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Lecture 2 Free Vibration of Single Degree of Freedom Systems. ERT 452 VIBRATION. MUNIRA MOHAMED NAZARI SCHOOL OF BIOPROCESS UNIVERSITI MALAYSIA PERLIS. COURSE OUTCOME. CO 2

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lecture 2 free vibration of single degree of freedom systems

Lecture 2Free Vibration of Single Degree of Freedom Systems

ERT 452

VIBRATION

MUNIRA MOHAMED NAZARI

SCHOOL OF BIOPROCESS

UNIVERSITI MALAYSIA PERLIS

ERT 452 SESION 2011/2012

course outcome
COURSE OUTCOME

CO 2

Ability to DEVELOP and PLAN the solutions to vibration problems that contain free and forced-vibration analysis of one degree of freedoms system.

ERT 452 SESION 2011/2012

course outline
COURSE OUTLINE
  • 2.1 Introduction
  • 2.2 Free Vibration of an Undamped Translational System
  • 2.3 Free Vibration of an UndampedTorsional System

ERT 452 SESION 2011/2012

2 1 introduction
2.1 Introduction
  • Free Vibration occurs when a system oscillates only under an initial disturbance with no external forces acting after the initial disturbance.
  • Undamped vibrations result when amplitude of motion remains constant with time (e.g. in a vacuum).
  • Damped vibrations occur when the amplitude of free vibration diminishes gradually overtime, due to resistance offered by the surrounding medium (e.g. air).

ERT 452 SESION 2011/2012

2 1 introduction1
2.1 Introduction
  • A spring-mass system in horizontal position.

ERT 452 SESION 2011/2012

2 1 introduction2
2.1 Introduction
  • Several mechanical and structural systems can be idealized as single degree of freedom systems, for example, the mass and stiffness of a system.

Equivalent spring-mass system for the cam follower system.

ERT 452 SESION 2011/2012

2 1 introduction3
2.1 Introduction
  • Modeling of tall structure as spring-mass system.

ERT 452 SESION 2011/2012

2 2 free vibration of an undamped translational system
2.2 Free Vibration of an Undamped Translational System
  • Equation of Motion Using Newton’s Second Law of Motion:
  • Procedure
    • Select a suitable coordinate to describe the position of the mass of rigid body in the system (linear or angular).
    • Determine the static equilibrium configuration of the system and measure the displacement of the mass or rigid body from its static equilibrium position.
    • Draw free body diagram.
    • Apply Newton’s second law of motion.

ERT 452 SESION 2011/2012

2 2 free vibration of an undamped translational system1
2.2 Free Vibration of an Undamped Translational System

where

If mass m is displaced a distance when acted upon by a resultant force in the same direction,

If mass m is constant, this equation reduces to

is the acceleration of the mass.

2 2 free vibration of an undamped translational system2
2.2 Free Vibration of an Undamped Translational System

where

is the acceleration of the mass.

For a rigid body undergoing rotational motion, Newton’s Law gives

where is the resultant moment acting on the body and and are the resulting angular displacement and angular acceleration, respectively.

2 2 free vibration of an undamped translational system3
2.2 Free Vibration of an Undamped Translational System

For undamped single degree of freedom system, the application of Eq. (2.1) to mass m yields the equation of motion:

2 2 free vibration of an undamped translational system4
2.2 Free Vibration of an Undamped Translational System
  • Equation of Motion Using Other Methods:
  • D’Alembert’s Principle.The equations of motion, Eqs. (2.1) & (2.2) can be rewritten as

The application of D’Alembert’s principle to the system shown in Fig.(c) yields the equation of motion:

2 2 free vibration of an undamped translational system5
2.2 Free Vibration of an Undamped Translational System
  • Principle of Virtual Displacements.“If a system that is in equilibrium under the action of a set of forces is subjected to a virtual displacement, then the total virtual work done by the forces will be zero.”

Consider spring-mass system as shown in figure, the virtual work done by each force can be computed as:

2 2 free vibration of an undamped translational system6
2.2 Free Vibration of an Undamped Translational System

When the total virtual work done by all the forces is set equal to zero, we obtain

Since the virtual displacement can have an arbitrary value, , Eq.(2.5) gives the equation of motion of the spring-mass system as

2 2 free vibration of an undamped translational system7
2.2 Free Vibration of an Undamped Translational System
  • Principle of Conservation of Energy.A system is said to be conservative if no energy is lost due to friction or energy-dissipating nonelastic members.

If no work is done on the conservative system by external forces, the total energy of the system remains constant. Thus the principle of conservation of energy can be expressed as:

or

2 2 free vibration of an undamped translational system8
2.2 Free Vibration of an Undamped Translational System

The kinetic and potential energies are given by:

or

Substitution of Eqs. (2.7) & (2.8) into Eq. (2.6) yields the desired equation

2 2 free vibration of an undamped translational system9
2.2 Free Vibration of an Undamped Translational System
  • Equation of Motion of a Spring-Mass System in Vertical Position:
2 2 free vibration of an undamped translational system10
2.2 Free Vibration of an Undamped Translational System

For static equilibrium,

where W = weight of mass m,= static deflectiong = acceleration due to gravity

The application of Newton’s second law of motion to mass m gives

and since , we obtain

2 2 free vibration of an undamped translational system11
2.2 Free Vibration of an Undamped Translational System

Notice that Eqs. (2.3) and (2.10) are identical.

This indicates that when a mass moves in a vertical direction, we can ignore its weight, provided we measure x from its static equilibrium position.

slide20

2.2 Free Vibration of an Undamped Translational System

The solution of Eq. (2.3) can be found by assuming

Where C and s are constants to be determined. Substitution of Eq. (2.11) into Eq. (2.3) gives

Since C ≠ 0, we have

slide21

2.2 Free Vibration of an Undamped Translational System

And hence,

Roots of characteristic equation or known as eigenvalues of the problem.

1/2

Where i= (-1) and

2 2 free vibration of an undamped translational system12
2.2 Free Vibration of an Undamped Translational System

Hence, the general solution of Eq. (2.3) can be expressed as

where C1 and C2 are constants. By using the identities

where A1 and A2 are new constants.

2 2 free vibration of an undamped translational system13
2.2 Free Vibration of an Undamped Translational System

Hence, . Thus the solution of Eq. (2.3) subject to the initial conditions of Eq. (2.17) is given by

2 2 free vibration of an undamped translational system14
2.2 Free Vibration of an Undamped Translational System
  • Harmonic Motion:

Eqs.(2.15),(2.16) & (2.18) are harmonic functions of time. Eq. (2.16) can also be expressed as:

where A0 and are new constants, amplitude and phase angle respectively:

and

2 2 free vibration of an undamped translational system15
2.2 Free Vibration of an Undamped Translational System

Note the following aspects of spring-mass systems:

1) Circular natural frequency:

Spring constant, k:

Hence,

2 2 free vibration of an undamped translational system16
2.2 Free Vibration of an Undamped Translational System

Hence, natural frequency in cycles per second:

and, the natural period:

2 2 free vibration of an undamped translational system17
2.2 Free Vibration of an Undamped Translational System

2) Velocity and the acceleration of the mass m at time t can be obtained as:

3) If initial displacement is zero,

If initial velocity is zero,

2 2 free vibration of an undamped translational system18
2.2 Free Vibration of an Undamped Translational System

4) The response of a single degree of freedom system can be represented in the state space or phase plane:

Where,

By squaring and adding Eqs. (2.34) & (2.35)

2 2 free vibration of an undamped translational system19
2.2 Free Vibration of an Undamped Translational System

Phase plane representation of an undamped system

example 2 2 free vibration response due to impact
Example 2.2 Free Vibration Response Due to Impact

A cantilever beam carries a mass M at the free end as shown in the figure. A mass m falls from a height h on to the mass M and adheres to it without rebounding. Determine the resulting transverse vibration of the beam.

example 2 2 solution
Example 2.2 Solution

Using the principle of conservation of momentum:

or

The initial conditions of the problem can be stated:

Thus the resulting free transverse vibration of the beam can be expressed as:

example 2 5 natural frequency of pulley system
Example 2.5 Natural Frequency of Pulley System

Determine the natural frequency of the system shown in the figure. Assume the pulleys to be frictionless and of negligible mass.

example 2 5 solution
Example 2.5 Solution

The total movement of the mass m (point O) is:

The equivalent spring constant of the system:

example 2 5 solution1
Example 2.5 Solution

By displacing mass m from the static equilibrium position by x, the equation of motion of the mass can be written as

Hence, the natural frequency is given by:

2 3 free vibration of an undamped torsinal system
2.3 Free Vibration of an UndampedTorsinal System

From the theory of torsion of circular shafts, we have the relation:

Shear modulus

Polar moment of inertia of cross section of shaft

Torque

Length shaft

2 3 free vibration of an undamped torsional system
2.3 Free Vibration of an UndampedTorsionalSystem

Polar Moment of Inertia:

Torsional Spring Constant:

2 3 free vibration of an undamped torsional system1
2.3 Free Vibration of an UndampedTorsionalSystem
  • Equation of Motion:

Applying Newton’s Second Law of Motion,

Thus, the natural circular frequency:

The period and frequency of vibration in cycles per second are:

2 3 free vibration of an undamped torsional system2
2.3 Free Vibration of an UndampedTorsionalSystem
  • Note the following aspects of this system:
  • If the cross section of the shaft supporting the disc is not circular, an appropriate torsional spring constant is to be used.
  • The polar mass moment of inertia of a disc is given by:
  • An important application: in a mechanical clock

where ρ is the mass density h is the thickness D is the diameter W is the weight of the disc

2 3 free vibration of an undamped torsional system3
2.3 Free Vibration of an UndampedTorsionalSystem
  • General solution of Eq. (2.40) can be obtained:

where ωn is given by Eq. (2.41) and A1 and A2can be determined from the initial conditions. If

The constants A1 and A2 can be found:

Eq. (2.44) can also represent a simple harmonic motion.

example 2 6 natural frequency of compound pendulum
Example 2.6 Natural Frequency of Compound Pendulum

Any rigid body pivoted at a point other than its center of mass will oscillate about the pivot point under its own gravitational force. Such a system is known as a compound pendulum (shown in Figure). Find the natural frequency of such a system.

example 2 6 solution
Example 2.6 Solution

For a displacement θ, the restoring torque (due to the weight of the body W ) is (Wd sin θ) and the equation of motion is

Hence, approximated by linear equation:

The natural frequency of the compound pendulum:

example 2 6 solution1
Example 2.6 Solution

Comparing with natural frequency, the length of equivalent simple pendulum:

If J0 is replaced by mk02, where k0is the radius of gyration of the body about O,

example 2 6 solution2
Example 2.6 Solution

If kG denotes the radius of gyration of the body about G, we have:

and

If the line OG is extended to point A such that

Eq.(E.8) becomes

example 2 6 solution3
Example 2.6 Solution

Hence, from Eq.(E.5), ωn is given by

This equation shows that, no matter whether the body is pivoted from O or A, its natural frequency is the same. The point A is called the center of percussion.

text book problems

Text book Problems

Let’s try!!!

problem 2 64
Problem 2.64
  • A simple pendulum is set into oscillation from its rest position by giving it an angular velocity of 1 rad/s. It is found to oscillate with an amplitude of 0.5 rad. Find the natural frequency and length of the pendulum.

ERT 452 SESION 2011/2012

problem 2 66
Problem 2.66
  • Derive an expression for the natural frequency of the simple pendulum shown in Fig. 1.10. Determine the period of oscillation of a simple pendulum having a mass, m = 5 kg and a length l = 0.5 m.

ERT 452 SESION 2011/2012

problem 2 77
Problem 2.77
  • A uniform circular disc is pivoted at point O, as shown in Fig. 2.99. Find the natural frequency of the system. Also find the maximum frequency of the system by varying the value of b.

ERT 452 SESION 2011/2012

problem 2 78
Problem 2.78
  • Derive the equation of motion of the system shown in Fig. 2.100, by using Newton’s second law of motion method.

ERT 452 SESION 2011/2012