Worked structural analysis example
Solve a 12 m Warren truss step by step, including reactions, axial member forces, zero-force diagonals and centre deflection.
Warren truss fundamentals
A Warren truss uses a repeating series of triangular panels between its top and bottom chords. The web members alternate direction rather than using the regular vertical-and-diagonal pattern of a Pratt or Howe truss. Under joint loading, the diagonals carry changing combinations of tension and compression as panel shear moves toward the supports.
Warren truss analysis converts the triangular geometry, support conditions and panel-point loads into support reactions, axial member forces and nodal deflections. A statically determinate pin-jointed Warren truss can be solved by equilibrium using the method of joints and method of sections; a direct-stiffness solver provides the complete force and displacement response.
In this symmetric example, the bottom chord is in tension and the top chord is in compression. The outer diagonals are in compression, while the next diagonals alternate between tension and compression. The two centre diagonals carry zero force for this particular symmetric load case.
The diagonal web transfers panel shear between the chords. As the shear changes from panel to panel and reverses across midspan, successive diagonals can switch between tension, compression and—at a symmetry point—zero force.
No. Analysis determines reactions, axial forces and displacements for an idealized model. Structural design must additionally check member yielding and buckling, connections, bracing, load combinations, serviceability and the applicable design code.
01 · Problem definition
A simply supported, four-panel Warren truss spans 12 m and is 3 m high. Four 12 kN downward loads act at the upper panel points. All members use the same steel modulus and cross-sectional area.
| Node | X | Y | Support | Restrained DOFs |
|---|---|---|---|---|
| N1 | 0.0 | 0.0 | Pin | UX, UY |
| N2 | 3.0 | 0.0 | Free | — |
| N3 | 6.0 | 0.0 | Free | — |
| N4 | 9.0 | 0.0 | Free | — |
| N5 | 12.0 | 0.0 | Vertical roller | UY |
| N6 | 1.5 | 3.0 | Free | — |
| N7 | 4.5 | 3.0 | Free | — |
| N8 | 7.5 | 3.0 | Free | — |
| N9 | 10.5 | 3.0 | Free | — |
| Load | Node | FX | FY | Moment |
|---|---|---|---|---|
| L1 | N6 | 0.000 kN | -12.000 kN | 0.000 kN·m |
| L2 | N7 | 0.000 kN | -12.000 kN | 0.000 kN·m |
| L3 | N8 | 0.000 kN | -12.000 kN | 0.000 kN·m |
| L4 | N9 | 0.000 kN | -12.000 kN | 0.000 kN·m |
Sign convention: Positive axial force is tension. Negative axial force is compression. Positive global Y is upward.
02 · Structural idealization
The analytical model isolates the in-plane axial response of an ideal pin-jointed truss. It is an analysis benchmark, not a completed member or connection design.
Use the calculated forces to inform design checks, but separately verify buckling, yielding, connections, bracing and code compliance for the real structure.
03 · Step-by-step hand calculation
The determinate truss can be solved from equilibrium alone. The selected joint and section checks below demonstrate the sign and magnitude of key member forces without reproducing every joint calculation.
The count m + r = 2j satisfies the necessary condition for static determinacy. Inspection of the repeating triangular geometry and pin-and-roller support arrangement confirms that the truss is stable and has no internal or external mechanism.
M11 and M12 are mirror-symmetric about midspan, so their axial forces are equal. At unloaded joint N3, their horizontal components cancel while their vertical components add. Because M2 and M3 are horizontal and no external vertical load acts at N3, ΣFy = 2F sin θ = 0 and both centre diagonals carry zero force.
Apply a 1 kN downward virtual load at N3. Using the real member forces Ni, virtual member forces ni, member lengths Li and EA = 1,000,000 kN, the virtual-work sum gives the displacement in the direction of the unit load.
The result is positive in the direction of the downward unit load; in global coordinates, uy = -0.453246 mm because global +Y is upward.
Symmetric members with identical Ni, ni and Li are grouped. The contribution includes every member listed in the row.
| Members | Ni (kN) | ni (kN) | Li (m) | Group contribution (mm) |
|---|---|---|---|---|
| M1, M4 | +12.000 | +0.250000 | 3.000 | 0.018000 |
| M2, M3 | +24.000 | +0.750000 | 3.000 | 0.108000 |
| M5, M7 | -18.000 | -0.500000 | 3.000 | 0.054000 |
| M6 | -24.000 | -1.000000 | 3.000 | 0.072000 |
| M8, M15 | -26.833 | -0.559017 | 3.354 | 0.100623 |
| M9, M14 | +13.416 | +0.559017 | 3.354 | 0.050312 |
| M10, M13 | -13.416 | -0.559017 | 3.354 | 0.050312 |
| M11, M12 | 0.000 | +0.559017 | 3.354 | 0.000000 |
| Total displacement in the direction of the unit load | 0.453246 | |||
04 · Optimal Beam Solver results
The Optimal Beam Solver uses the direct stiffness method to reproduce the determinate force solution and calculate the displacement response from the assigned EA values.
| Node | RX | RY |
|---|---|---|
| N1 | 0.000 kN | 24.000 kN |
| N5 | 0.000 kN | 24.000 kN |
| Node | UX | UY |
|---|---|---|
| N1 | 0.000 | 0.000 |
| N2 | 0.036 | -0.331 |
| N3 | 0.108 | -0.453 |
| N4 | 0.180 | -0.331 |
| N5 | 0.216 | 0.000 |
| N6 | 0.198 | -0.200 |
| N7 | 0.144 | -0.435 |
| N8 | 0.072 | -0.435 |
| N9 | 0.018 | -0.200 |
05 · Hand-versus-solver verification
The reaction, selected member-force and unit-load displacement calculations are independently stated above, then compared with the stored Optimal Beam Solver snapshot.
| Check | Hand calculation | Optimal Beam | Difference | Status |
|---|---|---|---|---|
| Left vertical reaction R_Ay | 24.000000 kN | 24.000000 kN | 0.000% | Pass |
| Right vertical reaction R_By | 24.000000 kN | 24.000000 kN | 0.000% | Pass |
| Bottom chord M2 | 24.000000 kN | 24.000000 kN | 0.000% | Pass |
| Top chord M5 | -18.000000 kN | -18.000000 kN | 0.000% | Pass |
| Diagonal M10 | -13.416408 kN | -13.416408 kN | 0.000% | Pass |
| Zero-force diagonal M11 | 0.000000 kN | 0.000000 kN | 0.000% | Pass |
| Centre vertical displacement N3 | -0.453246 mm | -0.453246 mm | 0.000% | Pass |
| Vertical equilibrium | 48.000 kN load | 48.000 kN reaction | 0.000% | Pass |
06 · Common modeling mistakes
These are not cosmetic differences. Each one changes the idealization, restraint count, load path or interpretation of the result.
Applying a distributed load directly to an ideal truss member instead of converting it to panel-point loads.
Assuming every Warren diagonal alternates force sign without checking the actual geometry and load positions.
Missing the two zero-force centre diagonals created by this symmetric load case.
Drawing crossing members without creating a shared node where a connection is intended.
Restraining both supports horizontally and unintentionally introducing an extra reaction.
Using the axial-force result as a strength, buckling or connection design check.
Continue exploring
Each variation starts from the same verified model. Change one modeling decision at a time so the resulting force and displacement changes remain explainable.
Apply a downward load at N3 and observe how the two zero-force centre diagonals become active.
Open base model → Editable variationOpen the Pratt benchmark and compare its vertical-and-diagonal web with the repeating Warren triangles.
Open base model → Editable variationKeep the determinate member forces unchanged and observe how axial displacement varies with EA.
Open base model →Use the exact benchmark
The page, editable calculator model, structural diagram, tables, downloadable JSON and regression test all use this same reviewed example record.