Dynamics of Machines - Part II - IFS.pdf

In order to calculate the stiffness matrix (k11, k12, k21, k22) of the 2 d.o.f. system,
Ky = f (16)
where
K =
k11 k12
k21 k22
y = { y1 y2 } T
f = { f1 f2 } T
one has to solve the following system of equations for case (I) and case (II):
k11 k12
k21 k22
y1
·
y2
=
Case (I): f = { F 0 } T
k11 k12
k21 k22
·
Case (II): f = { 0 F } T
k11 k12
k21 k22
f1
f2
F ·L 3 1
3·EI
F
6EI · (2L3 1 + 3L2 1 (L2 − L1))
F
6EI
·
· (2L31 + 3L21 (L2 − L1))
F ·L3 2
3·EI
=
=
F
0
0
F
It leads to a system of 4 equations, which can be written in a matrix form:
⎡
⎢
⎣
F ·L 3 1
3·EI
F
6EI · (2L31 + 3L 2 0
1(L2 − L1))
0
0
F ·L
0
3 1
3·EI
F
6EI · (2L31 + 3L 2 F
6EI
1(L2 − L1))
· (2L31 + 3L 2 1(L2 − L1))
F ·L 3 0
2
3·EI
0
0
F
6EI
0
· (2L31 + 3L 2 1(L2 − L1))
F ·L 3 ⎤
⎥
⎦
2
3·EI
·
⎧
⎪⎨
·
⎪⎩
⎫ ⎧
⎪⎬ ⎪⎨
F
0
=
⎪⎭ ⎪⎩
0
F
⎫
⎪⎬
⎪⎭
(21)
k11
k12
k21
k22
Solving this matrix system of order 4 by using the software MATHEMATICA, or by using
Cramer’s rule, one achieves the stiffness matrix:
EI
K =
(4L2 − L1)(L1 − L2) 2
⎡
12(L2/L1)
⎣
3 ⎤
6(L1 − 3L2)/L1
⎦ (22)
6(L1 − 3L2)/L1 12
Similar procedure can be made in order to get the stiffness matrix of the 3 d.o.f system. This is
the motivation of an exercise later on. The results (stiffness matrix with 9 stiffness coefficient)
will be presented in the section describing mechanical systems with 3 d.o.f.
8
(17)
(18)
(19)
(20)