If waves act on both sides of the structure, the maximum net horizontal 

 force will occur when the clapotis crest acts against one side when the 

 trough acts against the other. Hence the maximum horizontal force will be 



F 



conditions . 



with F and F, determined for the appropriate wave 

 Assuming for the example problem that the wave action is 



identical on both sides of the wall, then 



F^g^ = 0.63 (10) (3)2 - (-0.31)(10)(3)2 



^net " ^°'^^ "'' ^'-^^^ ^^^^ ^-^^^ " ^^'^ ^^^^ (5,800 lb/ft) 

 say 



\et = 85 kN/m 



(T = 6 s) 



Some design problems require calculation of the total force including the 

 hydrostatic contribution; e.g. seawalls. In these cases the total force is 

 found by using equation (7-76). For this example, 



F ^ ^ , = 0.5 (10) (3)2 + 56.7 = 101.7 kN/m (7,000 lb/ft) 

 a total 



F = 0.5 (10) (3)^ + (-27.9) = 17.1 kN/m (1,200 lb/ft) 



The total force acts against the seaward side of the structure, and the 

 resulting net force will be determined by consideration of static loads 

 (e.g., weight of structure), earth loads (e.g., soil pressure behind a 

 seawall), and any other static or dynamic loading which may occur. 



The moment about point A at the bottom of the wall (Fig. 7-89) may be 

 determined from Figure 7-92. The procedures are identical to those given 

 for the dimensionless forces, and again the moment caused by the hydrostatic 

 pressure distribution is not included in the design curves. The upper 

 family of curves (above M/wd = 0) gives the dimensionless wave moment when 

 the crest is at the wall, while the lower family of curves corresponds to 

 the trough at the wall. Continuing the example problem, from Figure 7-92, 

 with 



M M 



a t 



= 0.44; = -0.123 



3 3 



wd wd 



(T = 6 s) 



Therefore, 



M = 0.44 (10) (3)^ = 118.8 ^^^-^ (26,700 ^^zll ) 

 G m r t 



(T = 6 s) 



M = -0.123 (10) (3)^ = -33.2 ^^^ (-7,500 ^^~l^ ) 

 t m r t 



7-172 



