The bound from underneath

Guessing the shape found the two properties that make Rayleigh’s quotient a method rather than a guess. The load it returns is always too high, because an assumed shape is a constraint and a constraint stiffens. And the error in the load is the square of the error in the shape, because the quotient is stationary at the true mode. It ended by naming what the quotient does not do: it gives no second, better number from the same shape, and it gives no number from below at all.

Both are available, and neither takes more work than the quotient did. The first is a different way of dividing the same guess. The second is a property of the column that needs no guess whatever. Together they turn a single number that is always too high into a bracket with the answer inside it — and the bracket matters, because the place a critical load is most often used is inside the amplifier $1/(1 - P/P_{cr})$, where a load that is too high is on the unsafe side.

One guess, divided two ways #

A pin-ended column buckled into a shape $v(x)$ carries a bending moment $M = Pv$ at every section, because the axial load acts at the ends and the section is displaced $v$ from the line between them. Rayleigh’s quotient ignores that relation. It takes the guess’s own curvature, $v''$, and prices it as strain energy $\tfrac12\int EI (v'')^2$. But the guess’s curvature is only as good as the guess, and a guess wrong in its second derivative — the parabola, whose curvature is the same at the pins as in the middle — is priced wrongly where it is most wrong.

The alternative is to price the moment instead. The strain energy of a moment $M$ is $\tfrac12 \int M^2/EI$, and with $M = Pv$ that is $\tfrac12 P^2 \int v^2/EI$. Set it equal to the work the load does going down, $\tfrac12 P \int (v')^2$, and the load is

$$P_T = \frac{\displaystyle\int_0^L (v')^2\,dx}{\displaystyle\int_0^L v^2/EI\,dx},$$

Timoshenko’s quotient. It needs the guess and its slope and never its curvature. The curvature it uses is $Pv/EI$ — what the guess’s own moment would cause in the column as it is actually built — and that curvature is automatically zero at a pin, automatically large where the column is soft, and automatically discontinuous wherever the rigidity is.

The parabola is the telling case. Its Rayleigh quotient is 12 $EI/L^2$, and its Timoshenko quotient is exactly 10 — from a shape that needed no more thought than “zero at the ends, largest in the middle”. The mid-span sag improves from 10.000 to 9.882, and the uniform-load sag, already only an eighth of a per cent out, to a hundredth.

It is also still an upper bound, and it is never worse than Rayleigh’s quotient from the same shape. Both facts come from one inequality. Write the three integrals $a_0 = \int EI (v'')^2$, $a_1 = \int (v')^2$ and $a_2 = \int v^2/EI$; integrating $a_1$ by parts gives $\int (-v'')\,v$, and the Cauchy–Schwarz inequality applied to $\sqrt{EI}\,v''$ and $v/\sqrt{EI}$ says $a_1^2 \le a_0 a_2$. That rearranges to $a_1/a_2 \le a_0/a_1$ — Timoshenko’s quotient under Rayleigh’s — with equality only when $EI\,v''$ is proportional to $v$, which is the buckling equation itself.

Where each quotient puts its error #

The figure is the whole explanation, and it is worth reading before the algebra. The parabola’s own curvature is a flat line: as much bending at the pins as at mid-height. The true mode has none at the pins, because a pin carries no moment, and all of its curvature concentrated where the moment is. Rayleigh’s quotient charges the column for bending that does not happen, at exactly the places where the charge is largest in proportion. The implied curvature, $Pv/EI$, is the parabola itself turned over — zero at the pins, largest in the middle — and it is so close to the true mode’s curvature that the two lines nearly coincide.

This is what the essay on guessed shapes meant by weighting the error where the moment is. Stated more exactly, Rayleigh’s quotient asks the guess to be right twice differentiated, and Timoshenko’s asks it only to be right. A shape is always better than its second derivative, so an error that enters the load as the square of the curvature’s error enters it, in the second quotient, as roughly the square of the shape’s.

The sequence the two quotients start #

The two quotients are the first two terms of a sequence, and seeing the sequence is what turns them into a method. Start with the guess $F_1 = v$. The moment it implies, over its rigidity, is a curvature; integrate that curvature twice with the column pinned at both ends and the result, $F_2$, is the shape the guess’s own moment would bend the column into. Do it again for $F_3$, and so on. Each step is a deflection calculation of the kind a unit-load integral does, and each hands on a shape that is nearer the mode than the one it came from, because the operation that produces it — the column’s flexibility — amplifies the first mode by more than any other.

The quotients between consecutive integrals of the sequence, $\mu_k = a_{k-1}/a_k$, are Rayleigh’s for $k = 1$, Timoshenko’s for $k = 2$, and a further upper bound for every $k$ after that, each no larger than the last. These are Schwarz’s quotients, and iterating a guessed shape through the structure’s flexibility to find a mode is the Stodola–Vianello method — which is how buckling loads and natural frequencies of non-uniform members were computed by hand for the first half of the twentieth century.

The parabola’s sequence runs 12.000, 10.000, 9.882, 9.871, and the third term is exactly the Rayleigh quotient of the uniform-load sag shape. That is not a coincidence, and it explains the advice the essay on guessed shapes gave without deriving it — that the best free guess is the column’s own deflected shape under a transverse load. Feed a parabola through the column once and it becomes the deflection under a parabolic moment, which is the uniform-load sag. The sag shape is the parabola after one iteration, and it is good for the same reason the iteration converges.

A number from below #

Every quotient in the sequence is too high. To say how much too high, a number from the other side is needed, and the first one that comes is Temple’s inequality. From any two consecutive quotients $\mu_k$ and $\mu_{k+1}$ it gives

$$\lambda_1 \ \ge\ \mu_{k+1} - \frac{\mu_k - \mu_{k+1}}{\ell_2/\mu_{k+1} - 1},$$

where $\ell_2$ is any number known to be below the second critical load. The gap between two consecutive quotients measures how much of the other modes the current shape still contains, and $\ell_2$ says how far away those modes are; together they fix how far the next quotient could still fall.

For the uniform column $\ell_2$ is easy: the second mode of a pinned strut is two half-waves at $4\pi^2 = 39.48$ $EI/L^2$, four times the first. With the parabola’s first pair, 12 and 10, Temple gives 9.322; with its second pair, 9.843. One iteration of a parabola brackets the load between 9.843 and 9.882, a band of 0.4 per cent that contains $\pi^2$.

What Temple needs is the catch. A lower bound on the second load is a statement about the column that is as hard to make, in general, as a statement about the first — and a poor one makes the bound poor. For a non-uniform column the honest $\ell_2$ is the second load of a uniform column as soft as the softest section, $4\pi^2 EI_{min}/L^2$, and when the softest section is much softer than the rest that number can come down to meet the first load, at which point the formula divides by nearly nothing and returns nothing useful.

A lower bound with no guess in it #

There is a second route from below that needs neither a guess nor a second mode, and it is older than it looks.

A pin-ended column’s flexibility — the deflection at one point from a unit lateral load at another, over the column’s rigidity — is an operator whose eigenvalues are the reciprocals of the critical loads, $1/\lambda_1, 1/\lambda_2, \dots$, all positive. The trace of that operator, the sum of its diagonal, is the sum of all of them, and it is a single integral: $\int_0^L g(x,x)/EI(x)\,dx$, where $g(x,x) = x(L-x)/L$ is how far a pinned span deflects under its own unit load. Since the sum of all the reciprocals is at least the largest of them, $1/\lambda_1$, the trace’s reciprocal is a lower bound on the first critical load. The trace of the flexibility’s square is the sum of the reciprocal squares, a double integral, and its reciprocal square root is a tighter bound; the cube’s is tighter again.

For the uniform strut the first trace is $\int_0^L x(L-x)/L\,dx$ over $EI$, which is $L^2/6EI$, and the bound it gives is 6 $EI/L^2$. It is 39 per cent low, and it has a name: it is Dunkerley’s rule, the estimate that sums reciprocals and errs low, which the essay on periods met in its original setting, the whirling of shafts. Dunkerley’s rule is the first trace of a flexibility.

The surprise is in what the first trace equals. The critical loads of the uniform strut are $n^2\pi^2 EI/L^2$, so the trace is also $\sum L^2/(n^2\pi^2 EI)$. Setting the two expressions equal gives

$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6},$$

which is the Basel problem, solved by Euler in 1735 — nine years before he published the buckling load of a strut. The second trace gives $\sum 1/n^4 = \pi^4/90$ and a lower bound of $\sqrt{90}$ = 9.487, 3.9 per cent low; the third $\pi^6/945$ and 9.813; the fourth $\pi^8/9450$ and 9.859. The lower bounds on Euler’s column are Euler’s sums, and a structural engineer who computes the second trace of a pinned strut’s flexibility has evaluated $\zeta(4)$ without meaning to.

For the uniform column nothing here is needed, since the answer is known. What the traces offer is that the same integrals exist for every column, whatever its rigidity does along its length, and none of them assumes anything about the mode.

A column the smooth guess cannot see #

An average stiffness is not a safe stiffness found that buckling weights a column’s rigidity by the square of the mode’s curvature, so the middle of a pinned column decides almost everything and its ends almost nothing. That is true of the mode the column would have if its rigidity were uniform, and it is exactly the assumption a smooth guess makes. Make the ends soft enough and the mode itself changes, and a smooth guess cannot follow it.

Take a pin-ended column whose outer quarters keep a tenth of the middle’s rigidity — a column spliced to lighter sections at both ends, or a member whose ends have been cut down for a connection over a quarter of its length each. Its exact critical load is 3.225 $EI_0/L^2$, a third of the uniform column’s.

The true mode concentrates its curvature in the soft quarters, ten times more of it per unit moment than in the middle, and the half sine, whose curvature is smooth and largest where the column is stiff, charges the column for bending in the wrong place entirely. Rayleigh’s quotient from it is 8.256 $EI_0/L^2$. It is 156 per cent high, on a column where the half sine is a visually reasonable picture of the buckled shape — the mode is bent more sharply in the soft quarters, but it is still a single bow, zero at the pins and largest at mid-height.

Timoshenko’s quotient from the same half sine is 3.745. Its curvature is the moment over the rigidity, which jumps where the rigidity does, and the jump is the feature of the true mode that matters.

Every guess fails Rayleigh’s way and nearly recovers Timoshenko’s. The order also changes: on a uniform column the parabola is the worst of the four shapes and here, by both quotients, it is the best. The reason is the one already found on a tapered column, where the worst guess for a prismatic strut became the best. A parabola’s curvature is spread evenly along the column, which is less wrong when the column’s own curvature is spread toward its ends.

This is also where the rule that the load’s error is the square of the shape’s has to be read carefully. It is true near the answer. It is not a promise that a reasonable-looking shape is near the answer, and here a reasonable-looking shape is not.

The bracket on a column with no closed form #

The same column, with the sequence carried on and both lower bounds beside it:

The quotients still fall fast, but Temple’s bound is poor, and the reason is visible in the number it was given: a floor of 3.95 under the second load, when the first is 3.225, leaves almost no gap for the formula to work with. Temple is only as good as what is known about the next mode, and for a column with a very soft part what is honestly known is very little.

The traces do not have that problem. For this column the first six give 1.574, 2.833, 3.088, 3.171, 3.202 and 3.215 $EI_0/L^2$, rising steadily onto 3.225 with no second mode anywhere in the calculation. The fourth, 3.171, is 1.7 per cent low.

Across the whole range the bracket holds the answer, and it is narrow where it matters least and widest where the column is strangest. With half the rigidity gone from the outer quarters the load is caught between 8.246 and 8.285, a quarter of a per cent either side. With nine tenths gone the band has widened to 3.171–3.492, but it is still a statement with a floor, and its floor is 1.7 per cent under the answer. The half sine’s Rayleigh quotient, meanwhile, has hardly moved: it reads 8.97 when the column has lost half its end rigidity and 8.26 when it has lost nine tenths, because its curvature barely touches the quarters where the loss is.

The half sine has not been told about the column. That is the whole of what went wrong, and the Timoshenko quotient repairs it at no cost, because the rigidity enters it where the moment does.

Which end goes in the amplifier #

A critical load is rarely wanted for itself. It is wanted as the denominator of the amplifier a sway frame multiplies its moments by, $1/(1 - P/P_{cr})$, and there the direction of the error decides the sign of the mistake.

At 80 per cent of the true critical load the moments in this column are amplified five times. A designer who took the critical load from a half sine by Rayleigh’s quotient would compute an amplifier of 1.45 and design the column for less than a third of the moment it will carry. Timoshenko’s quotient, 8 per cent high, gives 3.83, still a quarter short. Every upper bound on a critical load is a lower bound on an amplifier. Only the floor of the bracket, 3.171, gives an amplifier that is safe: 5.36 against 5.00, 7 per cent conservative.

This is the practical reason the lower bound is worth the trouble. An energy method is usually trusted because its error has a known direction, and that is right — but the direction is the wrong one for the calculation that most often consumes it. A strength check against a buckling load wants the load low. So does a stability index, a sway classification, and every second-order moment. The upper bound serves none of them.

The integrals, by hand, for the uniform strut #

The whole argument can be checked on the uniform pin-ended strut with nothing but polynomial integrals. For the parabola $v = x(L-x)$: $v' = L - 2x$, so $a_1 = \int_0^L (L-2x)^2 dx = L^3/3$; $v'' = -2$, so $a_0 = 4EI\,L$; and $a_2 = \int_0^L x^2(L-x)^2 dx / EI = L^5/30EI$. Rayleigh’s quotient is $a_0/a_1 = 12\,EI/L^2$ and Timoshenko’s is $a_1/a_2 = 10\,EI/L^2$.

Temple’s bound from the pair, with $\ell_2 = 4\pi^2 = 39.48$: $10 - (12 - 10)/(39.48/10 - 1) = 10 - 2/2.948 = 9.322$. The traces: $\int_0^L x(L-x)/L\,dx = L^2/6$, so the first bound is 6; the second is the double integral of the square of the flexibility, $\int\!\!\int g(x,\xi)^2 = L^4/90$, and the bound $\sqrt{90}$ = 9.487. Between 9.487 and 10, from arithmetic that fits on one side of a sheet, lies $\pi^2$.

Pinned ends, a conservative load and an elastic column #

Both ends are pinned. Timoshenko’s quotient uses $M = Pv$, which is the moment only when nothing else acts on the column: with a built-in end the moment also carries the unknown end moments, which have to be found before the quotient can be written, and the sequence then needs the column’s flexibility with those ends. The traces carry over to any support, but the kernel $g(x, \xi)$ becomes that support’s own deflection function.

The rigidity is elastic and known. A step is drawn here as a clean change of $EI$ at the quarter points; a real splice is a short stiff region with bolts that can slip, and a cut-down end has a stress concentration the section rigidity does not see. The column is also taken as perfectly straight, so the critical load is a bifurcation and a bowed column never reaches it.

The load is conservative and axial. The quotients all come from a work balance, which a follower load does not have.

What the pictures cannot show #

That the traces are exact integrals and not approximations of one. Each bar in the trace figure is a definite integral of the column’s own flexibility, computed by quadrature to more figures than are printed. The fact that each is below the load is a theorem, not a trend.

The figures also cannot show how rarely Temple’s floor on the second mode is available in practice. The uniform column’s is known in closed form, and the non-uniform column’s is found only by comparing it with a uniform column as soft as its softest part — which is a bound, and a poor one exactly when the column is interesting.

Still open: whether a trace can be taken of a frame #

Every trace here belongs to one member, whose flexibility is a single function of two points. A sway frame’s stability is decided by all its columns and beams together, leaning columns included, and its flexibility is a matrix over its storeys whose entries are themselves deflections of a structure with rotating joints. The traces of that matrix would bound the frame’s elastic critical load factor from below with no assumed sway shape — which is the number a code’s sway classification actually asks for, and which is usually estimated from above by a single 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, is the question that comes after this one.