Vibration Chapter01 Edited Std

8/19/2019 Vibration Chapter01 Edited Std

1/23

MCT 4311 : Mechanical Vibration

“I n e d u c a t i o n , n o t h i n g w o r k s' ”

Dr. [email protected]

Mechanical Vibrations

Fifth Edition in SI Units

Singiresu S. Rao

Chapter Outline

1.1 Preliminary Remarks

. Br e H story o V rat on

1.3 Importance of the Study of Vibration

.

1.5 Classification of Vibration

1.6 Vibration Analysis Procedure

1.7 Spring Elements

1.8 Mass or Inertia Elements

. amp ng emen s

1.10 Harmonic Motion

.

3

1.1 Preliminary Remarks

•

• Examination of vibration’s important role

• Vibration anal sis of an en ineerin s stem

• Definitions and concepts of vibration

• Concept of harmonic analysis for general periodic motions

4


2/23


3/23

1.2 Brief History of Vibration

Recent contributions:

1950s – developed finite element method

– yapunov a e oun a ons o

modern stability theory which is

applicable to all types of dynamical

numerically detailed vibration analysis of

complex mechanical, vehicular, and

systems

1920 – Duffling and Van der Polof degrees of freedom with the aid of

computers

into the theory of nonlinear

vibrations and drew attention to its

• Turner, Clough, Martin and Topp

presented the finite element method as

importance in engineering

– Introduction of the correlation

function b Ta lor

nown o ay

1950 – advent of high-speed digital

computers

9

– genera e approx ma e so u ons

1.3 Importance of the Study of Vibration

Tacoma Bridge !

• Why study vibration?

• Vibrations can lead to excessive deflections and failure on themachines and structures

• To reduce vibration through proper design of machines and theirmountings

• To utilize profitably in several consumer and industrial applications

• , ,welding processes

• To stimulate earth uakes for eolo ical research and conduct

10

studies in design of nuclear reactors

1.4 Basic Concepts of Vibration

• =

• Vibratory System consists of:

1. spring or elasticity2. mass or inertia

3. damper

• Involves transfer of potential energy to kinetic

energy and vice versa

11


• . . . =

min. no. of independent coordinates required to determine

com letel the ositions of all arts of a s stem at an instant of

time

• Exam les of sin le de ree-of-freedom s stems:

12


4/23


• - -

13


• - -

14


• - - - - -

• Infinite number of degrees of freedom system are termed

continuous or istri ute systems

• Finite number of degrees of freedom are termed discrete or lumped

parame er sys ems

• More accurate results obtained by increasing number of degrees of

ree om

15


•

A system is left to vibrate on its own after an initial disturbance and

no external force acts on the s stem. E. . sim le endulum

• Forced Vibration:

A s stem that is sub ected to a re eatin external force. E. .

oscillation arises from diesel engines

• Resonance occurs when the fre uenc of the external force

coincides with one of the natural frequencies of the system

16


5/23


•

When n o energy is lost or dissipated in friction or other resistance

durin oscillations

• Damped Vibration:

When a n ener is lost or dissi ated in friction or other resistance

during oscillations

• Linear Vibration:

When a l l basic components of a vibratory system, i.e. the spring,

the mass and the damper behave linearly

17


•

If a n y of the components behave nonlinearly

•

If the value or magnitude of the excitation (force or motion) acting

on a vibrator s stem is known at an iven time

• Nondeterministic or random Vibration:

When the value of the excitation at a iven time cannot be

predicted

18

1.5 Classification of Vibration

•

19


Step 2: Derivation of Governing Equations

Step 3: Solution of the Governing EquationsStep 4: Interpretation of the Results

20


6/23


•

21


.

Mathematical Model of a Motorcycle

Figure shows a motorcycle with a rider. Develop a sequence of three

mathematical models of the system for investigating vibration in the

. ,

damping of the struts (in the vertical direction), masses of the wheels,

and elasticity, damping, and mass of the rider.

22


.


SolutionWe start with the simplest model and refine it gradually. When the

equiva ent va ues o t e mass, sti ness, an amping o t e system

are used, we obtain a single-degree of freedom model of the

. ,

equivalent stiffness (k eq) includes the stiffness of the tires, struts, and

rider. The equivalent damping constant (c e ) includes the damping of

the struts and the rider. The equivalent mass includes the mass of thewheels, vehicle body and the rider.

23


Example 1.1

Solution

24


7/23


.


Solution

This model can be refined by representing the masses of wheels,

e asticity o tires, an e asticity an amping o t e struts separate y,

as shown in Figure (c). In this model, the mass of the vehicle body

v r , v

mr. When the elasticity (as spring constant k r) and damping (as

damping constant c r) of the rider are considered, the refined model

shown in Figure (d) can be obtained.

25


.


Solution

26


.


SolutionNote that the models shown in Figure (b) to (d) are not unique. For

examp e, y com n ng e spr ng cons an s o o res, e masses

of both wheels, and the spring and damping constants of both struts

as sin le uantities the model shown in Fi ure e can be obtained

instead of Figure (c).

27

1.7 Spring Elements

•

to have negligible mass and damping

•

F = spring force,k = spring stiffness or spring constant, and x = deformation (displacement of one end

with respect to the other)

28


8/23

1.7 Spring Elements

•

energy is given by:

• W en an incrementa orce ∆F is a e to F:

* Δ+=Δ+ x xF F F

...)(!2

1

)()(

2

2

2*

** +Δ+Δ+= xdx

F d

xdx

dF

xF x

29

1.7 Spring Elements

30

1.7 Spring Elements

• Static deflection of a beam at the

W = mg is the weight of the mass m, E = Young’s Modulus, and I = moment of inertia of cross-sectionof beam

• Spring Constant is given by:

31

1.7 Spring Elements

•

1) Springs in parallel – if we have n spring constants k 1, k 2, …, k n in

, eq

• Combination of Springs:

2) Springs in series – if we have n spring constants k 1, k 2, …, k n in

series, then the equivalent spring constant keq is:

32


9/23

1.7 Spring Elements

33

1.7 Spring Elements

.

Torsional Spring Constant of a Propeller Shaft

Determine the torsional spring constant of the speed propeller shaft

shown in the figure.

34

1.7 Spring Elements

.

Torsional Spring Constant of a Propeller Shaft

SolutionWe need to consider the segments 12 and 23 of the shaft as springs in

combination. From the figure, the torque induced at any cross section

of the shaft (such as AA or BB) can be seen to be equal to the torque

, . ,

corresponding to the two segments 12 and 23 are to be considered as

series s rin s. The s rin constants of se ments 12 and 23 of the

shaft (k t12 and k t23) are given by

35

1.7 Spring Elements

Example 1.6

Torsional S rin Constant of a Pro eller Shaft Solution

)2(32

)2.03.0()1080(

32

)( 449

12

4

12

4

12

12

12

12

−×=

−==

π π

l

d DG

l

GJ k

t

)15.025.0()1080()( 44944 −×− π π d DGGJ

m/rad - N105255.25 6

×=

m/rad - N109012.8

)3(32326

2323

23

×=

===ll

t

Since the springs are in series

)109012.8105255.25()109012.8)(105255.25(

66

66

2312

2312

×+×××=

+=

t t

t t

t

k k

k k k eq

36

m/rad - N105997.6 6×=


10/23

1.7 Spring Elements

.

Equivalent k of a Crane

The boom AB of crane is a uniform steel

bar of length 10 m and x-section area of2 , .

the crane is stationary. Steel cable CDEBF

has x-sectional area of 100 mm2. Neglect

effect of cable CDEB, find equivalent spring

constant of system in the vertical direction.

37

1.7 Spring Elements

.


Solution

A vertical displacement x of pt B will cause

t e spring k 2 oom to e orm y

x2 = x cos 45º and the spring k 1 (cable)

= º – .

Length of cable FB, l1 is as shown.

38

1.7 Spring Elements

.


SolutionThe angle θ satisfies the relation: =−+ 10cos)3)((23

21

221 θ ll

The total potential energy (U):

°=∴= 0736.35,8184.0cos θ θ

2

2

2

1 )]90cos([1

)45cos(1

θ −°+°= xk xk U

N/m106822.1)10207)(10100( 6

9611 ×=

××==

− E Ak

0355.121l

N/m101750.5)10207)(102500( 7

9622

2 ×=××

==− E A

k

39

2

1.7 Spring Elements

.


SolutionPotential Energy of the equivalent spring is:

By setting U = U eq, hence: N/m104304.26 6

eq ×=k

40


11/23


•

E.g. In the figure below, the mass and damping of the beam

spring-mass system as shown.

41


•

E.g. Assume that the mass of the frame is negligible compared

.

represent the mass elements, and the elasticities of the vertical

members denote the s rin elements.

42


• Case 1: Translational Masses Connected by a Rigid Bar

Ve ocities o masses can e expresse as:

By equating the kinetic energy of the system :

43


•

meq = single equivalent translational mass

= translational velocity= rotational velocity

J = mass moment of inertia

x&θ &

44

Jeq = single equivalent rotational mass

1 8 Mass or Inertia Elements


12/23


•

1. E uivalent translational mass:

Kinetic energy of the two masses is given by:

Kinetic ener of the e uivalent mass is iven b :

Since and , equating T eq & T gives

45


•

.

Here, and , equating T eq and T givesθ θ && =eq R x θ && =

46


.

Equivalent Mass of a System

Find the equivalent mass of the system as shown, where the rigid link1 is attached to the pulley and rotates with it.

47


.


SolutionAssuming small displacements, the equivalent mass (meq) can be

systems. When the mass m is displaced by a distance , the pulley

and the rigid link 1 rotate by an angle . This causes the

x p pr x /

1==θ θ

rigid link 2 and the cylinder to be displaced by a distance

. Since the cylinder rolls without slippage, it rotates by p pr xll x /

112==θ

an angle . The kinetic energy o the system (T) canbe expressed (for small displacements) as:

c pcc

r r xr x //12

==θ

48


13/23


.


Solution

E.1111111

222222

xm J xm J J xmT &&&&&& +++++= θ θ θ

where Jp, J1, and Jc denote the mass moments of inertia of the pulley,

222222 22211 ccc p p

link 1 (about O), and cylinder, respectively, indicate the

angular velocities of the pulley, link 1 (about O), and cylinder,

c pθ θ θ &&& and ,

1

&&respec ve y, an represen e near ve oc es o e mass m

and link 2, respectively.

2

49


Example 1.11

qu va en ass o a ys em

Solution

Noting that , Equation (E.1) can be rewritten2/2

cccr m J = 3/and 2

111lm J =

as

.

system

50


.


Solution

We obtain the equivalent mass of the system as

11 21

2

1

2

12

2

11 c p llmlmlm J

.23 22222

p

c

p p p p

eq

r r r r r =

51


.

Cam-Follower Mechanism

-

shaft into the oscillating or reciprocating motion of a valve. The followersystem consists of a pushrod of mass mp, a rocker arm of mass mr, andmass moment of inertia about its C.G. a valve of mass m and avalve spring of negligible mass. Find the equivalent mass (meq) of thiscam-follower system by assuming the location of meq as (i) point A and(ii) point C.

52


14/23


.


53


.


Solution

The kinetic energy of the system ( T ) is:

If meq denotes equivalent mass placed at pt A, with , the kinetic

energy equivalent mass system T eq is:

x x && =eq

54


.


SolutionBy equating T and T eq , and note that

11

3

1

2 and ,,,l

x

l

x x

l

x x x x r r v p

&&

&&

&&&& ==== θ

23

22 ll J r

Similarl if e uivalent mass is located at oint C hence

.21

21

21

eq lll

r v p

x x && =

( )4.E11 2

eq 2eq eq eq v xm xmT && ==

55


.


Solution

Equating (E.4) and (E.1) gives

222

( )5.E21

3

2

122

eq ⎟⎟

⎜⎜⎝

+⎟ ⎠

⎜⎝

++=l

ml

ml

mm r pr

v

56


15/23

1.9 Damping Elements

• Damping force is proportional to the velocity of the vibrating bodyin a fluid medium such as air, water, gas, and oil.

• Coulomb or Dry Friction Damping:Damping force is constant in magnitude but opposite in direction tothat of the motion of the vibrating body between dry surfaces

• Energy is absorbed or dissipated by material during deformation

due to friction between internal planes

57


•

58


• τ

the fixed plate is:

where d u/ d y = v/h is the velocity gradient.

• Shear or Resisting Force (F) developed at the bottom surface of the

where A is the surface area of the moving plate and is the

damping constant

59


• ,

operating velocity (v*) and the equivalent damping constant is:

60


16/23


.

Equivalent Spring and Damping Constants of a Machine Tool

A precision milling machine is supported on four shock mounts, as

.

can be modeled as a spring and a viscous damper, as shown in Figure

(b . Find the e uivalent s rin constant k and the e uivalent

damping constant, c eq, of the machine tool support in terms of the

spring constants (k i ) and damping constants (c i ) of the mounts.

61


.


62


.


Solution

The free-body diagrams of the four springs and four dampers are shown

in Figure (c). Assuming that the center of mass, G, is located

,

that all the springs will be subjected to the same displacement, , and

all the dam ers will be sub ect to the same relative velocit where& xand denote the displacement and velocity, respectively, of the centerof mass, G. Hence the forces acting on the springs (F si ) and the dampers

x&

63

(F di ) can be expressed as


.


Solution

64


17/23


.


Solution 4,3,2,1; == i xk F isi

Let the total forces acting on all the springs and all the dampers be F s

E.1)(4,3,2,1; == i xcF idi&

and F d , respectively (see Figure d). The force equilibrium equations can

thus be expressed as

E.2

4321 sssss

F F F F F

F F F F F

+++=

+++=

65


.


Solution

where F s + F d = W , with W denoting the total vertical force (including

the inertia force) acting on the milling machine. From Figure (d), we

have

E.3)( xcF eqd

eqs

&=

66


.


Solution

Equation (E.2) along with Eqs. (E.1) and (E.3), yield

44321

k k k k k k eq

=+++=

where k i = k and c i = c for i = 1, 2, 3, 4.

.4321eq==

67


.


Solution

Note: If the center of mass, G, is not located symmetrically with respect

to the four springs and dampers, the i th spring experiences a

displacement of and the i th damper experiences a velocity of wherei i

x& i x

& &

of mass of the milling machine, G. In such a case, Eqs. (E.1) and (E.4)

need to be modified suitabl .

i

68


18/23


•

• Harmonic Motion: simplest type of periodic motion

• Displacement ( x): (on horizontal axis)

• Velocity:

• Acceleration:

69


•

The similarit between c clic harmonic and sinusoidal motion.

70


• ,

• Thus,22 , j j j

2,1;tan 1 =

⎟⎟ ⎞

⎜⎜⎛

= − ja

b

j

j

jθ

• Operations on Harmonic Functions:

v

e X =

vector,Rotat ng

71


•

where Re denotes the real part

• Displacement, velocity, and accelerations as rotating vectors

• Vectorial addition of harmonic functions

72



19/23


73

Example 1.18 Addition of Harmonic Motions

Find the sum of the two harmonic motions.

Method 1: By using trigonometric relations: Since the circular frequency

1 2 ,

74


.

Addition of Harmonic Motions

Solution

By equating the corresponding coefficients of cosωt and sinωt on both

sides, we obtain

2cos1510cos += A α ⎞⎛ = −

2sin15tan

1α

( ) )2sin15(2cos1510

s ns n

22

=

++=

=

A

α

°=

+

5963.74

2cos1510

75

.


.


Solution

Method 2: By using vectors: For an arbitrary value of ωt , the harmonic

motions x 1(t) and x 2(t) can be denoted graphically as shown in Figure.

By adding them vectorially, the resultant vector x(t) can be found to be

°= ...

76


20/23


.


Solution

Method 3: By using complex number representation: the two harmonic

motions can be denoted in terms of complex numbers:

[ ] [ ]10ReRe)()2()2(

11

++

=

≡=t it i

t it i

ee At xω ω

ω ω

The sum of x 1(t) and x 2(t) can be expressed as

.22

77

E.8)(Re)( )( α ω += t i Aet x


.


Solution

where A and α can be determined using Eqs. (1.47) and (1.48) as A =

14.1477 and α = 74.5963º

78


•

mp u e A s e max mum sp acemen o a v ra ng o y

from its equilibrium position

Period of oscillation (T ) is time taken to complete one cycle of

.

79


•

a ura requency s e requency w c a sys em osc a es

without external forces

Phase angle (φ ) is the angular difference between two synchronous

80

1 10 H i M ti

1 11 H i A l i


21/23


•

.

often used as a notation of various quantities such as

displacement, velocity, acceleration, pressure, and power

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

0

log10dBP

P

⎟⎟ ⎠

⎜⎜⎝

=0

log20dB X

X

reference voltage.

81

1.11 Harmonic Analysis

•

If t is a eriodic function with eriod its Fourier Seriesτ representation is given by

• Gibbs Phenomenon:An anomalous behavior observed from a periodic function that is

being represented by Fourier series.

82


• Gibbs Phenomenon:

As n increases, the approximation can be seen to improveeverywhere except in the vicinity of the discontinuity, P . The error

in amplitude remains at approximately 9 percent, even when ∞→

83


• Complex Fourier Series:

The Fourier series can also be represented in terms of complex

.

84


22/23

1 11 Harmonic Analysis 1 11 Harmonic Analysis


23/23


• Half-Range Expansions:

The function is extended to include the interval as shown in the

figure. The Fourier series expansions of x 1(t ) and x 2(t ) are known as

0toτ −

half-range expansions.

89


• Numerical Computation of Coefficients

If x (t ) is not in a simple form, experimental determination of the

amplitude of vibration and numerical integration procedure like the

trapezoidal or Simpson’s rule is used to determine the coefficients

an and bn. N 2

π i N

i

i

t n xa

x N

a

2cos2

1

0

=

=

=

π

τ

i

N

in

i

in

t n xb

N

2sin

2

1

∑

=

=

90

i 1=

1.11 Harmonic Analysis xamp e .

Fourier Series Expansion

Determine the Fourier series expansion of the motion of the valve in the

cam- o ower system s own in t e Figure.

91

Date post:	07-Aug-2018
Category:	Documents
Upload:	eiyba-adibah
View:	214 times
Download:	0 times

Vibration Chapter01 Edited Std

Documents