Structural Analysis Fundamentals / Pin-jointed trusses
Assumptions and determinacy
What we pretend about trusses, and whether equilibrium can solve 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 notesWhy this matters
Idealising a truss as pin-jointed members reduces its analysis to axial forces. Check the assumptions before applying that model to a real connection or loading arrangement.
By the end of this lesson you should be able to
- List the truss assumptions
- Apply the determinacy check
- Identify zero-force members
Truss analysis rests on three assumptions. The joints are frictionless pins, so they cannot carry moment. Loads are applied only at the joints. And the members are straight and light enough that self-weight can be ignored or split between the joints.
Put those together and each member can only pull or push along its own length. No bending, no shear — just axial force. That is why trusses are efficient and why they are easy to solve.
What it calculates: Whether equilibrium alone can find every member force and reaction.
- m
- Number of members
- r
- Number of reaction components
- j
- Number of joints
This assumes
- A plane, pin-jointed truss
- The members are arranged so the truss is actually stable
In plain terms: Each plane joint supplies two equilibrium equations. For a stable truss with independent equilibrium equations, m + r = 2j gives static determinacy and m + r > 2j gives redundancy. A deficient count implies a mechanism; an equal or higher count can still conceal instability.
Predict first
At an unloaded joint where exactly two members meet and they are not in line with each other, what are the member forces?
Members like that are zero-force members. They are not useless: they often brace another member against buckling, or they carry load only under a different load case. But for the load case in front of you they carry nothing, and spotting them first saves a lot of work.
Practice
A plane truss has 11 members, 7 joints and 3 reaction components. What is m + r − 2j?
Practice
A stable plane truss has 14 members, 8 joints and 3 reaction components. To what degree is it statically indeterminate?
Practice
A plane truss has 9 members, 7 joints and 3 reaction components. What is m + r − 2j, and what does that tell you? (Enter the number.)
Summary
- Pinned joints, loads at joints, light straight members → axial force only
- m + r = 2j is a necessary but not sufficient determinacy check
- Zero-force members carry nothing under this load case but may still be needed