Page 133 - Mechatronic Systems Modelling and Simulation with HDLs

122 6 MECHANICS IN HARDWARE DESCRIPTION LANGUAGES

These four equations are differentiated once with respect to time and then rear-
ranged to give:

4 4
1
˙

C ij ¨u ij + u ij = i i i = 1 ... 4 (6.40)
L ij
j=1 j=1
The four degrees of freedom of the beam element u yk ,u yl ,r zk and r zl should now
be represented by the potentials ϕ 1 ,ϕ 2 ,ϕ 3 and ϕ 4 , which for this reason are used
here for the branch voltages u ij . The following is true:
(i = j)
u ij = ϕ i − ϕ j
(6.41)
u ii = ϕ i
Substituting into the above formula yields:
1 1
˙
C ii ¨ϕ i + C ij ( ¨ϕ i − ¨ϕ j ) + ϕ i + (ϕ i − ϕ j ) = i i i = 1 ... 4 (6.42)
L ii L ij
i =j i =j
After rearranging this yields in vector notation:

C ¨ϕ + Lϕ = ˙ ı (6.43)

where:
¨ ϕ = [¨ϕ 1 , ¨ϕ 2 , ¨ϕ 3 , ¨ϕ 4 ] T

ϕ = [ϕ 1 ,ϕ 2 ,ϕ 3 ,ϕ 4 ] T
˙ i = [˙ ı 1 , ˙ ı 2 , ˙ ı 3 , ˙ ı 4 ] T
 
C 11 + C 12 + C 13 + C 14 −C 12 −C 13 −C 14
 
−C 12 C 12 + C 22 + C 23 + C 24 −C 23 −C 24
C =  
−C 13 −C 23 C 13 + C 23 + C 33 + C 34 −C 34
 
−C 14 −C 24 −C 34 C 14 + C 24 + C 34 + C 44
 1 1 1 1 1 1 1 
+ + + − − −
L 11 L 12 L 13 L 14 L 12 L 13 L 14
1 1 1 1 1 1 1
 
− + + + − −
 
L =  L 12 L 12 L 22 1 L 23 L 24 1 1 L 23 1 1 L 24 
1
1
 
− − + + + −
 
L 13 L 23 L 13 L 23 L 33 L 34 L 34
1 1 1 1 1 1 1
− − − + + +
L 14 L 24 L 34 L 14 L 24 L 34 L 44
th
Let us now return to the equations of the i mechanical, finite beam element:
M i ¨ u i + K i u i = p i (6.37)
where
u i = [u y0 , r z0 , u y1 , r z1 ] T