Structural Engineering & Statics
where the load goes
Structural form

The truss with no diagonals

A rectangle without a diagonal is a mechanism, so a Vierendeel girder makes its corners rigid instead. The shear a diagonal would have carried as axial force now travels as bending in the chords, and bending is a far more expensive way to move a force.

Assumes The triangle that cannot fold, and everything built out of it, The frame that leans, and what stops it and Neither pinned nor rigid, which is every real connection.

18 min read 10 figures The load must go somewhereGeometry beats material

The founding figure of the structural-form field is a triangle, and the argument for it is that a rectangle of pin-jointed bars is a mechanism: it folds over without any member changing length, so it has no stiffness at all and no amount of making the bars stronger helps. A Vierendeel girder is a row of rectangles, it does not fold over, and the reason is the one thing the triangle argument assumed away.

The shear goes round the corner instead of across itA 6-panel Vierendeel girder, 12 m by 1500 mm, under 100 kN at mid-span. There is no diagonal in it, so each panel's 50 kN of shear is carried as bending in the chords: the curves drawn along them are the chord moments, and every one passes through zero at the middle of its own panel. The local moment is the panel shear times the panel length over four, 25.0 kNm, and it adds to an axial force of 200 kN from the global moment at the same point. The girder deflects 6.20 mm against 3.18 mm for the same members triangulated — 1.95 times — and 68% of that movement is chord bending that a diagonal would have removed entirely.100 kNchord moment 25.0 kNm from the panel shear · axial 200 kN from the global moment6.20 mm against 3.18 mm triangulated — 68% of it is chord bending
Fig. 1 A six-panel Vierendeel girder, 12 m by 1,500 mm, under 100 kN at mid-span. There is no diagonal in it, so each panel’s 50 kN of shear is carried as bending in the chords: the curves drawn along them are the chord moments, and every one passes through zero at the middle of its own panel. It deflects 6.20 mm against 3.18 mm for the same members triangulated, and 68% of that movement is chord bending a diagonal would have removed entirely.

The joints are rigid. That is the whole of the difference, and it costs a factor of two.

Which free body produced the number

Cut the girder vertically through the middle of a panel and take the piece to the left.

Four things cross the cut: an axial force in each chord, and a shear in each chord. The two axial forces are equal and opposite and form the couple that carries the global moment: at mid-span, M/d=300/1.5=200M/d = 300/1.5 = 200 kN, and the frame solver returns 155 kN — the difference being the part of the global moment the chords are carrying as local bending rather than as a couple.

The two shears carry the panel shear, 50 kN, half in each chord. And a chord carrying 25 kN of shear over a panel of 2,000 mm with no load along it must be in double curvature: equal and opposite end moments, and a point of contraflexure at the middle.

Mchord=Vpanel s4=50×2.04=25 kNmM_{chord} = \frac{V_{panel}\,s}{4} = \frac{50 \times 2.0}{4} = 25\ \text{kNm}

which is where the contraflexure points in the figure come from. They are not drawn there for convenience; they are at the panel middles because a member in pure double curvature has its inflection at its middle, and the moment at the panel middle is genuinely zero.

Cutting there is therefore the clever cut. It is the one place in each panel where the free body has no chord moment on it, so the four unknowns reduce to two, and the whole girder can be analysed by hand — which is the classical method and the reason the arrangement was usable at all before computers. It is the method of sections with the cut chosen for the same reason: to make a question answerable without solving everything else.

The two things a chord is doing at once

At a panel end the two terms add. At a panel middle only the axial term survives. So the chord’s extreme-fibre stress varies along its own length in a sawtooth, peaking at every joint and dipping between them, and the peak is

σ=NA+MchordZ\sigma = \frac{N}{A} + \frac{M_{chord}}{Z}

with NN from the global moment and MM from the panel shear. The two follow different diagrams — NN tracks the moment and peaks at mid-span, MM tracks the shear and peaks at the ends — so the worst chord in a Vierendeel girder is not at mid-span and not at the support but somewhere between, and finding it is not a matter of inspection.

A Pratt truss of 6 panelsA Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 10 members came out in tension, 9 in compression and 2 carrying nothing.29.429.447.147.129.429.4-47.1-52.9-52.9-47.1-15.0-10.0-15.0-38.6-38.623.27.77.723.2tensioncompression2 carrying nothing
Fig. 2 The frame the comparison is against: the same geometry with one diagonal per panel. Every member carries axial force alone, the chords carry the moment as a couple, the diagonals and posts carry the shear, and no member in it has any bending at all.

That is the arrangement’s cost stated structurally rather than numerically. In a truss, a member carries force along its own axis, at a uniform stress over its whole section. In a Vierendeel, a chord carries force along its axis and a moment across it, so its material is used at a varying stress with the middle of the section barely loaded — which is the same inefficiency a beam has against a tie, imported into a member that was supposed to be a chord.

Where the movement goes

Splitting the deflection by virtual work — the frame’s own member forces against the forces from a unit load — gives an exact decomposition rather than a comparison:
from chord and post bending 4.25 mm
from chord extension 1.96 mm
total 6.20 mm
what the stiffness solver found 6.20 mm

The two routes agree to four figures, which is the check that the split is real rather than a fitted attribution. 68% of a Vierendeel girder’s deflection is bending that a single diagonal would have deleted.

Depth is worth far less to a Vierendeel than to a trussMid-span deflection against depth, for the same 6-panel girder solved twice: as a Vierendeel frame with rigid corners, and as a triangulated truss with the same members and one diagonal per panel. The truss improves as the reciprocal of the depth, because a chord force is M/d; the Vierendeel very nearly stops improving, because the part of its movement that comes from chord bending — 92% of it at the deepest section here — depends on the panel length and not on the depth at all. So the penalty grows with depth: 0.81 times at a depth of 5% of the span and 4.98 times at 25%.6%8%10%12%14%16%18%20%22%24%051015depth ÷ spanmid-span deflection (mm)Vierendeelthe same, triangulated
Fig. 3 And the property that makes the arrangement genuinely awkward: the penalty grows with depth. The truss improves as the reciprocal of the depth, because a chord force is M/d. The Vierendeel very nearly stops improving, because the chord-bending part of its movement depends on the panel length and not on the depth at all.

That figure is the finding. At 5% of the span the two frames are about equal; at 13% the Vierendeel is 1.95 times worse; at 25% it is 4.98 times worse and 92% of its movement is chord bending.

Deepening a Vierendeel girder does almost nothing after a point, because the axial term it reduces is not the term that dominates. What does help is shortening the panels: at three panels the penalty is 2.82 and at twelve it is 1.27, because Mchord=Vs/4M_{chord} = Vs/4 is linear in the panel length and the flexibility goes as its cube.

panels panel length chord moment deflection penalty
3 4,000 mm 50.0 kNm 10.77 mm 2.82
4 3,000 37.5 9.15 2.46
6 2,000 25.0 6.20 1.95
8 1,500 18.8 4.97 1.63
12 1,000 12.5 3.89 1.27

The design variable is the panel length, and it is the one the architecture is usually dictating — because the whole reason for leaving the diagonals out was to have a clear opening of a particular size.

What it is bought for

The count is necessary and not sufficientTwo pin-jointed frames, each satisfying m + r = 2j exactly. One of them folds anyway, because the equations are not independent; the ghosted outline is the motion that costs no member any change of length, drawn at an exaggeration of 0.55 of the span.one panel braced twice, the next not at allm 9 + r 3 = 2j 12 · rank 11a mechanismthe same count, properly arrangedm 9 + r 3 = 2j 12 · rank 12stands up
Fig. 4 The alternative, stated at its bluntest: a truss panel without its diagonal and with pinned joints is a mechanism, and moves under no load at all. Rigid corners are what stand between a Vierendeel girder and that drawing, and they have to be rigid in fact rather than in the analysis.

Nobody chooses this arrangement for its structural efficiency. It is chosen for the hole in the middle of every panel, and the list of places where that hole is worth a factor of two is short and consistent:

A rectangular opening through a storey-deep truss — the same problem a transfer structure has at a different scale — a corridor, a door, a plant route — where the diagonal would land exactly where the opening has to be.

A transfer structure over an entrance, where the whole point is that nothing crosses the void.

A bridge whose deck sits between the girders, where a diagonal would foul the traffic envelope.

And, at the smallest scale, a hole cut through a beam’s web, where the two tees above and below become a one-panel Vierendeel girder whether or not anybody meant them to.

A hole in a web is a Vierendeel panelA 400 × 300 rectangular opening in a 533 deep beam, 15% along a 9 m span carrying 20 per metre — where the moment is 103 kNm and the shear 63 kN. The moment is a couple on the two tees, 301 kN on a lever arm of 343 mm, which is 70.3 N/mm² of uniform stress. The shear has nowhere to go but through the tees, so each carries 32 kN over the opening and bends in double curvature: a Vierendeel moment of 6.3 kNm and 167.6 N/mm² on top. So 70% of the stress at the corner exists because the hole has a LENGTH, and only 30% of the section's second moment has gone.contraflexure at the middle of the holefrom the moment: 70.3 N/mm²from the shear: 167.6 N/mm²total 237.9 of 355second moment lost: 30%deflection up by 0.7%the shear the tees must carry: 63 kN, half each, over 400 mm
Fig. 5 That last case drawn: an opening in a beam web, with the two tees bending in double curvature over it and a point of contraflexure at the middle of the hole. It is the girder above with one panel, and 70% of the corner stress comes from the same Vs/4.

The joints have to be real

The joints are not pins, and this is what that costsA 4-panel Pratt truss solved twice on the same stiffness matrix: once with a moment release at every member end, which is the pin-jointed idealisation, and once with the joints continuous, which is what welding them produces. The axial forces are the same to within a per cent; the bending the second solution adds is worst in member 0, where the bending stress reaches 24.3% of the axial stress. Members are shaded by that ratio.worst secondary bending: 24.3% of the axial stress, in the member markedaxial force there 16.7 kN · end moment 0.20 kNm · slenderness of the member 18the same members, the same loads, the same solver — only the releases differ
Fig. 6 What rigidity is worth in a frame that has diagonals: the secondary moments a truss develops because its joints are not pins, which are a nuisance there. In a Vierendeel they are not a nuisance, they are the load path, and a joint that is 80% rigid is a girder that is 80% there.

Every argument on this page assumes the corners transmit moment fully. Real joints do not, and a Vierendeel girder is the structure most sensitive to that of any in this collection — a truss whose joints are not really pins has secondary moments it did not want, and a Vierendeel with semi-rigid joints has no load path at all.

So the arrangement moves the difficulty from the members to the connections. The chords and posts are ordinary; the joints have to develop the full plastic moment of the members meeting them, in both directions, at every panel point, and a Vierendeel girder is therefore an expensive thing to fabricate for reasons that have nothing to do with the tonnage.

The moment does not stop at the end of the beamA portal frame of 8 m by 4 m with fixed bases, carrying 20 kN/m on the beam. The bending moment is drawn on the tension side of every member, and it runs round the corner without a break: 65.2 kNm arrives at the end of the beam and 65.2 kNm leaves down the column, which is the same number, since joint rotational equilibrium is one of the equations the frame solve satisfied. Midspan carries 94.8 kNm, and the two add to 160.0 — the 160.0 kNm of a simply supported span, to 0.0e+0 kNm. The corner takes 61% of the wL²/12 a fully built-in beam would have carried, because the columns are springs rather than walls: the beam-to-column stiffness ratio is 1.27. The beam's moment crosses zero 0.92 m from the corner and the column's 1.33 m above its base.w = 20 kN/mcorner 65.2 kNmmidspan 94.8 kNm94.8 + 65.2 = 160.0 = wL²/8
Fig. 7 The joint that has to be made, at the scale of one corner. A moment crossing from a horizontal member to a vertical one has to pass through a panel zone that shears, and the panel zone’s own flexibility is a term nobody counts until the frame is a Vierendeel.

The thing it is unexpectedly good at

One property is worth setting against the whole of the cost, and it is the reason Vierendeel girders keep being built in places where a truss would fit.

Take that one away and the load finds another routeA 6-panel pratt truss under 20 kN at each top node, before and after member 2 is removed. The load redistributes. The worst-affected survivor now carries 2.03 times what it did, and four members that carried nothing before are now working. Whether that is survival depends on how much spare capacity was there, which is a different question from whether the frame was strong enough.intactmember 2 removedworst demand 2.03×
Fig. 8 A truss losing one member, and what happens to the rest. A determinate truss that loses a diagonal is a mechanism; the Vierendeel has no diagonals to lose, and every one of its panels is redundant several times over.

A Vierendeel girder is highly indeterminate, and every load path in it is duplicated: remove a post, and the two chords span the double panel; remove a length of chord, and the frame redistributes around it. Robustness is easy where the structure is a network and hard where it is a tree, and a Vierendeel is emphatically a network.

The price is the usual one for indeterminacy. The girder is sensitive to things statics cannot see — a support that settles puts moments through the whole frame, and a temperature difference between the chords does the same. Both are imposed deformations rather than loads, and both are invisible to the analysis that sized the members.

Chord force against truss depthThe force in a truss chord for a fixed bending moment, against the depth of the truss. The relationship is a reciprocal: the chords form a couple whose lever arm is the depth, so a shallow truss pays for it steeply.0.511.52050100150200250300depth of the truss2501671251007150the same moment, resisted by a longer lever arm
Fig. 9 And the comparison that makes the depth result so counterintuitive. A truss’s chord force is M/d and falls as a rectangular hyperbola with depth — the strongest scaling argument in the whole field. The Vierendeel keeps that benefit for its axial term and has a second term that ignores it entirely.

What it costs, counted properly

The deflection ratio understates the price, and the honest comparison is worth setting out.

A portal frame swaying under 20A portal frame pushed sideways, solved by the stiffness method because statics cannot divide the load between two columns. The base shears come out at 10.0 and 10.0 and add to the applied 20; the peak moment is 22.2. The sway is drawn hugely exaggerated, and the moment diagram is plotted on each member's tension face.20H 10.0 M 22.2H 10.0 M 22.2the two base shears add to the applied 20 — the split came from stiffness, not staticsthe sway is exaggerated; a real frame at this load moves a fraction of a millimetre
Fig. 10 The arrangement stood on end, which is what every unbraced building frame is. A portal frame resists sway by exactly this mechanism — column and beam bending in double curvature, with contraflexure points near the middles — and the whole of sway analysis is this page’s argument turned through ninety degrees.

At equal member sizes the Vierendeel deflects 1.95 times as far. But a truss designed for this load would not use these members: its chords carry 200 kN of axial force and nothing else, so they can be light, and its diagonals are the only members that see the shear. The Vierendeel’s chords carry 155 kN and 33.2 kNm, which for a section with a modulus of the order of 10⁶ mm³ is a stress of the same order again — so they have to be perhaps twice the size before anything is said about deflection.

The real cost is therefore closer to a factor of three or four in steel weight, plus a full-strength moment connection at every panel point, against a truss’s pinned or nominally pinned joints. Against that sits one thing, and it is not small: the panels are empty.

That is the shape of the trade in a sentence, and it is the shape of a great many structural trades. Depth is the cheapest strength there is and a clear opening is one of the most expensive, and a Vierendeel girder is the structure that buys the second with the first.

Where the model stops

The frame is elastic and the joints are perfect. Both assumptions flatter the arrangement, and the second flatters it a great deal.

Panel-zone shear is not in the model. The frame solver joins members at points; a real joint has a depth, the moment reverses across it, and the shear in the panel zone is large. On a Vierendeel with deep chords the panel zone can contribute an appreciable fraction of the total deflection again.

The comparison uses the same member sizes for both frames, which is fair for the deflection ratio and unfair as a design comparison: a truss designed for this load would use much lighter chords, so the real cost of the arrangement is larger than the ratio suggests.

Nothing here is a stability check. A Vierendeel chord is a beam-column with a point of contraflexure in it, and its effective length is a question the analysis above does not touch.

And the whole page is a two-dimensional frame. A Vierendeel bridge or a Vierendeel core is a three-dimensional assembly whose panels interact out of plane.

What the pictures cannot show

The chord moment curves are drawn along the chords at an exaggeration chosen to make them visible. They are moment diagrams, not shapes, so a reader who sees them as the deflected form of the chords has read a curvature as a displacement — and the two are not even in phase, since the moment is zero where the chord’s own curvature is zero and the chord’s displacement there is not.

The joints are drawn as circles, which is the site’s convention for a truss node and is exactly the wrong symbol here. A pin is what these joints must not be, and no drawing on this page shows the plates, welds and stiffeners that make them what they are.

And the depth sweep plots two curves crossing, which suggests a designer might choose the depth at which they cross. Nobody would: at that depth the Vierendeel and the truss deflect the same and the truss is far lighter, so the crossing is a curiosity rather than an optimum.

The ladder from here

Later rungs on this anchor: the classical hand method — cut at the contraflexure points, solve each panel, march along — which is one of the last hand methods for an indeterminate frame that a person can actually carry out. Vierendeel girders with unequal chords, where the shear does not split in half and the contraflexure points move off the panel centres. Panel-zone flexibility, and how much of a deep-chord girder’s movement it is. The Vierendeel as a stability problem, since its chords are beam-columns with moment reversal. Multi-storey Vierendeel frames, which is what every unbraced building frame actually is — the portal frame is a one-panel Vierendeel girder stood on end, and the whole of sway analysis is this page’s argument turned through ninety degrees. And the cellular beam, which is a rolled section turned into a Vierendeel girder by cutting and rewelding it.

Arthur Vierendeel patented the arrangement in 1896 and built a good many bridges with it in Belgium, at a moment when riveted moment connections had just become reliable enough to make it possible. The idea has not aged: it is still chosen for exactly the reason he chose it, which is that a panel with nothing across it is worth paying for.

Where this goes next

tensioncompression2 carrying nothing The triangle that cannot fold, and everything built out of it A square of pinned bars is a mechanism. A triangle is not, and that single fact is the reason trusses exist and the reason they look the way they do. worst secondary bending: 24.3% of the axial stress, in the member markedaxial force there 16.7 kN · end moment 0.20 kNm · slenderness of the member 18the same members, the same loads, the same solver — only the releases differ The joint that is not a pin Every truss on this site is analysed as though its joints were frictionless pins. Almost none are. The bending that follows is called secondary, which is a claim about size — and the claim is checkable. 20H 10.0 M 22.2H 10.0 M 22.2the two base shears add to the applied 20 — the split came from stiffness, not staticsthe sway is exaggerated; a real frame at this load moves a fraction of a millimetre The frame that leans, and what stops it A rectangle of pinned bars folds flat. Make the corners rigid instead of adding a diagonal and it does not — which buys an unobstructed opening and costs bending in every member of it. contraflexure at the middle of the holefrom the moment: 70.3 N/mm²from the shear: 167.6 N/mm²total 237.9 of 355second moment lost: 30%deflection up by 0.7%the shear the tees must carry: 63 kN, half each, over 400 mm The hole that costs nothing, and everything A service opening removes 30% of a beam's second moment and 0.7% of its deflection. What it costs is not that. Across the opening the shear has nowhere to go but through the two tees, and a tee carrying shear over a length bends.

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Chord forceJoint rigidityLoad pathMechanismMoment connectionPanel shearPoint of contraflexureSecondary bendingSpan to depth ratioStiffnessTriangulationVirtual work