I have the following load cases:
L1: W+T1+P1 (OPE)
L2: W+T1+P1+D1 (OPE)
L3: W+T2+P1 (OPE)
L4: W+T2+P1+D1 (OPE)
L5: W+P1 (SUS)
L6: W+D1
L7: W

Where T1 = -10 degr C, T2 = 30 degr C and ambient temp = 15 degr C.
D1 is a tank settlement of 150mm.

Now I have used the following combinations to get my expansion stress range (all combinations are combined using the ABS method):
L8: L3 - L1 (EXP)
L9: L4 - L2 (EXP)
L10: L6-L7 (to get the streses resulting from the tank settlement.

L11 = L8 + L10 (EXP) combined using ABS method, to get the total stress range of both temperature and tank displacement.

Is this correct? Am I using the correct combinations and the correct combination method (ABS)? Thanks in advance for your replies.

Hi Corne,

In my opinion D due to temperature should be separated with displacement due to settlement as I read from load case pdf file in our forum. I quote some as below:

Load Cases with thermal displacements and settlement:

For settlement, use a Cnode on any affected restraints. This Cnode must be a node number not used elsewhere in the model. Then place the settlement on the Cnode using a displacement vector not already used for thermal displacements. We will use D3 to describe restraint settlement in this example.

L1 W+T1+D1+D3+P1 (OPE)
L2 W+T2+D2+D3+P1 (OPE)
L3 W+P1 (SUS)
L4 L1-L3 (EXP) * effects of T1, D1, and settlement
L5 L2-L3 (EXP) * effects of T2, D2, and settlement
L6 L1-L2 (EXP) * full expansion stress range with settlement

Settlement is evaluated as an expansion load because it is strain-related with a half-cycle.

Note: For piping codes with no expansion stress computation simply add the thermal and settlement displacements to the operating cases as shown above.

* Use algebraic combination method under the Load Case Options tab.
** Use SRSS combination method under the Load Case Options tab.
*** Use ABS or Scalar combination method under the Load Case Options tab.

I'm kinda confused with this though. As L1-L2 has D3-D3 in it. Does this really incorporate the settlement?

In my opinion in this case T1 and T2 influence D1 and D2 respectively which we can calculate. But They don't influence D3. We can say D3 value which caused by settlement is constant along T1 to T2.

But the stresses resulting from D3 should be added to the expension stress imo. If we use L1-L2 we don't have the stresses resulting from D3. We only have the stresses resulting from temperature differences.

Hy Corne
Just try this:

L1: WW
L2: W+T1+P1
L3: W+T2+P1
L4: W+T1+D1+P1
L5: W+T2+D1+P1
L6: W+P1
L7: W+D1
L8: D1+T1
L9: D1+T2
L10: L4-L8
L11: L5-L9
L12: L2-L3
L13: L2-L6
L14: L3-L6
L15: L4-L2
L16: L5-L3
L17: L16+L6
L18: L17+L6

I guess you want to be more strict than usual, so the key would be the number of cycles/ the equivalent total cycles. "Settlement is evaluated as an expansion load because it is strain-related with a half-cycle"- not thousands of cycles, I would add.


regards

Hi Corne,

I did trial in the same one model with "This Cnode must be a node number not used elsewhere in the model. Then place the settlement on the Cnode using a displacement vector not already used for thermal displacements"

L1 W+P+T1+D1
L2 W+P+T2+D2
L3 W+P+T1+D1+D3
L4 W+P+T2+D2+D3
L5 L2-L1 *
L6 L4-L3 *

I got L5 and L6 have a different result. I am not sure I did differently.

I too saw in my model it gives different results. But the difference is just a couple of N/mm2 while the stresses from D3 only (W+D3) - (W) is much higher.
I'm still not completely satisfied. Maybe someone with a final answer?

The fact that "D3 only (W+D3) - (W)" is much higher makes sense if your model has non-linear boundary conditions (+Ys, gaps, friction). As soon as your model is non-linear you can't apply superposition.

You should really be addressing the EXP load case as an ALG combination, not ABS.

Thanks Richard, to check: the following should be correct?
L1: W+T1+P1 (OPE)
L2: W+T1+P1+D1 (OPE)
L3: W+T2+P1 (OPE)
L4: W+T2+P1+D1 (OPE)
L5: W+P1 (SUS)
L6: L3-L1 (EXP)
L7: L4-L2 (EXP)
where EXP is calculated using ALG method?

I believe the analysis should examine the maximum strain range acheivable through the life of the installation. I do not see how you can do that without that D1, maybe you can but you do not show that test here.

As Dave states, you will have to evaluate the other EXP cases.

So you mean I should do:
L1: W+T1+P1 (OPE)
L2: W+T1+P1+D1 (OPE)
L3: W+T2+P1 (OPE)
L4: W+T2+P1+D1 (OPE)
L5: W+P1 (SUS)
L6: L3-L1 (EXP)
L7: L4-L2 (EXP)
L8: L2-L1 (EXP)
L9: L4-L3 (EXP)
??
In my opion L6 and L7 will give the EXP stress due to thermal influences. Where L8 and L9 give the stresses due to D1 only.
If this is true I need L6+L8 etcetera too. Is this correct?
Or can I easily do L4-L1 and L2-L3 to get the total stress range?

Ignoring settlement for a moment I see three states - installed (W but W+P1 would work in most cases), operating #1 (W+T1+P1) and operating #2 (W+T2+P1). CAESAR II would recommend two expansion stress ranges: (Operating #1 - Installed) and (Operating #2 - Installed). It would your responsibility to create a third range: (Operating #1 - Operating #2). That might be the worst case here since T1 and T2 straddle ambient temperature.
Now add the settlement.
If settlement occurs immediately after the system goes into service, I suggest there are four states - installed (W or W+P1, as above), installed with settlement (W+D1)and the two operating cases (W+T1+P1+D1)(W+T1+P1+D1). Range calculations might then be between installed and installed with settlement; the two operating cases with settlement and the installed with settlement; and also between the two operating cases with settlement. If settlement is slow then the original two operating cases come into play. You should consider all these positions when you are looking for the maximum expansion stress range. Perhaps you can eliminate some of the ranges but you should consider all of these cases in your "State #1 - State #2" evaluation.
Most users stop there but to follow the Code you should include the damage of lesser stress range cycles in setting the number of cycles used in the allowable stress range calculation for that highest stress range.
Things are simpler for linear systems.

I see no distinction between pipe strain caused by settlement and that caused by thermal pipe strain - as you imply.

Dear Mr. Diehl,

As usual, your answer is sharp and complete, so there isn’t much room to comment about.
Indeed we are so lucky you can find out some time to share with us your knowledge. Thank you for so many valuable advices!

I have a question about the numerical results.

Presuming we are under the B31.3 code, the equation that must be applied is (1d).
This equation is clear explained in the Code and provides a calculation for equivalent number of full displacement cycles associated with the greatest computed displacement stress range.
In fact, the equation can be equally written as:
N*SE^5= NE*SE^5+ΣNi*Si^5
I would consider the "fifth power" law as more general than is shown in B31.3.Can it be used to calculate the equivalent number of cycles Nx associated with any displacement stress range Sx?
Nx*Sx^5= NE*SE^5+ ΣNi*Si^5
It is correct this interpretation?

Let me explain why I'm asking… Let’s suppose we consider only 2 calculation ranges. First, the stress range S1 due to settlement and we have one-half of cycle (N1=0.5) associated with. And we have S2 stress range due to OPE1- OPE2, associated with N2 (let’s say 7000 cycles). If S1>S2, the Code asks to consider S1 as base of calculation.
However, just for my curiosity I would ask how many cycles of S2 represent one-half of cycle of S1.
Presuming the "fifth power law" is still functioning also for one-half of cycle, the result would be
N2equiv*S2^5= N1*S1^5+N2*S2^5
or
N2equiv=0.5* (S1/S2)^5+N2
If S1=2*S2, the contribution of one-half cycle of S1 would be as 0.5*2^5=16 cycles of S2, in addition to N2 cycles.
If S1=3*S2, the contribution of one-half cycle of S1 would be as 0.5*3^5=121.5 cycles of S2, etc

It is correct this approach? Is working the "fifth power law" also in this case?

Thank you and my best regards

Equation (1d) is a form of Miner's Rule (see Wikipedia). The fifth power comes from Markl's fatigue curve.

Your math looks right but I don't know if you can work this with half cycles. Wikipedia mentions that Miner's may fall apart when there is a big excursion followed by small excursions.

Some Codes address one-time displacements (settlement) directly. The stress (range) from a single, one-time displacement is limited to 3Sh.

Thank you very much for your quick answer. Very appreciated it!
Yes, I've had the same concern about Miner's Rule (thinking that probably is quite forced in our case).
I think B31.3 is not so clear about what is permissive on the stress (range) from a single, one-time displacement, even some guidance is provided within. If a client insists to remain under B31.3, I have to follow (1d) "fifth power rule" also for this case, isn’t it?

Thank you again,
My best regards

It is a strain on the system and should be addressed. I suggest you take a look at B31.1 as I think that code itemizes such one-time displacement. You may be able to qualify this approach as "more rigorous" and thus acceptable in the view of B31.3 rules.

Dear Mr. Diehl,

Thanks a lot.
I hope also Corne is completely satisfied about the final answer.

My best regards

All problems and troubles solved in the mean time. Thanks all for your help.