1 Vibration Analysis Procedure - Kasetsart Universitypirun.ku.ac.th/~fengwtc/Teaching/208322...

208322 Mechanical Vibrations Lesson 2

1 Vibration Analysis Procedure The analysis of a vibrating system usually involves four steps: mathematical modeling, derivation of the governing equations, solution of the equations, and interpretation of the results.

Step 1: Mathematical Modeling Mathematical model represents all the important features of the system for the purpose of deriving the mathematical equations governing the system’s behavior. The mathematical model should include enough details to be able to describe the system in terms of equations without making it too complex. The mathematical model may be linear or nonlinear, depending on the behavior of the system’s components. Linear models permit quick solutions and are simple to handle; however, nonlinear models sometimes reveal certain characteristics of the system that cannot be predicted using linear models. First, we can use a very crude or elementary model to get a quick insight into the overall behavior of the system. Subsequently, the model is refined by including more components and details so that the behavior of the system can be observed more closely. ☻ Example 1: [1] Consider a forging hammer shown in Figure 1. The forging hammer consists of a frame, a falling weight known as the tup, an anvil, and a foundation block. The anvil is a massive steel block on which material is forged into desired shape by the repeated blows of the tup. The anvil is usually mounted on an elastic pad to reduce the transmission of vibration to the foundation block and the frame. Come up with two versions of mathematical model of the forging hammer.

Figure 1: A forging hammer. Solution

1 Copyright 2007 by Withit Chatlatanagulchai


☻ Example 2: [1] Consider a motorcycle with a rider in Figure 2. Develop a sequence of four mathematical models of the system. Consider the elasticity of the tires, elasticity and damping of the struts (in the vertical direction), masses of the wheels, and elasticity, damping, and mass of the rider.

Figure 2: A motorcycle with a rider.

Solution



Step 2: Derivation of Governing Equations Once the mathematical model is available, we use the principles of

dynamics and derive the equations that describe the vibration of the system. The equations of motion can be derived by drawing the free-body diagrams of all the masses involved. The equations of motion of a vibrating system are in the form of a set of ordinary differential equations for a discrete system and partial differential equations for a continuous system. The equations may be linear or nonlinear, depending on the behavior of the components of the system. Several approaches are commonly used to derive the governing equations. They are

Newton’s second law • of motion • Energy method • Equivalent system method (Rayleigh’s method)

• Virtual work method (d’ Alembert’s Method) • Lagrange’s method

We will study each method in details in the following lessons.

Step 3: Solution of the Governing Equations The equations of motion must be solved to find the response of the vibrating system. We can use the following techniques for finding the solution:

• Standard methods of solving differential equations • Laplace transform methods • Numerical methods • Matrix methods

In this course, we will focus on the first three methods; each method will be discussed in the following lessons. The last method, namely, matrix methods are mostly used in higher degree of freedom and can be studied once the students are familiar with linear algebra.

Step 4: Interpretation of the Results The solution of the governing equations gives the displacements, velocities, and accelerations of the various masses of the system. These results must be interpreted with a clear view of the purpose of the analysis and the possible design implications of the results.

2 Vibration Model: Mechanical Elements As discussed before, a vibratory system consists of three elements: mass, spring, and damper.

2.1 Spring Elements A spring is generally assumed to have negligible mass and damping. A force developed in the spring is given by

,F kx=

where F is the spring force, x is the displacement of one end with respect to the other, and k is the spring stiffness or spring constant.



The work done in deforming a spring is stored as strain or potential energy in the spring and is given by

21 .2

U kx=

Actual springs are linear only up to a certain deformation. Beyond a certain value of deformation, the stress exceeds the yield point of the material and the force-deformation relation becomes nonlinear as in Figure 3.

Figure 3: Nonlinearity beyond the yield point.

☻ Example 3: [1] Deflection of the cantilever beam in Figure 4 is given as

lx

y

Pa

Figure 4: A cantilever beam.

( )( )

( )

2

2

3, 0

6 ,3

,6

Px a xx a

EIy xPa x a

a x lEI

⎧ −≤ ≤⎪⎪= ⎨

−⎪ ≤ ≤⎪⎩

(1)

where of inertia of the cross section of the beam, and

E is Young’s modulus, I is moment P is external force. Find the spring constant of a

cantilever beam with an end mass m shown in Figure 5.

lx

y

m

Figure 5: A cantilever beam with end mass.



Solution From (1), we have that the static deflection of the beam at the free

ven by end is gi

( )2 33.

6 3st

mgl l l mgl−EI EI

δ = =

Therefore the spring constant is

3

3 .st

mgk = =EIlδ

be called an equivalent stiffnessk can of the cantilever beam.

Several springs are used in combination. They can be connected in s can be combined into a single

quival

2.1.1. Combination of Springs

parallel or in series or both. These springe ent spring as follows. Springs in Parallel

Consider two spring s connected in parallel, from the free-body

diagram we have

Figure 6: Springs in parallel.



1 2

.st st

eq st

W k kkδ δδ

= +

=

Therefore, the equivalent spring is

In general, if we have constants

parallel, then the equivalent spring constant can be obtained as

Springs in Series

1 2.eqk k k= +

n springs with spring 1 2, , , nk k k… in

keq

1 2 .eq nk k k k= + + +…

Next, we consider two springs connected in series in Figure 7.

The static deflection

Figure 7: Springs in series.

of the system stδ is given by

1 2st .δ δ δ= + (2)

Since both springs are subjected to the same force

,.eq st

W kW k

,W we have

1 1,W k

2 2

δδδ

===

Therefore,

11

eq stkkδ

δ = and 22

eq stkkδ

δ =

Substitute the equations above in , we get to (2)

1 2

,eq st eq stst

k kk kδ δ

δ+ =

therefore,

1 2

1 1 1 .eqk k k= +

In general, if we have constants

series, then the equivalent spring constant can be obtained as

n springs with spring 1 2, , , nk k k… in

eqk

1 2

1 1 1 1 .= + + +… eq nk k k k

☻ Example 4: [1] Figure 8 shows the suspension system of a freight truck

ith a parallel-spring arrangement. Find the equivalent spring constant of wthe suspension if each of the three helical springs is made of steel with a shear modulus 9 280 10 /G N m= × and has five effective turns, mean coil

diameter 20D cm= , and wire diameter 2 .d cm=



Figure 8: Parallel springs in a freight truck.

The equivalent spring constant of a helical spring under axial load is given by

4

3 ,8eqknD

= Gd

where wire diameter, d = D = mean coil diameter, er of active turns.

olution

n = numb

S The spring constant of each helical spring is given by

( )( )( )( )

494

33

80 10 0.0240,000 / .

8 8 5 0.2nDGdk N m

×= = =

Since there are three springs connected in parallel, we have

3 120,000 / .eqk k N m= =

☻ Example 5: [1] Determine the torsional spring constant of the steel propeller shaft shown in Figure 9. A hollow shaft under torsion has equivalent spring constant

( )4 4 ,32eq

Gk D dl

π= −

where , l = length D = outer diameter, ner diameter.

Figure 9: Propeller shaft.

d = in



olution

S There are two sections of torsional springs, section 1-2 and section -3. Since 2 12 23 ,stδ δ δ+ = we treat the two sections as springs connected series. The spring constant of each section is given by in

( )( ) ( )

( )( ) ( )

94 4

12

6

94 4

23

6

80 100.3 0.2

32 2

25.53 10 / ,

80 100.25 0.15

32 2

8.90 10 / .

k

Nm rad

k

Nm rad

π

π

×= −

= ×

×= −

= ×

Since the springs are connected in series, we have

12 23

1 1 1

eqk k k= +

6 6

6

1 1 ,25.53 10 8.90 106.60 10 / .eqk Nm rad

= +× ×

= ×

☻ Example 6: [1] Consider a crane in Figure 10(a). The boom AB has a cross-sectional area of 2,500 ble is made of steel with a cross-sectional area of 100 ecting the effect of the cable CDEB, find the equivalent spring constant of the system in the vertical direction.

2.mm The ca2.mm Negl




Figure 10: Crane lifting a load.


Solution

The equivalent system is given in Figure 10(c). First, we need to se trigonometry to find the length the angle

u 1l and .θ Using law of osine, we have c

( )( )2 2 21

1

3 10 2 3 10 cos135 ,12.31 .

ll m= + −

=

d using the law of cosine and the knowledge of 1l The angle θ can be foun

( )( )2 2 21 110 3 2 3 cos ,

35.07 .

l l θ

θ

= + −

=

x of point BA vertical displacement will cause the spring

eform by an amount of the spring o deform by an 2k to

d 2 cos 45x x= and 1k t

amount of ( )1 cos 90 .x x θ= −

th q The potential energy stored in e spring of the e uivalent system is

21 .2eq eqU k x=

The potential energy stored in the springs of the actual system is

( ) ( ) 22

2 11 1cos 45 cos 90 ,2 2

U k x k x θ⎡ ⎤= + −⎣ ⎦

where




( )( )

( )( )

6 961 1

11

100 10 207 101.68 10 / ,

0

A El

−

−

× ×= = = ×

6 9

72 2

12.31

2500 1 207 105.17 10 / .

k

A Ek× ×

= = = ×

Since the total potential energy stored in the springs of the equivalent system and the actual system must be equal we have

eq

U U=

2.2 Mass or Inertia Elements In many practical applications, several masses appear in

tion. For a simple analysis, we can replace these masses by a single equivalent mass. If the system has translational masses, we can replace them with a single equivalent mass. If the system has rotational masses, we can replace them with a single equivalent inertia. And if the system has both translational and rotational masses, we can choose to replacsingle equivalent mass or a single equivalent inertia. We can do this using the fact that the kinetic energy of the actual system and its equivalent system must be equal.

equivalent system is given in Figure 11(b). Find the equivalent mass

Figure 11: Translational masses connected by a rigid bar.

22 10l

N m

N m

,

626.43 10 / .eq

k N m= ×

,

combina

e them with a

☻ Example 7: [1] Figure 11(a) shows a system whose

.eqm


Solution Let e kinetic energy of the actual system and let the

inetic energy of the equivalent system. We have

T be th eqT be

k

2 2 21 1 2 2 3 3

2 22 3 12 1

1 1 2 31 1

21

1 1 12 2 2

1 1 1 ,2 2 21 .2eq eq

T m x m x m x

l xl xm x m ml l

T m x

= + +

⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

=

Figure 12: Translational and rotational

masses in a rack and pinion arrangement.

Therefore,

2 22 23 12 1

1 1 2 3 11 1

2 2

321 2 3

1 1

,eqT T=

1 1 1 1 ,2 2 2 2

.

eq

eq

l xl xm x m m m xl l

llm m m ml l

⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

☻ Example 8: [1] Consider translational and rotational masses coupled in

valent translational mass b) Find equivalent rotational inertia

Figure 12. a) Find equi



Solution ) Let T be the kinetic energy of the actual system and let e

inetic energy of the equivalent system. We have

a eqT be th

k

2 20

22

0 ,2 21

mx JR

= + ⎜ ⎟⎝ ⎠

2

1 12 21 1

.2eq eq

T mx J

x

T m x

θ= +

⎛ ⎞

=

Therefore,

22 2

0

02

,

1 1 1 ,2 2 2

.

eq

eq

eq

T T

xmx J m xR

Jm mR

=

⎛ ⎞+ =⎜ ⎟⎝ ⎠

= +

b) Let of the actual system and let e

kinetic energy of the equivalent system. We have

T be the kinetic energy eqT be th

( )

2 20

2 20

2

2 21 .T J

( )2 2 20

20

,

1 1 1 ,2 2 2

.

eq

eq

eq

T T

m R J J

J mR

=

θ θ θ

J

+ =

= +

☻ Example 9: [1] Find the equivalent mass of the system in Figure 13.

Figure 13: An example system. 1 1

2 21 1 ,

2eq eq

T mx J

m R J

θ

θ θ

= +

= +

Therefore,

θ=



Solution Let e kinetic energy of the actual system and let the

inetic energy of the equivalent system. Neglecting the rotational kinetic nergy of link 2, we have

T be th eqT be

ke

2 2 2 21 1 1 1

2 2 22 2 2

2 2 22

2 1 1 11

1 1 1 12 2 2 12 2 2p

x m l x xmx J mr r r

2 22

12 1

1 1 1 12 2 2 21 1 12 2 2

1 1 12 2 2 2

p p

c c c

p p p

c c

p p c

T mx J J m x

m x J m x

l

m rx x lm lr r r

θ θ

θ

⎛ ⎞= + + +⎜ ⎟⎝ ⎠

⎛ ⎞+ + +⎜ ⎟⎝ ⎠

⎡ ⎤⎞⎛ ⎞ ⎛ ⎞ ⎛⎢ ⎥= + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2

1

2

,

1 .2

cp

eq eq

xm lr

T m x

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

=

Therefore,

2 2 22

2 2 1 1 11

2 2 22

12 1 1

2 2 21 1 1 1 2 1

2 2 2

1 1 1 1 12 2 2 2 12 2 2

1 1 1 ,2 2 2 2

12 4

eq pp p p

c cc

p p c p

peq

p p p p

x m l x x lm x mx J mr r r

m rx x l xm l m lr r r r

J m l m l m lm mr r r r

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥= + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

= + + + +2 2

1 12 2 2 .

2c c

p p

m l m lr r

+ +



2.3 Damping Elements Damping is a mechanism by which the vibrating energy is gradually converted into other energy such as heat or sound. There are three common types of damping: viscous damping, Coulomb or dry friction damping, and hysteretic damping. Viscous damping dissipates energy by using the resistance offered by the fluid such as air, gas, water, or oil. Coulomb or dry friction damping is caused by friction between dry rubbing surfaces. It has constant magnitude but opposite in direction to the motion of the vibrating body. Hysteretic damping happens when some materials are deformed and the vibrating energy are absorbed and dissipated by the material.

Among the three types of damping, viscous damping is the most commonly used and can be modeled as Figure 14.

ous fluid in between.

From Newton’s law of viscous flow, we have

Figure 14: Parallel plates with a visc

,F du vA dy h

μτ μ= = = (3)

olute viscosity of the fluid.

Since in viscous damping, the damping force is proportional to the velocity of

the vibrating body, we have

where μ is the abs

,F cv= (4)

where led damping constant. Comparing (3) with (4), we have

c is cal

.Achμ

= (5)

Combination of dampers is done similar to that of springs. The inetic energy due to damping is given by k

21 .U cv= 2

☻ Example 10: [1] A bearing, which can be approximated as two flat plates separated by a thin film of lubricant, ofwhen the relative velocity between the platplates is 0.1 olute viscosity of the oil in-between is 0.3445 Pa-

termin

fers a resistance of 400 N es is 10 m/s. The area of the

2.m The abss. De e the clearance between the plates.



Area (A)

v

h

Figure 15: Flat plates separated by thin film

of lubricant.

olution

S From (4),

( ),

400 10 ,40 / .

F cvc

c Ns m

=

=

=

From (5),

( )

,

0.3445 0.140 ,

0.86 .

Ach

hh mm

μ=

=

=

☻ Example 11: [1] A milling machine is supported on four shock mounts

Figure 16(a). The elasticity and damping of each shock mount can be modeled as a spring and a viscous damper as shown in Figure 16(b). Find the equivalent spring constant ent damping

constant the milling machine support.

as shown in

eqk and the equival

eqc of



Figure 16: A milling machine.



Solution From the free-body diagram of the actual system in Figure 16(c), we have

1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4

,

.

s s s s s

d d d d d

F F F F Fk x k x k x k x

F F F F Fc x c x c x c x

= + + += + + += + + += + + +

From the free-body diagram of the equivalent system in Figure 16(d), we have

,

.s eq

d eq

F k x

F c x

=

=

Since the forces of the two systems are equal, we have

1 2 3 4

1 2 3 4

,

.eq

eq

k k k k k

c c c c c

= + + +

= + + +

One Degree of Freedom, Free Vibrations

3 Systems with Mass and Spring Consider a simple mass-and-spring system in Figure 17. Initially, the mass stretches the spring to an amount of Δ below the unstretched position. From the Newton’s second law, we have

.k mgΔ = (6)

m

Unstretchedposition

Δ

m

Staticequilibriumposition

m

kΔ

mg

x

( )k xΔ +

mgxx

Figure 17: A system with mass and spring.

With x chosen to point downward from the static equilibrium position and using (6), we have

( ) ,.

mg k x mxkx mx

− Δ + =

− =



Let the natural frequency be nω and let

2 ,nkm

ω = (7)

we have

2 0.nx xω+ =

The solution of the second-order differential equation above is

sin cos .n nx A t b tω ω= +

The constants A and are evaluated from initial conditions B ( )0x and

( )0x to be

( ) ( )0sin 0 cos .n n

n

xx t x tω ω

ω= +

One remark is that when x is chosen to be from the static equilibrium position, the weight mg is cancelled with kΔ and disappears from the differential equation.

4 Natural Frequency ☻ Example 12: [1] A 0.25 kg mass is suspended by a spring having a stiffness of 0.1533 N/mm. Determine its natural frequency in cycles per second. Determine its static deflection.

Solution From (7), we have

0.1533 0.783 / .0.25n

k rad sm

ω = = =

From (6), we have

( )0.25 9.8116 .

0.1533mg mmk

Δ = = =

☻ Example 13: [2] Determine the natural frequency of the mass on the end of a cantilever beam of negligible mass shown in Figure 18.

m

ml

x

Figure 18: A cantilever beam with end mass.



Solution A cantilever beam with end load as an equivalent spring constant

3

3 .eqEIkl

=

Therefore, the natural frequency is

3

3 / .nk EI rad sm ml

ω = =

☻ Example 14: [2] A rod in Figure 19 has 0.5 cm in diameter and 2 m long. When the disk is given an angular displacement and released, it makes 10 oscillations in 30.2 seconds. Determine the polar moment of inertia of the disk.

l

J

θ

Figure 19: A rotating disk.

Solution From the Newton’s second law, we have

.eqJ kθ θ= −

For a shaft under torsion, the equivalent spring constant is

( )( )( )

49 24 80 10 0.5 102.455 / .

32 32 2eqGdk Nm rad

lππ

−× ×= = =

From (7), we have

2

2

2

,

10 2.4552 ,30.2

0.567 .

nkJ

J

J kg m

ω

π

=

⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠=



☻ Example 15: [2] Consider a pivoted bar in Figure 20. The bar is horizontal in the equilibrium position with spring forces and Determine its natural frequency.

1P 2.P

a b

c

. .C G

k k

θ

1P 2P

O

Figure 20: A pivoted bar.

Solution From Newton’s second law, we have

( ) ( )0

2 2

0

2 2

0

,

0,

.n

J ka a kb b

ka kbJ

ka kbJ

θ θ θ

θ θ

ω

= − −

++ =

+=



Lesson 2 Homework Problems 1.3, 1.4, 1.6, 1.7, 1.8, 1.30, 1.32, 1.34, 1.35 Homework problems are from the required textbook (Mechanical Vibrations, by Singiresu S. Rao, Prentice Hall, 2004)

References [1] Mechanical Vibrations, by Singiresu S. Rao, Prentice Hall, 2004 [2] Theory of Vibration with Applications, by William T. Thomson and

Marie Dillon Dahleh, Prentice Hall, 1998


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1 Vibration Analysis Procedure - Kasetsart Universitypirun.ku.ac.th/~fengwtc/Teaching/208322...

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