ASCE 7 wind load for round irrigation tank

[Ricyteach]

Hi if anyone would be willing to help me with what I think it a simple question I would appreciate it. What I have is a group of 8 irrigation tanks, about 45 ft diameter and 18 ft tall, spaced at about 60 ft (so greater than 1.25 D but less than 2.0 D, so they are partway between isolated and grouped). These tanks have floating roofs, so I'm only looking at the wind load applied to the walls. The foundation is a concrete doughnut; the bottom is just sand with a liner on top.

Looking at Ch. 29 of ASCE 7, the sections for wind loading of isolated and grouped tanks are 29.4.2.1 (isolated) and 24.4.2.4 (grouped).

I think I understand that the Cf normal to the wind direction can be assumed to be 0.63 for an isolated tank and it is 1.3 for my tanks when grouped. At a 1.33D spacing the interpolation for C.f comes to about 1.23.

I think I have all that right. Here's my question: what do I use for Af in equation 29.4-1?:

F = qz GCf Af

The text says: "projected area normal to the wind except where Cf is specified for the actual surface area, in ft2 (m2)"

Does this mean that in order to compute drag and overturning moment on the tank I should only include the WINDWARD area, so that Af = HxD? Or does it mean I would include both the windward and leeward areas combined, or Af = 2 HxD?

Second question: when they are empty these tanks are open, effectively no roof. Should there be some kind of additional consideration taken into account in that case? I don't see anything in Chapter 29 about the case there is no roof on a tank.

 

[JStephen]

"Does this mean that in order to compute drag and overturning moment on the tank I should only include the WINDWARD area, so that Af = HxD?"- Yes
"Should there be some kind of additional consideration taken into account in that case?"- None that is required by ASCE 7-16 at least. There are provisions in API-650 and AWWA D100 for wind girders and blow-in resistance of tank shells.

There is a requirement for minimum wind load (I think 16 psf on projected area)- on roofed tanks, this will not control; on yours, it might.

There are some shortcomings in the wind design provisions. For example, what about a group of 2 tanks, or tanks where H/D is outside the range they give, etc.

 

[driftLimiter]

In this equation its the 'projected' area. Meaning you project the surface area to a single plan. My read is that there are no windward and leeward components to account for when using it.

Your second question is more interesting. You could treat it like wall that is open to atmosphere and compare.

 

[Ricyteach]

Thanks. No way that 16 psf minimum pressure controls; this is Indian River County in Florida.

I believe the product designer is taking some responsibility for the structural design and I'm only looking at foundation design, drag, OTM, and giving them the applied loads (they're in Europe so the client just needs help supplying them those loads). However since it sounds somewhat specialized to apply wind pressure to an open tank I think I will push harder for them to take more responsibility for the wind girders/reinforcement.

 

[Ricyteach]

Anyone care to check my math on this?

As a placeholder I'm assuming a corrugated section for the tank walls with moment of inertia of 0.0024 in^4/in. This is equivalent to a wall thickness of 0.307 in.

The unreinforced height equation (H1) per API 650 5.9.7.1 is:

H1 = 6000000 t (t/D)^(3/2) (120/V)^2

Where V is a service case wind speed, t is in inches, D is in feet, and H1 is feet. Based on Vult = 141 mph, V is about 110. So with a diameter of 46 ft for the tank I get:

H1 = 600000 * 0.307 * (0.307/46)^(3/2) * (120/110)^2 = 119.5 feet

This seems really high! Is it right...? Note that this is for an isolated tank; need to increase the wind speed for closely spaced tanks.

 

[HTURKAK]

I dont have any idea how you have calculated (t) . The subject formula is revised at 13 th ed.

V is the design wind speed (3-sec gust), in mph .
H1 = 600000 * 0.307 * (0.307/46)^(3/2) * (120/141)^2 = 72 feet

 

[Ricyteach]

Great this helps me understand that I'm in the right ballpark.

For t I've calculated an equivalent thickness, t, based on the moment of inertia, I, since the buckling resistance is a function of the moment of inertia in the circumferential direction.

t = (12 I)^(1/3)

Btw I'm pretty sure the 3 second gust windspeed in API 650 is based on the 50 yr return period. The 141 mph speed is the 300 yr return period (ASCE 7 risk category I). So I would not use 141 mph, I'd use a reduced speed to make it the 50 yr. Typically what I've done in the past to get the 50 yr is use the Risk II wind speed multiplied by square root of 0.60, or 0.78. So I guess in this case I should use 0.78 x 150 mph... I'll make that adjustment.

The API equation also doesn't know about the doubled wind pressure for closely spaced tanks that was added in ASCE 7 2016. So you'd also multiply V by square root of 2 to double the pressure. So I'll do that too.

 

[HTURKAK]

...So I guess in this case I should use 0.78 x 150 mph... I'll make that adjustment.

The following is the relevant clause from API 650 13th ED. 2020. You can see that API 650 follows ASCE 7 but with a ten years difference. I would like to remind also API 650 uses WSD.

5.2.1 k
k) Wind (W): The design wind speed (V) shall be either:
— the 3-sec gust design wind speed determined from ASCE 7-05 multiplied by √I, Figure 6-1; or
— the 3-sec gust design wind speed determined from ASCE 7-10 for risk category specified by the Purchaser
(Figure 26.5-1A, Figure 26.5-1B, or Figure 26.5-1C) multiplied by 0.78; or
— the 3-sec gust design wind speed specified by the Purchaser, which shall be for a 3-sec gust based on a 2 %
annual probability of being exceeded (50-year mean recurrence interval).
The 3-sec gust wind speed used shall be reported to the Purchaser

 

[Ricyteach]

Yup makes sense. So I would not multiply the category II by 0.78, I would multiply whatever the applicable category speed is, I, II, III, or IV, by 0.78. That makes more sense than just the II to get the 50 yr.