Type a branch, a subject or a topic — “round robin”, “paging”, “civil”.

Method of Joints

Solved members fill in one at a time, spreading outward from the only joint you were allowed to start at.

Skip to the animation

Isolate one pin at a time and apply ΣH = 0 and ΣV = 0, starting where only two members are unknown — each solved joint reduces the unknowns at its neighbours, so the solution propagates outward until every member force is known.

The procedure

  1. 1Find the reactions first, treating the whole truss as one rigid body. This is what hands you a joint with only two unknowns.
  2. 2Start at a joint with exactly two unknown members — two equations can solve no more than two unknowns.
  3. 3Assume every member is in tension. Resolve, and let a negative answer report compression.
  4. 4Move to an adjacent joint where the newly known force reduces the unknown count to two, and repeat.
  5. 5Use the last joint as a check. A determinate truss always leaves one spare equation.

Assuming tension consistently is what removes the guesswork: you never need to know the answer before starting, and the algebra reports the truth as a sign.

Why the starting joint matters

A pin gives exactly two equations. Start where three members are unknown and the joint is unsolvable — and it fails immediately rather than subtly, which is at least merciful.

The reactions are found first precisely so that a support joint qualifies. Choosing a starting joint is the only judgement the method requires; everything after it is mechanical.

Zero-force members

  • Two non-collinear members at an unloaded joint — both carry zero, since neither can balance the other's perpendicular component.
  • Three members at an unloaded joint, two of them collinear — the third carries zero.
  • A support reaction or an applied load at the joint invalidates both rules.

Spotting these first can halve the work on an exam truss. But a zero-force member is not removable: it braces its neighbours against buckling, and it carries force under other load cases — which is why the structure has one.

Sign convention

Draw every unknown member force pulling away from the joint — the tension assumption. A positive answer confirms tension; a negative one means compression, and the magnitude is still correct.

Carrying the sign forward to the next joint is where errors creep in. A member in compression pushes *into* the joint at its far end, so the arrow reverses while the value does not.

When to use it, and when not

Method of joints gives every member force, which is what a full design check needs. Its weakness is that it propagates from a support, so reaching one member in the middle of a fifty-panel truss means solving every joint before it.

When you want one specific member — usually the most heavily loaded one — the method of sections reaches it in a single cut. In practice a long truss is often solved by sectioning once to get three forces in the middle, then continuing with joints from there.

The numbers you will be asked for

Joint equilibrium

ΣH = 0 · ΣV = 0

Two equations per pin; no moment equation exists.

Equation count

2j equations for m + r unknowns

Equality means determinate.

Member components

H = F·cos θ · V = F·sin θ

θ measured from the horizontal.

Sign convention

positive = tension

Assume it everywhere; the algebra corrects you.

Advantages and disadvantages

Advantages

  • Gives every member force, which a full design check requires.
  • Uses only two equations at a time, so the arithmetic stays simple.
  • Zero-force members can be removed by inspection before starting.
  • The final joint provides a free check on everything upstream.

Disadvantages

  • Must propagate from a support, so a middle member is expensive to reach.
  • An early arithmetic error contaminates everything after it.
  • Requires a joint with only two unknowns to start, which is not always obvious.
  • Tedious for large trusses, which is what sections exists to fix.

Watch it work

loading visualisation…

Check yourself

question 1 / 4

One question at a time. Pick an answer to see why it is right or wrong, then move on — there is no score to keep and nothing is saved.

Why must you start at a joint with exactly two unknown members?
Why assume every member is in tension before you know the answer?
A member comes out with zero force. Can it be removed from the design?
After solving every member, one joint is left over. What is it for?

0 / 4

4 still unanswered — the dots above jump straight to them.