Structural Engineering & Statics
where the load goes
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Geometry beats material — page 7

Essays 145 to 168 of 329 on this thread, in the same order.
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.

 A weld is a force, and it is applied where the weld is. The bow a welded girder leaves the shop with, against how far its welds sit from the section's centroid. A weld cannot contract while the plate holds it, so it yields in tension and what is left when everything is cold is a locked-in force at about the yield stress: 312 kN for the 1.2 kJ/mm of heat drawn, over a shrinkage zone of 439 mm². Applied 210 mm off the centroid that is a moment, and a moment applied along a member is a curvature: the 12 m girder comes out bowed 12.5 mm, which is L/962 against a fabrication tolerance of L/1000. It also comes out 1.2 mm shorter. Welding symmetrically about the centroid puts the resultant on the neutral axis and the bow becomes 0.00 mm — the same heat, the same force, and no moment at all. Connections

The shape that came out of the shop

A weld cools by seven hundred degrees while the plate holds it, so it yields in tension and stays that way. What is left is a locked-in force of three hundred kilonewtons applied where the weld is, and if that is not on the centroid the member leaves the shop bent.

 A skew deck spans square, and the corner knows it. Plan of a 30-degree skew slab, 12 m along the road by 10 m wide, with the reaction per unit length of abutment drawn as a bar at each support point. The load takes the shortest route between the abutments, which is the square span of 10.4 m rather than the 12 m of carriageway — so the reaction runs to the two OBTUSE corners, where the abutments are closest, and drains away from the acute ones. Peak 2.15 times the average, least 0.16. Nothing about the loading is uneven; the geometry is. Internal forces

The deck that spans square

A slab bridge crossing a road at an angle is loaded uniformly and does not carry uniformly. Load takes the shortest route between the abutments, which is not the direction the carriageway runs, and the reaction piles up in two corners.

 Where a bar may stop, and how far past there it goes anyway. The tension the bottom steel must carry along a 9 m beam, and the resistance of the bars actually present, drawn as a staircase. The demand is the moment diagram divided by the lever arm and then SHIFTED 270 mm toward midspan, because the truss inside the beam delivers its shear diagonally — the two constructions agree to 0.36 per cent, which is the second-order term and nothing else. Each curtailed layer then runs a further 1150 mm to develop, so the outer layer stops at 245 mm rather than the 1665 mm the moment diagram allows. The tail is 1420 mm at each end — 16 per cent of the span — and it is what turns a 20 per cent saving into 2.2. Sections and stress

Where a bar may stop

The moment diagram falls away from midspan, so the steel midspan needs is not needed everywhere, and curtailing it saves real money. Then two things get in the way, and between them they take nine tenths of what the moment diagram promised.

 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.

 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.

 Cross the hangers and the chords stop bending. The same tied arch, the same sixteen hangers, the same load on half the span — hung vertically and hung as a network. Vertical hangers make the two chords a Vierendeel frame, which has no truss action at all, so a partial load is carried by bending: 3316 kNm in the tie and 6234 in the arch. Inclined hangers can carry the shear between the chords axially, and the same load gives 686 and 831 — factors of 4.8 and 7.5. The thrust is identical in both, because that is decided by the span and the rise and nothing else. Structural form

Cross the hangers and the bending goes

A tied arch with vertical hangers is a Vierendeel frame with a curved top chord — it has no truss action at all, so a load on half the span is carried by bending. Incline the hangers so they cross and the same two chords become a truss.

 Every level added is a longer span, not a shorter one. Steel per square metre of floor against the number of levels in the hierarchy, for a 12 m bay with a deck that can span 3.0 m. A bending level's weight per unit area is 3ρqrL/8σ — it contains the SPAN and not the spacing — so breaking a floor into more levels cannot make the members lighter by making them closer together. It adds one more system, and the last system always spans the whole bay: a second level costs 52 per cent more steel than one, and a third 107 per cent. The structural zone grows with it, 730 mm to 1346 mm. Hierarchy is not an economy, it is a way of reaching, and it is paid for in both currencies at once. Structural form

Every level is a longer span

A floor is a hierarchy — deck to joists to beams to girders — and the reason usually given is that breaking a long span into short ones saves material. A bending level's weight per square metre contains its span and not its spacing, so it does not.

 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 closed force polygon. The forces on a joint, laid tip to tail. Equilibrium is the statement that the polygon closes, and the gap when it does not is the out-of-balance force, to scale. Structural form

One drawing solves the whole truss

The method of joints solves a truss one joint at a time, and each solution is thrown away as soon as the next begins. Drawn instead of computed, the joints share their edges — every member's force appears once in a single figure, and the figure's own closure is the check.

 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.

 Most of a long pile is doing nothing. Head deflection against embedded length, both measured in the pile's own characteristic length T = 14.67 m. A pile shorter than about two T is a lever with nothing holding its foot and deflects several times as much; past four T the curve is flat to within a per cent, because the ground below that depth is never asked for anything. A lateral check is a check on the top four T of a pile, however deep it goes for its axial load — and adding length to fix a lateral deflection is the one remedy that does not work. Internal forces

The pile that is too short to bend

A laterally loaded pile is long or short in its own characteristic length, and the two are different structures. The slender pile bends and the ground below four characteristic lengths never hears about it; the monopile is two lengths deep, rotates about a point, and every tabulated coefficient written for the first case is wrong for the second.

 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.

 The studs are evenly spaced and the demand is not. The force per unit length the shear connection carries along half of a 12 m composite beam, from Newmark's solution. It is largest at the support — 282 N/mm — falls to nothing at mid-span, and averages 156: the end studs are asked for 1.81 times the mean. Studs are nevertheless placed at a uniform spacing, and the justification is the one the variable-angle truss uses for its stirrups — a ductile connector sheds what it cannot carry to its neighbours, so the uniform distribution is a plastic redistribution and not a description of the elastic state. Internal forces

The connection is busiest where the beam is not

A composite beam's studs are spaced evenly along it and the demand on them is not even at all. It peaks at the supports, where the bending stress is nothing, and falls to zero at mid-span, where the section is working hardest — so the connection is designed from a diagram nobody looks at.

 The second moment of area is a function of direction. Second moment of area of an equal angle against the angle of the axis it is taken about, with the product of inertia beneath it. The maximum is 5.943 × 10⁶ mm⁴ and the minimum 1.523 × 10⁶, a ratio of 3.90, and they occur where the product of inertia passes through zero — at 45.0° from the drawn axis. The value the drawing suggests, 3.733 × 10⁶, is neither of them. Sections and stress

The axis a column buckles about

A strut buckles about the axis with the smallest second moment of area, and for a section with no axis of symmetry that axis is neither of the two on the drawing. An angle used as a strut is 2.45 times weaker than the number a designer reads off its own dimensions.

 The shear centre of a channel. A channel of 100 by 250, with the shear flow in its flanges drawn. Those flows form a couple, so the load has to be applied 38.0 outside the web to leave the section untwisted — a point in the air, outside the material entirely. Sections and stress

The eccentricity a purlin cannot avoid

A channel's shear centre is outside the material, so a load applied anywhere on the section misses it. The distance is fixed by the proportions rather than by the detailing, it is 38 mm on an ordinary purlin, and the torque it produces is not an error anybody made.

 One point, every plane through it, one circle. A point carrying 180 N/mm² across one face, 90 across the other and 40 of shear. As the plane is turned, the pair (σ, τ) runs round a circle of radius 60.2 centred at 135.0 — and it goes round at twice the rate the plane does, which is the part always misremembered and the part that makes the picture work. The principal stresses are 195.2 and 74.8, on planes 20.8° from the face the 180 acts on; the largest shear on any plane is 60.2, exactly the radius, and it sits 45° from those — which is 90° round the circle. The von Mises stress that ranks this state against any other is 170.6. Sections and stress

The circle nobody draws

A plane stress state has three principal stresses and the third is zero. When the two on the drawing share a sign, the largest shear in the state involves the one that is not there — and the circle a designer has drawn is not the circle that governs.

 A short timber beam is a shear problem, and a steel one never is. Utilisation of the bending and shear checks on one beam, against span-to-depth. The two cross where the ratio equals f_m ÷ f_v exactly — no load, no width and no span survives the cancellation — which for this timber is 6.7 and for steel is 1.73. So a timber beam shallower than about six times its depth is governed by shear parallel to the grain, and a steel beam would have to be shorter than twice its own depth before the same thing happened, which is not a beam. The third check is bearing across the grain, which does not move with the span at all: on the beam drawn it is at 0.40, and it is the one that governs. Sections and stress

The shear that decides a timber beam

A steel beam is never governed by shear, because its bending strength is only 1.73 times its shear strength and no beam is that short. Timber's ratio is 6.7 along the grain and 23 across it, so shear governs at proportions people build every day.

 A 10 mm plate, and the width it can be. The elastic critical stress of a plate in compression against its width, with the yield stress drawn across it. Below 462 mm the plate reaches yield before it buckles; above it the plate ripples first, and the fraction of the width still carrying load falls away — at 900 mm only 46 per cent of it is still working. Stability

The coefficient that is not four

A plate's buckling stress carries a coefficient that looks like a constant and is not. It is 4 for an internal element, 0.43 for an outstand and 23.9 for a panel in shear — and the width a 10 mm plate may be runs from 152 mm to 1,130 across that range.

 Warping stiffens a short member and nothing at all a long one. The stiffening 1/[1 − tanh(κ)/κ] against kL, both axes logarithmic, over kL from 0.05 to 200. At the low end the curve is a straight line of slope −2, because for small kL the bracket is κ²/3 and the stiffening is 3/kL²: it reaches 1201 at kL = 0.05, falls to 1.005 at the top, and every open section ever rolled sits somewhere on it. The same three plates arranged three ways are marked: the 533 by 190 mm I-section at kL 2.76 and ×1.562, the tee at kL 37 and ×1.027, the angle at kL 34 and ×1.030. A tee's warping constant is 288 times smaller than the I-section's and an angle's 236 times, because their plates meet at a point and there is no pair of flanges to bend against each other — so they have no warping resistance to offer at all, and that is the reason an angle is a poor thing to twist. Stability

The restraint that beats the gradient

A moment-gradient factor is worth up to 2.7 on a beam's critical moment and is tabulated everywhere. Holding the ends against warping is worth more, is achieved by a detail rather than by a load case, and appears in no table at all.

 A channel has three critical loads, not one. The three critical loads of a channel in compression, against its length, with the load it actually buckles at drawn over them. At 3500 mm the flexural loads are 13970 kN about the major axis and 2306 kN about the minor, while twisting about the shear centre takes 1555 kN. The lowest root is 1484 kN, and the column twists. The shear centre sits 109.7 mm from the centroid, so the modes cannot happen separately: the lowest root of the coupled problem is 4.5 per cent below the lowest of the three, and the Wagner coefficient β is 0.60. The governing mode changes at 5813 mm: below that length the column twists, above it, it bends — because the torsional resistance keeps a term that does not grow when the member is shortened, and the flexural loads have none. Stability

The third root of the cubic

A column has three buckling loads and an Euler calculation finds two of them. The third is a twist about the shear centre, and for a section whose shear centre is not at its centroid the three cannot happen separately — so the answer is the lowest root of a cubic and can be a third below anything the two familiar modes report.

 Every member's share of the movement, and they are not the members expected. A Pratt truss of eight panels at a depth of 1, carrying 15 kN at each top node, with the movement of the bottom chord at mid-span attributed member by member. The unit-load sum δ = ΣF·f·L/EA gives 2019.41 at EA = 1: 49.4% from six top chords, 30.8% from eight bottom chords, 16.8% from eight diagonals, 3.0% from seven verticals. The single worst member is a top chord at mid-span at 11.9% of the whole. Each member is drawn at the width of its own share. The same deflection from a stiffness solution that shares none of this arithmetic is 2019.41, a relative residual of 1.4e-14. Deflection

The member that is not worth stiffening

A truss's deflection is a sum of one term per member, and a term is zero whenever either force in its product is. A vertical carrying the whole of a panel load can contribute nothing at all to the movement — which a total can never show and a per-member sum shows nothing else.

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