large overturning force in vertical tank 2

[EngineerofSteel]

I have a 52' vertical tank. I am designing the footing. I am using [(4*M)/(N*D) - W/N] from AWWA.

This gives me the max tensile for the legs. My question is, do I use this also for the resisting Dead Load? I have a 52,335 pound tensile load per bolt (6 anchor bolts resisting). This will require a very large resisting DL. And, 52K requires 1.25" ABs, but the existing anchors are 3/4", giving me more cause to doubt my results.

If the pad upon which this tank sits extends 5 feet from the AB placement, can I multiply the load at 5' times the actual DL?

Otherwise, I need 349 CUYD per each of 6 bolts. This gives me 16' x 16' x 8' thick. Seems large to me. But, I have not worked with forces like this before. Previous vertical tanks I have designed footings for have been 14' or shorter.

Any guidance, especially tips on common errors and frequently overlooked design requirements will be appreciated.

 

[Dinosaur]

What is AWWA? I never heard of it, but I'm a bridge guy.

What is the diameter of your 52' high tank? What is the design wind speed in this area?

Is this an elevated tank or a tall tank sitting on the ground? If it is elevated, how much trusswork is there in the tower supporting the tank?

 

[csd72]

If you have a single footing under the entire tank, then the leg loads have very little relevance to the overturning of the whole structure. In this case you would need to check 2 things for overall stability: 1. FOS against overturning is calculated simply as weight x lever arm from the front toe of the footing. This should be at least 1.5 depending on the code. 2. check the bearing pressure as an eccentric footing.

If the the legs are each on their own individual footings, then what you say is 100% correct, and the leg force is the same as the uplift on the footing.

When you are not sure, you need to go back to your first year statics and check it from first principles. Regardless of the load path the general equilibrium of moments and forces apply to the structure as a whole. If this doesnt match up then you have done something wrong.

PS: dont forget that half the bolts are pushing downwards with equal and opposite force under the bending! Well a little more actually due to dead load.

 

[civilperson]

Only the bolts in tension require an opposing gravity load. All six bolts will not be in tension at the same time. A cylindrical ring foundation with soil in the center can use the soil weight for stability. Is the controlling moment from wind or from seismic force?

 

[JStephen]

Is this some sort of elevated tank? (Deduced from the mention of "legs")? And where is the overturning force coming from?

Note that if this is seismic overturning, and if the weight of the contents is supported by the legs, then that weight of the contents would also be used to resist overturn. There are also some requirements for seismic loading that the structure has to be designed so that sway rods will yield, and this may affect foundation sizing.

If it's wind overturning, it should be with the tank empty.

Your 348 cubic yards should be cubic feet, but that checks out on the 16x16x8'. Normally, this wouldn't be a solid chunk of concrete, but would be a concrete footing with soil over the footing to resist uplift. And if the footings become too big, consider a single large slab with ringwall or some other configuration of foundation. Piers might be another approach.

Keep in mind that if you are in an area of high seismic forces, it may be very difficult to design an arbitrarily-shaped tank to suit the conditions.

 

[EngineerofSteel]

More data:
*AWWA is American Water Works Association. I found this equation in an old thread on eng-tips.
*The tank IS elevated on 10 legs around the 14.5' diameter circular tank. Each leg's base plate has two 3/4" anchor bolts.
*The design wind speed is 70 mph.
*I checked empty tank with WL and full tank with EQ load. EQ load governs.... as expected.
*The tank will hold unshelled walnuts. There is a slab under the tank at grade. The footings will be piers beneath each leg.
* the tension load of 52K already includes a deduction for the dead load of the contents and tank self weight.

Thanks!

 

[JStephen]

Just making a guess here, but looks like your tank would weigh around 400,000-500,000 lbs loaded. You seem to have 10 legs, and each leg needs 6 bolts per your design, and each bolt carries 52,335 lbs. Setting 4*M/(N*D) - W/N = 52,335 lbs with N = 60 and W = 450,000 lbs gives me 13,000,000 ft-lbs of moment, or a lateral load of around 100%-125% of the weight of the tank. Is this more or less what you have figured? I think this would correspond to setting the tank right near a California fault line.

Anyway, first thing to do is go through your seismic load in detail and make sure you're not missing the boat somewhere on the loading end of it. I think ASCE 7 has some requirements specifically for silos, but these are not generally covered in AWWA D100 or API-650 or other common specifications.

If your seismic calculations and structural design are correct, then you're stuck building a foundation for those loads. With that small of a tank, you'd be looking at a solid slab or maybe a slab with a pedestal.

You've mentioned the 3/4" anchor bolts as though this were an existing silo, but it seems you are building a new foundation for it? Anyway, if the silo itself was not originally designed for high seismic loads, it's quite possible that the legs or leg attachment details won't handle the seismic loading you have figured.

 

[EngineerofSteel]

the DL is 17,520 lbs, LL = 208,680 (black walnuts w/skin).

the center of gravity is at 28.8'. the total seismic (Fp/Wp=.41) gives me 2,670,800 ft-lbs. Plugging this into the AWWA equation gives me the 52K lbs max tension force per bolt.

There are only 2 bolts per footing. I am saying that 3 legs resist uplift. If it is legitimate to use 1/2 the legs to resist uplift, let me know.

In all my previous projects, and those done before my time, the DL exceeded the forces, and no footings were prepared.

In this project, I DO have resultant uplift, and I need some input before I proceed.

Thanks!

 

[EngineerofSteel]

Oh, yes: this is in seismic zone 3, in California, but not within 5 miles of a fault. The tanks are existing silos previously used also in zone 3 to hold grain, which is substantially heavier than walnuts. So, I am sure the legs are sufficient. Also, I found out the ABs in the previous arrangement were 7/8ths, not 3/4" as I had been told.

Thanks!

 

[dig1]

I just finished a project with similar (slightly higher) lateral coefficient plus a vertical seismic coefficent of about 0,37 for several large storage tanks.

In addition to tension the anchorge will have to withstand the seismic shear or you can add a shear key to the bottom of the base plates. Some codes are not allowing shear to be taken by friction on the bottom of the base plate and are requiring either the bolts or preferably shear keys be used.

You need to look at the seismic load acting in various (probably just two)directions relative to the legs. "Pressure Vessel Design Manual" by D.R. Moss gives some visual examples of what I'm talking about.

As mentioned in posts above, some legs and their anchor bolts will in compression from the overturning while others will be in tension.

Looking at the variation in tensile and compressive loads from OTM in the legs will tell you how many bolts are actively resisting the uplift load.

If your building code requirements include vertical seismic be aware that it acts both to cause compression and tension. So in addition to the variation in axial leg load around the perimeter due to OTM you may have (if required by the code) have an additional axial component. When designing the anchor bolts this load has to taken as tensile but for base plates and the buckling check of the leg itself the load needs to be considered compressive.

One thing to remember is that codes evolve over time and requirements for loads and structural ductility have changed. Just because a design was acceptable at one time does not mean it will be acceptable if re-built on the same site today. I understand the density differences in the material but a detailed analysis still needs to be performed.

 

[JStephen]

Plugging those numbers into [(4*M)/(N*D) - W/N], I get 42,547 lbs/bolt or 85,095 lbs per leg, which would require a footing maybe 10'x10'x6' thick- still large enough that you'd be looking at a single slab-type foundation in lieu of separate piers.

Normally, in a building, you would not assume that live load contributed to the stability of the structure. However, in tanks, when the seismic load is primarily due to the weight of the contents, then the contents weight is also included as part of resisting weight to overturning, to the extent that it is supported by the tank.

The seismic loading has been updated several times through the years, and a tank designed now may have design loads quite a bit higher than 20 or 30 years ago.

The equation referenced from AWWA is basically Mc/I - P/A. The distribution for I is taken as a thin ring of equivalent area to the total bolt pattern. If the construction of the leg supports is such that an Mc/I loading is reasonable, then it should still hold. But that could vary somewhat depending on the arrangements of bracing in a leg-type structure.