Structural Engineering & Statics
where the load goes
Concept

Arch — where it appears

A curved member that carries a transverse load principally as compression along its own line, at the price of a horizontal thrust at each springing. Its efficiency depends on how close its shape is to the funicular of the load it carries, and its design is usually decided by whatever receives the thrust.

Named by 11 essays across 4 fields — each of them below, with the objects they name alongside it.

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 wcr 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.

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.

stability · Arch buckling 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 2227 kN, within 1.0 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 2166 kN — 2.7 per cent of the flexibility is the tie — and whatever thrust the arch does not get, it carries as bending: 760 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 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.

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.

structures · Buttress 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 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. The loss is a little more than that ratio, because the released rib also shortens under its own shear, and at the 10 per cent rise drawn it is 0.11 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 0.11 per cent that the rib shortening removes from the thrust leaves 29 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 Two lines and a triangle: a three-hinged arch drawn. A three-hinged arch of 20.0 m span, springings at (0.0, 0.0) and (20.0, 0.0) m and the crown hinge at (10.0, 5.0), under 100.0 kN at 5.0 m. The right half carries no load, so it is a two-force member and its reaction lies along the line from its springing through the crown hinge. That line meets the load's line at K, 7.50 m up, and the left reaction must pass through K too. The triangle of the load and the two reaction directions gives the reactions as 90.1 kN at A and 55.9 kN at B, with a horizontal thrust of 50.0 kN — the values four equilibrium equations return, to 7e-15 kN. The shape of the rib entered nowhere.

The arch that is only its three hinges

A three-hinged arch's reactions come from three points and nothing else, so a parabola, a circle and a portal frame on the same hinges push on their abutments identically. Move a load across and the point where the reactions cross runs along two straight lines through the crown. That is the arch's influence line, drawn with a straightedge — and friction in the hinges it was built around blurs it.

equilibrium · Graphic statics Putting a funicular through three points. Four loads — 40.0 kN at 3.0 m, 60.0 kN at 7.0 m, 30.0 kN at 12.0 m, 50.0 kN at 16.0 m — and three points the polygon must pass through: A and B at the springings and C, 5.0 m above their chord at 10.0 m. A trial pole, dashed, draws a polygon from A that ends 6.39 m below B. The ray through the trial pole parallel to its own closing line cuts the load line at Q, 95.0 kN from the top, which is the left reaction of a simple beam on A and B and does not depend on the pole at all. Every pole whose polygon passes through A and B lies on the line through Q parallel to A B; the one whose polygon also reaches C is 98.0 kN from the load line, which is the moment at C of that simple beam, 490.0 kN·m, divided by C's height above the chord. That is the three-hinged arch's thrust — the four equilibrium equations give 98.0 kN.

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 Two pencil lines, and where they cross. Two lines drawn with a pencil 0.2 mm wide are two bands, drawn here much wider than a pencil so the shape can be seen, and they cross not at a point but in a parallelogram. At 60° apart the parallelogram's long diagonal is 2.0 pencil widths — 0.40 mm; at 12° apart the parallelogram's long diagonal is 9.6 pencil widths — 1.91 mm. The crossing's uncertainty along the bisector is the width divided by twice the sine of half the angle, so it grows without limit as the lines turn parallel, and a construction that finds a point by crossing two lines inherits it.

How wrong a drawing is

A pencil line is a band, and two bands cross in a parallelogram that grows as they turn parallel. The accuracy of a graphical construction is therefore a property of the angles it makes, not of the hand that made it — and the worst case is the shallow arch, the structure the method was most used on. Measured properly, the drawing's error there is the size of the builder's, and the check draughtsmen relied on cannot see it.

equilibrium · Graphic statics The rib yields long before it would buckle. The largest combined stress, axial plus bending, anywhere in a 60 m two-pinned parabolic rib at a 12 m rise (EI 250,000 kN·m²) with a 12,000 mm² section 1,000 mm deep, against a factor on 15 kN/m of dead load over the span and 6 kN/m of live load over the left half together. Dashed: first order; solid: second order; dotted: the 355 N/mm² yield stress. First order, the rib reaches yield at a factor of 1.73; second order, at 1.18. The same load pattern, spread uniformly, would bifurcate the rib at a factor of 2.97, the right-hand edge — which is the number the eigenvalue reports and a number the rib never reaches.

The weight that bends a rib it cannot bend

A parabolic rib carries its own uniform dead load as pure thrust, with no bending in it at all. Put a live load on half the span and the rib bends in the shape of its own buckling mode from the first kilonewton, and the whole thrust, most of it from the dead load, multiplies that bending. The rib never reaches its buckling load. It yields well short of it, at a load its first-order check says it can carry.

stability · Arch buckling The same rib, fixed, bends at both ends as well as the crown. The bending moment that axial shortening leaves in a 60 m parabolic concrete rib at a 6 m rise (EI 420,000 kN·m², EA 20,400,000 kN) under 60 kN/m, sagging upward, from a frame solve of the rib. Dashed: two-hinged, sagging throughout, 29.4 kN·m at the crown. Solid: fixed at the springings, 57.9 kN·m sagging at the crown and 112.4 kN·m hogging at each springing. The funicular load itself leaves no moment in either rib; all of this is the 0.11 and 0.63 per cent of the thrust the two ribs lose by getting shorter.

The springings that make shortening worse

Fixing an arch at its springings is the stiffer, cheaper and usual way to build one in concrete, and it makes the arch six times as sensitive to its own shortening. The thrust it loses acts at the elastic centre, two thirds of the way up, so the moment lands at the springings as well as the crown, twice as large and the other way round — at the section the fixed arch is designed at, not away from it.

deflection · Rib shortening How stiff an abutment has to be to count as fixed. The shortening moments in a 60 m parabolic concrete rib at a 6 m rise (EI 420,000 kN·m²) under 60 kN/m as shares of the fixed rib's, against the rotational stiffness of each springing as a multiple of EI/L on a logarithmic scale: at the springings (solid, −112 kN·m when fixed) and the crown (dashed, 58 kN·m fixed, 29 two-hinged). The springing moment is half the fixed value at kθ = 11.3 EI/L and nine tenths at 102 EI/L; at EI/L itself it is 8 per cent. The crown moment never falls below the two-hinged rib's 51 per cent of the fixed value.

The abutment that spreads before it turns

A fixed arch is fixed only if its abutments hold still, and they have two ways not to. Turning releases the springing moment, but slowly: an abutment has to be a hundred times the rib's EI/L to hold nine tenths of it, and a footing on stiff clay under a slender concrete rib already is. Spreading does the opposite. The rib's own shortening is worth 13.6 mm of spread, so an abutment that gives 7 mm a side under the thrust doubles every secondary moment in the arch — and a footing on dense sand gives 8.

deflection · Rib shortening

Named alongside it

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

FunicularThrustAxial shorteningThrust lineGraphic staticsIndeterminacyInfluence lineAntisymmetric modeCompatibilityElastic centreFlexibilitySecond-order

All concepts