pvc pipe under external pressure 5

[hyposmurf]

Hi all
How deep in water could I sink a 20" schedule 20 pipe, assuming the internal pressure of pipe is the atmosphere??
Of course it have not to collapse or be deformed.

http://www.cadtutorforum.net

 

[MikeHalloran]

Uh, not real deep.

It would make a very bad submarine, if that's what you're thinking.

What _are_ you thinking?



Mike Halloran
NOT speaking for
DeAngelo Marine Exhaust Inc.
Ft. Lauderdale, FL, USA

 

[Cockroach]

This is a very nice problem.

Assuming you are in pure water, the internal pressure is 14.7 psi and the 20 diameter PVC pipe of schedule 20 behaves as rigid plastic with properties 5218 psi yield and Young Modulus of 406106 psi, then you can go down to 544.6 feet to get the onset of failure.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[hyposmurf]

Hi Cockroach , nice nick name.
No one can beat you
Could you explain me how do yo get this values???
The 544.6 feet deep.

To Mikehallorman , no I do not want to use it as a sub .
I'm XXXLarge size.

My final task is to know if I can make a 1 inch HG absolute vacuum to a 40 inch lenght pipe of such dimension.
Do not worry by the caps , I had solve it yet.
My concern is with the pipe.

Thanks both .

http://www.cadtutorforum.net

 

[MikeHalloran]

Maybe I'm doing it wrong.

I'm getting about 50 feet for a perfectly round sch40 (.593" wt) pipe, and about 40 feet for one that's a half inch out of round.



Mike Halloran
NOT speaking for
DeAngelo Marine Exhaust Inc.
Ft. Lauderdale, FL, USA

 

[Cockroach]

Yeah, this forum does not lend itself well to the mathematics behind the theory, I will just give you the numbers and logic of the paradigm.

Given a 20 inch diameter pipe of schedule 20, the internal pressure is 14.7 psi, taken to represent atmospheric conditions. You could take this pipe and submerge it to a depth "h" in water of specific weight 62.5 lbf/ft^3 or:

h = P/(rho g)
h = (14.7 psig/62.5 lbf/ft^3)X(12 in/ft)^3 = 406.4 in.

At this depth, the wall is in neither compression or tension since internal pressure exactly balances external pressure subjected by the water. I would call this the neutral axis, wall stresses are exactly zero.

Continue to submerge the pipe. As the water gradient increases with depth, there shall be a point below this neutral axis to which pressure will crush the pipe. By Thick Wall Pressure Vessel Theory, an element of wall does not care if external pressure is crushing it or internal pressure is blowing it outwards. We can treat internal and external pressure by the same equation, I have chosen the Von Mises-Hencky Equation since it relates hoop, radial and longitudinal stresses as a vector gradient for a triaxial state of stress of an element representative of the wall under the loaded condition.

Grinding through the mathematics, the stress gradient applied to Thick Wall Pressure Vessel for hoop, radial and longitudinal stresses becomes:

S = sqrt(3) P [R^2 / (R^2 - 1)] for R = D/d
D = 20.0 in (OD), d = D-2t = 19.25 in (ID)

"t" is the wall thickness, commecial pipe schedule gives 0.387 inch for your 20.0 OD schedule 20 pipeline.

I set the stress equal to material yield, PVC rigid plastic listed at 5218 psi and a Young's Modulus of 2.8 GPa (metric value in Canada). By giggling around the terms in the Von Mises-Hencky Equation, you would find the required pressure to bring the wall to a yield point, hence the onset of failure to be:

R = D/d = 20.0 in / 19.25 in = 1.03896
P=[5218 psi/srt(3)][(1.03896^2 - 1)/1.03896^2] = 221.7 psi

This corresponds to a depth of h' BELOW the previously established neutral point.

h' = P'/(rho g) for P' = 221.7 psi
h' = (221.7 psig/62.5 lbf/ft^3)X(12 in/ft)^3 = 6130 in.

Therefore, total distance as measured from surface, hence the depth you seek is:

H = h + h' = 406.4 + 6130 in = 6536 in = 544.7 ft

This computation will be influenced by the physical values to the PVC material, dependent upon grade. For purposes of illustration and finding this an curiosity, I used generalized values from a DuPont listing. Be careful to check an MTR from your supplier and make the appropriate changes.

Hope this clarifies your apprehensions. This is a very cute problem.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[Cockroach]

Slight typo, 20.0 OD schedule 20 pipe has a wall thickness of 0.375 inches. I said 0.387 inches; this doesn't effect the calculation since the inner diameter is 19.25 inches as correctly stated.

544.7 feet. I was more worried of spelling mistakes, sorry!

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[TGS4]

Time out here!

Under external pressure, the failure mode is not going to be yield, but rather buckling. So, I'm going to completely disagree with Cockroach's approach on this one - you headed in the completely wrong direction.

 

[Cockroach]

TGS$, you failed to post your answer and method.

As for your comment, best to review the theory. I suggest Advance Mechanics of Materials, Boresi & SideBottom, pg 169 or Design Stress Analysis, Timoshenko & MacCullough, chapter 4. Also try the University of Tennessee at Martin, Lecture 15 in the course Engineering 473 touches on the method relative to interference fits. I would imagine that Roark or Mark deal with the subject, probably drops the mathematical theory for the practicioner however.

I stand by my analysis.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[jte]

Cockroach-

I'm looking at my copy of Advanced Mechanics of Materials 5th ed by Boresi, Schmidt, and Sidebottom. On page 169 they begin section 5.2, "Castigliano's Theorem on Deflections." Is that the section you were referring to? I suspect you have a different edition.

jt

 

[CRG]

I would agree with TGS4 that the correct that the failure mode will be from buckling. This has been discussed before on this site. Take a look at these two threads:

thread378-103771

Thread794-95854

 

[jte]

Cockroach-

Leaving the failure mode (buckling vs yield) aside for a moment, I believe you should revisit the "submerge it to 406 inches to achieve zero gage pressure" approach. I'd argue that as soon as the pipe is submerged to any depth, it is under external gage pressure. You start on the surface with the cylinder open which achieves 0 psig or 14.7 psia. Now close the cylinder and begin to submerge it. At a depth of 1 foot, you'd have the weight of the water, 62.4 #/ft^2 or 0.43 pounds per square inch. Now, if you are playing in the "gage" sandbox, you started at 0 psig and now have 0.43 psig external added for a grand total of 0.43 psig acting on the walls of the cylinder. If you prefer to play in the "absolute" sandbox then you started with 14.7 psia in the cylinder countered by the weight of the water and the air above the water for a total of 0.43 psi + 14.7 psi. So you wind up with 14.7 psi internal and 15.1 psi external for a differential pressure of 0.43 psi pressing in on the cylinder.

jt

 

[TGS4]

Unless someone is going to pay some consulting fees, I won't likely put answers or formulae on this forum. That said, I think that pointing hyposmurf in the right direction is the appropriate thing to do.

I thinkt hat jte has covered the gage vs. absolute thin well enough, so I will focus solely on the failure mode(s). In each design, there are many different failure modes that need to be considered. In no particular order (and not necessarily complete) they are:
1) Plastic collapse
2) Local failure (exceeding local strain limits)
3) Hydrostatic material failure
4) Buckling / Instability
5) Progressive distortion
6) Fatigue
7) Brittle fracture / crack growth
8) Material degradation over time (creep, radiation damage, hardening, etc)

So, I think that cockroach has adequately covered the first failure mode. However, there are many others (as shown in my list) that need to be considered.

Hyposmurf, what have you done to evaluate these other failure modes? If you need the forum to point you in the right direction regarding calcualtions, we will be glad to oblige.

 

[Cockroach]

JTE, for sure it is not the 5th edition. Sorry, I should of stated that with the title. It is probably not the 1st edition, perhaps second (1987-ish). Let me get back to you, it's not in my office as we speak.

Of course your comment and that of TGS4 would unnecessary complicate the mathematics. As stated in my analysis, with the absence of more information, like tension placed on the pipeline, sagging inbetween supports or laying on the bed (inelastic foundation), etc, etc, the problem would be unsolvable. I have made assumptions based on a reasonable approach to the problem and came up with 545 ft.

But how could you model the problem using fatigue, much less crack propagation? We don't have and would never hope to have material properties to assist us with modelling these phenonema.

I would hope that HypoSmurf begin with the computation set forth in this forum, research the discussion and direct his efforts accordingly. Is this not the point of the forum, healthy academic discussion?

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[TGS4]

As a part of the healthy academic discussion, all I want to do is point out that many different failure mechanisms MUST be examined. While one might, by applying engineering judgement, bypass a calculation regarding, say, fatigue, that does not alleviate the necessity of considering the other failure modes.

For example, in this particular case, if hyposmurf were to say that the pipe would not be repeatedly submerged, then I could say that, regardless of the state of stress, it would not fail due to fatigue in one cycle. Similarly, I would not be concerned about progressive distortion. Likewise, if material degredation is not a concern, then time-dependant properties are likely not significant.

However, all of the other failure must be dealt with. From my perspective and based on my engineering experience and judgement, the buckling failure mode would likely be the governing failure mode. B31.3 has formulae to address external pressure and the buckling/instability failure mode.

"But how could you model the problem using fatigue, much less crack propagation? We don't have and would never hope to have material properties to assist us with modelling these phenonema." To use a very legalise approach to this, ignorance is no excuse. Unless you can achieve positive confirmation that such failure modes will not occur (such as my discussion above), then as an engineer you are obligated to "get" the information necessary for you to make a sound judgement. To simply throw your hands up in the air and say that it is impossible ("never hope to have") is a complete cop-out. It's that kind of attitude that results in failures and deaths (Challenger and Columbia to name a few recent examples).

Finally, cockroach, please tell the forum why you neglected the buckling/instability failure mode. Do you have additional information on PVC that the rest of us may not be privy to?

 

[Cockroach]

I chose yield over collapse because of the API collapse equations giving pressures greater than that of Von Mises. The latter was 221.7 psi. Yield would occur before collapse, 232.4 psi.

In similar fashion, buckling instability in my judgement, is not a factor. My transition points between straight collapse, elastic and transitional, show pressures around the yield point. Again, without additional information I would suggest yield to be predominant mode of failure.

I do have PVC properties not typically available to most engineers in the field. To suggest ignorance as an excuse or believe another failure may occur, again, let's see your numbers complete with the model.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[TGS4]

Following the methodology of ASME Code Case 2286-1 (Article 3.1) for external pressure in a cylinder:

Mx = L / sqrt(Ro*t)
assuming an unsupported span of 20ft
Mx = 87.64
Therefore, Ch = 0.55*t/Do
Ch=0.0103125
Fhe=1.6*Ch*E*t/Do
Fhe=123.75 psi

Since Fhe/Fy<0.552, Fhs=Fhe/FS

Assuming a FS of 3, Fha=41.25 psi

Calculating Pa = 2 * Fha * t / Do
Pa=1.547 psi

------------------------------
Assumed values:
Do=20
t=0.375
E=4e5
------------------------------

That would mean that you could (with a design margin of 3) place the 20"NPS Sch. 20 pipe to a depth of 43 inches. Reducing the design margin to unity gives a depth of 129 inches (give or take a few since the pressure would not be uniform).

So, let's see you API collapse equations...

 

[stanier]

COckroach

You are of course assuming that the cylinder is perfectly round. If there is an ovality then buckling would override the other parameters.

The factor of safety for buckling is thus much higher.

Suggest Timoshenko Theory of Plates and Shells is referred to for the subject.

 

[hyposmurf]

Hi all you.
Thanks all , for pay attention to this task I have to do.
Hope we get at the end the proper result.
Please keep it on run.
Thanks again .

http://www.cadtutorforum.net

 

[MikeHalloran]

This is from a spreadsheet that implements the equations in ref. 2 below

20 20 in od of inner tube
sch 40 sch 20 tube gage
0.593 0.375 in wall thickness
PVC PVC material
4.00E+054.00E+05psi Elastic modulus
1.54E+051.54E+05psi Shear modulus
0.30 0.30 Poisson's ratio
5.22E+035.22E+03psi yield point for tube material

25.08 6.13 psi critical buckling pressure for a _perfect_ tube

per eqn 2.6 of ref 2

0.5 0.5 in radial deviation from circle at worst point
20.08 5.69 psi critical pressure for collapse of imperfect tube

ref 1: "Tubing Limits for Burst and Collapse", Tech Note, CTES, L.C., Conroe TX

http://www.ctes.com

ref 2: "Effect of Initial Eccentricity on Collapse Pressure of Circular Beam Tubes", S. Yadav
Fermi National Accelerator Laboratory, Batavia IL

http://www.flab.gov

Mike Halloran
NOT speaking for
DeAngelo Marine Exhaust Inc.
Ft. Lauderdale, FL, USA