Hung from the top, and nine per cent lighter

There is an argument for hanging a building that is completely correct and turns out to be much smaller than it sounds.

A tie is sized at $N/(f_y/\gamma_M)$. A strut of the same force is sized at $N/(\chi f_y/\gamma_M)$, where $\chi$ is the reduction it takes for being able to buckle — and $\chi$ is less than one always, and can be a great deal less. So a floor hung on a rod should need less steel than the same floor propped on a column, and a building whose floors all hang from a truss at roof level should need less steel than a conventional one.

It does. Here is how much.

The symmetry nobody expects #

The reason the saving is small is visible in that drawing before any arithmetic is done. The two structures carry the same total force through the same total length.

In a column building the load accumulates downward: the column below the top floor carries one floor, the one below the second carries two, and the one at the base carries all twenty. In a hung building the load accumulates upward: the hanger below the hat carries all twenty floors, the next one down carries nineteen, and the one holding the bottom floor carries one.

Those are the same twenty numbers in the opposite order, which is a coincidence worth being suspicious of and is not one: both structures have to deliver the weight of $j$ floors across $j$ storey heights, and which end they start from cannot change the product. It is the same conservation that makes the total static moment of a span immovable however the structure redistributes it — a quantity fixed by the loads and the geometry before any member has been chosen. Summing $N \times \ell$ over the height gives the same total in both cases, so the two structures would need identical steel if their allowable stresses were identical. The entire difference is $\chi$, and the saving is therefore

$$1 - \frac{\sum N_j \ell / f_d}{\sum N_j \ell / (\chi_j f_d)} = 1 - \bar\chi,$$

a weighted average of the buckling reductions. For the frame drawn that average is 0.9096 and the saving is 9.0 per cent. The two numbers agree because they are the same number.

Where the reduction actually lives #

$\chi$ is not uniform up the building, and where it is worst is the surprise.

The bottom column of the twenty-storey frame carries 18.0 MN, needs 53,262 mm², and is therefore an enormous stocky member: relative slenderness 0.333, $\chi = 0.952$. It gives away almost nothing.

The top column carries 900 kN, one floor’s worth. If it could be sized on strength alone it would be 2,535 mm² — a section about 60 mm square. At that size a 4 m storey height gives a relative slenderness of 1.11 and $\chi = 0.529$, so the area has to nearly double, and it settles at 4,788 mm². The top column is 89 per cent bigger than its load asks for, and the bottom column is 5 per cent bigger.

Length costs more than it looks, because a longer column both attracts the same load and resists it worse, and the second effect is the one that compounds. In a building the storey height is fixed and the load varies, so the relation is read from the other end: the lightly-loaded columns are the slender ones, and the lightly-loaded columns are all at the top. Shortening the building does not change that. It only changes how much of the building is top.

So the case for hanging is a case about the top of a building, which is exactly where there is least steel to save. That is why 9 per cent and not 40.

It is worth noticing that the top column is not slenderness-governed in the sense of failing by buckling. It is slenderness-governed in the sense of being bought by buckling: the load never comes near either capacity, and the member exists at that size because a smaller one would have an unacceptable $\chi$. Nearly half the steel in that column is doing nothing but making the rest of it less slender, which is the same purchase a lattice column makes when it spreads two chords apart and a rather worse deal.

There is also a floor below which the comparison stops being about buckling at all. Once the top column has been driven up to 4,788 mm², the section that provides it is a real rolled profile with a minimum available size, and below about six storeys the top columns of both arrangements are whatever the catalogue’s smallest sensible member is. The answer is continuous and the catalogue is not, and the 38 per cent at two storeys is a calculation about members nobody would specify.

And it gets worse with height #

The argument for hanging is strongest for the buildings nobody would ever hang — a two-storey shed, where a third of the steel is genuinely at stake and a hat truss makes no sense at all. By the time the building is tall enough for the arrangement to be an architectural idea, the material case has almost gone.

This is a scaling result of the same family as weight growing faster than strength, running the other way: the thing that makes tall columns economical is that they are fat, and being fat is what makes buckling stop mattering.

What it costs, which is movement #

The nine per cent is not free, and the price is in a quantity the material comparison does not contain.

A hanger is at its full design stress by construction — that is the entire point of it — so its strain is $f_d/E$ everywhere, and the elongation per storey is the same all the way down: 6.93 mm for a 4 m storey at 355 N/mm². The movements accumulate from the hat downward.

A column is at $\chi f_d$, which is less than full stress and is least near the top. Its movements accumulate from the ground upward.

Which end matters is not a symmetric question. The top of a building is where movement is cheapest to accommodate: the roof, the plant, a parapet, a joint nobody sees. The bottom is the entrance, the ground-floor glazing, the lift pit, the point where the cladding meets a pavement that is not moving at all. A hung building puts its largest movement exactly where it is most expensive.

The gradient matters as much as the total. A column building’s movement per storey is largest at the bottom, where $\chi$ is near one, and it tapers upward; a hung building’s is constant, at 6.93 mm every storey from the hat to the ground. So the differential movement between adjacent floors — the quantity a partition, a lift guide rail or a service riser actually feels — is uniform in the hung case and cannot be designed out of any one storey. A movement budget is a sum of terms and this one contributes the same term twenty times over.

And the movement is not only elastic. It arrives with the floors as they are built, so a hung building has to be built downward — the hat truss first, then the floors hanging from it — and each floor’s position depends on how many are already hung below it.

That is the same arithmetic a conventional tall building performs when its columns and its core shorten by different amounts and the floors have to be built out of level to arrive level, with the sign reversed and a larger number in it, because the hangers are worked harder than columns are permitted to be. And unlike the material saving, it grows straight with the height rather than shrinking.

The core, which is where the argument really lands #

The hangers are the members the comparison is about, and they are not the members that decide the scheme.

Every kilonewton of floor load in a hung building arrives at roof level and then has to come back down. The route is the core, and the core is therefore carrying the entire weight of the building in compression — 18.0 MN per hanger line, times however many lines there are — through a member that also has to be the whole of the lateral system, because there are no perimeter columns to help it.

That is a genuine reversal of the usual arrangement. In a conventional tall building the perimeter columns carry a large share of the gravity load, and an outrigger exists precisely to get some of the overturning couple out into them; the core takes what is left. A hung building has already spent that option: its perimeter is in tension under gravity, so it cannot be recruited into a compression couple without first cancelling the tension it is already carrying. The wind moment and the gravity load both arrive at the same core, and the core is not larger for it.

The nine per cent saved on the hangers is spent, several times over, on the element the hangers deliver to. That is not an argument the material comparison can make, because the material comparison never leaves the vertical member it is about — which is the standing hazard of comparing two structures by one of their parts.

One hole is larger than the whole saving #

A tie is checked twice, and the second check is the one that decides whether the nine per cent exists.

The gross section yields at $N/(f_y/\gamma_{M0})$, which is the number the whole comparison is built on. But wherever the tie is bolted, the net section ruptures at

$$N_{u,Rd} = \frac{0.9\,A_{net} f_u}{\gamma_{M2}}$$

and for S355 with $f_u = 490$ N/mm² and $\gamma_{M2} = 1.25$ that allowable stress is $0.9 \times 490/1.25 = 353$ N/mm² — within one per cent of the 355 the gross section is worked at. The two checks are, by construction, the same check. The factors were calibrated to make them so.

Which means the tie has no margin for the holes at all. The top hanger needs 2,535 mm² of gross section on yield and 2,551 mm² of net section on rupture, so with a bolt line removing a typical 15 per cent it needs

$$A_{gross} \ge \frac{2{,}551}{0.85} = 3{,}001\ \text{mm}^2$$

— 18 per cent more than the tie the saving was computed with. Apply that to the hung frame’s 0.412 m³ and it becomes 0.486, against the column building’s 0.453. The nine per cent saving has become a seven per cent penalty, and the whole of the difference is a row of holes.

The escape is real and it is the reason hung structures look the way they do. Either the splice is welded, so there are no holes and the gross section is the section everywhere; or the tie’s ends are upset — thickened before the thread or the hole is made, so the net area at the connection exceeds the gross area of the shank. Both are entirely standard, and both are only worth their cost because the alternative is to carry an 18 per cent penalty along the full length of a member whose entire justification was a nine per cent gain.

A column is under no such pressure. Its splice carries compression in bearing, the holes are for erection rather than for load, and its own net section is checked against a stress it is nowhere near. The tension member is the one whose connection is made of the same argument as the member, which is why a tie’s joint is the harder half.

A stronger steel makes the case for hanging better #

The saving is $1 - \bar\chi$, so anything that lowers $\chi$ raises it — and the most ordinary specification decision in the building lowers $\chi$ substantially.

Go from S355 to S460. The tie’s area falls in exact proportion: $900\,\text{kN}/460 = 1{,}957$ mm², a reduction of 23 per cent, because a tie’s capacity is $A f_y$ and nothing else. The top column’s does not. Its area appears in its own slenderness, so a smaller section is a more slender one, and the relative slenderness carries $\sqrt{f_y}$ in it as well:

$$\bar\lambda = \frac{\ell}{i}\cdot\frac{1}{\pi}\sqrt{\frac{f_y}{E}}$$

Both terms push the same way. At S355 the top column settled at 4,788 mm² with $\chi = 0.529$; at S460 it settles near 4,660 mm² with $\chi$ around 0.42 — a saving of under three per cent against the tie’s twenty-three.

The reason is one this collection has already priced: a stronger steel does not change $E$, and a slender column’s capacity is Euler’s, which contains $E$ and no yield stress whatever. Buying grade buys the tie everything it costs and buys the slender column almost nothing.

So the hung building’s advantage is not a fixed nine per cent; it is a number that grows with the grade specified, and it grows fastest at the top of the building, where the columns are most slender and the tie is smallest. It is still spent on the core, and it is still spent again at every hole.

The two things that actually decide it #

Material is not why hung buildings are rare. Two other properties are, and neither appears in the comparison above.

The first is robustness. Removing a member and asking what happens is a question a hung building answers badly. A hanger is a series load path: lose one segment and every floor below it loses its support at once, with nothing beneath to catch them. Lose a column and the floors above it have to find another route, which a frame with beams can sometimes provide. Tension chains fail wholesale and compression chains fail retail, and which of those a building does is decided by the direction its load accumulates rather than by any detail of the members.

There is a third that is neither structural nor thermal: the load has to go up before it comes down. Every kilonewton of floor load travels to roof level in a hanger and then back to the ground in the core or in perimeter compression members, so the total axial load path is roughly twice as long as a column building’s, and all of it converges on one element. Interrupting a load path that would otherwise run straight to the ground is expensive in proportion to how far the load is moved, and a hung building interrupts every load path in it by the full height of the building. The hat truss is a transfer structure carrying the whole weight of the building on one span, and it is sized by the deflection of that span rather than by its strength.

Where the model stops #

One hanger line and one column line. The comparison is between a single vertical member in each arrangement, carrying an identical accumulated load. A real hung building has a core taking everything the hangers deliver, and a real framed one has beams that tie the columns together and change their effective lengths — so $\chi$ is not a property of a member alone, and the column half of this calculation is the pessimistic one.

$\chi$ is computed at an effective length of one storey. That is right for a column continuous through a floor with a rigid diaphragm at each level, and it is generous where the floor plate is flexible or the column is spliced badly. Taking $K = 1.2$ instead moves the saving from 9.0 per cent to about 13.

The hanger has no fittings. Every hanger has connections at each floor, and a connection to a tension member is either a bolted splice that needs a net section considerably larger than the member, or a welded one that has to be inspected. The connection is where the material saving is spent, and a splice at every floor of a highly-stressed tie is not a small item.

Nothing here is about wind. A hung building’s floors are supported vertically by hangers and horizontally by nothing at all; the lateral system is entirely the core, and it has no perimeter columns to work with. That is a much larger design consequence than the nine per cent, and it points the other way.

The loads are static and permanent. In practice the hanger stress under permanent load alone is well below $f_d$ — the full stress is reached only under the full factored combination, and the movements quoted are the movements at that combination. The installed movement, which is the one construction has to chase, is the permanent-load fraction of it, perhaps 60 per cent.

And the drawings show two structures at one instant. They do not show what either looks like half-built, and that is where the arrangement is genuinely different: a column building is stable at every stage, and a hung building is a hat truss on temporary props with nothing under it until the last floor arrives.

The ladder from here #

Later rungs on this anchor: the hat truss itself, which is a transfer structure spanning the whole plan and carrying the whole building — sized by deflection rather than by strength, and the reason hung buildings tend to be short. Hangers that are cables rather than rods, where the axial stiffness is lower again and the movement problem doubles. Partially hung arrangements — the suspended floor at an atrium, the hung facade, the cantilevered floor hung back to a core — which are common and get the benefit without the fire and robustness problems, because only two or three floors depend on any one tie. Progressive collapse of a tension chain, and the tie-force rules that exist to prevent it. And the same comparison run on a bridge, where hanging is universal rather than eccentric: a suspension bridge’s deck hangs from cables for exactly the argument on this page, and the reason it wins there and loses here is that a bridge’s hangers are short and its main span is very long indeed.