Reduced forces and moments may be calculated from Equations 7-69 

 and 7-71 using the values of F and M found in the example problem 

 of the previous section; for T = 6 sec. 



F^ = 0.981 (8,000) = 7,850 Ibs./ft ; 



lb. -ft. 

 M' = 0.950 (48,300) = 45,900 — -- , 

 '^ ft. 



F; = 1.0 (1,860) = 1,860 Ibs./ft. ; 



Ib.-ft. 

 m; = 1.0 (5,120) - 5,120 — r- . 

 ' ft. 



Again assuming that the wave action on both sides of the structure is 

 identical, so that the maximiom net horizontal force and maximum over- 

 turning moment occurs when a clapotis crest is on one side of the 

 structure and a trough is on the other side 



(T = 6 sec.) 



say (T = 6 sec.) 



lb .-ft. 

 ^net = 41,000 -^ . 



A similar analysis for the 10-second wave gives, 



Ket = 6,065 Ibs./ft. , 



Ibs.-ft. 

 Ket - 44,700-^. 



(T = 10 sec.) 



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



7.325 Wall on Rubble Foundation . Forces acting on a vertical wall built 

 on a rubble foundation are shown in Figure 7-73, and may be computed in a 

 manner similar to computing the forces acting on a low wall if the comple- 

 ments of the force and moment reduction factors are used. As shown in 

 Figure 7-73, the value of b which is used for computing b/y is the 

 height of the rubble base and not the height of the wall above the 

 foundation. The equation relating the reduced force F" against the 

 wall on a nibble foundation with the force F which would act against 

 a wall extending the entire depth is. 



fl 



ry) F. (7-73) 



7-142 



