Structural form

The stiffness that comes from the shape

A cable has no bending stiffness whatever, and it still holds up a roof. What resists the load is the change of its own geometry, so its stiffness is a function of the tension already in it — and prestress buys stiffness that no change of material could.

Assumes The shape that carries itself, and the arch that is its reflection, The load put on backwards and The load that makes itself worse.

Cut a cable anywhere and look at what can cross the cut. A force along its own length, and nothing else. No moment, because a moment needs a section with depth and this one folds. No shear, for the same reason. Whatever holds the roof up has to be assembled out of that single component, and the material has no say in the matter.

So a cable pushed sideways cannot resist by getting stiffer in the way a beam does. It resists by moving — by taking up a shape in which the force along its length has a component pointing back against the load. The resistance is a property of the geometry it has adopted, not of the steel it is made of, and the geometry is different at every load.

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. 1 Total load against midspan sag for a 30 m cable of 1000 mm² prestressed to 500 kN under 5 kN/m. Solving the cubic at every point of the curve gives 0.740 m of sag at the full 150 kN, against the 1.125 m the flat-cable formula WL/8T₀ predicts — that formula is the dashed straight line, the tangent at the origin and nothing more. The tangent stiffness has climbed from 133.3 kN/m to 341.2 kN/m by the time the load is all on, and the horizontal component of the tension has gone from 500 kN to 760 kN.

That curve is the whole essay. It rises, it steepens, and everything on this site that bends gives a straight line instead.

The stiffness is in the shape, and the shape is in the load

The claim, stated once and plainly: a cable’s transverse stiffness is geometric, and it is proportional to the tension already in it. Put no tension in and there is no stiffness at all. Put tension in and the stiffness arrives, without a single kilogram of extra steel and without changing anything about the steel that is there.

This inverts the rule the rest of the collection runs on. Material far from the middle is how a beam gets stiff, and the rule is so reliable that it produces a factor of forty between two shapes cut from the same plate. A cable has no middle to be far from. What it has is a sag, and a sag is what turns an axial force into a transverse one — which is exactly the argument the funicular shape makes about strength, running here instead into stiffness.

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. 2 A uniformly loaded cable and its reflection. The identical curve carries the identical load, in tension one way up and compression the other. The shape and the force are one statement, which is why a cable that is asked to resist more load has no option but to change shape.

The free body, and the constant that comes out of it

Every number on this page comes from a cut, so here is the cut.

Take the cable between its lowest point and some station along it and keep that piece. Three things act on it — the horizontal pull HH at the low point, the tension along the cable at the far cut, and the weight hanging between. Nothing horizontal is applied anywhere in between, so horizontal equilibrium of that piece says the horizontal component of the tension is HH at every station. The horizontal force in a cable is constant along it, and that one result from one force sum is what makes the rest tractable.

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. 3 The funicular polygon for five point loads, with a vertex at every load. The horizontal component is 27.9 at every station, and the segments differ only in slope — 38.6 in the steepest against 28.0 in the flattest. The tension varies along a cable; the horizontal part of it does not.

Now take half the cable and sum moments about the support. That half carries wL/2wL/2, acting through the quarter point of the span, and HH arrives at midspan a distance δ\delta below the support, so

Hδ=wL2L4δ=WL8H.H\delta = \frac{wL}{2}\cdot\frac{L}{4} \quad\Longrightarrow\quad \delta = \frac{WL}{8H}.

That is the only equilibrium statement the problem contains, and it has two unknowns in it. The second equation is not an equilibrium statement at all — and that is the interesting part.

The equation that is not equilibrium

A parabola of sag δ\delta over a chord LL is longer than the chord by approximately 8δ2/3L8\delta^2/3L. If the cable is to sag further, it has to become longer, and the only thing that can lengthen it is elastic extension under the increase in tension, (HT0)L/EA(H - T_0)L/EA.

Setting the length the geometry needs equal to the length the steel supplies, and substituting δ=wL2/8H\delta = wL^2/8H, gives

H3T0H2w2L2EA24=0.H^3 - T_0 H^2 - \frac{w^2L^2 EA}{24} = 0.

A cubic, for what in any other member would be a division. It is bisected here rather than linearised, on a bracket widened by doubling — Newton from H=T0H = T_0 has a stationary starting point for a slack cable and a derivative of only T02T_0^2 otherwise, so it lands wherever the load happens to throw it.

Eliminating HH between the two equations gives the load–deflection law in closed form, which is the quotable version:

W(δ)=8T0δL+64EAδ33L3.W(\delta) = \frac{8T_0\delta}{L} + \frac{64\,EA\,\delta^3}{3L^3}.

Two terms, and they are two different mechanisms. The first is the prestress being turned toward the load by the sag. The second is the steel stretching. Differentiate and the tangent stiffness is 8T0/L+64EAδ2/L38T_0/L + 64EA\delta^2/L^3, so at zero deflection

k0=8T0Lk_0 = \frac{8T_0}{L}

and there is no material property in it whatsoever. A hemp rope and a steel strand of the same length, pulled to the same tension, have exactly the same initial transverse stiffness. They part company only in the cubic term, which is to say only after they have moved.

Prestress is the stiffness

The consequence is the reason cable structures are tensioned at all, and it is worth seeing at four levels of prestress rather than argued.

Prestress buys a stiffness no change of material canFour cables of identical steel — 30 m, 1000 mm², E = 160000 MPa — differing only in the tension put into them before the load arrived. The initial stiffness is 8T₀/L exactly: 0.0, 33.3, 133.3, 533.3 kN/m at T₀ = 0, 125, 500, 2000 kN, and no property of the steel appears in that expression. The slack cable leaves the origin flat — it has no stiffness whatever at zero load, and its sag grows as the cube root of the load, reaching 1.059 m under the same 150 kN that puts 0.276 m into the tightest of them. Four curves of one cable: the tightest starts 16 times stiffer than the slackest that has any stiffness at all, and every other property they share.00.20.40.60.81020406080100120140160midspan sag (m)total load on the cable (kN)T₀ = 0 kN · k₀ = 0.0T₀ = 125 kN · k₀ = 33.3T₀ = 500 kN · k₀ = 133.3T₀ = 2000 kN · k₀ = 533.3all at 5 kN/mat zero prestress the curveleaves the origin flat
Fig. 4 Four cables of identical steel — 30 m, 1000 mm², E = 160000 MPa — differing only in the tension put into them before the load arrived. The initial stiffnesses are 0.0, 33.3, 133.3 and 533.3 kN/m at T₀ = 0, 125, 500 and 2000 kN, which is 8T₀/L exactly. The slack one leaves the origin flat and reaches 1.059 m under the same 150 kN that puts 0.276 m into the tightest.

The slack cable makes the point. It has no stiffness at all at the origin, so an arbitrarily small load produces a disproportionate movement — with T0=0T_0 = 0 the law reduces to Wδ3W \propto \delta^3, and the sag grows as the cube root of the load. A washing line has this property and everybody has watched it happen. Adding steel does not fix it; adding tension does.

This is prestress used for a different purpose than a prestressed beam uses it. There, the tendon is a load put on backwards so that the material never has to work in tension. Here, nothing about strength is being bought at all. The tension is there to supply a stiffness the structure does not otherwise possess, and it is spent by relaxation and creep over the life of the building in exactly the way a prestressed beam’s is, with a different symptom — not cracking, but a roof that has gone soft.

The curve that bends the other way

Every load–deflection curve in this collection that departs from a straight line has so far departed downward.

The further it deflects, the harder it pulls backTotal load against midspan sag for a 40 m cable of 1000 mm² prestressed to 900 kN, carrying 8 kN/m. The cubic H³ − T₀H² − w²L²EA/24 = 0 was bisected at every point of the curve, so the sag at the full 320 kN is 1.228 m rather than the 1.778 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 = 180.0 kN/m; at the marked point the tangent has reached 421.5 kN/m, 2.34 times as stiff, and the horizontal component of the tension has risen from 900 kN to 1302 kN. Nothing about the steel changed. The geometry got better at the job.00.511.5050100150200250300350midspan sag (m)total load on the cable (kN)the design load, 320 kNsolved 1.228 m1.778 mtangent here 421.5 kN/mk₀ = 8T₀/L = 180.0 kN/mthe flat-cable law
Fig. 5 The same argument on a 40 m cable at 900 kN of prestress under 8 kN/m. The solved sag is 1.228 m against the flat-cable formula’s 1.778 m, and the tangent stiffness has risen from 180.0 to 421.5 kN/m — a factor of 2.34, against 2.56 for the shorter cable. The stiffening is not a quirk of one set of numbers.

A structure that has leaned carries its weight off the line it was acting on, so the extra moment bends it further, so it leans more. Its stiffness falls as the load rises, and it runs out entirely at the buckling load. A shallow frame that snaps through does the same thing more violently. Both are geometric nonlinearity, and both are the geometry working against the structure.

A cable is the same phenomenon with the sign reversed. Its deflection improves the geometry — more sag means a steeper cable at the ends, which means more of the tension is pointing the right way — so the structure gets better at its job the harder it is pushed. Two structures, one mechanism, opposite signs — and the sign is decided by whether the deflection lengthens or shortens the lever arm the internal force works on.

Chord force against truss depthThe force in a truss chord for a fixed bending moment, against the depth of the truss. The relationship is a reciprocal: the chords form a couple whose lever arm is the depth, so a shallow truss pays for it steeply.0.511.52050010001500depth of the truss1406760563500375281the same moment, resisted by a longer lever arm
Fig. 6 Chord force against depth for the hero cable’s own moment, wL²/8 = 562.5 kNm. A cable’s horizontal pull obeys the same reciprocal with the sag in place of the depth, so the two sags argued about above are two points on this curve — 760 at the solved 0.740 m and 500 at the flat formula’s 1.125 m. The difference from a truss is that the cable chooses which point it sits at.

The reciprocal is the argument depth makes everywhere, and the cable’s version of it is unusual only because the depth is a variable rather than a decision. A truss designed with a lever arm of 1 m has a lever arm of 1 m at every load it will ever see.

Two families, pulling against each other

A single cable is stiff only against load in the direction that increases its sag. Push it the other way and it goes limp, which makes it useless as a roof surface on its own.

The fix is to cross it with a second family curving the opposite way and pull the two against each other. Both are then in tension with no load on the roof at all, and any deflection adds curvature to both — the sagging family gains sag, the hogging family loses rise, and both resist, because transverse resistance is tension times curvature and the curvature the deflection adds has the same sign for both. That is a surface with stiffness in two directions built entirely out of things with none.

The free body is a unit square of the surface at the centre, cut on all four sides. Vertical equilibrium of the unloaded square requires nsfs=nhfhn_s f_s = n_h f_h, and that is not a design decision — the two pretensions are fixed relative to each other by equilibrium, and only their common scale is free. A net is a structure whose internal forces exist before anything is applied and are not arbitrary, which is what a self-stress state is.

The net is nearly linear right up to the moment half of it lets goLoad against centre deflection for a 30 m square net of cables at 2 m centres, a sagging family 1.5 m deep and a hogging family 1.5 m high, pretensioned to 400 kN. The tangent stiffness at the origin is 18.73 kN/m³ and the curve barely bends: at the design load of 1 kN/m² the centre has moved 53.4 mm. What ends the story is not a stress. At 407 mm the hogging family's tension has fallen to zero and it goes slack, which happens at 7.84 kN/m² — 7.8 times the design load. Past that point half the net has stopped working and the rest has to find the whole load by sagging, so the real limit on a cable roof is a loss of geometry rather than a want of strength.010020030040002468deflection at the centre of the net (mm)load on the roof (kN/m²)the hogging family goes slack: 7.84 kN/m²7.8× the design loaddesign: 1 kN/m² at 53.4 mmk₀ = 18.73 kN/m³ at the origin
Fig. 7 Load against centre deflection for a 30 m square net at 2 m centres, sagging family 1.5 m deep, hogging family 1.5 m high, pretensioned to 400 kN. The tangent stiffness at the origin is 18.73 kN/m³ and the curve barely bends over its whole range — at the design load of 1 kN/m² the centre has moved 53.4 mm. The straight line for comparison is that same origin stiffness extended.

The net’s near-linearity is worth a moment. A single cable stiffens by a factor of two and a half over its range; the net hardly stiffens at all, because most of its stiffness was there before the load arrived.

Most of it was not bought with pretension

Splitting k0k_0 into its two terms produces the number that reverses the usual account of what a cable net is.

Most of a deep net's stiffness is not bought with pretensionThe tangent stiffness of a 30 m cable net at the origin, split into the part the pretension provides — 8(H_s + H_h)/sL², which does not depend on the curvature at all — and the part the sag itself provides elastically. At the 1.5 m sag drawn elsewhere on this page the total is 18.73 kN/m³, of which only 3.56 is pretension and 15.17 is the shape: 81% of the stiffness comes from the geometry rather than from the jacks. The elastic part grows as the square of the sag, so the two are equal at a sag of 0.726 m and below that a net really is held up by how hard it was pulled. That is why a shallow net has to be tensioned so hard and a deep one hardly at all.0.511.52010203040sag and rise of the two families (m)stiffness at the origin (kN/m³)the shape's own15.17 kN/m³the pretension's3.56 kN/m³total 18.73equal at a sag of 0.73 m
Fig. 8 The origin stiffness of the 30 m net against the sag of its two families, split into the part the pretension provides — 8(H_s + H_h)/sL², which does not depend on curvature at all — and the part the shape provides elastically. At the 1.5 m sag drawn on this page the total is 18.73 kN/m³, of which only 3.56 is pretension and 15.17 is geometry. The two are equal at a sag of 0.726 m.

Eighty-one per cent of that net’s stiffness comes from its curvature rather than from its jacks. The pretension term is 8(Hs+Hh)/sL28(H_s + H_h)/sL^2 and contains no curvature at all; the elastic term goes as the square of the sag, so it overtakes the pretension term at a sag of 0.726 m — about a fortieth of the span — and everything deeper than that is mostly shape.

Which recovers the practical rule from the arithmetic rather than from experience: a shallow net has to be tensioned very hard and a deep one hardly at all, because a shallow net has almost none of the second term and has to buy the whole stiffness with the first. It also explains why the anchorages of a flat cable roof are so much more brutal than the surface above them suggests.

What actually ends it

The interesting failure of a cable net is not a failure of a cable.

One family is being wound up while the other is being let goThe tension in each family of a 30 m cable net as the centre is pushed down. Both start at the self-stress state — 400 kN in the sagging cables and 400 kN in the hogging ones, which is not a choice but the condition n_s·f_s = n_h·f_h for a surface in equilibrium under no load at all. Deflection adds sag to one and takes rise from the other, so by 407 mm the sagging family has climbed to 925 kN while the hogging family has reached zero. That crossing is the design case: below it the net has stiffness in both directions and above it the surface is a set of parallel cables with nothing holding them down.010020030040002004006008001000deflection at the centre of the net (mm)horizontal tension per cable (kN)sagging: 925 kNhogging: nothing leftboth start at 400 and 400 kNslack at 407 mm
Fig. 9 The tension in each family of the net as the centre is pushed down. Both start at the self-stress state — 400 kN sagging and 400 kN hogging, fixed by n_s·f_s = n_h·f_h rather than chosen. Deflection winds one up and lets the other go, so by 407 mm the sagging family has reached 925 kN and the hogging family has reached zero. That crossing happens at 7.84 kN/m², 7.8 times the design load.

At 407 mm the hogging family has nothing left in it. It goes slack, half the surface stops contributing in one step, and the remainder has to find the whole load by sagging — from a state in which it is already at 925 kN and its geometry has moved. The sagging cables are nowhere near any strength that matters; the structure has run out anyway.

So the design case for a cable roof is the loss of a self-stress state, and the quantity to compute is a displacement rather than a stress. This is unusual enough to be worth naming as a category: the limit is set by a member that stops working rather than by a member that breaks, in the same way that a brace’s job is a stiffness rather than a strength and serviceability rather than strength usually governs a floor. Ponding on a cable roof is the vicious version — water that will not run off collects in the sag it has just deepened, on a surface whose stiffness against that particular deflection is what the ponding is destroying.

Where the same argument turns up

Geometric stiffness is not a cable speciality. It is what a membrane roof does, what a spoked bicycle wheel does, and what a curved shell does from the other side of the sign — a shell is a surface that carries in its own plane because it is curved, and needs almost no thickness to manage it.

The same span, the same pressure, thirty times the thicknessA 3.0 m diameter carried two ways at 0.5 MPa, both drawn to the same scale across and both allowed 150 MPa. Curved, the wall is in pure tension: the free body is half the cylinder cut along its length, and N_θ = pR = 750 kN/m needs 5.0 mm of steel. Flat, the same width is a strip in bending: M = p(2R)²/8 = 563 kNm/m needs 150 mm, a factor of 30.0. That factor is √(3σ/p) = 30.0 and it is not a proportion but a change of exponent: the membrane thickness is linear in the pressure and the bending one is a square root of it, so the advantage grows as the load falls. Both wall thicknesses are drawn 10 times over, because at the scale of the span the curved one is a third of a pixel.curved — carried in the surfaceflat — carried in bending5.0 mm of wallN_θ = pR = 750 kN/m150 mm of plateM = p(2R)²/8 = 563 kNm/ma factor of 30.0, which is √(3σ/p) · wall thickness drawn 10× over
Fig. 10 A 3.0 m diameter carried two ways at 0.5 MPa, both allowed 150 MPa. Curved, the wall is in pure tension and 5.0 mm of steel does it. Flat, the same width is a strip in bending and needs 150 mm — a factor of 30.0, which is √(3σ/p). The shell and the cable are the same trade, made once in compression and once in tension.

The connection runs in the other direction too. The stiffest path takes the load whenever members share a displacement, and a cable net is the extreme case of that rule, because the stiffness deciding the split is itself a function of the deflection being shared. And a structure whose stiffness depends on its prestress is a structure whose natural frequency does too, which is why the period nobody chose is, for a cable roof, a number set by the tensioning sequence rather than by the mass and the section.

Where the model stops

The profile is a parabola, at every stage. The solver imposes that shape and computes only its amplitude, which reduces a continuum to one degree of freedom. It is a good approximation for shallow cables under distributed load and a poor one for a deeply sagging chain or a cable under a concentrated load, where the true shape has a kink the parabola cannot represent.

The load is uniform and stays uniform. A cable’s stiffness is directional in the deepest possible sense — it resists what deepens its sag and cannot resist anything else. An asymmetric load on a suspended structure produces a shape change the funicular argument never contemplated, which is what stiffening trusses exist to absorb.

The supports do not move. Every horizontal force here is delivered to something assumed rigid. Real anchorages give, and a support that draws inward reduces the tension, which reduces the stiffness, which increases the deflection — the same runaway loop as a leaning frame’s, this time with the geometry losing.

The prestress is the prestress that was intended. It is applied by jacking, measured indirectly, and then leaks away through relaxation of the strand and creep of whatever it is anchored to. Since k0k_0 is exactly proportional to T0T_0, a net that has lost a fifth of its pretension has lost a fifth of the first term of its stiffness, and there is nothing about the loss visible from underneath.

The state depends on the sequence. A net has no shape until it has been tensioned, and the order in which the cables are pulled decides the self-stress state it lands in — which makes it, like every structure whose analysis assumes it arrived complete, a case where the construction sequence is part of the answer.

What the figures cannot show

Every picture on this page is a plot. Not one of them is a cable, and that is deliberate, because the argument is entirely about a relationship between a load and a movement and a drawing of a cable would show a curve that looks the same at every load.

Which conceals two things. The first is that the sags being plotted are real and large — 740 mm on a 30 m cable is a sag ratio of one in forty, and 407 mm of net deflection is span over seventy-four. Nothing here is drawn exaggerated, because nothing here is drawn at all; these are movements a person standing underneath would see.

The second is that the axes hide how little of the structure is doing anything different. The steel is the same steel at every point of every curve, at a stress that never approaches anything interesting. The whole of the nonlinearity — the factor of 2.56 in stiffness, the cube-root behaviour of the slack cable, the collapse of one family of a net — is a change of shape, and a plot of stress against load would be a straight line with nothing to say.

The ladder from here

Later rungs on this anchor: the catenary’s own stiffness, which differs from the parabola’s by the term that made Galileo wrong. The cable-stayed system, where the stay’s apparent modulus falls with its own sag and the Ernst correction repairs it. Form-finding by force density, which makes the nonlinear problem linear by choosing the wrong unknown deliberately. The dynamic relaxation method, which solves a static net by giving it a fictitious mass and damping it. Cable-stayed against suspended, and why the two carry live load so differently. The saddle roof and the ring, where the two families’ thrusts close on themselves rather than on the ground. Wind uplift, which reverses the sign and asks the hogging family to do the sagging family’s job. And the prestress that has to be locked in during erection, where a sequence of jacking operations has to arrive at a state that equilibrium fixed in advance.

Named alongside this one

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

What links here

Every essay whose body links to this one.

The objects this essay names

Each one links to every other essay that touches it.

Cable netCompatibilityFunicularGeometric stiffnessHorizontal thrustPrestressSag ratioStiffness