Series

Curved surface pressure — the series

3 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. Every pressure points at the pin, so the water lifts nothing. A radial gate of radius 8 m holding 6 m of water, with its pivot 6 m above the sill. The pressure on a curved surface cannot be obtained by multiplying anything by anything, so it is integrated round the arc: the horizontal component comes to 176.6 kN/m and the vertical to 110.5. Both are recoverable without any integral at all — the horizontal is the pressure force on the surface's own vertical projection, γH²/2 = 176.6, and the vertical is the weight of the water standing above it, 110.5. They agree to 0.000 per cent. And because every pressure is normal to a circle, every one of them passes through the centre: the moment of the whole 208 kN/m about the pivot is -3.4e-15 kNm, against 353 for a flat gate on the same hinge.

    Every pressure points at the pin

    Pressure acts normal to a surface, so on a curved one every element pushes in a different direction and no multiplication gives the resultant. Two free bodies recover it without an integral — and on a circular surface a third observation makes the whole force disappear from the equation a hoist has to satisfy.

    part 1 · equilibrium
  2. The water pushes on the chord, and the abutments push back along the arch. One horizontal ring of an arch dam, one metre high, 50 m below the water, spanning a 100 m gap with a central angle of 133.6°, drawn in plan with the water on the convex side. The water presses 491 kPa normal to the face everywhere (arrows); by the projection rule its resultant is that pressure times the chord, 49,050 kN per metre of height, pointing straight downstream. Each abutment pushes back along the ring's tangent, at 66.8° to the chord, so the two thrusts are N = pw/2sin(θ/2) = 26,686 kN — the pressure times the ring's radius, 54.4 m, which is the hoop formula of a pipe, arrived at without a pipe.

    The angle that saves the most concrete

    Cut an arch dam into horizontal rings and each ring is a curved surface with water on one side. The projection rule gives the water's whole push on it without an integral — the pressure times the chord — and the two abutments, pushing back along the ring, turn that into a thrust equal to the pressure times the radius, the hoop formula of a pipe arrived at without a pipe. From there one line of calculus says that the ring using least concrete has a central angle of 133.6°, whatever the canyon, the depth or the concrete. That number explains why arch dams in narrowing canyons are drawn at a constant angle rather than a constant radius, and why the angle that suits the concrete is not the one that suits the rock.

    part 2 · equilibrium
  3. In a V-shaped canyon the rings carry the water, and the top rings more than theirs. The water pressure down the crown of a dam 100 m high in a V-shaped canyon 200 m wide at the crest, 5 m thick at the crest and 25 m at the base, with its rings at 133.6° (dashed), divided between the horizontal ring at each depth (solid) and the vertical cantilever at the crown (thin) so that the two deflect alike. At 4 m below the crest the ring carries 195% of the water there — the cantilever, loaded from below and free at the top, leans on it; at 49 m the cantilever carries 11%; at 89 m, 6%. Over the whole reservoir the cantilever carries 10% of the water and the rings the rest.

    The water the valley floor takes

    An arch dam is not a stack of rings. At its crown it is also a vertical cantilever fixed into the valley floor, and the water at each depth is shared between the ring there and the cantilever in whatever proportion makes them bend alike. In a V-shaped canyon the short lower rings are stiff and carry nearly everything, and the top rings carry up to twice their own water because the cantilever leans on them. In a U-shaped canyon the cantilever takes half the reservoir and bends the heel into tension that the ring formula never sees.

    part 3 · equilibrium

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