The column that stops

A column that runs from the roof to the foundation is nearly free. It is a stack of short compression members, each one sitting on the one below, and its cost per storey barely changes with how many storeys are above it — the load simply goes somewhere, straight down, by the shortest route there is.

Then the ground floor has to be a foyer, or a road passes underneath, or the car park below wants its columns on a 7.5 m grid where the flats above want them on 5.4 m. One column has to stop, and its load has to be carried sideways to a column that continues. The drawing of that is a beam, and the beam looks like any other beam on the sheet.

It is not any other beam. Thirteen and a half thousand kilonewton-metres is delivered in one member, and the depth that answers for it is more than half a storey — depth that is lost on every floor the member passes through. The transferred load is paid for once; the transfer member’s depth is paid for by everybody who wanted to stand where it is.

What the cut under the column says #

The number comes from one free body and one equation, and both are worth naming before anything is built on them.

Cut the transfer member vertically, immediately under the stopped column, and keep the left-hand piece. Acting on it are the left reaction $P(L-a)/L = 4500$ kN upward, and, on the cut face, the shear and moment the right-hand piece was applying. Moments about the cut give

$$M = \frac{P\,a\,(L-a)}{L} = \frac{6000 \times 3 \times 9}{12} = 13\,500\ \text{kNm},$$

and vertical equilibrium gives the shear on that face directly: 4500 kN on the short side of the load, 1500 kN on the long side. Nothing in that is special to a transfer; it is the point-load moment every textbook carries. What is special is the size of $P$, which is the accumulated weight of ten floors rather than a floor load, and the existence of $a$, which is a decision somebody made in plan.

The depth that answers for the moment follows from the section rather than from the frame. For a rectangle of width $b$ at an allowable stress $\sigma$, $M = \sigma b d^2/6$, so

$$d = \sqrt{\frac{6M}{\sigma b}} = \sqrt{\frac{6\,P\,a\,(L-a)}{\sigma\,b\,L}}.$$

Every term the design controls sits under the square root. Depth is the cheapest strength there is precisely because the moment goes as $d^2$, and here that generosity runs in reverse: the load has to quadruple before the depth doubles.

Four times the load is exactly twice the depth #

The ratio printed on that chart is 2.000 to three decimal places, and it is not a rule of thumb that happens to work near the middle of the range. It is the square root, arriving as arithmetic. Quadruple the load on a transfer member and its depth exactly doubles; double the load and the depth grows by 41 per cent.

The other term is the one worth carrying into a plan meeting, and it is not linear in $a$. Moving the stopped column from 3 m into the span to 1.5 m cuts the moment from 13 500 to 7875 kNm, a 42 per cent saving; moving it from 3 m to 6 m raises it to 18 000 kNm, a 33 per cent penalty. The depth follows at half the rate in each direction. The transfer is cheapest when the column that stops lands nearly on top of a column that continues, which is a statement about the plan rather than about the beam, and is the only part of this whole subject that costs nothing to act on.

Stiffness governs, and the deflection belongs to somebody else #

The dashed outline in the first figure is 1.94 m and the solid member is 2.55 m, thirty-one per cent deeper. The extra depth is bought by nothing the strength calculation can see, and the reason is a change of free body.

A beam’s deflection is normally the beam’s own problem. It is checked against a span fraction, it is a serviceability matter rather than a strength one, and the consequence of getting it slightly wrong is a door that binds. A transfer member’s deflection is not its own problem, because a column is standing on it. Whatever the transfer member does at that point, the column above does too, and so does every floor connected to that column. The deflection of the beam is the settlement of a support, ten times over.

The number is 57.6 mm in the long term, for the beam sized by its moment: 23.4 mm when the load first arrives and three times that when creep has finished with it, because concrete at ten thousand days has deflected roughly $1+\varphi$ times what it did on the day. Span over 208 — a deflection ratio no beam in the building would fail on, and a settlement no floor above it can absorb.

Three ways to move the same column #

The bending stiffness of a rectangle goes as $d^3$, so buying stiffness with depth is even more efficient than buying strength with it, and every way of making a transfer is a way of getting depth.

The truss is the striking column of that table. It carries the identical moment with one fifth of the material, and it does so by the oldest trick in the collection: resolving a moment into a pair of forces and then pushing them as far apart as the storey allows. At 3.6 m centres the chords carry 3750 kN each, against a lever arm the deep beam could only reach by being 3.6 m deep itself.

Its price is occupancy of a different kind. The beam takes 0.54 of a storey and leaves the rest; the truss takes the whole storey height and gives back the triangles between its diagonals — a storey that can be walked through but not planned freely. Its deflection was found by virtual work over the members after they were sized, which is one deflection obtained without solving the structure twice: the unit-load forces in a determinate truss are the real ones divided by $P$, so the sum $\Sigma F^2 L/EA$ needs no second analysis. And the truss is steel, so nothing about that 16 mm arrives three years late.

The wall is the extreme case and the honest one, because it is what a real podium usually does: the storey above the transfer is a party wall or a core wall, and a wall spanning 12 m over a 7.2 m depth is not a beam at all.

Forty-one per cent of the transfer wall’s 4.34 mm is shear deformation, and that fraction is not a correction — it is the deflection that is not bending, scaling as $(d/L)^2$ and therefore unavoidable in exactly the members the √ law has made deep. A transfer member is deep by construction. Beam theory is at its least reliable precisely where transfer structures live.

The lever arm used for the wall’s tension tie is 5.28 m against a 7.2 m depth, and that ratio is a curve fit rather than a derivation: 0.6 of the span while the wall is at least as deep as it is long, a straight line to an ordinary bent section by the time the span is twice the depth. It descends from the deep-beam testing done at Stuttgart by Leonhardt and Walther in the mid-1960s, which established what the figure above shows — below a span-to-depth ratio of about two, plane sections do not stay plane and the lever arm stops growing with the depth. Everything deeper is material that has stopped helping.

The floor above does not get a bigger moment. It gets a different one #

A floor continuous over several supports carries hogging over each support and sagging in each bay, and that redistribution is what continuity buys.

Now settle one of those supports. The second free body of this essay is the floor above the transfer, cut nowhere and loaded not at all: a continuous beam over two 8 m bays, with no load applied whatever, and its middle support pushed down 57.6 mm. That is the support that moved, at building scale, and the moments that come back are in equilibrium with nothing.

Two things in that diagram are worse than a larger number would have been. The first is the sign. The induced moment is sagging over the settled support, so it does not pile onto the 240 kNm of hogging — it cancels it and keeps going, and past the crossing the top face over the transferred column is in compression while the bottom face, carrying the light bottom steel of a support region, is in tension. A designer reading only the peak magnitude sees 346 kNm where 240 was expected, notes an increase of 44 per cent, and misses that a different face has become the critical one.

The second is that this happens on every floor. The settlement is imposed at the bottom of the stack and the column carries it up rigidly, so the tenth floor is displaced as much as the first, and each of the ten picks up its own 432 kNm. The transfer member has one design problem and has created ten.

A threshold with nothing in it but two shapes #

The question a designer actually wants answered is how much settlement is too much, and it has a closed form of unusual purity.

Push the middle support of a two-span continuous beam down by $\delta$ and the induced moment there is $3EI\delta/\ell^2$. Set that equal to the floor’s own hogging under a uniform load, $w\ell^2/8$, and the settlement that makes the two equal is

$$\delta_1 = \frac{w\,\ell^4}{24EI}.$$

That is 32 mm for these floors. Now compare it with the deflection those same floors would have if their spans were simply supported, $5w\ell^4/384EI$, which is 10 mm:

$$\frac{\delta_1}{\delta_{\text{ss}}} = \frac{384}{24 \times 5} = 3.2.$$

No load, no modulus, no second moment, no span. All four cancel, because both quantities are the same load acting on the same beam through two different bending shapes, and the ratio of two shapes is a pure number. A floor can absorb a support settlement of about three times its own simply supported deflection before the settlement is doing as much to it as its design load does — and since a floor’s own deflection is a number every designer already knows to within a factor of two, the threshold needs no analysis at all. It is the same species of result as span to the fourth: a coefficient that belongs to the geometry of bending rather than to any structure, and that survives every change of material and scale.

The floors hold the column up #

Everything so far has treated the settlement as imposed. It is not. The floors that are being bent by the settlement are pushing back on the column while they bend, and the column is passing that push down to the transfer member — so the transfer member never sees the whole 6000 kN.

That gap between 0.909 and 1.000 is a finding rather than an error, and it is the reason the curve is drawn at all. The closed-form depth is computed as though the transfer member carried the full load; the curve solves the coupling. Each floor resists the settlement of the column it is continuous over with a stiffness of 1875 kN per metre, ten floors give 18 750 kN/m, and the transfer member’s own flexibility decides how the 6000 kN divides between the two. The fixed point is exact and needs no iteration, because both relations are linear:

$$\delta = \frac{cP}{1 + c\,n\,k_f},$$

with $c$ the transfer member’s flexibility. It gives 57.6 mm where the uncoupled calculation gives 70.3, and 4920 kN reaching the transfer member where 6000 was applied. Eighteen per cent of the load never arrives.

This is the stiffest path taking the load, arriving in a place where nobody drew a second path. The floors above a transfer were designed to carry load horizontally to columns; they are also, whether anyone intended it or not, a set of springs holding one column up. The coupling makes the transfer member’s job slightly easier and makes the floors’ job substantially harder, which is a trade nobody chose and no drawing shows.

Where the model stops #

The transfer member is simply supported. It is drawn on a pin and a roller, and a real one is monolithic with the columns at its ends, which reduces its span moment and its deflection and introduces moments into those columns instead. The direction of the error is favourable, and its size is whatever the end restraint is — which is neither pinned nor rigid and is not usually established.

The load arrives all at once, on a finished structure. It does not. The transfer member is propped while it is cast and the stack is built one floor at a time, so the deflection develops as the building rises and each floor’s induced moment depends on when it was built. The structure was never complete while it was being loaded, and the ten identical floors of the figure are ten different problems.

Creep is one number. The factor of three used here compresses the whole of concrete’s time behaviour into a single multiplier applied to an elastic answer. The floors above are also creeping, which relaxes some of the induced moment they are being asked to carry, and the two effects work against each other on timescales the elastic analysis has no way to represent.

The column is axially rigid. Ten storeys of column shorten under 6000 kN by an amount comparable to the settlement being discussed, and the difference between that shortening and the shortening of the columns beside it is another imposed displacement on the same floors, from a completely separate cause.

Nothing here fails. The whole essay is elastic, and a reinforced floor that cracks over a settling support sheds stiffness exactly where the induced moment is highest, which reduces the moment. The elastic calculation is a conservative upper bound on the induced moment and says nothing about the crack widths that bought the relief.

The pictures on this page cannot show the one thing that makes a transfer structure worth arguing about, which is that it is a single member with no alternative route around it. Every figure here draws a load path that works. What none of them draws is the same building with that member removed — and a transferred column is a column whose survival depends on one beam, in a structure whose other columns each depend on the ground. That asymmetry is the subject of the structure that survives losing a member, and it is why transfer members are detailed to a standard the rest of the frame is not. The moment diagram is silent on it, because a moment diagram is drawn for a structure that exists.

The ladder from here #

Later rungs on this anchor: the strut-and-tie model of a deep transfer beam, where the load path is drawn explicitly as struts and ties and the plane-sections calculation is abandoned. Transfer plates, where the transfer is two-way and the moment has nowhere tidy to be cut. The construction sequence properly resolved, floor by floor, with the props struck at a stated stage. Differential column shortening in tall buildings, which is this argument with no transfer member in it at all. The outrigger, which is a transfer of moment rather than of load. Robustness and the key element, and what it means to design a member that is not permitted to fail. Transfer in seismic design, where a stiffness discontinuity at the podium is the soft-storey mechanism that has flattened more buildings than any other. And the economics of the whole thing: at what column offset it becomes cheaper to move the grid above than to build the transfer, which turns out to be a question about $\sqrt{P a}$ and about nothing else.