Structural Engineering & Statics
where the load goes
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Scale changes everything — page 3

Essays 49 to 72 of 164 on this thread, in the same order.
Deflection goes as the fourth power of the span. Deflection against span for a constant load intensity and section, with two slower relationships drawn faintly behind it for comparison: the load itself, which grows in proportion to the span, and the bending moment, which grows as its square. Doubling the span multiplies the deflection by sixteen, while the moment only quadruples. Equilibrium

The weight that has to be known before it can be found

Every other load arrives from outside and can be looked up. A structure's own weight depends on how big it is, and how big it is depends on the load — so the first calculation on any project is a fixed point, and the fraction of a member spent carrying itself turns out to be the square of its span as a fraction of a span it can never reach.

 Three ways to move the same column, and they are not close. The same 2000 kN moved 3 m across 14 m, built three ways and drawn to one scale. The deep beam is 1.14 m of concrete, 24.1 tonnes, and settles 60.6 mm in the long term. The storey-deep truss takes the same moment as a couple at 3.6 m centres, so its chords carry M/h and it weighs 2.6 tonnes — a fifth of the beam — while settling 15.8 mm, and it does not creep. The wall is 72.8 tonnes and hardly moves at all, 1.71 mm, of which 36% is shear rather than bending — which is what a member as deep as it is long always does, and is why beam theory does not describe one. A wall as a deep beam is the stiffest of the three by a factor of 35.4. Structural form

The same span, four ways

A beam, a truss, an arch and a cable can all cross the same gap under the same load, and the choice between them is usually described as a matter of judgement or of taste. It is neither. Each carries the load by a different mechanism, each mechanism has a different exponent, and an exponent decides the ordering at every span rather than at some spans.

 The characteristic strength, which nothing was measured at. A lognormal population of strengths with a mean of 30 N/mm² and a coefficient of variation of 0.15. The characteristic value is the 5% fractile — 23.2 N/mm², which is 77% of the mean, and which need not be the strength of any specimen that was tested. Dividing it by 1.50 gives 15.5, and the shaded sliver below that is the fraction of the population that would fail to reach it: 6.4e-6, or one in 155,818. A factor applied to a fractile is not covering the scatter, because the scatter has already been spent getting to the fractile. Materials

The strength no specimen had

A material property is written into a calculation as a number, and a material does not have one. It has a population of strengths with a mean and a spread, and the number used is a low fractile of that population — a value that need not have been measured, that most of the material exceeds, and whose distance below the mean is decided entirely by the scatter.

 A base plate, and when the bolts start working. A 500 × 500 mm plate carrying 600 kN and 180 kN·m, so the resultant sits 300 mm from the centre against a kern of 83.33 mm. The plate is in bolts engaged: bearing over 150.88 mm at a peak of 20 N/mm², with the holding-down bolts carrying 154.42 kN. The plate lifts at 50 kN·m and crushes at 126 kN·m, and the bolts are not needed until 150 kN·m. Connections

The failure that is in the concrete

An anchor bolt is a steel component and its capacity is usually decided by something else entirely — a cone of concrete pulled out around it, failing in tension, in a material every other calculation on the project has assumed cannot take tension at all. The exponent in the capacity says so: it is not the square the geometry implies.

 A broad tank sloshes and a tall one does not. The liquid's division into the part that moves with the wall and the part that sloshes, against the tank's proportion. The convective masses come from the potential-flow solution and the impulsive mass is whatever is left, so the two sum to the liquid's mass exactly at every proportion rather than approximately over part of the range. A tall tank at H/R = 3 is 84% impulsive and behaves almost like a solid; a shallow one at H/R = 0.5 is 72% convective and most of its contents never notice the earthquake. This tank sits at H/R = 1.33, which is 65% impulsive. Dynamics

The liquid has a period of its own

Shake a tank and its contents do not all go with it. Part of the liquid moves as though it were bolted to the wall and part sloshes at a period fixed by gravity and the radius, which the tank's stiffness has no influence over whatever. The split is decided by one proportion, and the two parts then take entirely different amounts of the earthquake.

 A strength with no mechanism in it, made of four. The shear a member carries with no links in it, split into the mechanisms that carry it, against the member's effective depth on a logarithmic axis. The three bands are calibrated to Taylor's measured shares at one 300 mm × 500 mm member and are then evaluated everywhere else, so the shape of the total is a prediction. Aggregate interlock is the band that dies: it depends on how tightly the crack faces are held together, crack width grows with member depth, and it falls from 62% of a shallow member's strength to 22% of a deep one's. That decay is the whole of the size effect, and the dashed line is the design code's fitted k = 1 + √(200/d), which knows nothing about interlock and falls by a factor of 1.52 where the model falls by 2.05 over the same twentyfold range. Dowel action is why the expression contains the flexural reinforcement ratio, which nothing in a truss analogy would predict. Internal forces

The strength with no mechanism in it

A concrete member with no links in it carries shear, and the expression that says how much is three variables raised to fitted powers with a size term in front. There is no free body anywhere in it. What it is fitting is a competition between four things that carry shear across a crack, and only one of them explains why a deeper member is worse at it.

 The dimension the capacity rides on is not a drawn dimension. Moment capacity of a 225 mm slab against the cover to the reinforcement. The line is very nearly straight, because capacity is A_s·f_yd·z and z is about 0.9d — so the capacity is proportional to a dimension that is not on the drawing. What is on the drawing is the overall depth, and the effective depth is what the cover, the link and half a bar diameter leave of it: 225 − 30 − 0 − 8 = 187 mm here. Every one of those three is a site tolerance rather than a design decision. Ten millimetres of bar position is 5.7% of this slab's capacity and 1.8% of a 600 mm beam's — the same workmanship costs 3.0 times as much in the shallow member, and the shallow member is the one whose steel is walked on before the pour. Sections and stress

The dimension nobody can measure

Every flexural capacity in reinforced concrete is proportional to the effective depth, and the effective depth is not on the drawing. It is what is left of the thickness after a cover, a link and half a bar diameter have been taken off it — and each of those is a site tolerance. In a slab, ten millimetres of workmanship is six per cent of the strength.

 The gap is a sum of five things and only one of them is computed. What a 30 mm movement joint is asked to accommodate, by three combination rules. The top bar is every term at its extreme, added: 33.2 mm, which assumes the hottest day, the fullest floor, the whole of the shrinkage and the worst-placed wall arrive together. The chance of that is about 1.5%. The bottom bar treats them as independent and asks for 16.0 mm. The middle bar is the rule used for actions and almost never for movements — one term at its full value and the rest at their coincidence factors — and gives 25.5 mm. The segments across the top bar are the terms themselves, and the ordering is the finding: the largest is tolerance at 10.0 mm, which is not a structural quantity at all, and the smallest is deflection at 3.2 mm — the only one anybody computes carefully, and 10% of the total. Deflection

The gap nobody computed

A movement joint is sized by adding up everything the structure will do to it, and the deflection calculation — the only term anybody computes carefully — is usually the smallest one in the list. The largest is a construction tolerance, which is not a structural quantity at all, and the sum of the extremes is nearly twice what treating them as independent would ask for.

 Ten decades of time for forty per cent of the strength. Strength as a fraction of the five-minute test value, against the length of time the load is held, over 12 decades of seconds. The relation is a straight line on this axis, which is the reason a single duration factor works at all: every decade of time costs about the same amount of strength. A load held for 60 years leaves 59% of the short-term strength, and the member need not have moved — this is not creep, and it is not fatigue, because nothing about the load varies. It is that a static stress slowly breaks fibres. The curve has an asymptote at 18% that nobody quotes: below about a fifth of its short-term strength a member has no time to failure at all, so there is a stress under which duration stops being a question rather than merely becoming a small one. Materials

The load that was left on too long

A timber beam that carries a load for fifty years fails at about fifty-nine per cent of the stress the same beam carries for five minutes in a testing machine. Nothing about the load varies, the member need not have deflected, and the relation between the two is a straight line on a logarithmic time axis over ten decades.

 A restoring moment that gets smaller the further it leans. Restoring moment against rotation for a block 0.90 m wide and 4.20 m tall, both normalised — the moment by its value at first uplift, the rotation by the angle α = 12.1° at which the block topples. It is mgR·sin(α − θ), which is a weight times a geometry with no material property in it at all, and it has two features an elastic system does not. There is a jump at the origin: the moment is whatever the ground demands until uplift and then it is mgR sinα, so the law is discontinuous where a spring's is steepest. And the slope is negative — the further it leans the less it pushes back — so the equilibrium at θ = 0 is stable only because of that jump, and there is no stiffness to divide into a mass. The straight line is what a spring of the same first-uplift strength would have done. Dynamics

The block that is safer for being bigger

A block resting on the ground lifts off at an acceleration that depends only on its shape and not at all on its size, and then falls over at one that depends strongly on its size. Two objects of identical proportion begin rocking at the same instant and only the smaller one topples — which is why the slender water towers stood in Chile and the squat tanks beside them did not.

 The edge, and the length over which it is forgotten. A cylinder of radius 4.00 m and wall 12 mm under 0.6 N/mm² of internal pressure, held at its base. Away from the base the wall carries the pressure as pure hoop tension and bends nowhere, which is why a pressure vessel is a cylinder. At the base the hoop force is zero, because the wall cannot grow there, and the difference is made up by a boundary layer of bending that dies out inward. The length it dies out over is 1/β = 170 mm — 0.778√(Rt), a geometric mean of the radius and the thickness — and the moment is under a twentieth of its edge value by 3.07 of them. Nothing in that length is the load. The base moment is p/2β², and the bending stress it produces is 1.82 times the membrane hoop stress the whole design is about, at every pressure, every radius and every thickness: the ratio is √3/√(1 − ν²) and contains none of them. The hoop force overshoots by 4.3% at 3.2 lengths in, which is the wall springing back past where it was going. Internal forces

The length a structure was never given

A disturbance applied at one place dies out over a distance, and the distance is not something anybody chose. A beam forgets a badly applied load over its own depth. A beam on the ground forgets a point load over the fourth root of its stiffness against the soil's. A shell forgets a held edge over the square root of the radius times the thickness — a geometric mean of two lengths three orders of magnitude apart, which is neither of them and is not near either.

 The steel that is sized by the concrete. Ultimate moment of a 1000 × 400 mm section against the area of tension steel in it, with the moment that cracks the section drawn across. The cracking moment is 77.2 kNm and contains no steel at all — it is f_ctm times the gross section modulus, 2.90 N/mm² times bh²/6 — so it is a horizontal line, and every section to the left of where the two meet is one whose first crack is its failure. The crossing is at 507 mm², and rearranging the two expressions gives 0.245·(f_ctm/f_yk)·bd against the 0.26 the codes print — the constant is a section modulus divided by a lever arm and not a fitted number. The rule as printed asks for 538 mm² here, which is 6% more than the derivation needs, and that margin is the whole of the safety in a check whose failure mode is sudden. Sections and stress

The steel the concrete asks for

Every other bar in a concrete member is there because of an action. This one is there because of the member itself — enough steel that the cracked section can carry more than the moment that cracked it, so that the first crack is not also the failure. The requirement contains no load, and both of its consequences run the wrong way round.

 The same steel, and a crack three times as wide. Calculated crack width against bar diameter, with the area of steel held at 1340 mm² per metre throughout — so the spacing changes with the square of the diameter and the amount of reinforcement does not change at all. The width runs from 0.173 mm at 8 mm bars to 0.372 at 25, a factor of 2.15 for identical steel. The reason is in the crack spacing: after a crack forms the bar has to re-anchor the concrete's tensile force before the next one can, and the length that takes is proportional to the bar's diameter. Of the 337 mm spacing drawn, 35% is the cover term and 65% is the bar term — and the cover term is the one that puts crack control and durability in opposition, because cover protects the bar and widens the crack that reaches it. Sections and stress

The same steel, and a wider crack

A crack's width is the distance between cracks times the strain the steel carries over that distance. Neither of those is decided by how much reinforcement there is. The spacing is a bond length, so it goes as the bar diameter; the strain is set by the stress in the steel. Two arrangements of identical steel can differ by a factor of two in crack width, and the one that wins is the one with more, smaller bars.

 The group is not weaker; it is very much softer. A 3 × 3 pile cap on the left, with each pile's share of 9.0 MN and 4.5 MNm in meganewtons — N/n plus M·y/Σy², the same three terms in the same order as a bolt group under an eccentric load and a section under biaxial bending. The corner piles take 1.25 times the average and a pile added at the centroid would change that by nothing at all, because it adds to neither second moment. On the right is the effect a bolt group cannot have: the piles share ground, so the stress bulbs overlap and the group settles 3.9 times as much as a single pile at the same load per pile, rising to 14.2 for 144 of them. The capacity check everyone makes — block failure against the sum of the piles — comes out at 4.54 here and does not govern at all. The check nobody tabulates is the one that does. Structural form

Nine piles, and four times the settlement

A pile cap divides its load between its piles by the same three terms a bolt group uses and a section under biaxial bending uses. What a bolt group does not have is neighbours it shares ground with — and the group effect that matters is not the strength check everybody makes, but a stiffness effect nobody tabulates.

 The average is not the answer, and it is unsafe. Critical load of a pinned column whose middle third has been given a different stiffness, against the whole-column Euler load, with the two numbers a hand check reaches for beside it. The eigenvalue is taken from K − P·Kg over 24 elements, so nothing here is a formula for a stepped column — it is the same computation the uniform case gets. At a middle third of 0.50 times the rest the true load is 0.612 of Euler's, the arithmetic average says 0.832 and the weakest segment says 0.496. The average is high by 36% and it is high on the unsafe side, because the third of the column it is averaging over is the third where the mode has all its curvature. The weakest-segment answer is safe everywhere and wasteful by about as much. Stability

An average stiffness is not a safe stiffness

Euler's load belongs to a column of one EI. Give the same column two, and the temptation is to average them — which is wrong, and wrong in the unsafe direction by a quarter. Buckling weights stiffness by the square of the curvature of the mode, so the middle of a pinned column decides everything and the ends decide almost nothing.

 The force nobody applied, and the speed it wins at. Lateral force per unit weight for a vehicle on a 400 m curve, against speed. The rising curve is what the free body demands — v²/gR, which is the body's own acceleration written on the other side of the equation — and the flat line is what 6.0° of cant supplies from the weight. They cross at 73 km/h, which is the speed the curve was set out for; below it the deficiency has the other sign and the rail is pushed the other way. The upper line is overturning, at b/2h = 0.399 — and there is no mass in that number, so a loaded vehicle and an empty one go over at the same 160 km/h and only the height of the load decides. At the 108 km/h drawn the deficiency is 0.124 of the weight, which is 49 kN on this 40 tonne vehicle. Equilibrium

The force that is really an acceleration

Every other load in this collection is applied by something. This one is applied by nothing at all — it is the body's own acceleration, written on the other side of the equation so that statics can be used on a problem statics has no business with. The move is legitimate, it is a hundred and eighty years old, and it is exactly half done more often than it is done.

 The check that depends on a date. Total deflection and the deflection occurring after the brittle finishes are built, for one 12 m beam, against the day those finishes go up. The total barely moves — the beam ends up where it ends up. The increment falls from 32 mm at a week to 14 mm at a year, because creep is fast at first and slow later and a partition built early inherits nearly all of it: 44% of the final creep has already happened by day 28. The span/500 limit is 24 mm and the span/250 limit is 48; this beam passes the first only after day 25. Camber subtracts from both terms of the difference and therefore changes the upper curve and not the lower one, which is the reason a cambered beam can satisfy every total-deflection check and still crack the wall. Deflection

The limit that depends on a date

Total deflection can nearly always be met, and on a long span it is met with camber. The limit that actually decides the member is the other one — the deflection occurring after the brittle finishes are built — and camber does nothing for it whatever, because it is subtracted from both terms of a difference. The same beam passes or fails on the day the partitions went up.

 The strength that does not keep pace. Concrete's mean tensile strength against its characteristic compressive strength, with the ratio of the two on the same axis, scaled. The tensile strength goes as f_ck^⅔, so it rises from 1.57 to 5.04 N/mm² over a sixfold rise in the compressive strength, and the ratio between them falls from 11.1% at C20 to 6.0% at C80 — a factor of 1.83. Nothing in a bending or a column calculation ever uses the lower curve, and everything that decides a transition does: when the section cracks, how much minimum steel it needs, how far a bar has to be lapped, what a member without links carries. So a stronger concrete needs more minimum reinforcement, longer laps and a bigger crack-control check, in a member whose ultimate capacity has barely moved. Its real scale is a length: E·G_F/f_ct² is 299 mm here, which is the size at which a member stops behaving plastically and starts behaving like a fracture problem. Materials

The strength that is never used

Concrete's tensile strength appears in no bending calculation, no column calculation and no shear calculation with links in it. The whole design philosophy is that it cracks and the steel takes over. And it decides where nearly every transition in the subject sits — when a section cracks, how much minimum steel it needs, how far a bar has to be lapped, what a member without links can carry, and how wide a crack opens.

 The same steel, brittle in January. Fracture toughness against temperature for a 25 mm ferritic plate, from Wallin's master curve — 30 + 70·exp(0.019(T − T₀)) in MPa√m, whose shape is the same for every ferritic steel and whose only free parameter is the reference temperature T₀ = -60 °C. Beside it, and on its own scale, is the crack length that toughness will tolerate at 200 N/mm², which goes as the square of it. At +20 °C this plate carries a 764 mm flaw and at -20 °C it carries 199 mm — a factor of 3.8 for a forty-degree change in the weather, in a steel that met its specification on both days. Thickness moves the curve as well, and the wrong way: constraint at a crack tip suppresses the yielding that would have blunted it, so a 100 mm plate of this steel tolerates 27% of the flaw a 10 mm plate does. Loading it in a millisecond shifts the whole curve another forty degrees. Materials

The same steel, brittle in January

Every other material property in this collection is a number. Toughness is a curve, and the axis it runs along is temperature. Between twenty degrees and minus twenty a structural steel does not get gradually weaker — it changes the mechanism by which it fails, and the flaw it will tolerate falls by a factor of four.

 The ground is a structure, and it has a period. The transfer function of 38 m of soil at 75 m/s over bedrock at 800 m/s: how much the surface moves for a given motion in the rock, at every frequency. It is a column fixed at the bottom and free at the top, so its resonances are the odd harmonics — the peaks stand at 1, 3, 5, 7 times the first, which is a fixed-free column and nothing else. The fundamental is at 0.493 Hz, a period of 2.03 s, and it is 4H/v_s exactly. The peak amplification is 7.5 against the bound 1/(α + πξ/2) = 7.5, which agrees to 0.2% — and the α in it is the impedance ratio, 0.0554 here. That is the term that keeps the answer finite: assume rigid bedrock and α is zero, the bound becomes 13 and the model is predicting an amplification set by damping alone. What limits the surface motion is that the energy can leave downward. Dynamics

The ground has a period of its own

An earthquake is measured on rock and felt on soil, and between the two is a layer that behaves exactly like a structure — a column fixed at bedrock, free at the surface, with a fundamental period of four times its depth over its shear wave velocity and a set of odd harmonics above it. The motion a building receives is the rock motion through that filter, and the filter is sharp.

 The resonance that ran out of time. The response of a 0.100 s oscillator at 2.0% damping while the driving frequency sweeps up through its own, plotted against the driving frequency rather than against time. The steady-state amplification is 1/2ζ = 25; this sweep reaches 23.6, which is 95% of it, because the time spent inside the half-power band is a limited number of build-up time constants. The whole answer depends on one group, β/ζ²ω², and over the range this figure's sibling sweeps the fraction falls from 100% to 41%. Two features fall out of the integration and neither is guessable from the steady-state picture: the peak arrives after the frequency has passed resonance, by 1.2% of it here, and the response beats afterwards at the difference between the two frequencies. A machine's instrument therefore reads its largest amplitude while it is already above its critical speed. Dynamics

The resonance that ran out of time

A resonance amplifies by one over twice the damping, which for a lightly damped structure is fifty or a hundred. That is a steady-state answer and it takes time to arrive — with a time constant containing the same small damping — so nothing that sweeps through a resonance ever collects all of it, and past a certain rate the peak reached stops depending on the damping at all.

 The worst force in a pile is not at the top of it. A 0.6 m pile 24 m long through ground that is settling, carrying 800 kN at its head. Above the neutral plane the soil moves down past the shaft and the friction acts downward, so the axial force grows with depth; below it the friction acts upward in the ordinary way and the force falls again to the 300 kN the base takes. The maximum is 1282.87 kN at 13.78 m — 1.60 times the load applied, and it is at a depth where nothing is applied, nothing is connected and nothing can be inspected. A pile section chosen for the head load is under-sized by that factor over the middle third of its length. Internal forces

The ground that hangs on instead of holding up

A pile is driven through fill that has not finished settling. The fill goes down past the shaft, the friction along that length turns round, and the pile is now carrying the soil rather than the other way about. The worst force is not at the head and not at the toe, and nothing at that depth can be seen.

 A tie is worth all of itself and a strut is not, which is worth nine per cent. The same 20 floors carried two ways, with every member drawn at the width its own force requires. Hung, the loads accumulate upward, so the largest hanger is at the top: 18.0 MN at 355 N/mm² with no buckling reduction of any kind. On columns they accumulate downward and the largest column is at the bottom, at the same force — but every column above it is understressed by its own slenderness, worst at the top where a 900 kN column still has to be 4.8 × 10³ mm² to reach χ = 0.529. Over the height the hangers total 0.002 m³ of steel against 0.002: a saving of 9.0 per cent, which is the average χ and nothing else. Structural form

Hung from the top, and nine per cent lighter

A tie is worth its full strength and a strut is not, so hanging the floors of a building from a hat truss ought to be an obvious economy. It is a real one, it is measurable, and it is nine per cent of the steel — shrinking as the building gets taller, which is the opposite of what the argument sounds like.

 The tube flattens because of the bending, and then cannot carry it. Moment against curvature for a long tube of radius 300 mm and wall 4 mm. Compression on one face and tension on the other are both directed along a curved line, so each produces an inward transverse pressure and the circle is squashed into an oval by the bending it is carrying. That reduces the second moment, so the curve bends over and reaches a limit point — no bifurcation, no imperfection, nothing to be sensitive to. It arrives at an ovalisation of exactly 2/9 for every tube of every size in every material, at 1018 kNm, where the secant stiffness has fallen to 67 per cent of the undeformed value and the tangent stiffness is zero. The relaxed path reproduces the closed form to 0.004 per cent. Stability

The tube that flattens itself

Bend a tube and the compression on one face and the tension on the other are both running along a curve, so both push inward. The circle becomes an oval, the second moment falls, and the moment–curvature curve turns over at a limit point that needs no imperfection, no bifurcation and nothing to be sensitive to.

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