1

ME/AE455

Mechanical Vibrations and Control

Chapter 1

Sections 1.1 to 1.3

Fall 2019

Prof. Ç. Çetinkaya

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Course Learning Objective:

The main objective of this course is to provide engineering juniors and seniors with conceptual and

analytical skills required for modeling and analyzing vibrating mechanical systems for design,

maintenance and testing purposes.

Prerequisite by Topic: ES 223 Rigid Body Dynamics, and MA330 Engineering Math

Textbook: D. Inman, Engineering Vibration (4th Edition),

Prentice-Hall, ISBN-10: 0132871696, ISBN-13: 978-0132871693

Official Course Website: http://people.clarkson.edu/~ccetinka/ME455/

Instructor: Prof. C. Cetinkaya, Dept. of Mech. and Aero. Eng., CAMP 241, cetin@clarkson.edu

Office Hours: Tue. 1:00-2:30pm | Wed. 1:30-3:30pm | Thur. 1:00-2:30pm | By Appointment |

Teaching Assistant: Xiaochi Xu xiaxu@clarkson.edu CAMP 242

Grader Information: TBA

TA Office Hours: Wed. 1:00-3:00pm | Wed. 1:00-3:00pm | Thur. 3:00-5:00pm | By Appointment |

TA Office Hour Meeting Place: Outside CAMP 250 (Table Reserved for Class)

Classroom/Class Hours: CAMP 177, Tuesday 4:00-5:15 and Thursday 4:00-5:15

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Fall 2019

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Policies:

Cell/Smart Phones/Computers with Internet access

in Classroom: No such devices are allowed in class

during lectures and exams except emergencies.

• Turn it off

• Put it in your bag

2

Extra Points Opportunity During Class:Added to Next Homework Assignment Grade

If you are the first to point out, you will receive:

• Error in Derivations – Minor Typo: 10 pts

• Error in Derivations – Major Typo: 20 pts

• Error in Derivations – Logic: 30 pts

• “Excellent Questions”: 15 pts

• “Excellent Comments”: 15 pts

Concept Map: Terminology

Source: https://courses.edx.org/static/content-mit-mrev~2013_Summer/handouts/Concept_Map.png 6/57

Examples: Structures and Buildings

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Examples: Mechanical Systems

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3

Examples: Aerospace Structures

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Examples of Vibrating Systems: Music

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Examples of Vibrating Systems: Organs

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Chapter 1

1.1 Basic of Mechanical Vibrations and Free

System Response

Vibration: a (mechanical) phenomenon, oscillations occur about an equilibrium

point.

Free Vibrational Motion: During vibration, no external effect is in place (except

the initial conditions/excitations)

Elasticity: A solid object deforms when forces are applied on it. If the material is

elastic, the object returns to its initial shape when the forces are removed.

Plasticity: The object does not return to its initial shape when the forces are

removed.

Material Fatigue: Structures may fail when subjected to cyclic loads. This type of

structural damage occurs even when the exerted stress range is far below the

static material strength. Fatigue is the most common source behind failures of

mechanical structures.

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4

Observation 01: Highway traffic light poles oscillating due to snow coverhttp://www.youtube.com/watch?v=xkOuPuawVeY

Observation 02: Vibrating flag pole

http://www.youtube.com/watch?v=bvLTErW5zNo

Observation 03: Vibration. See the unseen. (Fluke Corp.)

http://www.youtube.com/watch?v=W4s2UwKm7dc

Possible Questions:

1. What am I seeing/observing?

2. What is taking place? (engineering analysis)

3. Why is this occurring? (hypothesizing)

4. Am I missing something? (testing logical consequences of the hypothesis)

5. How can I apply this phenomenon to an engineering application? (commercial interest)

The scientific method requires observations of nature to formulate and test hypotheses.It consists

of these steps (wikipedia.org):

1. Asking a question about a natural phenomenon

2. Making observations of the phenomenon

3. Hypothesizing an explanation for the phenomenon

4. Predicting a logical consequence of the hypothesis

5. Testing the hypothesis by an experiment, an observational study, or a field study6. Creating a conclusion with data gathered in the experiment, or forming a revised/new hypothesis and

repeating the process

Direct Observations

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Linear

Behavior

Nonlinearity

Consider a spring-mass system and perform a static

experiment: The spring is used to model elasticity in a system.

A plot of force versus displacement:

Linear Spring: A Thought Experiment

FBD:

From strength of materials, we know that:

kf k x Experiment

k

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Free-body Diagram and an Equation of Motion of a System

0

0

( ) ( ) ( ) ( ) 0

(0)

(0)

k x t m x t m x t k x t

x x

x v

Newton’s Second Law of Motion (i.e. ΣF = m a) along x results in:

Second order ordinary

differential equation (ODE)

Generating an Equation of Motion

Free Body Diagram

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Stiffness and MassVibration is caused by the interaction of two types of forces:

(i) Restitution Force: Related to position (stiffness) and

(ii) Inertia Force: Related to acceleration (mass, or mass distribution in

space)

Mass

m

k

x

Displacement (x)

Spring

Stiffness (k): Restitution Force

Mass (m): Inertia Force

Proportional to displacement

Proportional to acceleration

( )kf k x t static

( ) ( )mf m a t m x t dynamicO

Reference

Displacement

Coordinate

x

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5

Single-Degree-of-Freedom Systems: SDoF

m

q

l =length

Pendulum

q(t) g

lq(t) 0

Gravity g

Shaft and Disk

q

( ) ( ) 0J t k tq q

Torsional

Stiffness

kMoment of

inertia

J = m r2/2

Restitution: Gravity

Inertia: Mass (m)

Restitution: Rotational Stiffness

Inertia: Moment of inertia (J)17/57

Solving Second Order ODEs

x(t) Asin(nt )

2 2

( ) cos( )

( ) sin( ) - ( )

n n

n n n

x t A t

x t A t x t

Assume a solution in this form: Φ, A and ωn=?

Differentiating it twice gives:

Substituting back into the equations of motion gives:

2

2

sin( ) sin( ) 0

0 or

n n n

n n

m A t k A t

km k

m

Natural

Frequency

t

x(t)

rad/s

( ) ( ) 0mx t kx t

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Example 1.1.1

Solution Steps:

1. Sketch the structure or part of interest

[contacts, geometry, mass, etc.]

2. Write down all the forces and make a “free

body diagram”

[internal/external forces + coordinates]

3. Apply Newton’s Law and/or Euler’s Law to find

the equations of motion

[Universal Law, fundamental relations, with

respect to 0 to eliminate reaction forces]

Free Body Diagram (FBD):

2

0 0 0( ) J t J mq M

Derive the equation of motion for the pendulum.

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The problem is one dimensional, thus a scalar

equation results as an equation of motion:

2

0 ( ) sin ( ) ( ) sin ( ) 0

restoring moment

J t mg l t m t mg tq q q q

Over-dots denote differentiation with respect to time t. This is a second

order, nonlinear ordinary differential equation (ODE).

2 ( ) ( ) 0 ( ) ( ) 0g

m t mg t t tq q q q

Requires knowledge of (0) and (0)q q the initial position and velocity at t =0.

Solution 1.1.1: Generating the Equation

of Motion

sinq qUsing the small-angle approximation:

sin ( )tq

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6

Initial Conditions (IC)

If a system vibrates, it means that, some energy into to the system has

been transferred (injected into) and caused it to move. (I. Newton)

0

0

(0) sin( 0 ) sin( )

(0) cos( 0 ) cos( )

n

n n n

x x A A

v x A A

From our earlier solution, we know that the form of the solution is as

follows:

For example, the mass in an SDOF could have been:

• Moved a distance x0 and then released at t = 0 (Potential Energy)

• Given an initial velocity v0 at t = 0 (Kinetic Energy)

• Some combination of the these two cases

x(t) Asin(nt )

and its initial conditions (IC):

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Initial Conditions (IC)

Solving this set of equations gives for the form of solution:

2 2 2 1 00 0

0

Amplitude Phase

1( ) sin( ) tan n

n n

n

xx t A t A x v

v

x0

Displacement x(0) = x0 and Velocity = v0 at t = 0( )x t

0 nv

x0

1

n

n

2x0

2 v0

2

t

x(t)

Slope

here is v0

n

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Simple Harmonic Motion

fn n rad/s

2 rad/cycle

n cycles

2 s

n

2Hz

Amplitude

t

x(t)

0x Slope

Here is v0

n

Period

T 2

n

Maximum

VelocityAn

x(t) Asin(nt )

Phase

Difference

A

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0

Minimum

Velocity

Total Solution

2 2 2

0 0 1 0

0

( ) sin tann n

n

n

x v xx t t

v

This is the solution to a simple harmonic oscillator, describing an

oscillatory motion (also refereed to as a simple harmonic motion).

At t = 0, this solution satisfies the initial condition (IC) for displacement:

2 2 2

0 0 00

2 2 2

0 0

(0)n n

n n

x v xx x

x v

Solution with the initial conditions (IC) x(t) Asin(nt )

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7

In MATLAB, use the atan2(x,y) function to determine the correct phase.

Calculating Arctangent: Phase

+

+

_

+

+

Calculating arctangent from a calculator requires some attention.

The argument atan( ̶ / + ) is in a different

quadrant from atan( + / ̶ ) .

Important: A typical calculator will return an

arctangent in between -/2 and +/2, reading

only the atan(-) for both of the above two cases.

_

__

1 0

0

Phase

tan n x

v

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Example 1.1.3: Tire Suspension Measured: m = 30 kg, f = 10Hz. What is k, the spring stiffness?

k

m

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Example 1.1.3: Solution

22 530 kg 62.8 1.184 10 N/mnk m

Obviously, there are more sophisticated models of suspension systems

(than this one-D model).

Such models will be covered latter in the course as our tools and

understanding develop.

m = 30 kg, f = 10Hz, k = ?

Tire suspension m = 30 kg, fn= 10 Hz = 10 cycles/sec

cycle 2 rad10 62.8 rad/sec

30 sec cyclen

k k

m

k

m

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Section 1.2 Harmonic Motion

2 rad 2 1 sec = =0.1 sec

rad/sn n n

Tf

The natural frequency in the SI units of Hertz (Hz = 1/sec):

The period is the time elapsed to complete one complete cycle:

rad/s cycles Hz =10 Hz

2 2 rad/cycle 2 s 2

n n n nnf

For the pendulum: rad/s 2 secn

gT

g

rad/s 2 secn

k JT

J k For the disk and shaft:

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8

Note how the relative magnitude of each increases with n for n > 1

Relationship between Displacement,

Velocity and Acceleration

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-200

0

200

Time (sec)

a

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1

0

1

x

A=1 m, n=12 rad/sec

x(t) Asin(nt )

x(t) nAcos(nt )

x(t) n

2Asin(nt )

Displacement

Velocity

Acceleration

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-20

0

20

v

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Example 1.2.1

A hardware store spring (k) and a bolt (m) with k = 857.8 N/m and m = 49.2 gr.

Exert IC of x0 =10 mm at t = 0. No initial velocity is prescribed.

Compute n and the max displacement of resulting vibrational motion.

k

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Example 1.2.1: Solution

-3

2 2 2

max 0 0 0

857.8 N/m132.04 rad/s

49.2 10 kg

21.01 Hz = 21.012 sec

2 1 10.0476 s

21.01sec

1( ) 10 mm

n

nn

n n

n

n

k

m

cyclesf

Tcyclesf

x t A x v x

0

Common

units are Hertz

(Hz)

To avoid serious

errors use fn when

working in Hertz and

n when in rad/s

Units depend on

system

m= 49.2x10-3 kg, k=857.8 N/m and x0 =10 mm

n = ? and max displacement A = ?

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2 2

( ) sin( ) ( )

( ) cos( ) ( )

( ) sin( ) ( )

n

n n n

n n n

x t A t x t A

x t A t x t A

x t A t x t A

Example 1.2.1: SolutionMax Amplitudes for Harmonic Motion: displacement, velocity

and acceleration

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9

max

2 3 2

max

2

1 0

( ) 1320.4 mm/s=1.32 m/s

( ) 174.35 10 mm/s

=174.35 m/s 17.8

tan rad0 2

( ) 10 sin(132.04 / 2) 10 cos(132.4 ) mm

n

n

n

v t A

a t A

g

x

x t t t

g = 9.8 m/s2

2.92 mph

90°

~0.4 in max = A

Example 1.2.1: SolutionMax Amplitudes

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Does Gravity Matter in Spring Problems?

Let be the deflection caused by hanging a mass on a spring

( = x1 - x0 in the figure)

Then from static equilibrium:

m g k

Next sum the forces in the vertical for some point x > x1 measured

From the equilibrium position

0

( ) ( ) 0

m x k x m g k x m g k

m x t k x t

So no, gravity does not have an effect on the vibration

(note that this is not the case if the spring is nonlinear)

Reference

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Example 1.2.2

A 2 m pendulum swings with a period of 2.893 s. What is the acceleration

due to gravity at that locality? Namely, g = ?

Pendulums and Measuring the gravitational acceleration g

m

q

l =length

Gravity g

Reference

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Example 1.2.2: Solution

2 2

2 2 2

2

4 42 m

2.893 s

= 9.434 m/s

gT

g

T = 2.893 sec, l = 2m

m

q

l =length

q(t) g

lq(t) 0

Gravity g 2

n

T

rad/s, 2 sn

gT

g

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10

Review: Complex Numbers and Complex

Exponential (See Appendix A)

eel

m

b

a

A

jc a j b Ae q

A complex number can be written with a real and imaginary

part or as a complex exponential

where

cos , sina A b Aq q

Multiplying two complex numbers:

1 2( )

1 2 1 2

jc c A A e

q q

Dividing two complex numbers:

1 2( )1 1

2 2

jc Ae

c A

q q

O

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Equivalent Solution Forms to 2nd ODEs

1 2

1 2

( ) sin( )

( ) sin cos

( ) n n

n

n n

j t j t

x t A t

x t A t A t

x t a e a e

All of the following solutions are equivalent:

The relationships between A and , A1 and A2, and a1 and a2 can be found

in Window 1.4 of the course textbook (page 19).

the Cartesian form

Called the magnitude-and-phase form

the polar form

• Each is useful in different situations

• Each represents the same information

• Each solves the equation of motion

( ) ( ) 0m x t k x t

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Derivation of the Solution: Polar

(Exponential) Form

2 2

1 2

1 2

Substitute ( ) into the ODE 0

0 0

( ) and ( )

( )

n n

n n

t

t t

n

j t j t

jt jt

x t a e m x k x

m a e k a e m k

k kj j

m m

x t a e x t a e

x t a e a e

Two solutions: Each satisfies the ODE

Linear combination is a solution,

too, as it satisfies the ODE.39/57

Is Frequency always positive?

x(t) a1en jt a2e

n jt

( ) sin nx t A t

From the preceding analysis, = ± n then

Using the Euler relations for trigonometric functions, the above

solution can be written as

It is in this form that we identify as the natural frequency n.

This is positive, the + sign being used up in the transformation from

exponentials to the sine function.

40/57

11

Calculating Average Values (Oscillations)

0

2 2

0

2

peak value

1lim ( ) = average value

1lim ( ) = mean-square value

= root mean square (rms) value

T

T

T

T

rms

A

x x t dtT

x x t dtT

x x

Proportional to

kinetic energy

Not very useful since for a

sine function the average

value is zero

May need to be limited due to

physical constraints

Also useful when the

vibration is random

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Example 1.2.1:

The acceleration magnitude is 20 log10(17.8) = 25dB relative to 1g.

The Decibel (dB) Scale

The dB scale is always relative to some reference value x0:

2

10 10

0 0

10log 20log x x

dBx x

For example: if an acceleration value was 19.6m/s2 , then relative to 1g (or

9.8m/s2), the level would be 6dB:

10log10

19.6

9.8

2

20log10 2 6dB

42/57

1.3 Viscous Damping

A real system dissipate energy as it vibrates.

We often model this effect as damping. The simplest form of damping is

called viscous damping (or dashpot) .

A viscous damper produces a force that is proportional to velocity.

Reaction force in a (viscous) damper (c):

( ) ( )cf c v t c x t

x

fc

x

x(t) Asin(nt )

( ) ( ) 0mx t kx t

Model:Diagram:

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Governing Equation Including Damping

m

kx

Displacement:

c

For this damped single DoF system, the reaction

force ( fm ) acting on the mass is due to the spring

(k) and the dashpot (c):

To solve this equation (governing ODE), it is useful

to assume a solution of the form:

tx(t) a e

Free Body Diagram:x

m k cf f f

( ) ( ) ( ) is re-arranged to:

( ) ( ) ( ) 0

m x t k x t c x t

m x t c x t k x t

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12

Solution to Governing ODE with Damping

2 t t tx(t) a e x(t) a e x(t) a e

2 0 ta e (m c k)

The velocity and acceleration can then be calculated as:

If these are substituted into the equation of motion:

Divide both sides of the equation by m, substitute for the natural

frequency and assume a non-trivial solution:

20 0t c ka e

m m

Characteristic Equation

2( ) ( ) ( ) 0 t t tmx t cx t kx t a e m a e c a e k

After collecting for the (common) exponential terms, we obtain:

45/57

Solution to ODE with Damping

2 2 20 2 0 2n n n

c k c

m m m

For convenience, define a new term, known as the damping ratio, as:

To represent the characteristic equation as:

Solving this quadratic equation for results in:

2

c=

k m

2

1 2 1, n n

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Critically Damped Motion: ζ =1

definition of criticaldamping coefficient

=1 2 22

cr n

c= c c k m m

k m

Critical damping occurs when ζ =1. The damping coefficient c, in

this case, is given by:

Solution then takes the form:

Solving for then gives,

2

1 2 1 1 1, n n n A repeated, real root

1 2( ) n nt tx t a e a t e

Needs two independent solutions,

hence the t in the second term

2

1 2 1, n n

tx(t) a e

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Critically Damped Motion: ζ = 1Calculate a1 and a2 are calculated from initial conditions (t=0),

1 2 1 0

1 2 2

0 1 2

2 0 0

( ) ( )

( ) ( )

n

n

t

t

n n

n

n

x t a a t e a x

v t a a t a e

v a a

a v x

0 1 2 3 4

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

Time (sec)

Dis

pla

cem

en

t (m

m)

x0=0.4mm v

0=1mm/s

x0=0.4mm v

0=0mm/s

x0=0.4mm v

0=-1mm/s

• No oscillation occurs

• Quickest return to equilibrium position

• Application: door mechanisms

• Application: analog gauges

k = 225 N/m m=100kg and ζ=1

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13

Over-damped Motion: ζ > 1

An over-damped case occurs when ζ > 1.

The two roots of the equation are real

(and different, not repeating).

2 2

2

1 2

1 1

1 2

1

n nn

, n n

t ttx(t) e (a e a e )

a1 and a2 are calculated from the ICs

(t=0):

2

0 01

2

2

0 0

22

1

2 1

1

2 1

n

n

n

n

v ( ) xa

v ( ) xa

0 1 2 3 4-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

Time (sec)

Dis

pla

cem

en

t (m

m)

k = 225 N/m m=100kg and ζ=2

x0=0.4mm v

0=1mm/s

x0=0.4mm v0=0mm/s

x0=0.4mm v

0=-1mm/s

Slower to respond than

critically damped case

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Under-damped Motion: ζ < 1When < 1, an under-damped motion occurs.

The roots of the equation are complex conjugate pairs. This is the only

case that yields oscillation.

1 2

2 2

2

1 2

1 2

1 1

1 2

1

=

sin

n nn

n

, n n

t t

j t j tt

t

d

j

x(t) a e a e

e (a e a e )

Ae ( t )

The frequency of oscillation d is called the damped natural frequency is

given by:

21 d n

2

c=

km

tx(t) a e

50/57

Under-damped Motion: ζ < 1Calculate A and f from the ICs (xo and vo) at t = 0

A 1

d

(v0 nx0)2 (x0d)

2

tan1 x0d

v0 nx0

0 1 2 3 4 5-1

-0.5

0

0.5

1

Time (sec)

Dis

pla

ce

me

nt

• Gives an oscillating response

with exponential decay

• Most natural systems vibrate

with and under-damped

response

• See Window 1.4 (page 19) for

details and other

representations( ) sinnt

dx t Ae ( t )

51/57

Example 1.3.1

3

3

49.2 10 kg, 857.8 N/m

2 2 49.2 10 857.8 12.99 kg/s

0.11 kg/s= 0.0085 1 thus, underdamped

12.99 kg/s

cr

cr

m k

c km

c

c

Consider the spring of Example 1.2.1, if c = 0.11 kg/s;

The motion is under-damped and the bolt will oscillate.

Determine the damping ratio of the spring-bolt system

52/57

14

Example 1.3.2The human leg has a measured natural frequency of

around 20 Hz when in its rigid (knee locked) position,

in the longitudinal direction (i.e., along the length of

the bone) with a damping ratio of

= 0.224.

Calculate the response of the tip if the leg bone to an

initial velocity of v0 = 0.6 m/s and zero initial

displacement and plot the response.

This would correspond to the vibration induced while

landing on your feet, with

your knees locked form a height

of 18 mm.

What is the maximum acceleration

experienced by the leg assuming

no damping?

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Example 1.3.2: Solution

22

2 2

-1

0

28.148

20 cycles 2 rad20 2 125.66 rad/s

1 cycles

1 125.66 1 .224 122.467 rad/s

0.6 0.224 125.66 0 0 122.4670.005 m

122.467

0tan 0

0

0.005

n

d n

d

n

t

s

A

v

x t e

sin 122.467t

20

0.224 1 underdamped

I.C.: 0, 0.6 / sec

n

o o

f Hz

x v m

A 1

d

(v0 nx0)2 (x0d)

2

tan1 x0d

v0 nx0

54/57

Solution (1.3.2): Obtain Max Acceleration

2

2 00 0 0

0

2 2 2 2

max

, 125.66, 0.6, 0 (no damping case, 0)

0.6m m

0.6max 0.6 125.66 m/s 75.396 m/s

n

n

n n

n n

n

vA x v x

vA

a x A

Maximum Acceleration with No Damping Approximation:

2

max 2

75.396 m/s7.68g

9.81 m/sa g

28.1480.005 sin 122.467tx t e t

A 1

d

(v0 nx0)2 (x0d)

2

tan1 x0d

v0 nx0

55/57

Solution (1.3.2): Displacement Response

28.1480.005 sin 122.47tx t e t

2 2

max 75.396 m/sna A

0.224 1

Underdamped

I.C.:

0

0.6 / sec

o

o

x

v m

56/57

15

Example 1.3.3

1 2

0

0 1 2 2 0

1 2 1 2

0

( ) sin( ) Use sin( ) sin sin cos cos

( ) sin( ) ( sin cos )

(0) ( sin(0) cos(0))

( sin cos ) ( cos sin )

n

n n

n n

t

d

t t

d d d

t t

n d d d d d

x t Ae t x y x y x y

x t Ae t e A t A t

x x e A A A x

x e A t A t e A t A t

v

0 01 0 1 0 1

0 00

( sin 0 cos 0) ( cos 0 sin 0)

( ) sin cosn

nn d

d

t nd d

d

v xA x A x A

v xx t e t x t

Compute the form of the response of an under-damped system using the

Cartesian form of the solution (given in Window 1.5.)

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