Structural Engineering & Statics
where the load goes
Sections and stress

The point that is not in the section

A channel loaded down its web twists. To stop it, the load must be applied through a point outside the steel entirely — in the air beside the section, where nothing can be attached.
15 min read 6 figures Geometry beats materialWhich failure arrives first

Load a channel section vertically, straight down the middle of its web, and it will twist. Not slightly and not because of an imperfection — twisting is what that section does under that load, and the amount is predictable.

To stop it, the load has to act through a particular point called the shear centre, and for a channel that point lies outside the section altogether: in the air, on the open side, some distance from the web. Nothing can be bolted there.

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. 1 A channel with the shear flow in its flanges drawn. Those flows form a couple, so a load applied through the web has a net torque on the section — and the point that leaves it untwisted sits outside the steel entirely.

The flanges are pulling in opposite directions

The explanation is the shear flow, followed to its conclusion in a thin-walled section.

In an open thin-walled member the flow runs along the walls rather than across the section. At a free edge — the tip of a flange — there is no material beyond, so QQ is zero and the flow is zero. Moving along the flange toward the web, QQ grows steadily, and the flow with it. So each flange carries a shear flow that starts at nothing at its tip and reaches a maximum where it meets the web.

Now look at the direction. The flow must be continuous through the section, running in one continuous path: in along the top flange, down the web, and out along the bottom flange. So the flow in the top flange points one way horizontally and the flow in the bottom flange points the other.

Two horizontal forces, equal, opposite, and separated by the depth of the section. That is a couple, and it has a moment whether or not anybody wanted one. The web’s flow carries the vertical shear, as expected. The flanges’ flows carry nothing vertically at all and produce a torque.

That torque exists whenever the section carries shear. The only question is what balances it — and if the applied load acts down the web, nothing does, so the section twists until something else stops it.

Where the point is

Locating the shear centre is a moment equation, and it is short.

Integrate the flow along one flange to get its horizontal force. For a channel of flange width bb, depth hh between the flange centrelines, and thickness tt, the flow is triangular along the flange and its total is

F=V t h b24I.F = \frac{V\,t\,h\,b^2}{4I}.

The two flange forces form a couple FhF h. For the section not to twist, the applied load VV must produce an equal and opposite moment about the web, so it must act at a distance

e=FhV=t h2b24Ie = \frac{F h}{V} = \frac{t\,h^2 b^2}{4I}

from the web centreline, on the side away from the flanges.

Two features of that expression are worth reading. It contains only geometry — no material, no load, no length of member — so the shear centre is a property of the cross-section in the same way the centroid is. And nothing constrains ee to be small, or inside the section: for an ordinary channel it comes out at something like forty per cent of the flange width beyond the web, which is well outside the material.

The shear centre of a channelA channel of 100 by 250, with the shear flow in its flanges drawn. Those flows form a couple, so the load has to be applied 38.0 outside the web to leave the section untwisted — a point in the air, outside the material entirely.web centrelineshear centree = 38.0no twisttwiststhe flange flows are equal, opposite, and separated — which is a coupleand nothing about the section's 32.90 × 10⁶ second moment predicts it
Fig. 2 A wider, deeper channel. The offset grows with the square of the flange width, so a broad shallow channel has its shear centre a long way outside the steel while a narrow deep one has it close to the web.

The centroid and the shear centre are different points, and confusing them is the standard error. The centroid is where the axial force acts if the section is not to bend. The shear centre is where the transverse load acts if the section is not to twist. They coincide for any section with two axes of symmetry — an I-section, a rectangle, a circular tube — and they do not coincide for a channel, an angle, a tee or a zed.

Which sections have the problem

The behaviour follows from symmetry, and the rule is short enough to carry.

Two axes of symmetry — I-sections, rectangles, tubes, cruciforms. The shear centre is at the centroid, and a load through the centroid produces no twist.

One axis of symmetry — channels, tees, zeds. The shear centre lies on the axis of symmetry and not at the centroid. For a channel it is outside the web; for a tee it is at the junction of the flange and stem, which is a point on the section but not its centroid.

No symmetry — angles, and most cold-formed shapes. The shear centre is somewhere that has to be computed, and for an equal angle it sits at the corner where the two legs meet.

The tee and the angle are worth separating out. For both, the shear centre lies at the point where the walls intersect — and the reason is neat: a shear flow running along a wall whose line passes through a point has no moment about that point, so a section made entirely of walls meeting at one point has its shear centre there. An angle, a tee and a cruciform all satisfy that condition, which is why all three have their shear centre at the intersection rather than at the centroid.

The same material, four waysFour cross-sections of identical area, so identical weight and cost, with the second moment of area computed from each profile's own geometry. Only the arrangement differs, and the stiffest is many times the flattest.tall rectangleI = 10.00 × 10⁶1.0× the firstteeI = 11.22 × 10⁶1.1× the firstI-sectionI = 24.29 × 10⁶2.4× the firstevery section here has an area of 3000 — only the shape differsthe bar is the second moment of area, to scale
Fig. 3 Three profiles of the same area. The rectangle and the I-section have shear centre and centroid coincident by symmetry; the tee does not, and its shear centre sits at the junction of flange and stem rather than at its centroid.

The number, and the check on it

For the channel in the hero figure — flanges of eighty, a depth of two hundred, walls of ten — the offset comes out at 31.731.7, which is thirty-nine per cent of the flange width and roughly a sixth of the section depth beyond the web.

Two properties of that number are worth testing rather than assuming.

It does not depend on the load. Doubling VV doubles the flange forces and doubles the moment they must balance, and ee is their ratio. So the shear centre is a fixed geometric point, and a member loaded through it stays untwisted at every load level — which is what makes it usable as a design rule rather than a load-case calculation.

The flow has to add up to the shear. The offset was computed from a flange force, which came from a flow, which came from QQ and II. Every one of those could be wrong and still produce a plausible-looking distance. What cannot be wrong quietly is the total: a distribution of shear stress has to integrate, over the whole section, back to the shear it is a distribution of. The generator behind these figures runs that integral before drawing anything and refuses if it misses by more than a fifth of a per cent — it returns 0.99960.9996 of the applied shear for the hero section, the shortfall being the sweep’s resolution rather than the theory.

That check is worth more than it looks, because it is a statement about forces summing on a chosen free body rather than about shear centres at all. The free body here is the cross-section itself, with the flange flows, the web flow and the applied load on it — an unusual body to draw, and the reason the twist gets noticed in the first place.

The shear centre of a channelA channel of 45 by 220, with the shear flow in its flanges drawn. Those flows form a couple, so the load has to be applied 13.9 outside the web to leave the section untwisted — a point in the air, outside the material entirely.web centrelineshear centree = 13.9no twisttwiststhe flange flows are equal, opposite, and separated — which is a coupleand nothing about the section's 17.66 × 10⁶ second moment predicts it
Fig. 4 A narrow deep channel. Since the offset goes as the square of the flange width, halving the flange brings the shear centre to within a fraction of its former distance — close enough to the web that loading through the web is a small error rather than a large one.

The square in b2b^2 is the practically useful part of the formula. It says the problem is a flange width problem: deep narrow channels barely twist, and broad shallow ones twist enormously. A designer who cannot avoid a channel can very often choose a proportion that makes the eccentricity tolerable, which is a cheaper answer than any of the four in the next section.

The same offset, buckling a column

There is a second place the shear centre appears, and it is one nobody meets it in first.

A column can buckle by bowing about either principal axis — that is ordinary flexural buckling. An open section can also buckle by twisting, rotating about its longitudinal axis while its ends stay put, because it is so feeble in torsion. And for a section whose shear centre and centroid do not coincide, the two modes are not independent: bowing displaces the section sideways, which moves the centroid away from the shear centre’s line, which produces a torque, which twists it — and twisting moves the centroid again.

The result is torsional-flexural buckling, a coupled mode at a critical load below either pure mode. The parameter that governs the coupling is exactly the distance between the centroid and the shear centre: coincident, and the modes decouple; far apart, and the coupled mode dominates.

That is why a plain channel or a lipped C-section used as a column is checked against a mode that an I-section never needs, and why cold-formed section design is so much more elaborate than hot-rolled. A cruciform section is the extreme case — its shear centre is at its centroid, so there is no coupling, and it is nonetheless so weak in pure torsion that it buckles by twisting alone at a load far below its flexural one.

The column curveFailure load against slenderness, as a fraction of the squash load. A stocky column crushes; a slender one buckles at the Euler load; the crossover is where the two curves meet, and real columns fall below both near it.5010015020000.20.40.60.811.2slenderness (effective length ÷ radius of gyration)they cross at λ = 75squashingEuler bucklingreal columns, which are neither
Fig. 5 The column curve, which is about flexural buckling. An open unsymmetric section has a second curve beneath this one for the torsional-flexural mode, and the gap between them is set by how far the shear centre sits from the centroid.

The general lesson is one this site keeps arriving at from different directions: efficiency in one property buys weakness in another. Opening a section out to put material far from one axis is what makes it a good beam, and it is precisely what makes it torsionally feeble and gives it a shear centre somewhere inconvenient. There is no shape that escapes the trade, only shapes where the weakness is somewhere it does not matter.

Why the twist is expensive

If the consequence were a small rotation nobody would care much. It is not, and the reason is that an open section is feeble in torsion.

A closed section resists twist by a shear flow circulating continuously round the tube. An open one has no such loop; each of its plates can only twist about its own mid-thickness, giving a torsion constant of roughly 13∑bt3\frac{1}{3}\sum bt^3 — a number containing the cube of a thickness, which for thin plate is very small. A hollow section is typically a hundred and fifty times stiffer in torsion than an open one of the same area.

So the torque produced by the flange couple is applied to a member with almost no capacity to resist it, and the rotation is large. Worse, twisting an open section makes it warp: the flanges displace along the member’s length, one forward and one back, so the cross-section does not stay plane. If that warping is restrained anywhere — at a fixed end, at a connection, at a stiffener — direct stresses develop that the ordinary torsion theory does not contain, and they can be a substantial fraction of the bending stress.

The combination is why a twisted channel is not a cosmetic problem. It is bending, plus torsion in a member bad at torsion, plus warping stresses at every restraint, all superimposed on a section that was designed for bending alone.

What is done about it

Since the shear centre cannot be attached to, four practical answers are in circulation, and all four are worth recognising because they show up in real details.

Use a symmetric section. The commonest answer by a wide margin. If a member will carry transverse load, an I-section or a hollow section is chosen and the problem does not arise. Channels are used as secondary members — purlins, rails, edge trimmers — where the twist is either small or restrained by what they carry.

Restrain the twist rather than avoid it. A purlin fixed to roof sheeting is prevented from rotating by the sheeting — a restraint that exists on the drawing and has to exist in the building — and the sheeting carries the torque back as a couple in its own plane. This is the usual arrangement for cold-formed zeds and channels, and it is why a purlin’s design is inseparable from the cladding attached to it.

Use them in pairs. Two channels back to back, or toe to toe with battens between, form a section with two axes of symmetry. The shear centre returns to the centroid, and the torque of one channel is balanced by the other’s.

Design for the torsion. Compute the eccentricity, take the resulting torque, and check the member for combined bending and torsion including the warping stresses. Correct, expensive, and reserved for cases where nothing else will do — a crane beam with an eccentric rail, a spandrel carrying a facade on one side, a curved beam. In each of those the torque is an applied couple rather than a by-product, and it has to be carried whatever the section.

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. 6 Shear stress across an I-section. Its two axes of symmetry put the flange flows in balance, so there is no residual couple, and the shear centre falls at the centroid — which is the whole reason this section can be loaded through its web without further thought.

Where the model stops

Thin walls. The derivation assumes the flow is uniform through the wall thickness and runs along the wall’s centreline. For a thick-walled or solid section the idea of a shear centre survives and the simple integral does not.

Open sections. For a closed section the flow has a circulating component that is statically indeterminate within the section itself, and locating the shear centre requires an extra condition — the twist being zero — rather than a direct integration.

Elastic behaviour. The shear centre is an elastic property. Past yield the flow redistributes and the point moves, which matters for a section being designed plastically.

No axial force. A member carrying axial load as well as shear has both a centroid and a shear centre in play, and eccentricity about either produces a moment. The two are different eccentricities and both have to be checked.

Prismatic members. A tapered or castellated member’s shear centre varies along its length, and the torque it generates is then distributed rather than applied at a point.

The figures on this page carry a distortion that cannot be avoided and should be stated. The shear centre is drawn as a dot in empty space beside the section, which invites the reading that something is there. Nothing is there. The point is a property of the section’s geometry, computed from an integral over the material, and it identifies a line of action rather than a location where anything exists — in the same way the resultant of a distributed load acts at a point where no force is applied.

The ladder from here

Later rungs on this anchor: shear flow in thin-walled open sections generally. Shear centres of closed and multi-cell sections. St Venant torsion and the torsion constant. Warping torsion and the warping constant. Combined bending and torsion. Cold-formed sections, where the shear centre governs routinely. Purlins and their restraint by cladding. Lateral-torsional buckling of monosymmetric sections, where the shear centre’s offset changes the answer. And the general theory of thin-walled beams, in which the shear centre, the centroid and the warping function are three aspects of one formulation.

Timoshenko identified the shear centre in 1913 and Vlasov built the general theory of thin-walled beams round it in the 1940s. The practical impetus was aircraft structures — thin-walled, open, unsymmetric, and unforgiving of a torque nobody had accounted for.

Where this goes next

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 The shear nobody draws A stack of loose planks slides at its ends when it is loaded. Glue them and the sliding stops — and whatever the glue is now carrying is a stress that no bending calculation contains. neutral axiscontribution of each striptotal I = 79.86 × 10⁶the outer strips do almost all of the work The material far from the middle does nearly all the work A strip of steel contributes to bending stiffness in proportion to the square of its distance from the centre. Move the same steel outward and the section gets stiffer for nothing. 2012.57.5shear 7.5moment 22.5the cut, at x = 5nothing was applied here — the internal forces are what the left-hand piece needs What a cut reveals, and why it was there all along Cut a beam anywhere and two quantities appear on the face — a shear force and a bending moment. Nothing was applied there. They are what the material was already doing.