The counter fitted after the load

A truss’s camber is a cutting list. Make each member a little shorter or longer than drawn, and the truss assembles into a shape that its dead load then pulls level; every millimetre of a member’s length arrives at mid-span multiplied by that member’s force under a unit load there. That was a determinate truss, where any set of lengths fits — the joints simply go where the members put them.

A truss with a counter-diagonal in every panel is not like that. Two diagonals in one panel are one more member than statics needs, and each such panel has a state of self-stress: a pattern of forces it can carry with nothing applied. A member of the wrong length in such a truss does not simply move a joint. It has to be forced into place, and the force of being built to the wrong length stays in the truss. The question the earlier essay left was whether a camber can be cut into a redundant truss without locking any of that force in.

The truss, and three cutting lists #

The truss is eight panels of 3 m, 3 m deep — 24 m of span — with a counter-diagonal beside the diagonal in each of its six interior panels, carrying 60 kN of dead load at each interior bottom joint. Its chords are about two and a half times as stiff axially as its webs, and under the dead load it sags 16.3 mm at mid-span. It has six self-stress states, one for each counter.

Three cutting lists, each a way a fabricator might set out a camber.

Every member cut short by its own dead-load elongation. The truss is level under its dead load, everywhere, and no member’s force changes at all. That is not luck. The list is the dead-load stretch of every member with its sign reversed, and the dead-load stretch is the elongation field of a real movement of the joints — the sag itself. A set of length changes that some movement of the joints could produce is compatible with every self-stress state, because a self-stress does no work on a movement the truss can make. The truss assembles into the reversed sag with nothing to fight.

The chords alone, cut by their own dead-load elongations. The webs are left at their drawn lengths, which is how a camber is often set out because the chords are spliced and the webs are not. The truss still sags 5.0 mm, since the webs’ stretch is part of the sag, and this list is not quite compatible: it locks 2 kN into a counter.

The top chord at one uniform strain, chosen so that mid-span comes level — the simplest list to fabricate, one percentage on one group of members. Mid-span is level and the rest of the chord within 1.3 mm. But a uniform strain on one chord is far from a real elongation field of this truss, and it locks in 91 kN.

What the incompatible list does #

The uniform list locks force into every panel, and the widest of it is in a vertical, 91 kN against a largest dead-load force of 471 in any member. The pattern is the self-stress states, added in whatever proportions the list’s incompatibility projects onto them: each counter-braced panel has been asked to close a gap its members do not fit, and closing it has put the panel’s diagonal and counter into opposite forces and its chords and verticals into the forces that balance them.

So the answer to the question as posed is reassuring. A redundant truss can be cambered with no locked-in force, by a list that is exactly as easy to compute as the determinate one; and the simple lists that a shop might prefer lock in tens of kilonewtons, not hundreds. The camber was never the problem.

The tolerance locks in more #

The cutting list is what the drawing asks for. What the shop delivers is that list plus its own errors, a millimetre or two on every member, and in a redundant truss those errors are misfits like any other.

With errors of a standard deviation of 1 mm on every member — a good shop’s tolerance — the median draw locks 174 kN into some member, and one draw in twenty more than 288 kN. At 2 mm it is twice that, 348 and 577, and the twentieth draw is beyond the largest force the dead load puts in any member.

The force is large because the panels are stiff. A counter 4.2 m long at 630 MN of axial stiffness resists a millimetre of misfit with about 150 kN, and the panel around it shares that out but does not soften it much. The redundant truss is a stiff structure against its own lengths, which is the same property that makes it stiff against a load — and stiffness is what turns a misfit into force.

So the comparison with the cutting lists is lopsided. The worst of them locks in 91 kN by design; the shop’s ordinary tolerance locks in twice that by chance, in members nobody can predict, and the cutting list cannot do anything about it.

Fit the counters last #

The remedy is an order of assembly. Leave the counters out, or loose, while the truss is erected and its dead load comes on. Without its counters the truss is a Pratt truss, which is determinate: every length error moves a joint and stresses nothing, and the dead load sags the truss to wherever its members’ real lengths put it. Then measure the gap each counter has to fill and cut it to fit.

The locked-in force is then exactly zero, at any tolerance. The counters fit the truss as built, not as drawn, so there is nothing to close. The force they take afterwards is only what loads beyond the dead load give them — live load, wind, temperature — which is what they were there for.

That is the procedure a bridge closure uses for the same reason. A structure that was never complete is assembled determinate and made redundant at the end, with the last members fitted to the gap the rest have left, and the self-stress it carries is the one somebody chose rather than the one the shop’s errors added up to.

And the camber is no worse #

The obvious worry is that a truss assembled without its counters is a truss whose camber is at the mercy of its errors, because the counters are what would have held it to shape. They are not. The scatter of the mid-span level from 1 mm errors is 5.2 mm with the counters fitted in the shop and 5.4 mm with them fitted last — a few per cent apart.

The counters stiffen the truss against a load. Against its own length errors they do almost nothing for the shape, because those errors are imposed deformations that the redundant truss resists only by generating force — the force the previous figure counted — and most of the misfit is still delivered as movement. Fitting them last gives up a few per cent of camber accuracy and saves every kilonewton of the tolerance’s locked-in force.

Why the every-member list fits #

The compatibility of the every-member list has a one-line reason, and it is worth having because it says which other lists are free too.

A self-stress state is a set of member forces $s$ in equilibrium with no load. A movement of the joints $u$ stretches the members by amounts $e$, and the work the self-stress does on that movement is the sum of each member’s force times its stretch, $\sum s_i e_i$. But a set of forces in equilibrium with nothing does no work on any movement of the joints — the principle of virtual work that lets any structure carry the unit load — because the work it does is the work the joint forces do, and the joint forces are zero. So for any stretch pattern that some movement could produce, $\sum s_i e_i = 0$ for every self-stress state — and a misfit locks in force exactly in proportion to how far it fails that.

The every-member list is the dead-load stretch reversed, and the dead-load stretch came from a movement, so it passes. Any list made from a real movement passes: a camber shaped like the sag under some other load, or a camber computed by cutting the truss’s members to the stretches of any displacement field at all. A list made by a rule of thumb — one strain on one chord, a fixed percentage on the webs — passes only by accident.

What the counters are for #

A counter-diagonal earns its place under loads that reverse the shear in a panel — a moving load, a load on half the span — where the Pratt diagonal alone would go into compression. Under this truss’s full dead load the counters carry at most 79 kN, in compression beside diagonals in tension, and they matter much more under the patterns of live load the truss was made redundant for.

That changes what their tolerance is worth. A counter fitted in the shop has to be sized for its share of the live load plus whatever the tolerance happens to lock into it — at the median draw, more than twice its whole dead-load force, and of either sign. A counter fitted last, to the gap, is sized for the live load alone. So the order of assembly is not only a way to keep the truss honest; it is a way to make the counters smaller, and their connections with them, since a connection sized for a force that might be either sign is the larger of two connections.

Two tolerances, not one #

A specification for a redundant truss therefore wants two numbers where a determinate truss needs one. The length tolerance on the members sets how far the built camber can be from the drawn one — 5 mm of scatter at mid-span for a millimetre per member here, the same whether the counters are fitted first or last. The fit-up tolerance on the counters sets how much force they lock in — none if they are cut to the gap, 174 kN at the median if they are cut to the drawing.

The first is a serviceability tolerance and can be generous. The second is a strength tolerance and, if the counters are fitted in the shop, has to be tight enough to keep the locked-in force inside the members’ margin — which at this truss’s stiffness means a fraction of a millimetre, a tolerance most shops would not promise on a 4 m member.

The truss carrying only its misfits #

The free body is the whole truss with its supports, carrying nothing but the forces its members’ misfits create. Equilibrium alone allows any combination of its six self-stress states; compatibility picks one, the combination whose member stretches, added to the imposed misfits, form an elongation field that some movement of the joints could produce.

Written as a stiffness problem, each misfit acts as a pair of forces at its member’s ends — the force the member would exert if it were stretched into place — and the joints move until those forces are balanced by the rest of the truss. The locked-in force in each member is its stiffness times the difference between its stretch and its misfit. A compatible misfit moves the joints until every member’s stretch equals its misfit, and the difference is zero everywhere.

The counter by hand #

A single counter-braced panel 3 m square, its diagonals 4.24 m long at 630 MN, its chords at 1,680 MN and its verticals at 630, has one self-stress state: a unit tension in both diagonals and $1/\sqrt{2}$ of compression in each chord and vertical. A misfit $e$ in one diagonal locks in a force

$$X = \frac{e}{\sum s_i^2 L_i/(EA)_i},$$

and the sum is $2 \times 6.7 + 2 \times 0.5 \times 1.8 + 2 \times 0.5 \times 4.8 \approx 20$ micrometres per kilonewton, so a millimetre’s misfit locks in about 50 kN in that panel alone. Thirty-five members each a millimetre out, in eight panels that share their chords, add up to the 174 kN the median draw found.

What the model assumes #

Members that are as they were cut. The errors are lengths. Joints that are not quite where the drawing put them, bolt holes with clearance and gusset plates that slip are additional sources of misfit, and a bolted truss with holes made bigger so the steel would fit takes up a millimetre or two before any force is locked in — which works like a partial version of fitting the counters last.

A truss that stays elastic. Locked-in force adds to the force the load puts in, and a member that yields under the sum relieves the misfit by yielding. For a truss designed to its elastic capacity that is not a comfort: the yielding member is the one the analysis did not expect.

And a dead load that is known. Fitting the counters to the gap under dead load works only for the load present when they are fitted. Finishes and services added afterwards load a truss that is now redundant, and their share of the sag is resisted by the counters — which is ordinary structural action and locks in nothing a load would not.

What the figures cannot show #

They cannot show the order of erection. Which members go in first decides which misfits are taken up as movement and which as force; fitting the counters last is one choice of order, and there are others between it and fitting everything at once.

They cannot show temperature on the day. A uniform change of temperature moves a truss of one material without stressing it, but a top chord in the sun over a bottom chord in shade is a misfit like any other, a redundancy only the sun can find — 15 degrees of difference is about 4 mm across a 24 m chord — and that one arrives after the counters are fitted and cannot be fitted away.

And they cannot show tension-only counters. If the counters are slender rods that cannot take compression, a locked-in compression simply slackens them, and the truss is a Pratt truss until the load puts them back into tension.

What it comes to #

A redundant truss can be cambered with no locked-in force. Cut every member short by its own dead-load elongation: that list is compatible with every self-stress state and levels the truss exactly.

The simple lists lock in little. Chords alone, 2 kN; one strain on the top chord, 91 kN.

The shop’s tolerance locks in far more. A millimetre per member puts 174 kN into some member at the median draw and 288 at one in twenty.

Fit the counters after the dead load is on, and it locks in nothing — for a few per cent of the camber’s accuracy.

Still open: the truss that is continuous over a support #

Every truss here is simply supported, and its redundancy is inside its panels. A truss continuous over an interior support is redundant in a second way: the reaction at the middle support is a redundant force, and the truss’s camber decides it. A camber cut to level the truss under its dead load on all three supports puts the middle reaction where the elastic analysis says; a camber cut a few millimetres too high at that support jacks the truss up off it and moves load to the ends. Whether a continuous truss can be cambered so that it lands on its middle support with exactly the reaction the design assumed — or whether, as with the counters, the only reliable answer is to set the middle bearing last, to the gap — is the question that takes this argument from the panel to the span.