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.

Assumes The hinge put in on purpose, The middle third and Weight is the only thing resisting it.

A masonry arch stands if some line of compression can be drawn inside the stonework. That is a complete account of the arch and it says nothing at all about what happens at the springing, where the line arrives at the abutment travelling at an angle and carrying a horizontal push it has to give to something.

This essay is about the something. It is the same question one free body further down, and the answer has a property the arch’s does not: weight added at the top makes it better.

The line, and the stone it has to stay insideA 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.T = 40 kN at 25°no pinnaclee = 0.503 moutside the middle thirdthe line stays in the stone · σ = 351 kPa at the toe
Fig. 1 A masonry pier nine metres high, taking a thrust of forty kilonewtons a metre at twenty-five degrees. The line drawn through it is the resultant on each horizontal cut; the dashed pair is the middle third. It stays in the stone, and it does not stay inside the third.

Which free body produced the number

Cut the pier horizontally and take everything above the cut. Three things act on it: the arch thrust arriving at the top, the weight of the masonry above the cut, and whatever is standing on top. Their resultant has to be carried by the cut face, and its position is

e=MNe = \frac{M}{N}

with MM the moment of all three about the pier’s centre-line at that level and NN their total vertical force. Trace ee down the pier and the locus is the thrust line.

For this pier the base cut carries 424 kN a metre of self weight plus 16.9 kN of vertical thrust component, against a moment from 36.3 kN of horizontal thrust acting nine metres up. That puts the resultant 0.503 m from the centre of a base 2.68 m wide.

Two thresholds matter, and they are geometric:

Inside the stone, e<w/2|e| < w/2. Beyond this there is no set of compressive forces in equilibrium with the load and the pier is a mechanism. Here w/2=1.34w/2 = 1.34 m, so there is a factor of nearly three in hand.

Inside the middle third, e<w/6|e| < w/6. Beyond this the joint opens on one side: a material that cannot be pulled cannot hold a resultant outside its kern, so part of the bed joint stops bearing. Here w/6=0.447w/6 = 0.447 m, and 0.503 is outside it.

The stone’s strength appears in neither. The toe stress at the base is 351 kPa, against a masonry compressive strength of several megapascals — a factor of twenty unused. What the pier is short of is not strength.

Why the joint opening is worth caring about

A resultant outside the middle third does not mean the pier falls down. It means the bed joint has opened, and three things follow.

The bearing area shrinks, so the stress at the toe rises — faster than linearly, because the resultant is moving towards the toe while the area is shrinking beneath it. That is the same non-linearity every eccentrically loaded no-tension section has, and it is why the stress runs away well before the resultant reaches the edge.

The pier becomes softer, because a section bearing on a third of its width has a fraction of the second moment of the full one. A softer pier deflects more under the thrust, which moves the resultant further out, which opens the joint more — a mild version of the load that makes itself worse, and the reason a slender masonry pier has a genuine second-order problem while a squat one does not.

And an open joint admits water, which is a durability problem that becomes a structural one over a few centuries.

The pressure runs away outside the middle thirdPeak bearing pressure under a 4 × 3 m base carrying 900 kN, against the eccentricity of the load. Inside the middle third the line is straight and the pressure has doubled by the time it reaches the edge of it: 75 kPa at the centre, 150 kPa at e = B/6. Beyond that the base lifts, the contact length shortens, and the curve turns upward without limit — at e = 1.54 m the peak is 433 kPa on 1.38 m of base.00.20.40.60.811.21.40100200300400eccentricity of the resultant (m)peak pressure (kPa)B/6: the base is on the point of liftinguniform: 75 kPa
Fig. 2 What happens to the bearing pressure as the resultant crosses the kern boundary. Inside it the whole width bears; outside it the width shrinks while the load moves towards its edge, and the peak stress rises far faster than the eccentricity does.

The pinnacle, and the shape of what it buys

Now add weight on top. A pinnacle sits on the pier’s own centre-line, so its weight contributes to NN and contributes nothing to MM — it has no lever arm about the axis the eccentricity is measured from.

e(P)=M0N0+Pe(P) = \frac{M_0}{N_0 + P}

which is a hyperbola. Two things follow from the shape rather than from the numbers.

The eccentricity falls as 1/N1/N, so doubling the weight on the joint exactly halves it. That is a strong return at first and a weak one later: the first tonne is worth ten of the tenth.

There is no threshold. Any pinnacle helps, and none of it is wasted; the improvement is smooth from zero. That is unusual — most structural interventions have a size below which they do nothing at all, as a brace does.

The first tonne is worth ten of the tenthThe eccentricity of the resultant at the base of the pier, against the weight added on top of it. The moment about the base does not change at all when a pinnacle is added — its own weight has no lever arm about the pier's centre — while the vertical force grows with it, so the eccentricity falls as exactly 1/N. That is a hyperbola, and it is why the shape of the answer matters: the pier starts at 0.503 m against a half-width of 1.34 m, and 55 kN brings it back inside the middle third at 0.447 m. Doubling that weight again buys a third as much as the first half of it did. A pinnacle is not strengthening the pier — nothing about the stone has changed. It is aiming it.02040608010012014016000.20.40.60.811.21.4weight added on top (kN per metre of pier)eccentricity at the base (m)the middle thirdthe edge of the stone55 kN brings it back
Fig. 3 Base eccentricity against the weight added on top. The moment does not change and the vertical force does, so the curve is one over N — and 55 kN a metre brings this pier back inside its middle third.

For this pier the answer is 55 kN per metre of run, which is a stone pinnacle of ordinary Gothic proportions. That is the whole of the mediaeval device: the pinnacle is not ornament with a structural excuse, and it is not ballast in the sense of dead weight resisting an uplift. It is a term in a denominator.

The batter, which is worth twelve times as much

There is a second way to move the same eccentricity, and on this pier it is far more powerful.

Take the batter away — make the pier a uniform 1.6 m rather than widening to 2.68 m at the base — and the pier does not stand at all. The base eccentricity goes from 0.503 m to 0.978 m against a half-width of 0.80 m: the resultant is outside the stone.

To recover it takes 74 kN per metre of pinnacle merely to get the line back inside the section, and 890 kN — twelve times the battered pier’s requirement — to get it back inside the middle third.

The batter does two things where the pinnacle does one. It adds self weight, which is the pinnacle’s effect. And it widens the section, which raises both thresholds at the level where the resultant is furthest out. The two multiply rather than adding, so widening the base is worth much more per tonne of stone than piling the same tonne on the top.

The line, and the stone it has to stay insideA masonry pier 9 m high, 1.6 m thick at the top and battered 0% 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 leaves the stone before it reaches the ground, which means no set of compressive forces inside this pier is in equilibrium with the thrust: it is a mechanism. Nothing about the strength of the masonry appears anywhere in this figure, and that is the point.T = 40 kN at 25°no pinnaclee = 0.978 moutside the middle thirdthe line leaves the stone: the pier is a mechanism
Fig. 4 The same pier with no batter. Same height, same thrust, same masonry, and the resultant leaves the stone before it reaches the ground — the only thing that changed is that the section did not grow on the way down.

Push the batter to 0.25 and the pier reaches a 3.85 m base, carries 540 kN of its own weight, and lands at an eccentricity of 0.116 m — comfortably inside the middle third with no pinnacle whatever.

Which is what a Gothic buttress actually looks like, and it is worth noticing what it does not look like. It is not a uniform pier. It is not a pier with a lump on top. It is a wedge that steps outwards as it descends and carries a weight at its head, and both features are doing the same arithmetic on the same equation.

The section that governs, which is not the base

The base carries the most vertical force and has the widest section, and it is easy to assume it is therefore the critical one. It is not, and the counterexample is sharp.

Take the same pier and make the thrust shallower — five degrees rather than twenty-five, which is what a flatter arch delivers. The base eccentricity falls to 0.594 m, well inside a 2.68 m base. And the pier does not stand.

The reason is that the line leaves the stone near the top, where the pier is only 1.6 m wide and almost none of its own weight has accumulated. Up there the moment from the thrust is small but the vertical force is smaller still, and the ratio is what matters.

So the check has to be made on every cut, and the governing one is wherever e/(w/2)|e|/(w/2) peaks. On a battered pier under a steep thrust that is usually near the base; on a uniform pier or under a shallow thrust it is near the top. The critical section moves with the angle of the load, which is not a property any single cut can report — the same trap an envelope sets in a different problem.

That is also the mechanical reason a flying buttress is placed where it is. A flatter arch delivers a flatter thrust; a flatter thrust is harder to turn downwards; and the fix is to deliver it high on a pier that has room to widen below — which is a flyer arriving at the top of a pier, rather than a thrust arriving at the bottom of a wall.

A line of thrust, and the masonry it has to stay insideAn 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.thrust anywhere from 3.85 to 5.23 fitsH = 3.85, leastH = 5.23, most
Fig. 5 The arch that delivers the thrust. Which of the admissible thrusts it actually takes is not decided by statics — so the pier below has to be able to take the whole range, not one member of it.

What a buttress is competing against, which is a tie

There is a completely different way to deal with an arch’s thrust, and comparing the two is the clearest statement of what a buttress is for.

Put a tie across the springings and the thrust never reaches the ground at all: the arch and its tie are a self-contained assembly delivering nothing but a vertical reaction, and the piers below can be as slender as their axial load allows. That is the tied arch, and it is structurally superior in every measurable way — less material, no eccentricity problem, no dependence on the ground.

It is also unusable in the building the buttress was invented for. A tie across the springing of a nave vault runs across the nave at the height of the triforium, and the whole point of the building is that the space is open. So the thrust has to be taken to the ground on the outside, through masonry, by a structure that cannot be pulled.

Seen that way the flying buttress is not a clever solution to carrying a thrust. It is what is left when the good solution has been ruled out by the brief, and every feature of it — the flyer’s angle, the pier’s batter, the pinnacle — is compensation for the absence of a tie. The comparison is worth making because it is the honest reason the device is elaborate.

The tie is a redundancy, so its stiffness decides the thrustThrust 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.0.10×0.32×10×00.20.40.60.81tie stiffness, relative to the one drawnthrust ÷ its rigid-tie value, and bending ÷ its worstthrustbending, to 7422 kNmas built: 2168 kN, tie stretches 68 mmfunicular wL²/8f = 2250 kN · rigid tie 2229 · the difference is the rib's own shortening
Fig. 6 The alternative, which was ruled out. A tie takes the thrust in tension across the span and the supports carry nothing horizontal at all — which is cheaper, simpler and in the wrong place.

The other way a pier goes

Everything above is about the resultant leaving the section. There is a second failure on the same free body, and it is checked on the same cut: sliding.

The horizontal thrust is 36.3 kN a metre. The friction available on the base joint at a coefficient of 0.6 is 264 kN. A factor of seven, and it is comfortable here — but it is not comfortable everywhere, and its behaviour is the opposite of the overturning check’s in one important respect.

Adding a pinnacle helps sliding too, because friction is proportional to the normal force. Widening the base does not help sliding at all beyond the weight it adds, because friction does not care how wide the joint is. So the two interventions that are nearly equivalent for the thrust line are not equivalent for sliding — and on a low pier under a very flat thrust, where sliding is the governing check, the pinnacle is the better purchase of the two.

The other place it bites is a joint that is not level. A bed joint sloping outwards reduces the friction available and adds a component of the weight to the demand at the same time, which is why a buttress’s joints are cut square to the load path rather than to the horizon on any pier of consequence.

The reaction lies inside the cone, so the block standsA block of 100 on a plane at 15°, against a coefficient of friction of 0.6. Resolving across and along the plane gives a normal force of 96.6 and a friction demand of 25.9, against a capacity of μN = 58.0 — a ratio of 0.45. Added together the two make one contact reaction leaning 15.0° from the normal, and the admissible reactions fill a cone of half-angle arctan μ = 31.0°. Equilibrium is possible exactly when the demanded reaction lies inside that cone, which here it does. The weight enters neither the cone nor the lean: a block of any weight on this slope leans its reaction by the same 15.0°, which is why the angle of repose is a material property and the size of a heap of sand is not.15°reaction, leaning 15.0° from the normalthe cone: half-angle arctan μ = 31.0°W = 100demand 25.9 against a capacity of 58.0 — F/μN = 0.45the weight appears nowhere in the cone — only the direction of the reaction is asked about
Fig. 7 The condition sliding actually is: the resultant has to lie inside a cone about the joint’s normal, whose half-angle is the friction angle. A check on the position of a line again, and again with no strength in it.

What a mediaeval builder was actually optimising

None of the above was available to the people who built these. There was no thrust line, no eccentricity, no kern; Coulomb’s analysis of the arch is 1773 and the buildings are three hundred years older.

What there was, is a rule of proportion — the pier’s width as a fraction of the vault’s span, the pinnacle’s height as a multiple of the pier’s width — passed on in lodges and refined by what fell down. Those rules encode the arithmetic above surprisingly well, and it is worth seeing why they could.

The thrust from a vault of given shape scales with its span times its load. The pier’s self weight scales with its width times its height. The eccentricity is a ratio of the two, so a proportional rule is dimensionally the right kind of rule: scale the whole building and the ratio holds. That is not true of a strength check, where the demand grows as a volume and the capacity as an area, and it is why proportional rules worked for masonry for centuries and failed for iron almost immediately.

The failures, when they came, were at the edges of the proportion set — the tallest naves, the widest vaults, the thinnest piers. Beauvais fell in 1284, and its choir was the tallest ever attempted.

Both failures are decided by the same two numbersFactors of safety against overturning and against sliding, for a body 4 m wide weighing 900 kN under a wind pressure of 1 kN/m², as its height grows. Overturning falls as the square of the height and sliding as the first power, so they cross: below 34.6 m the body overturns at a factor of one, and uplift at one edge has already begun at 20.0 m — a ratio of exactly √3, whatever the numbers are.051015200123456height of the body (m)factor of safetyuplift starts at 20.0 moverturningsliding
Fig. 8 The check underneath the proportional rules, in its simplest form. Whether a block stands is a ratio of a width to a height, which is a number that survives being scaled — and it is the property that lets a rule of proportion be right at every size.

Where this model stops

Three limits, and the first is the one that makes the whole approach honest rather than approximate.

The thrust is not a number. An arch with two pinned feet is indeterminate, and the thrust it delivers is anywhere in a range. The pier has to be safe for the whole range, and — this is the useful half — it only has to be safe for one line at each thrust. That is limit analysis: finding any admissible line inside the stone is a proof of safety, and no line has to be the real one. What the drawing above shows is a lower bound and is not a prediction.

The masonry is assumed to have no tension and infinite compressive strength. The first is nearly true and conservative. The second is not true and is why the toe stress was computed above: it is a check that has to be made and that almost never governs, which is a different thing from a check that can be skipped.

The pier is assumed rigid. Real ones deflect, the deflection moves the resultant, and on a slender pier that feedback matters. Squat piers — which is most of them — are unaffected.

What the picture cannot show

The drawing is a plane. A real buttress is a three-dimensional block, often stepped in two directions, and the thrust it takes arrives at an angle to its own plane wherever the vault is not a simple barrel. The thrust line is then a curve in space and the condition is that it stays inside a solid, which is the same statement and a much harder drawing.

Nor does the drawing show time. A masonry pier settles, its mortar creeps, and the arch above it relaxes; the thrust five hundred years after construction is not the thrust the day the centring came out. Settlement of one support would rewrite the internal forces of a redundant frame; here it rewrites nothing, because the analysis never claimed to know which line the structure had taken. What makes the structure survivable is precisely that it is a lower bound: a pier that can accommodate any thrust in a wide range does not need the range to stay put.

The generalisation

The habit worth carrying out of this is a way of reading a stability problem.

Every check in this essay has the same form: is a line inside a region? The thrust line inside the stone. The resultant inside the kern. The reaction inside the friction cone. None of them contains a material strength, all of them contain a geometry, and each is improved by moving either the line or the region.

That is a different kind of design variable from the ones most of this site uses. A beam is made safer by raising its capacity; a pier is made safer by moving a line, and the moves available — weight on top, width at the bottom, the angle the load arrives at — are all geometric. It is the reason the mediaeval builders got as far as they did with no theory of stress at all: the problem they were solving does not have stress in it.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

ArchBearing stressButtressEccentricityEquilibriumFree bodyFrictionKernLimit analysisMasonryNo tensionOverturningSelf weightStabilityThrust line