Uplift from wind and Overturning - Anchorage 2

[LeonEarle]

Gentlemen,

A. How to apply 5.2.1.j.2?

1. Calculate value for F.4.1
2. Convert F.4.1 * 1.6 to moment
3. Calculate moment resulting from pressure Mpi 5.11.2 (could be zero)
4. Calculate moment from horizontal wind
5. Calculate moment from vertical wind - but not greater than (moment from F.4.1 * 1.6 - Mpi)
6. Add moment from horizontal wind and lesser of vertical wind or (moment from F.4.1 * 1.6 - Mpi) for Mw

If this is not how to apply 5.2.1.j.2, then how is it applied?

B. What do you do with the negative number from F.4.2 on large tanks when all other values calculate positive?

Thanks!!!

 

[pmover]

it would be helpful if you identified the reference std or spec your are quoting. people can guess, but . . .

good luck!
-pmover

 

[LeonEarle]

Standard applicable to this question is API 650, 11th Edition.

 

[TankDude]

A. I would recommend forgetting about the overturning moments initially. Simply determine the combined uplift pressure from wind and design internal pressure. This total value must be less than or equal to 1.6P with P being calculated per F.4.1 Therefore, you should use the combined uplift pressure of wind and design internal pressure unless it exceeds 1.6P, at which time you should use 1.6P as the uplift pressure. From here you can calculate the contributing overturning moment and add it to that calculated from the wind pressure on the shell.
B. From F.4.2 it would appear that when this formula yields a negative result that the tank has no additional capacity for uplift due to internal pressure. In fact, it could be viewed to indicate that the tank requires anchorage simply due to the wind moment, and if the tank were to be anchored this would make F.4.2 n/a and you would be referred to F.7. I have not analyzed the corelation (if any) between F.4.2 and 5.11.2, but it could be an interesting exercise.

 

[LeonEarle]

TankDude,

I think we are essentially doing the same thing. If there is no pressure on the tank, F.4.2 does not enter into the calculation. Where there is pressure, if this limit is not applied, there is always a negative value. In the designs I have done recently using the limit, F.4.2 does not calculate a negative number although the amount of pressure allowed is extremely small. I think your method may be simpler though. Thanks