Depth of embedment of cantilever sheet pile walls in sandy soils
The active pressure acting on the back of the wall tries to move the wall away from the backfill.If the depth of embedment is adequate the wall rotates about a point O’ situated above the bottom of the wall sa shown in the following figure The types of pressure that act on the wall when rotation is likely to take place about O’ are:
-active earth pressure at the back of wall from the surface of the backfill down to the point of rotation O’,the pressure is designated as Pa1;
-passive earth pressure in front of the point of rotation O’ to the dredge line,this pressure is designated as Pp1;
-active earth pressure in front of the wall from the point of rotation to the bottom of the wall ;this pressure is designated as Pa2;
-passive earth pressure at the back of the wall from the point of rotation O’ to the bottom of the wall,this pressure is designated as Pp2 .
The active pressure acting on the wall are shown in the following figure.
If the passive and active pressures are algebrically combined, the resultant pressure distribution below the dredge line will be given as in the following figure.The various notations used are:
D-miminum depth of embedment with a factor of safety =1;
KA-Rankine active earth pressure coefficient;
Kp-Rankine passive earth pressure coefficient;
K=Kp-KA;
Pa-effective earth pressure acting against the sheet pile at the dredge line=yHKA;
Pp-effective passive earth pressure at the base of the pile wall and acting towards the backfill=yDKp
P’p-effective passive earth pressure at the base of the sheet pile wall acting against the backfill side of the wall =Pp”+yKD0 ;
P0p-effective passive pressure at level of 0=yyQK+yHKp;
y-effective unit weight of the soil assumed the same below and above the dredge line ;
y0-depth of point 0 below dredge line where the active and passive pressures are equal;
y—height of point of application of the total active pressures above Pa above point 0 ;
h-height of point G above the base of the wall ;
D0-height of point 0 above the base of the wall .
At the point 0, the passive pressure acting towards the right should equal the active pressure acting towards the left ,that is:
yy0Kp=y(H+y0)Ka ,therefore yy0(Kp-Ka)=yHKa, thus y0=(yHKa)/(y(Kp-Ka)=pˉa /yK.
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