Civil Engineering Reference
In-Depth Information
■
Solution
For the first mode, we have
S
11
=
A
(1)
T
[
M
]
A
(1)
=
[
11
m
00
0
m
0
00
m
1
.
4325
2
.
0511
]
.
1
4325
2
.
0511
4404
m
so the fi
rs
t modal amplitude vector is normalized by dividing each term by
√
S
11
=
3
.
3824
√
m
,
which gives the normalized vector as
=
11
.
0
.
2956
0
A
(1)
=
1
√
m
.
4289
0
.
6064
Applying the same procedure to the modal amplitude vectors for the second and third
modes gives
0
.
6575
0
0
.
6930
−
A
(2)
=
A
(3)
=
1
√
m
1
√
m
.
.
5618
0
7124
−
0
.
3550
0
.
0782
and the normalized modal matrix is
0
.
2956
0
.
6575
0
.
6930
1
√
m
[
A
]
=
.
.
−
.
0
4289
0
5618
0
7124
0
.
6064
−
0
.
3550
0
.
0782
To verify Equation 10.109, we form the triple product
0
.
2956
0
.
4289
0
.
6064
m
00
0
m
0
00
m
1
m
[
A
]
T
[
M
][
A
]
=
.
.
−
.
0
6575
0
5618
0
3550
0
.
6930
−
0
.
7124
0
.
0782
=
0
.
2956
0
.
6575
0
.
6930
100
010
001
×
.
.
−
.
0
4289
0
5618
0
7124
0
.
6064
−
0
.
3550
0
.
0782
as expected.
The triple product with respect to the stiffness matrix is
0
.
2956
0
.
4289
0
.
6064
−
20
3
k
m
[
A
]
T
[
K
][
A
]
=
.
.
−
.
−
−
0
6575
0
5618
0
3550
23
1
0
.
6990
−
0
.
7124
0
.
0782
0
−
11
0
.
2956
0
.
6575
0
.
6990
×
.
.
−
.
0
4289
0
5618
0
7124
0
.
6064
−
0
.
3550
0
.
0782
which evaluates to
0
.
1532
0
0
2
1
00
k
m
[
A
]
T
[
K
][
A
]
=
=
.
0
1
2912
0
2
2
0
0
0
0
5
.
0557
00
2
3
