2.3 Free Undamped Vibration of the Sdf System
25
comparison with total mass M of the girder. Determine the period of free horizontal
vibration.
Like the previous example, the equivalent stiffness of the columns is to be determined first. The columns CD and EF have one end hinged and the other end fixed.
For such cases, the stiffness for a unit displacement at the top is 3E I /L
3 , where L
is the length of the member. Therefore, the equivalent stiffness of the columns is
k eq =
12E I
(5a) 3 +
3E I
(4a) 3 +
3E I
(6a) 3 = 0.157
E I
a 3
The period of free horizontal vibration then is
T = 2π
M
k eq
= 2π
Ma 3
0.157E I
= 15.86
Ma 3
E I
Example 2.5 Find the natural frequency of the system shown in Fig. 2.10. The mass
of the beam is negligible in comparison with the suspended mass.
E = 2.1 × 10
5 N/mm
2
In addition to the flexible beam which provides a spring action, there is a spring
attached to the mass. In this, the deformation of the central mass is equal to the sum
of the deformations of the beam at the centre and the attached spring.
The beam can be replaced by a spring k 1 of stiffness 48E I /L
3 , where EI is the
flexural rigidity of the beam and L, its length. Two springs in this case are said to
be connected in series (Fig. 2.10c), which can be replaced by an equivalent spring
having stiffness k eq . Due to a unit force, the deformation in both the cases will be
Fig. 2.10 Example 2.5
25
comparison with total mass M of the girder. Determine the period of free horizontal
vibration.
Like the previous example, the equivalent stiffness of the columns is to be determined first. The columns CD and EF have one end hinged and the other end fixed.
For such cases, the stiffness for a unit displacement at the top is 3E I /L
3 , where L
is the length of the member. Therefore, the equivalent stiffness of the columns is
k eq =
12E I
(5a) 3 +
3E I
(4a) 3 +
3E I
(6a) 3 = 0.157
E I
a 3
The period of free horizontal vibration then is
T = 2π
M
k eq
= 2π
Ma 3
0.157E I
= 15.86
Ma 3
E I
Example 2.5 Find the natural frequency of the system shown in Fig. 2.10. The mass
of the beam is negligible in comparison with the suspended mass.
E = 2.1 × 10
5 N/mm
2
In addition to the flexible beam which provides a spring action, there is a spring
attached to the mass. In this, the deformation of the central mass is equal to the sum
of the deformations of the beam at the centre and the attached spring.
The beam can be replaced by a spring k 1 of stiffness 48E I /L
3 , where EI is the
flexural rigidity of the beam and L, its length. Two springs in this case are said to
be connected in series (Fig. 2.10c), which can be replaced by an equivalent spring
having stiffness k eq . Due to a unit force, the deformation in both the cases will be
Fig. 2.10 Example 2.5