Two cells, one equation, and a web with nothing in it

A single closed cell under torque is one of the few genuinely easy calculations in this subject. The shear flow $q$ — force per unit length round the wall — is constant all the way round, because a longitudinal cut anywhere shows it has nowhere to change; taking moments about any point in the plane gives $T = 2qA$, where $A$ is the area the wall encloses; and that is the whole thing. One unknown, one equation, no stiffness anywhere in it.

Now put a web down the middle. There are two cells, two shear flows, and $T = 2(q_1A_1 + q_2A_2)$ is still one equation.

The equation that is missing #

The section has become torsionally redundant, to a degree equal to the number of cells less one. What statics cannot supply, compatibility must, and the compatibility statement is the one thing that has to be true of two cells sharing a wall: they twist by the same amount. They are parts of the same cross-section, and a cross-section rotates as a unit.

Writing the rate of twist of cell $i$ as a circuit integral round its own walls,

$$2A_i\,G\theta' \;=\; \oint_i \frac{q_s}{t}\,ds,$$

where $q_s$ is the flow in each wall as seen going round cell $i$. On the outer walls that is $q_i$. On a wall shared with cell $j$ it is $q_i - q_j$, because the two cells’ flows run in opposite directions through the shared wall and what physically exists there is their difference.

That gives one equation per cell, plus the torque equation, and the system closes. It is worth noticing what has entered: the shear modulus, the wall thicknesses and the cell areas are all in the compatibility equations and none of them is in the torque equation. An indeterminate problem is one whose answer depends on stiffness, and a multi-cell box has just become one — which is the price every redundant structure pays for the redundancy, and it is paid here by a cross-section rather than by a frame.

The shear modulus turns out to cancel, because it multiplies every cell’s twist equally and the twists are being set equal. The thicknesses do not.

Why it is zero, without solving anything #

The result does not need the linear algebra. The box is symmetric about the internal web; the torque is symmetric about it; so the solution must be. Two identical cells under identical conditions carry identical flows, and $q_1 - q_2 = 0$ by the symmetry alone.

What is worth checking is the second half of the claim — that $J$ is then exactly the single-cell value rather than merely close to it. With the shared wall carrying nothing, cell 1’s circuit integral has no contribution from it, so

$$2A_1 G\theta' = q\sum_{\text{cell 1's outer walls}} \frac{s}{t},$$

and with $A_1 = A/2$ and cell 1’s outer walls being half the outer perimeter’s $\sum s/t$, this rearranges to $J = 4A^2 / \oint (ds/t)$ over the outer perimeter. Bredt’s single-cell formula, with the web absent from every term. The web is not merely unhelpful; it is not in the answer.

The consequence for design is direct and slightly startling. The web in the middle of a symmetric box girder — several hundred millimetres of concrete or a full-depth stiffened steel plate, running the length of the span — is doing nothing at all about torsion. Everything it is in the section for is something else.

It is also, stated carefully, not quite the same claim as the web is unstressed. The web carries no shear flow from the torque. It carries plenty from everything else, and the finding is about attribution rather than about the member. That distinction matters because it is the one a design check gets wrong in the safe direction and a section optimisation gets wrong in the other: nobody is going to under-design the web by believing this, and somebody might delete it.

The same shape of finding has appeared on this site before in a truss. A member can carry the full applied panel load and contribute exactly nothing to the deflection being measured, because the zero belongs to the pair — member and question — rather than to the member. The web here is idle with respect to torsion, and to nothing else.

What the web is actually for #

Three things, and none of them is on this page’s title.

The first is vertical shear. Under vertical load the webs carry the bending shear, and here the internal web is a full participant rather than a bystander: the shear from bending runs down all three webs roughly in proportion to their share of the section, so removing the middle one raises the stress in the surviving two by about half. That flow is not a loop. It starts at the free edge of the top slab, accumulates down the web and returns to zero at the bottom, because it is the change in bending stress being handed across the section. Nothing about it closes on itself, which is precisely why the web that is invisible to the torsion calculation is unavoidable in this one. A box with three webs is a box designed for shear, whatever its torsion does with them.

Plate buckling is the fourth reason and it is often the governing one: a 3 m wide compression flange with a web under its centre is two 1.5 m plates rather than one 3 m plate, and the buckling stress of a plate goes as the inverse square of its width. Halving the width quadruples the critical stress, which is a very large return for a member the torsion calculation values at nothing.

And there is a fifth that is not structural at all. A wide single-cell box has an internal void several metres across with no support to its soffit formwork during construction; a web under the middle of it is something to prop from. The structure that exists during construction is not the one being designed, and members put in for the first are checked in the second and then quietly credited with helping.

Move it and it starts working #

If the flow is the difference between two cells, then making the cells different should make the difference nonzero. It does.

The direction of the difference is worth reading. The smaller cell carries the larger flow, which is the opposite of what an area-based intuition suggests. A cell twists at a rate set by its flow divided by its area, so a small cell needs a large flow to twist as fast as a big one — and the shared wall then carries the surplus, in the direction that runs against the small cell’s circulation.

Three per cent is the honest size of the whole effect, and it needs a comparison to be read properly.

More cells #

Three cells is the first case where symmetry does not save the arithmetic, and it shows what the general answer looks like.

The middle cell’s circuit runs through two thin internal webs rather than one thin and one thick outer web, so its $\oint ds/t$ is smaller and it needs a larger flow to twist at the same rate. The flows are decided by the wall thicknesses and the cell areas, and by nothing about the material — $G$ cancels out of the ratios, exactly as it does for a single cell.

That cancellation is worth a sentence of its own, because it is not obvious and it is what makes the calculation usable. The compatibility equations each contain $G\theta'$ on one side; setting them equal to one another eliminates it, and the torque equation then fixes the scale. So a multi-cell box’s distribution of flow is a pure geometry problem, and only its twist per unit length needs a modulus — which is the same separation that lets a truss’s forces be found before anything is known about its members, arriving here in a section rather than a structure.

The general result generalises the two-cell one. A web between two cells carries the difference of their flows, and two cells carry equal flows when their circuit integrals divided by their areas are equal. Symmetry is the easy way to arrange that and not the only way: a small cell with thin walls and a large cell with thick ones can be tuned to carry the same flow, and the web between them would then also carry nothing. Nothing about the arrangement has to look symmetric for the result to hold.

Stated generally, this is the force method arriving inside a cross-section. A redundant structure needs as many compatibility statements as it has redundancies, and each statement is one deformation set equal to another. In a frame, choosing which quantity to release is the whole art of it, and two engineers may pick different releases and get the same answer by different arithmetic. In a multi-cell box nobody gets to choose: the cells are the releases, the equal twists are the compatibility conditions, and the number of redundancies is the number of cells less one. A four-cell box is three simultaneous equations before it is anything else, and it is the section’s own geometry that writes them.

The number a designer actually needs #

The reason any of this is computed is that a box girder on a curved alignment, or one carrying an eccentric load, has a torque to get to its bearings, and the check is a shear stress in a wall. That stress is the flow divided by the thickness, so the busiest wall is the thinnest one and not the one carrying most.

For the symmetric box under 4,000 kNm the flow is 444.4 N/mm everywhere on the outer loop, and the walls report 1.78 N/mm² in the 250 mm top slab, 2.02 in the 220 mm bottom slab, and 1.27 in the 350 mm webs. The soffit is the critical wall and it is the one nobody thinks of as a shear member. It is also the wall most likely to be thinned in a value-engineering exercise, because it carries no bending reinforcement over most of the span and looks like the cheapest place to save concrete.

Two more consequences follow from the flow being constant round the loop. Any wall can be checked in isolation once the flow is known, which is why a torsion check reduces to a list. And the sum of the wall thicknesses is irrelevant — what matters is the smallest one, exactly as a chain’s strength is its worst link and not its average. Thickening the webs of this box does nothing whatever for its torsional shear stress; thickening the soffit by 15 per cent removes the governing check.

The outline is the whole of the stiffness #

The multi-cell arithmetic answers how the flow divides. The question of how much torsional stiffness the section has at all is answered by one term, and it is worth putting the two side by side because they point in different directions.

Bredt’s constant is $J = 4A_m^2/\oint ds/t$, and the enclosed area is squared while the perimeter appears once. So a section’s torsional stiffness is dominated by its outline and barely notices anything inside it.

Take a box 4 m wide by 2 m deep: $A_m = 8$ m², perimeter 12 m. Widen it to 6 m: $A_m = 12$ m², perimeter 16 m. The area term rises by $(12/8)^2 = 2.25$ and the perimeter term by $16/12 = 1.33$, so

$$\frac{J_{6}}{J_{4}} = \frac{2.25}{1.33} = 1.7$$

A fifty per cent wider box is seventy per cent stiffer in torsion. Now add a central web to the original instead. The enclosed area of the outline has not changed at all, and this essay’s own finding is that the web carries almost no flow — so the torsional stiffness is essentially where it was.

Which gives the design instruction in one line. To make a box stiffer in torsion, make it bigger; adding cells does nothing. Cells are added for other reasons entirely — to carry the deck’s transverse bending, to shorten the panels for shear buckling, to provide a load path at a support — and every one of those is a reason that has nothing to do with the quantity the multi-cell equations were written to find.

And the torque has to be measured from the right point #

Everything above computes the flows produced by a torque. What produces the torque is a load times an eccentricity, and the eccentricity is measured from the shear centre rather than from the centroid.

For a doubly symmetric box the two coincide and nobody notices. For an asymmetric one they do not, and the difference is exactly the sort a bridge has: a longer cantilever on one side, a footway, a services bay, unequal cells. There the shear centre is displaced from the centroidal axis, and a load applied on the centreline of the deck produces a torque.

The consequence is easy to state and easy to get wrong. A designer who computes the torque as the load times its distance from the section’s centre of gravity has used the wrong reference for an asymmetric section, and the error is a pure offset — it does not vanish for a centred load, it changes sign for a load on the other side, and it is largest for the load case everybody assumes is torsion-free.

Locating the shear centre of a closed section uses the same machinery as this page: apply a transverse shear, solve for the flows with the twist set to zero, and find where the resultant of those flows acts. It is the same point defined two ways — the point a load causes no twist about, and the point the section twists about — and for a closed cell it is found from a compatibility condition rather than by integration, which is the one respect in which a box is harder than a channel.

The saving grace is that a box’s shear centre is usually inside the section, often close to the centroid, and never far outside it. A channel’s is outside the metal and an angle’s is at a corner; a box’s misbehaviour is measured in millimetres rather than in section depths, which is another of the quiet reasons boxes are used where a torque is expected.

Where the model stops #

Every wall is thin and every flow is uniform across it. Bredt’s assumption, and progressively wrong as walls thicken. A concrete box girder’s webs at 350 mm on a 1.5 m depth are not thin by any reasonable standard, and the shear flow across such a wall is not constant.

The section does not distort. The compatibility statement used is that the cells rotate together as a rigid cross-section, which is exactly the assumption distortion violates. For pure torque on a box with adequate diaphragms this is close to true; for an eccentric point load on a long unbraced box it is not, and the flows found here are the torsional part of a three-part answer.

Warping is free. Closed sections warp very little and the warping stresses are usually neglected, which is standard and is not always right — a box with a very wide, very thin cell warps more than the calculation admits, and a box restrained at a diaphragm generates axial stresses this has no term for.

It is elastic throughout. At collapse the flows redistribute: a wall that reaches its shear capacity sheds flow to the others, and the plastic torque of a multi-cell box is a mechanism problem rather than a compatibility one. The centred web that carries nothing elastically may carry a great deal at collapse, which is a good reason not to design it away.

The webs are vertical. A trapezoidal box — which is most of them, because a sloping web sheds rain and looks better — has webs whose length is not the depth, and both the circuit integral and the enclosed area change. The arithmetic is identical and none of the numbers on this page survive it.

And the drawing is of a torque alone. No box girder anywhere is loaded in pure torsion. Every real load case is bending plus torsion plus distortion, the webs carry all three at once, and the picture of a web with nothing in it is a picture of one third of one load case. What the calculation licenses is not removing the web — it is declining to claim credit for it in the torsion check.

The ladder from here #

Later rungs on this anchor: the shear-flow superposition properly done, with bending, torsion and distortion resolved on the same webs and the governing wall found from the sum rather than from any part of it. Cells that are not rectangles — a trapezoidal box, an aerofoil, a ship’s hull section — where the areas and perimeters have to be computed before the system can be assembled. Open cells connected to closed ones, where a cantilevering deck slab contributes $bt^3/3$ to a section otherwise dominated by a closed loop and the two contributions differ by three orders of magnitude. The plastic torque of a multi-cell section, from the sand-heap analogy with islands in it. Shear lag across a wide cell, which changes the effective wall a flow is running through. And the same redundancy in a structure rather than a section: a frame with more restraints than equations is solved by the identical move — count the redundancies, write one compatibility statement for each, and let the stiffnesses decide what statics could not.