Structural Engineering & Statics
where the load goes
Concept

Funicular — where it appears

The shape a cable takes under a given load, in which the only internal force anywhere is along the cable itself. It is funicular for one load case only, so a different arrangement puts bending into a structure whose whole efficiency was that it had none.

Named by 25 essays across 4 fields — each of them below, with the objects they name alongside it.

Three forces must meet at a point. A body held by two supports and one load. The reaction at the roller is vertical and the load's direction is given, so their lines of action fix a meeting point — and the pin reaction has to point at it.

Three forces must meet at a point, and a drawing can find it

A body held by exactly three forces has their lines of action concurrent. That is a theorem, it is enough to solve for direction and magnitude, and for a century it was done with a straightedge.

equilibrium · Graphic statics 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.

The shape that carries itself, and the arch that is its reflection

Hang a chain and it takes the one shape that carries its load in pure tension. Turn the shape upside down and it carries the same load in pure compression. That is what an arch is.

structures · Funicular 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 27.9 throughout. The end segments carry the most — 38.6 against 28.0 in the flattest one — because they are steepest.

The polygon that finds the shape

A hanging string under five loads has no smooth curve in it — it has five vertices and six straight segments, and every slope in it is a running sum divided by one number.

structures · Funicular A line of thrust, and the masonry it has to stay inside. An 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.

The line that must stay inside

A masonry arch does not stand because its shape is right. It stands because some line of compression can be drawn inside the stonework — any one will do, and there are infinitely many to choose from.

structures · Arch The further it deflects, the harder it pulls back. Total 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.

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.

structures · Cable stiffness The hoops change their mind at an angle no proportion chose. The two membrane forces of a spherical dome of radius 30 m under 3 kN/m² of surface, taken from the crown to a base at 60°. The free body for the meridional force is the cap above a cone of half-angle φ, and vertical equilibrium of it gives N_φ = −wR/(1 + cos φ) directly: -45.0 kN/m at the crown falling to -60.0 at the base, compression everywhere. Equilibrium normal to the surface then gives the hoop force, which starts at -45.0 kN/m and reaches 15.0 — it changes sign, and the angle at which it does was found here by bisecting N_θ rather than quoted: 51.827292°. Setting N_θ = 0 gives cos²φ + cos φ − 1 = 0, so cos φ is (√5 − 1)/2, the reciprocal of the golden ratio — an identity this site's solver gate checks against the bisection to nine decimals rather than asserting, because it is too pretty to be believed on sight. Below that parallel the hoops are in tension, which masonry has none of, and that is where every old dome is cracked.

The surface that carries by being curved

A flat plate spends its thickness on a lever arm of a few millimetres. Curve the same sheet and the load is carried in the surface itself, at a thirtieth of the thickness — and the entire bill is presented at the edges.

structures · Shell action 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.

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.

structures · Stiffening girder The force runs along the chord whatever route the member takes. A member pinned at two points 10 m apart, loaded only at those two points, and bowed 1.2 m off the line between them. Equilibrium leaves the two end forces no choice: equal, opposite, and along the chord. So at every section the axial force is P cos α, the shear is P sin α, and the bending moment is the force times the perpendicular offset — 240 kNm at the crown for 200 kN at 1.2 m, which is a multiplication rather than an analysis. Straighten the member and the moment diagram is identically zero; that is the case a truss member is in, and the reason it carries one number.

The member with only one direction

If a body is in equilibrium under forces applied at exactly two points, those forces are equal, opposite and along the line joining the points. It is three conclusions from two equations, it is the shortest real theorem in statics, and nearly everything that follows depends on it without saying so.

equilibrium · Two force member The split is where buckling puts it. A branching column carrying 400 kN to two points 3.0 m apart, over 9.0 m. Every member is drawn at the thickness it needs: the area is the larger of N/σ and what Euler asks of a strut of that length, and 3 of 3 members here are sized by buckling rather than by strength. The split sits at 57% of the height, which is where the total volume is least — 51% less than the fan of straight struts that carries the same load with the same stresses.

The tree that strength does not ask for

A branching column carries a roof on many points and reaches the ground on one. Size every member by its stress and the optimum tree turns out to have no trunk at all — the best answer is a fan of straight struts from the base. Put buckling in and the trunk appears, at 57 per cent of the height.

structures · Branching structure The tie is a redundancy, so its stiffness decides the thrust. Thrust and rib bending for a 60 m tied arch of 0.15 rise ratio, against the stiffness of its tie. Cut the tie and the structure is a curved simply supported beam, so the tie force is the one redundant and the force method gives it: with a rigid tie the answer is 2229 kN, within 0.9 per cent of the funicular wL²/8f, and the shortfall is the arch's own axial shortening. A real tie stretches 68 mm and returns 2168 kN — 2.7 per cent of the flexibility is the tie — and whatever thrust the arch does not get, it carries as bending: 739 kNm at 30 m. A tenth of the tie stiffness is not a tenth of the problem; it is a different structure.

The thrust that never reaches the ground

Every arch on this site has ended at the same sentence — the foundation is where an arch is really decided. A tie changes the sentence without changing the arithmetic: the horizontal force is still there, still the same size, and it now closes on itself through a bar at deck level.

structures · Tied arch A fan, and where its forces go. Half a cable-stayed bridge: a tower 70 m above a deck, 12 stays reaching out over 200 m, and a uniform 200 kN/m on the deck. Each stay is drawn at a weight proportional to the force in it, from 3427 kN at the innermost to 10090 kN at the outermost — the outer stay carries the same vertical share and is far flatter, so it carries far more. The deck's shading is its own accumulated compression, 61905 kN at the tower, which is 1.55 times the load being lifted and is the horizontal half of every stay force added up. Nothing in this drawing is a catenary: every stay is straight and every one of them is a spring.

The cable that is a spring

A suspension bridge's cable is a funicular — it takes the shape the load gives it, and the deck's job is to make the load one that shape is right for. A stayed deck is not that at all. Its cables are straight, each reaches the deck at one point, and every one of them behaves as an inclined spring.

structures · Cable-stayed Held up by a pressure nobody can feel. An air-supported roof of 60 m span and 9 m rise. The membrane has no bending stiffness whatever, so the only thing that can hold it in tension is a pressure difference, and the pressure has to exceed the load per unit plan area and nothing else: 0.25 kN/m² of fabric plus 0.6 of snow is 0.85 kN/m², so 1.19 kN/m² does it — 1190 pascals, which is 1.17 per cent of an atmosphere and 121 millimetres of water. Ears do not notice it. A door does: at 2.1 kN on an ordinary leaf, the building needs an airlock rather than a handle. And the whole of it arrives at the foundation as 3365 kN of uplift — 17.9 kN on every metre of perimeter — which is the bill the pressure's smallness conceals.

Held up by the air inside

A membrane has no bending stiffness at all, so the only thing that can hold it in tension is a pressure difference. The pressure needed to hold up a roof is smaller than the pressure a closed door makes — and the same pressure arrives at the foundation as hundreds of tonnes of uplift.

structures · Pneumatic structure Deflection goes as the fourth power of the span. Deflection against span for a constant load intensity and section, with two slower relationships drawn faintly behind it for comparison: the load itself, which grows in proportion to the span, and the bending moment, which grows as its square. Doubling the span multiplies the deflection by sixteen, while the moment only quadruples.

The weight that has to be known before it can be found

Every other load arrives from outside and can be looked up. A structure's own weight depends on how big it is, and how big it is depends on the load — so the first calculation on any project is a fixed point, and the fraction of a member spent carrying itself turns out to be the square of its span as a fraction of a span it can never reach.

equilibrium · Dead load Three ways to move the same column, and they are not close. The same 2000 kN moved 3 m across 14 m, built three ways and drawn to one scale. The deep beam is 1.14 m of concrete, 24.1 tonnes, and settles 60.6 mm in the long term. The storey-deep truss takes the same moment as a couple at 3.6 m centres, so its chords carry M/h and it weighs 2.6 tonnes — a fifth of the beam — while settling 15.8 mm, and it does not creep. The wall is 72.8 tonnes and hardly moves at all, 1.71 mm, of which 36% is shear rather than bending — which is what a member as deep as it is long always does, and is why beam theory does not describe one. A wall as a deep beam is the stiffest of the three by a factor of 35.4.

The same span, four ways

A beam, a truss, an arch and a cable can all cross the same gap under the same load, and the choice between them is usually described as a matter of judgement or of taste. It is neither. Each carries the load by a different mechanism, each mechanism has a different exponent, and an exponent decides the ordering at every span rather than at some spans.

structures · Form selection The tendon is a load, pointing the other way. A 14 m beam with a parabolic tendon dropping 260 mm to midspan, stressed to 1440 kN after losses. Its curvature pushes the beam up along its whole length with an intensity of 8Pe/L² = 15.28 kN/m, against an applied 17.63 kN/m — so 2.34 kN/m is left to bend anything, and the beam carries 57.4 kNm where an unstressed one carries 432 kNm. What the section then feels is 6.40 MPa of uniform compression and very little else.

The load that comes from changing direction

A force that travels in a straight line asks nothing of anything. Bend its path and it asks for a transverse load of F over R along every millimetre of the curve, and that load is real, is nowhere on the load schedule, and is the same statement behind a prestressing tendon, a hoop force, an arch thrust and a web that buckles with nothing applied to it.

internal-forces · Deviation force Two curvatures of opposite sign, which is what makes it a structure. A cable net over a 36 m square, drawn as the two families of cables that are also the two rulings of the surface. One family sags and carries downward load by hanging; the other rises and carries upward load — wind uplift, and a load reversal anywhere — by the same mechanism upside down. Neither can do anything alone. A single family of cables is a mechanism: it changes shape freely under any load pattern it was not tensioned for, and the shape it moves to is decided by the load rather than by the designer. Put the two together and each is the other's restraint, but only if they are pulled against one another first — the pretension of 520 kN in the sagging family and 715 in the hogging one is a self-equilibrating state that exists with no load on the roof at all, and it is what turns two mechanisms into one structure. The curvatures are drawn four times their true value: a real net of this span sags 2.2 m over 36, which is flatter than it looks anywhere.

Two curvatures of opposite sign

A single family of cables is not a structure. It is a mechanism that takes whatever shape the load asks for, and it will do that under any load pattern it was not tensioned for. Cross it with a second family curved the other way, pull the two against each other, and the pair becomes stiff — with no bending anywhere and no material property involved in the stiffness at all.

structures · Cable net 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.

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.

structures · Stressed ribbon A tenth of a per cent of the thrust is all of the moment. The thrust a two-hinged arch loses to its own axial shortening, against rise-to-span. The flexibility equation's denominator has two terms — ∫y²ds/EI for bending and ∫cos²θ ds/EA for shortening — and their ratio is about (15/8)(i/f)², the square of the radius of gyration over the RISE. At the 10 per cent rise drawn that is 0.10 per cent of the thrust, which sounds like a rounding error and is not: a parabolic arch under a uniform load is funicular, so the rigid solution has NO crown moment at all, and the 0.10 per cent that the rib shortening removes from the thrust leaves 28 kNm behind. The correction that is a tenth of a per cent of the thrust is a hundred per cent of the bending. At a two per cent rise the loss is 2.6 per cent, because a shallow arch's thrust is enormous and its lever arm is not.

The arch that gets shorter

A parabolic arch under a uniform load is funicular, so the perfect solution gives it no bending at all. Then the rib shortens under its own thrust by a tenth of a per cent, and every kilonewton-metre of moment the arch will ever carry comes from that.

deflection · Rib shortening

One drawing solves the whole truss

The method of joints solves a truss one joint at a time, and each solution is thrown away as soon as the next begins. Drawn instead of computed, the joints share their edges — every member's force appears once in a single figure, and the figure's own closure is the check.

structures · Truss

The tension that was left out

A suspension bridge's deck sits on a cable pulling hard along it, and a member with a large tension in it is stiffened by that tension. Leaving the term out of the deck's own equilibrium is what elastic theory does, and on a long span it asks for fourteen times the girder.

structures · Stiffening girder

The line that pairs four forces

Three forces in equilibrium meet at a point; four need not. But they pair off. The resultant of two passes through the point where their lines cross, the resultant of the other two through theirs, and the two resultants must share the line joining those points. Culmann's line turns a four-force body into two triangles — and the method of sections into a drawing.

equilibrium · Graphic statics

The pole decides the drawing, not the answer

Five forces will not pair off the way four do. They need a point that is nowhere on the structure — chosen freely, by whoever is holding the pencil — and the string of lines it generates. Every choice draws a different polygon and finds the same resultant, and the shape it draws turns out to be the beam's bending moment diagram.

equilibrium · Graphic statics

The centre that hangs in the air

Cross's analogy turns a closed frame into a short column and reads its redundant moments off as that column's stresses. Nothing in it requires the frame to be straight. Run round a fixed arch, the analogous column's centroid is a point two thirds of the way up the rise, on no part of the arch at all, and the bending stress about it is the arch's horizontal thrust. For a parabola under a uniform load the answer it returns is a diagram of nothing, which is exactly right.

deflection · Moment-area

Every space a point, every joint a polygon

A truss's force diagram is a second drawing of the truss in which the joints have become polygons and the spaces between members have become points. Maxwell showed in 1864 that the exchange runs both ways, so a designer can draw the forces first and ask what shape carries them. His theorem also says which frames have such a diagram at all, and the answer is a surprise, because it is about polyhedra.

equilibrium · Graphic statics

The polygon that runs out of freedom

A funicular polygon for given loads has exactly three freedoms, so it can be made to pass through three chosen points and no more. Three points is a three-hinged arch, and the drawing solves it. Take the crown hinge away and one freedom is left over. The drawing then offers a whole family of thrust lines and cannot say which one the arch uses — the rib's stiffness decides, and stiffness is not on the paper.

equilibrium · Graphic statics

Named alongside it

The objects these essays reach for when they reach for this one.

Load pathHorizontal thrustThrust lineEquilibriumFree bodyGraphic staticsMembrane actionPrestressCompatibilityForce polygonForm-findingArch

All concepts