322
8 Free Vibration Analysis of Continuous Systems
Therefore, equation to be solved is
pL
a
tan
pL
a
=
80.75
4.971
= 16.244
The first root of Eq. (e) is
pL
a
= 1.48
or
p =
1.48
L
G
ρ
=
1.48
1000
8.4 × 10 10
7800
= 4.85 rad/s
8.5 Free Flexural Vibration of Beams
Let us consider a beam sufficiently long in comparison with its cross-section, so
that the shear deformation is ignored. The effect of rotary inertia, which has been
discussed later, has not been considered in this deri-vation. The freebody diagram of
a beam segment of elemental length dx has been shown in Fig. 8.8.
Let A be the cross-sectional area of the beam,
I be the second moment of area of beam cross section,
ρ be the mass density for the material of the beam,
w (x, t) be the intensity of the external loading, acting on the beam per unit
length, and
V, M are the shear force and bending moment at a section, respectively.
Fig. 8.8 A beam and its freebody diagram
8 Free Vibration Analysis of Continuous Systems
Therefore, equation to be solved is
pL
a
tan
pL
a
=
80.75
4.971
= 16.244
The first root of Eq. (e) is
pL
a
= 1.48
or
p =
1.48
L
G
ρ
=
1.48
1000
8.4 × 10 10
7800
= 4.85 rad/s
8.5 Free Flexural Vibration of Beams
Let us consider a beam sufficiently long in comparison with its cross-section, so
that the shear deformation is ignored. The effect of rotary inertia, which has been
discussed later, has not been considered in this deri-vation. The freebody diagram of
a beam segment of elemental length dx has been shown in Fig. 8.8.
Let A be the cross-sectional area of the beam,
I be the second moment of area of beam cross section,
ρ be the mass density for the material of the beam,
w (x, t) be the intensity of the external loading, acting on the beam per unit
length, and
V, M are the shear force and bending moment at a section, respectively.
Fig. 8.8 A beam and its freebody diagram