10ft Retaining Wall With only 12" soil behind it

[StructuralEngineerTX]

Hi Everyone.
Here's the problem. I have approx 10ft tall retaining wall. Behind the wall, there's stable limestone stratum, does not need a wall because it can retain itself.

However, there will be a gap between the back of the wall and the face of the rock, about 12", which will be filled with gravel.

So, it's a 10ft tall wall which is only retaining a 12" wide x 10ft tall "column" of soil per foot of wall length.

I'm trying to figure out what the pressure will be behind this wall. I know it must not be gamma x H because it's only 12" of soil... so what Is the pressure behind the wall?

Thanks.

 

[msquared48]

Around 50 psf uniform.

Mike McCann
MMC Engineering

 

[fattdad]

I'd go with 50 also, or maybe a bit less.

f-d

¡papá gordo ain’t no madre flaca!

 

[EBSEngineering]

I'd also fill the void with angular clear stone (3/4") and put a drain to outlet at the bottom.

 

[StructuralEngineerTX]

OK guys, I appreciate the response... How are you coming up with 50 psf uniform?

And yes, thanks for bringing up the drainage, it will be gravel with a perforated pipe at the bottom.

 

[dgillette]

Yes, drainage could be important. If you have standing water behind the wall, the hydrostatic pressure is the same whether the fill is a mile or a millimeter wide.

 

[paddingtongreen]

The only way I know to calculate a value would be to use a pressure formula for deep bins, such as Janssen's.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[fattdad]

The Rankine active wedge is defined by an angle (measured from the horizontal) of 45+(phi/2). Enscribe that engle downward from the top of rock and toward the backside of the wall. You'll see that the line intersects the wall at the depth of 2 ft.

Calculate the active earth pressure at the depth of 2 ft.

For gravel use a phi=38 degrees, so Ka=0.238 and Eh(at 2 ft)= 59.5 pcf. That's the maximum value of earth pressure below the depth of 2 ft. You see when you say the rock is providing no earth pressure, below the depth of 2 ft the only affect on the Rankine wedge is that the wedge includes more rock, which provides no new horizontal load.

Considering the above posts, I'd now suggest that you use 60 psf as a uniform load below the depth of 2 ft (or for the whole wall), depnding on how refined you want to be.

Hope this helps. Try making a drawing, it may become clearer.

f-d

¡papá gordo ain’t no madre flaca!

 

[jeffhed]

StructuralEngineerTX,
I did a wall just like this a few years ago. The geotech on the project gave me this drawing showing what the wall pressure should be.

 

[VoyageofDiscovery]

Have a look at this link:

http://www.tgs.org.tw/jge/files/articlefiles/v2i2200709051462998162.pdf

 

[bookowski]

I had a similar situation and the geotech on the project told me to still use a traditional pressure diagram. I argued with him as best I could but he kept telling me that it was called the 'silo effect' and that the amount of soil behind the wall didn't matter. It sounds like he was wrong?

 

[StructuralEngineerTX]

Well that makes a lot of sense. The pressure must be related to the amount of fill. There's no way 1ft of soil pushes with the same pressure as 10ft of soil. If that would be the case, then to create energy all you would need would be a tall skinny pipe with water and a generator.

So, soil pressure must me related to the mass of it behind it...

 

[dgillette]

I believe the silo effect is real, but with only 1 foot of soil, I would expect friction on the sides (wall and rock) to provide a lot of support. Replace the rough rock with smooth teflon, and it would be different.

 

[Mflamer]

Hydrostatic pressure is independant of the diameter of the pipe. Pressure and energy are different. StructuralEngineerTX is correct! All you need to create energy is a tall skinney pipe and a generator, untill the pipe is empty! I know it's counter-intuitive, but correct. Review your undergrad fluids textbook.

 

[dgillette]

? This isn't a fluid mechanics problem. It's solid particles, with internal friction and friction against the sides.

 

[VoyageofDiscovery]

Have a look at this link:

http://www.nrcresearchpress.com/doi/pdf/10.1139/t01-063

Should help.

 

[fattdad]

I'm still going with my answer. This is not a complicated problem to solve using first principals.

f-d

¡papá gordo ain’t no madre flaca!

 

[paddingtongreen]

The problem with some of these answers is the method of fill. If the fill is placed in compacted layers, each should be considered separately with the thickness plus a surcharge for the compaction, but then no further lateral pressure from the next layers.

If the fill is tailgated, it becomes analogous to the narrowbunker in which case, Janssen's formula for narrow bins would apply. The result would not be too different. Janssen recognized that if the normal methods, say Rankine were used, the increasing lateral pressure and the friction would cause bridging of the fill material, but material does flow in bunkers and silos so he derived his formula.

Janssen's formula yields a curved line, rather than the straight line of Rankine, et al. See the attached link, below, for a diagram. There are other places on the web where it is explained, do a search.

The above is for anyone who wants to explore the behavior, my guess is that fifty or sixty lbs is about right.

Michael.
Timing has a lot to do with the outcome of a rain dance.