A row of frames is not a foundation

A half-through girder carries a deck at its bottom flange and has nothing across its top, because the top is where the traffic is. Its top chord is in compression along the whole span and is held sideways only by U-frames: a cross-girder under the deck and the two verticals rising from its ends, which together form a U that resists the chord’s sideways movement by bending. Every few metres along the span there is one.

Held everywhere, and it forgets its length analysed that chord as a strut on a continuous elastic foundation — a spring of stiffness $k$ per unit length along the whole of it — and found the property that makes the arrangement work: the chord chooses its own number of half-waves, each $\pi(EI/k)^{1/4}$ long, and its critical load $2\sqrt{kEI}$ stops depending on the span. The buckle that will not spread out gave the chord its real, parabolic force and found the buckle gathering at midspan.

Both made the substitution Engesser made in 1884 and that every design method since has kept: the U-frames are smeared. A frame of stiffness $K$ every $a$ metres is replaced by a foundation of stiffness $K/a$ everywhere. The essay before this one noted where that substitution must fail — when the buckle is short compared with the spacing — and did not compute it. This one does.

Two chords with the same restraint #

The chord is 24 m long, pinned at its ends, with the flexural stiffness of the earlier essays and a smeared restraint of 0.35 N/mm per mm of length. Smeared, it buckles at 1,897 kN in half-waves of 5.1 m.

Now deliver that restraint as frames. A frame every $a$ millimetres with a stiffness of $0.35a$ N/mm gives exactly the same smeared value at any spacing, so the smeared calculation cannot tell a frame every 2 m from one every 6 m. The chord can.

The picture at the top of this page shows it. With frames every 2 m the chord buckles in the same half-waves the smeared model predicts, running straight through the frames as though they were not there, at 1,892 kN — the smeared answer. With frames every 4 m it does something else entirely: it buckles in one half-wave per bay, with a node at every frame, at 1,548 kN, 18 per cent below the smeared answer.

Where the smeared answer stops being true #

Sweep the spacing and the chord’s behaviour has two regimes and a narrow crossing between them.

While the frames are close, the buckle is long compared with the spacing and cannot tell a row of springs from a continuous bed. The chord on frames matches the smeared answer to within two per cent up to a spacing of 3 m.

Once the frames are far apart, the chord stops trying to buckle through them. It buckles between them instead, each bay a pinned column of length $a$ with the frames as its nodes, at $\pi^2 EI/a^2$. The frames are still there and still stiff; they simply are not moving, because the chosen mode puts a node at each of them. At 6 m that load is 691 kN, 64 per cent below the smeared answer, and it falls as the square of the spacing from there.

Between the two, the chord on frames sits below both, because a mode that half-uses the frames is cheaper than either pure mode. The crossing is not gradual: the chord leaves the smeared answer at about two thirds of the half-wavelength and within a metre is on the bay’s Euler load.

The crossover is a fixed ratio #

The place where the two limits meet can be written down. The smeared load is $2\sqrt{kEI}$ and the load between frames is $\pi^2EI/a^2$; setting them equal gives

$$a = \frac{\pi}{\sqrt 2}\left(\frac{EI}{k}\right)^{1/4} = \frac{\lambda}{\sqrt 2}$$

where $\lambda = \pi(EI/k)^{1/4}$ is the smeared half-wavelength. The two limits cross when the frame spacing is 0.71 of the buckle’s own half-wavelength, for any chord and any restraint. The discrete answer peels away a little before that, at about 0.67, where the interaction of the two modes costs a few per cent.

Why it should be a fixed number is worth a moment, because it is the reason the rule is usable. The smeared problem has exactly one length in it that does not belong to the span: $\lambda$, the length over which bending and restraint balance. A row of frames adds one more, the spacing $a$. Any comparison between the two models can depend on those two lengths only through their ratio — there is nothing else to make a pure number from — so the ratio at which they part company cannot depend on the chord’s stiffness, the frames’ stiffness or the span. It can only be a constant, and the constant is found once.

The second sweep tests the rule. Quadruple the restraint and the smeared answer doubles, as $2\sqrt{kEI}$ says it must, and the half-wavelength shortens by the fourth root of four, from 5.1 m to 3.6 m. The spacing at which frames stop acting as a foundation shortens with it, and in the same proportion: the chord leaves the smeared answer at 0.66 of the new half-wavelength, against 0.67 of the old.

That gives a check a designer can make on the back of the drawing, with no eigenvalue at all. Compute the half-wavelength the smeared model implies. If the frames are closer than about two thirds of it, the smeared answer stands; if they are further apart, the chord’s capacity is the Euler load of one bay, and the smeared answer is unconservative by the ratio of the two. The spacing is compared not with the span, not with the chord’s depth, but with a length the restraint itself defines.

The stiffer the restraint, the shorter that length and the closer the frames have to be. That is the counter-intuitive half of the rule. A designer who makes the U-frames stiffer to raise the smeared capacity has shortened the half-wavelength the frames must be close enough to support, and a spacing that was adequate for the soft frames may not be for the stiff ones.

The frame stiffness past which only spacing matters #

When the chord buckles between frames, the frames are not deflecting. Their stiffness is doing nothing at all in that mode, and making them stiffer cannot change it. That is a statement the smeared model has no way to make, because in the smeared model every increase in stiffness raises the load.

Hold the spacing at 4 m and vary the frames’ stiffness, and the chord’s capacity follows the smeared curve while the frames are soft — the mode then runs through them, deflecting them — and then bends over and stops at the bay’s Euler load, 1,554 kN. The knee is at 1.40 kN/mm a frame, and the frames that give this chord its smeared 0.35 N/mm per mm are exactly that stiff. Double their stiffness and the smeared calculation reports 41 per cent more capacity; the chord gains nothing.

This is the result the brace that need not be strong found for a single brace on a column: a brace needs a threshold stiffness to force a node, and beyond it buys nothing. A row of U-frames is a row of such braces, and it has the same threshold. The smeared model is the one description of a braced member that does not have a threshold, because it has no nodes to force — and that is exactly the property that makes it wrong past the crossover.

The whole of it, once, for the 4 m frames #

The chord’s flexural stiffness is $EI = 2.52 \times 10^{12}$ N·mm², its smeared restraint $k = 0.35$ N/mm per mm. The half-wavelength is $\pi(EI/k)^{1/4} = \pi \times (7.2 \times 10^{12})^{1/4} = \pi \times 1{,}638 = 5{,}146$ mm, and the smeared load $2\sqrt{kEI} = 2\sqrt{0.35 \times 2.52 \times 10^{12}} = 1{,}878$ kN for an infinitely long chord; on 24 m, with a whole number of half-waves, 1,897.

Frames every 4 m are at $4{,}000/5{,}146 = 0.78$ of the half-wavelength, past the crossover. Each is $0.35 \times 4{,}000 = 1{,}400$ N/mm stiff. The bay’s Euler load is $\pi^2 \times 2.52 \times 10^{12}/4{,}000^2 = 1{,}554$ kN, and the eigenvalue of the chord on its five interior frames is 1,548 — within half a per cent of it. The design check that smears these frames overstates the chord by 1,897 against 1,548, or 23 per cent of the true capacity.

That is not a fine distinction. A half-through girder checked at a utilisation of 0.85 by the smeared method is at 1.04 by the discrete one.

Where a real U-frame puts the chord #

The two sweeps above used restraints chosen for round numbers. A real U-frame’s stiffness comes from its members, and working one out places a real girder on the figures.

A frame’s sideways stiffness at chord level is the inverse of its flexibility, and its flexibility has two parts: the vertical bending as a cantilever from the cross-girder, $h^3/3EI_v$, and the cross-girder bending under the equal and opposite moments the two verticals put into its ends, $h^2 b/2EI_q$. For a vertical 1.2 m tall with a second moment of $2 \times 10^7$ mm⁴ — a stiffener of ordinary proportions — on a cross-girder 6 m long with $5 \times 10^8$ mm⁴, the two terms are $1.37 \times 10^{-4}$ and $0.41 \times 10^{-4}$ mm/N, and the frame is about 5.6 kN/mm stiff.

At a spacing of 4 m that is a smeared restraint of 1.4 N/mm per mm — the stiffer of the two sweeps. Its half-wavelength is 3.6 m, and the crossover sits at two thirds of that, 2.4 m. Frames of ordinary proportions at an ordinary spacing are already past the crossover, because a stiff frame shortens the half-wavelength it has to be close enough to support. The chord on those frames buckles between them, at the Euler load of a 4 m bay, and the smeared calculation overstates it by the ratio of 3,783 kN to 1,554 — a factor of 2.4.

That example is built from assumed sizes, and a different vertical or cross-girder moves it. The point it makes does not depend on them: the stiffer the frames are made, the closer they must be to act as a foundation, and a designer who stiffens frames at a fixed spacing is walking toward the crossover rather than away from it.

Closer frames, stiffer frames, a heavier chord #

The check costs one line. Compute the smeared half-wavelength $\lambda = \pi(EI/k)^{1/4}$ with $k = K/a$. If the spacing $a$ is below about $0.67\lambda$, the smeared answer stands. If it is above, take the smaller of the smeared answer and $\pi^2EI/a^2$, and near the crossover allow a few per cent for the interaction.

Three moves are then available, and they behave differently.

Closer frames raise the bay’s Euler load as the square of the reduction in spacing, and leave the smeared answer unchanged. Past the crossover this is the only move that helps directly, and it is most effective where the chord’s force is largest: at midspan, where the parabolic force gathers the buckle, two extra frames can do what ten stiffer ones cannot.

Stiffer frames raise the smeared answer and shorten the half-wavelength. Past the crossover they do nothing; below it they help until they push the crossover down onto the spacing, and then stop.

A heavier chord raises both limits at once: the bay’s Euler load in proportion to $EI$, the smeared load in proportion to $\sqrt{EI}$, and the half-wavelength in proportion to $EI^{1/4}$, which moves the crossover outward. It is the one move that helps in both regimes, and it is the one most often reached for last, because the chord was sized for its axial force before its stability was examined.

The same reasoning applies wherever a member in compression is restrained at intervals by something flexible: purlins holding a rafter’s flange, a brace on the wrong flange of a beam, the floors of a tall building holding its columns. Each is a row of springs that may or may not act as a foundation, and each has a half-wavelength to measure its spacing against.

Why the error runs the unsafe way #

The smeared substitution is usually described as an approximation and left there, as though its error could fall on either side. It cannot. The smeared model allows the chord to buckle only in smooth half-waves of its own preferred length; the discrete model allows those and every other shape, including the shapes with nodes at the frames. A structure allowed more ways to buckle finds a cheaper one or the same one, never a dearer one. Smearing a discrete restraint removes buckling modes, and removing modes can only raise the computed critical load.

That is the same argument, from the other side, that an average stiffness is not a safe stiffness made about columns whose stiffness varies along their length: any simplification that constrains what the mode may look like is unconservative, and the only question is by how much. For U-frames the answer is: by nothing at all while the frames are closer than two thirds of a half-wavelength, and by the ratio of the two limits beyond that.

The chord cut from its frames #

The free body is the whole chord, pinned at its ends, cut from each U-frame at the point where the frame’s vertical meets it. Crossing each cut is a sideways force equal to the frame’s stiffness times the chord’s sideways movement there; everything else acting on the chord is its own axial force. The chord is divided into beam elements between frames, several to a bay, with the geometric stiffness of the axial force in each; the critical load is the smallest force at which the elastic and geometric stiffnesses together become singular, found by counting the negative pivots as the force is raised. The smeared figures use the closed form of the earlier essays, and the chord on frames converges to it as the spacing shrinks — to within a tenth of a per cent with frames every metre — which is the check that the two calculations are of the same chord. The other limit is checked the same way: past the crossover the chord’s eigenvalue sits within half a per cent of $\pi^2EI/a^2$, the load of one pinned bay, which it has to reach because that is the mode it has found.

A varying force, the frames’ strength, the deck and yielding #

A constant force. Every chord here carries the same force along its whole length. A real chord carries a parabola, and the buckle then gathers at midspan; the frames that matter are the middle ones, and a chord whose midspan frames are too far apart is in the between-frames regime even if the average spacing says otherwise.

The frames’ own strength. A U-frame has a stiffness and a strength, and near the crossover the frames deflect and so carry force. What that force is depends on the chord’s initial bow — an imperfection no eigenvalue contains — and it is what the frame is actually designed for.

The deck. A concrete deck on the cross-girders makes the U-frames’ bottoms continuous and changes their stiffness; a steel deck plate with longitudinal stiffeners can itself restrain the chord. Both change $K$ and neither changes the rule.

Inelastic buckling. At forces this large a stocky chord may yield before it buckles elastically, and then $E$ in every formula is a tangent modulus smaller than $E$, which lengthens the half-wavelength and makes a given spacing safer.

Frames that act at points #

That the frames act at points. A real U-frame vertical is a few hundred millimetres wide and is connected to the chord over that width, and it restrains twist as well as sideways movement. Neither matters to the crossover while the spacing is several metres; both would matter to a detail where frames are spaced at a fraction of a metre — which is the regime where the frames are a foundation anyway.

Still open: the force the frames have to carry #

Every number here is a critical load, the force at which a perfect chord can first move sideways. A real chord is not straight. It arrives with a bow of a few millimetres, and under its axial force that bow grows, as every imperfection does under the load that finds it, pressing on the frames with a force that rises steeply as the critical load is approached. The frames have to be strong enough to carry it, and the design rules give it as a percentage of the chord force that nobody derives. How large the frame force really is — and whether it is largest, as the eigenvalue would suggest, at the frames nearest the crossover where the chord is half using them — is a question about an imperfect chord, and it cannot be answered by an eigenvalue at all.