Hold a chain by its two ends and it falls into a particular curve. It is not a parabola, though it looks like one, and it is not an accident: it is the only shape in which a flexible thing can carry its own weight, because a chain cannot resist bending and so must find the shape where no bending is required. Robert Hooke worked out in 1675 what that means for builders and wrote it as an anagram to keep it secret. Translated: as hangs the flexible line, so but inverted will stand the rigid arch.
The catenary is aΒ·cosh(x/a). The parabola is the first two terms of its own Taylor series β cosh x = 1 + xΒ²/2 + xβ΄/24 + β¦ β so at small x they are almost the same curve and at large x they are not remotely. At x = 0.5 they differ by 0.3 per cent; at x = 2 the catenary is a quarter again as tall. On a real suspension bridge the sag is shallow, so the two shapes sit within half a per cent of each other, and β this is the part people get wrong β the cable really is a parabola anyway, because it carries a heavy uniform deck rather than only its own weight. Turn the sag up past thirty per cent and watch the two curves separate.
| x | cosh x | 1 + xΒ²/2 | difference | catenary is taller by |
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A chain can only pull. Whatever shape it settles into is therefore a shape in which every part is in pure tension with no bending anywhere. Reflect that shape and every force reverses: pure compression, still no bending. Stone is hopeless in tension and superb in compression, which is why this trick built cathedrals. GaudΓ hung bags of lead shot from strings for ten years to design the Sagrada FamΓlia upside down. The shaded band is the thrust line β where the compression actually runs. Change the load and watch it wander out of a circular arch, which is exactly where old arches crack.
Four bars pinned into a square fold over the moment you lean on them β the corners can rotate and nothing stops them. Add one diagonal and it becomes rigid, because a triangle is the only polygon whose shape is fixed by its side lengths alone. Maxwell counted it: a flat pin-jointed frame needs exactly 2j β 3 bars for j joints. Fewer and it is a mechanism; more and it is over-braced. The truss below is solved properly β every joint balanced β so drag the load and watch the forces redistribute. The bottom always pulls and the top always pushes, and that is the whole reason a truss is shaped the way it is.
A beam bends because its top squashes and its bottom stretches, and the further those two faces are from each other the harder that is. The measure of it is the second moment of area, bΒ·hΒ³/12 β and that cube is the single most useful number in structural engineering. Double the depth of a beam and you have not doubled its stiffness, you have multiplied it by eight. Turn a plank on edge and it gets four times stiffer for exactly no extra material, which is why floor joists stand upright and why an I-beam has almost nothing in the middle. Span is even more brutal: the sag goes as the fourth power, so doubling the span multiplies the sag by sixteen.
Every bridge type is an answer to the same question β how do I get the load to the ground without anything bending β and each answer works over a different range. Beams are simple and hopeless past about fifty metres because of that fourth power. Arches push outward and need something to push against. Suspension cables are the only thing that gets past a kilometre, because a cable in pure tension has no buckling problem at all and can simply be made thicker. Note the last column: the shallower you make a cable, the harder it pulls, and that is why a suspension bridge sags as much as it does.
| type | carries load by | typical span | longest built | what limits it |
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The curves are the real functions and the page checks itself against them: cosh(0.5) = 1.12763 against a parabola 1.12500, cosh(1) = 1.54308 against 1.50000, and cosh(2) = 3.76220 against 3.00000 β the gap being exactly the xβ΄/24 term and everything after it. The catenary is fitted to your chosen span and sag by solving sag = a(cosh(span/2a) β 1) numerically, so the two curves below always share their endpoints and their lowest point, which is the only fair comparison. The truss is not a drawing: it assembles the equilibrium equations at every joint and solves them, and the residual force left at the worst joint is about 4 Γ 10β»ΒΉΒ³ kN. It reproduces the textbook answers β a 200 kN load at midspan of a six-panel Warren truss gives reactions of exactly 100 and 100, bottom chords in tension rising to 312.5 kN and top chords in compression reaching β375 kN. Maxwell count is computed from the frame you have chosen, which is why the bare square is reported as a mechanism rather than merely drawn as one.
What is missing is most of engineering. Everything here is a pin-jointed frame in two dimensions with weightless bars, small deflections and perfectly linear elastic material β real joints are welded or bolted and carry moments, real bars weigh a great deal, and a real bridge is three-dimensional and has to resist wind sideways and twisting, which is what actually destroyed Tacoma Narrows rather than any of the forces on this page. Most importantly, nothing here can buckle. A compression member does not fail by being crushed; it fails by bowing sideways, at a load that depends on its length squared, and that single omission is the difference between the numbers here and a member that would actually survive. Nor is there any fatigue, corrosion, temperature movement, settlement of the foundations, or safety factor β and the last of those is not a fudge, it is the formal acknowledgement that the load and the strength are both uncertain.