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Clarification on Thick Walled Pressure Vessel Failure

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Unread postAuthor: Moonbogg » Mon May 24, 2010 5:18 pm

For all practical purposes and in any case I can think of, yes it will yield long before the pressure equals the material yield strength. From just my experience, and this is only 1 example, look at the venom's chamber. The material has a yield of at least 35kpsi, usually around 40kpsi. The cannon will yield at around 800psi, and increasing the wall thickness by even an inch won't get you close to 40kpsi. This is about 4" dia.
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Unread postAuthor: DYI » Mon May 24, 2010 6:49 pm

For all practical purposes and in any case I can think of...


I'm sorry, I think you've got the wrong thread. See, in this one I was asking a question about the practical implications of the asymptotic behaviour of a certain equation as one variable (ID) is held constant and another (OD) approaches infinity.

Responses to the following question would be appreciated:

btrettel's equation limits the yield pressure of any vessel to a value below that of the yield strength of the material the vessel is made of. Is this behaviour realistic, or a result of simplifying the situation?
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Unread postAuthor: Moonbogg » Mon May 24, 2010 7:14 pm

Its a result of simplifying the situation. If you have a 1"ID chamber with a 10ft wall thickness, your pressure vessel yield pressure will be greater than the material yield strength. Good enough? If not, then I am too stupid to talk to you.
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Unread postAuthor: btrettel » Mon May 24, 2010 8:42 pm

The asymptotic behavior doesn't surprise me. It makes intuitive sense to me. If the walls are infinitely thick, then the area where the pressure is applied is negligibly small. So think of the pressure as being applied in a two dimensional square (circles can be approximated by squares, especially when comparing against a far larger area). What'll be the stress in either direction? It'll be the same as the pressure. So the material yields when the pressure is the yield stress.

Another way to think about it is like this: Does the stress in a rod in axial tension change as the rod gets longer? No, it doesn't. A similar thing is happening here.

As for whether the result is physical or my interpretation is correct, I don't know. It makes sense to me, but I could be wrong. I'll know for certain after I take the failure theories class.
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Unread postAuthor: Moonbogg » Tue May 25, 2010 9:26 am

I did a test. I made a 1"ID tube with a 10ft wall thickness and entered the material yield psi for the internal pressure. It failed at .5 safety factor. Interesting. It seems it doesn't matter how thick the wall is after all. Oh well, it was a good guess.
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Unread postAuthor: DYI » Mon Apr 16, 2012 9:54 pm

It is my sad, sad duty to report that btrettel made a minor error in his yield pressure calculation. In his use of the Von Mises yield criterion he added the σ<sub>1</sub>σ<sub>2</sub> term, when he should have subtracted it.

The correct form of the equation (which I've checked very thoroughly) simplifies to:

P<sub>yield</sub>=σ<sub>yield</sub>*(r<sub>external</sub><sup>2</sup>-r<sub>internal</sub><sup>2</sup>)*(r<sub>internal</sub><sup>4</sup>+3r<sub>external</sub><sup>4</sup>)<sup>-(1/2)</sup>

Which, in the case of r<sub>external</sub> >> r<sub>internal</sub>, becomes P<sub>yield</sub>=σ<sub>yield</sub>/3<sup>1/2</sup>.

I'm so sorry...
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Unread postAuthor: Moonbogg » Tue Apr 17, 2012 12:23 am

DYI wrote:It is my sad, sad duty to report that btrettel made a minor error in his yield pressure calculation. In his use of the Von Mises yield criterion he added the σ<sub>1</sub>σ<sub>2</sub> term, when he should have subtracted it.

The correct form of the equation (which I've checked very thoroughly) simplifies to:

P<sub>yield</sub>=σ<sub>yield</sub>*(r<sub>external</sub><sup>2</sup>-r<sub>internal</sub><sup>2</sup>)*(r<sub>internal</sub><sup>4</sup>+3r<sub>external</sub><sup>4</sup>)<sup>-(1/2)</sup>

Which, in the case of r<sub>external</sub> >> r<sub>internal</sub>, becomes P<sub>yield</sub>=σ<sub>yield</sub>/3<sup>1/2</sup>.

I'm so sorry...


Well, I am surprised you haven't considered making the entire thing out of graphene. The cannon will weight 2 pounds and be good to 50x.
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Unread postAuthor: DYI » Tue Apr 17, 2012 9:54 am

Here's a little plot to illustrate the effect of my correction on the yielding formula:

Image

The purple line is btrettel's original equation, the blue line is my correction, and the brown line is Barlow's formula. The pressure vessel has an internal diameter of 1", with external diameter (the x-axis) varying from 1" to 3", which is getting near the limiting value for both thick-wall equations and covers the majority of practical vessels (by the time you start doubling the wall thickness to gain 3% yield pressure, it's time to take a new approach :wink: ).

It's apparent that they all behave very similarly in the thin-walled limit, hence the difficulty in recognizing the problem - they all give reasonable estimates for familiar pipes. However, simply plotting them all against each other illustrates a problem which we really should have noticed earlier: btrettel's formula predicted higher yield pressures than Barlow's formula which, intuitively, cannot be right, when one considers the assumptions made in Barlow's derivation.

The vessels which I thought were yielding internally at 100kpsi were actually doing so at about 62kpsi. I do have a few ideas as to reducing this problem, but I won't have time to do the necessary work on them for a while yet. I'll probably start a new thread on this after exams are done.
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Unread postAuthor: DYI » Sun May 13, 2012 6:13 pm

A correction to the above: the assumption of equal stress throughout the wall used in the thin-walled vessel derivation yields the formula P<sub>yield</sub> = σ<sub>yield</sub> * (D<sub>outside</sub> - D<sub>inside</sub>)/D<sub>inside</sub>, which for a given internal diameter increases linearly with external diameter. Barlow's formula substitutes D<sub>outside</sub> for D<sub>inside</sub> in the denominator to be conservative, in which case the intuitive conclusion mentioned above cannot be made.

It's not as easy to see the problem as I had originally suggested.
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Unread postAuthor: ramses » Wed May 16, 2012 8:22 pm

DYI wrote:I recently noticed that btrettel's equation approaches a limit of p<sub>yield</sub> = YS as the ID is held constant and the OD increases. Is this accurate? Will any arbitrarily thick walled pressure vessel yield when the internal pressure exceeds the yield strength of the material?


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Unread postAuthor: DYI » Wed May 16, 2012 8:52 pm

Ramses, please note that the location of the asymptote in the section you just quoted is incorrect - my correction can be seen in the post above yours. This is quite an old thread, and the update is very recent in comparison.


Also, knowing more about 3-D stress states and yield criteria now than I did when you posted your thread, I can say with greater certainty that my speculation as to how DACs contain such high pressure is likely correct. In particular, the anvils apply a compressive stress along the longitudinal axis of the "tube" which would tend to cause simple compressive rupture. The internal hydrostatic pressure tends to counter this effect, preventing the tube from collapsing. The same force is not applied to the entire surface of the gasket - it decreases out toward the edges, so the outer edge does not buckle either.

That's the intuitive explanation at least. If you use Lame's (missing accent, I know...) equations for an infinite cylinder and the Von Mises yield criterion you can get a basic idea of how the stress terms can be made to cancel, so long as the longitudinal compression increases in step with the internal pressure. Obviously end effects are *very* important here, so this is far from the full story, but at least it's analytical and "makes sense", as it were.
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