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

Essays 169 to 182 of 182 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: 3.40° at mid-span with the ends restrained against warping, against 9.09° if they are not, a factor of 2.67. What that angle does is move the flange tip sideways by 11.9 mm — 82 per cent of the member's own vertical deflection, and 37 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, 80 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 further sideways than four 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.

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