Abstract:
A retaining wall or a counterfort retaining wall incorporates a vertical stem wall section, a base section having a toe, a heel, and a number of shallow piles at the bottom of the base section to improve the sliding and overturning safety factor of the wall. To further improve the sliding and overturning safety factor of the wall, an edge of the heel is slanted with a bevel or a batter with an angle to pick up more resisting load.

Description:
CROSS-REFERENCE TO RELATED APPLICATION 
     This application claims the benefit of U.S. Provisional Application No. 61/036,859, filed Mar. 14, 2008, which is herein incorporate by reference. 
    
    
     FIELD OF THE INVENTION 
     The present invention relates to retaining walls and counterfort retaining walls. 
     BACKGROUND 
     In many wall designs, such as retaining walls or counterfort retaining walls or flood walls, the site parameters dictate the final design of the wall. In many cases there is also a property line that abuts the wall and moving the wall as close as possible to the property line is desirable. The retaining wall of the present invention gives the designer the ability of doing so. In addition, the embodiments of the retaining wall disclosed can be designed to have less concrete and backfill and thus may be more economical to construct. 
     SUMMARY 
     This summary is provided to introduce a selection of concepts in a simplified form that are further described below in the Detailed Description. This summary is not intended to identify key features of the claimed subject matter, nor is it intended to be used as an aid in determining the scope of the claimed subject matter. 
     To address the problems discussed above, a retaining wall or a counterfort retaining wall includes a vertical stem wall section and a base section at the bottom of the stem. The base section has a heel portion and a number of piles that extend downwardly from the bottom of the base. In one embodiment, the heel is beveled. In another embodiment, the piles are replaced with a key that runs continuously along the bottom of the base section. In yet another embodiment, the base section has a toe portion that extends outwardly from the stem wall. 
    
    
     
       DESCRIPTION OF THE DRAWINGS 
       The foregoing aspects and many of the attendant advantages of this invention will become more readily appreciated as the same become better understood by reference to the following detailed description, when taken in conjunction with the accompanying drawings, wherein: 
         FIG. 1  is an isometric view of an embodiment of a retaining wall in accordance with the present invention; 
         FIG. 2  illustrates a cross-section of the retaining wall shown in  FIG. 1 ; 
         FIG. 3  illustrates a front face elevation of retaining wall shown in  FIG. 1 ; 
         FIG. 4  illustrates an embodiment of retaining wall shown in  FIG. 2  with a key option instead of piles; 
         FIGS. 5A ,  5 B,  5 C, and  5 D illustrate different embodiments of a retaining wall with their corresponding location relative to a property line; 
         FIG. 6  is an isometric view of an embodiment of a counterfort retaining wall in accordance with the present invention; 
         FIG. 7  illustrates one embodiment of a retaining wall using a bevel on the heel portion and a number of shallow piles; 
         FIG. 8  illustrates another embodiment of a retaining wall using the bevel on the heel portion and a key; 
         FIG. 9  is a pressure diagram showing forces and weights encompassing the retaining wall shown in  FIG. 7 ; and 
         FIG. 10  is a pressure diagram showing forces and weights encompassing the retaining wall shown in  FIG. 8 . 
     
    
    
     DETAILED DESCRIPTION 
       FIGS. 1 ,  2  and  3  show one embodiment of a retaining wall according to the present invention. A retaining wall includes a non-prismatic stem wall  10  and a base section  20  that is positioned below the stem wall. The base section has a width that is wider than the width of the stem wall. In the embodiment shown, the base section also has a toe portion  80  and a heel portion  70 . Extending downwardly from the base section  20  is a number of shallow piles  30 . In one embodiment, the base section has a bevel  40  positioned on its upper outside edge that engages the backfill. The retaining wall supports a sloped backfill  50  behind the stem wall and a level backfill  60  is placed in front of the stem wall. In the embodiment disclosed, the piles have a length that is generally 10 feet long or less and that are approximately 2 to 24 inches in diameter, depending on soil conditions. 
       FIG. 4  shows another embodiment of the present invention with a continuous beveled key  90  extending along the bottom of the base section instead of the shallow piles. 
       FIGS. 5A ,  5 B,  5 C, and  5 D show several optional embodiments of the present retaining wall invention and their corresponding location relative to property line  100 . In  FIG. 5A , the toe portion is wider than the heel portion such that the wall stem is nearer the beveled edge of the heel portion than a front edge of the toe portion. In  FIG. 5B , the retaining wall lacks a heel portion and simply includes a toe portion that extends horizontally outwards from the stem wall section. A number of shallow piles extend downwardly from the toe portion. Conversely, in  FIG. 5C , the retaining wall lacks a toe portion but has a heel portion with a beveled edge. The exposed face of the stem wall is generally aligned with the property line. In  FIG. 5D , the retaining wall lacks both a toe portion and the shallow piles. 
       FIG. 6  shows one embodiment of a counterfort retaining wall according to the present invention. A counterfort retaining wall includes a non-prismatic stem wall  10  and a base section  20  that is positioned below the stem wall. The supporting intermediate non-prismatic column beams  110  and  120  support the stem and the base as shown in both sides of the stem wall. It is optional to use the supporting beams  110  only or  120  or both on either side of the stem. The base section has a width that is wider than the width of the stem wall. In the embodiment shown, the base section also has a toe portion  80  and a heel portion  70 . Extending downwardly from the base section  20  is a number of shallow piles  30 . In one embodiment, the base section has a bevel  40  positioned on its upper outside edge that engages the backfill. The retaining wall supports a sloped backfill  50  behind the stem wall and a level backfill  60  is placed in front of the stem wall. In the embodiment disclosed, the piles have a length that is generally 10 feet long or less and that are approximately 2 to 24 inches in diameter, depending on soil conditions. 
     The description provided below provides one exemplary methodology for calculating the wall dimensions. As will be appreciated by those skilled in the art, the dimensions are affected by soil type and expected loads on the wall. Therefore other methods may be used in calculating the wall dimensions. In the embodiments described, the retaining wall is formed of steel reinforced concrete. The retaining wall illustrations in  FIGS. 7-10  are not shown to scale. 
     In the exemplary calculations provided below, the principals of soil mechanics as described in the Concrete Reinforcing Steel Institute Design Handbook CRSI 2002 are used. A retaining wall with a prescribed earth pressure due to a sloped backfill is to be designed. The pressure due to the bevel has a different active coefficient and the passive pressure on the pile is assumed to increase from the passive wedge due to the friction and cohesion at the side of the wedge. These changes in the pressures are primarily what constitute the success of the retaining wall design disclosed. Historically these new pressure interactions along with the geometrical changes in this invention were not observed or, if assumed, they were then dismissed without actually having a close look on this phenomenon. Piles and battered piles have been used in the past with retaining walls for axial loads and moments. However, shallow piles were not used because designers preferred using a key not realizing the advantages of using shallow piles. 
     The given description is of wall backfilled with cohesive-less such as sand and gravel backfill that are commonly used. The stem wall is assumed prismatic. The description of the wall shown in  FIG. 7  is such that the dimension A base  is the base dimension and can be anything but zero. The dimension A stem  is the stem dimension and can be any distance but zero. The dimension B is the dimension of the toe and can be any distance including zero. The dimension C is the dimension of the heel and can be any distance including zero. The dimension D is the depth of the piles and can be any distance including zero. The dimension d1 is the diameter of the pile and can be any distance. Also, one can substitute the piles with a standard key  FIG. 8 , where K is the distance of the key and t is the width of the key. The location of the piles and of the key can be in anyplace at the bottom of the base. Additionally, the piles can be battered if more resistance is needed. The angle β′ can be anything smaller or equal to 90 degrees. For a counterfort retaining wall that includes columns or vertical beams that support the stem and incorporates the beveled angle β′ can be with or without a shallow pile can be analyzed similarly. The angle ε can be from −90 degrees to φ, where φ is the internal friction angle of the soil. For other types of backfill different analysis is needed. However, the following principals can be used as a guide for other conditions. 
       FIGS. 9 and 10  show the pressure diagram and the weights on the retaining walls of  FIGS. 7 and 8 , respectively. 
     The analysis can vary depending on how the pressure is treated at the bevel. Active pressure can be used at reset pressure can be used or uplift pressure can be used. In all conditions the stability can be shown to improve from a normal retaining wall. It will be demonstrated the case for active pressure for a pile and a key.
         β=180−β′=bevel angle   α=local slip surface at the bevel   φ=internal friction angle of the soil   δ=concrete friction angle with soil at bevel   γ=unit weight of soil   c=the cohesion below the base       

     Let: 
     
       
         
           
             
               n 
               m 
             
             = 
             
               
                 
                   sin 
                   ⁡ 
                   
                     ( 
                     
                       β 
                       - 
                       90 
                       - 
                       ϕ 
                     
                     ) 
                   
                 
                 
                   cos 
                   ⁡ 
                   
                     ( 
                     
                       β 
                       - 
                       90 
                       - 
                       ϕ 
                     
                     ) 
                   
                 
               
               = 
               
                 tan 
                 ⁡ 
                 
                   ( 
                   
                     β 
                     - 
                     90 
                     - 
                     ϕ 
                   
                   ) 
                 
               
             
           
         
       
         
         
           
             u=tan φ+tan δ 
             v=(tan φ+tan δ) tan(β−90−ε)+1−tan φ tan δ 
             w=(1−tan φ tan δ)tan(β−90−ε) 
           
         
       
    
               tan   ⁢           ⁢   α     =       -     n   m       +           (     n   m     )     2     +     w   u     -       (     n   m     )     ⁢     (       v   ⁢             u     )                         f   =         sin   ⁡     (     β   -   90     )       +       cos   ⁡     (     β   -   90     )       ⁢     tan   ⁡     (     α   -   ϕ     )                 [     1   +     tan   ⁢           ⁢     δtan   ⁡     (     α   -   ϕ     )           ]     ⁡     [       tan   ⁢           ⁢   α     +     tan   ⁡     (     β   -   90   -   β     )         ]       ⁢       cos   2     ⁡     (     β   -   90     )                       g   =         [       sin   ⁡     (     β   -   90     )       +       cos   ⁡     (     β   -   90     )       ⁢     tan   ⁡     (     α   -   ϕ     )           ]     ⁢   cos   ⁢           ⁢   ɛ               [     1   +     tan   ⁢           ⁢   δ   ⁢           ⁢     tan   ⁡     (     α   -   ϕ     )           ]     ⁡     [       tan   ⁢           ⁢   α     +     tan   ⁡     (     β   -   90   -   ɛ     )         ]                 cos   ⁢           ⁢     (     β   -   90   -   ɛ     )     ⁢     cos   ⁡     (     β   -   90     )                       
Then:
 
     
       
         
           
             
                 
             
             ⁢ 
             
               
                 K 
                 ba 
               
               = 
               
                 
                   0.5 
                   ⁢ 
                   
                     f 
                     ⁡ 
                     
                       [ 
                       
                         
                           
                             
                               
                                 sin 
                                 ⁢ 
                                 
                                   ( 
                                   
                                     β 
                                     - 
                                     90 
                                   
                                   ) 
                                 
                               
                               + 
                             
                           
                         
                         
                           
                             
                               cos 
                               ⁢ 
                               
                                 ( 
                                 
                                   β 
                                   - 
                                   90 
                                 
                                 ) 
                               
                             
                           
                         
                         
                           
                             
                               tan 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               δ 
                             
                           
                         
                       
                       ] 
                     
                   
                   ⁢ 
                   
                     
                       .075 
                       ⁢ 
                       
                         A 
                         base 
                       
                     
                     
                       
                         h 
                         1 
                       
                       - 
                       
                         0.375 
                         ⁢ 
                         
                           A 
                           base 
                         
                       
                     
                   
                 
                 + 
                 
                   g 
                   ⁡ 
                   
                     [ 
                     
                       
                         
                           
                             
                               sin 
                               ⁢ 
                               
                                 ( 
                                 
                                   β 
                                   - 
                                   90 
                                 
                                 ) 
                               
                             
                             + 
                           
                         
                       
                       
                         
                           
                             cos 
                             ⁢ 
                             
                               ( 
                               
                                 β 
                                 - 
                                 90 
                               
                               ) 
                             
                           
                         
                       
                       
                         
                           
                             tan 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             δ 
                           
                         
                       
                     
                     ] 
                   
                 
               
             
           
         
       
       
         
           
             
               K 
               a 
               ″ 
             
             = 
             
               
                 
                   
                     
                       
                         0.5 
                         ⁢ 
                         
                           f 
                           ⁡ 
                           
                             [ 
                             
                               
                                 
                                   
                                     
                                       cos 
                                       ⁢ 
                                       
                                         ( 
                                         
                                           β 
                                           - 
                                           90 
                                         
                                         ) 
                                       
                                     
                                     - 
                                   
                                 
                               
                               
                                 
                                   
                                     sin 
                                     ⁢ 
                                     
                                       ( 
                                       
                                         β 
                                         - 
                                         90 
                                       
                                       ) 
                                     
                                     ⁢ 
                                     tan 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     δ 
                                   
                                 
                               
                             
                             ] 
                           
                         
                         ⁢ 
                         
                           
                             ( 
                             
                               0.75 
                               ⁢ 
                               
                                 A 
                                 base 
                               
                             
                             ) 
                           
                           2 
                         
                       
                       + 
                     
                   
                 
                 
                   
                     
                       
                         g 
                         ⁡ 
                         
                           [ 
                           
                             
                               
                                 
                                   
                                     cos 
                                     ⁢ 
                                     
                                       ( 
                                       
                                         β 
                                         - 
                                         90 
                                       
                                       ) 
                                     
                                   
                                   - 
                                 
                               
                             
                             
                               
                                 
                                   
                                     A 
                                     r 
                                   
                                   ⁢ 
                                   
                                     sin 
                                     ⁡ 
                                     
                                       ( 
                                       
                                         β 
                                         - 
                                         90 
                                       
                                       ) 
                                     
                                   
                                   ⁢ 
                                   tan 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   δ 
                                 
                               
                             
                           
                           ] 
                         
                       
                       ⁢ 
                       
                         ( 
                         
                           
                             h 
                             1 
                           
                           - 
                           
                             .75 
                             ⁢ 
                             
                               A 
                               
                                 base 
                                 ⁢ 
                                 
                                     
                                 
                               
                             
                           
                         
                         ) 
                       
                       ⁢ 
                       
                         ( 
                         
                           0.75 
                           ⁢ 
                           
                             A 
                             base 
                           
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   0.5 
                   ⁢ 
                   
                     
                       ( 
                       
                         0.75 
                         ⁢ 
                         
                           A 
                           base 
                         
                       
                       ) 
                     
                     2 
                   
                 
                 + 
                 
                   0.75 
                   ⁢ 
                   
                     
                       A 
                       base 
                     
                     ⁡ 
                     
                       ( 
                       
                         
                           h 
                           1 
                         
                         - 
                         
                           .75 
                           ⁢ 
                           
                             A 
                             base 
                           
                         
                       
                       ) 
                     
                   
                 
               
             
           
         
       
       
         
           
             
               N 
               ⁡ 
               
                 ( 
                 
                   for 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   cohesiveless 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   soil 
                 
                 ) 
               
             
             = 
             
               1 
               + 
               
                 
                   ( 
                   
                     
                       4 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         K 
                         0 
                       
                       ⁢ 
                       tan 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       ϕ 
                     
                     3 
                   
                   ) 
                 
                 ⁢ 
                 
                   ( 
                   
                     D 
                     
                       d 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       1 
                     
                   
                   ) 
                 
               
             
           
         
       
       
         
           
             
               N 
               ⁡ 
               
                 ( 
                 
                   for 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   cohesiveless 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   soil 
                 
                 ) 
               
             
             = 
             
               1 
               + 
               
                 
                   [ 
                   
                     
                       2 
                       ⁢ 
                       
                         c 
                         
                           γ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           D 
                         
                       
                     
                     
                       1 
                       + 
                       
                         4 
                         ⁢ 
                         
                           ( 
                           
                             c 
                             
                               γ 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               D 
                             
                           
                           ) 
                         
                       
                     
                   
                   ] 
                 
                 ⁢ 
                 
                   ( 
                   
                     D 
                     
                       d 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       1 
                     
                   
                   ) 
                 
               
             
           
         
       
     
     Where,
         h 1 =h w +C tan ε+0.375 A base (1+tan ε/tan β′)   h″=h w +C tan ε+A base +0.75 A base  tan ε/tan β′   h 0 =h w +C tan ε+0.75 A base  tan ε/tan β′   h a =E+A base +0.5(h 0 −E)   γ c =unit weight of concrete   N=Number of diameters the pile passive can increase   μ=Coefficient of friction between soil and concrete at the bottom of the base   K a =Active pressure coefficient using standard soil mechanics pressure   K″ a =Active pressure coefficient at bevel at heel   K p =tan 2  (45+φ/2)=Passive pressure coefficient assumed cohesion-less   K 0 =1.06 (1−sin φ)=At rest pressure coefficient at level   p=Surcharge pressure   w0=Sloped Backfill=0.5γC 2  tan ε   w1=Wall=A stem h w γ c      w2=Backfill=Ch w γ   w3=Base=A base Cγ c      w4=(15/32)(A base ) 2 γ c  cot β′   w5=Base=A base (B+A stem )γ c      w6=Pile=(π/4)(d 2 )D(γ c )(d1/s)   w6=Key=K(t+K/10)γ c      wb=0.75 K ba γA base (h 1 )   Vp=ph″ tan ε   Va=0.5K a γ(h″) 2  tan ε   P1=K a γh″D 2 N(d1/s)(π/4)   P2=0.5 K a D 2 N(d1/s)(π/4)   P1′=[2K 0 γh a D(d1/s)(π/4)+P1/N]μ   P2′=[K 0 γD 2 (d1/s)(π/4)+P2/N)]μ   W T =Pile=w0+w1+w2+w3+w4+w5+w6+wb+Vp+Va+P1+P2+P1′+P2′   W T =Key=w0+w1+w2+w3+w4+w5+w6+wb+Vp+Va       

     Taking Moments at point A of w0, w1, w2, w3, w4, w5, w6, wb, Vp, Va, P1, P2, P1′, P2′ with respective moment arm gives M T . 
     Acting Force and Moment:
         H=0.5K a γ(h 0 ) 2 +0.75K″ 1 γh 1 A base +p(h 0 +A base )+0.25 K a γ(h 0 +0.875 A base )A base      M 0 =K a γ(A base +h 0 /3)(h 0 ) 2 /2+0.75K″         γh         A base (0.875 A base )+0.5p(h 0 +A base )+0.25 K a γ(h 0 +0.875 A base )A base (0.125 A base )       

     Passive Pressure:
         F1=K p γEA base      F2=0.5 K p γ(A base ) 2      F3=Pile=K p γ[(E+A base )DN(d1/s)   F3=Key=K p γ[(E+A base )K   F4=Pile=0.5 K p γK 2 N(d1/s)   F4=Key=0.5 K p γK 2      F=F1+F2+F3+F4   Sliding Safety Factor=SSF=[μW T +F]/H   Over turning safety Factor=OTSF=M T /M 0          

     For the case of at rest pressure at the bevel when assuming zero deflection at the base we have: 
     
       
         
           
             
               K 
               0 
               ″ 
             
             - 
             
               
                 ( 
                 
                   
                     
                       cos 
                       2 
                     
                     ⁢ 
                     β 
                   
                   + 
                   
                     
                       K 
                       
                         0 
                         ⁢ 
                         r 
                       
                     
                     ⁢ 
                     
                       sin 
                       2 
                     
                     ⁢ 
                     β 
                   
                 
                 ) 
               
               ⁡ 
               
                 [ 
                 
                   1 
                   + 
                   
                     cot 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       β 
                       ⁡ 
                       
                         ( 
                         
                           
                             tan 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             δ 
                           
                           - 
                           
                             
                               tan 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 β 
                                 ⁡ 
                                 
                                   ( 
                                   
                                     1 
                                     - 
                                     
                                       K 
                                       
                                         0 
                                         ⁢ 
                                         r 
                                       
                                     
                                   
                                   ) 
                                 
                               
                             
                             
                               1 
                               + 
                               
                                 
                                   K 
                                   
                                     0 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     r 
                                   
                                 
                                 ⁢ 
                                 
                                   tan 
                                   2 
                                 
                                 ⁢ 
                                 β 
                               
                             
                           
                         
                         ) 
                       
                     
                   
                 
                 ] 
               
             
           
         
       
       
         
           
             
               K 
               
                 b 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 0 
               
             
             = 
             
               
                 ( 
                 
                   
                     
                       cos 
                       2 
                     
                     ⁢ 
                     β 
                   
                   + 
                   
                     
                       K 
                       
                         0 
                         ⁢ 
                         r 
                       
                     
                     ⁢ 
                     
                       sin 
                       2 
                     
                     ⁢ 
                     β 
                   
                 
                 ) 
               
               ⁡ 
               
                 [ 
                 
                   
                     
                       - 
                       cot 
                     
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     β 
                   
                   + 
                   
                     tan 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     δ 
                   
                   - 
                   
                     
                       tan 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         β 
                         ⁡ 
                         
                           ( 
                           
                             1 
                             - 
                             
                               K 
                               
                                 0 
                                 ⁢ 
                                 r 
                               
                             
                           
                           ) 
                         
                       
                     
                     
                       1 
                       + 
                       
                         
                           K 
                           
                             0 
                             ⁢ 
                             r 
                           
                         
                         ⁢ 
                         
                           tan 
                           2 
                         
                         ⁢ 
                         β 
                       
                     
                   
                 
                 ] 
               
             
           
         
       
       
         
           
             
               K 
               
                 0 
                 ⁢ 
                 r 
               
             
             = 
             
               
                 
                   
                     cos 
                     2 
                   
                   ⁢ 
                   ϕ 
                 
                 
                   
                     z 
                     0 
                     2 
                   
                   - 
                   
                     z 
                     m 
                     2 
                   
                 
               
               ⁡ 
               
                 [ 
                 
                   
                     
                       A 
                       1 
                     
                     ⁡ 
                     
                       ( 
                       
                         1 
                         - 
                         
                           z 
                           m 
                           2 
                         
                       
                       ) 
                     
                   
                   - 
                   
                     4 
                     ⁢ 
                     
                       
                         A 
                         2 
                       
                       ⁡ 
                       
                         ( 
                         
                           1 
                           - 
                           
                             z 
                             m 
                           
                         
                         ) 
                       
                     
                   
                   + 
                   
                     2 
                     ⁢ 
                     ln 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       z 
                       0 
                     
                   
                   - 
                   
                     2 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       A 
                       3 
                     
                     ⁢ 
                     ln 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       z 
                       m 
                     
                   
                 
                 ] 
               
             
           
         
       
         
         
           
             A 1 [1+tan(φ−ε)tan φ] 2    
             A 2 =tan φ tan(φ−ε)[1+tan(φ−ε)tan φ] 
             A 3 =tan 2  φ tan 2 (φ−ε) 
             z 0 =1+sin φ√{square root over (1−cot φ tan ε)} 
             z m =sin φ√{square root over (1−cot φ tan ε)} 
           
         
       
    
     Where,
         K″ 0 =At rest pressure coefficient at bevel at heel   K b0 =Vertical at rest pressure coefficient at bevel at heel   K 0r =At rest coefficient for a sloped backfill       

     Thus replace K ba  by K b0  in wb above and H and M 0  becomes:
         H=0.5K a γ(h 0 ) 2 +0.75K″ 0 γA base h 1 +p(h 0 +A base )+0.25 K 0 γ(h 0 +0.875 A base )A base      M 0 =K a γ(A base +h 0 /3)(h 0 ) 2 /2+0.75K″ 0 γ(A base )(h 1 )(0.875 A base )+0.5p(h 0 +A base ) 2 +0.25 K 0 γ(h 0 +0.875 A base )A base (0.125 A base )       

     For the case of uplift analysis in the case of rotation: 
     
       
         
           
             
                 
             
             ⁢ 
             
               
                 Δ 
                 
                   Δ 
                   p 
                 
               
               = 
               
                 
                   
                     1 
                     - 
                     
                       K 
                       0 
                       2 
                     
                   
                   
                     ( 
                     
                       
                         K 
                         p 
                       
                       - 
                       
                         K 
                         0 
                       
                     
                     ) 
                   
                 
                 ⁡ 
                 
                   [ 
                   
                     
                       6 
                       ⁢ 
                       
                         ( 
                         
                           
                             K 
                             u 
                           
                           - 
                           
                             K 
                             0 
                           
                         
                         ) 
                       
                       ⁢ 
                       
                         ( 
                         
                           
                             h 
                             w 
                           
                           + 
                           
                             A 
                             base 
                           
                         
                         ) 
                       
                       ⁢ 
                       e 
                     
                     
                       
                         
                           ( 
                           
                             
                               K 
                               u 
                             
                             - 
                             
                               K 
                               0 
                             
                           
                           ) 
                         
                         ⁢ 
                         
                           X 
                           2 
                         
                       
                       - 
                       
                         
                           ( 
                           
                             1 
                             - 
                             
                               
                                 K 
                                 0 
                               
                               ⁢ 
                               
                                 K 
                                 u 
                               
                             
                           
                           ) 
                         
                         ⁢ 
                         
                           
                             ( 
                             
                               
                                 h 
                                 
                                   w 
                                   ⁢ 
                                   
                                       
                                   
                                 
                               
                               + 
                               
                                 A 
                                 base 
                               
                             
                             ) 
                           
                           2 
                         
                       
                     
                   
                   ] 
                 
               
             
           
         
       
         
         
           
             X=B+A stem +C+0.75 A base /tan β′ 
             x c =[M T −M0]/W T    
             e=X/2−x c   
           
         
       
    
     
       
         
           
             
                 
             
             ⁢ 
             
               
                 K 
                 u 
               
               = 
               
                 
                   
                     - 
                     
                       ( 
                       
                         
                           K 
                           p 
                         
                         - 
                         
                           K 
                           0 
                         
                       
                       ) 
                     
                   
                   ⁢ 
                   
                     
                       ( 
                       
                         Δ 
                         
                           Δ 
                           p 
                         
                       
                       ) 
                     
                     2 
                   
                 
                 + 
                 
                   2 
                   ⁢ 
                   
                     ( 
                     
                       
                         K 
                         p 
                       
                       - 
                       
                         K 
                         0 
                       
                     
                     ) 
                   
                   ⁢ 
                   
                     ( 
                     
                         
                     
                     ⁢ 
                     
                       Δ 
                       
                         Δ 
                         p 
                       
                     
                     ) 
                   
                 
                 + 
                 
                   K 
                   0 
                 
               
             
           
         
       
       
         
           
             
               N 
               q 
               l 
             
             = 
             
               
                 
                   [ 
                   
                     
                       
                         - 
                         cot 
                       
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       β 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         N 
                         q 
                       
                     
                     - 
                     
                       K 
                       
                         b 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         0 
                       
                     
                   
                   ] 
                 
                 ⁡ 
                 
                   [ 
                   
                     
                       
                         K 
                         u 
                       
                       - 
                       
                         K 
                         0 
                       
                     
                     
                       
                         K 
                         p 
                       
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                   ] 
                 
               
               + 
               
                 K 
                 
                   b 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   0 
                 
               
             
           
         
       
     
     And replace wb by
         wb=0.75N l   q γA base (h 1 )       

     If, for example, we take a 12-foot retaining wall with level backfill and use a key and design per standard practice we find the required sliding safety factor is SSF=1.5 and the required overturning safety factor is OTSF 2.0 where β′=90 degrees and β=90 degrees. The parameters used are:
         φ=30 degrees   ε=0.0   δ=24.23 degrees   γ=130   γ c =150 pcf   μ=0.45   Soil reaction=0.75   K a =0.333   K″ a =varies depending on β   K p =3.0   c=0.0   p=0.0   h w =12 feet   A stem =9 inch   E=14.4 inch   t=18 inch   K=12 inch   A=12 inch   B=24 inch   C=45 inch   X=78 inch       

     We find when varying the bevel angle β′ for the same retaining wall the following Table 1 for SSF and OTSF is obtained. 
     
       
         
               
               
             
               
               
               
               
             
               
               
               
               
               
               
               
               
               
               
               
               
               
             
               
               
               
               
               
               
               
               
               
               
               
               
               
             
           
               
                   
                   
               
               
                   
                 B′ 
               
             
          
           
               
                   
                 90 degrees 
                 60 degrees 
                 45 degrees 
               
             
          
           
               
                   
                 K α   
                 Uplift 
                 K 0   
                 Uplift 
                 K α   
                 Uplift 
                 K 0   
                 Uplift 
                 K α   
                 Uplift 
                 K 0   
                 Uplift 
               
               
                   
                   
               
             
          
           
               
                 SSF 
                 1.5 
                 1.5 
                 1.38 
                 1.38 
                 1.57 
                 2.47 
                 1.59 
                 2.37 
                 1.64 
                 3.08 
                 1.76 
                 3.08 
               
               
                 OTSF 
                 2.18 
                 2.18 
                 2.15 
                 2.15 
                 2.46 
                 5.59 
                 2.63 
                 5.37 
                 2.98 
                 7.42 
                 2.93 
                 7.48 
               
               
                   
               
             
          
         
       
     
     Thus, the stability increases with reducing β′. Where the base is extended 5.2 inches for β′=60 degrees making a concrete volume increase by 1.6% and extended 9 inches for β′=45 degrees making a concrete volume increase 2.8%. So for practically the same amount of concrete we have increased the stability. 
     From this conclusion, it would be wise to reduce the distance C for a given β′ and obtain the required SSF of 1.5 and OTSF of 2.0. This may require changing the key. Additionally, if it is required to reduce the concrete, then using piles is more suitable since the passive pressure increase by N diameters due to the side friction on the passive wedge. Since we are able to reduce C and use piles, we have minimized the volume of concrete and backfill. Furthermore, if the distance C is encroaching to a property line, then we are able to move the wall toward the property line and gain more real estate in front of the wall. In many design practices the property line is a crucial element to avoid in design. To show this finding, we redesign the wall two ways: with a key and with piles. 
     Using:
         f y =60,000 psi for steel reinforcement   f′ c =3000 psi concrete strength       

     The original wall has 0.655 cubic yard of concrete per foot and 62.53 lbs. of reinforcing steel per foot. For a redesign with key and K 0  condition at the bevel:
         β′=36 degrees   A stem =9 inch   A base =12 inch   K=12 inch   E=14.4 inch   B=24 inch   C=27 inch   X=72.39 inch   t=18 inch   SSF=1.59   OTSF=2.04   SSF Uplift=4.17   OTSF Uplift=8.80   0.602 cubic yard of concrete per foot   59.21 lbs. of reinforcing steel per foot   Savings in concrete=8.14%   Savings in steel=5.31%   Savings in backfill=4.52%       

     For a redesign with piles at 5 ft. on center and K 0 , the condition at the bevel for minimum C:
         β′=36 degrees   A stem =9 inch   A base =12 inch   d1=12 inch   D=47 inch   s=5 ft   N=2.6   E=14.4 inch   B=61 inch   C=0 inch   X=82.39 inch   Distance to property line at heel edge=21.39 inch   SSF=1.79   OTSF=2.00   SSF Uplift=3.16   OTSF Uplift=6.15   0.596 cubic yard of concrete per foot   68.48 lbs. of reinforcing steel per foot   Savings in concrete=9%   Additional steel=9.51%   Savings in backfill=21.32%       

     For a redesign with piles at 5 ft. on center and K 0 , the condition at the bevel for minimum concrete:
         β′=36 degrees   A stem =9 inch   A base =12 inch   d=12 inch   D=32 inch   s=5 ft   N=2.08   E=14.4 inch   B=14 inch   C=32 inch   X=67.39   SSF=1.79   OTSF=2.02   SSF Uplift=4.38   OTSF Uplift=8.31   0.542 cubic yard of concrete per foot   63.65 lbs of reinforcing steel per foot   Savings in concrete=17.18%   Additional steel=1.8%   Savings in backfill=0.4%       

     Finally we can observe the results. Note the piles does not need to be concrete piles they can be steel tubing or square tubing of I beams. They can be driven or installed in a hole. 
     While illustrative embodiments have been illustrated and described, it will be appreciated that various changes can be made therein without departing from the scope of the invention.