## Retaining Wall with inclined backfills

## Retaining Wall with inclined backfills

(OP)

Hi sifu. I have questions and require expertise inputs in regarding the geotechnical’s analysis approaches for a retaining wall with inclined backfills. The reasons are that I saw hand calculations that are different from what I normally read in Das or Bowles text book. The hand calculations (which i was told was based on older test book) appear to present lower forces acting against the retaining wall

I would like to discuss and check whether these alternative approaches are correct or not. I have attached the sketches for better understanding.

For simplification, let’s assume the inclined slope is infinity, no hydrostatic pressure acting against the wall, the backfills and the existing soils are of the same cohesionless soils, and all the extent and sizes are in the attached sketches. I hope that’s’ more than enough info.

Approach 1

The approach 1 is typical textbook formula. Calculate Ka based on the incline slope, friction angle of backfills, wall-soil friction, etc, and derive with a resultant force acting parallel with the inclined slope. The forces acting against the wall are basically the resultant forces time the cos (inclined slope angle).

Approach 2

The approach 2 is to draw a line in 1:1 from the heel and intersect with the horizontal line from the top of wall (at point 0). Then treat the inclined backfills within the triangle area (ABO) as surcharge, and then redistribute them as an equivalent uniform surcharge load along line AO. Since the triangle area (AB0) was treated as surcharge, then the backfills beneath the triangle was treated as normal active pressure. Thus, Ka (excluding the inclined slope factor) was much smaller than the approach 1. Subsequent analyses of forces acting horizontal against the walls were calculated.

Approach 3

The approach 3 is similar to approach 2 but with exception that line from heel shall intersect with inclined slope. Then treat the inclined backfills within the triangle area (ACO) as surcharge, and then redistribute them as an equivalent uniform surcharge load along line AD.

The approach 2 and 3 provide identical results but almost ½ smaller than approach 1.

Any ideas?

I would like to discuss and check whether these alternative approaches are correct or not. I have attached the sketches for better understanding.

For simplification, let’s assume the inclined slope is infinity, no hydrostatic pressure acting against the wall, the backfills and the existing soils are of the same cohesionless soils, and all the extent and sizes are in the attached sketches. I hope that’s’ more than enough info.

Approach 1

The approach 1 is typical textbook formula. Calculate Ka based on the incline slope, friction angle of backfills, wall-soil friction, etc, and derive with a resultant force acting parallel with the inclined slope. The forces acting against the wall are basically the resultant forces time the cos (inclined slope angle).

Approach 2

The approach 2 is to draw a line in 1:1 from the heel and intersect with the horizontal line from the top of wall (at point 0). Then treat the inclined backfills within the triangle area (ABO) as surcharge, and then redistribute them as an equivalent uniform surcharge load along line AO. Since the triangle area (AB0) was treated as surcharge, then the backfills beneath the triangle was treated as normal active pressure. Thus, Ka (excluding the inclined slope factor) was much smaller than the approach 1. Subsequent analyses of forces acting horizontal against the walls were calculated.

Approach 3

The approach 3 is similar to approach 2 but with exception that line from heel shall intersect with inclined slope. Then treat the inclined backfills within the triangle area (ACO) as surcharge, and then redistribute them as an equivalent uniform surcharge load along line AD.

The approach 2 and 3 provide identical results but almost ½ smaller than approach 1.

Any ideas?

## RE: Retaining Wall with inclined backfills

chfting- There is an error in your calculations for "Approach 2" and "Approach 3". The problem is for "Inclined Backfill" but the calculations use the "Level Backfill" equation. See the attached pages showing the Rankine / Coulomb method for "Equivalent Horizontal Liquid Pressure".Also attached are my calculations using the inclined backfill equation. A summary diagram is shown below. My calculations =

~~2580~~1290lb/ ft of wall length, which differs from your "Approach 1" value of 1904 lb/ft of wall length.www.SlideRuleEra.net

www.VacuumTubeEra.net

## RE: Retaining Wall with inclined backfills

SlideRuleEra.For your comments on "approach 2" and "approach 3",

exactly, that’s what I have in mind. The "approach 2" and "approach 3" follows the principles of the hand calculations that have been shown to me (which i should mentioned in my earlier post that are different from what I normally do). These principles are provided by seniors who are saying coulomb or rankine are too conservative when with deep slope. And 45 degree from heel is a conservative line for a rupture surface or active wedges. Of course, to be really exact, may need to do a cullmann’s graphical method to determine it. But that will be another story….My question is that have anybody (apart from my seniors) have used the approach 2 and 3 before? or similar principle like that. I couldn't find available info for those approaches.

For your comments on "approach 1",

I really liked your quick simple calculation, I will save in my data for future use. For information, my calculated Ka is 0,65, if based on some other charts, Ka may range from 0.55 to 0.67. I use 0.65 for severity. However, I do note that ur H is 6, mine is 8 at the heel area. Based on text book(seen as below), my H was determined at heel. And I noticed that ur result should be divided by 2 since it is a triangle area. So, if I use ur formula, force per length = ½*8*8*71.6= 2291.2 Ib/ft….which is still higher than mine, but much closer. What do u think?## RE: Retaining Wall with inclined backfills

chfting- Thank you for finding my error. I have corrected both the sketch and my post.To solve your example using the Rankine-Coulomb method for inclined backfill method I posted, the wall height is 6 ft., not 8 ft. See the image shown below.

The following is a link to a paper on "Some Simplifications of Coulomb's Active Earth Pressure Theory". The proposed changes improve accuracy of Rankine's inclined backfill method.

Also, I have attached a 13-page paper on

Krey's Method. This method uses better assumptions to provide lower, more accurate lateral loads than the Rankine-Coulomb Method. The inclined backfill case is discussed.www.SlideRuleEra.net

www.VacuumTubeEra.net

## RE: Retaining Wall with inclined backfills

## RE: Retaining Wall with inclined backfills

SlideRuleEra&OldestGuySorry for not replying earlier. Was busy

I have downloaded and went through ur information. Very interesting article. Yes, I agree with that Rankine solution, in general, over- and underestimate the active and passive pressures, respectively. In the passive case, however, the Coulomb theory tends to overestimate the pressure. Previously I have known that the reliable values are found from applications of the log spiral method; such as Kerisel & Absi . I have never use muller formula previously. I will implement it to check its effects.

As for the value of H, I think I know why we come with different values. Please correct if I’m wrong.

For geotechnical purpose, we tend to use heel’s H when checking for sliding, overturning, and bearing capacities. (H=8)

For structural purpose, we tend to use the stem wall’s H when checking for bending and shear at the stem wall. (H=6)

For structural purpose, my approach 1’s result will be 1070lb/ft (forces acting horizontal against the wall)

For geotechnical purpose, my approach 1’s result will be 1903lb/ft (forces acting horizontal against the wall). If based on chart 10.9 by OldestGuy, i calculate it to be 1856lb/ft. (1/2*58*8*8)

By comparing to the results (1073lb/ft) of the approach 2 and 3 in my first post, i noticed that my 1070lb/ft or yours 1290lb/ft are quite similar to the results derived from these 2 Unorthodox analysis approach. If based on chart 10.9, it will be 1/2*6*6*58=1044lb/ft. hmmm....maybe these analysis approaches 2 and 3 are not that unorthodox at all.

OldestGuy. Have u seen the analysis approach 2 and 3 before?## RE: Retaining Wall with inclined backfills

assumed"liquid" would flow over the top of the wall.www.SlideRuleEra.net

www.VacuumTubeEra.net