ASME sec III Div NC subpara 3658.3 gives the following limit foe checking flanges
Mfs ≤ 3,125(Sy/36,000)CAb

36000 appears as the min yield strength of material like SA 105. Can anyone explain the origin of 3125.
Regards
Siv

Dear Siv,

I can explain you the factor 3125, but I don’t understand very well the appearance of Sy/36000. The last one it is told to be a factor that compensates for the effect of temperature on the flange allowable stress. For bolts it appears to be an empirical correction or maybe has another meaning that makes more sense.

Now about your question.
The bolts are assumed to react to a bending moment M load.
Following the theory of circular plates, the bolts are not equally loaded, and the maximum “local” bolt load can be considered as 4M/C.
Now this load can be counted as giving the bolt stress 4M/(C*Ab)<S and that leads to M<(S/4)*C*Ab.
It is supposed that 12.5 ksi- one half of the allowable bolt stress 25ksi- is available to compensate the external load… At least is what you find in K. R. Rao book “Companion Guide to the ASME BPVC”

...so 12500/4 means 3125 psi…

You may note that the bolts are assumed to react to the external load, while the “Pressure Equivalent Kellogg Method” assumes exactly the opposite (by the way, Rao’s book gives an inadequate explanation on EQP method- in my opinion).

Funny story, isn’t it? Don’t like what we have as “A.D. 2008 flange engineering”?
Not good, not bad- I can say...because, as MoverZ concluded in another post “it certainly works, albeit with obvious technical limitations”!

Best regards,