The formula that reads four numbers

The shape of the diagram, and not its peak found why a beam’s lateral-torsional capacity depends on the shape of its moment diagram: the buckling condition weighs the square of the moment against the twist the beam wants to make, so a diagram that is large only over a short length destabilises less than a flat one of the same peak. The factor that records the benefit, the moment-gradient factor $C_1$, ran from 1.0 for uniform moment to about 2.7 for equal and opposite end moments. The restraint that beats the gradient then found a detail worth more than any gradient.

Both computed $C_1$ from the buckling problem. Almost no designer does. The factor is taken from a formula — in North American practice the one that reads the diagram’s peak $M_{\max}$ and the magnitudes $M_A$, $M_B$ and $M_C$ at the quarter point, mid-span and three-quarter point of the unbraced segment:

$$C_b = \frac{12.5\,M_{\max}}{2.5\,M_{\max} + 3M_A + 4M_B + 3M_C}$$

It is short, it handles any diagram, and it is a fit: its coefficients were chosen so that it reproduces the buckling result for the diagrams someone checked. The question this essay asks is the one every fit invites. What does it do on the diagrams nobody checked?

Four readings, and what they miss #

The figure below is the moment diagram of a beam fixed at both ends against rotation in its plane and carrying a single load at mid-span. Its end moments are $-PL/8$, its mid-span moment $+PL/8$, and it passes through zero at exactly the quarter points. The formula reads the peak, 1.0 on the scale drawn; zero at the quarter point; 1.0 at mid-span; zero at the three-quarter point. It returns $12.5/(2.5 + 0 + 4 + 0) = 1.923$.

The buckling problem for the whole diagram, for a beam 8 m between lateral restraints with forked ends, gives 1.723. The formula is 12 per cent too high, and too high is the unsafe side: it credits the beam with a critical moment it does not have.

The reason is in what the formula cannot see. A reading of zero at the quarter point tells it that the moment is small there, and it infers that the moment is small near there. For this diagram it is not: the moment rises linearly to its full value at the ends, a quarter of a span away. The two end regions each carry the peak moment, and they carry it with the sign that puts the bottom flange in compression — which the squared moment in the buckling condition treats exactly as it treats the top. The formula reads two zeros and believes in a diagram much kinder than the one that is there.

On the diagrams it was built around #

Start where the formula is expected to be good. A beam with end moments and no load between them has a straight-line diagram, described by the ratio $\psi$ of its end moments. For those diagrams the exact factor runs from 1.0 at $\psi = 1$ to 2.72 at $\psi = -1$, and the quarter-point formula is below it everywhere — by 5 per cent at $\psi = 0.5$, by 9 per cent at $\psi = 0$, by 21 per cent near $\psi = -0.8$. The older quadratic in $\psi$, $1.88 - 1.40\psi + 0.52\psi^2$ capped at 2.7, follows the exact curve more closely, because it was fitted to exactly this family and to nothing else.

So on linear diagrams the quarter-point formula is conservative, sometimes generously. Its reputation as a safe general-purpose fit is earned here, and it is earned on a family in which the moment at the quarter points is always between the moments at the ends. That is the property the fixed-ended beam lacks.

A load with end moments, growing #

The fixed-ended beam belongs to a family: a central point load with equal end moments added, from none — the simple beam — to end moments as large as the load’s own simple-span moment. Walk the family and the formula’s error moves through a pattern with a single cause.

With small end moments the diagram is still a triangle, the quarter points still read half the peak, and the formula is close. As the end moments grow, the quarter-point readings fall toward zero, and the formula, reading them, grows more optimistic faster than the true factor rises. It crosses above the truth, peaks at 12 per cent too high exactly when the end moment is half the simple-span moment — the fixed-ended beam, with its zeros at the quarter points — and then, as the end moments grow past it and the quarter-point readings rise again with the opposite sign, falls back below, becoming 39 per cent conservative at three quarters.

The formula is unsafe exactly where its sampling points land on the diagram’s zeros, and the fixed-ended beam under a central load puts them there precisely. It is not an unusual beam. It is the textbook’s second case after the simply supported one, and every continuous beam’s interior span under a concentrated load looks like it once its support moments have been shared out.

A uniform load makes the point from the other side. Its family has an unsafe band too, peaking at 6 per cent where the end moment is half the simple-span moment and the quarter-point readings are small. But the fixed-ended beam under a uniform load has its zeros at about a fifth of the span, not at the quarter points, so the formula’s readings there are small and non-zero, and it is 9 per cent conservative. The same boundary conditions give a safe answer or an unsafe one according to where the zeros fall relative to the sampling points — which is a fact about the formula, not about the beam.

Two diagrams, one reading #

The sharpest way to see what a sampled formula is blind to is to find two diagrams it cannot tell apart. The fixed-ended beam’s diagram is one. A smooth wave through the same four values — hogging at the ends, sagging at mid-span, zero at the quarter points — is the other. The formula reads identical numbers from both and gives both 1.923.

Their exact factors are 1.723 and 1.409. The smooth wave has more of its length at large moment than the triangle does — its magnitude is near its peak over a wider band around the ends and the middle — and the squared moment in the buckling condition charges for that. The information that separates them lies entirely between the points the formula reads, and it is worth 22 per cent of the critical moment.

No sampled formula can escape this. Any rule that reads a function at a fixed set of points will give one answer for every function through those points, and functions through the same points can have integrals — which is what the buckling condition is — that differ by as much as their shape allows. The quarter-point formula’s four readings were a good choice for the diagrams it was fitted to, because on those diagrams the readings determine the shape. They do not determine it in general.

A hundred and sixty diagrams #

The fixed-ended beam is the canonical case, and a design rule should be judged over the whole range of diagrams it will be given. The figure gives it one hundred and sixty — smooth curves through random values at the ends and the quarter points, some single-signed, some reversing, some peaked near the ends and some in the middle — and compares the formula with the buckling solution for each.

Most of them the formula handles conservatively, some very conservatively. But eighteen, 11 per cent, are more than 5 per cent on the unsafe side, and the worst is 21 per cent. Fourteen of those eighteen share the fixed-ended beam’s feature: at one of the three interior points the formula reads, the diagram is within 15 per cent of zero. Among the diagrams where the formula is safe, only 28 per cent have such a reading. A near-zero reading at a quarter point is the warning sign, and it is visible on the diagram before any calculation is made.

Why a fit, and what it replaced #

The formula exists because the thing it replaces could not be done by hand. The buckling condition for an arbitrary diagram is an eigenvalue problem, and before computers made that trivial a designer needed a number from a table or a line of arithmetic. The first widely used answer was a quadratic in the end-moment ratio — for straight-line diagrams only, because only they have a ratio. The quarter-point formula was published in 1979 to cover the rest: a beam carrying loads between its restraints, whose diagram is curved and has no single ratio to describe it.

Its authors did what a fit should do. They took a set of diagrams — linear ones, and a range of loaded spans — computed the true factor for each, and chose coefficients that matched them closely and erred to the safe side. On that set the formula is excellent, which is why it has survived four decades and a change of code. Nothing about the method is wrong. What it cannot do is extend its guarantee to a diagram outside the set, because a fit is a statement about the cases it was fitted to, and a formula that is conservative on every case its authors checked has made no promise about the cases they did not.

A beam that fails sideways is the one failure in steel design where the resistance depends on the load’s distribution as well as its size, and that is why the gradient factor needs a formula at all. Every other capacity check compares a peak demand with a resistance. This one compares a shape — and a shape cannot be reduced to four numbers without losing something, which the fixed-ended beam happens to put in exactly the wrong place.

The same blindness elsewhere #

Reading a function at a few points and inferring its integral is not peculiar to this formula; it is what every quadrature rule does. Simpson’s rule on the same five points — the ends, the quarter points and mid-span — would weight them 1, 4, 2, 4, 1 and would be exact for any cubic. The fixed-ended diagram is not a cubic; it has a kink at mid-span where the load is. A rule sampled at fixed points is exact for the functions its points can resolve and blind to the rest, and a kinked diagram with zeros at the sampling points is precisely what a rule of five points cannot resolve.

The same bargain has appeared twice before, in other clothes. Equivalent in work, not in resultant found that a load replaced by nodal forces reproduces some quantities exactly and others not at all, depending on what the replacement was chosen to preserve. And the answer that depends on how it was divided found a finite-element result that moved with the mesh, because a few sampled values stood in for a field. The quarter-point formula is the same bargain struck in 1979, with the sampling points fixed for ever.

Where the error comes from, in the buckling condition #

The buckling condition for a doubly symmetric beam with forked ends balances two integrals over the length: the twist’s own stiffness, warping and St Venant, against $\int M(x)^2 \varphi(x)^2\,dx / EI_z$, the squared moment weighted by the square of the twist the beam buckles in. The twist $\varphi$ is largest at mid-span and zero at the forked ends.

A formula that reads $M$ at four points is estimating that weighted integral from four samples. The weights it uses — 3, 4, 3 on the quarter points and mid-span, 2.5 on the peak — are sensible for a twist shaped like a half-sine and a diagram that varies slowly. What they cannot represent is a diagram that changes sign between samples, because a squared moment has no sign: the region near a zero reading contributes almost nothing if the diagram really is small there, and a great deal if it passes through zero and rises again steeply, as the fixed-ended beam’s does. The formula has to guess which, and for a diagram whose samples land on its zeros it guesses the kind one.

Look at the diagram before reading it #

The first remedy is to look at the diagram before using the formula on it. If the moment is near zero at any quarter point of the unbraced segment, the formula’s answer is suspect, and the suspicion runs toward the unsafe side. Fixed-ended and continuous spans under concentrated loads are the commonest cases; so is any segment containing a point of contraflexure near a quarter point.

The second is to use the answer that does not sample. The buckling eigenvalue for an arbitrary diagram is a small calculation — a handful of terms in a sine series, which is how every exact factor on this page was found — and it is what the formula was fitted to reproduce. Where software computes it, the formula is not needed. Where it does not, taking $C_1 = 1$ for a segment whose diagram reverses near a quarter point costs a few per cent and removes the question.

The third is to move a restraint. A lateral restraint at the point of contraflexure, or at the load, divides the segment into pieces whose diagrams are nearly linear — the home ground of every formula — and each piece is shorter. It is the move the restraint that beats the gradient recommended for a different reason, and here it has two — provided the restraint holds the compression flange, which a brace on the wrong flange found is the whole of what a beam’s brace is for, and at a point of contraflexure the compression flange changes sides.

The whole of it, once, for the fixed-ended beam #

The beam is fixed in its plane at both ends and carries $P$ at mid-span, so its end moments are $-PL/8$ and its mid-span moment $+PL/8$; the moment is linear between, passing through zero at $L/4$ and $3L/4$. Laterally the beam is restrained only at its ends, where it is free to rotate on plan and to warp.

The formula’s four readings, in units of $PL/8$, are 1, 0, 1, 0, and it gives $12.5 \times 1/(2.5 + 0 + 4 + 0) = 1.923$. The buckling problem, solved in twelve sine terms for a 457 mm universal beam spanning 8 m, gives 1.723. A design check using the formula would credit the beam with 1.923/1.723 = 1.116 times its true elastic critical moment. On a slender beam whose resistance is governed by that moment, that is an 11.6 per cent overestimate of capacity, from a formula whose inputs were read correctly off a correct diagram.

The buckling problem behind every exact factor #

The free body is the beam between its lateral restraints, forked at both ends — free to rotate on plan and to warp, held against twisting — carrying the in-plane moment diagram of each case. The buckling condition is the energy balance above, solved as a generalised eigenvalue problem with the twist written as a sum of twelve half-sine terms; the factor is the critical peak moment divided by the critical uniform moment of the same beam. The same routine reproduces the built-in factors for uniform load, 1.13, and central point load, 1.36, which every figure checks before it draws, and each diagram is scaled so its largest magnitude is one before the formula reads it.

Load height, monosymmetry and other formulas #

Load height. Every case here has its load at the shear centre, and a load applied above it moves with the twist. A load on the top flange is destabilising and one on the bottom flange stabilising, and the formula has a separate treatment for that which interacts with the gradient in ways this page does not test.

Monosymmetric sections. For a section with unequal flanges, the sign of the moment matters, because it decides which flange is in compression; the squared-moment argument above does not hold, and the fixed-ended beam’s hogging ends load the other flange.

Other formulas. Codes that use a formula based on a square root of the sum of squared readings are closer on the fixed-ended beam and farther on others; every sampled formula has diagrams it cannot see, and which ones depends on where it samples.

Forked ends, and the same ends for both answers #

That the segment is the whole beam between its end restraints, with forked ends. A fixed-ended beam is fixed in its plane; whether it is also fixed against lateral rotation and warping at its ends is a separate question, and if it is, the buckling problem changes and so does the exact factor. The comparison on this page is between the formula and the buckling problem for the same boundary conditions, which is the comparison the formula is used to make.

Still open: the cantilever, where the tables disagree #

Every segment here is held at both ends. A cantilever is held at one, and its critical moment depends on what the root allows — warping, lateral rotation, both or neither — and on whether the tip is restrained at all. There the gradient formulas are not merely inaccurate; the published tables disagree with one another by factors of two for what appear to be the same case, because each tabulates a different root. Which root a real cantilever has, and how far its critical moment moves between the boundary conditions the tables choose among, is the question after this one.