Deck Elevation 



To ensure that the deck is above the design crest elevation; thereby avoiding unnecessarily 

 large horizontal and vertical forces and damage to the platform base, the height of the lower 

 elevation of the deck will be: 



h" = h + n^,, + h' 



where h is the design water depth, Vmax ^^ ^^^ maximum displacement of the wave above 

 design stillwater level, and h' is the deck freeboard (say 10 feet for this problem). V^ax 

 will occur at zero phase angle (6 = 0°) and from any of the first nine tables, eta/height = 

 0.89 for e = 0°. Therefore, r?^„^ = 0.89 (H) =28.3 feet and h" = h + 17 + h' = 41 + 28.3 + 

 10 = 79.3 feet. The platform wiU be constructed so the lower deck elevation will be 79.3 

 feet above the bottom. 



In determining the forces and moments, it is assumed that the pihng are sufficiently far 

 apart to be considered isolated. First, the forces acting upon several structural members will 

 be determined. The total force, ¥^{6, S),wiU be a summation of the drag force, Fj){d,S), 

 and inertia force, Fj{d, S), components at any particular phase angle. Each component 

 wiU be presented graphically; the components will then be added to estabhsh the total force, 

 and the maximum force acting upon each member wiU be obtained from the graph. 



Forces on Member "a" 



In the case of the outrigger, Member a, the drag force is given by: 



C PD 



S 

 a 



ululdS' 



where D is the pihng diameter, S^(= 20.5') is the height of the outrigger above bottom, 

 and p = mass density of sea water, 1.99 sluggs/f oot^ . To determine F^(0, S^), select the 

 tabulated dimensionless drag value for the force, Fp(0,S^), at depth Sjh - 0.5 from 

 Table V and multiply the dimensionless force by : 



Cj^pD(H/T)^h 



Cj^pD(H/T)2h ^ 1.05(1.99 ) (6) (31.78/20)^41 ^ r648.9 lbs 



1.05(1.99) (6 



2 [0-6489 kips. 



60 



