vertical and lateral earth pressure - length req'd

[BurgoEng]

This may be the most basic of concepts, but these are the things that always leave me scratching my head.

Generally, lateral pressure is based on At rest, active or passive pressure, dependent upon depth and friction angle. So at a depth z, the lateral active pressure would be
pa=Ka g z, which would be in units psf or plf/ft of wall, etc... To my knowledge, this is based on an infinite amount of soil area (if looking down upon the ground from the air). In the case of a wall, say an infinite length of retained soil behind the wall.

The question I have, again sticking with a wall, what is the minimum amount of length behind the wall for the calculated pressure to be realized? Is it 1ft, such that as long as the wall is restraining a 1 ft length (as measured perpendicular from the wall) that the equations would be adequate.

This probably sounds silly to ask, but I actually have a case of a basement being constructed by excavating rock and I am trying to figure out the lateral loads applied to this wall if the space between the wall and the rock is filled with crushed stone, and the space between the rock and wall is only about 1-2ft deep.

 

[DRC1]

If a full failure wedge can develop behind the wall, then you have full active pressure. If the full wedge can not develop, you will have less. For a basement wall I would think you would be looking at at rst pressure. The stone has a high internal friction angle, so a foot width of stone will not add a lot of pressure, but if it is not well drained, any water building up in the stone will add full hydrostatic pressure.
If your basement is down 6.5 ft below grade (lowing for 8 ft basement and 18 in. above ground), and the failure wedge is 30deg from vert, it would day light about 5 feet out. Your set back is about 4 ft say about half. I would say 60% to 75%. As a more rigerous alternate, you could do a trial wedge diagram. Generally basement walls are a litt

 

[fattdad]

The "envelope" of soil is defined (according to Rankine earth pressure) by the angle (measured from the horizontal), 45+phi/2 (45-phi/2 from the vertical). If you have a soil with a friction angle of 30 degrees and a basement that's 8 ft deep, you'd go to the depth of 8 ft and strike a 30 degree angle from the basement wall (60 degrees from the horizontal). The soil within this "active wedge" would then be what's moving toward the wall. If part of the material in the "mass" is rock, you would then proportion the earth pressure by taking a combination of the rock and the soil properties.

The explanation above is for Rankine earth pressures. You would get a different answer if you used Coulomb's construction or a log spirial analysis. However, you'd also end up taking some combination of factors to develop the earth pressure - using both the rock and the soil properties.

Hope this helps.

f-d

 

[DRC1]

Don't know what happened to my last sentence. What I wanted to say was that generally basement walls (residential)seem to tolerate unaticipated loads pretty well, especially those during construction.

 

[BurgoEng]

The wall we have is part retaining wall, part bearing wall, and part non-load bearing. There are building columns supported on piers integrated into the wall. the rock excavation is pretty much solely for the basement of this building.

As for the angle of the failure wedge, if, for instance there was a wall , 8ft high but was only supporting 2ft of fill behind it (say there was another wall supporting beyond the 2 ft) than the pressure on the wall would need to be scaled down as the failure wedge would extend beyond what was actually needed to be suported?

 

[dgillette]

fattdad -You appear to have switched Rankine and Coulomb. Coulomb is based on sliding wedges; Rankine is based on shear failure within the mass of soil next to the wall. It's most easily visualized with a Mohr circle. In active pressure, the wall pressure is Sigma3 in the soil as it fails (meets the strength envelope), and in passive pressure, it's Sigma1. (You can derive it pretty easily from a Mohr circle with one pressure set equal to the vertical pressure.) If you had greased Teflon between the fill and the wall, and between the fill and the opposite side, you could, in theory, develop Rankine active with any thickness of fill.

But it's always more complicated in soil mechanics (or politics, marriage, or religion). You'd probably get something a little HIGHER than Coulomb or Rankine active if you do any compaction at all. Both Rankine active and Coulomb active require the wall to give, and/or the soil to be compressible (so the condition is active rather than at rest like Ko); furthermore, they assume there is no pressure induced against the wall by compaction. If there is compaction, the pressure just below the ground surface won't be zero like with Rankine or Ko. Design with a reasonable safety factor for Ka, and you'll probably be OK. Compact the backfill a little to keep the settlement small, but don't put it in like it was a highway subgrade or dam core.

Oh, and don't let the wall get too long without some sort of buttress or tieback. My basement wall is bowed in visibly, damaging some of the sheetrock, and the backfill was not compacted so it settled, leaving a shallow depression along the wall, correction of which is one of those projects I ought to get to sooner rather than later. Too many corners were cut in construction.

DRG

 

[fattdad]

The shear plane for Rankine earth pressure intercepts the toe of the failure surface at an angle of 45+phi/2 (measured from the horizontal - Terzaghi & Peck Figure 27.3). Log spiral and Coulomb are just other ways to derive earth pressures.

Getting back to the topic at hand, I've thought through the question of how to calculate (easily) the earth pressures of two walls that are 10 ft apart and 30 ft high (i.e., like a road grade with retaining walls on either side). Clearly, you'd never use Ka*gammaM*H for the pressure on the base of both opposing walls. However if you took the shear plane surface (i.e., 45+phi/2) at some point you'd intercept the opposing wall. Below that depth the earth pressures should then stay uniform (i.e., go to a rectangular distrubion, notwithstanding a surcharge pressure on the road).

On the matter of compaction, DRG is correct. You must consider that compacted soil backfill will "lock-in" additional earth pressure - especially in the upper interval of the backfill. One way to minimize this concern is to use open-graded aggregate adjacent to the wall.

I like these problems. . . .

f-d

 

[dgillette]

The first paragraph of Article 27 describes the Rankine states as plastic equilibrium throughout the soil mass. As such, it is conceptually different from Coulomb and log spiral (which is basically the same as Coulomb except with a spiral sliding surface instead of planar). All those diamonds in figures 27.1, 27.2, 27.3... represent the "infinite" number of shear surfaces within the soil mass at failure or plastic equilibrium. They are indeed oriented at 45 +/- phi/2, as you would get from a Mohr circle in contact with the strength envelope. Coulomb, on the other hand, comes from static equilibrium of a block on a distinct slide plane, which has the benefit of allowing you to do the static equilibrium calcs considering friction angle AND cohesion, slope of the backfill, friction on the wall, etc. The slide's orientation is also 45 +/- phi/2, which you can derive by making the force on the wall a function of the angle of the slide surface, then taking the derivative of it and setting the derivative equal to zero, so as to get the minimum value for sliding back up the surface for passive, and the maximum value for sliding down the surface for active.

Somewhere in my files is a copy of J. MacQuorn Rankine's original paper for the Royal Society.

I'm not clear on why the pressures below the slide plane-wall intercept would be uniform. Wouldn't you need to consider the whole thing above a sliding plane to the floor as a sliding block, going back to the derivation of the Coulomb method? It would be a triangle with the top corner away from the wall being analyzed, clipped off 10 feet from the wall.

This would be so much easier if we could draw pictures for each other. I'll try. It's not to scale of course, but the width is 10 feet, the height is 30 feet, and the slope angle is 60 degrees (assuming phi=30). Find the total force needed to maintain equilibrium against sliding with whatever wall friction you like, and I believe you will find it to be only slightly less than the force you'd get by assuming simple Coulomb active (the only difference being from the weight of the little piece clipped off in the top right corner). Wall friction might help significantly.

|<--wall
| .|<--other wall
| .|
| .|
| .
| . <--60 degrees
| .
| .
|._____ floor

Without a similar force on the other wall, the mirror image slide would occur, and horizontal equilibrium has to be maintained. Then there's that arching problem, which probably isn't really a problem with the 3:1 aspect ratio and compacted fill that wouldn't settle much as additional material is placed over it.

You’re right; these are fun, at least to a certain subset of engineers.
DRG

 

[BurgoEng]

I think this is why I wanted to learn site design.

 

[fattdad]

DRG, You have some interesting (and correct) points. Maybe uniform is not right, but it also can't continue to increase proportional to some "equivalent fluid density". I'm sitting at my desk with a calc sheet trying to figure out an appropriate "Rankine" solution and am somewhat flumoxed - ha.

Below the depth where the active wedge intercepts the "other wall", the lowest shear surface remains a constant length. With increasing depth the weight of the soils above the lowest shear surface increases. With an increased weight, there is an increase to the resisting shear force (i.e., W*tan phi). Now the question is how does this affect the resultant Eh (horizontal earth force) vector.

I'll continue to ponder. Nice to know that we all kept our T&P - final ha.

f-d

 

[dgillette]

Here's what I got for a simple example (feel free to check my math):

Assume Phi=30 degrees, c=0, frictionless walls (delta=0)

The slide mass weighs 26.67 kips (W). Fn is the normal force on the slide plane. Fw is the force on the wall.

Vertical forces:

W=Fn cos 60 + Fn tan 30 sin 60

Horizontal forces:

Fw=Fn sin 60 - Fn tan 30 cos 60

I get Fw = 15,390, which compares with a simple Rankine active of 18,750, which should be the same as Coulomb active with no wall friction. Adding wall friction would complicate matters a little because you'd have a term for Fw tan delta in the vertical force equation and you'd have to solve the equations simultaneously, but it's still the same principal. Rankine's model is out the window if there is wall friction because it assumes the ground is all at plastic yield and Sigma3 is horizontal.

If you move the wall farther back so you change the weight of the slide mass to 32,475, the wall force goes up to 18,750, which matches what both Rankine and Coulomb predict, so I think I have it right.

I now think you're right about the active pressure being constant below the wedge intercept. I didn't think that would be so, but played with the mechanics a little and change my mind. (At rest pressure would not be constant.)

In my office, I'm often the (please pardon me for saying this) Designated Deriver.

 

[cap4000]

BurgoEng

With only 1 to 2 feet of gravel between your basement wall and the rock wall your lateral pressure is reduced considerably. If you can, get the 1982 Spangler and Handy Soil Engineering Book 4th Edition. On page 572 it has the very unique calculation procedure. Good Luck.

 

[Panars]

I use a procedure outlined in a journal article in ASCE's journal of geotechnical engineering.

Frydman, S. and Keissar, I. (1987). "Earth Pressure on Retaining Walls Near Rock Faces," J. Geot. Engrg., ASCE, 113(6), 586-599.

If I remember correctly, the procedure is similar to one for the pressures from granular material (grain) in a silo.

 

[BurgoEng]

Thanks to everyone for their input. Not surprisingly, the wall location has changed, yet again, and I don't think I need to deal with this any more.

 

[cap4000]

Panars

I have that great article. The only thing is what K value do you use. At Rest or Active. I have another 2006 article by Zornberg and Yang that seems to indicate for rigid concrete walls th K value is between Ka and Ko. Their is a big difference if you have a MSE geogrid type wall versus rigid concrete wall.

 

[eric1037]

We usually use Ka if the wall is allowed to move at least 0.001 times the height of the wall for dense sand (eg compacted granular backfill). If the wall is restrained at the top (eg. with a basement floor) and sufficiently stiff, then you would use something closer to Ko. If the wall is a site retaining wall, I would use something closer to Ka.

 

[eric1037]

Another thing:

For this particular example, it is probably harder to use Ka if the foundation is bearing on rock. It is less likely to move enough to use Ka.

Also, dgillette has a good point on compaction. In addition, there have been experiments on granular soils by Matsuo, Kenmochi, and Yagi in 1978 in their paper entitled "Experimental Study on Earth Pressure of Retaining Wall by Field Tests", that indicates that even if the soils are not compacted, they will return to at-rest values. This is especially true if there are long-term vibrations, water level fluctuations, or temperature changes.