Truss Analysis & the Method of Joints

Pick a web pattern, a span, a depth and where the load stands. The page solves every member by the method of joints, marks the ones carrying nothing, and shows what the same bridge would do with its diagonals the other way round.

Worked example — Truss Analysis & the Method of Joints

Reaction at the roller	60.00 kN
Reaction at the pinned end	60.00 kN
First unknown at joint L0: L0–U0	-69.97 kN
Second unknown there: L0–L1	+36.00 kN
Heaviest tension, in L2–L3, L3–L4	+180.00 kN
Heaviest compression, in U2–U3	-216.00 kN
Members carrying nothing	0
Members + reactions − 2 × joints	0

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Truss Analysis & the Method of Joints

A truss is a frame of straight members joined at their ends, loaded only at those joints. Idealise the joints as frictionless pins and every member can carry only a force along its own length — tension if it is being pulled, compression if pushed — with no bending anywhere. That is what makes the whole thing solvable by arithmetic: at each joint the forces must add to nothing in two directions, which gives 2j equations for m member forces and r reactions. When m + r = 2j the truss is statically determinate and equilibrium alone fixes every force, with no material, no stiffness and no cross-section entering anywhere. The method of joints is those equations solved one joint at a time, starting wherever only two unknowns meet.

What each symbol means

m: the number of members. Each one carries a single unknown force along its own line
j: the number of joints. Each gives two equations, one for each direction, so the count to beat is 2j
r: the number of reaction components — three here, two at the pin and one at the roller
F: the force in a member, in kilonewtons. Positive is tension throughout this page, so the sign is half the answer
θ: the angle a diagonal makes with the bottom chord. It is the only trigonometry in the method

Assumes: Pin joints, straight members, and loads applied only at the joints. Every one is an idealisation that a real bridge breaks: the gusset plates are bolted or welded and do carry some bending, the members have weight of their own that is not applied at a joint, and a deck spreads its load along the chord rather than dropping it on a node. All three are small for a well-proportioned truss, which is why the idealisation survived a century and a half of building things that stood up. The analysis also assumes the truss does not move: every force here is what equilibrium requires, and nothing on this page says whether the bridge would deflect an acceptable amount.
Breaks when: The moment m + r stops equalling 2j. Add one more member and equilibrium is no longer enough — the forces then depend on how stiff each member is, which means on material and cross-section, and the whole method on this page stops working. That is the same wall Kirchhoff's Laws runs into when a bridge circuit cannot be reduced, and it has the same answer: more equations, from somewhere other than equilibrium. It breaks the other way too. Take a member away and the frame becomes a mechanism that folds up rather than a structure, and the determinacy row goes negative. And the pin-joint idealisation itself breaks where the members are stubby, where the joint is stiff enough to force real bending into them.

Three of these members carry nothing, and you can name them first 🖖

Press The same span, Pratt. Three members read exactly zero: L1–U1, L3–U3 and L5–U5. You did not need the solver for any of them. Look at joint U3: two top chords in a straight line, and the vertical. Nothing else pulls on that joint, so nothing can balance a force in the vertical, so there is none — and that holds whatever the load is and wherever it stands. The other two are the same argument at their own bottom joints. The tool finds these two entirely separate ways, once from the solved forces and once from the drawing alone with no arithmetic at all, and the two answers agreed across 23,814 different trusses and loadings. They are not spare: a member carrying nothing today is holding a member that is carrying something straight, which is the next lesson's whole subject.

Mirror the diagonals and the long members change sign 🖖

Press The same span, Pratt and then Diagonals mirrored. Identical nodes, identical member count, identical load. In the Pratt the four interior diagonals — 7.81 m each — are in tension at +93.72 kN, and the short 5 m verticals take the compression or nothing at all. Mirror them and those same four are in compression at −93.72 kN while every vertical is pulled. The two end diagonals are in compression either way, because they carry the reaction straight up into the truss and nothing can change that. The magnitude never moved; only the sign did. It matters because a long member in compression buckles at a load that has nothing to do with how strong its steel is, and a long member in tension does not care about its length at all. That is why a steel railway truss is nearly always a Pratt and why a timber one, where the connections favour compression, is often a Howe.

The same swap moves 72 kN between the two chords 🖖

It changes the chords as well, and the reason is worth following. In the Pratt the bottom chord carries +144.00 kN and the top −216.00 kN; in the Howe those swap to +216.00 and −144.00. Cut the truss between the third and fourth panel points and three members cross the cut: a chord, a chord and a diagonal. Take moments about the point where the other two meet and one chord force falls out on its own. In the Pratt the diagonal runs from U2 down to L3, so it meets the top chord at U2, twelve metres from the support. Mirror it and it runs from L2 up to U3, meeting the top chord at U3, eighteen metres out. Same reaction, same depth, a lever arm of 12 m against one of 18 m: 60 × 12 ÷ 5 = 144 against 60 × 18 ÷ 5 = 216.

Problem solved in full

Solving the first joint by hand, and then skipping five of them 6 steps
A 36 m Pratt truss, 5 m deep, six bays, with 120 kN standing at midspan. Find the force in the two members meeting at the left-hand support, and then find the bottom chord at midspan without solving the four joints in between. Press The same span, Pratt to follow along.

Open this problem in the tool
1. The load is at midspan, so the two reactions share it equally. Nothing about the web pattern enters this step.
  
  $R_{A} = R_{B} = \frac{120}{2} = 60 kN$
2. Start at joint L0, because it is the only joint in the truss with just two unknown members — everywhere else has three and the method has no foothold. The diagonal rises 5 m over 6 m, so sin θ = 0.6402 and cos θ = 0.7682.
  
  $sin θ = \frac{5}{6 ^{2} + 5 ^{2}} = 0.6402, cos θ = \frac{6}{6 ^{2} + 5 ^{2}} = 0.7682$
3. Vertical equilibrium at L0 has only the reaction and the diagonal in it, because the bottom chord is horizontal and cannot help.
  
  $Σ F_{y} = 0 : 60 + F_{1} sin θ = 0$
4. The sign is the answer: negative means the member is pushing back on the joint, so L0–U1 is in compression at 93.72 kN.
  
  $F_{1} = - \frac{60}{0 . 6402} = - 93.72 kN$
5. Now horizontal equilibrium at the same joint, with the diagonal's force known. The bottom chord comes out positive, which is tension.
  
  $Σ F_{x} = 0 : F_{2} + F_{1} cos θ = 0 \Rightarrow F_{2} = 93.72 \cdot 0.7682 = + 72.00 kN$
6. You could repeat that eleven more times. Instead, cut the truss between the third and fourth panel points and take moments about U2, where the two members you do not want both meet — the bottom chord falls out in one line, and the four joints in between never had to be solved.
  
  $Σ M_{U_{2}} = 0 : 60 \cdot 12 = F_{L_{2} L_{3}} \cdot 5 \Rightarrow F_{L_{2} L_{3}} = + 144.00 kN$
Answer

L0–U1 at −93.72 kN and L0–L1 at +72.00 kN, and the midspan bottom chord at +144.00 kN from a single moment equation. The two methods are not rivals: joints is systematic and gives you everything, sections gives you one member and does it in a line. An engineer sizing a bridge wants one member — the worst one — and reaches for a cut. The last step also shows where the 144 comes from, which is 60 kN of reaction with a twelve-metre lever divided by the five-metre depth. Halve the depth and it doubles, which is the flatter preset on this page and the reason a long bridge is a deep one.

$F_{1} = - 93.72, F_{2} = + 72.00, F_{L_{2} L_{3}} = \frac{60 \cdot 12}{5} = + 144.00 kN$

Learning path

Where the load goes

References (2)

Where the counting rule on this page comes from. Maxwell gives the condition a frame must satisfy for equilibrium alone to fix every force — the m + r = 2j the results table prints as a row: J. Clerk Maxwell, "On the calculation of the equilibrium and stiffness of frames." The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science 27(182), 1864, pp. 294-299. DOI 10.1080/14786446408643668
A standard statics text for the method of joints and the zero-force member rules the page claims a reader can apply by eye. R. C. Hibbeler, Engineering Mechanics: Statics, 14th ed. Pearson, 2016. ISBN 978-0-13-391892-6.

By the 42Math team · Reviewed against published sources. Corrections welcome.