Truss analysis calculator
A pin-jointed truss solved by the direct stiffness method: pick the geometry, and every member force, reaction and deflection comes out of the same global system a desktop package would assemble. Warren, Pratt and Howe patterns, any panel count.
| Member | Nodes | Length (m) | Force (kN) | Stress (MPa) |
|---|---|---|---|---|
| 12 | 11–6 | 3.61 | -36.1 | -20.0 |
| 11 | 0–7 | 3.61 | -36.1 | -20.0 |
| 9 | 9–10 | 2.00 | -36.0 | -20.0 |
| 3 | 2–3 | 2.00 | 36.0 | 20.0 |
| 5 | 4–5 | 2.00 | 32.0 | 17.8 |
| 8 | 8–9 | 2.00 | -32.0 | -17.8 |
| 4 | 3–4 | 2.00 | 32.0 | 17.8 |
| 2 | 1–2 | 2.00 | 32.0 | 17.8 |
| 18 | 1–8 | 3.61 | -21.6 | -12.0 |
| 21 | 5–10 | 3.61 | -21.6 | -12.0 |
| 7 | 7–8 | 2.00 | -20.0 | -11.1 |
| 10 | 10–11 | 2.00 | -20.0 | -11.1 |
| 6 | 5–6 | 2.00 | 20.0 | 11.1 |
| 1 | 0–1 | 2.00 | 20.0 | 11.1 |
Stress is force over area; compare it with the yield strength for a utilisation, or with a buckling curve for the compression members.
K·u = F · Assembled from each member’s axial stiffness.
k = EA/L · [c², cs; cs, s²] · c and s are the member direction cosines.
N ≈ W·L / (8h) · Simply supported, uniform load; a check on the solver, not a substitute.
A clear path from inputs to results.
A pin-jointed truss is the cleanest structural model there is: every member carries axial force only, and the whole thing can be solved from one global stiffness matrix. This calculator builds the geometry, assembles that matrix, solves for displacements and reads every member force back out — the same route a desktop package takes.
How to use
- Choose the truss pattern and set the span, depth and panel count.
- Enter the total roof load; it is applied at the top-chord panel points.
- Read the peak tension and compression, then check the member forces against yield and buckling.
The formula
Reference: Direct stiffness method for a two-dimensional pin-jointed truss.
Worked example
Six-panel Pratt truss, 12 m span
- Span 12 m, depth 3 m, 6 panels, 60 kN total roof load, 1800 mm² members, 200 GPa.
- 1
Build 8 nodes and 13 members; pin the left support and roller the right. Assemble K, apply 20 kN at each of the three top-chord nodes, solve for the 13 free degrees of freedom, then recover member forces from the elongations.
Peak compression in the end post, peak tension in the bottom chord, equilibrium satisfied to the kilonewton.
Interpreting the result
The chord force estimate W·L/(8h) is shown alongside the solved value: if the two disagree by more than a few per cent, the geometry is doing something the simple beam analogy cannot see.
Assumptions & limitations
- Pins at every joint: no bending in members.
- Linear elastic material with a constant section.
- Small displacements; no P-delta.
- Loads applied at joints only, not along members.
- Out-of-plane buckling is not checked here.
Frequently asked questions
Why does the solver refuse some geometries?
An untriangulated panel is a mechanism, so the stiffness matrix is singular. The calculator reports that rather than returning a number.
Which pattern is most efficient?
Pratt puts the diagonals in tension under gravity, which is why it is common in steel; Howe reverses that and suits timber.
Is the deflection realistic?
It is the elastic axial deflection of the truss; joint slip and connection flexibility usually add to it.
Can I model a non-uniform section?
Not in this form — every member shares one area. Use the section builder to size the chord, then check the peak force against it.
Methodology last reviewed 2026-10-22. Results are informational and not financial or engineering advice — see our disclaimer.