mechanic of machine-mechanical vibrations

VECTOR MECHANICS FOR ENGINEERS: DYNAMICSDYNAMICS

Ninth Ninth EditionEdition

Ferdinand P. BeerFerdinand P. Beer

E. Russell Johnston, Jr.E. Russell Johnston, Jr.

Lecture Notes:Lecture Notes:

J. Walt OlerJ. Walt Oler

Texas Tech UniversityTexas Tech University

CHAPTER

© 2010 The McGraw-Hill Companies, Inc. All rights reserved.

19Mechanical Vibrations


Vector Mechanics for Engineers: DynamicsVector Mechanics for Engineers: Dynamics

Nin

thE

ditio

n

Contents

19 - 2

Introduction

Free Vibrations of Particles. Simple Harmonic Motion

Simple Pendulum (Approximate Solution)

Simple Pendulum (Exact Solution)

Sample Problem 19.1

Free Vibrations of Rigid Bodies

Sample Problem 19.2

Sample Problem 19.3

Principle of Conservation of Energy

Sample Problem 19.4

Forced Vibrations

Sample Problem 19.5

Damped Free Vibrations

Damped Forced Vibrations

Electrical Analogues



Nin

thE

ditio

n

Introduction

19 - 3

• Mechanical vibration is the motion of a particle or body which oscillates about a position of equilibrium. Most vibrations in machines and structures are undesirable due to increased stresses and energy losses.

• Time interval required for a system to complete a full cycle of the motion is the period of the vibration.

• Number of cycles per unit time defines the frequency of the vibrations.

• Maximum displacement of the system from the equilibrium position is the amplitude of the vibration.

• When the motion is maintained by the restoring forces only, the vibration is described as free vibration. When a periodic force is applied to the system, the motion is described as forced vibration.

• When the frictional dissipation of energy is neglected, the motion is said to be undamped. Actually, all vibrations are damped to some degree.



Nin

thE

ditio

n


19 - 4

• If a particle is displaced through a distance xm from its equilibrium position and released with no velocity, the particle will undergo simple harmonic motion,

0

kxxm

kxxkWFma st

• General solution is the sum of two particular solutions,

tCtC

tm

kCt

m

kCx

nn cossin

cossin

21

21

• x is a periodic function and n is the natural circular frequency of the motion.

• C1 and C2 are determined by the initial conditions:

tCtCx nn cossin 21 02 xC

nvC 01 tCtCxv nnnn sincos 21



Nin

thE

ditio

n


19 - 5

txx nm sin

n

n 2

period

2

1 n

nnf natural frequency

20

20 xvx nm amplitude

nxv 00

1tan phase angle

• Displacement is equivalent to the x component of the sum of two vectorswhich rotate with constant angular velocity

21 CC

.n

02

01

xC

vC

n



Nin

thE

ditio

n


19 - 6

txx nm sin

• Velocity-time and acceleration-time curves can be represented by sine curves of the same period as the displacement-time curve but different phase angles.

2sin

cos

tx

tx

xv

nnm

nnm

tx

tx

xa

nnm

nnm

sin

sin2

2



Nin

thE

ditio

n

Simple Pendulum (Approximate Solution)

19 - 7

• Results obtained for the spring-mass system can be applied whenever the resultant force on a particle is proportional to the displacement and directed towards the equilibrium position.

for small angles,

g

l

tl

g

nn

nm

22

sin

0

:tt maF

• Consider tangential components of acceleration and force for a simple pendulum,

0sin

sin

l

g

mlW



Nin

thE

ditio

n

Simple Pendulum (Exact Solution)

19 - 8

0sin l

gAn exact solution for

leads to

2

022 sin2sin1

4

m

nd

g

l

which requires numerical solution.

g

lKn

2

2



Nin

thE

ditio

n

Sample Problem 19.1

19 - 9

A 50-kg block moves between vertical guides as shown. The block is pulled 40mm down from its equilibrium position and released.

For each spring arrangement, determine a) the period of the vibration, b) the maximum velocity of the block, and c) the maximum acceleration of the block.

SOLUTION:

• For each spring arrangement, determine the spring constant for a single equivalent spring.

• Apply the approximate relations for the harmonic motion of a spring-mass system.



Nin

thE

ditio

n

Sample Problem 19.1

19 - 10

mkN6mkN4 21 kk SOLUTION:

• Springs in parallel:

- determine the spring constant for equivalent spring

mN10mkN10 4

21

21

kkP

k

kkP

- apply the approximate relations for the harmonic motion of a spring-mass system

nn

n m

k

2

srad14.14kg20

N/m104

s 444.0n

srad 4.141m 040.0 nmm xv

sm566.0mv

2sm00.8ma 22

srad 4.141m 040.0

nmm axa



Nin

thE

ditio

n

Sample Problem 19.1

19 - 11

mkN6mkN4 21 kk • Springs in series:

- determine the spring constant for equivalent spring

- apply the approximate relations for the harmonic motion of a spring-mass system

nn

n m

k

2

srad93.6kg20

400N/m2

s 907.0n

srad .936m 040.0 nmm xv

sm277.0mv

2sm920.1ma 22

srad .936m 040.0

nmm axa

mN10mkN10 4

21

21

kkP

k

kkP



Nin

thE

ditio

n

Free Vibrations of Rigid Bodies

19 - 12

• If an equation of motion takes the form

0or0 22 nn xx the corresponding motion may be considered as simple harmonic motion.

• Analysis objective is to determine n.

mgWmbbbmI ,22but 23222

121

05

3sin

5

3 b

g

b

g

g

b

b

g

nnn 3

52

2,

5

3then

• For an equivalent simple pendulum,35bl

• Consider the oscillations of a square plate ImbbW sin



Nin

thE

ditio

n

Sample Problem 19.2

19 - 13

k

A cylinder of weight W is suspended as shown.

Determine the period and natural frequency of vibrations of the cylinder.

SOLUTION:

• From the kinematics of the system, relate the linear displacement and acceleration to the rotation of the cylinder.

• Based on a free-body-diagram equation for the equivalence of the external and effective forces, write the equation of motion.

• Substitute the kinematic relations to arrive at an equation involving only the angular displacement and acceleration.



Nin

thE

ditio

n

Sample Problem 19.2

19 - 14

SOLUTION:• From the kinematics of the system, relate the linear

displacement and acceleration to the rotation of the cylinder.rx rx 22

rra

ra • Based on a free-body-diagram equation for the equivalence of

the external and effective forces, write the equation of motion. : effAA MM IramrTWr 22

rkWkTT 2but21

02

• Substitute the kinematic relations to arrive at an equation involving only the angular displacement and acceleration.

03

8

22 221

21

m

k

mrrrmrkrWWr

m

kn 3

8

k

m

nn 8

32

2

m

kf nn 3

8

2

1

2



Nin

thE

ditio

n

Sample Problem 19.3

19 - 15

s13.1

lb20

n

W

s93.1n

The disk and gear undergo torsional vibration with the periods shown. Assume that the moment exerted by the wire is proportional to the twist angle.

Determine a) the wire torsional spring constant, b) the centroidal moment of inertia of the gear, and c) the maximum angular velocity of the gear if rotated through 90o and released.

SOLUTION:

• Using the free-body-diagram equation for the equivalence of the external and effective moments, write the equation of motion for the disk/gear and wire.

• With the natural frequency and moment of inertia for the disk known, calculate the torsional spring constant.

• With natural frequency and spring constant known, calculate the moment of inertia for the gear.

• Apply the relations for simple harmonic motion to calculate the maximum gear velocity.



Nin

thE

ditio

n

Sample Problem 19.3

19 - 16

s13.1

lb20

n

W

s93.1n

SOLUTION:• Using the free-body-diagram equation for the equivalence

of the external and effective moments, write the equation of motion for the disk/gear and wire.

: effOO MM

0

I

K

IK

K

I

I

K

nnn

2

2

• With the natural frequency and moment of inertia for the disk known, calculate the torsional spring constant.

22

221 sftlb 138.0

12

8

2.32

20

2

1

mrI

K

138.0213.1 radftlb27.4 K



Nin

thE

ditio

n

Sample Problem 19.3

19 - 17

s13.1

lb20

n

W

s93.1n

radftlb27.4 K

K

I

I

K

nnn

2

2

• With natural frequency and spring constant known, calculate the moment of inertia for the gear.

27.4293.1

I 2sftlb 403.0 I

• Apply the relations for simple harmonic motion to calculate the maximum gear velocity.

nmmnnmnm tt sinsin

rad 571.190 m

s 93.1

2rad 571.1

2 n

mm

srad11.5m



Nin

thE

ditio

n

Principle of Conservation of Energy

19 - 18

• Resultant force on a mass in simple harmonic motion is conservative - total energy is conserved.

constantVT

222

2212

21 constant

xx

kxxm

n

• Consider simple harmonic motion of the square plate, 01 T

221

21 2sin2cos1

m

m

Wb

WbWbV

2235

21

2232

212

21

2212

21

2

m

mm

mm

mb

mbbm

IvmT

02 V

00 22235

212

21

2211

nmm mbWb

VTVT

bgn 53



Nin

thE

ditio

n

Sample Problem 19.4

19 - 19

Determine the period of small oscillations of a cylinder which rolls without slipping inside a curved surface.

SOLUTION:

• Apply the principle of conservation of energy between the positions of maximum and minimum potential energy.

• Solve the energy equation for the natural frequency of the oscillations.



Nin

thE

ditio

n

Sample Problem 19.4

19 - 20

01 T 2

cos12

1

mrRW

rRWWhV

22

43

22

221

212

21

2212

21

2

m

mm

mm

rRm

r

rRmrrRm

IvmT

02 V

SOLUTION:• Apply the principle of conservation of energy between the

positions of maximum and minimum potential energy.

2211 VTVT



Nin

thE

ditio

n

Sample Problem 19.4

19 - 21

2211 VTVT

02

0 2243

2 m

m rRmrRW

2243

2

2 mnmm rRmrRmg

rR

gn

3

22g

rR

nn

2

32

2

01 T 221 mrRWV

2243

2 mrRmT 02 V

• Solve the energy equation for the natural frequency of the oscillations.



Nin

thE

ditio

n

Forced Vibrations

19 - 22

:maF

xmxkWtP stfm sin

tPkxxm fm sin

xmtxkW fmst sin

tkkxxm fm sin

Forced vibrations - Occur when a system is subjected to a periodic force or a periodic displacement of a support.

f forced frequency



Nin

thE

ditio

n

Forced Vibrations

19 - 23

txtCtC

xxx

fmnn

particulararycomplement

sincossin 21

222 11 nf

m

nf

m

f

mm

kP

mk

Px

tkkxxm fm sin

tPkxxm fm sin

At f = n, forcing input is in resonance with the system.

tPtkxtxm fmfmfmf sinsinsin2 Substituting particular solution into governing equation,



Nin

thE

ditio

n

Sample Problem 19.5

19 - 24

A motor weighing 350 lb is supported by four springs, each having a constant 750 lb/in. The unbalance of the motor is equivalent to a weight of 1 oz located 6 in. from the axis of rotation.

Determine a) speed in rpm at which resonance will occur, and b) amplitude of the vibration at 1200 rpm.

SOLUTION:

• The resonant frequency is equal to the natural frequency of the system.

• Evaluate the magnitude of the periodic force due to the motor unbalance. Determine the vibration amplitude from the frequency ratio at 1200 rpm.



Nin

thE

ditio

n

Sample Problem 19.5

19 - 25

SOLUTION:• The resonant frequency is equal to the natural frequency of

the system.

ftslb87.102.32

350 2m

ftlb000,36

inlb30007504

k

W = 350 lb

k = 4(350 lb/in)

rpm 549rad/s 5.5787.10

000,36

m

kn

Resonance speed = 549 rpm



Nin

thE

ditio

n

Sample Problem 19.5

19 - 26

W = 350 lb

k = 4(350 lb/in)

rad/s 5.57n

• Evaluate the magnitude of the periodic force due to the motor unbalance. Determine the vibration amplitude from the frequency ratio at 1200 rpm.

ftslb001941.0sft2.32

1

oz 16

lb 1oz 1

rad/s 125.7 rpm 1200

22

m

f

lb 33.157.125001941.0 2126

2

mrmaP nm

in 001352.0

5.577.1251

300033.15

1 22

nf

mm

kPx

xm = 0.001352 in. (out of phase)



Nin

thE

ditio

n


19 - 27

• With viscous damping due to fluid friction,

:maF 0

kxxcxm

xmxcxkW st

• Substituting x = et and dividing through by et yields the characteristic equation,

m

k

m

c

m

ckcm

22

220

• Define the critical damping coefficient such that

ncc m

m

kmc

m

k

m

c 2202

2

• All vibrations are damped to some degree by forces due to dry friction, fluid friction, or internal friction.



Nin

thE

ditio

n


19 - 28

• Characteristic equation,

m

k

m

c

m

ckcm

22

220

nc mc 2 critical damping coefficient

• Heavy damping: c > cc

tt eCeCx 2121

- negative roots - nonvibratory motion

• Critical damping: c = cc

tnetCCx 21 - double roots - nonvibratory motion

• Light damping: c < cc

tCtCex ddtmc cossin 21

2

2

1c

nd c

c damped frequency



Nin

thE

ditio

n

Damped Forced Vibrations

19 - 29

2

222

1

2tan

21

1

nf

nfc

nfcnf

m

m

m

cc

cc

x

kP

x

magnification

factor

phase difference between forcing and steady state response

tPkxxcxm fm sin particulararycomplement xxx



Nin

thE

ditio

n


19 - 30

• Consider an electrical circuit consisting of an inductor, resistor and capacitor with a source of alternating voltage

0sin C

qRi

dt

diLtE fm

• Oscillations of the electrical system are analogous to damped forced vibrations of a mechanical system.

tEqC

qRqL fm sin1



Nin

thE

ditio

n


19 - 31

• The analogy between electrical and mechanical systems also applies to transient as well as steady-state oscillations.

• With a charge q = q0 on the capacitor, closing the switch is analogous to releasing the mass of the mechanical system with no initial velocity at x = x0.

• If the circuit includes a battery with constant voltage E, closing the switch is analogous to suddenly applying a force of constant magnitude P to the mass of the mechanical system.



Nin

thE

ditio

n


19 - 32

• The electrical system analogy provides a means of experimentally determining the characteristics of a given mechanical system.

• For the mechanical system,

0212112121111 xxkxkxxcxcxm

tPxxkxxcxm fm sin12212222

• For the electrical system,

02

21

1

121111

C

qq

C

qqqRqL

tEC

qqqqRqL fm sin

2

1212222

• The governing equations are equivalent. The characteristics of the vibrations of the mechanical system may be inferred from the oscillations of the electrical system.

mechanic of machine-mechanical vibrations

Education