Internal forces

The internal force with no diagram

A cut through a member reveals four things, and this collection has drawn diagrams for three of them. The fourth is a torque, it obeys exactly the same rules, and whether it exists at all can depend on a decision the designer is free to make.

Assumes What a cut reveals, and why it was there all along, The point that is not in the section and Six equations, and the drawing shows three.

Cut a member and four quantities appear on the face: an axial force, a shear, a bending moment, and a torque. This collection has drawn diagrams for the first three about forty times, and has not once drawn the fourth.

That is not an oversight in the writing. It is a property of the drawings: a plane diagram has one axis to turn about, and a torque turns about the axis the drawing is looking along. So the internal force that this field is named for has, for ninety essays, been invisible — and it obeys precisely the same rules as the others.

A torque diagram is a shear diagram about a different axisA torque of 40 kNm applied 2 m along a member of 6 m held against twist at both ends. The two ends take 26.7 and 13.3 kNm, in inverse proportion to their distances, because the two halves are springs in parallel and torsional stiffness is GJ over length. The diagram steps at the applied torque and closes at the far end, exactly as a shear diagram does — the only difference is which axis the arrows turn about.40 kNm26.7 kNm13.3 kNmthe step at the load is 40 kNm, and the diagram closes
Fig. 1 A torque of 40 kNm applied 2 m along a 6 m member held against twist at both ends. The ends take 26.7 and 13.3 kNm, in inverse proportion to their distances, because torsional stiffness is GJ over length and the two halves are springs in parallel. The diagram steps by the applied torque and closes at the far end, exactly as a shear diagram does.

Read that figure with the axis label covered and it is a shear diagram: constant between load points, stepping at each applied action, closing at the end. Everything the relationship between load, shear and moment established carries across whole, because the derivation used nothing but a free body and a sum.

Where a torque comes from

A member is twisted when the line of action of its load misses a particular point in its cross-section. Not the centroid — the shear centre, which for an asymmetric section is somewhere else entirely and is frequently outside the material.

The shear centre of a channelA channel of 80 by 200, with the shear flow in its flanges drawn. Those flows form a couple, so the load has to be applied 31.7 outside the web to leave the section untwisted — a point in the air, outside the material entirely.web centrelineshear centree = 31.7no twisttwiststhe flange flows are equal, opposite, and separated — which is a coupleand nothing about the section's 20.19 × 10⁶ second moment predicts it
Fig. 2 A channel with its shear centre marked, outside the metal. A load applied through the web produces a torque about that point, and the section twists as well as bending. The distance between the two is a purely geometrical property, and it is the lever arm of every accidental torque a channel ever carries.

The everyday sources are all of that form: a beam supporting a slab on one side only, a crane runway with the wheel load offset from the web, an edge beam carrying a facade hung outside its face, a curved beam in plan where the load’s line of action cannot pass through any straight axis. In each case the torque is small compared with the bending, and in each case it is carried by a mechanism far weaker than bending — which is the whole difficulty.

Two mechanisms, and one of them is hundreds of times better

A closed section resists a torque by a shear flow that circulates round the walls: constant round the perimeter, equal to T/2AmT/2A_m where AmA_m is the area the walls enclose. An open section cannot do that — a shear flow going round would have to cross the slit — so it resists instead by a shear flow that runs up one face of each plate and back down the other, which is a far smaller lever arm.

One slit, and the torsional stiffness falls by a factor of hundredsA 300 by 500 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 6.11×10⁸ mm⁴; slit, the loop is broken and only each wall's own thickness resists, giving 8.94×10⁵ mm⁴. The ratio is 683 to one, so the same torque twists the slit section 683 times as far and raises a peak shear stress 45 times as high. Nothing about the material changed.closedJ = 6.11×10⁸ mm⁴twist 0.017° over 3.0 mpeak shear stress 1.5 N/mm²slit along its lengthJ = 8.94×10⁵ mm⁴twist 11.869° over 3.0 mpeak shear stress 67.1 N/mm²J closed ÷ J open = 683
Fig. 3 The same 300 × 500 × 12 tube, closed and slit along its length. Every millimetre of steel is present in both. The closed section’s torsion constant is 6.11 × 10⁸ mm⁴ and the slit one’s is 8.94 × 10⁵ — a factor of 683 — and for the same applied torque the shear stress is 45 times higher in the slit one.

That figure is the most extreme geometry-beats-material result in the collection, and it is worth comparing with the others. Rearranging the same steel changes a section’s bending stiffness by perhaps a factor of ten; a saw cut two millimetres wide, removing about a tenth of a per cent of the material — 24 mm² of 19,200 — changes its torsional stiffness by a factor of several hundred. Nothing else on this site is that sensitive to anything.

The arithmetic behind it is short. For a thin closed tube, Bredt’s formula gives

J=4Am2ds/tJ = \frac{4A_m^2}{\oint ds/t}

which contains the enclosed area squared — so it grows as the fourth power of the size. For an open section made of thin plates,

J=13bt3J = \frac{1}{3}\sum b t^3

which contains the thickness cubed and does not know how far apart the plates are at all. One formula is about the shape’s extent; the other is about how thick its pieces are. A closed section resists torsion with its outline and an open one resists it with its wall thickness, and no amount of making an open section deeper helps.

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 0.41 against a mean of 0.19 — a ratio of 2.15 — and it falls at the neutral axis, where the bending stress is zero.neutral axispeak 0.4stressflow, q = VQ ÷ Imean stress 0.19 — the value a shear divided by an area would givepeak 2.15× that, and in the place bending ignoresthe flow is continuous; the stress jumps wherever the width does
Fig. 4 Shear flow through an I-section under a transverse shear, sweeping the section and accumulating first moments. The same picture explains the torsional case: a closed tube lets this flow circulate round a loop enclosing a large area, and an open section leaves it nowhere to go but back along the plate it came from.

What holds the ends decides more than the section does

The formula J=13bt3J = \frac{1}{3}\sum bt^3 describes uniform torsion, in which every cross-section is free to warp — to move out of its own plane, so that the flanges of an I-section slide along the member relative to each other. Prevent that warping at a support and a second mechanism appears: the flanges bend in their own planes, in opposite directions, and the pair of flange shears is itself a torque.

Two mechanisms, and which one is working whereThe share of an applied torque carried by circulating shear and by bending of the flanges, along a member of 6 m held against warping at its left-hand end. The parameter λL is 10.91 and the decay length 1/λ is 0.55 m: at the restrained end every bit of the torque is carried by the flanges bending in opposite directions, and about three decay lengths along, none of it is. The restraint stiffens the member by a factor of 1.101 — which is worth having on a short member and nothing at all on a long one.012345600.20.40.60.81distance from the restrained end (m)share of the torqueone decay length, 0.55 mSaint-Venant:shear round the sectionwarping:flanges bending apart
Fig. 5 The share of the applied torque carried by each mechanism along a 6 m member held against warping at its left end. At the restrained end every bit of it is carried by the flanges bending apart; one decay length along — 0.55 m here — most of that has been handed over to the circulating shear, and past three of them there is none of it left. The parameter λL is 10.91 and the restraint stiffens the member by a factor of 1.101.

The decay length is 1/λ1/\lambda where λ=GJ/EIw\lambda = \sqrt{GJ/EI_w}, and it does the same job in this field that Saint-Venant’s principle does in the connections field: it says how far a disturbance at a boundary reaches into a member. For open sections it is a fraction of a metre, so a long member is almost entirely in uniform torsion and the restraint at its ends buys nothing.

The interesting case is a short one, where λL\lambda L is a small number and the whole member is inside its own boundary layer:

member length λL stiffening from warping restraint
1.5 m 2.73 1.571
3 m 5.45 1.224
6 m 10.91 1.101
12 m 21.82 1.048

A 1.5 m member is 57% stiffer in torsion than its torsion constant says, and a 12 m one is 5% stiffer. That is a large effect appearing and disappearing over ordinary lengths, and it is the reason torsional stiffness quoted as a section property is a partial answer: in torsion, the length is a property of the section too.

For closed sections the same calculation gives λL=109\lambda L = 109 for the box above, and the stiffening is 1.009. Warping restraint is a phenomenon of open sections, and the reason is the one already established — their uniform torsional stiffness is so small that anything else is comparable with it.

A torque that has to exist, and a torque that does not

The distinction this field turns on has no analogue in bending, and it is the reason a competent designer can make a torsion problem disappear rather than solve it.

Equilibrium torsion is a torque required by statics. A cantilevered canopy hung off one side of a beam delivers a torque that nothing else can carry; take the beam’s torsional stiffness to zero and the canopy falls off. There is no design decision here — the torque is as real as any bending moment and the member must be sized for it.

Compatibility torsion exists only because two members are joined and must rotate together. A floor beam framing into the side of an edge beam wants to rotate at its end; the edge beam resists, and the torque that arises is whatever it takes to make the two rotations equal. Nothing about equilibrium requires it.

A torque that can be declinedThe share of a joint's moment attracted into a torsional member, against that member's torsional stiffness measured in units of the bending stiffness it is competing with. The two are springs in parallel, so the share goes to zero with the stiffness: an open section of the same size attracts 70% where a closed one attracts 0.3%, a difference of 683 times in torsion constant. Where the torque is a matter of compatibility rather than of equilibrium, softening the member is a way of not having the problem — and nothing falls down.10⁻³10⁻²10⁻¹110¹00.20.40.60.81torsional stiffness ÷ bending stiffnessshare of the moment taken in torsionclosed box: 70%open section: 0.3%
Fig. 6 The share of a joint’s moment attracted into a torsional member, against its torsional stiffness measured against the bending stiffness it is competing with. The two are springs in parallel, so the share goes to zero with the stiffness: the closed box takes 25.5% and the same section slit takes 0.34%. In the second case the floor beam simply behaves as though it were simply supported, and nothing has failed.

That curve is the practical content of this whole essay. Faced with a compatibility torque, three responses are available: make the member strong enough to carry it, make it stiff enough that it attracts even more of it, or make it soft and let the torque leave. The third is usually the right answer, and it is the opposite of the instinct that adding stiffness is adding safety.

The one condition is that the load must have somewhere else to go. Softening the edge beam sends the moment back into the floor beam, which now behaves as a simply supported member and needs the midspan capacity to match. That is a redistribution, and it is legitimate exactly when the receiving member has the ductility to accept it — the same argument, in a different field, with the same condition attached.

The check that governs is usually the rotation

A member in torsion is rarely limited by its strength. The shear stresses a working torque produces in a closed section are modest — Bredt’s flow spread over the whole perimeter is an efficient way to carry anything — and what runs out first is the amount of twist the things attached to the member will tolerate.

A torque diagram is a shear diagram about a different axisA torque of 25 kNm applied 4.5 m along a member of 9 m held against twist at both ends. The two ends take 12.5 and 12.5 kNm, in inverse proportion to their distances, because the two halves are springs in parallel and torsional stiffness is GJ over length. The diagram steps at the applied torque and closes at the far end, exactly as a shear diagram does — the only difference is which axis the arrows turn about.25 kNm12.5 kNm12.5 kNmthe step at the load is 25 kNm, and the diagram closes
Fig. 7 A 25 kNm torque at the middle of a 9 m member restrained at both ends: 12.5 kNm to each end, by symmetry. The torque diagram is the least interesting drawing in the field and the rotation it implies is the number that decides the design — the twist is the torque times the length over GJ, and GJ is the quantity a section shape can change by three orders of magnitude.

The numbers make the case. An edge beam supporting a facade twists by TL/GJTL/GJ, and a rotation of a hundredth of a radian at the top of a 3 m cladding panel is 30 mm of movement at its head — a serviceability failure with no stress anywhere near a limit. The same beam as a closed box twists by 1/683 of that, which is nothing at all.

This is why the answer to a torsion problem is so often a change of section rather than a change of size. Deepening an open member increases its bending stiffness and leaves its torsional stiffness essentially unchanged; welding a plate across the open face of a channel to make a box multiplies the torsional stiffness by hundreds and the bending stiffness by a few per cent. The two stiffnesses respond to completely different changes, which is unusual — for most of this collection, making a member better at one thing makes it better at most things.

The one place plane analysis is completely blind

A plane frame analysis cannot report a torque, because in its coordinate system there is nowhere for one to be. Every commercial frame program will happily analyse a grillage of beams at right angles and report torsion in the members — but only if the model was built as a grillage, which is a decision made before any analysis ran.

This is the missing three equations at their most concrete. The moment at the end of a floor beam, in the plane analysis of that beam, is a number resisted by “the support”. In the real building it is resisted by the edge beam twisting, and the amount depends on a torsional stiffness that appears in neither model. Both analyses can be correct and the pair of them can still miss a member’s governing action, because the action lives in the joint between two models.

A moment about an axis, not about a pointThe same force, and the moment it makes about three different axes through the same point. In a plane drawing there is only one axis and the moment is a number; in three dimensions it is a vector, and what matters is its component along the axis the structure can actually resist. The component drawn faint is the one a plan view throws away, and it is the one that twists a beam rather than bending it.F100%82%34%the moment resisted is F·d·cos θ — the rest goes somewhere elseeach axis is drawn through the point, both ways
Fig. 8 The same force and its moment about three axes through one point. A member can only resist the component along an axis it has stiffness about; the rest is carried by something else or by nothing. In a plane analysis the other two components are assumed to be somebody’s problem, and torsion is the commonest way for that assumption to be wrong.

The history is a case of the tail wagging the dog

Torsion was solved for a circular shaft by Coulomb in 1784, and for anything else in 1855 by Saint-Venant, whose treatment introduced warping as the thing a non-circular section does that a circular one does not. The engineering pressure came from machinery: a drive shaft’s whole purpose is to carry a torque, so the theory was developed by people for whom torsion was the design case rather than a nuisance.

Structural engineering inherited it and needed the awkward half. Machine shafts are circular and closed; building members are open and thin-walled, which is exactly where the simple theory stops working and where the warping term Vlasov added in the 1930s is required. The order of events is worth noticing: the case that is easy to derive is the one structures never use, and the theory that structures need was the correction.

The practical consequence in the standards is a marked reluctance. Design codes treat torsion with more conservatism than anything else, and the recommended approach for compatibility torsion is usually to detail for it rather than to compute it — a piece of engineering judgement that the curve above justifies exactly.

What the picture cannot show

The whole treatment is elastic and thin-walled. Bredt’s formula assumes the shear flow is uniform through the wall thickness, which is a good approximation for a wall thinner than a tenth of the section and a poor one for a solid rectangle.

Combined actions are absent. A member carrying bending and torsion together has shear stresses from both adding on one face of a flange and subtracting on the other, and the check that matters is the combined one. The figures above draw each mechanism alone, which is what makes them legible and what makes them incomplete.

The torsional restraint at the ends is drawn as perfect or absent. Real end conditions are neither, and the torque a member attracts is decided by the same stiffness comparison as everything else here — so an assumption about the support decides the answer, and the assumption is rarely stated.

Where the ladder goes

The first rung is the one that has appeared twice already and has its own essay waiting: what happens to a member whose load misses its shear centre by an amount nobody intended, which is a question about tolerance rather than about analysis.

The second is the failure this field produces. A beam in torsion does not usually fail by twisting; it fails because the torsion has reduced its capacity in bending, or because the twist has moved its compression flange sideways and it has buckled laterally. Torsion is more often the mechanism of some other failure than a failure of its own.

The third is the one this essay opened with. A cut reveals four internal forces and this collection has now drawn all four — which means the next question is what happens when a section carries several of them at once, and how an interaction surface is drawn in more than two dimensions.

Named alongside this one

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

What links here

Every essay whose body links to this one.

The objects this essay names

Each one links to every other essay that touches it.

Compatibility torsionInternal forcesShear centreShear flowStiffnessTorsionTorsion constantWarping