The angle that saves the most concrete

Every pressure points at the pin. On a curved surface every element of a fluid’s pressure pushes in a different direction, so no multiplication gives the resultant, but two free bodies do: the horizontal component is the force on the surface’s vertical projection, and the vertical component is the weight of the fluid above it. On a radial gate, every element’s line of action passes through the trunnion as well, and the hoist that lifts it sees none of the water’s moment.

That essay worked in section, with the radial gate curved in a vertical plane. An arch dam is the same argument turned through ninety degrees. Its curvature is in plan: seen from above, the dam bows upstream into the reservoir, and each horizontal slice of it is a ring of concrete with water pressing on its convex face and rock at its two ends. The projection rule applies to each ring exactly as it applied to the radial gate, and it gives the whole design of the ring in two lines.

The water pushes on the chord #

Take one ring, a metre high, 50 m below the water surface, spanning a 100 m gap between the canyon walls. The water presses on it at $\gamma_w y$ = 491 kPa, normal to its face everywhere.

The resultant of that pressure, by the projection rule, is the pressure times the ring’s projection on a plane normal to the stream: the chord across the canyon, 100 m. So the water pushes the ring downstream with 49,050 kN for every metre of its height, and nothing about the ring’s curvature enters. A flat wall across the gap would receive the same push; a deeply curved one would receive the same push. The curvature does not change the load. It changes where the load goes.

A flat wall would have to carry it by bending, spanning 100 m across the canyon as a beam. A curved ring carries it by compression along its own length, and the compression arrives at the abutments along the ring’s tangent. With a central angle $\theta$ the tangent at each end makes $\theta/2$ with the chord, so the two abutment thrusts $N$ balance the water when

$$2N\sin(\theta/2) = p\,w.$$

The chord is $w = 2R\sin(\theta/2)$, so this is $N = pR$ — the hoop tension of a pipe under internal pressure, with the sign reversed because the water is outside the ring. The formula usually derived by cutting a cylinder in half has been reached here without any cylinder, by the projection rule and two forces. A shell carries its load by being curved in exactly this sense: the load is the same as on a flat plate, and the curvature turns it into membrane force.

A ring as thick as its thrust #

A ring of concrete allowed a compressive stress $\sigma$ needs a thickness $t = pR/\sigma$. At 5 N/mm² — a working stress of the kind used to proportion arch dams by this method — the 50 m ring with a 54.4 m radius is 5.34 m thick. The choice is the radius, or equivalently the central angle, and the gap is fixed by the canyon.

The three rings show the trade. The 60° ring is nearly flat: short, but with a radius of 100 m it carries a large thrust and is nearly 10 m thick. The semicircle has the smallest radius any ring across a 100 m gap can have, 50 m, so its thrust and its thickness are least, 4.91 m; but it is half again as long. Concrete is thickness times length, and between those two extremes is a ring that uses less than either.

One line of calculus #

The volume of a ring per metre of height is its thickness times its length:

$$V = \frac{pR}{\sigma}\cdot R\theta = \frac{p}{\sigma}\left(\frac{w}{2}\right)^2 \frac{\theta}{\sin^2(\theta/2)}.$$

Everything outside the last fraction is fixed by the water, the concrete and the canyon. The last fraction depends on the angle alone, and setting its derivative to zero gives

$$\tan\frac{\theta}{2} = \theta, \qquad \theta = 133.56°.$$

Whatever the canyon, the depth or the concrete, the ring that uses least concrete has a central angle of 133.6°. The thrust is least at 180°; the length is least for a flat ring; their product has a minimum between them, at an angle that is a property of the geometry of circles and nothing else. The minimum is also flat: anywhere from 118° to 150° costs within 2 per cent of the least, which is why real arch dams are drawn at angles in that range rather than at one exact figure, and why the abutments, not the concrete, usually decide where in the range.

This is an unusual kind of result for this subject. Most optimal forms depend on the loading — the line of thrust follows the load and changes when it does — but the water pressure here enters only as a scale, and so does everything else. The optimum is pure shape.

Where the thrust goes into the rock #

The abutment thrust leaves the ring along its tangent, at $\theta/2$ to the chord. Part of it pushes straight into the canyon wall, across the valley; the rest pushes along the valley, downstream, and has to be resisted by the rock’s strength in shear.

A flat ring pushes mostly into the rock: at 60°, 87 per cent of its thrust is directed across the valley, into the canyon wall, where rock is strongest. At the economical 133.6° only 39 per cent is, and 92 per cent of the thrust’s magnitude — $\sin 66.8°$ — is directed along the valley. A semicircle pushes its abutments straight downstream, parallel to the canyon walls, and holds on only by the rock’s shear and friction along the contact — the question of whether it slides, asked of a mountain.

The angle that suits the concrete is not the angle that suits the rock. In a canyon of sound rock the concrete wins and the dam is drawn near 133.6°. In a canyon whose walls are jointed parallel to the river, or weathered at the surface, the abutment governs and the rings are drawn flatter, with more concrete, so that more of their thrust goes into the rock rather than along it. Many of the historic failures of arch dams were failures of an abutment rather than of the arch, which is this division of the thrust being decided in the wrong direction.

Constant angle, or constant radius #

A dam is many rings stacked, and the canyon gets narrower with depth. That raises a question the single ring did not: as the gap shrinks towards the foot, what should the rings do?

The first arch dams kept a constant radius — every ring an arc of the same circle, the upstream face a vertical cylinder — because that is simple to set out and to build. As the canyon narrows, a ring of fixed radius spans a shorter chord, which means a smaller central angle, which means a flatter ring: the expensive end of the volume curve. And its thickness, $pR/\sigma$ with $R$ fixed, grows in proportion to the depth all the way to the foot.

The constant-angle dam keeps every ring at 133.6°, so each ring’s radius shrinks with its chord. Its thickness is then the pressure, which grows with depth, times a radius that shrinks with depth; in a V-shaped canyon the product is greatest half-way down and falls to nothing at the foot, where the gap closes. The cheapest dam in a V-shaped canyon is thickest half-way down, not at its foot, which is the opposite of every gravity structure and of every intuition about water pressure. Over the whole dam it needs 60 per cent of the constant-radius dam’s concrete.

The reason is visible ring by ring. At the crest the two dams are the same ring, 200 m across at 133.6°, with a radius of 108.8 m. Half-way down the V has narrowed to 100 m. The constant-angle ring there is the 54.4 m ring of the free-body figure; the constant-radius ring keeps its 108.8 m radius, so across a 100 m chord its central angle has closed to 54.7°, and by the volume curve a ring at 54.7° uses 64 per cent more concrete than one at 133.6° across the same chord. Lower still the constant-radius rings are flatter yet, while carrying the deepest water. The constant-radius dam is paying the steep side of the volume curve exactly where the pressure is greatest.

The constant-angle form is old: it is credited to Lars Jorgensen, whose Salmon Creek Dam in Alaska, finished in 1914, was the first built on the principle. Its upstream face is not a cylinder but a twisted surface, every ring centred on a different point, and its overhang near the crest — where the rings are largest and the dam leans upstream — is the visible signature of the arithmetic.

The saving depends on how fast the canyon narrows. In a canyon with vertical walls the gap is the same at every depth, the two forms are the same dam, and nothing is saved. In a U-shaped canyon that narrows slowly at first the constant-angle dam saves about a quarter; in a V, two-fifths; in a canyon that pinches towards a narrow gorge at its foot, half. A constant radius suits a canyon that does not narrow, and canyons that do not narrow are rare.

A pipe that is open on one side #

The hoop formula is usually met in a pipe or a tank, where a closed ring holds a pressure on its inside and the force in the wall is the pressure times the radius whatever the ring is made of. The arch dam’s ring is the same ring cut open and turned round: the pressure is outside, the force is compression rather than tension, and the two cut ends are held by rock instead of by the rest of the ring.

Cutting the ring open is what creates the optimum. A closed pipe has no choice of angle — it is 360° — and its only free variable is its radius, which is set by how much it has to carry. An open ring across a gap has a free angle, because the gap fixes the chord and not the radius, and the angle trades thrust against length. The pipe and the dam share the formula and differ in what is fixed.

The same open ring appears wherever pressure meets a curve that ends on supports. A curved retaining wall in plan, a cofferdam of sheet piles bowed against the water, a vault seen in section carrying its own weight — each is a ring of some angle spanning a fixed gap, and each has the same trade between a tight curve that is thin but long and a flat one that is short but thick. The number 133.6° is exact only for a circle under uniform pressure, but the shape of the trade is general, and so is the flatness of its minimum: a curve somewhat flatter or tighter than the optimum costs almost nothing, which leaves the designer free to let something else decide.

What the projection rule does here #

The two derivations in this essay — the push on the chord and the thrust in the ring — both come from the rule that the resultant of pressure on any curved surface equals the pressure on its projection. The rule works because pressure has no preferred direction: it pushes equally on every face of a fluid element, so any closed free body of water is in equilibrium under its own weight and the pressures on its faces, and the curved surface can be replaced by a flat one without changing the force.

That is also why the dam’s shape cannot reduce the water’s push. Every arch dam across a 100 m gap, at every angle, receives exactly 49,050 kN per metre of height at a depth of 50 m. What the shape chooses is the path: through bending of a flat wall, through compression of a curved ring, or — in a gravity dam — through the weight of a block of concrete large enough to resist the push by friction on its base. The arch dam is the cheapest of the three in a narrow canyon of good rock because compression is the most efficient way concrete can carry anything, and it can only use compression because the rock at its ends can push back.

One ring, by hand #

Fifty metres down, the water pressure is $9.81 \times 50 = 491$ kPa. Across a 100 m chord the push is $491 \times 100 = 49{,}050$ kN per metre of height.

At the optimum, $\tan(\theta/2) = \theta$; writing $\phi = \theta/2$, $\tan\phi = 2\phi$, which a few trials settle at $\phi$ = 1.1656 rad = 66.78°. The radius is $R = 50/\sin 66.78° = 54.4$ m and the thrust $N = pR = 491 \times 54.4 = 26{,}690$ kN; check: $2 \times 26{,}690 \times \sin 66.78° = 49{,}060$ kN, the push on the chord. At 5,000 kPa the thickness is $26{,}690/5{,}000 = 5.34$ m, and the length $54.4 \times 2.331 = 127$ m, so the ring holds $5.34 \times 127 = 677\ \text{m}^3$ for each metre of its height.

For the semicircle, $R$ = 50 m, $t$ = 4.91 m, length 157 m, 770 m³ — 14 per cent more concrete for a ring 8 per cent thinner.

Rings, and nothing else #

The calculation rests on choices that limit it.

The dam is a stack of independent rings. A real arch dam is also a set of vertical cantilevers fixed into the valley floor, and the water’s load divides between ring action and cantilever action by their relative stiffness — the stiffer path taking more. Near the foot the cantilevers are short and stiff and carry most of the load, which is why a real dam is not thin there even in a V-shaped canyon; the ring calculation is the preliminary method, and the division between the two was the subject of the trial-load analyses of the 1930s.

The ring is thin. The hoop formula takes the thrust on a single radius; a ring whose thickness is a tenth of its radius has a few per cent more stress on its upstream face, and the abutments’ fixity puts bending into rings that the thin-ring model treats as hinged.

The stress is a single allowable number. A real design checks the arch’s own buckling, temperature changes that shorten and lengthen the rings, and the uplift of water in the foundation, which a basement under water shows can be the governing load of a structure that sits in it.

The reservoir is full and still. A dam spends its life at many levels, and the rings near the crest are lightly loaded for most of it; an earthquake adds a hydrodynamic pressure that is greatest part-way down and acts in both directions, so that a ring designed only to be pushed must also survive being pulled, which a thin compression ring cannot. The ring calculation sizes the dam for the one load it carries every day.

The abutments do not move. Rock deforms under 27 MN per metre of height, and an abutment that gives lets the ring flatten and bend; the stiffness of the rock is as much a part of an arch dam as the concrete.

Still open: the ring that is not circular #

Every ring here is an arc of a circle, because the hoop formula is simplest for a circle and because a circle under uniform pressure carries it in pure compression. But the water pressure on a ring is uniform only in magnitude: on a ring whose ends are fixed into rock that gives, or one loaded by silt on its lower part, it is not the circle that carries the load in pure compression but some other curve, flatter at the crown or steeper at the ends. Whether a ring shaped to the funicular of its real loading — elliptical or parabolic arches are used in some dams for this reason — saves more than the 2 per cent margin around 133.6°, or whether the circle’s simplicity is worth more than whatever the funicular ring saves, is the question of what the optimal angle becomes once the ring is allowed a shape as well as an angle.