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Mechanics of Materials · Chapter 5 of 10 · Intermediate

Beams: Shear and Moment Diagrams

Before you can find a beam's stress, you need its bending moment, and where it peaks. Shear and moment diagrams map both along the whole beam and pinpoint the section that governs design.

01

Readiness check

This chapter rests on equilibrium and internal forces. Tick only what you can do closed-notes.

  • Find support reactions with ΣF = 0 and ΣM = 0.
  • Make a cut and find the internal force from a free body.
  • Integrate a constant and a linear function.
  • Recognise a uniformly distributed load as a resultant at its centroid.
  • Track sign conventions.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsReview reactions and internal forces in Statics.
3 or more weak itemsRevisit rigid-body equilibrium in Statics first.
02

The core idea

The shear force and bending moment vary along a beam; plotting them locates the maximum moment, which is the section that bending stress will govern.

dV/dx = −wdM/dx = Vmax M where V = 0

After finding the reactions, cut the beam at a section and the internal shear V and moment M follow from equilibrium of the piece. As you move along, V and M change according to the load: the slope of the shear diagram is minus the distributed load, and the slope of the moment diagram is the shear. So the moment is greatest where the shear crosses zero. The diagrams turn a whole beam into one critical number, Mmax, for the stress check of Chapter 4.

The skill works when: you find reactions first, then build V and M from the relationships, reading Mmax where V = 0.
The skill breaks down when: reactions are skipped, or sign conventions are mixed so the diagrams come out inconsistent.
The concept. Loads set the shear diagram; the shear sets the slope of the moment diagram. The moment peaks where the shear passes through zero, the section that governs bending.
03

The skills, taught in order

Diagrams are equilibrium applied along the beam. Five skills cover supports, reactions, the internal quantities, their relationships, and the diagrams.

5.1 Beams, supports, and loads

Common beams are simply supported (a pin and a roller), cantilever (one fixed end), and overhanging. Loads are point forces, uniformly or non-uniformly distributed loads, and applied moments. A distributed load of intensity w over a length acts, for equilibrium, as a resultant at its centroid.

5.2 Reactions

The first step is always the supports. Apply ΣF = 0 and ΣM = 0 to the whole beam to find the reactions, replacing each distributed load by its resultant. Everything downstream depends on getting these right.

5.3 Shear force and bending moment

Cut the beam at a distance x and consider one side. The shear V is the net transverse force on that piece; the bending moment M is the net moment. The standard sign convention takes sagging (concave up) moment as positive and the shear that rotates the segment clockwise as positive.

5.4 The load-shear-moment relationships

Across any span, dV/dx = −w and dM/dx = V. In words: the change in shear equals minus the area under the load, and the change in moment equals the area under the shear. These let you build the diagrams by adding areas instead of cutting at every point.

Beam and loadMax momentMax shear
Simply supported, central point PPL/4P/2
Simply supported, UDL wwL²/8wL/2
Cantilever, tip point PPLP
Cantilever, UDL wwL²/2wL

5.5 Drawing the diagrams and finding Mmax

Start at one end and step along, jumping at point loads, sloping under distributed loads. The maximum moment occurs where the shear is zero (or changes sign). That Mmax feeds straight into σmax = Mmax/S for the bending-stress check.

Engineering connection: every beam, joist, axle, and bridge girder is checked at its Mmax; the diagrams are the bridge from loads to the bending stress of Chapter 4.

04

Worked example 1: simply supported beam, off-center load

A simply supported beam spans 6 m and carries a 24 kN point load 2 m from the left support. Find the reactions and the maximum bending moment.

Figure 1. The shear steps down by 24 kN under the load, crossing zero there; the moment rises to its peak at that section, then falls back to zero at the right support.
  1. ProblemFind the reactions and Mmax for the beam in Figure 1.
  2. Given / findL = 6 m, P = 24 kN at a = 2 m from A. Find RA, RB, Mmax.
  3. AssumptionsRigid supports (pin and roller), weightless beam, static load.
  4. ModelReactions from moment equilibrium, then the moment at the load from the left-side free body.
  5. EquationsRA = P(L−a)/L, RB = Pa/L Mmax = RA·a
  6. SolveRA = 24(4)/6 = 16 kN, RB = 24(2)/6 = 8 kN (sum 24, checks). The shear is +16 from A to the load, then −8 to B, crossing zero at the load. Mmax = RA·a = 16 × 2 = 32 kN·m.
  7. CheckThe moment is zero at both simple supports and peaks under the load, where V changes sign, as expected. The larger reaction is on the side nearer the load.
  8. ConclusionReactions first, then the moment from a free body to one side. The peak under the point load is the section to carry into the bending-stress check.
Result. RA = 16 kN, RB = 8 kN; Mmax = 32 kN·m under the load.
05

Worked example 2: cantilever with a distributed load

A cantilever 3 m long carries a uniformly distributed load of 4 kN/m over its full length. Find the maximum shear and bending moment, and where they occur.

Figure 2. On a cantilever the shear builds linearly to the wall and the moment grows as a parabola, both greatest at the fixed end, the opposite of a simply supported beam.
  1. ProblemFind the maximum shear and moment for the cantilever in Figure 2.
  2. Given / findL = 3 m, w = 4 kN/m over the full span. Find Vmax, Mmax, and their locations.
  3. AssumptionsFixed support at the wall, uniform load, static.
  4. ModelThe total load wL acts at mid-span; shear and moment are largest at the fixed end.
  5. EquationsVmax = wL Mmax = wL²/2
  6. SolveVmax = 4 × 3 = 12 kN at the wall. Mmax = 4 × 3²/2 = 18 kN·m at the wall (hogging, tension on top).
  7. CheckThe moment is the shear-diagram area: a triangle of height 12 over length 3 gives ½ × 12 × 3 = 18 kN·m, matching. Both peak at the wall, where a cantilever always fails first.
  8. ConclusionA cantilever concentrates shear and moment at the support, unlike a simply supported beam that peaks at mid-span. Knowing where Mmax sits is half of beam design.
Result. Vmax = 12 kN and Mmax = 18 kN·m, both at the fixed wall.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Skipping reactionsDiagrams do not close to zero"Did I find the reactions first?"Solve equilibrium for reactions before cutting.
Max moment at the load alwaysWrong Mmax for distributed loads"Where does V cross zero?"Mmax is where shear is zero, not always under a point load.
Distributed load as a pointReactions or diagram shape wrong"Is w spread or concentrated?"Use the resultant for reactions, but slope the diagram across the span.
Sign convention driftV and M signs inconsistent"Sagging positive throughout?"Fix one convention (sagging M positive) and apply it everywhere.
07

Practice ladder

Level 1 · Direct skill

A simply supported beam, span 4 m, carries a central 10 kN load. Find the reactions and Mmax.

Show answer

R = P/2 = 5 kN each; Mmax = PL/4 = 10 × 4/4 = 10 kN·m at mid-span.

Level 2 · Mixed concept

A simply supported beam, span 6 m, carries a UDL of 5 kN/m. Find Mmax and where it occurs.

Show answer

R = wL/2 = 15 kN each; shear is zero at mid-span, so Mmax = wL²/8 = 5 × 36/8 = 22.5 kN·m at the centre.

Level 3 · Independent problem

Take the Worked Example 1 result (Mmax = 32 kN·m) and size a rectangular timber beam (allowable 12 MPa, h = 2b).

Show answer

Sreq = M/σ = 32×10⁶/12 = 2.67×10⁶ mm³. With S = 2b³/3, b³ = 4.0×10⁶, b = 159 mm, h = 318 mm; use about 160 × 320 mm. The diagram feeds straight into the Chapter 4 design.

Level 4 · Transfer to real engineering

Find a real beam with a known load (a shelf, a footbridge, a balcony). Sketch its shear and moment diagrams, locate Mmax, and estimate the bending stress.

What good work looks like

Reactions found, V and M diagrams sketched with correct shape, Mmax located where V = 0, and a bending-stress estimate from σ = M/S.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that my shear and moment diagrams close to zero at the ends."
"Give me five beams; I will say where Mmax occurs before computing."
"Draw the diagrams." Finding reactions and building V and M is the skill.
"What is the max moment?" Locating where V = 0 is the point.

Portfolio task

For one real beam, find the reactions, draw the shear and moment diagrams, identify Mmax, and carry it into a bending-stress check.

Must include: reactions, consistent V and M diagrams, Mmax at V = 0, and a σ = M/S result.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What is the first step in any beam problem?

Find the support reactions from equilibrium.

2. Write the load-shear-moment relationships.

dV/dx = −w and dM/dx = V.

3. Where is the bending moment maximum?

Where the shear is zero (or changes sign).

4. Give Mmax for a simply supported beam under a central point load and under a UDL.

PL/4 (point) and wL²/8 (UDL).

5. Where do shear and moment peak on a cantilever?

At the fixed end.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-draw both example diagrams from a blank page.
+3 daysOne simply supported and one cantilever diagram.
+7 daysCarry Mmax into bending stress, Chapter 4, and shear stress, Chapter 6.
+30 daysUse the diagrams again for deflection in Chapter 9.
10

Textbook mapping

ItemMapping
Primary sourceBeer, Johnston, DeWolf and Mazurek, Mechanics of Materials (6th ed), Chapter 5 (Analysis and Design of Beams for Bending)
Cross-referenceHibbeler, Ch. 6 · Gere and Goodno, Ch. 4
Core topics5.1 Supports and loads · 5.2 Reactions · 5.3 Shear and moment · 5.4 Load-shear-moment relations · 5.5 Diagrams and Mmax
Engineering connectionEvery beam, joist, axle, and girder is checked at its Mmax.
Read nextChapter 6: Shear Stress in Beams.