Structural Engineering & Statics
where the load goes
The collection

Every essay — page 12

Essays 265 to 288 of 728, in the same order.
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.

8 figures · Detail category 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.

8 figures · Rocking The hinge is at mid-height in exactly no storey. The height of the point of contraflexure in each column of a 4-storey, 3-bay frame under lateral load, as a fraction of the storey height, against the portal method's assumption that it is at the middle. The exact solution is a plane-frame stiffness analysis of the same frame. In the bottom storey the zero sits at 0.609 of the height, because a fixed base is stiffer than the joint above it and takes more of the column's moment; in the top storey it sits at 0.359, because there is no column above to share that joint. The average over the whole frame is 0.475, which is why the assumption survives — it is right on average and wrong everywhere. The worst error in the storey shear share is 25%, and the column shears still add to the storey shear to 0e+0 of it, because the method is exact statics applied to an assumed structure. Internal forces

The analysis that assumes the answer

A rigid frame is indeterminate, so statics cannot finish it. The hand methods finish it anyway, by assuming where the bending moment is zero and treating those points as hinges. That is not a shortcut around the analysis — it is a different kind of answer, exact in equilibrium and wrong in compatibility, and knowing which half is which is what makes the error a bound rather than a mystery.

7 figures · Portal method Two of these move and the third cannot. The first span of a 3-span beam under 5 kN/m, with the stiffness of the middle span swept over a factor of 25. The support moment and the mid-span moment both move — that is what redundancy does, and it is the whole reason a continuous beam has to be analysed rather than read off. Their combination does not: the mid-span ordinate plus the average of the two end moments is 30.6 kNm at every point on this axis, which is wL²/8 for this span and this load and contains nothing else. The largest departure anywhere on the sweep is 2.3e-16 of the value, which is the arithmetic of the stiffness solution rather than a property of the beam. Continuity buys a distribution and not a capacity, and this is the line that says so. Internal forces

Two of these move and the third cannot

Cut one span of a continuous beam free and add up the forces on it. What comes out is that the mid-span moment plus the average of the two end moments equals the free bending moment of that span, with nothing else in it — no stiffness, no support settlement, no analysis at all. Continuity moves moment about. It does not reduce the total, and it never has.

7 figures · Static moment 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.

8 figures · Edge disturbance 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.

9 figures · Minimum reinforcement 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.

8 figures · Equal-area axis 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.

9 figures · Crack width 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.

7 figures · Propped excavation 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.

8 figures · Pile group The lining that carries less for being weaker. Bending moment and hoop thrust in a circular lining, against the lining's own bending stiffness, both as fractions of the free-ring values. The ground arrives already stressed — 500 kPa vertically and 300 horizontally at K₀ = 0.6 — and the difference between them tries to squash the hole into an ellipse. A lining stiff enough to refuse absolutely collects the whole distortion pressure, p₂R²/3 = 300 kNm/m; one flexible enough to go with the ground collects nothing, because there is no curvature change left to resist. The thrust is the flat line: it comes from the mean stress rather than the difference, so it does not move at all. Putting 8 joints in this ring drops the moment to 34% of the solid one and leaves the thrust exactly where it was, which is why a segmental lining is jointed and why the intuition carried over from a beam is inverted here. Structural form

The lining that is stronger for being weaker

A tunnel lining is not loaded. The ground arrives already stressed and the hole wants to squash into an ellipse; the lining's only job is to refuse, and how much moment it collects depends entirely on how hard it refuses. Make it stiffer and it takes more. Make it flexible — put joints in it, make it thin — and it takes almost none, while the hoop thrust it carries does not move at all.

9 figures · Tunnel ring 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.

7 figures · Follower force 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.

8 figures · Stepped column 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.

8 figures · Centrifugal load 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.

7 figures · Soil arching The same curve, computed twice and from opposite ends. A cantilever of 6 m, with its deflected shape drawn from a double integration of M/EI and the bending moment of its conjugate beam drawn on top of it. The conjugate is the same span carrying M/EI as a load, with its supports transformed — a real fixed end becomes a free end and a real free end becomes a fixed one, because a fixed end has no slope and no deflection and the conjugate therefore needs no shear and no moment there. The two curves agree to 4.7e-6 of the largest deflection, which is the trapezium rule and not the method. The largest deflection is 36.00 mm against the closed form's 36.00. The reaction of the conjugate beam is -9.000 milliradians, which is the real beam's rotation at that support — so the whole of a slope calculation is one reaction. Deflection

The beam whose moment is a deflection

A beam's bending moment is the second integral of its load. Its deflection is the second integral of M/EI. They are the same problem, so a deflection can be found by loading a fictitious beam with M/EI and asking a statics question — and the only thing to remember is the supports, which are not remembered but derived, one boundary condition at a time.

8 figures · Conjugate beam 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.

8 figures · Incremental deflection 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.

8 figures · Tensile strength 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.

8 figures · Transition temperature 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.

8 figures · Site response 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.

8 figures · Transient resonance 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.

8 figures · Fit-up 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.

8 figures · Downdrag 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.

8 figures · Restraint cracking