Structural Engineering & Statics
where the load goes
The collection

Every essay — page 13

Essays 289 to 306 of 306, in the same order.
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.

9 figures · prestressed shear A soap film over a hole, and its volume is the torsion constant. The stress function for Saint-Venant torsion of a square of 100 mm, relaxed on a 97 by 97 grid until it stopped moving — 385 sweeps. Its contours are the lines the shear stress runs along, its slope is the magnitude of that stress, and twice its volume is the torsion constant: 1.4053e+7 mm⁴ against a closed form of 1.4058e+7, an error of -0.035 per cent. The steepest slope is 67.515 at the boundary, at the point on it nearest the centre — which is why the peak shear in a solid section is at the middle of the longest side and never at a corner, where the film comes down to zero from two directions and its slope vanishes. Sections and stress

Two volumes, and both of them are torques

Torsion of a solid section that is not a circle has no elementary answer, and for thirty years after Saint-Venant posed it the only way to get one was to blow a soap film over a hole cut in a plate and measure it. The film is not an illustration of the solution. It is the solution, and so is a heap of sand poured on the same hole.

9 figures · membrane analogy 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.

10 figures · multicell torsion 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.

9 figures · hung structure 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.

9 figures · stressed ribbon A buckling load with no compression anywhere in the member. A tube of radius 81.6 mm and wall 5 mm under pure torque — no axial load at all — buckles into a helix at T = 2πEI/L, which for the section drawn is 1877 kNm. Two things are absent from that expression and both are surprising: the length appears to the first power rather than the second, and the shear modulus does not appear at all, so the torsional stiffness of the member has nothing to do with the torque at which it buckles in torsion. The comparison is the torque that yields the same tube, 42.9 kNm — a factor of 43.8 below it. The mode arrives first only past a length of 263 m, which is πE/τ_y = 3219 radii and contains no thickness whatever. Stability

The buckling load with no compression in it

Twist a straight bar hard enough and it snaps into a helix, with no axial load on it anywhere. The load at which that happens is 2πEI/L — first power of the length, and no shear modulus in it at all, so how stiff the bar is in torsion has nothing to do with the torque that buckles it in torsion.

9 figures · torque buckling 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.

9 figures · stiffener rigidity 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.

7 figures · brazier buckling 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.

7 figures · slope stability 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.

8 figures · curved surface pressure 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. Everything on this site that adds 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.

8 figures · superposition 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.

8 figures · twist serviceability 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.

8 figures · thermal compatibility 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.

8 figures · gauge length 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.

9 figures · single lap 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.

10 figures · coped beam 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.

8 figures · pounding Evenly spaced modes, so one of them is always where the feet are. The first 6 modes of a 120 m stay under 3.50 MN. A taut string's frequencies are an arithmetic progression — every one of them 1.006 Hz above the last — where a beam's go as the square of the mode number and spread out. That difference is the whole of why a cable is a lively member and a beam is not: a beam has a first mode and then a gap, and a cable has a mode every 1.01 Hz for ever. The shaded band is ordinary walking, 1.6 to 2.4 Hz, and mode 2 sits inside it. Nothing about the tension can move a mode out of the band without moving another one in. Dynamics

The force read off a frequency

Nothing can measure the tension in a stay cable directly — there is no gauge, no accessible end and no place to put a load cell. What there is, is a member whose frequencies are an arithmetic progression whose spacing is the square root of its own tension, so a phone taped to it for thirty seconds returns the force.

8 figures · cable dynamics