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