Two volumes, and both of them are torques

Every other section property on this site is an integral anybody can do. The second moment of area is $\int y^2 \,dA$; the first moment above a cut is $\int y \,dA$; the shear flow round a thin wall is that first moment divided by two numbers. Hand a draughtsman a shape and a calculator and the answers come out.

The torsion constant of a solid section that is not a circle is not like that. There is no integral to do, because the quantity being integrated is the solution of a partial differential equation over the section, and that equation has a closed form for exactly three shapes: the circle, the ellipse, and the infinitely thin strip. For a square, a rectangle, a hexagon, a rail, a crane hook or a propeller shaft with a keyway there is nothing to write down.

What the equation says, and why it is hard #

Saint-Venant’s insight in 1855 was to stop assuming plane sections stay plane. A twisted circular shaft’s do; nothing else’s does, and the axial movement — the warping — is an unknown function over the cross-section that has to be found before anything else can be.

The trick that makes the problem tractable is to work with a stress function $\phi$ rather than with the warping, chosen so that the two shear stresses are its derivatives:

$$\tau_{zx} = \frac{\partial\phi}{\partial y}, \qquad \tau_{zy} = -\frac{\partial\phi}{\partial x}.$$

Equilibrium is then satisfied automatically, and compatibility reduces to one equation:

$$\nabla^2\phi = -2G\theta',$$

with $\phi = 0$ on the boundary — because no shear can cross a free surface, so the boundary must be a contour of $\phi$. From it, two statements follow that make the whole method:

The torque is twice the volume under $\phi$. $T = 2\iint \phi \,dA$.

The shear stress at a point is the slope of $\phi$ there, and it runs along the contour.

That is a complete solution and it is completely useless without a way of solving Poisson’s equation on an arbitrary domain, which in 1855 there was not.

It is worth naming what has just happened, because it is a move this collection makes over and over. A quantity that could not be computed — the warping — has been replaced by a quantity that can, chosen so that one of the two governing conditions is satisfied identically and only the other has to be worked at. The unit-load method does the same thing with a deflection, and the stress block does it with a distribution nobody can measure. The price is always the same: the new quantity has no physical meaning, so the answer comes out of it by a rule rather than by inspection.

The film #

Ludwig Prandtl’s observation of 1903 is that a thin membrane stretched over a hole and blown up from beneath satisfies exactly the same equation. Its deflection $w$ obeys $\nabla^2 w = -p/S$, with $w = 0$ on the rim. Same operator, same boundary condition, same shape of answer.

So the film’s height is $\phi$, up to a constant that comes out in the wash. The volume under it is the torque, the slope of it is the stress, and its contour lines are the shear trajectories. Cut a hole the shape of the section, stretch a soap film over it, put a small pressure underneath and measure the bubble.

The figures on this page are not photographs of soap films. They are the same equation relaxed numerically — successive over-relaxation on a grid, iterated until nothing moves — which is the twentieth century’s answer to the same difficulty. But the relaxation and the film are computing the same object, and the film got there first by fifty years.

Where the stress is, and where it is not #

The most useful thing the analogy gives is not a number. It is an instant, correct intuition about where a twisted section is in trouble, available to anyone who can picture a bubble.

The stress is largest where the film is steepest, which is on the boundary, at the point of the boundary nearest the middle of the section. On a rectangle that is the middle of the long side. On a shape with a re-entrant corner it is at the corner, where the film has to dive into a crevice.

The stress is zero at a convex corner. A film over a square hole meets the rim along both edges at a corner, so it arrives there flat: it has already come down to zero from both directions, and there is no slope left. The corner of a twisted square bar carries no shear at all, which is a claim nobody believes on first hearing and which the picture makes obvious in a second.

The stress is zero at the centre, because the film is at its peak there and a peak has no slope. So the material on the axis of a twisted bar is doing nothing whatever, which is the torsional restatement of why depth is worth more than area and is the entire argument for a hollow shaft. Torsion is the one action in this collection whose stress is nothing in the middle of a solid section — bending’s worst stress is at the extreme fibre and shear’s is at the neutral axis, and torsion’s is at the boundary but not at all of it.

The heap #

The second analogy takes the same picture to collapse, and it needs no relaxation at all.

When every point of a section has yielded in shear, the magnitude of the stress is $\tau_y$ everywhere, which means $|\nabla\phi| = \tau_y$ everywhere. A surface of constant slope over a given base is a heap of dry granular material at its angle of repose: pour sand onto a plate cut to the section’s shape and let it settle. The height at any point is $\tau_y$ times the distance to the nearest edge, and the plastic torque is twice the volume.

The circle’s heap is a cone, of volume $\pi\tau_y R^3/3$, so $T_p = 2\pi\tau_y R^3/3$. Against the first-yield torque $\pi\tau_y R^3/2$, the shape factor is

$$\frac{T_p}{T_y} = \frac{2\pi/3}{\pi/2} = \frac{4}{3}$$

exactly — and the relaxation returns 1.3334.

Both shape factors are larger than the same sections give in bending, and the reason is geometric rather than material: the elastic stress in torsion falls off from the boundary in two dimensions rather than one, so a larger fraction of the section is understressed at first yield and there is more left to recruit.

The reserve is also much less useful, and for a reason the heap makes plain. A section reaches its plastic torque only when every point of it has yielded, and the twist required to do that is large — the whole section has to be strained past yield, not just an outer layer. Ductility is being assumed on a scale nothing else on this site asks for, and a member twisted that far has usually failed at something else first: a bolt has slipped, a weld has torn, or the twist itself has become the limit state.

Why the circle wins #

Every solid section can now be compared on the same basis, and the ranking is not the one shape intuition supplies.

At equal area, a circle gives $J = 1.5914\times10^7$ mm⁴ and a square gives $1.4053\times10^7$ — the circle is 13.2 per cent stiffer for the same material. The film explains it in one sentence: the volume under a film is largest when the boundary is as far as possible from as much of the interior as possible, and the circle is the shape that maximises that. Every corner is material sitting where the film is nearly flat and therefore contributing almost nothing.

The rectangle’s approach to the thin-strip limit is worth reading off the same machinery. Writing $J = \beta b t^3$:

The thin-strip formula $bt^3/3$ is the film’s ridge running the length of a long narrow hole, where the ends stop mattering. At $b/t = 10$ it is 6 per cent optimistic and at $b/t = 2$ it is 46 per cent — which is why the sum $\sum bt^3/3$ used for an open section is written for thin plates and is not a general answer.

The thin-walled version is the same picture #

The analogy does not stop at solid sections; it explains the thin-walled result too, and rather better than the usual derivation does.

Bredt’s $q = T/2A$ is that plateau read as a volume: a flat roof of height $h$ over an enclosed area $A$ has volume $hA$, so $T = 2hA$, and the slope across a wall of thickness $t$ is $h/t$ — which is the shear stress. The famous formula is a rectangle’s area, seen sideways.

The contours of the film are shear flows, and it is worth being exact about what distinguishes them from the shear flow a bending moment produces. Both are flows in the same sense: a force per unit length running along the wall, constant through the thickness, with nowhere to leak away to. The difference is where they end. A bending shear flow starts at a free edge, builds along the wall and returns to zero at the far free edge, because it is accumulating the change in bending stress over the material it has passed. A torsional shear flow has no ends at all — the plateau’s contours are closed loops, and a loop that closes on itself never has to start anywhere. That is the whole of why an open section is hopeless in torsion and perfectly adequate in bending shear: the first mechanism needs a closed circuit and the second does not.

The ratio between the two volumes is a shape factor #

The film gives the elastic behaviour and the heap gives the plastic one, and the ratio between the torques they predict is the torsional twin of a quantity this collection has spent a good deal of time on.

For a circle: the elastic torque at first yield is $\tau_y J/r = \pi\tau_y r^3/2$, and the heap is a cone of height $\tau_y r$, so twice its volume is $2\pi\tau_y r^3/3$. The ratio is

$$\frac{T_p}{T_y} = \frac{2\pi/3}{\pi/2} = \frac{4}{3} = 1.333$$

For a thin rectangle the heap is a prism with a triangular section, and the same arithmetic gives 1.50. For a square it is a pyramid, and the ratio is $0.333/0.208 =$ 1.60.

Those are torsional shape factors, and they behave exactly like the bending ones: the reserve past first yield is the material that was under-stressed, being brought up. A circle has the least of it because its elastic shear stress is already proportional to radius and therefore already large over most of the section; a square has the most, because its corners carry no shear at all elastically and its edges carry it unevenly.

And the ordering is different from bending. A solid circle’s bending shape factor is $16/3\pi = 1.70$, well above a rectangle’s 1.50; its torsional one is 1.33, well below. The same two shapes, ranked oppositely by the same kind of ratio, because the two problems put the under-stressed material in different places — bending wastes what is near the neutral axis and torsion wastes what is near a corner.

The ridges on the heap are visible on the bar #

The film’s contours are the elastic shear trajectories. The heap has trajectories of its own, and they are worth naming because they can be seen.

In the fully plastic state the shear stress is $\tau_y$ everywhere — the heap has a constant slope by construction — so what varies is only the direction. The direction is along the contour of the heap, which means the shear runs parallel to the nearest boundary, and the lines of steepest descent run perpendicular to it.

That makes the heap’s ridges the interesting feature. A ridge is a line where two slopes meet, so it is a line where the shear direction changes discontinuously — sand arriving from two sides at once. For a rectangle the heap is a roof: a straight ridge down the middle over most of the length, with a 45-degree hip at each end.

Those ridges are not an artefact of the analogy. A steel bar twisted well past yield and then etched shows the same pattern, because the ridge is where the material has been sheared in two directions and its microstructure records it. The sand heap is a physical model whose geometry — not merely its volume — corresponds to something in the metal.

It also settles a question the elastic solution leaves open. Elastically the corners of a rectangle carry no shear at all, which is the free-surface argument and is exactly true. Plastically they carry $\tau_y$ like everywhere else, and the corner is simply where two 45-degree hips meet. The corner goes from carrying nothing to carrying as much as anything, which is the largest single redistribution in the section and is where most of the 1.60 comes from.

There is a practical residue of the ridge that is worth having. The hip lines meet the boundary at 45 degrees, so on a rectangle they arrive at the corners — and on a section with a re-entrant corner, such as a keyway or the junction of a flange and a web, the heap has a pit rather than a ridge and the film is infinitely steep. Elastically that is an unbounded stress concentration; plastically it is a small region that yields early and redistributes. Which is the same pairing as everywhere else in this collection: an elastic singularity and a plastic non-event, with a ductile material deciding which of the two the member experiences.

Where the model stops #

Saint-Venant torsion is uniform torsion. Every result here assumes the section is free to warp identically at every station along the member, so no axial stress is generated. Restrain that warping — at a fixed end, at a change of section, or by applying the torque non-uniformly — and a completely different mechanism appears alongside this one, with axial stresses the stress function knows nothing about.

The relaxation is a grid. The film is exact and the numerical film is not: the square comes in 0.035 per cent low and the 4:1 rectangle 0.247 per cent, and the difference between those two errors is the aspect ratio squeezing the grid rather than anything about the solver. The boundary treatment matters more than the mesh does — with the arms next to a curved boundary taken as full grid steps, the circle came out 2.3 per cent wrong; with the true distances used, seven parts in a hundred thousand.

The sand heap assumes rigid–plastic material and no hardening. The real collapse torque of a steel bar is higher, because steel hardens; the real usable torque is lower, because the twist required to reach a fully plastic state is large — of the order of five times the first-yield twist for a circle, and torsional deformation is not usually something a structure has to spare.

Neither analogy sees a hole. A section with an internal void has a film with an island in it, held at an unknown constant height rather than at zero, and that height is another unknown fixed by a circulation condition — which is the multi-cell problem and is a different calculation.

The comparison at equal area is one comparison. Sections are not usually chosen at equal area — they are chosen at equal depth, or equal outside dimension, or from what a catalogue happens to contain — and on those bases the ranking moves. At equal outside dimension the square beats the circle, because it has more material.

And the drawings are of a state. The contours are the shear trajectories at one instant of one loading; they do not move as the torque grows, because the problem is linear until it yields. What they cannot show is the warping itself, which is the out-of-plane movement that the whole formulation was invented to avoid having to draw.

The ladder from here #

Later rungs on this anchor: sections with holes, where the film gains a flat island and the island’s height is the extra unknown. The keyway and the fillet, where a re-entrant corner makes the film infinitely steep and the elastic stress concentration is unbounded — the practical reason shafts have generous radii. The narrow open section built of several plates, and where the sum $\sum bt^3/3$ stops being adequate. Composite and layered sections, where the film has a kink at every change of modulus. Numerical torsion in general, where this relaxation is the ancestor of the finite element method and Southwell’s relaxation tables of the 1940s are the same arithmetic done by hand. And the analogy’s cousin: the same Poisson equation, over the same domain, describes steady seepage, laminar pipe flow and the shape of a stretched drum — so a torsion constant is also a flow rate and a fundamental frequency, and the tables in an old handbook of one of them are tables of all of them.