The Chartership Forum

Structural Analysis Fundamentals / Pin-jointed trusses

Method of joints and method of sections

Two ways to get member forces, and how to choose between them.

Lesson review: expert verification remains open

This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

Checks needed for project use

  • An independent structural reviewer must check remaining derivations, diagrams, sign conventions and the interactive models across their full input ranges.
  • Independent review of diagrams, idealisations and full interactive input ranges remains required.

These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.

Concrete and steel calculations retain their stated first-generation Eurocode teaching basis. Use the standards and amendments specified for your project; do not mix generations without an appropriate design basis.

Read the lesson register and outstanding review notes

Why this matters

Use joint equilibrium when a sequence of joints has at most two unknown forces. A section through a suitable set of members can give a target force directly.

By the end of this lesson you should be able to

  • Solve a joint by resolving in two directions
  • Cut a section and take moments to isolate one member
  • Interpret the sign as tension or compression

In the method of joints, you isolate one joint at a time and resolve horizontally and vertically. Because a joint is a point, there is no moment equation — just two equations, so you can only solve a joint with two unknown member forces. Always assume every unknown member is in tension, pulling away from the joint. If the answer comes out negative, the member is in compression. Sticking to that habit removes almost all sign confusion.

Try it

Work joint by joint

Pick a joint, isolate it, and resolve. Start where only two member forces are unknown.

Isolate joint

Joint A has a support reaction and two members, so it is a sensible starting point.

Pin-jointed truss with two loaded top joints15 kN15 kNABCDE

Worked example

Member forces in a triangular truss

Member forces in a triangular truss24 kNABC

Given

  • Span A to B = 6.0 m, apex C is 4.0 m above the chord, 3.0 m from each support
  • Pin at A, roller at B
  • Vertical load of 24 kN downwards at C

Find

The forces in members AB and AC, stating tension or compression.

Assumptions

  • Pinned joints
  • Load applied at a joint
  • Member self-weight ignored

    The method of sections cuts an imaginary line right through the truss, including the member you want. You then treat everything on one side as a free body. Because you now have a whole body rather than a point, you get the moment equation back — and taking moments about the point where the other two cut members meet leaves your target member as the only unknown.

    Practice

    In the truss above, if the apex load is increased to 40 kN, what is the force in the bottom chord AB? Give the magnitude in kN (it stays in tension).

    Practice

    In the triangular truss above (span 6.0 m, apex 4.0 m high), if the apex load is 18 kN, what is the force in the sloping member AC? Give the magnitude in kN.

    Practice

    For that same truss under the 18 kN apex load, what is the force in the bottom chord AB? Give the magnitude in kN (it is in tension).

    Worked example

    One cut, one member force: the method of sections

    Given

    • Parallel-chord truss, span 24.0 m in six panels of 4.00 m, depth 3.00 m
    • Simply supported, with 60.0 kN hanging at each of the five interior bottom joints
    • Wanted: the force in the bottom chord of the third panel, between 8.00 and 12.0 m

    Find

    The bottom chord force, without working joint by joint across the truss.

      Practice

      A parallel-chord truss spans 24.0 m in 4.00 m panels with a depth of 3.00 m, carrying 60.0 kN at each of the five interior bottom joints. What is the support reaction, in kN?

      Practice

      For that truss, take a section through the third panel and moments about the top-chord joint at 8.00 m. What is the force in the bottom chord, in kN?

      Practice

      If the same truss were built 4.00 m deep instead of 3.00 m, what would that bottom chord force become, in kN?

      Summary

      • Method of joints: two equations per joint, so start where only two forces are unknown
      • Always assume tension; a negative answer means compression
      • Method of sections: cut through the target member and take moments about where the other cut members meet