3.3 Mathematical analysis
Fig. 3.1 (a): A cantilever beam having tip mass at free end
Fig. 3.1 (b): The beam under forced vibrations
Fig. 3.1(a) shows a cantilever beam which is fixed at one end and other end is carrying a lumped mass. Fig. 3.1(b) shows the cantilever beam which is subjected to forced vibration. An exciter is used to give excitation to the system. The exciter is capable to generate different type of forcing signal e.g. sine, swept sine, rectangular, triangular etc.
Single DOF Model - When we study the forced vibration of a cantilever beam, we can consider it as a discrete system in which the beam is mass less and one third of the whole mass is concentrated at the free end of the beam as shown in Fig. 3.1.(a). If the system is subjected to a harmonically excitation force, F 0 sin ωt , the equation of motion for the equivalent single-DOF system subjected to forced vibration can be given as,
(3.1)
Where, m is one third of the mass of the beam, c is the damping coefficient; k is stiffness of the beam, F 0 is the amplitude of the excitation force, ω is the circular frequency of excitation, y ( t ) is the displacement of the free end of the beam and t is time.
Fig. 3.2: An equivalent single-DOF spring-mass-damper system of the beam
Then the forced response is given as
(3.2)
Where, X is the amplitude of vibration and φ is phase difference of displacement with respect to exciting force
(3.3)
(3.4)
The above equations can be represented in terms of following quantities
So,
The equation (3.5) is termed as Magnification factor, which is the ratio of the amplitude of steady–state response to the static deflection under the action of force F 0 . The plot of magnification factor against the frequency ratio for different values of ς is shown in Fig. 3.3 (a). The curves show that as the damping ratio increases, the maximum value of magnification factor decreases. When there is no damping i.e.( ς =0 ), it reaches to infinity at the resonance i.e. when ω/ω n = 1. In practice the magnification factor cannot reach infinity owing to friction which tends to dampen the vibration.
Fig. 3.3 (a): Magnification factor versus frequency ratio for different values of damping ratio
Fig. 3.3 (b): Phase angle versus frequency ratio for different values of damping ratio
Continuous Beam Model - In actual case, the beam is a continuous system, i.e. the mass along with the stiffness is distributed throughout the beam. The equation of motion in this case will be ( Meirovitch, 1967),
(3.7)
where, E is the modulus of rigidity of beam material, I is the moment of inertia of the beam cross-section, y ( x , t ) is displacement in y direction at distance x from fixed end, m is the mass per unit length, m = ρ A(x) is the material density, A ( x ) is the area of cross-section of the beam, f(t) is the force applied to the system at x = L 1 .
Free Vibration Solution - We have following boundary conditions for a cantilever beam (Fig. 3.1).
(3.8)
(3.9)
For a uniform beam under free vibration from equation (3.7), we get
(3.10)
A closed form of the circular natural frequency ω nf , from above equation of motion for first mode can be written as
(3.11)
The natural frequency is related with the circular natural frequency as
(3.12)
where I , the moment of inertia of the beam cross-section, for a circular cross-section it is given as
(3.13)
Where, d is the diameter of cross section and for a rectangular cross section
(3.14)
Where b and d are the breadth and width of the beam cross-section as shown in the Fig. 3.4.
Fig. 3.4: Cross-section of the cantilever beam