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Why does glass break more easily than plastic. It seems like it should break less easily than plastic because it's stronger than plastic.

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Glass is brittle because it has many microscopic cracks in it which act as seeds for a fracture. If you can make glass without these cracks, as is done in fiberglass, then it is not so fragile.

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  • $\begingroup$ Even with the cracks, glass still has a higher observed strength than plastic. That's why I asked this question. I also added that information into my answer because of your answer. $\endgroup$ – Timothy Jan 16 '17 at 6:04
  • $\begingroup$ @user46757 But strength has virtually nothing to do with this process... $\endgroup$ – lemon Jan 16 '17 at 9:23
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Polymers that aren't brittle are glasses with long-chain or crosslinked long-chain molecules (or mixtures). Glass, on the other hand, is just a badly crystallized bunch of small not-very-sticky silicate molecules.

So, when a polyethylene sample fails under tensile stress, it stretches (and doesn't snap back), but only the linking between adjacent chains is broken, the chains may stay intact (and just straighten out a bit under stress). Because the chains are intact, there's stiffer resistance to tension when stretched (because the chains are now oriented axially by tension, having been previously randomly jumbled and disoriented).

Glass, having no such inner fibers, just opens up a crack and lets go. The high stress at the leading edge of a crack generally makes the first crack to open propogate all the way across a tensile strength specimen.

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Glass actually has a higher tensile strength than plastic but it still breaks more easily when it falls. Suppose you have a nanosmooth floor made of the hardest material diamond made by brittle fracturing a diamond crystal along its cleavage plane. When ever a sphere of any substance lands on it, it will dent when the area of contact has more strees than it can handle which is similar to its tensile strengh if it's a malleable material or form a circular crack around the area of contact if it's a brittle material. When ever a sphere of a material hits the floor, the size of the area of contact in the collision is nonzero because it's not perfectly rigid. The maximum stress is at the area of contact and is approximately the mass of the sphere times the collisional acceleration divided by the square of the size of the area of contact. A sphere made of a brittle material will form a crack when the maximum stress exceeds a certain value approximately equal to the strength of the material. We want to determine the speed a sphere must hit the diamond floor at in order to form a crack as a function of sheer modulus, strength, and density.

From the square cube law, the speed is independent of the size of the sphere for a given density, sheer modulus and strength. According to my answer at Why are so many forces explainable using inverse squares when space is three dimensional?, the square cube law is not exactly true but is very close to true. I think the theoretical strength of a smooth sphere of a given stable material actually approximately varies as the reciprocal of the log of how many times larger than an atom it is so the fraction of its theoretical strength lost from doubling its size also varies as the reciprocal of the log of how many times larger than an atom it is. According to the Wikipedia article Fracture mechanics, when a brittle material has cracks, it has a lower strength. The speed a sphere must hit the diamond floor at in order to crack is a function of its observed strength, density, and shear modulus and not its theoretical strength. Although the question of whether a crack will start in the first place follows the square cube law, the question of how far in will propagate when it does start doesn't. According to the square cube law, the size of the area of contact divided by the size of the sphere is independent of size of the sphere for a given shear modulus and strength. When you multiply the density, strength, and sheer modulus all by the same amount, the speed and what fraction of the size of the sphere the size of the area of contact is don't change either. Finally, when you multiply the strength by the same amount as the sheer modulus while leaving density and size constant, the speed multiplies by the square root of the amount you multiplied the sheer modulus by and the fraction of the size of the sphere that's the size of the area of contact doesn't change.

For a sphere of a given size, density, and speed it must go in order to form a crack, the stress varies as the collisional acceleration divided by square of the size of the area of contact but the collisional acceleration varies as the minus second power of the size of the area of contact so the stress varies as the minus fourth power of the size of the area of contact. The sheer modulus is the stress divided by the strain but the strain varies as the size of the area of contact so the sheer modulus varies as the minus fifth power of the size of the area of contact.

While leaving the size and density constant, you can multiply the strength by any positive real number r while leaving the shear modulus constant by first multiplying the sheer modulus by r^-5 and the strength by r^-4 then multiplying the sheer modulus and strength by r^5. The first operation doesn't change the speed then the second operation multiplies the speed by r^2.5 so in total, multiplying the strength by r multiples the speed by r^2.5. You can also multiply the sheer modulus by r by first dividing the strength by r then multiplying the sheer modulus and strength by r. The first operation divides the speed by r^2.5 and the second operation multiplies the speed by r^0.5 so in total, multiplying the sheer modulus by r divides the speed by r^2. The density can be multiplied by r by first multiplying the density, sheer modulus, and strength by r then dividing the sheer modulus and strength by r. The first operation doesn't change the speed and the second operation divides the speed by r^0.5. Finally, the speed is independent of the size of the sphere for a given strength, sheer modulus, and density. Therefore, the speed a sphere must hit the diamond floor in order to break is some constant times sheer modulus^-2 times strength^2.5 times density^-0.5. I don't know what power of strength, shear modulus, and density the speed required to create a crack that propagates all the way through breaking the sphere into more than one piece varies as.

I once read that the theoretical strength of glass is 31 GPa, almost as high as its shear modulus and that a tensile strength of 14 GPa was observed in wires of it drawn out from the molten state in a vacuum but can't find that web page anymore. According to https://en.wikipedia.org/wiki/Fracture_mechanics, the reason glass has a much lower observed strength is because it has been scratched creating a crack and any tension applied to the object gets magnified at the tip of the crack and once the tension at the tip of the crack gets magnified to the theoretical strength, the crack will grow making tension at the tip of the crack get magnified more easily causing a runaway effect of the crack growing at the speed of sound in the material. That's why glass appears to break instantly when the tension becomes too high. The deeper it gets scratched, the less tension is needed to make it fracture. It's the observed strength of the sphere from its surface cracks, not its theoretical strength that's used to predict whether those cracks will further propagate into cracks large enough to be visible. Even with the cracks, glass is still stronger than plastic. It also has a similar density to plastic. It requires a lower speed of collision with the diamond floor than plastic in order to break into more than one piece only because it's subject to more stress in the collision before any of its microscopic cracks propagate because it has a higher sheer modulus.

A fused quartz glass sphere etched nanosmooth dropped onto the diamond floor would be going nowhere near fast enough to break because the strength would be similar to the shear modulus but the speed required to break varies as the 2.5 power of strength and the -2 power of sheer modulus.

On the other hand, it could hit a sharp tetrahedral diamond tip made by fracturing it along its cleavage planes at a much lower speed and still form a crack. That's because diamond is much harder than fused quartz and the few atoms of the glass that contact the diamond tip are not strong enough to stop the whole glass sphere in its tracks. The slower it hits the diamond tip, the smaller the crack it will get. Hitting a diamond tip is actually far off from following the square cube law. A nanosmooth fused quartz sphere of twice the size hitting the diamond tip at the same speed will get a crack more than twice as big.

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protected by Qmechanic Jan 16 '17 at 13:21

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