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Homework answers / question archive / EXERCISE The geotechnical engineer in charge at a specific site has decided to use an Anchored sheet pile wall system penetrating clayey soil

EXERCISE The geotechnical engineer in charge at a specific site has decided to use an Anchored sheet pile wall system penetrating clayey soil

Earth Science

EXERCISE

The geotechnical engineer in charge at a specific site has decided to use an Anchored sheet pile wall system penetrating clayey soil.  The anchor tie rod is located at point O’ and will be contributing with a Force “F” against the lateral pressure from the backfill.

The lateral pressure distribution diagram above the dredge line is identical to that of a cantilever sheet pile wall penetrating clay (as shown).  However, the net pressure distribution diagram below the dredge line in simpler, since the theoretical depth of penetration of the sheet pile will be considerably reduced due to the addition of an anchor tie rod

For a given set of soil properties and sheet pile dimensions, as shown below:

  1. Determine the actual depth of embedment “D” using a 40% increase of theoretical depth of embedment as Factor of Safety.  (Note:  The sheet pile wall must satisfy static equilibrium condition, in terms of both summation of all horizontal forces and summation of moments, about an given point along the wall  
  2. Determine the magnitude of the anchor force, “F” using the actual depth of embedment (D) obtained in part A

 

 

 

Figure 9.22  Anchored sheet pile wall penetrating clay

  

 

 

 

Part (A)

Determine the actual depth of embedment “D” using a 40% increase of theoretical depth of embedment as Factor of Safety.   

    

Part (B)

Determine the magnitude of the anchor force, “F” using the actual depth of embedment (D) obtained in part A

 

 

Ka = Rankine active pressure coefficient  

  ?′1 = γ ?1 ??         Equation 1  where ,  

?′2 = (γ ?1 + γ′ ?2) ??             γ= γ ??? − γ ?            Equation 2 

                       Now,  the Active pressure from left to right ;

                                             ?? = [(γ ?1 + γ′ ?2) + γ ???(z − ?1 − ?2)] − 2?         Equation 3

Passive pressure from left to right ;  

                                              ?′? = γ ??? (z − ?1 − ?2) + 2?                                     Equation 4

Net lateral pressure ,      ?6 = ?? − ??

?6 =  ??? (z − ?1 − ?2) + 2?] − [(γ ?1 + γ′ ?2) + γ ???(z − ?1 − ?2)] − 2?

                                                 ?6 =  4? − (γ ?1 + γ′ ?2)                                             Equation 5

 

                       Now, summation of all horizontal forces per unit length of wall = 0

                                    Area of pressure diagram – Area below the dredge line – F  = 0

? − ?6D − F = 0        

                                                                ? = ? − ?6D                                                    Eq. 6  

 

                       Now, taking the Moment about  O’  (located at the level of the tie rod, give us ;  

2

 

 

                          −?[(?1 + ?2) − (?1 + ?1? )] + ?6D (?2 + ?1 + ?2 − ?1) = 0             

?6D2

                                −?[(?1 + ?2) − (?1 + ?1? )] + 2 + ?6D (?1 + ?2 − ?1) = 0        

 

                  ?6?2 + 2?6?(?1 + ?2 − ?1) − 2?1(?1 + ?2 − ?1 − ?1? ) = 0             Eq. 7 

 

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