Why thin metal foil does not break like a metal stick? 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?
 A: 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.
A: 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.
A: 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.
A: "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.
