6.3 About the experiment
A beam system with two lumped masses, and under simply-supported boundary conditions is shown in Fig.1. Free (or forced) vibrations of any system can be determined by the use of its mode shapes. Mode shape describes the relative vibration of different points in a system at any time. The mode shapes for the system shown in Fig. 6.1 (with the assumption of the mass less beam) are shown in Fig. 6.2.
Fig. 6.1: Two DOF System
Fig. 6.2: Mode Shapes (left) in-phase motion (right) out-of-phase (or anti-phase) motion
Mode of vibration of a beam is characterized from the mode shapes and its modal frequency. The number of half waves in the mode shape indicates the mode of vibration of a beam. For example, a beam with a mode shape as one-half sine wave, is said to be vibrating in first mode. Hence, Fig. 6.2 indicates the mode I and mode II vibrations of the beam shown in Fig. 6.1.
In mode I, the system vibrates with its lowest natural frequency (i.e., the fundamental natural frequency) with both the masses vibrating in-phase; with a phase difference of 0 degrees (for no or little damping case). Whereas, in mode II, the system vibrate with its second natural frequency with the masses vibrating in out-of-phase with a phase difference of 180 degrees.
6.3.1 Equations of motion
The equations of motion of a vibrating system can be derived from the Lagrange’s equations. Lagrange’s equations for an n degree of freedom system can be stated as,
where T and V are the kinetic and potential energies of the system and F i is the nonconservative generalized force corresponding to the i th generalized coordinate, is the time derivative of x i .
The kinetic and potential energies of a multi degree of freedom system can be expressed in matrix form as,
(6.2)
(6.3)
By substituting Eqs. (6.2) & (6.3) into Eq. (6.1), we obtain the desired equations of motion in matrix form as,
(6.4)
Since there are no nonconservative forces F i , so the equation of motion becomes,
(6.5)
Where [ m ] is the diagonal mass matrix and [ k ] is the stiffness matrix.
Let us assume the solution for the Eq. (6.5) as,
(6.6)
Then Eq. (6.5) can be written as,
(6.7)
Eq. (6.7) can be written as,
(6.8)
where [ α ] is the Influence coefficient matrix, and
(6.9)
For a nontrivial solution of Eq. (6.8), the determinant of coefficient matrix must be zero. That is,
(6.10)
On simplifying, we get,
(6.11)
Substituting Eq. (6.9) into Eq. (6.11), we get,
(6.12)
where m 1 and m 2 are the masses of the discs in the system shown in Fig. (6.1).
The influence matrix coefficients can be found out using the procedure described in [1]. For the system shown in Fig. 6.1, the influence matrix is obtained as,
where E is the Young’s modulus and I is the Moment of inertia of the beam.
The two mode shapes can be obtained by substituting the two values of [ ω ] in the differential equations of motion (6.8).