Truss analysis: finding the force in every member of a truss bridge

Truss analysis starts with a deliberate simplification

Put 20 kN on the centre joint of the 12 m Warren truss drawn above and the two supports each take 10 kN, which is the only part of the problem anyone can do in their head. The harder part is the framework in between: fifteen members, each pushing or pulling at its ends by an amount nobody has told you yet.

Truss analysis makes that question answerable by agreeing on a model first. Every joint is treated as a frictionless pin. Every load is applied at a joint, never in the middle of a member. The weight of the members is ignored. Under those three agreements a member can only be squeezed or stretched along its own length, so it carries exactly one number, its axial force. A truss with fifteen members has fifteen unknowns, and the job is to find them.

The idealisation is false in a useful way. Real bridge joints are gusset plates, bolts or welds that carry some bending, and real members have weight. But the pin-jointed forces are usually close to the real axial forces, which is why a truss can be analysed on paper at all.

Counting unknowns: is the truss determinate?

A truss can be solved by statics alone when the number of unknowns equals the number of equations. Each joint offers two equations (forces balance horizontally and vertically), so a truss with j joints offers 2j. The unknowns are the m member forces plus the r support reaction components. The test is therefore

m + r = 2j

Take a single triangle: three members, three joints, a pin (two reactions) and a roller (one reaction). That gives 3 + 3 = 6 = 2 × 3, and it solves. The Warren preset on this page has 15 members and 9 joints, with a pin and a roller supplying 3 reactions: 15 + 3 = 18 = 2 × 9. The five-joint truss on the method of joints page has 7 members: 7 + 3 = 10 = 2 × 5. Add one more diagonal to either and m + r exceeds 2j: the truss becomes indeterminate, and its forces then depend on member stiffness as well as geometry.

A short rule falls out of this. A simple truss built by adding two members for every new joint, starting from one triangle, has m = 2j − 3, which is the same count with r = 3.

The count is necessary but not sufficient. Four members arranged as a square have the right count for one support type and still fold flat, because a square is not a triangle. The count checks the arithmetic; the triangles check the shape.

Sign convention and the two hand methods

Tension is positive and compression is negative. A member in tension pulls on the joints at its ends, so on a free-body drawing its arrow points away from the joint. A member in compression pushes, and the arrow points toward the joint. Assume every unknown is tension at the start. A negative answer then means the member is in compression, and the sign corrects itself without any further thought.

The page on tension and compression explains what those two words mean for the steel, and why the answer decides how thick a member must be.

There are two hand methods, and they differ in what they isolate.

The method of joints isolates one pin at a time. At each joint ΣFx = 0 and ΣFy = 0, which allows two unknown forces per joint. The work starts at a support, where the reaction is already known, and work along the truss, each solved member feeding the next joint. It gives every member in order and it is the method to learn first.

The method of sections cuts the truss in two with an imaginary line through no more than three members, then balances the whole piece on one side. Taking moments about the point where two cut members meet leaves a single unknown, so one equation gives one member directly. It is the method to reach for when only the centre top chord of a long truss is wanted and the members at the ends do not matter. The worked case is on the method of sections page.

Where a computer solver fits

Past about a dozen members, hand truss analysis by joint-by-joint arithmetic stops being instructive and starts being error-prone. A solver sets up all 2j equations at once and solves them together. The truss bridge designer on this site does that with a direct-stiffness method each time a member moves, which is why the colours update while you drag.

For the Warren truss above, the solver reports the centre top chord G-H at 20.0 kN in compression, the bottom chords B-C and C-D at 15.0 kN in tension, the end bottom chords A-B and D-E at 5.0 kN in tension, and every diagonal at 11.18 kN, alternating push and pull. A moment equation about the bottom centre joint confirms the 20.0 kN in one line: 10 kN × 6 m divided by the 3 m height of the truss.

Truss bridge design calculations, in working order

Truss bridge design calculations follow one sequence, and the analysis above is its middle step. First the loads: the dead load of the deck and members, then the live load at each joint it reaches. Second the reactions, by taking moments about one support. Third the member forces, by joints or sections, with the sign convention above. Fourth the capacity of each member: the yield force in tension, the buckling load in compression. Last the margin, capacity divided by force, for every member in turn. The member with the smallest margin sets the strength of the bridge, which is why the design calculations are repeated after every change to a section or a panel length. The truss bridge designer repeats steps two to five for every move, and the worked example for a Pratt layout is on the Pratt truss bridge page.

What no hand method tells you

Analysis gives forces. It does not say whether a member survives them.

A long compression member can buckle at a fraction of the load that would yield the steel. Euler’s formula, P_cr = π²EI/L², says how much, and the bar that governs is almost always a compression bar. The factor of safety page turns the force list into a margin, and the same buckling reasoning is why a folded strip of card carries far more than a flat one.

Joints are the second blind spot. A glued or bolted joint can fail well below the strength of the members it joins; on a popsicle-stick model it is the usual point of failure. Self-weight is the third: a deck and its trusses weigh something, and it arrives as extra load on every joint. Imperfections complete the list, since no real member is perfectly straight.