Seventy-five per cent each way

A precast unit sits on a steel bearing plate, held in place by nothing but friction. It carries 1000 kN, the surfaces are good for μ = 0.4, and so the seat can resist 400 kN of horizontal force before anything moves. Braking and temperature ask it for 300 kN along the unit. Wind asks it for 300 kN across. Each check is written on its own line, each comes out at 75 per cent, and the unit slides off its seat.

Nothing in either calculation is wrong. What is wrong is that they are two calculations, and the contact only ever sees one force.

The law bounds a length, not two components #

Every earlier treatment of friction in these essays worked in a plane. A block on a slope, a leaning bar against a wall, the plates of a preloaded joint — the tangential force had one direction, up or down the surface, and the bound was an interval: anything from −μN to +μN, and nothing in statics prefers a point inside it.

A real contact has a surface, and a surface has two tangential directions. The Coulomb inequality does not say that each component of the friction force is at most μN. It says that the friction force is, and a force in a plane is a vector with a length:

$$\sqrt{F_x^2 + F_y^2} \;\le\; \mu N$$

The set of tangential forces that satisfies that is a disc of radius μN, centred on zero. The pair of checks

$$|F_x| \le \mu N \qquad |F_y| \le \mu N$$

describes the square that circumscribes the disc. The two agree along the axes, where one of the components is zero, and they disagree everywhere else. At the corners of the square the disagreement is largest: a demand at 45 degrees can reach $\mu N \sqrt 2$ before both checks fail, while the contact gave up at $\mu N$.

That is the whole of the defect, and it is not small. The square overstates the contact by 41 per cent at its corners. Put the other way round, a demand at 45 degrees is at the limit when each of its components is at $1/\sqrt 2$ of the capacity — seventy-one per cent, not a hundred.

The figure is worth reading as a statement about what a check is. A capacity check compares a demand with a resistance, and it is only as good as its agreement about what the demand is. The two one-direction checks each compare a component with a length, which is a comparison between two different kinds of object, and it happens to give the right answer on the axes because a component and a length coincide there.

In space it is a cone, and two checks draw a pyramid #

Add the normal force back and the picture becomes the one drawn in the treatment of friction as an inequality: the reaction the contact supplies is the normal force plus the tangential one, and the inequality caps how far it can lean from the normal. In a plane that lean is limited to a wedge of half-angle $\phi = \arctan\mu$. With two tangential directions it is limited to a circular cone of the same half-angle, because the cap is on the length of the tangential part and a circle is the set of equal lengths.

The two one-direction checks replace the cone with a square pyramid. Along its faces the pyramid leans exactly as far as the cone does. Along its diagonal edges it leans further — $\arctan(\mu\sqrt 2)$ — and the difference is the room those two checks give away.

The pyramid is not an invention for the sake of a figure. It is what a design calculation produces whenever it treats the two horizontal directions of a support as separate load cases, checks each against μN, and never forms the resultant — which is the natural thing to do when the two directions carry different actions with different factors, drawn on different sheets. Longitudinal forces on a bearing come from braking and temperature; transverse ones from wind and from the lateral load path of the deck. Nothing in either set of arithmetic mentions the other.

What the check should have been #

The repair is not a new coefficient or a larger margin. It is to form the resultant inside each load combination, before it is compared with anything:

$$\sqrt{H_{x,d}^2 + H_{y,d}^2} \;\le\; \mu N_d$$

where every term belongs to the same combination. That last condition is where the work is. The longitudinal and transverse actions on a support rarely come from one source, so in any given combination one of them is usually the leading action at its full factor and the other an accompanying one reduced by its combination factor. The combination that governs the resultant is therefore not necessarily the one that governs either direction alone. A case with braking leading and wind accompanying can produce a smaller longitudinal force than the braking-only case and a smaller transverse force than the wind-only case, and still a larger resultant than both.

The normal force belongs to the same combination too. A favourable permanent load that presses the contact together is taken at its lower value when friction is resisting, so the disc that each resultant is compared against is itself smaller in the combinations where the horizontal actions are largest.

When the two directions cannot be put into one combination — because they are checked by different parties, or at different stages — the straight-line rule described below is the safe substitute: add the two utilisations and hold the sum under one. It never overstates the contact, and it costs less than the square gives away.

A slope has already spent part of the disc #

The seat was flat, so every horizontal force it resisted was applied. Tilt the contact and the weight does some of the pushing. A block of weight $W$ on a slope of angle $\alpha$ presses on it with $N = W\cos\alpha$ and asks it for $W\sin\alpha$ straight down the fall line, before anything else is applied at all.

That demand sits inside the disc like any other. On a slope of 20 degrees with μ = 0.5, a block of 100 kN presses with 94.0 kN, the disc has a radius of 47.0 kN, and gravity is using 34.2 kN of it — 73 per cent. By the planar check the block stands with a comfortable margin, and it does.

What the planar check does not say is how much margin is left across the slope. The answer is the other leg of a right-angled triangle whose hypotenuse is the radius:

$$Q_{\max} = \sqrt{(\mu W\cos\alpha)^2 - (W\sin\alpha)^2}$$

which here is 32.2 kN — 69 per cent of μN, not all of it.

The block in that figure is the precast seat in another costume, and the costume matters. On the flat seat, the designer at least knew that two actions were being applied and chose to check them apart. On the slope, one of the two demands is not an action anyone wrote down. It is the weight, doing what weight does on a slope, and it has quietly used most of the contact before the sideways push arrives. A stair unit sitting on a sloping seat, a pipe resting on an inclined support, and a machine skid on a cambered slab are all in the same position, and in each of them the check that is usually made is the one along the slope.

What is used one way is not left the other #

Take the argument one step further and it stops being about a particular block. A contact using a fraction $u$ of its disc in one direction has $\sqrt{1 - u^2}$ of it available in the perpendicular direction, whatever the coefficient, the weight or the angle.

The quarter-circle is unforgiving near its ends. Using 50 per cent one way leaves 87 per cent the other, which sounds like almost everything, but using 90 per cent leaves 44 and using 95 leaves 31. A contact that is nearly exhausted in one direction is weak in every direction, and the two checks cannot see it, because each of them is looking at a line through the centre of the disc.

The curve also explains why the cautious alternative is cautious. A straight-line interaction — the two utilisations added, and the sum held below one — sits inside the quarter-circle everywhere. It gives nothing away at the ends and gives away most in the middle: $\sqrt 2 - 1$, or 0.414 of the whole capacity, at 45 degrees. That is the same $\sqrt 2$ the square overstates by, arriving from the safe side.

The shape is familiar from elsewhere in structural design, and the familiarity is the useful part. An I-section under axial force and moment has an interaction curve between two ways to fail, and the straight line inside it is conservative for exactly the reason it is conservative here. A fillet weld is stronger across than along because its throat resolves one force into two components against a criterion on their combination. And a yield criterion is the three-dimensional version of the same question, where von Mises draws a circle and Tresca a hexagon around the same stress states. Friction’s disc is the simplest member of the family: two components, one length, and no material behaviour to argue about.

There is a place most people have met this curve without calling it statics. A car’s tyre is a friction contact with a disc of available force, and braking uses some of it. A driver who brakes hard in a bend has spent the part of the disc that was holding the car on the curve, and the car leaves the road along the combination of the two demands — which is the next point, and the one with the most practical consequence.

It leaves along the demand, not along the push #

A plane friction problem knows which way the body will go if it goes: along the one tangential direction the plane has, one way or the other. With two directions available, the direction of slip is a result.

For a rigid block on a contact with the same coefficient in every direction, impending slip runs along the whole tangential demand, and the friction force points straight back along it. On a slope, that demand is the weight’s component down the fall line added to whatever is applied across. The block therefore leaves diagonally downhill, at an angle to the fall line that grows with the push.

The end of each curve has a closed form that is worth stating because it has nothing in it but the fraction already used. At the push that just starts the slide, the demand is exactly the radius of the disc, its fall-line part is $u$ times the radius, and so it leans from the fall line by $\arccos u$. A slope using 90 per cent of its grip, nudged sideways, leaves 25.8 degrees off straight downhill. A slope using a quarter of it has to be pushed hard, and leaves nearly — never quite — in the direction of the push.

That settles a detail question that looks as though it needs a test. A stop or a keeper provided to prevent a unit being pushed off a sloping seat sideways has to catch a body travelling mostly downhill, and a keeper placed only across the slope, in the direction the push came from, is placed for a movement that does not happen. The seating length that matters is measured along the diagonal the demand points in.

An unknown direction at every contact #

The unknown direction is the new thing in this problem, and on a single contact it is harmless because it is just the direction of the resultant demand. On several contacts it stops being harmless.

Consider a plate held to its support by friction at a number of points — a slip-resistant bolted joint is exactly this — and twist it. Each contact can supply a force of length up to μN in any direction, and none of the directions is given. The plate turns, if it turns, about some point; each contact slides perpendicular to its radius from that point; each friction force points back along its own slip. The problem is solved when those forces balance the applied load.

That is word for word the procedure a bolt group under eccentric shear is solved by at its ultimate load, where every bolt’s force is perpendicular to its own radius from the instantaneous centre and the centre is found by search. The bolt method was written for bearing bolts deforming in their holes. It applies to a slip-resistant joint for a reason that has nothing to do with bolts: friction at a point is a force of bounded length and unknown direction, which is precisely the property the instantaneous-centre method needs its fasteners to have.

The same unknown also explains why the three-dimensional problem is harder than the plane one in a way that has nothing to do with the arithmetic. In the plane there are three equations of equilibrium and each friction contact adds one unknown force with one inequality on it. In space there are six, each contact adds two tangential components with one inequality on their combination, and a body on several contacts can also turn about its normal — a twist that the cone at each point does not bound at all, because a single point supplies no moment about its own normal.

The free body the disc belongs to #

Cut the seat free on the plane of contact and three things cross the cut: a normal force and two tangential components. Every result above is a statement about those three numbers, and only one of them is constrained by the friction law.

The disc is the constraint on the two tangential components together. The cone is the same constraint written with the normal force included, as a bound on the direction of the whole reaction. The two one-direction checks are the same free body with the constraint applied to each tangential component as though the other were absent — which is a different constraint on the same forces, and a weaker one.

Nothing about the cut requires the contact to be small. A bearing plate, a skid, a precast seat and the plates of a friction joint all have a contact area, and the area does not enter the law any more in three dimensions than it did in two. What the area does allow is a distribution of friction forces over the surface, and that is where the rigid-body picture gives out.

What a single disc leaves out #

The coefficient is the same in every direction. Many real surfaces are not. A machined face, a ground plate with a lay, timber along and across its grain, and a sliding bearing with a guide all have a larger coefficient one way than the other, and the disc becomes an ellipse. The one-direction checks can then be right on both axes and wrong everywhere between them by a different amount.

The normal force does not depend on the tangential ones. On the flat seat it does not. On anything that tilts, rocks or carries a push with a vertical component, it does, and the disc changes size as it is being used — the coupling the tipping-or-sliding comparison had to set aside for the same reason.

The contact is a point, or a rigid plate that behaves like one. A long contact under a twist slips at its ends while its middle has not, and a flexible plate under a tangential load develops a distribution of friction that no single disc describes.

Nothing resists a twist about the normal. A point contact supplies no moment about its own normal, and an area contact supplies one that depends on how the pressure is distributed. A seat that is asked to resist a torque about the vertical is a different problem, and its answer is an integral over the area rather than a disc.

The coefficient is static and single. Whether the contact will stick or slide is decided by one number here, and real surfaces have a separate coefficient for starting and for continuing, which in two directions makes the disc a pair of discs with a gap between them.

Still open: whether the bound theorems survive friction #

A disc is a yield surface in all but name, and it invites the question that yield surfaces always invite: can a frictional structure be analysed by bounding its collapse from both sides, the way a plastic frame can?

The honest answer is that it cannot, and the reason is geometric. The bound theorems require that a structure at its limit moves in the direction normal to its limiting surface. For a friction cone, the outward normal to the surface has a component pointing away from the contact — it says a sliding body should also separate from what it slides on. Real contacts slide along their surface and do not lift. The flow is non-associated, exactly as it is for a sand whose friction angle is larger than its dilation, and the theorems that make a lower bound safe are not available for it.

What remains is weaker and still useful: the frictionless and the fully bonded versions of the same structure bracket it in some cases, and the set of equilibrium states that satisfy every contact’s disc can still be written down. Which of those states a real structure is actually in, and how it got there, is not a question about discs at all. It depends on the order in which the loads arrived.