Structural Engineering & Statics
where the load goes
Concept

Slip resistance — where it appears

The friction a preloaded bolted joint develops between its faying surfaces, which is the bolt preload times a surface treatment coefficient. It carries no load once it is exceeded, at which point the joint slips into bearing and behaves as a different connection.

Named by 10 essays across 3 fields — each of them below, with the objects they name alongside it.

A preloaded joint, before and after it slips. Two preloaded bolts at 137 kN each, on one friction face at μ = 0.5. The joint carries 137 kN by friction with the bolts in tension and not in shear at all; past that it slips into bearing and carries 188 kN with the bolts now in shear. Two different mechanisms, one joint.

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.

connections · Slip-critical 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.

equilibrium · Friction Two beams, or one beam four times as stiff. Two 200 × 150 planks spanning 4 m under 6 per millimetre. Loose, they have 112.5×10⁶ mm⁴ between them and deflect 16.2 mm, with the two faces at the interface sliding past one another. Bonded, the pair has 450.0×10⁶ — exactly 4 times as much, because doubling a depth cubes — and deflects 4.0 mm at half the extreme-fibre stress. Nothing was added but a restraint on slip. With connectors of stiffness 200 the same beam deflects 5.4 mm, which is 89% of the way from one bound to the other.

Two beams, or one beam four times as stiff

Stack two planks and they bend as two beams whose faces slide past one another. Bond the faces and the pair has one neutral axis, four times the second moment and half the stress. Nothing was added but a restraint on slip.

internal-forces · Composite action 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 Three fasteners, three completely different clocks. What each of three shear fasteners carries against how far the joint has moved. A 400 kN fillet weld is linear to 0.4 mm and then gone — it is stiff and it is not ductile. A 380 kN bolt in a hole 2 mm larger than itself carries nothing until the hole closes and then rises over several millimetres of hole elongation. A 250 kN preloaded bolt is at its slip resistance in under a tenth of a millimetre and holds it until it slips. Two of these in one joint are at the same displacement, so the one that gets there first carries the load — and the weld gets there 23 times sooner than the bearing bolt does.

Two fasteners that never arrive together

Every steel code forbids adding a weld's capacity to a bolt's in one shear joint, and states it as a rule rather than deriving it. It is derivable. Two fasteners in parallel are at the same displacement rather than the same force, and a weld has ruptured at four tenths of a millimetre while a bolt in a standard hole has not yet touched the side of it.

connections · Weld group 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 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

Named alongside it

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

FrictionCoefficient of frictionPreloadConnectionFree bodyLimit stateEquilibriumBearingBolt tensionDuctilityOverturningSelf-locking

All concepts