Structural Engineering & Statics
where the load goes
Structural form

The deck that keeps the hangers in the truss

Crossing the hangers of a tied arch turns it from a frame that bends into a truss that does not, as long as every hanger is pulling. A hanger cannot push, and under a load on half the span some of them would have to. Solved as it really is — with those hangers taken out, and the ones that follow them — the arch keeps the linear answer exactly until the live load on half its span is about equal to its own dead weight, and past that the tie's moment doubles and then quadruples. The arch's own weight is the pretension nobody had to apply, and a heavier deck is a stiffer network.

Assumes Cross the hangers and the bending goes, The thrust that never reaches the ground and Held by something that goes soft.

15 min read 6 figures The load that will not hold stillOne support too manyWhich failure arrives first

Cross the hangers and the bending goes. A tied arch with vertical hangers is a frame with a curved top chord, and a load on half its span is carried by both chords bending; incline the hangers so that they cross and the same chords become a truss, the shear passes through the hangers as axial force, and the chords’ moments fall by an order of magnitude. The price is that a hanger is a rod or a cable and can only pull. Under a half-span load, five of sixteen crossed hangers came out in compression, and the essay counted them as absent and argued for having many hangers, so that losing a third of them still leaves a web.

It counted them; it did not solve without them. A hanger that the linear solution puts in compression is not merely a member with nothing in it. It is a member that is not there, and the structure without it is a different structure, with different forces in every other member — including the hangers next to the one that went, some of which may now be in compression themselves. And it loaded the arch with live load alone, over half the span, with none of the arch’s own weight. The arch’s own weight turns out to be the most important load in the problem.

Eight slack where the linear answer has four

The arch here is 100 m long with a rise of 17 m, hung on 32 hangers inclined at 60° and crossing each other several times, arranged symmetrically so that the right half mirrors the left. Its deck weighs 60 kN/m over the whole span, and a live load of 120 kN/m — twice the dead load — stands on its left half.

A hanger that would push is a hanger that is not there. The force in each hanger of a tied arch of 100 m span and 17 m rise on 32 crossed hangers, carrying 60 kN/m of dead load over the span and 120 kN/m of live load over its left half, against the hanger's position along the tie. Solved as though every hanger could push (pale bars), 4 come out in compression, the worst at 185 kN. Solved with those hangers taken out until no remaining one is compressed (dark bars), 8 are slack — more than the linear solution has in compression, because each hanger that goes passes its share to its neighbours and some of them follow.
Fig. 1 The force in each hanger of the 100 m arch under 60 kN/m of dead load and 120 kN/m of live load on its left half. Solved as though every hanger could push (pale bars), four come out in compression, the worst at 185 kN. Solved with those hangers taken out until no remaining one is compressed (dark bars), eight are slack.

The pale bars are the linear answer. Most hangers are in tension, some heavily — the ones under the loaded half carry over a thousand kilonewtons — but four, all in the unloaded half, are in compression, the worst by 185 kN. The dark bars are the arch solved as it is: those four are taken out, the frame is solved again, any hanger now in compression is taken out too, and the process repeats until every hanger left is pulling and every hanger taken out would be shortened if it were put back. Eight hangers end up slack, not four. Each hanger that goes hands its share to its neighbours, and some of them follow it.

The slack hangers all lean the same way, and all but two of them are in the unloaded half; those two stand just inside the loaded half, beside its edge. A half-span load makes the arch want to sway — down on the loaded side, up on the other — and in the unloaded half the hangers leaning one way are stretched by that sway while those leaning the other way are shortened. The shortened ones are the ones that go. In a network the two families of hangers cross precisely so that one family can take the shear the other cannot, and slackness is the moment when only one family is left in a stretch of the span: still a truss, but a truss whose diagonals all point one way, which carries shear in one direction only.

This iteration is the same one that a footing on ground that cannot pull needs, and that a counter-braced truss needs for its diagonals: a member or a spring that only acts in one direction makes the structure’s stiffness depend on its own answer, and there is no way to the answer except by guessing the set of active members and checking it. It converges in four passes here.

The weight that comes free

Under the arch’s own weight alone, every one of the 32 hangers is in tension — the least by 29 kN, the most by 282. That is not a coincidence of these numbers; a uniformly loaded arch of this shape hangs its deck evenly, and every hanger, whichever way it leans, carries a share. The dead load is a pretension that nobody had to apply. Before any live load arrives, every hanger has been stretched by the deck it holds, and a live load can slacken a hanger only by putting more compression into it than the dead load put tension.

The dead load holds the hangers taut until the live load outweighs it. How many of the 32 crossed hangers of a tied arch of 100 m span and 17 m rise, carrying 60 kN/m of dead load over the span, are slack, against the live load on its left half as a multiple of the dead load: counted from the linear solution (dashed) and solved with the slack ones taken out (solid). At 0.50, 1; at 1.00, 1; at 1.50, 4; at 2.00, 8; at 3.00, 11. The arch's own weight stretches every hanger before any live load arrives, so the dead load is pretension nobody had to apply; slackening begins in earnest once the live load on half the span exceeds it, and once hangers start to go, the ones beside them carry the difference and follow.
Fig. 2 How many of the 32 hangers are slack against the live load on the left half as a multiple of the dead load: counted from the linear solution (dashed) and solved with slack hangers taken out (solid). At half the dead load, one; at the dead load, one; at one and a half times it, four; at twice, eight; at three times, eleven.

So the count of slack hangers is a function of one ratio: live load on half the span over dead load. Up to about one, only the end hanger with the least dead-load tension goes, and it carries little either way. Past one the count climbs — four at one and a half, eight at two, eleven at three — and the solid line pulls away from the dashed one, because the cascade adds hangers the linear solution never flagged.

Where the linear answer stops being right

The question that matters is not the count but the moment, since the whole reason for crossing the hangers was to keep the chords from bending.

The linear answer holds until the live load outweighs the dead. The worst bending moment in the tie of a tied arch of 100 m span and 17 m rise on 32 crossed hangers, carrying 60 kN/m of dead load over the span, against the live load on its left half as a multiple of the dead load, on a logarithmic scale: solved linearly as if hangers could push (dashed), solved with the slack ones taken out (solid), and for the same arch on vertical hangers (dotted). At 1.00, 291 kN·m against 291 kN·m linear and 3,296 kN·m vertical; at 1.50, 445 kN·m against 387 kN·m linear and 4,926 kN·m vertical; at 2.00, 998 kN·m against 491 kN·m linear and 6,557 kN·m vertical; at 3.00, 2,537 kN·m against 700 kN·m linear and 9,818 kN·m vertical. Up to about the dead load the two network answers agree; past it the slack hangers open gaps in the web, the chords bend across them, and the linear answer is out by a factor of 2.0 at twice the dead load and 3.6 at three times — still a fraction of what vertical hangers give.
Fig. 3 The tie’s worst bending moment against the live load on half the span as a multiple of the dead load, on a logarithmic scale: linear (dashed), with slack hangers taken out (solid), and for the same arch on vertical hangers (dotted). At the dead load, 291 kN·m both ways against 3,296 vertical; at one and a half times it, 445 against 387; at twice, 998 against 491; at three times, 2,537 against 700, with 9,818 vertical.

The two network answers are the same, to the kilonewton-metre, while the live load is no more than the dead load: 291 kN·m. Past it they separate. At one and a half times the dead load the real moment is 445 kN·m against the linear 387, fifteen per cent more; at twice, 998 against 491, a factor of two; at three times, 2,537 against 700, a factor of 3.6. The linear analysis of a network arch is exact until the live load on half the span outweighs the dead load, and then it fails fast. It never fails as badly as vertical hangers do — 9,818 kN·m at three times the dead load, still four times the slack-aware network — so the crossed arrangement keeps most of its advantage. But the advantage it keeps is not the one the linear answer reports.

Where a hanger goes slack, the tie bends. The bending moment along the tie of a tied arch of 100 m span and 17 m rise on 32 crossed hangers, carrying 60 kN/m of dead load over the span and 120 kN/m of live load over its left half: solved linearly (dashed) and with the 8 slack hangers taken out (solid). Linearly the tie's moment never exceeds 491 kN·m; with the slack hangers out it reaches 998 kN·m, in the stretches of tie where neighbouring hangers have gone and the tie spans across the gap like a beam between the hangers that remain.
Fig. 4 The bending moment along the tie under twice the dead load on the left half: linear (dashed) and with the eight slack hangers taken out (solid). The linear tie never exceeds 491 kN·m; the real one reaches 998 kN·m in the stretches where neighbouring hangers have gone.

The tie’s moment diagram shows where the extra moment comes from. The linear tie is a beam on closely spaced hangers, its moment a row of small humps between hanger points. Where hangers have gone slack the humps are longer and taller: the tie spans across the gap left by the missing hangers, and the panel of chord there is no longer part of a truss but a short stretch of a frame with no diagonals, carrying its shear by bending. The rest of the tie is untouched. A network arch with slack hangers is a truss with a few Vierendeel panels in it, and the panels are where the slack hangers were.

The rib pays as well

The tie is where the missing hangers show first, because it is the chord the load hangs from. The arch rib above it loses the same hangers, and it pays in the same currency.

Under the dead load and up to a live load equal to it, the rib’s worst moment is the linear one, 573 kN·m at the dead load. At one and a half times the dead load the two still agree, 716 against 713. At twice the dead load the rib carries 1,553 kN·m against a linear 852; at three times, 4,471 against 1,131 — a factor of four, larger than the tie’s. The rib is the more slender chord, and in a network arch it is sized almost entirely for axial thrust, on the understanding that the crossed hangers will keep its bending small. A rib designed to the linear moment at three times the dead load would be carrying four times the bending it was given.

The rib also loses something the moment does not show. Each hanger that pulls on it is, for the rib, a restraint against moving in its own plane: the crossing hangers tie every few metres of rib to the tie below, and a rib held at close intervals can carry its thrust without bending sideways in its own plane. A run of slack hangers is a run of rib with no such tie, and the in-plane buckling length of that stretch grows with every hanger that goes. The slack-aware analysis that finds the moment also finds the stretch, and a rib check that uses the linear hanger forces uses a buckling length the arch does not have.

There is a family resemblance here to a shallow arch that turns aside before it snaps. In both, the arch’s symmetric or linear behaviour is right up to a threshold and then wrong in a specific direction, because the structure has found a mode the simple analysis forbade it: there, a sideways shape; here, a set of hangers the analysis assumed were working. Neither shows up in the analysis that excludes it, and both are reached by a load that is not unusual.

Where road and rail sit

The ratio that decides everything — live load on half the span over the dead load on all of it — is set by what the bridge carries and what its deck is made of, and the two kinds of bridge that use network arches sit at opposite ends of it.

A road bridge with a concrete deck has a heavy dead load: the slab, the surfacing, the edge beams and the tie itself. Its live load is a lane load plus axle loads, spread across the width, and over half a 100 m span it is a modest fraction of the deck. The ratio is well under one, the hangers stay taut through every pattern, and the linear analysis is the right one. That is the bridge Tveit designed for.

A railway bridge carries a train that can weigh as much per metre as the deck it runs on, concentrated on one track and present on half the span for long stretches of a crossing. Its ratio can be one or two, and on a light steel deck more. That bridge is in the region where the cascade begins, where every passage of a train slackens some hangers and re-tensions them as it leaves. For a rail network arch the slack-aware analysis is the design, not a refinement of it, and the hangers’ fatigue under repeated slackening becomes a design case of its own — a load that arrives suddenly, each time a slack hanger snaps taut.

More hangers, wider error

The first essay’s answer to slackness was to have more hangers, so that losing some does not matter. Solved properly, that answer still holds, but less well than the linear analysis says.

More hangers lower the moment and widen the error. The worst moment in the tie of a tied arch of 100 m span and 17 m rise, with 60 kN/m of dead load and 120 kN/m of live load on half the span, against the number of crossed hangers: linear (dashed) and with slack hangers taken out (solid). 16 hangers: 1,878 kN·m, 2 slack, against 1,706 kN·m linear; 20 hangers: 1,180 kN·m, 4 slack, against 1,122 kN·m linear; 24 hangers: 1,035 kN·m, 5 slack, against 862 kN·m linear; 32 hangers: 998 kN·m, 8 slack, against 491 kN·m linear; 40 hangers: 788 kN·m, 10 slack, against 431 kN·m linear; 48 hangers: 802 kN·m, 11 slack, against 362 kN·m linear. More hangers lower the moment overall, though not at every step, since which hangers go slack depends on where each one lands; and the linear answer falls faster than the real one throughout, because it counts the work of hangers that are not there — by 48 hangers it is less than half the moment the tie actually carries.
Fig. 5 The tie’s worst moment against the number of crossed hangers, with 60 kN/m of dead load and 120 kN/m of live load on half the span: linear (dashed) and with slack hangers taken out (solid). With 16 hangers, 1,878 kN·m against 1,706 linear; with 32, 998 against 491; with 48, 802 against 362.

More hangers lower the real moment — from 1,878 kN·m with sixteen to about 800 with forty or more — though not at every step, since which hangers go slack depends on exactly where each one lands. The linear answer falls faster: from 1,706 to 362. With sixteen hangers the two differ by ten per cent; with 48 the linear answer is less than half the real moment. The more hangers a network has, the more of its apparent stiffness belongs to hangers that are not there, because the linear solution distributes the shear over all of them and the real arch only over the ones that are pulling.

That does not argue against many hangers; the real moment still falls. It argues that the benefit of the forty-eighth hanger cannot be taken from a linear analysis, and that the analysis which sizes the chords has to be the one that lets hangers go.

A heavier deck is a stiffer network

The ratio that governs is live load over dead load, and the dead load can be chosen.

A heavier deck lowers the moment the live load causes. The worst moment in the tie of a tied arch of 100 m span and 17 m rise on 32 crossed hangers, carrying 120 kN/m of live load on half its span, against the dead load over the whole span: linear (dashed) and with slack hangers taken out (solid). With 20 kN/m of dead load the tie carries 2,773 kN·m (15 slack); with 60, 998 kN·m (8); with 120, 582 kN·m (1); with 200, 781 kN·m. The linear answer rises slowly with the dead load, which adds its own small moments; the real one falls steeply until the hangers stay taut, because the deck's weight is what keeps the web in the structure, and only then rises with the linear one.
Fig. 6 The tie’s worst moment under 120 kN/m of live load on half the span, against the dead load over the whole span: linear (dashed) and with slack hangers taken out (solid). With 20 kN/m of dead load, 2,773 kN·m and fifteen hangers slack; with 60, 998 and eight; with 120, 582 and one; with 200, 781.

Under the same live load, a light deck of 20 kN/m leaves fifteen hangers slack and the tie carrying 2,773 kN·m. A deck of 60 kN/m leaves eight and 998. A deck of 120 kN/m leaves one, and the tie carries 582 kN·m. Past that the moment rises again, slowly, with the dead load’s own contribution, as the linear answer always does. Adding weight to the deck lowers the moment the live load causes, by a factor of nearly five, because the weight keeps the web in the structure.

That is the reason network arches are so often built with concrete decks. Per Tveit, who developed the form, paired it with a concrete tie that is heavy, stiff in its own plane and cheap, and a road bridge’s live load on half its span is a modest fraction of such a deck’s weight: the hangers stay taut and the linear analysis is right. A railway network arch, or one with a light steel deck, sits much further along the ratio. Its live load on half the span can be the deck’s weight or more, and it is the arch for which the slack-aware analysis is not a refinement but the design.

The other way to move along the ratio is to add tension that is not weight: pretension the hangers when they are installed. That is the same self-stress the first essay mentioned — a locked-in tension that shifts every hanger’s range upward — and its arithmetic is exactly the dead load’s, without the dead load’s cost in the arch and foundations. The hangers then have to be sized for the pretension plus their working range, which for a slender rod in fatigue is not free.

Thirty-two hangers, by hand

The onset of slackness can be read hanger by hanger from two linear solutions, because before anything goes slack the arch is linear and its hanger forces add. Under the dead load alone the hangers carry 208 kN on average — the deck’s 60 kN/m over a hanger spacing of 3.1 m, about 190 kN of vertical load, divided by sin 60° — and between 29 and 282 kN individually. Under a live load on the left half equal to the dead load, each hanger’s force changes by a fixed amount, and a hanger slackens when the live load reaches its dead-load force divided by that change:

Hanger at Dead-load force Change per unit of live over dead Slackens at a ratio of
97.9 m 29 kN −65 kN 0.45
60.4 m 238 kN −211 kN 1.13
54.2 m 210 kN −149 kN 1.41
66.7 m 259 kN −170 kN 1.52
79.2 m 273 kN −98 kN 2.79

The end hanger, nearly unloaded by the deck, goes first and matters least. The first hanger that matters — at 60.4 m, just past the edge of the live load, where the shear of a half-span load is largest — goes at a ratio of 1.13, which is why the two network curves part at about one. Once it has gone, the table no longer holds: its share passes to its neighbours, and the ones at 54.2 and 66.7 m go earlier than their own linear onsets say. That is the cascade, and it is the part a hand calculation cannot follow.

A static load, a perfect cable

The calculation rests on choices that limit it.

A slack hanger is exactly slack. A real hanger rod has a little bending stiffness and some sag, so it carries a small compression before it buckles and goes slack gradually; a cable is closer to the model. Neither changes the count much.

The load is static and the half-span pattern is the worst. A moving train is a sequence of patterns, and the hangers slacken and re-tension as it passes; a hanger that goes slack and snaps taut again is loaded dynamically, which is one reason slackening is avoided even where the chords could carry the moment. The pattern that slackens the most hangers is an envelope over arrangements, searched for a shape rather than a maximum.

The chords are elastic and the arch does not buckle. Slack hangers also leave the arch rib with longer unbraced lengths in its own plane, which lowers its in-plane buckling load; that is a second consequence of slackness the moment alone does not show.

Every hanger is the same. Real network arches vary the hanger slope along the span to keep the end hangers taut, and the end hanger here is the first to go for that reason.

Still open: the arch that lets its hangers slacken on purpose

Every design rule for network arches is written to keep the hangers taut, by dead load, pretension or geometry. The slack-aware solution says what happens if they are allowed to go: the tie’s moment doubles at twice the dead load, which a heavy tie might carry at no great cost, while the hangers that remain carry little more than they did. Whether a network arch with a light deck and a tie sized for the slack-aware moment is cheaper than one with a heavy deck or pretensioned hangers that never slacken — once the fatigue of a hanger that goes slack and snaps taut under every train is counted — is the question of whether slackness is a failure to be designed out or a load case to be designed for.

Where this goes next

Cross the hangers and the chords stop bending. The same tied arch, the same sixteen hangers, the same load on half the span — hung vertically and hung as a network. Vertical hangers make the two chords a Vierendeel frame, which has no truss action at all, so a partial load is carried by bending: 3316 kNm in the tie and 6234 in the arch. Inclined hangers can carry the shear between the chords axially, and the same load gives 686 and 831 — factors of 4.8 and 7.5. The thrust is identical in both, because that is decided by the span and the rise and nothing else. Cross the hangers and the bending goes A tied arch with vertical hangers is a Vierendeel frame with a curved top chord — it has no truss action at all, so a load on half the span is carried by bending. Incline the hangers so they cross and the same two chords become a truss. The shear goes round the corner instead of across it. A 6-panel Vierendeel girder, 12 m by 1500 mm, under 100 kN at mid-span. There is no diagonal in it, so each panel's 50 kN of shear is carried as bending in the chords: the curves drawn along them are the chord moments, and every one passes through zero at the middle of its own panel. The local moment is the panel shear times the panel length over four, 25.0 kNm, and it adds to an axial force of 200 kN from the global moment at the same point. The girder deflects 6.20 mm against 3.18 mm for the same members triangulated — 1.95 times — and 68% of that movement is chord bending that a diagonal would have removed entirely. The truss with no diagonals A rectangle without a diagonal is a mechanism, so a Vierendeel girder makes its corners rigid instead. The shear a diagonal would have carried as axial force now travels as bending in the chords, and bending is a far more expensive way to move a force. Fit the counters last, and the tolerance locks in nothing. The largest force random length errors lock into an eight-panel Pratt truss with a counter-diagonal in each of its six interior panels, 24.0 m long and 3.0 m deep, carrying 60 kN of dead load at each interior bottom joint, against the errors' standard deviation: the median draw (solid) and the 95th percentile (dashed) with every member fitted in the shop, and the same with the counters left loose until the truss carries its dead load and then cut to the gap (on the axis). With everything fitted, 174 and 288 kN at 1 mm, 348 and 577 at 2 — straight lines, because the force is proportional to the misfit. With the counters fitted last, nothing at any tolerance: the Pratt truss without its counters is determinate, and a determinate truss takes any length error by moving. The counter fitted after the load A truss with a counter-diagonal in every panel cannot take a change of length for free, so a camber cut into its members might seem to lock force into it. One cutting list locks in nothing — every member shortened by its own dead-load stretch — and the simpler lists lock in little. What locks in force is the shop: a millimetre of error per member puts 174 kN into some member of a truss whose largest dead-load force is 471. Leave the counters loose until the dead load is on and cut them to the gap, and the same errors lock in nothing at all.

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CableDead loadHangerLive loadLoad arrangementNetwork archNonlinearityTied arch