Information Technology Reference
In-Depth Information
FIGURE 5.1
A beam and its corresponding transfer matrices and state vectors.
3
2
b
w
1
−
−
/
6
EI
−
/
2
EI
·
w
w
. θ
2
0
b
0
1
/
2
EI
/
EI
V
V
.
V
b
0
0
1
0
=
(5.3)
.
M
b
M
···
1
0
0
1
M
···
1
.
···
···
···
···
···
.
0
0
0
0
1
b
a
U
i
z
b
=
z
a
Note that these relations, i.e., Eqs. (5.2) and (5.3), are equivalent.
Consider now the system of Fig. 5.1 formed of several beam elements joined end to end.
By definition of the transfer matrices,
U
1
z
a
z
b
=
(5.4a)
U
2
z
b
z
c
=
(5.4b)
U
3
z
c
z
d
=
(5.4c)
U
4
z
d
=
z
e
(5.4d)
U
5
z
e
=
z
f
(5.4e)
where
z
a
is the state vector at the origin, i.e., at the left end. As illustrated in Fig. 5.1,
U
1
is
the transfer matrix for the first element at the left end of the beam, i.e., use Eq. (5.3) with
equal to the length of the first element.
Vector
z
b
is the state variable at the right end of the first element. Matrix
U
2
represents the
second element and so on. Note in Eqs. (5.4a to e) that each of the state vectors
z
b
,
z
c
,
z
d
,
z
e
,
and
z
f
can be expressed in terms of the initial state vector
z
a
by progressively replacing
z
b
in Eq. (5.4b) by
z
b
of Eq. (5.4a),
z
c
of Eq. (5.4c) by
z
c
of Eq. (5.4b), etc. Thus, from Eq. (5.4b),
z
c
U
2
U
1
z
a
. In summary, the
state vector at locations, i.e., stations or nodes,
b, c, d, e
, and
f
, can be written as
U
2
z
b
. Equation (5.4a) permits this to be expressed as
z
c
=
=
U
1
z
a
=
z
b
(5.5a)
U
2
z
b
U
2
U
1
z
a
z
c
=
=
(5.5b)
U
3
z
c
=
U
3
U
2
U
1
z
a
z
d
=
(5.5c)
U
4
z
d
=
U
4
U
3
U
2
U
1
z
a
z
e
=
(5.5d)
U
5
z
e
=
U
5
U
4
U
3
U
2
U
1
z
a
z
f
=
(5.5e)
Search WWH ::
Custom Search