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Consider a metal stick, say iron or aluminum. From the experience, even if it's resilient, bend it forward and backward a couple of times, it would be broken.

However, consider a thin iron foil or thin aluminum foil. From the experience, we know that it could be bend forward and backward for almost as many time as time was permitted.

How to explain this in solid states? Why was it that the thin foil seemed to be much more deformable than stick?(Does it has anything to do with the fact that in the normal direction, the metallic bound was weak?) Why thin foil doesn't break?

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    $\begingroup$ You won't break a metal stick if you bend it with bend radius >> stick thickness. Same with the foil. $\endgroup$
    – Dmitry Grigoryev
    Commented Nov 12, 2020 at 7:39
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    $\begingroup$ The current answers focus on the bend radius, but that cannot be the only thing to consider. Some metals or alloys are brittle. Tin foil needs a heat treatment to become bendy. $\endgroup$
    – smcs
    Commented Nov 12, 2020 at 13:12

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Almost all solid metals are made up of individual small crystals called grains. A small stretching movement will simply stretch the crystal lattice of each grain a little, so the whole thing bends.

When you flex thin foil, it is so thin that the stretching distance is small and the grains can deform to match.

But with a thicker rod, the stretching is much bigger and the stress force it creates in the material is much higher. The outermost grain boundaries (the furthest stretched) will begin to pull apart, creating surface cracks in the metal. Each time you flex it, these cracks grow until they pass right through and the thing snaps in two. If you look closely at such a "fatigue" fracture with a magnifying glass, you can sometimes see the individual crystals forming a rough surface. Or, sometimes you can see the individual "waves" as the crack progressed at each stress peak.

The formation and behaviour of these grains, and the factors which control them, is the principal phenomenon studied by metallurgists.

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    $\begingroup$ This answer, as currently written, may suggest that the phenomenon is unique to crystalline materials. But the same thing happens with optical fibers of amorphous silica glass. Independent of the material, bending a thicker solid produces higher stresses on the surfaces for a given curvature. $\endgroup$
    – Chemomechanics
    Commented Nov 11, 2020 at 21:04
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    $\begingroup$ @Chemomechanics Really? "But with a thicker rod, the stretching is much bigger and the stress force it creates in the material is much higher." states a general principle and does not confine itself to crystal structures, I would suggest that the average reader will understand that. It seldom helps to distract to distract them with indirectly-related side-issues. $\endgroup$
    – Guy Inchbald
    Commented Nov 12, 2020 at 9:40
  • $\begingroup$ The actual crystals require a good microscope and chemical etchant. They are typicaly on the order of 100nm in size. You can often see beach marks through a magnifying glass but those are structures caused by the extension of a crack with each fatigue cycle, not the crystals themselves. $\endgroup$
    – BobTheAverage
    Commented Nov 12, 2020 at 17:18
  • $\begingroup$ What one can see depends very much on the grain size, which varies enormously. $\endgroup$
    – Guy Inchbald
    Commented Nov 12, 2020 at 19:05
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"Thin" is a relative term, but let's assume we're talking a foil that's 0.01 to 0.02 mm thick (i.e., kitchen aluminum foil). Let's also assume that our foil is a soft alloy -- i.e. nearly pure aluminum or iron.

If I take a piece of kitchen foil and I bend it with my hands, the bend radius is likely going to be no less than 5mm. So the ratio between bend radius and thickness is something like 500:1.

If I take a 1cm square bar of 1100 aluminum and I bend it on a 5m radius, it'll survive a lot of bending. Ditto a 1cm square bar of 1020 steel.

This is because the amount of stretch the material must undergo is small, and it's because the material is soft.

I know from experience that you see the same effect with wires. To break thin material, you must bend it using an instrument (like pliers that have a nice sharp edge) that makes the radius on the order of the material thickness. Do that, and it'll break. Keep the radius large with respect to the material thickness, and it won't.

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  • $\begingroup$ If I fold a piece of kitchen aluminum foil in half, the bend radius for that 180 degree bend is far less than 5mm or even 0.5mm- it's hard to tell the exact radius but it is on the order of the material thickness, and it does not break from that. $\endgroup$
    – Peteris
    Commented Nov 11, 2020 at 21:26
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    $\begingroup$ @Peteris it will break if you do that a couple of times back and forth, to induce fatigue. $\endgroup$
    – leftaroundabout
    Commented Nov 11, 2020 at 22:31
  • $\begingroup$ @leftaroundabout: I think it takes a few more times, and it's hard to get each bend to lie on top of the last. It's been long enough since I've done it that I couldn't describe the process to get a repeatable experiment -- hence I vectored off to talking about the equivalent operation using wire. If someone has some spare time they could try to quantify this -- it may even be a good science fair experiment if there's someone of school age out there that wants to give it a whirl. $\endgroup$
    – TimWescott
    Commented Nov 17, 2020 at 19:16
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    $\begingroup$ @TimWescott well, with wire it works just as well – in fact I've often resorted to fatigue when I needed to clip wire but didn't have pliers at hand. You just need to once bend it really tight (this can require squeezing it between two hard surfaces, but not necessarily pliers – I find usually my fingernails are enough), then back, the second time it'll already be much easier to get it in the tight bend. Vigorously repeat some 20 times, and you'll be able to snap it with bare hands. (Probably this doesn't work with all wire materials, but it did work with all I've tried it with.) $\endgroup$
    – leftaroundabout
    Commented Nov 17, 2020 at 22:13
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Let me use just a simple geometrical reason.

You can in principle bend a monodimensional row of bound particles to a wide angle affecting only the angle of the bond at about the pivotal point.

Conversely, if you bent a 2-D ensemble of particles (variously bound, it can be multiple rows as above slightly interacting or a thoroughly connected lattice, etc...) particles which are far from a point chosen as the pivot must be stretched apart.

Again I am sure things will depend on more defined solid state parameters, but at the core is the geometrical consideration above.

Bigger the sample, bending is effectively stretching.

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    $\begingroup$ Yes - the typical example is the old puzzler about how many times you can fold a sheet of paper, where the number of layers is quite clear. $\endgroup$
    – Carl Witthoft
    Commented Nov 11, 2020 at 15:31
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The phenomenon that causes a metal wire to break after being bent and straightened a couple of times is called work hardening. First the wire will become more strong at the point of deformation but at the same time more brittle, and then it will break as it can’t respond to the force by bending anymore.

Thinner materials are less susceptible to this because the same bend radius causes less deformation to the crystal lattice, and also because the deformation happens in different points each time. I’ve just destroyed some pieces of foil by bending them several times, rather gently, while holding them in pliers to localize the deformation, so the same principles are at work here as is the case with other metals.

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