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# Passive pressure on an inclined foreslope for permanent soldier beam wall

## Passive pressure on an inclined foreslope for permanent soldier beam wall

(OP)
Have been asked to design soldier beam for permanent cantilever wall (exposed height ~6').
Borings show silty-sand (report provides Phi = 29 deg, unit weight 125 PCF). Borings show there is fractured shale beneath the soil, but I would like to design the wall as if it was encountering only soil (which may be the case, we only have one boring to go on). Someone called for a rock socket of 5' minimum, which seems too little considering 1) we're on an angle, and 2) the shale is fractured. In any case:

The retained side of the proposed wall has a 27 deg backslope. The groundline is front of the proposed wall however, has an approximate downward slope of 51 deg.

Problem arises with calculating the passive pressure coefficient, Kp. Using Coulomb, (PHI = 29 deg, DELTA = PHI/2, BETA = 51), a Kp = 263 was calculated.

After this, tried looking for other methods of calculating passive pressure with inclined slope.

Using NAVFAC DM 7.02 (Figure 3 on Pg. 64), the formula for Kp yielded Kp = 46 (formula was dependent on PHI and BETA only).

Using NYDOT GDP-11 (Case 1, Infinite Slope Analysis, Pg. A-3), the formula for Kp (dependent on PHI and BETA again), doesn't even yield a number because the cosine square of 27 subtracted from cosine square of 51 yields a negative number, which there is obviously no square root for.

Does anyone have any recommendation for calculating passive pressure in a situation like this with an inclined foreslope? Is my assumption of treating it like it's all soil incorrect, given that the friction angle is less than the measured slope?

Thanks all.

### RE: Passive pressure on an inclined foreslope for permanent soldier beam wall

The negative numbers you're getting probably indicates that with a phi of 29 degrees it shouldn't even be standing at a slope of 51 degrees. By the numbers, you have no passive resistance because the soil shouldn't even be holding itself up, much less providing resistance to external loading. Sounds to me like a case for ignoring the soil and calculating the required embedment into the bedrock.

### RE: Passive pressure on an inclined foreslope for permanent soldier beam wall

(OP)
Thanks HotRod10, I was suspecting that was the case... supposing I wanted to create a soil-only design (for myself at least), maybe assume a foreslope of 29 deg or slightly less?

### RE: Passive pressure on an inclined foreslope for permanent soldier beam wall

Bayek949, when using the Coulomb equation for Kp, the downward slope in front of the wall needs to be a negative number. You should use -51 degrees. However, the equation Kp will not give a rational number if the slope is steeper than the phi angle of the soil.

To check this try using reasonable angles. Check a soil with 29 degrees = phi and no slope below the wall. Kp with wall friction being 1/ x phi = 14.5 degrees = 4.635. If you slope the ground down at -10 degrees in front of the wall, Kp reduces to 2.977. If you slope the ground up at +10 degrees, Kp rises to 7.437. If you slope down at -51 degrees, Kp = -0.225+0.508i. If you slope the ground upward +51 degrees in front of the wall, Kp = 262.992. If you slope down at -29 degrees, Kp = 0.790.

Summary: A downward slope in front of the wall needs to be a negative angle with an absolute value of at least phi.

### RE: Passive pressure on an inclined foreslope for permanent soldier beam wall

The angle of repose (the angle that soil will stand) for granular materials is generally similar to the phi angle. If the phi angle is truly around 29 degrees, then there obviously some cohesion also, a property of clay soils, which also creep under sustained load. That means you still wouldn't have much, if any, residual strength available to resist movement of the soldier piles in the long term.

### RE: Passive pressure on an inclined foreslope for permanent soldier beam wall

(OP)
@PEinc,
Thank you for the detailed response, I see how those numbers make sense. In that case, I'd design for the assumption there is rock below the subgrade.
Using AASHTO (1996 Standard Spec. for Highway Bridges), there seem some options:

1. For a "highly fractured, moderately weathered shale" with REC 92%, RQD 0% (seems pretty unreliable rock), could I use a medium-stiff clay value for shear strength (750 PSF?)

2. If I use a medium-clay value in lieu of a shale strength, would using Figure 5.6.2A (Embedment in Rock) or Figure 5.6.2C (Embedment in Cohesive Soil Retaining Granular Soil) be more appropriate to determine the passive pressure? The 5.6.2A equation is dependent on shear strength, foreslope inclination (-51 deg), and embedment depth, D. The 5.6.2C equation is dependent on b (wall element width), foreslope inclination, embedment depth, and also undrained shear strength of clay, effective saturated unit weight, and exposed wall height.

3. Would you still recommend using the measured foreslope angle (-51 deg), or another, if embedment is not considered to be in cohesionless soil now?

4. If there was really clay, I could use plasticity index to correlate friction angle. Since there's really shale rock, that I'm analyzing as clay, I suppose I can't correlate some friction angle to use in my usual Kp calculation?

@HotRod10,
Agreed, that makes sense. Guessing that in reality there's shallow shale rock underneath the soil on the slope, but with one boring to go on... it's not much.

### RE: Passive pressure on an inclined foreslope for permanent soldier beam wall

Yeah, you really need more borings to determine where your bedrock is.

With slopes that steep, a soldier pile wall would not be my first choice. I would be looking at an anchored wall of some kind, especially if the bedrock is sloped like the ground surface.

I also recommend using a newer version of the AASHTO spec. The knowledge base has expanded, the analysis tools are much more advanced, and the available solutions have changed drastically in the last 20 plus years. I don't recommend any of the first 3 editions of the AASHTO LRFD spec., though.

### RE: Passive pressure on an inclined foreslope for permanent soldier beam wall

If the rock is highly fractured with RQD = 0, maybe the rock should/could be considered a gravel? Also, maybe assume that the base of the exposed wall is deeper than 6' so that you can assume a slope that is much flatter than -51 degrees. Your 6' exposed height could theoretically increase and still allow use of a cantilevered wall - depending on how much of the steep slope you ignore. You would need to ignore enough slope so that you had significant "flat" rock (or gravel?) in front of the embedded soldier beam. Installing tieback anchors may be necessary but installation won't be easy working from a high, 51 degree slope.

For a permanent wall, I would assume no cohesion (c = 0). Just because your phi angle is 29 degrees, that doesn't mean there is cohesion. For example, loose clean sand could have phi = 29 with c = 0. Also, a phi angle of 29 degrees does not mean that you will have long term creep. You said that you have sand, not soft clay.

HotRod10 is right about getting a newer AASHTO. Much have changed in AASHTO for designing non-gravity retaining walls. AASHTO is expensive. So, try downloading a free, or at least more economical, Geotechnical Circular Design Manual from FHWA's geotechnical publications web site. It has the same stuff as AASHTO.

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