Structural Engineering & Statics
where the load goes
Concept

Coefficient of friction — where it appears

The ratio of the friction force available at an interface to the normal force pressing it together. It is a property of the two surfaces rather than of either material, it is not a designed quantity, and a slip-resistant connection's whole capacity is proportional to it.

Named by 9 essays across 2 fields — each of them below, with the objects they name alongside it.

The bearing that is drawn as a roller. The horizontal force a sliding bearing delivers, against the vertical load it is carrying, with its coefficient of friction on the same picture. The coefficient is not a constant: PTFE's falls as the contact pressure rises, and the standard fit is μ = 1.2/(10 + σ), so the bearing drawn is at 30.0 N/mm² and μ = 0.030 while the same bearing at a fifth of the load is at 0.075 — 2.5 times as much. The force curve is therefore strongly non-linear: a fifth of the load gives 50% of the force. Two readings follow and only one of them is usually taken. The largest force is at full load, 108 kN, and that is what the pier is designed for. The largest nuisance is at light load, where 54 kN of friction is 39% of the 140 kN of wind the bearing was put there to release the structure from. Cold makes it worse again: below about −5 °C the same bearing delivers 216 kN. A roller symbol on a drawing means this, and it is a pair of load cases rather than one, because friction opposes whichever way the deck happens to be going.

The roller that is not a roller

A sliding bearing is drawn as a roller and detailed as a sheet of PTFE, and it delivers a horizontal force of a few per cent of whatever it is carrying. The coefficient everybody quotes is the one at full design pressure, and PTFE's coefficient rises as the pressure falls — so the bearing is at its freest exactly where nobody checks it.

connections · Bearing friction 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.

equilibrium · Friction Weight is the only thing holding it down. A body 1.6 m wide and 4.5 m tall weighing 22 kN, under a wind pressure of 1 kN/m². The wind delivers 2 kN and an overturning moment of 4 kNm about the leeward toe; the weight restores 18 kNm, a factor of 4.35. The resultant lands 0.18 m from the centre against a middle third of ±0.27 m, so the base is still wholly in bearing.

Whether it tips or slides

A free body pushed sideways has two ways of leaving, and which one it takes is decided before any load is known. The condition is a width divided by a height set against a coefficient of friction, and the weight, the wind pressure and the depth of the body all cancel out of it.

equilibrium · Overturning 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.

equilibrium · Friction 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.

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.

connections · Slip-critical 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.

equilibrium · Friction 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.

equilibrium · Friction Nine-tenths of a tightening torque stretches nothing. Where the torque applied to the nut of an M20 grade 10.9 bolt goes, against the coefficient of friction in its thread and under its nut, taken as equal. The bottom band is the thread's lead — the only part of the work that stretches the bolt — the middle band is friction in the thread and the top band is friction under the nut. At μ = 0.14 the lead takes 11% of the torque, the thread 39% and the nut face 50%. At μ = 0.06 the lead's share is 22% and at 0.24 it is 6%, so a coefficient nobody measured decides how much of a specified torque arrives in the bolt as preload.

The torque that goes into the thread

A preload specified as a torque is a preload specified through two coefficients of friction that nobody measures. Nine-tenths of the torque on a bolt is spent turning against its own thread and the face of its nut, so a change in the grease moves the clamping force by half — and the one method that escapes it does so by yielding the bolt on purpose.

connections · Slip-critical An old joint passes the bolts that have lost their preload. The chance that an M20 grade 10.9 bolt holds the inspection torque of 647 N·m without its nut turning, against the tension actually in it as a fraction of the minimum preload, for a joint new and 1 year old, 5 years old and 20 years old. New, a bolt at half the minimum preload passes 0% of the time and one at the minimum 83%; 20 years old, the bolt at half passes 80% of the time and one at a quarter 9%.

The wrench that reads the rust

The obvious check on a preloaded joint that has been in service for years is a wrench on the nut, and it measures the wrong thing. The nut holds against the bolt's tension times a friction bracket, and twenty years of rust raise the bracket until a bolt with half its preload passes four times in five. The washer that could have said more remembers only the day it was crushed.

connections · Slip-critical

Named alongside it

The objects these essays reach for when they reach for this one.

FrictionSlip resistanceFree bodyPreloadLimit stateBearingBolt tensionConnectionEquilibriumInspectionSliding bearingTolerance

All concepts