Structural Engineering & Statics
where the load goes
Structural form

The triangle that cannot fold, and everything built out of it

A square of pinned bars is a mechanism. A triangle is not, and that single fact is the reason trusses exist and the reason they look the way they do.
9 min read 8 figures Geometry beats materialThe load must go somewhere

Four bars pinned into a square fold flat under the lightest push. Three bars pinned into a triangle do not, and cannot, without one of them changing length.

That is the entire idea. A triangle’s shape is fixed by the lengths of its sides — the side-side-side congruence from school geometry, doing structural work — and a frame assembled entirely from triangles inherits the property. Everything else about trusses is bookkeeping.

A Pratt truss of 6 panelsA Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 10 members came out in tension, 9 in compression and 2 carrying nothing.tensioncompression2 carrying nothing
Fig. 1 A Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; members shown in one colour came back in tension and the others in compression.

Solving one joint at a time

A pin joint transmits force and not moment, so every member is in pure tension or pure compression and carries one unknown number.

At a joint, all the forces pass through a point, so the moment equation is satisfied automatically and two equations remain. A joint where only two member forces are unknown can therefore be solved outright, and its answers become known quantities at the next joint along.

Start at a support, where typically two members meet and the reaction is already known from overall equilibrium. Solve it. Move to a neighbour. Repeat. The whole truss falls out in sequence, and the sequence is the method of joints.

Joint 0 of the truss, cut outOne joint of the truss with every force acting on it. Two equations — the horizontal and vertical sums — are enough for a joint with no more than two unknown member forces, which is the whole method.HV29.4-38.6reaction 0.0reaction 25.0ΣH = 0 and ΣV = 0, and nothing else is needed
Fig. 2 The support joint, cut out. Two members and a reaction, two equations, and no ambiguity — which is why the solution has to start somewhere like this.

The figures on this page are produced differently, and the difference is worth stating. Rather than working joint by joint, the generator assembles all the joint equations at once — two per joint, with one column per member force and one per reaction — and solves the whole system by elimination. The answers are identical. What the simultaneous approach avoids is needing to find a joint simple enough to begin at, which for an awkward geometry can be the hardest part.

Joint 1 of the truss, cut outOne joint of the truss with every force acting on it. Two equations — the horizontal and vertical sums — are enough for a joint with no more than two unknown member forces, which is the whole method.HV29.429.40.0ΣH = 0 and ΣV = 0, and nothing else is needed
Fig. 3 A joint further along the bottom chord. Four members meet here, which would be two too many to start with — but by the time the solution arrives, two of the four are already known.

Which members pull and which push

The interesting differences between named trusses are not about strength. They are about which members end up in tension.

A Howe truss of 6 panelsA Howe truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 10 members came out in tension, 10 in compression and 1 carrying nothing.tensioncompression1 carrying nothing
Fig. 4 A Howe truss of the same span and depth. The diagonals lean the other way, and the consequence is exact — the diagonals are now in compression and the verticals in tension, the reverse of the Pratt.

Under gravity, and by the same couple argument that governs a beam section, a simply supported parallel-chord truss always has its top chord in compression and its bottom chord in tension. That much is forced: the two chords form a couple resisting the bending moment, and the moment sags. What is not forced is the diagonals, and the two classic arrangements make opposite choices.

Pratt: diagonals slope down toward mid-span, and they carry tension. The verticals carry compression.

Howe: diagonals slope the other way and carry compression, with the verticals in tension.

The choice follows the material. Long members in compression buckle, so a long member is better in tension; short members in compression are fine. A Pratt puts the long diagonals in tension and the short verticals in compression, which suits steel. The Howe was the timber version, where the long timber diagonals took the compression and short iron rods took the tension — because timber is good in compression and fastening timber in tension is difficult.

Two trusses of identical count and identical geometry, differing only in the lean of some bars, chosen by what the members are made of.

A Warren truss of 6 panelsA Warren truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 10 members came out in tension, 11 in compression and 2 carrying nothing.tensioncompression2 carrying nothing
Fig. 5 A Warren truss, with no verticals at all. The diagonals alternate between tension and compression along the span, which is why every second one is doing the opposite job to its neighbour.

The members carrying nothing

Some members in a loaded truss have exactly zero force, and the solver returns them as zero rather than as something small.

They are not mistakes and they are not decoration. A zero-force member holds the geometry: it stops a long member buckling out of plane, or it braces a joint that would otherwise be free to move, or it becomes loaded under a different load case entirely. A truss designed for a single symmetric load case and stripped of its zero-force members will find a use for them the first time the load is asymmetric.

There are two patterns worth recognising by eye. A joint where exactly two non-collinear members meet, with no load applied, has zero force in both. A joint where three members meet, two of them collinear, with no load, has zero force in the third. Spotting those before starting the arithmetic removes a surprising number of unknowns.

Why the depth matters more than anything

The chords carry the bending, and they do it as a couple whose lever arm is the depth of the truss.

Chord force against truss depthThe force in a truss chord for a fixed bending moment, against the depth of the truss. The relationship is a reciprocal: the chords form a couple whose lever arm is the depth, so a shallow truss pays for it steeply.0.511.52050100150200250300depth of the truss2501671251007150the same moment, resisted by a longer lever arm
Fig. 6 Chord force against truss depth for a fixed bending moment. The relationship is a reciprocal, so halving the depth doubles the chord force and a shallow truss pays steeply.

For a given moment, the chord force is the moment divided by the depth. Depth is therefore the cheapest strength available: doubling it halves the chord forces without adding any material to the chords at all, at the cost of longer web members and a taller structure.

The limit is not structural but architectural. A roof truss can be as deep as the roof pitch allows; a floor truss competes with headroom; a bridge truss competes with approach gradients. Almost every truss in existence is as deep as something non-structural would permit.

Where the forces come from

A truss’s member forces are not arbitrary. They are the bending moment and shear of an equivalent beam, split between chords and web.

Load, shear and moment — a simple spanThe applied load, the shear force it produces and the bending moment that follows, drawn one above another to the same horizontal scale. Shear is the integral of the load and moment is the integral of shear.4 per unit lengthshear16.0moment32.0 at x = 4.00the moment peaks exactly where the shear passes through zero
Fig. 7 The moment and shear diagrams of a beam of the same span and loading. The chord forces of a truss follow the moment diagram and the diagonal forces follow the shear.

That correspondence is exact and useful. The moment at a station divided by the truss depth gives the chord force there, which is why the chords are heaviest at mid-span and lightest at the supports. The shear at a station divided by the sine of the diagonal’s angle gives the diagonal force, which is why the diagonals are heaviest at the supports and lightest in the middle.

Counting unknowns against equationsThree frames differing by one member. Two equilibrium equations per joint, one unknown per member and one per restraint: fewer unknowns than equations is a mechanism, equal is solvable by statics, more needs stiffness.m 4 + r 3 − 2j 8 = -1a mechanismm 5 + r 3 − 2j 8 = 0statically determinatem 6 + r 3 − 2j 8 = +1one member too manystatics can answer only the middle case
Fig. 8 The counting rule that decides whether the member forces can be found at all. Every truss on this page satisfies it exactly, which is why the solver returns one answer rather than a family of them.

A truss is therefore a beam with the material removed from where the stress was low, and the reason it works is depth — the chords are as far apart as the geometry allows.

Where the model stops

The pin-jointed idealisation is the foundation of everything above, and no real truss has pins in it.

Joints are not pins, whatever the counting rule assumed. Members are welded or bolted through gusset plates, and a rigid joint transmits moment. Real trusses therefore have bending in their members — secondary stresses — of perhaps ten to twenty percent of the axial values. The analysis is still done as pin-jointed because the primary forces dominate and the assumption is conservative for the chords.

Loads are not applied at joints. Real loads arrive along the chords, between the panel points. A chord loaded between joints bends as well as carrying axial force, and the design has to add the two.

Members have weight, and at long spans it dominates. Self-weight is distributed along each member, not concentrated at its ends, so the same objection applies to the truss’s own mass.

Compression members buckle, and the end restraint decides at what load. Nothing in the joint equations knows the difference between tension and compression. A compression member fails at a load set by its slenderness, often far below its squash load, and the solver returns a force with no comment on whether the member can carry it.

Plane behaviour. A truss analysed in its plane must be prevented from falling over out of plane by something else — purlins, bracing, a deck. That something is invisible in every figure here and is not optional.

The figures share one honest distortion. Member forces are drawn as line weights, so a heavily loaded chord is thicker than a lightly loaded diagonal. That encodes magnitude usefully and encodes nothing about whether the member is adequate, since a compression member’s capacity depends on its length as well as its force. The thickest line in a truss diagram is not necessarily the member in trouble.

The ladder from here

Later rungs on this anchor: the method of sections, which answers one member’s force without solving the rest. Bow’s notation and the Cremona diagram. Zero-force member rules. Space trusses. Secondary stresses from rigid joints. The K-truss and the Baltimore, and what subdividing a panel buys. Bridge truss types and their history. The Vierendeel, which has no diagonals at all and works entirely by joint rigidity. And the transition to the space frame, where the counting rule changes and so does everything else.

The Pratt patent is from 1844 and the Howe from 1840. Both were designed for wooden railway bridges, and the reason they are still built is that the argument about which members should be in tension has not changed.

Where this goes next

m 4 + r 3 − 2j 8 = -1a mechanismm 5 + r 3 − 2j 8 = 0statically determinatem 6 + r 3 − 2j 8 = +1one member too manystatics can answer only the middle case Counting the unknowns, and finding out whether statics can answer Two equations per joint, one unknown per member, one per restraint. Subtract, and the sign of the answer says whether the structure is a mechanism, solvable, or beyond what equilibrium alone can settle. 0.511.52050100150200250300depth of the truss2501671251007150the same moment, resisted by a longer lever arm Depth is the cheapest strength there is Doubling the depth of a truss halves its chord forces without adding a gram of material to the chords. Nothing else in structural design is that cheap, and almost every structure has already spent it. cable: pure tensionarch: pure compression The shape that carries itself, and the arch that is its reflection Hang a chain and it takes the one shape that carries its load in pure tension. Turn the shape upside down and it carries the same load in pure compression. That is what an arch is.