The section that will not keep its shape

Drive a lorry down one side of a box-girder bridge and the deck twists. That much everybody expects, and it is the reason the section is a box in the first place: a closed cell carries torque round itself as a shear flow, at a stiffness hundreds of times an open section’s, and the twist that results is negligible.

The twist is not the problem. The problem is that the rectangle stops being a rectangle.

Three load cases, and only two of them are checked #

Put a load on the axis and the box bends. Put it at an eccentricity $e$ and the standard move is to replace it by the same load on the axis plus a torque $m = p\,e$ per unit length — moving a force and paying for it with a couple. For the box drawn that torque is 60 kNm for every metre of span.

The torque is then applied to the section as a set of edge forces, and there is more than one set that does it. The obvious one is a pair of vertical forces on the two webs, $m/b$ apart. The useful one is the set that drives Bredt’s uniform shear flow round the cell: vertical forces of $m/2b$ on the webs and horizontal forces of $m/2h$ on the flanges, arranged to circulate. Both are statically equivalent to $m$; they are not equivalent to each other.

Subtract the second from the first and what is left is a third set — the same magnitudes with the flange forces reversed. That set has zero resultant torque: the vertical pair contributes $m/2$ and the horizontal pair contributes $-m/2$. It is self-equilibrating in exactly the sense that Saint-Venant’s principle is about, and it is the reason this essay exists.

For the box here the distortional set is 10 kN/m up one web and down the other, with 15 kN/m pushing the flanges the other way. It does the one thing a rigid-section analysis has no coordinate for: it changes the shape of the cross-section.

Nothing in that decomposition depends on how far off the axis the load happens to be, and the quickest way to see it is to move the load and do the subtraction again.

That invariance is what makes the next few sections possible. The distortional load is always $m/2$, so the whole problem can be solved once for a cross-section and then scaled by whatever eccentric loading arrives.

Half the torque, and the arithmetic is exact #

The natural coordinate for that shape change is $\gamma$, the change in the corner angles — the rectangle becoming a parallelogram of the same side lengths. Under the distortion mode the corners move by $\pm h\gamma/4$ horizontally and $\pm b\gamma/4$ vertically, which is what the ghosted rhombus in the first figure is drawn from.

Work out what the distortional force set does through those displacements and a clean number arrives. Each corner takes a horizontal force $m/4h$ through a displacement $h\gamma/4$ and a vertical force $m/4b$ through $b\gamma/4$, so each contributes $m\gamma/8$, and the four together give $m\gamma/2$.

The generalised load driving the distortion is exactly $m/2$. A box girder spends half of an eccentric load’s torque on changing its own shape, and the half that is left is the only half a torsion calculation sees.

Which free body produced the number #

Take a strip of the girder one millimetre long and look at its cross-section as a closed four-member frame: two flange plates, two web plates, rigidly jointed at the corners.

Apply the distortional force set to that frame and solve it. The answer is a corner-angle change per unit torque, and its reciprocal — doubled, because the generalised load is $m/2$ — is the transverse stiffness $K_d$. For the box here it comes to 238 kN·mm per radian per millimetre of girder.

That number is solved rather than quoted, and there is a case where it can be checked against a closed form. Make the box square with all four plates the same thickness and the mode has four-fold symmetry: every joint rotation is zero, every wall bends in pure double curvature with a chord rotation of $\gamma/2$, and the strain energy per wall is $6EI\psi^2/L$. Four of them give $U = 6EI\gamma^2/L$, so

$$K_d = \frac{12EI}{L}$$

with $I = t^3/12(1-\nu^2)$ per unit width. The frame solve returns that to the last digit, which is the only reason the general case is worth believing.

The load path is worth saying out loud, because it is the whole of the transverse design of a box girder. The plates are not carrying the distortional forces in their own planes; they are carrying them by bending out of their planes, about their own transverse axes, at a stiffness that contains $t^3$. That is a thousand times less stiffness than the same plates offer along the girder, and it is why a deformation that no longitudinal calculation contains can dominate the answer.

It is the same plate-bending stiffness that a folded plate spans on, used for a different purpose. A folded plate has depth because its folds hold their angle; a box girder’s cross-section holds its shape for exactly the same reason, and the corner is the fold. Where the two differ is what happens when the fold has to work over a long length with nothing holding it — which is the next section.

The equation is Winkler’s, with an angle where the deflection goes #

Distortion is resisted twice, by two quite different actions.

The cross-section frame resists $\gamma$ locally, wherever it occurs, with the stiffness just computed. And the walls resist a change of $\gamma$ along the girder, because a varying distortion warps them longitudinally: the corner on one diagonal goes forward while the corner on the other goes back, at a warping function of $\pm bh/8$ — three quarters of a square metre for this box — and the longitudinal stress that produces is a real stress in a real flange.

Integrate $w^2$ over the walls and the distortional warping constant comes out as

$$I_{dw} = \frac{b^2h^2(b\,t_f + h\,t_w)}{96}$$

which for the box here is $2.70 \times 10^{16}\,\mathrm{mm^6}$. Put the two together and the governing equation is

$$E I_{dw}\,\gamma'''' + K_d\,\gamma = \frac{m}{2}$$

That is Winkler’s equation with $\gamma$ where the deflection goes. A box girder distorting is a beam on an elastic foundation — the beam being the walls’ longitudinal warping, and the foundation being the cross-section’s own transverse frame.

Which means it has a decay length, and the decay length is the number the whole design turns on.

The number the spacing is chosen against #

$\lambda = (K_d/4EI_{dw})^{1/4}$ gives a decay length of 17.5 metres for this box. Not 17.5 metres of a 40 m span, or a fortieth of anything: a length fixed by the cross-section, the same on a thirty-metre span and a ninety-metre one.

That is what a diaphragm is placed against. A diaphragm holds $\gamma = 0$ at a point, and it helps its neighbours only if they are inside its own reach.

Left alone, the box drawn here distorts by 0.084 radians, which is 4.8 degrees of corner-angle change and is not a structure. The transverse bending stress in the webs reaches 210 N/mm², the longitudinal warping stress at the corners reaches 75, and the section’s bending stress — the one the girder was sized for — is 71. The stress nobody computed is larger than the stress everybody did.

That is not a warning about a marginal case; it is the reason box girders have diaphragms at all, and the reason the drawing above is a hypothetical rather than a design.

What the transverse frame is really made of #

$K_d$ contains $t^3$, so it is the most sensitive quantity in the whole calculation to something a designer usually decides for other reasons. Taking the web from 12 mm to 20 mm — a change made for shear, or for a plate-buckling check — nearly doubles the frame stiffness, takes the decay length from 17.5 m to 15.6, and takes the distortional stress from 106 per cent of the bending stress to 71.

It also means that the honest way to control distortion is not always a diaphragm. Thicker plates work. So do transverse stiffeners on the webs, which are a frame of their own in parallel with the plates. And so does making the box narrower, since $I_{dw}$ goes as $b^2h^2$ and the driving torque goes as the eccentricity the width allows.

What does not work is anything longitudinal. Adding flange area raises the section’s bending stiffness and leaves $K_d$ where it was, so the distortional stress as a share of the bending stress goes up.

None of that is visible in the shear flow the box was chosen for, which is the pure-torsion half of the decomposition: uniform round the cell, the thing that makes a closed section stiff, and entirely absent from the distortional case, where the four edge forces circulate in opposite senses on the flanges and the webs and add to no torque at all.

Distortion is not warping, and the difference is a plane #

The section next door to this one on the site is the open section that cannot stay flat, and the two are easy to confuse because both are about a cross-section that stops behaving as a rigid figure.

They are perpendicular. Warping is out-of-plane: an I-section twisted about its axis has its flanges bend in opposite directions and the cross-section dishes out of its own plane, while remaining, in projection, exactly the shape it started as. Distortion is in-plane: the box’s cross-section stays flat and changes shape.

A box does both, and the second is the larger. It also has a companion the open section does not: distortional warping, the longitudinal displacement that follows from $\gamma$ varying along the member, which is what produces the corner stresses plotted above. Three deformations, then, and a designer who has computed torsion has computed one.

The mechanisms are the same in outline and at right angles in geometry. Both are a deformation the rigid-section theory has no coordinate for; both are harmless while they are free and expensive once something restrains them; both produce a longitudinal stress that no shear check contains. What differs is the plane the deformation happens in, and therefore what stops it: an open section is restrained by an end plate or a symmetry condition, and a box is restrained by its own cross-section acting as a frame.

The diaphragm is a structure of its own #

The analysis above treats a diaphragm as a point where the corner angle is held at zero. Something has to do the holding, and what it costs is worth naming.

A diaphragm carries the distortional force set as an in-plane load: the two vertical forces on the webs and the two horizontal forces on the flanges arrive at its edges and have to be carried across it. For a plate diaphragm that is a shear panel, and the check is straightforward. For a cross-braced one it is a truss, and the diagonal carries the whole of it.

What complicates it is the access hole. A box girder has to be inspected, so every diaphragm has an opening in it large enough for a person, and an opening in the middle of a shear panel removes the material where the shear is highest. The opening is therefore framed, the framing is welded to the diaphragm, and the diaphragm is welded to four plates it meets at right angles — which is why a diaphragm costs several times what its weight of steel suggests, and why the spacing curve above is a spending decision rather than a drawing one.

The decay length has no material in it #

Collect the scaling and something falls out that explains why concrete boxes and steel boxes are detailed so differently.

The decay length is the beam-on-elastic-foundation one, $\ell \sim (4EI_w/K_d)^{1/4}$. For a thin-walled box the longitudinal warping stiffness $EI_w$ goes as $E t$ — linearly in the wall thickness, as every thin-walled section property does. The transverse frame stiffness $K_d$ goes as $E t^3/b^3$, because it is a frame of plates bending. So

$$\ell \;\propto\; \left(\frac{E t}{E t^3/b^3}\right)^{1/4} = b^{3/4}\,t^{-1/2}$$

and the modulus cancels exactly. The distance a distortion reaches along a box depends on the box’s proportions — its width and its wall thickness — and not at all on what it is made of.

Put the two materials through it. A steel box with 12 mm walls and a concrete box of the same outline with 300 mm walls differ by a factor of $\sqrt{300/12} = 5$ in decay length. So the 17.5 m of the steel box becomes about 3.5 m in concrete, on a member that is very much larger.

That is the whole of why the two are detailed as they are. A steel box needs intermediate diaphragms every few metres, at a fabrication cost the previous section describes; a concrete box of the same span needs them only at the piers, because a distortion introduced at mid-span has died away long before it reaches anything. Nobody chose that; it is a consequence of concrete being thick and of the modulus dropping out of the arithmetic.

An internal web is a diaphragm with no spacing #

The whole of this essay treats the transverse frame as four walls and a diaphragm as a device inserted at intervals to hold their corners. There is a second way to hold them, and it is continuous.

Add a web. A two-cell box has a wall down the middle, and that wall prevents the rhombus mode outright — the top and bottom flanges cannot move relative to one another in the direction the mode requires, because something is joining them everywhere. A multi-cell box has essentially no distortion problem at all, and the diaphragms it carries are there for load introduction at supports rather than for shape.

So the design choice is between adding material at points and adding it along the length, and the comparison is a straightforward one. A diaphragm every 4 m on a 40 m span is ten plates, each fabricated, fitted and welded inside a closed box through an access hole. An internal web is one plate 40 m long, welded on a line, with none of the access difficulty and with a stiffening effect on the shear and the flange buckling as well.

Which is why boxes above a certain width acquire an internal web rather than more diaphragms, and why the crossover is a fabrication argument rather than a structural one. Nothing on the decay-length curve says where it is; the curve says what a diaphragm has to achieve, and the cost of achieving it decides which of the two ways it is done.

Where the model stops #

The frame is solved with straight walls and rigid corners. Real box girders have haunches, fillets and longitudinal stiffeners, all of which stiffen the transverse frame and none of which appears here. The effect is in the right direction — the model is conservative — but a heavily stiffened box can be several times stiffer transversely than the bare plates, which moves the decay length by the fourth root of that.

The distortion mode assumes the walls are inextensible in their own planes. They are not quite, and the frame solve above uses their real axial stiffness rather than a large number standing in for rigidity, which is why the square-box check agrees with the closed form to the last digit rather than approximately.

Diaphragms are treated as rigid. A plate diaphragm with an access opening in it is not, and a cross-braced diaphragm is a frame with a stiffness of its own; both belong in the model as a spring rather than as a support, which the force method above would take without modification and which nothing here does.

And the load is a line load on all of the span. A wheel load is a point, and a point distortional load on a foundation of this stiffness produces a local peak the sine series here resolves only approximately. The qualitative answer does not move — it still decays over 17.5 m — but the peak does.

What the pictures cannot show #

The rhombus in the first figure is drawn at an amplitude a reader can see. The real corner-angle change on a well-diaphragmed box is of the order of $10^{-3}$ radians, which at the scale of the drawing is thinner than the line the section is drawn with. What is being illustrated is a mode shape, not a movement.

Nor can any of these figures show the thing that makes distortion expensive on a real bridge, which is that a diaphragm is a piece of steelwork somebody has to fabricate, fit and weld inside a closed box, through an access hole, at a rate of one every few metres. The curve of stress against spacing is smooth; the cost of moving along it is not.

The assumption the figure rests on #

The transverse frame is taken to have four rigid corners and nothing else. That is the assumption every number above depends on, and it is the one worth naming, because a real box girder’s corner is a welded joint between a 16 mm plate and a 12 mm plate — usually with a longitudinal stiffener running past it, sometimes with a fillet, occasionally with a bolted splice.

If that corner is softer than rigid, $K_d$ falls, the decay length rises as the fourth root, and diaphragms that were inside each other’s reach are no longer. If it is stiffer, the whole problem shrinks. Nothing in the longitudinal analysis of a box girder — not the moment, not the shear, not the torsion — is sensitive to that detail at all, and the deformation this essay is about is sensitive to almost nothing else.

That is the sense in which a diaphragm belongs with a deck whose job is to distribute rather than to carry, one dimension down. Neither member is there to resist the load. Each is there to make the structure behave the way the analysis of the load assumed it did, and each is therefore invisible to every calculation that assumption appears in.