Lecture 2 Free Vibration of Single Degree of Freedom Systems

Lecture 2Free Vibration of Single Degree of Freedom SystemsERT 452VIBRATION

MUNIRA MOHAMED NAZARISCHOOL OF BIOPROCESSUNIVERSITI MALAYSIA PERLIS1ERT 452 SESION 2011/2012CO 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 2011/20122COURSE OUTCOME2.1 Introduction2.2 Free Vibration of an Undamped Translational System2.3 Free Vibration of an Undamped Torsional System

COURSE OUTLINE

3ERT 452 SESION 2011/2012Free 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 2011/201242.1IntroductionA spring-mass system in horizontal position.ERT 452 SESION 2011/201252.1Introduction

Several mechanical and structural systems can be idealized as single degree of freedom systems, for example, the mass and stiffness of a system.

ERT 452 SESION 2011/201262.1Introduction

Equivalent spring-mass system for the cam follower system.Modeling of tall structure as spring-mass system.

ERT 452 SESION 2011/201272.1Introduction

ProcedureSelect 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 Newtons second law of motion.ERT 452 SESION 2011/201282.2Free Vibration of an Undamped Translational SystemEquation of Motion Using Newtons Second Law of Motion:92.2Free Vibration of an Undamped Translational SystemIf 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

where

is the acceleration of the mass.102.2Free Vibration of an Undamped Translational Systemwhere

is the acceleration of the mass.For a rigid body undergoing rotational motion, Newtons Law gives

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

112.2Free Vibration of an Undamped Translational SystemFor undamped single degree of freedom system, the application of Eq. (2.1) to mass m yields the equation of motion:

122.2Free Vibration of an Undamped Translational SystemEquation of Motion Using Other Methods:

The application of DAlemberts principle to the system shown in Fig.(c) yields the equation of motion:

DAlemberts Principle.The equations of motion, Eqs. (2.1) & (2.2) can be rewritten as13Principle 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.2.2Free Vibration of an Undamped Translational System

Consider spring-mass system as shown in figure, the virtual work done by each force can be computed as:14Since the virtual displacement can have an arbitrary value, , Eq.(2.5) gives the equation of motion of the spring-mass system asWhen the total virtual work done by all the forces is set equal to zero, we obtain2.2Free Vibration of an Undamped Translational System

15Principle of Conservation of Energy.A system is said to be conservative if no energy is lost due to friction or energy-dissipating nonelastic members.2.2Free Vibration of an Undamped Translational SystemIf 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:

or162.2Free Vibration of an Undamped Translational SystemThe kinetic and potential energies are given by:

or

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

172.2Free Vibration of an Undamped Translational SystemEquation of Motion of a Spring-Mass System in Vertical Position:

182.2Free Vibration of an Undamped Translational SystemFor static equilibrium,

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

The application of Newtons second law of motion to mass m gives

and since , we obtain

19This indicates that when a mass moves in a vertical direction, we can ignore its weight, provided we measure x from its static equilibrium position.2.2Free Vibration of an Undamped Translational SystemNotice that Eqs. (2.3) and (2.10) are identical.202.2Free Vibration of an Undamped Translational SystemThe 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

212.2Free Vibration of an Undamped Translational SystemAnd hence,

Where i = (-1) and1/2

Roots of characteristic equation or known as eigenvalues of the problem.222.2Free Vibration of an Undamped Translational System

Hence, the general solution of Eq. (2.3) can be expressed aswhere C1 and C2 are constants. By using the identities

where A1 and A2 are new constants.

232.2Free Vibration of an Undamped Translational SystemHence, . Thus the solution of Eq. (2.3) subject to the initial conditions of Eq. (2.17) is given by

242.2Free Vibration of an Undamped Translational SystemHarmonic Motion:

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

and

Eqs.(2.15),(2.16) & (2.18) are harmonic functions of time. Eq. (2.16) can also be expressed as:251) Circular natural frequency:2.2Free Vibration of an Undamped Translational SystemNote the following aspects of spring-mass systems:

Spring constant, k:

Hence,

262.2Free Vibration of an Undamped Translational System

Hence, natural frequency in cycles per second:and, the natural period:272) Velocity and the acceleration of the mass m at time t can be obtained as:2.2Free Vibration of an Undamped Translational System

3) If initial displacement is zero,

If initial velocity is zero,

284) The response of a single degree of freedom system can be represented in the state space or phase plane:2.2Free Vibration of an Undamped Translational System

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

292.2Free Vibration of an Undamped Translational System

Phase plane representation of an undamped system30Example 2.2Free Vibration Response Due to ImpactA 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.

31The initial conditions of the problem can be stated:Example 2.2 Solution

Using the principle of conservation of momentum:orThus the resulting free transverse vibration of the beam can be expressed as:

32Example 2.2 Solutionwhere

33Example 2.5 Natural Frequency of Pulley SystemDetermine the natural frequency of the system shown in the figure. Assume the pulleys to be frictionless and of negligible mass.

34Example 2.5 Solution

The total movement of the mass m (point O) is:The equivalent spring constant of the system:35Example 2.5 Solution

By displacing mass m from the static equilibrium position by x, the equation of motion of the mass can be written asHence, the natural frequency is given by:

Free Vibration of an Undamped Torsinal System36

372.3Free Vibration of an Undamped Torsinal SystemFrom the theory of torsion of circular shafts, we have the relation:

Shear modulusPolar moment of inertia of cross section of shaftLength shaftTorque

382.3Free Vibration of an Undamped Torsional System

Polar Moment of Inertia:Torsional Spring Constant:392.3Free Vibration of an Undamped Torsional SystemEquation of Motion:

Applying Newtons Second Law of Motion,Thus, the natural circular frequency:

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

402.3Free Vibration of an Undamped Torsional SystemNote 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 is the mass density h is the thickness D is the diameter W is the weight of the disc41where n is given by Eq. (2.41) and A1 and A2 can be determined from the initial conditions. If2.3Free Vibration of an Undamped Torsional SystemGeneral solution of Eq. (2.40) can be obtained:

The constants A1 and A2 can be found:

Eq. (2.44) can also represent a simple harmonic motion.42Example 2.6Natural Frequency of Compound PendulumAny 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.

43For a displacement , the restoring torque (due to the weight of the body W ) is (Wd sin ) and the equation of motion isExample 2.6 Solution

Hence, approximated by linear equation:The natural frequency of the compound pendulum:

44Comparing with natural frequency, the length of equivalent simple pendulum:Example 2.6 Solution

If J0 is replaced by mk02, where k0 is the radius of gyration of the body about O,

45If kG denotes the radius of gyration of the body about G, we have:Example 2.6 Solution

If the line OG is extended to point A such thatandEq.(E.8) becomes

46Hence, from Eq.(E.5), n is given byExample 2.6 Solution

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 ProblemsLets try!!!47A 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 2011/201248Problem 2.64Derive 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 2011/201249Problem 2.66A 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 2011/201250Problem 2.77Derive the equation of motion of the system shown in Fig. 2.100, by using Newtons second law of motion method.ERT 452 SESION 2011/201251Problem 2.7852