Reinforced concrete deep walls help 2

[derecha1]

I have been trying to find some information out about a deep reinforced concrete rectangular vault I am designing. The dimensions are 26' x 30' x 29' deep. Using "Rectangular Concrete Tanks" I am getting a wall thickness between 22" and 28" based on shear. Does this seem reasonable? It sounds a little big but I don't know how I would determine it otherwise. I am using a lateral force of 135 pcf x 0.6 (at-rest coefficient) based on geotechnical.

Any help with this would be appreciated.

 

[whymrg]

2 feet thick wall does not seem to be too much for 29 feet deep vault.

 

[derecha1]

I appreciate your comment. I understand that with retaining walls it would not be unusual and I just want to make sure that I am not missing something since this is a rigid box as opposed to just a retaining wall.

 

[JAE]

What are you using for phiVc?

 

[derecha1]

phiVc for 28" wall is 37,504 lb. Vu is 37,074 lb based on coeff. and analysis from "Reinforced Concrete tanks" using fixed base conditions. Hinged base is less but shear at edge controls under hinged base.

 

[Focht3]

To me, the wall seems too light - particularly if groundwater could be an intermittent problem. Deflection, not strength, should control this design. The critical design case would be with the vault empty of water, with groundwater at the ground surface. I'd expect wall thicknesses to be more than 30 inches for the GWT below the vault, and substantially larger for the GWT at the surface.

I participated in the forensic evaluation of an above ground sewage treatment plant - aeration basin walls had failed. The tank had been designed using ultimate strength design techniques (a precursor to LRFD) instead of ASD. The as-built walls were 14 inches, while ASD would have dictated 20+ inches. Oh, and the basin walls were only about 15 feet high, and the fluid unit weight [γ]_sewage = 63 pcf. That's considerably less than the [γ] = 81 pcf you have assumed, and your wall is twice as high. Your design [γ] = ( (135-63)*0.6 + 63 ) = 106 pcf if the GWT is at the ground surface.

Proceed with caution...

Perform an ASD design to check your current calculations. If there's a big difference, you have a problem...



Please see FAQ731-376 for great suggestions on how to make the best use of Eng-Tips Fora.

 

[derecha1]

Thanks FOCHT3 for the reply. GW is not an issue here based on drilling and testing. The top of the vault is restrained at the top with a slab at the surface. (Sorry not included previously. This vault is only being analyzed empty because it is only a vault to house valving for an adjacent tank.

Could you clarify what ASD Design is and where I can find it?

 

[Focht3]

ASD is the abbreviation for Allowable Stress Design - the "old" way of doing concrete design.

How are you performing the design - by "hand", or computerized design package?



Please see FAQ731-376 for great suggestions on how to make the best use of Eng-Tips Fora.

 

[derecha1]

Sorry, just not familiar with the acronym.

I am performing the computations by hand using PCA's resource "Rectangular Concrete Tanks" to analyze the structure, which gives factors based on finite element analysis for these types of structures.

 

[tincan]

The shear plate coefficent is in the ballpark order of 0.33 for a wall fixed at the base and both sides, top free.
The working stress would put you at approximately 22,500#/ft of shear at the mid-span (worst point). Without the factored loads, and f'c = 3500psi & 65psi shear,
22500/12(65) = 28.8 inch d. 28.8 + 1/2 bar thickness + 2" cover==> 31+. Your numbers are not out of whack, I did a 75' x 100' x 25' deep recently, had the same order of magnitude. Focht3 has good advice, Best of luck
Tincan.

 

[derecha1]

Thank you all, this has been helpful to just check where I was at.

 

[JedClampett]

An interesting thing I've noticed in reviewing LRFD (ultimate) strength design vs. ASD when live load is the predominate load. Shear is slightly more conservative using LRFD. Allowable stress shear is 1.1*f'c^.5. LRFD shear is (.85*2*f'c^.5)/1.7 which simplifies to f'c^.5 which is 10% less.

 

[pmkPE]

ASD in concrete is known as working stress design, LRFD in concrete is known as ultimate strength and utilizes load factors which I did not notice in your calculations above.