Structural Engineering & Statics
where the load goes
Concept

Optimisation — where it appears

Choosing a structure's proportions so that two competing limits arrive together, which is where the best member of a family usually sits. Driving two limits together is also what makes a structure imperfection-sensitive, because coincident failure modes interact — so an optimised member needs a larger allowance rather than a smaller one.

Named by 15 essays across 6 fields — each of them below, with the objects they name alongside it.

Where to put the supports. Peak sagging and hogging moment for a uniformly loaded beam, against how far the supports are moved in from the ends. The best arrangement is where the two curves cross, and it is nowhere near the ends.

Where to put the supports, which is not at the ends

Moving the supports of a uniformly loaded beam inward by about a fifth of its length halves the worst bending moment. The load has not changed and nor has the beam.

internal-forces · Support layout 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.

structures · Folded plate 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.

structures · Outrigger 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.

structures · Branching structure Not where the two loads meet. How much a column loses below the weaker of its two single-mode capacities, against the ratio of its local critical load to its global one. The received claim is that the worst place is where the two coincide; the arithmetic says otherwise. The erosion is largest at a ratio of 0.47 — 23% — sits within a per cent of that for every ratio below about a half, and at exact coincidence is only 2%. What the curve does say is the useful half of the folk claim: once the plates are stocky enough that the local critical load is twice the global one, the interaction is nothing at all, and the section is worth thickening only up to there.

Two ways of buckling at once

A thin-walled column can bow as a whole or ripple in its plates, and each has its own critical load. The received advice is that the worst arrangement is the one where the two are equal. The arithmetic says the opposite — at coincidence the interaction costs two per cent, and the expensive region is where the plates go first.

stability · Mode interaction Half the beam does nearly all of the deflecting. The virtual-work integrand M·m/EI along the member, normalised to its own peak, with the running share of the answer beside it. The integrand is a density: it says how much of the deflection each millimetre of the beam produced. For this case the half nearest the root supplies 87.5 per cent of it, and the rest of the member supplies the remainder. Stiffening the busy 50 per cent by 1.5 times takes the deflection down by 29.2 per cent; the same material spent on the quiet end takes it down by 4.2 — a factor of 7.0 for the same steel. The map of what is contributing is not the map of where the moment is largest, and the second is the one that gets drawn.

Where a deflection comes from

The unit-load method gives a deflection as an integral, and this collection has treated that integral as a number to evaluate. It is not a number. It is a density, and it says which millimetres of the member produced the answer — which is not the same map as where the moment is largest.

deflection · Deflection distribution The answer is continuous and the catalogue is not. Capacity bought against capacity required, over a real rolled series. The straight line is what a continuous section would give — exactly the moment asked for, and nothing can be bought on it. The staircase is what a catalogue gives: each tread is one section, each riser is the step to the next, and the vertical gap between the two is steel that is paid for and does nothing. The steps in this series run from 23% to 59% in plastic modulus, so the average waste is 15.0% and the worst is 46% — just above a riser, where the section below has been missed by a kilonewton-metre. The 1200 kNm marked buys a 686×254×125 at 1418 kNm, which is 85% utilised. Two things follow that a continuous treatment cannot see: the sensitivity of a design to an assumption is zero over most of a tread and enormous at a riser, and an optimisation that returns three significant figures is answering a question with twelve answers in it.

The answer is continuous and the catalogue is not

Every optimisation in this subject returns a number with three significant figures in it, and nothing with three significant figures can be bought. What can be bought is a rolled series whose steps are a quarter to a half apart, so the member that goes on the drawing is on average a tenth stronger than the one that was calculated and can be a third stronger for no reason at all.

sections · Available sections The average is not the answer, and it is unsafe. Critical load of a pinned column whose middle third has been given a different stiffness, against the whole-column Euler load, with the two numbers a hand check reaches for beside it. The eigenvalue is taken from K − P·Kg over 24 elements, so nothing here is a formula for a stepped column — it is the same computation the uniform case gets. At a middle third of 0.50 times the rest the true load is 0.612 of Euler's, the arithmetic average says 0.832 and the weakest segment says 0.496. The average is high by 36% and it is high on the unsafe side, because the third of the column it is averaging over is the third where the mode has all its curvature. The weakest-segment answer is safe everywhere and wasteful by about as much.

An average stiffness is not a safe stiffness

Euler's load belongs to a column of one EI. Give the same column two, and the temptation is to average them — which is wrong, and wrong in the unsafe direction by a quarter. Buckling weights stiffness by the square of the curvature of the mode, so the middle of a pinned column decides everything and the ends decide almost nothing.

stability · Stepped column The best design is where two failures arrive together. A fixed area of steel rolled into tubes of every proportion, with the three things that can end each one. Euler's load goes as r² because I = A r²/2; the local buckling stress goes as 1/r² because the wall thins as the tube grows; squashing does not care. The capacity is the lowest of the three, so it has a maximum — and the maximum is exactly where the two buckling curves cross, at r/t = 129 and 2364 kN, which the closed form r*, the fourth root of αAL² over π³β√3, reproduces to 0.52 per cent. That is the general result and it is not about tubes: the optimum of a minimum of a rising and a falling curve is always their intersection, so optimising a design against two failure modes puts both of them at the design point — which is the one configuration imperfections hurt most.

The best design is the most sensitive one

Take a fixed area of steel and roll it into a tube. Euler's load rises with the radius and local buckling falls with it, so the capacity has a maximum — and the maximum is exactly where the two failure modes arrive together, which is the one configuration imperfections hurt most.

stability · Wall optimum 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.

structures · Hierarchy 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.

structures · Outrigger Where the span moment and the support moment cross. Span moment and support moment against the joint's stiffness, for a 6 m beam under 30 kN/m. They move in opposite directions because they add to a constant — the simple-span 135 kN·m is fixed by statics and the joint only decides how it is split. They cross at 68 kN·m, where the joint is delivering 75 per cent of the fixed-end moment, and neither the crossing nor the fraction depends on the beam, the span or the load: it is the point where f·wL²/12 equals wL²/8 − f·wL²/12, which is f = 0.75 for every beam there has ever been.

The joint that was chosen

A joint's stiffness decides how a beam's moment divides between its span and its supports, and the two add to a constant. So there is a stiffness at which they are equal, the beam is sized by the smaller of two numbers rather than the larger of one, and the design moment is half what a simple connection leaves behind.

connections · Joint classification One volume of steel, divided three ways. A Pratt truss of eight panels at a depth of 1, carrying 10 kN at each top joint, drawn three times with every member as wide as its area, the total volume the same in each. With equal areas the mid-span deflection is 43503; fully stressed, with area in proportion to force, 32702; with area in proportion to the square root of the product of each member's real and virtual forces — the division that makes mid-span as stiff as this steel can make it — 31318. The fully stressed truss is 4 per cent short of the stiffest; the equal-area truss is 39 per cent short. No member is allowed less than 10 per cent of the equal area; the deflections are in units of load × length / (E × volume).

The truss that is stiff by accident

Give a truss a fixed volume of steel and ask how to divide it among the members. Sized for strength — every member at the same stress — its mid-span deflection comes within four per cent of the stiffest that steel can make, although stiffness was never asked about. The reason is an inequality, and the same inequality says where the accident stops: at the quarter point the strength design is sixty-nine per cent short of the best.

deflection · Truss deflection The strength design, and the least steel that is stiffer. A Pratt truss of eight panels at a depth of 1, carrying 10 kN at each top joint, each member drawn as wide as its section from a catalogue whose sections step by 25 per cent in area, the smallest 10 per cent of the largest member's need. Above, every member at the smallest section that carries its force. Below, the least steel that makes mid-span 1.50 times as stiff with every member still strong enough — the discrete optimum — with the 24 members it made larger drawn in the second colour: six of six top chord members, eight of eight bottom chord members, six of eight diagonals, four of seven verticals. The optimum uses 44 per cent more steel than the strength design, and it puts it where a member's real and virtual forces are both large; members either force leaves small are not touched.

The calculus answer, rounded, is the worst one

A real truss is built from a catalogue, and every member is rounded up to the next section. That rounding costs its stiffness almost nothing: the few per cent that separate the strength design from the stiffest survive it. What does cost is the next step. When a deflection limit governs, the obvious move — take the continuous optimum and round it up — needs more steel than any other way of stiffening the truss, and the exact discrete answer is within one per cent of a bound no catalogue can beat.

deflection · Truss deflection Fully stressed, with a choice of diagonals. Two trusses sized so that every member is at the allowable stress under the full load, each member drawn as wide as its area, tension and compression in two colours. Above, an eight-panel Pratt truss as deep as a panel is long, with a counter-diagonal in each of its six interior panels, loaded by 10 at every bottom joint, found by resizing and reanalysing until nothing changes; below, the same truss without its counters. The counters in the two middle panels have shrunk to nothing (dotted) and the four nearer the supports have stayed, working in compression beside the diagonals in tension. The truss that kept them needs 1,200 units of steel against the Pratt's 1,220, and deflects 24.0 at mid-span against 26.0.

The truss whose forces follow its sections

In a determinate truss each member's force is fixed before its section is chosen, so sizing for strength and sizing for stiffness can be done in either order. Put a counter-diagonal in every panel and they cannot. The fully stressed design becomes an iteration that starves some counters to nothing and keeps others, lands on a different truss from every start, and — whichever it lands on — weighs the same and deflects the same. Then enlarge one group of members to stiffen it, and a vertical nobody touched is overloaded by 58 per cent.

deflection · Truss deflection

Named alongside it

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

ServiceabilitySlendernessDeflectionFully stressed designVirtual workBucklingCritical loadImperfection sensitivityLocal bucklingSecond momentSelf-weightStiffness

All concepts