For clay, then 



F^Q^ = (1.09) [(4 lb/in. ^) (8,640 in.^) + (4 lb/in. ^) (10) (144 in.^)] 



+ y (3 lb-sec^/ft^)(1 ft^) (1.5) (1.69 ft/sec)^ 



= (1.09) (34, 560 + 5,760) + 6 



^TOT " ^^'°°° ^^ 



Achieving a 95% reduction in force due to upward soil cutting vibration 

 gives 



'tot , , = 2.200 1b 

 reduced 



In cohesionless soils, (sand) the situation is much the same, but F is 

 given by 



Y, N A^ D Yi, A u D 

 „ b q f . 'b s ^ 



where y = Soil buoyant unit weight {'2d 40 lb/ft ) 

 b 



N = Dimensionless parameter {y 90 for friction angle (j) = 40 ) 



D = Depth of embedment (ft) 



p = Friction coefficient (Q; 0.5) 



Therefore, we have 



^TOT = ^v ^s + ^I 



= (1.09) 



(40 lb/ft^)(90)(1 ft2)(3 ft) ^ (40 lb/ft^)(60 ft2)(0.5)(3 ft) 



+ 6 



= 1.09(5,400 + 1,800) + 6 



A 95% force reduction due to vibration gives 



"TOT 



= 400 lb 



reduced 



57 



