Vacuum collapse -- why do strong metals implode but glass doesn't?

Question

This question has been puzzling me lately.

I'm sure you've seen demonstrations of metal containers imploding when evacuated. Here, for example, are two videos of vacuum collapse: experiment 1, experiment 2.

However, when the same experiment is conducted with a material as fragile as glass, nothing implodes or shatters. Two videos of the same experiment conducted with glass: experiment 3, experiment 4.

Nothing is special about the quality of the glass used in experiments 3 and 4. The glass is also not very thick. Yet, evacuating the glass, almost to 100% vacuum, doesn't so much as put a scratch on it, but the metal containers implode with great force. What is the reason for this difference?

My guesses are:

1. The total surface area of the glass used in experiments 3 and 4 are much smaller compared to the surface area of the metal in experiments 1 and 2. Greater surface area equates to greater force absorbed by the entire structure, even though the force per unit area remains the same.
2. Ductile deformation (metal) is fundamentally different from brittle fracture (glass), involving different mechanisms of atomic displacement.

(Blaise Pascal famously conducted several experiments with vacuum in a glass test tube in the 17th century.)

Please share your thoughts on this. Thank you!

Edit: I don't mean to say that glass never implodes/shatters in such experiments, obviously.

Answer 1

For a cylindrical pressure vessel loaded in compression (that is, vacuum inside), failure occurs by buckling instability in which a random and small inward perturbation of the stressed wall grows without bound at and beyond a certain critical load value. This is analogous to buckling instability in a thin column loaded in compression.

The characteristic which resists buckling instability is not the yield strength but the stiffness of the wall, which depends on its thickness and on its elastic modulus. The thicker the wall and the higher the modulus, the more resistant to buckling the cylinder will be.

The elastic modulus of common glass is about $48 \times 10^6$ psi compared to that of steel at $29 \times 10^6$ psi. The glass cylinder will hence be more resistant to implosion than a steel cylinder of identical size and wall thickness.

Note also that putting a scratch or microcrack in glass renders it weak in tension. In compression, however, the applied stress tends to press microcracks shut, so scratching a glass vacuum vessel does not cause the sort of catastrophic failure you expect to get when the glass is in tension.

Answer 2

Niels's answer is right at pointing the difference of stiffness as the main reason for the different behaviour of glass and steel. However, there is an additional element at play: relative wall thickness. In the videos, glass containers are a lot smaller than steel ones, but their walls are very thick when compared with their size - if we scaled the tubes in the last video to truck size, we would get a some inches thick wall instead of just a few mm, like the steel container.

Glass is a lot more brittle than metals under tension (as Niels said), a lot cheaper (by mass) and usually lighter. That combine to make usual glass containers a lot ticker than metal containers made for the same purpose. That's the reason we can easily crush an empty beer can with bare hands but crushing an empty beer bottle is a lot harder. The same difference holds if we replace bare hands with external air pressure.

Answer 3

There are two failure modes that need to be considered for a vacuum tube/vessel:

The first is tensile/compressive failure, caused by the applied stress exceeding the material's ultimate tensile stress. This is largely governed by the wall thickness and ultimate tensile stress. However, note that in a ductile material, this is preceeded by a period of plastic deformation.

The second is buckling (also known as elastic instability failure). In this case, the structure of the vessel becomes unstable to small perturbations, which can cause a failure below the load of the first case. This case is largely governed by wall thickness and elastic modulus.

Vacuum vessels made from either metal or glass can fail by either of the two modes, depending on their design and the relative values of the parameters mentioned.

Another thing to bear in mind though is that most metals are much more ductile than glass, which is very brittle. So, a failure in the two cases will look very different. A metal tube will tend to deform quite a lot before it ruptures and may 'crumple up'. However, glass cannot deform much before it breaks, so if it fails it will tend to shatter very suddenly. This tends to be more dangerous and is probably why many glass vessels have thicker walls than metal ones - because they need a higher factor of safety.

Answer 4

"Strong" is a vague and overloaded term.

Does it refer to?

• tensile strength (resistance to stretching).
• compressive strength (resistance to squishing)
• hardness (resistance to cutting)
• fracture strength (resistance to fracturing).

You then additionally have the difference in failure types:

• elastic (it stretches but then returns to original)
• plastic (it starts to deform slowly and then stays deformed)
• brittle (when it starts to fail it fails immediately).

Glass and steel have very different behavior and failure modes. Drop a metal can and it will dent or deform, glass on the other hand will chip or fracture. This doesn't mean one is stronger or weaker when under load, it just means they have different failure thresholds and behaviors. Metal has reasonably high fracture strength but plastic deformation - it is much easier to dent or bend than shatter. Glass on the other hand will either show no damage at all or be fractured/chipped. You can't usually dent it.

Answer 5

Expanding on Pere's answer about the relative wall thickness being the criteria for failure.

The document below describes the derivation of the stress profile in a thick walled cylinder (known as Lame's equations)

http://courses.washington.edu/me354a/Thick%20Walled%20Cylinders.pdf

For an external pressure and zero internal pressure, like in your video examples, the stress is largest at the inner surface of the wall

$$\sigma_\theta = \left[ \frac{-2P_o r_o^2}{r_o^2-r_i^2}\right]$$

If we were to only consider pressure vessels of the same wall thickness $r_o = r_i + t$, then the expression for stress is

$$\sigma_\theta = \left[ \frac{-2P_o \left( r_i^2 +2 r_i t + t^2\right)}{2r_i t + t^2}\right]$$

Mathematically, the numerator grows larger than the denominator for large $r_i$ (quadratic vs linear behavior). Physically, that means that the stress is greater for larger containers, like your metal examples. The stress is lower for smaller containers, like your glass examples.

Answer 6

The failure mechanism for a cylindrical container under vacuum is buckling.

If we look at the equation predicting buckling pressure we can figure out which variables are most responsible for the differences you see in the videos:

$$P_b=\frac14 \frac{E}{1-\nu^2}\left(\frac{t}{r}\right)^3$$

Where $P_b$ is buckling pressure, $E$ and $\nu$ are the modulus of elasticity and poisson's ratio of the material, and $t$ and $r$ are the wall thickness and radius of the cylinder. This is neatly split into a material factor and a geometry factor.

For glass: $$\frac{E}{1-\nu^2}=\frac{70 \, \text{GPa}}{1-0.22^2}= 74\text{GPa}$$ And Mild Steel: $$\frac{E}{1-\nu^2}=\frac{200 \, \text{GPa}}{1-0.3^2}= 220\text{GPa}$$

So from this it seems your intuition that steel is the superior material is still definitely true. So what's going on?

It must be the geometry. The glass containers were designed to handle vacuum and their thickness to radius ratio is much better suited for it.

So let's find out approximately how thin that unreinforced tanker wall must have been:

$$ 1\text{ATM}=\frac14\;220 \, \text{GPa}\;\left(\frac{t}{1\,\text{m}}\right)^3$$ $$ t \approx 1 \text{cm}$$

You may be wondering, well why are the steel containers so thin? Wouldn't they rupture under pressure then?

For comparison, let's look at the equation for rupture pressure:

$$P_r=\sigma_y \frac{t}{r}$$

Where $P_r$ is rupture pressure, $\sigma$ is the stress limit of the material, and $t$ and $r$ are still the wall thickness and radius of the cylinder. This is also neatly split into a material factor and a geometry factor.

For Mild Steel: $$\sigma=200 \text{MPa}$$

So that tanker could handle a pressure of:

$$P_r=\sigma_y \frac{t}{r}=200\text{MPa} \frac{1 \text{cm}}{1 \text{m}}= 2\text{Mpa}$$

So it could handle a pressure of roughly 20 atmospheres before rupturing. That's probably way more pressure than it would ever need. So then the wall is actually much thicker than it would really need to be. It's designed that thick so that it won't collapse under its own weight and so it won't buckle (extremely) easily. This brings us to the reinforced tanker. The reason it has support structure in bands around it is to allow the metal to be thinner since it doesn't actually need to be that thick to hold the pressure of the fluid.

Then you might be asking if the wall was so much thinner for the reinforced tanker, why didn't it collapse at a much weaker vacuum? For that we need to examine the buckling pattern. The unreinforced tanker just folded with two portions moving out and two portions moving in. This is the simplest buckling mode and has the lowest pressure threshold. The reinforcements on the other tanker prevented that mode from occurring by providing additional rigidity, so the buckling that occurred followed a much more complex pattern.