I noticed that I have been bending my book all along, when I was reading it with one hand.


This also works for plane flexible sheets of any material.

Illustration using an A4 sheet

Without bending the sheet: No bend

With a bend along perpendicular axis With bend

How do you explain this sturdiness, that comes only when the object is bent along the perpendicular axis ? I feel that this is a problem related to the elastic properties of thin planes. But any other versions are also welcome.

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    $\begingroup$ Numberphile answered a version of this with a pizza rather than a book: The Remarkable Way We Eat Pizza - Numberphile $\endgroup$
    – MannyC
    Commented Apr 16, 2019 at 14:32
  • 3
    $\begingroup$ @MannyC: Whatever benefits gained by watching Numberphile videos are more than outweighed by the wrong mathematics that pervade them. Sad to say, but true. $\endgroup$
    – user21820
    Commented Apr 17, 2019 at 14:04
  • 1
    $\begingroup$ The answer below is long, so I will just suggest the Wikipedia article on Theorema Egregium if you already have a good understanding of how surfaces work, as it explains it in terms of differential geometry. This question was likely one that must have occurred to Gauss at one point when he had one of his notebooks in his hand and realised he was keeping it straight by bending it. Basically rigidity is created perpendicular to the direction of the folding, which can be explained rigorously using differential geometry on surfaces. $\endgroup$
    – Tom
    Commented Apr 17, 2019 at 16:07
  • 1
    $\begingroup$ You re-invented corrugated iron, which allows you to adjust stiffness and strength separately. Stiffness is detemined by the thickness of the structure, including corrugations, while strength depends on the thickness of the sheet. $\endgroup$
    – Rainer P.
    Commented Apr 17, 2019 at 17:15
  • $\begingroup$ See Gauss's Theorema Egregium en.wikipedia.org/wiki/Theorema_Egregium $\endgroup$
    – R. Rankin
    Commented Apr 18, 2019 at 10:21

8 Answers 8


Understanding why this works turns out to be quite deep. This answer is kind of a long story, but there's no maths. At the end ('A more formal approach') there is an outline of how the maths works: skip to that if you don't want the story.

Insect geometry

Consider a little insect or something who lives on the surface of the paper. This insect can't see off the paper, but it can draw straight lines and measure angles on the paper.

How does it draw straight lines? Well it does it in two ways: either it takes two points, draws lines between them on the paper, and finds the shortest line between them, which it calls 'straight'; or alternatively it draws a line in such a way that it is parallel to itself and calls this 'straight'. There is a geometrical trick for constructing such 'parallel-to-themselves' lines which I won't go into. And it turns out that these two sorts of lines are the same.

I'm not sure how it measures angles: perhaps it has a little protractor.

So now our insect can do geometry. It can draw various triangles on the paper, and it can measure the angles at the corners of these triangles. And it's always going to find that the angles add up to $\pi$ ($180^\circ$), of course. You can do this too, and check the insect's results, and many people do just this at school. The insect (let's call it 'Euclid') can develop an entire system of geometry on its sheet of paper, in fact. Other insect artists will make pictures and sculptures of it, and the book on geometry it writes will be used in insect schools for thousands of years. In particular the insect can construct shapes out of straight lines and measure the areas inside them and develop a bunch of rules for this: rectangles have areas which are equal to $w \times h$ for instance.

I didn't specify something above: I didn't tell you if the paper was lying flat on a desk, or if it was curved in your hand. That's because it does not matter to the insect: the insect can't tell whether we think the paper is curved, or whether we think it's flat: the lines and angles it measures are exactly the same. And that's because, in a real sense, the insect is right and we're wrong: the paper is flat, even when we think it's curved. What I mean by this is that there is no measurement you can do, on the surface of the paper which will tell you if it is 'curved' or 'flat'.

So now shake the paper, and cause one of the insects to fall off and land on a tomato. This insect starts doing its geometry on the surface of the tomato, and it finds something quite shocking: on a small scale everything looks OK, but when it starts trying to construct large figures things go horribly wrong: the angles in its triangles add up to more than $\pi$. Lines which start parallel, extended far enough, meet twice, and there is in fact no global notion of parallelism at all. And when it measures the area inside shapes, it finds it is always more than it thinks it should be: somehow there is more tomato inside the shapes than there is paper.

The tomato, in fact, is curved: without ever leaving the surface of the tomato the insect can know that the surface is somehow deformed. Eventually it can develop a whole theory of tomato geometry, and later some really smart insects with names like 'Gauss' and 'Riemann' will develop a theory which allows them to describe the geometry of curved surfaces in general: tomatoes, pears and so on.

Intrinsic & extrinsic curvature

To be really precise, we talk about the sheet of paper being 'intrinsically flat' and the surface of the tomato being 'intrinsically curved': what this means is just that, by doing measurements on the surface alone we can tell if the rules of Euclidean geometry hold or not.

There is another sort of curvature which is extrinsic curvature: this is the kind of curvature which you can measure only by considering an object as being embedded into some higher-dimensional space. So in the case of sheets of paper, the surfaces of these are two dimensional objects embedded in the three dimensional space where we live. And we can tell whether these surfaces are extrinsically curved by constructing normal vectors to the surfaces and checking whether they all point in the same direction. But the insects can't do this: they can only measure intrinsic curvature.

And, critically, something can be extrinsically curved while being intrinsically flat. (The converse is not true, at least in the case of paper: if it's intrinsically curved it's extrinsically curved as well.)

Stretching & compressing

There's a critical thing about the difference between intrinsically flat and intrinsically curved surfaces which I've mentioned in passing above: the area inside shapes is different. What this means is that the surface is stretched or compressed: in the case of the tomato there is more area inside triangles than there is for flat paper.

What this means is that, if you want to take an intrinsically flat object and deform it so that it is intrinsically curved, you need to stretch or compress parts of it: if we wanted to take a sheet of paper and curve it over the surface of a sphere, then we would need to stretch & compress it: there is no other way to do it.

That's not true for extrinsic curvature: if I take a bit of paper and roll it into a cylinder, say, the surface of the paper is not stretched or compressed at all. (In fact, it is a bit because paper is actually a thin three-dimensional object, but ideal two-dimensional paper is not.)

Why curving paper makes it rigid

Finally I can answer the question. Paper is pretty resistant to stretching & compressing: if you try and stretch a (dry) sheet of paper it will tear before it has streched really at all, and if you try and compress it it will fold up in some awful way but not compress.

But paper is really thin so it is not very resistant to bending (because bending it only stretches it a tiny tiny bit, and for our ideal two dimensional paper, it doesn't stretch it at all).

What this means is that it's easy to curve paper extrinsically but very hard to curve it intrinsically.

And now I will wave my hands a bit: if you curve paper into a 'U' shape as you have done, then you are curving it only extrinsically: it's still intrinsically flat. So it doesn't mind this, at all. But if it starts curving in the other direction as well, then it will have to curve intrinsically: it will have to stretch or compress. It's easy to see this just be looking at the paper: when it's curved into a 'U' then to curve it in the other direction either the top of the 'U' is going to need to stretch or the bottom is going to need to compress.

And this is why curving paper like that makes it rigid: it 'uses up' the ability to extrinsically curve the paper so that any further extrinsic curvature involves intrinsic curvature too, which paper does not like to do.

Why all this is important

As I said at the start, this is quite a deep question.

  • The mathematics behind this is absolutely fascinating and beautiful while being relatively easy to understand once you have seen it. If you understand it you get some kind of insight into how the minds of people like Gauss worked, which is just lovely.
  • The mathematics and physics behind it turns out to be some of the maths that you need to understand General Relativity, which is a theory all about curvature. So by understanding this properly you are starting on the path to understanding the most beautiful and profound theory of modern physics (I was going to write 'one of the most ...' but no: there's GR and there's everything else).
  • The mathematics and physics behind it also is important in things like engineering: if you want to understand why beams are strong, or why car panels are rigid you need to understand this stuff.
  • And finally it's the same maths: the maths you need to understand various engineered structures is pretty close to the maths you need to understand GR: how cool is that?

A more formal approach: a remarkable theorem

The last section above involved some handwaving: the way to make it less handwavy is due to the wonderful Theorema Egregium ('remarkable theorem') due to Gauss. I don't want to go into the complete detail of this (in fact, I'm probably not up to it any more), but the trick you do is, for a two dimensional surface you can construct the normal vector $\vec{n}$ in three dimensions (the vector pointing out of the surface), and you can consider how this vector changes direction (in three dimensions) as you move it along various curves on the surface. At any point in the surface there are two curves which pass through it: one on which the vector is changing direction fastest along the curve, and one along which is changing direction slowest (this follows basically from continuity).

We can construct a number, $r$ which describes how fast the vector is changing direction along a curve (I've completely forgotten how to do that, but I think it's straightforward), and for these two maximum & minimum curves we can call the two rates $r_1$ and $r_2$. $r_1$ & $r_2$ are called the two principal curvatures of the surface.

Then the quantity $K = r_1r_2$ is called the Gaussian curvature of the surface, and the theorema egregium says that this quantity is intrinsic to the surface: you can measure it just by measuring angles et cetera on the surface. The reason the theorem is remarkable is that the whole definition of $K$ involved things which are extrinsic to the surface, in particular the two principal curvatures. Because $K$ is intrinsic, our insects can measure it!

Euclidean geometry is true (in particular the parallel postulate is true) for surfaces where $K = 0$ only.

And we can now be a bit more precise about the whole 'stretching & compressing' thing I talked about above. If we're not allowed to stretch & compress the sheet of paper, then all the things we are allowed to do to it don't alter any measurement that the insects can do: lengths or angles which are intrinsic, that is to say measured entirely in the surface of the paper, can't change unless you stretch or compress the paper. Changes to the paper which preserve these intrinsic properties are called isometries. And since $K$ is intrinsic it is not altered by isometries.

Now consider a sheet of paper which is flat in three dimensions. It's obvious that $r_1 = r_2 = 0$ (the normal vector always points in the same direction). So $K = 0$.

Now fold the paper in a 'U' shape: now it's clear that $r_1 \ne 0$ -- if you draw a curve across the valley in the paper then the normal vector from that curve changes direction. But this folding is an isometry: we didn't stretch or compress the paper. So $K$ must still be $0$: the paper is still intrinsically flat. But since $K = r_1r_2$ and $r_1 \ne 0$ this means that $r_2 = 0$.

And what this means is that the other principal curvature must be zero. This principal curvature is along the line that goes down the valley of the 'U'. In other words the paper can't bend in the other direction without becoming intrinsically curved ($K \ne 0$), which means it needs to stretch.

(I have still handwaved a bit here: I have not defined how you compute $r$, and I've not shown that there is not some other curve you can draw along the paper which has $r = 0$ apart from the obvious one.)

One of the reasons that this is all quite interesting is that this maths is the beginning of the maths you need to understand General Relativity, which also is about curvature.

Failure and folding

Of course, if you take the U-shaped bit of paper and try to bend it in the other direction at some point it will fail suddenly and become folded in some complicated way. I think there's a whole area of study which thinks about that. I suspect that when this happens (during the sudden failure, not after it I think) there must be, locally, non-zero intrinsic curvature at places on the paper. I'm sure there is a lot of interesting maths about this (apart from anything else it must be very interesting for engineered structures), but I don't know it.

  • $\begingroup$ Um, I think you made a mistake. If you bend the paper, you have $r_1 ≠ 0$ at some point, and hence $r_2 = 0$ at that point. But you can't from that alone conclude that you have $r_2 = 0$ along a straight line. I think it can be fixed, by observing that $r_1 ≠ 0$ in some open region around that point and hence $r_2 = 0$ everywhere in that region, and then show that it extends indefinitely, but I know very little in this area and can't tell if my idea can be easily made to work. $\endgroup$
    – user21820
    Commented Apr 17, 2019 at 14:18
  • $\begingroup$ @user21820 My understanding is, everywhere that the paper is bent (has extrinsic curvature), you can show that $r_1 \neq 0$. Therefore $r_2 = 0$ at all of those points. Then you can show by continuity that it's true in the whole interval. $\endgroup$
    – Draconis
    Commented Apr 17, 2019 at 18:23
  • 2
    $\begingroup$ @user21820 Note that the paper is curved into a 'U' shape, not a 'V': any line crossing the valley of the 'U' has $r \ne 0$ along much of its length. Find the line which maximises $r$ amongst these (by symmetry this is the line which crosses the valley at right angles). This line defines $r_1$. Translate this line up and down the line of the valley and it is now clear that $r_1 \ne 0$ everywhere in the curved part of the 'U'. I think you must have assumed I was folding the paper in a 'V': $r$ is horribly discontinuous (zero almost everywhere, not finite at a single point) for that shape. $\endgroup$
    – user107153
    Commented Apr 18, 2019 at 15:30
  • 2
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented Apr 18, 2019 at 16:41
  • $\begingroup$ @tfb: You don't seem to get the point. If you assume the entire paper has the U-shape with a straight valley, you are already assuming the conclusion you claim. If you want to claim an explanation for the effect in question, which is that curving the paper at a single point is enough to 'keep it straight', then your answer fails to do so, and I've already sketched how it may be possible to do it properly in my first comment. It's not sufficient to start from the assumption that $r_1 ≠ 0$ everywhere on the paper, since that does not explain why it is stable. $\endgroup$
    – user21820
    Commented Apr 24, 2019 at 12:47

You have essentially discovered principles behind bending moments and structural engineering.

As another poster stated, physically the structure you made is stronger, because to bend something (for example, a beam loaded at the top) layers at the top are compressed whereas layers at the bottom are stretched. This is simply due to geometry and the physical nature of materials. In short, the load (force) is being transformed from a direction normal to the beam, to an internal force—longitudinal stress. More specifically, the applied load (from weight, gravity, whatever) results in a bending moment in the member, this bending moment manifests as internal stresses (tensile and compressive forces) inside the member that resists the bending of equal magnitude.

enter image description here

Some primers on forces: compression and tension are the same thing, just different "directions" ie: if compression is -1 or -2, then tension will be 1 or 2. Knowing that, and knowing that the top of the member is in compression and the bottom is in tension, we can reason that there is a force distribution across the member. And I think the most important part to your question is that since the force distribution goes from -x to +x across the member, there must be some point where x = 0 (the neutral surface). In the picture below the stress (green arrows) at some point cross 0.

enter image description here

Therefore, we can observe that maximum stresses occur at the edges, top and bottom of the beam in our example. This principle is precisely how and why I-beams work. The strength of the member comes from the material properties of the material (its ability to resist compression or tension (stretching)). That means something like a steel beam will be limited in its ability to resist bending by the calculation of tensile load at the surface. Physically that equation is (for the $x$ direction):


Where $c$ is the neutral surface (the imaginary plane where $\sigma_{x}=0$) and $y$ is the distance from the neutral surface, and $\sigma_{m}$ is the maximum absolute value of the stress in the member.

In layman's terms, the height of the beam is the driving factor in it's strength, not the thickness. But in the plane that is experiencing maximum loads (tension and compression) thickness there will give you more strength. This results in the classic I-beam shape.

enter image description here

What does all of this have to do with paper?

When the OP orients the paper horizontally (flat), the height of the paper relative to the neutral surface is basically 0. IE we can consider that the entire paper IS a neutral surface. That means it literally cannot resist any bending. Flip the paper 90 degrees and now all of the paper is height, and all of the paper can resist bending and it cannot be bent. It will usually buckle or tear before bending.

enter image description here

The curved shape OP creates, is exploiting the all the concepts we covered here. Instead of making an I shape, OP makes a C-shape which leads to the idea of exploiting thin materials by using corrugation to add incredible strength while keeping weight low. For example, the inside layers of a cardboard box are corrugated or folded into little curved shapes to resist bending. So we can use less material to achieve much higher strengths.

enter image description here

enter image description here

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    $\begingroup$ To me, this doesn't really explain anything about why this, or I-beams work. Why does a force distribution across the sheet allow it to reduce bending moments in a different direction? You've explained what bending moments are; but you didn't explain why bending paper adds stability in another direction, or how this relates to I-beams. Your explanation for how I-beams "work" is seriously lacking. All you've said is that your bending the top and compressing the bottom. I would argue that this makes the I-beam weaker than the unstressed I-beam, in a general sense. $\endgroup$
    – JMac
    Commented Apr 16, 2019 at 18:13
  • 1
    $\begingroup$ The point isn't that the moment is reduced, so much as the force (stress). Roughly speaking moments are $r \times F$, so increasing $r$ from 'thickness of paper' to 'height of curve in paper' reduces the force in the paper needed to counteract its own weight. engineering.stackexchange.com/questions/68/… is the same principle. $\endgroup$
    – patstew
    Commented Apr 17, 2019 at 9:49
  • 1
    $\begingroup$ @patstew I don't think the "height of curve" is really that relevant. It's really more geometrical, as shown in tfb's answer. Consider that you could get the same height by sandwiching two pieces of cardboard with straight pieces of cardboard between, and still get the same $r$. It's important how the corrugation effects stability in other directions as well. $\endgroup$
    – JMac
    Commented Apr 17, 2019 at 11:15
  • 1
    $\begingroup$ I was going to +1 this answer as I was reading it, up until I noticed that you did not explain why the corrugations give increased strength, which is the essence of the OPs question. If you can amend your answer with that explanation you get the +1. $\endgroup$
    – Alopex
    Commented Apr 17, 2019 at 11:56
  • 2
    $\begingroup$ @patstew The wavyness is important to its structure. The "corrugated" plastic that you're talking about introduces another failure mode. It's a lot more susceptible to buckling in the vertical members because they have no curvature at all. It's fairly easy to get the squares to buckle compared to actual corrugation. $\endgroup$
    – JMac
    Commented Apr 17, 2019 at 12:31

When you bend a piece of material, the resistance is provided by stretching the material on the outside part of the bend, and compressing the material on the inside of the bend.

A thin flat sheet bends easily because, physically, not of lot of stretching or compressing occurs when it bends.

When you give your book a fold, like a trough, that shape can not physically bend without a lot of stretching along the top edges, and a lot of compressing along the bottom of the trough. A very tiny bend would create a lot of stretching and compressing, and so the shape has a lot of bending resistance.


The other answers so far are technically correct, but none of them really seem to give a commonsense/intuitive and simple answer. So I'll have a go at one.

Imagine very slightly bending some kind of object downwards at one end, while holding the other end firmly horizontal. (It could be almost any object, could be paper, a branch off a tree, some plastic pipe, a long thin block of rubber, even a concrete block!) But only bending the object very slightly from one end, so you aren't breaking or fracturing it.

To bend at all, the top of the object has to stretch more than the bottom, because it is on the "outside" of the "curve" that forms when you bent the object.

(The bottom is squashed, or "compressed" as well, but it's easier to visualise if we ignore that and focus on what happens at the top side of the object)

Almost all materials and objects will resist stretching and compression, at least within some small limits. Some resist it massively (try stretching a steel bar). Others don't resist it much (try pulling some nylon cord or elastic, or a spring). Some will quickly break or tear (concrete and paper don't stretch well at all, they quickly break or tear instead). Other materials will stretch for quite a bit (steel is one, which is why it's used to reinforce concrete structures, unlike concrete it will keep resisting a stretching action).

That difference between how much the "top" and the "bottom" must bend, and the fact that if the object is bent even slightly, they must both bend and their curves will have different radii, is what determines the outcome, whether the object is your piece of paper, an entire notepad of paper, a tree branch, or a steel beam.

Back to your paper.

If your paper is flat, the top and bottom surfaces of the sheet are extremely close vertically. So it can bend or flop down, with almost no stretching of the top. The top surface does actually stretch a bit, which is why even the flopped sheet flops in a curved shape - there comes a point where if it bent more, the top surface would have to stretch enough more than the bottom surface, that the fibres of paper resist it, so it doesn't easily bend any more (without you creasing it or something).

But now suppose you bend the sheet along its length, even slightly. Now the "top" and "bottom of the curve aren't the two surfaces of the sheet, a tiny distance apart. They are the "valley" of the bent sheet and the two edges that are higher (the two sides of the sheet which bend upward). Those are a * lot * further apart vertically than the two surfaces were. So the sheet of paper still tries to flop, but it can't flop at all (or only microscopically or at the corners) because the "top" would now have to stretch a lot, just to have the sheet bend a tiny bit. Paper fibres don't stretch well (they are bonded to each other and resist stretching beyond a small amount; they will eventually tear instead). Gravity also isn't pulling down the far end of the sheet enough to force the end down even at the "cost" of tearing some of the fibres.

The end result is that now, the fibres at the "top" edges would have to stretch a lot to allow the paper to "flop" - so they can't stretch enough to flop - and they also aren't being pulled downward enough to tear (or to bend in other ways). So the sheet simply stays where it is. So now, the sheet acts a lot more rigidly.

You can see this by imagining trying the same thing, but with a sheet of silicone, or something else really floppy and flexible, instead of paper. Now your bending the sheet along its length doesn't work well, because the material itself doesn't resist its "top" surface or edges stretching much at all, so it can still find a way to flop down.

( ** I've simplified a fair bit. The main areas that I have simplified are - if the object is long and slender enough, it may end up finding other ways to bend, such as curving diagonally with one diagonal half up, and the other down. So if you try and hold a metal tape-measure out too far, this is what happens. It'll happen to your sheet of paper too, if it can. So there are other ways of bending. In engineering, where bending of a beam or column is usually a failure they're called "modes of failure", so steelworks much be designed with their 3d shape in mind, to prevent this kind of thing. Also many objects are complex or are not "elastic" beyond a small amount, for example your paper is made up of fibres bonded together, and how that bonding affects the fibres plays a big role too. Living wood from trees is made of different parts and those also interact so it splinters after a while, but doesn't fully break. But this should give you a good idea what's going on. Just be aware it's a simplified version)


"Curving" the paper increases the second moment of area, beacuse it effectively increases the distance of the paper cross-sectional area from the section centroid.

Stiffness of a cross-section is proportional to the square of distance from the centroid (also see parallel axis theorem), so curving the paper effectively multiplies it's stiffness by several orders of magnitude, thus the curved paper exhibits minimal displacement (= remains straight).

Here's another example of the same principle. A paper held horizontally bends under it's own weight. A perfectly flat paper held perfectly vertically is perfectly capable to carry it's own weight with minimal displacement. It is the same principle, radical increase in stiffness along the bending direction through increase of distance from the centroid.

Note: I'm using "curving" here as a verb even though it's probably not correct, so as not to confuse the action with the effect of paper bending due to gravity.

  • 1
    $\begingroup$ I don't understand your comments about paper bending. A paper held both horizontally or vertically will bend very easily, with literally the same amount of force. You definitely need to be more clear about what you're trying to get at there. More importantly... this seems to repeat what several other answers are talking about, but that really omits most of the complex and interesting parts of why this actually works. $\endgroup$
    – JMac
    Commented Apr 17, 2019 at 19:53
  • $\begingroup$ I clarified the vertical paper part. Hope it helps. Sorry to not have a "story" to tell here, but this is a very basic topic in structural mechanics and this is the correct scientific answer. $\endgroup$
    – insys
    Commented Apr 17, 2019 at 21:18
  • $\begingroup$ I guess I get this approach, I just think it over-simplifies it at the core compared to how interesting it is. I still think the paper standing vertical isn't a good example though, since structurally it will still fail under it's own weight in any realistic circumstance. Anyways, I retracted the downvote since I get what you were getting at with the paper example. $\endgroup$
    – JMac
    Commented Apr 17, 2019 at 21:44

All of these other answers are way too long and complicated (although probably more technically correct than my answer). When you bend the paper you're essentially creating a one-ended suspension bridge. Think about a suspension bridge that doesn't fully span a valley/canyon/river, i.e. one end hangs in the air. As you fold the sheet of paper, the sides that are more vertical become the suspension that holds the 'deck' up. Take off the suspension and the deck doesn't have sufficient strength to hold it's own weight.


I'm going to try another intuitive answer, since we seem to have quite a few technical answers here. As you say, it's all about the elastic properties.

When you hold the paper without a crease, you are giving a boundary condition to the surface - horizontal, in this case. Each point on the rest of the paper feels a downwards force of gravity, as well as parallel contact (electrostatic) forces to the surface. However, these forces are entirely in the direction of the curve, because the boundary condition you've set does not include any components along the translation direction of the cylinder (see the figure).

However, when you do induce those components, by changing the boundary conditions, you create forces in all (parallel to the surface) directions at every point. These forces are essentially there because the paper cannot be discontinuously changed (that's part of the elastic properties you mentioned). If the paper is long enough, the gravitational force may eventually win and the paper might fall (or rip, or crease).

enter image description here


I think it is because of structure of the paper. The fibres in the pulp that it is build from are aligned in one direction. This is also reason why it is much easier to tear the sheet in one direction (with the fibres), then in the other (across them).

  • 7
    $\begingroup$ The orientation of the fibers doesn't have much effect on this. It happens regardless. $\endgroup$
    – JMac
    Commented Apr 17, 2019 at 11:09
  • 7
    $\begingroup$ This incorrectly predicts that rotation of the sheet by 90 degrees would prevent the bending. $\endgroup$
    – user191954
    Commented Apr 18, 2019 at 15:14

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