From Figure 7-85, 



Cp = 0.7 

 and from equation (7-53), with Rg > 5 x 10 , 



%= 1.5 



Therefore, 



^.-r. = Cw pg ^ HK.„ 



2 

 F.= (1.5) (1025.2) (9.8) "^^^'^^^ 



^^ ,..., ^.^^^.^, w.-/ ^ (10.0) (0.40) = 74.0 kN (16,700 lb) 



F^ = (0.7)(0.5)(1025.2)(9.8)(1.25)(10.0)^ (0.35) = 153.8 kN (34,600 lb) 

 Mv^ = F. dS.„ = (74,000)(26)(0.59) = 1,135 kN-m (0.837 x 10^ ft-lb) 

 ^Dm " ^Dm^^Dm " ( 153,800)(26)(0.79) = 3,160 kN-m (2.33 x 10^ ft-lb) 



From equation (7-41), 



V ^ (1.5) (1.25) ^ 



CpH (0.7) (10.0) ^' ' 



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

 = 0.0183 , 



d)„ = 0.20 



Therefore, from equation (7-42), 



F„ = (J)„wCnH^D 



F^ = (0.20) (10,047) (0.7) (10.0)^ (1.25) = 175.8 kN (39,600 lb) 

 Interpolating between Figures 7-81 and 7-82 gives 



Therefore, from equation (7-43), 

 \ = cv^wC^H^Dd 



7-148 



