f. The new velocity and embedment depth are calculated analogously 

 to step (c) (Equations 20 and 21) . 



g. This iterative procedure continues until the velocity either 

 equals or becomes negative. The embedment depth corresponding to 

 this situation is assumed to be the ultimate embedment depth. 



SAMPLE PROBLEM 



Consider a 10-foot diameter cylinder 30 feet in length with conical 

 ends striking the seafloor vertically with the axis of the cylinder 

 perpendicular to the seafloor. The altitudes of the conical ends are 

 10 feet. The entry velocity is 20 ft/sec, and it is assumed that this 

 is the object terminal velocity. The object weighs 1000 pounds in 

 water and has a mass of 5576 slugs. The remolded strength profile is 

 given in Figure 20. (This is suggested as a typical profile for 

 seafloor sites containing weak, cohesive soils.) 



It is desired to calculate the ultimate depth of penetration. 



1. From geometry, the distribution of horizontal and vertical 



area components are as given in the table below (taking A£ = 0.5 ft). 



The values of horizontal area. A, , A , and A„ as functions of I 



nv 6 An ASL 

 correspond to the section of object between Z + ^±2L and Si - . 



For example, the second increments of area ^ 2 



(corresponding to I = 0.5) are calculated as follows: 



The section is represented by a frustrum of a cone with the cross 

 section: 



,25 ft 



22 



