What the frames hold is the bow

A row of frames is not a foundation took the compression chord of a half-through girder, held sideways by U-frames at intervals, and asked when it would buckle. The frames act as a smeared foundation only while they are closer together than about two-thirds of the half-wavelength the smeared model predicts; beyond that the chord buckles between frames, and frame stiffness buys nothing. Every number there was a critical load, the force at which a perfect chord first moves.

A frame is not sized by a critical load. It is sized by a force, and the force it carries comes from the chord not being straight: it arrives with a bow of a few millimetres, the compression grows the bow as every imperfection grows under the load that finds it, and the frames push back. The design rules give that force as a percentage of the chord force. How large is it really, and where along the range of frame stiffness is it largest? The earlier essay guessed the crossover, where the chord is half using the frames. The calculation says the opposite.

The chord, the frames and the bow #

The chord is the one the earlier essay used: 24 metres between the ends of the girder, flexural rigidity $2.52 \times 10^{12}$ N·mm² sideways, held by five U-frames at 4-metre spacing, each a spring of 1,400 N/mm. It carries 930 kN. Its critical load with these frames is 1,548 kN, 1.66 times the working force.

An eigenvalue has one shape. An imperfection can have any, so the calculation tries many: bows of one to twelve half-waves along the chord, each as high as a five-hundredth of its own half-wavelength — 48 mm for a single arch over the whole 24 m, 12 mm for a bow of four half-waves, 4 mm for one of twelve. For each, the chord is solved to second order: the bow is stress-free, the frames are fitted to it, and under the compression the chord moves further, by an amount that grows as the compression approaches the load at which that shape would buckle. Each frame then carries its stiffness times the chord’s extra movement at the frame.

The worst bow for these frames has four half-waves, each six metres long, and the compression roughly doubles it. Four of the five frames stand near a crest and push back with 13.6 kN each. The middle frame stands at a node of this particular bow and carries nothing — though another bow, of three half-waves, puts a crest on it and loads it to 12.5 kN. The frame force is 1.47 per cent of the chord force.

Which bow the frames feel #

The bows separate into three kinds. A long bow — one or two half-waves over the whole chord — is large but gentle: its curvature is small, the frames can hold it with little force, and it produces well under one per cent. A short bow of seven or more half-waves, shorter than a bay, lives mostly between the frames: the chord bends between its supports and the frames feel only a little of it.

And one bow produces no frame force at all, whatever the compression. A bow of six half-waves has its nodes exactly at the five frames. The chord moves, and grows, and could buckle in that shape, but at every frame its movement is zero, so no frame is asked for anything. It is the shape in which the chord buckles once the frames are stiff enough — the between-frames mode of the earlier essay — and it is invisible to the frames by construction.

The worst bows are the ones in between: a few half-waves, each longer than a bay, so that the frames stand on their flanks and crests. For soft frames the worst has three; as the frames stiffen it shortens, to four and then five, the longest bow short of the one the frames cannot see.

The obvious bow is the harmless one #

The calculation a designer would most naturally do is with one bow: the chord bent into a single arch over its whole length, 48 mm at mid-span, the shape a column curve assumes. It gives the frames 0.34 per cent of the chord force. The two-wave bow gives 0.68, the three-wave 1.34, and the worst, four waves, 1.47 — more than four times the single arch.

The single arch is harmless for the frames for the same reason it is dangerous for an unrestrained column: it is long. Its curvature is small, so the compression presses on the frames gently, and the frames have raised the load at which that shape would buckle so far above the working force that it hardly grows. The shapes that load the frames are the ones the frames make likely — shorter waves, near the half-wavelength the frames themselves select. The imperfection that matters for a restraint is the one shaped like the buckle the restraint allows, not the one shaped like the buckle it prevents, and a single-bow calculation looks at the wrong one.

The force is least where the eigenvalue is most interesting #

Sweep the frame stiffness and the envelope has a simple shape with one surprise in it. Frames that are only just stiff enough carry the most: at a stiffness that gives the chord a critical load 1.13 times its working force, 5.55 per cent of the chord force. Stiffen them and the force falls quickly, to a minimum of 1.38 per cent at about 1,900 N/mm — which is almost exactly the stiffness at which the chord’s buckled shape switches from waves spanning several frames to waves between them. Stiffen them further and it rises again, slowly, toward 1.64 per cent.

So the crossover is not where the frames are most loaded. It is where they are least. The eigenvalue makes the crossover look critical because it is where frame stiffness stops buying critical load; the frame force does not care about that. It is governed by two different things at the two ends of the range, and the crossover is simply where one hands over to the other.

Two regimes, two reasons #

At the soft end the force is amplification. A bow in the shape the chord buckles in grows as one over one minus the ratio of the compression to the critical load, and frames that barely hold the chord leave that ratio close to one. The frames are soft, so each carries only its stiffness times the chord’s movement — but the movement is large, and grows without limit as the frames soften toward the stiffness at which the chord cannot carry its force at all.

At the stiff end the force is the bow itself. Rigid frames do not let the chord move at them, so nothing is amplified there, and each frame carries the force needed to hold the chord’s bow in place against the compression. That force can be written down by hand. A bow of height $e$ over a half-wavelength $\ell$ has a curvature at its crest of $\pi^2 e/\ell^2$, and a compression $P$ acting on a curved member presses sideways with $P$ times its curvature per unit length. A frame collects a bay’s worth of that, so

$$F \approx P\,\frac{\pi^2 e}{\ell^2}\,a = P\,\frac{\pi^2}{500}\,\frac{a}{\ell}, \qquad e = \frac{\ell}{500}.$$

The worst bow the frames can feel is the shortest one whose nodes are not all on frames — five half-waves in six bays, $\ell = 1.2a$ — so the stiff-frame limit is $\pi^2/500 \times 5/6 = 1.64$ per cent of the chord force, whatever the chord’s stiffness, the frames’ spacing or the force itself. With many frames the fraction $a/\ell$ tends to one and the limit to $\pi^2/500$: 1.97 per cent.

That is a number with a history. Rules for the strength of a brace have for a long time been written as one or two per cent of the force in the member braced, stated rather than derived. Here two per cent arrives from one assumption — a bow of a five-hundredth of its own length — and the geometry of curvature. The rule is the stiff-frame limit of this calculation, and it is right for stiff frames and wrong, by up to a factor of three on the unsafe side, for frames just stiff enough.

The four-wave bow, by hand #

The drawn case can be reproduced with a pencil, and doing it shows which parts of the calculation matter. Treat the frames as the smeared foundation of the first essay on this chord, $k = K/a = 0.35$ N/mm per millimetre. A shape of $m$ half-waves then buckles at

$$P_m = EI\left(\frac{m\pi}{L}\right)^2 + k\left(\frac{L}{m\pi}\right)^2,$$

the chord’s own bending resistance plus the foundation’s, one rising with $m$ and the other falling. For four half-waves the two terms are 691 kN and 1,277 kN, so $P_4 = 1{,}968$ kN. The working force of 930 kN is 0.47 of that.

A bow already in that shape grows under the compression by the factor $(P/P_m)/(1 - P/P_m)$, which here is 0.90: the 12 mm bow gains 10.8 mm at its crests. The frames do not stand exactly on the crests. A four-wave bow has its crests at 3, 9, 15 and 21 metres, and the frames at 4, 8, 16 and 20 metres, where the bow is 0.87 of its crest height; there the chord has moved 9.3 mm further, and a frame of 1,400 N/mm pushes back with 13.0 kN. The full calculation gives 13.6.

The difference is the smearing. Discrete frames are a little less stiff against a four-wave shape than the smeared foundation says — the chord bends between them — so the true $P_4$ is lower and the amplification a little higher. Everything else in the hand calculation is the full calculation’s own logic: a bow, a buckling load for that bow’s shape, an amplification, and a spring.

It also shows why the soft end climbs so steeply. As the frames soften, $P_m$ falls toward the working force for the worst shape, the amplification factor $1/(1 - P/P_m)$ climbs without bound, and the frame force follows it while the frames’ own stiffness falls only in proportion.

A share that depends on how hard the chord works #

The same point seen from the other axis. Keep the frames at 1,400 N/mm and vary the chord force. At a fifth of the chord’s critical load the frames carry 0.95 per cent of it — nearly the bow-holding force, since the compression amplifies little. At half, 1.20 per cent. At eight-tenths, 2.35 per cent, and rising without limit toward the critical load.

A frame force quoted as a fixed percentage of the chord force is therefore a statement about how close to buckling the chord is allowed to work, made without saying so. A chord designed to work at a large fraction of its critical load needs frames sized for much more than a chord that works at half of it, and a single percentage covers one of those cases and not the other. The quantity that governs is the ratio of the critical load to the working force, which the bracing essays met as the brace stiffness demanded; here it reappears in the brace strength.

What to size the frames for #

The picture suggests a way of choosing the frames that the eigenvalue alone does not. The earlier essay showed that stiffness beyond the crossover buys no critical load, since the chord then buckles between frames at a load no frame can raise. This one shows that stiffness beyond the crossover buys no reduction in frame force either: the force is at its least there and creeps back up toward the rigid-frame limit. The crossover stiffness is where both the critical load and the frame force are as good as they will get, so there is nothing to be gained by stiffer frames and a great deal to be lost by softer ones.

At that stiffness the frames carry about 1.4 per cent of the chord force, a little under the two per cent of the stiff limit. Softer frames save material in the frames and pay for it twice: in critical load, which falls, and in frame force, which climbs — to 1.65 per cent at half the crossover stiffness, 2 per cent at three-eighths and 5.5 per cent at two-ninths. A frame designed for two per cent of the chord force is therefore safe only if it is also at least about three-eighths as stiff as the crossover demands, and the stiffness and the strength have to be checked together, since a percentage rule for one says nothing about the other. It is the same pairing that decides whether a beam’s brace works: a restraint is a stiffness and a strength, and a member can be strong enough and still fall over when the first is missing.

None of this needs anything more elaborate than a designer already has. The envelope is one factorisation of the chord’s stiffness and a dozen right-hand sides, one per bow; the hand version is one buckling formula and one amplification factor per shape. What it needs is the decision to try more than one shape, and to read the frame force off the worst of them rather than off the one that looks most like a column.

Every frame is somebody’s worst #

A last consequence for design. No single bow loads every frame to its worst. The four-half-wave bow loads four frames and misses the middle one; the three-half-wave bow loads the middle one and the two nearest the ends less. Each frame’s design force comes from its own worst shape, and they come out nearly equal along the chord: 13.6 kN at four frames, 12.5 at the middle. The frames nearest the girder’s ends, where the chord force of a real girder is smaller and a designer might reduce them, are as heavily loaded here as the middle ones, because a bow of several half-waves puts a crest beside every frame.

The second-order chord, by elements #

The chord is divided into eight beam elements per bay, each with the cubic stiffness of a beam and the consistent geometric stiffness of a compressed one; the frames are springs at their nodes; the ends are pinned. For each bow, the compression acting on the bowed shape is an equivalent sideways load, and the chord’s extra movement follows from one linear solution of the stiffness less the compression times the geometric stiffness — the matrix is factorised once and every bow is a new right-hand side. Each frame’s force is its stiffness times the chord’s extra movement at its node. The critical load comes from the same matrices as an eigenvalue. The stiff-frame limit above is a check on the whole: at 56,000 N/mm the calculation gives 1.54 per cent, approaching the hand value from below as it should.

A pinned chord, a constant force, and a bow nobody measured #

The chord force is constant. In a real girder the compression in the top chord follows the bending moment, largest at mid-span and falling toward the supports, and the buckle then localises toward the middle. The frames near the ends would carry less than they do here.

The bow’s size is assumed. A five-hundredth of the half-wavelength is a design convention, not a measurement, and every frame force here is proportional to it. The shape is not assumed — every shape was tried — which is the part a single-bow calculation gets wrong.

The chord is a column between frames. In a truss girder the chord is also a continuous beam over its panel points, carrying local bending that adds to the bow’s.

The frames are springs. A U-frame is a cross-girder and two verticals with its own flexibility and its own strength, and it is loaded by the chord’s bow while also carrying the deck; its design combines the two.

Still open: the frame that is also a beam #

Every frame here is a spring that does one job. A real U-frame is the end of a cross-girder that carries the deck, and the deck’s load bends the cross-girder and rotates its ends — which rotates the verticals and pushes the chord sideways before any bow is involved. The frame’s stiffness against the chord is then partly spent before the chord asks for it, and the deck load itself becomes an imperfection with a shape set by the cross-girders’ spacing. Whether that deck-induced push, amplified like any bow, adds to the frame force computed here or dominates it, is a question about a restraint that is also loaded.