Worked structural analysis example
Solve a 12 m Howe truss step by step, including support reactions, axial member forces, a zero-force member and deflection.
Howe truss fundamentals
A Howe truss is a triangulated structure with top and bottom chords connected by vertical and diagonal web members. Its characteristic interior diagonals slope up toward the centre of the span. Under typical downward gravity loading, those longer interior diagonals usually act in compression, while the bottom chord and some vertical members act in tension. The exact force pattern still depends on the geometry, supports and load positions.
Howe truss analysis converts the truss geometry, support conditions and joint loads into support reactions, axial member forces and, when stiffness properties are included, nodal deflections. A statically determinate pin-jointed Howe 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.
For this symmetric gravity-load example, the bottom chord and outer verticals are in tension, while the top chord, end posts and centre-sloping interior diagonals are in compression. The centre vertical is a zero-force member. Reversing or relocating loads can change that pattern.
The interior diagonals slope up toward midspan and transfer panel shear between the chords. For the downward joint loading used here, joint equilibrium puts both centre-sloping diagonals into compression. A different load direction or arrangement can reverse their force sign.
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 Howe truss spans 12 m and is 3 m high. Three 16 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 | 3.0 | 3.0 | Free | — |
| N7 | 6.0 | 3.0 | Free | — |
| N8 | 9.0 | 3.0 | Free | — |
| Load | Node | FX | FY | Moment |
|---|---|---|---|---|
| L1 | N6 | 0.000 kN | -16.000 kN | 0.000 kN·m |
| L2 | N7 | 0.000 kN | -16.000 kN | 0.000 kN·m |
| L3 | N8 | 0.000 kN | -16.000 kN | 0.000 kN·m |
Sign convention: Positive axial force is tension. Negative axial force is compression. Positive global Y is upward.
02 · 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 triangulated geometry and pin-and-roller support arrangement confirms that the truss is stable and has no internal or external mechanism.
At N3, the two bottom-chord members M2 and M3 are collinear and no external load acts at the joint. The non-collinear centre vertical M10 is therefore a zero-force member.
Apply a 1 kN downward virtual load at N7. 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 = -0.631529 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 | +24.000 | +0.500000 | 3.000 | 0.072000 |
| M2, M3 | +32.000 | +1.000000 | 3.000 | 0.192000 |
| M5, M6 | -24.000 | -0.500000 | 3.000 | 0.072000 |
| M7, M13 | -33.941 | -0.707107 | 4.243 | 0.203647 |
| M8, M12 | +8.000 | +0.500000 | 3.000 | 0.024000 |
| M9, M11 | -11.314 | -0.707107 | 4.243 | 0.067882 |
| M10 | 0.000 | 0.000000 | 3.000 | 0.000000 |
| Total displacement in the direction of the unit load | 0.631529 | |||
03 · 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 | 0.072 | -0.468 |
| N3 | 0.168 | -0.632 |
| N4 | 0.264 | -0.468 |
| N5 | 0.336 | 0.000 |
| N6 | 0.240 | -0.444 |
| N7 | 0.168 | -0.632 |
| N8 | 0.096 | -0.444 |
04 · 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 | 32.000000 kN | 32.000000 kN | 0.000% | Pass |
| Top chord M5 | -24.000000 kN | -24.000000 kN | 0.000% | Pass |
| Diagonal M9 | -11.313708 kN | -11.313708 kN | 0.000% | Pass |
| Zero-force vertical M10 | 0.000000 kN | 0.000000 kN | 0.000% | Pass |
| Centre vertical displacement N7 | -0.631529 mm | -0.631529 mm | 0.000% | Pass |
| Vertical equilibrium | 48.000 kN load | 48.000 kN reaction | 0.000% | Pass |
05 · 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.
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.
Drawing the interior diagonals in the Pratt direction while intending to model a Howe truss.
Assuming the Howe diagonals are in tension under this gravity load instead of checking their force sign.
Missing the zero-force centre vertical M10 at the unloaded bottom joint.
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.
Open the matching Pratt benchmark and compare its reversed diagonals under the same geometry and loading.
Open base model → Editable variationApply a downward load at N3 and observe how the zero-force centre vertical M10 becomes active.
Open base model → Editable variationKeep the determinate member forces unchanged and observe how axial displacement varies with EA.
Open base model →Truss guide library
Move between verified examples built to the same span, depth, supports and loading basis.
Use the exact benchmark
The page, editable calculator model, structural diagram, tables, downloadable JSON and regression test all use this same reviewed example record.