Add joints and members to solve.
How to use the truss calculator
Each panel on the left holds one part of a truss. Open a panel, fill in its fields, and Apply to commit. The diagram redraws as you type.
- Add the joints. Open Nodes and enter the coordinates of every joint in the figure. The origin can be placed anywhere; only the distances between joints affect the answer.
- Connect them into members. Open Members and join the joints into bars, either by clicking two of them on the grid or by typing their IDs. One bar per line in the drawing.
- Add the supports. Open Supports and place a pin or a roller on each restrained joint. A pin gives two reaction components, a roller one.
- Apply the loads. Open Point Loads and enter each force as components: a 10 kN weight is a y magnitude of −10. A load spread along a bar goes in Distributed Loads.
- Read the answer. Member forces are solved as the truss is built. The diagram labels each bar with its magnitude and a T or a C, with thicker bars carrying more, and the tables below list the forces and the support reactions.
Fields that name a joint or a bar can be filled by clicking the diagram instead of typing, so a member takes two clicks. Joints can also be dragged to change the geometry directly.
How it solves: the method of joints
Each joint gives two equations, requiring that the horizontal and vertical forces on it sum to zero. The unknowns are the axial force in every member plus each restrained reaction component. That is a square system exactly when m + r = 2j, with m members, r reaction components and j joints, the standard determinacy condition.
Solving all the joint equations together, rather than one joint at a time, removes the need for a starting joint with only two unknowns. The result is identical to the hand method.
When a truss cannot be solved, and why
If the determinacy count does not work out, or the arrangement cannot carry load, the calculator reports which of the three cases applies rather than returning a result.
- Mechanism: not enough constraints (m + r < 2j)
- Part of the structure would move under load. Check for an untriangulated panel, a joint attached by a single bar, or a missing support.
- Statically indeterminate (m + r > 2j)
- More unknowns than equilibrium equations. Statics alone cannot find these forces; they depend on the stiffness of each member. Remove a redundant bar or support and it solves.
- Unstable, despite the count (m + r = 2j)
- Three collinear bars, a joint free to swing, or supports that let the whole truss slide or rotate. The equations come out singular.
Sign convention, tension and compression, units
A positive member force is tension: the bar pulls its end joints towards each other, and the diagram marks it T. A negative force is compression, marked C: the bar pushes its joints apart. Zero-force members are drawn plain and reported as zero.
Loads are entered as components rather than a magnitude and an angle, with positive x to the right and positive y up, so a 10 kN weight is a y magnitude of −10. Lengths and forces carry whichever unit pair is selected and are never converted, so one problem should use one consistent pair.
Distributed loads, and why there is no moment input
A deck, a roof sheet or a member’s own weight arrives spread along a bar rather than at a joint. Since a pin-jointed truss carries no bending, a uniform load w over a bar of length L is replaced by half of its total at each end joint, wL/2. This is the standard treatment before the method of joints is applied. The equivalent joint loads are drawn on the diagram and listed in the results, so the conversion can be checked.
There is no moment input for the same reason: a frictionless pin transmits no couple, and that assumption is what leaves two equations and only axial unknowns at every joint. A structure with applied moments or rigid connections is a frame, not a truss.
Built-in examples
The Examples panel loads a truss that already solves. All of them are statically determinate, and opening a panel afterwards shows the values that produced the diagram.
- Warren truss, four panels. Nine joints and fifteen bars on a pin and a roller.
- King post roof truss. A ridge load carried down a pair of rafters and hung from the tie beam.
- Single triangle. The smallest determinate truss, useful for checking the sign convention.
- Cantilever bracket. Two bars off a wall, one in tension and one in compression.
- Deck load on a Pratt panel. A uniform 6 kN/m along the bottom chord, resolved to the joints.
Frequently asked questions
Is this truss calculator free?
Yes. It runs in the browser, requires no sign-up, and no input is uploaded.
What method does it use to solve the truss?
The method of joints, assembled as one linear system rather than joint by joint. Every joint contributes two equilibrium equations, horizontal and vertical, and the unknowns are the axial force in each member plus each restrained reaction component. Solving them together gives the same answer as working joint to joint by hand, without needing a starting joint that has only two unknowns.
Why does it say my truss is statically indeterminate?
It has more unknowns than equations: m + r > 2j, where m is the number of members, r the number of reaction components and j the number of joints. Statics alone cannot find those forces, because they depend on the stiffness of each member. Removing a redundant bar or support will make it solvable.
Why does it say there are not enough constraints?
The opposite count: m + r < 2j, so part of the structure is a mechanism and would move under load rather than carry it. Check for an untriangulated panel, a joint attached by a single bar, or a missing support.
What does it mean that the counts match but the truss is unstable?
The determinacy count is necessary but not sufficient. Three collinear bars, a joint free to swing, or supports that let the whole truss slide or rotate all give a singular set of equations even when m + r = 2j. These cases are reported rather than solved.
What do T and C mean, and what is the sign convention?
A positive member force is tension (T): the bar pulls its two end joints towards each other. A negative force is compression (C): the bar pushes them apart. For loads and reactions, positive x points right and positive y points up, so a downward load is negative.
Can it handle distributed loads?
Yes, on a member. A pin-jointed truss cannot carry bending, so a uniform run of load w along a bar of length L is replaced by half of its total at each end joint, wL/2. This is the standard treatment of a deck or roof load before the method of joints is applied. The equivalent joint loads are listed in the results.
Can it apply a moment at a joint, or analyse a frame or a beam?
No. A frictionless pin transmits no couple, which is the assumption that leaves two equations and only axial unknowns at each joint. A structure with applied moments, rigid connections or members carrying bending is a frame or a beam, and requires a different analysis.
Which units does it use?
Metres, millimetres, centimetres, feet or inches for length, and kN, N, lb or kip for force. Units are labels only and are not converted, so enter one problem in one consistent pair and read the answer in the same pair.
My answer differs from the textbook. What should I check?
The solve is independent of any published answer, so a difference usually means the input differs: a coordinate with the wrong sign, a missing diagonal, a roller where the problem has a pin, or a downward load entered as positive. The counts in the rail, joints against members plus reactions, are the quickest way to spot a missing bar.