From the previous example problem, L"L^=171m for d = 26 m and T = 



6 



X Tl 



12 s . Then, from the expression — = -=— , 



L 2ir 



e = ^ = M^sOi ^ 



n L 171 



or 



. _ 360° (26.0) _ ^, ,0 



Values in Table 7-6 can be shifted by 55 degrees and represent the variation 

 of force on the second pile with the phase angle. The total horizontal 

 force is the sum of the two individual pile forces. The same procedure can 

 be applied for any number of piles. Table 7-6 can be used by offsetting the 

 force values by an amount equal to 55 degrees (preferably by a graphical 

 method). The procedure is also applicable to moment computations. 



The maximum force is about 183.0 kN when the wave crest is about 8 degrees 

 or [(8° /360°) 171] « 3.5 m (11.5 ft) in front of the reference 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 structures 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 example, the forces on a 

 single pile are (from preceding example problems). 



[F ) 



m -^finite-amplitude 



= 175.9 kN (39,600 lb) 



[f ),. = 102 kN (22,930 lb) 



Therefore, the total force on the two-pile group, corrected for the finite- 

 amplitude design wave, is given by, 



fF 1 



^ m^ finite-amplitude 



Total 2 piles ~ Wl 



'^^ Airy 



(corrected for 



finite-amplitude 



design wave) 



Total 



2 piles 



(computed from 

 Airy theory) 



r Total 2 piles 



175.9 

 102.0 



(183.0) = 315.6 kN (71,000 lb) 



*1s*********is***ic1tis**-k1c*is*************** 



7-154 



