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Statics · Module 7 of 11 · Intermediate

Internal Forces

Cut the structure to see what it carries inside.

Readiness check

From Module 5. Tick only what you can do closed-notes.

Find beam support reactions in under five minutes, with a verification.
Replace a distributed load with a resultant at its centroid.
Take moments about an arbitrary point, not just supports.
Keep a consistent sign convention through a multi-step solution.
Sketch a simple function (linear, parabolic) from its slope behavior.

0 or 1 weak itemsContinue with this module.

2 weak itemsRedo Levels 1 to 3 of the Module 5 ladder; internal forces start where reactions end.

3 or more weak itemsStep back to Module 5; every section cut is a mini reaction problem.

The core idea

An imaginary cut exposes three internal resultants: N, V, and M.

N (axial)V (shear)M (bending)

Whatever the cut face carries, the removed piece must have been providing. Solve either side of the cut; pick the simpler one. Convention: positive V rotates the segment clockwise; positive M sags the beam (smile).

The model works when: external reactions are already known and the member is rigid: every cut is just another equilibrium problem.

The model breaks down when: you need stresses or deflections; N, V, M are the inputs to Mechanics of Materials, not the final answer.

The concept. Slice the member anywhere: the exposed face must carry exactly the axial force, shear, and moment the removed piece was providing.

The method

1Look

Reactions first: solve the whole beam as a rigid body.

2Simplify

Choose the cut location the question asks about.

3Draw

Keep one side; expose N, V, M on the cut face.

4Solve

Equilibrium of the kept segment; verify from the other side.

Worked example: shear and moment at a section

A 6 m simply supported beam (pin at A, roller at B) carries 12 kN down at midspan. Find the internal shear and bending moment at x = 2 m from A, and the maximum bending moment.

Figure 1. Problem setup. Reactions by symmetry: A_y = B_y = 6 kN.

Figure 2. Left-segment free-body diagram. Solution: V = +6 kN, M = +12 kN·m, N = 0.

applied loadsupport reactioninternal N, V, Mdimensions

ProblemFind V and M at the cut marked in Figure 1, then the maximum bending moment.
Given / findL = 6 m, P = 12 kN at x = 3 m. Find V and M at x = 2 m, and M_max.
AssumptionsBeam weight negligible; 2D; standard sign convention (positive M sags).
ModelReactions by symmetry: A_y = B_y = 6 kN. Cut at x = 2 m and keep the left segment (Figure 2): it carries only one force, which is simpler.
EquationsΣF_y = 0: 6 − V = 0 ΣM_cut = 0: M − 6(2) = 0
SolveV = +6 kN, M = +12 kN·m at x = 2 m. The maximum moment occurs under the load: M_max = 6 × 3 = 18 kN·m at midspan.
CheckRight-segment cross-check at x = 2 m: M = 6(4) − 12(1) = 12 kN·m. The slope relation dM/dx = V = 6 kN matches the 6 kN·m rise per metre.
ConclusionThe beam's critical station is midspan with 18 kN·m. That number, not the loads themselves, is what sizes the beam cross-section in Mechanics of Materials.

Result. At x = 2 m: V = +6 kN, M = +12 kN·m. Maximum: M = 18 kN·m at midspan.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Never actually cutting the beam	External reactions found, then nothing to write	"Where is my section, and which side am I keeping?"	Draw the kept segment as a brand-new FBD with V, M (and N) on the cut face.
Forgetting the reaction in the kept segment	V and M wrong by exactly the reaction's contribution	"Does my segment FBD include every force on that piece?"	Supports inside the kept piece contribute. Recount the arrows.
Sign convention improvised per problem	Correct magnitudes, random signs; diagrams flip	"What does positive M physically look like?"	Fix it once: positive V is a clockwise pair, positive M is sagging. Use it forever.
Misusing the slope relations	Moment diagram with kinks in the wrong places	"What are dV/dx and dM/dx equal to?"	dV/dx = −w, dM/dx = V. Point loads jump V; couples jump M; M is extreme where V = 0.

Practice ladder

Level 1 · Direct skill

For the worked-example beam, find V and M at x = 4 m (right of the load).

Show answer

The left segment now includes the load: V = 6 − 12 = −6 kN; M = 6(4) − 12(1) = +12 kN·m. Note the shear sign flip across the load.

Level 2 · Mixed concept

Sketch the full V and M diagrams for the worked-example beam. Where is V zero, and what happens there?

Show answer

V: constant +6 kN from A to midspan, jumps to −6 kN, constant to B. M: rises linearly 0 to 18 kN·m at midspan, falls linearly back to 0. V crosses zero at midspan, exactly where M peaks.

Level 3 · Independent problem

A 4 m cantilever (fixed at the wall) carries a uniform 3 kN/m load. Find V(x) and M(x) from the free end, and the values at the wall.

Show answer

Measuring x from the free end: V(x) = 3x kN, M(x) = −3x²/2 kN·m (hogging). At the wall: V = 12 kN, M = −24 kN·m. Check: ½wL² = ½ × 3 × 16 = 24.

Level 4 · Transfer to real engineering

Model a person (70 kg) standing at the tip of a 1.8 m diving board (cantilever). Draw V and M diagrams, find the wall moment, and explain in one paragraph why diving boards are thickest at the clamp and thin at the tip.

What good work looks like

W ≈ 687 N; M_wall ≈ 687 × 1.8 ≈ 1236 N·m; M tapers linearly to zero at the tip, so the required cross-section tapers too. Material follows the moment diagram: that is structural shaping in one sentence.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"I will describe my V and M diagrams in words (shapes, peaks, zeros). Tell me which feature is inconsistent with dM/dx = V, but not the fix."

"Give me a beam with one point load and one UDL; I will predict diagram shapes before computing. Grade the prediction first."

"Draw the shear and moment diagrams for this beam." Diagram fluency is the core deliverable of this module.

"What's the max moment?" Locating the critical section yourself is the design skill.

Portfolio task

Extend your Module 5 reaction calculator to output V(x) and M(x) arrays and plot both diagrams. Validate against the worked example (V = ±6 kN, M_max = 18 kN·m) and your Level 3 cantilever. Screenshot the plots into your portfolio with the validation table.

Must include: both validation cases, the correct jump at the point load, and a stated limitation (for example point loads and one UDL only).

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What do N, V, and M each represent physically?

N: axial push or pull along the member. V: transverse sliding force across the section. M: bending, the tendency to curve the member.

2. State the positive sign conventions for V and M.

Positive shear rotates the element clockwise; positive moment makes the beam sag (concave up, "smile").

3. What are the differential relations between w, V, and M?

dV/dx = −w(x) and dM/dx = V(x). Areas under one curve give changes in the next.

4. Where can the bending moment reach its maximum?

Where V = 0 or changes sign, at points of applied couples, or at a fixed support; check all candidates.

5. Why do later courses care about M more than the applied loads?

Bending stress σ = Mc/I is set by the internal moment at the critical section; the loads matter only through M.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRedraw both diagrams of the worked example from memory.

+3 daysOne full cantilever V-M analysis with new numbers.

+7 daysMixed set: reactions, cut, diagrams, end to end in one sitting.

+30 daysFeed an M_max into a stress estimate when you reach Mechanics of Materials.

Textbook mapping

Item	Mapping
Main textbook	R.C. Hibbeler, Engineering Mechanics: Statics, Chapter 7, Internal Forces
Core sections	7.1 Internal Loadings in Structural Members · 7.2 Shear and Moment Equations and Diagrams · 7.3 Relations between Distributed Load, Shear, and Moment
Recommended problems	Fundamental Problems F7-1 onward (partial solutions in the back). Draw at least six full V-M diagram pairs by hand before automating anything.
Skip on first pass	7.4 Cables: elegant but rarely examined first; return before Dynamics if your program uses it.
Read next	Chapter 8, sections 8.1 to 8.2 before opening Module 8.