The strip that becomes a plate

The one number a stronger steel does not change took Young’s modulus as the thing a designer can’t buy, and it listed the other elastic constants as later business. Poisson’s ratio is the first of them, and in a beam it is usually treated as having no job. A fibre stretched along a beam shrinks across it, by $\nu$ times as much, and there’s nothing across the beam to mind. It is what makes the stiffness of a bent beam $EI$, with no $\nu$ in it anywhere.

A plate has a job for it. In a wide sheet, the fibre that wants to shrink sideways cannot, because its neighbours are in the same state and the plate’s width stops them. The material is held at a transverse strain of zero, and to bend it to the same curvature takes $E/(1 - \nu^2)$. For steel that is 1.099 times as much. Engineering tables hold both numbers, and it seems to be a matter of shape which of them to use: a beam is narrow, and a plate is wide. The only difficulty is that this makes the choice depend on the drawing, and a strip of metal doesn’t look at drawings. If width were the switch, somewhere between a ruler and a deck plate the bending stiffness would have to jump by a tenth, and there is no such jump. What really happens is more interesting, and it depends on how far the strip has been bent.

Why a beam’s cross-section is a saddle #

Take a long strip of width $b$ and thickness $t$, and bend it along its length to a curvature $\kappa = 1/R$. The top fibres get shorter along the length and the bottom ones longer. In a beam with nothing to stop it, the top then wants to widen and the bottom to narrow, by Poisson’s ratio. A cross-section thus curves the other way, with curvature $\nu\kappa$, and the strip takes up a shallow saddle. This is what’s called anticlastic curvature, and anyone can see it by bending a rubber eraser: its top surface curves one way along its length and the opposite way across it.

The saddle is what lets a beam off the plate’s stiffness. Its fibres are all free to take their Poisson strain, and no transverse stress forms. That only holds as long as a saddle is cheap, and the sting is in what a saddle does on a line that is already curved. The edges of a saddled strip stand further from the strip’s mean surface than the middle does. In a strip bent along its length, being further from the neutral surface means a longer or shorter arc. So every fibre at a distance $u$ above the mean has been given a membrane strain $\kappa u$, in addition to its bending strain. The saddle has to pay for that stretch, and for a large enough curvature it cannot afford it.

The force per unit width from that strain is $Et\kappa u$. It acts along a line with curvature $\kappa$, so it pushes back towards the mean surface with a force per unit area $Et\kappa^2 u$. That is an elastic foundation, in the form that the beam that sits on the ground solved. Across the width of the strip, the cross-section behaves as a beam of stiffness $D = Et^3/12(1 - \nu^2)$ sitting on a bed of springs whose stiffness is set by the curvature of the strip:

$$D\,u'''' + Et\kappa^2\,u = 0, \qquad \text{or} \qquad u'''' + 4\beta^4 u = 0, \quad \beta^4 = \frac{3(1 - \nu^2)\kappa^2}{t^2}.$$

At the free edges there is no transverse moment, which sets $u'' = -\nu\kappa$, and no shear, which sets $u''' = 0$. The rest is the same closed form, a sum of $\cosh\beta y \cos\beta y$ and $\sinh\beta y \sin\beta y$, that a beam on soil has. The characteristic length $1/\beta$ is about $0.78\sqrt{Rt}$, and it is the distance over which the curl dies away. Both the width and the curvature are in the problem only as a ratio of the width to that length, and the square of the ratio is a number named after G. F. C. Searle, who measured elastic constants by bending strips at the start of the twentieth century:

$$\mathcal{S} = \frac{b^2}{Rt}.$$

The curve #

The stiffness of the strip comes from integrating the moment across its width, and it has two contributions: the bending of each fibre in its own thickness, and the membrane stress on its lever arm $u$.

So the transition isn’t a jump in the width, it is a smooth change in one combined number, and it is spread over a couple of decades. Below a Searle number of about 1, the strip is a beam, whatever its width. Around 10 it has moved most of the way. Above about 30 it is within 2 per cent of the plate’s value, and it approaches it slowly from then on, because the curled edge zones are a smaller and smaller share of the width.

For the small end, the check is short. At small curvature the membrane strain is negligible, so $u'' = -\nu\kappa$ everywhere and the section is one parabola, and the edge rises above the middle by $\nu\kappa b^2/8$. Put $\mathcal{S} = 1$ for a strip 100 mm wide and 1 mm thick, so $R = 10$ m: the rise is $0.3 \times 10^{-4} \times 10^4/8 = 0.0375$ mm, and the solution here gives 0.037. The stiffness is the beam’s, plus a term $E\kappa\int u^2$ from the second contribution, which is what brings in the 0.3 per cent.

What the strip looks like at each end #

This is the picture that explains the rest. At a Searle number of 1, the section is a single parabola, the beam’s saddle. At 10 it has flattened in the middle, with the transverse curvature concentrated towards the edges. At 100, the middle of the strip is dead flat, and the only thing left of the saddle is a narrow curl at each edge, of a width $\sqrt{Rt}$ or so. Press a metal rule against a curved surface, tightly, and look along its edges. They curl, and its middle is flat. It’s a common thing to see, and nobody who has seen it needs the equation to believe it.

The middle is a plate because it cannot do otherwise. It is flat, so its transverse curvature is zero, and the material there is held at zero transverse strain, by the same geometry that makes a plate. The strip has not been given a different property by its width. Its middle has been turned into a plate by being bent far enough.

The stress tells the same story, and it has a consequence that doesn’t appear in the stiffness. A plate has a transverse stress $\nu\sigma$ that a beam does not have. In this strip that stress is zero at the free edge, as it must be, and it rises across the edge zone to Poisson’s ratio times the longitudinal stress. So a wide plate that’s been bent hard has a biaxial stress state in the middle of its width, and a uniaxial one at its edge. Those two states have different yield conditions, as the shear strength nobody measured sets out. By von Mises, the plate’s middle in plane strain yields at $2/\sqrt{3}$ of the uniaxial stress, and the edge does not get that benefit. So the first yield of a plate in bending is at its edges, with its plastic moment gain coming from the middle. When a thick plate is cold-bent, cracks are found both at a sheared edge and at mid-width, where the stress is biaxial and ductility is lower. Choosing between those is a matter of edge finish and metallurgy, but the biaxial state that drives the second comes from here.

The last stress is the one that appears out of nothing and is usually forgotten. The fibres that have moved away from the mean surface carry a longitudinal membrane stress, of one sign at the edges and of the other just inside them. It is a self-equilibrating field with no resultant, and it is a large one: nearly a fifth of the bending stress at the surface, once the Searle number has passed about 10. It doesn’t go on growing, as the bigger curvature concentrates it near the edges without raising it. An edge in compression at a fifth of the fibre stress may not matter in a section of thick steel, but it is of direct interest in a thin one, with the flange that curls away from its stress and the flange bent hardest where it curls least looking at the same shape in a beam’s own flange, driven by the same force on a curved line. In a beam’s flange, the curvature comes from the beam, and the flange’s own curl is limited by its width in the same way as the strip’s.

Which side of the curve real plates are on #

This is the most useful reading of the curve, and it is at odds with habit.

The ordinary structural cases are at the bottom. A curve for curvature says that a wide plate is plate-like because it is bent hard, and structures aren’t bent hard. A 1,000 mm wide strip of 12 mm deck plate that spans 2.5 m and has sagged by 10 mm, at span over 250, has a radius of about 78 m and a Searle number of 1.07. A free-edged strip in that state is a beam, with a stiffness 1.003 times $EI$, and its cross-section is a saddle with a rise of about half a millimetre.

The plate modulus appears nonetheless in tables for that same deck. The reason is that a deck plate doesn’t have free edges. Edges that are held do the job of curvature. A plate supported on its sides, or running continuously across several ribs, or restrained by its neighbours, has its transverse strain suppressed by those supports, at any curvature whatever. That is the true reason the plate formula holds for an actual plate, rather than its being wide. So the formula isn’t wrong for a real deck; it is right for a reason not given in the table, and it is therefore wrong for the free-edged cases that look like real decks. These are a plank bridge, the free flange of a thin section, or a strip test specimen cut from a plate.

The specimen is the case worth stopping at. To measure the stiffness of a plate material in a four-point bend, a coupon is cut out, with free edges, of a size chosen for convenience, and its modulus is reported. A modest load gives a small Searle number, so the test reads $E$. Load the coupon heavily, with a modestly large width and it reads 3 per cent higher, and if the reader doesn’t know the curve, that’s a stiffness that looks like it rises with load. In laminated materials and timber panels, where the effective Poisson’s ratio can be large, the gain is much greater and the error is larger. A careful test standard limits width to thickness for this reason, without necessarily saying why. Searle’s own use of the strip ran the other way. His 1908 book explained a way of finding Poisson’s ratio by measuring the saddle’s two curvatures directly, in a strip kept at small enough curvature to remain in the beam regime.

On the other side of the curve is rolling. Plates rolled to a radius have a large Searle number and are fully in the plate regime. A 1,200 × 6 mm plate rolled to a 2 m radius, staying elastic, stands at 120 and 1.088. The calculation of its spring-back with $E$ would be wrong by the tenth that the curve says it should be. A bent steel rule stands at 9, in the middle, which is part of why it feels as though it stiffens the more it is pressed.

The gain is Poisson’s ratio and nothing else #

A material with no Poisson’s ratio would have one curve, flat at 1. Cork, famously, comes close, which is why it goes into a bottle without bulging, and a sheet of cork has no difference between its beam and plate stiffness. Concrete, at about 0.2, gains at most 4 per cent, which is why reinforced concrete slab design has never been very concerned. Steel, at 0.3, gains 10. A rubber-like material at 0.4 and above gains 19 or more, up to a limit of a third more at 0.5, where a fully incompressible plate has $4/3$ of the beam’s stiffness. So the transition has its largest effect in the soft polymer sheets and bearing layers where a designer doesn’t usually have a plate formula in mind, and that is where the bearing Engesser says has buckled has its rubber layers.

The argument has a surprising relative in the buckling of shells. The same bed of springs that the curvature gave here is the one in a cylinder, on a much greater scale, where the curvature is built in. In a cylinder, $\beta$ has the same form, and the decay length $0.78\sqrt{Rt}$ is exactly the bending boundary layer of a tank wall or a dome, which the surface that carries by being curved gave as the whole price of its edge. A strip bent to a radius $R$ and a shell built to one are the same equation, and the only difference is whether the curvature came from the load or from the design. This also means that a beam’s lateral stiffness can be thought of as what remains after this boundary layer has been allowed for. Three kilopascals that halve a tube in bending is the same effect going the other way, with a closed section whose flattening costs it strength.

What a long uniform strip assumes #

The strip is long. This model has no ends. Near the end of a real strip the curvature falls away, and the transition to the plate state happens over a length of about the width. A short specimen’s middle section may never reach it.

The bending is uniform along the length. A real beam has a moment gradient, so its Searle number changes from section to section, and a three-point bend test has a zone of plate near the load and of beam near the supports. The stiffness read from such a test is a blend.

The material is elastic. Rolling a plate to a useful radius usually yields it, and a plastic strip has its own, larger effective Poisson’s ratio of one half, so its plate gain is $2/\sqrt{3}$, a 15 per cent increase, rather than the 10 per cent elastic one.

The deflections across the width are small. The equation assumes shallow transverse slopes. For a very wide strip bent very hard, an edge can begin to buckle into waves, which is an entirely different way of relieving its compression, and it happens at large Searle numbers on thin sheet.

The material is isotropic. Rolled plate has a small anisotropy, and timber has a very large one, with different ratios in the two directions. In the material that has a direction, transverse properties would change the numbers on this curve substantially and also its point of transition.

Nothing is applied across the width. A pressure or an axial load changes the bed of springs. The picture’s Searle number stands for a strip in pure bending and only that.

Still open: the rib that holds an edge partway #

A free edge and a held edge are the two ends of a range. A plate with a light edge stiffener, a rolled lip or a lightly stiffened flange holds its edge partly, as a spring would, so its section takes only part of its saddle. There should then be a third parameter beside $\nu$ and Searle’s number: the transverse stiffness of the edge, and how large that needs to be before a strip at small Searle number takes on its plate value. Since many light sections are exactly that case, with thin wide plates at low curvature and lips or returns, the size of a lip needed to get the extra tenth for free is a real design question. So far nothing has been computed for it, as the boundary condition used here allows the edge no restraint at all.