When we design a pipe or pressure vessel, we look at hoop stress and longitudinal stress and then look at the material properties. We assume that the material reacts to the load as a solid. However This answer and comments claim:

Wall thickness doesn't matter if you try to contain a pressure that's greater than the yield strength. The material will permanently deform no matter how thick it is. It's like i'm trying to contain a gas bubble in a soft mud. No matter how much mud I add around it - i can't contain it.

Is this true? Do metals behave significantly different when subjected to pressures beyond their yield strength?

This could be tested experimentally: To arrive at a very thick walled pipe we could drill a hole into a block of gold (ultimate tensile strength 220 Mpa). Build the block large enough and the pipe wall thickness should withstand the immense pressure - So when we pressurize the hole to something > 220 MPa (a pressure common in water cutting, but at much other places.)

If the statement cited above is correct, we should see a different failure mode than a thin walled bursting vessel - for example creep. Or not, if material behaviour is dictated by the stress in the vessel material, not in the medium contained.

If we look at the pressures involved for even a soft metal like gold, we see that this a mostly academic question - for the vast majority of applications, the pressures encountered are well below the yield strneghts of the materials so we simply don't need to consider additional failure modes.

The above experiment is just an example, to show that such a situation could indeed be engineered. I'm sure other experiments can (and have) been done. My intution is that the material will start to flow in some way (towards the least resistance) but I'd like to know for certain.

  • $\begingroup$ In practice it will be almost guaranteed to fail at the interface between the feed pipes from the pump and the lead block, which won't be very interesting. You might be able to somehow enclose some explosive material entirely within a lead block and then detonate it, or something similar. $\endgroup$
    – alephzero
    Commented Aug 6, 2020 at 10:25
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    $\begingroup$ Bomb calorimeter anyone? $\endgroup$
    – Solar Mike
    Commented Aug 6, 2020 at 10:26
  • $\begingroup$ cambridge.org/core/journals/journal-of-materials-research/… - an experiment with gold at pressures higher then it's yield strength. The answer to the question if additional failure modes pop up is probably in there. $\endgroup$
    – mart
    Commented Aug 7, 2020 at 5:20
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    $\begingroup$ I've deleted my previous answer as I realize now it is incorrect. Having gone back and done the arithmetic, an internal pressure greater than the material yield strength will always result in a hoop stress greater than material yield strength. My apologies. $\endgroup$
    – jko
    Commented Aug 7, 2020 at 10:16
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    $\begingroup$ Reminds me of a lawsuit where I was a witness; One attorney never could understand how the internal gas pressure was in psi and the steel strength was in psi , and there was NO relationship. $\endgroup$ Commented Aug 10, 2020 at 20:25

4 Answers 4


A rupturing pressure vessel consists of an interplay between rupture of the walls, that causes a reduction in pressure via venting, while rupture of the walls also weakens the ability of the structure to sustain more pressure. Which one declines faster, and to what extent, determines what happens next.

If the rupture on the wall is small and does not propagate, or the pressure vessel has a relief valve, then excess pressure inside it is vented out in a controlled manner.

As a rupture forms on the pressure vessel, the structural integrity of the vessel decreases. For example, if the material is brittle such as glass, cracks in the pressure vessel would propagate quickly due to the low fracture toughness. Unless the internal pressure is relieved equally quickly (unlikely, because the speed of sound in solids is usually higher than gases), then rupture of the pressure vessel will accelerate, possibly leading to a mechanical explosion.

If the material from which the pressure vessel is made of is ductile, such as lead, rupture of the vessel would probably leave you with large pieces and bent metal. There are studies done on the fragmentation pattern of black powder driven bombs, and the metal pieces are usually fewer in number and larger.

Lastly, if the rise in pressure inside the pressure vessel was near instantaneous, for example due to detonating explosives, regardless of the material properties of container, you would end up with numerous, fine bits of container.

  • $\begingroup$ those are the failure modes I specifically did not ask about. Unless you can somehow show that the statement I quoted is wrong, of course. $\endgroup$
    – mart
    Commented Aug 6, 2020 at 10:59

As jko points out in a comment, the hoops tress is always higher than the medium pressure, so the case as presumed in the question cannot happen and we don't see "ineresting" new failure modes.

I threw away my notes since but this can be shown using the formula for hoop stress found here - hoop stress approaches medium pressure as the inner diameter goes towards zero, so it can never be lower.


Short answer

No. Although material may yield locally, it does not have to lead to failure. There is also a design with multiple pre-stressed walls, which may contain fluid at a higher pressure than material yield strength without any yielding.

Additional information

Elastic approach

Stresses in your answer are calculated based on fully elastic behaviour. When you increase the internal fluid pressure, the highest stress is the hoop on internal surface, but there is also radial stress component on the internal surface, which is basically minus fluid pressure. So if you apply Tresca yield criterion, elastic fluid pressure limit $P_{max,elastic,Tresca}$ for the cylindrical shell is just a half of its yield strength $R_e$ even for infinitely thick wall (you can see this from the following formula by taking internal diameter $D_i\rightarrow 0$). $$P_{max,elastic,Tresca} = R_e\frac{D_e^2-D_i^2}{2D_e^2}$$ Although theoretically correct, I have never seen this formula in the practice. It is interesting, that formula for minimum thickness design based just on maximum hoop stress is used in practice, e.g., in EN 13480-3 for thick-walled pipes and possibly also in ASME standards (it was definitely used in the past, I am not sure about now).

Plastic limits

That being said, the vessel should not break when internal surface is yielding, because rest of the thickness is just fine. When you want to break it, you have to accomplish yielding in the whole thickness. Limiting pressure can theoretically go higher than yield strength of the material and the limiting factor would be allowable contraction of the material (I think this criterion can be found in ASME BPVC alternative rules): $$P_{max,plastic,Tresca} = R_e\cdot \ln\left(\frac{D_e}{D_i}\right)$$ Since the second principal stress in cylindrical vessel is half of the first, the stress state is such that there is maximum difference between Tresca and Mises yield criteria. So the maximum pressure would be actually using Mises criterion (this one is used in EN 13480-3): $$P_{max,plastic,Mises} = \frac{2}{\sqrt{3}}P_{max,plastic,Tresca}$$ If you need really high pressures, which would be difficult to achieve with single wall, you can use multiple walls which are carefully prestressed. Pressure limit of such a design is approaching plastic limit with number of walls going up. Note that plastic formula do not take into account hardening of the material, which may in some cases lead to significant increases of limit pressures.

Usual practice

In practice, most pressure vessel and piping design is done using formulae based on thin shell theory and maximum principal stress criterion, which leads to simple formula for limit pressure: $$P_{max,simple} = 2R_e\frac{D_e-D_i}{D_e+D_i}$$

This lies somewhere between fully elastic and fully plastic states of the wall and is much easier to work with than other formulae. With slight modifications, it is used in many pressure vessel, boiler and piping standards.

Some additional remarks

Although formulae for thickness design are not based on elastic limit, there is a safety factor involved, so there may be no yielding. However, partial yielding can happen during hydrotest, where the test pressure is higher than design pressure, for example in EN 13445, safety factor may be 1.5, but then the test pressure may be 1.43 of design pressure. With sufficient material and fabrication quality, this is not a problem, on the contrary the vessel may actually be improved due to beneficial pre-stress and material hardening. This procedure is taken to its extreme in cryogenic pressure vessels (see slide 7), where the hydrotest does basically cold stretching and yield strength of special austenitic steels is increased significantly.


Maybe; Like most things it depends on specifics. Even at pressure below that which would cause general yield , there is local yield at stress concentrations and imperfections. This is shown when using acoustic emission to monitor a vessel during hydrotest; Typically 1.5 x design pressure. Sounds are produced by local plastic strain. Then methods such as ultrasound (UT) and radiography (RT) are used to evaluate the source. Any case I know of did not reveal a significant problem. There has probably been at least one case of failure of the vessel caused by overpressure. Over 60 years ago (long before I was an engineer) a pressure vessel disintegrated in a refinery in Whiting IN, (Standard Oil , IN) It was some kind of hydrocracker and had an internal explosion; I do not recall ever seeing an estimated peak pressure. Unfortunately it had low toughness even at the elevated operating temperature. Pieces were blown many hundreds of feet. I happen to have a fist size shard that was being disposed of many years later which reminded me of the accident.


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