Structural Analysis Methods I: Beam Elements - Mechanics of Structures: Variational and Computational Methods - page 219

Solve these four equations for c 0 ,c 1 ,c 2 , and c 3 .
c 0 = n 2 cosh n 1 +

n 1 cos n 2 / n 1 + n 2

c 1 = n 2 / n 1 sinh n 1 + n 1 / n 2 sin n 2 / n 1 + n 2

c 2 = cosh n 1 − cos n 2 /

n 1 +

n 2

(4.104)

2

/

n 1 +

n 2

3

c 3 = [ ( 1 / n 1 ) sinh n 1 − ( 1 / n 2 ) sin n 2 ]

Finally, the transfer matrix is given by Eq. (4.98) as

U i

2 3

= c 0 I + c 1 ( A ) + c 2 ( A ) + c 3 ( A ) (4.105)

or, after some algebraic manipulations

3 3

EI

c 1 + c 3 η c 3

c 0 +

2 c 2 η

− c 1 −

3 c 3 (η − ζ)

− −

2 c 2 /

EI

k s GA

3

2 c 2
EI

c 1 − c 3 ζ

3 2 c 2 ζ

λ c 3 c 0 −

EI

U i

=

3 c 3 ) 2 2 3 c 3 λ

λ EI ( c 1 + η − λ EIc 2 c 0 + c 2 η −

2 3 2 3 2

λ EIc 2 EI [ − c 1 ζ + c 3 (ζ − λ) ] c 1 + c 3 (η − ζ) c 0 − c 2 ζ

(4.106)

It is convenient, from the standpoint of installing this transfer matrix as a subroutine in
a computer program, to redefine c 0 ,c 1 ,c 2 , and c 3 in terms of new funtions e 0 ,e 1 ,e 2 ,e 3 ,
and e 4 , where

3 [ 2

e 0 = (η − ζ) c 1 + (ζ − λ) + η(η − ζ) ] c 3

2

e 1 = c 0 + (η − ζ) c 2

3

e 2 = c 1 + (η − ζ) c 3

2 c 2

e 3 =

3 c 3

e 4 =

or

e 0 = n 1 sinh n 1 − n 2 sin n 2 / n 1 + n 2

e 1 = n 1 cosh n 1 + n 2 cos n 2 / n 1 + n 2

n 2 sin n 2 )/ n 1 + n 2

(4.107)

e 2 = ( n 1 sinh n 1 +

2

e 3 = c 2

3

e 4 = c 3

The above transfer matrix then becomes

e 1 + ζ e 3 − e 2 − e 4 / EI + ( e 2 + ζ e 4 )/ k s GA − e 3 / EI

λ e 4 e 1 − η e 3 e 3 / EI ( e 2 − η e 4 )/ EI

U i =

λ EI ( e 2 + ζ e 4 ) − λ EIe 3 e 1 + ζ e 3 − λ e 4

λ EIe 3 EI ( e 0 − η e 2 ) e 2 e 1 − η e 3

(4.108)

This transfer matrix is presented in Table 4.3, where the functions e 1 , e 2 , e 3 , and e 4 are
given in a form suitable for use in a computer program. This table yields the same transfer
matrix as in Eq. (4.86) if k, k ∗ , 1 / k s GA, and N are set equal to zero.