Pmax F.4.2 Evaluation of Equation

Dear All,

as per clause F.4.2 API650. we have the following equation:

Pmax=0.000849*DLS/D^2 + 0.00127*DLR/D^2 - 0.00153*MW/D^3

First Part of the equation = 0.000849*DLS/D^2

Second Part of the equation = 0.00127*DLR/D^2

3rd part of the Equation = 0.00153*MW/D^3

can any one derive/evaluate these equations?

In Addendum 3 of API 650, the expression for "the maximum design pressure limited by uplift" given in F.4.2 is mathematically equivalent with 5.11.2 condition (1)- i.e 0.6Mw + MPi < MDL /1.5 + MDLR.

You can check easily that in case pressure is expressed in [Pa], coefficients are 4/(π*1.5)=0.8488 in the first part, 4/π=1.2732 in the second part and 0.6*2*4/π=1.5278 in the third part.
Above π is 3.14159..

Because the pressure is considered in kPa, they are multiplied by 10^(-3) so they would be 0.0008488, 0.0012732 and 0.0015278.

Best regards.

Thanks Mariog,

it is clear for me that the how to calculate resisting uplift pressure P=DLR/pi*(D^2/4).
could you please define the factor 1.5 in as resisting uplift pressure P=DLS/1.5*pi*(D^2/4) & 0.6 in wind case as increasing uplift by P=0.6Mw*(2/D)/pi*(D^2/4).

Dear Jamil,

I don't know from where there is the conclusion the resisting uplift pressure is P=DLR/pi*(D^2/4)

I attach a pdf document with the calculation.

Basically your question is why API 650 does not correlate various sections, for example 5.11.2 and Table 5-21a—(SI) Uplift Loads. I cannot answer, I just expect they will do it.

Best regards.

Thanks Mariog,

you really send helpful document.