The bearing that does not know what it carries

Made weaker on purpose priced isolation in general, as a period bought with a displacement, and gave the friction pendulum two sentences on the way past. A slider sits in a spherical dish; the restoring force comes from gravity; the period is set by the radius alone, $T = 2\pi\sqrt{R/g}$, with no mass in it. That removes the uncertainty about what the building weighs, which was treated as the end of the matter.

It is not the end. Having no mass in the period is the visible consequence of something larger, which also settles a different problem, and that problem is one that the floor that twists in its own time and the essays after it spent a long time on. On an ordinary building, torsion comes from a gap between the centre of mass and the centre of stiffness, a gap that is usually unplanned and that seismic codes make a whole subject of. A building on friction pendulums does not have that gap, whatever is done to its mass. The reason takes one line, and it is a statics argument.

Stiffness is weight over radius, one bearing at a time #

A slider displaced by $u$ on a dish of radius $R$ rises slightly, and the bearing’s weight $W$ wants to run it back down. For small slopes the horizontal restoring force is $W u/R$. So the bearing’s stiffness is

$$k = \frac{W}{R}.$$

Its friction force is $\mu W$. Both of the bearing’s forces are proportional to the load it happens to carry. That is the whole of the mechanical difference from a rubber bearing, whose stiffness is a property of the rubber, $GA/h$, with a slight correction if the load gets near its buckling load.

Applied to the whole structure, with mass $m$, the total stiffness is $\sum W_j/R = mg/R$, and the mass cancels:

The figure is the familiar part, and it is familiar as a convenience. A building’s weight is uncertain by perhaps a tenth, and a rubber bearing’s period moves by half of that, which matters little. The convenience is much larger for a structure whose load changes greatly: a storage tank that may be full or empty, a warehouse, or a building whose fit-out comes much later. The dish doesn’t need to know any of that, and the same bearing gives the same period at any filling.

The important step is to read the same line locally. Each bearing’s stiffness follows its own load. That is what the next figure shows, and it is a much stronger statement than saying the total is right.

The centre goes where the mass goes #

Consider a plan, 30 m by 18 m, on twelve bearings in a four-by-three grid, with the mass 3.6 m off centre across the shorter dimension. Statics gives the load on each bearing. It is a plane through the reactions whose resultant must pass through the mass, so the row nearest the mass carries more, and here the bearing shares run from 3 per cent to 13 per cent of the weight across the plan’s width. For rubber bearings, all alike, the centre of stiffness is the centre of the grid, and the building has a lever arm of 3.6 m. For pendulums the centre of stiffness is

$$y_c = \frac{\sum k_j y_j}{\sum k_j} = \frac{\sum (W_j/R)\,y_j}{\sum W_j/R} = \frac{\sum W_j y_j}{\sum W_j} = y_m.$$

The radius cancels, and what is left is the definition of the resultant of the bearing loads, which statics has already placed at the mass.

The same holds for friction: a total friction force of $\mu W$, shared in proportion to load, also has its resultant at the mass. There is therefore nothing for a lever arm to work with. A single sentence does the same job for every load case at once. For a plate that has not yet twisted, all the bearings share one displacement vector and one velocity vector, so each bearing’s force is its own load times one vector that all of them have in common, and the resultant of a set of parallel forces in proportion to the loads goes through the centre of the loads. Since the plate does not twist at the start, and has no torque to make it, it never twists. The result does not depend on the record, on the amount of friction, or on how far the slider has travelled.

That is not a claim about small errors. There is no approximate step in the argument, as long as the plate is rigid and the friction is proportional to the load.

The two long edges through one record #

A computed record is a sharp test of that claim, and it is a fair one because the integration knows nothing of the argument. Rubber bearings with 15 per cent damping and the same 3 s period carry the two long edges apart. They have a torsional mode close to the translational one, as they would on any symmetric plan with bearings spread across it. The pendulums show one line for both edges.

The picture does not say the pendulums are simply better: they differ from the rubber bearings in more than one way. Here the friction takes more energy out than the 15 per cent of viscous damping does, so the two peak travels shouldn’t be compared directly. That comparison belongs to a later figure. The separation of the edges is the point here.

Over a set of records, the rubber plan’s corner bearing moves 10 per cent further than the mass at 1.8 m of offset and 17 per cent further at 3.6, and that number holds up with changes in the damping: it stays between 1.16 and 1.20 for damping anywhere between 5 and 30 per cent. The pendulum’s reads 1.000 at every offset. Seventeen per cent is not an alarming number by the standards of a fixed-base building, whose torsional amplification can be far worse. But isolators are the most expensive items in the structure, and their stroke is sized to the travel, so each percentage point goes straight into the size of the dish and the width of the moat. A designer of rubber bearings can arrange a layout with the stiff bearings under the heavy part, and that does work, but it has to be recalculated each time the mass moves, whereas the pendulum does it automatically. It also takes away much of what a fixed-base plan has to answer for, including the cross term a modal combination adds when a torsional mode sits close to a translational one, since a mode that is never excited has nothing to combine.

What rocking does to it #

There is an obvious objection. The load on each bearing is the static share only when the building stands still. During shaking, the inertia force acts at the mass’s height $h$ above the bearings and overturns the building, adding load on the side the building has moved towards and taking load from the other. So the load centre is on the move, and it is reasonable to expect the centre of stiffness to go with it, and with it the proof.

The centre of stiffness does move, and the first half of the objection is right. Take the shear $\mathbf{F}$ acting on the building at height $h$. The overturning moment it causes shifts the resultant of the bearing loads by

$$\mathbf{d} = \frac{h}{W}\,\mathbf{F}.$$

Now look at what that gives. The shift is parallel to the force. The bearing force resultant, $\mathbf{F}$, acts through a point moved along its own line of action. A torque about the mass is $\mathbf{d} \times \mathbf{F}$, and the cross product of a vector with a multiple of itself is zero. The building can rock as much as it pleases, and a force pushing along its own line, no matter where on that line it acts, has no lever arm about a point on the same line.

The computation includes the overturning and shakes both directions at once, so the shear’s direction wanders through the record, and that is the case that should break it. With the mass central, the twist’s contribution at the plan’s corner is under a tenth of a millimetre up to 20 m and half a millimetre at 40 m, which is a remarkably small number for a building whose height is more than twice its width. With the mass 3.6 m off centre it remains under a millimetre up to 20 m, and then starts to climb: 3 mm at 30 m, 8 mm at 40 m.

The argument above explains that departure too. The proof needs every bearing to be able to take its share of the shift, so it depends on all of them staying in compression. In a plan with the mass 3.6 m off centre, the bearing row on the light side begins with only 3 per cent of the weight, so the first tilt takes it to nothing, the slider lifts, and the distribution stops being a plane. In the computation, a bearing’s load is cut off at zero. Below the height at which the lightest bearing lifts, the twist is what the integration’s own small errors give. So the failure is a contact failure, and the mechanism that protects the plan from torsion goes on working until the bearing has physically left the dish.

It would be a strange result for a rubber bearing, whose stiffness would not respond to any of this. Here, the fact that the bearing has no opinion of its own about its stiffness has turned out to be its strength.

What friction buys, and where the argument ends #

Friction is the second half of what a pendulum does. A bearing with a coefficient of 5 per cent starts to slide when the base shear reaches 5 per cent of the weight, and until then it behaves as a fixed base. Once sliding, it dissipates $4\mu W u$ a cycle, and the loop is a parallelogram of height $2\mu W$ on a sloping line of $W/R$.

The travel falls steadily with friction: from 344 mm at a coefficient of 0.02 to 112 mm at 0.09. The base shear falls to a low point near that same value and then rises, because beyond there the force in the bearing is nearly all friction, whatever the record does, and a larger friction simply is a larger force. At that point the curve for the isolator has moved towards a stiff structure with a fuse. This curve is a cleaner account of the choice than a general preference for low friction, and a real dish’s coefficient comes from its material and contact pressure and is not free to go anywhere. The roller that is not a roller found the same trouble: a PTFE coefficient falls with pressure, and it doubles in the cold.

There is also a price in what the building does afterwards. A slider stops where the restoring force no longer exceeds the friction, at any displacement inside a band of $\pm\mu R$. At 5 per cent and 2.24 m that band is 112 mm either way. The figure’s records end with a mean residual of about 20 mm, well inside it, but after a big earthquake the building can be left noticeably off-centre, and it stays there, with no restoring force available to return it. Rubber bearings come back almost entirely by themselves: 4 mm of residual in the same set of records. The codes check this by demanding a minimum of restoring stiffness against friction, and that is why real pendulums cannot have an arbitrarily large radius, even though a longer period is otherwise attractive.

That leaves the last and most interesting way for torsion to come back, since friction is also what the argument depends on. The proof required the friction force at each bearing to be its load times one coefficient. Suppose the coefficient varies from one bearing to another. The friction force’s resultant is then no longer at the mass, and a torque appears, independent of where the mass is.

So the identity has a limit, and it isn’t small: a linear variation of ±20 per cent across the plan, which a batch of bearings can easily have between one end of the plan and another, or a bearing that has run cold beside one in the sun, gives 4 per cent, with no eccentricity at all. At ±40 per cent it is 8 per cent. That is still less than half of what the rubber plan had for a large offset. It is nevertheless a real and unavoidable twist, and a surprising thing to find: on rubber, the source of torsion is where the mass is, and on a pendulum, it is what the contact surfaces did. The friction’s dependence on pressure raises a further issue. A coefficient that falls as the pressure rises makes the heavy bearings relatively slippery, so friction is no longer proportional to load, and the mass offset’s contribution comes back in a small way through a gap that statics cannot close. Bearing tests are done to narrow this, and EN 15129 sets a tolerance on the friction of a set. Now it is clear what that tolerance is for.

What the stroke is worth #

Put the ratio into hardware. A dish is sized for the travel of its most-travelled bearing, plus the residual a previous earthquake may have left, plus a margin, and the whole plan’s bearings are normally made alike. On the rubber plan in the set of records above, 3.6 m of offset takes the corner bearing from 371 mm to 434 mm, a 63 mm increase on the radius of the stroke, so 126 mm across the diameter of a bearing whose plan area grows with the square of it. The moat around the building, the flexible services crossing it, and the stair and lift connections all pick up the same 63 mm. A designer can claw that back, by moving stiff bearings towards the heavy side, or by adding lead cores where they are needed, and all of those moves are valid as long as the mass stays where it was assumed to be.

The pendulum’s corner bearing stands at 227 mm and it has no such term. Its travel is lower here because 5 per cent of friction takes more energy out of these records than 15 per cent of viscous damping does at this level, which is a separate choice from the one being examined. The point is the offset’s share of the stroke, and that is nothing, whatever the offset turns out to be. For a tank farm, an archive or a hospital whose equipment is rearranged every few years, the mass is a moving target through the building’s life, and it is the one quantity that does not have to be recalculated when it moves.

What the rigid plate takes as given #

The superstructure is rigid. A real building above a pendulum has its own modes, and a flexible floor doesn’t share one displacement across its bearings. The proof’s first line, about one common displacement vector, then holds only in an average sense. A stiff diaphragm at the bearing level is therefore a working requirement, and the floor is a beam lying down is the check on whether one is stiff enough to count. The same goes for the mass itself: the mass here sits at one level, and a real building’s mass is at many.

The slope is small. $k = W/R$ is the small-angle form, and at 300 mm of travel on a 2.24 m dish the slope is 0.13, so its error is about 1 per cent. At strokes near a metre, it matters.

The dish has one radius. Double and triple pendulums use several surfaces and several friction values, to let the effective stiffness and damping change with the size of the earthquake. Those are in effect pendulums whose coefficient changes with travel, and the one-line proof still applies to them only if each stage is reached by all the bearings at once, which needs a load-independent activation that real bearings approach but don’t have exactly.

Vertical shaking is left out. A vertical acceleration changes the weight on all the bearings at once. Since both the stiffness and the friction follow weight, it modulates the whole system by one factor and does not disturb the proof, but it does change the period and the friction through the record. A fair check would need both.

The picture is a mean of a few records. The ratios above are means over four synthetic filtered records, and an individual record will depart from them. A real site’s records have long-period pulses that favour a much larger travel, and the ground has a period of its own on that side of the argument.

Still open: when one bearing has been replaced #

Bearings are inspected and sometimes exchanged. A new slider next to bearings that have run for thirty years has a different surface, and so a different coefficient, in a pattern that is local rather than linear across the plan, and an old bearing has probably gone stiffer in the cold while a new one has not. Whether a single replacement in a twelve-bearing plan gives any twist worth measuring, whether its effect is largest at the corner or at the centre of an edge, and whether a programme of replacing bearings should go corner by corner or side by side, comes down to how the local friction anomaly is balanced by the plan’s torsional stiffness, which itself is set by every other bearing’s weight over the radius. The one-way linear pattern used here doesn’t have enough detail to answer that question.