Introduction

1

Dynamics II Dr. Jorge A. Olórtegui Yume, Ph.D.

INTRODUCTION TO MECHANICAL VIBRATIONS

Lecture No. 1

Mechanical Engineering School

National University of Trujillo

Introd. to Mech. Vibrations Dr. Jorge A. Olortegui Yume, Ph.D.2

The Millenium Bridge

PLACE :London, England

OPEN TO PUBLIC:June 10th, 2000

CLOSED:3 days after!!!

CAUSE:Vibrations felt by pedestrians”The Wobbly Bridge”

COST:28.4 Million Dollars

RE-OPEN TO PUBLIC:Feb. 22nd, 2002

ANALYSIS, REINFORCING, TESTING COST:7.8 MILLION Dollars

ANALYSIS:Based upon Tacoma Narrows Bridge case


THE TACOMA NARROWS BRIDGE

PLACE:Tacoma - Gig Harbor, Washington, USA

OPEN TOPUBLIC:July 1st, 1940

COLLAPSE:Nov 7th, 1940

CAUSE:RESONANCEWind-induced oscillation frequency coincided with one of the bridge´s natural frequencies


TODAY´S TACOMA NARROWS BRIDGE

BEFORE AFTER AFTER

BEFORE


BEFORE WE CAN SOLVE SUCH COMPLEX PROBLEMS ….

NEED MECHANICAL VIBRATIONS FUNDAMENTALS

CRAWL, WALK, RUN AND THEN…FLY


WHAT DO WE DO IN THE MECH.

VIBRATIONS COURSE?

Physical

System



VIBRATIONS COURSE?

Physical

System

Engineering

Model

x

2k 1k

2l 1l



VIBRATIONS COURSE?

Physical

System

Engineering

Model

Mathematical

Model

0

0

0

02

22

2

112211

221121

2

x

lklklklk

lklkkkx

mr

m



VIBRATIONS COURSE?

Physical

System

Engineering

Model

Mathematical

Model

Mathematical

Solution

22

2

211

1

1 sin1

sin1

tAtA

t

tx


VIBRATION: Any motion that repeats itself after an interval of time

Examples:• Motion of a plucked guitar string• Ground motion in an earthquake• Beating of your heart• Oscillation of mass attached to spring & damper• Pendulum

MECH. VIBRATIONS FUNDAMENTALS

Vibration



•Mean of Epotential storage• Spring (k) or elasticity in general

•Mean of Ekinetic storage• Mass (m) or inertia in general

•Mean of gradual Energy disipation •Damper (c), friction (m)

ViIBRATION

VIBRATORY SYSTEM PARTS

2

2

1kxEpotential

2

2

1mvEkinetic

2cvdt

dEP

dissipated

dissipated


DEGREE OF FREEDOMMín. # of independent coordinates needed to define position of all parts of a system at any instant.

ONE DEGREE OF FREEDOM SYSTEM (1-DOF)

Mass-Spring System


TWO DEGREE OF FREEDOM SYSTEM (2-DOF)

Torsional system

THREE DEGREE OF FREEDOM SYSTEM (3-DOF)


“N” DEGREE OF FREEDOM SYSTEM (N-DOF)


A cantilever beam showing an infinite number of DOF

N CONTINUOUS

SYSTEM

N = Finite #

DISCRETE OR LUMPED SYSTEM

Model as


MECH. VIBRATIONS FUNDAMENTALSVIBRATION CLASIFICATIONSeveral criteria

Free Vibration: System disturbed initially thenleft to oscillate on its ownForced Vibration: System subject to an externalforce usually oscillating

Undamped Vibration: If no energy is dissipated by friction or othe resistanceDamped Vibration: If energy is gradually lost


MECH. VIBRATIONS FUNDAMENTALSVIBRATION CLASIFICATION

Linear Vibration: Basic components of system behave linearly

Non-linear Vibration: Real components are non linear

Equation Of Motion (EOM) Equation Of Motion (EOM)

0

kxxcxmIF m, c, k are constantsTHENDiff. Equation is LINEAR

0

xxkxcxm

Diff. Equation is NON-LINEAR

Great!. Soln. To EOM is usually known and relatively easy to use

Ooops!. Soln. To EOM is usually not known and/or difficult to obtain and difficult to use Num. Methods


MECH. VIBRATIONS FUNDAMENTALSSOME CONCEPTS IN VIBRATIONS MODELLING



Equation Of Motion (EOM)

0

kxxcxm

xmF

Newton´s 2nd Law (Dynamics)

teXtx d

tn sin11



Which part of the modelling

procedure is this?


MECH. VIBRATIONS FUNDAMENTALSExample: The figure shows a motorcycle and its rider. Develop a

sequence of three mathematical models of the system to investigate vibration in the vertical direction. Consider the elasticity of the tires, struts, and rider; damping of the struts (vertically) and the rider; masses of the wheels, body vehicle and rider.


MECH. VIBRATIONS FUNDAMENTALSSolution:

Physical System: Rider-Motorcycle Model Rider-Motorcycle using basic components

Model 1• 1-DOF

• keq = stiffness of tires, struts, and rider

• ceq = includes damping of struts, and rider

• meq = mass of wheels, vehicle body, and rider

• Elasticity k (stiffness) of tires, struts, and rider

• Energy dissipation c( damping) of struts and rider• Inertia m (mass) of wheels, vehicle body, and rider



Physical System: Rider-Motorcycle Model Rider-Motorcycle using basic components

Model 2• 2-DOF • Stiffness of tires & struts separately. Rider stiffness 0• Struts damping separately and rider damping 0• Mass of vehicle body and rider together

• Elasticity k (stiffness) of tires, struts, and rider

• Energy dissipation c( damping) of struts and rider• Inertia m (mass) of wheels, vehicle body, and rider


MECH. VIBRATIONS FUNDAMENTALSSolution:Physical System: Rider-Motorcycle Model Rider-Motorcycle using

basic components

Model 3• 3-DOF • Stiffness of tires, struts and rider considered• Struts and rider damping considered• Masses of tires, vehicle body and rider considered


MECH. VIBRATIONS FUNDAMENTALSSolution:Physical System: Rider-Motorcycle Model Rider-Motorcycle using

basic components

Model 4• 2-DOF • Struts Stiffness together. Rider stiffness 0• Struts damping together and rider damping 0• Masses of vehicle body and rider together• Masses of wheels together



Exercise: A study of the

vibratory response of a human body subjected tovibration/shock is importantin many applications. In a standing posture, themasses of head, uppertorso, hips, and legs and the elasticity and dampingof the neck, spinal column, abdomen, and legsinfluence the vibratoryresponse characteristics. Develop a sequence of three improvedapproximations formodeling the human body

Now the ball is on your court !!!

HEAD

UPPER TORSO

HIPS

ARM

LEGS

NECK

SPINAL COLUMN

http://images.google.com.pe/imgres?imgurl=http://www.inglaner.com/images/equipos/plancha_compactadora_01.jpg&imgrefurl=http://www.inglaner.com/equipo.html&usg=__Scon-Bq4NcfuUf3p99nfiYV0sn4=&h=590&w=650&sz=123&hl=es&start=19&sig2=kCt3IxcXEASu2cEIauuf2g&um=1&itbs=1&tbnid=jJzO9Hfx6qveMM:&tbnh=124&tbnw=137&prev=/images?q=compactadora&um=1&hl=es&lr=&rlz=1G1GGLQ_ENUS247&tbs=isch:1&ei=5ny_S_GCCIS0lQe56IHnBw

http://images.google.com.pe/imgres?imgurl=http://www.inglaner.com/images/equipos/plancha_compactadora_01.jpg&imgrefurl=http://www.inglaner.com/equipo.html&usg=__Scon-Bq4NcfuUf3p99nfiYV0sn4=&h=590&w=650&sz=123&hl=es&start=19&sig2=kCt3IxcXEASu2cEIauuf2g&um=1&itbs=1&tbnid=jJzO9Hfx6qveMM:&tbnh=124&tbnw=137&prev=/images?q=compactadora&um=1&hl=es&lr=&rlz=1G1GGLQ_ENUS247&tbs=isch:1&ei=5ny_S_GCCIS0lQe56IHnBw



HEAD

UPPER TORSO

HIPS

ARM

LEGS

NECK

SPINAL COLUMN



HEAD

UPPER TORSO

HIPS

ARM

LEGS

NECK

SPINAL COLUMN

SHOULDER


MECHANICAL VIBRATIONS FUNDAMENTALSBASIC ELEMENTS OF A VIBRATING SYSTEM

Spring Elements

• Linear • Mass and Damping negligible• Restoring Force opossed todeformation

Assume : x1 > x2

Fext FextFint Fint

Fext = Fint

Deforming Internal


MECHANICAL VIBRATIONS FUNDAMENTALSBASIC ELEMENTS OF A VIBRATING SYSTEMSpring Elements

Free Body Diagrams (FBD´s)

Strectching(“Coming out”)

Shrinking(“Coming in”)

Lo

Lo x

Fs

Lf

Fs

xLf

Fs Fs

Spring FBD

Shrinking

Fs

W

N

(“Coming in”)

Stretching

Fs

W

N

(“Coming out”)

FBD of Body attached to spring

Fs : Spring Force (in N)Epot : Potential Energy (in J)x : Spring elongation (in m)k : Spring Constant or Stiffness (in N/m)

Spring Force Potential Energy stored in spring

kxFs 2

2

1kxEpot


MECHANICAL VIBRATIONS FUNDAMENTALSBASIC ELEMENTS OF A VIBRATING SYSTEMSpring Combinations

In Parallel

• Equivalent spring can replace original system

• All elongations are equal

• Forces in each spring are different

• Equilibrium

keq=

21 st

111 kF stkF 22 stkF 11 222 kF

221121 kkFFW

stststeq kkFFk 2121 21 kkkeq


MECHANICAL VIBRATIONS FUNDAMENTALSBASIC ELEMENTS OF A VIBRATING SYSTEMSpring Combinations

In Series

• Equivalent spring can replace original system

• Total elongation is summation of elongations

• Forces in each spring are equal because of equilibrium

=

21 st

21 FFW

111 kF 12kW

22kW 222 kF

1

1

k

W

2

2

k

W

21 st

21 k

W

k

W

k

W

eq 21

111

kkkeq

keq


MECHANICAL VIBRATIONS FUNDAMENTALSBASIC ELEMENTS OF A VIBRATING SYSTEMSpring Combinations in general

neq kkkk ...21

eqn ...21

eqeqnn

seqsnss

kkkk

FFFF

...

...

2211

21In Parallel

nkkeq Special case

kkkk n ...21

neq kkkk

1...

111

21

kkkk n ...21n

kkeq

2n21

21

kk

kkkeq

eqn ...21

eq

seq

n

snss

k

F

k

F

k

F

k

F ...

2

2

1

1In Series

Special case

Special case



Solution:

Example: Find the equivalent stiffness k of the following system diameter d = 2 cm

Springs in parallel and series:

k1k2

k3

k4

m

k5

k3

k4

m

k1+k2+k5



Solution:

m

43

43521

kk

kkkkkkeq

m

k1+k2+k5

1

1

k31

k4

k3k4k3 k4

k3

k4

m

k1+k2+k5



Exercise: Determine the equivalent spring constant of the system shown

Your turn !!!



Solution:

keq=


MECHANICAL VIBRATIONS FUNDAMENTALSBASIC ELEMENTS OF A VIBRATING SYSTEMMasa suspendida al final de una viga en voladizo (Flexión):

De Resistencia de Materiales

Deflección Estatica al final de una viga en voladizo debido a masa “m” en el extremo.Asumir que masa de barra << “m”

EI

lmg

EI

Wlst

33

33

kF

Fk

Ley de Hooke Analogía

3

3

l

EIWk

st

Sistema Real o Situación Física

(a) Modelo de 1GDL (asume que no hay amortiguamiento c=0)


MECHANICAL VIBRATIONS FUNDAMENTALSBASIC ELEMENTS OF A VIBRATING SYSTEMMasa suspendida al final de una barra (Torsión):

(a) Sistema Real o Situación Física

(b) Modelo de 1GDL (asume que no hay amortiguamiento )

De Resistencia de Materiales

Desplazamiento angular quasi-estatico al final de una barra redonda debido a torque “M” en el extremo.

44

32

32

dG

ML

dG

ML

GI

ML

p

st

tkM

Mkt

Ley de Hooke Analogía

L

dG

dG

ML

MMk

st

t3232

4

4

t

(a) (b)

M

LG

L

dGkt

32

4

d


MECHANICAL VIBRATIONS FUNDAMENTALSBASIC ELEMENTS OF A VIBRATING SYSTEMConstantes de Rigidez para otros Tipos Elementos Simples (Ejemplo)



Solution:

Example: The figure shows the suspension system of a freight truck with a parallel spring arrangement . Find the equivalent spring constant of the suspension if each of the three helical springs is made of steel (G=80x109 N/m2) and has five effective turns, mean coil diameter D =20 cm, and wire diameter d = 2 cm

The stiffness of each helical spring is:

mNnD

Gdk /000,40

520.08

02.01080

83

49

3

4

Parallel spring arrangement:

mNmNkkeq /000,120/000,4033



Solution:

Ejemplo: Determine the torsional spring constant of the steel propeller shaft shown

• Consider shaft by parts: 12 y 23• Induced torque in any cross section of the shaft equal to the applied torque “T” (draw imaginary sections at A-A y B-B)• Segments 12 y 23 regarded as series springs

radmN

l

dDG

l

GJkt /1053.25

232

2.03.01080

32

6

449

12

4

12

4

12

12

12

12



Solution: (cont´d)

radmN

l

dDG

l

GJkt /109.8

332

15.025.01080

32

6

449

23

4

23

4

23

23

23

23

Series spring

radmNkk

kkk

tt

tt

eqt /106.6 6

2312

2312

2312

111

tteqt kkk


BIBLIOGRAPHY

BÁSICA:•Thomson, W.T., Dahleh, M.D., 1997, “Teoria de Vibraciones con Aplicaciones”, Prentice HallIberoamericana, 5ta Edición, México.•Inman, D., 2007, “Engineering Vibration”, Prentice Hall, 3rd Edition, USA.•Moore, H., 2008, “Matlab for Engineers”, Prentice Hall, 2nd Edition, USA.

COMPLEMENTARIA:•Balachandran, B., Magrab, E., 2006, “Vibraciones”, Thomson, 5ta Edición, México•Rao, S.S., 2004, “Mechanical Vibrations”, Ed. Prentice Hall, 4th Edition, USA.

ESPECIALIZADA:•Hartog, D., 1974, “Mecánica de las Vibraciones”, Cecsa, Mexico.•Harris, C., Piersol, A., 2001, “Harri´s Shock and Vibration Handbook”, McGraw Hill Professional,

5th Edition. USA.

Introduction

Documents

Transcript of Introduction