Therefore from equation (7-39) 



M. = F. d S. = 1619 (4.5) (0.82) = 5,975 N-m (4,407 ft-lb) 

 ^m tm ^m 



and from equation (7-40) 



Mn- = Fn d Sn_ = 6741 (4.5) (1.01) = 30.6 kN-m (22,600 ft-lb) 



Lm Dm urn 



The value of a is found by interpolation between Figures 7-81 and 7-82 

 using W = 0.29 '", H/gT^ = 0.0031 , and d/gT^ = 0.0046 . 



Figure 7-81: W = 0.1 ; a = 0.33 



m 



Interpolated Value W = 0.29 ; a « 0.34 



m 



Figure 7-82: W = 0.5 ; a = 0.35 



m 



The maximum total moment about the mud line is found from equation (7-43). 



2 

 M = a wC H Dd 



m m D 



M = 0.34 (10,047) (0.7) (3)^ (0.3) (4.5) = 29.1 kN-m (21,500 ft-lb) 



m 



The moment arm, measured from the bottom, is the maximum total moment M 

 divided by the maximum total force F ; therefore, 



/ = ^|4§T= ^-2 n. (13.8 ft) 

 m ' 



If it is assumed that the upper 0.6 m (2 ft) of the bottom material lacks 

 significant strength, or if it is assumed that scour of 0.6 m occurs, the 

 maximum total horizontal force is unchanged, but the lever arm is increased 

 by about 0.6 m . The increased moment can be calculated by increasing the 

 moment arm by 0.6 m and multiplying by the maximum total force. Thus the 

 maximum moment is estimated to be 



[m 1 0.6 m below mud line = (4.2 + 0.6) F = 4.8 (6,931) = 



33.3 kN-m (24,500 ft-lb) 

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



*************** EXAMPLE PROBLEM 21************** 



GIVEN : A design wave with height H = 3.0 m (9.8 ft) and period T = 10 s 

 acts on a vertical circular pile with a diameter D = 0.3 m (1.0 ft) in a 

 depth d = 30.0 m (98.4 ft) . Assume C = 2.0 and C = 1.2. 



FIND : The maximum total horizontal force and the moment around the mud line 

 of the pile. 



7-129 



