Barlow vs. Roark Pressure Vessel EQs 1

[DBCox]

Hello everyone,

I am working on some thick walled pressure vessels and would like to apply the Roark equation for a thick walled for this case and noticed something odd.

Until now, we have always used the Barlow equation with a hefty safety factor, thick walled or not, but now we are moving on to more advanced methods. It is my understanding that the Barlow equation assumes a uniform hoop stress (or an average stress through the thin wall). The Roark eq does not and can be used to find the stress at any point, so I assumed the Roark equation result for max hoop stress would be greater since the sum of all the stresses would need to equal the Barlow result (or be close). This is not the case. The Roark resultant stress is around 3/4 that of the barlow equation for a thick walled tube.

What is wrong here? Are my assumptions and theory wrong, or am I making an algebraic mestake?

Thanks!

 

[JStephen]

Are you checking stress at the inside of the wall? (IE, Roark's "Maximum stress"?)

 

[DBCox]

JStephen,

Yes, I am checking at the inside.

This brings up another question. I thought the max stress occured at the outer wall, not the inner wall. The outer wall seems reasonable because the strain has to be greater on the outside wall.

 

[iainuts]

I'd assume that when you use the Barlow equation, you're using the OD or outside radius as opposed to the ID or inside radius. The use of the OD is debatable, since it could be considered overly conservative.

If you do a freebody on the cross section (ie: draw a pair of annular circles on a paper) and then show the X component of the pressure (ie: show arrows pointing horizontal to represent the X component of pressure) then you probably realize that you're only putting those arrows on the ID of the tube. The equation for the freebody you just created would be the Barlow equation with one difference, the OD would become the ID.

Try that and see if it doesn't clear up the discrepancy between the thick wall equation and Barlow equation.

Also, because pressure causes compressive stress on the ID and not the OD, the ID is going to have the higher total stress, not OD.

 

[JStephen]

The limiting case of a thickwalled cylinder would be a hole drilled in a semi-infinite body, which still has stress at the hole, decreasing to zero at distance.

If a thick-wall cylinder is moved radially outward by a uniform amount, the change in circumference is the same at inside and outside. But the inside circumference is less to begin with, so strain is higher there. Additionally, due to radial strain through the wall, the radial deformation at the inside should be less than the radial deformation at the outside surface.

 

[athomas236]

Is there any reason why you are not using the formula from a pressure vessel design code such as ASME Section VIII ?

Regards,

athomas236

 

[DBCox]

iainuts,

I am not sure I agree that the stress at the inside wall would be greater because that is where the pressure acts. I agree that the compressive forces due to the pressure would be higher there, but, typically the highest stress is along the circumference of the tube; hoop stress.

JStephen's point about the limit with a hole in an infinite body makes sense to me, but again, only for the compressive stresses due to the pressure, not the hoop stresses. I am afraid I don't follow you on radial strain.

athomas236, I am using Roark because I have it here. I will eventually have several methods, including ASME's and compare them all. I am not the type to blindly accept a regulatory body's equations without verifying them. Not that they are wrong, but they tend to derive them for their purposes and the Roark seems like it is a more universal method. I try not to use regulatory body equations unless I am submitting calculations for ASME approval, or whoever the body is who would approve it. I mostly deal with DNV and ABS, and use there equs for the jobs they approve, but I don't trust the calcs without seeing a derivation, or at least some sort of explanation. We run our own calcs first, then verify they meet ABS or DNV requirements.

 

[CoryPad]

blakrapter,

Could you explain your loading conditions?

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[iainuts]

Just a clarification. For cylindrical vessels, there is a radial component of stress and a tangential component of stress which are orthogonal to each other. The tangential component is the hoop stress. Orthogonal to the hoop stress is the compressive radial stress created by the pressure pushing outward on the material which is equal to internal pressure on the ID and zero on the OD. The principal stress can then be found to be a combination of these two stresses, which is a maximum on the ID and a minimum on the OD.

Note that hoop stress also varies across the thickness depending on wall thickness, it is not uniform even for thin wall vessels.

I'd agree with athomas that the equations provided by the regulatory body should be used. Using Roark for vessels is fine as long as you're also ensuring the equations such as ASME are met. You may always exceed those factors of safety, but as a minimum, you need to ensure the applicable codes are met.

 

[DBCox]

iainuts,

Thank you for the clarification. I understand now what you are referring to, but, I still do not understand why the hoop stress is greater on the ID than OD. I have always been under the impression that the hoop stress is the greatest at the outside. I may very well be wrong, but I am curious why it is greater on the inside than outside. Once I get that figured out, I can figure out/understand why or if the Barlow equation is the conservative approach to determining the stress in a thick walled vessel.

I agree about the reg. bodies. Of course we meet or exceed their requirements, I just find it amazing that each body has its own equations, yet the results vary quite a bit.

 

[RebelBrill]

blakrapter,

I believe the hoop stress is greater on the ID than it is on the OD because the strain is greater on the ID. This is so because of Poisson's ratio. The inner diameter of a cylinder under pressure expands more than the outer diameter because of the effect characterized as Poisson's ratio.

 

[JStephen]

Check a quick example. Suppose you take a cylinder 10" ID X 12" OD with 2,000 PSI pressure. Assuming the stress is uniform through the thickness, you can cut a cross section and sum forces, and get average hoop stress = 2,000 psi x 5"/1" = 10,000 psi. This should be the Barlow result.

Using the equations in Table 32, Case 1(b) of 5th Ed. of Roark, a=6", b=5", sigma-2 = 11,090 psi at the inside face, and 9,091 psi at the outside face.

From ASME Section VIII, t=PR/(SE-0.6P) or re-arrange to SE= 0.6P + PR/t = 11,200 psi, or just slightly different from the Roark equation.

As to why the stress is higher at the inside: Suppose you take this same cylinder, and expand it outwardly by 1" in radius, so that is now 12" ID, 14" OD. The change in circumference is the same at both faces, but the strain is based on original circumference, which is smaller at the inside- so hoop strain at inside = 1/5, hoop strain at outside = 1/6. Hoop stress is based primarily on hoop tension and follows a similar trend. This neglects some obvious effects, but seems the clearest way to explain it.

 

[corus]

The stress will be greater on the inside as there is less material.

iainuts is almost correct, but the radial stress, and the hoop stress are both principal stresses. The Von Tresca stress intensity is largest on the inside as the radial stress is negative whereas the hoop is positive.

The stresses differ from ASME as this assumes a thin walled vessel. Expressions for a thick walled vessel will differ.

corus

 

[prex]

Can't really follow tha debate here.
My point is that for pressure vessel wall thickness calculations what matters is not the peak local value of stress, but the average stress across the wall.
And the formula PD/2t (don't know if this is what is referred here as the Barlow formula), if D is the inner diameter (and for internal overpressure), will always give the average hoop stress irrespective of thick or thin wall.
So to answer first the original question of blakrapter, there must be a mistake in their calculations, as the average hoop stress from exact (Roark) equations must exactly give the stress from the above formula, otherwise the laws of equilibrium would be violated.
Concerning code formulae, that blakrapter shows to have little confidence in, these, as recalled by JStephen, add a fraction of P to stress and this is to account for the radial stress in the wall. The added fraction of P varies from code to code: the theoretical value for thin walls would be 0.5 (and indeed this is the value used by Div.2 that is officially based onto Tresca criterion). I suppose that such variations are primarily due to tradition, but also may represent a corrective factor when, like in piping codes such as B31.3, the diameter to be used in the formula is the outside diameter.

prex

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Online tools for structural design

 

[BarryEng]

On the subject of the Barlow formula, I would like to quote from 'Tubular Steel Structures' by Troitsky (one of the Linclin Welding series of books).

9.3.1.2 - The Barlow formula for internal fluid pressure gives results on the side of safety for all practical thickness ratios. This formula is similar to the 'common formula' (which uses the mean diameter instead of the outside diameter). In the common formula, it is assumed that the hoop stress is uniformly distributed across the cylinder wall. This condition does not hold, except in the case of pipes having walls of infinitesimal thickness.

While Barlow's formula is widely used because of its convenience of solution, it was not generally considered to have any theoretical justification until formulae based on the maximum energy of distortion theory showed that for thin walled pipe with no axial tension, Barlow's formula is actually theortically correct. (See Lester, C. B., "Hydraulics for Pipelines", Oldom Publishing Co., Bayonne, New Jersey, 1958, p 61.)

Since most commercially important pipes have a ratio of wall thickness to outside diameter less than 0.10, Barlow's formula for thin walled pipes is of great significance.

Comprehensive bursting tests on commercial steel pipe have demonstrated that the formula predicts the pressure at which the pipe will rupture with an accuracy well within the limits of uniformity of commercial pipe thickness. In general, failure occurred at a pressure about 3% higher than predicted.

BarryEng

 

[GBor]

There is an FEA code from a company that used to advertise here...perhaps they still do...called Paulin Research (

http://www.paulin.com,

I think). I'm not necessarily proposing their code, but it develops reports to meet various pressure vessel codes and, if I'm not mistaken, shows the equations that they used to get there. You may be able to ask for their documentation. I don't know if it just does ASME, but it has a great deal of useful information that may be pertinent to this discussion.

 

[unclesyd]

Here is how LLNL approaches PV design. There is a lot of very good information in this document. I like the Chart that shows the relationship between the MAWP and MOP.

http://www.llnl.gov/es_and_h/hsm/doc_18.02/doc18-02.html

 

[israelkk]

One issue that is missing in the previous posts is the fatigue and fracture mechanics of the vessel. When doing fatigue or crack propagation analysis the stress in the inside diameter (especially for thick walled vessels) should be the one to use.

 

[DBCox]

Jstephen,

Thank you for the information. Your strain example makes sense to me. I could not for the life of me figure out why the ID would be higher stressed than the OD, but I have it now.

Thanks!

 

[Cockroach]

The Barlow Equation is wrong, one dimensional (Hoop only). Roarks is Von Mises-Hencky which presupposes triaxial stresses in the wall.

There are a lot of threads in this problem.

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