The shear the chords take

Every shear calculation on this site so far has begun with a cut and a vertical force. Cut the beam, look at what is left of the applied load on one side, and whatever does not balance has to cross the cut face as shear. The chords carry the moment as a pair of horizontal forces; the web carries the shear.

That division of labour holds for a prismatic member, and it is so nearly universal in the drawings here that it has never needed stating. It stops holding the moment the two chords stop being parallel — and the reason is not subtle once it is drawn. The chord force is $M/z$, it acts along the chord, and if the chord slopes, part of that force is vertical.

Which free body produced the number #

Cut a tapered member at some station and take the piece towards the tip. On the cut face there are three things: a compression $C$ along the top chord, a tension $T$ along the bottom one, and whatever shear crosses the web.

Horizontal equilibrium makes the two chord forces equal in their horizontal components; call that common value $H$. Taking moments about a point on the line of action of one of them gives $M = H z$, so $H = M/z$ exactly as for a parallel-chord member. The new statement is the vertical one. Each chord carries a vertical component equal to $H$ times its own slope, and summing them gives

$$V_{web} = V - \frac{M}{z}\frac{dz}{dx}$$

with $z$ the lever arm and $x$ measured along the member. That is Résal’s result, published in 1899, and it is the whole of this essay in one line.

Three things about it are worth pausing on. It contains the moment, not the shear, which is why the correction is largest where the moment is largest rather than where the shear is. It contains $dz/dx$ and not the individual chord slopes, so a beam haunched symmetrically about its centreline and one haunched on the soffit alone behave identically as long as the lever arm changes at the same rate. And the sign is everything: where the depth grows in the same direction as the moment, the chords take work off the web, and where it grows the other way, they add to it.

A closed form with nothing in it #

Take the simplest case there is: a cantilever of length $L$ carrying a tip load $P$, tapering linearly from depth $d_0$ at the tip to $d_1$ at the root. Then $M = Px$, $V = P$, and $z = k d$ with $d$ linear in $x$, so $dz/dx$ is constant. At the root, $x = L$ and $z = k d_1$:

$$V_{web} = P\left(1 - \frac{L \cdot k(d_1 - d_0)/L}{k d_1}\right) = P\,\frac{d_0}{d_1}$$

and the answer is a ratio of two depths. No length appears, no load, no material, no lever-arm factor. A three-to-one haunch leaves the web a third of the applied shear at the section where the moment is largest and the shear check would otherwise be hardest.

Turn the same member round — put the deep end at the tip and the shallow end at the root, which is what a fish-bellied cantilever is — and the same derivation gives $P \, d_1/d_0$. The web carries three times the applied shear, at the root, where it is thinnest. The two answers are reciprocals of one another, and the whole difference is which way the taper runs.

The limiting case, which is a truss #

Push the taper until the depth is proportional to the moment. For the tip-loaded cantilever that means $d \propto x$ — a wedge whose two chords meet at the point where the load is applied — and the arithmetic gives

$$V_{web} = P\left(1 - \frac{x \cdot k c}{k c x}\right) = 0$$

exactly, everywhere along the member. A cantilever whose depth follows its own moment diagram has no web shear at all.

That is not a curiosity, it is a whole family of structures. A crane jib is a wedge. A cable-stayed pylon head is a wedge. The tapered brackets under a mill floor, the triangulated cantilever of a Warren bridge, the tapered flange plate on a plate girder support — all of them are the same idea, and all of them have one property in common: they are two force paths and a spacer. The web of a wedge has nothing to do but hold the chords apart, which is exactly what the diagonals of a truss with no diagonals to carry shear do not have the option of doing.

Read the other way round, the result explains something about trusses. A parallel-chord truss puts real force in its diagonals; a triangular one, loaded at its apex, does not. The difference is entirely the same $M/z \cdot dz/dx$ term, arrived at from the geometry of the chords rather than from a lever arm.

Where the shear check actually goes #

The practical consequence is a migration. In a prismatic cantilever the worst shear is at the root and everyone knows it. In a tapered one it is not.

Under a tip load the applied shear is constant, so the web shear falls monotonically towards the root and the critical section for shear moves to the tip — the shallowest part of the member, which is also where the web is thinnest. That is a genuinely awkward result, because the tip is where nobody looks: it has almost no moment, its flanges are small, and it is drawn as the unimportant end.

Under a uniform load the applied shear rises towards the root and the correction rises with it, so the two effects fight and the worst station lands somewhere in between. Sweeping it is a two-line calculation and guessing it is not, which is the same finding the tapered member’s bending check produces: a member whose capacity moves has to be checked everywhere rather than at the obvious place.

The corner of a portal, which is the case that matters #

The single most common haunch in construction is the one at the eaves of a portal frame, and it is worth looking at because the sign there is not obvious from the drawing.

A pitched portal has its largest moment at the eaves, and the haunch is put there to provide the depth for it. Reading along the rafter from the apex towards the eaves, both the moment and the depth are increasing — so the chords are helping and the web shear falls, which is convenient because the eaves is also where the shear is largest. The haunch does two jobs and the second is free.

What is true of the shear along such a member is true of the bending as well, and for the same reason: two quantities are climbing at different rates along one length.

So the section that decides the bending is not at either end either, and there is no reason for it to sit at the same station as the one that decides the shear. A haunched member has two migrating checks, and they migrate independently.

Read along the column from the base upwards, however, and the picture inverts on one common detail. Where a haunch is fabricated as a straight cut off a rolled section and welded on, the inner flange of the haunch is inclined and the outer one is not, so $dz/dx$ is set by one chord alone; and if the frame is designed with a pinned base and a stiff rafter, the moment in the column can be falling over part of the haunched length while the depth is still rising. Over that stretch the correction reverses sign and the web is asked for more than the applied shear.

The general rule is not “a haunch helps”. It is: compare the direction the moment is growing in with the direction the depth is growing in, over the specific length being checked. Two identical haunches on two ends of the same member can be doing opposite things.

The web that is not a web #

There is a third case, and it is where the effect stops being a correction and becomes the design.

In a reinforced concrete beam the truss model assumes a horizontal tension chord and a horizontal compression chord, and computes a strut angle from the shear the web has to carry. Taper the member and the tension chord follows the soffit. The vertical component of the chord force then subtracts from the shear the struts have to carry, and the links needed fall in proportion — often by a third in a haunched transfer beam, which is a substantial saving in a place where congestion is the real constraint.

The trap in that calculation is the one this essay opened with: the shear used for the link design has to be $V_{web}$ and not $V$, and every shear diagram a designer is handed plots $V$. Nothing on the diagram says that the member under it is not prismatic.

The man it is named after, and the bridge behind it #

Jean Résal was an engineer of the French state bridge service, and the reason his name is attached to a one-line correction is that he was designing the Pont Alexandre III when he needed it.

That bridge is a very flat steel arch of 107 metres, and its ribs are deep at the springings and shallow at the crown. Read along a rib from the crown outwards, the moment and the depth grow together — so the inclined chords take a large share of the shear, and the web plate needed is far lighter than a prismatic analysis would ask for. On a structure whose whole design case was to be as thin as possible over the Seine, that was not a refinement. It was the reason the section worked.

The correction reached the German literature as the Résal-Effekt and the English as, mostly, nothing at all: it appears in bridge codes as a clause about inclined flanges and in very few textbooks as a mechanism. The result is that generations of engineers have derived it from scratch at a haunch, usually correctly and occasionally with the sign the wrong way round.

There is a small historical irony in it. The same effect had been in plain sight for a century in the form every engineer already understood — a triangular truss carries no web force — and what Résal did was to notice that a solid tapered web is the same statement with the diagonals smeared out. The truss analogy has run in that direction ever since: a continuum result recognised because somebody drew the discrete version of it first.

The plate girder, where it is worth money #

The place the correction earns most is a welded plate girder over a support, and the arithmetic is worth doing because it is not marginal.

Take a two-span continuous girder haunched over its internal pier from 1,200 mm at midspan to 2,400 at the support. At the pier the moment is at its largest and so is the shear, which is the combination that decides the web thickness for every girder of this shape. With parallel chords the web carries the whole reaction’s share of the shear; with the haunch running the right way it carries half.

The straightness of the taper is itself a decision with a price on it, and it is the one a plate girder is built around.

Eleven per cent of a web is a real quantity of steel, and it buys something a curved cut does not: a flame-cut line a machine can follow in one pass, and an offcut that is the next girder’s flange.

Halving the web shear does not halve the web plate, because a web is usually sized for buckling rather than for yielding and the buckling stress goes as the square of the thickness over the depth — but the haunch has also made the panel deeper, which pushes the other way. The two effects have to be taken together, and the net is typically a plate one or two millimetres thinner over a length of twenty metres, or one fewer transverse stiffener every panel. On a bridge that is tonnes.

What it never buys is a shallower girder. The lever arm still has to be there, the chords still carry $M/z$, and the haunch is providing that as its first job. The shear relief is a second-order benefit of a decision already taken for a first-order reason, which is why it so rarely gets designed for deliberately and so often gets discovered afterwards.

The step where the haunch stops #

A haunch has an end, and the end is a station where $dz/dx$ changes abruptly from a finite value to zero. The chord force does not change there — it is $M/z$ and both are continuous — but its direction does, and a force that changes direction at a point applies a concentrated load there.

The size of it is the term this essay has been carrying all along, evaluated at the cut-off. For the tapered cantilever above, at the station where the depth has reached 800 mm the moment is 360 kNm and the lever arm 0.72 m, so $H = 500$ kN; the taper is $dz/dx = 0.12$; and terminating it delivers

$$H\,\frac{dz}{dx} = 500 \times 0.12 = 60\ \text{kN}$$

into the web at a point — half the applied tip load, arriving at a station where nothing was applied. It is the same $F\Delta\theta$ a kinked force path always produces, and it is why the end of every welded haunch carries a transverse stiffener.

The web shear steps by the same amount, in the direction that makes it worse: inside the haunch the chords were taking 60 kN off the web and past the cut-off they are not, so the web’s demand jumps from a third of the applied shear to all of it, at a section where the depth has just stopped growing. The worst web panel in a haunched member is very often the first prismatic one, and the shear diagram handed to the designer is flat across that station.

Where the model stops #

The derivation assumes the chord force acts along the chord. For a genuinely trussed member — a truss, a cable-stayed deck, a haunched girder with distinct flanges — that is exact. For a solid tapered member it is an idealisation, because the compression in a tapered beam is a stress field rather than a chord force, and it does not all act at the same inclination.

A finite-element solution of a tapered beam shows the correction arriving at somewhat less than the simple formula gives, because the extreme fibre is inclined at $dz/dx$ and the fibres nearer the middle are inclined at less. The usual reconciliation is to compute $dz/dx$ from the centroids of the chords rather than from the extreme fibres, which is what the lever arm $z$ means anyway. That is right in principle and it depends on knowing where the chord centroids are, which for a solid section with a varying stress distribution is not a fixed fraction of the depth. Nothing here compares the formula against a plane-stress solution of the same wedge.

And the picture cannot show what happens after the web buckles. Everything above is an elastic force division, and a slender tapered web that has gone into a tension field redistributes its shear along the diagonal the buckle sets rather than the one the geometry sets. The Résal correction survives — the chords still resolve — but the panel geometry that decides the tension-field capacity is a trapezium rather than a rectangle, and the standard treatment of it does not exist.

The generalisation #

The habit worth carrying away is smaller than the formula and more useful. A force that is not perpendicular to the cut is not entirely in the plane of the cut, and every internal force diagram on this site is drawn as though it were.

The same slip appears elsewhere. A curved beam’s chord forces are inclined by definition and the same term appears as a radial force. An arch’s thrust is the same statement carried to the point where the whole shear has been resolved away. A sloping column carries part of the floor shear as an axial component. In each case the picture that gets drawn — a vertical cut, a shear force, a moment — carries an assumption about direction that nobody wrote down and nothing on the diagram records.

The way to catch it is the free body. Draw the cut face, draw every force actually crossing it in its own direction, and resolve. A diagram that arrives with its shear already separated from its moment has done part of the work already, and it may have done it for a beam whose chords were parallel.

A shear-flow diagram is the clearest instance of the slip, because it is drawn as though the flanges were horizontal and is used on members whose flanges are not. Incline them and part of that diagram is carried by the flange force instead, with nothing in the drawing changing. The same is true one level further back, at the cut every result on this site descends from: what crosses it is decided by the geometry of what is being cut, and a diagram that arrives with its shear already separated from its moment has assumed an answer to that question.