Structural Engineering & Statics
where the load goes
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One support too many — page 6

Essays 121 to 144 of 188 on this thread, in the same order.
A frame bent by nothing at all. A portal of 8.0 m span and 5.0 m columns at EI = 20000.0 kN·m², with its right foot settled 10.0 mm and no load on it anywhere. The moment diagram is drawn on the members: −3.95 kN·m at the left foot, −3.95 at the left knee, −0.00 at the crown, 3.95 and 3.95 on the right. The settlement is drawn hugely magnified; at true scale it is 10.0 mm on an 8.0 m frame. The diagram is antisymmetric, the vertical force the settlement develops is 0.99 kN, and every one of those numbers is proportional to EI. Deflection

The load that is not a load

Settle one foot of a portal frame by ten millimetres and the frame develops moments with nothing applied to it anywhere. In the analogous column the case is simpler than a load case, not harder — the section carries no direct stress at all, and the whole answer is one bending stress. And it scales the wrong way: the moments are proportional to EI, so the stiffer the frame, the more a settlement costs it.

 Every shape the analogy reaches, which is every single ring. Four frames of 10.0 kN/m on an 8 m span, each solved by the column analogy and each checked against a stiffness solution of the same frame: a portal, largest moment 40.6 kN·m, agreeing to 0.24 per cent; a pitched portal, largest moment 35.6 kN·m, agreeing to 0.15 per cent; a stepped frame, largest moment 46.6 kN·m, agreeing to 0.16 per cent; splayed legs, largest moment 16.9 kN·m, agreeing to 0.09 per cent. The diagrams are drawn normal to each member. Nothing in the method asks what shape the frame is: it needs an area, a centroid and three second moments of the centreline, and a polyline has all five however it bends. Deflection

Three, and what three is a property of

The column analogy works because a closed ring cut once has three redundants and a plane section has three stress resultants. That match is the whole method, and it is topological rather than geometric — a portal, a pitch, a step, a splay and a polygonised arch are all one ring and all exact. Add a second bay and the frame has six redundants with nothing to be a drawing of, and the outer ring on its own is out by 190 per cent.

 Two lines and a triangle: a three-hinged arch drawn. A three-hinged arch of 20.0 m span, springings at (0.0, 0.0) and (20.0, 0.0) m and the crown hinge at (10.0, 5.0), under 100.0 kN at 5.0 m. The right half carries no load, so it is a two-force member and its reaction lies along the line from its springing through the crown hinge. That line meets the load's line at K, 7.50 m up, and the left reaction must pass through K too. The triangle of the load and the two reaction directions gives the reactions as 90.1 kN at A and 55.9 kN at B, with a horizontal thrust of 50.0 kN — the values four equilibrium equations return, to 7e-15 kN. The shape of the rib entered nowhere. Equilibrium

The arch that is only its three hinges

A three-hinged arch's reactions come from three points and nothing else, so a parabola, a circle and a portal frame on the same hinges push on their abutments identically. Move a load across and the point where the reactions cross runs along two straight lines through the crown. That is the arch's influence line, drawn with a straightedge — and friction in the hinges it was built around blurs it.

 Putting a funicular through three points. Four loads — 40.0 kN at 3.0 m, 60.0 kN at 7.0 m, 30.0 kN at 12.0 m, 50.0 kN at 16.0 m — and three points the polygon must pass through: A and B at the springings and C, 5.0 m above their chord at 10.0 m. A trial pole, dashed, draws a polygon from A that ends 6.39 m below B. The ray through the trial pole parallel to its own closing line cuts the load line at Q, 95.0 kN from the top, which is the left reaction of a simple beam on A and B and does not depend on the pole at all. Every pole whose polygon passes through A and B lies on the line through Q parallel to A B; the one whose polygon also reaches C is 98.0 kN from the load line, which is the moment at C of that simple beam, 490.0 kN·m, divided by C's height above the chord. That is the three-hinged arch's thrust — the four equilibrium equations give 98.0 kN. Equilibrium

The polygon that runs out of freedom

A funicular polygon for given loads has exactly three freedoms, so it can be made to pass through three chosen points and no more. Three points is a three-hinged arch, and the drawing solves it. Take the crown hinge away and one freedom is left over. The drawing then offers a whole family of thrust lines and cannot say which one the arch uses — the rib's stiffness decides, and stiffness is not on the paper.

 Two flanges, one bolt, one prying force. An end plate 18 mm thick with its bolt 45 mm from the beam web, bolted to a column flange 14 mm thick with the same bolt 30 mm from the column web; their tips are 35 mm beyond the bolt and bear on each other, so the prying force there is one force acting on both. Checked as two separate tee stubs on rigid bases, the end plate carries 109.3 kN per bolt (mode 2, web hinge and bolt) and the column flange 104.4 (mode 1, both hinges), so the component method gives the joint 104.4. The pair carries 92.3 kN, 12 per cent less, with the prying force at 44.7 kN and hinges in the end plate at its web and the column flange at the bolt line — the end plate's web hinge working with the column flange's bolt-line hinge, a mechanism that exists in neither tee stub on its own. Connections

The hinge that forms in the other flange

An end plate bolted to a column flange is two tee stubs sharing one bolt, and the force at their tips is one force pressing on both. Checked separately, as the component method checks them, the end plate carries 109 kN a bolt and the column flange 104. Together they carry 92 — by a mechanism with one hinge in each flange that neither tee stub has on its own.

 The same restraint, spread and gathered. The buckled shape of a 24.0 m compression chord with the same smeared restraint, 0.35 N/mm per mm, delivered by U-frames at two spacings. With frames every 2.0 m the chord buckles in half-waves of about 5.1 m that ignore the frames — the smeared shape — at 1892 kN, matching the smeared answer. With frames every 4.0 m it buckles between them, with a node at every frame, at 1548 kN: 18 per cent below the smeared 1897 and at the Euler load of one bay. The dots are the frames. Stability

A row of frames is not a foundation

The top chord of a half-through girder is held sideways by U-frames, and the standard calculation smears them into a continuous elastic foundation. That is right while the frames are close and quietly wrong once they are not. The crossover sits at seven-tenths of the buckle's own half-wavelength, whatever the frames' stiffness — and beyond it the chord buckles between frames at a load no stiffening of the frames can raise.

 The forces the girder can hold with nothing on it. The one self-stress state of a K-braced girder of eight 3 m panels, 3 m deep: member forces in equilibrium at every joint with no load and no reaction. It lives entirely in the two middle panels — twelve members: the four diagonals meeting at the middle joint of the central vertical in tension, and the four central chord members and the half-verticals at the outer side of each middle panel in compression, at 0.89 and 0.45 of the diagonal force. It is symmetric about mid-span, so a symmetric load can call on it as freely as any other: symmetry does not supply the missing equation. Every other member of the girder is idle in it, drawn faint. Equilibrium

The redundancy only the sun can find

A K-braced girder that is symmetric about mid-span has one member more than statics can resolve, and it is at the centre, where two K's meet at one joint. Symmetry does not remove it, because the forces it allows are symmetric themselves. Loads barely touch it — a section calculation that ignores it is exact for a load at almost any joint. What finds it is a millimetre of misfit or a sunlit top chord, which put forces into the middle of the girder that no load calculation contains.

 The date the joint is cast decides the sign. The moment at the pier after thirty years, for two 12 m pretensioned beams, 300 mm wide and 700 mm deep with a 160 mm slab, made continuous over the pier by a joint cast with the slab, against the beams' age when the joint and slab are cast; sagging positive, with its three parts dashed. Cast at 7 days the joint ends at +320 kN·m; at 28, +214; at 90, +59; at a year, −164. It exceeds the joint's cracking moment of 159 kN·m for any joint cast before about 45 days, and it changes sign at about 130 days. From 7 days to a year the prestress's share falls from +555 to +204, because an old beam has made most of its upward creep before it is joined; the differential shrinkage's grows from −57 to −285, because an old beam has finished its own shrinking and the slab's is then all difference; the dead load's eases from −178 to −83. The first two move the joint the same way as the beams age. Materials

The pier that bends the wrong way

Two precast beams are made continuous over a pier by a joint cast with the deck, and the joint is designed for the hogging moment a continuous beam has there. Thirty years later it is sagging, by more than the moment that cracks its underside, because the beams were still cambering upward when they were joined. Whether that happens is decided by two dates — when the beams were cast and when the joint was — and the deck's shrinkage, which pulls the other way, is not enough to stop it.

 At the tip it is the tension flange that swings. The buckled shape of a 3.00 m cantilever of 457 mm universal beam carrying a load at its tip at the shear centre, root built in: slope and warping held, seen from above: the sideways movement of the top flange and of the bottom flange along the length, scaled so the larger is one. The cantilever hogs, so its bottom flange is in compression, and that flange bows out furthest 0.58 of the way from the root, where it moves 0.15 — and has come back to 0.00 by the tip. The top flange, in tension, swings out to the full 1.00 at the tip. The section twists as it goes, so the flange that is doing the buckling is not the flange that moves most at the end. Critical moment at the root: 1266 kN·m. Stability

The flange that swings is in tension

A cantilever hogs, so its bottom flange is the one in compression and the one that buckles. At its tip the bottom flange barely moves: the section turns about it, and the flange that swings out is the top one, in tension. So the restraint that matters at a cantilever's tip is on the flange every rule for spans calls the wrong one — and the root, which a drawing shows as a single line, moves the critical moment by a factor of thirty-four.

 A load on one span moves the other. The change in level along two 100 m spans of stressed ribbon at a sag of 2.00 per cent, carrying 35 kN/m, when 20 kN/m of live load is added to the left-hand span, with a rigid pier (dashed) and with a pier of 200,000 kN/m (solid). With a rigid pier the loaded span sags a further 195 mm and the other does not move. With a pier of 200,000 kN/m the pier top leans 25 mm towards the load, the loaded span sags 358 mm and the unloaded span rises 191 mm; the pier carries 4,953 kN of the 9,440 kN a rigid pier would. Structural form

The pier that carries only the difference

A stressed ribbon continued over a pier pulls on it from both sides, and under its own weight the two pulls cancel. Load one span and they do not: the pier is asked for the difference, nearly ten meganewtons for a crowd on one of two hundred-metre spans. Whether it gives it depends on its stiffness against a number the ribbon supplies itself, and whatever the pier does not carry, the ribbon turns into movement — a third of a metre of extra sag in the loaded span and a fifth of a metre of rise in the empty one.

 Fully stressed, with a choice of diagonals. Two trusses sized so that every member is at the allowable stress under the full load, each member drawn as wide as its area, tension and compression in two colours. Above, an eight-panel Pratt truss as deep as a panel is long, with a counter-diagonal in each of its six interior panels, loaded by 10 at every bottom joint, found by resizing and reanalysing until nothing changes; below, the same truss without its counters. The counters in the two middle panels have shrunk to nothing (dotted) and the four nearer the supports have stayed, working in compression beside the diagonals in tension. The truss that kept them needs 1,200 units of steel against the Pratt's 1,220, and deflects 24.0 at mid-span against 26.0. Deflection

The truss whose forces follow its sections

In a determinate truss each member's force is fixed before its section is chosen, so sizing for strength and sizing for stiffness can be done in either order. Put a counter-diagonal in every panel and they cannot. The fully stressed design becomes an iteration that starves some counters to nothing and keeps others, lands on a different truss from every start, and — whichever it lands on — weighs the same and deflects the same. Then enlarge one group of members to stiffen it, and a vertical nobody touched is overloaded by 58 per cent.

 One quantity, three names. The middle braced bay's share of the storey force (solid) and one end bay's (dashed), for a 48 by 12 m plate, 200 mm of concrete, spanning two 24 m bays between three braced bays of 100,000 kN/m each, under 40 kN/m, against a factor on the plate's stiffness as built, on a logarithmic scale. The share moves continuously from 61 per cent, a continuous plate on unyielding supports, to 33 per cent, the rigid plate. Left of 0.02 (shaded) ASCE 7 calls the plate flexible and assigns tributary shares, 50 per cent to the middle; the plate there actually gives it 59 per cent at the boundary. Right of 1.54 (shaded) EN 1998-1 calls it rigid and assigns stiffness shares, 33 per cent; at that boundary the plate gives 37. The plate as built, factor 1, gives 38 per cent to the middle and 31 to each end. Structural form

Rigid by one code and not by the other

A floor plate between shear walls shares its storey force out in proportions that move continuously with its stiffness, from a continuous beam's to a rigid body's. The codes put it in one of two boxes and use a different model in each. ASCE 7 and EN 1998-1 draw their boxes' edges a hundred times apart in plate stiffness, each at a point where its own licensed model already misstates the middle wall by about a tenth — and an ordinary 200 mm slab sits in the gap between them.

 The same rib, fixed, bends at both ends as well as the crown. The bending moment that axial shortening leaves in a 60 m parabolic concrete rib at a 6 m rise (EI 420,000 kN·m², EA 20,400,000 kN) under 60 kN/m, sagging upward, from a frame solve of the rib. Dashed: two-hinged, sagging throughout, 29.4 kN·m at the crown. Solid: fixed at the springings, 57.9 kN·m sagging at the crown and 112.4 kN·m hogging at each springing. The funicular load itself leaves no moment in either rib; all of this is the 0.11 and 0.63 per cent of the thrust the two ribs lose by getting shorter. Deflection

The springings that make shortening worse

Fixing an arch at its springings is the stiffer, cheaper and usual way to build one in concrete, and it makes the arch six times as sensitive to its own shortening. The thrust it loses acts at the elastic centre, two thirds of the way up, so the moment lands at the springings as well as the crown, twice as large and the other way round — at the section the fixed arch is designed at, not away from it.

 How stiff an abutment has to be to count as fixed. The shortening moments in a 60 m parabolic concrete rib at a 6 m rise (EI 420,000 kN·m²) under 60 kN/m as shares of the fixed rib's, against the rotational stiffness of each springing as a multiple of EI/L on a logarithmic scale: at the springings (solid, −112 kN·m when fixed) and the crown (dashed, 58 kN·m fixed, 29 two-hinged). The springing moment is half the fixed value at kθ = 11.3 EI/L and nine tenths at 102 EI/L; at EI/L itself it is 8 per cent. The crown moment never falls below the two-hinged rib's 51 per cent of the fixed value. Deflection

The abutment that spreads before it turns

A fixed arch is fixed only if its abutments hold still, and they have two ways not to. Turning releases the springing moment, but slowly: an abutment has to be a hundred times the rib's EI/L to hold nine tenths of it, and a footing on stiff clay under a slender concrete rib already is. Spreading does the opposite. The rib's own shortening is worth 13.6 mm of spread, so an abutment that gives 7 mm a side under the thrust doubles every secondary moment in the arch — and a footing on dense sand gives 8.

 Fit the counters last, and the tolerance locks in nothing. The largest force random length errors lock into an eight-panel Pratt truss with a counter-diagonal in each of its six interior panels, 24.0 m long and 3.0 m deep, carrying 60 kN of dead load at each interior bottom joint, against the errors' standard deviation: the median draw (solid) and the 95th percentile (dashed) with every member fitted in the shop, and the same with the counters left loose until the truss carries its dead load and then cut to the gap (on the axis). With everything fitted, 174 and 288 kN at 1 mm, 348 and 577 at 2 — straight lines, because the force is proportional to the misfit. With the counters fitted last, nothing at any tolerance: the Pratt truss without its counters is determinate, and a determinate truss takes any length error by moving. Deflection

The counter fitted after the load

A truss with a counter-diagonal in every panel cannot take a change of length for free, so a camber cut into its members might seem to lock force into it. One cutting list locks in nothing — every member shortened by its own dead-load stretch — and the simpler lists lock in little. What locks in force is the shop: a millimetre of error per member puts 174 kN into some member of a truss whose largest dead-load force is 471. Leave the counters loose until the dead load is on and cut them to the gap, and the same errors lock in nothing at all.

 One span hung, the other still bare tendon. The level along two 100 m spans at a sag of 2.00 per cent, 35 kN/m finished and 3.0 kN/m of bare tendon, the tendons at 1100 N/mm² when finished, with the deck hung on the left span only (solid) and as finished (dashed), on a pier of 200,000 kN/m. The hung span sags 2.04 m and pulls 21,442 kN; the bare tendon beside it sags 0.20 m and already pulls 18,363 kN, because it was cut 455 mm shorter than the span so as to carry its share when finished. The pier is asked for the difference, 3,080 kN, and leans 15 mm towards the hung span — against 4,953 kN under 20 kN/m of crowd on one finished span. Structural form

The tendon that pulls before the deck arrives

A stressed ribbon is hung one span at a time, and the obvious fear is the stage at which one span carries its deck and the next does not: the pier between them asked for a whole span's thrust. It is not asked for that, because tendons cut to the finished length are already stretched across the empty span and pulling. For two 100 m spans at a fiftieth, the rigid pier's stage force is 4,109 kN against a crowd's 9,440 — unless the tendons were sized generously, when the stage overtakes the crowd.

 The flange over the support works over a third of its width. The effective width of the flange as a share of its overhang along two continuous 20 m spans with a flange overhang of 3.0 m each side of the web, the support bearing over 1.0 m, under a uniform load, left out where the moment is under three tenths of its peak (solid), with EN 1993-1-5's factors over the regions they apply to (dashed) and the simply supported span's mid-span value (dotted). At the sagging peak, 7.6 m from the end, the flange works over 0.81 of its width; over the support, over 0.33. EN 1993-1-5 gives 0.83 in the span and 0.34 over the support, from effective lengths of 17.0 and 10.0 m; a single span of 20 m has 0.88 at mid-span. Sections and stress

The reaction that is made of short waves

Run a wide-flanged girder continuously over a support and its worst moment moves to the support. That moment is made by a reaction, a reaction is made of short waves, and a short wave puts its stress next to the web. Over the support of two 20 m spans with a 3 m overhang the flange works over a third of its width, against four fifths at the sagging peak, so the stress at the web there is 4.2 times the sagging peak for a moment only 1.7 times as large — and how much flange works depends on what the support bears on.

 The shear is read from the pressure line, not the tendon. The first 9.6 m of a two-span beam of 16 + 16 m, 400 × 900 mm, prestressed with 2,400 kN on a parabolic tendon and carrying 45 kN/m, with depths drawn from the top face (−450 mm) to the bottom (+450 mm): the tendon's centre (dashed) and the pressure line, where the resultant of the prestress actually acts once the supports have pushed back (solid). 0.8 m from the support the tendon falls 87.00 mm per metre, which is the slope the rule P·sin θ reads, worth 209 kN; the pressure line falls 76.39 mm per metre, worth 183 kN. The difference is the secondary shear, 26 kN, which the supports put back. Internal forces

The relief that belongs to the pressure line

An inclined tendon carries part of a prestressed beam's shear, and the rule is to subtract its vertical component, P times the slope of the tendon. In a continuous beam the tendon also provokes reactions, and the reactions make a shear of their own — a constant in each span, too small to look at. It is not small where it matters. Beside the end support of a two-span beam it doubles the shear left for the links, and moving the tendon without changing the beam at all can make the tendon's slope report that there is none.

 Five ways to share one deflection among four members. The sway at the top of the left column of a portal 12 m wide and 6 m high with fixed bases, an IPE 500 beam on HEB 300 columns, carrying 25 kN/m on the beam and 40 kN sideways at the top of the left column, 10.38 mm, found five times by the unit load: on the frame itself and on four releases of it. Each bar is split into what each member's integral contributes — left column, left half of the beam, right half, right column — and the red line is the total, which every bar reaches. In millimetres, member by member: the frame itself −3.72, 2.16, 0.66, 11.28; left column alone 10.38, 0.00, 0.00, 0.00; right column alone 0.00, 0.17, 0.17, 10.04; three hinges −15.78, 4.90, 1.41, 19.84; pin and roller −31.43, 27.10, 14.55, 0.15. A member's share can be negative, and on one structure it is the whole answer. Deflection

One column is enough to find the sway

The unit load that finds a deflection may be carried on any structure that can carry it, so in a fixed-base portal the sway can be found from one column's two end moments and nothing else. Carry it on a three-hinged frame instead and the same sway comes out of all four members, one of them contributing a negative amount. Every structure gives the same total and a different account of where it came from — which means "how much of the sway is axial shortening" has no answer until a structure is named, and only one structure's answer is worth having.

 Two formulas that agree until the shear is soft. The buckling load over the Euler load, by Engesser (dashed) and Haringx (solid), against the shear stiffness over the Euler load on a logarithmic scale. Marked: double lacing, 23.48: Engesser 0.959, Haringx 0.961; battens, 0.30: Engesser 0.232, Haringx 0.419; rubber bearing, 6.6e-4: Engesser 0.001, Haringx 0.025. They agree within 5 per cent while the shear stiffness is more than about 4 times the Euler load, and for a shear stiffness a thousandth of it Haringx's is 31 times Engesser's. Stability

The bearing Engesser says has buckled

A member that is flexible in shear buckles below its Euler load, and there are two classical formulas for how far below. For a laced column they agree to a fraction of a per cent. For a battened column one gives nearly twice the other. For a laminated rubber bearing, Engesser's says it buckles at a fourteenth of the load it carries every day and Haringx's says it is safe by a factor of nearly three. They are not two theories. Put the axial load into Engesser's shear stiffness and it becomes Haringx's, exactly — so the whole disagreement is about which shear stiffness was measured.

 The slab carries what the collector does not, beside the bay. The plate's shear flow just beside the same brace line, along it: the rule's uniform 40 kN/m (dashed); with a 150 mm concrete slab (solid), concentrated beside the bay — 132 kN/m on average over the bay's 6 m, 3.3 times the rule's, with peaks at the bay's ends that grow as the mesh is refined, because a stiff support ending in a plate is a stress singularity — and little elsewhere; with a steel deck of shear stiffness 10,000 kN/m (dotted), close to uniform. Each curve adds up to the line's 960 kN. Structural form

The collector the slab does not need

A floor delivers its storey force to a short braced bay through a collector, a member along the brace line that gathers the floor's shear and carries it to the bay. The rule that sizes it assumes the floor hands over its shear uniformly along the whole line. A concrete slab does not: it is so stiff in shear that it sends most of its load straight into the bay, the collector carries about a quarter of the rule's force, and the slab beside the bay works at more than three times the rule's shear. A steel deck is the other way round. The rule is right for one floor, and safe for the other only while the slab stays uncracked.

 Where the pile and the ground settle alike. For a pile 24 m long and 600 mm across, carrying 800 kN, in ground whose top 18 m settles — by 100 mm at the surface, less with depth — over a toe that resists by moving: on the left, how far the ground (dashed) and the pile (solid) settle down the pile's length; on the right, the force in the pile. Above 15.9 m the ground settles more than the pile and hangs on it; below, the pile settles more and the ground holds it up. The force peaks there, at 1,407 kN against the 800 kN applied, and the toe, settling 10.6 mm, mobilises 623 kN; the head settles 13.6 mm. Internal forces

The plane where the pile and the ground agree

Downdrag is usually found by equilibrium: the ground hangs on the pile above a neutral plane, holds it up below, and the plane is where those balance the load and the toe's resistance. But the toe's resistance is not a number the designer can supply. It is whatever the toe's own settlement mobilises, and the toe settles because the ground drags the pile down. Solved by compatibility instead — the plane where the pile and the ground settle by the same amount — the drag, the neutral plane, the toe's force and the pile's own settlement all turn out to grow with how far the ground goes, and each kilonewton put on the head costs the pile three quarters of one, not half.

 What leaves the interior panel lands in the span. A frame of two 8 m bays and two 4 m storeys with fixed bases, 457 mm beams and 254 mm columns, with 20 kN/m on the left bay's beams only, against the stiffness of every beam-to-column joint as a multiple of the beam's EI/L, on a logarithmic scale. The shear across the floor interior joint's column-web panel (solid): 154 kN with rigid joints, 146 at EN 1993-1-8's rigid limit of 25 EI/L, 104 at 3.0 EI/L, 23 MN·m/rad, 63 at EI/L. The loaded beam's largest sagging moment (dashed), in kN·m: 71, 76, 102 and 126, against 160 for a simply supported span. The faint lines are EN 1993-1-8's boundaries for a frame that sways: nominally pinned below 0.5 EI/L, rigid above 25. Internal forces

The panel a soft joint empties into the span

Under a load on one bay, an interior column's web panel carries the difference between the two beams' moments. Make the beam-to-column joints semi-rigid and the panel is relieved — by 5 per cent at the stiffness EN 1993-1-8 still calls rigid, by a third at three times the beam's EI/L — because a softer joint takes less moment. The moment it does not take lands at mid-span, where the beam was not designed for it, and in sway it lands in the columns. And soften only some of a frame's joints, and the sway goes looking for the ones still stiff: their panels carry up to a quarter more than they would in an all-rigid frame.

 However heavy the chord, ordinary posts cap its share. The heavy chord's share of the shear in a panel of uniform shear in a Vierendeel girder of 2 m panels, 1.5 m deep, against how many times stiffer it is than the light chord, from the one-panel model (lines) and from a whole girder (dots). Splitting by second moment (dashed) runs from 0.5 towards 1. With posts as stiff as the light chord the share is 0.54 at a ratio of 4 and still only 0.55 at 30; posts 4, 16 and 100 times stiffer reach 0.65, 0.80 and 0.93 at 30. The posts, not the chord, set the ceiling. The dots are a twenty-panel girder at its sixth panel, far from both support and load: with ordinary posts 0.50, 0.53, 0.56, 0.60, 0.65, 0.69 at ratios of 1, 2, 4, 10, 20, 30. They follow the model until the heavy chord's own second moment becomes a noticeable part of the girder's — 31 per cent at a ratio of 30 — and it starts to carry the global moment by bending as a beam. Structural form

The heavy chord the posts will not let work

A Vierendeel girder carries its shear as bending in its two chords, and when one chord is four times as stiff as the other the obvious rule gives it four fifths of the shear. With posts as stiff as the light chord it takes 0.56. The heavy chord's end moments have to be carried round each joint by a post, the post bends, and the heavy chord turns with its joints instead of bending between them. No amount of extra steel in that chord takes its share past 0.55 in the middle of a long girder; only the posts can, and they have to be about 150 times as stiff as the light chord to make the rule come true.

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