8.11 Free Vibration of the Shear Beam
345
Fig. 8.15 Freebody diagram
of a beam
The general solution of the equation becomes
y s = sin αx(C 3 sin pt + C 4 cos pt)
(8.165)
8.12 Effect of Axial Force on the Free Flexural Vibration
of Beams
In practical problems, a beam undergoing flexural vibration may be subjected to an
axial tension or compression.
Freebody diagram or an element of length dx of a beam has been shown in
Fig. 8.15. The beam is subjected to an axial tension N, which is supposed to
be constant throughout the length for small deflection of the beam. The shear
deformation and rotary inertia effects have not been considered in the present
derivation.
Considering the dynamic equilibrium of the forces acting on the element, the
following equation results in
ρ A
∂
2 y
∂ t 2 dx = −
V +
dV
∂ x
dx
+ V + N
θ +
∂θ
∂ x
dx
− N θ
or
ρ A
∂
2 y
∂ t 2 = −
∂ V
∂ x
+ N
∂θ x
∂ x
or
ρ A
∂
2 y
∂ t 2 = −
∂ V
∂ x
+ N
∂
2 y
∂ x 2
(8.166)
345
Fig. 8.15 Freebody diagram
of a beam
The general solution of the equation becomes
y s = sin αx(C 3 sin pt + C 4 cos pt)
(8.165)
8.12 Effect of Axial Force on the Free Flexural Vibration
of Beams
In practical problems, a beam undergoing flexural vibration may be subjected to an
axial tension or compression.
Freebody diagram or an element of length dx of a beam has been shown in
Fig. 8.15. The beam is subjected to an axial tension N, which is supposed to
be constant throughout the length for small deflection of the beam. The shear
deformation and rotary inertia effects have not been considered in the present
derivation.
Considering the dynamic equilibrium of the forces acting on the element, the
following equation results in
ρ A
∂
2 y
∂ t 2 dx = −
V +
dV
∂ x
dx
+ V + N
θ +
∂θ
∂ x
dx
− N θ
or
ρ A
∂
2 y
∂ t 2 = −
∂ V
∂ x
+ N
∂θ x
∂ x
or
ρ A
∂
2 y
∂ t 2 = −
∂ V
∂ x
+ N
∂
2 y
∂ x 2
(8.166)