Statics · Module 5 of 11
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
Learning objectives
- Draw a free-body diagram showing every reaction each support can carry.
- Solve three equilibrium equations for three unknown reactions.
- Recognise when a missing applied couple makes a problem unsolvable.
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.
Worked example 2: diagnose a missing applied couple
Problem. A 4 m simply supported beam carries an 8 kN downward force 1 m from pin A and a 2 kN·m clockwise applied couple. A student calculates By = 2 kN from the point load alone.
- Audit the free-body diagramThe applied couple is an external action and must appear in the moment equation even though it has no line of action.
- Take moments about AUsing counterclockwise positive: 4By - (8)(1) - 2 = 0.
- SolveBy = 2.5 kN. Then ΣFy = 0 gives Ay = 8 - 2.5 = 5.5 kN.
- CheckMoments about B give -4(5.5) + 8(3) - 2 = 0, so the reactions are consistent.
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 original test case that you create yourself.
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.