Rigid by every test, and still folding

Counting the unknowns asked whether a frame has as many equations as unknowns, and the count that does not see it found frames that pass the count and fold anyway, because some of their equations say the same thing twice. The real test, it concluded, is the rank of the equilibrium matrix: the joint equations are independent or they are not, and when they are not the mechanism sits in the matrix’s null space.

That test has one property the count shares. It is binary. A frame has full rank or it does not. And a question about a physical structure that has only two answers is a question that will be asked, sooner or later, of a structure that is nearly one and nearly the other.

That structure is the subject here: the frame a few degrees from a critical form. It has as many unknowns as equations. Its matrix has full rank and a determinant that is not zero. By every test on the two essays below, it is a determinate, rigid frame. It is also, in three distinct senses that arrive in a definite order, nearly a mechanism — and the sense that the tests measure is the last of the three.

Two bars, and how flat they are #

The simplest frame with a critical form is two bars between two pinned supports, meeting at a joint in the middle. If the joint lies on the line between the supports, the bars are collinear, and the joint can move sideways — perpendicular to them — without either bar changing length. That is a mechanism with a full count: two bars and four reactions against three joints’ six equations. It is the flat pair of the essay below.

Raise the joint by an angle $\theta$ and the frame is sound. The rank is full; the forces follow from equilibrium of the joint, each bar carrying $P/2\sin\theta$ under a load $P$ there. At 15 degrees that is 97 kN for a load of 50. At 6 degrees, 239. At 3 degrees, 478.

The figure below follows four quantities as the angle falls from 30 degrees to half a degree, on logarithmic axes so that each power law is a straight line. The rest of the essay is an account of why the four lines have the slopes they have, and why their order matters more than any one of them.

The measure that is not yes or no #

The equilibrium matrix $A$ of a determinate frame is square: one row per joint equation, one column per unknown force. The forces are $t = A^{-1}(-f)$. Whether $A$ can be inverted is the rank test. How well it can be inverted is measured by its smallest singular value, $\sigma_{\min}$, which is the smallest factor by which $A$ can shrink a vector of unit length — zero exactly when there is a mechanism, and small when there nearly is one.

For the two bars, $\sigma_{\min}$ is proportional to $\sin\theta$. Everything else follows from that. No force can exceed $|f|/\sigma_{\min}$, so the bar forces grow as $1/\sin\theta$. The condition number $\sigma_{\max}/\sigma_{\min}$ — the factor by which a small relative error in the matrix can grow into a relative error in the answer — grows as $1/\sin\theta$ too. At 2 degrees the bars carry 14.6 times the load and the condition number is 54.

That is the continuous version of the rank test, and it is the answer to the question the essay on graphic accuracy left open. How wrong a drawing is found that a pencil line’s width is amplified by $1/\sin$ of the angle between two lines, and ended by asking what the same amplification does to a computer, whose pencil is sixteen significant figures wide. The condition number is the computer’s version of that angle, and it costs $\log_{10}$ of itself in figures: at 2 degrees, 1.7 of sixteen; at half a degree, 2.3.

The measure that is not about arithmetic #

The bars are not rigid. Each is a steel member of 1,000 mm² section, 200 MN in axial stiffness, and each stretches under its force. The joint’s vertical stiffness is what those stretches allow: $2EA\sin^2\theta/l$, which has two factors of $\sin\theta$ in it, because a bar at a shallow angle both carries more force for a given load and converts its stretch into more vertical movement.

So the sag under load goes as $1/\sin^2\theta$, and the rise it is measured against goes as $\sin\theta$. Their ratio goes as $1/\sin^3\theta$:

$$\frac{\delta}{h} = \frac{P}{2EA\sin^3\theta}$$

At 15 degrees the joint sags 8 mm on a rise of 1,072 — nothing. At 6 degrees it sags 46 mm on 420, a ninth of the rise. At 3 degrees it sags 183 mm on a rise of 210: the load has flattened the frame to an eighth of the height its forces were computed on. The forces were found by resolving at a joint 210 mm above the line; the joint is now 27 mm above it, where the same load needs bars carrying eight times as much. The linear analysis has not become inaccurate. It has become an answer about a different frame.

What happens next is the shallow arch’s snap-through: as the joint approaches the line, the bars’ force rises faster than the load, the vertical stiffness falls, and past a critical load the joint passes through the line to the far side. A frame that the rank test calls rigid, analysed by the method that test belongs to, is predicting the equilibrium of a structure that has already jumped.

Four limits, and the order they arrive in #

Put a practical limit on each measure and ask at what angle it is passed, and the order is fixed by the exponents.

The sag goes first. It reaches a tenth of the rise at 6.2 degrees, because it grows as the cube of $1/\theta$ and starts at a value set by the load over the axial stiffness — $50 \text{ kN}/2 \times 200 \text{ MN}$, about one in eight thousand, which the cube eats quickly.

The force goes next. Ten times the load arrives at 2.9 degrees, as $1/2\sin\theta$ says it must.

Fabrication tolerance goes third. A millimetre’s error in any joint’s position changes the bar force by one per cent below 1.4 degrees, because the force depends on the rise and a millimetre is a larger fraction of a smaller rise. This is the real-world counterpart of the condition number: the geometry is the matrix, and a fabricator’s tolerance is the relative error in it.

The arithmetic never goes. Three lost digits needs a condition number of a thousand, which this frame does not reach above half a degree, and a computer with sixteen figures to lose would not produce a wrong answer until the frame was flatter than any fabricator could make it.

The ratio between the first and the third is 4.4. The frame stops being the structure its forces describe at an angle four times larger than the one at which its fabrication starts to matter, and at an angle at which the matrix is perfectly well-conditioned. The ordering is not a property of these particular bars; it is a property of the exponents, and every frame near a critical form has the same ones, because every such frame has a stiffness that vanishes as the square of its distance from the form and forces that grow as its inverse.

The same order in other frames #

The two bars are one critical form. The concurrent triangle of the count that does not see it is another: a triangular frame whose three reactions pass through one point, so that the whole frame can turn about it. Tilt the roller’s track slightly away from the line through the pin and the frame is sound, and it approaches its critical form as the tilt goes to zero.

The measures behave in the same order. At 8 degrees of tilt the largest force is 5.2 times the push and the frame moves 5.7 per cent of the distance its track is offset from the pin’s line; at 2 degrees, 18.6 times and 387 per cent. The frame’s rotation about the nearly-concurrent point is resisted only by the roller’s small lever arm, and its movement grows faster than its forces for the same reason as the bars’: the stiffness that resists the mechanism vanishes as the square of the offset.

A shallow truss is the everyday case. At zero depth its chords are collinear and it is a mechanism, and every shallow truss is somewhere on the road there. At a depth of a tenth of its span, the Pratt truss drawn is an ordinary truss: its largest force is 1.7 times the load it carries and it sags 1.3 per cent of its depth. At a fiftieth of its span its largest force is 8.4 times the load and its computed sag is one and a half times its depth — a truss analysed as rigid that the analysis itself says has turned inside out.

Why the tests cannot see it, and what can #

The count asks whether there are enough equations. The rank asks whether they are independent. Neither can ask how independent, because independence is a yes-or-no property of a set of vectors, and the frame near a critical form has independent equations by a margin. What measures the margin is a number, and the useful numbers here are two.

The smallest singular value of the equilibrium matrix measures how far the frame is from a mechanism in the forces: its reciprocal bounds the largest force a unit load can produce. It needs no material properties; it is a property of the geometry alone, like the rank it refines.

The stiffness of the frame in the direction of its nearest mechanism measures how far it is from a mechanism in the displacements, and it goes as the square of the first. It needs the members’ axial stiffness, which is why no test of equilibrium can find it — and it is the one that decides whether the analysis means anything, because a structure whose load moves it by a large fraction of its own geometry is not described by any linear analysis, rigid by count or not.

A designer checking a shallow frame therefore has a simple sequence. Compute the sag under the design load on the actual members and compare it with the geometric quantity that keeps the frame away from its critical form — the rise, the depth, the offset. If the ratio is more than a few per cent, the frame must be analysed on its deflected geometry, and its forces will be larger than the linear analysis says. The rank test and the condition number, which a computer program reports, will not have noticed.

Where the nearly-critical frame is met #

Nobody designs two bars at three degrees on purpose. The nearly-critical frame arrives by other routes, and each of them is ordinary.

A shallow roof. A truss whose depth is set by the roof’s pitch rather than by its span is a shallow truss, and a low-pitched roof over a long span is a truss a long way down the road to its critical form. The forces are large and the designer sees that; the sag is large compared with the depth and nothing in a linear analysis reports it as anything but a deflection to check against a serviceability limit.

A brace at a shallow angle. A triangle that cannot fold is rigid because its three sides are not collinear, and a diagonal brace set nearly parallel to the member it braces — to clear a door, a duct, a window — makes a triangle that is nearly flat. The brace carries a force of $1/\sin$ of its angle, and the frame’s stiffness against the mechanism the brace was meant to prevent goes as the square of that sine.

A frame whose geometry changes during construction. A three-hinged arch lowered onto its crown hinge, a roof whose supports spread under load, a truss whose bearings slide: each can move a frame toward a critical form after it was analysed in a sound one. A three-hinged arch is the clearest: the hinge put in on purpose makes it determinate, and a flat three-hinged arch is the two bars of this page with a rib instead of a bar — its thrust grows as one over its rise and its crown drops as the arch shortens.

A computer model built from a drawing. A stiffness program inverts the same matrix that this essay has been measuring, and it will report forces for a nearly critical frame with every digit it has. The warning it can give — a condition number, a small pivot — is the arithmetic measure, which is the last to arrive. The measure that arrives first needs one extra line: compare the joint movements it has already computed with the geometry that keeps the frame from its critical form.

In every case the same fact decides the danger, and it is the one the tests of the essays below cannot contain: nearness to a mechanism is measured in the frame’s stiffness, and stiffness goes as the square of the distance. A frame twice as close to its critical form carries twice the force and moves four times as far for each unit of load, on a geometry half as large — eight times the relative movement. That cube is the same steep path just past a critical point that makes some structures imperfection-sensitive, met here before any buckling has happened.

The whole of it, once, at 6 degrees #

The two bars span 8 m, so each is $4{,}000/\cos 6^\circ = 4{,}022$ mm long, and the joint is $4{,}000 \tan 6^\circ = 420$ mm above the line between the supports. Under 50 kN each carries $50/(2 \sin 6^\circ) = 239$ kN.

Each stretches by $239{,}000 \times 4{,}022 / 200 \times 10^6 = 4.8$ mm. A bar at 6 degrees converts a stretch into vertical movement of the joint by a factor of $1/\sin 6^\circ = 9.6$, so the joint drops 46 mm — 11 per cent of its rise. On the flattened geometry the angle is 5.3 degrees and the bars need about 269 kN, 12 per cent more than the linear answer. At 3 degrees the same arithmetic puts the joint at an eighth of its rise and the linear answer is short by a factor of eight.

The joint equations, and the matrix they make #

The free body is each joint of the frame, with the members and reactions cut: two equations of equilibrium per joint, assembled into the square equilibrium matrix whose columns are the unknown member forces and reactions. Its singular values are found from the eigenvalues of $A^{T}A$; the forces by solving the system directly; the fabrication sensitivity by moving every joint coordinate by a millimetre in turn and solving again; the sag by giving each member its extension under its force and solving the compatibility equations, which use the transpose of the same matrix. For the two bars the forces and the sag are checked against their closed forms, $P/2\sin\theta$ and $Pl/2EA\sin^2\theta$, and agree to the precision of the arithmetic.

Snap-through, buckling bars and stiff joints #

The large-displacement answer. Every sag here is the small-displacement estimate, and when it is a large fraction of the rise the estimate is itself wrong — the real sag under that load is larger, and past a critical load there is no equilibrium near the original shape at all. The figures show where the linear analysis stops, not what replaces it.

Buckling of the bars. The bars at a shallow angle carry large compressions if the load is reversed, or tensions as drawn; a compressed slender bar near a critical form buckles long before the frame snaps.

Joint stiffness. Real joints are not pins. A welded joint resists the rotation the mechanism needs, and a nearly critical frame with stiff joints behaves as a frame in bending — a different structure with a different, usually much greater, stiffness.

A determinate frame, with nothing to prestress #

That the frame is determinate. A redundant frame near a critical form has self-stress states as well as a nearly-mechanism, and a prestress can stiffen the mechanism — which is how a cable net or a tensegrity stands up at all. For a determinate frame nothing can be prestressed, the nearly-mechanism has only the members’ axial stiffness to resist it, and the cube law is exact.

Still open: the frame that is a mechanism and stands up anyway #

A flat pair of bars with nothing applied is a mechanism, and the rank test says so. Pull the two supports apart until the bars are in tension, and the same two bars resist a sideways load at the joint: the tension acts as a stiffness, because a joint moved sideways turns the tensioned bars and their tension acquires a component that pushes it back. That is the flat pair’s state of self-stress stiffening its mechanism, and it is how every guitar string and every cable net carries load across its own line. A frame can therefore be a mechanism by every test on this page and on the two below it and still be a structure, and when a self-stress stiffens a mechanism — and when, like the kite of the essay below, it does not — is the question after this one.