The stiffness a row of braces shares out

A torsional brace’s demand is a quadratic in the moment: a rotational spring at midspan resists the twist of a beam buckling sideways, the stiffness it needs to hold a stated moment rises as the square of that moment, and the curve crosses zero at the beam’s own unbraced capacity. That essay held one brace and found its ceiling — the moment at which the brace is stiff enough to be a node, the beam stops twisting it, and the buckle jumps to two half-waves either side of it. No stiffer brace raises the moment past that.

Real girders are not braced once. A plate girder in a bridge is held by cross-frames every five or six metres, a rafter by purlins and fly braces at every other purlin, a crane girder by a horizontal truss at intervals. The question this essay asks is what the arithmetic of one brace becomes when there are several, and the answer depends, more than anything else, on what the braces are being asked to do.

A girder, held at its quarter points #

The beam is the plate girder of the essay on one brace — 1,225 mm between flange centres, $I_z = 2.67 \times 10^8$ mm⁴, $J = 4.57 \times 10^6$ mm⁴, $I_w = 1.00 \times 10^{14}$ mm⁶ — over 24 m rather than 12, in uniform moment. Unbraced it buckles laterally at 837 kN·m. Its plastic moment is 5,627 kN·m, which is what a designer would like it to reach, and asking for an elastic critical moment equal to that is asking for a girder whose slenderness for lateral–torsional buckling is one: still reduced, but no longer governed by twisting.

The braces are torsional — cross-frames or diaphragms between this girder and its neighbour, which resist the section’s rotation rather than the movement of either flange. Each is a rotational spring of stiffness $\beta_T$, and the row of them is equally spaced along the span.

The figure is the whole essay in one picture, and it carries two readings. Read along the horizontal axis and the striking thing is how little the count matters. At 4,336 kN·m/rad in all, two braces carry the girder to 3,815 kN·m, three to 3,726 and five to 3,578. The beam is responding to the total stiffness spread along it and barely notices whether it arrives in two lumps or five.

Read upward and the count is everything. Each row stops at its own ceiling, and the ceilings are far apart: one brace at 2,637 kN·m, two at 5,585, three at 9,704, five at 21,466. The 5,627 kN·m line runs just above the two-brace ceiling, so one or two braces cannot get the girder there at any stiffness — two braces fall short by less than one per cent, and that one per cent is a wall. Three braces can, and so can five. The total sets how far up its curve a row has climbed; the count sets where the curve ends.

The beam stops twisting at the braces #

What happens physically at the ceiling is easiest to see in the buckled shape.

With braces of 542 kN·m/rad each the girder buckles much as it would with none: one half-wave of twist over the whole 24 m, the braces dragging a little at each tick. The critical moment has tripled to 2,646 kN·m, which is what the springs bought, but the shape is unchanged.

At 2,710 kN·m/rad each — the stiffness that takes four braces to 5,627 kN·m — the shape is still one half-wave. It is flatter across the middle, because the braces are resisting hardest where the twist is largest, but the girder still turns at every brace. The braces are working as springs, not as supports. This is the regime real braces live in, and it is why bracing design is a stiffness calculation and not a question of where the nodes are.

Only at stiffnesses an order of magnitude higher does the shape change. At 20,897 kN·m/rad each the twist has broken into lobes, and at four times that the braces are close to being nodes: the girder twists in five half-waves, one in each bay, at 14,998 kN·m, which is the critical moment of a 4.8 m length. That is the ceiling, and it is the moment of the bay between braces rather than of the girder.

Asking each brace for its ceiling #

Bracing rules were first written around the ceiling, because a brace stiff enough to reach it lets the designer treat the beam as a set of short beams between braces — the idea behind the ideal brace stiffness and the knee past which more stiffness buys nothing.

The demand per brace rises steeply with the count. One brace needs 1,615 kN·m/rad to reach 95 per cent of its ceiling, three need 11,007 each, eight need 118,954 each — seventy-four times as much per brace, and nearly six hundred times as much in total. The reason is two effects multiplied together. Each brace added shortens the bay, and the ceiling, being the critical moment of a bay, rises nearly as the square of the count on a girder whose resistance to twist is mostly warping. And the stiffness a torsional brace needs rises as the square of the moment it holds. A ceiling growing as the count squared, demanded through a stiffness growing as the moment squared, is a demand growing as roughly the fourth power of the count, shared among the count. That is the steep staircase in the figure.

If the ceiling were what a row of braces was for, more braces would be a bad bargain. But the ceiling for eight braces is 47,925 kN·m on a girder whose plastic moment is 5,627. Nobody wants a critical moment eight times the section’s capacity; the section yields long before. Asking each brace to deliver its ceiling is asking it to protect the girder against a moment it can never carry.

Asking the row for a moment instead #

The question a designer actually has is different: here is the moment the girder must reach, what must the braces be?

Asked for 5,627 kN·m, the picture reverses. One and two braces cannot get there at any stiffness, because their ceilings are below it — marked with crosses. Three braces get there with 3,510 kN·m/rad each, four with 2,710, five with 2,233, eight with 1,477. Each brace added makes every brace softer.

The total is the line across the top, and it barely moves: 10,530 kN·m/rad for three, 10,841 for four, 11,813 for eight, a spread of 12 per cent across a count that more than doubles. That is the first reading of the hero figure turned into a design statement. Below the ceiling, the beam sees the row’s total stiffness, so the total is what the moment demands, and the count only divides it.

The small rise in the total with the count is real and worth a sentence. A brace close to a support sits where the twist is small and does less than one in the middle; more braces put more of the total near the ends, and the row needs a little more stiffness in all to compensate. It is a second-order effect, and it runs the opposite way to the intuition that more braces should be more efficient.

The count has one job in this reading, and it is a sharp one: to put the ceiling above the target. Two braces at 5,585 kN·m miss by 42 kN·m and cannot be rescued by any stiffness; a third brace lifts the ceiling to 9,704 and the stiffness needed falls from infinite to 3,510. Once the ceiling is comfortably clear, adding braces is a way to make each brace cheaper, not a way to make the girder stronger.

The continuous restraint the rules assume #

The design rules for torsional bracing are built on a result from 1966. Taylor and Ojalvo solved a beam with a rotational restraint spread continuously along its length, of stiffness $\bar\beta$ per unit length, and found that the critical moment is

$$M_{cr}^2 = M_0^2 + \bar\beta\,E I_z$$

with $M_0$ the unbraced value. Replace $\bar\beta$ by the row’s total divided by the span, $\Sigma\beta_T / L$, and the formula describes a row of braces as though it were smeared. The design rules then drop $M_0$, write $n\beta_T$ for the total, and solve for the stiffness each brace needs. That is why the rules have the number of braces in the denominator.

For four braces the smeared formula reads between 90 and 93 per cent of the row’s critical moment all the way up to the ceiling. It is close, and it is on the safe side. Asked for the stiffness to reach 5,627 kN·m it gives 13,264 kN·m/rad in all — 26 per cent more than three braces actually need, 12 per cent more than eight. The safe side comes from where the smeared restraint puts its stiffness: some of it at the supports, where the beam is not twisting and the restraint does nothing. A row of braces keeps its stiffness off the supports, so it does a little better than the formula assumes.

This is a contrast worth drawing with a row of U-frames holding a girder’s compression chord, where the same smearing errs the other way once the frames are spaced at more than about two-thirds of the chord’s own half-wavelength. There the chord buckles between frames below the smeared answer. Here the corresponding failure is the ceiling, and the smeared formula has no ceiling in it at all — it keeps climbing in the figure long after the row has stopped. The formula is safe below the ceiling and silent about the ceiling, so a design that uses it must check separately that the bays are short enough for the target moment. With three braces on this girder that check is passed by a factor of 1.7, with two it is failed by one per cent, and the formula alone would not have said which.

One brace lost #

The total-stiffness reading has a corollary that is wrong, and it is the one that matters for a bridge in service: if the beam sees only the total, losing one brace out of four should cost a quarter of the total and no more.

The smeared formula, given three-quarters of the stiffness, predicts that the girder keeps 4,891 kN·m, a loss of 13 per cent. Removing a brace from the actual row costs 34 per cent: the critical moment falls to 3,787 kN·m without one of the outer braces and to 3,704 without one of the inner ones. The beam does markedly worse than its total says, because a missing brace is not a quarter of the restraint gone evenly; it is a bay twice as long, 9.6 m with no restraint in it, and the buckle gathers there. The twist that was spread over five bays concentrates in the long one, and the remaining braces, all of them at the edges of the long bay or beyond it, are poorly placed to stop it.

This matters because cross-frames are lost. They are removed for an inspection, cut out to route a service, fail at a bolted connection that has worked loose, or are left out of an erection sequence for a week because the next girder has not arrived. A row designed exactly to its total has no margin for any of that, and the margin it needs is not the brace’s share of the total but rather more. A designer who wants a girder to tolerate one missing brace has to check the girder with that brace missing, which is a different calculation from adding a quarter to the stiffness.

The count during erection #

The rows above are a finished bridge. A girder spends part of its life with fewer braces than that, and the count is then the number that decides everything.

Steel girders are usually lifted in pairs, braced together on the ground by some of their cross-frames, and set on their bearings; the remaining cross-frames go in afterwards, and the deck’s wet concrete is often the heaviest load the bare girders ever carry. A pair lifted with one cross-frame at midspan has a ceiling of 2,637 kN·m, whatever that cross-frame is made of. A pair with two, at the third points, has one of 5,585. If the wet concrete and the formwork bring the moment to more than either, no stiffening of the cross-frames present can save the girder, and the only remedy is another cross-frame. The count is a strength the girder either has or does not, and erection is when it is most often short.

This is also why a missing brace in service and a brace not yet fitted are the same calculation. Both shorten the row by one and lengthen a bay, and both have to be checked as the beam that is actually standing rather than as a fraction of the beam that was designed.

The torsional brace has one virtue in all of this that the lateral brace lacks. A lateral brace works only on the flange the buckle moves; a torsional brace resists rotation, and rotation is the same at every height in the section, so a cross-frame fitted in a hurry cannot be fitted to the wrong flange. That is what makes cross-frames the bracing of choice for girders, and it is why their count and stiffness, rather than their position in the depth, are the whole of their design.

The numbers, by hand #

The two numbers that the essay turns on can be had without a computer.

The ceiling of a row of $n$ braces is the critical moment of one bay of length $L/(n+1)$:

$$M_{ceiling} = \frac{\pi}{L_b}\sqrt{E I_z\left(G J + \frac{\pi^2 E I_w}{L_b^2}\right)}, \qquad L_b = \frac{L}{n+1}.$$

For two braces, $L_b = 8$ m, and with $E I_z = 5.60 \times 10^{13}$ N·mm², $G J = 3.70 \times 10^{11}$ N·mm² and $\pi^2 E I_w / L_b^2 = 3.24 \times 10^{12}$ N·mm², the product under the root is $2.02 \times 10^{26}$, its root $1.42 \times 10^{13}$, and times $\pi/8000$ it is $5.585 \times 10^{9}$ N·mm — 5,585 kN·m, just short of the 5,627 asked for. For three braces, $L_b = 6$ m, the warping term rises to $5.76 \times 10^{12}$ and the ceiling to 9,704 kN·m.

The total stiffness to reach a moment is the smeared formula solved for $\Sigma\beta$:

$$\Sigma\beta_T = \frac{(M^2 - M_0^2)\,L}{E I_z} = \frac{(5{,}627^2 - 837^2) \times 10^{12} \times 24{,}000}{5.60 \times 10^{13}}$$

which is $1.33 \times 10^{10}$ N·mm/rad, 13,264 kN·m/rad. Divided among three braces that is 4,421 each, against the 3,510 the row actually needs; among eight it is 1,658, against 1,477.

A beam in uniform moment, with perfect braces #

The calculation rests on choices that limit it.

The moment is uniform. A girder under distributed load has its largest moment at midspan and much less near the supports, and the braces near the supports are then doing less than those at midspan. The design rules carry a moment-gradient factor for this; the row’s behaviour is the same in kind and the braces near the ends matter even less.

The braces are elastic springs with no slack. A cross-frame connected with bolts in clearance holes has a small rotation before it engages, and the girder twists that much before any brace resists. The stiffness that matters is that of the cross-frame, its connections and the girder’s web in series, as the essay on one brace found, and the web is usually the softer.

The load acts at the shear centre. A deck’s weight arrives on the top flange, above the shear centre, and a load applied above it moves sideways with the twist and adds to the buckling; every brace then has to hold more. A deck that has hardened and is fixed to the top flange restrains it continuously, and the girder then forgets its length altogether.

The girder is straight. A real girder starts out slightly twisted, and the braces then carry a force from the first load, growing as the moment approaches the critical. That force is what the strength half of a bracing rule is for, and it grows with the same ratio — moment over critical moment — so a row designed to the minimum stiffness carries large forces. Rules ask for twice the ideal stiffness for exactly this reason.

The section does not distort. A deep girder’s web can bend between its flanges, so that the compression flange rotates more than the brace does. Web stiffeners at the braces prevent it; without them the brace’s stiffness is partly wasted on bending the web.

Still open: the brace near the support #

The total stiffness rises slightly with the count because braces near the supports do less. The natural design response is not to space braces equally at all, but to concentrate them where the twist is largest. Whether a row of three braces, spaced more closely about midspan, reaches the same moment with less total stiffness — and whether the bays near the supports, now longer, lower the ceiling enough to cancel the gain — is a question of where a girder buckles first when its braces are not evenly spread. It is the same trade this essay found between the total and the count, set this time between the total and the spacing.