Generator

A three-pinned arch, rise 2.6 on span 9

Rendered here at the parameters it defaults to, with every essay that calls it — which is the same list as the blast radius of changing it.

A three-pinned arch, rise 2.6 on span 9. A three-pinned arch under a uniform load. One moment equation about the crown hinge gives a horizontal thrust of 23.37, with no stiffness and no assumption about the section. The thrust line lands on the axis everywhere, so there is no bending anywhere in the arch.

11 essays call arch-thrust. The drawing above is what it returns with no arguments at all; every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about.

Where it is called

Changing this generator changes every one of these figures.

A beam, its loads and its reactions. A free body: the beam cut away from everything it touches, with the forces that were doing the touching drawn on it. The reactions are computed from the loads, so their arrows are to scale relative to each other. The moment of everything on the body is summed about three marked points, and every sum comes to zero. Nothing here is a new equation: with only vertical forces on the body, the moment about a point does not depend on that point's height, so every centre anywhere on the plane returns the same equation — and the third equilibrium equation, the horizontal sum, reads nothing equals nothing.

Equilibrium

The equation that is not new, and the three that are

A plane free body yields exactly three independent equations. Most attempts at a fourth are one of the first three wearing different clothes — and on a beam under vertical load, one of the three is already saying nothing.

Structural form

The hinge put in on purpose

An arch with two pinned feet cannot be solved by statics. Add a third hinge at the crown — deliberately weakening it — and the whole structure falls out of one moment equation.

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.

Structural form

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.

A load with a maximum in it, and nothing bifurcates. Load against apex movement for a two-bar frame of half-span 1000 mm and rise 150 mm. The load rises to 133.4 kN at a movement of 64 mm — well short of the 150 mm that would bring the apex level — and then falls. Past that point the frame can only be held by taking load away, so under a dead weight it goes: 260 mm of movement at constant load, arriving inverted and in tension. The minimum on the path is -133.4 kN, the exact negative of the maximum, because the geometry is symmetric about the flat position and the arithmetic knows it.

Stability

The roof that jumps

Every stability failure in this collection so far has been a bifurcation — a straight thing discovering it can be bent. A shallow frame does something else entirely. It stays perfectly symmetric, deforms steadily, and at some point the load it can carry starts to fall while it is still moving in the direction it was pushed.

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.

Structural form

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.

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.

Structural form

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.

The arch does not squash; it leans. The first two buckling modes of the same rib, drawn against its undeformed shape at an exaggeration of a few hundred. The first, at w_cr L³/EI = 49.7, is antisymmetric: one half rises while the other falls and the crown moves sideways. The second, at 117.1, is symmetric — the whole rib settling. The two differ by a factor of 2.36, which is why an arch is braced against sideways movement of its crown rather than against the load it is carrying, and why a tied arch with a single hanger at midspan is doing nothing for the mode that governs it.

Stability

The arch that leans instead of squashing

A masonry arch is asked whether a line of thrust fits inside it. A steel rib is asked a different question entirely: it is a column carrying an axial force along its whole length, and the mode it buckles in puts one half up and the other half down while the crown moves sideways.

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.

Structural form

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.

The line, and the stone it has to stay inside. A masonry pier 9 m high, 1.6 m thick at the top and battered 12% on its outer face, taking a thrust of 40 kN per metre of run at 25° to the horizontal. The line drawn through it is the locus of the resultant on each horizontal cut: everything above the cut is the free body, and the resultant's position is the moment divided by the vertical force. The dashed pair is the middle third, inside which no tension is implied anywhere on the joint. The line stays inside the stone throughout and reaches the base at 0.503 m from the centre, against a half-width of 1.34 m — but outside the middle third, so part of the base joint is open and the toe is carrying a triangle. Nothing about the strength of the masonry appears anywhere in this figure, and that is the point.

Structural form

The weight that makes it safer

Every load in this collection makes a structure worse. A pinnacle does not. A masonry pier fails when the line of compression leaves the stonework, and adding weight at the top rotates that line back towards the vertical without adding anything the pier cannot carry — so the stone is not being strengthened, it is being aimed.

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.

Deflection

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.

Cross the hangers and the chords stop bending. The same tied arch, the same sixteen hangers, the same load on half the span — hung vertically and hung as a network. Vertical hangers make the two chords a Vierendeel frame, which has no truss action at all, so a partial load is carried by bending: 3316 kNm in the tie and 6234 in the arch. Inclined hangers can carry the shear between the chords axially, and the same load gives 686 and 831 — factors of 4.8 and 7.5. The thrust is identical in both, because that is decided by the span and the rise and nothing else.

Structural form

Cross the hangers and the bending goes

A tied arch with vertical hangers is a Vierendeel frame with a curved top chord — it has no truss action at all, so a load on half the span is carried by bending. Incline the hangers so they cross and the same two chords become a truss.

The library, page 1 of 7 — where arch-thrust sits