Mechanical vibration

1. Types of Damping:
Damping can be classified into the following types:
I. Viscous damping
II. Hysteretic or Material Damping
III. Dry Friction of Coulomb Damping
IV. Damping using Electromagnetic Fields
Refer to section 3.7, page 70, for more information.

2. The particular integral ˲ (ˮ) under the excitation force
˘ ˮ = ˘J˩J ˮ in Eq.2.1 is of the form
˲ (ˮ) = IJ˩J( ˮ − ∅) Eq.2.17
It can be shown that the amplitude I of the steady-state
response is
I =
( )
Eq.2.20
Dividing both sides by ˫
I =
9
# 9 ( )
Eq.2.21
and ∅ = tan #
or ∅ = tan #
#
Eq.2.22

3. I is the amplitude of the steady-state response and −∅ is
the phase angle of ˲ (ˮ) relative to the excitation ˘J˩J ˮ.
The last two equations can be written using $ = ,
= 2ξ and J =
as
9
= =
#
# ($ )
= H Eq.2.23
and ∅ = tan # $
#
, Eq.2.24
where H is called the magnification factor and J the frequency
ratio of the excitation frequency to the natural frequency
of the system.
Eq.2.23 and 2.24 are plotted below with damping factor ξ as
a parameter.

4. Fig. Phase Angle Φ Versus
Frequency Ratio r
Fig. Magnification Factor K Versus
Frequency Ratio r

5. Note:
See comments on these 2 diagrams given in
Textbook of “Machine Vibration Analysis” by
Prof. Dr. Abdul Mannan Fareed
1st Edition, 2007

6. The general solution of Eq. 2.1 represents the system
response to a harmonic excitation and the initial conditions.
Substituting Eq.2.16 and 2.19 into Eq.2.2, the general
solution becomes
˲ ˮ = ˲ (ˮ)+ ˲ (ˮ)
˲ ˮ = ˓˥
sin ˮ + +IJ˩J( ˮ − ∅), Eq.2.25
where I=
( )
and ∅ are calculated from
Eq.2.23 and 2.24.

7. Find the transient response and the steady-state
response of the system in Example 1, if the excitation
force of ˘ ˮ = 2ŸJ˩JŵŹˮ N is applied to the mass in
addition to the given initial conditions.
Example 3: Forced Damped Vibrations
Solution: The displacement of mass ˭ is
obtained by direct application of Eq.2.25. The
system parameters are identical to those
calculated in Example 1.
The steady-state response from Eq.2.23 is
˲ = ҒJ˩J( ˮ − ∅)

8. ˲ = ҒJ˩J( ˮ − ∅) or
˲ =
˘
˫
ŵ
˫ − $˭ $ +( I)$
J˩J( ˮ − ∅)
Substituting the values of all parameters
˲ =
$
[# (#'/$) ] [$(.# $')(#' $)]
J˩J(ŵŹˮ − ∅),
=6.0J˩J(ŵŹˮ − ∅) mm,
where ∅ = tan # $
#
=
$(.# $')(#' $)
# (#'/$)
=29.1°.

9. The general solution becomes from Eq.2.25
˲ ˮ = ˓˥ sin ˮ + + HJ˩J( ˮ − ∅)
˲ ˮ = ˓˥ %.$'
sin ŵ9.7ˮ + + 6.0J˩J(ŵŹˮ − 29.ŵ°).
Applying the initial conditions at ˮ = 0,
˲ 0 = 0 = ˓J˩J + 6.0sin(−29.ŵ°)
˲Ӕ 0 = ŵ00 = ˓ −3.2ŹJ˩J + ŵ9.7IJJ + 6.0
× ŵŹcos(29.ŵ°).
Solving for ˓ and , we obtain = tan # ŵ.87 = 6ŵ.8°
and ˓ = 3.3ŵ mm.
Thus ˲ = 3.3ŵ˥ %.$' sin ŵ9.7ˮ + 6ŵ.8° + 6.0sin(ŵŹˮ
− 29.ŵ°) mm.
The above equation can now be plotted as shown next.

10. Plot the harmonic motions of steady-state response ˲
and the general solution ˲ ˮ = ˲ + ˲ respectively in
Example 2 for at least three consecutive cycles. Use for
this purpose size A4 graph papers only.
Assignment 4

11. 2.5 Comparison of Rectilinear Rotational
Systems:
Earlier discussions and problems were centred mostly on
systems with rectilinear motion. The theory and
interpretations given are equally applicable to systems
with rotational motion.
The analogy between the two types of motion and the
units normally employed are tabulated below.
Extending this analogy concept, it may be said that
systems are analogous if they are described by equations
of the same form.
Thus the theory developed for one system is applicable
to its analogous system.

12. Table 1: Analogy between Axial and Rotational
Systems

13. Table 2: Analogy between Responses of the Two
Systems