The arm that makes the columns work

A core alone is a vertical cantilever, and a cantilever’s tip drift under a uniform load goes as $wH^4/8EI$. Past about thirty storeys that fourth power is what decides the core’s thickness — not its strength, which was settled long before, but the number of millimetres the top of it moves.

Meanwhile the perimeter columns are standing there doing nothing about it. They are as far from the centre of the building as anything can be, which is exactly where material is worth most against an overturning moment, and they take almost none of it — because a floor slab connects them to the core through a diaphragm that transmits in-plane shear and no moment at all.

An outrigger is a storey-deep arm that changes that, and it does so by a mechanism that is worth being precise about: it restrains the core’s rotation.

Which free body produced the number #

Cut the core just below the outrigger and take the piece above.

The wind above that level pushes it sideways; the core below it holds it; and the outrigger applies a couple. The couple is not a shear — the arm delivers equal and opposite vertical forces at its two ends, and their resultant is zero — so the shear diagram of the core is untouched by the whole arrangement. Only the moment diagram changes, and it changes by a step.

The couple’s size follows from one compatibility statement. The core wants to rotate at that level by $\theta_{load}$; the couple $M_1$ takes some of that rotation back out of the core; and the outrigger with its columns allows a rotation of its own. Equate them:

$$\theta_{load}(x) - \frac{x}{EI}M_1 = \frac{M_1}{K}$$

$$M_1 = \frac{\theta_{load}(x)}{x/EI + 1/K}$$

with $\theta_{load}(x) = \dfrac{w}{6EI}\left[H^3 - (H-x)^3\right]$ for a uniform wind.

One equation, one unknown. The outrigger is a singly redundant structure and its whole analysis is that line.

What $K$ is made of #

The rotational stiffness on the right-hand side is the arm and its columns in series, and the columns dominate.

A couple $M_1$ across a bay of width $2c$ is a force pair $M_1/2c$. Each column stretches or shortens by $(M_1/2c)\,L_c/EA$, where $L_c$ is the height of column below the outrigger. The rotation that produces is the difference of those movements over the bay:

$$\frac{1}{K_{cols}} = \frac{L_c}{2c^2 EA}$$

For a 200 m building with the outrigger at 150 m, columns of 1.5 m² concrete at $\pm7.5$ m, that is about $1.8\times10^8$ kNm per radian. The arm itself — a storey-deep truss or wall spanning from core to column — adds its own flexibility in series, which typically brings the pair down to something around $10^8$.

Two things fall out of that expression and both are counter-intuitive.

The column stiffness depends on the height below the outrigger, not above. A column shortening over 150 m is much softer than one shortening over 50, so an outrigger placed high has softer columns to work against — which is one of two effects pulling its optimum height around.

And $c$ enters squared. Doubling the bay width quadruples the rotational stiffness. Position beats area by a wide margin, which is the same statement plan torsion makes about walls and the second moment of area makes about flanges.

That is depth as the cheapest strength applied to a plan dimension instead of a section depth: the outrigger’s lever arm is the distance from core to column, and everything it can do is proportional to that distance squared.

The optimum, and the two questions it answers differently #

Maximising the drift reduction over the height at which the arm is placed gives a stationary point, and it is worth deriving rather than quoting.

For a rigid arm ($K \to \infty$), $M_1 = EI\theta_{load}(x)/x$, and the drift it removes is $M_1 x(H - x/2)/EI$. Substituting and writing $u = x/H$:

$$\Delta \propto \left[1 - (1-u)^3\right]\left(1 - \tfrac{u}{2}\right)$$

whose derivative vanishes at $u = 0.549$. Not a half; a little above it. The reduction there is 87.9%.

Give the arm a realistic stiffness and the optimum climbs, because a soft arm needs the largest rotation it can find and the core’s rotation grows with height:

And asking a different question moves it the other way. The base moment is $wH^2/2 - M_1$, so minimising it means maximising $M_1$ alone with no regard for where it acts — and $M_1$ is largest at 0.49 H, where the core’s rotation and the arm’s leverage are balanced differently. At that height the base moment falls 26.5% and the drift only 39.1%.

There is no best height for an outrigger. There is a best height for a stated objective on a stated building, and the folk number of “about mid-height” is the rigid-arm drift answer rounded.

Two arms, and where they go #

A second outrigger is a second redundant, so the compatibility becomes a 2 × 2 system: each couple relieves the core’s rotation at its own level and at the other one, so they compete.

Searching the pair on the same building puts them at 0.555 H and 0.805 H, with couples of 111 and 97 MNm and a combined drift reduction of 60.9% against 45.6% for one. Neither sits where the single one did, and the lower of the two is close to it while the upper one has moved well up.

The rule of thumb that two outriggers go at roughly a third and two thirds of the height is the rigid-arm answer again; at realistic stiffness both migrate upward, for the same reason a single one does.

The diminishing return is steep. One arm buys 46 points of drift; the second buys 15 more; a third would buy about six. Against the cost of a storey of usable floor at 150 m, that is where the sequence stops.

Two arms are a two-by-two system of redundants competing for the same rotation, which is the arrangement solved by passing it around — each couple changes the rotation the other one sees, so neither can be sized without the other.

The moment diagram acquires a step #

The clearest way to see what an outrigger does to the core is to look at the core’s own bending-moment diagram.

Without one it is a parabola: zero at the top, $wH^2/2$ at the base, curving smoothly. With one, the couple appears as a discontinuity — the diagram jumps by $M_1$ at the outrigger level and continues parallel to where it was.

Everything about the arrangement is visible in that step. It is a step rather than a kink because the arm delivers a couple rather than a force. It is a downward step because the couple opposes the wind. And the base moment is reduced by exactly the height of the step, whatever level it happens at — which is why the base-moment optimum is simply wherever $M_1$ is largest, and has nothing to do with leverage.

The drift, by contrast, is the area of the diagram weighted by distance, so it cares very much where the step is: a step near the top removes a small area, a step near the bottom removes a large one but has less rotation to work with. The competition between those two is the whole of the 0.549.

What it costs the columns #

The couple is carried by the perimeter columns as a tension–compression pair, and 4,877 kN of tension on a gravity column is a serious number. Three consequences follow, and all of them are why outriggers are harder than the arithmetic suggests.

The windward columns can go into net tension. A column carrying 20,000 kN of gravity load and 4,877 of outrigger tension is fine; one near the top of the building, carrying only the few floors above it, is not. This is one reason the arm is not placed higher than the optimum even where the optimum says it could be.

The foundations under those columns see the couple too. It travels down the columns to the ground, so the outrigger has converted part of the core’s base moment into a pair of vertical forces on two pad foundations a bay apart — which may be a better arrangement or a much worse one, depending on the ground.

And the arm is fighting differential shortening for its whole life. The core and the perimeter columns are at different stress levels and shorten by different amounts under gravity and creep; the outrigger connects them and resists the difference; and the force that generates has nothing whatever to do with wind.

Why it exists at all, which is a fourth power #

None of this would be worth a storey of floor area if drift went as anything gentler.

A cantilever’s tip deflection under a uniform load is $wH^4/8EI$. Take a building from 40 storeys to 50 — a factor of 1.25 in height — and the drift goes up by $1.25^4 = 2.44$ while the wind load itself has grown too. Meanwhile the drift limit is generally stated as a fraction of the height, so the target has only risen by 1.25.

So the demand grows as the fourth power and the allowance as the first, and the gap has to be closed by $EI$. Doubling the core’s wall thickness roughly doubles $EI$ and costs floor area on every storey. An outrigger buys 46% of the drift for one storey of area, once — and on a 50-storey building that is a very much better trade.

The whole family of devices tall buildings use — outriggers, belt trusses, tubes, diagrids, megaframes — exists because of that exponent, and each is a different way of getting material further from the centre without paying for it on every floor.

The exponent is the same one a beam meets in its span, with the span turned on its end: nothing else in the subject grows fast enough to make a plan dimension worth this much.

Against thickening the core, which is closer than it sounds #

The trade the previous section asserted deserves the arithmetic, because on floor area alone it is very nearly a draw.

Buying the outrigger’s 45.6 per cent by stiffness alone means $EI \times 1.84$, since drift goes as $1/EI$. For a core of thin walls $I$ is roughly proportional to the wall thickness, so a 600 mm core becomes 1,104 mm. Round a 60 m core perimeter that is

$$60 \times 0.504 = 30\ \text{m}^2\ \text{of floor lost on every storey}$$

and over 50 storeys, 1,500 m² — which is about one floor plate. The outrigger costs one floor plate. On lettable area the two options are level.

Three things break the tie and all of them point the same way.

The material is not level at all. The thicker core is $30 \times 200 = 6{,}000$ m³ of extra concrete, some 15,000 tonnes, against a few hundred tonnes of steel in two arms. That is a factor of fifty, and it is paid for again in the foundations.

The outrigger’s storey was never lettable. It goes at a plant level, where the floor is already occupied by air handling and lift machinery, so its marginal cost is close to zero — while every square metre of core wall is taken from space that would have been sold. The comparison that matters is not one floor against one floor; it is one floor already spent against fifty floors each losing 30 m².

And a heavier core attracts more force. Mass is the load in an earthquake and the extra concrete is at the middle of the plan where the lateral system is, so thickening the core raises the demand it was thickened to meet. The outrigger adds almost nothing to the mass and takes the force out through columns that were there anyway.

The general form is worth keeping, because it recurs whenever a stiffness problem is solved by section rather than by arrangement: material added at the centre is paid for everywhere and works only where it is, while material that changes the arrangement is paid for once. That is the same reading plan torsion gives about walls, and the same one a section gives about its flanges.

The outrigger has no strength job #

There is a question underneath all of this that decides how the whole arrangement is checked, and it is not a question about numbers: is the outrigger allowed to fail?

The core alone carries the wind. Without the arm it drifts 500 mm rather than 272, and 500 mm over 200 m is $H/400$ — outside a serviceability limit and nowhere near a strength one. So the outrigger is a stiffness device, and losing it is a serviceability event.

That is only true if the core was designed for the moment it would carry alone. Two positions are available and they are genuinely different structures.

Design the core for $wH^2/2$, unrelieved. The outrigger then reduces a base moment the core could have taken anyway, its loss costs drift and nothing else, and it needs no more robustness than any other member. This is the expensive option and the one that keeps the arrangement honest.

Or take the 26.5 per cent relief into the core’s design. The core is then thinner, and the outrigger has become a member whose loss is a collapse rather than a deflection — a single arm, at one level, carrying a 4,877 kN couple through two connections into a concrete wall, with no alternative path. A structure that survives losing a member is a question worth asking of any element carrying that much with no spare, and the answer here is decided by an economic choice made much earlier.

There is a second reason the first position is the safer one, and it is about the nature of the force rather than its size. The couple is deformation-driven: it exists because the core wants to rotate and the arm will not let it, so it is bounded by the deformation rather than by the load. If the columns yield, the connection slips, or the delayed splice is made late, the couple simply does not reach the value computed — and the structure is fine, because the core was carrying the moment regardless. A structure relying on a deformation-driven force for its strength is relying on a quantity that its own construction sequence can change.

Where the model stops #

The core is a prismatic cantilever. Real cores step, change wall thickness, lose walls at podium level and are perforated by door openings — every one of which changes $EI$ with height, and the compatibility equation then needs the real flexibility rather than $x/EI$.

The columns are axial springs and nothing else. A perimeter frame with real beams is also a frame, so it takes some shear directly, and the outrigger’s couple is then not the only thing connecting the two systems.

And everything is elastic and first-order. A 200 m building at 500 mm of drift has a P-delta amplification of several per cent, which multiplies the drift the outrigger was there to reduce and therefore makes the outrigger worth slightly more than this calculation says.

What the pictures cannot show #

The deflected shape is drawn at an enormous exaggeration. 272 mm over 200 m is one part in 735, which at the scale of the figure is a fraction of the line width, and the kink at the outrigger — which is the whole visual signature of the arrangement — is smaller still.

Nor can the figures show the belt. In practice an outrigger is accompanied by a belt truss running right round the perimeter at the same level, whose job is to bring the other perimeter columns into the couple as well. Without it only the two columns the arm reaches are working, and the arithmetic above is for exactly those two.

The assumption the figure rests on #

The outrigger’s stiffness $K$ is a single number, constant, known. It is neither: it is the arm in series with the columns, the columns’ contribution depends on where the arm is, and the arm’s own contribution depends on how it is connected to a core that is a concrete wall rather than a node. The optimum height computed above depends on a stiffness that itself depends on the height, and the table’s migration from 0.549 to 0.854 is a sensitivity to exactly that number.

The other thing a stiff level does #

An outrigger is a stiff plane inserted into a building that is otherwise uniform, and stiffness discontinuities have consequences beyond the one they were put there for.

The storey immediately above and below the outrigger sees a large change in the way load is shared, and the columns at those levels carry a jump in axial force that has to be delivered by the slab or by the arm itself. On a concrete building the transfer is a real detailing problem: several thousand kilonewtons arriving into a column over one storey, in a region where the arm’s own reinforcement is already congested.

And the outrigger level is a discontinuity in the dynamic properties too. A building’s mode shapes are the shapes its stiffness distribution admits, and inserting a stiff plane at three quarters of the height changes them — usually by flattening the top of the first mode, which reduces the modal participation of the upper floors and therefore the accelerations people feel there. That is a side effect rather than a design intent, and it is generally a welcome one.

The ladder from here #

Later rungs on this anchor: the belt truss, and how many columns a couple can actually reach. The outrigger as a damper location, since a level with a large relative rotation is also a good place to put something dissipative. Optimum placement for a triangular wind profile rather than a uniform one, which moves both optima downward. The construction-sequence problem in full, with creep and shrinkage and a delayed connection. Outriggers in steel, where the arm is a truss two storeys deep and its own shear flexibility is not negligible. And the comparison this anchor eventually has to make: an outrigger against a tube, against a diagrid, against simply making the core bigger — four ways of buying the same fourth power, priced.