Truss Analysis: Method of Joints Tutorial

Turn a whole truss into a sequence of two-equation joint problems — and learn the one sign convention that keeps it all straight.

StaticsMechanical Engineering Year 1Free preview
⏱️ About 18 min

A bridge truss looks like a tangle of steel, but almost every member in it is doing one very simple job: pulling or pushing straight along its own length. That is the secret that makes truss analysis so clean. Under the right idealizations, a truss is not a mystery at all — it is a chain of pin joints, each of which yields to nothing more than ΣF<sub>x</sub> = 0 and ΣF<sub>y</sub> = 0. Solve one joint, walk to the next, repeat. This lesson builds that walk from the ground up, and the simulator lets you make the tension-or-compression call at each joint before the solver confirms it.

💡
The big idea: A <strong>truss</strong> is a structure of straight, slender members connected at pin joints, with loads applied only at the joints. Under those idealizations every member is a <em>two-force member</em>, so its internal force is purely axial — pure tension or pure compression along the member's axis. The <strong>method of joints</strong> isolates one joint at a time and applies the two force-balance equations; because only two members can be unknown at any solvable joint, the whole truss unravels joint by joint.
🎯 By the end, you'll be able to
  • State the three idealized-truss assumptions and why each makes members two-force
  • Adopt a single tension/compression sign convention and apply it consistently
  • Recognize which joint to solve first (the one with at most two unknowns)
  • Carry out a full method-of-joints solution on a king-post truss
📎 Helpful to know first

What makes a truss a truss

A truss is a structure built from straight, slender members joined at their ends to form triangles. The triangle is the key: unlike a four-sided frame, a triangle cannot be deformed without stretching or shortening a member, so a triangulated assembly is rigid. Real roof trusses, bridge trusses, and crane jibs are all elaborations on this triangulated skeleton.

For analysis we replace each real joint with an idealized pin and make three simplifying assumptions. These are what let us treat every member as a clean two-force member — and they are also the assumptions whose violation moves a structure out of "truss" territory and into the frames-and-machines lesson.

🔑 The three idealized-truss assumptions
  1. Pinned joints. Every connection is a frictionless pin that transmits force but no moment. (A real bolted or welded joint does transmit some moment, but for slender members the effect is small.)
  2. Loads applied only at the joints. No force acts in the middle of a member. If a load must be applied between joints, a secondary member or a joint is added so the load still lands at a node.
  3. Weightless, straight members. Each member's own weight is either negligible or lumped onto the joints. Members carry load only along the straight line between their two pins.

With all three holding, each member is loaded at exactly two points (its pins) and nowhere else — the definition of a two-force member. Equilibrium then forces the two pin forces to be equal, opposite, and collinear along the member. There is no bending, no shear — only axial tension or compression.

A king-post truss with joints A, B, C, D. Pin support at A (bottom left), roller at B (bottom right), apex C, mid-bottom joint D. A 12 kN load acts downward at C; reactions of 6 kN act upward at A and B. Members AC and BC slope up to the apex, AD and DB form the bottom chord, and CD is the vertical king post. AC BC AD DB CD A B C D 12 kN 6 kN 6 kN
The worked truss: a king-post with pin at A, roller at B, 12 kN at the apex C, and a mid-bottom joint D. Reactions are 6 kN each by symmetry.

Tension or compression? Pick one convention and keep it

At every joint you must guess the sense of each unknown member force before writing equations. It does not matter which sense you assume — but you must be consistent, and you must interpret the sign of the answer correctly. This lesson uses the convention used by the simulator:

  • Assume every unknown member force is tension — i.e. draw it pulling away from the joint.
  • If the solved value is positive, the member really is in tension (it pulls on the joints).
  • If the solved value is negative, the member is in compression (it pushes on the joints). The magnitude is still correct; only the assumed sense was wrong.

A negative answer is never a mistake to fix — it is information. Compression members are the ones that can buckle, so flagging them matters for design (a later module). In the simulator below, tension members turn red and compression members blue.

✨ Solve joints with two unknowns first

A pin joint gives only two independent equations (ΣFx, ΣFy), so it can solve for at most two unknown member forces. The whole method hinges on ordering: find a joint with two or fewer unknowns, solve it, and that newly-known member reduces the unknown count at the neighbouring joint. Start at a support (its reactions are found first from overall equilibrium), and let the solution propagate. In the simulator, joints that are ready to solve glow green.

🎮 Interactive: method-of-joints step-through LIVE
Predict first: Before you solve joint A: member AC slopes up to the load at C — do you think it is in tension (stretching) or compression (being squeezed)? Make the call, then reveal.

An interactive truss on a canvas. Solvable joints (those with at most two unknown members) glow green; clicking one opens a panel to predict each member's state before the solver reveals the axial force and colours the member.

Click a glowing (green) joint, predict Tension / Compression / Zero for each unknown member, then reveal. Tension turns red, compression blue, zero-force grey. Switch presets to try a 7-member Pratt-style truss.

Reactions first, then joint A

Before touching the joints, solve the truss as a single rigid body for its support reactions. For the king-post: sum moments about A to get By, then ΣFy for Ay. By symmetry each is half the apex load.

📝 Worked example: For the king-post truss above (pin at A, roller at B, apex load P = 12 kN at C, rafters AC and BC with slope 3 up for every 4 across), find the support reactions and then the force in member AC using joint A.
  1. Overall equilibrium, sum moments about A: B_y (8 m) - P (4 m) = 0, so B_y = (12*4)/8 = 6 kN.
  2. ΣFy = 0: A_y + B_y - P = 0 -> A_y = 12 - 6 = 6 kN. (ΣFx = 0 gives A_x = 0.)
  3. Joint A: members AC and AD. Assume both in tension (pulling away). The slope of AC is a 3-4-5 triangle, so its unit vector is (4/5, 3/5) = (0.8, 0.6).
  4. ΣFy at A = 0: A_y + F_AC*(3/5) = 0 -> 6 + 0.6*F_AC = 0 -> F_AC = -10 kN.
  5. Negative means compression: member AC carries 10 kN in compression (it pushes on A and C).
  6. ΣFx at A = 0: F_AD + F_AC*(4/5) = 0 -> F_AD + 0.8*(-10) = 0 -> F_AD = +8 kN (tension).
  7. Sanity check: the rafter pushing up-and-in on A balances the 6 kN reaction pushing up; the bottom chord AD pulls outward to balance the rafter's inward push. Physically right.
✓ Reactions A_y = B_y = 6 kN. Member AC = 10 kN compression; member AD = 8 kN tension.
⚠️ These are idealizations — they have limits

The two-force-member result is a consequence of the three assumptions, not a universal truth. The moment any assumption fails the analysis changes:

  • A rigid (moment-resisting) joint lets a member carry bending, so it is no longer pure axial — that is a frame, not a truss (Lesson 4).
  • A load applied between joints puts a transverse force on the member, giving it bending and shear in addition to axial force.
  • A heavy member whose weight is not negligible bends under its own weight, again adding bending.

Engineers add joints or secondary members precisely to keep real loads landing at nodes, preserving the truss idealization where it is useful.

✏️ Practice: Continuing the king-post: at joint A we found F_AC = -10 kN. Using ΣF<sub>x</sub> = 0 at joint A with member AC's horizontal component (4/5)·F_AC and member AD horizontal, the relation is F_AD + (4/5)·F_AC = 0. Compute F_AD, in kN.
kN
Solution
  1. F_AD = -(4/5)*F_AC = -0.8*(-10) = +8 kN.
  2. Positive -> tension. Member AD (bottom chord) is stretched by 8 kN.
✏️ Practice: By symmetry, member BC equals member AC: F_BC = -10 kN (compression). At the apex joint C, members AC and BC each push with vertical component (3/5)·10 = 6 kN upward (compression pushes the joint away). The applied load is 12 kN down and member CD is vertical. From ΣF<sub>y</sub> = 0 at C, what is F_CD? (Recall at joint D, members AD and DB are collinear and D is unloaded.)
kN
Solution
  1. Joint D is unloaded and members AD, DB lie on the same straight line (collinear). By Rule 2, the third member at D — the vertical CD — must be a zero-force member, so F_CD = 0 kN.
  2. Check at C: the two rafters push up 6 + 6 = 12 kN, exactly balancing the 12 kN load, so CD contributes nothing. Confirmed: F_CD = 0.

Check your understanding

1. Under the idealized-truss assumptions, every member is a two-force member. Which condition is NOT one of those assumptions?
The three assumptions are pinned joints, loads at joints, and weightless straight members. The material (steel, timber, aluminium) is irrelevant to the two-force-member result.
2. You assume a member force is tension and solve to get F = -7 kN. The member is:
A negative answer to a tension assumption means the member is in compression with magnitude 7 kN. The magnitude is correct; only the assumed sense was wrong.
3. Why must you start the method of joints at a joint with at most two unknown members?
A pin joint provides only ΣFx=0 and ΣFy=0 — two equations — so it can resolve at most two unknown member forces. Starting at a two-unknown joint (usually a support) lets the known values propagate outward.
🔎 Related lessons
Trusses: Method of Sections · soon Zero-Force Members & Truss Design Intuition · soon Two-Force & Three-Force Members · soon 2D Equilibrium: ΣF=0, ΣM=0
✅ Key takeaways
  • A truss idealizes joints as pins, loads at joints, and weightless straight members; under these, every member is a two-force member carrying pure axial force.
  • Adopt one sign convention (assume tension; negative answer = compression) and apply it at every joint.
  • Method of joints solves one joint at a time using ΣFx=0 and ΣFy=0; start at a joint with at most two unknowns (usually a support).
  • For the king-post: reactions 6 kN each; AC = BC = 10 kN compression; AD = DB = 8 kN tension; CD = 0 (zero-force).
➡️ Method of joints walks the truss one joint at a time — fine, but slow if you only need the force in one member deep inside the structure. The method of sections cuts straight to it: slice the truss and apply three equilibrium equations to one piece.
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