Structural Engineering & Statics
where the load goes
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Series — page 6

A field says what an essay is about. A series follows one idea essay by essay — from the question that introduces it to the one that assumes all the others.
Four guesses at one buckling mode. A pin-ended column, with four assumed shapes and the load each of them gives. The reference is a ten-term Ritz expansion solved as an eigenvalue problem, at 9.8696 EI/L² — which is π², as it must be. a half sine gives 9.870, its own sag shape gives 9.882, a mid-span sag gives 10.000, a parabola gives 12.000. Every one of them is high and none of them is low, because an assumed shape is a constraint on the column and a constraint can only stiffen it.

Stability energy

  1. 1 Guessing the shape, and getting the load anyway
  2. 2 The bound from underneath
2 essays · stability Three times the stress, and it does not matter how big the hole is. The hoop stress around a circular hole in a wide plate pulled at 100 N/mm², from Kirsch's exact solution. At the sides of the hole it is 3.0 times the applied stress — 300 N/mm² — and the factor is the same for a hole of any radius, because the radius cancels. At the top and bottom of the hole it is -1.0 times the applied stress, which is compression in a plate that nothing is pushing. The disturbance dies quickly: the stress is within 5% of the applied value by 3.5 hole radii, which is Saint-Venant's principle with a number on it.

Stress concentration

  1. 1 The hole that multiplies the stress by three
  2. 2 The notch a crack does not feel in full
2 essays · materials 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.

Stressed ribbon

  1. 1 The deck that is its own cable
  2. 2 The pier that carries only the difference
2 essays · structures A truss drawn inside a solid, and solved as one. A deep member 4000 mm between bearings and 2000 mm deep, carrying 1200 kN at mid-span. The model is two struts and one tie, on a lever arm of 1600 mm, and it is solved by the truss solver rather than by a formula: the tie comes back at 750 kN and each strut at 960 kN, at 38.7° to the horizontal. Spread over a strut width of 812 mm the compression is 3.0 N/mm² against a limit of 15.8 for concrete cracked across its own strut, and the tie needs 1724 mm² of steel. A beam calculation on the same member would have asked the tie for 702 kN, which is 7% less than the model does.

Strut-and-tie

  1. 1 When there is no section to design
  2. 2 The node is the part that is checked
2 essays · internal-forces Two curves climbing together, and the one that catches up first. A 6 m member tapering from 200 to 600 mm, with the moment it carries and the moment it can carry drawn on the same scale below it. The demand rises linearly and the capacity as the square of the depth, so the gap between them closes and then opens again. It is narrowest at 3.00 m from the free end, where the member is 400 mm deep and 79% used, against 70% at the root where the moment is largest.

Tapered member

  1. 1 The section that changes along the span
  2. 2 The section that governs is inside the haunch
2 essays · sections A buckled panel is a truss that nobody drew. A 1000 × 1000 panel of 6 mm web, at d/t = 167. It buckles in shear at 63.8 N/mm², which is 383 kN — and it then carries 696 kN, 1.82 times as much, because the tension diagonal takes over from the compression one that has gone. The band runs at 22.5° with a membrane stress of 252 N/mm² over a width of 541 mm, and it pulls on the flange at 221.3 N per millimetre of its length. A web that never buckled at all would have reached 953 kN, so the panel ends at 73% of a stocky web's capacity on a fraction of its steel.

Tension field

  1. 1 The panel that carries more after it has failed
  2. 2 The tension has to pull on something
2 essays · stability One slit, and the torsional stiffness falls by a factor of hundreds. A 200 by 200 box of 8 mm wall, drawn closed and then slit along its length. Closed, the torque runs round the wall as a shear flow and the torsion constant is 5.66×10⁷ mm⁴; slit, the loop is broken and only each wall's own thickness resists, giving 1.31×10⁵ mm⁴. The ratio is 432 to one, so the same torque twists the slit section 432 times as far and raises a peak shear stress 36 times as high. Nothing about the material changed.

Torsional constant

  1. 1 The slit that costs a factor of six hundred
  2. 2 The corner the rolling mill rounds
2 essays · sections Where a beam's load comes from. A 8 × 6 m panel carrying 5 kN/m², divided at 45° from the corners. The long beams take a trapezoid of 15.0 m² each and the short beams a triangle of 9.0 m²; the four areas sum to 48.0 m², which is the panel, so no load has been invented or lost. The line loads quoted are the uniform equivalents; the real distributions peak at 15.00 kN/m at midspan.

Tributary area

  1. 1 The load a beam is given is a decision
  2. 2 The column given more than its rectangle
2 essays · equilibrium 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.

Twist serviceability

  1. 1 The movement with no limit against it
  2. 2 A deck is a spring, not a wall
2 essays · deflection The shear goes round the corner instead of across it. A 6-panel Vierendeel girder, 12 m by 1500 mm, under 100 kN at mid-span. There is no diagonal in it, so each panel's 50 kN of shear is carried as bending in the chords: the curves drawn along them are the chord moments, and every one passes through zero at the middle of its own panel. The local moment is the panel shear times the panel length over four, 25.0 kNm, and it adds to an axial force of 200 kN from the global moment at the same point. The girder deflects 6.20 mm against 3.18 mm for the same members triangulated — 1.95 times — and 68% of that movement is chord bending that a diagonal would have removed entirely.

Vierendeel

  1. 1 The truss with no diagonals
  2. 2 The frame is a girder stood on end
2 essays · structures The deflection at x = 4, by virtual work. Three diagrams: the moment from the real load, the moment from a unit load placed where the answer is wanted, and their product. The area under the third, divided by EI, is the deflection — 213.33 here. No standard case was consulted, so the method works for any load pattern at all.

Virtual work

  1. 1 One deflection, without solving everything
  2. 2 Any structure will carry the unit load
2 essays · deflection Lock-in: the frequency the wind sheds at, and what it does to the chimney. A 1.2 m cylinder at 0.9 Hz. Vortices leave it at St·U/D, so the critical speed at which they match the structure is 6 m/s — a breeze, met many times a year rather than once in fifty. The upper panel shows the shedding frequency locking on to the structure across a band from 5.1 to 7.8 m/s; the lower shows the amplitude that results. At a Scruton number of 11.2 the peak amplitude is 180.94 mm, which is 15% of the diameter.

Vortex shedding

  1. 1 The wind that brings its own frequency
  2. 2 The breeze that uses up a chimney
2 essays · dynamics A hole in a web is a Vierendeel panel. A 400 × 300 rectangular opening in a 533 deep beam, 15% along a 9 m span carrying 20 per metre — where the moment is 103 kNm and the shear 63 kN. The moment is a couple on the two tees, 301 kN on a lever arm of 343 mm, which is 70.3 N/mm² of uniform stress. The shear has nowhere to go but through the tees, so each carries 32 kN over the opening and bends in double curvature: a Vierendeel moment of 6.3 kNm and 167.6 N/mm² on top. So 70% of the stress at the corner exists because the hole has a LENGTH, and only 30% of the section's second moment has gone.

Web opening

  1. 1 The hole that costs nothing, and everything
  2. 2 Four corners and a mechanism
2 essays · sections The damping a wind leaves behind, and the speed that uses it up. Total damping ratio against wind speed, for a structure with 0.60% of its own and a modal mass of 25 kg per metre at 1.2 Hz. The aerodynamic damping of a section whose lift falls with angle of attack is negative and grows with speed, so at 4.14 m/s what is left is nothing. Below the threshold a disturbance dies away. Above it, the structure feeds itself and the motion grows out of nothing at all — which is what makes this a different mechanism from resonance rather than a severe case of it.

Aeroelastic instability

  1. 1 The motion that feeds itself
1 essay · dynamics A base plate, and when the bolts start working. A 500 × 500 mm plate carrying 600 kN and 180 kN·m, so the resultant sits 300 mm from the centre against a kern of 83.33 mm. The plate is in bolts engaged: bearing over 150.88 mm at a peak of 20 N/mm², with the holding-down bolts carrying 154.42 kN. The plate lifts at 50 kN·m and crushes at 126 kN·m, and the bolts are not needed until 150 kN·m.

Anchor breakout

  1. 1 The failure that is in the concrete
1 essay · connections The force spreads, and the spreading needs a tie. The end block behind an anchorage of 1200 kN on a 200 mm plate, in a section 700 mm deep. Half the force enters at the quarter point of the plate and leaves at the quarter point of the section, so a strut between the two rises 125 mm and needs a transverse tie to turn it. Placing the tie 0.5 depths from the face makes that tie force 214 kN — and at exactly half a depth this reproduces Guyon's 0.25P(1 − a/h) to the digit, which makes that famous coefficient a lever arm somebody chose rather than a property of concrete. The bearing stress under the plate is 20.0 N/mm² against 5.7 once the force has spread.

Anchorage zone

  1. 1 The force that splits what it pushes on
1 essay · internal-forces Strength at an angle, and the straight line that is not it. Compressive strength against the angle between the load and the grain. Hankinson's formula — f₀f₉₀ ÷ (f₀sin²α + f₉₀cos²α) — is an interpolation rather than a failure theory, and what makes it worth having is how far it sits from the straight line anyone would otherwise draw between 21 and 2.5 N/mm². At forty-five degrees it gives 4.5 N/mm² against the line's 11.8: 38 per cent of it, and 21 per cent of the strength along the grain. The curve drops away in the first twenty degrees because the weak direction starts governing as soon as it has any component at all, which is the same arithmetic as a section's weak axis and the reason a skewed bearing detail is a real loss rather than a small one.

Anisotropy

  1. 1 The material that has a direction
1 essay · materials Two drawings of one deck, and they are not the same structure. A 112 m viaduct on five supports, articulated two ways. Above, the fixed point is at the left abutment: the far end has to be given 45 mm of movement, and the friction of every sliding bearing runs one way, so the fixed support takes 660 kN before any wind or braking is applied. Below, the fixed point is at the middle pier: the largest joint halves to 22 mm and the friction now cancels across the fixed point, leaving 0 kN. The movement arrows are drawn at 900 times the scale of the deck, because a 45 mm movement on a 112 m span is thinner than the line the deck is drawn with. Nothing about the deck, the loads or the ground has changed between the two.

Articulation

  1. 1 Where the structure is allowed to move
1 essay · structures A section modulus for each face, and only the smaller one is a strength. Four profiles of equal area with the second moment divided by BOTH distances to an extreme fibre rather than by the larger of them. A symmetric section has one section modulus and an asymmetric one has two, differing here by as much as 1.00 to one — so the same member has two bending strengths, and which of them applies is decided by the sign of the moment rather than by anything about the section. The bar is the smaller of the two, which is the one that governs when the moment can go either way.

Asymmetric section

  1. 1 Two strengths, depending which way up
1 essay · sections The answer is continuous and the catalogue is not. Capacity bought against capacity required, over a real rolled series. The straight line is what a continuous section would give — exactly the moment asked for, and nothing can be bought on it. The staircase is what a catalogue gives: each tread is one section, each riser is the step to the next, and the vertical gap between the two is steel that is paid for and does nothing. The steps in this series run from 23% to 59% in plastic modulus, so the average waste is 15.0% and the worst is 46% — just above a riser, where the section below has been missed by a kilonewton-metre. The 1200 kNm marked buys a 686×254×125 at 1418 kNm, which is 85% utilised. Two things follow that a continuous treatment cannot see: the sensitivity of a design to an assumption is zero over most of a tread and enormous at a riser, and an optimisation that returns three significant figures is answering a question with twelve answers in it.

Available sections

  1. 1 The answer is continuous and the catalogue is not
1 essay · sections

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