The column that leans on its neighbours

Take a single column, pin it at both ends, and stand it up with a vertical load on top. It falls over. Not by buckling — its Euler load may be enormous — but because a pin at each end permits a rigid-body sway, and nothing resists it. Its critical load in sway is exactly zero.

Now put three more of them in a row and connect the tops with a beam. Still zero. Add a fourth column with a fixed base, and the whole row stands.

That fourth column is doing all of the work, and the question this essay is about is what it costs it.

Which free body produced the number #

The whole storey, cut just below the beams.

Give it a sway $\Delta$. Each column that has lateral stiffness pushes back with $s_i\Delta$; a pinned-pinned column pushes back with nothing. Each column carrying axial load $P_i$ contributes a destabilising moment because its own load is now acting through an offset, which at storey level is a horizontal force of about $1.2P_i\Delta/L$ — the consistent value from a cubic shape function, which reproduces the textbook $K = 2$ for an isolated cantilever column to within one and a half per cent where the naive $P\Delta/L$ is 22% out.

Equilibrium of the storey gives

$$\left(\sum s_i\right)\Delta = 1.2\frac{\sum P_i}{L}\Delta$$

and buckling is where a non-zero $\Delta$ satisfies it:

$$P_{cr,\text{storey}} = \frac{L\sum s_i}{1.2}$$

Two sums, and they are over different things. The stiffness sum runs over the columns that resist sway. The load sum runs over every column in the storey. Nothing requires the two sets to be the same.

What it costs the column that is doing the work #

Once the storey’s buckling load is known, each column’s own effective length follows from the load it is carrying at that instant:

$$K_i = \frac{\pi}{L}\sqrt{\frac{EI_i}{\lambda P_i}}, \qquad \lambda = \frac{P_{cr,\text{storey}}}{\sum P_i}$$

Run it for a single column with a fixed base and nothing leaning on it and the answer is $\pi/\sqrt{2.5} = 1.99$, which is the 2.0 of strong enough and still falls over and of every textbook — arrived at here by a route that never mentions a chart, and a useful check that the geometric stiffness above is the right one.

Now add the leaners. $\sum P$ rises, $\sum s$ does not, $\lambda$ falls in proportion, and

$$K_{braced} = 1.99\sqrt{\frac{\sum P}{P_{braced}}}$$

The effective length rises as the square root of the load ratio. A column providing all of a storey’s stiffness while carrying a quarter of its gravity load is at $K = 4$. Alignment charts stop at 3.

And what it costs the leaners #

The second half of the result is the one that catches people, because it runs the opposite way to intuition.

A pinned-pinned column has no critical load of its own in sway. So what should it be designed for?

The storey buckles at a load factor $\lambda$, and at that instant the leaning column is carrying $\lambda P_{lean}$. The effective length that reproduces that as an Euler load is 2.60 on the storey drawn — for a member a designer would otherwise have taken at $K = 1.0$ without hesitation.

That is not a subtlety. It is a factor of 2.6 on the length and therefore 6.8 on the slenderness ratio’s effect, on every gravity column in the building, and it comes entirely from the fact that they are in a frame that sways.

The reason it is so easy to miss is that a gravity column looks like a member with nothing to do with the lateral system. It has pinned connections, it is not on the bracing drawing, it is not in the wind analysis, and it is sized by a spreadsheet from an axial load. All of that is correct and none of it contains the storey.

There is a way of putting the whole result in one sentence that is worth having, because it makes the arithmetic unnecessary. A storey buckles as a unit, and every column in it fails at the same load factor. Effective length is then not a property of a member at all; it is a way of expressing one member’s share of a collective event, translated into the units a member check happens to use. Read that way, the square root stops being surprising: a column carrying a quarter of the load that will cause the collective failure is at a quarter of the load factor, and a load factor of a quarter is a length factor of two, because the Euler load goes as the inverse square of the length.

It also explains why the concept behaves so oddly. The ends decide the length is exactly right for a braced member, whose ends really are held by something independent of it, and it is a category error for a sway member, whose ends are held by an average over the storey. The same phrase is doing two different jobs and only one of them is a statement about ends.

The two ways designers deal with it #

Design the storey. Compute $\lambda$ for the storey and use it for every column in it, which is what the storey-stiffness method does and what almost every commercial frame program now does internally.

Or move the problem into the loads. Apply a notional horizontal force proportional to the vertical load, run a second-order analysis, and let the amplification take care of it. The load that is really a lean is that substitution, and its virtue here is that a leaning column’s contribution appears automatically: it has vertical load, so it generates notional force, so it loads the bracing.

The second method is now the usual one, and it has a property worth noticing: it never mentions effective length at all. A second-order analysis with imperfections included computes the moments in the deformed geometry directly, and the member check that follows is against a column of its actual length. The effective length has not disappeared — it has been converted into an amplified moment.

Which of the two is used decides where the answer appears. In the first, a gravity column’s problem shows up as a large $K$. In the second, it shows up as a large moment on the bracing and a slightly larger axial force everywhere. The physics is identical.

There is a practical asymmetry between them that is worth naming. The effective-length method makes the member look wrong, and a member that looks wrong gets attention. The notional-load method makes the system look slightly worse everywhere, and a system that is uniformly five per cent worse gets none. So the second method, which is the better one, is also the one that hides the finding — and a designer using it can pass a frame with six bays of gravity columns leaning on one braced bay without ever seeing a number that says so. The count that does not see it is about a different arithmetic missing a mechanism; this is an arithmetic that finds it and then reports it in a form nobody reads as a warning.

Why the amplifier is the more useful reading #

The storey’s critical load has a second life as the denominator of the sway amplifier:

$$\text{amplification} = \frac{1}{1 - \sum P/P_{cr}}$$

which is the load that makes itself worse written at storey level. It gives a much better feel for the problem than an effective length does, because it is a number that can be looked at and judged: below about 1.1 the second-order effects are ignorable, above about 1.4 the frame is in trouble, and the range in between is where most real buildings sit.

It also makes the leaning columns’ contribution vivid. Adding a bay of gravity columns adds $\sum P$ and leaves $P_{cr}$ alone, so it moves the frame to the right along that hyperbola. A frame at an amplifier of 1.15 with two bays of gravity columns is at 1.35 with six, and nothing about the bracing has changed.

Where the stiffness actually comes from #

Everything above treats $\sum s_i$ as given, and it is worth asking what supplies it, because the answer is often less than the drawing suggests.

A column’s sway stiffness depends on its base fixity and on the beams framing into its top. A base drawn as fixed and built on four bolts in a thin plate is a spring rather than a restraint, and held and not held is the essay about the difference. A beam drawn as rigidly connected and detailed with a flexible end plate contributes a fraction of what the model gives it.

That matters more here than in most stability problems, because the storey’s stiffness is a sum over few terms. In a frame where one bay is braced and eight are not, the whole of $\sum s$ comes from one bay, and an error in that bay’s stiffness is an error in the storey’s critical load in full proportion — while the same error in a member’s own second moment would have been diluted.

A worked reading of the storey drawn #

It is worth putting the four numbers side by side, because the arithmetic is short and the conclusion is not obvious from any one of them.

The braced column has a fixed base, a second moment of 3.2×10⁸ mm⁴ and 4.2 m of height, so its sway stiffness is $3EI/L^3 = 2{,}720$ N/mm. It is the only column in the storey with any. The storey’s critical load is $sL/1.2 = 9{,}520$ kN.

The gravity load on the storey is 1,100 kN on the braced column and 820 on each of three leaners: 3,560 kN in total. So $\lambda = 2.67$ — the storey buckles at two and two thirds times the load it is carrying, which sounds comfortable and is the number a stability check would report.

Now read what that means member by member. The braced column’s effective length factor is 3.63, so its buckling length is 15.2 metres for a column 4.2 metres long. Each leaner’s is 2.60, giving 10.9 metres. A designer checking those columns at their true lengths — 4.2 metres, $K = 1$ — would find every one of them enormously adequate, and would be wrong about all four by the same cause.

And the amplifier is $1/(1 - 1/2.67) = 1.60$: the sway moments in the frame are sixty per cent larger than a first-order analysis gives. That is the number to look at, and it is the number that would have made the problem obvious, because 1.6 is well past any threshold anybody would ignore.

Three ways of expressing one fact, and only one of them is legible at a glance. Drawing as calculation usually means a figure; here it means choosing which of three equivalent numbers to write down.

Where the model stops #

The storey was treated in isolation. A multi-storey frame’s columns are continuous, so a stiff storey helps a soft one and the buckling mode may involve several floors. The storey method is a good approximation when the storeys are similar and a poor one when they are not — and the classic failure is a soft storey at ground level, whose critical load the method computes correctly and whose consequence it understates.

All the columns were assumed to sway together. They do, provided the floor is a rigid diaphragm in plan. Where it is not — a long narrow floor, a plate with a large opening — different parts of a storey sway by different amounts and the sharing above does not hold.

Imperfections were left out. A perfectly straight, perfectly plumb frame buckles at $P_{cr}$; a real one has an out-of-plumb of about one in two hundred and begins to sway from the first kilonewton. The column that was never straight is the member-level version, and at storey level it is the reason a notional force exists at all.

The geometric stiffness is linearised. $1.2P/L$ is the consistent first-order term and it is excellent up to about 60% of the critical load, which is where frames live. Nearer the critical load the true relation curves away and the amplifier under-predicts.

The columns were assumed to buckle in sway and not in the plane at right angles to it. A column in a frame braced in one direction and sway in the other has two different effective lengths about its two axes, and the one that governs is not always the sway one — a column with a large minor-axis slenderness and a short braced length can be governed by the braced direction even when the sway factor is 3. Both have to be computed, which means the storey argument has to be run twice on two different sets of stiffnesses.

And nothing here is about the columns’ own axial shortening. A tall building’s differential shortening changes the geometry the second-order analysis is run on, and the two effects are computed separately by nearly everyone.

The generalisation #

The habit is to ask, of any stability check, what the free body is — and to notice how often the honest answer is larger than the member being checked.

A member-by-member stability check is a check on a free body consisting of one member with assumed end conditions. That is exact when the assumptions are exact, and the assumptions are exact only when the member’s ends are genuinely held by something that does not itself depend on the member. In a braced frame they are. In a sway frame they are not, and the free body has to grow until it closes — which for sway means the whole storey, and for a slender building means the whole building.

There is a diagnostic worth carrying from all of this, and it needs no calculation. On any frame drawing, count the columns and then count the ones that appear on the bracing drawing. If the two numbers differ by a lot — and in a typical commercial building the ratio is five or ten to one — then the columns that do appear are carrying a stability problem several times the size of their own, and their effective lengths are not the ones on any chart. The stiffness the load takes away is the mechanism; the count is how to notice it is happening.

The second reading is about accounting. A leaning column is a member that consumes capacity and produces none, and there is no line in any calculation where that shows up as a cost. It appears instead as a slightly larger number in somebody else’s check, several drawings away, and the connection between the two is a sum that nobody’s spreadsheet contains. One support too many is usually about redundancy being a benefit. This is the same bookkeeping run the other way: a member that shares the load without sharing the resistance is a redundancy that costs.