Hi all,

I've some experience in load cases, think of hot sustained, all kinds of OCC combinations and EXP cases.
But now the following issue.
I have a system where the highest secondary stresses in the + direction are from a load case with W (for example W+T1+P1+D1). The highest secondary stresses in the - direction are from a load case with WNC (for example WNC+T2+D2).
How to make a good EXP load case to get the full stress range?
Let's assume:
L1 = W+T1+P1+D1 (OPE)
L2 = WNC+T2+D2 (OPE)
L3 = L1-L2 (EXP)
This is not true cause actually L3 = (W-WNC) + P1 + (T1-(T2+D2)), so combining primary and secondary stresses.

How can I best make load combinations to get the actual secondary stress range from L1 to L2?

Thanks in advance for your replies.
Corné

A change in weight distribution is considered a secondary stress (for example the redistribution when a +Y lifts off). Therefore your case 3 above is correct.

Hi Richard, I can understand this in case of the uplift of a +Y support, as small plastic deformation can bring the pipe back on its rest.
In my case the stresses occur after a bend in a long line. Due to friction from this long line on its supports I need to combine the load cases. In such a case the difference between W and WNC is not a secondary stress, although the resulting stresses after the bend are thermal stresses of course.
I'm now puzzling myself. I'll overthink it once again.

Thanks for your reply.

Hmmmmm ...

Corne,

Just so that things don't get off track, would you please edit your post and change:
L3=L1-L3 (EXP)
to:
L3=L1-L2 (EXP)

Thanks.

By the way, the piping codes do not address this sort of issue directly.
B31.3 para. 319.2.3(b) implies that the expansion stress range is:
"the algebraic difference between strains in the extreme
displacement condition and the original (as-installed) condition (or any anticipated condition with a greater differential effect)".

If L1 & L2 are the extremes, that's what you should use.

Hi Dave,
I can't find an edit button. How to edit my opening post?
[edit]
I see there is an edit button on this last post, but not on my other posts in this topic. Apperently a post that has been replied on can't be edited.[/edit]

On topic:
My "concern" is that in L3 case I also have primary stresses (that can't be solved by plastic deformation) in the main run between the supports, as is W-WNC (so content only). These are primary stresses I'm also seeing in my EXP case when I use load cases as mentioned in my opening post. That's why I'm in doubt. However these two are the extremes indeed.

My humble opinion:

L1 = W+T1+P1+D1 (OPE)
L2 = WNC+T2+D2 (OPE)
L3 = W+P1 (SUS)
L4 = WNC (SUS)
L5 = L1-L3 (EXP)
L6 = L2-L4 (EXP)
L7 = L5-L6 (EXP)

and the reason is that you always have a sort of "check point" which is installation temperature.

I hope that I've made myself clear..

In order to go from W+P to WNC, first you have to shut down and empty the line. If, for example, you have a system that will always operate at instalation temperature W+P-WNC will never be secondary.

If not, there are a lot of lines full of liquid that have also the "steam out" condition. I think that nobody made a check W+P-WNC.


Best regards,

L7=L5-L6?

I would propose L7=L1-L2.

Then, to keep things simple, I'd make sure all three stress ranges can survive the total number of cycles expected. Otherwise you should be using that adjusted Miner's Rule to accummulate damage from each set.

I'm afraid I do not understand.

If L7=L1-L2, and at limit D2=0 and T2 is installation temperature, mean that we should always do the following:

L1=W+P1+D1+T1 (OPE)
L2=W+P1 (SUS)
L3=WNC (SUS)
L4=L1-L3 (EXP) =[(W-WNC)+P1]+D1+T1

which we don't.

Or do I miss something?

Best regards,

I was thinking there were two operating cases L1 & L2. I see now that I was incorrect. My teaching point is inappropriate in this case.
But what is the purpose of your L5-L6?

I was trying to void the inclusion of P.

Regards,