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.

� The particular integral ��(�) under the excitation force

� � = �����in Eq.2.1 is of the form

��(�) = �(�� − ∅) Eq.2.17

� It can be shown that the amplitude �of the steady-state

response is

=�

����� ��(��)�Eq.2.20

� Dividing both sides by �

=�

��

�������

��(�

�! )�Eq.2.21

and ∅ = tan����

�����or ∅ = tan��

�� �!

����� �!Eq.2.22

� is the amplitude of the steady-state response and −∅ is

the phase angle of ��(�) relative to the excitation ����.

� The last two equations can be written using �%& =

�

�,

��

�= 2ξ� �%! and ) =

�

�*as

+�

��=

+

+,=

�

��-� ��(&.-)�= / Eq.2.23

and ∅ = tan��&.-

��-�, Eq.2.24

where / is called the magnification factor and ) 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.

Fig. Phase Angle Φ Versus

Frequency Ratio r

Fig. Magnification Factor K Versus

Frequency Ratio r

Note:

See comments on these 2 diagrams given in

Textbook of “Machine Vibration Analysis” by

Prof. Dr. Abdul Mannan Fareed

1st Edition, 2007

� 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

� � = ��(�)+ ��(�)

� � = 01�.�*2 sin �5� + 7 + �(�� − ∅), Eq.2.25

where =�

����� ��(��)�and ∅ are calculated from

Eq.2.23 and 2.24.

� Find the transient response and the steady-state

response of the system in Example 1, if the excitation

force of � � = 28�9:��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

�� =�

�<�(�� − ∅)

�� =�

�<�(�� − ∅) or

�� =�

�

9

� − �&; & +(�=)&�(�� − ∅)

Substituting the values of all parameters

�� =&> ?@@@!

[��(�B/&@)�]��[&(@.�F&B)(�B &@! )]���(9:� − ∅),

=6.0��(9:� − ∅) mm,

where��∅ = tan��&.-

��-�=&(@.�F&B)(�B &@! )

��(�B/&@)�=29.1°.

� The general solution becomes from Eq.2.25

� � = 01�.�*2 sin �5� + 7 +�

�/�(�� − ∅)

� � = 01�I.&B2 sin 99.7� + 7 + 6.0�(9:� − 29.9°).

� Applying the initial conditions at � = 0,

� 0 = 0 = 0�7 + 6.0sin�(−29.9°)

�M 0 = 900 = 0 −3.2:�7 + 99.7=O�7 + 6.0× 9:cos�(29.9°).

� Solving for 0 and 7, we obtain 7 = tan�� 9.87 = 69.8°and 0 = 3.39 mm.

� Thus � = 3.391�I.&B2 sin 99.7� + 69.8° + 6.0sin�(9:�− 29.9°) mm.

� The above equation can now be plotted as shown next.

� Plot the harmonic motions of steady-state response ��and the general solution � � = �� + �� respectively in

Example 2 for at least three consecutive cycles. Use for

this purpose size A4 graph papers only.

Assignment 4

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.

Table 1: Analogy between Axial and Rotational

Systems

Table 2: Analogy between Responses of the Two

Systems