areas of the pressure diagrams below that point. Thus this shear 



^F 5 5~" (103) 



728 X 3.U 505 X 2.36 

 _ ^ _2 



= 1235 - 595 = 6U0 Tjounds per foot of wall, 



338. Above this point, the shear would decrease as oressure diagram area 

 is added. The quantits'- ^8 ^7 = 26.7 pounds per square- foot per foot of wall 



\ 



would be rate of change of pressure over hi , and calling this rate R, the equation 

 for finding the point of aero shear irhen this point is within h> is 



2 



R Z 



Sj, = (pg - R^ Z^) Z^ + -^ (lOli) 



or R^ Z^ - 2pg Z^ + 2Sp, = 



where Z, , is the distance of this ^ero shear point above the dredgsd bottom F. 

 Solving for Z, 



L 



2 



rU 



= 1.31 feet 

 The pressure at Z would be p„ = pg - R. Z = h70 pounds per square foot, 



339. The b ending moment M^ at the dredged bottom is the svm of the moments 

 of the individual areas pg HG and pg FG 



2 



M =1^ (h. + I h,) - % P8 = "^^6^ r2h. Oh.) - ^5 P8 (106) 



fNote: R is the rate of change of pressure between pg and p ) 



II, = 21U X (3.1i)^ (2.36 + 2.26) - (236)^(505) 

 T g g 



16U 



