Structural Engineering & Statics
where the load goes
Concept

Friction — where it appears

A contact force that is whatever it needs to be up to a limit proportional to the normal force pressing the surfaces together. It is a reaction rather than an action: it takes whatever value equilibrium requires until it reaches its limit, and only then does the surface slide.

Named by 20 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 Five loads behind one wall, and the water is the biggest. The horizontal pressure on a 6 m wall retaining soil at 18 kN/m³ with a friction angle of 30°, a surcharge of 10 kPa and the water table 2 m down, drawn once as the profile the wall feels and then once per term. The terms are surcharge 20.0 kN/m at 3.00 m, soil above water 12.0 kN/m at 4.67 m, soil at the water table 48.0 kN/m at 2.00 m, submerged soil 27.2 kN/m at 1.33 m, water 78.5 kN/m at 1.33 m, and they sum to 185.7 kN/m — matched to 7e-8 by integrating the drawn profile numerically. The largest single term is the water, at 78.5 kN/m: water has no shear strength, so its coefficient is exactly one where the soil's is 0.333, and it acts on top of the soil's effective stress rather than instead of it. The combined resultant sits at 1.90 m above the base, 0.317 of the height rather than the third point at 2.00 m that a pure triangle would give.

The load that depends on what carries it

Every other load in this collection is a number the structure is given. Retained soil is not — it pushes with a fraction of its own weight, and the fraction is decided by how far the wall moves. Six millimetres of retreat on a six-metre wall takes a third off the load, and being held still puts it back.

equilibrium · Lateral pressure A tank grows without limit; a silo stops. Vertical pressure against depth in a 8 m silo of a solid weighing 9 kN/m³, beside the straight line a liquid of the same weight would have produced. The free body is a slice: its own weight in, the wall friction out, and the friction is μK times the vertical pressure that generates it. The result saturates at γR/μK = 89 kN/m² and reaches 63% of it at one characteristic depth, R/μK = 9.9 m. At the base the pressure is 85 kN/m² against a liquid's 270 — 69% less — and the wall has taken 69% of the stored weight down with it. The exponent is the capstan's, and for the same reason.

The pressure that stops growing

A tank of liquid presses harder the deeper it gets, without limit. A silo of grain does not. Wall friction carries part of the weight, the pressure that generates the friction is proportional to the pressure being carried, and the equation that follows is the one that describes a rope round a bollard.

equilibrium · Silo pressure The line, and the stone it has to stay inside. A masonry pier 9 m high, 1.6 m thick at the top and battered 12% on its outer face, taking a thrust of 40 kN per metre of run at 25° to the horizontal. The line drawn through it is the locus of the resultant on each horizontal cut: everything above the cut is the free body, and the resultant's position is the moment divided by the vertical force. The dashed pair is the middle third, inside which no tension is implied anywhere on the joint. The line stays inside the stone throughout and reaches the base at 0.503 m from the centre, against a half-width of 1.34 m — but outside the middle third, so part of the base joint is open and the toe is carrying a triangle. Nothing about the strength of the masonry appears anywhere in this figure, and that is the point.

The weight that makes it safer

Every load in this collection makes a structure worse. A pinnacle does not. A masonry pier fails when the line of compression leaves the stonework, and adding weight at the top rotates that line back towards the vertical without adding anything the pier cannot carry — so the stone is not being strengthened, it is being aimed.

structures · Buttress Two drawings of one deck, and they are not the same structure. A 112 m viaduct on five supports, articulated two ways. Above, the fixed point is at the left abutment: the far end has to be given 45 mm of movement, and the friction of every sliding bearing runs one way, so the fixed support takes 660 kN before any wind or braking is applied. Below, the fixed point is at the middle pier: the largest joint halves to 22 mm and the friction now cancels across the fixed point, leaving 0 kN. The movement arrows are drawn at 900 times the scale of the deck, because a 45 mm movement on a 112 m span is thinner than the line the deck is drawn with. Nothing about the deck, the loads or the ground has changed between the two.

Where the structure is allowed to move

One drawing decides how big every movement joint on a bridge is and where every horizontal force goes, it takes an afternoon, and it appears on no calculation sheet. Move the fixed point from an abutment to the middle pier and the largest joint halves and the horizontal force on that support drops from the whole of the friction to none of it.

structures · Articulation 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 force nobody applied, and the speed it wins at. Lateral force per unit weight for a vehicle on a 400 m curve, against speed. The rising curve is what the free body demands — v²/gR, which is the body's own acceleration written on the other side of the equation — and the flat line is what 6.0° of cant supplies from the weight. They cross at 73 km/h, which is the speed the curve was set out for; below it the deficiency has the other sign and the rail is pushed the other way. The upper line is overturning, at b/2h = 0.399 — and there is no mass in that number, so a loaded vehicle and an empty one go over at the same 160 km/h and only the height of the load decides. At the 108 km/h drawn the deficiency is 0.124 of the weight, which is 49 kN on this 40 tonne vehicle.

The force that is really an acceleration

Every other load in this collection is applied by something. This one is applied by nothing at all — it is the body's own acceleration, written on the other side of the equation so that statics can be used on a problem statics has no business with. The move is legitimate, it is a hundred and eighty years old, and it is exactly half done more often than it is done.

equilibrium · Centrifugal load Two identical pipes, and one carries three times the other. Load per metre on a buried conduit against the depth of cover, in trench widths, with the weight of the prism of soil directly above it drawn between them. A conduit laid in a narrow trench is stiffer than nothing and softer than the sides: the backfill settles relative to the undisturbed ground, the friction on the trench walls acts upward, and the conduit gets 64% of the prism. Lay the same conduit on the ground and build an embankment over it and it is now stiffer than the fill beside it, the interior prism settles less, the friction acts downward, and it gets 172% — a factor of 2.71 between two pipes with nothing different but which way the ground moved. The equation is Janssen's, the same one a silo wall obeys, with a trench for a silo; both curves start on the prism line, because with no depth there is no shear to redistribute anything. This is why a flexible pipe is buried rather than a rigid one: making the conduit weaker moves it down the page.

The pipe decides what the soil weighs

A buried conduit is not loaded by the soil above it. It is loaded by whatever share of that soil the relative movement leaves it — and which way the shear on the sides of the prism acts depends on whether the conduit settles more or less than the ground beside it. Two identical pipes under identical fill, one carrying two thirds of the prism and one carrying nearly twice it.

equilibrium · Soil arching The circle is searched for, and the first guess is 39 per cent optimistic. The same slope with 81 trial circles evaluated, each one through the toe and each one giving its own factor of safety. There is no equation whose solution is the answer: the slip surface is a shape the ground chooses, so the calculation is a search over shapes and the answer is the smallest number found — 1.191 against 1.650 for the circle a first guess puts through the toe from above the middle of the slope, which is 39 per cent optimistic. A slope analysis that reports one circle has reported nothing.

The surface that has to be searched for

Every other check in this collection is made at a section somebody drew. A slope has no section — the failure surface is a shape the ground chooses, so the calculation is a search over shapes, and the answer is the smallest number found rather than the solution of anything.

equilibrium · Slope stability 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 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.

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

The pin that is not a point

Every line of action drawn so far passes exactly through a pin's centre, which is true of a frictionless pin and of nothing else. A real one carries its force tangent to a small circle instead, so a link's line is a band, a construction's answer is a range, and a support drawn as a hinge hands a couple of a hundred kilonewton-metres to whatever it is pinned to.

equilibrium · Graphic statics

The pull that Johansen left out

Johansen's mechanisms treat a timber fastener as a beam in a bed of crushing wood, bent and pushed sideways and nothing else. A real fastener that tilts across the joint is also pulled along its own axis, and if a washer or a thread resists the pull, it clamps the two members together and adds to what the joint can carry. The addition is a quarter of the axial resistance, it goes only to the modes in which the fastener tilts, and it is capped by fastener type — nothing for a dowel, a quarter for a bolt, all of it for a screw — which is how a screw keeps gaining strength past the thickness at which Johansen's capacity stops.

connections · Dowel yield

Named alongside it

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

Free bodyCoefficient of frictionSlip resistanceBearingEquilibriumOverturningLimit statePreloadConnectionRestraintSliding bearingThermal movement

All concepts