Trans RINA, Vol 161, Part A4, Intl J Maritime Eng, Oct-Dec 2019


The normalization integrals are: 1


N Y x x


N X x x 2


kk ( )d


ii ( )d 1


=  = 


2 0 0


(13a) (13b)


The eigenvalue problems (10a) and (10b) with the boundary conditions (11a) and (11b) are now analytically solved to yield:


X x = x Y x = x


ii kk


( ) sin( ) ( ) sin( )


where the eigenvalue is obtained: ,


i i =


(14a) (14b)


Xx Xx = N


i() k() k i


Yx Yx = N


i


The next step is to define the integral transform pair – the integral transform itself and the inversion formula. For the transverse displacement of the free span:


z t X x z x t x i ( ) =  i =1, 2, 3,   , 1, 2, 3, k =kk= 


(15a) (15b)


And by introducing Equations (14) and (15) to Equation (13), the normalization integrals are evaluated as


Ni = N = k


1 , 2 1 , 2


i =1, 2, 3,  k =1, 2, 3,  (16a) (16b)


Therefore, in this case, the normalized eigenfunction correlates with the original eigenfunction through the following function:


1 0


z x t X x z t 


i i i=1


For the wake variable: 1


q t Y x q x t x k ( ) =  0


q x t Y x q t 


k k k=1 i ( ) ( , )d , transform ( , ) = ( ) ( ) , inversion (18a) (18b)


() 1/2


(17b)


i() 1/2


(17a)


k ( ) ( , )d , transform ( , ) = ( ) ( ) , inversion


(19a) (19b)


The third step is the transformation of the governing partial differential equations into a system of ordinary differential equations with respect to the time t, by employing the definition of ()i


Equations (18a) and (19a). By multiplying both sides of equation system (8a) by


k ()


respectively, integrating on x from 0 to 1, and then using Equations (18b) and (19b), the following ordinary differential equation system is yielded:


i()


    


 


    





 z t( ) ( + 2


4 i + i + s + f ) p


miU L T L −


(r r L d ( ) t


d q ( ) dt


2 k


22 1 1 1


tww q t


+ f E q t q t    klrs l r s = = = l ( ) ( ) r


d ( ) dt


s − f


d ( ) dt


q t k


+w q t( ) =Fki 2


 f k i=1


ppj= 4


mm Bij a


22 2 PAi


EI EI EIm dd EID


z tji pi) sin  D q t() 2


=1 +


(m mi me ) d ( ) 2


p + + mp


z t Cii k t


+ (m mL + g


=11 dt 4


ij ij j −C


egAe sin L EID


d ( )zti dt


2


where the coefficients are analytically determined by the following integrals:


A X x( ) ij i i


C X x x D X x Y x x E Y x Y x Y x Y x x F Y x X x x


== ==


 


11 2


d ( ) 2


00 11


i ( )d , l ik 00 klrs== ( ) ( )d 


11 00


k ( ) ( ) ( ) ( )d , r s ki A-326 k i k


dd d , ij


X x


xx x d ,x B X x ( )


jj i


i ( ) ( )d , d ( ) X x (20b) k


The initial conditions are also transformed with spatial coordinate being eliminated, yielding


z (0) 0 ,==t i


d (0) z


q (0)==t k ( ) ( ,0)d


 Y x q x x , 1 0 d (0) q


d k


d i


0 0 For computation efficiency, the expansions for (21a) (21b) z( , )x t ©2019: The Royal Institution of Naval Architects = k 1


 =


ik )A z t( ) +


jj pi


 j +


(m m gL) cos  EI


3  B z t ( ) + 2miU  d ( )ztj (20a)


zt and qt given by Xx and Yxk()


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