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The load must go somewhere — page 10

Essays 217 to 240 of 450 on this thread, in the same order.

Prying against flange thickness. The ratio of bolt force to applied force, for a tee stub carrying 140 kN per bolt, as the flange thickness varies. Prying disappears above 31.91 mm and the flange has become a mechanism below 22.75 mm, where the shaded region begins and the bolt has stopped being the thing that decides.

The thickness that decides who fails

A bolt in a tee stub carries more than the load applied to it, because the flange bends and levers against its own edge. How much more, and whether the bolt or the flange is the thing that gives way, are both decided by one dimension — and the two regimes it separates fail in completely different ways.

A preloaded joint, before and after it slips. Four preloaded bolts at 172 kN each, on one friction face at μ = 0.5. The joint carries 344 kN by friction with the bolts in tension and not in shear at all; past that it slips into bearing and carries 362 kN with the bolts now in shear. Two different mechanisms, one joint.

Connections

The hole made bigger so the steel would fit

A preloaded joint carries load by friction, and the friction is reduced by the shape of the hole the bolt passes through — not by how much steel the hole removes, but by a coefficient in a table. An oversize hole costs fifteen per cent of the resistance; a long slot costs thirty-seven. Both are provided because the steel would not otherwise line up.

Block shear: the metal between the holes. Three bolts in a 9 mm plate end connection. The shaded block tears out along a shear plane 180 mm long and a tension plane 55 mm long. Shear ruptures first, and the capacity is the sum of two different strengths on two different planes: 524.79 kN, of which the shear plane carries 63.03%.

Connections

The end that is only a plate

Cut one flange off a beam's end and what is left is a tee. Cut both and what is left is a plate with holes in it — no flanges, no section modulus worth the name, and none of the checks the beam was selected by. Three plate checks replace them, and the one that governs depends on dimensions that appear in no section table.

A check made on a perimeter, not on a section. One bay of a flat slab, 7.2 m square, on a 400 × 400 mm column. The heavy closed line is the control perimeter, 2d from the column face with its corners rounded at that radius — 4427 mm long against 1600 mm round the column itself. The shaded area inside it delivers no shear across it and is subtracted from the load; everything outside arrives through the perimeter. At 12 kN/m² that is 604 kN across 4427 × 225 mm, a shear stress of 0.697 N/mm² against a resistance of 0.658.

Internal forces

Turn the column, and the slab passes

A flat slab that is comfortable under gravity fails its punching check the moment a moment arrives at the column, and nothing about the load has changed. The fix is not more concrete. It is the column's plan shape and, at equal area, which way round it is turned — worth more than adding half again as much column.

The worst force in a pile is not at the top of it. A 0.75 m pile 30 m long through ground that is settling, carrying 1200 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 600 kN the base takes. The maximum is 2331.39 kN at 18.86 m — 1.94 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 coating that takes the resistance with it

The cure for downdrag is to make the pile slippery, and it works — a bitumen slip layer takes the drag on this pile from 1,131 kN to 34. It also removes the shaft friction that was holding the pile up, in the same proportion and over the same length, and past a certain smoothness there is no neutral plane to find because there is no equilibrium.

The shear goes round the corner instead of across it. A 6-panel Vierendeel girder, 21 m by 6000 mm, under 300 kN at mid-span. There is no diagonal in it, so each panel's 150 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, 131.3 kNm, and it adds to an axial force of 263 kN from the global moment at the same point. The girder deflects 109.09 mm against 6.53 mm for the same members triangulated — 16.70 times — and 98% of that movement is chord bending that a diagonal would have removed entirely.

Structural form

The frame is a girder stood on end

Every unbraced building frame is a Vierendeel girder turned through ninety degrees, and the identification is not an analogy — it is the same equations with the axes swapped. Which means the frame inherits results that read as absurd for a building — more bays is stiffer, a wider building is not, and doubling one section property halves the sway.

$A buckled panel is a truss that nobody drew. A 1500 × 2000 panel of 8 mm web, at d/t = 188. It buckles in shear at 41.0 N/mm², which is 492 kN — and it then carries 1187 kN, 2.41 times as much, because the tension diagonal takes over from the compression one that has gone. The band runs at 18.5° with a membrane stress of 348 N/mm² over a width of 788 mm, and it pulls on the flange at 280.2 N per millimetre of its length. A web that never buckled at all would have reached 2460 kN, so the panel ends at 48% of a stocky web's capacity on a fraction of its steel.$ Stability

The tension has to pull on something

A buckled web carries its shear on a diagonal band of membrane tension, and the band pulls sideways on the flanges and stiffeners that bound it. That pull is the design output nobody plots — it runs from 72 to 603 newtons per millimetre across ordinary panel proportions, it is largest exactly where the panel is most efficient, and at the end of the girder there is nothing beyond to take it.

One of these two curves is a stiffness and the other is a statement of statics. The torque a spandrel beam carries, against how much of its torsional stiffness is left. The rising curve is compatibility torsion — a floor beam framing into the side of the spandrel, which shares its fixed-end moment of 197 kNm between the spandrel's torsional stiffness and its own flexural one. Uncracked, the spandrel takes 51% of it, or 100 kNm; at a quarter of that stiffness it takes 21%, or 41 kNm, and the floor beam picks up what was shed. The flat line is equilibrium torsion — a canopy cantilevering 2.2 m off the same spandrel, whose 116 kNm is fixed by statics and contains no stiffness at all. The first can be designed away by accepting a rotation. The second cannot be designed away by anything.

Internal forces

The torque that should not be shed

A compatibility torque can be let go, because the load has somewhere else to go. What the rule does not say is what it costs the somewhere else — a twenty per cent rise in a floor beam's midspan moment, a crack width nobody limits, and a rotation the spandrel has to actually deliver. There is a size of torque past which shedding is the wrong answer, and no code states it.

The distortion runs the length of the span, and a diaphragm stops it. Longitudinal stress at a corner of the box from distortional warping, along a 60 m span carrying 45 N/mm at 2.0 m off the axis. With no interior diaphragm it peaks at 73 N/mm², which is 83 per cent of the bending stress the girder was designed for. Two interior diaphragms take it to 18. The governing length is Winkler's: the distortion decays over 23.6 m, so a diaphragm helps its neighbours only if it is closer than that, and past it the spacing stops mattering.

Sections and stress

One diaphragm is nearly none

A box girder's distortion decays over a length the section decides, and on a sixty-metre span that length is twenty-four metres. So a single diaphragm at midspan sits further from each end than the distortion can reach and removes a third of the problem; three diaphragms remove nine tenths. The spacing rule is not span over five — it is a property of the plates.

Removing each member in turn. Every member of a 8-panel pratt truss removed one at a time, with the worst demand on the survivors plotted against the member removed. Four of the 35 leave a mechanism — the bars drawn to the top of the frame — and for those there is no redistribution to compute, because there is no structure left. The rest redistribute, and the worst of them asks a survivor for 2.04 times what it carried before. A single number for robustness does not exist: it depends on which member goes.

Structural form

A determinate truss has no robustness at all

Remove any one of a Warren truss's thirty-one members and what is left is a mechanism. Not weakened — gone, with no set of forces that holds the load in any position. Robustness is not a property a structure has by degree; it is bought by adding members that carry nothing until something else stops carrying, and a truss without them has none of it to measure.

The stub column a bearing stiffener makes, in plan. A plan through the girder at the bearing. The 8 mm web runs across; the pair of 100 × 12 stiffeners stands off it; and the shaded strip of web either side — 15ε t_w, or 98 mm each way — is the width that buckles with the stiffeners rather than independently of them. Together they are an area of 3962 mm² with a second moment of 9.01·10⁶ mm⁴ about the web's centreline, a radius of gyration of 48 mm over a buckling length of 900 mm — 0.75 of the depth, because the flanges hold the ends. That is a slenderness of 0.25, at which the column curve returns 0.98: the stub column reaches 98 per cent of its squash load, and the section's own strength is very nearly the whole answer.

Stability

A column nine hundred millimetres long

The patch-load check asks how much of a web a flange can spread a wheel over, and answers in a plate-buckling reduction that throws seven tenths of it away. A pair of stiffeners does not improve that answer. It replaces the question with a different one, from a different family, with a different failure in it.

The same chord, buckling under a constant force and under a varying one. Two buckled shapes of the same 40 m compression chord on the same continuous restraint of 0.35 N/mm per mm, at the same scale. Under a constant force the buckle fills the member — 8 half-waves over 98 per cent of the length — and the critical force is 1881 kN, which is the length-free answer the closed form gives. Under the parabolic force a uniformly loaded deck delivers to it, the buckle LOCALISES: 5 half-waves over 56 per cent of it, gathered where the force is largest, and the peak force at buckling is 2113 kN. The shaded curve is the force distribution the second shape is buckling under.

Stability

The buckle that will not spread out

A compression chord on a continuous restraint chooses its own number of half-waves and forgets how long it is. Give it the force it actually carries — a parabola, largest at midspan — and the number barely moves while the shape changes completely, which is the half that decides where the restraint has to be.

What a crack makes of a notch, against how sharp the notch is. The fatigue notch factor against the root radius, at a fixed elastic factor of 3 in a 430 MPa steel. K_t is a property of the shape and does not move along this axis at all — the dashed line — while the factor fatigue actually feels climbs toward it from below. At a 1 mm root the answer is 2.40, which is 30 per cent of the notch relieved; at 0.1 mm it is 1.38, and a notch that concentrates by 3 elastically is barely felt. Nothing has changed about the stress field: what has changed is that the peak is confined to a smaller volume than the material's own process size, so the crack starts against an average rather than against a maximum.

Materials

The notch a crack does not feel in full

The elastic concentration factor is a property of shape and knows nothing about size, which is what makes it so useful and so misleading. A fatigue crack starts against an average over a volume the material owns, so two notches with the same factor and different radii have different fatigue strengths — and the stronger the steel, the less of that relief it gets.

Two webs of one box, carrying different amounts of the same load. The equivalent stress on each web of a 3.0 × 2.0 m box along the half span, under 40 N/mm at an eccentricity of 1500 mm, with the bending stress alone drawn for comparison. The two webs carry the same bending and opposite torsion, so one gets the sum and the other the difference: at the quarter point the loaded web is at 116 N/mm² and the far web at 11, a ratio of 10.9. A check made on the section rather than on the wall reports the average of two numbers that differ by that much.

Sections and stress

Three actions on one web

A load applied off the centreline of a box girder is three actions at once, and every textbook decomposes it into them. What the decomposition does not say is that all three land on the same piece of plate — added on one web and subtracted on the other, so the two walls of one section differ by a factor of ten.

The worst section of a haunched rafter is inside the haunch. Utilisation along a 15.3 m portal rafter carrying 8 kN/m, with an eaves moment of 500 kNm and an apex moment of 150, haunched over 3 m from 906 mm deep down to the rafter's own 453. The moment is largest at the eaves and the depth is largest there too, so the eaves is at 0.42; the apex is at 0.31. The peak is 0.46 at 2.98 m — the haunch tip, where the section has just become the bare rafter and the moment is still 226 kNm. The dashed curve is the same rafter with no haunch, which reaches 1.02 and does not pass.

Sections and stress

The section that governs is inside the haunch

A tapered cantilever has its worst section somewhere along it because the moment grows linearly and the modulus quadratically. A haunched rafter has the same competition with a step in it, and the step is where the check lands — at neither end of the member, at a station no formula names.

Damping that is computed rather than measured. Radiation damping of a 2.5 m block on soil with a shear wave speed of 178 m/s, against how heavy the block is made — the horizontal axis is a multiple of the 150 tonne block drawn. At that mass the three modes are at 47, 29 and 13 per cent of critical. None of this is dissipation: the energy leaves as waves travelling away into the half-space, so the quantity is a geometrical coupling and it can be computed from the size of the footing, the density of the soil and its shear modulus. Every curve falls as the block gets heavier, because a heavier block moves less for the same wave field — which is the one counter-intuitive thing here: mass buys frequency and costs damping.

Dynamics

The damping that is radiated

Every response in this field is quoted with a damping assumption attached, because damping is measured rather than designed and the measurement has a factor of two in it. A machine block on the ground is the exception: its damping is not dissipation at all, and it can be computed from three numbers none of which is a material property of anything that dissipates.

Dynamics

The floor that arrives at a column

The gap between two buildings is computed from their roof displacements, which is where each of them moves most. It is not where they touch, and it is not what is there when they do — a slab edge meeting a column part way up its height is a different event from two slabs meeting, and it is the one that appears in the photographs.

The same bridge by the two theories it might have been designed by. Deck moment along a 500 m suspended span with half of it loaded, computed twice. Elastic theory treats the deck as a beam and applies the cable's extra tension as an upward load: 80.1 MN·m. Deflection theory keeps the cable's total tension acting on the deck's own deflected shape — a geometric stiffness — and returns 56.3, which is 30 per cent less. The extra cable tension is very nearly the same in both (6.20 against 6.11 MN), so nothing about the cable is in the difference: it is entirely the H·v″ term the older theory drops.

Structural form

The tension that was left out

A suspension bridge's deck sits on a cable pulling hard along it, and a member with a large tension in it is stiffened by that tension. Leaving the term out of the deck's own equilibrium is what elastic theory does, and on a long span it asks for fourteen times the girder.

What the second, third and fourth arms are worth. Top drift removed against the number of outriggers, each arrangement at its own optimum levels, on a 40-storey core 200 m tall. One arm at 59 per cent of the height removes 82.3 per cent of the drift. A second, with both moved to 35 and 71, takes it to 91.6 — a gain of 9.3 points, which is half of what was left. The third is worth 3.0 and the fourth 1.5, and each one costs a storey of the building's most valuable height.

Structural form

What the second arm is worth

One outrigger at its best height removes five sixths of a tall core's drift, which sounds like the end of the argument. A second removes half of what is left, a third half of that, and each of them costs a storey of the most valuable floor area in the building — so the question is not where to put an outrigger but how many the arithmetic still justifies.

Two restraints, and only one of them cares how much was imposed. Crack width against the restrained strain, for the same 20 m wall restrained two ways. Edge restraint — a wall cast on a base — gives a width proportional to the strain, because the concrete has to accommodate the movement and the cracks are where it does. End restraint — a bay cast between two that have hardened — gives 0.32 mm at every strain on the axis, because the crack opens only until the steel can push the cracking force back into the concrete, and that force is a property of the section. At the 304 microstrain this wall is asked for, the two are 0.17 and 0.32 mm, and the movement is divided into 35 cracks and 19.

Internal forces

The bay that is cast last

A wall cast on a base cracks in a way the steel controls: more movement gives wider cracks, and reinforcement decides how many. A bay cast between two walls that have already hardened cracks in a way that has nothing to do with how much movement there was — and the threshold that separates a controlled crack from a single wide one is a quantity of steel rather than a limit on anything.

Two load sets with the same resultant and different work. The two ways of putting a uniform load of 10 kN/m onto a beam element 6.00 m long. Both put 30.0 kN at each node, so both have the same resultant and the same moment about any point — they are equivalent for a rigid body. The consistent set adds a couple of 30.0 kN·m at each end, in opposite senses, which is what makes it do the same virtual work over the element's shape functions as the real load does. The couples cancel in the resultant, which is exactly why the resultant cannot see them, and they are the whole difference between an exact answer and one that is a third out.

Equilibrium

Equivalent in work, not in resultant

Two force systems with the same resultant and the same moment about every point are interchangeable — for a rigid body. A finite element is not a rigid body, and substituting one for the other on a beam element leaves the tip of a cantilever a third too low with no warning of any kind.

The part of a base plate that is delivering anything. A plan of the plate: the 260 × 260 column in the middle, and the shaded region it can reach — its own outline grown by 43 mm, which is t√(f_y/3f_jd) for a 20 mm plate on grout at 20 N/mm². That is 78176 mm² of a 200000 mm² plate, or 39 per cent of it. The corners are outside it and are carrying nothing: a plate cantilevers from the column's perimeter and runs out of bending capacity at c, so making it larger in plan beyond that changes nothing at all, and making it thicker changes everything.

Connections

The plate that is only as big as it is thick

A base plate's design is presented as a bearing calculation: an area, a bearing strength, and a check that the pressure fits. The area in that calculation is the plate, and the plate cannot deliver it — a 20 mm plate on a 500 square reaches 43 mm past the column and the corners carry nothing at all.

Force and capacity round a weld group, which do not vary together. Utilisation round the c shape group under 100 kN at 150 mm, walked from one end of the weld to the other, with the force scaled onto the same axis for comparison. The capacity is 1460 N/mm where the force runs along the weld and 1789 where it runs across it — √1.5 more — so the utilisation is not simply the force in different units. The worst point is at 47 degrees to the weld, where the capacity is 1612 N/mm, and the group carries 182 kN against the 165 the along-the-weld value would allow.

Connections

The weld that is stronger where it is pulled

A fillet weld pulled across its axis is √1.5 stronger than the same weld pulled along it. On a group under an eccentric load the direction of the pull varies from point to point, so the capacity does too — and the gain that follows is worth twenty per cent at one eccentricity and nothing at all at another.

$Where the span moment and the support moment cross. Span moment and support moment against the joint's stiffness, for a 6 m beam under 30 kN/m. They move in opposite directions because they add to a constant — the simple-span 135 kN·m is fixed by statics and the joint only decides how it is split. They cross at 68 kN·m, where the joint is delivering 75 per cent of the fixed-end moment, and neither the crossing nor the fraction depends on the beam, the span or the load: it is the point where f·wL²/12 equals wL²/8 − f·wL²/12, which is f = 0.75 for every beam there has ever been.$ Connections

The joint that was chosen

A joint's stiffness decides how a beam's moment divides between its span and its supports, and the two add to a constant. So there is a stiffness at which they are equal, the beam is sized by the smaller of two numbers rather than the larger of one, and the design moment is half what a simple connection leaves behind.

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