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The load must go somewhere — page 2

Essays 25 to 37 of 37 on this thread, in the same order.
100 kNe = 150centroidworst bolt 50.37 kNSix bolts · direct shear 16.67 kN eachelastic vector method Connections

The connection is not a point, and every diagram on this site says it is

Every free body drawn here has joined its members at points. Real structures fail at the joints far more often than in the members, and the reason is that a joint is exactly the region the theory behind every other page explicitly excludes.

100 kNe = 150centroidworst bolt 50.37 kNSix bolts · direct shear 16.67 kN eachelastic vector method Connections

The bolt that carries more than its share

Six bolts, one hundred kilonewtons, and a worst bolt carrying fifty. The load is shared equally and the torque is not, and the second one is invisible on any drawing where the connection is a point.

100 kN appliedbolt 150.63 kNprying 50.63 kNm = 45n = 40flange 20 mm · one-hingebolt force is 1.51 times the applied load Connections

The force the bolt never saw applied

Pull a tee stub with a hundred kilonewtons and its bolt carries a hundred and fifty. The extra comes from the flange bending and pressing its own edge against the thing it is bolted to, and no free body of the connection as a point contains it.

pullshear plane, 180 mmtension plane, 40 mmcapacity 421.7 kN0.6 fu Anv = 322.5 kN · 0.6 fy Agv = 297 kN · fu Ant = 124.7 kNthe yield value governs the shear plane Connections

The metal between the holes, which comes out as a block

A bolted end connection can fail without a single bolt breaking and without the plate reaching its tensile strength anywhere. A block of metal simply comes out, bounded by two surfaces with two different strengths on them.

00.511.522.533.544.55050100150200displacement, mmload, kNfriction 137 kNbearing 188 kNslipthe rising branch is drawn, not solved: it is elastic shear of the plates Connections

The joint that carries nothing until it slips

Tighten the bolts hard enough and the plates are clamped together with a force nothing applied. The joint then carries shear by friction, the bolts are in tension and not in shear at all, and the load path has nothing in common with the joint it looks identical to.

x̄ = 19.77connected legoutstanding legLc = 150net areaU = 0.87 of it worksU = 1 − x̄ / Lc = 0.87both halves are geometry — where the centroid sits, and how long the connection is Connections

The angle that uses half of itself

Bolt an angle through one leg and the other leg is not fully working. The correction is one over a length — both halves of it are geometry, neither involves a material, and a two-bolt connection throws away a quarter of the section.

flexibility contributed by each componentthey add, so the softest dominates — Sj = 25227.71 kN·m/radwhat doubling it buyscolumn web in shear21.89%×1.12column web in compression11.55%×1.06column flange in bending39.62%×1.25end plate in bending18.09%×1.1bolts in tension8.85%×1.05 Connections

A joint made of springs in series

A connection's stiffness is not a property of the connection. It is the series combination of the flexibilities of everything the force passes through — so the softest component decides, and stiffening any of the others changes almost nothing.

180 kNrow 1 · 420 mm180 kNrow 2 · 340 mm180 kNrow 3 · 200 mm540 kN compressionPlastic distribution — M = 172.8 kN·mthe other distribution is drawn faintly: 1.21× between them Connections

Making a moment cross a gap

A moment is not a thing that can be handed across a joint. It has to be turned into a pair of forces, carried separately, and reassembled — and the whole design is a question of how the tension is shared between bolt rows that are not equally able to take it.

600 kN90 kN·mresultant at e = 150middle thirdbolt carries nothingPartial contactbearing over 300 mm at 10 N/mm² · 50% of 20 Connections

Where the structure meets the ground, and when the bolts start working

Push a base plate off its middle third and it lifts off the foundation. The holding-down bolts then carry exactly nothing, and go on carrying nothing until the plate has crushed the concrete underneath it.

012345-30-20-10102030time (s)displacement (mm)20 kN applied at once and heldthe load arrives in no time at allstatic, 12.67 mmpeak 24.56 mm at 0.250 s Dynamics

Twice the deflection, for the same load

A weight placed gently on a beam deflects it by one amount. The same weight let go from rest, a millimetre above the same beam, deflects it by twice as much — and the factor of two is exact, for every structure ever built.

eight modes · 2400 ttwo modes reach 90% of itmode 1 · 0.93 s85.6%85.6% cumulativemode 2 · 0.31 s9.1%94.7% cumulativemode 3 · 0.19 s3.0%97.7% cumulativemode 4 · 0.14 s1.3%99.0% cumulativemode 5 · 0.12 s0.6%99.6% cumulativemode 6 · 0.1 s0.3%99.9% cumulativemode 7 · 0.09 s0.1%100.0% cumulativemode 8 · 0.09 s0.0%100.0% cumulative Dynamics

Most of the mass moves together

A sixteen-storey frame has sixteen modes, and the first one carries eighty-four per cent of the mass. That is why a hand calculation on one mode gets the base shear right to within a tenth — and why the same calculation gets the acceleration at roof level wrong by a factor of two.

three structures · one recordT = 0.3 speak 12.54 mmT = 0.8 speak 46.2 mmT = 1.8 speak 108.95 mmpeak displacementperiod Dynamics

The spectrum is not a load

A response spectrum looks like a load curve and is not one. Every point on it is the peak of a complete time integration of one particular structure, and the curve is what you get by doing that again for every structure there could be.

05101520253035400246810drop height ÷ static deflectionpeak ÷ static deflectiontwice the static answer — a weight placed, not droppedthe static answera weight dropped from a height Dynamics

The weight that was dropped

A half-tonne load lowered onto a beam produces 5 kN. The same load dropped one metre onto the same beam produces 160 kN — and onto a beam ten times softer, 54 kN. The stiff structure is the one that suffers, which is the opposite of every other rule on this site.

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