Okay, I know this subject has been beat to death, but after searching and searching I could not find an exact match to my question.

Anyways, I am trying to determine what load cases I need to run for temperature. Basically I have a line designed for say 200 F. I have an ambient temperature of 70 F. I have a winter temperature at worse of -20 F. This is to be installed in Oklahoma, where the last 61 years the lowest temp. recorded was -14F.

Of course under normal operation, the line will never get close to even ambient, if there is flow always going through. If the pipe happened to be empty and it was during a record winter, then yeah maybe it could get down below 0 F. So, would load cases be set up such as this?

CASE 1 (OPE) W+T1+P1
CASE 2 (OPE) WNC+T2
CASE 3 (SUS) W+P1
CASE 4 (SUS) WNC
CASE 5 (EXP) L5=L1-L3
CASE 6 (EXP) L6=L2-L4
CASE 7 (EXP) L7=L5+L6

Where T1 = 200 F and T2 = -20 F.

In Example 1 of Appendix S of ASME B31.3 I set up this exact example in Caesar, and used these same load cases as listed above, and my CASE 7 stress results almost matches exactly with the results of Table S301.7.

If for my case, where we may hit minimum temperature once maybe TWICE in the pipe's lifetime, is it still required per code to design to that stress range?

I have attached the B31.3 Caesar model that I made. (Interestingly enough, it wouldn't allow me to attach my .cfg file!)

You may consider eq.(1d)/ B31.3/302.3.5 (d) that gives the equivalent number of full displacement cycles during the expected service life of the piping system.

In this particular case will not be very helpfull because SE is associated with 200F not with -20F.

But you can try Appendix P.

Regards,

mariog, even for N=1,your f is still at most, 1.2. We typically only design to 7000 cycles (Caesar default) anyways. So that is not helping us here unless I am not seeing something.

danb, not sure what you are talking about with your first sentence. Appendix P appears to have much higher allowables, but of course calculates stresses different. I will look into that.

You may have 7000 cycles from T3>T2 (i.e. from a normal low operating temperature) to T1 and 2 cycles from install to T2, or from T2 to T1, etc. Anyway you may know/imagine such assumptions and maybe you are interested to know how many eq cycles (e.g. from T3 to T1) are?

I think Dan remark is about the possibility to count effectively the equivalent number of cycles. B31.3 (1d) provides a calculation for equivalent number of full displacement cycles associated with the greatest computed displacement stress range.

About this subject, I would add that (1d) is a particular form of Miner's Rule and may be equally written as:
N*SE^5= NE*SE^5+SUM(Ni*Si^5)

If is of interest (is it?) one would consider the fifth power law - Miner's Rule - as more general than the particular form shown in B31.3 and consequently such form can be used it to calculate the equivalent number of cycles Nx associated with any displacement stress range Sx:
Nx*Sx^5= SUM(Nj*Sj^5), where Nj&Sj are all cycles (includes NE&SE and all Ni&Si). Just my opinion, of course.

One further tidbit. After reading the suggested load case method for expansion ranges, I noticed that the preferred method according to the Caesar manual, as well as moderators here to do, as for in this case, L1-L2 (Algebraic)..so for my Caesar model above, I added this load case for LOAD CASE 8 = L1-L2. It turns out, my original LOAD CASE 7 = L5+L6 gives numbers much more closely resembling the results from Example 1 out of Appendix S of 31.3, Table S301.7, than this new LOAD CASE 8 = L1-L2.

So, what is the correct method to use? It appears that adding expansion cases (abs) like in my LOAD CASE 7 is more in line with ASME 31.3 no? I have attached my new Caesar model.

Well, it seems like this works fine and dandy for this particular case, but when you start dealing with temperatures on the same side of ambient, adding the expansion cases doesn't work...so I guess I will revert back to the way the Caesar User Guide recommends and just algbraically subtract operating cases from each other for my expansion cases.

That Example 1 is linear, right? In that case adding the two excursions ( one above and the other below ambient) will equal the stress range calculation based on the difference between the two states.
The algebraic difference follows Code intent for linear and nonlinear support situations.

One of the topics at CAU2012 is proper use of Miner's Rule and (1d) in CAESAR II.

L1-L2 can not be used in this particular case:

L1-L2= (W+T1+P1)-(WNC+T3)= (T1-T3)+(W-WNC)+P1 (EXP)??

My opinion is that you can not include fluid weight and pressure in this loadcase.

Ah, danb...maybe that is why our company was adding the two expansion cases for that case...makes sense.

It is my interpretation that displacement stress range ("...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...")should not contain stresses from sustained loads. Otherwise we should use

L1=W+T1+P1(OPE)
L2=W+P1(SUS)
L3=WNC(SUS).... but is also an operating state
L4=L1-L2(EXP)
L5=L1-L3(EXP)

As far as I remember nobody use L5

But if you use Appendix P, your loadcase L8 in your example and L5 in my example are valid because are "operating stress range"

Regards,

Back to original post of threeouts, in my opinion the load cases should be written:
(OPE) L1=W+T1+P1
(OPE) L2=WNC+T2
(SUS) L3=W+P1
(SUS) L4=WNC
(EXP) L5=L1-L3 [=T1-70F=130F; disp range from instal. to max metal temp]
(EXP) L6=L2-L4 [=T2-70F=-90F; disp range from instal. to min metal temp]
(EXP) L7=L5-L6 [=T1-T2=220F; full disp range max to min metal temp]
L5, L6, L7 are Algebraic combinations.

Another method:
(OPE) L1=W+T1+P1
(OPE) L2=W+T2+P1 [to use only for cases 6, 8 below]
(SUS) L3=W+P1
(SUS) L4=WNC
(EXP) L5=L1-L3 [=T1-70F=130F; disp range from instal. to max metal temp]
(EXP) L6=L2-L3 [=T2-70F=-90F; disp range from instal. to min metal temp]
(EXP) L7=L5-L6 [=T1-T2=220F; full disp range max to min metal temp]
(EXP) *L8=L1-L2 [*optional case for full disp range max to min metal temp]
L5, L6, L7, L8 are Algebraic combinations.
This method can be used for T2 at above or below zero deg C.

Both methods above (linear/or non-linear analysis) will provide results almost exactly the same in example 1 of B31.3 appendix-s.

I disagree with "L7", your taking a range of ranges this way. My preference is case "L8", which is much more obvious.

If you want to combine "L5" and "L6" to obtain the full displacement range, you want "L5+L6", not "L5 - L6". This provides the sum of the ranges, instead of the difference between the ranges.

But it seems to me that summing the ranges does not work when you are on the same side of ambient.

In Yhebostress's first example, I would just do L1-L2. I think you end up with slightly higher stresses since you have weight and pressure added, but..at least it is conservative.

Agreed, you can't sum the ranges if the components are on the same side of ambient.

Both L7 and L8 of my previous post will result to exactly the same code stress values if the combination method is "ALGEBRAIC" following the code rule. Just to explain:

on METHOD-1: USE ALGEBRAIC
L7=L5-L6; substitute L5, L6
L7=(L1-L3)-(L2-L4); substitute L1,L2,L3,L4
L7=[(W+T1+P1)-(W+P1)]-[(WNC+T2)-WNC]
= W+T1+P1-W-P1-WNC-T2+WNC
L7= T1-T2 (THERMAL STRESS RANGE)

on METHOD-2: USE ALGEBRAIC
L7=L5-L6; substitute L5, L6
L7=(L1-L3)-(L2-L3); substitute L1,L2,L3
L7=[(W+T1+P1)-(W+P1)]-[(W+T2+P1)-(W+P1)]
= W+T1+P1-W-P1-W-T2-P1+W+P1
L7= T1-T2 (THERMAL STRESS RANGE)

hence, if L5= +130F, and L6= -90F
L7= +130F - (-90F); ALGEBRAIC COMBINATION
L7= +220F (THERMAL STRESS RANGE)

also,
L8=L1-L2; (USE ALGEBRAIC) substitute L1, L2
=(W+T1+P1)-(W+T2+P1)
=W+T1+P1-W-T2-P1
L8=T1-T2 (THERMAL STRESS RANGE)
then, if T1= +200F, and T2= -20F
L8= +200F - (-20F); ALGEBRAIC COMBINATION
L8= +220F (THERMAL STRESS RANGE)

***FOR L7=L5+L6, the combination method to use shall be "ABSOLUTE". This means;
L7=|L5| + |L6| ; for L5= +130F, and L6= -90F
=|130F| + |-90F| ; ABSOLUTE COMBINATION
L7= +220F (THERMAL STRESS RANGE)
Note that using the ABSOLUTE combination if both T1,T2 are above zero will result into unacceptable stress range value because it will force to get the SUM (not the ALGEBRAIC difference). So, say T1=+200F, T2=+20F will have L7=|200F|+|20F|=220F (Not ok). However, for L7=L5-L6 using ALGEBRAIC combination, L7=(200F)-(20F)=180F (correct stress range between T1 & T2). This means that the latter is better to use than the former method for any +/- values of T2.

No. L 7=L5+L6(ABS) will not sum the response to absolute loads. Instead, it will sum the absolute response to those loads.
A subtle but important difference.