The trace that adds up every storey

The bound from underneath found a way to catch a column’s buckling load from below without guessing its shape. A column’s flexibility under its own load — how far each point moves when a unit force acts at another — is a kernel, and the traces of its powers are sums over all its buckling loads: the first is $\sum 1/P_k$, the second $\sum 1/P_k^2$, and so on. Each sum is at least its largest term, so each trace bounds the smallest buckling load from below, and the bounds close in as the power rises: 6, $\sqrt{90}$, 9.81 and 9.86 for a pinned strut whose load is $\pi^2 = 9.87$.

The essay ended on whether the same can be done for a sway frame, whose stability belongs to all its columns and beams at once and whose flexibility is a matrix over its storeys, with rotating joints in every entry. The traces of that matrix would bound the frame’s elastic critical load factor — the number a design code’s sway classification actually asks for — with no assumed sway. Whether the first two traces are close enough to be useful for a real frame, where the storeys interact and the modes crowd together, was the question.

A frame’s flexibility, and its traces #

The frame is one bay, six storeys of 3.5 m, 6 m wide, fixed at its base; its columns have a bending stiffness of $40{,}000\ \text{kN·m}^2$ and its beams 60,000, and every floor carries 400 kN of gravity load. Its lateral flexibility $F$ is a six-by-six matrix: the sway of each floor under a unit horizontal force at each floor, from six first-order analyses. Its gravity load’s P-Δ effect is a geometric stiffness $K_g$: each storey couples its two floors through $N_i/h$, the load above it over its height — the stiffness the load takes away.

The frame buckles in sway at λ times its load when $F K_g$ has an eigenvalue $1/\lambda$. The matrix $B = F K_g$ is not symmetric, but it is similar to one that is, so its eigenvalues $\mu$ are real and none is negative, and the critical factor is one over the largest. The traces are $\operatorname{tr} B^k = \sum \mu^k$, and need nothing but matrix products.

Two bounds follow from each, for any positive numbers. The sum is at least its largest term, so $\mu_{\max} \le (\operatorname{tr} B^k)^{1/k}$ and

$$\lambda \ge (\operatorname{tr} B^k)^{-1/k}.$$

And $\sum\mu^{k+1}/\sum\mu^k$ is an average of the $\mu$ weighted by $\mu^k$, which cannot exceed the largest, so

$$\lambda \le \frac{\operatorname{tr} B^k}{\operatorname{tr} B^{k+1}}.$$

The column essay had only the first. A matrix gives both, from the same traces.

The two sequences close on the critical factor from either side. The first trace gives 5.12 from below and nothing useful from above; the second, 10.82 and 19.23; by the fourth the bracket is 14.07 to 16.39, which brackets 14.96 within 6 and 10 per cent. Every number in it came from the flexibility and the loads, with no sway shape assumed at any point. That is the frame version of what the column essay found, and it answers the first half of the question: the traces do bound a frame’s critical factor, from both sides, and they converge.

The second half — whether the first two are close enough to be useful — is the harder one. Here they are not. The first trace is a third of the answer.

The first trace is a sum over the storeys #

What the first trace is becomes clear on a frame with rigid beams, whose storeys sway independently. Each storey then has its own critical factor $\alpha_i = k_i h/N_i$, its sway stiffness against the P-Δ of the load above it, and $B$’s eigenvalues are exactly $1/\alpha_i$. So

$$\operatorname{tr} B = \sum_i \frac{1}{\alpha_i},$$

and the first bound is $\lambda \ge 1/\sum(1/\alpha_i)$. That is Dunkerley’s formula — the reciprocal of the frame’s critical factor is at most the sum of the reciprocals of its parts’ — with the storeys as the parts. The true answer for a rigid-beamed frame is the weakest storey, $\min \alpha_i$, because the storeys do not help each other; Dunkerley adds them all, as if every storey’s sway were contributing to one mode.

The spectrum says when that hurts. With beams of ordinary stiffness the eigenvalues fall off — 1.00, 0.67, 0.49, 0.38 — and the first trace adds a tail worth about twice the first term. With rigid beams they fall off more slowly, because the storey factors of a uniform frame rise only gently up the height. And with columns sized so that every storey has the same factor, all six are equal, and the first trace is six times the first eigenvalue.

The better the frame is balanced, the worse its first trace. A frame designed by sizing each storey’s columns for that storey’s load — which is how frames are designed — has storey factors that are nearly alike, and its first trace bounds it at about one over the number of storeys. The modes crowd together exactly in the frames engineers build, and crowding is what a sum of reciprocals cannot see past. The second trace does better, $1/\sqrt{n}$ in the worst case, 0.41 for six storeys. The $k$-th, $n^{-1/k}$.

The figure has a second line that runs the other way. EN 1993-1-1’s storey estimate takes each storey’s drift under horizontal loads, computes $(H/V)(h/\delta)$ storey by storey, and keeps the smallest. With rigid beams that is exactly $\min \alpha_i$, and it is exact for every frame on the figure — including the balanced frame on which the first trace is a sixth of the answer. The two methods fail in opposite frames.

Flexible beams couple the storeys #

Rigid beams are the case the storey estimate was built for. Real beams bend, a storey’s sway rotates the joints above and below it, and the storeys stop being independent.

With stiff beams the storey estimate is exact and Rayleigh’s quotient on the sway is 24 per cent high; the second-trace bracket is wide, 20 to 40 around 32. As the beams soften the picture inverts. The storey estimate falls below the true factor, 11.00 against 12.93 with beams as stiff as the columns, because a storey’s drift under its own shear now includes joint rotations that the neighbouring storeys partly resist — the storeys help each other, and a storey-by-storey count gives that help away. And the bracket narrows, to 2.29 to 2.85 around 2.52 with beams a tenth of the columns, because coupled storeys have a spread spectrum, one mode clearly first, and a sum of powers is dominated by its largest term.

So the traces are best exactly where the storey estimate is worst: in frames whose storeys interact. That is not a coincidence. Both are reading the same spectrum. The storey estimate is exact when the eigenvectors are single storeys, which is when the eigenvalues are free to be anything, including all alike. The traces are good when one eigenvalue stands clear of the others, which needs the storeys coupled into a frame-wide mode.

More storeys #

The first trace degrades with height steadily, because every storey adds a term: a third of the answer at six storeys, an eighth at twenty. The fourth trace holds up far better, 0.94 at six and 0.82 at twenty, and on the six-storey frame it beats the storey estimate. On the twenty-storey frame the storey estimate wins again, 0.92 against 0.82: a tall uniform frame has many nearly equal storeys near its base, its modes crowd, and the storey estimate is close to exact for the same reason the traces are not.

What the traces never do is overstate. Every lower bound on every figure is below the true factor, by construction, and that is a guarantee Eurocode 3’s estimate does not carry. EN 1993-1-1’s storey formula happens to be conservative on every frame drawn here, but the storey that cannot see the building lean found where it goes wrong: a building whose sway is an overall bending, where the least storey misreads the building’s factor and, worse, the ground storey reads an amplification of 1.02 where the truth is 1.20. A trace has no such blind spot, because it is a sum over every mode the flexibility contains, the bending one among them.

The sway shape is not the mode #

Rayleigh’s quotient is the classical way to estimate a frame’s critical factor quickly: take the first-order sway under horizontal loads in proportion to the floor loads, and divide the work the loads do through it by the P-Δ work. It is an upper bound, and it is as good as the sway shape is like the buckling mode.

A soft bottom storey shows how far apart they can be. The buckling mode concentrates its sway in the soft storey: the first floor moves 0.65 of the roof’s sway and the floors above move almost as a block. The first-order sway under storey shears is spread up the height, 0.31 at the first floor, because every storey’s shear drives its own drift. Rayleigh’s quotient on that shape gives 11.70 against 9.91, 18 per cent high, on the unsafe side.

The traces need no shape and so have no such error to make; with four traces the soft-storey frame’s bracket is 9.60 to 10.50. The storey estimate is 9.01 here, 9 per cent low and safe, because the soft storey is where it looks.

The traces are a power method #

The upper bound $\operatorname{tr} B^k/\operatorname{tr} B^{k+1}$ is worth looking at once more, because it is not new. Multiplying a vector by $B$ again and again turns it into the first mode, and the ratio of successive lengths into the first eigenvalue: the power method, which is how a frame’s buckling mode was found by hand before the stiffness matrix replaced the hand methods, and how Stodola and Vianello found a column’s. The traces are the power method run on every starting vector at once and summed, so the ratio of two of them is an eigenvalue estimate that cannot depend on a lucky or unlucky start.

That is also why they converge the way they do. The power method converges at the rate of the second eigenvalue over the first, and so do the traces: on the frame with ordinary beams, whose second eigenvalue is 0.67 of the first, each step closes the bracket by about a third; on the balanced frame with rigid beams, whose second is the same as the first, no number of steps separates them, and the bounds stop improving at $n^{-1/k}$. The crowding that defeats the first trace is the crowding that slows the power method, seen from the other side.

And it says what the bracket is for. A designer who wants a sway frame’s critical factor to decide whether second-order effects can be counted rather than checked needs to know that it is above 10, not what it is. The second-trace lower bound answers that for the frame of ordinary beams — 10.82 — in two matrix products and no eigenvalue routine, and it answers it with a number that cannot be above the truth.

The six-storey frame, by hand #

The first-order sway under a unit force at every floor gives storey drifts, and each storey’s own factor is its shear over its load times its height over its drift. For the six-storey frame those are 17.4, 12.9, 15.1, 19.9, 29.7 and 56.0 from the bottom up. The storey estimate is the smallest, 12.86, from the second storey, which carries five floors’ load and is held by beams that bend at both its ends.

The first trace is close to the sum of the reciprocals of those factors — close rather than equal, because with flexible beams a storey’s drift under its own shear is not quite the diagonal of $B$. The reciprocals sum to $0.057 + 0.078 + 0.066 + 0.050 + 0.034 + 0.018 = 0.30$, which would give 3.3; the true first trace is 0.195, which gives 5.12, because the diagonal of $B$ is each storey’s drift under its own shear alone, and that is smaller than its drift when every storey’s shear acts at once and rotates the joints it shares with its neighbours. Either way the bound is a third of the answer, and the reason is visible in the list: six storeys whose factors lie within a factor of four of each other add up to more than three times the worst of them.

Checking a number a program prints #

A designer today does not compute a frame’s critical factor by hand; a frame program reports it, from an eigenvalue routine run on a stiffness matrix nobody looks at. What the designer can check is whether the number is plausible, and the usual check is the storey estimate, worked from the first-order drifts the program also prints. On a frame of racking storeys that check is excellent, and on a frame whose storeys lean on each other through their beams — the case the column that leans on its neighbours describes within one storey — it is conservative by the amount the figures show. Neither is a check in the strict sense: the storey estimate can agree with a wrong program and disagree with a right one.

The traces can be a check in the strict sense. The flexibility matrix is six first-order analyses with a unit force at each floor, which any program will run; $K_g$ is the loads and the storey heights; and the bracket from the fourth traces is two numbers between which the true factor must lie. A program that reports 14.96 for the six-storey frame is inside 14.07 to 16.39; one that reports 12, or 18, has modelled something other than the frame — a pinned base where a fixed one was meant, beams released at their ends, a load case with the gravity doubled. The bracket cannot say which, but it cannot be talked out of the fact that something is wrong.

That use needs the upper bound as much as the lower one, which is why the matrix version of the argument is more than the column version with more numbers in it. A column’s traces only bound its buckling load from below. A frame’s bound its critical factor on both sides, from the same products.

A frame, linear, with its load where it is #

The frame is one bay and two-dimensional. A building’s sway in plan includes twist, and a building of many frames has frames that interact through its floors; both make a larger flexibility matrix with more modes, and the traces’ lower bounds only get further from the answer as modes are added. The upper bounds do not.

The gravity load is concentrated at the floors and the columns are axially rigid. Columns that shorten add an overall bending to the sway, which the traces include automatically once it is in the flexibility; leaning columns add load without stiffness and change only $K_g$. Neither changes the argument.

The analysis is elastic. An elastic critical factor is what Eurocode 3’s sway classification is defined by — whether second-order effects may be neglected — and that is the use the bounds are for. A frame near its strength has a lower critical factor than its elastic one, and none of these estimates sees that.

What the traces are for #

The traces of a frame’s flexibility are a lower bound that cannot be wrong and an upper bound beside it, with no mode shape to guess and no eigenvalue routine to run: products of a matrix that a first-order analysis already produces, one column per floor. On a frame whose storeys interact they converge quickly and bracket the critical factor within a few per cent by the fourth trace. On a frame whose storeys are independent and alike — the frame that a storey-by-storey design produces when it succeeds — the first trace is Dunkerley’s sum over the storeys and bounds it at one over their number, while the storey estimate is exact.

So the first two traces are not, on their own, a design tool for a balanced frame. What they are is a check that cannot be fooled: where the storey estimate is exact the traces are poor and safe, and where the storey estimate goes wrong — a core that bends as a whole, storeys that lean on each other — the traces are at their best.

Still open: the frame that buckles while it yields #

Every bound here is on an elastic critical factor, the number Eurocode 3’s classification uses. A frame carrying its design loads has some members near yield, and a member at yield has a tangent stiffness far below its elastic one, so the frame’s real critical factor falls as its load rises — the inelastic second-order problem, where the load makes itself worse through a stiffness that is itself being spent. The traces would bound that factor too, given the tangent flexibility, but the tangent flexibility depends on the state of every member and so on the load path. Whether a frame’s inelastic critical factor can be bracketed at its design load from the traces of a tangent flexibility computed once, or whether the bracket has to be recomputed at every step of the load, decides whether the elastic number on the classification has any meaning near the load the frame is designed for.