66
3 Forced Vibration of Single Degree of Freedom System
the damping is equivalent to a force acting at the middle of the beam proportional to
the velocity and equal to 500 N at a velocity of 2.5 cm/s.
The frequency of forcing function is given by
ω = 2π ×
600
60
= 20π rad/s
The mass of the vibrating body is
m =
5000
9.8
kg.
The static deflection is
δ st =
0.025
100
m
The stiffness is given by
k =
F 0
δ st
=
5000 × 100
0.025
= 20 × 10
6 N/m
Based on the information given, the damping constant is
c = 500 ×
100
2.5
= 20,000 Ns/m
The natural frequency of the system is given by
p
2
=
k
m
=
20 × 10
6
× 9.8
5000
= 39,200
Therefore, p = 198 rad/s.
The damping constant is
n =
c
2m
=
20,000
2 × 5000
× 9.8 = 19.6
η =
ω
p
=
20π
198
= 0.32
ζ =
n
p
=
19.6
197
= 0.099
The magnification factor is
μ =
1
(1 − η 2 ) 2 + ( 2ηζ ) 2
3 Forced Vibration of Single Degree of Freedom System
the damping is equivalent to a force acting at the middle of the beam proportional to
the velocity and equal to 500 N at a velocity of 2.5 cm/s.
The frequency of forcing function is given by
ω = 2π ×
600
60
= 20π rad/s
The mass of the vibrating body is
m =
5000
9.8
kg.
The static deflection is
δ st =
0.025
100
m
The stiffness is given by
k =
F 0
δ st
=
5000 × 100
0.025
= 20 × 10
6 N/m
Based on the information given, the damping constant is
c = 500 ×
100
2.5
= 20,000 Ns/m
The natural frequency of the system is given by
p
2
=
k
m
=
20 × 10
6
× 9.8
5000
= 39,200
Therefore, p = 198 rad/s.
The damping constant is
n =
c
2m
=
20,000
2 × 5000
× 9.8 = 19.6
η =
ω
p
=
20π
198
= 0.32
ζ =
n
p
=
19.6
197
= 0.099
The magnification factor is
μ =
1
(1 − η 2 ) 2 + ( 2ηζ ) 2