Shear and Moment in Beams

Learning Objectives

  • Determine support reactions before constructing internal-force diagrams.
  • Apply a consistent sign convention for internal shear and bending moment.
  • Relate distributed load, shear, and moment through differential and area relationships.
  • Construct and verify shear-force and bending-moment diagrams.
  • Use singularity functions to represent piecewise loading in a compact form.

Internal Shear Force

Internal shear force VV is the transverse resultant exposed by cutting a beam and enforcing equilibrium on either side of the cut.

Internal Bending Moment

Internal bending moment MM is the resultant couple exposed at a beam cut that balances the moments of external actions on the isolated segment.

Physical Support Contacts

The pin and roller hardware show the two boundary contacts that anchor a simple beam; exact reaction components and values come from equilibrium, not from the image.

Text-free technical view of a continuous steel beam resting on a pin support at one end and a roller support at the other.

Interpreting Support Contacts

This contextual view separates physical bearing hardware from idealized support symbols. Use the connected pin and roller to locate the boundary contacts, then solve their reaction components with equilibrium; the raster carries no arrows, values, or sign convention.

Load-Shear-Moment Differential Relations

Differential relationships among distributed load, shear force, and bending moment.

dVdx=−w(x),dMdx=V(x)\frac{dV}{dx}=-w(x), \qquad \frac{dM}{dx}=V(x)
Floor-to-Beam Load Path

The slab and secondary members meet the main beam across a broad contact zone, providing physical context for distributed loading; no load intensity or diagram is encoded.

Text-free cutaway of a concrete floor slab and secondary steel members resting on a main steel beam.

Interpreting Broad Floor Loading

Observe how the broad floor assembly and secondary members touch the beam across a zone rather than at one isolated point. This illustrates physical load transfer only; tributary width, load intensity, and shear or moment ordinates must be defined in the analysis.

Diagram Shape Rules

A concentrated force produces a jump in the shear diagram. A concentrated couple produces a jump in the moment diagram. Constant distributed load gives linear shear and quadratic moment. Where V=0V=0 and remains continuous through the point, the bending moment typically has a local extremum.

Local Hanger Attachment

A single connected hanger concentrates an attachment at one station along the beam; use the equations and diagram rules for the resulting shear discontinuity.

Text-free close-up of a steel I-beam with one bolted underside hanger supporting a short secondary member.

Interpreting a Local Attachment

The connected hanger is a physical source of a localized action on the main member, but the raster gives no force magnitude or idealized point location. Treat any exact jump in the shear diagram as a deterministic analysis result.

Beam Section-Cut Context

The supported beam and clean interior section plane show where internal shear and bending moment are imagined; use the deterministic simulator for exact signs, ordinates, and diagram shapes.

Text-free technical illustration of a simply supported steel beam with an exposed section plane between pin and roller supports.

Reading the Section-Cut Context

The beam remains an intact supported member, while the translucent plane marks a conceptual section at an interior station. Use that plane as the physical location of the isolated segment; the image contains no load values, force arrows, sign convention, or shear/moment diagram, so rely on the simulator and equations for quantitative construction and verification.

Interactive Exploration

Move the point load along the simply supported beam and vary its magnitude. Track the reaction redistribution and the corresponding shear and moment diagrams.

Shear and Moment Diagram Explorer

Concept and model scope

Move one point load along a simply supported 10 m beam and track reactions, the shear jump, and the bending-moment peak.

Controls

Point load

Downward concentrated load. The support reactions and internal-force diagram ordinates scale linearly with its magnitude.

Range: 5–60 kN. Step: 1 kN.

20 kN

Load position

Distance from the left support to the point load. Moving the load redistributes the support reactions and moves the maximum moment location.

Range: 0.5–9.5 m. Step: 0.1 m.

5.0 m
20 kNspan = 10 m · load at x = 5.0 mV+10.0-10.0M50.0 kN·m
Left reaction
10.00 kN

RA = P(L-a)/L

Right reaction
10.00 kN

RB = Pa/L

Maximum moment
50.00 kN·m

Occurs beneath the point load where the shear changes sign.

Analysis Workflow

Use the process below whenever a beam requires a full shear-force and bending-moment construction.

Shear and Moment Diagram Workflow
Shear and Moment Diagram WorkflowStart → Draw beam FBD and choose sign convention; Draw beam FBD and choose sign convention → Solve support reactions from equilibrium; Solve support reactions from equilibrium → Partition beam at load/discontinuity points; Partition beam at load/discontinuity points → Determine V(x) and M(x) or apply area rules; Determine V(x) and M(x) or apply area rules → Equilibrium, jumps, slopes, and end values consistent?; Equilibrium, jumps, slopes, and end values consistent? — Yes → Finalize SFD and BMD; Equilibrium, jumps, slopes, and end values consistent? — No → Correct reactions, signs, or segment equations; Correct reactions, signs, or segment equations → Solve support reactions from equilibrium

Start → Draw beam FBD and choose sign convention; Draw beam FBD and choose sign convention → Solve support reactions from equilibrium; Solve support reactions from equilibrium → Partition beam at load/discontinuity points; Partition beam at load/discontinuity points → Determine V(x) and M(x) or apply area rules; Determine V(x) and M(x) or apply area rules → Equilibrium, jumps, slopes, and end values consistent?; Equilibrium, jumps, slopes, and end values consistent? — Yes → Finalize SFD and BMD; Equilibrium, jumps, slopes, and end values consistent? — No → Correct reactions, signs, or segment equations; Correct reactions, signs, or segment equations → Solve support reactions from equilibrium

  • Start: terminator
  • Draw beam FBD and choose sign convention: process
  • Solve support reactions from equilibrium: process
  • Partition beam at load/discontinuity points: process
  • Determine V(x) and M(x) or apply area rules: process
  • Equilibrium, jumps, slopes, and end values consistent?: decision
  • Correct reactions, signs, or segment equations: process
  • Finalize SFD and BMD: terminator
Varied Span Zones

Different neighboring attachments along one continuous beam suggest why analysis may be partitioned at changing load conditions; the actual segments and equations remain deterministic.

Text-free architectural cutaway showing one continuous steel beam with a slab zone, open span, and localized hanger zone.

Reading Varied Span Zones

The same beam passes through a slab-supported region, a clear region, and a localized attachment region. These physical changes motivate segment boundaries, but they do not supply exact load functions, discontinuity locations, or solution values.

Singularity Function

A singularity or Macaulay function activates a load term only after the coordinate passes the load location.

Macaulay Bracket Definition

Piecewise definition used in singularity-function beam equations.

⟨x−a⟩n={0,x<a(x−a)n,x≥a\langle x-a\rangle^n= \begin{cases} 0, & x<a \\ (x-a)^n, & x\ge a \end{cases}

Using Singularity Functions

Singularity notation can replace multiple piecewise equations with one expression and integrates naturally from load to shear, moment, slope, and deflection. Constants of integration still require the correct support and continuity conditions.

Do Not Sketch Before Solving Reactions

A visually plausible diagram built from incorrect reactions will remain internally inconsistent. Always establish global equilibrium first and then verify the completed diagrams against the total applied loading.

Key Takeaways
  • Support reactions are the starting point for correct internal-force diagrams.
  • dV/dx=−wdV/dx=-w and dM/dx=VdM/dx=V control diagram slope and curvature.
  • Point forces jump shear; point couples jump moment.
  • Locations where continuous shear crosses zero are important candidates for moment extrema.
  • Singularity functions provide a compact alternative to many piecewise expressions.