Structural Engineering & Statics
where the load goes
Series

Friction — the series

6 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. The reaction lies inside the cone, so the block stands. A block of 100 on a plane at 15°, against a coefficient of friction of 0.35. Resolving across and along the plane gives a normal force of 96.6 and a friction demand of 25.9, against a capacity of μN = 33.8 — a ratio of 0.77. Added together the two make one contact reaction leaning 15.0° from the normal, and the admissible reactions fill a cone of half-angle arctan μ = 19.3°. Equilibrium is possible exactly when the demanded reaction lies inside that cone, which here it does. The weight enters neither the cone nor the lean: a block of any weight on this slope leans its reaction by the same 15.0°, which is why the angle of repose is a material property and the size of a heap of sand is not.

    The force that is whatever it needs to be

    Every other force in statics has a value the equations produce. Friction has an inequality instead, so it takes whatever value equilibrium demands and the bound only ever says no — which means a problem with friction in it has a range of answers rather than one.

    part 1 · equilibrium
  2. The reaction lies inside the cone, so the block stands. A block of 48 on a plane at 22°, against a coefficient of friction of 0.6. Resolving across and along the plane gives a normal force of 44.5 and a friction demand of 18.0, against a capacity of μN = 26.7 — a ratio of 0.67. Added together the two make one contact reaction leaning 22.0° from the normal, and the admissible reactions fill a cone of half-angle arctan μ = 31.0°. Equilibrium is possible exactly when the demanded reaction lies inside that cone, which here it does. The weight enters neither the cone nor the lean: a block of any weight on this slope leans its reaction by the same 22.0°, which is why the angle of repose is a material property and the size of a heap of sand is not.

    The area that is not in the equation

    Friction is proportional to the force pressing two surfaces together and independent of how large they are, which sounds like an approximation and is not. The area is absent because the contact that carries the load is a tiny fraction of the contact that is drawn, and that fraction grows in exact proportion to the load.

    part 2 · equilibrium
  3. A preloaded joint, before and after it slips. Eight preloaded bolts at 100 kN each, on two friction faces at μ = 0.35. The joint carries 560 kN by friction with the bolts in tension and not in shear at all; past that it slips into bearing and carries 900 kN with the bolts now in shear. Two different mechanisms, one joint.

    The force that is capped on purpose

    Everywhere else in this collection friction is a nuisance whose value nobody controls, checked with a coefficient known to one figure. In a friction damper the inequality is the design intent — the device is specified so that a member behind it can never be asked for more than a stated force.

    part 3 · equilibrium
  4. Both checks pass, and the contact lets go. A contact pressed together by 1000, with μ = 0.4. Every tangential force the contact can supply lies inside a disc of radius μN = 400.0, because the friction law bounds the length of the force and not its components. The contact is asked for 300.0 one way and 300.0 the other way — 75% and 75% of the radius taken one at a time — and 424.3 together, 106% of it. Each one-direction check passes and the force does not fit: the square those checks describe reaches √2 times further at its corners than the contact can.

    Seventy-five per cent each way

    A contact asked for friction in two directions at once can supply a force of a certain length pointing any way it likes, so its limit is a disc and not a square. Two checks made one direction at a time, each passing at seventy-five per cent, describe a contact that has already let go.

    part 4 · equilibrium
  5. The pier grips, gives, and grips again. A sliding bearing carrying 3000 kN on a pier head of 20 kN/mm, dragged by a deck expanding at 1.7 mm an hour, with a static coefficient of 0.05 and a kinetic one of 0.03. The force in the pier climbs while the bearing grips, reaches 150 kN, and falls in a fraction of a second to 30 kN: the pier springs back under only the kinetic friction, overshoots the 90 kN that friction would hold it at, and grips again. The swing is 120 kN — 2.00 times the 60 kN between the two coefficients — and the pier head jumps 6.00 mm each time, three times in 12 hours.

    The pier that moves in jumps

    A sliding bearing whose static friction is larger than its kinetic friction does not release a slow thermal movement as a drift. It grips, gives and grips again, and each time the force in the pier swings by twice the difference between the two coefficients — whatever the pier is made of.

    part 5 · equilibrium
  6. Same deck, same load, and two pier forces. A deck bearing on a pier of 20 kN/mm with μ = 0.03, taken to the same final state two ways: the deck moves 4 mm over the pier, and the bearing's load rises from 2000 to 4000 kN. Moved first, while the bearing carries 2000 kN, the pier force reaches the limit of 60 kN and the bearing slides for the rest of the movement; the load arriving afterwards raises the limit and changes nothing, and the pier is left carrying 60 kN. Loaded first, the limit is 120 kN before the deck moves, the bearing grips throughout, and the pier carries 80 kN. Both states are at the same displacement under the same load, and both satisfy equilibrium and the friction bound; the order is the only difference, and it appears in neither.

    The order the loads arrived in

    Statics allows a contact with friction a whole range of forces and has no way to choose between them. A real structure does choose, and what it chooses by is the order in which things happened to it — so the force in a pier under a sliding bearing is a record of its history, not a function of its loads.

    part 6 · equilibrium

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