Shear and Moment in Beams

Learning Objectives

  • Understand the definitions and concepts of Shear Force and Bending Moment in beams.
  • Identify different types of beams and their support conditions.
  • Apply the Method of Sections to determine internal shear and moment.
  • Utilize the Area Method to establish differential relationships and draw shear and moment diagrams.
  • Locate the Point of Contraflexure and understand its significance in design.
  • Analyze moving loads to determine the Absolute Maximum Bending Moment.
  • Apply Macaulay's Method (Singularity Functions) to represent load, shear, and moment algebraically.

Beams are structural members designed primarily to resist transverse (perpendicular) loads. These external loads, along with support reactions, create internal forces that vary along the longitudinal axis. To safely design a beam, engineers must plot these variations to locate the points of maximum internal stress.

Shear Force (VV)

The algebraic sum of the transverse forces acting on either side of a section of a beam. It represents the internal force resisting the tendency of one part of the beam to slide past the other.

Bending Moment (MM)

The algebraic sum of the moments of all forces acting on either side of a section of a beam. It represents the internal rotational effect resisting the tendency of the beam to bend or flex.

Types of Beams and Supports

Beam Types & Support Conditions

  • Simply Supported: Pinned at one end, Roller at the other. Statically Determinate.
  • Cantilever: Fixed at one end, Free at the other. Statically Determinate.
  • Overhanging: Supported at two points with one or both ends extending beyond the supports. Statically Determinate.
  • Continuous: Supported at more than two points. Statically Indeterminate.
  • Propped Cantilever: Fixed at one end, Roller at the other. Statically Indeterminate.

Method of Sections (Equation Method)

The most fundamental way to determine internal shear and moment at any point in a beam is the Method of Sections. This involves cutting the beam at a specific point and applying equilibrium equations to the isolated segment.

Sign Convention for Method of Sections

Standard Sign Convention:

  • Positive Shear (VV): Tends to rotate the cut segment clockwise. (e.g., pointing DOWN on the right face of a left-hand segment).
  • Positive Moment (MM): Tends to bend the segment concave upward, creating a "smiling" shape. This puts the top fibers in compression and bottom fibers in tension.

Calculation Steps

  1. Calculate Reactions: Draw a Free Body Diagram (FBD) of the entire beam and use static equilibrium (∑F=0,∑M=0\sum F = 0, \sum M = 0) to find support reactions.
  2. Section the Beam: Make an imaginary cut at a distance xx from the origin. Make a new cut whenever the loading condition changes (e.g., after a point load or at the start of a uniform load).
  3. Draw FBD of Segment: Draw the isolated segment (usually the left side for convenience). Place the unknown internal shear V(x)V(x) and moment M(x)M(x) at the cut face acting in their positive directions.
  4. Apply Equilibrium:
    • ∑Fy=0\sum F_y = 0 yields the equation for V(x)V(x).
    • ∑Mcut=0\sum M_{\text{cut}} = 0 yields the equation for M(x)M(x).

Load, Shear, and Moment Relationships (Area Method)

Writing equations for every segment is tedious. Because load, shear, and moment are mathematically related by differential calculus, we can construct the diagrams graphically using the Area Method.

Differential Relationships

The relationship between Load (ww), Shear (VV), and Moment (MM) is defined by differential calculus:

1. Slope of Shear Diagram for Downward-Positive Loading

Slope of Shear Diagram

For this lesson, downward distributed load intensity is positive, so it reduces the signed shear from left to right.

dVdx=−w(x)\frac{dV}{dx} = -w(x)

Variables

SymbolDescriptionUnit
VVshear force-
wwdownward-positive distributed load intensity under this sign convention-
xxdistance along the beam-

Sign convention for distributed load

Here w(x)w(x) is positive downward. If a different load sign convention is used, change the sign in the differential and area relations consistently.

Moment and Shear Differential Relationship

2. Slope of Moment Diagram = Shear Force

This means the shear value at any point is the slope of the moment curve at that point.

Slope of Moment Diagram

Relates bending moment to shear force.

dMdx=V(x)\frac{dM}{dx} = V(x)

Variables

SymbolDescriptionUnit
MMbending moment-
VVshear force-
xxdistance along the beam-

Area Method Rules

The Area Method provides a graphical and rapid way to construct shear and moment diagrams by calculating the areas under the respective curves:

With downward-positive ww, the signed shear change is the negative area under the load diagram. The signed moment change equals the area under the shear diagram.

Curve Degrees The relationship dictates that each successive diagram increases in polynomial degree:

  • If load w(x)w(x) is degree nn.
  • Then Shear V(x)V(x) is degree n+1n+1.
  • Then Moment M(x)M(x) is degree n+2n+2.

For example: A Uniform Load (degree 0, flat line) →\rightarrow Linear Shear (degree 1, sloped line) →\rightarrow Parabolic Moment (degree 2, curve).

Area relation from load to shear

For downward-positive distributed load, the shear change equals the negative signed area under the load diagram.

V2−V1=−∫x1x2w(x) dxV_2 - V_1 = -\int_{x_1}^{x_2} w(x) \, dx

Variables

SymbolDescriptionUnit
V1,V2V_1, V_2shear values at the interval endpoints-
w(x)w(x)downward-positive distributed load intensity-
x1,x2x_1, x_2interval endpoints along the beam-

Area relation from shear to moment

The moment change equals the signed area under the shear diagram over the interval.

M2−M1=∫x1x2V(x) dxM_2 - M_1 = \int_{x_1}^{x_2} V(x) \, dx

Variables

SymbolDescriptionUnit
M1,M2M_1, M_2bending moments at the interval endpoints-
V(x)V(x)signed shear force along the interval-
x1,x2x_1, x_2interval endpoints along the beam-
Shear and Moment Diagram Construction

Build signed shear and bending-moment diagrams from a verified beam model, then check extrema and equilibrium.

Use one consistent sign convention and split the beam wherever a concentrated force, applied couple, or distributed-load expression changes. The chart summarizes statically determinate construction; resolve indeterminate actions with the appropriate compatibility method before drawing final diagrams. It does not select a code load combination or design resistance.

Shear and Moment Diagram ConstructionBuild signed shear and bending-moment diagrams from a verified beam model, then check extrema and equilibrium.. Use one consistent sign convention and split the beam wherever a concentrated force, applied couple, or distributed-load expression changes. The chart summarizes statically determinate construction; resolve indeterminate actions with the appropriate compatibility method before drawing final diagrams. It does not select a code load combination or design resistance.. Define supports and applied loads → Solve support reactions from equilibrium; Solve support reactions from equilibrium → Mark load, couple, and distribution-change stations; Mark load, couple, and distribution-change stations → Construct V using jumps and dV/dx = −w; Construct V using jumps and dV/dx = −w → Construct M using jumps and dM/dx = V; Construct M using jumps and dM/dx = V → Locate V=0 and moment extrema / contraflexure; Locate V=0 and moment extrema / contraflexure → Verify end conditions and global equilibrium; Verify end conditions and global equilibrium → Use governing V and M

Define supports and applied loads → Solve support reactions from equilibrium; Solve support reactions from equilibrium → Mark load, couple, and distribution-change stations; Mark load, couple, and distribution-change stations → Construct V using jumps and dV/dx = −w; Construct V using jumps and dV/dx = −w → Construct M using jumps and dM/dx = V; Construct M using jumps and dM/dx = V → Locate V=0 and moment extrema / contraflexure; Locate V=0 and moment extrema / contraflexure → Verify end conditions and global equilibrium; Verify end conditions and global equilibrium → Use governing V and M

  • Define supports and applied loads: terminator
  • Solve support reactions from equilibrium: process
  • Mark load, couple, and distribution-change stations: process
  • Construct V using jumps and dV/dx = −w: process
  • Construct M using jumps and dM/dx = V: process
  • Locate V=0 and moment extrema / contraflexure: process
  • Verify end conditions and global equilibrium: process
  • Use governing V and M: terminator

Locating the Maximum Bending Moment

Because the shear force is the mathematical derivative of the bending moment (dMdx=V(x)\frac{dM}{dx} = V(x)), the bending moment curve will reach a local maximum or minimum value exactly at the points where the shear diagram crosses the zero axis (V=0V = 0).

In structural design, finding these "points of zero shear" is the most critical step, as the beam must be sized to resist the maximum moment occurring at these locations.

Point of Contraflexure

The location on a beam where the bending moment is exactly zero (M=0M = 0). Also known as an Inflection Point.

Significance of Point of Contraflexure

At this point, the beam's curvature changes from concave upward (positive moment, sagging) to concave downward (negative moment, hogging).

Significance in Design:

  • At this point, flexural stress is zero.
  • In reinforced concrete, this point dictates where top longitudinal reinforcement (for negative moment) can be terminated or shifted to the bottom (for positive moment).
  • It is a common location to place splices or hinges in continuous structures, as they will only need to transmit shear forces, not bending moments.

Moving Loads and Absolute Maximum Bending Moment

Moving Loads

While static loads are fixed in position, moving loads (like trucks on a bridge or crane hoists on a runway beam) change position over time. To design structures carrying moving loads, engineers must find the absolute maximum internal forces (shear and moment) that occur as the load system traverses the entire span.

Influence lines show how a response at a specified section changes as a unit load moves. On a simply supported span, the absolute maximum shear occurs near a support; find its value and sign from the appropriate shear influence line for the actual moving load system.

Absolute Maximum Bending Moment

For one downward point load on a simply supported span, the maximum positive moment occurs when the load is at midspan. For a train of downward concentrated loads on a simply supported span, the absolute maximum positive moment occurs beneath one of the loads. Identify which load controls by comparing plausible candidates; the heaviest or nearest load is not guaranteed to control.

For each candidate load PkP_k, the trial placement puts that load and the resultant RR of the active on-span loads equidistant on opposite sides of midspan. With xkx_k and xRx_R measured from the left support, this condition is xk+xR=Lx_k + x_R = L. It locates candidate arrangements; the moment under every feasible candidate still must be calculated and compared.

Candidate position for absolute maximum moment

For one candidate wheel on a simply supported span, the candidate wheel and the active load resultant are symmetric about the span midpoint.

xk+xR=Lx_k + x_R = L

Variables

SymbolDescriptionUnit
xkx_kcandidate load position measured from the left support-
xRx_Rresultant position of the concentrated loads currently on the span, measured from the left support-
LLdistance between the simple supports-

Moving-Load Absolute-Moment Search

  1. Verify the support model, span LL, point-load magnitudes and spacings, travel range, and the stated static load basis.
  2. For each plausible candidate load PkP_k, form the resultant and position of the loads currently on the span, then solve xk+xR=Lx_k + x_R = L for that trial placement.
  3. Retain trial placements that fit the span and travel range; calculate reactions and the moment directly beneath each feasible candidate load.
  4. Evaluate partial-train, load-entry, and load-exit positions with the actual influence-line ordinates, including changes in which loads lie on the span.
  5. Compare all feasible candidate and boundary moments, verify equilibrium and signed conventions, and report the maximum value, controlling load, and its position.
  6. If the support system differs from a simply supported single span, use its actual influence-line envelope. If inertia or transient response matters, use an appropriate dynamic analysis.
Absolute Maximum Moment under a Moving Point-Load Train

Find the maximum positive bending moment for a static train of downward concentrated loads on a simply supported single span. For each feasible candidate under a load, the load and the load-train resultant are equidistant from midspan; compare candidates instead of assuming the heaviest load controls.

The bisection condition is a candidate-position rule for this structural model, not a load-rating criterion. Verify all span, support, load, and spacing data. Include only loads on the span when forming a partial-span resultant and include load-entry/exit boundary cases. For another support system or moving distributed loads, use the actual static influence-line envelope; dynamic response requires a dynamic model. The representative figure and any sample values are illustrative, not measured loads or an AASHTO load specification.

Absolute Maximum Moment under a Moving Point-Load TrainFind the maximum positive bending moment for a static train of downward concentrated loads on a simply supported single span. For each feasible candidate under a load, the load and the load-train resultant are equidistant from midspan; compare candidates instead of assuming the heaviest load controls.. The bisection condition is a candidate-position rule for this structural model, not a load-rating criterion. Verify all span, support, load, and spacing data. Include only loads on the span when forming a partial-span resultant and include load-entry/exit boundary cases. For another support system or moving distributed loads, use the actual static influence-line envelope; dynamic response requires a dynamic model. The representative figure and any sample values are illustrative, not measured loads or an AASHTO load specification.. Define span, supports and moving load train → Are span, support, load and spacing data verified?; Are span, support, load and spacing data verified? — No → Hold: resolve missing or conflicting model data; Are span, support, load and spacing data verified? — Yes → Static point-load train on a simply supported single span?; Static point-load train on a simply supported single span? — No → Use the actual influence line; use dynamic analysis for transient response; Use the actual influence line; use dynamic analysis for transient response → Continue with the model-specific response envelope; Static point-load train on a simply supported single span? — Yes → For each active load set, find resultant R and location x_R; For each active load set, find resultant R and location x_R → For each plausible P_k, solve x_k + x_R = L; retain feasible trials; For each plausible P_k, solve x_k + x_R = L; retain feasible trials → Compute reactions and moment beneath every feasible P_k; Compute reactions and moment beneath every feasible P_k → Include partial-span and load-entry/exit positions by influence line; Include partial-span and load-entry/exit positions by influence line → Compare every feasible candidate and boundary moment; Compare every feasible candidate and boundary moment → Check equilibrium, signed convention and candidate coverage; Check equilibrium, signed convention and candidate coverage → Report maximum M, controlling load and position

Define span, supports and moving load train → Are span, support, load and spacing data verified?; Are span, support, load and spacing data verified? — No → Hold: resolve missing or conflicting model data; Are span, support, load and spacing data verified? — Yes → Static point-load train on a simply supported single span?; Static point-load train on a simply supported single span? — No → Use the actual influence line; use dynamic analysis for transient response; Use the actual influence line; use dynamic analysis for transient response → Continue with the model-specific response envelope; Static point-load train on a simply supported single span? — Yes → For each active load set, find resultant R and location x_R; For each active load set, find resultant R and location x_R → For each plausible P_k, solve x_k + x_R = L; retain feasible trials; For each plausible P_k, solve x_k + x_R = L; retain feasible trials → Compute reactions and moment beneath every feasible P_k; Compute reactions and moment beneath every feasible P_k → Include partial-span and load-entry/exit positions by influence line; Include partial-span and load-entry/exit positions by influence line → Compare every feasible candidate and boundary moment; Compare every feasible candidate and boundary moment → Check equilibrium, signed convention and candidate coverage; Check equilibrium, signed convention and candidate coverage → Report maximum M, controlling load and position

  • Define span, supports and moving load train: terminator
  • Are span, support, load and spacing data verified?: decision
  • Hold: resolve missing or conflicting model data: terminator
  • Static point-load train on a simply supported single span?: decision
  • Use the actual influence line; use dynamic analysis for transient response: process
  • Continue with the model-specific response envelope: terminator
  • For each active load set, find resultant R and location x_R: process
  • For each plausible P_k, solve x_k + x_R = L; retain feasible trials: process
  • Compute reactions and moment beneath every feasible P_k: process
  • Include partial-span and load-entry/exit positions by influence line: process
  • Compare every feasible candidate and boundary moment: process
  • Check equilibrium, signed convention and candidate coverage: process
  • Report maximum M, controlling load and position: terminator

Macaulay's Method (Singularity Functions)

Singularity functions (also called Macaulay functions) provide a single mathematical expression for the load, shear, and bending moment over the entire length of a beam, regardless of discontinuous loads.

Macaulay's Method (Singularity Functions)

A singularity function is defined by pointed brackets. It effectively "turns on" only when x>ax > a.

Definition:

  • If x<ax < a: ⟨x−a⟩n=0\langle x - a \rangle^n = 0
  • If x≥ax \ge a: ⟨x−a⟩n=(x−a)n\langle x - a \rangle^n = (x - a)^n

Macaulay Integration

Integration of a standard singularity function.

∫⟨x−a⟩ndx=⟨x−a⟩n+1n+1\int \langle x - a \rangle^n dx = \frac{\langle x - a \rangle^{n+1}}{n+1}

Variables

SymbolDescriptionUnit
xxposition along beam-
aalocation of load application-
nnpower indicating type of load-

For concentrated moments (n=−2n = -2) or forces (n=−1n = -1), integration increments the power by 1 but without dividing by the new exponent:

Macaulay Integration (Moment)

Integration for a concentrated moment.

∫⟨x−a⟩−2dx=⟨x−a⟩−1\int \langle x - a \rangle^{-2} dx = \langle x - a \rangle^{-1}

Variables

SymbolDescriptionUnit
xxposition along beam-
aalocation of concentrated moment application-

Macaulay Integration (Force)

Integration for a concentrated force.

∫⟨x−a⟩−1dx=⟨x−a⟩0\int \langle x - a \rangle^{-1} dx = \langle x - a \rangle^0

Variables

SymbolDescriptionUnit
xxposition along beam-
aalocation of concentrated force application-

Interactive Simulation

Experiment with different beam supports and loads to see how the Shear and Moment diagrams change in real-time.

Shear and Moment Diagram Explorer

Concept and model scope

Superpose a downward point load and full-span uniform load on a simply supported beam. Positive shear is upward on the left cut face; positive moment is sagging. A point load produces a shear jump; distributed loading gives dV/dx = −w and dM/dx = V. Diagram axes auto-fit independently and show engineering units. End-point loads pass directly into the support.

Controls

Point load

Downward concentrated force. Zero removes the point-load arrow and shear jump.

Range: 0–60 kN. Step: 1 kN.

20 kN

Uniform load

Constant downward load over the complete span; creates linear shear and parabolic moment between concentrated loads.

Range: 0.0–15.0 kN/m. Step: 0.5 kN/m.

0.0 kN/m

Span

Distance between the pin and roller. Load location is held as a fraction of this length.

Range: 2.0–12.0 m. Step: 0.5 m.

10.0 m

Load position / span

0 is the left support, 1 is the right support. The corresponding position in metres appears in results.

Range: 0.00–1.00. Step: 0.01.

0.50
P = 20 kN · w = 0 kN/mL = 10 m · a = 5.00 mV (kN)+10.00-10.00MPeak 50.00 kN·m at 5.00 m0x = 10 m
Left reaction
10.00 kN
Right reaction
10.00 kN
Peak moment
50.00 kN·m
Peak location
5.000 m
Shear just left of load
10.00 kN
Shear just right of load
-10.00 kN

Point-load jump: −20 kN. At a support, combine it with the support reaction to obtain the net boundary jump.

Point and distributed load comparison

Set either load to zero to isolate its contribution, then combine them. A point load creates a shear jump; a full-span uniform load creates linear shear and parabolic moment. The maximum moment may occur away from the point load when distributed loading is present.

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.

Beam shear and bending sign convention

A cut section identifies the positive internal shear and sagging bending directions used consistently by the lesson diagrams and equations.

Beam shear and bending sign conventionA cut section identifies the positive internal shear and sagging bending directions used consistently by the lesson diagrams and equations.+V+M saggingsection cut

Load, shear, and moment differential chain

Distributed load controls the slope of the shear diagram, and shear controls the slope of the bending-moment diagram.

Load, shear, and moment differential chainDistributed load controls the slope of the shear diagram, and shear controls the slope of the bending-moment diagram.w = −dV/dxV = dM/dxarea relations

Concentrated actions create diagram discontinuities

A point force creates a finite jump in shear while an applied couple creates a finite jump in bending moment at its station.

Concentrated actions create diagram discontinuitiesA point force creates a finite jump in shear while an applied couple creates a finite jump in bending moment at its station.point P → ΔVcouple M₀ → ΔMpiecewise diagrams

Bending-moment extrema occur where shear crosses zero

Because V = dM/dx, a smooth local maximum or minimum in the bending-moment diagram occurs at a station where the shear diagram passes through zero.

Bending-moment extrema occur where shear crosses zeroBecause V = dM/dx, a smooth local maximum or minimum in the bending-moment diagram occurs at a station where the shear diagram passes through zero.V > 0 → M risesV = 0M extremum
Key Takeaways
  • Shear (VV) is the algebraic sum of the vertical forces to one side of the section.
  • Moment (MM) is the algebraic sum of the moments of all forces to one side of the section.
  • Sign Convention: Shear is positive when it causes clockwise rotation; Moment is positive when it causes sagging.
  • Differential Relationships: For downward-positive distributed load, dV/dx=−wdV/dx=-w; shear is the derivative of moment.
  • Area Method: The shear change is the negative area under downward-positive load, and the moment change is the area under shear.
  • Zero Shear: The point where the shear diagram crosses zero corresponds to a local maximum or minimum bending moment.
  • Point of Contraflexure: The point where the bending moment is zero (M=0M=0) and curvature reverses, which is critical for reinforcement detailing.
  • Moving Loads: For a simply supported span, test plausible candidate point loads using xk+xR=Lx_k+x_R=L, then compare the resulting moments and boundary positions using the actual load arrangement. Other support conditions need their own influence-line envelope.
  • Singularity (Macaulay's) Functions: Allow a single equation to define shear or moment across an entire beam, handling discontinuous loads systematically by "turning on" terms only when x≥ax \ge a.

References