to compute the maximum horizontal force on the two-pile group. Compute 

 the phase difference between the two piles by Equation 7-49, 



^n " K '^°s"n " 100 (cos 30°), 



x^ = 86.6 ft. 



From the previous example problem, L =*= L^ = 554 ft. for d = 85 ft. 

 and T = 12 sees. Then from the expression. 



L 277 



27rx 2ir (86.6) 



d = = ^^ = 0.982 radians, 



" L 554 



360° (86.6) 



6„ = \ = 56.3° . 



" 554 



Figure 7-60 can be shifted by 56.3° to represent the variation of force 

 on the second pile with phase angle as shown in Figure 7-61. The total 

 horizontal force is the sum of the two curves (F^ = 42,000 lbs). The 

 same procedure can be used for any number of piles with one curve for 

 each pile. Table 7-4 can be used similarly simply by offsetting the 

 force values by an amount equal to 56.3°. The procedure is also appli- 

 cable to moment computations. 



Figure 7-61 shows the maximum force to be about 42,000 lbs. when the 

 wave crest is about 8° or [(8°/360°) 554] » 12 ft. in front of the refer- 

 ence pile. 



Because Airy theory does not accurately describe the flow field of 

 finite amplitude waves, a correction to the computed maximum force as 

 determined above could be applied. This correction factor for struc- 

 tures of minor importance might be taken as the ratio of maximum total 

 force on a single pile for an appropriate finite- amplitude theory to 

 maximum total force on the same pile as computed by Airy theory. For 

 the example, the forces on a single pile are (from preceding example 

 problems) , 



and 



P«,^ r. . 1- J = 43,900 lbs. , 



"• / jtntte ampbtude 



] = 25,000 lbs. 



'Airy 



7-119 



