This appvoaah. is cm approximation and should he limited to rough aalaulations 

 for cheeking purposes only. The use of tables of finite-amplitude wave 

 properties ( Skj'elbreia et al., 1960 and Dean, 1974) is reeommended for design 

 aalaulations . 



As the distance between piles becomes small relative to the wavelength, 

 maximum forces and moments on pile groups may be conservatively estimated by 

 adding the maximum forces or moments on each pile. 



The assumption that piles are unaffected by neighboring piles is not valid 

 when distance between piles is less than three times the pile diameter. A few 

 investigations evaluating the effects of nearby piles are summarized by Dean 

 and Harleman (1966). 



h. Calculation of Forces on a Nonvertical Cylindrical Pile . A single, 

 nonvertical pile subjected to the action of a two-dimensional design wave 

 traveling in the -he direction is shown in Figure 7-67 . Since forces are 

 perpendicular to the pile axis, it is reasonable to calculate forces by 

 equation (7-20) using components of velocity and acceleration perpendicular to 

 the pile. Experiments (Bursnall and Loftin, 1951) indicate this approach may 

 not be conservative, since the drag force component depends on resultant 

 velocity rather than on the velocity component perpendicular to the pile 

 axis. To consider these experimental observations, the following procedure is 

 recommended for calculating forces on nonvertical piles. 



For a given location on the pile (x^-, , y^ , z^^ in Figure 7-87), the force 

 per unit length of pile is taken as the horizontal force per unit length of a 

 fictitious vertical pile at the same location. 



*************** EXAMPLE PROBLEM 30*************** 



GIVEN : A pile with diameter D = 1.25 m (4.1 ft) at an angle of 45 degrees 

 with the horizontal in the x-z plane is acted upon by a design wave with 

 height H = 10.0 m (32.8 ft) and period T = 12 s in a depth d = 26 m (85 

 ft) . 



FIND ; The maximum force per unit length on the pile 9.0 m (29.5 ft) below the 

 SWL (z = -9.0 m). 



SOLUTION : For simplicity. Airy theory is used. From preceding examples, C^ 

 = 1.5 , C^ = 0.7 , and L = L^ = 171 m. 



From equation (7-25) with sin (-2ir/T) = 1.0 , 



_ „ ID^ hIjL cosh [2Tr(d + z)/L] 1 

 ^im - ^M^^ 4 "JL cosh [2Tid/L] I 



2 

 f^^ = 1.5 (1025.2) (9.8) ^^^^^^^ (10.0) ^ (0.8) = 2,718 N/m (186 lb/ft) 



7-155 



