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Truss & Frame Solver (Stiffness Method)
Forces, moments, reactions and deflections of 2-D/3-D trusses, beams and frames.
Try an example: Simply supported steel beam with a uniform load (5wL⁴/384EI) · Propped cantilever: statically indeterminate (prop reaction 3wL/8) · Hibbeler Statics Example 6.1: three-bar truss with a 500 N load · Warren roof truss, 12 m span, with buckling check · Portal frame with a hinge and wind load · 3-D tripod: three legs under a vertical load
How it works
Describe a truss, a continuous beam or a 2-D frame line by line — nodes, a steel or timber section, members with optional hinges, pin, roller and fixed supports, springs, point loads and distributed loads — and get the node displacements, the support reactions, the axial force, shear and bending moment of every member, the stress and a utilisation check against the yield strength, an Euler buckling check of the compression members and an equilibrium check. Diagrams show the member forces in colour, the deformed shape and the bending moments. Solved with the direct stiffness method and a sparse Cholesky solver; part of the Proofline math notebook. A checking and teaching model, not certified design.
- Direct stiffness method: bar elements (axial stiffness EA/L along the member direction) for trusses in 2-D and 3-D; Euler–Bernoulli beam-column elements (EA/L, 12EI/L³, 6EI/L², 4EI/L, 2EI/L) for 2-D frames, transformed to global axes and assembled into a sparse matrix.
- Uniform member loads enter as consistent fixed-end forces (wL/2, wL²/12). Hinges release the end moment by static condensation of the element; the released rotation is recovered afterwards. A node where only hinged or pin-ended members meet has an undefined rotation, which is held fixed.
- The system is reordered with reverse Cuthill–McKee and solved by an envelope Cholesky factorisation; a failing pivot means the structure is a mechanism and names the free direction.
- Member forces from the element end forces; shear and moment along each member from equilibrium, deflections from Hermite shape functions plus the exact fixed-end solution of the uniform load — exact for Euler–Bernoulli beams.
- Checks (not design-code verifications): stress |N|/A + |M|/W against fy, Euler load Pcr = π²EI/(KL)² for compression members, span deflection against L/limit, and the sum of reactions against the applied loads.
- Verified against Hibbeler Statics Example 6.1 (forces 500 N T, 707.1 N C, 500 N T), PL³/3EI and 5wL⁴/384EI, the propped cantilever (3wL/8, wL²/8), the symmetric three-bar truss P/(1 + 2cos³θ), a tripod and mechanism detection.
Not certified engineering. A linear-elastic, first-order model for checking and teaching: no safety factors, second-order (P-Δ) effects, imperfections, lateral-torsional or local buckling, shear deformation, connection design or code combinations. Have structures that matter designed by a qualified engineer.
Sources
- W. McGuire, R. H. Gallagher, R. D. Ziemian: Matrix Structural Analysis, 2nd ed. (Wiley, 2000), chapters 2–4 and 7 (Wiley / Cornell eCommons)
- R. C. Hibbeler: Structural Analysis (Pearson), chapters 14–16 (stiffness method), and Engineering Mechanics: Statics, Example 6.1 (method of joints) (Pearson)
- A. George, J. W. H. Liu: Computer Solution of Large Sparse Positive Definite Systems (Prentice-Hall, 1981) — reverse Cuthill–McKee and envelope Cholesky (Prentice-Hall)
- S. P. Timoshenko, J. M. Gere: Theory of Elastic Stability, 2nd ed. (McGraw-Hill, 1961), §2.1–2.5 (Euler buckling, effective length) (McGraw-Hill)
Part of Proofline
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Proofline Math notebook
Solve, graph, test and visualise in one notebook — with data, findings and a report.
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Solve trusses, beams and frames with the stiffness method
Describe a truss, beam or frame line by line and get displacements, reactions, member forces, bending moments, deflections and a buckling check.
What it is for
A roof truss with fifteen bars, a beam fixed at one end or a portal frame with a hinge quickly becomes a page of algebra. This tool solves them with the direct stiffness method, the method structural analysis programs use.
You write the structure one statement per line: node A 0 0, material steel E=210 fy=235, section IPE200 A=28.5 I=1943 W=194, member M1 A B, support A pin and loads such as load B Fy=-20 or udl M1 qy=-5. Three types are available: 2-D trusses with pin-jointed bars, 3-D space trusses, and 2-D frames whose members also carry bending. Members can have hinges, supports can be springs.
It is a linear-elastic, first-order model for checking: no design code, no second-order effects, no plasticity.
Worked example
Take a 6 m steel IPE 200 beam under 5 kN/m. With support A pin and support B roller the tool reports a largest displacement of 20.68 mm at 3 m, exactly 5wL⁴/384EI. Now change the left support to support A fixed: the beam becomes a propped cantilever. The largest displacement drops to 8.576 mm at 3.6 m, and the reactions are 18.75 kN plus a 22.5 kN·m fixed-end moment at A and 11.25 kN at B. Those are the textbook values 5wL/8, wL²/8 and 3wL/8. The bending stress at the fixed end, 22.5 kN·m ÷ 194 cm³, is 116 MPa, a utilisation of 0.494 against 235 MPa. The span ratio L/δ is 699.6, well inside an L/250 limit.
The Warren truss example (12 m span, four 10 kN panel loads, steel tubes with A = 7.07 cm²) gives 20 kN at each support, 30 kN tension in the middle bottom chords and 30 kN compression in the top chord member GH. Its Euler load is 64.9 kN at a slenderness of 150, so N/Pcr = 0.46. The two verticals near midspan carry zero force.
How to read the result
The member table lists the axial force (positive is tension), the largest shear and moment for beams, the largest stress N/A ± M/W and the utilisation against the yield strength. The Euler ratio N/Pcr is only filled in for compression members.
The equilibrium check adds the applied loads and the reactions in x, y and moment. All three should be near zero; a larger number means a support or a load was entered wrong.
The diagram draws tension members solid and compression members dashed, with an exaggerated deformed shape. Frames get a bending moment diagram on the tension side and a table of span deflections with L/δ.
Tips and common mistakes
- A truss needs enough supports: in 2-D one
pinplus oneroller. Too few raise a "mechanism" error that names the free direction. - Use
trusson a member in a frame for bracing that carries only axial force. - Section values are in cm² and cm⁴ as printed in steel tables; lengths are in metres and loads in kN.
- A hinge in the middle of a span is a hinge at the end of one member:
member BEAM B C hinge=end. - The buckling check uses the member length times K. Reduce K for braced members only if the bracing really exists.
FAQ
Is this enough to design a steel structure?
No. Design codes add load factors, lateral-torsional buckling, connection checks and more, so have a qualified engineer check anything that matters.
Why does my truss show a mechanism?
The stiffness matrix is singular: some joint or the whole structure can move without straining a bar. Add a support or a diagonal, or check that every node is connected.
How accurate are the beam results?
Each frame member is an exact Euler–Bernoulli beam element, so point loads and uniform loads give exact displacements and moments at the stations, matching the textbook formulas to rounding.
Can it handle 3-D frames?
3-D trusses yes, 3-D frames with bending no. Model a plane frame per direction or use specialist software.
Guide last updated:
For scripts and AI agents
- JSON API:
POST https://prooflinetrussframesolver.usefulweb.site/api/v1/calculate(same calculation as this page) - Interface description: OpenAPI 3.1 · AI guide: llms.txt
- MCP server (Streamable HTTP):
https://prooflinetrussframesolver.usefulweb.site/mcp· tool nameproofline_truss_frame_solver
Example request
curl -s -X POST https://prooflinetrussframesolver.usefulweb.site/api/v1/calculate \
-H 'content-type: application/json' \
-d '{"type":"frame2d","model":"# IPE 200, 6 m span, 5 kN/m\nnode A 0 0\nnode B 6 0\nmaterial steel E=210 fy=235\nsection IPE200 A=28.5 I=1943 W=194\nmember M1 A B\nsupport A pin\nsupport B roller\nudl M1 qy=-5"}'