Hi, Would you please advise methods (not the only one as I suppose there are more...) for user SIF values calculation, e.g. weldolets, non orthogonal tees, reducing tees etc. Users are requesting more and more information about that, there are some research studies available. Codes are very weak, e.g. in EN13480 there is less than in ASME. I find NB, NC codes much more detailed and the question is if we could apply SIF calculated according these much more detailed rules (of NB NC) for other codes?
EN 13480-3 app. H, issue 2, provides more examples, and I believe (even) more detailed calculations, for some very specific and complex geometries compared to B31.3:2012.
I agree that EN13480 has cases than B31.3 does not have, but ASME NB and NC may have some additional cases. The ASME NC lists a formula to convert stress indices in ASME NB to SIF factors. I recall it like this:
SIF = C2*K2 / 2
Also there are ways to estimate SIF using finite element analysis of the component. Finite elements calculates the stress concentration factor SCF. The SIF can be theoretically estimated using the simple formula.
SIF = SCF / 2
This is based on the fact that the SIF uses a welded pipe as the reference while SCF uses an un-welded pipe. The weld has an SCF of about 2.
Real SIF values are estimated based on fatigue tests as was done by the father of SIF, Markl.
Some software like FE-PIPE or NozzlePro can estimate the SIF for you and make the process much easier than using general purpose programs like ANSYS.
Karim Rinawi
Thank you. The problem is that users claim, and they ar right that e.g. simple reducing tee (pipe DN50 welded into pipe DN350) should have different value of SIF than DN300/DN300. Of course we need to follow the code :-)
Another issue is that it is difficult to find out where are these mentioned stress concentrations located, that refer to flange. Is it in the 'body' of flange or this is stress in the pipe, nearby the welding. In Markls' tests it was weld that was tested for leaking, so I guess this is the weakest point. How about bend? There is ovalization that causes flexibility change, but SIF refers to bend shell or ... welding connection?
Overall misunderstandings come from lack of knowledge what exactly is the measurable result of tests leading to SIF value estimation... Would you advise, please?
Regarding what Karim described...
SIF = SCF/2 but what about 0.75 factor ;-) i.e. you mean the SIF value for thermal stress range or 0.75i for static analysis, please?
I am aware this is complex problem, however I suffer from not having one common resource that would allow catch it and apply. I understand it piece by piece, but... exceptions are responsible for stress analysis quality, like Y-tee, reducing tee etc. This is why I am asking you to share what you know about it please.
Maciej Rydlewicz
Best regards,
Maciej Rydlewicz, PhD Eng.
maciej.rydlewicz@softdesk.pl
+48 512206994
Although most earlier tests are made on equal tees, reduced branches give very reasonable results. To my knowledge equal tee and reduced tee will have very different SIF values per the code. The fatigue crack in bending for the bend will occur at the middle lower point for horizontal bend. The PIPE STRESS ENGINEERING book by Peng has excellent discussion on the bend SIF and on the 0.75 factor for sustained stress.
For tees the failure point is assumed to be at the header junction weld for B31.3.
Thank you. The problem is that codes like ASME B31, EN13480 don't provide formulas for reducing tees and thus what we have in AutoPIPE seems to be inacurate, 'fortunately' for AutoPIPE this is due to the reason the code is kind of improper. I wonder if there is a chance Bentley gets kind of solution for that, what would be a good step forward, in my opinion.