When you need many member forces
Isolate one joint at a time. Each planar joint provides ΣFₓ = 0 and ΣFᵧ = 0, so start at a joint with no more than two unknown member forces.
Best for: tracing the entire truss joint by joint.
Principles of Engineering • Unit 2 • Lesson 2.9
Advanced Truss Analysis
Engineering question: How can we find forces in selected members of a larger truss without calculating every joint first?
Your learning pathway
What you'll be able to do
Follow the learning modules in order. This lesson adds strategy and larger structures—not a new set of physical fabrication requirements.
01 · RecognizeFind zero-force members and understand when the rules apply.
02 · ChooseUse joint analysis or a section cut to solve selected member forces.
03 · VerifyCheck force signs, support reactions and independent equilibrium equations.
Suggested 45–50 minute learning sequence
Zero-force inspection5–7 min
Watch & model10–12 min
Worked section cut15 min
Interact & practice15 min
The second video and additional practice may be assigned outside of class.
01 / Connect previous lessons
Two tools for different jobs
Both analysis methods use static equilibrium and ideal pin-jointed, joint-loaded, weightless two-force members.
When you need a few selected members
Imagine a cut through a small set of members. Isolate one side, replace each cut member with an unknown axial force and use ΣFₓ = 0, ΣFᵧ = 0 and ΣM = 0.
Best for: finding a key member without solving all the intermediate joints.
Whole truss: 3 equilibrium equationsOne 2D joint: 2 equationsOne 2D cut: 3 equations
Important nuance: A section cut crossing three unknown member forces is the standard starting point, but three equations do not guarantee a solution. If the unknown lines of action are arranged so that the equations are dependent, choose another cut or combine methods.
02 / Inspect before calculating
Two zero-force-member rules
These rules apply to ideal pin-jointed trusses with no applied external load or support reaction at the examined joint. The result is specific to that loading arrangement.
Two noncollinear members at an unloaded joint
If a joint connects only two noncollinear members and has no external load or reaction, their forces must both be 0 N. Otherwise, they could not balance in two perpendicular directions.
FJA = FJB = 0 N
Three members; two lie on one line
At unloaded J, JA and JB lie on the same horizontal line; JC is vertical. The collinear forces cannot balance a nonzero vertical component.
ΣFᵧ = FJC = 0 → JC = 0 N
Do not remove real structural members: A member with zero calculated force under one ideal load case may still prevent buckling, brace the structure, or carry forces when a vehicle moves. The rules do not apply if an external force or support reaction acts at that joint.
Reasoning check: does a support reaction count as an external load?
Yes. An unloaded joint for these rules must have no external forces, including support reactions. A joint with a pin reaction is not automatically eligible even if no separate hanging weight is shown.
03 / Plan the solution
Five steps for a section cut
Use this method when the truss has more members than you want to solve individually.
1 · InspectDetermine support types, loads and zero-force members.
2 · ReactSolve whole-truss reactions when needed.
3 · CutCross selected members—usually no more than 3 unknowns.
4 · IsolateDraw one side with external loads and cut-member forces.
5 · Solve and cross-check
Assume each cut member is in tension (arrows point outward from the isolated section along the member axis). Choose a moment center to eliminate unknowns, then solve horizontal and vertical equilibrium. A negative result means compression.
Cutting isn't removing support: Your free-body diagram includes all external loads and support reactions acting on the chosen side. Member forces appear only along members crossed by the cut. If you analyze the opposite side, their directions reverse.
04 / Engineering videos
Two worked statics tutorials
These videos are from Jeff Hanson's statics course. The first demonstrates how and where to cut a truss; the second teaches zero-force-member shortcuts on a much larger truss.
Jeff Hanson · Core tutorial
Statics Lesson 49 — Trusses: Method of Sections
Pause and try: At about 04:57, pause and trace the section cut. At about 07:52, explain how the moment center eliminates two unknown forces.
Watch on YouTube ↗Jeff Hanson · Zero-force rules
Statics Lesson 50 — Finding Zero-Force Members
Pause and try: At about 01:55 and 03:24, pause on each shortcut and check the joint load conditions before removing any members from the analysis.
Watch on YouTube ↗Videos contain longer university-level examples. You can complete this POE lesson using the simpler worked truss below.
05 / Full worked example
Find forces in three selected members
Analyze an ideal five-joint, seven-member truss. The bottom joints A–B–C are at (0,0), (2,0) and (4,0) m. Top joints D and E are at (1,2) and (3,2) m. Pin A, roller C; downward loads 120 N at D and 80 N at E. Neglect member weight.
Use the entire truss
Take moments about A (CCW positive) so reactions at A disappear from that equation.
ΣMA = 4Cᵧ −120(1) −80(3) = 0
Cᵧ = (120 + 240)/4 = 90 N upward.
ΣFᵧ = Aᵧ +90−120−80 = 0
Aᵧ = 110 N upward. Aₓ = 0 N.
Both reactions add to 200 N, the total downward load.
Cut DE, BE and BC
The vertical yellow line crosses DE, BE and BC—three members. Keep the right side, with joints E and C, because it contains only one applied load (80 N) and one known reaction (90 N).
Assume unknown axial forces are tensile, pulling away from the right-hand section and toward the removed left-hand material.
Coordinates: E = (3,2), C = (4,0), B = (2,0). For member BE, the vector from E toward B is (−1, −2), of length √5 m.
Step 3 · Draw the right-hand section FBD
There are three unknown axial forces: FDE points left, FBE points down-left, and FBC points left under the assumed-tension convention. The isolated right-hand portion also has its downward 80 N load at E and upward 90 N reaction at C.
Member BE: Fₓ = −FBE(1/√5); Fᵧ = −FBE(2/√5)
Solve BC first
Forces DE and BE pass through E, so their moments about E are zero. The 90 N reaction at C has a 1 m horizontal lever arm; the leftward force BC acts 2 m below E.
ΣME = +90(1) −2FBC = 0
FBC = +45 N tension.
The positive result means the originally assumed tensile direction is correct.
Solve BE next
ΣFᵧ = 90−80−(2/√5)FBE = 0
FBE = 5√5 ≈ +11.18 N tension.
Its vertical component is −10 N, while its horizontal component is −5 N.
Step 6 · Horizontal forces
ΣFₓ = −FDE −(1/√5)FBE −FBC = 0
Substitute the two previously calculated results:
−FDE −5−45 = 0 → FDE = −50 N
DE is in compression, magnitude 50 N. The result is negative because the assumed leftward tensile force is actually a rightward compressive force on this isolated right-hand portion.
| Member | Signed result under tension-positive convention | Final classification |
|---|---|---|
| BC | +45.00 N | Tension |
| BE | +11.18 N | Tension |
| DE | −50.00 N | Compression |
06 / Check without repeating every calculation
Verify results at another joint
A section solves selected forces efficiently, but a separate joint balance is a valuable way to catch directional mistakes.
Use known whole-truss reactions
Joint A is easy because there are only two unknown members, AB and AD. Member AD rises 2 m over a 1 m run, so its component ratios are 2/√5 vertically and 1/√5 horizontally.
ΣFᵧ = 110+(2/√5)FAD = 0
Thus FAD = −55√5 ≈ −122.98 N compression.
ΣFₓ = FAB +(1/√5)FAD = 0
Thus FAB = +55.00 N tension.
Continue at joint B to verify the section results.
Joint B: a consistency check
At B, the remaining diagonals have equal magnitudes but opposite classifications: FBD = −11.18 N compression and FBE = +11.18 N tension.
ΣFₓ = −55+45+5+5 = 0 N
ΣFᵧ = −10+10 = 0 N
Determinacy check: j = 5 joints, m = 7 members, r = 3 external reaction components, so m + r = 7 + 3 = 10 = 2j. This is a necessary count for an ordinary determinate planar truss; it does not alone prove stability or detect every geometric mechanism.
07 / Interactive truss investigation
Move the loads. Watch the members change.
Keep geometry fixed and change either downward joint load. Select stages to see the section cut and examine the selected member-force calculation. Pay special attention to BE, which can change from tension to compression as the load distribution changes.
110.00 NLeft reaction Aᵧ (up)
90.00 NRight reaction Cᵧ (up)
+45.00 N (T)Member BC
+11.18 N (T)Member BE
−50.00 N (C)Member DE
0.00 NVertical force residual
Load at D is larger, so BE is in tension.
Live truss and section cut
1 · Whole-truss equilibrium
The two downward joint loads balance the two upward support reactions. Take moments about A to find Cᵧ.
4Cᵧ − P_D(1) − P_E(3) = 0
Ideal pin-jointed truss with no member self-weight, loads only at D and E, and small static deformation. These are force calculations—not predictions of bending, failure or safe bridge capacity.
Investigation A · Swap the loads
Set D = 80 N and E = 120 N. Predict the force in BE before checking the explorer.
Expected result
BE becomes −11.18 N (compression). The sign switches because the right side now has a larger downward applied load than its upward support reaction; the member force reverses direction.
Investigation B · Make the loads equal
Set D = E = 100 N. What happens to BE under the ideal model? Does that mean removing it is a good design decision?
Expected result
FBE = 0 N in this particular load case, by equilibrium. That does not establish that BE is unnecessary under different loading, out-of-plane stability demands, buckling prevention or a real bridge's load cases.
08 / Independent checks
Try the diagram and math first
Use sketches and equations in your engineering notebook or 2.9 worksheet. Check your answer only after you have a reason for it.
Two members at an unloaded joint
Joint J connects exactly two noncollinear members. No applied load or reaction exists at J. What can you infer?
Show solution and reasoning
Both member forces equal 0 N for this idealized loading situation. Each noncollinear vector direction has no balancing partner.
A joint with a support
Two noncollinear members meet at a pin support, and the pin carries a nonzero reaction. Can you immediately declare both members zero force?
Show solution and reasoning
No. The support reaction is an external force at that joint, so the unloaded-joint condition required by the shortcut is not satisfied.
Choose an efficient cut
You need one member force in a large, planar, statically determinate truss. Why might a cut crossing three members be better than solving ten other joints first?
Show solution and reasoning
The method of sections exposes only the cut-member forces on one isolated part. You can use the three independent planar equilibrium equations, if the geometry makes them sufficient.
80 N at D and 120 N at E
Use the five-joint truss geometry from the worked example. Find Aᵧ and Cᵧ.
Show solution and reasoning
ΣMA = 4Cᵧ−80(1)−120(3)=0 ⇒ Cᵧ = 110 N. ΣFᵧ = Aᵧ+110−200=0 ⇒ Aᵧ = 90 N.
Solve BC and BE for swapped loads
For D = 80 N, E = 120 N on the same truss, use the right-hand section cut through BC, BE and DE. State BC and BE with tension/compression.
Show solution and reasoning
Cᵧ=110 N. Taking moments about E: 110(1)−2FBC=0 ⇒ BC=+55 N tension. Vertical balance: 110−120−(2/√5)FBE=0 ⇒ BE=−5√5≈−11.18 N compression.
Interpret DE
A force DE was assumed leftward on the isolated right-hand section, and the calculated answer was −50 N. What does the minus sign mean?
Show solution and reasoning
DE is in compression of magnitude 50 N. The actual force acts opposite the assumed tensile arrow on that side of the cut.
Count members and joints
A 2D truss has j = 5 joints, m = 7 members and r = 3 independent external reactions. Does m + r = 2j guarantee the truss is stable?
Show solution and reasoning
The count 7 + 3 = 2(5) is satisfied. That is only a necessary simple counting condition; geometry or constraints may still create a mechanism or dependent equilibrium equations.
Zero under one load
Your truss model predicts BE = 0 N when D and E both carry 100 N. Should BE automatically be deleted from your physical design?
Show solution and reasoning
No. A zero-force result is specific to the idealized load case and assumptions. BE can be engaged when loads shift, and it may help with real bracing, stability or buckling resistance.
Reference library
Study resources and additional truss problems
These open engineering texts explain the same approaches using larger examples and interactive illustrations.
Method of sections
Open statics textbook, complete procedure and a cut-truss interactive illustration.
Engineering Statics — Section 6.5 ↗Zero-force members
Two inspection rules with figures and a worked larger-truss example.
Engineering Statics — Chapter 6 ↗Method of joints
Return to the previous method and compare when it is more efficient than a section cut.
Engineering Statics — Section 6.4 ↗Key vocabulary
Section cut: imaginary boundary exposing internal forces. Moment center: point used to simplify rotational equilibrium. Collinear: lying on the same straight line. Zero-force member: axial force equal to zero for a specified ideal load case.
Student submission
Exactly what to submit
Submit the POE 2.9 Advanced Truss Analysis worksheet or teacher-directed notebook equivalent, showing your complete reasoning rather than copying a solved model from the site.
Required evidence
- Two small joint FBD sketches illustrating zero-force-member rules and a sentence explaining when each applies.
- A complete whole-truss FBD with A, B, C, D, E labels; geometry; both applied joint loads and support reactions.
- Support-reaction calculations using ΣM = 0 and ΣFᵧ = 0 with N and m units.
- One section line crossing exactly three members and a separate, clearly labeled FBD of the chosen section.
- Section equations leading to at least three signed member forces, each classified as tension or compression.
- A force-equilibrium check at another joint or an equivalent independent equation.
- Short reflection on why a zero-force member may still matter in a real bridge.
Before turning it in
- The cut matches the three members you identified.
- Your isolated FBD includes all loads and reactions on the retained side.
- Every moment uses a perpendicular distance.
- Assumed force directions and negative compression results are interpreted consistently.
- Numerical force values include N and moments include N·m.
- All assigned independent and homework questions are included.
Due date: Check the Daily Agenda and your instructor's submission instructions.
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