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The load must go somewhere — page 10

Essays 217 to 226 of 226 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.

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