Structural form

The deck that is its own cable

Every other cable structure hangs something from the cable. A stressed ribbon hangs nothing — the walking surface is the catenary, laid at a fiftieth of the span rather than a tenth, because a footbridge has to be walkable. That one decision hands the abutments six and a quarter times the entire weight of the bridge.

Assumes The shape that carries itself, and the arch that is its reflection, The stiffness that comes from the shape and The deck is not there to carry the load.

A suspension bridge has a cable and a deck, and the two are different objects doing different jobs: the cable carries and the deck distributes. A cable-stayed bridge has stays and a deck, and again they are different. A stressed ribbon has neither arrangement. The deck is the cable — a prestressed concrete slab, three hundred millimetres thick, hanging in a catenary between two abutments with nothing above it and nothing below it — and people walk on the curve.

Everything about the structure follows from one decision, and the decision is not structural.

The abutment force is the sag turned upside down. A 100 m ribbon carrying 35 kN/m at a sag of 2.0 per cent of its span. H = wL²/8f, so the horizontal force at each abutment is 21875 kN — 6.25 times the entire weight of the deck, and five times what a suspension bridge of the same span and weight at a tenth would have needed. The curve is a reciprocal and it has no flat part: halving the sag doubles the force, at any sag. What stops a designer flattening it further is not the ribbon, which is in tension and cannot buckle. It is what the ground at each end will take, and at 6.25 deck-weights that is usually rock or a very large anchor block.
Fig. 1 A 100 m ribbon carrying 35 kN/m at a sag of 2 per cent of its span, against the same structure at every other sag. H = wL²/8f is a reciprocal with no flat part: at a tenth of the span the abutment force is 1.24 times the whole weight of the deck, and at a fiftieth it is 6.25. The dashed comparison is the sag a suspension bridge would use.

The sag is set by a ramp gradient

A parabola of sag ff over a span LL leaves each end at a slope of 4f/L4f/L. That is not a small number at the sags cables like: at f/L=1/10f/L = 1/10 the deck arrives at the abutment at 40 per cent — a 22° slope — which is a roof, not a footpath.

Accessibility limits for a pedestrian route sit around 8 per cent, or 1 in 12.5. Setting 4f/L0.084f/L \le 0.08 gives

fL0.02,\frac{f}{L} \le 0.02,

exactly. The sag of a stressed ribbon is a wheelchair gradient rearranged, and it comes out at a fiftieth of the span for reasons that have nothing to do with any force in the structure. The 100 m span drawn therefore sags 2 m, arrives at 4.57°, and hands over the rest of the design as a consequence.

The funicular polygon for five loads. The shape a string takes under 5 point loads, with a vertex at every load and a constant horizontal component of 306.7 throughout. The end segments carry the most — 307.8 against 306.7 in the flattest one — because they are steepest.
Fig. 2 The funicular shape for the load it carries, which is what the ribbon takes up on its own the moment it is built. Under uniform load that is a parabola, and the ribbon has no bending in it anywhere except where something stops it being that curve.

What the abutment is asked for

H=wL28f=35×10028×2=21,875 kNH = \frac{wL^2}{8f} = \frac{35 \times 100^2}{8 \times 2} = 21{,}875\ \text{kN}

against a total deck weight of wL=3,500wL = 3{,}500 kN. The horizontal pull at each end is 6.25 times the weight of the entire bridge, and it is permanent, and it does not go away when the bridge is empty.

That number is what makes stressed ribbons rare rather than common, because it is a foundation problem rather than a structural one. It is also a number that gets no smaller when the bridge gets lighter: H/wL=L/8fH/wL = L/8f contains no load at all, so the abutment force in deck-weights is a pure function of the sag ratio. Making the deck out of carbon fibre would halve the force and leave the ratio exactly where it was, and the anchorage would still be sized at 6.25 times whatever the deck weighs. Twenty-two meganewtons of horizontal thrust has to be resisted by something: rock anchors into a competent stratum, a very large gravity block, or a tie under the ground connecting the two abutments — which is the same three answers an arch’s thrust gets, because it is the same problem with the sign of the curvature reversed.

The cable and the arch are the same curve. The shape a cable takes under a uniform load is a parabola, and it carries that load in pure tension. Reflected, the identical curve carries the same load in pure compression, which is what an arch is. A catenary of the same span and sag is drawn faintly against it: that is the shape of a cable carrying its own weight rather than a load spread evenly along the horizontal, and the two are close but not the same curve.
Fig. 3 The duality stated plainly. A cable in tension and an arch in compression under the same load take the same curve and deliver the same horizontal force to the same abutments. A stressed ribbon is a very flat inverted arch that people walk on, and the shallower an arch is the harder it pushes — the identical reciprocal, arrived at from the other side.

Two things soften it slightly and neither by much. The full live load raises HH from 21,875 to 34,375 kN, a factor of 1.57, so the anchorage is sized for the loaded case and lives most of its life at two thirds of it. And a ribbon can be built with its two abutments tied to one another below ground when the geometry permits, which converts the whole structure into a self-anchored system and the thrust into an internal force — at the cost of a tie 100 m long carrying 22 MN.

The end that cannot be clamped

The ribbon has bending stiffness. It is a concrete slab, not a chain. But that stiffness is almost entirely irrelevant, and where it is not irrelevant it is catastrophic.

Away from the ends, the deck’s own EIEI competes with the geometric stiffness the tension provides, and it loses badly: the length over which bending matters at all is

=EI/H=324,000/21,875=3.85 m,\ell = \sqrt{EI/H} = \sqrt{324{,}000/21{,}875} = 3.85\ \text{m},

which is 3.85 per cent of the span. Over the other 96 per cent the ribbon is a cable and its flexural stiffness contributes nothing.

A cable alone goes to a kink, and a kink is not a road. A point load of 1000 at mid-span of a 900 m suspended deck. The upper shape is the cable with no girder at all: two straight lines meeting under the load, because a cable takes the funicular shape of whatever is on it and the funicular of a point load is a kink — 0.0083 radians of it here. The lower shape is the same cable with the girder present, peaking at 1.125 against the bare cable's 1.873. The girder is not carrying the load — it takes only 17% of it — it is spreading it, over a characteristic length of √(EI/H) = 183 m, and what reaches the cable is spread over that length rather than arriving at a point.
Fig. 4 The same decay length from the other direction. A beam on a tensioned string spreads a local disturbance over √(EI/H) and no further, so a stiff deck on a slack cable is a beam and a flexible deck on a taut one is a cable. A stressed ribbon is at the extreme end of that scale — the tension is enormous and the deck is thin.

At the abutment the deck has to be turned from its 4.57° end slope to whatever the approach requires. If it were clamped, the moment would be

M=θEIH=0.08324,000×21,875=6,735 kNm,M = \theta\sqrt{EI\,H} = 0.08\sqrt{324{,}000 \times 21{,}875} = 6{,}735\ \text{kNm},

which is 4.21 times the deck’s own capacity. The clamped detail is not conservative or expensive; it is impossible.

A three-pinned arch, rise 2.6 on span 9. A three-pinned arch under a uniform load. One moment equation about the crown hinge gives a horizontal thrust of 23.37, with no stiffness and no assumption about the section. The thrust line lands on the axis everywhere, so there is no bending anywhere in the arch.
Fig. 5 What has to happen instead. The end curves over a saddle rather than meeting the abutment at a point — radius EI/M_cap = 203 m, turned through the end slope over an arc of 16.2 m, so that the curvature the deck is asked for is one it can actually take. The saddle is not a refinement of the detail. It is the detail, and it is a sixth of the span long.

The alternative, used where a saddle will not fit, is to haunch the deck locally — two or three times the depth over the last few metres — which raises McapM_{cap} faster than it raises \ell, since the first goes as d2d^2 and the second as d3/2d^{3/2}.

Which free body produced the number

Every quantity above comes from one of two cuts, and it is worth naming them because the second is the one people skip.

The whole ribbon, cut at mid-span. The half-bridge is in equilibrium under half the weight, the reaction at one abutment, and the horizontal force HH acting through the mid-span point. Taking moments about the abutment gives Hf=wL2/8H f = wL^2/8 directly, which is the entire derivation of the headline number. Nothing in it is about concrete, prestress, or the deck’s depth; it is the same moment equation a funicular polygon draws with a ruler.

A short length at the abutment. This one has the deck’s bending in it. Cut a metre of ribbon just inside the support and the forces on the cut face are the tension along the deck, a shear, and a moment — and the moment exists only because something is stopping the deck being the curve it wants to be. Away from the ends nothing does, so the moment is zero and stays zero. That is why the whole of the flexural design of a stressed ribbon happens in the last five metres at each end.

The contrast between the two cuts is the structure’s whole character. One free body decides everything about the forces and knows nothing about the deck; the other decides everything about the deck and is 4 per cent of the span long.

Temperature, which is the reason it is prestressed

A cable’s arc length is longer than its chord by

sL8f23L,s - L \approx \frac{8f^2}{3L},

107 mm for the ribbon drawn. That is a small number and it depends on f2f^2, which is what makes the ribbon so sensitive: differentiating, a change of arc length dsds produces

df=3L16fds,df = \frac{3L}{16f}\,ds,

and at f=L/50f = L/50 the multiplier 3L/16f3L/16f is 9.375. A 30 °C rise lengthens 100 m of concrete by 30 mm, and 30 mm of extra arc becomes 281 mm of extra sag — 14 per cent of the sag itself, and enough to change the horizontal force from 21,875 kN to 19,178.

The stress has no length in it and the movement is nothing but length. Stress and movement against member length, for a 30 °C change. Held rigidly, the stress is 10.8 MPa at every length there is — E·α·ΔT, with no L, no A and no I anywhere in it. Held by a spring of 100 kN/mm the answer climbs with length rather than falling, because a longer member hands the same spring more movement to absorb: 83% of full restraint at 10 m and 97% at 60 m. The free movement, plotted to its own scale, reaches 18 mm.
Fig. 6 The general statement behind that arithmetic: an axial strain is a movement proportional to length, and a flat cable is a geometry that multiplies a small movement into a large one. The multiplier is 3L/16f, and it is the reason a stressed ribbon’s shape is a function of the weather.

This is why the deck is prestressed rather than merely tensioned by its own weight. Prestress raises HH above what equilibrium requires, which does two things at once: it stiffens the geometry against sag change, and it keeps the whole section in compression so that the curvature reversals at the ends — and the tension a live load on half the span produces — never crack it.

The further it deflects, the harder it pulls back. Total load against midspan sag for a 100 m cable of 1000 mm² prestressed to 500 kN, carrying 5 kN/m. The cubic H³ − T₀H² − w²L²EA/24 = 0 was bisected at every point of the curve, so the sag at the full 500 kN is 4.536 m rather than the 12.500 m the flat-cable formula WL/8T₀ gives — the straight dashed line, which is the tangent to this curve at the origin and nothing more. Its slope is the initial stiffness 8T₀/L = 40.0 kN/m; at the marked point the tangent has reached 250.7 kN/m, 6.27 times as stiff, and the horizontal component of the tension has risen from 500 kN to 1378 kN. Nothing about the steel changed. The geometry got better at the job.
Fig. 7 Why prestress buys stiffness rather than strength. A cable’s stiffness is geometric: it resists a load by changing shape until the shape suits the load, and prestress reduces how much shape change is needed. In a ribbon that stiffness is the only kind there is over 96 per cent of the span.

Why it is lively

The last consequence of hanging a thin slab on a very large tension is the frequency, and it is the property that has closed more of these bridges than any structural check.

The ribbon behaves as a taut string, so its modes are

fn=n2LHμ=n×0.427 Hz.f_n = \frac{n}{2L}\sqrt{\frac{H}{\mu}} = n \times 0.427\ \text{Hz}.

Evenly spaced. Not merely low — a beam’s modes go as n2n^2 and spread out, so a beam has a first mode and then a gap; a string has modes at 0.427, 0.854, 1.281, 1.708, 2.135 Hz and onward, one every 0.427 Hz forever.

The first three modes of a simply supported beam. Three modes of a simply supported beam of 100 m span, drawn from the general solution with the constants fixed by the support conditions rather than assumed to be sines. Mode 1 is at 0.03 Hz with βL = 3.1416; Mode 2 is at 0.13 Hz with βL = 6.2832; Mode 3 is at 0.28 Hz with βL = 9.4248. The frequencies go as the square of βL, so the threeth mode is 9.0 times the first. The marked points are the nodes.
Fig. 8 The contrast. A beam’s modes climb as the square of the mode number and its spectrum is sparse at the bottom, so a designer can put the first mode above the pacing band and be finished. A string’s are an arithmetic progression, so there is always a mode within 0.21 Hz of any pacing rate a pedestrian can adopt, and the fourth and fifth modes here sit at 1.71 and 2.14 Hz, in the middle of ordinary walking.

The damping of a prestressed concrete ribbon is around 1 per cent, and there is very little mass per unit length to hide a crowd in. Every stressed ribbon of any span built since about 1990 has a tuned mass damper on it, and what a damper does is buy the only thing that stops a resonance.

What it is for

It is worth being clear about what this arrangement buys, because the list above is nearly all cost.

It is the lightest way to cross a gap. There is no separate cable, no tower, no stay, no hanger, no truss and no pier. The structure is a slab three hundred millimetres thick and nothing else, and for a 100 m footbridge that is a smaller quantity of material than any alternative by a wide margin. Ranking materials and forms is a question about the load case, and for a long light pedestrian span with good rock at both ends this one wins outright.

It needs almost no maintenance. A prestressed concrete slab in permanent compression has no bearings, no expansion joints, no cables to inspect and no coating to renew. The two details that do need attention — the saddles and the anchorages — are at the ends, on land, and reachable.

It is very difficult to make ugly. The shape is the funicular of its own weight, which is to say the structure is showing exactly what it is doing, and there is nothing else in the picture. That is not a structural argument and it is the reason most of these bridges got built.

The abutment force is the sag turned upside down. A 40 m ribbon carrying 35 kN/m at a sag of 2.0 per cent of its span. H = wL²/8f, so the horizontal force at each abutment is 8750 kN — 6.25 times the entire weight of the deck, and five times what a suspension bridge of the same span and weight at a tenth would have needed. The curve is a reciprocal and it has no flat part: halving the sag doubles the force, at any sag. What stops a designer flattening it further is not the ribbon, which is in tension and cannot buckle. It is what the ground at each end will take, and at 6.25 deck-weights that is usually rock or a very large anchor block.
Fig. 9 And the reason there are not more of them, drawn at a shorter span. The abutment force in deck-weights does not fall — it cannot, since it is L/8f and the sag ratio is fixed by the gradient — so a 40 m ribbon still pulls 6.25 times its own weight on ground that a 40 m beam would have loaded only vertically. Below about 30 m nobody builds one, because a beam needs no anchorage at all.

Where the model stops

The cable is a parabola. Under uniform load along the span it is; under uniform load along the arc, which is what self-weight actually is, it is a catenary. At a sag ratio of 1 in 50 the two differ by less than a millimetre and the parabola is the honest choice, but nothing in the arithmetic here would survive a deeper sag.

The deck is one member with one EIEI. A real ribbon is precast segments post-tensioned together, so it has a joint every few metres, and its bending stiffness depends on whether those joints are open — which depends on the prestress, which is the quantity being designed. The stiffness is a function of the load case.

Half-span loading is quoted as a force and not as a shape. A cable under an unsymmetric load changes shape rather than merely stretching, and the deflected form is not a scaled version of the dead-load one. Getting it right needs the cable equation solved with the geometry updated, which is a nonlinear problem the numbers here approach only through HH.

The abutment is a number. The 21,875 kN has to be resisted by ground, and ground moves: a few millimetres of abutment spread lengthens the chord, which flattens the sag, which raises HH, which spreads the abutment further. Whether that converges is a question about the foundation’s stiffness rather than the bridge’s, and it is the check a ribbon actually lives or dies by.

The live load is uniform. It is not — a crowd is patchy, and half the span loaded is the case that moves a cable most, because it is the case that asks the shape to change rather than merely to stretch. The 1.57 ratio quoted for HH under full load is the easy case.

And no drawing here shows a person on it. The dynamic behaviour that decides whether the bridge is usable is a crowd, correlated in time, on a structure with 3,000 kg per metre and 1 per cent damping — and the picture of that is a lock-in rather than a load.

The ladder from here

Later rungs on this anchor: the nonlinear cable analysis proper, with the shape updated as the load goes on and the tangent stiffness recomputed at each step. The multi-span ribbon over intermediate piers, where each span’s thrust nearly cancels its neighbour’s and the pier carries only the difference — which is how the long ones are built. Erection, which is the hardest part: the ribbon is a catenary of bare tendons before it is a deck, its sag at that stage is very different, and the segments are hung on it in an order that determines the finished geometry. The saddle detail resolved as a beam on a curved support with slip. Ribbon-and-arch hybrids, where an arch below the deck takes the thrust and the ground takes almost nothing. And the anchorage itself, which is a rock-mechanics problem rather than a structural one and is the reason these bridges appear in gorges and almost nowhere else.

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Cable stiffnessFunicularGeometric stiffnessLoad pathNatural periodPrestressServiceabilityThrust