How do I know whether a truss can be solved by hand?

Count the members m, the reaction components r and the joints j. If m + r equals 2j the truss is statically determinate and the method of joints will solve it from equilibrium alone. If m + r is less than 2j it is a mechanism and will move. If m + r is greater than 2j it is indeterminate and the member forces depend on stiffness (EA), so hand methods need compatibility as well. The count is necessary but not sufficient: a truss can satisfy m + r = 2j and still be unstable if the members are badly arranged.

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

When you only need one or two member forces rather than all of them. The method of joints is serial — every force depends on the one before it — so reaching a member in the middle of a long truss means solving everything up to it. The method of sections cuts straight through the member you want, crossing no more than three members, and one moment equation taken about the point where the other two intersect gives it directly.

What is a zero-force member and can I just delete it?

A member carrying no force in the load case being analyzed. You should not delete it. It is usually there to brace a compression member, reducing its buckling length, and in a different load case — wind uplift, an unbalanced snow load, a moving vehicle — it may be one of the working members. StructureCalcs draws zero-force members in gray rather than hiding them.

Why does my truss member force come out negative?

Because it is in compression. StructureCalcs uses the tension-positive convention throughout, in the truss, frame, beam and section tools alike, so a member reported as minus 45 kN is carrying 45 kN of compression. This is also why the method of joints is taught with every unknown assumed to be in tension: a negative answer is then simply the answer, and no direction has to be guessed.

Can I apply a load between the joints of a truss?

Not without leaving truss theory behind. The pin-joint idealisation only holds when load is applied at the joints; a load landing mid-panel puts bending into the chord, which a truss analysis cannot see. Either add a joint at that point, model the structure as a frame instead, or check the chord separately as a beam spanning between its panel points.

How to analyze a truss: joints, sections and the stiffness method

A visual reference. Every truss on this page is solved by the same engine that runs the Truss Calculator, so the member forces and colors are the tool’s own answers. For full hand derivations, see the ten truss worked examples.

What a truss is, and what makes it efficient

A truss is a set of straight members meeting at joints. Load it only at those joints and every member carries pure axial force — pulled apart (tension) or pushed together (compression) — with no bending anywhere. That is the whole efficiency argument: bending uses a section unevenly, with the material near the neutral axis barely working, while axial force uses all of it at once.

Figure 1.The smallest stable truss: three members, three joints, 20 kN at the apex. Red members are in tension, blue in compression. The bottom chord is pulled apart (6.67 kN tension) while the two rafters are squashed — which is the behavior of every roof truss ever built, in miniature.

The pin-joint assumption is what makes this work. Real joints are welded or bolted and do carry some moment, but the members are slender and the analysis is not sensitive to it; the idealization is standard and, for member sizing, conservative. Members that carry only axial force are called two-force members, and that restriction is exactly why a truss is so much easier to solve than a frame.

One consequence catches people out: a load applied between joints breaks the assumption. Purlin loads hung mid-panel put bending into the chord, and that has to be checked separately or modeled as a frame instead.

The vocabulary, on a drawing

Figure 2.A four-panel Warren truss spanning 12 m, with three 30 kN loads on the bottom chord. Node numbers in circles, member numbers on the members — the same labeling the calculator uses.

Chords — the members running along the top and bottom. In a simply supported truss the top chord is normally in compression and the bottom in tension.
Web members — everything between the chords. Diagonals slope; verticals (posts) do not.
Panel — the bay between two adjacent joints on a chord.
Node / joint — where members meet, and where load must be applied.
Zero-force member — a member carrying nothing in this load case, kept for stability or to shorten a buckling length. The calculator draws these gray.

The named truss types differ only in how the web is arranged. A Warren alternates diagonals with no verticals; a Pratt puts its diagonals in tension and its verticals in compression; a Howe is the reverse; a Fink subdivides for roof pitches. Which one is right depends on which members you would rather have in tension — steel is happy in tension and buckles in compression, so a Pratt is common in steel bridges.

Determinacy: can this be solved by hand?

Count the members (m), the reaction components (r) and the joints (j). Each joint gives two equilibrium equations in a plane truss, so:

m + r < 2j — a mechanism. It moves. No amount of analysis will fix it, and the solver will refuse it rather than return numbers.
m + r = 2j — statically determinate. Equilibrium alone gives every force; the method of joints works.
m + r > 2j — indeterminate. The forces now depend on member stiffness (EA), so hand methods need compatibility as well.

The Warren truss above has m = 15, r = 3 (a pin and a roller) and j = 9, so m + r = 18 and 2j = 18. Determinate — it can be worked by hand joint by joint.

⚠ The count is necessary, not sufficient. A truss can satisfy m + r = 2j and still be unstable if the members are badly arranged — three collinear members at a joint, or a panel with no diagonal. Geometry has to be checked as well as arithmetic.

The method of joints

Isolate one joint at a time and apply ΣFx = 0 and ΣFy = 0. Two equations means you can only start at a joint with at most two unknown members — usually a support — and then work along the truss as each solved member becomes a known force at the next joint.

Find the support reactions from global equilibrium first.
Pick a joint with no more than two unknowns.
Assume every unknown member is in tension — pulling away from the joint. A negative answer then simply means compression, and you never have to guess a direction.
Resolve horizontally and vertically; solve the two equations.
Move to the next joint that now has two or fewer unknowns, and repeat.

Figure 3.The three-member truss with its solved forces. Starting at the pinned joint 1 there are exactly two unknown members, which is what makes it the right place to begin. Signs follow the tension-positive convention, so the bottom chord comes out as +6.67 kN (tension) and each rafter as -12.02 kN — the minus sign IS the answer “compression”.

The method is exact but serial: every force depends on the one before it, so an arithmetic slip early on propagates silently to the end. That is the practical argument for checking against a solver, and the worked example does exactly that, line by line.

The method of sections

When you only need one or two member forces — the maximum bottom-chord tension, say, in the middle of a long bridge truss — the method of joints makes you solve everything up to it. The method of sections cuts straight there instead.

Cut an imaginary line through the truss, crossing the member you want and no more than three members in total.
Discard one side and treat the other as a rigid body.
Apply the three plane equilibrium equations: ΣFx, ΣFy, ΣM.
Take moments about the point where two of the three cut members intersect — that eliminates both in one line and gives the third directly.

The moment trick is the whole method. For the Warren truss above, one cut through the center panel and one moment equation about the top-chord joint gives the peak bottom-chord tension of 65.6 kN without touching any other member. The peak compression in the structure is 75.0 kN. Both are read straight off the figure below.

Figure 4.The Warren truss solved: tension red, compression blue, with the deflected shape dashed. The bottom chord is in tension throughout, the top chord in compression, and the diagonals alternate — the signature of the type.

What the stiffness method adds

The classical methods stop at determinate trusses and give forces but not displacements. The matrix stiffness method — what this site’s engine actually runs — has neither limit. Each member contributes a stiffness EA/L along its own axis, rotated into global coordinates; those contributions assemble into a global stiffness matrix K; and the whole structure is solved at once from K d = F, giving joint displacements first and member forces from them.

Three things become possible that hand methods cannot reach:

Indeterminate trusses — extra members and extra supports are no harder than the determinate case.
Support settlement — a support that moves induces forces in an indeterminate truss even with no load applied. See the settled pier example.
Fabrication errors and temperature — a member made 5 mm too short, or a chord in the sun, forces the structure into a state before any load arrives. See lack of fit and thermal stress.

All three need member areas, not just geometry — in a determinate truss the forces are independent of EA, but the moment the structure is indeterminate, stiffness decides how load is shared. Pick sections from the built-in library and the areas come with them.

Mistakes worth knowing about

Loading between joints. The pin-joint idealization only holds for joint loads. A mid-panel load bends the chord and needs a frame model or a separate local check.
Forgetting the sign convention. Tension is positive throughout this site. A compression member reported as −45 kN is not an error.
Ignoring buckling. Analysis returns the compression force; it does not tell you the member survives it. A long slender compression member fails well below its squash load, and that check is a separate design step.
Dismissing zero-force members. They carry nothing in this load case. Change the load and they may be the ones working — and they often exist to brace a compression member anyway.
Under-restraining. Two rollers and no pin leaves the truss free to slide sideways. The solver cannot invert a singular matrix and will say so rather than invent an answer.

Where to go next

Warren truss and Pratt bridge truss — the method of joints and sections, worked in full.
Fink roof truss with gusset joints — how joint detailing is drawn and what it means.
Truss Calculator — draw your own; forces update as you type.
Frame Calculator — for structures whose joints carry moment, or which are loaded between joints.
How to analyze a beam — the other half of the fundamentals.

Frequently asked questions

How do I know whether a truss can be solved by hand?: Count the members m, the reaction components r and the joints j. If m + r equals 2j the truss is statically determinate and the method of joints will solve it from equilibrium alone. If m + r is less than 2j it is a mechanism and will move. If m + r is greater than 2j it is indeterminate and the member forces depend on stiffness (EA), so hand methods need compatibility as well. The count is necessary but not sufficient: a truss can satisfy m + r = 2j and still be unstable if the members are badly arranged.
When should I use the method of sections instead of the method of joints?: When you only need one or two member forces rather than all of them. The method of joints is serial — every force depends on the one before it — so reaching a member in the middle of a long truss means solving everything up to it. The method of sections cuts straight through the member you want, crossing no more than three members, and one moment equation taken about the point where the other two intersect gives it directly.
What is a zero-force member and can I just delete it?: A member carrying no force in the load case being analyzed. You should not delete it. It is usually there to brace a compression member, reducing its buckling length, and in a different load case — wind uplift, an unbalanced snow load, a moving vehicle — it may be one of the working members. StructureCalcs draws zero-force members in gray rather than hiding them.
Why does my truss member force come out negative?: Because it is in compression. StructureCalcs uses the tension-positive convention throughout, in the truss, frame, beam and section tools alike, so a member reported as minus 45 kN is carrying 45 kN of compression. This is also why the method of joints is taught with every unknown assumed to be in tension: a negative answer is then simply the answer, and no direction has to be guessed.
Can I apply a load between the joints of a truss?: Not without leaving truss theory behind. The pin-joint idealisation only holds when load is applied at the joints; a load landing mid-panel puts bending into the chord, which a truss analysis cannot see. Either add a joint at that point, model the structure as a frame instead, or check the chord separately as a beam spanning between its panel points.