Sections and stress

The section that will not keep its shape

A box girder is closed, so torsion costs it almost nothing. What an eccentric load actually does to it is something a torsion calculation contains no term for — the rectangle becomes a parallelogram, in its own plane, along the whole length of the span.

Assumes The internal force with no diagram, The section that cannot stay flat and The beam that sits on the ground.

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.

An eccentric load is three load cases, and only two of them are checkedA line load of 40 N/mm at 1.5 m from the axis of a 3.0 by 2.0 m box, replaced by the three cases it is equivalent to. Bending is the load on the axis. The torque 60 kNm per metre then splits into a set of edge forces that drives Bredt's shear flow and distorts nothing, and a set with the flange forces reversed — 10.0 kN/m up one web and down the other, 15.0 kN/m across the flanges — which carries no torque at all and squashes the rectangle into the rhombus drawn behind it. Its generalised load is exactly half the torque, so a box girder spends half of an eccentric load's torsion on changing its own shape, and no torsion calculation contains that half.40 N/mmas appliedbendingpure torsiondistortion=++torque 60 kNm per metre · the distortional half is 30the third and fourth cases add to the second: same torque, and one of them has none of it
Fig. 1 A line load of 40 kN/m running 1.5 m off the axis of a 3.0 by 2.0 m box, replaced by the three load cases it is equivalent to. The first two are the ones a designer computes. The third carries no torque at all, and it is the one that squashes the section into the rhombus drawn behind it.

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 ee and the standard move is to replace it by the same load on the axis plus a torque m=pem = 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/bm/b apart. The useful one is the set that drives Bredt’s uniform shear flow round the cell: vertical forces of m/2bm/2b on the webs and horizontal forces of m/2hm/2h on the flanges, arranged to circulate. Both are statically equivalent to mm; 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/2m/2 and the horizontal pair contributes m/2-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.

Half the torque, and the arithmetic is exact

The natural coordinate for that shape change is γ\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 ±hγ/4\pm h\gamma/4 horizontally and ±bγ/4\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/4hm/4h through a displacement hγ/4h\gamma/4 and a vertical force m/4bm/4b through bγ/4b\gamma/4, so each contributes mγ/8m\gamma/8, and the four together give mγ/2m\gamma/2.

The generalised load driving the distortion is exactly m/2m/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.

One slit, and the torsional stiffness falls by a factor of hundredsA 300 by 200 box of 12 mm wall, drawn closed and then slit along its length. Closed, the torque runs round the wall as a shear flow and the torsion constant is 1.48×10⁸ mm⁴; slit, the loop is broken and only each wall's own thickness resists, giving 5.48×10⁵ mm⁴. The ratio is 270 to one, so the same torque twists the slit section 270 times as far and raises a peak shear stress 28 times as high. Nothing about the material changed.closedJ = 1.48×10⁸ mm⁴twist 0.153° over 4.0 mpeak shear stress 6.2 N/mm²slit along its lengthJ = 5.48×10⁵ mm⁴twist 41.279° over 4.0 mpeak shear stress 175.1 N/mm²J closed ÷ J open = 270
Fig. 2 Why the box was chosen in the first place: a closed cell against an open one of the same material, at a torsional stiffness ratio in the hundreds. Every word of that comparison is true and none of it is about distortion, which is a different deformation with a different stiffness and a different governing equation.

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/2m/2 — is the transverse stiffness KdK_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 γ/2\gamma/2, and the strain energy per wall is 6EIψ2/L6EI\psi^2/L. Four of them give U=6EIγ2/LU = 6EI\gamma^2/L, so

Kd=12EILK_d = \frac{12EI}{L}

with I=t3/12(1ν2)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 t3t^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.

The same sheet, twice, and a factor of ten thousandA 2400 mm developed width of 4 mm sheet, covering 1800 mm in plan — so the legs sit at 41.4° and the fold is 265 mm deep. Flat, its second moment about its own mid-plane is 12800 mm⁴, which spans nothing. Folded, it is 56.00×10⁶ — 4375 times as much, which is exactly the depth in thicknesses squared. The material is identical, the plan cover has fallen by 25%, and the only thing that changed is where the material sits. What limits it is buckling of the leg: at this leg length the flat between the folds goes at 76 N/mm², well below the steel's 275.flat: I = 12800 mm⁴folded: I = 56.00×10⁶ mm⁴265gain 4375× = (66)²limited by buckling of the leg: the leg goes at 76 N/mm²
Fig. 3 The same plate-bending stiffness in a structure that depends on it. A folded plate spans because its folds give it depth; 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 γ\gamma locally, wherever it occurs, with the stiffness just computed. And the walls resist a change of γ\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 ±bh/8\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 w2w^2 over the walls and the distortional warping constant comes out as

Idw=b2h2(btf+htw)96I_{dw} = \frac{b^2h^2(b\,t_f + h\,t_w)}{96}

which for the box here is 2.70×1016mm62.70 \times 10^{16}\,\mathrm{mm^6}. Put the two together and the governing equation is

EIdwγ+Kdγ=m2E I_{dw}\,\gamma'''' + K_d\,\gamma = \frac{m}{2}

That is Winkler’s equation with γ\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.

Everything a beam on the ground does is a function of βxDeflection, moment and shear along a beam on an elastic foundation, each divided by its own value immediately under the load and drawn against βx. Contact pressure is k times deflection, so it is the same curve as the first. The stations are exact and none of them depends on the load or on the beam: the moment crosses zero at βx = π/4, which is 2.52 m here; hogging peaks at βx = π/2 at 20.8% of the sagging moment; the deflection crosses zero at 3π/4, or 7.57 m, past which the beam lifts; and by βx = π the uplift is 4.32% of the settlement. One characteristic length along, the deflection is already down to 51% of its peak, and by three it is 4.2%. That is what it means for a raft to stop being a beam: past two or three of these lengths, nothing knows the load happened.00.511.522.53-0.200.20.40.60.81βx — distance in characteristic lengthseach quantity ÷ its value under the loadπ/4: the moment reversesπ/2: hogging peaks at 20.8%3π/4: the beam lifts offπ: uplift is 4.32%deflection, and pressure with itbending momentshear1/β = 3.21 m
Fig. 4 The equation this one is, in the setting it was written for. A point load on a beam on a Winkler foundation dies away over a characteristic length that belongs to the beam and the soil together and to nothing else — not to the length of the beam. Everything below is that sentence with a box girder’s cross-section in place of the soil.

The number the spacing is chosen against

λ=(Kd/4EIdw)1/4\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 γ=0\gamma = 0 at a point, and it helps its neighbours only if they are inside its own reach.

The distortion runs the length of the span, and a diaphragm stops itLongitudinal stress at a corner of the box from distortional warping, along a 40 m span carrying 40 N/mm at 1.5 m off the axis. With no interior diaphragm it peaks at 75 N/mm², which is 106 per cent of the bending stress the girder was designed for. Two diaphragms take it to 14. The governing length is Winkler's: the distortion decays over 17.5 m, so a diaphragm helps its neighbours only if it is closer than that, and past it the spacing stops mattering.010203040-80-60-40-2020406080along the span (m)corner stress from distortion (N/mm²)no diaphragmtwo diaphragmsbending stress 71 N/mm² · decay length 17.5 m · half-wave 54.8 mdiaphragms
Fig. 5 Corner stress from distortional warping along the span, with no interior diaphragm and with two. The undiaphragmed case peaks at 75 N/mm² — 106 per cent of the bending stress the girder was designed for — and two diaphragms take it to 14. The distortion is not a local effect near the load; it runs the length of the member, because that is what a beam on an elastic foundation does with a load that is on all of it.

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.

A diaphragm spacing is a decay length in disguiseThe peak distortional stress, as a share of the bending stress, against the spacing of the diaphragms that hold the section square. The curve is flat at the right — past the decay length of 17.5 m a diaphragm is too far away to help its neighbour — and falls steeply once the spacing comes inside it. At a spacing of one decay length the distortional stress is 55 per cent of the bending stress; halving that spacing again takes it to 7. The number a designer needs is not a rule of thumb about span over five; it is this length.01020304000.20.40.60.81diaphragm spacing (m)distortional ÷ bending stressdecay length 17.5 m3.0 × 2.0 m box · frame stiffness 238 kN·mm per radian per mm
Fig. 6 Peak distortional stress as a share of the bending stress, against diaphragm spacing. The flat right-hand end is the decay length doing its work: past it, one diaphragm cannot reach the next and adding them one at a time buys nothing until the spacing comes inside 17.5 m. Two diaphragms take the share to 19 per cent and five take it to 5.

What the transverse frame is really made of

KdK_d contains t3t^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 IdwI_{dw} goes as b2h2b^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 KdK_d where it was, so the distortional stress as a share of the bending stress goes up.

Shear stress across a sectionThe distribution of shear stress over an I-section, computed as VQ/It by accumulating the first moment of the area above every height. The peak is 0.38 against a mean of 0.21 — a ratio of 1.85 — and it falls at the neutral axis, where the bending stress is zero.neutral axispeak 0.4stressflow, q = VQ ÷ Imean stress 0.21 — the value a shear divided by an area would givepeak 1.85× that, and in the place bending ignoresthe flow is continuous; the stress jumps wherever the width does
Fig. 7 The shear flow the box was chosen for, which is the pure-torsion half of the decomposition. It is uniform round the cell, it is what makes the closed section stiff, and it is 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 γ\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.

Two mechanisms, and they add up to the torque at every sectionSaint-Venant torque and warping torque along a 400 by 200 mm I-section of 5 m, twisted by 2 kN·m with the ends fixed-fixed. J is 6.69×10⁵ mm⁴ and I_w 7.86×10¹¹ mm⁶, so k = √(GJ/EI_w) gives kL = 2.86 and a decay length of 1.75 m — 35% of the member. At each held end the shearing mechanism is exactly zero and all 1 kN·m is carried by the flanges bending in opposite directions; a decay length along, that share has fallen to 82%, and at the far end it is 100.0%. The two curves sum to the flat line at 1 kN·m at every one of the 161 stations, to the last bit of the arithmetic, which is the equilibrium of a slice of the member and is the only reason the split may be believed.012345-1-0.500.51distance along the member (m)torque (kN·m)1/k = 1.75 msum = ±1 kN·mone half each waysum = −1 kN·mSaint-Venantcirculating shearwarpingthe flanges bending
Fig. 8 The out-of-plane version, for comparison. The mechanism is the same in outline — a deformation the rigid-section theory has no coordinate for, restrained at the ends, producing longitudinal stress — and the geometry is at right angles to it.

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.

Shear stress across a sectionThe distribution of shear stress over an I-section, computed as VQ/It by accumulating the first moment of the area above every height. The peak is 0.47 against a mean of 0.26 — a ratio of 1.83 — and it falls at the neutral axis, where the bending stress is zero.neutral axispeak 0.5stressflow, q = VQ ÷ Imean stress 0.26 — the value a shear divided by an area would givepeak 1.83× that, and in the place bending ignoresthe flow is continuous; the stress jumps wherever the width does
Fig. 9 The action a plate diaphragm has to carry. The distortional force set arrives at its four edges and crosses it as in-plane shear, which is the one thing a plate is good at — until somebody cuts a hole in the middle of it.

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 10310^{-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, KdK_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.

How much of a point load the cable ends up takingThe fraction of a mid-span point load that reaches the cable, against μ = L√(H/EI) — how many characteristic lengths of girder fit in the span. A stiff girder gives a small μ and takes most of the load itself; a limp one gives a large μ and hands nearly all of it over. At μ = 17.0 the cable has 100% of it. What the curve does not show, and the shapes view does, is that the girder's real job is not on this axis at all: even where it carries almost nothing it is still the thing that turns a kink into a curve.0204060801001201401600%20%40%60%80%100%μ = L√(H/EI)share the cable takesthe whole loadwhat the cable takesleft is a stiff girder,right a limp one
Fig. 10 Where the box girder goes next: a deck whose job is to distribute rather than to carry. A diaphragm does the same thing one dimension down — it is not there to resist the load, it is there to make the cross-section behave as the analysis assumed it did.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

Beam on elastic foundationBox girderCross sectionDecay lengthDiaphragmDistortionEccentricityPlate bendingSecond moment of areaSelf equilibratingShear flowStiffnessSuperpositionTorsionWarping