Halving the panel buys a shorter strut

A truss is usually chosen by its depth, and depth is genuinely the cheapest strength there is. The second choice — how many panels to cut the span into — is made far more casually, often to suit a purlin spacing, and it changes something quite different from what most people assume it changes.

Take one truss and subdivide it. Span 24 m, depth 3.4 m, a total load of 60 kN arriving at the top panel points, and nothing altered but the number of panels.

The chord force is fixed by the moment and the depth #

The two chords are a couple. At any section the moment is carried by a tension in one and an equal compression in the other, separated by the depth, so

$$F_{\text{chord}} = \frac{M}{d}$$

and neither $M$ nor $d$ has anything to do with how many panels the truss has. The mid-span moment of a 24 m truss under 60 kN spread over its top is fixed by the span and the load; the depth is fixed by the drawing. The chord force is therefore fixed before the panel count is chosen.

The numbers in the figures confirm it, and the small residual movement is worth explaining rather than ignoring. The largest bottom chord force goes 52.9, 56.5, 56.7, 57.7 kN as the panels go 4, 6, 8, 12. It rises, by 9 per cent over a threefold subdivision.

The reason is that a coarse truss does not sample the moment diagram at its peak. With four panels the bottom chord runs from x = 0 to x = 6 m and its force is the moment at the middle of that run divided by the depth; with twelve panels the middle panel sits astride mid-span and picks up almost the full parabolic maximum. Subdividing does not raise the moment; it stops the coarse truss from getting away with missing it.

So the honest statement is stronger than “the chord force is unchanged”. Subdividing a truss makes the chord force slightly worse, converging on $M_{\max}/d$ from below, and anyone expecting an improvement there has misread which term the panel count is in.

What does change, and it changes fast #

Every compression member in a truss has a buckling length, and that length is a panel dimension.

The end diagonal is the truss’s hardest-working compression member, and across the four cases it carries −60.8, −46.3, −40.0 and −34.8 kN over lengths of 6.90, 5.25, 4.53 and 3.94 m. Both quantities fall, and the one that governs its size falls as their product with the length squared.

For a pin-ended strut the Euler load is $\pi^2 EI/L^2$, so the second moment of area needed to carry a force $F$ over a length $L$ is proportional to $F L^2$. In consistent units that is

A factor of 5.4 on the stiffness the strut has to have, for a truss whose chord force moved by 9 per cent in the wrong direction. That is the whole trade, and it is why the panel count is a stability decision rather than a strength one.

The top chord does even better, because its unbraced length is the panel itself rather than a diagonal. It carries −70.5, −63.5, −60.5 and −57.7 kN over 6.0, 4.0, 3.0 and 2.0 m, giving $F L^2$ of 2,538, 1,016, 545 and 231 — a factor of eleven between the coarsest and the finest.

A member strong enough and still falling over is the failure the panel count is aimed at, and it is aimed at it by the only means that works against an inverse square: shortening the length rather than strengthening the member.

One more consequence follows from the chord force being fixed. Because $M/d$ does not care how the web is arranged, every one of these four trusses needs the same chord section, and the choice of panel count changes only what happens between the chords. A designer comparing arrangements is therefore comparing web solutions to a fixed problem, which is a much smaller search than it looks — and it is the reason the named truss types differ in their diagonals and never in their chords. The load a member is given is decided before the member is, and here it is decided by two numbers that a panel count cannot reach.

Which free body produced the number #

Two free bodies, and they are the two halves of the argument.

For the chord force, cut the whole truss through a panel with a vertical line and take everything to the left. Crossing the cut are the two chord forces and one diagonal. Taking moments about the point where the diagonal meets the other chord eliminates two of the three unknowns and leaves the chord force as the moment at that point divided by the depth. The panel count enters only through where the cut can be taken, which is why the chord force is $M/d$ evaluated at a panel point rather than at mid-span.

For the strut, the free body is one member with a pin at each end, and the question asked of it is not equilibrium at all. It is whether the straight configuration is the only equilibrium available, which is a different question with a different kind of answer — an eigenvalue rather than a force. The panel count sets $L$ in that eigenvalue and nothing else about it.

That is the sharpest way to state why the two effects are independent. One free body is a section through the structure and the other is a member on its own; the panel count is a length in the second and a position in the first.

What it costs, counted #

Nothing about subdivision is free, and the cost is joints.

Read the two columns together. Going from four panels to twelve multiplies the members by 3.5 and the joints by 3. In a welded or bolted truss the joints are most of the fabrication cost and nearly all of the drawing office cost, and they are also where the fatigue lives — so the trade is a light structure that is expensive to make against a heavy one that is cheap.

There is a second cost that the count does not show, and it is the one that decides whether the trade is worth making at all. A member has a minimum size that has nothing to do with its force. A diagonal carrying 34.8 kN over 3.94 m is not sized by 34.8 kN; it is sized by a slenderness limit, by what can be handled without bending it, by the bolt group needed at each end, and by the smallest angle or hollow section the fabricator stocks. Past a certain fineness the members stop shrinking and only the joints keep multiplying, and every panel added beyond that point is pure cost.

That is the practical end of the argument. The benefit calculated above is real over the first doubling and largely spent by the second, because $F L^2$ falls fast while the member size hits a floor. The optimum is not where the arithmetic stops improving; it is where the member stops being sized by its force, and on an ordinary roof truss that happens between six and ten panels.

Where the balance sits has moved over time and is not a fact about mechanics. When steel was expensive relative to labour, trusses were finely subdivided; the nineteenth-century lattice girder with dozens of small diagonals is the extreme. When labour became expensive relative to steel the panels got longer, the members got fatter, and the modern equivalent of that girder is a plate girder with no web members at all.

The other way to subdivide, which adds no chord joints #

There is a version of this trade that gets most of the benefit without most of the cost, and it is what the named subdivided trusses are for.

A K-truss puts a node at mid-height of each vertical and runs two diagonals to it from the chord panel points, halving the diagonal’s buckling length without cutting the chord into shorter pieces. A Baltimore or Petit truss adds sub-diagonals that brace the mid-point of a long member while leaving the primary triangulation alone.

Both work on exactly the term identified above. They shorten $L$ for the member that is governing without multiplying the chord connections, which are the expensive ones — and they add members that carry very little force, since a brace’s job is to be stiff rather than strong. That is the same economy as a brace that need not be strong, applied inside a truss instead of to a beam.

The price is that the sub-members are lightly loaded and therefore slender, and a lightly loaded slender member is exactly the kind that is damaged in transport and left slightly bent on site.

What subdivision does to the web pattern #

Panel count also decides the angle the diagonals run at, and there is a shape argument hiding inside the arithmetic.

A diagonal at 45° carries a shear of $V$ with an axial force of $V\sqrt2$; one at 30° to the horizontal carries $2V$. Since the panel length and the depth set that angle, choosing the panel count is also choosing how efficiently the web carries shear — and the efficient range, between about 40° and 60°, is what fixes the panel length at roughly the truss depth.

That is where the conventional rule comes from. A panel about equal to the depth gives a diagonal near 45°, and a truss designed to it is close to optimal on both counts at once: the web forces are small and the buckling lengths are short. The 24 m truss here has a depth of 3.4 m, so its natural panel count is seven, and the eight-panel version is the one a designer would actually draw.

The length that subdivision does not shorten #

Everything above is about buckling in the plane of the truss, and the compression chord has a second buckling length at right angles to it that the panel count does not touch.

Out of plane the top chord is held only where something braces it — a purlin with a tie, a rafter bracing bay, a deck. Adding panels adds nodes, and a node is not a restraint: a node is a place where web members meet the chord, and web members lie in the truss’s own plane and offer nothing at all against movement out of it. A twelve-panel truss braced at three points has the same out-of-plane length as a four-panel truss braced at three points, and the chord force is the same in both.

That asymmetry is easy to lose, because a drawing of a truss shows the in-plane length and hides the other one. It is the reason a compression chord is usually governed by the out-of-plane case, the reason the plan bracing is drawn on a separate sheet, and the reason an effective length is a property of what holds a member rather than of the member.

It also changes what subdivision is worth. The benefit calculated above lands almost entirely on the diagonals, which buckle in plane and have no other length; the chord’s eleven-fold improvement is available only if the bracing keeps up, and a bracing system is a structure with its own economics.

The secondary effect that arrives with the joints #

Every joint added is a place where the pin-jointed idealisation is not true, and a finer truss has more of them.

Secondary bending is the price of a joint that is not a pin, and its size depends on the member’s slenderness rather than on its force. Subdividing produces shorter, stubbier members, and a stubbier member rotated through the same joint angle picks up more curvature and therefore more secondary moment.

So the fine truss wins on buckling and loses on secondary bending, and the two are not measured in the same units. The usual resolution is that secondary bending does not govern strength — it relaxes as soon as anything yields — but it does govern fatigue, which is why the finely subdivided lattice girders of the nineteenth century are the ones with cracked joints.

The limit of subdivision is not a truss #

Push the subdivision far enough and the structure stops being one.

Halve the panels again and again and the diagonals become shorter, more numerous, more lightly loaded and closer together, until what is between the chords is less a set of members than a continuous field carrying shear. That limit is the plate girder, and it is not an approximation to a very fine truss — it is the same structure with the discretisation removed. Its web carries shear as a diagonal tension and compression field distributed over the whole panel, which is why a web that has buckled goes on carrying load as a tension field rather than failing when the compression diagonal does.

Going the other way is equally instructive. Reduce a truss to two panels and the single diagonal in each half becomes very long and very heavily loaded, and its required stiffness runs away as the square. That is why a two-panel truss is almost never built and why a very long single-span truss is always finely subdivided: at 24 m the choice is a judgement, and at 120 m the arithmetic has already made it.

The panel count is therefore a scale-dependent decision rather than a proportion. A rule expressed as a ratio — panels equal to the depth — happens to work over the range of spans buildings use, and it works because the depth is itself chosen as a fraction of the span. Outside that range the rule stops being a rule and the $F L^2$ term takes over.

Where the model stops #

The loads stay at the panel points. Every figure here loads the top nodes only. Subdividing a truss usually means the purlins land on nodes that did not exist before, and if they do not, the chord carries bending between them and the whole comparison changes.

Self-weight is absent. The 60 kN is applied load. A finer truss weighs less in its web and more in its connections, and the crossover is a fabrication question rather than a structural one.

The buckling length is taken as the panel length. In plane that is roughly right and out of plane it is not: the top chord’s out-of-plane length is set by the bracing, which is a different structure entirely and does not change when the panels do.

Every member is pin-ended in the buckling calculation. The end restraint from continuous chords raises the real capacity by a factor that is itself larger in a fine truss, so the table above understates the benefit.

The comparison holds the depth constant. Real subdivision often accompanies a change of depth, and then both terms move at once — which is why the effect described here is so often attributed to the wrong variable. Depth is the cheapest strength and panel count is the cheapest stability, and a design that changes both and measures the result has learned nothing about either.

And nothing here is an optimisation. Four cases were drawn because four cases make the trend visible. The actual minimum-cost panel count depends on the price of a joint, which is not a structural quantity and varies by a factor of several between fabricators.

The ladder from here #

Later rungs on this anchor: the chord that runs through several panels, which is a continuous beam whatever its joints do. Counters and tension-only diagonals, where a member is present under one load case and absent under the other. The counting rule in three dimensions. The K-truss and the Baltimore worked properly, with the sub-member forces solved rather than asserted. Out-of-plane buckling of the compression chord, where the relevant length belongs to the bracing system and not to the truss. And the transition to the plate girder, which is the limit of subdivision — a web with no panels at all, where the shear is carried as a field rather than by members.

The Pratt patent of 1844 and the Howe of 1840 were both for timber railway bridges, and both fix a panel length close to the truss depth for the reason above. The subdivided types arrived a generation later with wrought iron, when spans grew past the point where a 45° panel could be crossed by a single member of sensible size — which is to say that the K-truss exists because a diagonal got too long, not because anybody wanted more joints.