• Author / Uploaded
• Aravind Drogba

• Categories
• Decibel
• Fuerza
• Amplitud
• Funciones trigonométricas
• Aceleración

Citation preview

MEMB343

Mechanical Vibrations

FUNDAMENTALS OF VIBRATIONS BASIC CONCEPTS

MEMB343

Mechanical Vibrations

LEARNING OBJECTIVES  Upon completion of this lecture, you should be able to:  Understand the concepts of degree-of-freedom, and discrete and continuous systems.  Compute the stiffness of some simple spring elements.  Determine the equivalent mass or inertia, and equivalent spring and damping constants of vibrating systems.  Understand the definitions of free and forced vibration, undamped and damped vibration, linear and non-linear vibration, and deterministic and random vibration.

MEMB343

Mechanical Vibrations

VIBRATION IN ENGINEERING PRACTICE Good  Machine condition monitoring.  Vibrating sieves, mixers and tools.  Electric massaging units, dentist toothbrushes.

drills,

Bad  Noise, vibration and harshness (NVH).  Machinery and structural failures.  Motion sickness, white finger syndrome, etc.

electric

MEMB343

Mechanical Vibrations

TACOMA NARROWS BRIDGE This photograph shows the twisting motion of the center span just prior to failure.

The nature and severity of the torsional movement is revealed in this picture taken from the Tacoma end of the suspension span. When the twisting motion was at the maximum, elevation of the sidewalk at the right was 28 feet (8.5m) higher than the sidewalk at the left.

MEMB343

Mechanical Vibrations

TACOMA NARROWS BRIDGE (cont.)

MEMB343

Mechanical Vibrations

POSITIVE DISPLACEMENT BLOWER

Shearing of shaft attributed to high torsional vibrations. Rotor operating speed was within the vicinity of a torsional natural frequency.

MEMB343

Mechanical Vibrations

POSITIVE DISPLACEMENT BLOWER (cont.)

MEMB343

EXAMPLES OF VIBRATING SYSTEM

 Masses attached to springs.  Flexible Rods.  Pendulums.

Mechanical Vibrations

MEMB343

Mechanical Vibrations

BASIC CONCEPTS OF VIBRATION  When a particle or a rigid body in stable equilibrium is displaced by the application of an additional force, mechanical vibration will result.  Some important concepts in mechanical vibration theory can be categorized into the followings:  Elementary parts of vibrating systems.  Degree of freedom.  Discrete and continuous system.

MEMB343

Mechanical Vibrations

ELEMENTARY PARTS OF VIBRATING SYSTEMS  To be subject to vibration, a system must be able to store energy in two different forms and allow energy to be transferred from one to the other.  In particular, for vibration to exist, there must be a transfer of energy from potential to kinetic and vice-versa.  Potential energy is due to either gravity or the elasticity of the system, whilst the kinetic energy is due to the motion of the mass.  The simplest mechanical oscillators are the pendulum and the spring-mass system. The corresponding simplest electrical oscillator is the capacitor-inductor system.

MEMB343

Mechanical Vibrations

SPRING ELEMENTS Linear Spring

FS  kx Torsional Spring

M S  kT  A spring is a mechanical link that is generally assumed to have negligible mass and damping.  A force is developed in a spring whenever there is a relative motion between two ends of the spring.  Work done in deforming a spring is stored as potential energy in the spring.

MEMB343

Mechanical Vibrations

SPRING ELEMENTS (cont.)  A spring element is generally made of an elastic material.  The stiffness in a spring element can be related more directly to its material (elastic modulus) and geometric properties.  A spring-like behavior results from a variety of motion configurations, including:  Longitudinal motion (vibration in the direction of the length).  Transverse motion (vibration perpendicular to the length).  Torsional motion (vibration rotating around the length).

MEMB343

Mechanical Vibrations

STIFFNESS OF SPRING ELEMENTS

 Stiffness associated with the longitudinal vibration of a slender prismatic bar.

MEMB343

Mechanical Vibrations

STIFFNESS OF SPRING ELEMENTS (cont.)

 Stiffness associated with the torsional vibration of a shaft.

MEMB343

Mechanical Vibrations

STIFFNESS OF SPRING ELEMENTS (cont.)

 Stiffness associated with a helical spring.

MEMB343

Mechanical Vibrations

STIFFNESS OF SPRING ELEMENTS (cont.)

 Beam stiffness associated with the transverse vibration of the tip of a beam.

MEMB343

Mechanical Vibrations

EQUIVALENT SPRING CONSTANT Springs in Parallel

Springs in Series

keq  k1  k2  ....  kn

1 1 1 1    ....  keq k1 k2 kn

MEMB343

Mechanical Vibrations

MASS OR INERTIA ELEMENTS Linear Motion

 F mx

Rotational Motion

 M  I  

 The mass or inertia element is assumed to be a rigid body.  A rigid body’s inertia is responsible for the resistance to acceleration of a system.  Work done on a mass is stored in the form of kinetic energy of the mass.

MEMB343

Mechanical Vibrations

EQUIVALENT MASS OF A SYSTEM Translational Masses Connected by a Rigid Bar. Original system

x1

Pivot point m1

m2

l1 l2 Equivalent system

l3

xeq meq

l1

x3

x 2 m3

From trigonometric relationship l2 x2  xeq l1

l3 x3  xeq l1

Assume

xeq  x1

MEMB343

Mechanical Vibrations

EQUIVALENT MASS OF A SYSTEM (cont.) Translational Masses Connected by a Rigid Bar. Equate the kinetic energy of the three masses to that of the equivalent system’s mass

1 1 1 1 2 2 2 2     m1 x1  m2 x2  m3 x3  meq xeq 2 2 2 2 The equivalent system’s mass is therefore obtained 2

2

 l2   l3  meq  m1    m2    m3  l1   l1 

MEMB343

Mechanical Vibrations

EQUIVALENT MASS OF A SYSTEM (cont.) Coupled Translational and Rotational Masses.

J0

From kinematics, the relationship between the linear and the angular velocity is x

 

r

m

r

To obtain the equivalent translational mass x  x eq

x

To obtain the equivalent rotational mass    eq

MEMB343

Mechanical Vibrations

EQUIVALENT MASS OF A SYSTEM (cont.) Coupled Translational and Rotational Masses. Equivalent Translational Mass 2

1 2 1  x  1 2  mx  J 0    meq xeq 2 2 r 2

J0 meq  m  2 r

Equivalent Rotational Mass

 

1  2 1 2 1 2 m r  J 0  J eqeq 2 2 2

J eq  mr 2  J 0

MEMB343

Mechanical Vibrations

DAMPING ELEMENTS Linear Damper

FD  cx Torsional Damper

M D  cT  A damper is generally assumed to have negligible mass and stiffness.  A force is developed in a damper whenever there is a relative velocity between two ends of the damper.  The damper dissipates energy from a system in the form of heat or sound.

MEMB343

Mechanical Vibrations

DAMPING ELEMENTS MODELS  Viscous Damping The damping force is proportional to the velocity of the vibrating body.

 Coulomb Damping The damping force is constant in amplitude but opposite the direction to that of the motion of the vibrating body.

 Hysteretic Damping The energy dissipated per cycle is proportional to the square of the vibration amplitude.

MEMB343

Mechanical Vibrations

EQUIVALENT DAMPING CONSTANT Dampers in Parallel

Dampers in Series

ceq  c1  c2  ....  cn

1 1 1 1    ....  ceq c1 c2 cn

MEMB343

Mechanical Vibrations

DEGREE OF FREEDOM  The minimum number of independent coordinates required to determine completely the positions of all part of a system at any instant of times.

y y

x z

z

x Unconstrained rigid body with 6 d.o.f.

MEMB343

Mechanical Vibrations

DISCRETE AND CONTINUOUS SYSTEM  Systems with a finite number of degrees of freedom are called discrete or lumped parameter systems, and those with an infinite number of degrees of freedom are called continuous or distributed systems.

Discrete System

Solution: 2nd Order Ordinary Differential Equation

Continuous System

Solution: Partial Differential Equation

MEMB343

Mechanical Vibrations

CLASSIFICATION OF VIBRATION

   

Free and Forced Vibration. Undamped and Damped Vibration. Linear and Nonlinear Vibration. Deterministic and Random Vibration.

MEMB343

Mechanical Vibrations

FREE AND FORCED VIBRATION  Free Vibration Oscillation occurring at a natural frequency, after an initial force input.

mx  kx  0 mx  cx  kx  0

 Forced Vibration Oscillation occurring at the frequency of a driving force input.

mx  kx  F (t ) mx  cx  kx  F (t )

MEMB343

Mechanical Vibrations

UNDAMPED AND DAMPED VIBRATION  Undamped Vibration No energy is lost or dissipated in friction or other resistance during oscillation.

mx  kx  0 mx  kx  F (t )

 Damped Vibration Energy is lost or dissipated during oscillation.

mx  cx  kx  0 mx  cx  kx  F (t )

MEMB343

Mechanical Vibrations

LINEAR AND NONLINEAR VIBRATION  Linear Vibration The cause (force) and effect (response) are proportionally related. Principle of superposition holds.

 Nonlinear Vibration Relationship between cause and effect is no longer proportional.

MEMB343

Mechanical Vibrations

DETERMINISTIC AND RANDOM VIBRATION

 Deterministic Vibration The instantaneous values of the vibration amplitude at any time (t) can be determined from mathematical expressions.

 Random Vibration Future instantaneous values of the vibration amplitude cannot be predicted in a deterministic sense.

MEMB343

Mechanical Vibrations

EXAMPLE 1  A composite propeller shaft, which is made of steel and aluminium, is shown below. Determine the torsional spring constant of the shaft The shear modulus G of steel is 80 GPa and for aluminium 26 GPa.

MEMB343

Mechanical Vibrations

EXAMPLE 1 (cont.)  The polar area moment of inertia for a hollow shaft is given by the following equation where D and d are the outer and inner diameters, respectively.

IP 

 D 4  d 4  32

 Torsional stiffness for the shaft is:

GI P G D  d kt   l 32l 4

4



MEMB343

Mechanical Vibrations

EXAMPLE 1 (cont.)  For the steel shaft, torsional stiffness is:

ktS 

  80 109 0.254  0.154  32  5

 5.34 106 Nm/rad

 For the aluminium shaft, torsional stiffness is:

ktA 

  26 10 0.15  0.1 9

4

4

32  5

  0.207 10

6

Nm/rad

 Shafts are in parallel, therefore:

keq  ktS  ktA  5.547 106 Nm/rad

MEMB343

Mechanical Vibrations

FUNDAMENTALS OF VIBRATIONS SIMPLE HARMONIC MOTION

MEMB343

Mechanical Vibrations

LEARNING OBJECTIVES  Upon completion of this lecture, you should be able to:  Determine if a vibration motion can be classified as simple harmonic.  Represent simple harmonic motion in both trigonometric and complex forms.  Understand some basic terminology that are used to define a vibration signal.

MEMB343

Mechanical Vibrations

INTRODUCTION  Oscillatory motion may repeat itself regularly as in the case of a simple pendulum, or it may display considerable irregularity, as in earthquakes.  When the motion is repeated in equal intervals of time  , it is called periodic motion.  The repetition time  is called the period of oscillation, and its reciprocal 1 is called the frequency. f    If the motion is designated by the time function xt  , then any periodic motion must satisfy the relationship: xt   xt   

MEMB343

Mechanical Vibrations

SIMPLE HARMONIC MOTION

 The simplest type of periodic motion is called a simple harmonic motion.  It is essentially a periodic motion with a single frequency.  It can be demonstrated by a mass suspended from a light spring.  If the mass is displaced from its rest position and released, it will oscillate up and down.

MEMB343

Mechanical Vibrations

SIMPLE HARMONIC MOTION (cont.)

 The motion of the mass can be expressed by the equation t  x  A sin 2     A is the amplitude of oscillation, measured from the equilibrium position of the mass.   is the period, and the motion is repeated when t   .

MEMB343

Mechanical Vibrations

SIMPLE HARMONIC MOTION (cont.)

 Harmonic motion is often represented as the projection on a straight line of a point that is moving on a circle at constant speed.  With the angular speed of the line OP designated by  , the displacement x can be written as x  A sint  .

MEMB343

Mechanical Vibrations

SIMPLE HARMONIC MOTION (cont.)  The quantity  is measured in radians per second, and is referred to as the circular or angular frequency.  Since the motion repeat itself every 2 radians, we have the relationship . 2   2f    and f are the period and frequency of the harmonic motion, usually measured in seconds and cycles per second (Hz), respectively.  The velocity and acceleration of harmonic motion can be simply determined by differentiation of x  A sint  . x  A cos t  A sin t  2  x   2 A sin t   2 A sin t   

MEMB343

Mechanical Vibrations

SIMPLE HARMONIC MOTION (cont.)

 Thus the velocity and acceleration are also harmonic with the same frequency of oscillation but lead the displacement by  / 2 and  radians, respectively.

MEMB343

Mechanical Vibrations

SIMPLE HARMONIC MOTION (cont.) 2  The acceleration can be written as x   x .  Therefore, in simple harmonic motion, the acceleration is proportional to the displacement and is directed towards the origin.  As Newton’s 2nd law states that the acceleration is proportional to the force, harmonic motion can be expected with linear springs with force varying as kx .

MEMB343

COMPLEX FORM HARMONIC MOTION

Mechanical Vibrations

OF

SIMPLE

 The trigonometric functions of sine and cosine are related to the exponential function by Euler’s equation: ei  cos   i sin   A vector of amplitude z rotating at constant angular speed  can be represented as a complex quantity A in the Argand diagram.

MEMB343

Mechanical Vibrations

COMPLEX FORM OF SIMPLE HARMONIC MOTION (cont.)

 The following relationship holds: z  Ae it  A cos t  iA sin t  x  iy  The quantity z is referred to as the complex sinusoid with x and y as the real and imaginary components.

MEMB343

Mechanical Vibrations

COMPLEX FORM OF SIMPLE HARMONIC MOTION (cont.)

 The above figure shows z and its conjugate z * which is rotating in the negative direction with angular speed   .  From this diagram, it is seen that the real component x is expressed in terms of z and z * by the equation: x  12 z  z *  A cos t  Re Ae it  Re stands for the real part of the quantity z .

MEMB343

Mechanical Vibrations

COMPLEX FORM OF SIMPLE HARMONIC MOTION (cont.)  Some of the rules of exponential operations between z1  A1ei1 and z2  A2ei 2 are:  Multiplication

z1 z2  A1 A2ei 1  2 

 Division

z1  A1  i 1  2    e z2  A2 

 Powers

z n  An ein 1 n

1 n

z A e

i n

MEMB343

Mechanical Vibrations

VIBRATION TERMINOLOGY

 Since the position, velocity and acceleration change continually with time, several other quantities are used to discuss vibration.  The peak value, defined as the maximum displacement, or magnitude A , usually indicates the maximum stress that the vibrating part is undergoing.

MEMB343

Mechanical Vibrations

VIBRATION TERMINOLOGY (cont.)  Another quantity useful in describing vibration is the average value, denoted as x , and defined by:

1 T x  lim  xt dt T  T 0  The average value indicates a steady or static value somewhat like the DC level of an electrical current.  For example, the average value for a complete cycle of a sine wave Asin t  is zero.  Its average value for a half-cycle is, however:

x

A





0

sin t dt 

2A



 0.637 A

MEMB343

Mechanical Vibrations

VIBRATION TERMINOLOGY (cont.)  Since the square of displacement is associated with a system’s potential energy, the average of the displacement squared is sometimes a useful vibration property to discuss.  The mean square value of a time function xt  , denoted 2 by x is found from the average of the squared values, integrated over some time interval T : 1 T 2 x  lim  x t dt T  T 0 2

 If xt   A sin t , its mean square value is: A2 x  lim T  T 2

T



0

1 1 2 1  cos 2t dt  A 2 2

MEMB343

Mechanical Vibrations

VIBRATION TERMINOLOGY (cont.)  The root mean square (rms) value is the square root of the mean square value.  From the previous example for xt   A sin t , the rms of the sine wave of amplitude A is:

Arms

A   0.707 A 2

 Vibration instrumentation generally measures root mean square vibration amplitudes (displacement, velocity or acceleration).

MEMB343

Mechanical Vibrations

VIBRATION TERMINOLOGY (cont.)  Since the peak value of the velocity and acceleration are multiples of the circular frequency times the displacement amplitude, these three basic quantities often differ in value by an order of magnitude.  For systems with circular frequency  larger than 1 rad/s, the relative amplitude of the velocity response is larger than that of the displacement by a multiple of  , and the acceleration response is larger by a multiple of  2 .  For systems with circular frequency less than 1 rad/s, the velocity and acceleration have smaller relative amplitudes than the displacement.

MEMB343

Mechanical Vibrations

VIBRATION TERMINOLOGY (cont.)  A common unit of measurement for vibration amplitudes and rms values is the decibel (dB). As the decibel is a logarithmic unit, it compresses or expands the scale.  The decibel was originally defined in terms of the base 10 logarithm of the power ratio of two electrical signals, or as the ratio of the square of the amplitudes of two signals. 2

 P1   x1   x1  dB  10 log10    10 log10    20 log10    P2   x2   x2 

 The equation based on the ratio of the square of the amplitudes of two signals results from the fact that power is proportional to the square of the amplitude or voltage.

MEMB343

Mechanical Vibrations

VIBRATION TERMINOLOGY (cont.)

 Cycle is the movement of a vibrating body from its equilibrium position to its extreme position in one direction, then to the equilibrium position, then to its extreme position in other direction, and back to equilibrium position. 1 revolution (angular displacement of 2 radians) of the pin P or one revolution of the vector OP constitute a cycle.

MEMB343

Mechanical Vibrations

VIBRATION TERMINOLOGY (cont.)  Amplitude, denoted by A , is defined as the maximum displacement of a vibrating body from its equilibrium position.  Period of oscillation, denoted by  , is defined as the time (second) taken to complete one cycle of motion.  It is equal to the time required for the vector OP to rotate through an angle of 2 and therefore: 2    Frequency of oscillation, denoted by f , is defined as the number of cycles per unit time. f is measured in cycles/second (Hz). 1  f    2

MEMB343

Mechanical Vibrations

VIBRATION TERMINOLOGY (cont.)

 Phase angle (  ) is the difference in time between two events such as the zero crossing of two waveforms.   is expressed in radians as the time between two events divided by the period, times 2. t   2 

MEMB343

Mechanical Vibrations

EXAMPLE 1  The maximum displacement and the maximum acceleration of the foundation of a centrifugal pump were found to be 0.25 mm and 0.4g. Find the operating speed of the pump. Assume the motion of the foundation is harmonic.

MEMB343

Mechanical Vibrations

EXAMPLE 1 (cont.)  For harmonic motion, the following relationship between displacement and acceleration is valid:

xmax    2 xmax  From the above equation, we solve for the angular frequency:



xmax 0.4  9.81   125.28 rad/s 3 xmax 0.25 10

MEMB343

Mechanical Vibrations

EXAMPLE 1 (cont.)  The operating speed of the pump is related to the angular frequency by the following equation:

60 60 125.28 N   1196.3 rpm 2 2

MEMB343

Mechanical Vibrations

FREE VIBRATION OF SINGLEDEGREE-OF-FREEDOM SYSTEMS UNDAMPED FREE VIBRATIONS

MEMB343

Mechanical Vibrations

LEARNING OBJECTIVES  Upon completion of this lecture, you should be able to:  Determine the equation of motion of an undamped singledegree-of-freedom system.  Determine the undamped natural frequency of a singledegree-of-freedom system.  Solve the equation of motion of the undamped free response.  Understand the relationship between various solutions to the equation of motion.

MEMB343

Mechanical Vibrations

INTRODUCTION  The vibration analysis procedure generally involves:    

Mathematical modeling. Derivation of governing equations. Solution of the governing equations. Interpretation of the results.

 This lecture deals with the undamped free vibration response of a single-degree-of-freedom system.

MEMB343

Mechanical Vibrations

UNDAMPED FREE VIBRATIONS  A free vibration is produced and maintained by forces such as elastic and gravitational forces.  These forces depend only on the position and motion of the body.  When forces that oppose the restoring force (friction, air resistance, etc.) are negligible, the vibration is called undamped.  An undamped free vibration will repeat itself indefinitely.  In real systems, the frequency and period of vibration obtained for a freely vibrating system are very close to the values obtained for a system that has a small amount of damping.

MEMB343

Mechanical Vibrations

UNDAMPED FREE VIBRATIONS (cont.)

 Consider a block of mass m sliding on a frictionless horizontal surface.  Vibration is induced by displacing the block a distance x0 and then releasing it with an initial velocity of x0  v0 .

MEMB343

Mechanical Vibrations

UNDAMPED FREE VIBRATIONS (cont.)  The elastic restoring force of the spring Fs  kx is always directed toward the equilibrium position, whereas the acceleration a  d 2 x / dt 2  x acts in the direction of positive displacement.  It is important to remember that since acceleration is the second time derivative of displacement, both the displacement x and the acceleration x must be measured positive in the same direction.  Applying Newton’s 2nd law to the block gives the differential equation of motion of the block:  kx  mx or mx  kx  0

MEMB343

Mechanical Vibrations

UNDAMPED FREE VIBRATIONS (cont.)  The general solution for the previous equation based on physical observation is given by: xt   A sin nt     A is the amplitude, or maximum value of the function.   n , the angular natural frequency measured in rad/s, determines the interval in time that the function repeats itself.   , the phase measured in radians, determines the initial value of the sine function.  It is standard to measure time t in seconds.

MEMB343

Mechanical Vibrations

UNDAMPED FREE VIBRATIONS (cont.)  The velocity and acceleration of the block are obtained by differentiating xt   A sin  nt    with time.  The velocity of the block is

x t    n A cos nt    .

 The acceleration of the block is xt    n2 A sin nt    .  Substitution of these equations into the equation of motion given by mx  kx  0 yields:

 m 2 A sin nt     kAsin nt     0

MEMB343

Mechanical Vibrations

UNDAMPED FREE VIBRATIONS (cont.)

 Dividing the previous equation by A and m yields the fact that this equation is satisfied if 2 k . n  m  The constant  n characterizes the spring-mass system, as well as the frequency at which the motion repeats itself, and hence is called the system’s natural frequency.

MEMB343

Mechanical Vibrations

UNDAMPED FREE VIBRATIONS (cont.)  The constants A and  are determined by the initial stage of motion of the mass-spring system. If no energy is imparted to the mass, it will stay at rest.  If, however, the mass is displaced to a position x0 at time t  0 , the potential energy in the spring will result in motion. Also, if the mass is given an initial velocity of v0 at time t  0 , motion will result.  These are called initial conditions and when substituted into the previous equations for displacement and velocity yield: x0  x0  A sin n 0     A sin 

v0  x0   n A cos n 0      n A cos 

MEMB343

Mechanical Vibrations

UNDAMPED FREE VIBRATIONS (cont.)

 Solving these two equations simultaneously for the two unknowns A and  yields:

 n2 x02  v02 A n

  tan

1

 n x0 v0

MEMB343

Mechanical Vibrations

UNDAMPED FREE VIBRATIONS (cont.)

 Thus the solution of the equation of motion for the springmass system is given by:

 n2 x02  v02  1  n x0   xt   sin  nt  tan n v0  

MEMB343

Mechanical Vibrations

UNDAMPED FREE VIBRATIONS (cont.)  Since the solution repeats whenever their argument increases by an angle 2 , the period of the oscillation is given by: 2 n 

n

 The natural frequency of the oscillation in Hz (cycles per second) is then: 1 n fn    n 2  The results above may be used to analyze the vibrational motion of a particle or rigid-body whenever the equations of motion reduce to the form which characterizes simple harmonic motion given by x   2 x  0 . n

MEMB343

Mechanical Vibrations

ALTERNATIVE SOLUTIONS  Following the theory of elementary differential equations, we can assume the solution for the equation of motion as:

xt   aet  a and  are nonzero constants to be determined.  Upon successive differentiation of xt  , we obtain:

x t   aet xt   2 aet  Substituting the assumed exponential form into the equation of motion yields: m2 aet  kaet  0

MEMB343

Mechanical Vibrations

ALTERNATIVE SOLUTIONS (cont.)  Since the term ae t is never zero, the previous expression can be divided by ae t to yield:

m2  k  0  Solving this algebraically results in:

  

k k  j   n j m m

j   1 is the imaginary number and  n is the natural frequency.

MEMB343

Mechanical Vibrations

ALTERNATIVE SOLUTIONS (cont.)  Note that there are two values for  :    n j    n j  This results because the equation for  is of second order, which implies that there must be two solutions of the equation of motion. t  Substituting these two values of  into xt   ae yields: xt   ae  jnt xt   ae  jnt  The principal of superposition for linear systems states that the sum of two solutions is also a solution, hence: xt   a1e jnt  a2e jnt  a1 and a2 are complex-valued constants of integration.

MEMB343

Mechanical Vibrations

ALTERNATIVE SOLUTIONS (cont.)  Another alternative solution for the equation of motion is:

xt   A1 cos  nt  A2 sin  nt  A1 and A2 are real-valued constants of integration to be determined from the initial conditions: xt   x0 and xt   v0 when t  0  The velocity and acceleration of the block are obtained by differentiating xt  with time.

xt    n A1 sin  nt   n A2 cos  nt

xt    n2 A1 cos  nt   n2 A2 sin  nt

MEMB343

Mechanical Vibrations

SOLUTIONS RELATIONSHIP  The solution of mx  kx  0 subject to non-zero initial conditions can be written in three equivalent ways.  First, the solution can be written as follows where A and  are real-valued constants. xt   A sin nt     Second, the solution can be written as follows where a1 and a2 are complex-valued constants.

xt   a1e jnt  a2e jnt  Third, the solution can be written as follows where A1 and A2 are real-valued constants. xt   A1 cos  nt  A2 sin  nt

MEMB343

Mechanical Vibrations

SOLUTIONS RELATIONSHIP (cont.)  Each set of two constants is determined by the initial conditions.  Following trigonometric identities and the Euler’s formulas, the various constants are related by:

A A  A 2 1

2 2

A1  a1  a2 a1 

A1  A2 j 2

 A1    tan    A2  A2  a1  a2  j -1

a2 

A1  A2 j 2

MEMB343

Mechanical Vibrations

DISPLACED EQUILIBRIUM POSITION

 The deformation of the spring in the static equilibrium position is  , and the spring force k is equal to the gravitational force W acting on the mass m .

k  W  mg

MEMB343

Mechanical Vibrations

DISPLACED EQUILIBRIUM POSITION (cont.)  Measuring the displacement x from the static equilibrium position, the forces acting on m are k and W .  With x chosen to be positive in the downward direction, so are all quantities (force, velocity and acceleration).  Applying Newton’s 2nd law of motion to the mass m : mx   F  W  k   x   Since k  W , we obtain k x  x  0 m  The choice of the static equilibrium position as reference for x has eliminated W , the force due to gravity, and the static spring force k , from the equation of motion.

MEMB343

Mechanical Vibrations

DISPLACED EQUILIBRIUM POSITION (cont.)  The equation of motion may be written as x   n2 x  0 where  n2  k / m .  The natural period of oscillation and the natural frequency is respectively established from: m and 1 k  n  2

k

fn 

2

m

  n , f n and  n depend only on the mass and stiffness of the system, which are properties of the system.  For this case and any case where vibration is measured from gravity equilibrium position, no gravity terms have to be included in the equations of motion.

MEMB343

Mechanical Vibrations

EXAMPLE 1  A heavy ring of mass moment of inertia 1 kgm2 is attached at the end of a two-layered hollow shaft of length 2 m. If the two layers of the shaft are made of steel and brass, repectively, determine the natural time period of torsional vibration of the heavy ring. Shear modulus for steel and brass are 80 GPa and 40 GPa, respectively.

MEMB343

Mechanical Vibrations

EXAMPLE 1 (cont.)

 Torsional stiffness for a hollow shaft is:

GI P G D 4  d 4  kt   l 32l

MEMB343

Mechanical Vibrations

EXAMPLE 1 (cont.)  For the steel shaft, torsional stiffness is:

ktS 

  80 109 0.054  0.044  32  2

 14490.6 Nm/rad

 For the brass shaft, torsional stiffness is:

ktB 

  40 10 0.04  0.03 9

4

32  2

4

  3436.1 Nm/rad

 Shafts are in parallel, therefore:

keq  ktS  ktB  17926.7 Nm/rad

MEMB343

Mechanical Vibrations

EXAMPLE 1 (cont.)  The torsional natural frequency is:

n 

keq

17926.7   133.89 rad/s J0 1

 The torsional natural time period is:

2

2 n    0.047 s n 133.89

MEMB343

Mechanical Vibrations

FREE VIBRATION OF SINGLEDEGREE-OF-FREEDOM SYSTEMS ENERGY METHOD

MEMB343

Mechanical Vibrations

LEARNING OBJECTIVES  Upon completion of this lecture, you should be able to:  Derive the equation of motion of a single-degree-of-freedom system using the energy method.  Determine the natural frequency of a single-degree-offreedom system using the energy method.  Understand that the energy method is based on the principle of conservation of energy and therefore is only valid for undamped vibration systems.

MEMB343

Mechanical Vibrations

INTRODUCTION  Dynamic systems can be characterized in terms of one or more natural frequencies.  The natural frequency is the frequency at which the system would vibrate if it were given an initial disturbance and then allowed to vibrate freely.  There are many available methods for determining the natural frequency. The most utilized include:    

Newton’s Law of Motion Rayleigh’s Method Energy Method Lagrange’s Equation.

MEMB343

Mechanical Vibrations

ENERGY METHOD  The total energy of a conservative system (damping and external forces equal zero) is constant. Thus

d T U   0 dt

T:

kinetic energy

U:

potential energy

 Kinetic energy is the energy of motion, as calculated from the velocity.  Potential energy has several forms. One is strain energy. Another is the work done against a gravity field.

MEMB343

Mechanical Vibrations

RAYLEIGH’S METHOD  The principle of conservation of energy in the context of an undamped vibrating system can be restated as

T1  U1  T2  U 2

1,2 :

denotes two different instants of time

 Subscript 1 denotes the time when the mass passes through its static equilibrium position; therefore U1=0.  Subscript 2 denotes the time corresponding to the maximum displacement of the mass; therefore T2=0.  For a system undergoing harmonic motion T1 and U2 denote the maximum values of T and U, respectively.

Tmax  U max

MEMB343

Mechanical Vibrations

EXAMPLE 1 x x

 A manometer used in a fluid mechanics laboratory has a uniform bore of cross-section area A.  If a column of liquid of length l and density r is set into motion as shown in the above figure, find the frequency of the resulting motion.

MEMB343

Mechanical Vibrations

SOLUTION TO EXAMPLE 1  Energy Method.

1 2 1 2   T  mx  lAr x 2 2 1 2 1  2 Axrg  2 U  kx   x 2 2 x  d T  U   lArxx  2 Argxx  0 dt  2g   x   x  0  l 

n 

2g rad / s l

MEMB343

Mechanical Vibrations

SOLUTION TO EXAMPLE 1 (cont.)  Newton’s Law.

 F mx

 2 Axg r  lArx  2g  x   x  0  l 

n 

2g rad / s l

MEMB343

Mechanical Vibrations

EXAMPLE 2 x r

k

m

 A circular cylinder of mass m and radius r is connected by a spring of modulus k.  If it is free to roll on the rough horizontal surface without slipping, finds its natural frequency.

MEMB343

Mechanical Vibrations

SOLUTION TO EXAMPLE 2  Energy Method.

1 2 1 2 3 2 T  mx  J 0  mx 2 2 4 1 2 U  kx 2 d 3 T  U   mxx  kxx  0 dt 2 3 mx  kx  0 2

1 2 J 0  mr 2 r  x r  x

2k n  rad / s 3m

MEMB343

Mechanical Vibrations

SOLUTION TO EXAMPLE 2 (cont.)  Newton’s Law.

 F mx

kx

mx  kx  F f (1)



Ff

M J 0

J 0   F f r  1 2  x   mr     F f r 2  r  1 F f   mx (2) 2

Insert (2) into (1)

3 mx  kx  0 2

2k n  rad / s 3m

MEMB343

Mechanical Vibrations

EXAMPLE 3  Determine the effect of the mass of the spring on the natural frequency of the spring-mass system. k:

spring stiffness

m:

mass density of the spring (mass/length)

l:

length of spring

y k

l dy m x

MEMB343

Mechanical Vibrations

SOLUTION TO EXAMPLE 3  Kinetic energy of the mass:

1 2 Tm  mx 2

 Potential energy of the spring:

1 2 U  kx 2

 Kinetic energy of the spring:

1 Ts  mlx 2 6

1 1  2  Total kinetic energy of the system: T   m  ml  x 2 3 

MEMB343

Mechanical Vibrations

SOLUTION TO EXAMPLE 3 (cont.)  Assume a harmonic motion where X is the maximum amplitude and n the natural frequency: xt   X cos  t n  The maximum potential and kinetic energies are:

1 1  2 2 Tmax   m  ml  X  n 2 3  1 U max  kX 2 2  Equating Tmax and Umax, we have

n 

k rad / s 1 m  ml 3

MEMB343

Mechanical Vibrations

RAYLEIGH’S METHOD: EFFECTIVE MASS  Rayleigh’s method can be used for multi-mass or for distributed mass systems, provided the motion of every point in the system is known.  In multi-mass systems which are joined by rigid links, levers or gears, the motion of the various masses can be expressed in terms of the motion of some specific point, and thus reducing the system to 1 DOF.  Kinetic energy can then be written as 1 2 .

T

2

meff x

 meff is the effective mass or an equivalent mass at the specified point where motion of the various masses are referred to.

MEMB343

Mechanical Vibrations

RAYLEIGH’S METHOD: EFFECTIVE MASS (cont.)  If the stiffness at the specified point is also known, the natural frequency can be calculated from:

n 

k meff

MEMB343

Mechanical Vibrations

RAYLEIGH’S METHOD: EFFECTIVE MASS (cont.)  In distributed mass systems such as springs and beams, a knowledge of the distribution of the vibration amplitude becomes necessary before the kinetic energy can be calculated.  Rayleigh showed that with a reasonable assumption for the vibration amplitude, it is possible to take into account previously ignored masses and arrive at a better estimate for the fundamental frequency.

MEMB343

Mechanical Vibrations

EXAMPLE 4  A simply supported beam of total mass m has a concentrated mass M at mid-span.  Determine the effective mass of the system at mid-span and find its natural frequency.

l/2 m

M

x

l

M:

mass of concentrated load

m:

total mass of beam

l:

length of beam

MEMB343

Mechanical Vibrations

SOLUTION TO EXAMPLE 4  Assume the deflection of the beam to be due to a concentrated load at mid-span:

 3x  x  3  y  y max   4    l    l  The velocity is therefore:

 3x  x  3  y  y max   4    l    l

MEMB343

Mechanical Vibrations

SOLUTION TO EXAMPLE 4 (cont.)  The maximum kinetic energy of the beam is given by:

Tmax Tmax

l

1 2 m 2   2 y x  dx 2 0 l 1 2  0.4857 m  y max 2



MEMB343

Mechanical Vibrations

SOLUTION TO EXAMPLE 4 (cont.)  If meq denotes the equivalent mass of the beam acting at mid-span, its maximum kinetic energy is:

 

1 2 Tmax  meq y max 2 meq  0.4857m  The effective mass at mid-span is expressed as:

meff  M  0.4857m

MEMB343

Mechanical Vibrations

SOLUTION TO EXAMPLE 4 (cont.)  The natural frequency is:

n 

n  n 

k meff

k M  0.4857 m 48 EI

l 3 M  0.4857 m 

MEMB343

Mechanical Vibrations

FREE VIBRATION OF SINGLEDEGREE-OF-FREEDOM SYSTEMS DAMPED FREE VIBRATIONS

MEMB343

Mechanical Vibrations

LEARNING OBJECTIVES  Upon completion of this lecture, you should be able to:  Determine the equation of motion of a damped singledegree-of-freedom system.  Determine if the system is underdamped, critically damped or overdamped.  Solve the equation of motion of the damped free response.  Determine the damped natural frequency of an underdamped single-degree-of-freedom system.  Determine the logarithmic decrement of an underdamped single-degree-of-freedom system, and relate it to the damping ratio.

MEMB343

Mechanical Vibrations

INTRODUCTION  The analysis of undamped free vibrations is only an idealization of real systems because it does not account for the energy lost to friction.  Once set into motion, such idealized systems vibrate forever with a constant amplitude.  All real systems, however, lose energy to friction and will eventually stop unless there is a source of energy to keep them going.  When the amount of energy lost in the system is small, the results of undamped free vibrations are often in good agreement with real systems.

MEMB343

Mechanical Vibrations

DAMPED FREE VIBRATIONS  Common types of friction forces that can remove mechanical energy from vibrating systems include:  Fluid friction (viscous damping force), which arises when bodies move through viscous fluids;  Dry friction (coulomb friction), which arises when a body slides across a dry surface; and  Internal friction, which arises when a solid body is deformed.

 Damping caused by fluid friction is quite common in engineering work, and only linear viscous damping is considered in this lecture.

MEMB343

Mechanical Vibrations

THE LINEAR VISCOUS DAMPER  Viscous damping is usually represented by a dashpot, which consists of a piston moving in a cylinder filled with viscous fluid.  The viscous dampers considered are linear; the magnitude of the viscous damping force is directly proportional to the velocity with which the damper is being extended or compressed F  cx .  The constant of proportionality c is called the coefficient of viscous damping. Units in the SI system are Ns/m .  The direction of the viscous damping force is always opposite to the direction of the velocity.

MEMB343

Mechanical Vibrations

VISCOUS DAMPED FREE VIBRATIONS

 Consider a block of mass m sliding on a frictionless horizontal surface.  Vibration is induced by displacing the block a distance x0 and then releasing it with an initial velocity of x 0  v0 .

MEMB343

Mechanical Vibrations

VISCOUS DAMPED FREE VIBRATIONS (cont.)  The elastic restoring force of the spring Fs  kx is still directed toward the equilibrium position (negative coordinate direction).  Since the positive directions for the velocity x and acceleration x are the same as the positive coordinate direction, the damping force Fd  cx also acts in the negative coordinate direction.  Applying Newton’s 2nd law to the block gives the differential equation of motion of the block or  kx  cx  mx mx  cx  kx  0  The above equation is a second-order linear differential equation with constant coefficients.

MEMB343

Mechanical Vibrations

VISCOUS DAMPED FREE VIBRATIONS (cont.)  From the theory of ordinary differential equations, the solution of any linear, ordinary differential equation with constant coefficients is always of the form: xt   aet .

 a and  must be chosen to satisfy the differential equation and the initial conditions.  Substituting the solution into the governing equation gives:





aet m2  c  k  0  If the constant a is zero, the trivial solution x  0 is obtained, which is of no interest. Since the exponential e t is never zero, we have the characteristic equation: m2  c  k  0

MEMB343

Mechanical Vibrations

VISCOUS DAMPED FREE VIBRATIONS (cont.)  The characteristic equation has roots

 c  c 2  4mk 1, 2  2m  The displacement of the block is then given by xt   a1e1t  a2e2t  The constants are determined from the initial conditions. t  0; x  a1  a2  x 0 and x  a11  a22  v0  The roots can be rewritten in terms of a non-dimensional parameter  known as the damping ratio. c c   2 mk 2m n

MEMB343

Mechanical Vibrations

VISCOUS DAMPED FREE VIBRATIONS (cont.)  In terms of the damping ratio  and the natural frequency  n , the roots of the characteristic equation become:

1, 2   n   n  2  1  The behavior of the system depends on whether the quantity under the radical is positive, zero or negative.  The value of c that makes the radical zero is called the critical damping coefficient cc . cc  2m n  2 mk  The solution will have three distinct types of behavior depending on whether the actual system damping c is greater than, equal to, or less than cc .

MEMB343

Mechanical Vibrations

OVERDAMPED SYSTEMS  When the damping coefficient c is greater than cc , then the damping ratio   1 , and the radical is real.  The two roots 1 and 2 are both real and are unequal, and since  2 1   , both roots will be negative.

1   n  n  2  1

2   n  n  2  1  The solution then becomes:     n  nt  xt   e a1e  

 2 1  t 

 a2 e

   2 1  t  n   

   

MEMB343

Mechanical Vibrations

OVERDAMPED SYSTEMS (cont.)  The real-valued constants of integration a1 and a2 are determined by the initial conditions.

a1 

a2 





v0     2  1  n x0 2 n  2  1





 v0      2  1  n x0 2 n  2  1

MEMB343

Mechanical Vibrations

OVERDAMPED SYSTEMS (cont.)

 Therefore the displacement simply decreases to zero as t increases, and the motion is non-vibratory.

MEMB343

Mechanical Vibrations

CRITICALLY DAMPED SYSTEMS  When the damping coefficient c is equal to cc , then the damping ratio   1 , and the radical is zero.  The two roots 1 and 2 are equal and negative. 1  2   n  The solution in this case has the special form: xt   a1  a2t ent  The real-valued constants of integration a1 and a2 are determined by the initial conditions. a1  x0

a2  v0   n x0

MEMB343

Mechanical Vibrations

CRITICALLY DAMPED SYSTEMS (cont.)

 The displacement simply decreases to zero as t increases, and the motion is non-vibratory.  Qualitatively the motion for critical damping is the same as the motion for overdamped system.

MEMB343

Mechanical Vibrations

CRITICALLY DAMPED SYSTEMS (cont.)  Critical damping is of special importance because it is the dividing point between non-vibratory motions and damped oscillatory motions.  Critical damping is the smallest amount of damping for which a system will not oscillate.  A critically damped system will come to rest in less time than any other system starting from the same initial conditions (i.e. the value of damping that provides the fastest return to zero without oscillation).

MEMB343

Mechanical Vibrations

UNDERDAMPED SYSTEMS  When the damping coefficient c is less than cc , then the damping ratio   1 , and the radical is imaginary.  The two roots 1 are 2 complex conjugates.

1   n  j d 2   n  j d 2  The constant  d   n 1   is called the damped natural circular frequency.  The solution then becomes: xt   e nt a1e jd t  a2e jd t





MEMB343

Mechanical Vibrations

UNDERDAMPED SYSTEMS (cont.) j

 By making use of the Euler formula e  cos   j sin  , the equation for the displacement can be written as:

xt   e  nt a1  a2  cos d t  j a1  a2 sin d t   e  nt  A1 cos d t  A2 sin d t   Ae  nt sin d t   

 The constants A and  are evaluated using the initial conditions. x0  x0  A sin 

x0  v0   n x0  x0d cot 

MEMB343

Mechanical Vibrations

UNDERDAMPED SYSTEMS (cont.)  The constants A and  can be expressed as: x x0 d A 0 tan   sin  v0   n x0  With this value of  , the sine becomes:

sin  

x0 d

v0   n x0 2  x0 d 2

 Thus value of A and  are determined to be:

A

v0   n x0 2  x0 d 2  d2

x0 d   tan v0   n x0 1

MEMB343

Mechanical Vibrations

UNDERDAMPED SYSTEMS (cont.)

 Motion of the underdamped system is called time-periodic.  The motion oscillates about the equilibrium position, but the amplitude Ae  nt decreases because the exponent is negative.

MEMB343

Mechanical Vibrations

UNDERDAMPED SYSTEMS (cont.)  Since the amplitude decreases monotonically with time, the oscillation will never repeat itself exactly, unlike the case of a undamped free vibration.  The constant  d   n 1   2 is called the damped natural circular frequency.  Since 0    1 for underdamped vibration , the damped natural frequency  d will always be less than the undamped natural circular frequency  . n

MEMB343

Mechanical Vibrations

UNDERDAMPED SYSTEMS (cont.)  By analogy with the undamped free vibration, a damped natural frequency  d and a damped natural period  d may be defined as d n fd   2 2 d   d n

1 2 2 2 1 2

 It is interesting to note that the damped natural period and the damped natural frequency are constant (independent of time) even though the amplitude is not.

MEMB343

Mechanical Vibrations

EXAMPLE 1  A torsional pendulum has a natural frequency of 200 cycles/min when vibrating in vacuum. The mass moment of inertia of the disc is 0.2 kgm2. It is then immersed in oil and its natural frequency is found to be 180 cycles/min. Determine the damping constant. If the disc, when placed in oil, is given an initial displacement of 2º, find its displacement at the end of the first cycle.

MEMB343

Mechanical Vibrations

EXAMPLE 1 (cont.)  Undamped and damped torsional natural frequencies are:

200 n  2f n  2   20.94 rad/s 60 180 d  2f d  2   18.85 rad/s 60  Damping ratio is obtained from:

 d   1    n

2

 18.85     1     0.4355  20.94   2

MEMB343

Mechanical Vibrations

EXAMPLE 1 (cont.)  Torsional damping coefficient is obtained from:

ct  2 J 0n  2  0.2  20.94  0.4355  3.65 Nms/rad  Initial conditions are:

 0  2  0.035 rad 0  0  Damped natural period is:

2

2 d    0.33 s d 18.85

MEMB343

Mechanical Vibrations

EXAMPLE 1 (cont.)  Value of  is determined from:







2 2   0   n 0   0d 

d2

0  0.4355  20.94  0.035  0.035 18.85 2



 0.039 rad

18.852

2

MEMB343

Mechanical Vibrations

EXAMPLE 1 (cont.)  Value of  is determined from:

  0d     tan     0   n 0  0.035  20.94  1   tan    0  0.4355 18.85  0.035   68.6  1.2 rad 1

MEMB343

Mechanical Vibrations

EXAMPLE 1 (cont.)  At the end of 1 cycle, t   d  0.33 s . The displacement is obtained from:

 t   e

 n t

sin d t   

 0.039  e 0.435520.940.33 sin 18.85  0.33  1.2  0.0017 rad  0.097

MEMB343

Mechanical Vibrations

LOGARITHMIC DECREMENT  Viscous damping in an underdamped 1-DOF system causes the vibration to decay exponentially.  The exponent is a linear function of the damping ratio  .  A convenient measure of the amount of damping is provided by the extent to which the amplitude xt  has fallen during one complete cycle of vibration.  is defined as the natural  Logarithmic decrement logarithm of the ratio of two successive amplitudes, and is defined as follows, where  d is the damped natural period: xt    ln xt   d 

MEMB343

Mechanical Vibrations

LOGARITHMIC DECREMENT (cont.)  Substitution of the analytical form of the underdamped response yields: Ae  nt sin  d t      ln  n t  d  Ae sin  d t   d d   

 d d  2 ,  As the expression for the logarithmic decrement reduces to:   ln e n d   n d  The circular damped natural frequency is substituted into the above equation to give: 2 2    n  2 n 1  1 2

MEMB343

Mechanical Vibrations

LOGARITHMIC DECREMENT (cont.)  Solving the expression for  yields:

 



4 2   2  Logarithmic decrement can be determined from a record of the displacement response of an underdamped system.  Let t1 and t2  t1   d  represent the times corresponding to consecutive displacements x1 and x2 measured one cycle apart.  The logarithmic decrement is given by: x1   ln x2

MEMB343

Mechanical Vibrations

LOGARITHMIC DECREMENT (cont.)

 The peak measurements can be used over any integer multiple of the period to increase the accuracy over the measurements taken at adjacent peaks ( m is an integer which corresponds to the number of complete cycles). 1 x1   ln m xm1

MEMB343

Mechanical Vibrations

LOGARITHMIC DECREMENT (cont.)

 For   0.3 or   2 , the log decrement and the damping ratio can be approximated as:

  2

  2

MEMB343

Mechanical Vibrations

EXAMPLE 2  The free vibration response of an electric motor of weight 500 N mounted on a flexible foundation is shown in the figure below. Find  the undamped and damped natural frequencies of the electric motor - foundation system, and  the spring constant and damping coefficient of the foundation.

MEMB343

Mechanical Vibrations

EXAMPLE 2 (cont.)

MEMB343

Mechanical Vibrations

EXAMPLE 2 (cont.)  From the time-response plot, the damped natural period is 0.2 s. The damped natural frequency is:

d 

2

d

2   31.416 rad/s 0.2

 Log decrement is also obtained from this plot from:

1 x1 1 8   ln  ln  0.693 m xm1 3 1

MEMB343

Mechanical Vibrations

EXAMPLE 2 (cont.)  Since   2 , damping ratio is obtained from:

 0.693     0.11 2 2  Undamped natural frequency is then computed from:

n 

d 1  2



31.416 1  0.112

 31.608 rad/s

MEMB343

Mechanical Vibrations

EXAMPLE 2 (cont.)  Spring constant of the foundation is obtained from:

500 k  m   31.6082  50920.8 N/m 9.81 2 n

 Damping constant of the foundation is:

500 c  2mn  2   31.608  0.11  354.4 Ns/m 9.81

MEMB343

Mechanical Vibrations

REVIEW OF FREE VIBRATIONS  There are four cases of interest with regards to the free vibrations of a single degree-of-freedom-system.  Undamped system (0) s   j 1, 2

 Underdamped system (0