» 14.26 pounds per square foot 



= pressure due to surcharge Wo exerted by submerged fill 

 and p^ = W3 ^^3 . ^ ) tan^ {^2^3 ) (9!,) 



= (75 X 3 + 1050) tan2 (32,50) 



a 518 pounds per square foot 



= pressure due to surcharge Wo and submerged fill 



Using W|^ = W3 (h^ + _3_) ^ 3^275 pounds per square foot 



h ^ 



= 1275 tan2 (30°) 

 s Ii.20 pounds per square foot 



= pressure due to surcharge W^ exerted by original ground, 



W| o 90 - 01 



and P8 =^k i\* Ji) tan^ / ^) (96) 



W|^ 2 



= (80 X 3 + 1275) tan^ (30°) 



- 505 pounds per square foot 



- pressure due to surcharge W|, and original ground to the 

 dredged bottom at 6 feet below the low water line. 



329. At the point of application of pg, both active pressure on the 



pile back and passive pressirre on the pile face woiald be applied. The 



active pressure on the pile back would increase in the same manner as 



it increased from py to pg, because there would be no change of material 



at the level of pg. ' From the expressions for p^ and pg, this rate of 



increase of active pressure with increase in depth would be Rg = wk tan 



90 - 01, 



(- = — -) per unit increase in hi . Similarly, the maximum passive pressure 



on the pile face would increase with depth at the rate Rp s w]_^ tan*^ (___^_Lii) 



per unit increase in hL, since at any point h below the level of the 

 dredged bottoni, this passive pressure pg = wl h tan2 90 + 0[j^ ^ Yhe rate 

 of increase 2 



161 . . 



