Structural Engineering & Statics
where the load goes
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The load must go somewhere — page 8

Essays 169 to 192 of 450 on this thread, in the same order.
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.

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

The steel decides how many, not how much

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

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

The crack that never reached forty-five degrees

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

 The web down the middle carries no torsion at all. A 2-cell box 3.0 m wide and 1.5 m deep under 4000 kNm of torque. Statics gives one equation, T = 2 Σ q_i A_i, and there are 2 unknown flows — so the section is torsionally redundant and the missing statements are that every cell twists by the same amount. Solving that system gives 444.4 and 444.4 N/mm in the cells, so the internal web carries 0.00 N/mm — 0.00 per cent of the outer wall's. J is 2.3679e+12 mm⁴ against 2.3679e+12 for the same outline with no internal web at all, a ratio of 1.000000. The same walls slit open would give 7.413e+10, so closure is worth 32 times and the internal web is worth what the ratio says. Sections and stress

Two cells, one equation, and a web with nothing in it

Bredt's formula answers a single closed cell because a single closed cell has one unknown and one equation. Put a web down the middle and there are two unknowns and still one equation — and the answer, when the missing statement is supplied, is that the new web carries exactly nothing.

 A tie is worth all of itself and a strut is not, which is worth nine per cent. The same 20 floors carried two ways, with every member drawn at the width its own force requires. Hung, the loads accumulate upward, so the largest hanger is at the top: 18.0 MN at 355 N/mm² with no buckling reduction of any kind. On columns they accumulate downward and the largest column is at the bottom, at the same force — but every column above it is understressed by its own slenderness, worst at the top where a 900 kN column still has to be 4.8 × 10³ mm² to reach χ = 0.529. Over the height the hangers total 0.002 m³ of steel against 0.002: a saving of 9.0 per cent, which is the average χ and nothing else. Structural form

Hung from the top, and nine per cent lighter

A tie is worth its full strength and a strut is not, so hanging the floors of a building from a hat truss ought to be an obvious economy. It is a real one, it is measurable, and it is nine per cent of the steel — shrinking as the building gets taller, which is the opposite of what the argument sounds like.

 The abutment force is the sag turned upside down. A 100 m ribbon carrying 35 kN/m at a sag of 2.0 per cent of its span. H = wL²/8f, so the horizontal force at each abutment is 21875 kN — 6.25 times the entire weight of the deck, and five times what a suspension bridge of the same span and weight at a tenth would have needed. The curve is a reciprocal and it has no flat part: halving the sag doubles the force, at any sag. What stops a designer flattening it further is not the ribbon, which is in tension and cannot buckle. It is what the ground at each end will take, and at 6.25 deck-weights that is usually rock or a very large anchor block. Structural form

The deck that is its own cable

Every other cable structure hangs something from the cable. A stressed ribbon hangs nothing — the walking surface is the catenary, laid at a fiftieth of the span rather than a tenth, because a footbridge has to be walkable. That one decision hands the abutments six and a quarter times the entire weight of the bridge.

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

The rib that is a boundary condition

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

 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.

 Every pressure points at the pin, so the water lifts nothing. A radial gate of radius 8 m holding 6 m of water, with its pivot 6 m above the sill. The pressure on a curved surface cannot be obtained by multiplying anything by anything, so it is integrated round the arc: the horizontal component comes to 176.6 kN/m and the vertical to 110.5. Both are recoverable without any integral at all — the horizontal is the pressure force on the surface's own vertical projection, γH²/2 = 176.6, and the vertical is the weight of the water standing above it, 110.5. They agree to 0.000 per cent. And because every pressure is normal to a circle, every one of them passes through the centre: the moment of the whole 208 kN/m about the pivot is -3.4e-15 kNm, against 353 for a flat gate on the same hinge. Equilibrium

Every pressure points at the pin

Pressure acts normal to a surface, so on a curved one every element pushes in a different direction and no multiplication gives the resultant. Two free bodies recover it without an integral — and on a circular surface a third observation makes the whole force disappear from the equation a hoist has to satisfy.

 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 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.

 Independence, and the floor it never goes below. The reduction factor a column is allowed, two ways. The Eurocode storey rule falls to 0.70 and stops. The independence argument — n bays each with a mean and a standard deviation, summed — gives (1 + zv/√n)/(1 + zv), which falls faster and stops at 0.503: a floor with no n in it at all, decided only by how variable the load is and how far out the fractile is drawn. The mean is never reduced away, because every bay really does carry its mean. At ten storeys the two differ by 10.0 points. Equilibrium

The load that is never all there at once

A column at the bottom of twenty storeys is designed for the imposed load of twenty floors added up, and twenty floors do not reach their own worst day together. The reduction that follows is not a discount on the safety margin. It is the central limit theorem, and it has a floor it never goes below.

 The moment a joint makes out of geometry. Bending stress in the chord divided by the axial stress already in it, against how far the diagonal's working line misses the node. A truss analysed as pin-jointed is drawn with its members as lines through their own centroids meeting at a point, because that is what makes the joint a pin with nothing but forces in it. A fabricated joint is not obliged to oblige, and the resultant then has a moment about the node — 16 kNm at the 60 mm drawn. It divides between the members meeting there in proportion to their rotational stiffnesses, 4EI/L or 3EI/L, which is the moment-distribution rule applied to an unbalanced moment that came from geometry rather than from load. The chord takes 48 per cent of it and carries 25 per cent of its axial stress again in bending. Ten per cent arrives at 28 mm, which on a chord of any depth is a detailing decision rather than a mistake. Connections

The joint that is not where it was drawn

A truss is drawn as lines through the centroids of its members meeting at points. A fabricated joint is under no obligation to oblige, and when the working lines miss each other the resultant has a moment about the node that has to go somewhere.

 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 pile has no length until the ground gives it one. Deflection, bending moment and soil reaction down a 0.6 m pile carrying 150 kN at a free head, in ground whose modulus grows by 0.005 N/mm³ per millimetre of depth. The one length in the problem is T, the fifth root of EI over n_h, which is 1.89 m here; the head moves 20.5 mm, the worst moment of 219 kNm is at 2.50 m — 1.32 T — and below about four T nothing happens at all. The classical coefficients come out of the finite differences rather than a table: 2.430 against Matlock and Reese's 2.435, and 0.772 against their 0.772. Internal forces

A pile has no length until the ground gives it one

Almost every other structural member is handed a length by the drawing. A pile goes into the ground until it stops, and what decides how much of it is working is a fifth root of the ratio between its own stiffness and the soil's.

 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.

 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.

 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 summer that is worse than the one before it. The earth pressure behind an integral abutment, summer by summer, as a multiple of the at-rest value it started at. A 60 m deck expands by 10.8 mm at each end and pushes the abutment into the backfill. Granular soil under cyclic strain densifies, so the same movement next year needs a higher pressure to achieve, and K climbs from 0.38 toward 0.96 — a factor of 2.49 on the force, reached after about a century. The design load on an integral abutment describes the bridge's whole life rather than a load case, and it is the only load in this collection that gets larger because time has passed rather than because something was added. Structural form

The summer that is worse than the last

An expansion joint is a hole in a deck that leaks salt water onto the bearings underneath it. Remove it and the thermal movement does not go away — it goes into the soil behind the abutment, twice a day for a hundred and twenty years, and granular soil under cyclic strain gets denser.

 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.

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