The pier's share is cut into the members

The counter fitted after the load cambered a truss that could not take a misfit. Its counter-diagonals made it redundant inside its panels, so a cutting list that was not compatible with some shape of the truss locked force into it, and the shop’s ordinary tolerance on member lengths locked in more than the dead load put into the counters. The answer that essay reached was to leave the counters loose, camber the truss as a determinate one, and fit the counters to the gap once the dead load was on.

Every truss in that essay was simply supported. Its redundancy was inside its panels, where a loose counter could remove it. A truss that runs continuously over a middle support is redundant in a way no counter can be left loose from: the reaction at the middle bearing is a force statics cannot find, because three bearings are one more than a beam in a plane needs. Its value comes from how the truss deforms, and a camber is a deliberate deformation. The question is how much of that reaction the camber decides.

Three bearings, and one reaction statics cannot find #

The truss is two equal spans of a Pratt truss, eight panels of 3 m each and 3 m deep, continuous over a middle bearing: 48 m end to end, its diagonals running down towards the middle of each span. A dead load of 60 kN hangs at each of its fifteen interior bottom joints, 900 kN in all. The chords have an axial stiffness of 1,680 MN and the web members 630 MN.

On three level bearings an elastic analysis puts 572 kN on the middle one and 164 kN on each end. Simply supported, each span would have put 240 kN on each of its bearings, so the middle would have carried 480 kN — the two spans’ inner reactions added — and the ends 210 each. The 92 kN difference is what continuity moves to the middle: the truss hogs over the pier, and the hogging lifts the ends’ share onto it.

How large that difference is depends on how stiff the truss is against bending relative to shear. A uniform beam of the same span would put five eighths of each span’s load on the pier, 600 kN. The truss puts 572, because its web members shorten and lengthen under shear and a truss is softer in shear than a solid beam; the deflection that is not bending found the same thing in a girder with a thin web.

Two cutting lists for one truss #

A camber is made by cutting members to lengths other than their drawn ones, so that under the dead load the truss comes to the shape intended. The camber that lives in the member lengths found the cutting list that does this exactly: shorten every member by the amount its own dead-load force stretches it, and lengthen every one the dead load shortens. The dead load then takes each member back to its drawn length and the truss to its drawn shape.

For a continuous truss that list depends on which dead-load forces it is computed from, and there are two plausible ones. The continuous list takes the forces of the truss on three bearings. The simple-span list takes each span on its own, as if it were simply supported — which is how each span’s camber would be computed by anyone who thought of the spans separately, or who copied the camber from a simply supported design of the same span.

Lay each truss on its end bearings before any load is on it. The continuous list gives two humps 15.7 mm high and a valley that just touches the middle bearing: that is the dead-load deflection of the truss on three bearings, upside down, and a deflection on three bearings is nothing at the middle one. The simple-span list gives two taller humps, 42 mm, and a middle node standing 35 mm above its bearing. Each span’s camber is right for that span alone, but a simply supported span’s camber leaves its ends free to turn, and two spans sharing a vertical over the pier cannot each turn their own way: the shared vertical has to stand up, and the pier node rises.

Both trusses are level once the dead load is on, if they are carried as their lists assumed. The simple-span truss can only be carried as two simple spans if the middle bearing stays out of the way until its spans have sagged 35 mm — and the middle bearing does not move.

The list chooses the moment diagram #

Land the simple-span truss on its three bearings and load it. Its dead load closes the 35 mm gap and then pushes the pier node onto its bearing — and the reaction it arrives at is exactly 480 kN, the two simple spans’ share. The truss carries its dead load as two simple spans with no moment over the pier, although it is continuous and sitting on all three bearings. The continuous list lands with 572 kN and carries it as a continuous truss: 715 kN·m of hogging in the panels beside the pier and 915 kN·m of sagging in the spans, against no hogging and 1,395 kN·m of sagging.

That is not two trusses. It is one truss whose redundant reaction has been set two ways, and both are in equilibrium with the same load. A redundant structure carries its load in whatever way its compatibility selects, and compatibility is a statement about lengths. Change the lengths and it selects another way. The support that moved found the same freedom in a continuous beam whose middle support settled: a settlement is a length too, and it moves the reactions without changing the load. A cutting list is a settlement built into the members.

So a camber in a continuous truss is a choice of moment diagram, made in the fabrication shop. Either list makes the truss level under its dead load. Only one of them makes it carry its dead load the way the member sizes were calculated for, and the drawing of the finished truss — level, on three bearings — is the same for both.

The simple-span list’s chords carry half as much again in the spans as the continuous analysis expected. If the members were sized from the continuous analysis, its top chords at mid-span are over-stressed by half and its chords beside the pier are carrying sagging forces of the wrong sign. If they were sized from simple spans, the truss is safe and the continuity, for its dead load, is doing nothing.

The pier’s share is a dimension #

The two lists are two points on one line. Whatever list a truss was cut to, the middle bearing’s height above the line through the end bearings decides the middle reaction, and it does so through a single stiffness: how much force it takes to lift the middle of the truss a millimetre while it spans end to end.

For this truss the stiffness is 2.6 kN per millimetre. A middle bearing set 10 mm high takes 598 kN; set 10 mm low, 545. The two lines are parallel and 35 mm apart along the axis: the simple-span list is exactly the continuous list with its middle bearing 35 mm too low, which is the gap the first figure showed. The pier’s share of a continuous truss’s dead load is set by a dimension, and 2.6 kN per millimetre is the exchange rate.

At 2.6 kN per millimetre a bearing set to ordinary site tolerances — a few millimetres of grout, a shim pack that is one plate thick or thin — moves the reaction by a few per cent. That is the easy part. The harder part is that the bearing’s height is not the only length in the problem, and the others are not measured on site at all.

A millimetre of fabrication on every member #

Every member of a fabricated truss is cut to its list within a tolerance. The counter-braced truss found that a 1 mm tolerance locked in forces larger than the dead load put into its counters. In a continuous truss the same errors have nowhere to lock in except the redundant reaction: the truss is determinate on its end bearings, so the errors bend it freely into some shape, and that shape puts the middle node above or below its bearing.

At a tolerance of 1 mm on each of its 65 members, the truss’s free shape puts its middle node anywhere within about ±14 mm of where the list intended, one standard deviation, and the pier’s share scatters by 36 kN, 6 per cent. One truss in forty lands with 78 kN more than the analysis, 14 per cent, and one in forty with 78 kN less. At 2 mm the scatter doubles, exactly, because the errors are the same draws scaled.

Why a millimetre per member becomes fourteen at the pier is the arithmetic of curvature. A bottom chord member near the middle of a span that is 1 mm too long bends the truss by $1/3000$ of a radian over its own depth — the chord’s extra length divided by the 3,000 mm lever arm to the top chord — and that kink, a quarter of the way along a 48 m truss, moves its middle node by about 4 mm. Thirty-two chord members each contributing a few millimetres of random sign add, as random errors do, to the square root of the sum of their squares, and the web adds a little more.

The band on the figure is what a pair of hydraulic jacks with load cells would set the reaction to: within 2 per cent, ±11 kN. Every truss cut to a 1 mm tolerance is outside it more often than not.

A deeper truss holds its shape and loses its reaction #

Making the truss deeper might seem to help. A chord error kinks the truss by the error over the depth, so a deeper truss kinks less and its middle node strays less. But a deeper truss is also stiffer at its middle bearing, and the reaction’s error is the node’s stray multiplied by that stiffness.

The two trends do not cancel. The stray falls roughly as the inverse of the depth, 34 mm at 1.2 m deep and 9 mm at 4.8, and the stiffness rises roughly as the square of it, from 0.5 to 4.9 kN per millimetre. Their product rises with the depth: the reaction’s scatter is 3 per cent at 1.2 m and 8 per cent at 4.8. A deeper truss is built more accurately in millimetres and less accurately in kilonewtons.

It is the general rule for an imposed deformation in a redundant structure, which the support that moved put in one sentence: the force an imposed deformation causes is proportional to the stiffness of the structure it is imposed on. A stiffer structure is better at carrying loads and worse at tolerating misfits, and a fabrication error is a misfit by another name.

The sun on a continuous top chord #

The members are cut to length once. They do not stay that length: a truss in the sun has a top chord warmer than its bottom chord through the middle of the day, and a chord that is warmer is longer.

A top chord 10 °C warmer than the bottom one arches the truss upward, lifts it off its middle bearing by the equivalent of 12 mm, and takes 31 kN off the pier. A difference of 20 °C, which a dark-painted steel truss under a summer sun reaches easily, takes 61 kN, 11 per cent. A clear night with the top chord radiating to the sky reverses it.

That has two consequences. The first is that the pier’s share is not one number but a daily cycle, and a design that depends on its exact value is depending on the weather. The second is practical: any attempt to set the reaction on site — by jacking, by shimming, by measuring — has to be made with the chords at the same temperature, which in practice means early in the morning before the sun is on the truss, or the setting is wrong by however many degrees the top chord was warm. The movement nobody applied found the same daily cycle in a frame with nothing restraining it but its own continuity.

Set by force, not by geometry #

The counters in the earlier essay were fitted to the gap after the dead load was on, which removed the misfit by construction. The middle bearing of a continuous truss cannot be treated that way. If it were set to the gap — the truss landed on its ends, sagging, and the bearing packed up to meet it — the middle reaction under dead load would be nothing, and the truss would carry its dead load as one 48 m simple span, which is the one way it was certainly not designed to carry it.

What can be done is to set the reaction rather than the height. Land the truss on its end bearings and on jacks at the middle, jack until the load cells read the design reaction, and pack the bearing to the height the jacks have found. The reaction is then whatever the jacks were set to, within their own accuracy, whatever the members’ errors and the camber list were; the 65 members’ tolerances go into the bearing’s height, where they belong, instead of into the reaction. That is the move the jack that creep carries past the mark made for a concrete beam whose staged construction had left its support moments wrong, and it is older than either: it is how continuous girders were set on their bearings before anyone could compute their compatibility.

It changes what the camber is for. Cut to the continuous list, the truss is level under its dead load when it is jacked to the design reaction, because that list was computed with that reaction. Cut to any other list and then jacked to the design reaction, it carries the design forces but is not level: its shape is out by however far the list was from the continuous one. Jacking makes the forces right; only the list makes the shape right; it takes both to have both.

The pier’s stiffness by hand #

The truss’s chords, 3 m apart with 1,680 MN of axial stiffness each, act as a beam of bending stiffness $EA\,d^2/2 = 1{,}680 \times 3^2/2 = 7{,}560\ \mathrm{MN\,m^2}$, and a beam on two end bearings 48 m apart deflects at its middle under a point load $P$ by $PL^3/48EI$. Its stiffness there is

$$k = \frac{48EI}{L^3} = \frac{48 \times 7{,}560}{48^3} = 3.28\ \text{MN/m},$$

or 3.3 kN per millimetre. The truss’s computed stiffness is 2.6, a fifth less, and the difference is the web: the diagonals and verticals stretch under the shear the middle load puts through each span, and a beam with no web deformation has none of that softness. The same web softness is why the truss puts 572 kN on its pier where a uniform beam would put 600.

The simple-span list’s gap follows from the same stiffness. The two lists’ dead-load reactions differ by $572 - 480 = 92$ kN, and $92/2.6 = 35$ mm, the gap the first figure showed. A designer who wanted to know whether a camber was cut for continuous or simple spans would not need the cutting list: lay the truss on its ends, unloaded, and measure the middle node.

What the continuous truss model leaves out #

The bearings are rigid. A real middle bearing on a pier compresses under its reaction, and the pier and its foundation compress further. Both are springs in series with the truss’s own stiffness at the bearing, and a softer support makes every number here less sensitive — the scatter in kilonewtons falls in proportion. A pier on piles a few millimetres soft under 600 kN would take a useful share of the tolerance off the reaction.

The members are pinned and straight. Real joints are gusseted and the chords are continuous through them, so the truss has some bending stiffness in its members as well as its triangulation. That stiffens it a little at the middle bearing and makes the tolerance a little worse; it does not change the finding that the list chooses the reaction.

The tolerance is independent from member to member. A shop that cuts every chord member slightly long because its tape reads short makes a correlated error, which a symmetric truss partly absorbs — a uniform elongation of both chords changes the length of the truss, not its curvature — and a correlated error in one chord only is the thermal case again, with the shop as the sun.

The live load is not here. Traffic and snow are carried by the truss as it stands, with whatever reaction its dead load left, and the continuity carries them elastically whatever the camber was. The camber’s choice of moment diagram applies to the dead load alone, and to whatever was on the truss when it was jacked.

Still open: the truss continuous over two piers #

Here there is one middle bearing and one redundant, and a pair of jacks sets it. A truss continuous over three or four spans has two or three redundants, coupled: jacking one pier changes the reactions at the others, as the beam built towards its own middle found for a continuous beam’s support moments. Whether the reactions of a multi-span truss can be set one pier at a time, in some order that does not undo the earlier settings, or whether they must be jacked together to a set of forces solved for at once — and whether a tolerance-scattered truss of four spans lands farther from its design reactions at the inner piers or the outer ones — is the question the second pier asks of the first.