Help with API650 - 3.11 - Wind load on tanks 2

[pdosreis]

Hi,
I'm calculating a storage tank, of D=13,5m and H=10,85m.
When calculating wind load on tanks (overturning stability), section 3.11, I've followed all the steps in this section and reached the conclusion of no need for anchorage, because M<2/3x(WxD/2).
Now, I really don't know (understand) what to do with 3.11.4 that refers to sliding due to wind, and gives me a "maximum allowable sliding friction of 0,40 multiplied by the force against the tank bottom".
What am I suppose to due with this factor, where do I apply it? How do I calculate if I need anchorage because of sliding? Can anyone help me?

 

[JStephen]

Take the total weight of the tank, including the floor and structure, and multiply by 0.4 to find the sliding resistance. Compare to the total lateral wind force.

Addendum 4 to API-650 revises the wind loading considerably by adding a wind uplift requirement- you need to look into it if this tank is not specified to be built by Addendum 3.

 

[pdosreis]

Thanks! That really helped.
Can you just tell me if I should consider also 2/3.

Like this:
MaxOverturningMomentwithoutAnch.=2/3x((Wtank+Wfloor+Wroof)xD/2)
And then compare this to:
OverturningMomentfromWindPress.=Wind PressurexDxH/2

Did I understood it right?

 

[JStephen]

Generally the floor would be considered for the sliding resistance, but not for the overturning. Or at least that's the way I would work it.

Your second formula would be M = pressure x D x H x H/2. But in the new Addendum 4, you have a term for uplift on the roof as well, and some additional equations to calculate uplift resistance.

 

[pdosreis]

Sorry, but maybe I didn't explained myself very well.
Forget the "Overturning" term I used, I'm only talking about sliding.
The overturning issue it's already taken care of, and I agree with you, I didn´t consider the floor when calculating overturning.
So:
MaxSlidingwithoutAnch.=2/3x((Wtank+Wfloor+Wroof)xD/2)

And then compare this to:

MomentfromWindPress.=Wind PressurexDxH/2

It's this what you were saying in your first answer?

 

[JStephen]

If you're considering sliding, then it's not a moment. I don't think there's any requirement to use a 2/3 factor, for that matter.

You'd have maximum sliding force = (Wtank+Wfloor+Wroof) x 0.4

And compare to Wind Pressure x D X H