The bearing Engesser says has buckled

The column made of two columns found that a built-up column has two ways to deflect — its chords bend, and its lattice shears — and that the two flexibilities add. A column whose bending would buckle it at the Euler load $P_E$ and whose lattice has a shear stiffness $S$ buckles at Engesser’s

$$\frac{1}{P} = \frac{1}{P_E} + \frac{1}{S},$$

which is why a battened column reaches only 23 per cent of its own Euler load — a column strong enough in its parts and still able to fall over as a whole. In a paragraph about where its model stops, that essay mentioned a second formula, Haringx’s, which “gives a higher answer” and differs in “what it assumes about how the shear force follows the deformed member”, and noted that the difference matters most exactly where the answer matters most.

That paragraph deserves an essay of its own, because the case where the two formulas disagree most is not a steel column at all. It is the laminated rubber bearing under a base-isolated building, where one of them says the bearing has buckled long before it is loaded and the other says it is safe.

Two readings of one slice #

Both formulas start from the same member — a bending stiffness and a shear stiffness in series — and both write the same equilibrium of a slice of it once it has deflected. They differ in one sentence about that slice.

In a member that shears, the axis and the cross-sections stop being perpendicular. The axis leans by its slope $\theta$; the section rotates by less, $\theta - \gamma$, where $\gamma$ is the shear strain. The axial load is vertical. Engesser resolves it against the axis: the shear force on the section is the load times the axis’s slope, $P\theta$. Haringx resolves it against the section: the load is carried across the section, normal to it, so the force that shears the section is the load times the section’s rotation, $P(\theta - \gamma)$. The two readings differ by $P\gamma$ — the axial load times the shear strain — which is nothing when the member barely shears and everything when it shears a great deal.

Carried through to a buckling load, Engesser’s reading gives his formula above. Haringx’s gives a quadratic,

$$P^2 + S\,P - S\,P_E = 0, \qquad P = \frac{S}{2}\left(\sqrt{1 + \frac{4P_E}{S}} - 1\right).$$

Where they agree, and where they part #

Plotted against the ratio of the shear stiffness to the Euler load, the two curves lie together on the right and fan apart on the left. Where the member is stiff in shear — more than about four times its Euler load — they agree within five per cent, and both tend to the Euler load as the shear stiffness grows. Where it is soft, they part completely: Engesser’s tends to the shear stiffness itself, $P \to S$, and Haringx’s to the geometric mean of the two, $P \to \sqrt{S\,P_E}$, which is much larger.

The three members marked on the curve sit in three different places. The double-laced column of the earlier essay has a shear stiffness 23.5 times its Euler load: the formulas give 0.959 and 0.961 of it, and nobody need care which is used. Its battened version has a shear stiffness of 0.30 of its Euler load: Engesser gives 0.232, Haringx 0.419 — a factor of 1.8. And a laminated rubber bearing has a shear stiffness of a few ten-thousandths of its Euler load, where the two formulas differ by a factor of forty.

The battened column, by both #

The battened column makes the disagreement concrete. With battens every 1.2 m, Engesser’s formula gives it 312 kN and Haringx’s 562 — the difference between a column that needs its chords doubled and one that does not. Spread the battens to 3 m and the gap opens further: 27 kN against 178, a factor of six, because the batten frame’s shear stiffness falls with the cube of the spacing and the formulas diverge as it does.

Design rules for built-up steel members use Engesser’s form. EN 1993-1-1 combines the Euler load and the lattice’s shear stiffness in series, exactly as the first equation does, and so did the rules before it. If Haringx’s were right for a battened column, every battened column designed in the last century would be nearly twice as strong as its designer believed — which would be a comfortable error, but an error.

One formula, two shear stiffnesses #

The disagreement dissolves under one line of algebra. Take Engesser’s formula and give it a shear stiffness of $S + P$ instead of $S$:

$$\frac{1}{P} = \frac{1}{P_E} + \frac{1}{S + P} \;\Longrightarrow\; P(S + P + P_E) = P_E(S + P) \;\Longrightarrow\; P^2 + S\,P - S\,P_E = 0.$$

That is Haringx’s quadratic, term for term. The two formulas are the same formula, and they differ only in what they mean by the shear stiffness: Haringx’s $S$ is Engesser’s less the axial load the member is carrying. The figure shows it as a check: Engesser’s curve, recomputed with $S + P$, lands on Haringx’s at every point.

That is the resolution Bažant gave in 2003 to what had been called a paradox for half a century. The shear stiffness of a member under axial load is not a single number until it says what it is the stiffness of — a force across the deformed axis, or a force across the rotated section — and the two definitions differ by exactly the axial load, because the axial load itself has a component across the section once the section has rotated relative to the axis. Neither formula is wrong. Each is right with its own shear stiffness, and wrong with the other’s.

So the question “which formula?” becomes “which stiffness did the measurement or the calculation produce?” — and that has a different answer for different members.

Why a battened column’s stiffness is Engesser’s #

The battened column shows how large “the axial load” is beside “the shear stiffness” when it matters. Its shear stiffness comes from the frame the chords and battens make: shear the column sideways and each panel racks like a portal, the chords bending between battens and the battens bending across the column. For battens every 1.2 m that frame has a shear stiffness of 406 kN — computed, as every design rule computes it, from the frame with no axial load in it at all. That is Engesser’s $S$, the stiffness against a force across the axis, and his formula with it gives 312 kN.

Haringx’s formula with the same 406 kN is Engesser’s with $406 + P$, and at buckling $P$ is 562 kN: it is quietly using a shear stiffness of 968 kN, more than twice what the frame has. The difference is not small because $P$ and $S$ are of the same size, which is exactly the battened case. In a double-laced column the lattice’s stiffness is 31.5 MN against a buckling load of 1.3 MN, and adding one to the other changes the answer by a fifth of a per cent.

The bearing, where the choice is a factor of forty #

A laminated bearing is layers of rubber vulcanised between thin steel plates. The plates stop the rubber bulging, which makes the bearing very stiff vertically — for this one, 600 mm across with twenty 10 mm layers, the vertical stiffness is 1,350 times the horizontal — and leaves it soft in shear, which is what isolates the building. Treated as a column 257 mm tall, it has an enormous Euler load, 440 MN, because its bending stiffness comes from the confined rubber, and a tiny shear stiffness, 0.29 MN, because shearing it is its job.

Engesser’s formula then says the bearing buckles at 0.29 MN — its shear stiffness, since the Euler load is so much larger that it drops out. Such a bearing carries perhaps 4 MN of building in service. Haringx’s says 11.2 MN, nearly three times the service load, and almost exactly the geometric mean of the two stiffnesses. Bearing design uses Haringx’s, in the form $\sqrt{P_S P_E}$, and tests on bearings bear it out; a building does not stand on bearings that Engesser’s formula says have buckled.

The bearing’s construction says why. Its cross-sections are the steel shims, which stay flat, and the axial load passes from shim to shim through the rubber, normal to the shims — across the rotated sections, which is Haringx’s reading of the slice. A lattice column is the other way round: its axial load is carried by the chords, which follow the deflected axis, and its lattice carries whatever force crosses the axis, which is Engesser’s reading. And the shear stiffness a designer computes for a lattice, from the frame action of its battens or the stretch of its diagonals with no axial load in the calculation, is Engesser’s $S$. Used in Engesser’s formula it is right; used in Haringx’s it overstates the column by the factor the battened case showed.

A test that can tell them apart #

A buckling load is not something anyone measures on a bearing that is going into a building, but a stiffness is, and the two formulas disagree about it long before either load is reached. A column approaching its buckling load loses its lateral stiffness: the axial load, acting through any sideways displacement, pushes the displacement further, and at the buckling load the net stiffness is zero. So the formula that says a bearing buckles at 0.29 MN also says that under 4 MN it has no lateral stiffness at all, and the formula that says 11.2 MN says it has most of its stiffness left.

Haringx’s equations give the bearing’s horizontal stiffness under load directly. Unloaded it is the rubber’s own, $GA/T_r$ = 1,131 N/mm. Under 4 MN it is 995 N/mm, 88 per cent of that, close to the approximation $1 - (P/P_{cr})^2$ = 87; under 8 MN, about half; and at 11.2 MN, nothing, exactly where the buckling load says it should vanish. Bearings are tested in combined compression and shear before they are installed, at their design load, and what those tests find is a modest reduction of the lateral stiffness of the size Haringx’s equations predict — not a bearing that has collapsed sideways. Every isolated building is, in this sense, a test of the identity: its isolation period, the period nobody chose for a fixed-base building and the one chosen deliberately here, is computed from a lateral stiffness that only Haringx’s reading leaves positive.

Where Haringx’s formula came from #

The history explains which member each formula belongs to. Engesser derived his in 1891 for latticed steel columns, when built-up columns had been failing at loads their second moment of area said were safe, and his members were exactly the kind in which the chords carry the axial load along the deflected axis. Haringx derived his in the late 1940s at the Philips laboratories in Eindhoven, for helical springs and rubber rods used as vibration-free mountings — members whose axial load passes through soft material from one rigid end to the other, across sections that rotate as the member shears. Each formula was written for its own member, by someone who knew which way the load went through it, and each was correct there. The paradox appeared only when they were treated as rival theories of a single abstract “member flexible in shear”, and it took the observation that the shear stiffness itself changes meaning with the axial load to put them back where they belong.

The bearing also buckles in tension #

Haringx’s quadratic has a second root, and it is negative:

$$P = -\frac{S}{2}\left(1 + \sqrt{1 + \frac{4P_E}{S}}\right).$$

A negative load is tension, so the equation predicts that a member soft enough in shear can buckle while being pulled. Engesser’s formula, with only one root, cannot say so. For the bearing the tensile root is 11.5 MN — almost the same magnitude as the compressive one, because the shear stiffness is so small against the Euler load that the quadratic is nearly symmetric. It is not a mathematical curiosity: an isolated building’s bearings at its corners can be pulled by overturning in an earthquake, and buckling of laminated bearings in tension has been studied and observed for that reason.

It also says something about the physics the identity hides. In Haringx’s reading, a section that has rotated relative to the axis turns part of a tensile load into a shearing force on itself, and in a member whose resistance to shear is small enough, that force can drive a lateral mode exactly as compression does. A stiff lattice, where Engesser’s reading applies and the rotations are negligible, has nothing of the kind.

The bearing by hand #

The bearing’s numbers come from its rubber in five lines. The shape factor, loaded area over free area of one layer, is $D/4t = 600/40 = 15$. The confined compression modulus is $6GS^2 = 6 \times 0.8 \times 225 = 1{,}080$ N/mm², and the bending modulus a third of it, 360. With $A = 282{,}700$ mm², $I = 6.36 \times 10^9$ mm⁴, 200 mm of rubber in a 257 mm bearing:

$$S = G A \frac{h}{T_r} = 0.8 \times 282{,}700 \times 1.285 = 0.291 \text{ MN}, \qquad P_E = \frac{\pi^2 \times 360 \times 6.36 \times 10^9 \times 1.285}{257^2} = 440 \text{ MN}.$$

Then $\sqrt{S\,P_E} = \sqrt{0.291 \times 440} = 11.3$ MN, against Haringx’s exact 11.2, and Engesser’s $1/(1/440 + 1/0.291) = 0.290$ MN. The ratio of the two formulas here is 38.

A bearing is not quite a column #

The bearing’s model is a beam. A laminated bearing is treated here as a Timoshenko column of equivalent stiffnesses, which is the usual design model and the one Haringx’s formula was derived for. A bearing’s buckling load also falls as it is sheared sideways — an isolated building’s bearings carry their load while displaced — and the reduced-area rules for that are a separate correction.

The rubber is linear. Natural rubber stiffens at large shear strain and its shear modulus depends on the strain amplitude; the buckling load uses the small-strain modulus.

And the lattice’s shear stiffness is elastic. A battened column’s batten joints slip and its battens yield, and either lowers the shear stiffness the formula uses; the choice between Engesser and Haringx is the smaller question there.

Sandwiches, springs and slipping joints #

They cannot show the members in between. A sandwich panel whose core shears, a helical spring, a timber column of nailed boards, a fibre composite with a soft matrix: each is flexible in shear, and for each the right formula depends on how the axial load passes from one part to the next — along the deformed axis or across rotated sections. The identity says what question to ask; it does not answer it for a member that has not been examined.

They cannot show imperfection. Both formulas are bifurcation loads of a straight member. A real built-up column was never straight, and its lacing carries a shear that exists only because of that, amplified through the same shear flexibility; the amplification follows whichever formula’s stiffness is right.

And they cannot show the shear deflection’s own share before buckling. Long before it buckles, a member flexible in shear deflects more than its bending says, by a term that is not bending at all; that is a serviceability question both formulas leave alone.

The stiffness has to say what it is of #

For a member stiff in shear the two formulas agree — within five per cent while the shear stiffness exceeds four times the Euler load — and for a laced column the choice is immaterial.

For a soft one they part. A battened column buckles at 312 kN by Engesser and 562 by Haringx; a rubber bearing at 0.29 MN by one and 11.2 by the other.

They are one formula. Engesser’s with $S + P$ is Haringx’s with $S$, so the dispute is about whether the shear stiffness was defined across the deformed axis or across the rotated sections.

A lattice is Engesser’s member and a bearing is Haringx’s, because a lattice’s chords follow the axis and a bearing’s load crosses its shims. And Haringx’s equation has a root in tension that Engesser’s cannot produce.

Still open: the bearing that is already displaced #

Every bearing here is upright. An isolated building’s bearings do their work displaced — sheared sideways by a large fraction of their own height during an earthquake — and while displaced the area through which the load passes from top plate to bottom plate shrinks to the overlap of two offset circles. Design reduces the buckling load in proportion to that overlap, and tests show bearings carrying their load at displacements where the overlap is small. Whether a bearing displaced sideways is still a Haringx member with a smaller area, or whether the offset itself changes which reading of the slice applies, is the question that decides whether the buckling check of an isolated building can be made with the bearing at rest.