Structural Engineering & Statics
where the load goes
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One support too many — page 2

Essays 25 to 48 of 133 on this thread, in the same order.
One restraint, and several times the load. The same portal — the same columns, the same beam, the same steel — buckling with its head held against sway and with its head free to sway. The braced frame's critical load is 16.46 EI/L² and the swaying one's is 5.69 EI/L², a factor of 2.89, and the effective length factor that comes out of each eigenvalue is 0.774 against 1.317. Both are eigenvalues of the assembled frame at a beam-to-column stiffness ratio of G = 1.00; the buckled shapes are the mode vectors themselves, drawn at 18 per cent of the storey height so that the movement can be seen. Stability

Held, and not held

One horizontal restraint at the head of a storey, carrying no vertical load whatever, moves the critical load of the columns beneath it by a factor of 2.89. Effective length is a property of the frame, not of the member.

 The depth is decided by how far it moves, not by what it can carry. A column carrying 6000 kN landing 3 m into a 12 m transfer member. The free body is the member itself, cut under the column: M = P·a(L − a)/L = 13500 kNm, with 4500 kN of shear on one side of the cut and 1500 on the other. At an allowable stress that moment asks for 1.94 m of depth — the dashed outline — and keeping the settlement it causes inside the floors' own bending asks for 2.55 m, which is the member drawn solid. 31% more depth is bought by nothing the strength calculation can see. The depth grows as √(P·a), so four times the load is exactly twice the depth, and depth in a transfer member is a storey nobody occupies. Structural form

The column that stops

A load path that runs straight to the ground costs almost nothing. Interrupting one costs depth in proportion to the square root of the load times the distance it is moved — and the interruption's own deflection becomes the settlement of everything standing on it.

 The same restraint, twice, with opposite signs. A 4 m strip of 200 mm slab whose ends cannot move apart, against deflection measured in its own thicknesses. The flat line is what a yield-line calculation gives, which is what the same strip would carry if its ends were free: 30.0 per unit width. The rising branch is compressive membrane action — the deflected strip is forced into an arch — and it peaks at 116.6, which is 3.89 times the yield-line load, at a deflection of 0.24 of the thickness. Past that the arch runs out of depth and the load falls back to the flexural one; past a deflection of one thickness there is no arch left and the reinforcement starts carrying the strip as a cable. It gets back to the arch's load at 2.17 thicknesses, which is one part in 9 of the span — a sag nobody would design for and exactly what a floor does instead of falling. Internal forces

The force nobody put in the model

A slab strip whose ends cannot move apart is not the strip in the yield-line calculation. Deflecting shortens the chord between its ends, the ends do not come in, and the strip is forced into an arch — worth four times the load it was designed for, at a movement nobody would see.

 Two centres, and the distance between them is a torque. A storey 30 by 18 m with its walls drawn heavy, pushed in one direction by 1000 kN. The force acts through the centre of mass and the storey turns about the centre of rigidity — the stiffness-weighted centroid of the walls, at x = 15.0 m — and the distance between the two is an eccentricity of 0.00 m before the 5% that has to be assumed anyway. The table below the plan splits each wall's force into its direct share and its torsional one. Torsion relieves the walls near the centre of rigidity and loads the far ones, so the wall in trouble is not the wall carrying the most: west wall is asked for 8% more than its direct share, and the walls at right angles to the push carry 19 kN each with nothing applied along them at all. Structural form

The corner that moves most

A lateral force is shared out in proportion to stiffness only if it passes through the centre of rigidity, which is not the centre of the plan and not the centre of mass. The distance between the two is a torque, and the wall that pays for it is the one furthest away and carrying least.

 Whether the floor shares the load out by stiffness or by area. The share of a uniform storey force taken by each of three equally stiff walls, against the stiffness of the floor plate that spans between them. A plate far stiffer than the walls translates almost rigidly, every wall deflects the same and the share is the ratio of stiffnesses — 60% to the middle wall here, the same as everyone else. A plate far softer than the walls behaves as a continuous beam over them and the middle wall takes 33%. Neither end is the tributary-area answer of 50%, which assumes a plate that is both soft and discontinuous over the wall, and which nothing here ever reaches. Structural form

The floor is a beam lying down

A floor plate spans horizontally between the walls that resist a lateral load, carries a distributed inertia load, and has chords, a web and a span-to-depth ratio like any other beam. Its stiffness decides whether the walls share the load by their stiffness or by the area of floor nearest them — and the familiar tributary answer turns out to be neither limit.

 A built-up column has a second way to bend. A 12 m column of two chords 300 mm apart, joined by double lacing. On the left it buckles the way a solid column does, by bending; on the right the chords stay straight and the lattice racks, which a solid column cannot do at all. Neither happens alone, and the two flexibilities add rather than the two stiffnesses — so the critical load is 1287 kN against an Euler load of 1341 kN, which is 96% of it, and the column behaves as though its slenderness were 80 rather than 79. Stability

The column made of two columns

A solid column buckles when its bending stiffness runs out. A laced one has a second way to go — the lattice shears, the chords stay straight — and the two flexibilities add rather than the two stiffnesses. A battened column reaches 23% of its own Euler load and behaves as though its slenderness were twice what it is.

 How stiff a brace has to be before the frame stops swaying. The effective length factor of a swaying portal against the stiffness of a horizontal spring at its head. The curve starts at k = 1.317, the unbraced value, and falls to 0.774 — the factor for the same frame with its head held — at a brace stiffness of 23.2 EI/L³. Past that point the frame buckles in the non-sway mode, which the brace does not restrain, and further stiffness buys nothing at all. The threshold is worth stating as 1.41 N꜀ᵣ/L, which is the form the number is memorable in: for a storey carrying a thousand kilonewtons over four metres it is about 0.35 kN per millimetre of sway. Against the frame's own lateral stiffness of 12.0 EI/L³ it is a factor of 1.93. Stability

The most dangerous day is before it is finished

A structure is analysed once, complete, with every restraint present. It spends weeks in states nobody drew — a beam landed with no deck on it holds 17% of the moment its section is worth, a frame not yet braced buckles at a third of the load it will, and a bolt not yet tightened is a pin where the analysis assumed a fixity.

 Two differences up the same building, peaking in different places. Differential shortening between a perimeter column and the core of a 40-storey building, plotted up the height. The part driven by load peaks at level 20 — exactly half way up, because a floor near the top has almost nothing built above it to shorten what is beneath, and a floor near the bottom has almost nothing beneath it to shorten. The part driven by shrinkage does not care what is above it at all and accumulates all the way to the roof. Their sum is worst at level 40, at 43 mm, which across a 9 m bay is a floor out of level by one in 208. Deflection

The columns are shorter than the core

Every column in a tall building gets shorter as the building is built on top of it, and the core beside it gets shorter by a different amount. The floors between them tilt by the difference — and the difference is largest exactly half way up, because a floor near the top has almost nothing built above it and a floor near the bottom has almost nothing beneath it.

 How much of a deflection belongs to the beam. The share of the total deflection that is the beam's own bending, against the stiffness of what it sits on. A 8 m beam on two supports under a uniform load: on rigid supports every millimetre is the beam's, and the share falls away as the supports soften until almost none of it is. The beam drawn beside this figure sits at 51% — so 49% of what it does is happening somewhere a beam calculation never looks. The two flexibilities are in series, which means the softer one governs and stiffening the other buys nothing. Deflection

The deflection that belongs to the support

A beam calculation answers a question about a beam sitting on things that do not move. Real ones sit on bearings, on other beams and on columns that shorten, and every one of those is a spring in series with the member — so a deflection is the sum of two things and only one of them is a property of the beam.

 One of these two curves is a stiffness and the other is a statement of statics. The torque a spandrel beam carries, against how much of its torsional stiffness is left. The rising curve is compatibility torsion — a floor beam framing into the side of the spandrel, which shares its fixed-end moment of 197 kNm between the spandrel's torsional stiffness and its own flexural one. Uncracked, the spandrel takes 51% of it, or 100 kNm; at a quarter of that stiffness it takes 21%, or 41 kNm, and the floor beam picks up what was shed. The flat line is equilibrium torsion — a canopy cantilevering 2.2 m off the same spandrel, whose 116 kNm is fixed by statics and contains no stiffness at all. The first can be designed away by accepting a rotation. The second cannot be designed away by anything. Internal forces

The torsion that goes away if you let it

A spandrel beam attracts a torque in proportion to its own torsional stiffness. Crack it and the stiffness falls by a factor of four, the torque falls with it, and nothing has failed — because the floor beam it was competing with picks up exactly what was shed. A canopy hung off the same spandrel is a different animal entirely.

 A reaction with no load, and the moment it bends the beam with. The prestress moments in a 2-span beam. The primary moment is −P·e, the tendon acting on its own section, and it reaches 540 kNm over the middle support. The secondary moment is what is left when the primary is taken off the total, and it is 306 kNm — 57% of the primary, with the same sign, so it does not cancel anything. It comes from the middle support refusing to let the beam lift: 51.0 kN pressing down there and 25.5 kN lifting at each end, a reaction set that sums to -2e-13 because nothing external was applied. Its diagram is straight between supports to 2.0e-13% of its own peak, which it has to be: reactions are point forces and a point force puts no curvature in a span. Internal forces

The prestress that pushes back

On a simply supported beam a tendon is an internal matter and changes no reaction. Put the same beam on three supports and the tendon lifts it off the middle one, the support refuses, and the force it takes to hold the beam down is a reaction produced with no load applied at all.

 A brace on the wrong flange never gets there, however stiff it is. The critical moment of an 8 m beam against the stiffness of a single midspan brace, drawn three times for the three heights the brace could sit at. On the compression flange it climbs from 143 kNm to the two-half-wave plateau of 447 — the beam braced into two 4.0 m beams — and reaches 99% of it at 447 kN/m. At the shear centre it needs 2252 kN/m, 5.0 times as much. On the tension flange it never arrives at all: at the stiffness that would have done the job on the other flange it has bought a factor of 1.068, and a stiffer brace in the same place buys the same nothing. Past the plateau the beam stops using the brace, which is where the idea of an ideal stiffness comes from. Stability

The brace on the wrong flange

A brace on a column has one property that matters, and it is stiffness. A brace on a beam has two, and the second decides whether the first is worth anything: put the identical restraint on the tension flange and it does not reach the answer at any stiffness whatever.

 Two null spaces of one matrix, and the count is their difference. Two pin-jointed frames, each with the forces it can carry with nothing applied to it drawn on its bars — tension one colour, compression the other, thickness in proportion. That force set is the null space of the equilibrium matrix; a mechanism is the null space of its transpose; and Maxwell's count b + r − 2j is the difference of their dimensions and knows neither of them separately. A square with both diagonals has s = 1 and m = 0. Two bars in a straight line has s = 1 and m = 1 with a count of 0, so the count is satisfied by a frame that both folds and can be prestressed — and the prestress stiffness is positive, which is why a tensioned pair of collinear bars is stiff at all. Structural form

The forces that are there with nothing applied

Maxwell's count is the difference between two dimensions, and it knows neither of them separately. A frame can satisfy it exactly and still both fold and be prestressable — and when it does, the second of those is what stops the first.

 A couple applied to the core, and two columns to make it. A 20-storey core with one outrigger at 59% of its height. The arm is stiff in bending and the perimeter columns are stiff in tension and compression, so between them they resist the core's rotation at that level — a couple of 35283 kNm here, carried as a 294 kN pair in the columns at 120 m centres. The compatibility is one equation: the core's rotation at that level, less what the couple takes back out of it, equals the rotation the arm and its columns allow. The top drift falls from 360 mm to 74, which is 80% of it, and the base moment from 73500 to 38217 kNm. The deflected shape is drawn hugely exaggerated: the real top drift is about one five-hundredth of the height. Structural form

The arm that makes the columns work

The perimeter columns of a tall building are already there, already carrying gravity, and already the furthest thing from the centre. They take almost none of the overturning, because a floor slab transmits shear and not moment — and one storey-deep arm at the right height changes that by nearly a half.

 A derivative taken with a ruler, and the step that makes it worst. Castigliano's theorem says the deflection is ∂U/∂P, and the derivative here is taken numerically — two solves at ±dQ and a central difference. Against the unit-load answer of 1.720635e-2 it agrees to 1.6e-13, which for a linear structure it must: ∂N/∂P is exactly the force a unit load produces, so the two expressions are the same sum written twice. The error against step size is the classic pair of straight lines — truncation falling as the step shrinks, round-off rising as the difference of two nearly equal energies loses its digits — meeting near dQ = 1.2e+1. For a linear structure the truncation term is exactly zero, so what is drawn here is round-off alone. Deflection

The deflection that is a derivative

A structure's strain energy is one number. Differentiate it with respect to a load and out comes the displacement under that load — and the trick that makes it a method rather than an identity is that the load does not have to be there.

 The building does not care how far it went down; it cares how much it tilted. Five footings on soil that is 35% as stiff under one of them, carrying 60 kN/m. They settle between 12 and 54 mm, and the number that matters is neither of those: it is the angular distortion between neighbours, 4.32 per thousand, or one in 231 — against a limit of one in 500 for cracking in finishes, which this does not. A building that went down half a metre uniformly would be undamaged and would need a new front step; this one has moved a twentieth as far and has cracked. Deflection

The settlement that matters is the difference

A building that goes down half a metre uniformly is undamaged and needs a new front step. One that goes down a twentieth as far, unevenly, has cracked. The superstructure can even the difference out — and the only way it can do so is by carrying the difference itself, as a force.

 The same load, two diagrams, both in equilibrium. One span of a pair of 7 m spans under 5 kN/m, drawn twice. The elastic solution puts 31 kNm over the support and 17 in the span. Reducing the support moment by 30% and taking what statics then gives leaves 21 and 21: the section the beam needs falls from 31 kNm to 21, a saving of 30%. Both curves are in equilibrium with the same load — the mid-span ordinate plus half the support moment is the free moment 31 kNm for either — and the second is legitimate for that reason alone. What it costs is 1.0 milliradians of rotation at the support, which the section has to be able to deliver. Internal forces

The moment that was moved on purpose

The elastic analysis of a continuous beam gives one set of moments. It is not the only set the beam is allowed to have, and taking a smaller one at the support is legal, cheaper, and paid for in a rotation that has to be delivered before the design exists.

 Between two beams and one, and much nearer one. How composite a beam is, against the one dimensionless group that decides it: αL, where α² = K·EI∞/(EA*·EI₀). At αL = 0 the layers slide freely and the beam is two beams; past about 20 the connection is stiff enough that the last per cent is unbuyable. The beam drawn sits at αL = 15.8 and is 96% composite, deflecting 36.7 mm against 33.9 for full interaction and 110 for none. The curve is steep where a real design sits, which is why halving the number of studs does not halve anything. Internal forces

Half the studs, and most of the beam

Bonding two layers together quadruples the stiffness of the pair. A real connection is a row of studs that deform, so the layers slide a little and the beam sits between the two answers — but not halfway between, and the number that decides where is a single dimensionless group.

 Four inequalities, and the wedge between them. The Magnel diagram: every limit on a prestressed section, plotted as a bound on 1/P against the eccentricity. Two of the four come from transfer, when the force is largest and the only moment is the beam's own weight, and two from service, when 20% of the force has been lost and the moment is 640 kNm. Each is linear in 1/P, which is the substitution that makes the problem a picture rather than a search. The shaded region is every force-and-eccentricity pair the section will accept: it is a wedge opening to the right, so the cheapest prestress is always at the largest eccentricity the cover allows — 1029 kN at e = 400 mm here. The section's kern is 241 mm, and every useful answer is outside it. Sections and stress

Four inequalities and a wedge

A prestressed section has to satisfy two stress limits when the force is largest and the load smallest, and two more when the force has relaxed and the load has arrived. Each is linear in one over the force — which turns a search for a prestress into a region on a page, and turns an impossible section into an empty one.

 The envelope is not a state of the structure. Every arrangement of the imposed load on three spans — 8 of them, since each span is loaded or not — drawn faintly, with the greatest sagging and greatest hogging at each station drawn over them. Each faint curve is a real state of equilibrium and satisfies the free-moment identity exactly: mid-span ordinate minus the mean of the end moments is wL²/8, to 3e-16 of it. The envelope satisfies it nowhere, missing by up to 23% — because it is assembled from different load cases at different stations and no arrangement of load produces it. seven of the 8 arrangements are needed to build it; the rest never govern anywhere. Equilibrium

The envelope is not a structure

A continuous beam whose imposed load may sit on any span has eight load cases, and every one of them is a genuine state of equilibrium. The curve the design is made against is not one of them — it is assembled from different cases at different stations, and it fails the identity all eight satisfy exactly.

 The tie is a redundancy, so its stiffness decides the thrust. Thrust and rib bending for a 60 m tied arch of 0.15 rise ratio, against the stiffness of its tie. Cut the tie and the structure is a curved simply supported beam, so the tie force is the one redundant and the force method gives it: with a rigid tie the answer is 2229 kN, within 0.9 per cent of the funicular wL²/8f, and the shortfall is the arch's own axial shortening. A real tie stretches 68 mm and returns 2168 kN — 2.7 per cent of the flexibility is the tie — and whatever thrust the arch does not get, it carries as bending: 739 kNm at 30 m. A tenth of the tie stiffness is not a tenth of the problem; it is a different structure. Structural form

The thrust that never reaches the ground

Every arch on this site has ended at the same sentence — the foundation is where an arch is really decided. A tie changes the sentence without changing the arithmetic: the horizontal force is still there, still the same size, and it now closes on itself through a bar at deck level.

 The two theorems close on the answer from opposite sides. A pinned-base portal frame under 200 kN at mid-span and 80 kN at the eaves, with a plastic moment of 200 kNm. The three horizontal lines are mechanisms — work equations, each an upper bound: the beam mechanism gives 1.000, sway 1.250 and the combined one 0.714. The curve is the lower bound: for each value of the one redundant, the largest load factor whose moment field stays inside the plastic moment everywhere. Its peak is 0.714, and it touches the lowest mechanism exactly — so the collapse load is known rather than bracketed. Using the beam mechanism instead would have claimed 1.40 times the real capacity, and the sway one 1.75 times, both of them on the wrong side. Equilibrium

Two ways of being wrong

Plastic analysis has two theorems and they point in opposite directions. Any equilibrium field that nowhere exceeds the plastic moment gives a load at or below the collapse load; any mechanism gives one at or above it. Only one of those errors is safe.

 The bond stress is crowded against the loaded end. A 20 mm bar embedded 806 mm, with the force in it and the bond stress on it plotted along the embedment. Uniform bond — the assumption behind every development length ever tabulated — is a flat stress and a straight line of force. An elastic bond of the same peak strength is neither: the slip is largest where the bar is pulled and dies away over 1/α = 471 mm, so the far end of the bar is doing almost nothing. At the design rule's length of 40 diameters the elastic bond is 55 per cent used. The uniform answer is what the bond looks like after it has yielded along the whole length, which is a statement about ductility rather than about strength. Internal forces

The force that arrives along a length

A bolt takes its force at a hole and a weld along a line. A reinforcing bar has no such place — it is a smooth cylinder in a hole of its own shape, and the only thing stopping it sliding out is a stress smeared over its surface. So the force in it is not a number, it is a function of position.

 A fan, and where its forces go. Half a cable-stayed bridge: a tower 70 m above a deck, 12 stays reaching out over 200 m, and a uniform 200 kN/m on the deck. Each stay is drawn at a weight proportional to the force in it, from 3427 kN at the innermost to 10090 kN at the outermost — the outer stay carries the same vertical share and is far flatter, so it carries far more. The deck's shading is its own accumulated compression, 61905 kN at the tower, which is 1.55 times the load being lifted and is the horizontal half of every stay force added up. Nothing in this drawing is a catenary: every stay is straight and every one of them is a spring. Structural form

The cable that is a spring

A suspension bridge's cable is a funicular — it takes the shape the load gives it, and the deck's job is to make the load one that shape is right for. A stayed deck is not that at all. Its cables are straight, each reaches the deck at one point, and every one of them behaves as an inclined spring.

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