"Correcting" an SN curve for a different % probability of survival?

[JBlack68]

To all

I am looking for some suggestions / explanation on the following. I came across a simple fatigue estimate in a report. What I don't really understand is the "correction" of the SN curve from the data in MMPDs for 84% probability of survival. MMPDS is giving the SN curve for 50% probability of survival (I believe) and it was somehow corrected for a higher probability of survival. The infinite life is therefore much lower (and so is the life at N=1000cycles)

1. How is this done? Is there a formula for that?
2. why 84%? that's 16% failure I suspect

Any ideas

Thanks
Regards

 

[rb1957]

sounds "odd" to me ... the typical SN curve is a mean life prediction, the mean of a bunch test specimens.

to change to an 84% life curve, you'd need to assume a normal (or some other) distribution of the population, and assume a standard deviation (or some other parameter to define the population distribution. Once you've done that I think it's an easy statistics exercise to determine the 84% curve (lower than the 50% curve).

again, seems an "odd" thing to do ...

another day in paradise, or is paradise one day closer ?

 

[rb1957]

the next "interesting" question would what safe life factor to apply ? typical safe life factor is 5 on mean life sn.

another day in paradise, or is paradise one day closer ?

 

[JBlack68]

safe life factor = 4 was used it seems

 

[JBlack68]

Thanks for the reply. Did some search and assuming a Normal distribution I think one can "correct" the SN curve using the formula

S_corrected = S_50%*10^SE*SD
where
SE = Standard Error
SD = Standard Deviation

I am no sure but I think 84% probability of survival corresponds to -1 SD from the mean

Taking MMPDS data as an example, the SN curves are 50% (survivability!) and the doc provides some info on SE & SD. See attached snapshot. Not sure how to read the data though! Any ideas?

 

[rb1957]

Good catch, I had forgotten those statistics with the curve fit.

My sense (and I'm no expert on this) would be to calc the mean life (well, log (life)) and factor log(life) (divide by 1.18 to get -1 SD ?) and then life (=10^log(life)) and then safe life=life/4 ??


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[JBlack68]

see attached for my interpretation of your thought/sense

then what does one do with the Standard Error (SE) quoted with the data?

 

[rb1957]

Maybe this is meant to be a "rational" safe life factor ... ie do you need to /4 later ?
Maybe what they're saying is they can accept 1 in 6 failing in-service ?
Notice how similar the results are ...
with SF = 4 applied to mean life = 12,500 cycles
applying 84% survivable factoring, life = 9,856 cycles

Maybe you're looking at someone else's calcs, and this is what they did ?? seems overly conservative (to apply the SF on top of the 84% reduction).

The only thing I'd change is "N factored" should be IMHO "N reduced" or "N 84%" or "N 5/6"

another day in paradise, or is paradise one day closer ?

 

[JBlack68]

I think I need to understand the "correction" of the 50% SN curve to take into account 84% probability of survavibility (& SSE?). With the "new" SN curve then n, damage, etc is straight fwd. Whether one then need to apply the safe life factor will be the final point to decide

 

[rb1957]

the correction would be straight-forward, knowing the SD ... from the mean curve, reduce lives by the factor ...

new life = mean_life^(1/1.18) ... I think !?

so this'd work into the logNf equation ... by dividing the constant (20.68) and the factor (9.84) by 1.18

more importantly, I wouldn't apply a safe life factor to the resulting 84% life ... I think 84% life equates to safe life, and explains why to do it in the first place !

another day in paradise, or is paradise one day closer ?

 

[JBlack68]

If one considers the equation log(N)= 20.68-m*Log(S)
the "shifted" SN curve is log(N)= 20.68-m*Log(S) - Z*SD
where SD is provided (1.18) and Z is the Normal quantile (Z = 1.0 for about 84% survival rate)

for 2 selected lives N0 & N1 get the S0shift & S1shift

did the exercise with N0=1000 & N1=10^8 (with R = -1) and the shift seems reasonable (the slope of the SN curve is conserved)

 

[rb1957]

why should the slope be preserved ?

another day in paradise, or is paradise one day closer ?

 

[JBlack68]

I got the same b (or m if you prefer) value with the calculated S0shift & S1shift (N0=1000 & N1=10^8) being unchanged