Structural form

The deck is not there to carry the load

A cable takes the shape of whatever is on it, which is exactly the problem — under a point load its shape is a kink, and a kink is not a road. The stiffening girder exists to spread the load until what reaches the cable is something the cable's own shape is right for.

Assumes The shape that carries itself, and the arch that is its reflection, The stiffness that comes from the shape and The beam that sits on the ground.

A cable under load takes the funicular shape of that load, and the fact is usually stated as a virtue. It is the founding result of the structural-form field: a shape that carries itself does so in pure tension, with no bending anywhere, and the arch is its reflection. Stated as a limitation instead, the same sentence reads: the cable’s shape is decided by the load, so change the load and the shape changes. Under a single point load the funicular shape is two straight lines meeting at a corner.

A cable alone goes to a kink, and a kink is not a roadA 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.the cable alonekink of 0.0083 radwith the stiffening girderthe cable takes 83% of the load; the girder spreads it over 183 mμ = L√(H/EI) = 4.93
Fig. 1 A point load 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, with a kink of 0.0083 radians. The lower is the same cable with the girder present, peaking at 1.125 against the bare cable’s 1.873. The girder takes 17% of the load and spreads it over 183 m.

A kink is not a road. Whatever else the deck of a suspension bridge is for, it is for turning that corner into a curve.

The system, which is a beam on a string

Take the deck as a beam of flexural rigidity EIEI and the cable as a string under a horizontal force HH that the dead load has already established. A load applied to the deck is shared between them, and the governing equation is one line:

EIvHv=p(x)EI\,v'''' - H\,v'' = p(x)

The first term is the beam and the second is the string. A string resists a change of shape with HvHv'' — its restoring force is proportional to its curvature rather than to its displacement, which is what distinguishes it from a beam on an elastic foundation, where the restoring force is kvkv and the equation is EIv+kv=pEIv'''' + kv = p.

Both are fourth-order and both have a characteristic length, and the two lengths behave completely differently. A Winkler foundation gives =(4EI/k)1/4\ell = (4EI/k)^{1/4}; a tensioned string gives

=EIH\ell = \sqrt{\frac{EI}{H}}

which for this deck is 183 m — a fifth of the span. That is the distance over which the girder spreads anything applied to it, and it is the single number the whole subject turns on.

Which free body produced the number

Cut the deck just to one side of the load and take the piece to the left.

Three things cross the cut. The girder’s shear, EIv-EIv'''; the vertical component the hangers have delivered, which is the integral of HvHv'' over the length; and, at the far end, the reaction. The load is shared between the second of those and the first, and the split at every wavelength is the ratio of the two terms in the equation.

That last observation is what makes the problem exactly solvable. Expand the load as a sine series, p=pnsin(nπx/L)p = \sum p_n \sin(n\pi x/L); then every term is uncoupled, because sin\sin is an eigenfunction of both operators:

vn=pnEI(nπ/L)4+H(nπ/L)2v_n = \frac{p_n}{EI\,(n\pi/L)^4 + H\,(n\pi/L)^2}

and the cable’s share of the $n$th harmonic is

Hk2EIk4+Hk2=11+(k)2,k=nπL\frac{H k^2}{EI k^4 + H k^2} = \frac{1}{1 + (k\ell)^2}, \qquad k = \frac{n\pi}{L}

Long-wavelength load goes to the cable; short-wavelength load goes to the girder. For this deck the first harmonic has k=0.64k\ell = 0.64 and the cable takes 71% of it; the third has k=1.91k\ell = 1.91 and the cable takes 21%; the ninth has k=5.7k\ell = 5.7 and the cable takes 3%.

A point load is all harmonics at once, and its sharp features are exactly the short-wavelength content — so the girder takes the corner and the cable takes the rest, which is the whole mechanism in a sentence.

The funicular polygon for five loadsThe shape a string takes under 5 point loads, with a vertex at every load and a constant horizontal component of 27.9 throughout. The end segments carry the most — 38.6 against 28.0 in the flattest one — because they are steepest.10148126H = 27.9, the same at every stationeach vertex is a load; each slope is the running vertical sum ÷ H
Fig. 2 The shape the cable would take on its own, found by drawing rather than by algebra. Every vertex of that polygon is a load, and a cable with a load at one point has one vertex — which is the kink the deck is there to remove.

What the girder actually buys

How much of a point load the cable ends up takingThe fraction of a mid-span point load that reaches the cable, against μ = L√(H/EI) — how many characteristic lengths of girder fit in the span. A stiff girder gives a small μ and takes most of the load itself; a limp one gives a large μ and hands nearly all of it over. At μ = 9.9 the cable has 99% of it. What the curve does not show, and the shapes view does, is that the girder's real job is not on this axis at all: even where it carries almost nothing it is still the thing that turns a kink into a curve.0204060800%20%40%60%80%100%μ = L√(H/EI)share the cable takesthe whole loadwhat the cable takesleft is a stiff girder,right a limp one
Fig. 3 The fraction of a mid-span point load the cable ends up taking, against μ = L√(H/EI) — how many characteristic lengths of girder fit in the span. A stiff girder gives a small μ and keeps most of the load; a limp one gives a large μ and hands nearly all of it over.
EIEI μ\mu \ell cable’s share peak deflection
10⁷ 98.6 9.1 m 100.0% 1.837
10⁸ 31.2 28.9 100.0% 1.755
4×10⁹ 4.93 183 83.1% 1.125
10¹¹ 0.99 913 11.0% 0.138

Read down that table and the girder’s contribution to carrying rises from nothing to almost everything. Read the last column and something else happens: the deflection falls by a factor of thirteen, and most of the fall has happened by the time the cable is still taking 83%.

So the girder is worth having long before it is carrying much. At μ=4.93\mu = 4.93 it has removed 40% of the deflection and 83% of the kink while accepting 17% of the load, which is the regime a real suspension bridge is built in — the Humber’s deck is a 4.5 m box on a 1,410 m span, and its μ\mu is of this order.

The further it deflects, the harder it pulls backTotal load against midspan sag for a 30 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 150 kN is 0.740 m rather than the 1.125 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 = 133.3 kN/m; at the marked point the tangent has reached 341.2 kN/m, 2.56 times as stiff, and the horizontal component of the tension has risen from 500 kN to 760 kN. Nothing about the steel changed. The geometry got better at the job.00.20.40.60.811.2020406080100120140160midspan sag (m)total load on the cable (kN)the design load, 150 kNsolved 0.740 m1.125 mtangent here 341.2 kN/mk₀ = 8T₀/L = 133.3 kN/mthe flat-cable law
Fig. 4 Where the cable’s own stiffness comes from, which is not the same question. A cable resists a load by changing shape, and its stiffness against that change is geometric rather than material — so the H in this page’s equation is a force the dead load put there, and the live-load behaviour depends on it entirely.

The load case that sizes it

Half a span loaded, which is the case that sizes the girderHalf of a 900 m span loaded at 10 per metre — the arrangement a suspension bridge is designed by, because it is the one the cable's own shape is least like. This model holds the cable force constant, so it does not credit the cable with the extra tension a full-span load would give it, and the two cases therefore come out closer together here than deflection theory puts them. The girder's moment peaks at 198438, against 1012500 for the same girder spanning alone — 19.6% of it. The reduction is not the cable carrying the load in the ordinary sense: it is that the cable holds the deck's shape close enough to a straight line that the deck barely has to bend. The stiffness parameter here is μ = 4.93, and the length over which the girder spreads anything is √(EI/H) = 182.6 m — about 20% of the span.0200400600800050000100000150000200000along the deck (m)moment in the girderpeak 198438 — 19.6% of the bare girder's
Fig. 5 Half a span loaded, which is the arrangement a suspension bridge is designed by — the one the cable’s own shape is least like. The girder’s moment peaks at 198,438 against 1,012,500 for the same girder spanning alone, so it carries 19.6% of what a plain beam would.

A uniform load over the whole span is very nearly what the cable’s shape already suits, and it produces an almost uniform additional sag with little curvature change — the cable simply deepens, in the way a funicular polygon redraws itself. Half a span loaded is antisymmetric, its content is concentrated in harmonics the cable is poor at, and it twists the cable into an S — so the girder has to supply the difference over the whole span.

Pattern loading is not a refinement here, it is the design case. The same is true of a continuous beam, and for the same reason: a structure whose response depends on the shape of the load has a worst shape, and it is rarely the one that is easiest to imagine.

The moment left in the girder once the cable has taken its shareA point load of 1000 at mid-span of a 900 m deck. The girder's moment peaks at 89749, against 225000 for the same girder spanning alone — 39.9% of it. The reduction is not the cable carrying the load in the ordinary sense: it is that the cable holds the deck's shape close enough to a straight line that the deck barely has to bend. The stiffness parameter here is μ = 4.93, and the length over which the girder spreads anything is √(EI/H) = 182.6 m — about 20% of the span.0200400600800020000400006000080000100000along the deck (m)moment in the girderpeak 89749 — 39.9% of the bare girder's
Fig. 6 The moment the girder is left with under a point load: 89,749 against 225,000 for the same girder spanning alone, or 39.9%. The reduction is not the cable taking the load in the ordinary sense — it is the cable holding the deck’s shape close enough to a straight line that the deck barely has to bend.

The generalisation worth carrying

The deck’s job — turn a concentrated action into a distributed one before handing it on — is a job many structures do, and naming it makes several of them the same structure.

A railway sleeper distributes a wheel load along the rail’s own beam-on-elastic-foundation length before it reaches the ballast.

A pad footing does the same for a column onto soil, over the characteristic length of a beam on the ground.

A floor slab distributes a point load to several beams over a width set by its own stiffness against theirs, which is the stiffest path taking the load in its most literal form.

And a raft does it for a whole building.

In every case the distributing member is not the load-carrying member, its stiffness sets a length rather than a capacity, and the useful question about it is “over what distance?” rather than “how much?”. The suspension bridge is the clearest example because the two functions are carried by two visibly different objects.

The ground pushes back hardest where the beam has gone down furthestA strip 16.4 m long and 1 m wide on ground of subgrade modulus 50 × 10³ kN/m³, carrying 1000 kN at its centre. The beam settles 3.90 mm under the load and the ground pushes back in proportion — the arrows are k times the settlement above them, peaking at 195 kN per metre — so the pressure diagram is the settlement bowl and not an assumed distribution. The characteristic length 1/β is 2.56 m: the bowl crosses zero at 6.04 m, which is 3π/4 of it, and beyond that the arrows reverse because the beam has lifted off. By 8.05 m — one π/β — the disturbance is 4.3% of what it was, which is why the moment 641 kNm and the peak pressure 195 kN/m contain no length at all. The settlement is drawn 217 times full size — the real bowl is 3.90 mm deep over 16.4 m, about 1 in 4207 — and at true scale the beam would be a straight line.1000 kN1/β = 2.56 mthe beam lifts off at 6.04 msettlement 3.90 mm · contact pressure 195 kN/m · moment 641 kNm, none of which contains the length of the beam
Fig. 7 The same argument with the restoring force proportional to displacement instead of curvature. The characteristic length is a fourth root here rather than a square root, and the physical statement is identical: a stiff member spreads a concentrated action, and how far it spreads it is what its stiffness buys.

The comparison the two shapes make

Setting the three limiting systems side by side is the fastest way to see what the arrangement is.

The cable and the arch are the same curveThe 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.cable: pure tensionreflected herearch: pure compression
Fig. 8 The cable and its reflected arch, carrying the same load by the same shape with the sign reversed. Both are efficient for the load they were shaped for and both are helpless against any other — which is why a masonry arch needs mass to keep its thrust line inside it, and a suspension bridge needs a girder to keep its cable’s shape near the one it was built with.

The cable alone deflects 1.873 units, has a kink, and takes 100% of the load. It is perfectly efficient and unusable.

The girder alone deflects 3.797 — twice as far — and takes 100% of the load. It is a plain 900 m beam, which is not a structure.

Together they deflect 1.125, which is 60% of the better of the two on its own, and share the load 83 to 17. The system is stiffer than either component because they resist different things: the girder resists curvature and the cable resists displacement, and the load’s content is split between the two.

A line of thrust, and the masonry it has to stay insideAn arch ring of 9% of the span in thickness, rising 28% of the span, under its own weight as a uniform load. Any horizontal thrust between 3.85 and 5.23 puts a line of compression entirely inside the masonry, so the arch stands — and which of them it actually takes is not decided by statics. The two extremes are drawn: the minimum-thrust line, which rides high at the crown and low at the haunches, and the maximum-thrust line, which does the opposite.thrust anywhere from 3.85 to 5.23 fitsH = 3.85, leastH = 5.23, most
Fig. 9 The arch version of the same requirement, and the reason masonry needs to be thick. A thrust line has to stay inside the material, and a change of loading moves it — so an arch’s depth is doing the job a stiffening girder does, by giving the funicular shape somewhere to be.

That pairing generalises. A structure that carries load by shape needs some way of tolerating a change of shape, and there are only two of them: mass, which is what a masonry arch uses, or bending stiffness, which is what a stiffening girder is. A prestressed cable net uses a third — opposing curvature — and is worth naming because it is the one that does not need either.

Reading the parameter

μ = L√(H/EI) is the only number in the problem, and it is worth translating into things a reader can picture.

A three-pinned arch, rise 2.6 on span 9A 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.crown hinge — no moment here, by constructionH = 23.4H = 23.427.027.0thrust line and axis coincide — the definition of funicular
Fig. 10 Where H comes from in the first place: a cable or an arch of sag f under a uniform load w develops a horizontal force wL²/8f, so a shallow cable pulls harder. The sag ratio is therefore inside μ as well, and a bridge with a shallow cable has a stiffer system for the same girder.

HH is set by the dead load and the sag: H=wL2/8fH = wL^2/8f for a parabolic cable, so a sag of a tenth of the span gives H=1.25wLH = 1.25\,wL. Substituting,

μ=LwL28fEI\mu = L\sqrt{\frac{wL^2}{8f\,EI}}

which says three things at once. Longer spans give larger μ — the cable dominates more, the deck matters less for carrying and more for spreading. Shallower cables give larger μ, because a shallow cable pulls harder for the same load. And a stiffer deck gives smaller μ, in the obvious way.

For the deck drawn here, μ is 4.93 and the characteristic length is a fifth of the span. A 19th-century bridge with an enormous stiffening truss might have μ near 1; a modern box-girder deck on a 2 km span has μ of 10 or more, and its deck is spreading a load over a few per cent of the span rather than a fifth of it.

Where the model stops

HH is held constant, and it is not. In real deflection theory the cable’s horizontal force increases when live load is added, and that increase is what makes a full-span uniform load so cheap — the cable simply deepens. This model cannot represent it, so it under-credits the cable for full-span loading and the two load cases come out closer together here than deflection theory puts them.

Everything is linear. Suspension-bridge behaviour is geometrically nonlinear by nature: the stiffness depends on the tension and the tension depends on the deflection. The linearised equation is the first term of that and is excellent for small live-load ratios.

The hangers are treated as a continuous connection. They are discrete, they can only pull, and a bridge under a severe antisymmetric load can slacken the hangers on the unloaded half — at which point the deck is a plain beam over that length.

Nothing here is dynamic. A suspension bridge’s most famous problem is aeroelastic, and the stiffening girder’s torsional stiffness — not touched anywhere above — is what decides it. Tacoma Narrows had an adequate vertical stiffening girder and a catastrophic torsional one.

And the cable’s own weight is absent. It sets the dead-load shape, it sets HH, and it is the reason a real bridge’s cable is a catenary rather than a parabola — a distinction that matters for the geometry and not for anything on this page.

What the pictures cannot show

The deflected shapes are drawn at an exaggeration and the kink is drawn as a sharp corner, which is the model’s answer rather than a bridge’s. A real cable has bending stiffness of its own, small but not zero, and the corner is rounded over a length of a few metres — which is negligible structurally and is not negligible if the reader is being invited to believe a picture.

Nothing in the figures shows a hanger, and the hangers are what make the whole arrangement a single system. Drawn as a continuous connection, they suggest the deck is glued to the cable; drawn discretely, at 20 m centres on a 900 m span, they would show that the “distributed” transfer is 45 point loads.

And the sharing curve has μ\mu on its horizontal axis, which is a number no reader has an intuition for. The useful translation is the last column of the table: a bridge whose girder spreads a load over a fifth of its span is a bridge with a real stiffening girder, and one that spreads it over a hundredth has a deck rather than a girder.

The ladder from here

Later rungs on this anchor: the deflection theory proper, with HH as an unknown determined by the cable’s own extensibility, and the difference between it and the elastic theory that preceded it — Melan’s equation and the reason the Brooklyn Bridge was designed by a method that made it several times heavier than it needed to be. Torsional stiffness of the deck and the flutter speed, which is the criterion that actually governs a long-span bridge. The self-anchored suspension bridge, where HH is delivered into the deck rather than into the ground and the deck is in compression. Cable-stayed bridges, where the stays are inclined and the deck is a beam on a set of elastic springs rather than a string. Hanger slackening and the nonlinear analysis it requires. And the cable-stiffened roof, where the same equation governs and the load that matters is uplift.

The thing worth carrying away is the reframing rather than any number. A structure’s most important member is not always the one carrying the most force, and asking “what fraction does it carry?” of the stiffening girder gets an answer — 17% — that is close to useless. Asking “over what length does it spread?” gets 183 m, and that is the number the bridge was designed around.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

CatenaryCharacteristic lengthDeflectionElastic foundationForm findingFunicularGeometric nonlinearityHorizontal thrustLoad sharingPattern loadingSag ratioStiffness