The maximum force can be assumed to be given by 



f = f — ^ 



where F^ and Fj^ are given by Equations 7-35 and 7-31. Substituting 

 these equations into the above gives 



From Equation 7-34, 



^M^ 1.5 (4) 



W = = —^ = 0.245. 



C^H 0.7 (35) 



Interpolating between Figures 7-49 and 7-50 with H/gT^ = 0.00755 and 

 d/gT^ = 0.0183, it is found that 4)„ = 0.20 



From a preceding problem, 



H 



= 0.54 . 



Enter Figure 7-44 with d/gT^ = 0.0183 and using the curve labeled 

 1/2 % read 



Therefore, 



C^„ = 0.35 . 



2^ 

 Dm K 



say 



Dm 

 '" 0.35 



f = 300lb./ft. 



The maximum horizontal force per unit length at z = -30 ft. on the 



fictitious vertical pile is f^ = 300 lbs. /ft. This is also taken as 



the maximum force per unit length perpendicular to the actual inclined 

 pile. 



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7-123 



