Structural Engineering & Statics
where the load goes
The collection

Every essay — page 11

Essays 241 to 264 of 728, in the same order.
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.

8 figures · Characteristic strength The middle third, computed. The kern of a 300 × 600 mm rectangle, computed by asking, for every direction, how far the resultant can move before the far face would be pulled. It reaches ±100.0 mm vertically and ±50.0 mm horizontally, which are h/6 and b/6 exactly, and the region between is a rhombus rather than the ellipse the two numbers suggest — a resultant on a diagonal has less room than either axis allows. Materials

The strength thrown away on purpose

Masonry, concrete and soil are all analysed as though they had no tensile strength whatever. Each of them has some. The decision to set it to zero is the single most consequential modelling assumption in the subject, it is safe for one kind of check and unsafe for another, and almost nothing that uses it says which.

8 figures · No tension 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.

8 figures · Anchor breakout The end bolts do the work and the middle ones very nearly nothing. A lap of 10 bolts at 75 mm pitch transferring 800 kN between two plates, with the force each bolt actually carries drawn above it and the flat line a division by the bolt count would have given drawn behind. The end bolts carry 1.16 of their nominal share and the middle ones 0.89. The reason is not in the bolts: at the leading end the first plate is carrying everything and the second nothing, so the two strain at different rates and the slip between them is largest there. In the middle they strain alike, there is almost no slip, and a bolt with no slip across it transfers almost no force. The mean over the worst is 0.861, and the end bolt has to slip 1.16 mm before the rest catch up. Connections

The joint that has to be as good as the member

A splice exists because members come in lengths and structures do not. It has to deliver the same force, at the same stiffness, in the same distribution across the section, through a discontinuity — and each of those three requirements is met by a different feature of the detail, with the third one usually left to look after itself.

8 figures · Splice What the shape of a load in time is worth, for two load shapes. The peak displacement as a multiple of the static deflection, against the load's duration divided by the structure's natural period, for two load shapes: a load that rises linearly, then stays; a rectangular pulse, then nothing. The lines are closed forms and eight dots are the peak of a complete time integration of an oscillator of 0.300 s period under that load, agreeing with the line to within 0.19% everywhere. Dynamics

The load that is over before it has moved

A blast delivers an enormous pressure for a few milliseconds. Everything else in this field asks what force a structure can carry; a load that has come and gone before the structure has travelled any distance is not asking that question, and the answer turns out to depend on the mass and the ductility with the strength barely in it.

8 figures · Blast 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.

8 figures · Sloshing 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.

8 figures · Concrete shear An enhanced strength that is the strength of a tie. Bearing strength as a multiple of the design cylinder strength, against how far the load is allowed to spread, with the bursting tension the spread creates on the same axis. The enhancement is √(A₂/A₁) and it reaches 2.80 for the 250 mm pad on a 700 mm block drawn — 47.6 N/mm² against a design strength of 17.0. There is no material property in that statement beyond the one being enhanced, and the reason is on the second curve: a load that spreads does so along inclined struts, a pair of inclined struts has a horizontal component, and that component is 16.1% of the load. It has to be tied. 1099 mm² of steel is what the enhancement actually is, and the cap of three is not a property of concrete — it is the angle past which nobody believes the strut. Internal forces

Three times as strong under a smaller pad

Press a small plate onto a large block of concrete and it will carry three times the stress a cylinder of the same concrete fails at. The enhancement is a ratio of areas with no material property in it, which should be a warning: what has actually been measured is not the concrete's strength but the strength of a tie holding it together.

8 figures · Bearing stress The tendon is a load, pointing the other way. A 14 m beam with a parabolic tendon dropping 260 mm to midspan, stressed to 1440 kN after losses. Its curvature pushes the beam up along its whole length with an intensity of 8Pe/L² = 15.28 kN/m, against an applied 17.63 kN/m — so 2.34 kN/m is left to bend anything, and the beam carries 57.4 kNm where an unstressed one carries 432 kNm. What the section then feels is 6.40 MPa of uniform compression and very little else. Internal forces

The load that comes from changing direction

A force that travels in a straight line asks nothing of anything. Bend its path and it asks for a transverse load of F over R along every millimetre of the curve, and that load is real, is nowhere on the load schedule, and is the same statement behind a prestressing tendon, a hoop force, an arch thrust and a web that buckles with nothing applied to it.

8 figures · Deviation force 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.

7 figures · Effective depth The answer is continuous and the catalogue is not. Capacity bought against capacity required, over a real rolled series. The straight line is what a continuous section would give — exactly the moment asked for, and nothing can be bought on it. The staircase is what a catalogue gives: each tread is one section, each riser is the step to the next, and the vertical gap between the two is steel that is paid for and does nothing. The steps in this series run from 23% to 59% in plastic modulus, so the average waste is 15.0% and the worst is 46% — just above a riser, where the section below has been missed by a kilonewton-metre. The 1200 kNm marked buys a 686×254×125 at 1418 kNm, which is 85% utilised. Two things follow that a continuous treatment cannot see: the sensitivity of a design to an assumption is zero over most of a tread and enormous at a riser, and an optimisation that returns three significant figures is answering a question with twelve answers in it. Sections and stress

The answer is continuous and the catalogue is not

Every optimisation in this subject returns a number with three significant figures in it, and nothing with three significant figures can be bought. What can be bought is a rolled series whose steps are a quarter to a half apart, so the member that goes on the drawing is on average a tenth stronger than the one that was calculated and can be a third stronger for no reason at all.

8 figures · Available sections A section has two areas and the tables give one of them. Peak shear stress divided by the mean, for four sections of exactly the same gross area and depth. The mean is V/A and is the number a first calculation uses; the peak is what the material actually sees, and the ratio between them is a property of shape alone. A rectangle's is 1.5 — the parabola's peak over its average — and it is one of the few numbers in this subject that is exactly derivable and universally ignored. An I-section's is near 1.98, and the reason is on the second bar: 97% of the shear is inside a web that is 56% of the area. So the flanges carry the moment and almost none of the shear, and the web carries the shear and almost none of the moment — which is why a shear check on an I-section uses the web area and a moment check uses the whole section, and why the two checks are about two different pieces of steel. Sections and stress

The section has two areas

A shear force divided by the area of the section is not the shear stress anywhere in it. A rectangle's peak is exactly one and a half times that number and an I-section's web carries nearly all of the shear over a fifth of the area, which is why a moment check and a shear check on the same member are checks on two different pieces of steel.

7 figures · Shear area Two curvatures of opposite sign, which is what makes it a structure. A cable net over a 36 m square, drawn as the two families of cables that are also the two rulings of the surface. One family sags and carries downward load by hanging; the other rises and carries upward load — wind uplift, and a load reversal anywhere — by the same mechanism upside down. Neither can do anything alone. A single family of cables is a mechanism: it changes shape freely under any load pattern it was not tensioned for, and the shape it moves to is decided by the load rather than by the designer. Put the two together and each is the other's restraint, but only if they are pulled against one another first — the pretension of 520 kN in the sagging family and 715 in the hogging one is a self-equilibrating state that exists with no load on the roof at all, and it is what turns two mechanisms into one structure. The curvatures are drawn four times their true value: a real net of this span sags 2.2 m over 36, which is flatter than it looks anywhere. Structural form

Two curvatures of opposite sign

A single family of cables is not a structure. It is a mechanism that takes whatever shape the load asks for, and it will do that under any load pattern it was not tensioned for. Cross it with a second family curved the other way, pull the two against each other, and the pair becomes stiff — with no bending anywhere and no material property involved in the stiffness at all.

9 figures · Cable net Two drawings of one deck, and they are not the same structure. A 112 m viaduct on five supports, articulated two ways. Above, the fixed point is at the left abutment: the far end has to be given 45 mm of movement, and the friction of every sliding bearing runs one way, so the fixed support takes 660 kN before any wind or braking is applied. Below, the fixed point is at the middle pier: the largest joint halves to 22 mm and the friction now cancels across the fixed point, leaving 0 kN. The movement arrows are drawn at 900 times the scale of the deck, because a 45 mm movement on a 112 m span is thinner than the line the deck is drawn with. Nothing about the deck, the loads or the ground has changed between the two. Structural form

Where the structure is allowed to move

One drawing decides how big every movement joint on a bridge is and where every horizontal force goes, it takes an afternoon, and it appears on no calculation sheet. Move the fixed point from an abutment to the middle pier and the largest joint halves and the horizontal force on that support drops from the whole of the friction to none of it.

7 figures · Articulation The two braces balance until one of them buckles. An inverted-V brace after the compression member has gone. While both braces are elastic they carry equal and opposite forces and their vertical components cancel on the beam above, which is why the beam in a chevron bay is usually sized for gravity alone. The compression brace buckles at 445 kN and then sheds most of what it was carrying — 30% is left here — while the tension brace goes on to yield at 1065. The difference between the two vertical components is 659 kN, applied at the middle of the span with no help from either brace, and it asks the beam for 1317 kNm against the 200 kNm the gravity load asks for — 6.6 times as much. The beam drawn does not: 1517 kNm against a capacity of 731. The force is not a load case anybody applies; it is what the frame leaves behind on its way to the state it will actually be in. Structural form

The force the brace leaves behind

Two braces meeting under a beam carry the storey shear as a tension and a compression whose vertical components cancel, so the beam above sees nothing. They cancel only while both braces are elastic. Once the compression brace buckles it sheds most of its force, the tension brace goes on to yield, and the difference is a point load at midspan that nobody applied.

7 figures · Chevron brace Effective length is a property of the storey. The effective length factor of the one column that resists sway, against the total gravity load on the storey as a multiple of its own. At the left-hand end it carries the storey alone and its K is 1.99 — the 2.0 every chart gives a column fixed at the base and free to sway at the top, reproduced here by a route that never mentions a chart. Then columns are added that have pinned bases and therefore no lateral stiffness whatever. They contribute load and nothing else, so they cannot buckle on their own and they lower the load at which everything buckles together. K rises as the square root of the load ratio, exactly, and at the storey drawn — three leaning columns carrying 69% of the gravity load — it is 3.57. That is off the end of every published alignment chart, and the leaning columns themselves, which a designer would take at K = 1.0 for pinned ends, are at 2.54. Stability

The column that leans on its neighbours

A column with a pinned base and a pinned top has no lateral stiffness at all and cannot stand up alone, and yet thousands of them do. What holds them is the rest of the storey, and what it costs is paid by whichever columns do have stiffness — whose effective length rises as the square root of the load being leaned on them, straight off the end of every chart.

7 figures · Storey buckling A transverse load with nothing applied. Web slenderness against web thickness, with the limit the flange's own curvature sets. A flange carrying 6213 kN and curved to a radius of 592 m needs 10.5 N per millimetre of radial force to stay on its curve, and the only thing available to supply it is the web. Nothing has been applied to the girder: the load comes from the deflected shape, which is why a straight beam has none of it and a beam at a plastic hinge has a great deal. Setting the radial force against the web's own plate-buckling resistance gives, in four lines, h_w/t_w ≤ k·(E/f_yf)·√(A_w/A_fc) — the form the codes use, arrived at without them. The constants differ: an elastic flange strain gives k = 1.34 and the rule uses 0.3, a factor of 4.5, and the gap is the curvature assumed. k goes as the inverse square root of the flange strain, so 0.3 is a flange strained to 3.4% — which is what a plastic hinge does to it. The rule is not conservative; it is written about a different beam. Stability

The web that is crushed from inside

A plate girder's compression flange is curved by the beam's own deflection, and a curved force needs a transverse load to stay on its curve. The only thing available to supply it is the web. So a deep girder can buckle its web vertically with nothing applied to it at all, and the rule that prevents it is the only clause in the codes about a load no load case contains.

7 figures · Flange induced A bearing capacity is a mechanism, and here it is. Prandtl's collapse mechanism under a 3.0 m footing in a soil of 32° friction. A rigid wedge is driven down with the footing at 61° to the horizontal; a fan of radial shear turns the stress through exactly ninety degrees on a logarithmic spiral whose growth rate is tanφ; and a passive wedge at 29° has to be pushed up and out of the way. Nothing here is empirical — every angle is a function of φ alone — and the mechanism reaches 15.9 m from the centre, which is 10.6 times the footing's half width. That is why two footings closer together than about four widths do not have separate bearing capacities. Equilibrium

The ground is a mechanism

Bearing capacity is met as a formula with three terms and a table of coefficients, and that presentation hides what it is. Underneath is a plastic collapse mechanism — a rigid wedge, a fan of radial shear on a logarithmic spiral, and a passive wedge that has to be pushed up and out of the way — and every coefficient in the table is a property of that one drawing.

9 figures · Bearing capacity The check that everything adds up, and the error it cannot see. Four versions of the same 3-bay, 4-storey frame, with the global equilibrium residual each one produces — the sum of the reactions against the sum of the applied loads, as a fraction of the applied total. It is the first thing every analysis prints and it is worth having: a lost restraint and a load entered in the wrong unit both show up immediately, at 8% and 32%, because both change what the structure is carrying. The fourth bar is the point. A member whose stiffness is wrong by a factor of ten redistributes the internal forces completely — the second bar shows the change in the member forces, 24% — and the global residual is exactly zero, because the wrong answer is still in equilibrium with the same loads. Equilibrium is one equation per degree of freedom of the whole body, and a stiffness error lives entirely in the many equations underneath it. A model can satisfy every equilibrium check ever devised and be a model of a different structure. Equilibrium

The check that cannot see the error

Every analysis prints a global equilibrium residual, and it is the first thing anybody looks at. It catches a lost restraint and a load entered in the wrong unit immediately. It is structurally incapable of catching a member whose stiffness is wrong by a factor of ten, because the wrong answer is still in equilibrium with the same loads.

8 figures · Equilibrium check 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.

8 figures · Movement budget 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.

7 figures · Measured stiffness 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.

8 figures · Duration of load 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.

7 figures · Heat-affected zone 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.

7 figures · Bearing friction