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Clarification on Thick Walled Pressure Vessel Failure

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Clarification on Thick Walled Pressure Vessel Failure

Unread postAuthor: DYI » Sat May 15, 2010 3:02 pm

From what I can tell, the maximum shear stress theory relates the maximum shear stress (before failure) to the material's yield strength by
MSS = YS/2, and that the maximum stress in the case with internal pressure only is found at r = r<sub>internal</sub>.

Based on this and some simple manipulation, I'm lead to believe that:

p<sub>failure</sub> = [YS*(r<sub>o</sub><sup>2</sup> - r<sub>i</sub><sup>2</sup>)]/[2*(r<sub>o</sub><sup>2</sup> + r<sub>i</sub><sup>2</sup>)]

Where:
p<sub>failure</sub> = internal pressure causing the vessel to yield
YS = yield strength of the material
r<sub>o</sub> = outside radius of the vessel
r<sub>i</sub> = inside radius of the vessel

Is this reasoning correct?

Some sample calculations show what seem to be unrealistically low failure pressures. It estimates the failure pressure of 2" SCH 40 steel pipe at ~2800psi, whereas the thin-walled approximation yields ~6000psi.
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Unread postAuthor: jrrdw » Sat May 15, 2010 5:54 pm

I can't tell you about the math, but is it not possible that processing the thin walled pipe differently make it stronger? Carbon percentage (higher or lower)??? Case hardening??? Different blends of iron ore???
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Unread postAuthor: Moonbogg » Sat May 15, 2010 6:53 pm

I duno, looks like you might want to take that post to a physics forum. Looks a little too brainy for a lot of us, especially me.
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Unread postAuthor: Gun Freak » Sat May 15, 2010 8:01 pm

Agree with Moonbbog, although Rag may have a clue...
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Unread postAuthor: btrettel » Sat May 15, 2010 10:21 pm

Disclaimer: I'm not a practicing engineer, but I can say what I remember from my mechanics classes. This might be wrong, so take it for what it's worth.*

Yielding means that the material deforms permanently. I don't know if you want to use that stress because the pressure vessel won't "fail" as most would imagine here. You likely want what's called "ultimate failure". I think the ultimate stress can be substituted in for the yield strength. I'm not certain because the stress-strain curve becomes non-linear beyond about the yield point and I can't recall exactly how all these failure theories were derived and if that is important.

No one uses the maximum shear stress theory. Switch to something like distortion energy theory. The distortion energy theory is less conservative, if I remember correctly.

It's also important to know precisely what material is used and its material properties. Many pressure vessels are made with processes that make the materials very hard, so textbook figures for yield and ultimate strengths might be wrong. I'm not familiar with the exact processes used for pipe, but I think pipes are drawn out, which should have the same hardening effect.

All of the things I mentioned in the above paragraphs contribute to the low estimate for ultimate failure pressure.

The thin-wall stuff is plainly wrong for thick-walled vessels. The thick-wall equations literally reduce to the thin-wall ones for very thin walls. I wouldn't expect the thin-wall equations to return similar numbers for "thick" pressure vessels.

The equations for pressure vessel stress in that file seem to be correct from my first glance. Be aware that longitudinal tension also contributes. You might have to do what's called a Mohr's circle to find the principle stresses to use in the failure theories.

* It's worth noting that I'll be taking a class on material failure next semester. If you have any questions after Dec., I'd probably be a good person to ask.
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Last edited by btrettel on Sun May 16, 2010 7:24 am, edited 1 time in total.
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Unread postAuthor: DYI » Sat May 15, 2010 10:58 pm

Thanks for the response, btrettel. I was aware that maximum shear stress theory gave conservative results, but I did not know how conservative they were (and I'd really like to know). In my application, measurable plastic deformation is equivalent in results to actual rupture, apart from how much noise is made in the process.

I saw a bit of distortion energy theory in my research this afternoon, but not enough to come up with anything useful based around it. I'll be sure to take a closer look at it when I get some time.

I have almost no background in material deformation and failure calculations, so any research I do into the matter is necessarily slow going.
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Unread postAuthor: btrettel » Sun May 16, 2010 7:49 am

I think how conservative the maximum shear stress theory is depends greatly on the situation.

Another thing I forgot to note was that there also is the radial component of the stress that contributes. Now, you probably realize that "contributing" could in fact increase the safety factor if it reduces total stress. Residual stresses could do similar stuff.

Below is an equation derived using the distortion energy theory. Let me know if this fits your intuition better. I'll recheck my math later today.

Pyield = YS * (do^2 - di^2) / sqrt((di^2 + do^2)^2 + (di^2 - do^2)^2 + (di^2 +do^2) * (di^2 - do^2))
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Unread postAuthor: D_Hall » Sun May 16, 2010 2:38 pm

I'm not particularly in the mood to check your math, assumptions, etc., but I will say that your link is good. I've played around with those sets of equations in the past and had very good luck predicting the failure of die sets in hydraulic presses.

No, I've not gotten into strain energy and such (although perhaps I should). I've done nothing more than find principle stresses (not to be confused with radial, tangential, or longitudinal stresses although obviously those three are the data required to find principles) and compare them to known material properties.

Note that I am I big fan in using yield stress not ultimate, however. Why? Well, in most real world applications a bent part is as useful as a broken part. Once your pressure vessel starts deforming you're not likely to maintain integrity of the overall system. I mean, just because nothing goes "bang" doesn't mean that you don't start seeing leaks in threaded connections and such. So even if it didn't FAIL, it is no longer functioning as it should be. Thus, I'm a big fan of yield stress.
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Unread postAuthor: axi0m » Sun May 16, 2010 7:46 pm

Here is everything you need to know:

http://www.physicsforums.com/showthread.php?t=397893

I can sum it up for you by linking you to this calculator: http://www.mechengcalculations.com/jmm/ ... rocess.jsp

It provides the hoop stress, which is the stress that the ID of the pressure vessel experiences.
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Unread postAuthor: DYI » Mon May 17, 2010 7:28 pm

@axi0m: I've already seen that calculator, and the forum link provided significantly less than everything I need to know. As btrettel mentioned, the maximum shear stress approach is skewed to the conservative side for estimating failure pressures.

@btrettel: Your equation seems to agree with Barlow's formula in the thin-walled limit, and gives yield pressures more congruent with my intuition and experience for thicker vessels. I'd be interested to see your derivation if time permits.
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Unread postAuthor: axi0m » Mon May 17, 2010 8:09 pm

Contrarily, it is everything you should need to know. You need to establish a safety margin, because the actual failure point of a given vessel will never be calculated with complete accuracy. From that you can satisfactorily calculate the geometry given that the calculator I linked provides the max hoop stress on the ID. There is no shearing involved. The axial stress, that which is felt by the end of the vessel is significantly less than the hoop stress.
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Unread postAuthor: btrettel » Tue May 18, 2010 8:32 am

axi0m, there is always shearing involved; draw a Mohr's circle to see the 2D stress state as a function of angle.

The derivation below is okay if the axial stress is negligible. It probably isn't neglegible, but I don't remember how to find the principal stresses in 3D. I used both the hoop and radial stresses here.

Apologies if this seems rough; I don't have time to explain everything.

St = P (do^2 + di^2) / (do^2 - di^2) (hoop stress)
Sr = -P (radial stress)
tauxy = 0

Save = (St + Sr) / 2 = P di^2 / (do^2 - di^2)

taumax = sqrt(((St - Sr) / 2)^2 + tauxy^2) = P do^2 / (do^2 - di^2)

S1,2 = Save +- taumax = (P / (do^2 - di^2)) * (di^2 +- do^2)

http://en.wikipedia.org/wiki/Von_Mises_ ... conditions

Scroll down to the part about plane stresses.

YS = (P / (do^2 - di^2)) * sqrt((di^2 + do^2)^2 + (di^2 - do^2)^2 + (di^2 +do^2) * (di^2 - do^2))

Again, this was rough, but this would take a while to explain to someone who's not familiar with this sort of analysis. Reading about Mohr's circle should help.
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Unread postAuthor: DYI » Mon May 24, 2010 12:03 pm

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: Moonbogg » Mon May 24, 2010 1:01 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?


If I understand you right, then no. The pressure in the vessel will certainly almost never need to be anywhere near the yield PSI of the material in order to make it fail. It depends on how much surface area is exposed to pressure and the thickness of the wall. The wall thickness will always allow more pressure to be held, at least as far as I am aware (NOT AN EXPERT!!)
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Unread postAuthor: DYI » Mon May 24, 2010 2:52 pm

If I understand you right, then no. The pressure in the vessel will certainly almost never need to be anywhere near the yield PSI of the material in order to make it fail. It depends on how much surface area is exposed to pressure and the thickness of the wall. The wall thickness will always allow more pressure to be held, at least as far as I am aware (NOT AN EXPERT!!)


Ah, I should have phrased the question differently: Is any arbitrarily thick-walled pressure vessel certain to yield before the internal pressure exceeds the material's yield strength?

While, for a given ID, a thicker wall will increase a vessel's yield pressure, the equation derived by btrettel implies that this yield pressure asymptotically approaches a limit of the material's yield strength as OD increases.
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