Structural Engineering & Statics
where the load goes
Depth

Series — page 14

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.
A column that has nothing on it but itself. A 30 m column carrying no load except its own weight, with its buckled shape and the axial force that produced it. The force is zero at the top and largest at the base, which is why the answer is not Euler's: the eigenvalue is a load intensity and comes out as q_cr L³/EI = 7.835, a number with no π in it, computed here as the smallest eigenvalue of the same stiffness and geometric-stiffness matrices that give a tip-loaded column its Euler load. It is equivalent to a tip load of 3.18 times as much total weight — a column carries its own weight better than it carries somebody else's, because most of the weight is near the base where the buckle is not. The height limit that follows is a cube root, so a section of radius of gyration 80 mm falls over on its own at 52 m and one of twice that reaches only 82.

Self-weight buckling

  1. 1 Too tall for nothing but itself
1 essay · stability What is left after the first fibre yields, which is a property of shape. The shape factor — plastic modulus over elastic — for six sections, computed by finding each one's equal-area axis and summing ±f_y over it. The numbers contain no dimension, no stress and no material: a rectangle is exactly 3/2 whatever its size, a diamond exactly 2, a circle 16/3π. The spread is the argument. An I-section keeps only 13 per cent in reserve past first yield, because nearly all its material is already at the extreme fibre and there is nothing further in to recruit; a diamond keeps 100 per cent, because most of its material is near the middle and doing very little elastically. So the section shapes that are best at elastic bending are the ones with the least left afterwards, which is exactly backwards from the way the reserve is usually described.

Shape factor

  1. 1 What is left after the first fibre yields
1 essay · sections A section has two areas and the tables give one of them. Peak shear stress divided by the mean, for four sections of exactly the same gross area and depth. The mean is V/A and is the number a first calculation uses; the peak is what the material actually sees, and the ratio between them is a property of shape alone. A rectangle's is 1.5 — the parabola's peak over its average — and it is one of the few numbers in this subject that is exactly derivable and universally ignored. An I-section's is near 1.98, and the reason is on the second bar: 97% of the shear is inside a web that is 56% of the area. So the flanges carry the moment and almost none of the shear, and the web carries the shear and almost none of the moment — which is why a shear check on an I-section uses the web area and a moment check uses the whole section, and why the two checks are about two different pieces of steel.

Shear area

  1. 1 The section has two areas
1 essay · sections An angle bolted through one leg. A 100 × 75 × 10 angle connected through its 100 mm leg with three bolts at 75 mm pitch. The centroid sits 19.77 mm from the connected face over a connection 150 mm long, so U = 1 − 19.77/150 = 0.87 and 13.18% of the net area is not working.

Shear lag

  1. 1 The angle that uses half of itself
1 essay · connections Nothing happens, and then everything happens. The moment capacity left to a section already carrying shear, against the shear as a fraction of what the web can take. The web holds 26.1% of this section's plastic modulus and the flanges hold the rest, and only the web's share is reduced — by the factor √(1 − v²) that von Mises leaves it. So the curve is flat for most of its length: the first per cent of moment is not lost until v = 0.27, half the shear capacity costs 3.5%, and 15% is not reached until v = 0.9. The tangent at v = 1 is vertical, which is why the last tenth of the shear range costs more than the first eight.

Shear moment interaction

  1. 1 Both at once, and neither matters until it does
1 essay · sections The cut that severs the stirrups is the cut that counts them. A cracked web drawn as the truss it has become: a tension chord along the bottom, a compression chord along the top, concrete struts between the cracks and stirrups crossing them. The free body is a cut parallel to the cracks, which over a lever arm of 495 mm severs z·cot θ/s = 8.3 stirrups at cot θ = 2.5. Every one of them is at yield, so the shear is their number times their strength, and flattening the crack raises the count rather than the strength. The same cut passes through the bottom chord, which is where the chord force the moment diagram does not contain comes from.

Shear truss analogy

  1. 1 The beam that becomes a truss
1 essay · internal-forces The hoops change their mind at an angle no proportion chose. The two membrane forces of a spherical dome of radius 30 m under 3 kN/m² of surface, taken from the crown to a base at 60°. The free body for the meridional force is the cap above a cone of half-angle φ, and vertical equilibrium of it gives N_φ = −wR/(1 + cos φ) directly: -45.0 kN/m at the crown falling to -60.0 at the base, compression everywhere. Equilibrium normal to the surface then gives the hoop force, which starts at -45.0 kN/m and reaches 15.0 — it changes sign, and the angle at which it does was found here by bisecting N_θ rather than quoted: 51.827292°. Setting N_θ = 0 gives cos²φ + cos φ − 1 = 0, so cos φ is (√5 − 1)/2, the reciprocal of the golden ratio — an identity the bisection reproduces to nine decimals, which is worth seeing done rather than taken on trust, because it is too pretty to be believed on sight. Below that parallel the hoops are in tension, which masonry has none of, and that is where every old dome is cracked.

Shell action

  1. 1 The surface that carries by being curved
1 essay · structures The least reliable number in the material decides the answer, briefly. What a twenty per cent error in the concrete's tensile strength does to a computed deflection, against how far past cracking the beam is. Well past the cracking moment it does almost nothing — at 1.9 times M_cr the spread is 56 per cent — because the section is nearly fully cracked and the interpolation has run out. Just above cracking it does everything: at 1.19 times M_cr the same twenty per cent moves the deflection by 3658 per cent. Tensile strength is the property with the widest scatter and the least direct test, and a beam designed to sit near its cracking moment has put the answer on it.

Shrinkage curvature

  1. 1 The curvature nobody applied
1 essay · deflection The ground is a structure, and it has a period. The transfer function of 38 m of soil at 75 m/s over bedrock at 800 m/s: how much the surface moves for a given motion in the rock, at every frequency. It is a column fixed at the bottom and free at the top, so its resonances are the odd harmonics — the peaks stand at 1, 3, 5, 7 times the first, which is a fixed-free column and nothing else. The fundamental is at 0.493 Hz, a period of 2.03 s, and it is 4H/v_s exactly. The peak amplification is 7.5 against the bound 1/(α + πξ/2) = 7.5, which agrees to 0.2% — and the α in it is the impedance ratio, 0.0554 here. That is the term that keeps the answer finite: assume rigid bedrock and α is zero, the bound becomes 13 and the model is predicting an amplification set by damping alone. What limits the surface motion is that the energy can leave downward.

Site response

  1. 1 The ground has a period of its own
1 essay · dynamics A strength that is a property of the specimen. Nominal strength against size for geometrically similar specimens of one material. On the left the specimen is too small for a crack to run and the strength is a plateau — a plastic limit, and the regime laboratory specimens sit in. On the right a crack releases more energy than it consumes as soon as it starts and the strength falls as the inverse square root of size, which is the regime real structures sit in. The turn happens at D₀ = 120 mm. A 100 mm specimen reads 3.10 N/mm² and a 1500 mm member of the same material carries 1.14: the test overestimates the structure by a factor of 2.71.

Size effect

  1. 1 The bigger one is the weaker one
1 essay · materials 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.

Skew

  1. 1 The deck that spans square
1 essay · internal-forces The snow that left the roof is standing against the wall. A roof in section with a 1.0 m obstruction at its downwind end, drawn with the vertical scale exaggerated 4 times because a metre of snow on twenty metres of roof is thinner than the line it would be drawn with. The balanced layer is 0.48 kN/m² everywhere; the wedge against the wall holds the snow that 30% of the 20 m upwind gave up, so its area is fixed by conservation of mass rather than chosen. That makes it 0.85 m deep and 3.4 m long, with a peak load of 2.60 kN/m² — 5.4 times the balanced value, and still only 27% of the snow on the roof. An intensity several times the design load, made of a quantity nobody would notice had moved.

Snow drift

  1. 1 The load that arrives where the wind stops
1 essay · equilibrium Two identical pipes, and one carries three times the other. Load per metre on a buried conduit against the depth of cover, in trench widths, with the weight of the prism of soil directly above it drawn between them. A conduit laid in a narrow trench is stiffer than nothing and softer than the sides: the backfill settles relative to the undisturbed ground, the friction on the trench walls acts upward, and the conduit gets 64% of the prism. Lay the same conduit on the ground and build an embankment over it and it is now stiffer than the fill beside it, the interior prism settles less, the friction acts downward, and it gets 172% — a factor of 2.71 between two pipes with nothing different but which way the ground moved. The equation is Janssen's, the same one a silo wall obeys, with a trench for a silo; both curves start on the prism line, because with no depth there is no shear to redistribute anything. This is why a flexible pipe is buried rather than a rigid one: making the conduit weaker moves it down the page.

Soil arching

  1. 1 The pipe decides what the soil weighs
1 essay · equilibrium The force falls, the drift rises, and the damping goes the wrong way. What a compliant foundation does to a 0.6 s building on a 8 × 8 m footing, against the stiffness of the ground under it. Three curves, all normalised to the fixed-base answer. The period lengthens — 1.31 times at 200 m/s — because the swaying and rocking of the foundation are flexibilities in series with the structure's own, and the rocking term carries an h², so it is the tall building that feels it. The base shear falls with the period, which is why a fixed base is usually called conservative. The displacement rises, by 51% here, and that is what breaks the cladding, the services and the gap to the building next door. And the effective damping falls rather than rises: the structure's own is divided by the cube of the lengthening — 2.2% of an original 5% — while a slender building's foundation radiates only 0.28% back, because rocking radiates almost nothing at these frequencies.

Soil-structure

  1. 1 The ground is a spring
1 essay · dynamics A frame with no plane to be drawn in. The tetrahedron, solved: three equations at every free joint, one axial force in every member. Members drawn heavy carry more; tension and compression are separated by the sign that came back from the solve rather than by inspection. The count is m + r = 12 against 3j = 12, which makes it exactly determinate, and joint equilibrium closes to 1.8e-15.

Space frame

  1. 1 Three equations at every joint
1 essay · structures Deflection goes as the fourth power of the span. Deflection against span for a constant load intensity and section, with two slower relationships drawn faintly behind it for comparison: the load itself, which grows in proportion to the span, and the bending moment, which grows as its square. Doubling the span multiplies the deflection by sixteen, while the moment only quadruples.

Span scaling

  1. 1 Span to the fourth, which is why spans are short
1 essay · deflection Chord force against truss depth. The force in a truss chord for a fixed bending moment, against the depth of the truss. The relationship is a reciprocal: the chords form a couple whose lever arm is the depth, so a shallow truss pays for it steeply.

Span-to-depth

  1. 1 Depth is the cheapest strength there is
1 essay · structures The end bolts do the work and the middle ones very nearly nothing. A lap of 10 bolts at 75 mm pitch transferring 800 kN between two plates, with the force each bolt actually carries drawn above it and the flat line a division by the bolt count would have given drawn behind. The end bolts carry 1.16 of their nominal share and the middle ones 0.89. The reason is not in the bolts: at the leading end the first plate is carrying everything and the second nothing, so the two strain at different rates and the slip between them is largest there. In the middle they strain alike, there is almost no slip, and a bolt with no slip across it transfers almost no force. The mean over the worst is 0.861, and the end bolt has to slip 1.16 mm before the rest catch up.

Splice

  1. 1 The joint that has to be as good as the member
1 essay · connections The props decide where the stress ends up. Bottom-fibre stress in the steel of a 12 m composite beam carrying 6 kN/m of wet concrete and 9 kN/m afterwards. Unpropped, the bare steel takes the first stage alone and reaches 146 MPa; propped, the finished composite section takes everything and reaches 93 MPa — a ratio of 1.57. 62% of the unpropped beam's final stress was locked in before the slab was structural at all. The deflections differ by 1.73 times for the same reason, and no drawing of the finished beam distinguishes the two.

Staged section

  1. 1 The section that changed while it was being loaded
1 essay · sections Two of these move and the third cannot. The first span of a 3-span beam under 5 kN/m, with the stiffness of the middle span swept over a factor of 25. The support moment and the mid-span moment both move — that is what redundancy does, and it is the whole reason a continuous beam has to be analysed rather than read off. Their combination does not: the mid-span ordinate plus the average of the two end moments is 30.6 kNm at every point on this axis, which is wL²/8 for this span and this load and contains nothing else. The largest departure anywhere on the sweep is 2.3e-16 of the value, which is the arithmetic of the stiffness solution rather than a property of the beam. Continuity buys a distribution and not a capacity, and this is the line that says so.

Static moment

  1. 1 Two of these move and the third cannot
1 essay · internal-forces

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