The floor the columns lean on

The braces that have to hold each other put several braces along one column and found how stiff each had to be, and how a row of braces fails together in a zigzag rather than one at a time. It ended on the brace that is not a bar: a floor. A floor ties every column in a storey to the bracing system, and it is a nodal brace for the columns near the core and a relative brace for the ones in the middle of a long plate, whose restraint has to pass through the floor’s own shear before it reaches anything that holds.

The arithmetic that is usually done treats the floor as rigid. The column that leans on its neighbours is the classic form of it: a pin-ended column carries no sway stiffness of its own and adds its load, divided by its height, to the softening the bracing system has to overcome; a storey of such columns needs a core whose stiffness exceeds the sum. The load that makes itself worse supplies the amplification, $1/(1 - 1/\lambda)$, and the column that was never straight supplies the lean that starts it. This essay puts a floor between the columns and the core that shears in its own plane, and asks what the sum misses.

A storey that leans on its floor #

The storey is a single line of eleven column lines along a floor plate ten bays long, with the core at one end. Each column carries 3,000 kN over a storey of 3.6 m and is pinned at both ends, so it contributes no sway stiffness: it only leans. Each is built a three-hundredth out of plumb, so even before it sways it pushes its floor line sideways with $P\theta_0 = 10$ kN, and once the floor line has swayed by $u$ the push is $P(\theta_0 + u/h)$. The term $P/h$, 833 kN per metre of sway, is the column’s sway softening: the stiffness the bracing has to supply before it can supply any more.

The core is sized for a buckling factor of 4 if the floor were rigid: a stiffness four times the eleven columns’ total softening. The floor carries the columns’ pushes to the core as a shear beam lying in its own plane — the floor is a beam lying down — with a shear stiffness between adjacent column lines of $k_f$. Write the floor lines’ sways as $u_0$ at the core to $u_{10}$ at the far end, and every line is in equilibrium between the floor’s shear on either side, the core’s spring if it has one, and its columns’ push:

$$(K_{\text{floor}} + K_{\text{core}})\,u = P\theta_0\,\mathbf{1} + \frac{P}{h}\,u .$$

The storey buckles at the load factor at which $K_{\text{floor}} + K_{\text{core}} - \lambda (P/h) I$ first loses its positive definiteness — the least eigenvalue of a tridiagonal matrix, divided by $P/h$. The only new number is the ratio $\beta = k_f h/P$: how many times stiffer each bay of floor is than one column’s softening.

How stiff a floor is, measured in columns #

A concrete slab is very stiff in its own plane. A 200 mm slab 15 m deep across the line of columns, with 6 m bays, has a shear stiffness per bay of $G t D/s \approx 12{,}500 \times 200 \times 15{,}000/6{,}000 = 6{,}250$ kN/mm, which is several thousand times a 3,000 kN column’s 0.83 kN/mm: off the right of the figure, where every curve has reached 4.0 and the floor is as good as rigid. Light floors are not. A timber floor, a metal deck without its concrete, a precast floor whose planks are not yet tied by a topping, a slab cut by a stair and service openings all the way across — each has an in-plane stiffness orders of magnitude lower, and the storey above a construction floor or a refurbished building often stands on exactly such a floor.

Below a thousand times the curves fall away. With the core at one end and the floor a hundred times as stiff as a column’s softening per bay, the storey buckles at 1.52 times its working load, not 4.0. The floor has not failed and the core has not changed. The far columns lean on the floor before they reach the core, and a floor that shears gives them somewhere to lean that moves.

Putting the core in the middle halves the distance the far columns’ push has to travel, and recovers 2.80. Splitting the same total core stiffness between the two ends recovers 3.02. Both are the floor’s argument rather than the core’s: the core is no stiffer, it is only nearer the columns that need it.

The far columns sway more #

With a stiff floor the storey sways as a block: every line moves about a third of the columns’ lean, which is the core’s own flexibility amplified by $1/(1 - 1/4)$. As the floor softens the sway fans out. At a hundred times, the line at the core sways 0.69 of a lean and the farthest line 2.39: the far columns have leaned on a floor that moved, leaned further because it moved, and taken the floor further with them. The second-order effect that a rigid-floor calculation applies once to the whole storey is applied here bay by bay, and it compounds along the plate.

That is the sense in which a floor is a relative brace for the far columns. The brace that need not be strong distinguished a nodal brace, which holds a point to the ground, from a relative brace, which holds one point to its neighbour, and found that a relative brace needs a stiffness set by its own segment’s length. A floor bay is a relative brace between two column lines, and the far columns are braced through ten of them in series.

What the core is asked for #

The core’s force is the floor’s answer to the same question, and it is the number a designer actually uses: the bracing system is sized for the sum of the columns’ notional horizontal forces, amplified. With a rigid floor that is 110 kN times 1.33, 147 kN. Through a floor a hundred times as stiff as P/h per bay it is 2.75 times the sum, 303 kN — twice what the rigid-floor sum promised. Through a floor at seventy-five times, 4.64 times: the storey is close to its own buckling load and every force in it is close to unbounded.

The extra force has not come from anywhere new. It is the same eleven columns’ push, each larger because each has swayed further than the rigid floor allowed, and the floor has to carry all of it. The figure for the floor’s largest shear tells the same story at the bay next to the core.

The bay next to the core carries every column’s push but the first, amplified: 2.60 times the first-order sum at a hundred times P/h. Its connections — the collector into the core, the chords along the plate’s edges, the fixings of a timber or deck diaphragm — are designed for the rigid-floor force in the ordinary calculation, and see twice it in the storey that actually stands. The collector the slab does not need dealt with the collector’s force under wind; under the columns’ own lean the collector’s force depends on the floor’s stiffness as much as on the load.

A longer plate, a shorter future #

The core in this comparison grows with the plate — it is always sized for a factor of four with a rigid floor — so every curve would sit at 4.0 if the floor were rigid. It does not. With two bays the floor barely matters; with ten and the core at one end the factor is 1.52; with fourteen bays it has fallen below one, and the storey cannot carry its working load at all. A longer plate has more columns pushing, which the core is sized for, and a longer path for their push to travel, which nothing is sized for. The floor’s own buckling load — the factor the storey would reach with a perfectly rigid core — falls as the square of the plate’s length, like any shear beam’s, and a long enough plate meets it.

Cores at both ends keep the plate’s longest path to half, and the factor stays near three out to twenty bays. That is the structural reason long buildings put their stability elements at both ends rather than at one, and it is independent of wind: the columns’ own lean is enough to need it.

No core is enough #

Turn the question round. Given a floor, how stiff must the core be to give the storey a factor of 2? With a rigid floor, twice the columns’ softening. With a floor a thousand times P/h per bay, 2.15 times: the floor costs a little. With a hundred times, 15.2 times — seven times the rigid-floor answer. Below about ninety, no core is stiff enough. A stiffer core buys a stiffer storey only while the floor can deliver to it; past that the storey’s stability belongs to the floor, and the core is a rigid wall at the end of a plate that buckles in its own plane.

The threshold is the floor’s own critical load. Behind a perfectly rigid core, the floor plate is a chain of ten shear springs held at one end, each joint carrying a column that softens it by $P/h$. The chain’s lowest mode is a quarter-wave — nothing at the core, most at the free end — and for a chain of $N$ springs its eigenvalue is $4k_f\sin^2\!\big(\pi/(2(2N+1))\big) \approx k_f(\pi/(2N+1))^2$. Divided by $P/h$, the floor alone buckles at

$$\lambda_{\text{floor}} \approx \beta\left(\frac{\pi}{2N+1}\right)^2 = \beta\left(\frac{\pi}{21}\right)^2 = 0.0224\,\beta$$

for ten bays. That is 2.0 at $\beta = 90$, which is where the figure’s curve leaves for infinity, and 1.0 — the storey unable to carry its working load even with a core of infinite stiffness — at $\beta = 45$. The formula also explains the bays figure: the floor-only factor falls as $1/(2N+1)^2$, so doubling the plate’s length quarters it.

What that means physically depends on the floor. A concrete slab at several thousand times P/h has a floor-only factor of a hundred or more, and the floor never matters. A plate whose in-plane stiffness is a few per cent of that — a plywood or metal-deck diaphragm, a precast floor before its topping, a slab cut almost through by an atrium — can sit near $\beta = 100$, and there the floor, not the core, decides the storey.

A check that sees the floor #

The two limits — the core with a rigid floor, and the floor with a rigid core — combine in a way a designer can use without the matrix. Two flexibilities in series add, and a buckling factor is a stiffness divided by the load’s softening, so the storey’s factor is bounded below by the Föppl–Dunkerley sum

$$\frac{1}{\lambda} \le \frac{1}{\lambda_{\text{core}}} + \frac{1}{\lambda_{\text{floor}}}.$$

For the ten-bay storey with its core at one end, $\lambda_{\text{core}} = 4$ and $\lambda_{\text{floor}} = 0.0224\beta$. At a hundred times P/h the sum gives $1/(0.25 + 0.446) = 1.44$, against the exact 1.52; at three hundred, 2.51 against 2.65; at a thousand, 3.39 against 3.49 — always a little below the exact factor, as a Dunkerley sum is, and within six per cent across the whole range. The rule’s content is the second term. A storey’s stability check that does not compute the floor’s own factor has assumed it infinite, and the sum says how large it has to be before the assumption costs less than a tenth: $\lambda_{\text{floor}}$ ten times $\lambda_{\text{core}}$, which for this plate is a floor four hundred times as stiff per bay as one column’s softening.

The rule also says what to do when the floor fails it. A stiffer core helps only through the first term, which is why the core-stiffness figure climbs without limit. Moving the core nearer the far columns, or adding a second one, changes $\lambda_{\text{floor}}$ itself, by shortening the chain: with cores at both ends the chain is half as long and the floor’s factor four times as large.

The same storey, by hand #

Two bays are enough to see every effect by hand. Take one bay with the core at one end: two column lines, the first on the core $k_c$, the second joined to it by the floor $k_f$. The stiffness matrix is

$$\begin{pmatrix} k_c + k_f & -k_f \\ -k_f & k_f \end{pmatrix},$$

and its least eigenvalue is $\mu = \tfrac12\left[(k_c + 2k_f) - \sqrt{(k_c + 2k_f)^2 - 4k_c k_f}\right]$. With a rigid floor it tends to $k_c/2$ — the core shared by two columns. With $k_c = 8\,P/h$ (a rigid-floor factor of 4 for two columns) and $k_f = 10\,P/h$, it is $\tfrac12[28 - \sqrt{784 - 320}] = \tfrac12(28 - 21.5) = 3.23\,P/h$: a factor of 3.23 instead of 4, from a floor only ten times as stiff as a column. The far column’s sway, with its neighbour held by the core, is amplified by its own spring: it leans on $k_f$ in series with $k_c$ shared, and the series stiffness is what it sees.

The ten-bay storey is the same matrix grown to eleven rows, and its least eigenvalue is what every figure above is drawn from.

How a design standard sees the floor #

A design standard asks whether a diaphragm may be treated as rigid, and asks it about lateral load: whether the floor distributes wind or earthquake forces to the stability elements in proportion to their stiffness or to their tributary areas. Rigid by one code and not by the other found how different the tests can be. None of them asks the question this essay asks — whether the floor is stiff enough not to lower the storey’s buckling load — and the answers need not agree. A floor can distribute wind loads closely enough to rigid to pass a code’s test and still be the storey’s weakest brace under the columns’ own lean, because the two questions are measured against different stiffnesses: the stability elements’ for the first, the columns’ softening for the second.

The floor that feeds two bays found a floor’s shear flexibility deciding how wind is shared between two brace lines. Here the same flexibility decides whether there is a storey to share anything at all.

What a single storey on one spring leaves out #

The columns are pinned and identical. Columns continuous through the floor carry some sway stiffness of their own, which the storey can use; columns of different loads lean by different amounts, and a heavily loaded line far from the core is the worst case. The figures treat every column as a pure leaner with the same load.

The floor is a shear beam. A deep floor plate also bends in its own plane, and a long narrow one bends more than it shears; its flexibility then grows faster with length than the shear beam’s. The essay’s conclusions about length are therefore optimistic for narrow plates.

The leans are all one way. Every column is out of plumb by the same amount in the same direction, which is the worst pattern for the core and the one codes assume for notional forces. Random leans partly cancel along the plate, which reduces the first-order force but not the buckling factor, since that does not depend on the leans at all.

The core is a spring at one storey. A real core is a cantilever whose stiffness at a given floor depends on every storey above and below it, and whose own sway adds the storeys’ leans together up the building. The figures hold one storey on one spring; a tall core with soft floors at several levels has each floor’s chain hanging off a core that is itself moving, and the factors multiply rather than repeat.

The floor is elastic and its connections are rigid. A diaphragm whose fixings slip, or whose collector has a gap, adds flexibility at exactly the place this essay finds the largest force.

Still open: the storey whose floor is not yet built #

The storey most likely to stand on a soft floor is one under construction. The columns of the floor above are erected before the slab that will tie them is cast; the deck is laid but not yet filled; the planks are placed but not yet topped. For a few days each storey leans on whatever temporary bracing and partial diaphragm the erection sequence has provided, and the core may itself be only a few floors further on. Whether the stage at which a storey’s buckling factor is lowest is the stage at which its floor is softest — or the stage at which the next storey’s columns add their load to it before its own topping has cured — is the question that turns this storey into the most dangerous day of a building’s construction.