An interesting thing has occurred with the design in API-650 11th Edition Addendum 1 with regard to Appendix F. With the changes, the design is now based on the yield strength of the material. When you look at the formula for P, the internal pressure, you will see that you have AFy which is the area times the yield strength. At first glance you would think that this will help you when you high strength materials. And in reality it does IF the shell, roof, and compression ring are all high strength material. However, if the shell and roof are high strength material and the compression ring is A36 material, then you have the same pressure as before. The yield strength, Fy, in the formula is based on the lowest yield strength of the compression area. Therefore, if the compression ring is the lowest yield strength, then that is the value you use.
An interesting thing occurs when you start looking at the area, A. The amount of area available in the shell and the roof have not changed. Those formulas are not based on the yield strength of the material. What did change is that the amount of area in the horizontal projection of the compression ring is based on the yield strength of the compression ring when the horizontal leg is not stiffened. In the previous editions this value was 16t. Now (in US units) it is the thickness of the material times 3000/(Fy)^0.5. Here is a list of what happens to the horizontal projection allowed (Le) for a thickness of 1/4".
16t = 4"
A36 material (Fy = 36000 psi) = 3.95"
A678A material (Fy = 50000 psi) = 3.35"
A240-304L SS (Fy = 25000 psi) = 4.74"
As you can see, the higher the yield strength of the compression ring, the smaller Le can be which reduces the area, A. Thus, if you used a high strength steel for the compression ring and the shell but used a low strength steel for the roof you will end up with a lower allowed internal pressure because the compression ring will not contribute as much area and the yield strength of the roof will be the governing value for Fy in the pressure formula.
If you really want to scratch your head about something, think about this. I could build a tank using a material with a yield strength of 50000 psi for the shell, roof, and compression ring and have it meet the frangibility requirements of 184.108.40.206. However, that same tank with the same thicknesses using a material with a yield strength of 36000 psi for the shell, roof, and compression ring would not necessarily meet the frangibility requirements of 220.127.116.11 (area of the compression area could be greater for the lower strength material).