From equation (7-41), 



u = ^ = 1«5(1.25) ^ ^ 97 

 " C^H 0.7(10.0) ^'^' 



Interpolating between Figures 7-77 and 7-78 with H/gT^ = 0.0075 and d/gT^ 



= 0.0184 , it is found that d) = 0.20 . 



m 



From a preceding problem. 



Enter Figure 7-72 with d/gX^ = 0.0183 and, using the curve labeled 1/2 

 H, , read 



Therefore, 



m Dm K 



Dm 



f = 3,394.1 ^^J^\V = 3,879 N/m (266 lb/ft) 

 m 0.35 



say 



f = 3,900 N/m (267 lb/ft) 



m 



The maximum horizontal force per unit length at z = -9.0 m (-29.5 ft) on 

 the fictitious vertical pile is f = 3,900 N/m . This is also taken as the 

 maximum force per unit length perpendicular to the actual inclined pile. 



*************************************** 



i. Calculation of Forces and Moments on Cylindrical Piles Due to Breaking 

 Waves . Forces and moments on vertical cylindrical piles due to breaking waves 

 can, in principle, be calculated by a procedure similar to that outlined in 

 Section III,l,b by using the generalized graphs with H = Hi . This approach 

 is recommended for waves breaking in deep water (see Ch. 2, Sec. VI, BREAKING 

 WAVES) . 



For waves in shallow water, the inertia force component is small compared 

 to the drag force component. The force on a pile is therefore approximately 



F "F^=C^|pgDH^K^ (7-62) 



m Dm D 2 ° Dm 



7-157 



