The bearing is not a smaller bearing

The bearing Engesser says has buckled set two formulas for a shear-flexible member beside each other on a laminated rubber bearing, 600 mm across with twenty 10 mm layers, and found 0.29 MN by Engesser and 11.2 MN by Haringx, with the bearing carrying 4 MN in service. Haringx’s, it argued, was the one whose assumptions a bearing’s steel shims actually satisfy, and it is the one the isolation industry uses. Every bearing in that essay was upright.

A bearing is not upright when it matters. An isolator’s whole purpose is to shear sideways during an earthquake, by a large share of its own diameter, and carry the building’s weight while it is doing so. At that moment the top plate is no longer over the bottom one: the two circles overlap in a lens. Design codes deal with this by reducing the buckling load in proportion to the lens’s area. The question left over was whether a bearing displaced sideways is really still a Haringx member with a smaller area, or whether the offset itself changes what the slice means, and so whether the buckling check of an isolated building can be made with the bearing at rest.

The lens has two different sizes #

Two circles of diameter $D$ offset by $u$ share a lens of area

$$A_r = \frac{D^2}{2}\left(\arccos\frac{u}{D} - \frac{u}{D}\sqrt{1 - \frac{u^2}{D^2}}\right),$$

and this is the area every design guide uses. A bearing’s resistance to bending, though, comes from the pair of faces on either side of its axis, and its stiffness goes with the second moment. The lens gets narrow in the direction of the offset, which is exactly the direction that counts, so its second moment falls off far faster than its area. At half the diameter the lens has 39 per cent of the area left and only 9 per cent of the second moment. By 0.8 the area is still a tenth and the second moment has practically gone.

That gap matters because Haringx’s buckling load comes from two stiffnesses. With $S$ the shear stiffness $GA\,h/T_r$ and $P_E = \pi^2 EI_s/h^2$ the Euler load of the bending stiffness, the load is the positive root of

$$P^2 + SP - SP_E = 0.$$

For a laminated bearing, $S \ll P_E$, so that $P \approx \sqrt{S\,P_E}$: it is the geometric mean of the two. Reducing $A$ reduces $S$. Reducing $I$ reduces $P_E$. Whichever is taken as following the lens, the result must be read through that square root, which is the whole of what follows.

What the rule is close to #

The obvious interpretation of the rule is that a displaced bearing is a smaller bearing, a lens-shaped one, with everything reduced. That reading gives $S$ and $P_E$ both following the lens, and it reaches 6.6 MN at a fifth of the diameter, 2.1 MN at a half and 0.2 at 0.8. At half the diameter the design rule’s load is twice that. The rule is not that of a smaller bearing. Had anyone derived it that way they would have had to explain a load in proportion to $A_r$, when the derivation gives the square root of $A_r I_r$.

Keep the bending on the lens and leave the shear on the whole area, however, and the load comes out as 7.6, 3.3 and 0.6, which is the rule’s shape at 90, 75 and 50 per cent. There is a physical reason to think this is the right pairing. Rubber shears across its whole plan area whether or not anything lies directly above it: a layer sheared sideways by the plate on top of it is held at every point by the shim immediately below, and those shims are not offset from their neighbours by anything like the whole displacement, since the shear is shared through twenty layers. A column of rubber at the outer edge of the bearing is part of the shear path even when it no longer lies between the plates. Bending is different. A bending stiffness relies on compression through the rubber from one end plate to the other, so a part of the section with no load path through it does not count. It is reasonable, then, for shear to stay whole and for bending to move to the lens.

Run the same reasoning backwards and ask what the rule implies. If $P = P_0 A_r/A$ and $P \approx \sqrt{SP_E}$, then what the rule assumes about bending depends on what it assumes about shear. If shear follows the lens the rule needs $P_E$ to fall with $A_r/A$: 0.39 at half the diameter. If shear stays whole it needs $(A_r/A)^2$: 0.15. The lens’s own second moment, at 0.09, is less than either. So the rule cannot be derived from any section at all. It is always a little more than the lens has to give, and nearer when the shear is left whole: this is the case in which the two sit nearest over the whole of the range. Its position above the geometry is the source of its appeal to testing, and of a concern as well: reported tests on bearings have in many cases carried more than the rule says at large displacement, for reasons this lens does not contain, including load carried by the shims overhanging one another and the rubber’s own rollover. Where the rule has been checked, it has mostly been found safe. That is a statement about the tests, and it is not an explanation.

Why a margin at rest reads so large #

The load at rest is a comfortable number partly because a bearing is unusual. Made weaker on purpose is the idea of isolation in three words: the bearing is laid out to be soft sideways, and that softness comes from thin layers of rubber whose shape factor makes them very stiff vertically. It is the same shape factor, 15 here, that lifts the bending modulus into the hundreds of N/mm² and gives a stack only 257 mm tall a Haringx load nearly three times its working load, with a shear stiffness two thousand times smaller than its Euler load. Few compression members have a square-root formula standing between their load and their stiffness in this way, and because of it a margin of three at rest looks more secure than it is. It is a margin against a geometric mean, and one of the things being averaged falls as the fourth power of a width.

That is also why a larger bearing helps so much. The block that is safer for being bigger found a scale effect in the other direction for a rocking body, whose stability comes from its width in much the same way. In both, a designer has one dimension which improves nearly everything, at a cost in the mass and the money and in whatever else depended on that width staying small.

The two-spring model, and what it asks of a push #

The figures so far have frozen a section at each displacement and asked for its buckling load. Yet a real bearing arrives at its displacement by being pushed. Koh and Kelly’s two-spring model is the simplest thing that follows that path. Two rigid plates a height $h$ apart are joined by a shear spring $K_s = S/h$ and a rotational spring $K_\theta = P_E h$, and the core can shear by $s$ and tilt by $\theta$ while the top stays level. The movements of the top are $u = s\cos\theta + h\sin\theta$ sideways and $v = s\sin\theta + h(1 - \cos\theta)$ down. Under an axial load $P$ and a shear force $F$, stationarity gives

$$K_s s = P\sin\theta + F\cos\theta, \qquad K_\theta\theta = P(s\cos\theta + h\sin\theta) + F(h\cos\theta - s\sin\theta).$$

At small movements this has exactly Haringx’s quadratic as its buckling condition, and it is a nice property of the model that it does. As $u$ grows it carries the geometry along, including the moment of the load across the offset. Now let the push go on at a fixed load and see when it needs no more force to go further.

If the springs take no notice of the overlap, there is no peak at all. A bearing whose springs stay whole goes on rising, 407 kN at 420 mm, a nearly linear curve, because geometry alone costs very little at these tilts. The offset itself does not buckle the bearing. The moment of the load across the offset, the thing that sounds as though it should matter most, is resisted by a tilt of about a milliradian per 27 mm of movement. It is only by being carried on a smaller lens that the bearing’s resistance falls. That answers the second half of the question: the offset does not change what the slice means. It is the overlap, through the rotational stiffness, that does the damage.

Once the bending stiffness follows the lens, the curve bends over, at 168 mm and 109 kN. By 0.28 of the diameter a bearing with a margin of nearly three at rest has run out, along a push. That is much earlier than the frozen section’s reading, because the rotational stiffness falling as the bearing moves is itself a softening, adding to whatever Haringx’s frozen load would say. And it is lower than the rule at every displacement: 1.5 MN at half the diameter against 4.4. Taken at face value, this would say that the rule is unsafe by a factor of nearly three. It should not be taken at face value. The rotational spring here falls at once with the lens’s second moment, in a reading that treats anything beyond the lens as gone, when tests suggest something remains there, and there is a real question about how much.

The margin, read each way #

The service load decides how far a bearing may go, and the answer is much more sensitive to the reading than the load is. At 4 MN, a third of the bearing’s load at rest, the rule says it is stable to 0.55 of the diameter, 330 mm. Haringx with the lens’s bending says 0.45, 270 mm, and a push in the two-spring model says 0.28, 170 mm. Halving the load to 2 MN extends these to 0.75, 0.65 and 0.44. All agree that a displaced bearing has a much smaller margin than its load at rest would suggest, so the claim in the question has a plain answer. The buckling check of an isolated building cannot be made with the bearing at rest, on any reading but the one in which nothing changes when the plates slide apart. That reading is the only one with no physical argument for it.

The earthquake’s demand is what these are compared against. The earthquake asks for a displacement makes the point that an isolator’s displacement comes from the ground and its period and is little affected by the bearing’s strength. A bearing that has 0.45 of its width in margin and is being asked for 0.5 is not one with a little too little capacity. It is one with no further resistance, and it is a load path for the building’s whole weight. Unlike a brace or a joint, there is no reserve behind it.

What it does to a diameter #

The design consequence is one of size. A larger bearing has more room for the same displacement, and that is what is available to a designer who wants more margin, since neither the load nor the demand can change. Holding everything else fixed, a demand of 300 mm needs a 600 mm bearing by the rule and 700 mm with the bending following the overlap, an increase of 36 per cent in plan area. That is a substantial difference in rubber, with an effect on the very period the isolation was chosen for: a larger bearing with the same rubber height is stiffer sideways. The column made of two columns worked through a similar trade for a built-up member: the capacity that a shear-flexible section gains from a greater width comes with a greater stiffness in the direction it would prefer to be soft.

The figure suggests a narrow and useful conclusion. The reading matters more at small demand than one might think, at 13 per cent in diameter at 200 mm, and about 20 per cent over the rest of the range. No site can settle between them by a calculation. The argument the whole of this essay comes to is that a test of one bearing at the design displacement and load is worth more than any of them, and this is what prototype testing in the isolation standards requires.

The hand check #

$S = GA\,h/T_r$: $0.8 \times 282{,}700 \times 257/200 = 290{,}600$ N, or 0.29 MN. $P_E = \pi^2 E_r I\,(h/T_r)/h^2$, with shape factor 15, compression modulus $6 \times 0.8 \times 225 = 1{,}080$ N/mm² and bending modulus a third of it: $P_E = 439.8$ MN. Then $P = 0.145(\sqrt{1 + 4 \times 439.8/0.29} - 1) = 11.2$ MN, the earlier essay’s figure.

At half the diameter, with $P_E$ falling to 9.3 per cent, 40.9 MN, the positive root with the whole shear is $0.145(\sqrt{1 + 564} - 1) = 3.3$ MN. With $S$ reduced as well to 39 per cent, 0.113 MN, it is $0.057(\sqrt{1 + 1448} - 1) = 2.1$ MN, both as drawn. And the rule is $11.2 \times 0.391 = 4.4$ MN. Both figures can be had on a pocket calculator. So the reason for the difference between the rule and the smaller bearing is in the arithmetic: a factor of $\sqrt{0.39}$ in shear, $\sqrt{0.093}$ in bending and their product of 0.19 for the smaller bearing, against the rule’s 0.39.

What the slice cannot say #

The rubber is linear. Rubber softens with shear strain, and a bearing at large displacement has a tangent shear stiffness much lower than its initial one, stiffening again at very large strain. Nagarajaiah and Ferrell’s work found that softening matters to the critical load of a displaced bearing, and the model here does not include it, though it could.

Top plate level. Both end plates are held flat, as they are when bolted to stiff construction. A flexible support changes the effective length of the bearing and has a direct effect on its buckling, and the pinned base that is not pinned is a reminder of how far a support’s real stiffness can be from its drawn one. A bearing that is not dowelled but merely recessed can also roll out at a lesser displacement, which is another limit altogether, and one more related to the column that creeps under the arch in being a question about what a support really does.

One direction, a monotonic push. Earthquake motion is two-dimensional and cyclic. A circular bearing’s lens does not care which direction the offset goes. A square one’s does, and a bearing’s rubber accumulates damage from repeated strain. A bearing’s stiffness at its second peak is not that of its first.

No rate. The rubber’s behaviour at the rates of a real earthquake, and the heating of a lead core, are outside this model. A real isolator is usually carrying a lead core or a high-damping compound, both of which change its shear curve.

Rollover and shims. Part of the load past the lens passes through the shims overhanging one another. It is real and it is the reason measured capacities are larger at very large displacement than this lens would say. No part of the argument here can measure it. It is exactly the thing that a prototype test is there to find, and exactly why the frozen readings on the second figure should be read as lower bounds rather than predictions.

Still open: the bearing that is displaced while the building rocks #

The axial load has been held constant at 4 MN on every figure, and in an isolated building it does not stay there. The overturning of the superstructure in the same earthquake raises the load on the bearings at one edge of the plan and lowers it at the other, at the same instant that the displacement is at its largest, and it may go as far as lifting the far-edge bearings into tension. The load at the corner bearing is then highest just when its lens is smallest. Whether a bearing at a corner of a slender isolated building has its worst moment before its peak displacement or at it, and whether a bearing that has been pulled into tension and has cavitated returns to its Haringx load at all when the load swings back, is what a wall that is allowed to lift would ask of an isolation system.