Field

Structural form

Trusses, cables and arches: the shapes that carry load by geometry rather than by bulk.

A Pratt truss of 6 panels. A Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 10 members came out in tension, 9 in compression and 2 carrying nothing.

The triangle that cannot fold, and everything built out of it

A square of pinned bars is a mechanism. A triangle is not, and that single fact is the reason trusses exist and the reason they look the way they do.

The cable and the arch are the same curve. The shape a cable takes under a uniform load is a parabola, and it carries that load in pure tension. Reflected, the identical curve carries the same load in pure compression, which is what an arch is.

The shape that carries itself, and the arch that is its reflection

Hang a chain and it takes the one shape that carries its load in pure tension. Turn the shape upside down and it carries the same load in pure compression. That is what an arch is.

Chord force against truss depth. The force in a truss chord for a fixed bending moment, against the depth of the truss. The relationship is a reciprocal: the chords form a couple whose lever arm is the depth, so a shallow truss pays for it steeply.

Depth is the cheapest strength there is

Doubling the depth of a truss halves its chord forces without adding a gram of material to the chords. Nothing else in structural design is that cheap, and almost every structure has already spent it.

The funicular polygon for five loads. The shape a string takes under 5 point loads, with a vertex at every load and a constant horizontal component of 27.9 throughout. The end segments carry the most — 38.6 against 28.0 in the flattest one — because they are steepest.

The polygon that finds the shape

A hanging string under five loads has no smooth curve in it — it has five vertices and six straight segments, and every slope in it is a running sum divided by one number.

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.

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 portal frame swaying under 20 kN. A portal frame pushed sideways, solved by the stiffness method because statics cannot divide the load between two columns. The base shears come out at 10.0 and 10.0 and add to the applied 20; the peak moment is 22.2. The sway is drawn hugely exaggerated, and the moment diagram is plotted on each member's tension face.

The frame that leans, and what stops it

A rectangle of pinned bars folds flat. Make the corners rigid instead of adding a diagonal and it does not — which buys an unobstructed opening and costs bending in every member of it.

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.

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.

The joints are not pins, and this is what that costs. A 4-panel Pratt truss solved twice on the same stiffness matrix: once with a moment release at every member end, which is the pin-jointed idealisation, and once with the joints continuous, which is what welding them produces. The axial forces are the same to within a per cent; the bending the second solution adds is worst in member 0, where the bending stress reaches 24.3% of the axial stress. Members are shaded by that ratio.

The joint that is not a pin

Every truss on this site is analysed as though its joints were frictionless pins. Almost none are. The bending that follows is called secondary, which is a claim about size — and the claim is checkable.

Take that one away and the load finds another route. A 6-panel pratt truss under 20 kN at each top node, before and after member 2 is removed. The load redistributes. The worst-affected survivor now carries 2.03 times what it did, and four members that carried nothing before are now working. Whether that is survival depends on how much spare capacity was there, which is a different question from whether the frame was strong enough.

The structure that survives losing a member

Every check in this collection asks what a structure carries. None of them asks what is left when part of it is gone — and two frames with the same members, the same weight and the same factor of safety can answer that question completely differently.

The slab that spans both ways

A panel supported on four sides sends its load in two directions at once, and the share is decided by a fourth power — so a panel a third longer than it is wide has already stopped being a two-way slab in any useful sense. What it does at collapse is a different calculation with a different answer.

Two shapes that are the wrong way up for each other. Deflected shapes of a 20-storey building under a uniform wind, drawn to the same scale. The wall alone bends: its shape is flattest at the base and steepest at the top, reaching 146 mm. The frame alone shears: it is steepest at the base where the storey shear is largest, reaching 140 mm. Tied together at every floor they reach 58 mm — less than a quarter of either, and less than the 72 mm two springs in parallel would give, because each is stiff exactly where the other is not.

How a tall building stands still

A shear wall bends and a framed tube shears, and the two deflected shapes are the wrong way up for each other. Tie them together at every floor and the pair is stiffer than the sum of their stiffnesses — because near the base the wall holds the frame back and near the top the frame holds the wall.

The props decide where the stress ends up. Bottom-fibre stress in the steel of a 12 m composite beam carrying 12 kN/m of wet concrete and 18 kN/m afterwards. Unpropped, the bare steel takes the first stage alone and reaches 292 MPa; propped, the finished composite section takes everything and reaches 186 MPa — a ratio of 1.57. 62% of the unpropped beam's final stress was locked in before the slab was structural at all. The deflections differ by 1.73 times for the same reason, and no drawing of the finished beam distinguishes the two.

The structure that was never complete

Every analysis in this collection is of a finished structure loaded once. Real ones are built in pieces, and each piece carries whatever was present at the moment it became structural — so the stress in a member depends on when it arrived, which appears nowhere on any drawing.

The further it deflects, the harder it pulls back. Total load against midspan sag for a 30 m cable of 1000 mm² prestressed to 500 kN, carrying 5 kN/m. The cubic H³ − T₀H² − w²L²EA/24 = 0 was bisected at every point of the curve, so the sag at the full 150 kN is 0.740 m rather than the 1.125 m the flat-cable formula WL/8T₀ gives — the straight dashed line, which is the tangent to this curve at the origin and nothing more. Its slope is the initial stiffness 8T₀/L = 133.3 kN/m; at the marked point the tangent has reached 341.2 kN/m, 2.56 times as stiff, and the horizontal component of the tension has risen from 500 kN to 760 kN. Nothing about the steel changed. The geometry got better at the job.

The stiffness that comes from the shape

A cable has no bending stiffness whatever, and it still holds up a roof. What resists the load is the change of its own geometry, so its stiffness is a function of the tension already in it — and prestress buys stiffness that no change of material could.

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.

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.

The depth is decided by how far it moves, not by what it can carry. A column carrying 6000 kN landing 3 m into a 12 m transfer member. The free body is the member itself, cut under the column: M = P·a(L − a)/L = 13500 kNm, with 4500 kN of shear on one side of the cut and 1500 on the other. At an allowable stress that moment asks for 1.94 m of depth — the dashed outline — and keeping the settlement it causes inside the floors' own bending asks for 2.55 m, which is the member drawn solid. 31% more depth is bought by nothing the strength calculation can see. The depth grows as √(P·a), so four times the load is exactly twice the depth, and depth in a transfer member is a storey nobody occupies.

The column that stops

A load path that runs straight to the ground costs almost nothing. Interrupting one costs depth in proportion to the square root of the load times the distance it is moved — and the interruption's own deflection becomes the settlement of everything standing on it.

The shear goes round the corner instead of across it. A 6-panel Vierendeel girder, 12 m by 1500 mm, under 100 kN at mid-span. There is no diagonal in it, so each panel's 50 kN of shear is carried as bending in the chords: the curves drawn along them are the chord moments, and every one passes through zero at the middle of its own panel. The local moment is the panel shear times the panel length over four, 25.0 kNm, and it adds to an axial force of 200 kN from the global moment at the same point. The girder deflects 6.20 mm against 3.18 mm for the same members triangulated — 1.95 times — and 68% of that movement is chord bending that a diagonal would have removed entirely.

The truss with no diagonals

A rectangle without a diagonal is a mechanism, so a Vierendeel girder makes its corners rigid instead. The shear a diagonal would have carried as axial force now travels as bending in the chords, and bending is a far more expensive way to move a force.

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.

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 same sheet, twice, and a factor of ten thousand. A 3000 mm developed width of 3 mm sheet, covering 2400 mm in plan — so the legs sit at 36.9° and the fold is 300 mm deep. Flat, its second moment about its own mid-plane is 6750 mm⁴, which spans nothing. Folded, it is 67.50×10⁶ — 10000 times as much, which is exactly the depth in thicknesses squared. The material is identical, the plan cover has fallen by 20%, and the only thing that changed is where the material sits. What limits it is buckling of the leg: at this leg length the flat between the folds goes at 27 N/mm², well below the steel's 275.

Folded until it spans

A flat sheet has a second moment of area of B·t³/12 and will not span anything. Folded, the same material has B·t·h²/12, and the gain is exactly the fold depth over the thickness, squared — a ratio with no material in it and no width in it.

A frame with no plane to be drawn in. The tetrahedron, solved: three equations at every free joint, one axial force in every member. Members drawn heavy carry more; tension and compression are separated by the sign that came back from the solve rather than by inspection. The count is m + r = 12 against 3j = 12, which makes it exactly determinate, and joint equilibrium closes to 1.8e-15.

Three equations at every joint

A plane truss is determinate when m + r = 2j. A space frame needs 3j, and that one changed digit is why a cube of twelve bars is six mechanisms short while looking perfectly solid — and why every three-dimensional frame ever built is made of triangles in several planes at once.

The corner that moves most

A lateral force is shared out in proportion to stiffness only if it passes through the centre of rigidity, which is not the centre of the plan and not the centre of mass. The distance between the two is a torque, and the wall that pays for it is the one furthest away and carrying least.

The floor is a beam lying down

A floor plate spans horizontally between the walls that resist a lateral load, carries a distributed inertia load, and has chords, a web and a span-to-depth ratio like any other beam. Its stiffness decides whether the walls share the load by their stiffness or by the area of floor nearest them — and the familiar tributary answer turns out to be neither limit.

Two null spaces of one matrix, and the count is their difference. Two pin-jointed frames, each with the forces it can carry with nothing applied to it drawn on its bars — tension one colour, compression the other, thickness in proportion. That force set is the null space of the equilibrium matrix; a mechanism is the null space of its transpose; and Maxwell's count b + r − 2j is the difference of their dimensions and knows neither of them separately. A square with both diagonals has s = 1 and m = 0. Two bars in a straight line has s = 1 and m = 1 with a count of 0, so the count is satisfied by a frame that both folds and can be prestressed — and the prestress stiffness is positive, which is why a tensioned pair of collinear bars is stiff at all.

The forces that are there with nothing applied

Maxwell's count is the difference between two dimensions, and it knows neither of them separately. A frame can satisfy it exactly and still both fold and be prestressable — and when it does, the second of those is what stops the first.

A couple applied to the core, and two columns to make it. A 20-storey core with one outrigger at 59% of its height. The arm is stiff in bending and the perimeter columns are stiff in tension and compression, so between them they resist the core's rotation at that level — a couple of 35283 kNm here, carried as a 294 kN pair in the columns at 120 m centres. The compatibility is one equation: the core's rotation at that level, less what the couple takes back out of it, equals the rotation the arm and its columns allow. The top drift falls from 360 mm to 74, which is 80% of it, and the base moment from 73500 to 38217 kNm. The deflected shape is drawn hugely exaggerated: the real top drift is about one five-hundredth of the height.

The arm that makes the columns work

The perimeter columns of a tall building are already there, already carrying gravity, and already the furthest thing from the centre. They take almost none of the overturning, because a floor slab transmits shear and not moment — and one storey-deep arm at the right height changes that by nearly a half.

Every path to the ground goes through the link. A braced bay 8 m by 4 m whose two diagonals stop 800 mm apart instead of meeting. The storey shear reaches the ground through the diagonals, and the vertical components they deliver to the beam have to pass through the segment between them: the link carries 47% of the applied shear as a shear force, at a lever arm short enough that its ends reach 0 kNm while the rest of the beam carries 0. The deflected shape drawn is the solved one, magnified — the real drift under this load is 0.008 mm. Everything outside the link is designed to stay elastic while the link is yielding, which is what makes the mechanism a choice rather than a hope.

The part that is meant to be weak

A braced frame is stiff and has nowhere to yield. A moment frame yields everywhere and is soft. Move the two diagonals a metre apart along the beam and the whole storey shear has to pass through the segment between them — which keeps most of the stiffness and puts every yielding in one member the designer chose.

The split is where buckling puts it. A branching column carrying 400 kN to two points 3.0 m apart, over 9.0 m. Every member is drawn at the thickness it needs: the area is the larger of N/σ and what Euler asks of a strut of that length, and 3 of 3 members here are sized by buckling rather than by strength. The split sits at 57% of the height, which is where the total volume is least — 51% less than the fan of straight struts that carries the same load with the same stresses.

The tree that strength does not ask for

A branching column carries a roof on many points and reaches the ground on one. Size every member by its stress and the optimum tree turns out to have no trunk at all — the best answer is a fan of straight struts from the base. Put buckling in and the trunk appears, at 57 per cent of the height.

The corner columns take what the middle ones did not. Axial stress in the columns across one flange face of a 30 by 40 m framed tube, at the base. Plane sections says the flat line: every column on the face at the same distance from the neutral axis, therefore at the same stress. The solved distribution is the curve — 69.4 N/mm² at the corner against 18.2 in the middle, a ratio of 3.81. The middle columns lag because the only route the axial force has into them is the in-plane shear of the spandrel frame, bay by bay from the corner. The face is carrying its resultant on an effective width of 51 per cent, and the tube deflects as though its second moment were 72 per cent of the gross.

The corner columns take more than their share

A framed tube is a hollow cantilever, and a hollow cantilever's flange ought to be uniformly stressed. It is not, and the reason is that the only route the axial force has into a column in the middle of a face is the in-plane shear of the frame — one bay at a time, from the corner inwards.

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.

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 windward guy tightens, the leeward one gives way. A 120 m mast on three guy levels, at a wind of 3 N/mm, with the deflection drawn 0.54 times its true size. The guys start at 160 kN each and end at 259 against 95, 299 against 83, 221 against 112 kN. The leeward guys still carry a real force — the lowest keeps 28 per cent of its partner's tension — and supply almost none of the restraint, because their tangent modulus has fallen to 54 per cent of the steel's. The mast top moves 100 mm, its worst bending moment is 554 kNm at 40 m, and it is carrying 801 kN of axial load that nothing but the guys put there.

Held by something that goes soft

A guy is a cable, so it has no stiffness of its own — what resists a mast's movement is the guy's geometry changing, and how much of that there is depends on the tension already in it. Wind pushes the mast towards the leeward guy, which is the one losing tension.

The cable that is a spring

A suspension bridge's cable is a funicular — it takes the shape the load gives it, and the deck's job is to make the load one that shape is right for. A stayed deck is not that at all. Its cables are straight, each reaches the deck at one point, and every one of them behaves as an inclined spring.

Held up by a pressure nobody can feel. An air-supported roof of 60 m span and 9 m rise. The membrane has no bending stiffness whatever, so the only thing that can hold it in tension is a pressure difference, and the pressure has to exceed the load per unit plan area and nothing else: 0.25 kN/m² of fabric plus 0.6 of snow is 0.85 kN/m², so 1.19 kN/m² does it — 1190 pascals, which is 1.17 per cent of an atmosphere and 121 millimetres of water. Ears do not notice it. A door does: at 2.1 kN on an ordinary leaf, the building needs an airlock rather than a handle. And the whole of it arrives at the foundation as 3365 kN of uplift — 17.9 kN on every metre of perimeter — which is the bill the pressure's smallness conceals.

Held up by the air inside

A membrane has no bending stiffness at all, so the only thing that can hold it in tension is a pressure difference. The pressure needed to hold up a roof is smaller than the pressure a closed door makes — and the same pressure arrives at the foundation as hundreds of tonnes of uplift.

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.

Four ways to make a cell resist being racked, and one that is not one. One cell of a grid shell under the membrane shear it has to carry, by the four mechanisms available for carrying it, with the racking each produces over a 30 m span under 1.2 kN/m² of asymmetric load. A serviceability limit of span/250 is 120 mm. Four pin-jointed bars in a quadrilateral have no in-plane shear stiffness whatever — the cell folds, and the answer is not a large deflection but a mechanism. Rigid nodes carry the shear by bending the members over a cell, which smears to 12EI/s³ and comes to 0.35% of what a continuous sheet of the same stretching stiffness gives: 1205 mm, ten times the limit. One diagonal per cell, or a third member direction, carries it axially instead and lands within a factor of two of the sheet. That is the whole difference between a grid shell and a row of arches.

A shell only if the grid takes shear

A curved surface carries load in its own plane at a fraction of the material a flat one needs, and every gridshell ever built is an attempt to buy that with members instead of with a surface. The attempt succeeds or fails on one property nobody draws — whether four bars meeting at a corner can resist being racked — and a pinned quadrilateral grid cannot resist it at all.

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.

The corner columns take what the middle ones did not. Axial stress in the columns across one flange face of a 36 by 36 m framed tube, at the base. Plane sections says the flat line: every column on the face at the same distance from the neutral axis, therefore at the same stress. The solved distribution is the curve — 78.5 N/mm² at the corner against 13.4 in the middle, a ratio of 5.88. The middle columns lag because the only route the axial force has into them is the in-plane shear of the spandrel frame, bay by bay from the corner. The face is carrying its resultant on an effective width of 45 per cent, and the tube deflects as though its second moment were 64 per cent of the gross.

The columns that lean

A framed tube carries its wind load by bending the spandrel beams between its columns, and it does it badly — the corner columns take nearly six times what the middle ones do. Tilt the columns instead, so the perimeter is triangulated, and the same shear is carried axially. The concentration falls to 1.21 and the tube recovers most of the stiffness the plan said it had.

Two curvatures of opposite sign

A single family of cables is not a structure. It is a mechanism that takes whatever shape the load asks for, and it will do that under any load pattern it was not tensioned for. Cross it with a second family curved the other way, pull the two against each other, and the pair becomes stiff — with no bending anywhere and no material property involved in the stiffness at all.

Two drawings of one deck, and they are not the same structure. A 112 m viaduct on five supports, articulated two ways. Above, the fixed point is at the left abutment: the far end has to be given 45 mm of movement, and the friction of every sliding bearing runs one way, so the fixed support takes 660 kN before any wind or braking is applied. Below, the fixed point is at the middle pier: the largest joint halves to 22 mm and the friction now cancels across the fixed point, leaving 0 kN. The movement arrows are drawn at 900 times the scale of the deck, because a 45 mm movement on a 112 m span is thinner than the line the deck is drawn with. Nothing about the deck, the loads or the ground has changed between the two.

Where the structure is allowed to move

One drawing decides how big every movement joint on a bridge is and where every horizontal force goes, it takes an afternoon, and it appears on no calculation sheet. Move the fixed point from an abutment to the middle pier and the largest joint halves and the horizontal force on that support drops from the whole of the friction to none of it.

The two braces balance until one of them buckles. An inverted-V brace after the compression member has gone. While both braces are elastic they carry equal and opposite forces and their vertical components cancel on the beam above, which is why the beam in a chevron bay is usually sized for gravity alone. The compression brace buckles at 445 kN and then sheds most of what it was carrying — 30% is left here — while the tension brace goes on to yield at 1065. The difference between the two vertical components is 659 kN, applied at the middle of the span with no help from either brace, and it asks the beam for 1317 kNm against the 200 kNm the gravity load asks for — 6.6 times as much. The beam drawn does not: 1517 kNm against a capacity of 731. The force is not a load case anybody applies; it is what the frame leaves behind on its way to the state it will actually be in.

The force the brace leaves behind

Two braces meeting under a beam carry the storey shear as a tension and a compression whose vertical components cancel, so the beam above sees nothing. They cancel only while both braces are elastic. Once the compression brace buckles it sheds most of its force, the tension brace goes on to yield, and the difference is a point load at midspan that nobody applied.

The prop load is decided by the digging, not by the hole. Prop forces in a 12 m excavation propped at 3 levels, with the force each prop reaches at any stage of the sequence drawn thick and the force the finished arrangement gives it drawn thin. The pale dots are the individual stages. The middle prop reaches 214 kN while the dig is at 9.5 m and finishes at 88 — a factor of 2.44 between the two, and the larger one is not in the final analysis anywhere. Terzaghi and Peck's apparent pressure diagram, a rectangle of 49.4 kN/m², reproduces the total of the staged maxima to 2% — which is what it is: an envelope of measured prop loads, back-figured into a pressure, and a shape nothing on a wall is ever loaded with.

Every prop has its own worst day

A braced excavation has no finished state worth analysing. It is dug in stages, a level of props goes in at each stage, and a prop's force is largely fixed the moment it is installed — so the force to design it for is the largest it sees during a sequence that appears on no calculation sheet, and which for the middle prop here is nearly two and a half times what the finished arrangement gives.

Nine piles, and four times the settlement

A pile cap divides its load between its piles by the same three terms a bolt group uses and a section under biaxial bending uses. What a bolt group does not have is neighbours it shares ground with — and the group effect that matters is not the strength check everybody makes, but a stiffness effect nobody tabulates.

$The lining that carries less for being weaker. Bending moment and hoop thrust in a circular lining, against the lining's own bending stiffness, both as fractions of the free-ring values. The ground arrives already stressed — 500 kPa vertically and 300 horizontally at K₀ = 0.6 — and the difference between them tries to squash the hole into an ellipse. A lining stiff enough to refuse absolutely collects the whole distortion pressure, p₂R²/3 = 300 kNm/m; one flexible enough to go with the ground collects nothing, because there is no curvature change left to resist. The thrust is the flat line: it comes from the mean stress rather than the difference, so it does not move at all. Putting 8 joints in this ring drops the moment to 34% of the solid one and leaves the thrust exactly where it was, which is why a segmental lining is jointed and why the intuition carried over from a beam is inverted here.$

The lining that is stronger for being weaker

A tunnel lining is not loaded. The ground arrives already stressed and the hole wants to squash into an ellipse; the lining's only job is to refuse, and how much moment it collects depends entirely on how hard it refuses. Make it stiffer and it takes more. Make it flexible — put joints in it, make it thin — and it takes almost none, while the hoop thrust it carries does not move at all.

A tie is worth all of itself and a strut is not, which is worth nine per cent. The same 20 floors carried two ways, with every member drawn at the width its own force requires. Hung, the loads accumulate upward, so the largest hanger is at the top: 18.0 MN at 355 N/mm² with no buckling reduction of any kind. On columns they accumulate downward and the largest column is at the bottom, at the same force — but every column above it is understressed by its own slenderness, worst at the top where a 900 kN column still has to be 4.8 × 10³ mm² to reach χ = 0.529. Over the height the hangers total 0.002 m³ of steel against 0.002: a saving of 9.0 per cent, which is the average χ and nothing else.

Hung from the top, and nine per cent lighter

A tie is worth its full strength and a strut is not, so hanging the floors of a building from a hat truss ought to be an obvious economy. It is a real one, it is measurable, and it is nine per cent of the steel — shrinking as the building gets taller, which is the opposite of what the argument sounds like.

The abutment force is the sag turned upside down. A 100 m ribbon carrying 35 kN/m at a sag of 2.0 per cent of its span. H = wL²/8f, so the horizontal force at each abutment is 21875 kN — 6.25 times the entire weight of the deck, and five times what a suspension bridge of the same span and weight at a tenth would have needed. The curve is a reciprocal and it has no flat part: halving the sag doubles the force, at any sag. What stops a designer flattening it further is not the ribbon, which is in tension and cannot buckle. It is what the ground at each end will take, and at 6.25 deck-weights that is usually rock or a very large anchor block.

The deck that is its own cable

Every other cable structure hangs something from the cable. A stressed ribbon hangs nothing — the walking surface is the catenary, laid at a fiftieth of the span rather than a tenth, because a footbridge has to be walkable. That one decision hands the abutments six and a quarter times the entire weight of the bridge.

The summer that is worse than the one before it. The earth pressure behind an integral abutment, summer by summer, as a multiple of the at-rest value it started at. A 60 m deck expands by 10.8 mm at each end and pushes the abutment into the backfill. Granular soil under cyclic strain densifies, so the same movement next year needs a higher pressure to achieve, and K climbs from 0.38 toward 0.96 — a factor of 2.49 on the force, reached after about a century. The design load on an integral abutment describes the bridge's whole life rather than a load case, and it is the only load in this collection that gets larger because time has passed rather than because something was added.

The summer that is worse than the last

An expansion joint is a hole in a deck that leaks salt water onto the bearings underneath it. Remove it and the thermal movement does not go away — it goes into the soil behind the abutment, twice a day for a hundred and twenty years, and granular soil under cyclic strain gets denser.

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.

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.

Every section is hogged and sagged before the bridge exists. The bending moment envelope of a launched deck, section by section along its own length, taken over every position of the launch. In service each section has one sign; during the launch 100 per cent of them see both, because each passes over every pier and through every span on its way out. The worst launch moment is 42568 kNm against 40500 in service, and with no launching nose at all it would be 185977. That is why a launched bridge is a constant-depth box with symmetric flanges: the design case is not a load, it is a history.

Every section was somewhere else

A bridge pushed out over its piers subjects each of its cross-sections to a history rather than to a load case. Every one passes over every support and through every span, so the design envelope is the envelope of envelopes — and no in-service condition produces it.

Every level added is a longer span, not a shorter one. Steel per square metre of floor against the number of levels in the hierarchy, for a 12 m bay with a deck that can span 3.0 m. A bending level's weight per unit area is 3ρqrL/8σ — it contains the SPAN and not the spacing — so breaking a floor into more levels cannot make the members lighter by making them closer together. It adds one more system, and the last system always spans the whole bay: a second level costs 52 per cent more steel than one, and a third 107 per cent. The structural zone grows with it, 730 mm to 1346 mm. Hierarchy is not an economy, it is a way of reaching, and it is paid for in both currencies at once.

Every level is a longer span

A floor is a hierarchy — deck to joists to beams to girders — and the reason usually given is that breaking a long span into short ones saves material. A bending level's weight per square metre contains its span and not its spacing, so it does not.

A closed force polygon. The forces on a joint, laid tip to tail. Equilibrium is the statement that the polygon closes, and the gap when it does not is the out-of-balance force, to scale.

One drawing solves the whole truss

The method of joints solves a truss one joint at a time, and each solution is thrown away as soon as the next begins. Drawn instead of computed, the joints share their edges — every member's force appears once in a single figure, and the figure's own closure is the check.

A Pratt truss of 8 panels. A Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 14 members came out in tension, 13 in compression and 2 carrying nothing.

Halving the panel buys a shorter strut

Subdividing a truss into more panels of the same span and depth barely changes the chord forces, because the couple that carries the moment has not moved. What it changes is the length of every compression member, and a buckling capacity goes as the inverse square of a length.

3 continuous spans against 3 simple ones. The bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 28.1 to 18.0, and a hogging moment of 22.5 appears over the supports where there was none.

The chord is a continuous beam

A truss analysis reports one number per member, and for a chord running through eight panels that number is an axial force with no bending in it. The chord is a single piece of steel spanning between its own nodes, and anything landing between them bends it.

Influence line for the shear force at x = 10.5. The shear force at one fixed station, plotted against the position of a unit load walking across the span. The beam was re-solved at 301 load positions. The worst position is x = 10.56, giving 0.560.

Two diagonals, one of which is absent

A truss diagonal is sized for the shear in its panel, and near mid-span that shear changes sign depending on where the load stands. A member that can only pull cannot carry the reversed case, so the panel gets a second diagonal — and at any instant one of the pair is not there.

The shear goes round the corner instead of across it. A 6-panel Vierendeel girder, 21 m by 6000 mm, under 300 kN at mid-span. There is no diagonal in it, so each panel's 150 kN of shear is carried as bending in the chords: the curves drawn along them are the chord moments, and every one passes through zero at the middle of its own panel. The local moment is the panel shear times the panel length over four, 131.3 kNm, and it adds to an axial force of 263 kN from the global moment at the same point. The girder deflects 109.09 mm against 6.53 mm for the same members triangulated — 16.70 times — and 98% of that movement is chord bending that a diagonal would have removed entirely.

The frame is a girder stood on end

Every unbraced building frame is a Vierendeel girder turned through ninety degrees, and the identification is not an analogy — it is the same equations with the axes swapped. Which means the frame inherits results that read as absurd for a building — more bays is stiffer, a wider building is not, and doubling one section property halves the sway.

Removing each member in turn. Every member of a 8-panel pratt truss removed one at a time, with the worst demand on the survivors plotted against the member removed. Four of the 35 leave a mechanism — the bars drawn to the top of the frame — and for those there is no redistribution to compute, because there is no structure left. The rest redistribute, and the worst of them asks a survivor for 2.04 times what it carried before. A single number for robustness does not exist: it depends on which member goes.

A determinate truss has no robustness at all

Remove any one of a Warren truss's thirty-one members and what is left is a mechanism. Not weakened — gone, with no set of forces that holds the load in any position. Robustness is not a property a structure has by degree; it is bought by adding members that carry nothing until something else stops carrying, and a truss without them has none of it to measure.

The same bridge by the two theories it might have been designed by. Deck moment along a 500 m suspended span with half of it loaded, computed twice. Elastic theory treats the deck as a beam and applies the cable's extra tension as an upward load: 80.1 MN·m. Deflection theory keeps the cable's total tension acting on the deck's own deflected shape — a geometric stiffness — and returns 56.3, which is 30 per cent less. The extra cable tension is very nearly the same in both (6.20 against 6.11 MN), so nothing about the cable is in the difference: it is entirely the H·v″ term the older theory drops.

The tension that was left out

A suspension bridge's deck sits on a cable pulling hard along it, and a member with a large tension in it is stiffened by that tension. Leaving the term out of the deck's own equilibrium is what elastic theory does, and on a long span it asks for fourteen times the girder.

What the second, third and fourth arms are worth. Top drift removed against the number of outriggers, each arrangement at its own optimum levels, on a 40-storey core 200 m tall. One arm at 59 per cent of the height removes 82.3 per cent of the drift. A second, with both moved to 35 and 71, takes it to 91.6 — a gain of 9.3 points, which is half of what was left. The third is worth 3.0 and the fourth 1.5, and each one costs a storey of the building's most valuable height.

What the second arm is worth

One outrigger at its best height removes five sixths of a tall core's drift, which sounds like the end of the argument. A second removes half of what is left, a third half of that, and each of them costs a storey of the most valuable floor area in the building — so the question is not where to put an outrigger but how many the arithmetic still justifies.

All essays