F gt = A N c s u f(0.84 + 0.16 B/L) (5-1) 



where s = 20.7 kPa (3.0 psi or 432 psf) (Figure 2-7) 



f = 0.7 (paragraph 5.1.1) 



B = 0.9 m 



L = 0.9 m 



A = B x L = 0.81 m 2 



D/B = 11/0.9 = 12 



N = 15 (Figure 5-1) 



Therefore , 



F = 0.81(15)(20.7)(0.7)[0.84 + 0.16(0.9/0.9)] 



= 176 kPa-m 2 = 176 kN (39,600 lb) 



Because the anchor will be in service for several years under 

 sustained loading' from the subsurface buoy, the long-term static capa- 

 city must be checked. The long-term capacity is calculated from 

 Equation 5-3 in paragraph 5.2.1. Because the soil is loose (soft), the 

 drained strength parameters are used in Equation 5-3. 



F_ = A (c' N' + y, D N ) (0.84 + 0.16 B/L) (5-3) 



Jtt V c d qy 



where (2/3)c' = (2/3) (3.5 kPa) = 2.3 kPa (48 lb/ft 2 ) 



(Figure 2-7 and paragraph 5.2.1) 



N* = 9 (Figure 5-3) 



V = 380 kg/m 3 (24 lb/ft 3 ) (Figure 2-7) 



tan -1 (2/3 tan <J>) = tan _1 (2/3 tan 35°) = 25 degrees 

 (Figure 2-7 and paragraph 5.2.1) 



N = 6 (Figure 5-2) 



63 



