2 retaining walls close to each other 1

[Vipul19]

I have a situation where there is a basement wall 10ft (3000mm) high. on the interior side of the wall approx 900mm behind it, is another wall to support a fireplace (also 10ft high). The small 900mm space between the two walls will be backfilled. While designing the exterior for retaining is fine, I feel like the interior wall which is technically retaining the 900mm wide portion of backfill doesn't need to be designed for the same pressure as the exterior wall. The soil is confined and in a small space.

Any thoughts on what pressure this interior wall should be designed for? (lateral)

 

[oldestguy]

Why do you suppose that designers use the words "equivalent fluid" when discussing lateral earth pressure? If you filled the cavity with water, then what? However,in-this case the "silo hang-up effect" probably applies to both "walls". I trust you are taking into account the vertical compressive stresses from loads on top also.

 

[Vipul19]

Yes I have, the slab above will actually be structural/suspeneded. This "silo hangup" effect is what I am wondering about. Can you use this?

 

[dcarr82775]

I don't agree that it needs to full equivalent fluid pressure. Ignoring silo affects. The EFP is based on a failure plane that extends up to the surface and includes all the weight of soil within that failure plane. If you don't have the full 'area' of soil inside the failure wedge, the load is smaller.

Water is a different animal altogether. EFP is a 'easy' way to think of it, but it is equivalent not equal.

 

[BigH]

You say you have 900 mm between the walls - just wondering if the owner might like to have this space for storage? As the interior wall is only supporting the fireplace, is it really the same "length" as the basement wall? I find that any "storage" space one can find in a house is a plus . . .

 

[fattdad]

Here's how I'd address this question.

The narrow slot is 35 inches.
Let's use open-graded backfill with a friction angle of 40 degrees and a unit density of 130 pcf.
Ka= tan2(45-phi/2)= 0.22
The angle that enscribes the active wedge is 45-phi/2 as measured from the vertical (i.e., 25 degrees)
At the depth of 75 inches the active wedge hits the opposing wall. Below that depth there will be no greater active soil contriubiton.

Using an equivalent fluid density of 130*0.22, you get 28.6 pcf. So, the active earth pressure would range from 0 at the top to 179 psf at the depth of 75 inches and then stay at 179 for the remainder of the depth.

I'd multiply all these values by 1.5 'cause I'd be using at-rest earth pressures.

f-d

¡papá gordo ain’t no madre flaca!

 

[Vipul19]

Thanks fatdad - You have explained the intuitive behaviour perfectly!

 

[RFreund]

I have discussed and used fattdads approach before. You could also do a trial modified trial wedge as explained in the complex geometries section of FWHA -NHI-10-024 and FHWA-NHI-10-025 (I beleive it is that reference)and Stable Features behind SMSE Walls - FHWA-CFL/TD-06-001.

And some more discussion.

http://files.engineering.com/getfil...f_Failure_Plane_and_Design_Considerations.pdf

I should point out though that all the above info is for flexible wall systems.

EIT

http://www.HowToEngineer.com