API 650 Section Modulus Calculation Help

[takiyasamsama]

I need to know the formula and calculation steps on how API got the values for Table 5.20a for Section Modulus of stiffeners based on the shell thickness.

Take example for shell thickness of 6 mm. Using "Detail C" for 125 x 75 x 8 the value in the table is 93.71 cm3. However when I refer to 125 x 75 x 8 section modulus is 29.6 cm3.

I would really appreciate help and guide in this calculation.

 

[IFRs]

The section modulus includes a portion of the tank shell, which adds to the cross section area. The Figure shows the distance used, it is related to the shell thickness.

 

[takiyasamsama]

As always IFRs thanks for your answer. What I wanted to know, these area is the governing area of 16t above and below including the area of the angle bar, correct?

If area it's not a problem to calculate, but in Table 5.20a is section modulus. I need to learn and understand how API calculated these values based on thickness and angle detail used.

 

[IFRs]

Please review the basic properties of sections - area, moment of inertia, etc. Perhaps a wiki or your college introductory structural engineering book. From the geometric properties of the section ( length, width, distance from a common axis ) you can determine the center of gravity ( neutral axis ), the moment of inertia and the distance from the neutral axis location to the furthest edge ( extreme fibers ) and thence calculate the section modulus. Basically, Section Modulus is the Moment of Inertia divided by the distance to the extreme edge from the neutral axis. A non-symmetrical part will have different Section Modulii to the compression flange as compared to the tension flange. The moment of inertia of the whole section can be quickly and easily calculated by considering simple rectangular pieces of the cross section, summing each part's moment of inertia about it's own neutral axis and adding to that the sum of the square of each part's distance to the overall neutral axis times the part's area. In equation it would be Itotal = I1 + A1 x D1^2 + I2 + A2 x D2^2 + I3 + A3 x D3^2...etc. The moment of inertia of a rectangular section about is's own neutral axis is width x height^3 / 12. This is a relatively straightforward set of calculations that you will be pleased to master.

 

[takiyasamsama]

Thanks for the explanations. I believe I gotta opened up my college book reference back. Thanks again IFRs