How do I know whether a truss is statically determinate?

A plane truss gives two equilibrium equations per joint, 2j in total, and the unknowns are the m member forces plus r support reactions. m + r = 2j is necessary for determinacy, m + r > 2j means it is indeterminate to degree m + r − 2j, and m + r < 2j is always a mechanism. Counting alone is not enough, though: if the three reactions are parallel or meet at one point, or a joint is held only by collinear members, the truss can still move. This tool computes the rank of the equilibrium matrix directly and flags the truss as unstable, with the number of free motions, whenever the rank is below 2j.

When should I use the method of joints versus the method of sections?

The method of joints isolates one joint at a time and has only two equations per joint, so you must start where at most two member forces are unknown and work outwards; it is the natural way to get every member force. The method of sections cuts through the truss and uses three equations on one side (usually a moment about a clever point), which is better for a few members in the middle, or for trusses like the K-truss where every joint has three or more unknowns and the method of joints gets stuck. The tool tells you when that happens, and the table still shows the stiffness-method values.

What is a zero-force member, and can I remove it?

A zero-force member carries no axial force under the current loads. Two common cases: an unloaded joint with only two non-collinear members makes both of them zero-force, and an unloaded joint with three members, two of them collinear, makes the third one zero-force. They carry nothing for this load case, but they brace compression members against buckling and pick up force as soon as the loads change, so real structures usually keep them.

Why check compression members for Euler buckling?

A slender compression member often bows sideways and fails long before the material yields. For pinned ends the critical load is P_cr = π²EI/L², inversely proportional to length squared and proportional to the second moment of area. That is why a tube resists buckling far better than a solid bar of the same area, and why long top chords of bridges are usually governed by stability rather than strength. The tool computes both the yield and buckling safety factors for each compression member, takes the smaller one and says which failure comes first.

What simplifications does the calculation make?

It is an ideal truss: frictionless pinned joints, loads applied only at joints, members carrying axial force only, self-weight ignored (it is listed separately for reference) and no secondary bending from rigid joints; the material is linear elastic with small deflections. Buckling uses pinned ends with an effective length factor of 1 and no reduction for initial crookedness or residual stress, so real designs should apply the stability factors from a design code and keep a proper safety margin.

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🌉Truss Solver - Member Forces, Reactions, Method of Joints, Stability

Q: When should I use the method of joints versus the method of sections?

The method of joints isolates one joint at a time and has only two equations per joint, so you must start where at most two member forces are unknown and work outwards; it is the natural way to get every member force. The method of sections cuts through the truss and uses three equations on one side (usually a moment about a clever point), which is better for a few members in the middle, or for trusses like the K-truss where every joint has three or more unknowns and the method of joints gets stuck. The tool tells you when that happens, and the table still shows the stiffness-method values.

Q: What is a zero-force member, and can I remove it?

A zero-force member carries no axial force under the current loads. Two common cases: an unloaded joint with only two non-collinear members makes both of them zero-force, and an unloaded joint with three members, two of them collinear, makes the third one zero-force. They carry nothing for this load case, but they brace compression members against buckling and pick up force as soon as the loads change, so real structures usually keep them.

Q: Why check compression members for Euler buckling?

A slender compression member often bows sideways and fails long before the material yields. For pinned ends the critical load is P_cr = π²EI/L², inversely proportional to length squared and proportional to the second moment of area. That is why a tube resists buckling far better than a solid bar of the same area, and why long top chords of bridges are usually governed by stability rather than strength. The tool computes both the yield and buckling safety factors for each compression member, takes the smaller one and says which failure comes first.

Q: What simplifications does the calculation make?

It is an ideal truss: frictionless pinned joints, loads applied only at joints, members carrying axial force only, self-weight ignored (it is listed separately for reference) and no secondary bending from rigid joints; the material is linear elastic with small deflections. Buckling uses pinned ends with an effective length factor of 1 and no reduction for initial crookedness or residual stress, so real designs should apply the stability factors from a design code and keep a proper safety margin.

Build a plane truss with a few taps: place nodes, connect members, add pin or roller supports and drag out load arrows. The tool solves it with the direct stiffness method and shows every member in tension (blue) or compression (red) with thickness proportional to the force, along with the support reactions and an exaggerated deformed shape. It checks the rank of the equilibrium matrix to tell you whether the truss is determinate, indeterminate or a mechanism. Choose steel, aluminium or timber and a section area to get stresses, Euler buckling loads and safety factors, and for determinate trusses it writes out the method of joints step by step, just like a textbook.

Free · No sign-up · Runs in your browserLast updated: 2026-10-29

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How to use Truss Solver (2D Truss Analysis)

1Pick a preset (Pratt, Howe, Warren or K-truss bridge, Fink roof, crane boom, cantilever) or clear the canvas and draw your own
2Edit with the node, member, support and load tools; in select mode you can drag nodes
3Read the coloured member forces, reactions, deformed shape and determinacy check
4Tap a member to set its material, section and area, then check stress, buckling and safety factor in the table
5Open the method-of-joints steps to compare with hand calculations, or copy a share link or JSON to save the design

FAQ

How do I know whether a truss is statically determinate?: A plane truss gives two equilibrium equations per joint, 2j in total, and the unknowns are the m member forces plus r support reactions. m + r = 2j is necessary for determinacy, m + r > 2j means it is indeterminate to degree m + r − 2j, and m + r < 2j is always a mechanism. Counting alone is not enough, though: if the three reactions are parallel or meet at one point, or a joint is held only by collinear members, the truss can still move. This tool computes the rank of the equilibrium matrix directly and flags the truss as unstable, with the number of free motions, whenever the rank is below 2j.
When should I use the method of joints versus the method of sections?: The method of joints isolates one joint at a time and has only two equations per joint, so you must start where at most two member forces are unknown and work outwards; it is the natural way to get every member force. The method of sections cuts through the truss and uses three equations on one side (usually a moment about a clever point), which is better for a few members in the middle, or for trusses like the K-truss where every joint has three or more unknowns and the method of joints gets stuck. The tool tells you when that happens, and the table still shows the stiffness-method values.
What is a zero-force member, and can I remove it?: A zero-force member carries no axial force under the current loads. Two common cases: an unloaded joint with only two non-collinear members makes both of them zero-force, and an unloaded joint with three members, two of them collinear, makes the third one zero-force. They carry nothing for this load case, but they brace compression members against buckling and pick up force as soon as the loads change, so real structures usually keep them.
Why check compression members for Euler buckling?: A slender compression member often bows sideways and fails long before the material yields. For pinned ends the critical load is P_cr = π²EI/L², inversely proportional to length squared and proportional to the second moment of area. That is why a tube resists buckling far better than a solid bar of the same area, and why long top chords of bridges are usually governed by stability rather than strength. The tool computes both the yield and buckling safety factors for each compression member, takes the smaller one and says which failure comes first.
What simplifications does the calculation make?: It is an ideal truss: frictionless pinned joints, loads applied only at joints, members carrying axial force only, self-weight ignored (it is listed separately for reference) and no secondary bending from rigid joints; the material is linear elastic with small deflections. Buckling uses pinned ends with an effective length factor of 1 and no reduction for initial crookedness or residual stress, so real designs should apply the stability factors from a design code and keep a proper safety margin.

Features

Direct stiffness solver with rollers and loads in any direction
Determinacy, degree of indeterminacy and mechanism detection from the equilibrium matrix rank
Tension and compression colouring with force labels and an adjustable deformed shape
Steel, aluminium and timber with three section types: stress, Euler buckling, slenderness and safety factor
Step-by-step method of joints for determinate trusses (whole-truss equilibrium first, then joint by joint)
Seven classic presets plus import and export by URL or JSON

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