0, \theta=0. Conjugate must have M=0,V=0M=0, V=0. Only a free end has zero moment and shear).
  3. Real: Free End →\rightarrow Conjugate: Fixed End Support (Reason: Real beam has y≠0,θ≠0y \neq 0, \theta \neq 0. Conjugate must have M≠0,V≠0M \neq 0, V \neq 0. Only a fixed support can resist both moment and shear).
  4. Real: Internal Simple Support (Roller) →\rightarrow Conjugate: Internal Hinge (Reason: Real beam has y=0y=0, but slope is continuous. Conjugate must have M=0M=0, but shear is continuous. An internal hinge transmits shear but not moment).
  5. Real: Internal Hinge →\rightarrow Conjugate: Internal Simple Support (Roller) (Reason: Real beam has y≠0y \neq 0 but the slope changes abruptly. Conjugate must have M≠0M \neq 0 but an abrupt change in shear).

Virtual Work (Unit Load Method)

Unit Load Method

The Virtual Work method, or Unit Load method, is an energy-based technique to calculate the deflection or slope at a specific point on a structure. It is highly versatile and can be applied to beams, trusses, and frames. To find the deflection δ\delta at a point, a fictitious "virtual" unit load (11) is placed at that point in the direction of the desired deflection.

The external virtual work done by the unit load is equated to the internal virtual strain energy stored in the body due to the real loads.

Virtual Work Equation for Beam Deflection

Relates real bending moments and virtual moments to calculate deflection at a specific point.

1⋅δ=∫0LmMEIdx1 \cdot \delta = \int_{0}^{L} \frac{m M}{EI} dx

Variables

SymbolDescriptionUnit
δ\deltaDesired real deflection at the point of interest.-
mmInternal virtual moment equation due to the applied unit virtual load.-
MMInternal real bending moment equation due to actual applied loads.-
EEModulus of Elasticity.-
IIMoment of Inertia.-

Macaulay's Method (Singularity Functions)

Overview of Macaulay's Method

When a beam has multiple different loads (point loads, partial uniform loads), writing a single moment equation M(x)M(x) across the entire length is impossible with standard algebra. Macaulay's Method uses singularity functions (indicated by angled brackets ⟨x−a⟩n\langle x-a \rangle^n) to write a single, continuous equation for the entire beam, making integration much simpler.

Rules of Singularity Functions

  • ⟨x−a⟩n=(x−a)n\langle x-a \rangle^n = (x-a)^n if x≥ax \ge a
  • ⟨x−a⟩n=0\langle x-a \rangle^n = 0 if x<ax < a
  • Integration rule: ∫⟨x−a⟩ndx=⟨x−a⟩n+1n+1\int \langle x-a \rangle^n dx = \frac{\langle x-a \rangle^{n+1}}{n+1} (for n≥0n \ge 0)

Implementation

When using Macaulay's method, you do not expand the bracket ⟨x−a⟩\langle x-a \rangle. You treat it as a single variable during integration. If a uniform load starts at aa and ends at bb, it is mathematically modeled as a continuous load starting at aa, with a counteracting negative continuous load starting at bb.

Interactive Tool: Beam Deflection

Interactive Simulation

Use the tool below to visualize the deflection shape for different beam types and loads. Note how the deflection (yy) curve corresponds to the double integration of the moment (MM) curve.

Beam Deflection and Superposition

Concept and model scope

Compare simply supported and cantilever elastic curves under a full-span uniform load plus a point load. The point force is at midspan for a simple beam and at the free end for a cantilever. Downward deflection and downward-positive slope are used here; this reverses the sign of the lesson's upward y coordinate. Constant EI, linear elasticity and small deflections are assumed. Axes use separate graph scales; this is not a physical-scale bent beam.

Controls

Support condition

Simple supports enforce zero end deflection. A left clamp enforces zero left deflection and slope.

Point load

Downward force at midspan or the cantilever tip, according to support selection.

Range: 0–50 kN. Step: 1 kN.

0 kN

Uniform load

Full-span downward uniform load. One kN/m equals one N/mm.

Range: 0–25 kN/m. Step: 1 kN/m.

8 kN/m

Span

Physical beam length. Uniform-load deflection varies with L to the fourth power.

Range: 2.0–10.0 m. Step: 0.1 m.

5.0 m

Elastic modulus

Young's modulus. Deflection varies inversely with E.

Range: 20–210 GPa. Step: 5 GPa.

200 GPa

Moment of inertia

Constant section second moment of area about the bending axis.

Range: 10–500 ×10⁶ mm⁴. Step: 5 ×10⁶ mm⁴.

100 ×10⁶ mm⁴

Inspection x / L

Station measured from the left support or clamp.

Range: 0.00–1.00. Step: 0.01.

0.50
0.000.01.632.53.265.0Deflection (mm), downward positive · independently scaled graphPosition x (m)
Maximum deflection
3.255 mm

At x = 2.50 m (unless unloaded).

Deflection at station
3.255 mm
Slope at station
0.000 mrad
Flexural rigidity
20000.00 kN·m²

Linear small-deflection model. No material strength or code serviceability check is implied.

Mechanics Visual Synthesis

Use these deterministic figures to connect the lesson equations to the corresponding free-body, section, transformation, or response geometry. Each visual addresses a separate mechanics question and is intentionally static so it can be compared directly with the interactive model above.

Moment, curvature, slope, and deflection

In Euler–Bernoulli bending, moment over EI defines curvature; integration gives slope and then transverse deflection subject to boundary conditions.

Moment, curvature, slope, and deflectionIn Euler–Bernoulli bending, moment over EI defines curvature; integration gives slope and then transverse deflection subject to boundary conditions.M/EIθ = y′y

Beam support boundary conditions

Fixed, pin, roller, and free boundaries impose different combinations of displacement, rotation, moment, and shear conditions that determine integration constants.

Beam support boundary conditionsFixed, pin, roller, and free boundaries impose different combinations of displacement, rotation, moment, and shear conditions that determine integration constants.fixed: y=0, θ=0simple: y=0free: M=V=0

Linear-elastic deflection superposition

When material and geometry remain linear elastic, the total deflected shape is the algebraic sum of the shapes produced by individual load cases.

Linear-elastic deflection superpositionWhen material and geometry remain linear elastic, the total deflected shape is the algebraic sum of the shapes produced by individual load cases.case 1+ case 2= total y(x)
Beam Deflection Method Selection
Beam Deflection Method SelectionDefine loading, EI, supports, desired displacement → Single simple load case with closed form?; Single simple load case with closed form? — Yes → Use standard elastic-curve solution; Single simple load case with closed form? — No → Many discontinuous loads?; Many discontinuous loads? — Yes → Use singularity / Macaulay integration; Many discontinuous loads? — No → Use moment-area, conjugate beam, or unit-load method; Use standard elastic-curve solution → Apply support boundary and continuity conditions; Use singularity / Macaulay integration → Apply support boundary and continuity conditions; Use moment-area, conjugate beam, or unit-load method → Apply support boundary and continuity conditions; Apply support boundary and continuity conditions → Check slope/deflection signs and extrema; Check slope/deflection signs and extrema → Report elastic deflection

Define loading, EI, supports, desired displacement → Single simple load case with closed form?; Single simple load case with closed form? — Yes → Use standard elastic-curve solution; Single simple load case with closed form? — No → Many discontinuous loads?; Many discontinuous loads? — Yes → Use singularity / Macaulay integration; Many discontinuous loads? — No → Use moment-area, conjugate beam, or unit-load method; Use standard elastic-curve solution → Apply support boundary and continuity conditions; Use singularity / Macaulay integration → Apply support boundary and continuity conditions; Use moment-area, conjugate beam, or unit-load method → Apply support boundary and continuity conditions; Apply support boundary and continuity conditions → Check slope/deflection signs and extrema; Check slope/deflection signs and extrema → Report elastic deflection

  • Define loading, EI, supports, desired displacement: terminator
  • Single simple load case with closed form?: decision
  • Use standard elastic-curve solution: process
  • Many discontinuous loads?: decision
  • Use singularity / Macaulay integration: process
  • Use moment-area, conjugate beam, or unit-load method: process
  • Apply support boundary and continuity conditions: process
  • Check slope/deflection signs and extrema: process
  • Report elastic deflection: terminator
Key Takeaways
  • Stiffness (EIEI): The product of Modulus of Elasticity (EE) and Moment of Inertia (II) is the Flexural Rigidity.
  • Double Integration: y′′=M/EIy'' = M/EI, y′=Slopey' = \text{Slope}, y=Deflectiony = \text{Deflection}.
  • Maximum Deflection Location: Compare interior zero-slope points with interval endpoints, including a cantilever's free end.
  • Superposition: Deflections are additive. Break complex loading into standard cases.
  • Area-Moment Method: Utilizes the area of the M/EIM/EI diagram and its moment to find slope (θ=Area\theta = \text{Area}) and deflection deviations (t=Area⋅xˉt = \text{Area} \cdot \bar{x}).
  • Conjugate Beam Method: Transforms slope into shear and deflection into bending moment on a fictitious beam loaded with the M/EIM/EI diagram. Requires strict boundary condition mapping (e.g., a Real Fixed Support becomes a Conjugate Free End).
  • Virtual Work (Unit Load Method): A powerful energy method applicable to trusses, beams, and frames.
  • Macaulay's Method: Uses singularity functions ⟨x−a⟩n\langle x-a \rangle^n to represent discontinuous loads with a single, continuous mathematical expression.