Structural Engineering & Statics
where the load goes
Theme

Which failure arrives first — page 7

Essays 145 to 168 of 495 on this thread, in the same order.
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.

 Every reason a building is stiffer than its model, added up. The computed natural frequency of a floor, and the same frequency after each source of stiffness that was deliberately left out is put back. Not one of them is a modelling error. Cladding and partitions are stiffness nobody is allowed to rely on for strength; a nominally pinned connection is never pinned; a slab acts with its beam whether or not shear connectors were provided; and concrete between the cracks is stiffer than a cracked section assumes. Together they multiply the stiffness by 1.83 and the frequency by 1.35, because a frequency is the square root of a stiffness, and every factor is halved on the way through. The asymmetry is the finding: leaving stiffness out makes a deflection conservative and a vibration check unconservative in the direction that matters, since a stiffer floor has a higher frequency and sits further from the footfall range. The model here reads 4.40 Hz against a 5.2 Hz criterion and fails it; the floor reads 5.95 Hz and passes. The correction that would have got it right is exactly the stiffness nobody is willing to count on. Deflection

Stiffer than the model said

Measured natural frequencies of finished buildings come out between ten and sixty per cent above the values computed for them, consistently and in one direction only. Nothing on the list of reasons is a modelling error: every one is a real source of stiffness deliberately left out — and leaving stiffness out is conservative for deflection and unconservative for vibration.

 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.

 The strength was bought in a furnace and the welder gives it back. Proof stress as delivered and beside a weld, for four aluminium alloys. The heat-treated alloys lose half of it: the strength of a 6xxx extrusion is in precipitates formed by an ageing treatment, and the arc dissolves them for 32 mm either side of the weld, permanently. The work-hardened tempers lose nearly as much, because the heat undoes exactly the work. The annealed ones lose nothing at all, because there is nothing left in them to anneal. The consequence is the crossover: 5083-H22 is 1.04 times 6061-T6 as delivered and 0.92 times it once welded, so the stronger alloy is the weaker member. The 240 mm member drawn, with two longitudinal welds, keeps 87% of its parent capacity — a weld along a member softens a strip and leaves a section, and the same weld across it softens the whole of one. Materials

The strength the welder gives back

A 6082-T6 extrusion is twice as strong as a 5083-H111 plate and, welded across, the two are within a few per cent of each other. The heat of the arc anneals the metal for thirty millimetres either side, permanently, and the strength that was bought in a furnace is given back at the first joint.

 The bearing that is drawn as a roller. The horizontal force a sliding bearing delivers, against the vertical load it is carrying, with its coefficient of friction on the same picture. The coefficient is not a constant: PTFE's falls as the contact pressure rises, and the standard fit is μ = 1.2/(10 + σ), so the bearing drawn is at 30.0 N/mm² and μ = 0.030 while the same bearing at a fifth of the load is at 0.075 — 2.5 times as much. The force curve is therefore strongly non-linear: a fifth of the load gives 50% of the force. Two readings follow and only one of them is usually taken. The largest force is at full load, 108 kN, and that is what the pier is designed for. The largest nuisance is at light load, where 54 kN of friction is 39% of the 140 kN of wind the bearing was put there to release the structure from. Cold makes it worse again: below about −5 °C the same bearing delivers 216 kN. A roller symbol on a drawing means this, and it is a pair of load cases rather than one, because friction opposes whichever way the deck happens to be going. Connections

The roller that is not a roller

A sliding bearing is drawn as a roller and detailed as a sheet of PTFE, and it delivers a horizontal force of a few per cent of whatever it is carrying. The coefficient everybody quotes is the one at full design pressure, and PTFE's coefficient rises as the pressure falls — so the bearing is at its freest exactly where nobody checks it.

 Four details, and no material anywhere on the plot. Stress range against cycles to failure for four detail categorys — 160, 112, 71, 36 N/mm² at two million cycles. The lines are parallel because they share a slope of three, and the spread between them is a factor of 4.4 in stress and therefore 88 in life. Nothing on this plot depends on the strength of the steel: the same detail in a grade twice as strong lies on the same line. At a stress range of 62 N/mm² the lives are 160: unlimited, 112: 2.1e+7, 71: 3.0e+6, 36: 3.9e+5 cycles. The knee in each line is the constant-amplitude limit, past which the slope becomes five. Connections

The detail decides and the steel does not

A fatigue check contains no material strength anywhere. The same detail in a steel twice as strong lies on exactly the same line, because a fatigue life is decided by the geometry of a weld and by the stress range it sees — and two per cent of the traffic does most of the damage, because life goes as the inverse cube of the range.

 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 axis moves when the section yields. Six sections, each drawn to its own scale, with their elastic neutral axis — the centroid, dashed — and their plastic neutral axis, the equal-area axis, solid. For the symmetric ones the two lines are the same line and the distinction never arises, which is why it is so easily missed. For the tee they are 23% of the depth apart, because the axis that makes the first moment of area vanish is not the axis that makes the two areas equal. The shape factors run from 1.144 to 1.800 across these six, and they are ratios of moduli taken about two DIFFERENT axes — which is also why an asymmetric section has two elastic section moduli, one to each extreme fibre, and only one plastic modulus. The tee's two elastic moduli differ by a factor of 2.78; a fully plastic section does not care which fibre reached yield first, so it has nothing to be two of. Sections and stress

The axis that moves when the section yields

An elastic section bends about its centroid. A fully plastic one bends about the axis that halves its area, and for anything symmetric those are the same line — which is why the distinction is almost never met. For a tee they are a fifth of the depth apart, and three things follow that the elastic calculation gives no warning of.

 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 prop load is decided by the digging, not by the hole. Prop forces in a 12 m excavation propped at 3 levels, with the force each prop reaches at any stage of the sequence drawn thick and the force the finished arrangement gives it drawn thin. The pale dots are the individual stages. The middle prop reaches 214 kN while the dig is at 9.5 m and finishes at 88 — a factor of 2.44 between the two, and the larger one is not in the final analysis anywhere. Terzaghi and Peck's apparent pressure diagram, a rectangle of 49.4 kN/m², reproduces the total of the staged maxima to 2% — which is what it is: an envelope of measured prop loads, back-figured into a pressure, and a shape nothing on a wall is ever loaded with. Structural form

Every prop has its own worst day

A braced excavation has no finished state worth analysing. It is dug in stages, a level of props goes in at each stage, and a prop's force is largely fixed the moment it is installed — so the force to design it for is the largest it sees during a sequence that appears on no calculation sheet, and which for the middle prop here is nearly two and a half times what the finished arrangement gives.

 Two frequencies that meet, and a determinant that never moves. The two natural frequencies of Ziegler's two-bar column against the follower load Pℓ/k, with the determinant of its stiffness matrix drawn along the top. The determinant is k² at every load — it varies over this whole axis by 1.1e-16 of itself, which is round-off — so a static buckling analysis of this structure finds no critical load whatever and reports it as stable everywhere. The frequencies say otherwise: they approach, meet at Pℓ/k = 2.0858, the closed form (7 − 2√2)/2, and become a complex pair, which is oscillation that grows. With no damping that merge is also where the column goes unstable. Adding any internal damping at all drops the load at which that happens to 1.4643, which is 41/28 and 30% below the undamped value; the limit of the damped system is not the undamped system, which is the paradox Ziegler found in 1952 and which was taken for an arithmetic error for a decade. Stability

The load it cannot buckle under

Every stability calculation on this site rests on an assumption nobody states: that the load has a potential, so a critical load is where a total potential energy stops being a minimum. A load that turns with the structure it is pushing has no potential, and the static analysis of such a column returns no critical load at all — a determinant that never vanishes, for a column that fails at a perfectly finite one.

 Two identical pipes, and one carries three times the other. Load per metre on a buried conduit against the depth of cover, in trench widths, with the weight of the prism of soil directly above it drawn between them. A conduit laid in a narrow trench is stiffer than nothing and softer than the sides: the backfill settles relative to the undisturbed ground, the friction on the trench walls acts upward, and the conduit gets 64% of the prism. Lay the same conduit on the ground and build an embankment over it and it is now stiffer than the fill beside it, the interior prism settles less, the friction acts downward, and it gets 172% — a factor of 2.71 between two pipes with nothing different but which way the ground moved. The equation is Janssen's, the same one a silo wall obeys, with a trench for a silo; both curves start on the prism line, because with no depth there is no shear to redistribute anything. This is why a flexible pipe is buried rather than a rigid one: making the conduit weaker moves it down the page. Equilibrium

The pipe decides what the soil weighs

A buried conduit is not loaded by the soil above it. It is loaded by whatever share of that soil the relative movement leaves it — and which way the shear on the sides of the prism acts depends on whether the conduit settles more or less than the ground beside it. Two identical pipes under identical fill, one carrying two thirds of the prism and one carrying nearly twice it.

 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.

 Five millimetres short, and a hundred kilonewtons in every member. An X-braced bay 6 m by 4 m in which one diagonal was fabricated 5 mm short, with the force that leaves in every member. Nothing is applied to this frame. The forces are the self-stress state the frame's one redundancy supports, scaled so that the diagonal is pulled back to the length it should have been: tension in both diagonals at 100 kN, compression in the four members round the outside, and the whole set in equilibrium with nothing. That is 25% of the force the diagonal was sized to carry, and it is there for the life of the structure. Take one diagonal out and the frame becomes determinate: the short member then simply puts the joint somewhere else, and the structure is in the wrong place instead of under stress. Redundancy is bought, and this is the price. Connections

Built to the wrong length

A redundant structure's members do not have independent lengths. Choose all but one and geometry decides the last, so a member made a different length has to be pulled or pushed into place — and the force required stays in the structure for as long as the structure does. Nothing has been applied to it, there is no load case and no factor, and the members are carrying real force.

 Restraint is a fraction, and the length decides how far up it reaches. A 20 m wall 3.0 m high cast against a base that has already hardened — a length-to-height ratio of 6.67. The base holds the bottom of the wall at R = 0.50 and the top of it at 0.304, decaying as 0.609 to the power of the height in wall heights. The free contraction is 380 microstrain, of which 84 per cent is the wall cooling from its own hydration peak and the rest is drying; the concrete's own strain capacity is 50. Everything to the right of the dashed line cracks, which here is the bottom 3.00 m of it. Nothing has been loaded. Internal forces

The steel decides how many, not how much

A wall cast on a base that has already set cools, tries to contract, and is not allowed to. What follows is not a stress problem with a strength on the other side of it. The movement is going to happen; the only question the reinforcement gets to answer is how many pieces it is divided into.

 Prestress buys shear as a square root, not as a sum. The shear stress an uncracked web can take before the principal tension reaches the concrete's tensile strength, against the axial compression the prestress put there. With no prestress it is 1.35 N/mm², the tensile strength itself, because pure shear has a principal tension of exactly its own magnitude at forty-five degrees. Adding compression gives √(f_ct² + σ_cp·f_ct), which is a square root and therefore flattens: the first newton of prestress is worth far more than the last. At the 7.14 N/mm² drawn the limit is 3.39 N/mm², a gain of 2.51, and doubling the prestress from there takes it only to 4.59. The straight line is what a rule that simply added the two strengths would have promised. Internal forces

The crack that never reached forty-five degrees

Every shear expression for reinforced concrete is a curve fitted to tests, because a cracked section has no free body worth drawing. An uncracked prestressed web has one — a single point, a Mohr's circle and a principal tension — and it is the only shear check in the subject that is derived rather than measured.

 A soap film over a hole, and its volume is the torsion constant. The stress function for Saint-Venant torsion of a square of 100 mm, relaxed on a 97 by 97 grid until it stopped moving — 385 sweeps. Its contours are the lines the shear stress runs along, its slope is the magnitude of that stress, and twice its volume is the torsion constant: 1.4053e+7 mm⁴ against a closed form of 1.4058e+7, an error of -0.035 per cent. The steepest slope is 67.515 at the boundary, at the point on it nearest the centre — which is why the peak shear in a solid section is at the middle of the longest side and never at a corner, where the film comes down to zero from two directions and its slope vanishes. Sections and stress

Two volumes, and both of them are torques

Torsion of a solid section that is not a circle has no elementary answer, and for thirty years after Saint-Venant posed it the only way to get one was to blow a soap film over a hole cut in a plate and measure it. The film is not an illustration of the solution. It is the solution, and so is a heap of sand poured on the same hole.

 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.

 A buckling load with no compression anywhere in the member. A tube of radius 81.6 mm and wall 5 mm under pure torque — no axial load at all — buckles into a helix at T = 2πEI/L, which for the section drawn is 1877 kNm. Two things are absent from that expression and both are surprising: the length appears to the first power rather than the second, and the shear modulus does not appear at all, so the torsional stiffness of the member has nothing to do with the torque at which it buckles in torsion. The comparison is the torque that yields the same tube, 42.9 kNm — a factor of 43.8 below it. The mode arrives first only past a length of 263 m, which is πE/τ_y = 3219 radii and contains no thickness whatever. Stability

The buckling load with no compression in it

Twist a straight bar hard enough and it snaps into a helix, with no axial load on it anywhere. The load at which that happens is 2πEI/L — first power of the length, and no shear modulus in it at all, so how stiff the bar is in torsion has nothing to do with the torque that buckles it in torsion.

 A stiffener is a boundary condition, and it is bought at a threshold. The buckling stress of a 2400 × 12 mm plate with one longitudinal stiffener, against how rigid that stiffener is. Below γ the stiffener rides on the buckle and the plate takes the whole-width mode; at γ the stiffener stays straight and the plate buckles between stiffeners at 74 N/mm², 4.0 times the bare plate's 18.5. Above γ nothing further happens at all, because the sub-panel mode does not know the stiffener is there. The curve is a ramp and then a horizontal line, so a stiffener at twice γ is exactly as good as one at γ. Here γ = 31.5, which asks for an outstand of 144 mm; the 150 mm one drawn gives γ = 35.5, a margin of 1.13. Stability

The rib that is a boundary condition

A rib on a plate is not a member carrying load. It is a line the buckle is not allowed to cross — and it becomes one at a threshold. Below the required rigidity it rides on the buckle and buys a fraction; at the threshold it stays straight and the plate buckles between stiffeners; above it, nothing further happens at all.

All themes