Civil Engineering Reference
In-Depth Information
and this is the
eigenvector
corresponding to the
eigenvalue
1
. Proceeding identically
with the values for the other two frequencies,
2
and
3
, the resulting amplitude vectors
are
1
A
(2)
=
A
(2)
2
.
0
8544
−
0
.
5399
1
A
(3)
=
A
(3)
2
−
.
1
0279
0
.
1128
This example illustrates that an
N
degree-of-freedom system exhibits
N
natural
modes of vibration defined by
N
natural circular frequencies and the correspond-
ing
N
amplitude vectors (mode shapes). While the examples deal with discrete
spring-mass systems, where the motions of the masses are easily visualized as
recognizable events, structural systems modeled via finite elements exhibit
N
natural frequencies and
N
mode shapes, where
N
is the number of degrees of
freedom (displacements in structural systems) represented by the finite element
model. Accuracy of the computed frequencies as well as use of the natural modes
of vibration to examine response to external forces is delineated in following
sections.
10.4 BAR ELEMENTS: CONSISTENT
MASS MATRIX
In the preceding discussions of spring-mass systems, the mass (inertia) matrix
in each case is a lumped (diagonal) matrix, since each mass is directly attached
to an element node. In these simple cases, we neglect the mass of the spring
elements in comparison to the concentrated masses. In the general case of solid
structures, the mass is distributed geometrically throughout the structure and the
inertia properties of the structure depend directly on the mass distribution. To
illustrate the effects of distributed mass, we first consider longitudinal (axial)
vibration of the bar element of Chapter 2.
The bar element shown in Figure 10.7a is the same as the bar element intro-
duced in Chapter 2 with the very important difference that displacements and ap-
plied forces are now assumed to be time dependent, as indicated. The free-body
diagram of a differential element of length
d
x
is shown in Figure 10.7b, where
cross-sectional area
A
is assumed constant. Applying Newton's second law to the
differential element gives
d
x
A
2
u
+
∂
∂
A
d
x
)
∂
−
=
(10.51)
A
(
x
∂
t
2
