Structural Engineering & Statics
where the load goes
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Which failure arrives first — page 8

Essays 169 to 192 of 495 on this thread, in the same order.
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.

 The circle is searched for, and the first guess is 39 per cent optimistic. The same slope with 81 trial circles evaluated, each one through the toe and each one giving its own factor of safety. There is no equation whose solution is the answer: the slip surface is a shape the ground chooses, so the calculation is a search over shapes and the answer is the smallest number found — 1.191 against 1.650 for the circle a first guess puts through the toe from above the middle of the slope, which is 39 per cent optimistic. A slope analysis that reports one circle has reported nothing. Equilibrium

The surface that has to be searched for

Every other check in this collection is made at a section somebody drew. A slope has no section — the failure surface is a shape the ground chooses, so the calculation is a search over shapes, and the answer is the smallest number found rather than the solution of anything.

 Two answers added, and the answer to the two together, drawn on top of each other. A 8 m beam under a 60 kN point load at mid-span (152.38 mm), under 12 kN/m of uniform load (152.38 mm), and under both at once (304.76 mm). The sum of the first two is 304.76 mm, and the residual between it and the third is zero — not small, zero, to the last bit of the arithmetic. That exactness is not a numerical accident: the governing equation is linear in the load, so the response is a linear operator applied to it, and a linear operator distributes over addition by definition. Every calculation that adds one load case to another is standing on that one line. Deflection

The addition everything else rests on

Influence lines add, the unit-load method adds, moment distribution adds, load combinations add, and a stiffness matrix is linear by construction. All of it stands on one sentence with three hypotheses in it — and when they fail, two of the failures point in opposite directions.

 The deformation with no limit against it. A 8 m open section carrying 12 kN/m at an eccentricity of 75 mm from its shear centre. The torque is small — 900 Nmm per mm — and the twist is not: 2.55° at mid-span with the ends restrained against warping, against 9.09° if they are not, a factor of 3.57. What that angle does is move the flange tip sideways by 8.9 mm — 62 per cent of the member's own vertical deflection, and 28 per cent of the span/250 that vertical deflection is checked against. The same load on a closed section of the same depth moves it 0.15 mm, 60 times less. No code gives a limit for this quantity, so it is the one movement in the collection that is computed only after somebody has complained about it. Deflection

The movement with no limit against it

Every code in the world gives a deflection limit. None gives a twist limit — and a beam loaded off its shear centre twists. On an open section a modest eccentricity moves the flange tip sideways by three fifths of the sag that does get checked, and nothing anywhere says whether that is acceptable.

 Steel and concrete happen to match, and nothing else on the list does. The mismatch strain a 40 degree change produces in seven pairs of materials that engineering bonds together, which is the difference of their coefficients of expansion times the temperature. Steel against concrete is 80 microstrain — 17 per cent of the larger coefficient, and by far the smallest on the list. It puts 0.223 N/mm² of tension into the concrete, 7.7 per cent of its tensile strength and 1.9 per cent of the 12 N/mm² a fully restrained member would have carried. Reinforced concrete works because of a coincidence in the third significant figure of two numbers nobody chose, and the same bar in aluminium would put in two and a third times as much. Materials

The coincidence reinforced concrete stands on

Steel expands at twelve microstrain per degree and concrete at ten. Nobody chose either number, they are not equal, and the seventeen per cent between them is the smallest mismatch of any pair of materials engineering bonds together — which is the reason the most-used structural material on earth does not tear itself apart every summer.

 The ductility number depends on the ruler. Elongation after fracture against the gauge length it was measured over, in units of √S₀. A tensile specimen extends uniformly until the ultimate load and then localises into a neck, so the total is a strain (16 per cent here) plus a length (7.2 mm), and dividing a length by the gauge length is what makes the curve fall. At the two standard gauges the same steel reports 27.5 per cent over 5.65√S₀ and 21.8 over 11.3√S₀ — a ratio of 1.264, and 42 per cent of the first number is a property of the specimen rather than of the material. Materials

The ductility that depends on the ruler

Percentage elongation after fracture is the most quoted ductility measure in the subject and one of the least well defined. A specimen stretches uniformly until the ultimate load and then localises, so the number is a strain plus a length — and dividing a length by the gauge length makes the answer a property of the specimen.

 The load path has a kink in it, and the kink is a plate thick. Two 10 mm plates lapped over 60 mm and pulled with 60 kN. The two load paths are offset by the thickness of a plate, so the joint carries a moment nobody applied: P × 10.0 mm / 2. Taken at face value that gives a peak stress 4.00 times the mean. The joint rotates under load and the moment falls to 86 per cent of it, leaving 3.57 times — a saving of 11 per cent and not, on a plate this thick, a rescue. The bolt is bent as well as sheared: 382 N/mm² of bending against 191 of shear. Connections

The joint that is crooked by construction

Lap two plates and fasten them and the two load paths are offset by the thickness of a plate. The joint carries a moment nobody applied, the peak stress is four times the mean, and the rotation that is supposed to straighten it out saves eleven per cent — because the rescue works for thin sheet with a long lap and a bolted structural joint is neither.

 The section that is checked is not the section that was chosen. A 457 mm beam coped 50 mm deep over 120 mm to frame into a girder. What is left is a tee with a section modulus of 3.836e+5 mm³ against the whole section's 1.438e+6 — 27 per cent. The moment at the end of the cope is the reaction on a lever arm of 130 mm: 23.4 kNm, giving 61 N/mm² and a flexural utilisation of 0.17. The web now has a free edge along the cope, so its buckling coefficient collapses from 4 to 0.425 — a factor of 9.4 — and the re-entrant corner has a stress concentration of 5.5 on a 10 mm radius. Connections

The section that is checked is not the one chosen

A beam framing into a girder has its top flange cut away so the two can sit at the same level. What is left is a tee with a quarter of the section modulus, a web with a free edge, and a re-entrant corner — and the beam was selected on a table entry that describes none of it.

 Buildings that sway alike need almost no gap between them. The separation two adjacent buildings need, against the ratio of their periods. The obvious answer is the sum of what each can do — 340 mm — and it is wrong, because the two peaks do not occur at the same instant. The right combination is the one modal responses use, √(u₁² + u₂² − 2ρu₁u₂), with ρ the cross-correlation coefficient of the two responses. ρ depends on the period ratio and behaves the opposite way to intuition: at a ratio of one the two buildings sway together, ρ = 1, and the gap collapses to the difference of the two, 100 mm. At the 0.57 drawn ρ is 0.029 and the gap is 248 mm — 27 per cent less than the sum, and 99 per cent of the square root of the sum of squares. Dynamics

The gap between two buildings

Two towers side by side in an earthquake need a gap. The obvious answer is the sum of what each can move, and it is wrong — because the two peaks do not happen at the same instant. What decides the answer is the ratio of the two periods, and buildings that sway alike need almost no gap at all.

 Steel has a third direction and it is not as good. Through-thickness strain demand against weld size, on a 30 mm plate, with the ductility the plate can supply in each of its three directions drawn across it. Rolling stretches the plate's inclusions into flat stringers, so a bar cut along the rolling direction, one cut across it and one cut THROUGH it are three different specimens of one steel — 60, 45 and 15 per cent reduction of area, which converts exactly to a true fracture strain of ln(1/(1 − Z)): 0.916, 0.598 and 0.163. A factor of four in the reported percentage is 5.6 in the strain the material can take. The demand goes as the deposited area over the square of the thickness, so doubling the weld size quadruples it: the 12 mm throat drawn asks for 5.4 per cent, which an ordinary plate supplies and a plate with a bad inclusion cluster does not. Connections

The direction a plate was never tested in

A rolled plate is not one material. Rolling stretches its inclusions into flat stringers lying in the plane, so a bar cut along it, one cut across it and one cut through it are three different specimens of one steel — and every mill certificate reports the first.

 One plate, three systems, one stress. The stress at the rib-to-deck weld of an orthotropic deck, split by which system produced it. The main girder contributes 120 N/mm², the crossbeam 34 and the trough 13, and they add to 167 because the deck plate is the top flange of all three at once. A calculation that treats the crossbeam as a support rather than a structure understates the total by 26 per cent. The stress RANGE at that weld is 47 N/mm², which is the number the whole deck is designed by, because the weld is a low fatigue category running the entire length of the bridge. Sections and stress

One plate and three structures

An orthotropic deck is a single steel plate stiffened by troughs, sitting on crossbeams, sitting on main girders. Nothing about that is unusual until it is noticed that the plate is the top flange of all three, so a wheel standing on it loads every one of them at once.

 It is the square of the diagram that destabilises. Four moment diagrams normalised to the same peak, and the buckling factor each one earns. Eliminating the lateral displacement from the coupled buckling equations leaves one functional in the twist, and its destabilising side is ∫M(z)²φ²/EI_z — the SQUARE of the moment, weighted by where the beam wants to twist. A diagram with a peak over a short length has a much smaller weighted square than a flat one of the same maximum, so it buckles at a higher peak: uniform 1.00, uniformly distributed load 1.13, central point load 1.36, cantilever 1.71. The root-mean-square of each diagram, printed beside it, very nearly predicts the order — which is as close to an intuition for C₁ as the subject has. Stability

The shape of the diagram, and not its peak

A beam's lateral-torsional capacity is quoted against uniform moment, which is the one case a beam carrying a load never has. Change the shape of the moment diagram without changing its peak and the buckling moment moves by a factor of nearly three.

 The best design is where two failures arrive together. A fixed area of steel rolled into tubes of every proportion, with the three things that can end each one. Euler's load goes as r² because I = A r²/2; the local buckling stress goes as 1/r² because the wall thins as the tube grows; squashing does not care. The capacity is the lowest of the three, so it has a maximum — and the maximum is exactly where the two buckling curves cross, at r/t = 129 and 2364 kN, which the closed form r*, the fourth root of αAL² over π³β√3, reproduces to 0.52 per cent. That is the general result and it is not about tubes: the optimum of a minimum of a rising and a falling curve is always their intersection, so optimising a design against two failure modes puts both of them at the design point — which is the one configuration imperfections hurt most. Stability

The best design is the most sensitive one

Take a fixed area of steel and roll it into a tube. Euler's load rises with the radius and local buckling falls with it, so the capacity has a maximum — and the maximum is exactly where the two failure modes arrive together, which is the one configuration imperfections hurt most.

 The confinement that slenderness switches off. The capacity of a concrete-filled tube against slenderness, divided by the plain sum of its two materials. Below about λ̄ = 0.5 the concrete is confined and the section is worth more than its parts — up to 29 per cent for a stub. Above it the bonus is gone, because confinement needs the concrete to dilate, dilation needs strain, and a slender column buckles before it gets there. The column drawn is at λ̄ = 0.42 and has 0.3 per cent of a bonus, which is to say none. What does not switch off is the other half: at d/t = 80 an empty tube buckles locally at 286 N/mm², below its own yield of 355, and the filled one reaches 508 because the wall cannot go inward. That is worth more than the confinement ever was, and it applies at every slenderness. Materials

Each one stops the other failing

A concrete cylinder crushes by splitting outward and a thin steel tube fails by rippling inward. Put one inside the other and each material's failure mode requires a movement the other one prevents, which is a much stronger statement than composite action.

 Where the demand crosses the capacity. Base shear against roof displacement for a eight-storey frame pushed with a triangular load pattern. It is elastic to 1994 kN, where the first storey reaches its shear capacity, and flattens as each of the others follows — 7 yield events in the order 3, 4, 2, 1, 5, 6, 7. The demand is a displacement rather than a force: the elastic spectrum at the building's own period of 0.93 s gives 173 mm, and the participation factor of 1.26 makes that 219 mm at roof level. Against an idealised yield of 70 mm that is a ductility demand of 3.10. Dynamics

Pushed over until it will not stand

A structure in an earthquake is asked for a displacement rather than a force. A pushover answers a different question cheaply — how much base shear the frame has against how far its roof moves — and the whole art is in what happens when the two are laid over each other.

 The loop a brace has when it cannot buckle. Force against axial deformation for two braces with the same core area, cycled six times at a storey drift of 2 per cent. An ordinary brace yields at 900 kN in tension and buckles at 482 in compression — 54 per cent of it — and the buckled shape leaves a plastic hinge that does not straighten, so the compression side loses capacity every cycle and is at 12 per cent of its first value by the last. A restrained brace has a casing that carries no axial force at all and only holds the core straight, which decouples axial capacity from flexural stiffness — the coupling that makes a strut weaker than a tie — so it yields at the same force both ways and hardens instead. The energy dissipated is 2.07 times as much over the six cycles, and the casing has to satisfy one inequality: π²EI/L² above the fully hardened core force, 2.56 here, which is a buckling check on a member carrying nothing. Dynamics

The brace that yields both ways

An ordinary diagonal yields in tension at its full strength and buckles in compression at half of it, and the buckle leaves a hinge that does not straighten. Stop it buckling with a sleeve that carries no load at all and the loop becomes symmetric.

 Weight is the only thing holding it down. A body 1.6 m wide and 4.5 m tall weighing 22 kN, under a wind pressure of 1 kN/m². The wind delivers 2 kN and an overturning moment of 4 kNm about the leeward toe; the weight restores 18 kNm, a factor of 4.35. The resultant lands 0.18 m from the centre against a middle third of ±0.27 m, so the base is still wholly in bearing. Equilibrium

Whether it tips or slides

A free body pushed sideways has two ways of leaving, and which one it takes is decided before any load is known. The condition is a width divided by a height set against a coefficient of friction, and the weight, the wind pressure and the depth of the body all cancel out of it.

 A preloaded joint, before and after it slips. Eight preloaded bolts at 100 kN each, on two friction faces at μ = 0.35. The joint carries 560 kN by friction with the bolts in tension and not in shear at all; past that it slips into bearing and carries 900 kN with the bolts now in shear. Two different mechanisms, one joint. Equilibrium

The force that is capped on purpose

Everywhere else in this collection friction is a nuisance whose value nobody controls, checked with a coefficient known to one figure. In a friction damper the inequality is the design intent — the device is specified so that a member behind it can never be asked for more than a stated force.

 A Pratt truss of 8 panels. A Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 14 members came out in tension, 13 in compression and 2 carrying nothing. Structural form

Halving the panel buys a shorter strut

Subdividing a truss into more panels of the same span and depth barely changes the chord forces, because the couple that carries the moment has not moved. What it changes is the length of every compression member, and a buckling capacity goes as the inverse square of a length.

 Influence line for the shear force at x = 10.5. The shear force at one fixed station, plotted against the position of a unit load walking across the span. The beam was re-solved at 301 load positions. The worst position is x = 10.56, giving 0.560. Structural form

Two diagonals, one of which is absent

A truss diagonal is sized for the shear in its panel, and near mid-span that shear changes sign depending on where the load stands. A member that can only pull cannot carry the reversed case, so the panel gets a second diagonal — and at any instant one of the pair is not there.

 A truss drawn inside a solid, and solved as one. A deep member 5000 mm between bearings and 2500 mm deep, carrying 2400 kN at mid-span. The model is two struts and one tie, on a lever arm of 2000 mm, and it is solved by the truss solver rather than by a formula: the tie comes back at 1500 kN and each strut at 1921 kN, at 38.7° to the horizontal. Spread over a strut width of 1031 mm the compression is 3.7 N/mm² against a limit of 18.1 for concrete cracked across its own strut, and the tie needs 3448 mm² of steel. A beam calculation on the same member would have asked the tie for 1404 kN, which is 7% less than the model does. Internal forces

The node is the part that is checked

A strut-and-tie model is safe if every part of it can carry what the model asks. The struts are wide and lightly stressed, the ties are steel and easy, and the whole of the difficulty collects at the points where they meet — in a volume of concrete a few hundred millimetres across that no drawing shows.

 Two diagrams for one load, and the second one has no straight-beam ancestor. Bending moment and torsion round a 90° arc of radius 6 m under a uniform load, built in at one end. The bending peaks at 576 and the torsion at 329, 57% of it. Both are zero at the free end and largest at the support, which is where a curved cantilever's bearing has to hold a torque it was probably not asked for. Internal forces

The torque that has nowhere to go

A curved beam on two supports splits its torsion between them, and the two halves cancel at mid-span. A curved cantilever has one end, so every increment of torque accumulates toward it — and the largest action at the root of a curved balcony is one that a straight beam does not have at all.

 More steel across the crack, until the roughness runs out. Shear resistance of the interface against the reinforcement crossing it. The steel clamps rather than carries, so the resistance is the clamping stress times the interlock coefficient and rises in a straight line — until the asperities crush at 5.50 N/mm², which happens at a reinforcement ratio of 0.79%. Past that the line is flat and every further bar is decoration. The dashed line is what the clamping alone would give if the concrete were unbreakable. Internal forces

Two models of one bracket

A corbel can be designed as a plane that has to be clamped or as a truss that has to be drawn, and the two are not approximations of each other. They describe different failures, they ask for steel in different places, and the honest answer is that both are checked because neither bounds the other.

 The same load, two diagrams, both in equilibrium. One span of a pair of 9 m spans under 30 kN/m, drawn twice. The elastic solution puts 304 kNm over the support and 171 in the span. Reducing the support moment by 30% and taking what statics then gives leaves 213 and 207: the section the beam needs falls from 304 kNm to 213, a saving of 30%. Both curves are in equilibrium with the same load — the mid-span ordinate plus half the support moment is the free moment 304 kNm for either — and the second is legitimate for that reason alone. What it costs is 13.0 milliradians of rotation at the support, which the section has to be able to deliver. Internal forces

The moment that was shed has to land

Redistribution takes a moment off a beam's support and pays for it with rotation. On a beam that is the whole story. In a frame the support is a column, the shed moment does not vanish, and it arrives at a member whose section was chosen from the diagram it has just left.

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