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

☻ 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.

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

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

2

1

.

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

l x

y

P a

Figure 4: A cantilever beam.

( )

(

)

(

)

2 2

3

,

0

6

,

3

,

6

Px

a

x

x

a

EI

y x

Pa

x

a

a

x

l

EI

⎧

−

≤ ≤

⎪⎪

= ⎨

−

⎪

_{≤ ≤}

⎪⎩

(1)

where of inertia of the cross section

of the beam, and

E

is Young’s modulus,

P

is external force. Find the spring constant of a

I

is moment

cantilever beam with an end mass

m

shown in Figure 5.

l x

y

m

Solution

From (1), we have that the static deflection of the beam at the free ven by end is gi

(

)

2

₃

3

.

6

3

st

mgl

l

−

l

mgl

EI

EI

δ

=

=

Therefore the spring constant is

3

3

.

st

mg

k

=

=

EI

l

δ

be called an equivalent stiffness

k

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 spring e ent spring as follows.

Springs in Parallel

Consider two springs connected in parallel, from the free-body diagram we have

1 2

.

st st eq st

W

k

k

k

δ

δ

δ

=

+

=

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

.

eq

k

= +

k

k

n

springs with spring

k k

₁

,

₂

,

…

,

k

_n in

k

_eq

1 2

.

eq n

k

= +

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

δ

_st is given by

1 2

st

.

δ

= +

δ δ

(2)

Since both springs are subjected to the same force

,

.

eq st

W

k

W

k

,

W

we have 1 1

,

W

k

2 2

δ

δ

δ

=

=

=

Therefore, 1 1 eq st

k

k

δ

δ

=

and ₂ 2 eq st

k

k

δ

δ

=

Substitute the equations above into (2), we get

1 2

,

eq st eq st st

k

k

k

k

δ

δ

δ

+

=

therefore, 1 2

1

1

1

.

eq

k

=

k

+

k

In general, if we have constants

series, then the equivalent spring constant can be obtained as

n

springs with spring

k k

1

,

2

,

…

,

k

n in eq

k

1 2

1

1

1

1

.

=

+

+ +

…

eq n

k

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 w

the suspension if each of the three helical springs is made of steel with a

shear modulus 9 2

80 10

/

G

=

×

N m

and has five effective turns, mean coil

Figure 8: Parallel springs in a freight truck.

The equivalent spring constant of a helical spring under axial load is given by 4 3

,

8

eq

k

nD

=

Gd

where

d

= wire diameter,

D

= mean coil diameter, er of

active turns.

olution

n

= numb

S

The spring constant of each helical spring is given by

(

)

(

)

( )( )

4 9 4 3 3

80 10

0.02

40, 000

/ .

8

nD

_{8 5 0.2}

Gd

k

=

=

×

=

N m

Since there are three springs connected in parallel, we have

3

120, 000

/ .

eq

k

=

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

)

,

32

eq

G

k

D

d

l

π

=

−

where

l

= length,

D

= outer diameter, ner diameter.

Figure 9: Propeller shaft.

olution S

There are two sections of torsional springs, section 1-2 and section -3. Since

2

δ

₁₂

+

δ

₂₃

=

δ

st

,

we treat the two sections as springs connected

series. The spring constant of each section is given by in

(

)

( )

(

)

(

)

( )

(

)

9 4 4 12 6 9 4 4 23 6

80 10

0.3

0.2

32 2

25.53 10

/

,

80 10

0.25

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

eq

k

=

k

+

k

6 6 6

1

1

,

25.53 10

8.90 10

6.60 10

/

.

eq

k

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 ca 2

.

mm

Negl

Solution

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

u

l

₁ and

θ

.

Using law of

osine, we have c

( )( )

2 2 2 1 1

3

10

2 3 10 cos135 ,

12.31 .

l

l

m

=

+

−

=

d using the law of cosine and the knowledge of

l

₁

The angle

θ

can be foun

( )( )

2 2 2 1 1

10

3

2

3 cos ,

35.07 .

l

l

θ

θ

= +

−

=

x

of point B

A vertical displacement will cause the spring eform by an amount of the spring o deform by an

2

k

to

d

x

₂

=

x

cos 45

and

k

₁ t

amount of

x

₁

=

x

cos 90

(

−

θ

)

.

th q

The potential energy stored in e spring of the e uivalent system is 2

1

.

2

eq eq

U

=

k x

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

(

)

2

(

)

2 2 1

1

1

cos 45

cos 90

,

2

2

U

=

k

x

+

k

⎡

_⎣

x

−

θ

⎤

_⎦

where

(

)(

)

(

)(

)

6 9 6 1 1 1 1

100 10

207 10

1.68 10

/ ,

0

A E

l

− −

×

×

=

=

=

×

6 9 7 2 2

12.31

2500 1

207 10

5.17 10

/ .

k

A E

k

=

=

×

×

=

×

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 replac

single 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. 2 2

10

l

N m

N m

, 6

26.43 10

/ .

eq

k

=

×

N m

,

combina e them with a

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

.

eq

Solution

Let e kinetic energy of the actual system and let the inetic energy of the equivalent system. We have

T

be th

T

eq be k 2 2 2 1 1 2 2 3 3 2 2 2 2 1 3 1 1 1 2 3 1 1 2 1

1

1

1

2

2

2

1

1

1

,

2

2

2

1

.

2

eq eq

T

m x

m x

m x

l x

l x

m x

m

m

l

l

T

m x

=

+

+

⎛

⎞

⎛

⎞

=

+

_⎜

_⎟

+

_⎜

_⎟

⎝

⎠

⎝

⎠

=

Figure 12: Translational and rotational masses in a rack and pinion arrangement.

Therefore, 2 2 2 2 1 3 1 2 1 1 2 3 1 1 1 2 2 3 2 1 2 3 1 1

,

eq

T

=

T

1

1

1

1

,

2

2

2

2

.

eq eq

l x

l x

m x

m

m

m x

l

l

l

l

m

m

m

m

l

l

⎛

⎞

⎛

⎞

+

_⎜

_⎟

+

_⎜

_⎟

=

⎝

⎠

⎝

⎠

⎛ ⎞

⎛ ⎞

=

+

_{⎜ ⎟}

+

_{⎜ ⎟}

⎝ ⎠

⎝ ⎠

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

valent translational mass b) Find equivalent rotational inertia Figure 12.

Solution

) Let

T

be the kinetic energy of the actual system and let e

inetic energy of the equivalent system. We have

a

T

_eq be th k 2 2 0 2 2 0

,

2

2

1

mx

J

R

=

+

_{⎜ ⎟}

⎝ ⎠

2

1

1

2

2

1

1

.

2

eq eq

T

mx

J

x

T

m x

θ

=

+

⎛ ⎞

=

Therefore, 2 2 2 0 0 2

,

1

1

1

,

2

2

2

.

eq eq eq

T

T

x

mx

J

m x

R

J

m

m

R

=

⎛ ⎞

+

_{⎜ ⎟}

=

⎝ ⎠

= +

b) Let of the actual system and let e

kinetic energy of the equivalent system. We have

T

be the kinetic energy

T

eq be th

( )

2 2 0 2 ₂ 0 2

2

2

1

.

T

J

( )

2 2 2 0 2 0

,

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

2

1

1

,

2

eq 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

T

eq be k e 2 2 2 2 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1

1

1

1

1

2

2

p

2 12

2

2

x

m l

x

x

mx

J

m

r

r

r

2 2 2 1 2 1

1

1

1

1

2

2

2

2

1

1

1

2

2

2

1

1

1

2

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 r

x

x l

m

l

r

r r

θ

θ

θ

⎛

⎞

=

+

+

_⎜

+

_⎟

⎝

⎠

⎛

⎞

+

+

_⎜

+

_⎟

⎝

⎠

⎡

_⎞

⎤

⎛ ⎞

⎛ ⎞

⎛

⎢

⎥

=

+

⎜ ⎟

_{⎜ ⎟}

+

⎜ ⎟

_{⎜ ⎟}

+

⎜

_⎜

⎟

_⎟

⎢

⎥

⎝ ⎠

_⎣

⎝ ⎠

⎝

⎠

_⎦

⎛

⎞

⎛

⎞

+

⎜

_⎜

⎟

_⎟

+

⎜

_⎜

⎟

_⎟

+

⎝

⎠

⎝

⎠

2 1 2

,

1

.

2

c p eq eq

x

m

l

r

T

m x

⎡

_⎛

_⎞

⎤

⎢

_⎜

_⎟

⎥

⎜

⎟

⎢

_⎝

_⎠

⎥

⎣

⎦

=

Therefore, 2 2 2 2 2 2 1 1 1 1 2 2 2 2 1 2 1 1 2 2 2 1 1 1 1 2 1 2 2 2

1

1

1

1

1

2

2

2

2 12

2

2

1

1

1

,

2

2

2

2

12

4

eq p p p p c c c p p c p p eq p p p p

x

m l

x

x l

m x

mx

J

m

r

r

r

m r

x

x l

x

m

l

m

l

r

r r

r

J

m l

m l

m l

m

m

r

r

r

r

⎡

⎤

⎛ ⎞

⎛ ⎞

⎛

⎞

⎢

⎥

=

+

⎜ ⎟

_{⎜ ⎟}

+

⎜ ⎟

_{⎜ ⎟}

+

⎜

_⎜

⎟

_⎟

⎢

⎥

⎝ ⎠

_⎣

⎝ ⎠

⎝

⎠

_⎦

⎡

⎤

⎛

⎞

⎛

⎞

⎛

⎞

⎢

⎥

+

⎜

_⎜

⎟

_⎟

+

⎜

_⎜

⎟

_⎟

+

⎜

_⎜

⎟

_⎟

⎢

⎥

⎝

⎠

_⎣

⎝

⎠

⎝

⎠

_⎦

= +

+

+

+

12 12 2 2 2

.

2

c c p p

m l

m l

r

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

v

A

dy

h

μ

τ

=

=

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

c

is called damping constant. Comparing (3) with (4), we have

.

A

c

h

μ

=

(5)

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

k 2

1

.

U

=

cv

2

☻ Example 10: [1] A bearing, which can be approximated as two flat plates separated by a thin film of lubricant, of

when the relative velocity between the plat

plates 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 abs

Area (A)

v

h

Figure 15: Flat plates separated by thin film of lubricant. olution S From (4),

( )

,

400

10 ,

40

/ .

F

cv

c

c

Ns m

=

=

=

From (5),

( )

,

0.3445 0.1

40

,

0.86

.

A

c

h

h

h

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

eq

k

and the equival

eq

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

F

k x

k x

k x

k x

F

F

F

F

F

c 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

,

eq

k

= +

k

k

+ +

k

k

= + + +

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

Δ =

mg

(6) m Unstretched position Δ m Static equilibrium position m kΔ mg x

(

)

k Δ +x mg x x

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

mx

kx

mx

− Δ +

=

− =

Let the natural frequency be

ω

_n and let 2

,

n

k

m

ω

=

(7) we have 2

0.

n

x

+

ω

x

=

The solution of the second-order differential equation above is

sin

n

cos

n

.

x

=

A

ω

t

+

b

ω

t

The constants

A

and are evaluated from initial conditions

B

x

( )

0

and

( )

0

x

to be

( )

0

_{( )}

sin

_n

0 cos

_n

.

n

x

x

ω

t

x

ω

t

ω

=

+

One remark is that when

x

is chosen to be from the static

equilibrium position, the weight

mg

is cancelled with

k

Δ

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

/ .

0.25

n

k

rad s

m

ω

=

=

=

From (6), we have

(

)

0.25 9.81

16

.

0.1533

mg

mm

k

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

m l

x

Solution

A cantilever beam with end load as an equivalent spring constant

3

3

.

eq

EI

k

l

=

Therefore, the natural frequency is

3

3

/ .

n

k

EI

rad s

m

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

.

eq

J

θ

= −

k

θ

For a shaft under torsion, the equivalent spring constant is

(

)(

)

( )

4 9 2 4

_{80 10}

_{0.5 10}

2.455

/

.

32

32 2

eq

Gd

k

Nm rad

l

π

π

×

×

−

=

=

=

From (7), we have 2 2 2

,

10

2.455

2

,

30.2

0.567

.

n

k

J

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.

1

P

P

₂

.

a

_b

c

. .

C G

k

k

θ

1

P

P

₂

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

kb

J

ka

kb

J

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