Structural Engineering & Statics
where the load goes
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The load must go somewhere — page 4

Essays 73 to 96 of 320 on this thread, in the same order.
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.

 A basement is a boat. A 20 by 30 m substructure dug 6 m into ground whose water table stands 2 m down. The head on the underside of the base slab is 4.0 m, so the pressure there is 39.2 kN/m² over the whole plan — 23.5 MN of it, pushing upward. Nothing about the structure changes that number. What resists it is weight: 18.7 MN of concrete and whatever is built above, giving a factor of 0.80. The structure floats if the water reaches 2.82 m below the ground, and a base slab alone would have to be 1.64 m thick to hold it down. Equilibrium

A basement is a boat

Every load in this collection presses down and is resisted by strength. Hydrostatic uplift presses up, is resisted by weight, and does not care what is built on it — so the check contains no material property at all. It is a ratio of two weights, and one of them is water.

 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.

 The cut that severs the stirrups is the cut that counts them. A cracked web drawn as the truss it has become: a tension chord along the bottom, a compression chord along the top, concrete struts between the cracks and stirrups crossing them. The free body is a cut parallel to the cracks, which over a lever arm of 495 mm severs z·cot θ/s = 8.3 stirrups at cot θ = 2.5. Every one of them is at yield, so the shear is their number times their strength, and flattening the crack raises the count rather than the strength. The same cut passes through the bottom chord, which is where the chord force the moment diagram does not contain comes from. Internal forces

The beam that becomes a truss

Once a web has cracked in shear there is no shear stress field in it any more. There are concrete struts, two chords and whatever crosses the cracks, and the angle of those cracks is not a property of the material — it is something the designer chooses, and every quantity in the beam moves when it changes.

 The strain it wants, the strain it is allowed, and the difference. A bridge deck 1.40 m deep with 18 °C at the top face falling away over 10% of the depth. The left curve is the free thermal strain αT(y); the straight line beside it is what a plane section will actually take, ε₀ + κy with ε₀ = 32.0 microstrain and κ = 0.063 per km. The right-hand block is E times the difference, and it reaches -3.98 N/mm² of compression at the surface and 1.83 of tension 140 mm below it. Its resultant force is 8.3e-14 kN and its resultant moment 3.0e-12 kNm, which is what self-equilibrating means: the field is invisible to every equilibrium check that could be made on the member. Sections and stress

The stress nobody restrained

A bridge deck lying loose on its bearings, with nothing holding it anywhere, develops four newtons per square millimetre when the sun comes out. The stress is not caused by restraint. It is caused by plane sections, and it is invisible to every equilibrium check that could be made on the member.

 Nothing happens, and then everything happens. The moment capacity left to a section already carrying shear, against the shear as a fraction of what the web can take. The web holds 26.1% of this section's plastic modulus and the flanges hold the rest, and only the web's share is reduced — by the factor √(1 − v²) that von Mises leaves it. So the curve is flat for most of its length: the first per cent of moment is not lost until v = 0.27, half the shear capacity costs 3.5%, and 15% is not reached until v = 0.9. The tangent at v = 1 is vertical, which is why the last tenth of the shear range costs more than the first eight. Sections and stress

Both at once, and neither matters until it does

A section carrying shear has less moment capacity, and the reduction is the web's share of the plastic modulus times one minus the root of one minus the shear ratio squared. On a rolled beam that share is a quarter, so half the shear capacity costs three and a half per cent — and then the last tenth costs more than the first eight.

 The arch does not squash; it leans. The first two buckling modes of the same rib, drawn against its undeformed shape at an exaggeration of a few hundred. The first, at wcr L³/EI = 49.7, is antisymmetric: one half rises while the other falls and the crown moves sideways. The second, at 117.1, is symmetric — the whole rib settling. The two differ by a factor of 2.36, which is why an arch is braced against sideways movement of its crown rather than against the load it is carrying, and why a tied arch with a single hanger at midspan is doing nothing for the mode that governs it. Stability

The arch that leans instead of squashing

A masonry arch is asked whether a line of thrust fits inside it. A steel rib is asked a different question entirely: it is a column carrying an axial force along its whole length, and the mode it buckles in puts one half up and the other half down while the crown moves sideways.

 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.

 The force runs along the chord whatever route the member takes. A member pinned at two points 10 m apart, loaded only at those two points, and bowed 1.2 m off the line between them. Equilibrium leaves the two end forces no choice: equal, opposite, and along the chord. So at every section the axial force is P cos α, the shear is P sin α, and the bending moment is the force times the perpendicular offset — 240 kNm at the crown for 200 kN at 1.2 m, which is a multiplication rather than an analysis. Straighten the member and the moment diagram is identically zero; that is the case a truss member is in, and the reason it carries one number. Equilibrium

The member with only one direction

If a body is in equilibrium under forces applied at exactly two points, those forces are equal, opposite and along the line joining the points. It is three conclusions from two equations, it is the shortest real theorem in statics, and nearly everything that follows depends on it without saying so.

 A tank grows without limit; a silo stops. Vertical pressure against depth in a 8 m silo of a solid weighing 9 kN/m³, beside the straight line a liquid of the same weight would have produced. The free body is a slice: its own weight in, the wall friction out, and the friction is μK times the vertical pressure that generates it. The result saturates at γR/μK = 89 kN/m² and reaches 63% of it at one characteristic depth, R/μK = 9.9 m. At the base the pressure is 85 kN/m² against a liquid's 270 — 69% less — and the wall has taken 69% of the stored weight down with it. The exponent is the capstan's, and for the same reason. Equilibrium

The pressure that stops growing

A tank of liquid presses harder the deeper it gets, without limit. A silo of grain does not. Wall friction carries part of the weight, the pressure that generates the friction is proportional to the pressure being carried, and the equation that follows is the one that describes a rope round a bollard.

 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.

 The force spreads, and the spreading needs a tie. The end block behind an anchorage of 1200 kN on a 200 mm plate, in a section 700 mm deep. Half the force enters at the quarter point of the plate and leaves at the quarter point of the section, so a strut between the two rises 125 mm and needs a transverse tie to turn it. Placing the tie 0.5 depths from the face makes that tie force 214 kN — and at exactly half a depth this reproduces Guyon's 0.25P(1 − a/h) to the digit, which makes that famous coefficient a lever arm somebody chose rather than a property of concrete. The bearing stress under the plate is 20.0 N/mm² against 5.7 once the force has spread. Internal forces

The force that splits what it pushes on

A prestressing tendon delivers its whole force through a plate a fraction of the section deep. One depth further along the stress is uniform, and the spreading in between requires a transverse tension nobody applied — the force that splits end blocks, and the only number in the design that no equilibrium equation on the member can see.

 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.

 Every path to the ground goes through the link. A braced bay 8 m by 4 m whose two diagonals stop 800 mm apart instead of meeting. The storey shear reaches the ground through the diagonals, and the vertical components they deliver to the beam have to pass through the segment between them: the link carries 47% of the applied shear as a shear force, at a lever arm short enough that its ends reach 0 kNm while the rest of the beam carries 0. The deflected shape drawn is the solved one, magnified — the real drift under this load is 0.008 mm. Everything outside the link is designed to stay elastic while the link is yielding, which is what makes the mechanism a choice rather than a hope. Structural form

The part that is meant to be weak

A braced frame is stiff and has nowhere to yield. A moment frame yields everywhere and is soft. Move the two diagonals a metre apart along the beam and the whole storey shear has to pass through the segment between them — which keeps most of the stiffness and puts every yielding in one member the designer chose.

 The split is where buckling puts it. A branching column carrying 400 kN to two points 3.0 m apart, over 9.0 m. Every member is drawn at the thickness it needs: the area is the larger of N/σ and what Euler asks of a strut of that length, and 3 of 3 members here are sized by buckling rather than by strength. The split sits at 57% of the height, which is where the total volume is least — 51% less than the fan of straight struts that carries the same load with the same stresses. Structural form

The tree that strength does not ask for

A branching column carries a roof on many points and reaches the ground on one. Size every member by its stress and the optimum tree turns out to have no trunk at all — the best answer is a fan of straight struts from the base. Put buckling in and the trunk appears, at 57 per cent of the height.

 The restraint chooses the buckling length, and it is not the member's. A compression flange 12 m long held sideways not at points but everywhere, by a restraint of 0.35 N/mm per mm of length. Unrestrained it would buckle at 173 kN in a single half-wave, drawn faintly. Restrained it buckles at 1968 kN — 11.4 times as much — in two half-waves, because the sum n²π²EI/L² + kL²/n²π² has its minimum there and every other n is worse. The effective length that answer implies is 3555 mm, which is 0.30 of the member and is a property of the restraint rather than of the span. Stability

Held everywhere, and it forgets its length

A brace at a point divides a member's buckling length. A restraint spread along the whole member does something else — the member chooses its own number of half-waves, and past a few of them the critical load stops depending on the length at all.

 Balanced, and four times as heavy on the bearing. A bascule leaf of 900 kN whose centroid is 9 m from the trunnion, balanced by 2700 kN at 3 m on the other side. What balancing achieves is exactly one thing: the moment about the pivot is zero at every opening angle, because both terms carry the same cosine. What it costs is two things that are not zero. The reaction on the trunnion becomes 4.0 times the leaf's own weight, since both weights are still there. And the rotational inertia rises by 33%, so the balanced leaf is the hardest one to start and to stop — which is why the counterweight is put as close to the pivot as it will fit, at the price of being heavy: the same balance at twice the radius weighs 1350 kN and carries 1.25 times the inertia. Equilibrium

Balanced, and four times as heavy

A counterweight cancels a moment about a pivot, and that is the only thing it cancels. The bearing beneath carries both weights, the inertia rises as the square of the radius, and a load that moves cannot be balanced at more than one position at all.

 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.

 One coefficient, and nothing else in it. The deflected shapes of one beam under four load cases, each scaled so that its mid-span deflection is the same, with the tangent at the left-hand support drawn on each. The end rotation is that deflection times a coefficient that depends only on the shape of the load: 3.20 for a uniform load, 3.00 for a load at mid-span, 2.99 for a triangular load, 3.60 for a load on half the span. Every material property, every second moment and the span itself cancel out of the ratio θL/δ, so a beam at any deflection limit has an end rotation that is known before anything about it is: at L/360 it is 8.89 milliradians, or 0.51 of a degree. Deflection

The angle nobody limits

Every serviceability rule in this collection limits a displacement. What a bearing, a joint and a cladding gap actually have to accommodate is an angle — and the angle is locked to the displacement by a coefficient that contains no material, no section and no span.

 One drift, two motions, opposite curvatures. The sideways movement of a 120 m building under a uniform wind, drawn as the sum of the two mechanisms that produce it. The bending curve is a cantilever's: flat at the base, steepening upward, concave one way. The racking curve is a stack of parallelograms: steepest at the base and flattening, concave the other. They add to 366 mm at the roof, of which 61% is bending. The one group that decides the split is αH = H√(GA/EI) = 2.48: below one the racking dominates and the building behaves as a frame, above about six the bending does and it behaves as a cantilever, and everything interesting is in between. Deflection

Two motions with one name

A tall building's sway is two movements added. A frame racks like a stack of parallelograms, worst at the bottom; a cantilever bends about its base, worst at the top. The total at roof level says nothing about which storey is worst, and on this building it is neither.

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