Every level is a longer span

A floor is almost always a hierarchy. A slab or a metal deck spans a short distance to joists; the joists span further to beams; the beams span further still to girders; the girders reach the columns. Each level spans further than the one it carries and is deeper for it.

The reason usually given is that a long span is expensive and breaking it into short ones is cheap. That reason is wrong, and the arithmetic that shows it is three lines long.

The result the whole thing turns on #

Take a bending member of span $L$ at spacing $a$, carrying $q$ per unit area. Its moment is

$$M = \frac{q a L^2}{8},$$

so the section modulus it needs is $M/\sigma$. For a section of depth $d$, the area is roughly $A \approx 3Z/d$ — the flanges are most of it and they are half a depth apart — and with a span-to-depth ratio $r$ so that $d = L/r$,

$$A = \frac{3 M r}{\sigma L} = \frac{3 q a L r}{8\sigma}.$$

Its weight per unit area of floor is $\rho A / a$, and the spacing cancels:

$$g = \frac{3\,\rho\, q\, r\, L}{8\,\sigma}$$

A bending level’s weight per square metre contains its span and not its spacing. Halving the spacing halves each member’s size and doubles the number of them, exactly.

That is not an approximation for a particular section. It follows from the moment being proportional to the spacing and the required area being proportional to the moment, which are both exact, and it is why the result is so clean.

Which means hierarchy cannot save weight #

If a level’s weight depends only on its span, then adding a level to a floor cannot make the existing levels lighter by making their members smaller — it makes them lighter only by making their spans shorter.

And it adds a level of its own, whose span is the longest yet, because the last level always reaches the columns.

For a 12 m bay with a deck that can span 3 m:

A second level costs 52 per cent more steel than one; a third costs 108 per cent more. The structural zone grows with it, because the levels stack. And the piece count rises from four to twenty-one, which is where the fabrication and erection cost is.

More hierarchy is heavier, deeper and more expensive to build, on every axis at once.

Which free body produced the number #

The free body is one member of one level, cut at midspan, and the interesting part is what defines its loading.

Crossing the cut are a moment and a shear. The moment is $qaL^2/8$, in which $a$ — the spacing — is the width of floor the member has been given, which is a decision rather than a measurement. Halve it and the member’s moment halves.

The step that makes the spacing cancel is the second one: the number of members per unit width of floor is $1/a$. So the weight of the level, per unit area of floor, is the member’s weight per unit length divided by $a$ — and the member’s weight per unit length is proportional to $a$ through its moment.

That cancellation is the whole result, and it holds for any member whose area is proportional to its required section modulus. It fails only where something else sets the member’s size: a minimum practical section, a deflection limit that does not scale the same way, or a stability requirement.

So what is hierarchy for #

If it costs weight, depth and pieces, something must be buying it. Three things are.

The deck cannot reach. A composite metal deck spans three metres, a precast plank six or seven, a flat slab eight or nine. A 12 m bay with a 3 m deck needs something at 3 m centres, and if that something spans 12 m directly then there is one level; if it cannot, there are two. Hierarchy exists to bridge the gap between what the deck can do and what the grid demands, and its depth is set by the ratio between those two lengths.

The members cannot be made. A single level spanning 12 m at 3 m centres is a perfectly good design and needs a 600 mm deep beam every three metres — four per bay. That is available and it is used, in the form of long-span composite beams and cellular beams. Where it stops working is where the span is 18 or 24 m and the beam becomes something a fabricator does not make.

The services have to fit. A deeper hierarchy has more, shallower zones and a shallower one has fewer, deeper members. Ducts and pipes have to pass, and passing them through a 900 mm beam once is a different problem from passing them under two 400 mm ones.

The deck is most of the weight #

The steel is not what a floor weighs, and the numbers are not close.

For the two-level floor above, the slab weighs 3,250 N/m² and the two levels of steel weigh 442 between them. The deck is 88 per cent of the structure.

So every decision about the steelwork is a decision about an eighth of the weight. And the deck’s own weight is proportional to its span — a slab’s thickness follows its span through its own span-to-depth ratio, exactly as a beam’s depth does — so the single most effective way to lighten a floor is to shorten the deck’s span, which is done by adding steel.

That is the real trade, and it is the opposite of the one the hierarchy argument states. Adding a level makes the steel heavier and the slab lighter, and which wins depends on where the deck’s span was and on whether it is at its minimum thickness.

For the case drawn, the slab is at its minimum thickness of 130 mm already, so shortening its span buys nothing at all and every level added is pure cost. On a floor where the slab is thicker than its minimum, the arithmetic can go the other way.

Where the cancellation fails, which is where the interest is #

The spacing cancels exactly, and the two situations in which it does not are the two that decide real floors.

A minimum member size. No fabricator makes a section lighter than about twenty kilograms a metre and no site handles one usefully. Once the computed section falls below that, the level’s weight per unit area is $m_{min}/a$ — inversely proportional to the spacing, so closer spacing now costs weight. That puts a floor on how fine a hierarchy can usefully get and it is the reason the degenerate answer of infinitely many infinitely light members does not appear.

Deflection rather than strength. A deflection limit of $L/360$ requires $I \propto q a L^3$, and $I \propto A d^2$, so $A \propto q a L r^2$ — the spacing still cancels, but the span now enters to the first power with an $r^2$ in front of it. A deflection-governed floor therefore rewards a deeper member much more strongly than a strength-governed one, and the span-to-depth ratio stops being a free choice.

The second of those is the more important because most floors are deflection-governed, and it changes the conclusions above quantitatively without changing their direction: hierarchy is still heavier, and the penalty is larger.

Depth is the currency, not weight #

The other axis is the one that decides buildings, and it is barely about structure at all.

The structural zone — the total depth from the top of the floor to the bottom of the lowest member — sets the floor-to-floor height, which multiplied by the number of storeys is the building’s height. That height is facade, cladding, cores, lifts, columns and, in many jurisdictions, planning permission.

Three hundred millimetres per floor over twenty floors is six metres of building, and six metres of facade on a large plan is a substantially larger sum than the weight difference computed above.

That is why depth is the cheapest strength there is and simultaneously the most expensive dimension in the building: the structure wants depth and the building cannot afford it, and every hierarchy decision is negotiating between them.

The one-level floor is 730 mm deep and the three-level one 1,346, which is 616 mm a floor. Over twenty storeys that is twelve metres of building, from a decision made on the grounds that shorter spans are cheaper.

Where the levels’ spans should sit #

Given a number of levels and the two lengths at the ends — the deck’s span and the bay — the intermediate spans still have to be chosen.

The weight is the sum of the spans, since each level’s weight is proportional to its own span and the loads accumulate slowly. So minimising the weight means minimising the sum, which for a fixed number of levels between fixed endpoints means making the intermediate spans as short as possible — pushing them all toward the deck’s span.

That is degenerate, and what stops it is the same minimum-member effect that stops everything else in this subject: a level whose span is barely longer than the one below it is a set of members carrying almost nothing, at whatever the smallest section anybody makes weighs.

The geometric progression used in the table above — each span the same multiple of the last — is the arrangement that keeps every level doing comparable work. It is not the lightest, and it is what gets built, because it is the one in which no level is trivial.

What the piece count is really costing #

The steel weight is one currency and the piece count is another, and on a building frame the second is usually larger.

The table above counts four members per bay for one level and twenty-one for four. Each of those members has two end connections, so the connection count goes from eight to forty-two — and a connection costs an order of magnitude more per kilogram than the member it joins, because it is fabricated rather than rolled.

That reverses the usual instinct about where to look for savings. A designer optimising the members is optimising the cheap part; a designer reducing the number of pieces is optimising the expensive one. A one-level floor with four heavy beams per bay is cheaper than a three-level floor with sixteen light ones, even though the light ones weigh less in total, because most of the cost is in the ends rather than the middles.

It is also why the trend in steel floor construction over several decades has been toward fewer, longer, deeper members rather than toward the finely graded hierarchies that older buildings show. The mechanics did not change. The relative price of steel and labour did.

What this says about long spans #

The result has an implication for the perennial question of whether long spans are expensive, and it is a more precise answer than the usual one.

A level’s weight per square metre is linear in its span, not quadratic. Doubling a span doubles the weight of that level and nothing else — the moment goes up four times and the depth goes up twice, and the two combine to a factor of two in area.

Long spans are therefore much less expensive in material than the moment diagram suggests, and their cost is elsewhere: in depth, which goes up linearly too; in deflection, which goes as the fourth power and usually governs; and in stability, since a deeper member is a more slender one.

The material cost of span is mild and the serviceability cost is severe, and that is a much better summary of the subject than any statement about hierarchy.

There is a caveat that saves the linear result from being too comfortable. It holds at fixed $r$ — fixed span-to-depth — which means a 24 m beam is twice as deep as a 12 m one. Hold the depth instead, because the building will not give any more, and the required area goes as $L^2$ rather than $L$. Span is linear in weight when depth is free and quadratic when it is not, and in a building it is never free.

The hierarchy that is not a hierarchy #

The whole of this page describes a one-way system: each level spans in one direction and hands its load to the next. Two-way systems do not work this way, and the difference is worth naming because it is the alternative that the arithmetic above quietly assumes away.

A flat slab, a waffle slab or a two-way grillage carries load in two directions at once. There is no hierarchy: the same material spans both ways, and the depth is one depth rather than a stack. The result is a floor with a very small structural zone and a great deal of material in it — heavy, shallow, and with no beams to route services around.

The comparison between the two is not a comparison of weight. A flat slab weighs more than a steel one-level floor and occupies a third of the depth, and which of those matters depends on the building.

What the two-way system buys is the elimination of the stacking, which is the mechanism by which hierarchy costs depth. It buys it by making one element do everything, which is why it is heavy, and by having no clear load path, which is why its reinforcement is arranged by a rule about statics rather than by a diagram.

The measurement that would settle it #

The claim that a level’s weight per square metre is independent of its spacing is a strong one and it can be checked against any building whose steel tonnage is known.

Take a floor with joists at spacing $a$ spanning $L$, and compare with a comparable building at a different spacing and the same span. The prediction is that the kilograms per square metre attributable to that level are the same, within the granularity of the section catalogue.

That granularity is what makes the test imperfect rather than the theory. Sections come in steps, so a member sized for a 2 m spacing is rounded up to a real section and one at 4 m is rounded up to a different one, and the two roundings do not preserve the ratio. The prediction holds on average over many bays and is noisy on any one of them.

Which is itself worth knowing: the exact result above is exact about a continuous variable, and the members it describes come in discrete sizes. Every conclusion drawn from it is a statement about a trend rather than about a bay.

Where the model stops #

The span-to-depth ratio is held fixed. In practice it varies with the member type, and a long-span cellular beam is a different animal from a short joist.

Deflection is not in it. The sizing above is a strength one. Most floor members are governed by deflection or by vibration, and both scale differently with span.

The section is idealised. $A \approx 3Z/d$ is a good approximation for an I-section and a poor one for a truss, a joist or a slab, and a truss’s weight scales differently again.

Stability is not in it. The levels of a hierarchy also brace one another — the deck restrains the joists, the joists restrain the beams — and a floor with fewer levels has fewer restraints, so its members are more slender and their capacity depends on the diagram they carry rather than on their section alone.

And the columns are absent. A denser column grid shortens every span in the hierarchy at once, which is the one move that beats all of the above — and it is a decision about the building’s plan rather than about its structure.

Where the ladder goes #

Later rungs on this anchor: the weight of a truss against a beam of the same span, where the exponent changes. Deflection-governed and vibration-governed floors, where the span exponent is not one. Two-way systems, which are not hierarchies at all and which the arithmetic here does not describe. The column grid as the outer variable. Integrated systems, where the services pass through the structure rather than under it, and the zone that saves. Long-span composite and cellular beams, which are the one-level answer. And the wider question this is an instance of: how much of a structure’s arrangement is decided by mechanics and how much by what can be made and moved.