Chapter (2): Vibration of Single-degree of Freedom

Systems (SDOF)

2.1 Degree of Freedom

2.2 Differential Equations of Motion in Time Domain:

- Newton’s Law of Motion

- Energy Method

2.3 General Solution of Equation: Transient and Steady-

state Response

2.4 Frequency Response Method in Frequency Domain:

Impedance Method

2.5 Comparison of Rectilinear & Rotational Systems

2.1 Degree of Freedom:

� The One-degree-of-freedom system is the keystone for

more advanced studies in vibration.

� The system is representation by a generalized model

shown below.

� Examples of one-degree-of-freedom systems are shown

in next figure. Though they differ in appearance, they can

be represented by the same generalized model.

� The model serves:

�To unify a class of problems commonly

encountered.

�To bring into focus the concept of vibration.

Fig. Generalized Model of One-degree-

of-freedom Systems SDOF

Fig. Examples of Systems with

One-degree-of-freedom SDOF

Definition of Degree of Freedom:�The number of degrees of freedom of a vibratory system is the

number of independent spatial coordinates required to uniquely

define the configuration of a system.

�To every degree of freedom there is an independent spatial

coordinate.

�To every degree of freedom there is an equation of motion.

�To every degree of freedom there is a natural frequency.

�Number of natural frequencies equal number of degree of

freedom.

�A rigid body will have 6 DOF to describe its motion – 3

translation and 3 rotation. Such a System has here 6

natural frequencies and hence 6 independent coordinates

and 6 equations.

Examples of Multi-degree-of-freedom Systems:

Fig. ? DOF Dynamic SystemFig. ? DOF Dynamic System

Animated Motion of Multi-degree-of-freedom

Systems

Fig. Generalized Model

of ? DOF

Fig. Generalized Model

of ? DOF

Quarter Car Model:

Fig. Generalized Model

of ?DOF

Human Body Dynamic Model:

Fig. Generalized Model

of ? DOF

2.2 Differential Equations of Motion in Time

Domain:

� The equation of motion of a dynamic system can be

written using the following methods:

�Force Method using Newton’s Law of Motion (Time

Domain)

�Energy Method using Kinetic & Potential Energies

(Time Domain)

�Frequency Response Methods (Frequency Domain)

�Equation of Motion using Laplace transform (Frequency

Domain)

�At first we consider the equations of motion of dynamic

systems in time domain.

Types of Vibratory Motions for SDOF Systems:

Fig. (a) Undamped Free

Vibration (b) Damped Free

Vibration and (c) Forced

Damped Motions

2.2.1 Equation of Motion - Newton’s Law of

Motion/Force Method:� This method states simply that the inertia force on a

dynamic system is equal to the sum of all forces and/or

moments acting on that system. In general we can write

��� = ������� ������������� ����� �.

���� = ������� ��������������� ����� � .

� To simplify solution, we can neglect in this analysis all the

forces and moments from Statics and keeping only the

dynamic ones.

� Following examples take into account these above

principles.

Example 1: Spring-mass System

� To derive equation of motion using

Newton’s Law of Motion.

� The inertia force is equal to sum of all

forces,

�+↓ ���= ������� ������������� ����� �

��� = �� − �( !" + �).�Neglecting forces from Statics, then

��� = −��,Or ��� + �� = 0, Differential Eq. of Motion.

� Rewriting �� + %& � = 0, where '() = %

& is the natural

frequency in (radians/s)²

Example 2: Simple Pendulum

� To derive equation of motion using

Newton’s Law of Motion.

� The inertia moment is equal to sum of all

moments acting on the system.

�+* ����= ������� ��������������� ����� ����� = −��+����.

� Since vibration is very small, then ���� , �,���� = −��+�-

Or ���� + ��+� = 0- Differential Eq. of Motion.

� �� + &./&/0 � =, where '() = .

/ is the natural frequency in (radians/s)².

Example 3: Torsional Pendulum

� The inertia moment is equal to sum

of all moments acting on the system.

�+1 ���� = ������� �������� = −�"�2 �" = 345 +6 .

� Since vibration is very small, then

���� , �,���� = −�"�.

Or ���� + �"� = 0- Differential Eq. of Motion.

� Rewriting �� + %789 � = �� + %7

:;0

0 � = 0, where '() = %7:;0

0 is

the natural frequency in (radians/s)².

Assignment 1

1. Calculate the natural frequency

<( using Force Method.

2. A motor weighing 150 kg is supported

by 4 springs, each having a stiffness of

1000 N/m. Determine the speed in

rpm at which resonance will occur.

3. Calculate the natural frequency <( of the

above system if �=10kg, �=10000 N/m and mass of pulley is 15 kg with radius 0.30 m.

� The equation of motion of a conservative system can be

established from energy conservations.

� The total mechanical energy in a conservative mechanical

system is the sum of the kinetic energy T and the potential

energy U.

� The kinetic energy T is due to the velocity of the mass.

� The potential energy U is due to the strain energy of the

spring by virtue of its deformation.

� For a conservative system, the total mechanical energy is

constant and its time derivative is zero.

Total Mechanical Energy= = + > = ?������. @@" = + > = 0.

2.2.2 Equation of Motion - Energy Method:

Fig. Potential Energy due to

Deformation of a Spring

Example 1: Mass-spring System

� To derive equation of motion using Energy Method.

�The displacement �(�) of the mass � is measured from its static equilibrium position.

�Let �(�) be positive in the downward direction.�The spring is massless.

� The kinetic energy of the system is = = A)��)B .

� The net potential energy of the entire system is the

algebraic sum of the strain energy of the spring and the

that due to the change in elevation of the mass and is >= C �� + �� D�� − ���E

F = > = A) ��).

� Total Mechanical Energy == + > = A)��B) + A

) ��)= ?������.� Therefore,

@@"

A)��B) + A

) ��) = 0 = ��� + �� �B = 0� �B(�) can not be equal to zero for all values of time �, then

��� + �� = 0.�Above is called Differential Equation of Motion of free

undpamed system and by rewriting:

�� + %& � = 0-

where '() = %& is the natural frequency of the system in

(radians/s)².

Example 2: Simple Pendulum

� To derive equation of motion using

Energy Method.

�The displacement �(�) of the mass �is measured from its static equilibrium

position.

�Let �(�) be positive to the right direction.

�The cord is massless.

� The kinetic energy of the system is = = A) ���)B .

� The net potential energy of the entire system is due to

the change in elevation of the mass and is > = ��(+− + GHI J).

� Total Mechanical Energy == + > = A) ���)B + ��(+

− + GHI J) = ?.� Therefore,

@@" (

A) ���)B + �� + − + GHI J) = 0.

� �B �(���)� + ��+����) = 0- since vibration is very small and

�B(�) can not be equal to zero for all values of time �, then���)� + ��+� = 0.�

�Above is called Differential Equation of Motion of free

undamped system and by rewriting:

�)� + &./89 � = 0.

where '() = &./&/² =

./ is called the natural frequency in

(radians/s)².

Example 3: A Cantilever Beam

� To derive equation of motion using Energy Method.

� The kinetic energy of the system is = = A)��)B .

� The potential energy of the entire system is the strain

energy of the spring and is > = A) ��).

Fig. Original Dynamic

System

Fig. Equivalent Dynamic

System

� Total Mechanical Energy= = + > = A)��B) + A

) ��) = ?� Therefore,

@@"

A)��B) + A

) ��) = 0 = ��� + �� �B= 0.

� �B(�) can not be equal to zero for all values of time �, then��� + �� = 0

�Above is called differential Equation of Motion of free

undamped system and by rewriting:

�� + %& � = 0-

where '() = %& is the natural frequency of the system in

(radians/s)².

Consideration of Equivalent Mass of a Spring:

� Consider again example 1. Let Lbe the mass of spring per unit

length and�+� length of the spring when the mass is at static

equilibrium position .

Example 4: A Simple Mass-spring

System

� Using Energy method, the kinetic energy is due to the

rigid mass ��and the spring �. � Consider an element DM at a distance M. By giving a displacement � � to the mass, the displacement at this

element is N/ � � . Thus, � � defines the system

configuration and the total kinetic energy is

�= = A)��B ) + A

)C L/F (N/ �B � ))DM = A)��B ) + A

O L+�B � ).

� The total potential energy is > = A) ��).

� The first derivative of the total mechanical system is zero.

Hence, Total Mechanical Energy= = + > = ? and

@@" (

A)��B ) + A

O L+�B � ) + A) ��)) = 0.

�B can not be zero for all values of time, Hence,��� + P/Q ��

+ �� = 0.� Rewriting above eq. yields � + P/

Q �� + �� = 0, or �� + %

&RSTU

� = 0 and the nat. frequency becomes

'() = %&RST

U.

Springs in Parallel:

- Determine the spring constant �VWfor equivalent spring

- Apply the approximate relations

for the harmonic motion of a

spring-mass system.

• '( = %XY& = AFZ

)F = 22. 36

rad/s.

Springs can be arranged as shown

below, either in parallel or series.

For each spring arrangement,

determine the spring constant

�VW for a single equivalent spring.

Springs in Arrangements in Parallel or Series:

[ = �A + �) δ�VW = [

δ = �A + �)=10 kN/m=10^ N/m

Fig. Springs in Parallel and Series

Connections with � = 20 kg

Calculation of Equivalent

Spring Stiffness `ab

- Determine the spring constant

�VW for equivalent spring.

- Apply the approximate relations

for the harmonic motion of a

spring-mass system.

Springs in Series:

VW = A + )-�where�� = ghg

%XY =g%i +

g%0=

A%i +

A%0

A%XY = A

^FFF+A

OFFF�VW=�2400�N/m.

Fig. Springs in Series Connections

with � = 20 kg

Calculation of Equivalent Spring

Stiffness `ab

• '( = %XY& = )^FF

)F= 10.95 rad/s.

• p = )qr = 0.574 s.

Equivalent Inertia Mass �VW:� The equivalent mass �VW of a vibration system may be

better explained with the help of the following Fig.3.1.

�Mass �VW is obtained using the kinetic energy = of the

system as illustrated below, Fig.3.1b.

�Assuming the spring to be of negligible mass, then,

Fig.3.1a Vibration System in

Rotation

Fig.3.1b Equivalent Spring

-mass System

= = A) ���)B =

A)�VW�)B ,

where �� = �t. +�(/^))= �+) AA) +

AAO = u

^v�+) andsubstituting �B = Q

^ +�B , then �VW = u)u�.

Equivalent Mass Moment of Inertia �VW� The equivalent mass moment of inertia �VW may be obtained

using the pinion-and-gear assembly, Fig.3.2.

� To find the �VW of the assembly referring to the motor shaft

1.

� The speed ratio between the motor shaft 1 and shaft 2 is

given by � = wA/w). Let �A and �) be the angular rotations of the pinion �A and the gear �) respectively.

Fig.3.2 Equivalent Mass Moment of Inertia

� The kinetic energy = of the assembly referred to the motor shaft 1 is

= = A) �A�BA) +

A) �)�B)) =

A) �VW�BA)

and substituting �B) = ��BA in above eq. of =, then�VW = �A + �)�)

�Hence the equivalent mass moment of inertia of �) referring to motor shaft 1 is �)�).

Effects of Orientation:� Consider the three figures and write down their equations of motion respectively.

� Eq. (a) and (c) are affected by gravity while (b) is not.

� In eq.(a) the natural frequency is increased, in (b) remains the same and in (c) is decreased.

Assignment 2

k

1. Calculate the natural frequency <(using Energy Method.

2. Calculate the natural

frequency <( using Energy Method.

3. Calculate the natural

frequency <( using Energy Method.

2.3 General Solution of Equation:

� The eq. of motion using Force

Method can be written as:

��� = ������� �.� Referring to the FBD, the forces

acting on the mass ��are: 1. Gravitational force ��,2. The spring force � � + !" , Fig. Model of Systems

with 1 DOF3. The damping force ��B and 4. The excitation force x���'�.

�Using Force Method and the attached figure, we can

write as follows:

∴ � @0@"0 � + !" = −� � + !" − � @

@" � + !" +��+ x���'�, Or ��� + ��B + �� = x���'�. Eq.2.1

General Solution:

� The general solution �(�) of Eq.2.1 is the sum of the

complementary function �t(�) and the particular integral �5(�), i.e.

� � = �t(�)+ �5(�) Eq.2.2

2.3.1 Complementary Function (Transient Response)

� This function satisfies the corresponding homogeneous

equation:

��� + ��B + �� = 0. Eq.2.3

� The solution is of the form �t � = � !", Eq.2.4

where � and � are constants. Substituting Eq.2.4 into Eq.2.3 gives ��) + �� + � � !" = 0.

� Since the constant � !"can not be equal to zero, then we deduce that ��) + �� + � = 0.

� This is called the auxiliary equation whose roots are:

�A-) = A)& (−� ± �) − 4��). Eq.2.5

� Since there are two roots, the complementary function is

�t = �A !i" + �) !0", Eq.2.6

where �A and �) are arbitrary constants to be evaluated using initial conditions.

� To write above eq. in a more convenient form, one uses

%& = '(),

t& = 2ξ'( or ξ = t

) %&, Eq.2.7

where '( is the natural frequency and ξ is the damping factor.

�Above equation can be rewritten as

�� + 2|'(�B + '()� = 0 Eq.2.8

�) + 2|'(� + '() = 0 Eq.2.9

�A-) = −|'( ± |) − 1'(. Eq.2.10

� Referring to Eq.2.10, value�Hf�|�in� |) − 1 may lead to

the following cases:

� Case 1: If | > 1, Eq.2.10 will have real, distinct and negative roots, |) − 1 < |.

�t = �A �5i" + �) �50" . Eq.2.11

� Thus, no oscillation is expected from Eq.2.11.

� Since both the roots are negative, the motion diminishes

with increasing time and it is aperiodic.

� Case 2: If | = 1, Eq.2.10 shows that both the roots are equal to −'(. Thus, the complementary function is of the

form: �t = (�Q + �^�) �r�", Eq.2.12

where �Q and �^ are constants. The motion is again

aperiodic and will diminish to zero as time increases. No

motion is expected.

� Case 3: If now | < 1, the roots in Eq.210 become complex conjugate and are of the form

�A-) = −|'( ± � 1 − |)'(- Eq.2.13

where � = −1, Define now '@ = 1 − |)'(.�Using Euler’s Formula ±�� = ��� ± �����, the complementary function �t in Eq.2.6 becomes:

�t = �� !i" + �O !0" = ��r�" �� �r�" + �O ��r�" -or �t = ��r�"( �� + �O ��'@� + � �� − �O ���'@�).

Eq.2.14

� Since �t is a real physical quantity, the coefficients �� + �Oand � �� − �O in Eq.2.14 must also be real. This requires that��� and �O be complex conjugates. Hence Eq.2.14 becomes

�t = ��r�"(�u��'@� + �v���'@�), Eq.2.15

�t = � ��r�" Iin '@� + � , Eq.2.16

where �u and �v are real constants to be evaluated by the initial

conditions, �= �u) + �v) and � = tan�A( �u �v)6 .

� The motion described by Eq.2.16 is a harmonic function decreases as time increases.

| > 1 – Over damped Eq.2.11, | = 1 – Critically damped Eq.2.12,

| < 1- Under damped Eq.2.16

Fig. Complementary Functions �t of SDOF Under Different Damping Factors

�Note that

� �t(�) is vibratory only if the system is underdamped, |< 1.

� The frequency of oscillation '@ is less than the natural

frequency '(.� In all cases, �t(�) will eventually die out regardless of the initial conditions. Hence complementary function

gives transient response.

� The critical damping �t is the amount of damping

necessary for a system to be critically damped, i.e. | = 1. From Eq.2.7, this value amounts

�t = 2 ��. Eq.2.17

�Damping factor | can be defined as | = tt� Eq.2.18

�A machine of 20 kg mass is mounted on springs

and dampers. The total stiffness of springs is 8

kN/m and the total damping is 130 N.s/m. If

the system is initially at rest and a velocity of 100

mm/s is imparted to the mass, determine:

a. The displacement and velocity of mass as a

time function

b. The displacement at � = 1.0 s.

Example 1: Free Damped Vibrations

� Solution:

The displacement �t(�) is obtained by the direct application of Eq.2.16. The parameters of the equation are:

'( = % &6 = vFFF )F6 = 20 rad/s, ξ = t) %& = AQF

) vFFF�)F= 0.1�25 and '@ = 1 − 0.1�25) � 20 = 19.7 rad/s.

a. Substituting these values into Eq.2.16, we obtain

�t = �Q.)�"(�A��19.7� + �)���19.7�) and�Bt = −3.25 �Q.)�" �A��19.7� + �)���19.7� +

19.7 �Q.)�"(−�A���19.7� + �)��19.7�)Applying the initial conditions gives � = 0- � 0 = 0 ,

→ �A = 0 and

� = 0- �B 0 = 100, → �) = AFF A�.u6 = 5.07 mm

�t = 5.07 �Q.)�"���19.7� mm and

�Bt = �Q.)�" −1�.47���19.7� + 100��19.7�= 101.3 �Q.)�"�� 19.7� + 9.5° mm/s.

b. The displacement at � = 1.0 s is �t= 5.07 �Q.)��A.F���19.7 � 1.0 = 0.1477�mm.

� Plot the harmonic motions of displacement �t and velocity �Bt developed by the solutions (a) respectively in example 1 for at least three consecutive cycles. Use

for this purpose size A4 graph papers only.

Assignment 3