Structural Engineering & Statics
where the load goes
Generator

A preloaded joint, before and after it slips

Rendered here at the parameters it defaults to, with every essay that calls it — which is the same list as the blast radius of changing it.
A preloaded joint, before and after it slips. Two preloaded bolts at 172 kN each, on one friction face at μ = 0.5. The joint carries 172 kN by friction with the bolts in tension and not in shear at all; past that it slips into bearing and carries 250 kN with the bolts now in shear. Two different mechanisms, one joint.

A preloaded joint, before and after it slips. Two preloaded bolts at 172 kN each, on one friction face at μ = 0.5. The joint carries 172 kN by friction with the bolts in tension and not in shear at all; past that it slips into bearing and carries 250 kN with the bolts now in shear. Two different mechanisms, one joint.

14 essays call slip-curve. The drawing above is what it returns with no arguments at all; every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about.

Where it is called

Changing this generator changes every one of these figures.

Bearing and tear-out against end distance. A 20 mm bolt in a 10 mm plate. Below 165 mm of end distance the bolt tears a channel out to the end and the capacity is proportional to that distance; above it the plate crushes in front of the bolt and the end distance stops mattering. At 40 mm the capacity is 52.12 kN and the mode is tear-out. Connections

The hole that goes oval, and the one that tears to the edge

A bolt pressing on the side of its hole either crushes the plate in front of it or shoves a channel of metal out to the end. Which one happens is decided entirely by a distance that is usually set by a minimum in a table.

 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. 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.

 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. Equilibrium

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.

 The coefficient is a slope, and that is why it can exceed one. The crack magnified: two rough faces, drawn as a sawtooth at 54° to the plane. Sliding one over the other cannot happen without lifting it, so a shear displacement forces a separation in fixed proportion — the tangent of that angle, which is the number written down as a coefficient of friction and here is 1.40. The bars crossing the plane are stretched by the separation and clamp the faces back together; they are not carrying the shear, they are supplying the normal force that lets the roughness carry it. A bar that is not anchored on both sides supplies nothing. Internal forces

Shear across a crack that is already there

Every shear calculation in this collection starts from an uncracked solid — a principal stress, a shear flow, a diagonal tension. This one starts after the crack, on a plane with no tensile strength at all, and the coefficient it uses is not a coefficient of friction. It is the slope of the roughness.

 Between two beams and one, and much nearer one. How composite a beam is, against the one dimensionless group that decides it: αL, where α² = K·EI∞/(EA*·EI₀). At αL = 0 the layers slide freely and the beam is two beams; past about 20 the connection is stiff enough that the last per cent is unbuyable. The beam drawn sits at αL = 15.8 and is 96% composite, deflecting 36.7 mm against 33.9 for full interaction and 110 for none. The curve is steep where a real design sits, which is why halving the number of studs does not halve anything. Internal forces

Half the studs, and most of the beam

Bonding two layers together quadruples the stiffness of the pair. A real connection is a row of studs that deform, so the layers slide a little and the beam sits between the two answers — but not halfway between, and the number that decides where is a single dimensionless group.

 The end bolts do the work and the middle ones very nearly nothing. A lap of 8 bolts at 70 mm pitch transferring 800 kN between two plates, with the force each bolt actually carries drawn above it and the flat line a division by the bolt count would have given drawn behind. The end bolts carry 1.09 of their nominal share and the middle ones 0.94. The reason is not in the bolts: at the leading end the first plate is carrying everything and the second nothing, so the two strain at different rates and the slip between them is largest there. In the middle they strain alike, there is almost no slip, and a bolt with no slip across it transfers almost no force. The mean over the worst is 0.918, and the end bolt has to slip 1.36 mm before the rest catch up. Connections

The bolts that do not share

Every bolted connection in this collection has divided a force by a number of bolts. That is right for a short joint and wrong for a long one, and the reason has nothing to do with the bolts — it is that the plates they join are elastic, and stretch by different amounts at different points along the lap.

 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. Connections

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.

 Five millimetres short, and a hundred kilonewtons in every member. An X-braced bay 6 m by 4 m in which one diagonal was fabricated 5 mm short, with the force that leaves in every member. Nothing is applied to this frame. The forces are the self-stress state the frame's one redundancy supports, scaled so that the diagonal is pulled back to the length it should have been: tension in both diagonals at 100 kN, compression in the four members round the outside, and the whole set in equilibrium with nothing. That is 25% of the force the diagonal was sized to carry, and it is there for the life of the structure. Take one diagonal out and the frame becomes determinate: the short member then simply puts the joint somewhere else, and the structure is in the wrong place instead of under stress. Redundancy is bought, and this is the price. Connections

Built to the wrong length

A redundant structure's members do not have independent lengths. Choose all but one and geometry decides the last, so a member made a different length has to be pulled or pushed into place — and the force required stays in the structure for as long as the structure does. Nothing has been applied to it, there is no load case and no factor, and the members are carrying real force.

 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. Equilibrium

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.

 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. Equilibrium

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.

 The studs are evenly spaced and the demand is not. The force per unit length the shear connection carries along half of a 12 m composite beam, from Newmark's solution. It is largest at the support — 282 N/mm — falls to nothing at mid-span, and averages 156: the end studs are asked for 1.81 times the mean. Studs are nevertheless placed at a uniform spacing, and the justification is the one the variable-angle truss uses for its stirrups — a ductile connector sheds what it cannot carry to its neighbours, so the uniform distribution is a plastic redistribution and not a description of the elastic state. Internal forces

The connection is busiest where the beam is not

A composite beam's studs are spaced evenly along it and the demand on them is not even at all. It peaks at the supports, where the bending stress is nothing, and falls to zero at mid-span, where the section is working hardest — so the connection is designed from a diagram nobody looks at.

 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. Connections

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.

 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. Connections

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.

 The bolt force is the larger of two lines. What an M20 bolt in a 25 mm tee flange actually carries, against the tension applied to the flange. Preloaded to 171 kN it starts there and climbs at Φ = 0.185 — the bolt's own stiffness over the bolt's plus the clamped plates', 857 against 3781 kN/mm — so 18 per cent of every kilonewton applied reaches it and the rest is unloading the contact. At 138 kN the contact runs out and the line joins the one an ordinary bolt has followed from the start, climbing at 2.12. The two lines meet, so the strength is the same either way; what differs is the slope, by a factor of 11.5. The flange's own mechanism is at 172 kN, comfortably past the crossing. Connections

The bolt that was already stretched

Prying is a lever that needs the flange to lift before it can act, and a preloaded bolt does not let it. The bolt carries a fifth of every kilonewton applied until the plates part, and everything after that is the ordinary calculation — which is why preload changes the fatigue answer by a factor of a thousand and the strength answer by nothing at all.

The library, page 5 of 7 — where slip-curve sits