Structural Engineering & Statics
where the load goes
Depth

Series — page 16

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.
The same strain, two moduli, and a width multiplied to say so. A timber section with a steel plate in it, carrying 20.0 kNm. Plane sections stay plane, so the strain at a height is the same in both materials; Hooke's law then puts the stresses in the ratio of the moduli, which here is 19.09. Multiplying the stiffer material's WIDTH by that ratio gives a fictitious section of one material with the same neutral axis and the same forces — 595.2×10⁶ mm⁴ of it, against 351.0 for the same shape with the moduli ignored. The steel plate is 3.8% of the area and carries 43% of the moment, at 96 N/mm² against the timber's 5.0. The transform is not an approximation: it is compatibility and Hooke's law written down.

Transformed section

  1. 1 A section made of two materials, one of them pretended away
1 essay · sections The resonance that ran out of time. The response of a 0.100 s oscillator at 2.0% damping while the driving frequency sweeps up through its own, plotted against the driving frequency rather than against time. The steady-state amplification is 1/2ζ = 25; this sweep reaches 23.6, which is 95% of it, because the time spent inside the half-power band is a limited number of build-up time constants. The whole answer depends on one group, β/ζ²ω², and over the range this figure's sibling sweeps the fraction falls from 100% to 41%. Two features fall out of the integration and neither is guessable from the steady-state picture: the peak arrives after the frequency has passed resonance, by 1.2% of it here, and the response beats afterwards at the difference between the two frequencies. A machine's instrument therefore reads its largest amplitude while it is already above its critical speed.

Transient resonance

  1. 1 The resonance that ran out of time
1 essay · dynamics The same steel, brittle in January. Fracture toughness against temperature for a 25 mm ferritic plate, from Wallin's master curve — 30 + 70·exp(0.019(T − T₀)) in MPa√m, whose shape is the same for every ferritic steel and whose only free parameter is the reference temperature T₀ = -60 °C. Beside it, and on its own scale, is the crack length that toughness will tolerate at 200 N/mm², which goes as the square of it. At +20 °C this plate carries a 764 mm flaw and at -20 °C it carries 199 mm — a factor of 3.8 for a forty-degree change in the weather, in a steel that met its specification on both days. Thickness moves the curve as well, and the wrong way: constraint at a crack tip suppresses the yielding that would have blunted it, so a 100 mm plate of this steel tolerates 27% of the flaw a 10 mm plate does. Loading it in a millisecond shifts the whole curve another forty degrees.

Transition temperature

  1. 1 The same steel, brittle in January
1 essay · materials The lining that carries less for being weaker. Bending moment and hoop thrust in a circular lining, against the lining's own bending stiffness, both as fractions of the free-ring values. The ground arrives already stressed — 500 kPa vertically and 300 horizontally at K₀ = 0.6 — and the difference between them tries to squash the hole into an ellipse. A lining stiff enough to refuse absolutely collects the whole distortion pressure, p₂R²/3 = 300 kNm/m; one flexible enough to go with the ground collects nothing, because there is no curvature change left to resist. The thrust is the flat line: it comes from the mean stress rather than the difference, so it does not move at all. Putting 8 joints in this ring drops the moment to 34% of the solid one and leaves the thrust exactly where it was, which is why a segmental lining is jointed and why the intuition carried over from a beam is inverted here.

Tunnel ring

  1. 1 The lining that is stronger for being weaker
1 essay · structures The force runs along the chord whatever route the member takes. A member pinned at two points 10 m apart, loaded only at those two points, and bowed 1.2 m off the line between them. Equilibrium leaves the two end forces no choice: equal, opposite, and along the chord. So at every section the axial force is P cos α, the shear is P sin α, and the bending moment is the force times the perpendicular offset — 240 kNm at the crown for 200 kN at 1.2 m, which is a multiplication rather than an analysis. Straighten the member and the moment diagram is identically zero; that is the case a truss member is in, and the reason it carries one number.

Two force member

  1. 1 The member with only one direction
1 essay · equilibrium A two-way slab is a one-way slab as soon as it is not square. The share of the load carried by the strips spanning the short way, against the ratio of the sides. The two families of strips cross at the centre and must deflect equally there, and a strip's deflection goes as the fourth power of its span — so at a ratio of 1.33 the short strips already take 76% and at 2 they take 94%. The panel drawn here is 6 × 8 m, a ratio of 1.33, and its short strips take 76.0%. Two-way action is worth having at a ratio of one and worth almost nothing by two.

Two-way spanning

  1. 1 The slab that spans both ways
1 essay · structures Two cantilevers, or one wall, and the beams decide which. The deflected shape of a coupled pair of 6 m walls, drawn against the two limits it lies between. Release the coupling beams entirely and the pair is two independent cantilevers, deflecting 111 mm. Make them rigid and it is one composite wall of the full width, deflecting 16 mm — 6.8 times stiffer, because the lever arm between the wall centroids is 8.40 m and everything inside either wall is smaller than that. Real beams of 600 × 350 mm over a 2.4 m opening land at 23 mm and carry 63% of the base overturning as an axial couple rather than as wall bending. The degree of coupling never reaches one, because a beam of finite depth cannot suppress the walls' curvature entirely.

Wall coupling

  1. 1 Two walls that agreed to be one
1 essay · internal-forces 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.

Wall optimum

  1. 1 The best design is the most sensitive one
1 essay · stability A straight line, and the comfortable case is already two thirds down it. The capacity of a 215 mm masonry wall as a fraction of its squash load, against the eccentricity of the resultant. Nothing in this figure is a buckling calculation. A material that cannot be pulled bears on a strip of width 3(t/2 − e) under a triangular stress block, so the capacity is exactly 1.5f(t − 2e) — a straight line, zero when the resultant reaches the face, and already at 67% at the edge of the kern. The middle third is treated everywhere as the comfortable case; a wall loaded there has given away a third of its capacity before slenderness has been mentioned. The lower line is the same wall with slenderness in it, which enters as an ADDITIONAL eccentricity of 17.4 mm rather than as a reduced stress — h_ef²/2400t, for h_ef = 3000 mm. Euler's load for this wall is 9.6 times what the eccentricity rule allows, which is why no masonry calculation contains it.

Wall slenderness

  1. 1 It does not buckle, it runs out of width
1 essay · stability The flanges go opposite ways, and the pair of them is the bimoment. A 305 by 165 mm I-section held against warping and twisted by 0.5 kN·m, with the section on the left and the two flanges seen in plan on the right. At the built-in end each flange bends in its own plane, one way at the top and the other at the bottom, through 29.3 mm at the free end — drawn 20 times its true size against the 6 m length. The pair of flange shears is 1.69 kN each, and 1.69 × 295 mm is 0.500 kN·m — the whole torque at that section, carried by two forces neither of which is a torque. The pair of flange moments is 3.04 kN·m each, and 3.04 × 295 mm is 0.897 kN·m², which is the bimoment. It puts 67.0 N/mm² into two diagonally opposite flange tips and takes the same out of the other two, so its net force and its net moment about every axis are zero — which is exactly why no member diagram has a place for it.

Warping

  1. 1 The section that cannot stay flat
1 essay · sections 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.

Weld distortion

  1. 1 The shape that came out of the shop
1 essay · connections The wind pushes on one face and pulls on three. A 30 × 20 m building in plan, with the measured pressure coefficient on each face and the arrows drawn in the direction the pressure acts. Only the windward face is pushed; the other three are sucked, and the side faces are sucked hardest of all at c_p = -0.7. The horizontal resultant is 842 kN at a velocity pressure of 0.9 kPa, and the arithmetic of it is the whole point: the leeward suction pulls the building downwind, so it ADDS, supplying 38% of the answer, while the two side faces cancel each other exactly and supply none of it. The coefficients are wind tunnel data; what is computed is the free body they are applied to.

Wind pressure

  1. 1 Most of it is suction
1 essay · equilibrium One criterion inside the other, touching at six points. The two yield criteria in principal stress space with the third principal stress zero, both normalised by the yield stress. Von Mises is the ellipse — σ₁² − σ₁σ₂ + σ₂² = f_y², which is a circle seen at an angle — and Tresca is the hexagon inscribed in it, touching at the six points where one principal stress is zero or the two are equal. Everywhere else Tresca is the smaller, by up to 15.5 per cent, and the widest gap is at pure shear, where σ₁ = −σ₂ and the two answers are 205 and 178 N/mm². The ratio there is exactly 2/√3, computed rather than quoted, and it is the whole reason a web is checked against f_y over root three.

Yield criterion

  1. 1 The shear strength nobody measured
1 essay · materials

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