Statics · Module 5 of 11 · Intermediate
Equilibrium of a Rigid Body
Rigid bodies can translate and rotate, so both forces and moments must balance. This module is the center of the whole course.
Readiness check
This module combines everything from Modules 2 to 4. Tick only what you can do closed-notes, right now.
- Draw a complete particle free-body diagram without prompting.
- Compute a moment about any point using components (M = x·Fy − y·Fx).
- Replace a distributed load with its resultant at the centroid.
- Solve three linear equations with three unknowns.
- Recall what a couple is and why it is a free vector.
The core idea
Three equations, three unknowns: that is 2D rigid-body statics.
ΣFx = 0ΣFy = 0ΣM = 0Each support type supplies known reaction directions. A roller gives one force perpendicular to its surface. A pin gives two components. A fixed support gives two components plus a moment. Count the unknowns before you solve anything.
The method
What object are we studying, and what touches it?
Particle, rigid body, beam, truss, or area?
FBD: forces, supports, axes, distances, unknowns.
Only now write the equations and calculate.
Worked example: support reactions of a loaded beam
A 6 m beam rests on a pin at A (left end) and a roller at B (right end). It carries 8 kN down at 1.5 m from A and 4 kN down at 4.5 m from A. Find the reactions.
- ProblemFind the support reactions of the beam in Figure 1.
- Given / findSpan 6 m; loads 8 kN at 1.5 m and 4 kN at 4.5 m from A. Find Ax, Ay, By.
- AssumptionsBeam weight negligible compared to the loads; rigid beam; 2D; loads purely vertical.
- ModelFigure 2: the pin at A gives Ax and Ay; the roller at B gives By only. Three unknowns, three equations: determinate.
- EquationsΣMA = 0: By(6) − 8(1.5) − 4(4.5) = 0 ΣFy = 0: Ay + By − 12 = 0 ΣFx = 0: Ax = 0
- SolveMoments about A first (this removes Ax and Ay in one stroke): 6By = 12 + 18 = 30, so By = 5 kN. Then Ay = 12 − 5 = 7 kN and Ax = 0.
- CheckIndependent moment check about B: Ay(6) − 8(4.5) − 4(1.5) = 42 − 36 − 6 = 0. Reasonableness: more load sits nearer A, so A carries more (7 over 5).
- ConclusionThe pin anchors must carry 7 kN, the roller seat 5 kN. These reactions become the input loads for the internal-force analysis in Module 7.
Misconceptions and diagnostics
Read each row as a practical debugging check: the mistake, how it shows up, what to ask yourself, and how to correct it.
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Wrong reactions for the support type | Too many or too few unknowns; unsolvable or trivial system | "What motion does this support prevent?" | Roller: 1 force. Pin: 2 components. Fixed: 2 components + 1 moment. Memorize the table cold. |
| Forgetting a force in the moment equation | ΣM check about a second point fails | "Did every FBD force appear once in ΣM?" | Tick each FBD arrow as you write its moment term. Always verify with a second moment point. |
| Treating an indeterminate beam as solvable | 4 unknowns, 3 equations, circular algebra | "How many reaction unknowns did my supports create?" | If unknowns exceed 3 in 2D, statics alone cannot solve it; flag it, do not force it. |
| Panic at a negative reaction | A correct answer gets "fixed" into a wrong one | "What does the minus sign mean here?" | Nothing is wrong: the reaction points opposite to your assumed direction. Keep the sign and move on. |
Practice ladder
A 4 m simply supported beam (pin left, roller right) carries 10 kN at midspan. Find the reactions.
Show answer
By symmetry Ay = By = 5 kN, Ax = 0. Verify with ΣMA: 4By = 10(2), so By = 5.
A 6 m beam (pin A, roller B) carries a uniform load of 2 kN/m across the full span. Find the reactions.
Show answer
Resultant 12 kN at the 3 m midpoint. ΣMA: 6By = 12(3), so By = 6 kN and Ay = 6 kN. Symmetric load, symmetric reactions.
A 3 m horizontal arm is fixed to a wall at A. It carries 1.2 kN down at the free end and a 0.8 kN/m uniform load over its full length. Find all reactions at the wall.
Show answer
UDL resultant 2.4 kN at 1.5 m. ΣFy: Ay = 1.2 + 2.4 = 3.6 kN up. Ax = 0. ΣMA = 1.2(3) + 2.4(1.5) = 7.2 kN·m counterclockwise, resisting the load moment.
Model a real diving board, balcony, or cantilevered shelf: measure or estimate dimensions and load (a person at the tip), compute the support reactions, and identify which connection element fails first if overloaded.
What good work looks like
A labeled FBD with the support modeled honestly (fixed, or a pin and roller pair), reactions computed including the moment, and a sentence connecting the wall moment to the most-stressed bolt or weld.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Write a reaction calculator (spreadsheet or Python) for a simply supported beam: inputs are span, up to three point loads with positions, and one uniform load; outputs are both reactions. Validate it against the worked example (7 kN and 5 kN) and one textbook Fundamental Problem.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write the three equilibrium equations for a rigid body in 2D.
ΣFx = 0, ΣFy = 0, ΣM about any point = 0.
2. What reactions do a roller, a pin, and a fixed support provide?
Roller: one force perpendicular to the rolling surface. Pin: two force components. Fixed: two components plus a reaction moment.
3. What is a two-force member, and what does it imply?
A member loaded at only two points: the forces must be equal, opposite, and directed along the line joining the points, so the force direction is known by geometry.
4. Why take moments about a support point first?
The unknown reactions through that point have zero arm, so they vanish, leaving one equation with one unknown.
5. How do you recognize a statically indeterminate 2D problem?
More than three unknown reaction components, or improper constraints. Statics alone cannot solve it.
Textbook mapping
| Item | Mapping |
|---|---|
| Main textbook | R.C. Hibbeler, Engineering Mechanics: Statics, Chapter 5, Equilibrium of a Rigid Body |
| Core sections | 5.1 Conditions · 5.2 Free-Body Diagrams · 5.3 Equations of Equilibrium · 5.4 Two- and Three-Force Members |
| Recommended problems | Fundamental Problems F5-1 through F5-12 (partial solutions in the back). The support-reaction table in 5.2 is worth memorizing verbatim. |
| Skip on first pass | 5.5 to 5.7 (3D equilibrium, constraints and statical determinacy in 3D); return after Module 6. |
| Read next | Chapter 6, sections 6.1 to 6.4 before opening Module 6. |