Hi, all,

I have question about the Sustained Stress and I attempt to find out the solution. But I still be confusing.

As per B31.3, The system is acceptable if the SUS stress shall be less than Sh. SUS stress is consisted of hoop stress, axial stress and bending stress.

In my case, the allowable stress is 29500 psi. Code stress is 20500 psi. but Hoop stress is 34000 psi. From this case, I'm surprise that why code stress is less than hoop stress and overall stress is acceptable. But it shouldn't be like this if we refer from B31.3 Code.

The other question that I have try to input the wall thickness less than the designed pipe wall thickness (Tmin + CA + Mill Tolerance) as per B31.3 requirement by substract the mill tolerance. The hoop stress is larger than the allowable (Sh). But code stress (W+P1)is acceptable. Why?

The final question that why the mill tolerance (12.5%) wasn't taken into account to check the pipe stress analysis. In fact that the mill tolerance shall occure in the mass production piping. This condition have difference stress result between them.

Please verify.
Thank you

You need to read the Code. Sustained stress combines longitudinal stresses, mainly weight and pressure (axial) stress. Axial pressure stress is by definition half of hoop stress in a pipe.

For example, if your weight only stress were 5,000 psi and axial pressure stress 15,000psi you would be within a sustained stres limit of 20,000 psi, but hoop stress would be unacceptable at 30,000psi.

Mill tolerance and corrosion allowance are taken into account in sustained stress equations. Weights are based on nominal wall as are all stiffness equations by default, although the set-up can alter this.

In B31.3 (other Code not see yet :)), the basis to calculate minimum WT is hoop stress with a little bit modified equation. So if your WT is enough, you don't worry about hoop stress. If your WT is not enough, there is a warning at error checker. To see, just input a very small WT. Hoop stress is not code stress, so there is no code stress check as per hoop stress. In this way it doesn't mean that we will ignore the warning regards the compute min. WT > user input WT.

BTW, dear MoverZ, I think mill. tolerance is not used for SUS stress calculation instead of min. WT... How do you think?

Samsul,

You have raised an interesting point there ... B31.3 para 302.3.5 states that ... The thickness of pipe used in calculating SL shall be the nominal thickness, T, minus mechanical, corrosion, and erosion allowance, c, for the location under consideration. The loads due to weight should be based on the nominal thickness of all system components unless otherwise justified in a more rigorous analysis.

I'm pretty sure this is how Caesar used to operate ... sustained (SL) stresses based on nominal - (corrosion + Tolerance), weights based on nominal. But the mill tolerance now seems to have no effect on any B31.3 stresses ???

One for Rich Ay ?

There is a different of nominal thickness, T (symbol Tbar-with line above T) and pipe wall thickness (measured or minimum per purchase specification), T I think. Correct me if I'm wrong.

The minimum thickness T for the pipe selected, considering manufacturer’s minus tolerance, shall be not less than tm.

The T-bar and T symbols in B31.3 are indeed different .... 'T' from para 304.1.1 gives you the figure to compare with commercial walls when selecting a purchase thickness for example. T-bar in para 302.3.5 is the nominal wall. My question about inclusion or not, of mill tolerance in SL calculations remains.

In appendix S301.2 Design condition:
tm = t + c = 6.1 mm + 1.59 mm = 7.69 mm

Select 16" sch.30:
T bar = Nominal thickness without minus mill tolerance = 9.53 mm
T = T bar - minus mill tolerance = 9.53 * (1.00-0.125) = 8.34 mm

tm must < T
7.69 mm < 8.34 mm ; (this is what Caesar check regards the warning in error checker)

Regarding to SUS stress:
"SL shall be the nominal thickness, T bar, minus mechanical, corrosion, and erosion allowance, for the location under consideration"

Wall thickness used in calc. of SL = T bar (without minus mill tolerance) - allowance.

I think Caesar follows the Code. The Q is raised to the Code author this way. Someone experience in fabrication/manufacture may explain. But honestly dear MoverZ, I just thinking the same w/ you, If we purchase T bar pipe, we mill the pipe so then reach T. The erosion, corrosion, and mechanical allowance occurs, now we got T-allowance (not T bar - allowance).

Regarding to piping input I think Caesar should breakdown all allowance not only listed corrosion allowance. Even this assumed the sum of all allowance? And I believe the value "t = pressure thickness" will give the most conservative stress for longitudinal pressure stress as part of SL which not Caesar taken just because follow the Code.

Except for IGE/TD/12, CAESAR II uses mill tolerance only in checking the minimum wall thickness calculation for pressure.

The pressure check for wall thickness monitors hoop stress and this is sensitive to changes in thickness around the pipe.

The stress analysis performed by CAESAR II is, for the most part, a longitudinal stress and mainly due to bending. The wall thickness check for pressure monitors pipe burst which can be quite localized. The bending stress is not as sensitive to local imperfections as the entire cross section must reach yield before the "plastic hinges" can form to allow collapse.

I see no reason to include mill tolerance in the stress calculations. If you want to see what happens if you do, instead simply reduce the wall thickness value that you enter in CAESAR II.

Thank you for the comments.
I've attempt to read and understand the Code about the longitudinal stress (SL) that considered the part of bending moment (only). This bending moment is determined from pressure and weight in piping system.
I just difficult to reply my client in case of the actual pressure is larger than the design pressure and the sustain stress is acceptable during the hoop stress is larger than allowable as I mention above.

The common senses of design engineer, not stress engineer, they understand that if pressure is higher than the pressure rating of piping system, the piping will be bursted and stress result must be over allowable but the fact is not like that eventhough the warning show the input thickness is less than min. pipe wall thickness requirement.

And I've try to input the wall thickness substracted Mill. at the begining of project. The result show that the hoop stress increase and overall stress still acceptable.

This's just my experience.
TH Engineer.

Caesar is not the tool you should be use for pipe wall thickness calcs. Those should have been done and the wall thickness required for the pressure (governed by the hoop stress) should have been done long before a system is given to a piping design to layout, much less for a stress engineer to review.

Your allowable doesn't change. If you've got an A106 B carbon steel at less than 400F, your allowable stress is 20,000 psi. That applies to the hoop stress as well as to the longitudinal stresses.

Your initial post suggested that you think hoop stress, and longitudinal stress due to pressure and external load are all combined and compared to Sh. Hopefully the other responses have made it clear that this is not the case. Hoop stress is checked on its own against Sh. And, since the hoop stress for pressure is, by definition/derivation, twice the longitudinal pressure stress, it is going to govern the wall thickness calculation for pressure.

Even if you use the full 20,000 psi in your hoop stress check, then your longitudinal pressure stress can be no greater than 10,000 psi, leaving another 10,000 psi for other longitudinal stress sources such as bending loads on the pipe.

Now, it is the case that, if, for the wall thickness available, you get a calculated hoop stress of 25,000 psi, you have exceeded the allowable and Caesar will give you a warning. But, this means that your longitudinal stress from pressure is only at 12,500psi. If your other bending stresses are less than 7,500psi, it will pass in Caesar, as that is the stress it is designed to check for.

If you are getting a warning that you don't have enough wall for the hoop stress, your first step should be to make sure you didn't make an error in the input. After that, you need to talk to the design folks who made the wall thickness calcs and find out why there is such an issue.

Well said.

David,

Going back to your comments on exclusion of mill tolerance from sustained stress calcs above.... I'm doubting myself on this issue now.

So it's ok to ignore mill tolerance you say. What about corrosion ? That is also commonly a localised effect, more of a concern in hoop stress due to pitting, risk of bursting etc. Fair enough, when corrosion is more general as it is sometimes, it is often worst in the upper or lower (or both ?) quadrants of a pipe periphery and that is where bending resistance lies. But surely if a negative tolerance is at its worst, maybe -12.5%, that may also be in the same areas of the pipe wall and thus would cause a serious reduction in bending resistance ? Should it not then be considered in a sustained stress equation as part of the 'mechanical' allowance ?

Interestingly I answered a very similar question this week. Here is my response.

Typically (most Piping Codes) mill tolerance is only considered in the initial determination of the pipe’s wall thickness, for hoop considerations. Mill Tolerance is not (typically) to be considered in the Flexibility Analysis. This is why most Piping Codes include something similar to B31.3 Paragraph 319.3.5 ( “… nominal thickness is to be used …” ).

Having said that, there are exceptions. For example, B31.8 Chapter VIII wants you to deduct mill tolerance when computing the “combined stress”. Similarly the British Gas Code (TD/12) considers load cases with both +mill and –mill values. (I’m always amazed at why people only assume mill tolerance is a deduction.) There could be a few other instances where mill tolerance is expressly dealt with, but typically once you determine “t”, you put mill tolerance behind you.

Corrosion is a different matter. B31.3 wants the Sustained and Occasional load cases corroded, and is silent on the Expansion load case. This is why we have that (configuration) directive “ALL_STRESS_CASES_CORRODED”. Many people feel that the Expansion case should be corroded. This is actually a “Client” or “Project” call. Personally, I think everything should be corroded.

Note that reducing the wall thickness (either for mill tolerance or corrosion) at the input level is incorrect. Reducing “t” lowers the system stiffness, i.e. it makes the system more flexible – which is non-conservative. Lowering “t” also reduces the weight load of the system, which again is non-conservative. This is why the Codes say "compute the stress using a corroded section modulus".

Dear all,

Actually if we purchase OD, ID, T bar pipe (as per ASME B36.10M), regarding to mill tolerance, does it mean that it is possible that they will deliver us with size OD', ID', T pipe? someone experience in fabrication/manufacturing please explain.

Yesterday I made calculation based on my own assumption that they will deliver us with size OD', ID', T. If I am wrong, I am afraid that many also do the same. So I try to put my opinion here even maybe not correct.

Starting from that assumption, If We calculate the Slp as part of Sl

Considering mill tolerance: Slp= 8,298 psi
Not considering mill tolerance: Slp= P*D/4*(Tbar-allowance)= 550 psi*16 in/4*(0.375-0.063 in)= 7,051 psi
We loss= 8,298-7,051= 1,247 psi
Sl= 14,370 psi (in appendix S); using Caesar not try yet

% loss = 1,247/ (14,370+1,247)= 8%

Dear Sir Richard Ay,
When I said that t=pressure thickness used for calculation, I mean how if that's only used in Slp as part of SUS stress case not for EXP case as you explained caused piping system more flexible.

P.S: I could not attach the excell calc. this time...system problem.

I try using pdf type file now...

You try to get rigorous results, so you really need to consider more accurate formulas for hoop/ longitudinal stress. In my opinion, the formulas that the Codes are using are a little bit conservative, so there is a margin that can partially compensate the effect you’ve observed. For example, the hoop stress is not constant over the thickness; for internal pressure load, there is a maximum on the "internal edge". Eventually, each Code has his own formula, which is a compromise between many variables.

My suggestion is to make the same comparison with a more rigorous formula for stress.

Best regards

Dear Mariog,

Yes in c2quick.pdf file, the default Slp = P*ID/(OD^2-ID^2); code exact equation, Caesar default. We can find also in Popov's book this Eq. I will try to recalculate.

But my point is: "Actually if we purchase OD, ID, T bar pipe (as per ASME B36.10M), regarding to mill tolerance, does it mean that it is possible that they will deliver us with size OD', ID', T pipe? someone experience in fabrication/manufacturing please explain"

If it is possible we have OD', ID', T, from manufacturer does Caesar anticipate it?

CAESAR II uses your input values in the analysis. The software doesn't guess, estimate, or assume anything.

I think I need to stop due to uncertainty (many possibilities pairs of OD' and ID' here) as said by Sir Dave Diehl. But I want to attach it once again...:)

Dear Samsul,

I don’t know how it’s the Excel file, but you’ve lost ID^^2 in the formula posted (Slp and P must have the same units dimension).
If you need to calculate more accurate you would consider the Caesar "B31 case 178 option". In fact it’s the same formula, but is more clear what thickness you need to consider (nominal pipe thickness less the sum of the mechanical -thread or groove depth-, corrosion and erosion allowances)

Yes, it is possible to have in field dimensions as OD', ID', T pipe…..- long time ago, a Client forced us to make such measurement in field…
Your remark is correct; however we must accept what the Code is asking for. Also, I think Caesar hasn’t the target to be more conservative than the Code is. Fortunately, I would say!

Best regards,

Dear Mariog,

Apologize for the miss typing, yes, Slp = P*ID^2/(OD^2-ID^2), in excel I put that formula...I revised the pair of OD' and ID' in my last calc. Thanks for remind.

Please find attached file, this may clear your doubt.

Hi,
I need a piece of information, please.

Consider A106 Gr. B pipe featuring 19999 psi sustained stress (design temperature is less than 200°C) and no warning as regards hoop stress.
Caesar tells that sustained stress (W+P1) complies B31.3 code.
Is my pipe safe?

Or must I reduce sustained stress to 6666,3 psi?
(6666.3 psi sustained stress) + (13332.7psi hoop stress) = 19999 psi.

Thanks.

Follow the Code rule: The sum of logitudinal stress due to sustained loads must be less than Sh (20ksi in your case). There is no hoop term here.
Burst due to pressure is prevented by the component specification (e.g. B16.9) referencing the line's design temperature & design pressure.

Thank You very much, Mr. Diehl.
Have a nice day.