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

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.

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.

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.

The joints further along make the point. At the first interior bottom node four members meet — two chords, a vertical and a diagonal — which is two unknowns too many for a joint to be solved on its own. It is only solvable in sequence, once the support joint has handed it two of the four. The simultaneous assembly never needs that ordering, and on an awkward geometry finding a place to begin is the part that fails.

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.

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.

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.

For a given moment, the chord force is the moment divided by the depth. The mid-span moment of this span is 45 whatever the truss under it looks like, so the top chord is 45/0.4 = 112.5 shallow and 45/0.85 = 52.9 as first drawn — a reciprocal, and a reciprocal is steep at the shallow end.

Take the depth the other way and watch the same relation run out.

That last observation is the whole economics of depth in one line. Depth is the cheapest strength available: doubling it halves the chord forces without adding any material to the chords at all, and it leaves the verticals exactly where they were. What it costs is longer web members — the diagonals lengthen with the depth — 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.

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.

None of this would be available if the counting rule came out differently. Every truss on this page has exactly as many member forces and reactions as there are joint equations to fix them, which is why the counting rule matters before any arithmetic starts: one short and the frame is a mechanism with no equilibrium solution at all, one over and there is a family of solutions rather than an answer, and the elimination that produced every number above would have nothing to return.

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.

The joint that started it, with its numbers #

The figures on this page all come from one solved system, so it is worth taking a single joint out of it and showing that the number is arrived at rather than asserted.

The Pratt truss at the top has six panels of unit width, a depth of $0.85$, and a load of $10$ at each of the five interior top nodes. Total load $50$; by symmetry each support carries $25$.

The free body is the support joint. Cut it out. Three forces act on it: the reaction of $25$ upward, the bottom chord pulling horizontally, and the end diagonal running up to the first top node, whose direction is fixed by the geometry — a rise of $0.85$ over a run of $1$, so a length of $1.312$, giving $\sin\theta = 0.648$ and $\cos\theta = 0.762$.

Vertical equilibrium has only two contributors, since the chord is horizontal:

$$25 + F_d(0.648) = 0 \quad\Longrightarrow\quad F_d = -38.6.$$

Negative, so the end diagonal is in compression, which is the standard convention on this site. Horizontal equilibrium then gives the chord:

$$F_c + (-38.6)(0.762) = 0 \quad\Longrightarrow\quad F_c = +29.4,$$

positive and therefore tension. Two equations, two answers, and the next joint along now has only two unknowns left.

The whole-truss solver returns $-38.60$ and $+29.41$ for those two members, which is the point of doing it by hand once: the elimination is not doing anything the joint was not already doing.

One more check is available, and it ties the truss back to the beam. The top chord at mid-span comes out at $-52.94$. The bending moment of an equivalent beam at mid-span is $25 \times 3 - 10 \times (1+2) = 45$, and $45 / 0.85 = 52.94$. The couple argument is not an approximation to what the solver did — for a parallel-chord truss it is exactly what the solver did.

The same solved system also settles the zero-force claim made earlier without any rule of thumb. Two members in this truss come back at exactly zero: the verticals at the first and last interior panel points, where three members meet, two of them collinear along the chord, and no load is applied. The rule predicts it; the elimination confirms it; and the value returned is zero rather than a small residue, which is worth noticing, because a numerical method that returned $0.003$ instead would leave a reader unable to tell a structural fact from a rounding error.

Half a truss’s deflection is its web #

The correspondence with a beam was offered for the forces and it does not extend to the deflection, which is where a truss stops behaving like the beam it replaces.

A truss’s deflection is the sum of its members’ extensions, weighted by virtual work: $\delta = \sum N n L/EA$. Split that sum into chords and web and the two halves are different animals. The chords give a bending deflection, with an effective second moment

$$I_{\text{eff}} = 2 A_c \left(\frac{d}{2}\right)^2 = \frac{A_c d^2}{2}$$

The web gives a shear deflection, because a panel whose diagonal stretches racks into a parallelogram, and the truss’s equivalent shear rigidity is a property of the diagonal and the vertical in series:

$$\frac{1}{GA} = \frac{1}{EA_d \sin^2\!\alpha\cos\alpha} + \frac{1}{EA_v \cot\alpha}$$

Put a real truss through both. Twelve metres, six panels, 2 m deep so the diagonals are at 45°; chords of 3,000 mm² and web members of 1,500; five loads of 10 kN. Then $EI = 1.26\times10^6$ kNm² and $GA = 82{,}400$ kN, and under the equivalent 4.17 kN/m:

$$\delta_b = \frac{5wL^4}{384EI} = 0.89\ \text{mm}, \qquad \delta_s = \frac{wL^2}{8GA} = 0.91\ \text{mm}$$

The web contributes 51 per cent of the deflection. A solid beam of the same proportions would put five per cent into shear; a truss puts half, because its “web” is two thin bars per panel rather than a continuous plate.

The ratio is $9.6\,EI/GAL^2$, and $I_{\text{eff}}$ carries $d^2$ while $GA$ does not — so the fraction goes as $(d/L)^2$ and a deeper truss is more web-dominated, not less. At $L/d = 10$ it falls to 18 per cent; at $L/d = 4$ it is two thirds.

Two design consequences follow, and both bite on the members the force analysis says least about.

Web members sized on strength alone make a truss that sags. The diagonals near mid-span carry almost nothing — the shear is small there — so a strength-driven design makes them minimum-section, and a minimum section is exactly what the deflection sum is most sensitive to.

And $I_{\text{eff}}$ on its own is not a deflection check. Using $A_c d^2/2$ in a beam formula underestimates a deep truss’s deflection by a factor of two or three. The sum over members is the calculation; the effective second moment is a way of describing half of its answer.

What the idealisation costs #

The pin joint is an assumption made for solvability, and it is a good one. The cost of it does not show up in the member forces at all; it shows up in the drawing office and the fabrication shop.

Connections dominate the price. A truss is lighter than the beam it replaces, often by half, and it is very frequently more expensive. Steelwork is priced by the fabricated tonne, and the rate per tonne for a member with two simple end connections is a fraction of the rate for a node where six members meet on a gusset plate with forty bolts through it. A truss trades material for joints, and joints are where the labour is. The crossover — the span past which the material saving finally beats the connection cost — sits somewhere around fifteen to twenty metres for ordinary building work, which is exactly why rolled beams are used below it and trusses above.

Secondary bending is real and is not analysed. Because the joints are rigid, a member that is forced to rotate with its joint bends. The resulting stresses are typically ten to twenty per cent of the axial ones, and they are habitually ignored — legitimately, because they are self-limiting: a little local yielding relieves them and they do not accumulate. That is a ductility argument rather than a statics one, and it fails for a brittle material or a fatigue-loaded member, which is why railway and crane trusses are detailed far more carefully than roof trusses.

Eccentricity has to be designed out. The analysis assumes every member’s centroidal axis passes through one point at each joint. If the setting-out is careless and the axes miss, the offset multiplies the axial force into a real moment at the node — and unlike secondary bending, this one does accumulate with load. The rule that member centrelines must be concurrent at a node is the single most consequential detailing rule in truss design, and it exists entirely to protect an assumption made three steps earlier in the analysis.

The member was fine #

There is a failure that makes the previous point sharply, because everything the analysis on this page produces was correct at the time and the bridge fell down anyway.

The I-35W bridge in Minneapolis was a steel deck truss carrying eight lanes over the Mississippi. On 1 August 2007 it collapsed during the evening rush, killing thirteen people. The investigation found no fractured member and no overloaded chord. What failed was a gusset plate — one of the plates at a node that the members bolt onto — which had been detailed at half the thickness the forces required, an error made in 1964 and carried unnoticed through forty-three years of inspections. Additional dead load from two resurfacings, and construction plant parked on the deck that afternoon, took it past what it could hold.

A truss analysis returns one number per member. It says nothing whatever about the plate that joins them, and the plate is not a member, has no line in the diagram, and appears in none of the figures on this page. The collapse changed gusset-plate checking from something assumed adequate to something calculated explicitly, and it is the clearest possible statement of where the pin-jointed model’s boundary actually lies: not at the ends of the members, but a little way inside the node, at a component the model does not represent.

The general form of the lesson recurs everywhere on this site. A model returns answers about the things it contains. The things it does not contain do not appear as large numbers or as warnings — they appear as nothing at all, and nothing is very hard to notice.

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.