Method of jointsA whole truss, joint after joint
Every joint = a small puzzle of two equations: ΣFx = 0 and ΣFy = 0.
The only rule: enter only a joint with at most 2 unknown members.
- 1 Reactions — ΣM about the pinned support
- 2 Zero-force members — mark them in advance
- 3 Joint with ≤ 2 unknowns → ΣFx, ΣFy
- 4 Move on to the neighbouring joint
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01
The trick that saves most mistakes
- Every unknown member — assume tension: the arrow leaves the joint.
- Got plus → tension. Got minus → compression. No guessing — the sign decides.
- A member already found in compression enters the next joint with an inward arrow and a positive magnitude. Never an inward arrow and a minus together.
- Every component points where the arrow points: right and up — plus, left and down — minus. A tension member to the left of the joint: the arrow leaves to the left → minus.
- An inclined member is decomposed: horizontal = N · cos α, vertical = N · sin α. Here α = 45° → 0.707 for both.
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02
A whole truss — 6 joints: 5 to solve, F to check
A truss of 6 m and height 2 m, 4 t at D and 1 t at F. The order: A → C → D → E → B.
N3 dashed — zero-force member (joint C) · scroll the drawing sideways - ReactionsΣMA: 4·2 + 1·4 − BY·6 = 0 → BY = 2 t, and from ΣFy: AY = 3 t
Joint A — two members, two unknowns Joint D — N1 in compression enters the joint - Joint AΣFy: 3 + N1·0.707 = 0 → N1 = −4.243 compression
ΣFx: N2 + N1·0.707 = 0 → N2 = 3 - Joint CN2 and N6 on one line, no load → N3 = 0 · ΣFx: −3 + N6 = 0 → N6 = 3
- Joint DN1 in compression enters D: 4.243 · 0.707 = 3 to the right and 3 up → ΣFy: 3 − 4 − N5·0.707 = 0 → N5 = −1.414
ΣFx: 3 + N4 + N5·0.707 = 0 → N4 = −2 - Joint EN5 in compression enters E: 1 to the right, 1 down → ΣFy: N7 − 1 = 0 → N7 = 1 · ΣFx: −3 + 1 + N9 = 0 → N9 = 2
- Joint BΣFy: 2 + N8·0.707 = 0 → N8 = −2.828 compression, and ΣFx comes to zero check ✓
That's it. 9 members, 5 joints — and at every joint only two short equations.
All the results (for checking)
Reactions: AY = 3 t, BY = 2 t Member Between joints Force State N1 A–D 4.243 t Compression N2 A–C 3 t Tension N3 C–D 0 Zero-force member N4 D–F 2 t Compression N5 D–E 1.414 t Compression N6 C–E 3 t Tension N7 E–F 1 t Tension N8 F–B 2.828 t Compression N9 E–B 2 t Tension - Reactions
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03
4 pitfalls everyone falls into
- Entering a joint with 3 unknowns. At D there are four at first — so we start at A.
- Lever arms from different supports in the same ΣM. All lever arms — from the same point.
- Inward arrow + minus. The sign is counted twice → at D you get N5 = −9.9 instead of −1.414.
- Swapping sin and cos. At 45° it doesn't show — at another angle it ruins everything.
Bonus: Finished? Substitute into the last joint — everything must come to zero. A free check.
Trusses binder · BST · beamsolvertool.com/en/learn/trusses/method-of-joints/
Solve a truss in BST BST guides you joint after joint, with a free body for every joint — or waits for you to solve on your own
All pages as a print-ready PDF — in the Binders to download of BST PRO