Truss Analyzer: Method of Joints Solver for Member Forces

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How to use this truss analyzer

Start by picking a truss type and setting the span, the height and the number of panels. Press Generate truss and the tool lays out every joint, every member and a set of joint loads for you. That is usually enough for coursework and for a quick sanity check on a real frame.

If your frame is not a textbook shape, treat the generated truss as a starting point and edit the three tables directly. The node table controls position and support type. The member table connects any two nodes. The load table applies a horizontal and a vertical force at any joint. Every edit re solves the whole frame immediately, so you can watch a member swing from tension into compression as you drag a support or move a load.

Support types matter more than people expect. A pin holds the joint in both directions and supplies two reaction components. A vertical roller supplies one upward or downward reaction and lets the joint slide horizontally. A horizontal roller does the opposite. A normal simply supported truss needs one pin and one vertical roller, which is three reaction components in total.

Sign convention is the usual one. A positive member force is tension, meaning the member is being pulled and is trying to stretch. A negative force is compression, meaning the member is being squashed. Loads follow the same axes as the geometry, so a downward load is a negative vertical force.

What the method of joints is actually doing

A pin jointed truss carries load through axial forces only. No member bends, because a frictionless pin cannot transmit a moment. That single assumption is what makes hand analysis possible.

Cut any joint free from the rest of the frame and you are left with a small set of forces meeting at a point: the axial force in each member attached to that joint, plus any applied load and any support reaction. Forces meeting at a point cannot produce a moment about that point, so the moment equation is automatically satisfied and you are left with two useful equations, sum of horizontal forces equals zero and sum of vertical forces equals zero.

Two equations solve at most two unknowns. So the method works by finding a joint with no more than two unknown member forces, solving it, then moving to the next joint where the newly known forces have reduced the count to two again. The tool shows that march in the step by step panel, in the same order you would do it on paper.

Whether the march can finish at all is a counting question. If a truss has m members, r reaction components and j joints, then the joint equations number 2j and the unknowns number m plus r. When m plus r equals 2j the truss is statically determinate and the method of joints alone gives a unique answer. When m plus r is greater than 2j the truss is indeterminate and equilibrium is no longer enough, because the load shares out according to how stiff each member is. When m plus r is less than 2j you have a mechanism, and no amount of algebra will help until you add a member or a support.

Reading tension, compression and the members that carry nothing

Tension and compression are not just signs on a page. A tension member can be slender, because a cable in tension does not care how thin it is until the steel yields. A compression member has to survive buckling long before the material reaches its strength, and buckling depends on length and on the shape of the cross section rather than on area alone. That is why in a real truss the compression chords and struts are chunky while the tension ties can be light.

This is also why the largest compression force in the results table usually deserves more attention than the largest tension force. Take that member length and its end conditions into a buckling check before you size it.

Zero force members show up in almost every real truss and they confuse people constantly. A member carrying no force under the current load case is not a mistake and it is not spare metal. Two situations produce them. First, at a joint where only two members meet and no load is applied, both members must be zero unless they happen to be in line. Second, at a joint where three members meet, two of them in line and no load applied, the odd member out is zero. Those members are there to brace the compression chord against buckling out of plane, to hold the geometry during erection, and to carry load under a different load case such as wind from the other side. Change the loading in the table and watch them wake up.

A worked check you can repeat on paper

Use the default setup: a Pratt truss with a 12 m span, 4 m height, four panels and 10 kN at each of the three loaded bottom joints. Total load is 30 kN and the frame is symmetric, so each support carries 15 kN.

Now take joint L0 at the left support on its own. Only two members reach it, the bottom chord L0L1 which runs horizontally and the end diagonal L0U1 which rises 4 m over 3 m, so its direction cosines are 0.6 horizontal and 0.8 vertical. Vertical equilibrium at that joint reads 15 plus 0.8 times the force in L0U1 equals zero, which gives 18.75 kN compression. Horizontal equilibrium then reads the force in L0L1 plus 0.6 times minus 18.75 equals zero, giving 11.25 kN tension. Both numbers appear in the results table.

Move to joint L1. Three members meet there, two collinear bottom chords and one vertical, and a 10 kN load hangs from it. Vertical equilibrium immediately gives 10 kN tension in the vertical. Horizontal equilibrium says the two bottom chord forces are equal, so L1L2 is also 11.25 kN tension.

The centre vertical L2U2 comes out as exactly zero, which is the classic result for this arrangement and a good check that the solver is behaving. Add a load at the centre top joint U2 instead and that vertical picks up force at once.

You can verify any chord force independently with the method of sections. Cut the truss vertically between L1 and L2, keep the left hand piece, and take moments about the top joint U1. The 15 kN reaction acts 3 m away and the 10 kN load sits directly under U1 with no lever arm, so the bottom chord force times the 4 m height equals 45, giving 11.25 kN. The two methods agree because both are simply equilibrium applied to different free bodies.

Mistakes that produce wrong member forces

The most common error is treating a real frame with continuous chords and welded joints as a pin jointed truss and then being surprised by bending. The truss idealisation is still the right starting point, since axial force dominates, but a welded chord will attract secondary bending moments that a pin model cannot show.

The second is loading a member at midspan rather than at a joint. Every load in this tool goes to a joint, which is the correct model. If in reality a purlin sits halfway along a top chord, that chord also acts as a beam and needs a bending check on top of the axial force from this analysis.

The third is over restraining the supports. Two pins on a simple span look safer but give four reaction components against three equations, which makes the truss externally indeterminate. The tool will say so rather than quietly returning a wrong answer. In a real structure that extra restraint also means thermal expansion has nowhere to go.

The fourth is forgetting that the answer changes with the load case. A truss designed only for gravity load can find its wind uplift case reverses the sign in the diagonals, turning slender ties into struts that then buckle. Run each load case separately and size every member for the worst force it ever sees in either direction.

Related Calculators and Reference

• Section Properties Calculator: Moment of Inertia and Section Modulus, for sizing the chords once you know the forces
• Euler Buckling Calculator: Critical Load and Slenderness, for checking every compression member the analyzer reports
• Beam Deflection Calculator: Moment, Shear and Four Standard Cases, for chords that also carry load between joints
• A36 Steel Yield Strength and S355 Steel Yield Strength, for the capacity side of the check
• AISC Steel Construction Manual resources, for design rules on axially loaded members