CHAPTER 8 TRANSVERSE VIBRATIONS-IV: MULTI-DOFs ROTOR SYSTEMS - iitg.ac.in · CHAPTER 8 TRANSVERSE VIBRATIONS-IV: MULTI-DOFs ROTOR SYSTEMS Transverse vibrations have been considered

CHAPTER 8

TRANSVERSE VIBRATIONS-IV: MULTI-DOFs ROTOR SYSTEMS

Transverse vibrations have been considered previously in great detail for mainly single mass rotor

systems. The thin disc and long rigid rotors were considered with various complexities at supports; for

example, the rigid disc mounted on flexible mass less shaft with rigid bearings (e.g., the simply

supported, overhung, etc.), the flexible bearings (anisotropic and cross-coupled stiffness and damping

properties), and flexible foundations. Higher order effects of the gyroscopic moment on the rotor, for

most general case of motion, was also described in detail. However, in the actual case, as we have

seen in previous two chapters for torsional vibrations, the rotor system can have several masses (e.g.,

turbine blades, propellers, flywheels, gears, etc.) or distributed mass and stiffness properties, and

multiple supports, and other such components like coupling, seals, etc. While dealing with torsional

vibrations, we did consider multi-DOF rotor systems. Mainly three methods were dealt for multi-DOF

systems, that is, the Newton’s second law of motion (or the D’Alembert principle), the transfer matrix

method, and the finite element method. We will be extending the idea of these methods from torsional

vibrations to transverse vibrations along with some additional methods, which are suitable for the

analysis of multi-DOF rotor system transverse vibrations. In the present chapter, we will consider the

analysis of multi-DOF rotor systems by the influence coefficient method, transfer matrix method, and

Dunkerley’s method. The main focus of these methods would be to estimate the rotor system natural

frequencies, mode shapes, and forced responses. The relative merit and demerit of these methods are

discussed. The continuous system and finite element transverse vibration analysis of multi-DOF rotor

systems will be treated in subsequent chapters. Conversional methods of vibrations like the modal

analysis, Rayleigh-Ritz method, weighted sum approach, collocation method, mechanical impedance

(or receptance) method, dynamic stiffness method, etc. are not covered exclusively in the present text

book, since it is readily available elsewhere (Meirovitch, 1986; Thomson and Dahleh, 1998).

However, the basic concepts of these we will be using it directly whenever it will be required with

proper references.

8.1 Influence Coefficient Method

In transverse vibrations due to coupling of the linear and angular (tilting) displacements the analysis

becomes more complex as compared to torsional vibrations. A force in a shaft can produce the linear

as well as angular displacements; similarly a moment can produce the angular as well as linear

displacements. Influence coefficients could be used to relate these parameters (the force, the moment,

and the linear and angular displacements) relatively easily. In the present section, the influence

421

coefficient method is used to calculate natural frequencies and forced responses of rotating machines.

Up to three-DOF rotor systems the hand calculation is feasible, however, for more than three-DOF the

help of computer routines (e.g., MATLAB, etc.) are necessary. The method is described for multi-

DOF i.e., n number of discs mounted on a flexible shaft (Figure 8.1) and supported by rigid bearings;

which can be extended for the multi-DOF rotor system with flexible supports.

Figure 8.1 A multi-DOF rotor system mounted on rigid bearings

8.1.1 The static case: Let f1, f2 , …, and fn are static forces on discs 1, 2, …, and n respectively, and

x1, x2, …, xn are the corresponding shaft deflections at discs. The reference for the measurement of the

shaft displacement is from the static equilibrium and the system under consideration is linear, so that

gravity effect will not appear in equilibrium equations. If a force f is applied to the disc of mass m1,

then deflection of m1 will be proportional to f, i.e.

or (8.1)

where α11 is a constant, which depends upon the elastic properties of the shaft and support conditions.

It should be noted that we will have deflections at other disc locations (i.e., 2, 3, …, n) as well due to

force at disc 1 due to the coupling. Now the same force f is applied to the disc of mass m2 instead of

mass m1, then the deflection of m1 will still be proportional to the force, i.e.

or (8.2)

where, α12 is another constant (the first subscript represents displacement position and the second

subscript represent the force location). In general we have α12 ≠ α11. Similarly, if force f is applied to

the disc of mass mn, then the deflection at m1 will be

1 1nx fα= (8.3)

1x f∝ x fα=1 11

1x f∝ fx 121 α=

422

where, α1n is a constant. If forces f1, f2, ..., and fn are applied at the locations of all the masses

simultaneously, then the total deflection at m1, will be summation of all the three displacements

obtained above by the use of superposition theorem, as

1 11 1 12 2 1n nx f f fα α α= + + +� (8.4)

In the equation, it has been assumed that displacements are small so that a linear relation exists

between the force applied and corresponding displacement produced. Similarly, we can write

displacement at other disc locations as

2 21 1 22 2 2n nx f f fα α α= + + +� (8.5)

and

1 1 2 2n n n nn nx f f fα α α= + + +� (8.6)

Here α2j, …, αnj, where j =1, 2, …, n are another sets of constants and can be defined as described for

α1i above. Hence, in general αij is defined as a displacement at ith station due to a unit external force at

station jth and keeping all other external forces to zero. Equations (8.4) to (8.6) can be combined in a

matrix form as

1 11 12 1 1

2 21 22 2 2

1 2

n

n

n n n nn n

x f

x f

x f

α α α

α α α

α α α

=

�

�

� � � � � �

�

(8.7)

It should be noted that due to the transverse force actually both the linear and angular displacements

take place, i.e., a coupling exists between the linear and angular displacements. We have already seen

such coupling in Chapter 2 due to bending of the shaft and due to gyroscopic couples, respectively.

Moreover, we have seen in Chapters 4 and 5 couplings between the horizontal and vertical plane

linear motions (x and y) due to dynamic properties of fluid-film bearings and between the horizontal

and vertical plane angular motions (ϕx and ϕy), respectively.

Similarly, a moment gives the angular displacement as well as the linear displacement. The method

can be extended to account for the angular displacement (tilting), ϕy, of the disc, and for the

application of the point moment, M, at various disc locations along the shaft a part of loading on discs.

Then, the equation will take the following form

(8.8) { } [ ]{ }d fα=

423

with

{ }1

1

n

n

y

y

x

xd

ϕ

ϕ

=

�

�

; { }

1

1

n

n

f

ff

M

M

=

�

�

and

11 12 1(2 )

21 22 2(2 )

(2 )1 (2 )2 (2 )(2 )

n

n

n n n n

α α α

α α α

α α α

�

�

� � � �

�

(8.9)

which gives

(8.10)

where αij, with i, j = 1, 2, …,n are influence coefficients. The first subscript defines the linear (or

angular) displacement location and the second subscript defines the force (or moment) location. The

analysis so far has referred only to static loads applied to the shaft. When the displacement of the disc

is changing rapidly with time, the applied force has to overcome the disc inertia as well as to deform

the shaft.

8.1.2 The dynamic case: In Figure 8.2 free body diagrams of a disc and the shaft is shown. Let

and (not shown in free body diagrams for brevity) be the external force and moment on the disc

whereas and are the reaction force and moment transmitted to the shaft (which is equal

and opposite to the reaction force and moment of the shaft on the disc). From the force and moment

balance of disc 1, we have

and 1 11 1 d y

M M I ϕ′ − = �� (8.11)

where Id is the diametral mass moment of inertia the disc.

Figure 8.2 Free body diagrams of (a) a disc and (b) the shaft

1{ } [ ] { }f dα −=

1f ′

1M ′

1m 1f 1M

1111 xmff ��=−′

424

Similarly at other disc locations, we can write

and 2 22 2 d y

M M I ϕ′ − = �� (8.12)

and

n n n nf f m x′ − = �� and n nn n d y

M M I ϕ′ − = �� (8.13)

Substituting for 1 2 1 2, , , , , ,nf f f M M� �

and

nM from equations (8.11)-(8.13) and remembering that

for the simple harmonic motion of discs 2

x xω= −�� and

2

y yϕ ω ϕ= −�� (when no external excitation is

present then ω is the natural frequency of the system ωnf , and when there is an external excitation

then it is equal to the excitation frequency, ω), equation (8.8) gives

{ } [ ]1 1

2

1 1 1

2

2

1

2

n

n n n

d y

n d y

f m x

f m xd

M I

M I

ω

ωα

ω ϕ

ω ϕ

′ + ′ +

= ′ +

′ +

�

�

(8.14)

which can be expanded as

{ } [ ]{ } [ ]1

1 1

2

1

n

n n

nfd

d n

m x

m xd f

I

I

α ω αϕ

ϕ

′= +

�

�

with { }

1

1

n

n

f

ff

M

M

′ ′

′ = ′

′

�

�

(8.15)

In view of equation (8.9), equation (8.15) can be rearranged as

{ } [ ]{ } { }

1

1

1

1

11 1 1 1( 1) 1(2 )

1 1 ( 1) (2 )2

( 1)1 1 ( 1) ( 1)( 1) ( 1)(2 )

(2 )1 1 (2 ) (2 )( 1) (2 )(2 )

n

n

n

n

n n n d n d

n nn n n n d n n d

nf

n n n n n n d n n d

n n n n n n d n n d

m m I I

m m I Id f d

m m I I

m m I I

α α α α

α α α αα ω

α α α α

α α α α

+

+

+ + + + +

+

′= +

� �

� � � � � �

� �

� �

� � � � � �

� �

(8.16)

2222 xmff ��=−′

425

which can be written in more compact form as

[ ] [ ] { } [ ]{ }2 2

1 1A I d fα

ω ω

′− = −

(8.17)

with

[ ] { }

1

1

1

1

11 1 1 1( 1) 1(2 )

1 1 ( 1) (2 )

( 1)1 1 ( 1) ( 1)( 1) ( 1)(2 )

(2 )1 1 (2 ) (2 )( 1) (2 )(2 )

n

n

n

n

n n n d n d

n nn n n n d n n d

n n n n n n d n n d

n n n n n n d n n d

m m I I

m m I IA d

m m I I

m m I I

α α α α

α α α α

α α α α

α α α α

+

+

+ + + + +

+

=

� �

� � � � � �

� �

� �

� � � � � �

� �

(8.18)

Disc displacements x and ϕy can be calculated for known applied loads (e.g., the unbalance forces and

moments) as

(8.19)

with

[ ] [ ] [ ] [ ]1

2 2

1 1R A I α

ω ω

−

= − −

(8.20)

where R represents the receptance matrix and for the present case it contains only real elements. For

free vibrations the right hand side of equation (8.17) will be zero, i.e.

[ ] [ ] { } { }2

10A I d

ω

− =

(8.21)

which only satisfy when

[ ] [ ] { }2

10A I

ω− = (8.22)

and it will give the frequency equation and system natural frequencies could be calculated from this.

Alternatively, through eigen value analysis of matrix [A] system natural frequencies and mode shapes

could be obtained directly. In general, the receptance matrix, [R], may contain complex elements

when damping forces also act upon the shaft, in which case applied forces and disc displacements will

not all be in phase with one another. Hence, a more general form of equation (8.19) would be that

{ } [ ]{ }d R f ′=

426

which indicates both the real and imaginary parts of x, and R. In such case the real and imaginary

parts of each equations need to be separated and then these can be assembled again to into a matrix

form, which will have double the size that with complex quantities. Some of the steps are described

below

{ } { } [ ] [ ]( ) { } { }( )j j jr i r i r id d R R f f′ ′+ = + +

where r and i refer to the real and imaginary parts, respectively. Above equation can be expanded as

{ } { } [ ]{ } [ ]{ }( ) [ ]{ } [ ]{ }( )j jr i r r i i i r r id d R f R f R f R f′ ′ ′ ′+ = − + +

Now on equating the real and imaginary parts on both sides of equations, we get

{ } [ ]{ } [ ]{ }r r r i id R f R f′ ′= − and { } [ ]{ } [ ]{ }i i r r id R f R f′ ′= +

Above equations can be combined again as

[ ] [ ][ ] [ ]

r r i r

i i r i

d R R f

d R R f

′ − =

′

(8.23)

It can be observed that now the size of the matrix and vectors are double as that of equation (8.19).

Now through simple numerical examples (for the two or more DOFs) some of the basic concepts of

the present method will be illustrated.

f ′

427

Example 8.1 Obtain transverse natural frequencies of a rotor system as shown in Figure 8.3. Take the

mass of the disc, m = 10 kg and the diametral mass moment of inertia, Id = 0.02 kg-m2. The disc is

placed at 0.25 m from the right support. The shaft has a diameter of 10 mm and a span length of 1 m.

The shaft is assumed to be massless. Take the Young’s modulus E = 2.1 × 1011

N/m2

of the shaft.

Consider a single plane motion only.

Figure 8.3 A rotor system

Solution: Influence coefficients for a simply supported shaft are defined as

11 12

21 22

y

yzx

fy

M

α α

ϕ α α

=

with

and

It should be noted that subscript 1 represents corresponding to a force or a linear displacement, and

subscript 2 represents a moment or an angular displacement. To obtain natural frequencies of the rotor

system having a single disc, from equation (8.21), we have

[ ] [ ]11 122

2

21 22 2

1

1

1

d

nf

nf

d

nf

m I

A I

m I

α αω

ωα α

ω

−

− = −

(a)

Hence, the determinant of the above matrix would give the frequency equation as

(b)

2 2 2 3 2

11 12

3 2;

3 3

a b a l a al

EIl EIlα α

− −= = −

2 2

21 22

( ) 3 3;

3 3

ab b a al a l

EIl EIlα α

− − −= = −

( ) ( )4 2 2

11 22 12 11 22 1 0d nf nf dmI m Iω α α α ω α α− − + + =

428

For the present problem, we have

m/N; m/N; m/N

Equation (a) becomes,

which gives two natural frequencies of the system, as

rad/sec and rad/sec

It should be noted that the linear and angular motions are coupled for the present transverse vibrations

since the disc is offset from its mid span; however, when the disc is at the mid span then the linear and

angular motions will be decoupled (i.e., α12 = α21= 0). Natural frequencies for such case would be

rad/s

for the pure translation motion of the disc, and

rad/s

for the pure rotational motion of the disc. These expressions also of course can be obtained from

frequency equation (a) for α12 = α21= 0.

Example 8.2 Find the transverse natural frequency of a stepped shaft rotor system as shown in Figure

8.4. Consider the shaft as massless and is made of steel with the Young’s modulus, E = 2.1 (10)11

N/m2. The disc could be considered as a point mass of 10 kg. The circular shaft is simply supported at

ends (In Figure 8.4 all dimensions are in cm).

Figure 8.4 A simply supported stepped shaft

4

11 1.137 10α −= × 4

12 21 3.03 10α α −= = − × 3

22 1.41 10α −= ×

4 4 2 78.505 10 7.3 10 0nf nfω ω− × + × =

129.4nfω =

2290nfω =

1 4

11

1 122.244

10 2.021 10nf

mω

α −= = =

× ×

2 3

22

1 1248.697

0.02 0.8084 10nf

dI

ωα −

= = =× ×

429

Solution: To simply the analysis let us consider that the angular displacement of the disc is negligibly

small. From equation (8.23), we have

[ ] [ ] 112 2

1 1

nf nf

A I mαω ω

− = −

from which the natural frequency is given as

Hence, now the next step would be to obtain the influence coefficient, . Using the energy method

this influence coefficient is obtained as follows

Figure 8.5 A free body diagram of the rotor system

For a load F when it acts at the disc, reaction forces at bearings can be obtained as (Figure 8.5)

and

Figure 8.6 Free body diagram of the shaft section 0 0.6z≤ ≤

From Figure 8.6, the bending moment in the shaft can be obtained as

1 1

0 0.4 0 0.4C x xM M Fz M Fz= ⇒ − = ⇒ =∑ (a)

11

1nf

mω

α=

11α

0 1 0.6 0 0.6A B BM F F F F+ = ⇒ × − × = ⇒ =∑

0 0.4A B AF F F F F F+ = ⇒ + = ⇒ =∑

430

Figure 8.7 Free body diagram of the shaft section 0.6 1.0z≤ ≤

From Figure 8.7, the bending moment in the shaft can be obtained as

2 20 ( 0.6) 0.4 0 0.6 (1 )D x xM M F z Fz M F z= ⇒ + − − = ⇒ = −∑ (b)

The strain energy stored in the shaft from bending moments can be obtained as

From the Castigliano’s theorem the linear displacement can be obtained as

(c)

On substituting equations (a) and (b) into equation (c), we get

{ }{ }0.6 1.0

0 0.61 2

0.6 (1 ) 0.6(1 )(0.4 )(0.4 )dz dz

EI EI

F z zFz zδ = +

− −∫ ∫

The stiffness of the beam given as

From above in fact it can be observed that two shaft segment can be thought as connected parallel to

each other at disc location.

1 2

2 20.6 1.0

0 0.61 22 2

x xM dz M dz

UEI EI

= +∫ ∫

1 2

1 20.6 1.0

0 0.61 2

x x

x x

M MM M

F Fdz dzEI EI

U

Fδ

∂ ∂

∂ ∂= +∂

=∂ ∫ ∫

2 20.6 1.0

0 0.61 2 1 2

0.16 0.36 ( 2 1) 0.01152 0.00768dz dz

EI EI

Fz F z z F F

EI EI= + =

− ++∫ ∫

1

11 1 2

1

δ

0.01152 0.00768Fk

EI EIα

−

=

= = +

431

With , we have

Hence the natural frequency is given as

rad/sec

The influence coefficient can be also obtained by the singular function approach and for more details

readers are referred to Timoshenko and Young (1968).

Example 8.3 Find transverse natural frequencies and mode shapes of a rotor system shown in Figure

8.8. B is a fixed end, and D1 and D2 are rigid discs. The shaft is made of the steel with the Young’s

modulus E = 2.1×1011

N/m2 and a uniform diameter d = 10 mm. Shaft lengths are: BD2 = 50 mm, and

D1D2 = 75 mm. The mass of discs are: m1 = 2 kg and m2 = 5 kg. Consider the shaft as massless and

neglect the diametral mass moment of inertia of discs.

Figure 8.8

Solution: For simplicity of the analysis, the shaft is considered as massless and disc masses are

considered as point masses (i.e., the diametral mass moment of inertia of discs are neglected). The

first step would be to obtain the influence coefficients corresponding to two disc locations acted by

concentrated forces. Basically, we need to derive linear displacements at two disc locations due to

forces F1 and F2 acting at these locations as shown in Fig. 8.9. These deflection relations are often

available in a tabular form in standard handbooks (e.g., Young and Budynas, 2002). However, for the

present problem the calculation of influence coefficients is explained by the energy method and by the

singularity function to clarify the basic concept for the completeness.

11 2 4 6 4 3 4 4

1 2

π π2 10 N/m ; 0.1 4.907 10 m ; 0.3 3.976 10 m

64 64E I I

− −= × = = × = = ×

7

11

18.45 10 m/Nk

α= = ×

7

11

1 8.452906.89

10

10n

mω

α= = =

×

432

Figure 8.9 A shaft with two concentrated forces

Fig. 8.10 The free-body diagram of a shaft segment for 0 ≤ z ≤ a

Energy method: In this method, we need to obtain the strain energy due to bending moments in the

shaft. Bending moments at different segments of the shaft can be obtained as

(i) Shaft segment for 0 ≤ z ≤ a

From Fig. 8.10, the bending moment can be written as

1 1yz yM F z= − (a)

(ii) Shaft segment for a ≤ z ≤ L

From Fig. 8.11, the bending moment can be written as

2 1 2 ( )yz

M F z F z a= − − − (b)

Fig. 8.11 The free-body diagram of a shaft segment for a ≤ z ≤ L

The strain energy is given by

433

1 2

2 2

0 2 2

a Lyz yz

a

M dz M dzU

EI EI= +∫ ∫ (c)

Using the Castigliano theorem, the linear deflection at station 1 can be written as

1 2

1 2

1 1

1

10

yz yz

yz yza L

y y

ay

M MM dz M dz

F FUy

F EI EI

∂ ∂

∂ ∂∂= = +

∂ ∫ ∫ (d)

On substituting equations (a) and (b) into equation (d), we get

( )( ) { }( )

( )( ) ( )

1 1 2

1

1 1 2 1 2

10

3 3 2 23 3 3 23 3 3

0

( )

3 3 3 2 3 3 2

a Ly y y

ay

La L

y y y y y

a a

F z z dz F z F z a z dzUy

F EI EI

L a a L aF F F F Fz z z aza L a

EI EI EI EI EI

− − − − − −∂= = +

∂

− − = + + − = + − + −

∫ ∫

which finally takes the form

( ) ( )1 2

3 3 2 23

13 3 2

y y

L a a L aLy F F

EI EI EI

− − = + −

(e)

Equation (e) has the following form

1 1 1 1 2 21 y f y y f yy F Fα α= + (f)

Hence, for a = 0.075 m, L = 0.125 m, N-m2, we have

1 1y fα = 6.31×10

-6 m/N

and

1 2y fα = 1.32 ×10

-6 m/N

(g)

The deflection at station 2 can be obtained as

1 2

1 2

2 2

2

20

yz yz

yz yza L

y y

ay

M MM dz M dz

F FUy

F EI EI

∂ ∂

∂ ∂∂= = +

∂ ∫ ∫ (h)

On substituting equations (a) and (b) into equation (h), we get

103.1EI =

434

( )( ) { }( )1 1 2

2

1 2

20

3 2 3 22

0 ( )

20

3 2 3 2

a Ly y y

ay

L L

y y

a a

F z dz F z F z a z a dzUy

F EI EI

F Fz az z aza z

EI EI

− − − − − +∂= = +

∂

= + − + − +

∫ ∫

which finally takes the form

( ) ( ) ( ) ( ) ( )1 2

3 3 2 2 3 3 2 2 2

23 2 3

y y

L a a L a L a a L a a L ay F F

EI EI EI EI EI

− − − − − = − + − +

(i)

Equation (i) has the following form

2 1 1 2 2 22 y f y y f yy F Fα α= + (j)

Hence, for a = 0.075 m, L = 0.125 m, N-m2, we have

2 1y fα = 6.07×10

-6 m/N

and

2 2y fα = 1.92×10

-7 m/N

(k)

Method of the singularity function: Now the influence coefficients are obtain by the method of

singularity function for illustration. The singularity function (< >) is defined as

< f > = f(z) if f(z) > 0

=0 if f (z) < 0 (l)

The bending moment can be written as

1 20.075

y yEIy F z F z′′ = < > + < − > (m)

On integrating twice equation (m), we get following expressions

1 2

1 12 2

12 20.075y yz zEIy F F c< > < − >′ = + + (n)

and

2

1

3

1 2

3

0.0756 6

y

yz z

FzEIy F c c< − >

< >= + + + (o)

103.1EI =

435

where the integration constants c1 and c2 are obtained by the boundary conditions, and are give as

(i) 0.125

0z

y=

′ = from equation (n), we have

1 2

2 2

1

0.125 0.050

2 2y yF F c

< > < >+ + =

which gives

1 21 0.0078 0.0013y yc F F= − − (p)

(ii)0.125

0z

y=

= from equation (o), we have

1 2 1 2

3 3

2

0.125 0.050.125 0.0078 0.0013 0

6 6y y y yF F F F c

< > < >+ + − − + =

which gives

1 2

4 4

2 6.51 10 1.354 10y yc F F− −= × + × (q)

Finally equation (o) becomes

( )2

1 1 2 1 2

3 4 4

3

0.075 0.0078 0.0013 6.51 10 1.354 106 6

y

y y y y yEIy z F F z F FFz

F − −= < − > − − × + ×< >

+ + +

(r)

On evaluating the deflection at station 1 for z = 0 (i.e., at the free end), we have

{ }1 2

4 4

0

16.51 10 1.354 10y yz

y F FEI

− −

== × + ×

which has the following form

1 1 1 1 2 21 y f y y f yy F Fα α= +

with

1 1y f

α = 6.316×10-6

m/N

and 1 2y f

α = 1.314×10-6

m/N

The deflection at station 2 for z = 0.075 m, we get

( )1 2

34 4

0.075

1 0.0750.075 0.0078 6.51 10 0.0013 0.075 1.354 10

6y yz

y F FEI

− −

=

= − × + × + − × + ×

(s)

which has the following form

436

2 1 1 2 2 22 y f y y f yy F Fα α= +

with

2 1y fα = 1.314×10

-6 m/N

and

2 2y fα = 0.404×10

-6 m/N

which is same as obtained by the energy method. For free vibrations from equation (8.17), we have

1 1 1 2

2 1 2 2

1 22

1

2

1 2 2

1

0

01

y f y f

nf

y f y f

nf

m my

ym m

α αω

α αω

−

=

− (t)

For a non-trial solution, it gives

1 1 1 2

2 1 2 2

1 22

1 2 2

1

01

y f y f

nf

y f y f

nf

m m

m m

α αω

α αω

−

=

−

(u)

which give the frequency equation as

( ) ( )1 1 2 2 1 2 1 1 2 2

4 2 2

1 2 1 2 1 0nf y f y f y f nf y f y f

m m m mω α α α ω α α− − + + = (v)

For the present problem, we have

1 1y fα = 6.316×10

-6 m/N,

1 2 2 1y f y fα α= = 1.314×10

-6 m/N

and

2 2y fα = 0.404×10

-6 m/N

1 2m = kg and 2 5m = kg

Equation (u) becomes,

4 6 2 111.772 10 1.209 10 0nf nfω ω− × + × =

which gives two natural frequencies of the system for the pure translator motion, as

1266.67

nfω = rad/sec and

21304.0

nfω = rad/sec

437

B D2 D1

Shaft length (mm)

Figure 8.12 Mode shapes; (1) - first mode shape, (2) - second mode shape

The mode shapes corresponding to these natural frequencies are plotted in Fig. 8.12. While plotting

mode shapes the maximum displacement corresponding to a particular mode has been chosen as unity

and other displacements have been normalized accordingly. It should be noted that for the case when

discs have appreciable diametral mass moment of inertia, then along with the forces at stations 1 and

2, the moments also need to be considered while deriving influence coefficients. In that case, we will

have sixteen influence coefficients, i.e., the size of the influence coefficient matrix would be 4×4,

however, the symmetry conditions of influence coefficients will still prevail. Correspondingly, we

would have four natural frequencies and corresponding mode shapes.

Example 8.4 Find transverse natural frequencies and mode shapes of the rotor system shown in

Figure 8.13. B is a fixed bearing, which provide fixed support end condition; and D1, D2, D3 and D4

are rigid discs. The shaft is made of the steel with the Young’s modulus E = 2.1 (10)11

N/m2 and the

uniform diameter d = 20 mm. Various shaft lengths are as follows: D1D2 = 50 mm, D2D3 = 50 mm,

D3D4 = 50 mm and D4B2 = 150 mm. The mass of discs are: m1 = 4 kg, m2 = 5 kg, m3 = 6 kg and m4 = 7

kg. Consider the shaft as massless. Consider the disc as point masses, i.e., neglect the diametral and

polar mass moment of inertia of all discs.

0 20 40 60 80 100 120-1

-0.5

0

0.5

1

(1)

(2)

Rel

ativ

e li

nea

r dis

pla

cem

ents

438

Figure 8.13 A multi-disc overhung rotor

Solution: The first step would be to obtain influence coefficients. We have the linear deflection, y, and

force relations from the strength of material for a cantilever shaft with a concentrated force at any

point (Fig. 8.14) as

2

(3 ) for 06

yf zy a x x a

EI= − < < and

2

(3 ) for a6

yf ay x a x L

EI= − < <

Figure 8.14 A cantilever shaft with a concentrated force at any point

Let us take stations 1, 2, 3, and 4 at each of the disc locations. Figure 8.15 shows a cantilever shaft

with forces at stations 1 and 2. Station 1 is at free end and it has an intermediate station 2 between

station 1 and the fixed support. Lengths L, a, and b are shown in Fig. 8.15 and for this case we have L

= a + b. Between two stations, it will have the following influence coefficients (which relates the

force to linear transverse displacement)

1

3

1

113

y

y L

f EIα = = ;

2

3

2

223

y

y a

f EIα = = ; ( )

1 2

2

2 1

21 123

6y y

y y aL a

f f EIα α= = = = − (a)

similar relations would also be valid between stations (1, 3) and (1, 4).

Figure 8.15 A cantilever shaft with a force at the free end and another at the intermediate point

439

Figure 8.15 shows a cantilever shaft with a force at station 2, and has a station 3 between the force

and the fixed end. Between two stations, it will have the following influence coefficients

( )3

223

a b

EIα

+= ;

3

333

a

EIα = ; ( )

2

32 232 3

6

aa b

EIα α= = + (b)

These relations could also be used between stations (2, 4) and (3, 4).

Figure 8.16 A cantilever shaft with two forces intermediate stations

Table 8.1 A calculation procedure of influence coefficients in a multi-discs cantilever shaft

S.N.

Disc

location* iL (m)

ija (m)

ijb (m) iiα (m/N) jiij αα = (m/N) jjα (m/N)

1 (1, 1) 0.30 0.30 0.00 5.4567×10-6

5.4567×10-6

5.4567×10-6

2 (2, 2) 0.30 0.25 0.00 3.1578×10-6

3.1578×10-6

3.1578×10-6

3 (3, 3) 0.30 0.20 0.00 1.6168×10-6

1.6168×10-6

1.6168×10-6

4 (4, 4) 0.30 0.15 0.00 6.8209×10-7

6.8209×10-7

6.8209×10-7

5 (1, 2) 0.30 0.25 0.05 5.4567×10-6

4.1052×10-6

3.1578×10-6

6 (1, 3) 0.30 0.20 0.10 5.4567×10-6

2.8294×10-6

1.6168×10-6

7 (1, 4) 0.30 0.15 0.15 5.4567×10-6

1.7052×10-6

6.8209e-007

8 (2, 3) 0.30 0.20 0.05 3.1578×10-6

2.2231×10-6

1.6168×10-6

9 (2, 4) 0.30 0.15 0.10 3.1578×10-6

1.3642×10-6

6.8209×10-7

10 (3, 4) 0.30 0.15 0.05 1.6168×10-6

1.0231×10-6

6.8209×10-7

* Numbers in bracket represent the force station numbers

It should be noted that in Fig. 8.16 the shaft segment from the force 2y

f to the free end would act as

rigid shaft. It will not contribute in the deformation of the shaft at station 2, it will act as if it is not

present at all (this is true only for mass-less shaft assumption).

440

When disc 1 and 2 is present, we have Li = L1 = 0.3 m, aij = a12 = 0.25 m. Hence the influence

coefficients takes the following form: α11 = 5.4567×10-6

m/N, α12 = α21 = 4.1052×10-6

m/N, α22 =

3.1578×10-6

m/N. On the similar lines other influence coefficients can be calculated and it is

summarized in Table 8.1

Now for free vibrations, the governing equation has the following form

1 1 1 2 1 3 1 4

2 2 2 3 2 4

3 3 3 4

4 4

1 2 3 41

2 3 4 2

2

33 4

44

1 0 0 0 0

0 1 0 0 01

0 0 1 0 0

0 0 0 1 0sym

y f y f y f y f

y f y f y f

y f y f nf

y f

m m m m y

m m m y

ym m

ym

α α α α

α α α

α α ω

α

− =

(c)

which can be written as for the data of the present problem, as

[ ] [ ] { } { }2

10

nf

A I yω

− =

(d)

with

1 1 1 2 1 3 1 4

2 2 2 3 2 4

3 3 3 4

4 4

1 2 3 4

2 3 4 4

3 4

4

0.2183 0.2053 0.1698 0.1194

0.1642 0.1579 0.1334 0.0955[ ] 10

0.1132 0.1112 0.0970 0.0716

0.0682 0.0682 0.0614 0.0477sym

y f y f y f y f

y f y f y f

y f y f

y f

m m m m

m m mA

m m

m

α α α α

α α α

α α

α

−

= =

The eigen value and eigen vector of the above matrix is obtained as

1

2

3

4

2

2

2 4

2

2

1 / 0.5076

1 / 0.0121[1 / ] 10

0.00101 /

0.00021 /

nf

nf

nf

nf

nf

ω

ωω

ω

ω

−

= =

;

and

31 2 4

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

0.7082 0.6740 0.5085 -0.2763

0.5434 0.0317 -0.5467 0.6740[ ]

0.3836 -0.4402 -0.3758 -0.6371

nf nf nf nf

Y Y Y Y

Y Y Y YY

Y Y Y Y

Y Y Y Yω ω ω ω

= =

0.2368 -0.5925 0.5489 0.2519

Hence, the transverse natural frequencies are

1nf

ω = 140.36 rad/s; 2nf

ω = 909.09 rad/s; 2nf

ω = 3162.28 rad/s, 4nf

ω = 7071.07 rad/s. The

corresponding eigen vectors (mode shapes) can be plotted.

441

8.2 Transfer Matrix Method

In the influence coefficient method, governing equations are derived by considering the equilibrium

of the whole system. Difficulties with such an approach are that for a large system it becomes very

complex and we need to evaluate influence coefficients for each case. While in the transfer matrix

method (TMM), which is also called the Myklestand & Prohl Method, the shaft is divided into a

number of imaginary smaller beam elements and governing equations are derived for each of these

elements in order to determine the overall system behavior. The TMM has an advantage over the

method of influence coefficient in that the size of matrices being handled does not increase with the

number of DOFs. It demands less computer memory and associated ill-conditioning problems (i.e.,

nearly singular matrices) are less. Also this method is relatively simple and straightforward in

application. For the present case discs are considered as point masses. In the case when the mass of

the shaft is appreciable then it is divided up into a number of smaller masses concentrated (or lumped)

at junctions (or stations) of beam elements so that concentrated masses and the shaft may be modeled

as shown in Figure 8.17. The station number can be assigned wherever there is change in the state

vector (i.e., the linear and angular displacements, shear forces, and bending moments) for example at

step change in shaft diameter, at discs, coupling, and support positions, etc. Station numbers are given

from 0 to (n+1).

Figure 8.17 Modeling a real rotor with discrete elements

Figure 8.18 A free body diagram of an ith shaft segment Figure 8.19 A cantilever beam

8.2.1 A field matrix: Figure 8.18 shows the free body diagram of the ith shaft segment, which is

between stations (i - 1)th

and ith. Let Sy be the shear force in y-direction, Myz

is the bending moment in

442

y-z plane, y is the linear displacement in y-direction, and is the angular displacement in y-z plane.

The back-subscript R refers to the right of a particular mass and L refers to the left of a particular

mass. Directions for the force, the moment, and displacements are chosen according to the positive

sign conversion of the strength of material (Timoshenko and Young, 1968).

The displacement (linear or angular) at ith

station will be equal to the sum of the relative displacement

between ith and (i - 1)

th stations and the absolute displacement of (i - 1)

th station. As a first step for

obtaining the relative displacement, (i - 1)th station can be considered to be fixed end as shown in

Figure 8.19. Hence, the displacement and the slope at the free end are related to the applied moment

and the shear force at free end by considering the beam as though it were a cantilever (Timoshenko

and Young, 1968). Then as a second step, the displacement and the slope of the fixed end is

considered by considering the beam as a rigid. On assumption of small displacements, hence, finally

above two steps could be superimposed to get the total displacement and slope of the ith shaft segment

at the left of ith station (i.e. i

th mass), as

1

2 3

12 3

i i

i

L yz L y

L i R i R x

M l S ly y l

EI EIϕ

−−= − − + (8.24)

and

1

2

2

i i

i i

L yz L y

L x R x

M l S l

EI EIϕ ϕ

−= + − (8.25)

where l is the length of shaft segment, and EI is the flexural rigidity of the shaft segment. It should be

noted that the first two terms in right hand side of equation (8.24) and first term in the right hand side

of equation (8.25) are related to the second step in which shaft is considered as rigid. Last two terms

on the right hand side of equations (8.24) and (8.25) are from the first step in which the shaft is

considered to be cantilevered. From the free body diagram (Figure 8.18) the shear force and the

bending moment at either ends of the ith shaft segment are related as

1i iL y R yS S

−= (8.26)

and

1i i iL yz R yz L yM M S l

−= + (8.27)

On substituting for iL y

S and iL yz

M from equations (8.26) and (8.27) into equations (8.24) and (8.25),

these equations could be rewritten as

xϕ

443

( )1 1 1

1

2 3

12 3

i i i

i

R yz R y R y

L i R i R x

M S l l S ly y l

EI EIϕ

− − −

−−

+= − − +

(8.28)

( )1 1 1

1

2

2

i i i

i i

R yz R y R y

L x R x

M S l l S l

EI EIϕ ϕ

− − −

−

+= + −

(8.29)

1i iL y R yS S

−=

(8.30)

and

1 1i i iL yz R yz R yM M S l

− −= +

(8.31)

These equations can be rearranged and expressed in a matrix form as

(8.32)

with 2 3

2

1

1

12 6

0 1{ } ; [ ] ; { }2

0 0 1

0 0 0 1

x x

L i i R i

yz yz

y yL i R ii

l lly yEI EI

l lS F SEI EI

M Ml

S S

ϕ ϕ−

−

− −

= = =

(8.33)

where is the field matrix for the ith shaft segment, and is the state vector at i

th mass. The

filed matrix, , transforms the state vector from the left of a shaft segment to the right of the shaft

segment. Equation (8.32) is for the motion in the vertical plane. Similar set of equations may be

written for motion in the horizontal plane. In general, state vectors will have both real and imaginary

components due to two plane motion (and/or due to the damping and gyroscopic moments, however,

the damping is not considered in the present analysis and because of this field matrix remains real.

The effect of gyroscopic moments will be considered subsequently). Equation (8.32) is expanded to

give a more general form as

(8.34)

with

1{ } [ ] { }L i i R i

S F S −=

[ ]i

F { }i

S

[ ]i

F

1{ } [ ] { }L i i R iS F S∗ ∗ ∗

−=

444

(8.35)

where is the modified state vector of the size 17×1 and is the modified field matrix of the

size 17×17. Subscripts h and v refer to the horizontal and vertical directions respectively; and r and j

refer to real and imaginary parts, respectively. The last equation of an identity has been added to

facilitate inclusion of the unbalance in the analysis, as will be made clear later (it was used for the

torsional vibrations also in Chapter 6). It should be noted that in the simplest case for the single plane

motion and without damping (i.e., imaginary components will be zero) the size of the modified field

matrix *

iF will be 5×5 and is given as

2 3

2

* *

1

1

1 02 6

0 1 02

{ } ; [ ] ; { }0 0 1 0

0 0 0 1 01 1

0 0 0 0 1

x x

L i i R iyz yz

y y

L i R i

i

l lly yEI EI

l lEI EI

S F SM Ml

S S

ϕ ϕ

−

−

− −

= = =

(8.36)

8.2.2 A point matrix: Figure 8.20 shows the free body diagram of the point mass mi, where iy

u is the

unbalance force at ith location. The relationship between forces and displacements at the concentrated

mass is given by its equations of motion

2

i i iR y L y y i i i iS S u m y m yω− + = = −�� (8.37)

where ω is the spin speed of the shaft. It should be noted that a synchronous whirl is considered here.

1

1

{ } { }[ ] 0 0 0 0

{ } { }0 [ ] 0 0 0

{ } ; [ ] ; { }{ } 0 0 [ ] 0 0 { }

0 0 0 [ ] 0{ } { }

0 0 0 0 11 1

r r

j j

r r

j j

h h

h h

L i i R iv v

v v

iL i R i

S SF

S SF

S F SS F S

FS S

∗ ∗ ∗

−

−

= = =

{ }S∗ [ ]F

∗

445

Figure 8.20 Forces, moments and linear & angular displacements on a concentrated mass

Similarly, for the moment and the angular displacement, we have the following relationship

2

i i i i i iR yz L yz d x d xM M I Iϕ ω ϕ− = = −�� (8.38)

where Id is the diametral mass moment of inertia of the disc. Moreover, since the linear and angular

(slope) displacements on each side of mass are equal. Hence, we can rewrite all governing equations

for a concentrated mass as

R i L iy y− = −

(8.39)

i iR x L xϕ ϕ=

(8.40)

2

i i i iR yz d L x L yzM I Mω ϕ= − + (8.41)

and

2

i i iR y i R i L y yS m y S uω= − + − (8.42)

Hence, we can combine these conditions including in a matrix form as

(8.43)

with

2

2

01 0 0 0

00 1 0 0{ } ; [ ] ; { } ; { }

00 1 0

0 0 1

x x

R i i L i i

yz yzd

y y yiR i L i i

y y

S P S uM MI

S S um

ϕ ϕ

ω

ω

− − = = = =

− −

(8.44)

{ } [ ] { } { }R i i L i i

S P S u= +

446

where is the point matrix and is the unbalance vector of the ith mass. The point matrix,

, transforms the state vector from the left of a disc to the right of the disc. Equation (8.43) can be

written for the horizontal plane also. In general since unbalance will also have both the real and

imaginary components, equation (8.43) can be expanded to a more general form as

(8.45)

with

1

{ } [ ] 0 0 0 { } { }

{ } 0 [ ] 0 0 { } { }

{ } ; [ ] ; { }{ } 0 0 [ ] 0 { } { }

{ } 0 0 0 [ ] { } { }

1 0 0 0 0 1 1

r r r

j j j

r r r

j j j

h h h

h h h

R i i L i

R i i L i

S P u S

S P u S

S P SS P u S

S P u S

ν ν ν

ν ν ν

∗ ∗ ∗

−

= = =

(8.46)

where is the modified point matrix and the size of the matrix is 17×17. By putting 1 in the last

row of the state vector, unbalance force components have been accommodated in the last column of

the modified point matrix itself. It should be noted that in the simplest case for a single plane motion

the size of the modified point matrix will be 5×5 and is given as

* 2

2

1 0 0 0 0

0 1 0 0 0

[ ] 0 1 1 0

0 0 1

0 0 0 0 1

i d

y

i

P I

m u

ω

ω

= −

− (8.47)

8.2.3 An overall transfer matrix: Once we have the point and field matrices, we can use these to

form the overall transfer matrix to relate the state vector at one extreme end station (say left) to the

other extreme end (right), provided we do not have intermediate supports that will be dealt

subsequently. Equations (8.34) and (8.45) can be combined as

(8.48)

with

iP][ iu}{

iP][

{ } [ ] { }R i i L iS P S∗ ∗ ∗=

[ ]iP∗

[ ]iP∗

1 1{ } [ ] { } [ ] [ ] { } [ ] { }R i i L i i i R i i R iS P S P F S U S∗ ∗ ∗ ∗ ∗ ∗ ∗

− −= = =

[ ] [ ] [ ]i i iU P F∗ ∗ ∗=

447

where is the modified transfer matrix for ith segment, which transforms the state vector from the

right of (i-1)th station to the right of i

th station. The transfer matrix for all (n+1) stations (i.e., from 0

th

to nth

as shown in Fig. 8.21) in the system may be obtained in the similar manner as equation (8.48) as

follows

{ } { }

{ } { } { }

{ } { } { }

* * *

1 1 0

* * * * * *

2 2 2 2 1 0

* * * * * * *

3 3 2 3 2 1 0

R R

R R R

R R R

S U S

S U S U U S

S U S U U U S

=

= =

= =

�

(8.49)

where is the overall modified transfer matrix for the complete rotor system and the size of the

matrix is 17×17.

Now the transfer matrix method is illustrated for a very simple case when there is no coupling

between the vertical and horizontal planes (e.g., from gyroscopic effects or from bearings, then only

single plane motion can be considered) and no damping in the system (i.e., in a single plane

displacements and forces are in phase with each other), then the overall transfer matrix [T] will take a

size of 5×5. Equation (8.49) can be written in the expanded form as

(8.50)

To determine the system characteristics it is first necessary to define the system boundary conditions,

which describe the system support conditions.

Figure 8.21 A simply supported multi-DOF rotor system

[ ]iU∗

{ } { } { }* * * * * * * * *

1 1 01 0[ ] [ ] [ ] [ ] [ ] { }

n n n RR n R n RS U S U U U S T S−−

= = =…

*[ ]T

1,1 1,2 1,3 1,4 1,5

2,1 2,2 2,3 2,4 2,5

3,1 3,2 3,3 3,4 3,5

4,1 4,2 4,3 4,4 4,5

0

1 0 0 0 0 1 1

x x

yz yz

y y

n

y t t t t t y

t t t t t

M t t t t t M

S t t t t t S

ϕ ϕ

− −

=

448

For illustration of the TMM, let us consider a simply supported multi-DOF rotor system as shown in

Figure 8.21, for which linear displacements and moments are zero at supports (i.e., and

). These states are corresponding to 1st and 3

rd rows in the state vector {S} and above

equation take the following form

1,1 1,2 1,3 1,4 1,5

2,1 2,2 2,3 2,4 2,5

3,1 3,2 3,3 3,4 3,5

4,1 4,2 4,3 4,4 4,5

0

0 0

0 0

1 0 0 0 0 1 1

x x

y y

n

t t t t t

t t t t t

t t t t t

S t t t t t S

ϕ ϕ

=

(8.51)

On considering only 1st and 3

rd set of equations from the matrix equation (8.51), since they have zero

on the left hand side vector, hence, we get

1,2 1,4 1,5

3,2 3,4 3,50

x

y

t t t

St t t

ϕ − = −

(8.52)

The remaining two equations, i.e., 2nd

and 4th set of equations from the matrix equation (8.51) can be

written as

2,2 2,4 2,5

4,2 4,4 4,50

x x

y yn

t t t

S St t t

ϕ ϕ = +

(8.53)

8.2.4 Free vibrations: For free vibrations, the frequency of vibration, ω, in equation (8.44) will be the

natural frequency, , of the system. For the present illustration of the simply supported shaft,

equation (8.52) becomes a homogeneous equation since right hand side terms (i.e., terms of t’s with

second subscript as 5) are related to unbalance forces, which are zero in free vibrations. Hence, we

have

1,2 1,4

3,2 3,4 0

0

0

x

y

t t

St t

ϕ =

(8.54)

For non-trivial solution the determinant of the above matrix will be zero, hence

(8.55)

0 0ny y= =

00

nyz yzM M= =

nfω

( ) , , , ,nff t t t tω = − =1 2 3 4 3 2 1 4 0

449

Equation (8.55) is a frequency equation in the form of a polynomial and it contains natural

frequencies as unknown. In case of simple systems, the frequency can be obtained in an explicit form.

As by hand calculations it is feasible to obtain roots of a polynomial of only third degree with the help

of closed form solutions. However, for complex systems any convenient root-searching techniques of

the numerical analysis (e.g., the incremental method, the bisection method, the Newton-Raphson

method, etc.) could be used. In such cases there is no need to multiple the field and point matrices in

the symbolic form, whilst these matrices should be multiplied in the numerical form by choosing

suitable guess value of the natural frequency and condition of equation (8.55) may be checked with

certain acceptable limits after the final multiplications. Based on the residue of equation (8.55) or its

derivatives, the next guess value of the natural frequency may be decided and the process may be

repeated till the residue is reduced to the desired level of accuracy. Since equation (8.55) has, in

general, several roots the procedure may be repeated to obtain the remaining roots. Care should be

exercised in obtaining all the roots (natural frequencies) without stepping over any of them. Torsional

vibrations by using the TMM may be referred for more details of the root searching algorithm.

To obtain mode shapes, from equation (8.54) we have

(8.56)

Let us take a reference value of the angular displacement at station 0 as . On substituting the

second term (the third term can also be used) of equation (8.56) into equation (8.53) for free

vibrations, we get

2,2 2,4

4,2 4,4 1,2 1,4

1

/

x

y n

t t

S t t t t

ϕ = −

(8.57)

Now the state vector at nth station is known completely. By back substitution of the state vector at n

th

station into equations (8.49), we can get state vectors at other stations. It should be noted that t’s are

function of the natural frequency, hence from the procedure described above we will get a mode shape

corresponding to a particular natural frequency. The procedure can be repeated for each of the system

natural frequencies to get the corresponding mode shapes. In general, for a system of n number of

degrees of freedom, it will have n number of natural frequencies; and that many number of mode

shapes. Moreover, the mode shape for a particular natural frequency has the unique relative

, ,

, ,

y x x

t tS

t tϕ ϕ= − = −

0 0 0

1 2 3 2

1 4 3 4

xϕ =0

1

450

amplitudes. For other kinds of boundary conditions Table 8.2 provides frequency equations and

equations to obtain mode shapes.

Table 8.2 Equations for the calculation of natural frequencies and mode shapes.

S.N. Boundary

conditions

Station

numbers

Equations to get natural

frequencies

Equations to get mode shapes

1 Simply

supported

(Pinned-roller

support)

0: Pinned

end,

n: roller

support

1,2 1,4

3,2 3,4 0

0

0

x

y

t t

St t

ϕ =

2,2 2,4

4,2 4,4 0

x x

y yR n

t t

S St t

ϕ ϕ =

2 Cantilever

(Fixed-free)

0: Fixed

end,

n: free end

3 Fixed-fixed 0, n: Fixed

ends

1,3 1,4

2,3 2,4 0

0

0

yz

y

Mt t

St t

=

3,3 3,4

4,3 4,4 0

yz yz

y yR n

M Mt t

S St t

=

4 Free-free 0, n: Free

ends

3,1 3,2

4,1 4,2 0

0

0z

t t y

t t ϕ

− =

1,1 1,2

2,1 2,2 0z zn

t ty y

t tϕ ϕ

− − =

5 Fixed-pined 0: Fixed

end,

n: pinned

end

1,3 1,4

3,3 3,4 0

0

0

yz

y

Mt t

St t

=

2,3 2,4

4,3 4,4 0

x yz

y yn

Mt t

S St t

ϕ =

8.2.5 Forced response: It should be noted that the extreme right hand side vector of equations (8.52)

and (8.53) are unbalance forcing terms (i.e., terms of t’s with second subscript as 5). Other terms of t’s

which contain ω is the spin speed of the shaft. Equation (8.52) can be used to obtain the state vector

at station 0. Subsequently, state vectors at the nth and other locations can be obtained by equations

(8.53) and (8.49), respectively. These state vectors contain both the angular and linear displacements,

and the reaction force and the moment. Hence, unbalance responses and reaction forces are known,

which can be obtained for various speeds. Equations (8.52) and (8.53) are tabulated in Table 8.3 along

with similar equations for other standard boundary conditions for a ready reference.

33 34

43 44 0

0

0

yz

y

Mt t

St t

=13 14

23 24 0

yz

yxR n

My t t

St tϕ

− =

451

Table 8.3 Governing equations for forced vibration with different boundary conditions

S.N. Boundary

conditions

Station

numbers

Equations to get state vectors

1 Simply

supported

(pinned-

roller

supports)

0: pinned

end,

n: roller

support

1,2 1,4 1,5

3,2 3,4 3,50

x

y

t t t

St t t

ϕ − = −

2,2 2,4 2,5

4,2 4,4 4,50

x x

y yR n

t t t

S St t t

ϕ ϕ = +

2 Cantilever

(fixed-free)

0: fixed

end,

n: free

end

3,5

4,5

33 34

43 44 0

yz

y

M t

t

t t

St t

− −

= 1,513 14

2,523 24 0

yz

yxR n

M ty t t

S tt tϕ

− = +

3 Fixed-fixed 0, n:

fixed

ends

1,3 1,4 1,5

2,3 2,4 2,50

yz

y

Mt t t

St t t

− = −

3,3 3,4 3,5

4,3 4,4 4,50

yz yz

y yR n

M Mt t t

S St t t

= +

4 Free-free 0, n: free

ends 3,1 3,2 3,5

4,1 4,2 4,50z

t t ty

t t tϕ

−− = −

1,1 1,2 1,5

2,1 2,2 2,50z zn

t t ty y

t t tϕ ϕ

− − = −

5 Fixed-

pined

0: fixed

end,

n: pinned

end

1,3 1,4 1,5

3,3 3,4 3,50

yz

y

Mt t t

St t t

− = −

2,3 2,4 2,5

4,3 4,4 4,50

x yz

y yn

Mt t t

S St t t

ϕ = −

8.2.6 Gyroscopic effects: If gyroscopic effects are allowed for, in equation (8.45) the modified point

matrix, [P*], will get some extra terms, however, the modified field matrix, [F

*], will not be affected.

For free vibrations, the formulation would have more general form since equations will contain both

the natural whirl frequency, ν (while considering gyroscopic effects we tried to distinguish the whirl

natural frequency symbol by ν in place of nfω ; while the former is speed-dependent ( )ν ω , the latter

is speed independent so that ( 0)nf

ν ω ω= = ), and the spin speed, ω , of the shaft. This leads to speed

dependency of the natural whirl frequency.

Equilibrium equations for the moment in the y-z vertical and z-x horizontal planes, considering

gyroscopic moments also, are given as (noting equation (8.38) and for gyroscopic terms refer Chapter

5)

i i i i i iR yz L yz d x p yM M I Iϕ ωϕ− = +�� � (8.58)

and

i i i i i iR zx L zx d y p xM M I Iϕ ωϕ− = −�� � (8.59)

where is the polar mass moment of inertia of the disc. In the modified matrix , following rows

and columns element will be affected: (i) Equations (8.58) and (8.59) are equations of moments, and

pI [ ]P∗

452

in the state vector {S}, 3rd

, 7th, 11

th and 15

th rows are for moment equations. Hence, only these rows in

the modified matrix will be affected. Now in equations (8.58) and (8.59), additional terms for

slopes, and (see Figure 8.22), are appearing because of gyroscopic effects. In the modified

state vector {S*} the angular displacements are at 2

nd, 6

th, 10

th and 14

th rows. Hence in the modified

point matrix [P*] columns 2

nd, 6

th, 10

th and 14

th will be affected. For more the clarity of the above

explanation let us express angular displacements and moments as

( ) jj ;tr j e

νϕ = Φ + Φ ( ) jj tr jM M M e

ν= + (8.60)

where ν be the natural whirl frequency with gyroscopic effects, and subscripts r and j represent the

real and imaginary parts of a complex quantity, respectively (for brevity the subscript i corresponding

to ith mass has been dropped). On substituting equation (8.60) into equation (8.58), we get

( ) ( ) ( ) ( )2j j j j jr j r j i r j i r jR yz R yz L yz L yz d x x p y yM M M M I Iν ων+ − + = − Φ + Φ + Φ + Φ (8.61)

Separating the real and imaginary parts from equation (8.61), we get

2

r r i r i jR yz L yz d x p yM M I Iν ων− = − Φ − Φ (8.62)

and

2

j j i j i rR yz L yz d x p yM M I Iν ων− = − Φ + Φ (8.63)

It should be noted that equation (8.62) corresponds to row in equation (8.45) and gyroscopic

effects term (i.e., ip

I ων− , which is the coefficient of jyϕ that is in the 6

th row of the modified state

vector {S*}) in equation (8.62) will corresponds to 6

th column in equation (8.45). Similarly, from

equation (8.63), it can be seen that gyroscopic effects will introduce another term (i.e., ip

I ων ) at 15th

row and 2nd

column corresponding to jyzM and

ryϕ , respectively. Hence, we have

11,6 ipP I ων= − and

15,2 ipP I ων= (8.64)

where Pi,j represent the additional element at the ith row and the j

th column of the modified point

matrix.

Similarly, using equation (8.59), we get the additional element of the modified point matrix as

[ ]P∗

yϕx

ϕ

11th

453

3,14 7,10 pP P I ων= − = (8.65)

Figure 8.22 Coordinate axes and positive conventions for angular displacements (a) Rectangular 3-

dimensional axis system (b) tilting of the shaft axis in z-x plane (c) tilting of shaft axis in y-z plane

Example 8.4 Obtain the unbalance response and transverse critical speeds of an overhang rotor

system as shown in Figure 8.23. End B1 of the shaft is fixed. The length of the shaft is 0.2 m and the

diameter is 0.01 m. The disc is thin and has 1 kg of mass with the radius of 3.0 cm. Neglect the mass

of the shaft and the gyroscopic effect. Take an unbalance of 3 gm at a radius of 2 cm. Choose the shaft

speed suitably so as to cover two critical speeds in the unbalance response. 112.1 10E = × N/m

2.

Figure 8.23

Solution: Let station 0 be the fixed end and station 1 be the free end. Consider only a single plane

motion. The overall transformation can be written as

{ } { }* * *

1 0RS T S = (a)

with

* * *

1 1T P F = (b)

454

2

1

2

1 0 0 0 0

0 1 0 0 0

[ ] ;0 1 0 0

0 0 1

0 0 0 0 1

d

y

P I

m u

ω

ω

∗

= −

−

2 3

2

1

1 02 6

0 1 02

[ ]0 0 1 0

0 0 0 1 0

0 0 0 0 1

l llEI EI

l lEI EI

Fl

∗

= (c)

{ }*;

1

x

yz

y

y

S M

S

ϕ

−

=

(d)

2 3

2

2

2

1 01 0 0 0 0 2 6

0 1 0 0 0 0 1 02

[ ][ ] 0 1 0 00 0 1 0

0 0 10 0 0 1 0

0 0 0 0 10 0 0 0 1

d

y

l llEI EI

l lEI EI

P F Il

m u

ω

ω

= − −

2 3

2

2 2 22

2 2 2 32 2

1 02 6

0 1 02

0 1 02

12 6

0 0 0 0 1

d dd

y

l llEI EI

l lEI EI

I l I lI l

EI EI

m l m lm m l u

EI EI

ω ωω

ω ωω ω

− − −=

+ −

(e)

Boundary conditions for the present case are that all the linear and angular displacements at station 0

are zero and all the moments and shear forces are zero at right of station 1. Hence, the state vector at

the station 0 and right of 1 will have the following form

{ } { }*

00 0 1

T

yz yS M S= (f)

and

{ } { }*

10 0 1

T

xRS y ϕ= (g)

From free vibrations, unbalance, yu , is zero. Hence, 3

rd and 4

th rows will give the eigen value problem

of the following form

455

2 2 2

2 2 2 3

0

12 0

01

2 6

nf d nf d

yz

ynf nf

I l I ll

EI EI M

Sm l m l

EI EI

ω ω

ω ω

− −

= +

(h)

which gives a polynomial of the following form

1 2 2 24 23

3 4

1212 0

d

nf nf

d d

ml I E IEI

ml I ml Iω ω

+− + = (i)

For the present problem, we have following data

m = 1 kg, r = 0.03 m, 1 2 4

42.25 10dI mr

−= = × kg-m2, l = 0.2 m,

112.1 10E = × N/m

2, d = 0.01 m, and 4 10

/ 64 4.91 10I dπ −= = × m4

Hence from equation (i), we get

( ) ( )4 6 2 119.3176 10 3.5421 10 0nf nfω ω− × + × =

which gives natural frequencies as

1195.4

nfω = rad/s

23046.2

nfω = rad/s

Now for the unbalance response from equations (a), (e), (g) and (f); we have

( )

2 3

2

2 2 22

2 2 2 32 2

0 1

1 02 6

00 1 0

2 0

00 1 02

0

1 1 12 6

0 0 0 0 1

x

d dyzd

y

y

l llEI EI

yl lEI EI

I l I l MI lEI EI

S

m l m lm m l uEI EI

ϕω ω

ω

ω ωω ω

− = − − −

+ −

(j)

The first two rows and the last three rows will give

456

2 3

2

2 6

2

yz

y x

l lM yEI EI

Sl lEI EI

ϕ

− =

(k)

and

( )

2 2 2

2 2 2 3

1 02

0

1 02 6

1 10 0 1

d d

yz

y y

I l I ll

EI EIM

m l m l u SEI EI

ω ω

ω ω

− −

+ − =

(l)

The form of equations (k) and (l) is similar to that given in Table 8.3. Equation (k) can be written as

( )

2 2 2

2 2 2 3

1 02

12 6

d d

yz

y y

I l I ll MEI EI

S um l m l

EI EI

ω ω

ω ω

− − =

+

(m)

or

( )

12 2 2

2 2 2 3

1 02

12 6

d d

yz

y y

I l I llM EI EI

S um l m l

EI EI

ω ω

ω ω

−

− − = +

(n)

Equation (n) can be used to get the bending moment and the shear force due to unbalance force.

Hence, on substituting equation (n) into equation (k), we get

( )

12 2 2

2 3

22 2 2 3

1 022 6

12 2 6

d d

yx

I l I ll l l

y EI EIEI EI

ul l m l m lEI EI EI EI

ω ω

ϕ ω ω

− − − − = +

(o)

Equation (o) can be solved for a particular spin speed to get the unbalance response (y and φx). Then

in the similar way the spin speed can be varied to get the variation of the unbalance response with the

spin speed. A plot of the y and φx with respect to the spin speed of the rotor is given in Figures 8.24(a

and b). The resonant condition can be seen as large amplitudes of vibration and it indicate critical

speeds. A similar plot can be obtained from equation (n) for and yz y

M S and are shown in Figs. 8.24(c

and d). It can be observed from all four plots that critical speeds are same as natural frequencies

457

obtained by free vibration analysis (1

195.4nf

ω = rad/s and 2

3046.2nf

ω = rad/s). In the plots of the

shear force and the bending moment at support, anti-resonances can be seen in between the two

critical speeds. This indicates that two modes of vibrations have cancelling effects on the shear force

and the bending moment. The frequency at which the anti-resonance occurs for the shear force is not

same as that of the bending moment. It indicates that it is not a system characteristics and its location

may change for different unbalance force, however, critical speed have fixed frequencies since it is

system characteristics.

Spin speed, ω, (rad/s)

Figure 8.24(a) Variation of the unbalance linear response with the spin speed

Spin speed, ω, (rad/s)

Figure 8.24(b) Variation of the unbalance angular response with the spin speed

0 500 1000 1500 2000 2500 3000 350010

-15

10-10

10-5

100

0 500 1000 1500 2000 2500 3000 350010

-10

10-5

100

105

Lin

ear

dis

pla

cem

ent,

y,

(m)

An

gu

lar

dis

pla

cem

ent,

ϕx r

ad/s

458

Spin speed, ω, (rad/s)

Figure 8.24(c) Variation of the support bending moment with the spin speed

Spin speed (rad/s)

Figure 8.24(d) Variation of the support shearing force with the spin speed

Example 8.5 Obtain transverse natural frequencies and corresponding mode shapes of a rotor system

as shown in Figure 8.25. Take the mass of the disc, m = 10 kg, the diametral mass moment of inertia,

Id = 0.02 kg-m2 and the disc is placed at 0.25 m from the right support. The shaft has the diameter of

10 mm and the span length of 1 m. The shaft material has the Young’s modulus E = 2.1 × 1011

N/m2

and consider the shaft as mass-less. Neglect gyroscopic effects and take one plane motion only.

Compare the results by the influence coefficient method.

0 500 1000 1500 2000 2500 3000 350010

-10

10-5

100

105

0 500 1000 1500 2000 2500 3000 350010

-10

10-5

100

105

1010

Ben

din

g m

om

ent,

M,

(Nm

) S

hea

rin

g f

orc

e, S

, (N

)

459

Figure 8.25 A simply supported rotor system

Solution: Station numbers 0, 1 and 2 can be assigned at the left support, at the disc and at the right

support; respectively. Then, the overall transformation of states is given as

(a)

which can be expanded as

(b)

with

1 1 /l EIβ = and 2 2 /l EIβ = .

Subscripts 1 and 2 outside the matrices belong to the following parameters: , , and d

l Iβ . On

multiplication of matrices, we get

which finally takes the following form

(c)

{ } [ ] [ ] [ ] { }2 02 1 1R S F P F S=

{ } { }

1 12 2

6 6

22 0

2

2 1 1

1 0 0 01 0.5 1 0.5

0 1 0 00 1 0.5 0 1 0.5

0 1 00 0 1 0 0 1

0 0 10 0 0 1 0 0 0 1

R

nf d

nf

l l l l l l

l lS S

Il l

m

β β β β

β β β β

ω

ω

= −

{ } { }

1

1

1

1 12 2 2 21 2

1 2 2 2 2 2 2 2 2 26 66

2 2

1 2 2 2 2 2 2

2 02 2

1 2 2

211

1 0.5 0.5 1 0.5

0.5 1 0.5 0 1 0.5

0 0 11

0 0 0 10 0 1

nf d nf

nf d nf

R

nf d nf

nf

m l l I l l l l l l

m l I l lS S

lm l I l

m

β ω β ω β β β β

β ω β ω β β β β

ω ω

ω

+ − − = −

11 12 13 14

21 22 23 24

31 32 33 34

41 42 43 442 0

x x

yz yz

y yR

y yt t t t

t t t t

M Mt t t t

S St t t t

ϕ ϕ

− − =

460

with

(d)

The following boundary conditions are applied for the present case (Figure 8.25)

(e)

On application of boundary conditions in equation (c), the following set of equations is obtained

and (f)

From the first set of equations (e), the frequency equation takes the following form

(g)

On substituting equation (e) into equation (g), we get

which simplifies to

( ) ( )

( ) ( )

( ) ( ) ( )

1

1

1

1 12 2 2 2 2

11 1 2 2 12 1 1 2 2 2 2 26 6

1 2 2 2

13 1 1 1 2 2 1 2 2 2 2 26

1 1 12 2 2 2 2

14 1 1 1 2 2 1 1 2 2 2 1 2 2 2 26 6 6

21 1

1 ; 1 0.5 ;

0.5 1 0.5 0.5 ;

1 0.5 0.5 0.5 ;

0.5

nf nf d nf

nf d nf

nf d nf

t m l t l m l l I l

t l m l l I l l

t l m l l l I l l l l

t m

β ω β ω β ω

β β ω β β ω β

β β ω β β ω β β

β

= + = + + −

= + + − +

= + + − + +

= ( ) ( )( ) ( )

( ) ( ) ( )

1

1

1

2 2 2

2 2 22 1 1 2 2 2

2 2

23 1 1 1 2 2 1 2 2

1 2 2 2

24 1 1 1 2 2 1 1 2 1 2 2 26

; 0.5 1 ;

0.5 0.5 1 ;

0.5 0.5 1 0.5 ;

nf nf d nf

nf d nf

nf d nf

l t l m l I

t l m l I

t l m l l I l l

ω β ω β ω

β β ω β β ω β

β β ω β β ω β β

= + −

= + − +

= + − + +

( ) ( ) ( )

( ) ( )( ) ( )

( ) ( )

1

1

1

1

1 2 2 2

24 1 1 1 2 2 1 1 2 1 2 2 26

2 2 2

31 1 2 32 1 1 2

2 2

33 1 1 1 2 1

1 2 2 2 2

34 1 1 1 2 1 1 1 2 41 16

42 1 1

0.5 0.5 1 0.5 ;

; ;

0.5 1;

0.5 ; ;

nf d nf

nf nf d nf

nf d nf

nf d nf nf

t l m l l I l l

t m l t l m l I

t l m l I

t l m l l I l l t m

t l m

β β ω β β ω β β

ω ω ω

β ω β ω

β ω β ω ω

ω

= + − − +

= = + −

= + − +

= + − + + =

= ( ) ( ) ( )12 2 2 2

43 1 1 1 44 1 1 16; 0.5 ; 1;nf nf nft l m t l mβ ω β ω= = +

0 0 2 2 0;R R

y M y M= = = =

12 14

32 34 0

0

0

x

y

t t

St t

ϕ =

22 24

42 442 0

x x

y yR

t t

S St t

ϕ ϕ =

12 34 14 32 0t t t t− =

( ) ( ){ } ( ) ( ){ }( ) ( ) ( ){ } ( ) ( ){ }

1 1

1 1

1 12 2 2 2 2 2

1 1 2 2 2 2 2 1 1 1 2 1 1 1 26 6

1 1 12 2 2 2 2 2 2

1 1 1 2 2 1 1 2 2 2 1 2 2 2 2 1 1 26 6 6

1 0.5 0.5

1 0.5 0.5 0.5 0

nf d nf nf d nf

nf d nf nf d nf

l m l l I l l m l l I l l

l m l l l I l l l l l m l I

β ω β ω β ω β ω

β β ω β β ω β β ω ω

+ + − + − + +

− + + − + + + − =

461

(h)

After substituting numerical values in equation (h), we get natural frequencies of the rotor system as

rad/sec and rad/sec

The relative linear and angular displacements, and the bearing and shaft reaction forces:

(i) For rad/sec: Let us assume 0

1z

ϕ = rad, from equation (f), we have

N

rad

N

Now state vectors at stations 0 and 2 are completely known. Hence, state vectors at the left and right

of station 1 can be obtained as

and

It should be noted that for relative amplitudes of displacements, reaction forces and moments have no

quantitative significance. In fact, estimation of unbalance responses would give exact value of loads at

various stations of the shaft.

(ii) For rad/sec: Let us again assume 0

1z

ϕ = rad. For this case also, on same lines as for

the previous case, we get following results

( ) ( ) ( ){ } ( )1 1

22 2 4 2 2 2 2 2

1 1 2 1 2 1 1 1 2 1 2 1 2 1 1 23 9 0

d nf d nfm I l l l l I l l m l lβ β ω β β β β ω− + + + + + + =

129.45nfω =

2289.23nfω =

129.45nfω =

( )( )0

12 2

14

6.12 10nf

y

nf

tS

t

ω

ω= − = − ×

2 0 0

24 1222 24 22

14

1.39R x x y

t tt t S t

tϕ ϕ= + = − = −

2 0 0

344 1242 44 42

14

1.93 10R y x y

t tS t t S t

tϕ= + = − = ×

1 12 2

6 6

12 14 21 1 0 1

0 0.2991 0.5 1 0.5

1 0.8050 1 0.5 0 1 0.5

0 496.1970 0 1 0 0 1

/ 661.5960 0 0 1 0 0 0 1

x x

yz yz

y yL

y yl l l l l l

l l

M Ml l

S S t t

β β β β

ϕ ϕβ β β β

− − − = = = − − −

2 2

2 2

1 1 1 1

1 0 0 0 1 0 0 0 0.299 0.299

0 1 0 0 0 1 0 0 0.805 0.805

0 1 0 0 1 0 496.197 482.234

0 0 1 0 0 1 661.596 1928.

x x

yz nf d yz nf d

y nf y nfR L

y y

M I M I

S m S m

ϕ ϕ

ω ω

ω ω

− − − − = = = − − − − − 937

2289.23nω =

462

N; rad; N;

and

Example 8.6 Find transverse natural frequencies and mode shapes of a two-disc rotor system shown

in Figure 8.26. B is a fixed end, and D1 and D2 are rigid discs. The mass of discs are: m1 = 5 kg and m2

= 2 kg, and the diametral mass moment of inertia are: kg-m2 and kg-m

2. The

shaft is made of the steel with the modulus of elasticity E = 2.1 (10)11

N/m2, and of the uniform

diameter d = 10 mm. Shaft lengths are: BD1 = 50 mm, and D1D2 = 75 mm. Consider the shaft as mass-

less.

Figure 8.26 An overhung two rotor system

Solution: Figure 8.27 shows station numbers with 0 at the fixed end, and 1 and 2 at the subsequent

discs.

Figure 8.27 The overhung rotor with the shaft and disc properties

We have the following data

m4, N-m

2

The transformation of the state between stations 0 and 1 can be written as

3

0 1.116 10Q = − ×2

1.088R xϕ =2

41.033 10R yS = − ×

1

0.011

2.044

836.783

1115.7

x

yL

y

M

S

ϕ

− − −

= −

− 1

0.011

2.044

2580

1033

x

yR

y

M

S

ϕ

− − −

= −

10.03dI =

20.01dI =

4 4 10

64 640.01 4.91 10I d

π π −= = = × 103.1EI =

463

(a)

with

(b)

(c)

(d)

Similarly, the transformation of the state between stations 1 and 2 can be written as

(e)

with

, (f)

(g)

On substituting equation (a) into equation (e), we get the overall transformation of the state from

station 0 to 2, as

{ } [ ] [ ] { } [ ]{ }2 0 02 1R S U U S T S= = (h)

{ } [ ] [ ] { } [ ] { }1 0 01 1 1R

US P F S S= =

2 21

2 2

1

1 0 0 0 1 0 0 0

0 1 0 0 0 1 0 0[ ]

0 1 0 0 0.03 1 0

0 0 1 5 0 0 1

d nf nf

nf nf

PI

m

ω ω

ω ω

= = − −

2 35 7

4 52

1

1

1 1 0.05 1.21 10 2.02 102 6

0 1 4.85 10 1.21 100 1[ ] 20 0 1 0.05

0 0 10 0 0 1

0 0 0 1

l llEI EI

l lF EI EI

l

− −

− −

× × × × = =

21

2

1 0 0 0

0 1 0 0[ ] ?

0 1 0

0 0 1

d nf

nf

UI

m

ω

ω

= −

{ } [ ] [ ] { } [ ] { }2 1 12 2 2RR R

US P F S S= =

22

2

1 0 0 0

0 1 0 0[ ]

0 0.01 1 0

2 0 0 1

nf

nf

Pω

ω

= −

5 7

4 5

2

1 0.075 2.73 10 6.82 10

0 1 7.27 10 2.73 10[ ]

0 0 1 0.075

0 0 0 1

F

− −

− −

× ×

× × =

22

2

1 0 0 0

0 1 0 0[ ] ?

0 1 0

0 0 1

d nf

nf

UI

m

ω

ω

= −

464

with

(i)

The expanded form of equation (i) has the following form

(j)

Boundary conditions are (i) y and ϕx at station 0 is zero, and (ii) at the free end shear force, Sy, and

bending moment, Myz are zero. On substituting these boundary conditions in equation (j), we get

(k)

From the last two expressions of equation (k), we have

(l)

Hence, on taking the determinant of equation (l) equal to zero, the frequency equation is

= 0 (m)

From which transverse natural frequencies can be obtained by the by root searching numerical method

described earlier. For the present problem it has a simple form as

??

Natural frequencies obtained are

? rad/s and ? rad/s

[ ] [ ] [ ] [ ] [ ] [ ]1 2 2 2 1 1

[ ] ?T U U P F P F= =

11 12 13 14

21 22 23 24

31 32 33 34

41 42 43 442 0

x x

yz yz

y yR

y yt t t t

t t t t

M Mt t t t

S St t t t

ϕ ϕ

− − =

11 12 13 14

21 22 23 24

31 32 33 34

41 42 43 442 0

0

0

0

0

x

yz

yR

t t t ty

t t t t

Mt t t t

St t t t

ϕ

−

=

33 34

43 44 0

0

0

yz

y

Mt t

St t

=

( ) ( ) ( ) ( )33 44 34 43 0nf nf nf nft t t tω ω ω ω− =

( ) ( )4 2 0nf nfω ω+ + =

1nfω =1nfω =

465

For obtaining the mode shape, from the first two equations of (k), we have

(n)

For a given natural frequency, the first equation of equation (l) gives

(o)

By choosing as a reference value of the displacement, the first equation of equation (n) gives

(p)

On substituting equation (o) into equation (p), the shear force at station 0 can be obtained as

or (q)

The bending moment at station 0 can be obtained now from equation (o). Now, we have obtained the

state vector at station 0 completely, and using the transformation matrices the state vectors at all other

station can be obtained and are given as

(i) For first natural frequency

; ?

13 14

23 242 0

yz

yxR

My t t

St tϕ

− =

340 0

33

yz y

tM S

t=

31

Ry =

13 14 1yz yt M t S+ =

3413 0 14 0

33

1y y

tt S t S

t+ = 33

0

13 34 14 33

y

tS

t t t t=

+

1

x

yz

yR

y

M

S

ϕ

−

= 2

x

yz

yR

y

M

S

ϕ

−

=

466

Figure 8.28 of mode shape for y and

(ii) For second natural frequency

; ?

Figure 8.29 of mode shape for y and

0

1

2

y

y

y

zϕ

1

x

yz

yR

y

M

S

ϕ

−

= 2

x

yz

yR

y

M

S

ϕ

−

=

0

1

2

y

y

y

zϕ

467

Example 8.7 Find transverse natural frequencies and mode shapes of the rotor system shown in

Figure 8.30. B1 and B2 are bearings, which provide simply supported end condition and D1, D2, D3 and

D4 are rigid discs. The shaft is made of the steel with the Young’s modulus E = 2.1 (10)11

N/m2 and

uniform diameter d = 20 mm. Various shaft lengths are as follows: B1D1 = 150 mm, D1D2 = 50 mm,

D2D3 = 50 mm, D3D4 = 50 mm and D4B2 = 150 mm. The mass of discs are: m1 = 4 kg, m2 = 5 kg, m3 =

6 kg and m4 = 7 kg. Consider the shaft as mass-less. Consider discs as thin and take diameter of discs

as d =1 8 cm, d =2 10 cm, d =3 12 cm, and

d =4 14 cm, however, neglect the gyroscopic effects.

Figure 8.30 A multi-disc rotor system

Solution: The discs have the following data

m1 = 4 kg, m2 = 5 kg, m3 = 6 kg, m4 = 7 kg

.d =1 0 08 m, .d =2 0 1m,

.d =3 0 12 m, .d =4 0 14 m,

1

1 12 2

1 14 44 0.04 0.0016dI m r= = × × = kg-m

2,

2

1 2

45 0.05 0.003125dI = × × = kg-m

2,

3

1 2

46 0.06 0.0054dI = × × = kg-m

2,

4

1 2

47 0.07 0.008576dI = × × = kg-m

2,

The shaft has EI = 1649.34 N-m2 and following dimensions according to station numbers

l1 = 150 mm, l2 = 50 mm, l3 = 50 mm, l4 = 50 mm, l5 = 150 mm

Now the overall transformation of the state vector can be written as

{ } [ ]{ }5 0R S T S= (a)

with

[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]5 4 4 3 3 2 2 1 1

T F P F P F P F P F= (b)

468

2

2

1 0 0 0

0 1 0 0[ ]

0 1 0

0 0 1

i

inf d

i nf

PI

m

ω

ω

= −

;

2 3

2

12 6

0 1[ ] 2

0 0 1

0 0 0 1

i

i

l llEI EI

l lF EI EI

l

=

; { }x

yz

y

y

SM

S

ϕ

−

=

(c)

From Table 8.2, we have the eigen value probelm for the simply supported boundary conditions, as

1,2 1,4

3,2 3,4 0

0

0

x

y

t t

St t

ϕ =

(d)

which gives the frequency equation as

( ) ( ) ( ) ( ) ( ){ }1,2 3,4 1,4 3,2 0nf nf nf nf nff t t t tω ω ω ω ω= − = (e)

On solving the roots of above function by the root searching method, it gives the following natural

frequencies

1nfω = 215 rad/s,

2nfω = 1076 rad/s,

2nfω = 2470 rad/s,

4nfω = 3847 rad/s,

5nfω = 6398 rad/s,

6nfω = 8885 rad/s,

7nfω = 10776 rad/s,

8nfω = 13282 rad/s.

From Table 8.2, the eigen vector can be obtain from the following equation

2,2 2,4

4,2 4,45 0

x x

y yR

t t

S St t

ϕ ϕ =

(f)

For example for a particular natural frequency from equation (d) and (f), we have

1,2 1,4

3,2 3,4 0

0

0

x

y

t t

St t

ϕ =

and 2,2 2,4

4,2 4,45 0

x x

y yR

t t

S St t

ϕ ϕ =

(g)

Now on choosing φx = 1 as reference value, from equations (g) and noting the boundary condition, we

get the state vector at 0th and 5

th station as

469

0

0

1{ }

0

66905.27

S

= −

; 5

0

1.033338{ }

0

72938.64

R S

−

=

At other stations also the state vectors can be obtained as

[ ] [ ]1 01 1

0.127182

0.543645{ } { }

10076.00

43389.27

R S P F S

= = − −

; [ ] [ ]2 12 2

0.146180

0.205305{ } { }

12275.12

9603.40

R RS P F S

= = − −

and

[ ] [ ]3 22 2

0.147021

0.174094{ } { }

12711.83

31172.87

R RS P F S

−

= = −

; [ ] [ ]4 33 3

0.129076

0.535830{ } { }

10940.80

72938.64

R RS P F S

−

= = −

These state vectors for the fundamental natural frequency are summarized in Table 8.4(a). The mode

shape for the linear and angular displacements are extracted for Table 8.4(a) and are drawn as shown

in Fig. 8.31(a) and (b), respectively. On the same lines other state vectors corresponding to various

natural frequencies can be obtained and are given in Table 8.4(b-h) and corresponding mode shapes

from 2nd

to 5th modes are shown in Fig. 8.31(c-d).

Table 8.4(a) Relative values of state vectors at various stations for the fundamental natural frequency

State vectors at first natural frequency, 215 rad/s

S0 S1 S2 S3 S4 S5

y 0 0.127182 0.146180 0.147021 0.129076 0

ϕx 1.00 0.543645 0.205305 -0.174094 -0.535830 -1.033338

Myz 0 -10076.00 -12275.12 -12711.83 -10940.80 0

Sy -66905.27 -43389.27 -9603.40 31172.87 72938.64 72938.64

Table 8.4(b) Relative values of state vectors at various stations for the second natural frequency

State vectors at second natural frequency, 1076 rad/s

S0 S1 S2 S3 S4 S5

y 0 0.08201 0.04415 -0.01806 -0.06734 0

ϕx 1.00 -0.35964 -1.10908 -1.26954 -0.62340 0.98042

470

Myz 0 -29234.01 -16196.41 13547.38 35270.03 0

Sy -199334.91 180497.74 436132.02 310671.00 -235133.53 -235133.53

Table 8.4(c) Relative values of state vectors at various stations for the third natural frequency

State vectors at third natural frequency, 2470 rad/s

S0 S1 S2 S3 S4 S5

y 0 0.05172 -0.009865 -0.03457 0.02368 0

ϕx 1.00 -0.96553 -1.25193 0.43362 1.7472 -1.11239

Myz 0 -33799.42 38772.86 58144.09 -62887.40 0

Sy -288162.68 974074.79 673141.85 -592459.95 419249.34 419249.34

Table 8.4(d) Relative values of state vectors at various stations for the fourth natural frequency

State vectors at fourth natural frequency, 3847 rad/s

S0 S1 S2 S3 S4 S5

y 0 0.0323 -0.03034 -0.0055 0.008635 0

ϕx 1.00 -1.3525 -0.7590 1.5790 -1.30565 0. 57196

Myz 0 -19708.29 93968.80 -65909.34 41291.063 0

Sy -344903.01 1571465.29 -673757.87 -1169875.76 -275273.75 -275273.75

Table 8.4(e) Relative values of state vectors at various stations for the fifth natural frequency

State vectors at fifth natural frequency, 6398 rad/s

S0 S1 S2 S3 S4 S5

y 0 0.01159 -0.01634 0.01017 -0.002194 0

ϕx 1.00 -1.7681 1.0278 -0.43533 0.103774 -0.010208

Myz 0 54931.80 -1947.42 1647.72 -2506.62 0

Sy -405838.58 1491974.8 -1852691.57 645435.50 16710.77 16710.77

Table 8.4(f) Relative values of state vectors at various stations for the sixth natural frequency

State vectors at sixth natural frequency, 8885 rad/s.

S0 S1 S2 S3 S4 S5

y 0 0. 6159 2.78 8.48 -9.555 0

ϕx 1.00 10.31 125.4 0.0000004289 19.06 1.01

Myz 0 -1098548.67 -22258104.4 -140562945.6 -4168712.59 0

Sy 1366381.12 195881901.58 1290696326.5 5308048574.2 27791417.25 27791417.25

471

Table 8.4(g) Relative values of state vectors at various stations for the seventh natural frequency

State vectors at seventh natural frequency, 10776 rad/s.

S0 S1 S2 S3 S4 S5

y 0 0.01552 0.2102 -0.2306 0.038953 0

ϕx 1.00 -1.6894 11.200 3.7243 -0.976793 0.035605

Myz 0 254747.85 -3468821.24 640227.57 22263.86 0

Sy -394294.80 6818067.54 128888160.07 -31812064.91 -148425.72 -148425.72

Table 8.4(h) Relative values of state vectors at various stations for the eighth natural frequency

State vectors at eighth natural frequency, 13282 rad/s.

S0 S1 S2 S3 S4 S5

y 0 -0.03468 0.02929 -0.001166 -0.0007505 0

ϕx 1.0 -2.69371 -1.06703 0.05671 -0.1437 8.9769

Myz 0 679094.86 16463.86 3645.74 200574.14 0

Sy -541528.33 -25017431.96 824200.49 -410329.07 -1337160.94 -1337160.94

Fig. 8.31(a) The fundamental mode shape of linear displacements

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

Axial position on the shaft

Dis

pla

cem

ent

ampli

tud

e

472

Fig. 8.31(b) The fundamental mode shape of angular (slope) displacements

Fig. 8.31(c) Higher mode shapes (2nd

to 5th) of linear displacements

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45-1.5

-1

-0.5

0

0.5

1

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

2

3

4 5

Axial position on the shaft

Dis

pla

cem

ent

ampli

tud

e

Axial position on the shaft

Dis

pla

cem

ent

ampli

tud

e

473

Fig. 8.31(d) Higher mode shapes (2nd

to 5th) of angular (slope) displacements

Example 8.8 Obtain bending critical speeds of a rotor system as shown in Figure 8.32. Take the mass

of the disc, kg and its diametral mass moment of inertia, kg-m2. The length of the

shaft segments are a = 0.3 m and b = 0.7 m; and the diameter of the shaft is 10 mm. Neglect the

gyroscopic effects. N/m2.

Figure 8.32 An overhang rotor system

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

5=m 0.02dI =

112.1 10E = ×

2

3

4

5

Axial position on the shaft

Dis

pla

cem

ent

ampli

tud

e

474

Solution: Figure 8.33 shows the station numbering and free body diagram of various segments and

supports.

Figure 8.33 Free body diagrams of rotor segments

For shaft segment (1) as shown in Figure 8.33(b), the state vector can be related as

(a)

with

For second branch, the state vector can be written as

(b)

with

End conditions for the overhang rotor as shown in Figure 8.33(a) can be written as

{ } 1 1 0 1 01[ ] [ ] { } [ ] { }

LS F P S U S= =

[ ] 1 11[ ] [ ]U F P=

{ } { } { }2 22 1 1[ ] [ ]

L R RS F S U S= =

2 2[ ] [ ]U F=

(a) Rotor system (b) FBD of segment (1)

(c) FBD of segment (2) (d) Free body diagram of a small shaft

segment at support A

475

At node 0: (c)

At node 1: 1 1 1 1 10; 0 and 0

L R L R ay M M S S R= = ≠ − + = (d)

where, Ra is support reaction at bearing A.

At node 2: (e)

From the first shaft segment, i.e., equation (a), on application of end conditions of equation (c), we

have

(f)

with

[ ]

1 211 12 13 14 6

21 22 23 24

2131 32 33 34

2

41 42 43 44 1 1

1 0 0 01 0.5

0 1 0 00 1 0.5

0 1 00 0 1

0 0 10 0 0 1

nf d

nf

u u u u l l l

u u u u lU

Iu u u u l

mu u u u

α α

α α

ω

ω

= = −

1 12 2 2 2

6 6

2 2

2 2

2

1

1 0.5 0.5

0.5 1 0.5

1

0 0 1

nf d nf

nf d nf

nf d nf

nf

m l l I l l l

m l I l

m l I l

m

ω α ω α α α

ω α ω α α α

ω ω

ω

+ −

− = −

(g)

and

(h)

From equation (f), the first set of equation gives

(i)

On using equation (i) into equation (f), we get state vectors in the left of station 1 as

0 00S M= =

2 2 20 and 0

R Ly M M= = =

11 12 13 14

21 22 23 24

31 32 33 34

41 42 43 441 01

0

0

0L

u u u u y

u u u u

u u u uM

u u u uS

ϕ ϕ

−

=

l

EIα =

1111 0 12 0 0 0

12

0 u

u y u yu

ϕ ϕ= − + ⇒ =

476

(j)

Now noting equations (d) and (j) we can have state vectors in the right of station 1 as

22 111 1 21 0

12

R L

u uu y

uϕ ϕ

= = − +

32 111 1 31 0

12

R L

u uM M u y

u

= = − +

42 111 1 41 0

12

R L a a

u uS S R u y R

u

= + = − + +

(k)

Equation (k) can be written as

021

31

411

0 0

0

0

1

a

R

y

yu

RuM

uS

ϕ

− =

(l)

with

Equation (l) can be written in a standard form as

(m)

with

10

Ly =

11 22 111 21 0 22 0 21 0

12 12

L

u u uu y u y u y

u uϕ

= − + = − +

32 11111 31 0 32 0 31 0

12 12

L

u uuM u y u y u y

u u

= − + = − +

11 42 111 41 0 42 0 41 0

12 12

L

u u uS u y u y u y

u u

= − + = − +

1 1 0R L

y y= =

32 1122 11 42 1121 21 31 31 41 41

12 12 12

; andu uu u u u

u u u u u uu u u

= − = − = −

{ } [ ]{ }111 SUSR =

477

{ }

0

21

1 131

41

0 0 0 0

00 0 0; and

00 0 0

0 0 1 a

y

uU S

u

Ru

− = =

For second branch, i.e., equation (b) and noting equation (m), we have

{ }

1 206

21

2 1 1231

412

0 0 0 01 0.5

00 0 00 1 0.5[ ][ ]{ }

00 0 00 0 1

0 0 10 0 0 1

L

a

yl l l

ulS U U S

ul

Ru

α α

α α

−

= =

(n)

On simplifying equation (n), we get

( )( )

( )

.

. .

( )

x

yz

y aL

y lu lu l u l y

u u lu l

M u lu l

S Ru

α α α

ϕ α α α

+ + −

+ += +

1 12 2

21 31 41 06 6

21 31 41

31 41

2 41 2

0 5 0 0

00 5 0 0 0 5

00 0

0 0 1

(o)

On expanding equation (o), we have

( )2 2 21 2 31 2 2 41 0 2 20.5 ( ) 0.5R L au u l u y l Rϕ ϕ α α α= = + + − +

2 2 31 2 41 0 20 ( )( )R L aM M u l u y l R= = = + − +

(p)

From equation (p) the first equation, we have

(q)

( )1 12 2

2 2 2 21 2 2 31 2 2 41 0 2 26 60 0.5 ( )R L ay y l u l u l u y l Rα α α= = = + + − +

2 2 41 0( )R L b a

S S R u y R= = = − +

( )1 2

0 2 21 2 2 31 2 2 416

1 2

2 26

0.5a

y l u l u l uR

l

α α

α

+ +=

478

On substituting equation (q) in the third equation of set of equations (p), and noting that since ,

we get

( )1 2

2 2 21 2 2 31 2 2 416

31 2 41 1 2

2 26

0.5( ) 0

l l u l u l uu l u

l

α α

α

+ +− + + = (r)

On simplification of equation (r), we get

1

21 2 3130u uα+ = (s)

which is the frequency equation. On substituting variables defined in equation (l) into equation (s), we

have

32 1122 1121 2 31

12 12

10

3

u uu uu u

u uα

− + − =

(t)

In view of variables defined in equations (g) and (h), the frequency equation (t) becomes

( )( )

3 32 2 2 21 1 1

22 21 2 2

12 22 21 1

1 1

1 1 16 6

02 3 3

2 2

nf nf d nf d nf

nf nf

nf d nf d

l l lm I I m

EI EI EIl l lm m l

EI EI EIl ll I l I

EI EI

ω ω ω ω

ω ω

ω ω

+ − − +

− + − =

− −

which can be simplified as

( ) ( )3 4 3 2 2 2

1 1 2 1 1 2 1 23 4 6 2 6 2 2 36( ) 0d nf d d nfmI l l l EI ml I l I l ml l EIω ω+ − + + + + = (u)

For the present problem, we have

; d = 0.01 m

On substituting in equation (u), we get

00y ≠

4; 1 20.02 m 0.3 m; 0.7 m; 5 kgdI l l m= = = =

4 10 4 6 2(0.01) 4.909 10 m ; 1.031 10 Nm

64I EI

π −= = × = ×

479

4 4 2 75.97 10 3.83 10 0nf nfω ω− × + × =

which gives

125.47nfω = rad/sec and

2nfω = 243.00 rad/sec

The mode shapes can be obtained by using transfer matrices between various intermediate stations

derived in equations (q), (p), (o), (m), (b) and (a). For a reference y0 = 1 may be chosen. This is left to

readers as an exercise (Fig. 8.34).

Alternative method for the above example is explained now. A transformation matrix to transform the

state vector from the left of a support to the right of the support can be developed as follows (refer

equation (d)):

1 1

1 0 0 0 0

0 1 0 0 0

0 0 1 0 0

0 0 0 1

1 0 0 0 0 1 1

A

R L

y y

M M

S R S

ϕ ϕ

− − =

or { } { }* * *

1 1AspR LS U S = (v)

where now all vectors and matrices are modified to accommodate the reaction force from the support.

Modifying all other transformation in equations (a) and (b), we get the modified overall

transformation as

{ } { } { }* * * * * * *

2 12 0 0spLS U U U S T S = = (w)

which can be expanded as

11 12 13 14 15

21 22 23 24 25

31 32 33 34 35

41 42 43 44

2 01 0 0 0 0 1 1

A

A

A

A

L

y t t t t t R y

t t t t t R

M t t t t t R M

S t t t t R S

ϕ ϕ

− − =

(x)

Boundary conditions given in equations (c) and (e), can be applied in equation (x) to get

480

11 12 13 14 15

21 22 23 24 25

31 32 33 34 35

41 42 43 44

2 0

0

0 0

0

1 0 0 0 0 1 1

A

A

A

A

L

t t t t t R y

t t t t t R

t t t t t R

S t t t t R

ϕ ϕ

− =

(y)

which gives

11 12 15

31 32 35

0

0A

t t tyR

t t tϕ

− = +

and

2521 22

4541 422 0

A

L

tt t yR

tt tS

ϕ

ϕ

− = +

(z)

The first of equation (z) can be rearranged as

11 12 15

31 32 35

0

0A

yt t t

t t tR

ϕ

−

=

(a1)

or

2 2

11 31 11 31 11 12 31 32 11 15 31 35 0

11 12 15 2 2

12 32 11 12 31 32 12 32 15 12 35 32 0

31 32 35 2 2

15 35 11 15 31 35 15 12 35 32 15 35

0

0A A

t t y t t t t t t t t t t yt t t

t t t t t t t t t t t tt t t

t t R t t t t t t t t t t R

ϕ ϕ

− + + + − = = + + + + + +

(b1)

For the non-trial solution of equation (b1), we should have

2 2

11 31 11 12 31 32 11 15 31 35

2 2

11 12 31 32 12 32 15 12 35 32

2 2

11 15 31 35 15 12 35 32 15 35

0

t t t t t t t t t t

t t t t t t t t t t

t t t t t t t t t t

+ + +

+ + + =

+ + +

It can be verified that equations (b1) and (u) will be identical. It should be noted that the

transformation between the right and left side of the support B can be written as

2 2

1 0 0 0 0

0 1 0 0 0

0 0 1 0 0

0 0 0 1

1 0 0 0 0 1 1

B

R L

y y

M M

S R S

ϕ ϕ

− − =

or { } { }* * *

2 2BspR LS U S = (a1)

which will be used for getting the state vector in the right of support B.

481

Example 8.9 Obtain the variation of the transverse natural frequency with the shaft speed (i.e.,

obtained the Campbell diagram) of an overhang rotor system as shown in Figure 8.35. From such

Campbell diagram obtain critical speeds. The end B1 of the shaft is having fixed end conditions.

Length of the shaft is 0.2 m and diameter is 0.01 m. The disc is thin and has 1 kg of mass and the

radius of the disc is 3.0 cm. Consider gyroscopic effects, however, neglect the mass of the shaft. Take

the range of the shaft speed such that it covers at least two critical speeds in the Campbell diagram.

Use the TMM.

Figure 8.35

Solution: Because of gyroscopic effect now the motion in the two orthogonal planes will be coupled.

This requires considering the modified field and point transfer matrices as given by equations (8.34)

and (8.45). In additional, the gyroscopic effect will introduce the following new terms in the point

matrix: * * * *

3,14 7,10 11,6 15,2 pP P P P I ων= − = − = = , where subscripts represent the row and column,

respectively, in the modified point matrix and ω is the spin speed of the shaft. Let station 0 be the

fixed end and station 1 be the free end. For free vibrations, then the overall transformation can be

written as

{ } { }* * *

1 0RS T S = (a)

with

* * *

1 1T P F = (b)

sym

skew

sym

skew

[ ] 0 0 [ ] 0

0 [ ] [ ] 0 0

[ ] ;0 [ ] [ ] 0 0

[ ] 0 0 [ ] 0

0 0 0 0 1

i

P G

P G

P G P

G P

∗

=

1

[ ] 0 0 0 0

0 [ ] 0 0 0

[ ] 0 0 [ ] 0 0

0 0 0 [ ] 0

0 0 0 0 1i

F

F

F F

F

∗

=

(c)

2

2

1 0 0 0

0 1 0 0[ ]

0 1 0

0 0 1

d

PI

m

ν

ν

= −

2 3

2

12 6

0 1 [ ] 2

0 0 1

0 0 0 1

l llEI EI

l lF EI EI

l

=

(d)

482

0 0 0 0

0 0 0 0[ ]

0 0 0

0 0 0 0

sym

p

GI ων

=

,

0 0 0 0

0 0 0 0[ ]

0 0 0

0 0 0 0

skew

p

GI ων

= −

(e)

{ }

{ }

{ } ;{ }

{ }

1

r

j

r

j

h

h

v

v

S

S

S S

S

∗

=

{ } ;y

h

xz

x

x

SM

S

ϕ

−

=

{ } ;x

v

yz

y

y

SM

S

ϕ

−

=

0

0{ }

0

0

u

=

(f)

sym

skew

* * *

sym

skew

1

[ ][ ] 0 0 [ ] [ ] 0

0 [ ][ ] [ ] [ ] 0 0

0 [ ] [ ] [ ][ ] 0 0

[ ] [ ] 0 0 [ ][ ] 0

0 0 0 0 1

P F G F

P F G F

T P F G F P F

G F P F

= =

(g)

( )

2 3

2 3

2

2

2 2 22 2

2

2 2 2 32 2

12 6

11 0 0 0 2 60 1

20 1 0 0 0 1[ ][ ] 20 1 0 0 1

0 0 1 20 0 1

0 0 0 11

2 6

d dd d

l llEI EIl ll

EI EI l lEI EIl l

P F EI EI I l I lI I ll EI EI

m

m l m lm m lEI EI

ν νν ν

νν νν ν

= = − − − − +

(h)

2 3

2

2sym

0 0 0 010 0 0 0 2 60 0 0 0

0 0 0 0 0 1[ ] [ ] 20 0 0 00 0 1 20 0 0 0

0 0 0 00 0 0 1

p pp p

l llEI EI

l lG F EI EI I l I lI Il EI EI

ων ωνων ων

= =

(i)

2 3

2

2skew

0 0 0 010 0 0 0 2 60 0 0 0

0 0 0 0 0 1[ ] [ ] 20 0 0 00 0 1 20 0 0 0

0 0 0 00 0 0 1

p pp p

l llEI EI

l lG F EI EI I l I lI Il EI EI

ων ωνων ων

= = − − − −

(j)

483

Boundary conditions for the present case are that all the linear and angular displacements at station 0

are zero and all the moments and shear forces are zero at right of station 1. Hence, the state vector at

station 0 and 1 have the following form

{ } { }*

00 0 0 0 0 0 0 0 1

r r i i r r i i

T

xz x xz x yz y yz yS M S M S M S M S=

and

{ } { }*

10 0 0 0 0 0 0 0 1

r r r i

T

r y i y r x i xRS x x y yϕ ϕ ϕ ϕ= − − − −

Following rows: 3, 4, 7, 8, 11, 12, 15, 16 will give the eigen value problem of the following form

484

( )

( )

22 2 2

2 2 2 3

22 2 2

2 2 2 3

2 2 2 2

1 0 0 0 02 2

1 0 0 0 0 0 02 6

0 0 1 0 02 2

0 0 1 0 0 0 02 6

0 0 12 2

p pd d

p pd d

p p d d

I l I lI l I ll

EI EI EI EI

m l m lEI EI

I l I lI l I ll

EI EI EI EI

m l m lEI EI

I l I l I l I ll

EI EI EI EI

ων ωνν ν

ν ν

ων ωνν ν

ν ν

ων ων ν ν

− −

+

− − − −

+

− −

( )

( )

2 2 2 3

2 2 2 2

2 2 2 3

0 0

0 0 0 0 1 0 02 6

0 0 0 0 12 2

0 0 0 0 0 0 12 6

r

r

i

i

r

r

i

i

xz

xz

xz

xz

yz

yz

yz

yzp p d d

M

S

M

S

M

S

m l m l MEI EI

SI l I l I l I l

lEI EI EI EI

m l m lEI EI

ν ν

ων ων ν ν

ν ν

+

− − − −

+

0

0

0

0

0

0

0

0

=

(k)

We need to find ν for which the determinant of the above matrix is zero. It should be noted that the above matrix is function of spin speed of the shaft, hence

these solutions have to be obtained for a particular speed at a time. For different operating speed the solutions will help in plotting the Campbell diagram (Fig.

8.36).

485

8.3 Dunkerley’s Formula

Dunkerley’s formula can be used to calculate the fundamental transverse natural frequency without

the help of the numerical methods. This method gives very rough estimation of natural frequency.

From the influence coefficient method, the natural frequency of the system is obtained by the

following conditions

0A = (8.66)

For two degrees of freedom system by considering only the linear displacements, the above equation

will be of the following form

1 11 2 122

1 21 2 22 2

1

01

nf

nf

m m

m m

α αω

α αω

−

=

−

1 11 2 22 1 2 12 212 2

1 10

nf nf

m m m mα α α αω ω

⇒ − − − =

which can be rearranged as

( )2

1 11 2 22 1 2 12 212 2

1 10

nf nf

m m m mα α α αω ω

− + − =

(8.67)

but for a polynomial whose first coefficient is unity, the second coefficient is equal to minus of the

sum of the roots of the equation (Scarborough, 1966)

( )1 2

1 11 2 222 2

1 1

nf nf

m mα αω ω

+ = + (8.68)

Now define

1 112

11

1m α

ω= and

2 222

22

1m α

ω=

(8.69)

where ω11 and ω22 are natural frequencies of single-DOF system, respectively, when mass m1 alone

is present and when mass m2 alone is present. Hence, we have

486

1 2

2 2 2 2

11 22

1 1 1 1

nf nfω ω ω ω+ = + (8.70)

It can be proved that in general for multi-DOF systems, we can write

1 2

2 2 2 2 2 2

11 22

1 1 1 1 1 1

nf nf nfN NNω ω ω ω ω ω+ + + = + + +� �

(8.71)

where N is the total DOFs of the rotor system. In most cases the fundamental frequency 1nf

ω will be

much lower than the all other higher natural frequencies, so the above equation may be written as

1

2 21

1 1N

inf ii

εω ω=

+ =∑

(8.72)

where ε is a small positive quantity. The above equation can be rearranged as

1

2

21

1

1nf N

i ii

ω

εω=

=

− +

∑

(8.73)

On neglecting ε, we can find the fundamental natural frequency of the rotor system, as

1

2

21

1

1nf N

i ii

ω

ω=

= ∑

(8.74)

It should be noted that since ε is a positive quantity, the we will get a lower bound of the fundamental

frequency from the present method as opposed to FEM in which we always get higher bounds of

natural frequencies.

1

2 2 2 2

11 22

1 1 1 1

nf NNω ω ω ω≅ + + +� (8.75)

Dunkerley first suggested this. Equation (8.75) always gives a value for fundamental frequency,

which is slightly lower than the true value, by virtue of the approximation involved. There are other

487

approximate methods are available that can be used to obtain bounds of natural frequencies of the

rotor system for example the Rayleigh's quotient (Meirovitch, 1986).

Example 8.10 Find transverse natural frequencies and mode shapes of a rotor system shown in Figure

8.37. B is a fixed end, and D1 and D2 are rigid discs. The shaft is made of the steel with the Young’s

modulus E = 2.1×1011

N/m2 and a uniform diameter d = 10 mm. Shaft lengths are: BD2 = 50 mm, and

D1D2 = 75 mm. The mass of discs are: m1 = 2 kg and m2 = 5 kg. Consider the shaft as massless and

neglect the diametral mass moment of inertia of discs.

Figure 8.37

Solution: For Figure 8.37, we have

N-m2, l1 =0.125 m, l2 =0.05 m,

11 3 3

1

3 3 103.1

0.125

EIk

l

×= = = 0.158×10

6 N/m,

22 3 3

2

3 3 103.1

0.05

EIk

l

×= = = 2.47×10

6 N/m,

62 61111

1

0.158 100.079 10

2

k

mω

×= = = × ,

62 62222

2

2.47 100.494 10

5

k

mω

×= = = ×

(a) Case 1 (b) Case 2

Figure 8.38 Overhang rotor systems with a single disc

The system natural frequency from the Dunkerley’s formula is given as

103.1EI =

488

1

6 6

2 2 2

11 22

1 1 1 1 110 14.68 10

0.079 0.494nfω ω ω− −

= + = + × = ×

or 1nf

ω = 260.10 rad/sec

Using the TMM the value of the fundamental natural frequency was 1

266.67nf

ω = rad/sec. Hence, the

Dunkerley’s formula estimates reasonably good estimate of the fundamental natural frequency, and it

gives the lower bound.

Example 8.11 Find fundamental transverse natural frequency of the rotor system shown in Figure

8.38. B1 and B2 are bearings, which provide simply supported end condition and D1, D2, D3 and D4 are

rigid discs. The shaft is made of the steel with the Young’s modulus E = 2.1 (10)11

N/m2 and uniform

diameter d = 20 mm. Various shaft lengths are as follows: B1D1 = 150 mm, D1D2 = 50 mm, D2D3 = 50

mm, D3D4 = 50 mm and D4B2 = 150 mm. The mass of discs are: m1 = 4 kg, m2 = 5 kg, m3 = 6 kg and

m4 = 7 kg. Consider the shaft as massless. Consider the discs as point masses.

Figure 8.38 A multi-disc rotor system

Solution: The influence coefficient for a simply support shaft with a disc is given as

2 2

3

a b

EIlα = with l = a + b = 0.45 m (a)

where l is the span of the shaft, and a and b are the disc position from the left and right bearings. The

natural frequency of a single-DOF rotor system can be obtained as

2 1

i i

ii

i y fmω

α= , 1,2,3,4i = (b)

We have d = 0.02 m, l = 0.45 m, EI = 1649.34 N-m2. Hence, EIl

3 =450.89 N-m

5. Table 8.5

summarises the calculation of the fundamental natural frequency.

489

Table 8.5 Calculation procedure of the fundamental natural frequency using the Dunkerley’s formula

S.N. ai

(m)

bi

(m)

Influence

coefficient,

αii (m/N)

Mass of

the disc,

mi, (kg)

For a single-DOF,

21 /i iii i y fmω α=

The fundamental

natural frequency,

( )1

21 / 1 /nf iiω ω= ∑ ,

(rad/sec)

1 0.150 0.300 4.49×10-6

4 1.80×10-5

95.18

2 0.200 0.250 5.54×10-6

5 2.77×10-5

3 0.250 0.200 5.54×10-6

6 3.32×10-5

4 0.300 0.150 4.49×10-6

7 3.14×10-5

Σ21/ iiω =10.93×10

-5

Exercise 8.12 Find transverse natural frequencies and mode shapes of the rotor system shown in

Figure 8.39. B is a fixed bearing, which provide fixed support end condition; and D1, D2, D3 and D4

are rigid discs. The shaft is made of the steel with the modulus of rigidity E = 2.1 (10)11

N/m2 and the

uniform diameter d = 20 mm. Various shaft lengths are as follows: D1D2 = 50 mm, D2D3 = 50 mm,

D3D4 = 50 mm and D4B2 = 150 mm. The mass of discs are: m1 = 4 kg, m2 = 5 kg, m3 = 6 kg and m4 = 7

kg. Consider the shaft as massless. Consider the disc as point masses, i.e., neglect the diametral and

polar mass moment of inertia of all discs.

Figure 8.39 A multi-disc overhung rotor

Solution: The influence coefficient for a cantilever shaft with a disc at free end is given as

3

3yf

L

EIα = (a)

where L is the span of the shaft. The natural frequency of a single-DOF rotor system can be obtained

as

2 1

i i

ii

i y fmω

α= , 1,2,3,4i = (b)

490

We have d = 0.02 m, and EI = 1649.34 N-m2. Table 8.6 summarises the calculation of the

fundamental natural frequency.

Table 8.6 Calculation procedure of the fundamental natural frequency using the Dunkerley’s formula

S.N. Li

(m)

Influence

coefficient,

αii (m/N)

Mass of

the disc,

mi, (kg)

For a single-DOF,

21 /

i iii i y fmω α=

The fundamental

natural frequency,

( )1

21 / 1 /nf iiω ω= ∑

, (rad/sec)

1 0.30 5.46×10-6

4 2.183×10-5

138.55

2 0.25 3.16×10-6

5 1.579×10-5

3 0.20 1.62×10-6

6 9.701×10-6

4 0.15 6.82×10-7

7 4.774×10-6

Σ21/ iiω = 5.209×10

-5

Concluding Remarks:

To summarise, in the present chapter we studied methods of calculation of natural frequencies and

forced responses. Two main methods namely, the influence coefficients and transfer matrix methods,

are given detailed treatment. The Durkerley’s formula for the approximate estimation of the

fundamental frequency is presented. All three methods are illustrated with simple examples keeping

calculation complexity to a minimum, while retaining various basic features of the solution method

for a variety of cases. The application of these methods for larger system is then straight forward;

however, it requires help of computer. The influence coefficient method is simple in application;

however, it requires calculations of influence coefficients with the help of load-deflection relations,

which are different for different systems. Moreover, with number of DOF of the system the matrix

size increases, so it requires higher computational time for large rotor systems to solve the eigen value

problem. The transfer matrix method is quite systematic and effective even for multi-DOF systems. It

does not require any a prior calculation of the system matrix element as in the influence coefficient

method. The overall size of the matrix remains the same, and it does not increase with the DOF of the

system. However, for the calculation of natural frequencies, this method requires roots searching

numerical method, which is time consuming and there is risk of missing some roots. The dynamic

matrix method (not described here) is similar to the TMM in that it relates to the state vectors at

different stations, however, while assembling the components equations the size of the matrix no

longer remains the same and it increases with the DOF of the system. In fact, the finite element

method is the improvised version of this method as we will see in the next chapter.

491

Appendix 8.1 Load deflection relations for various boundary conditions of the shaft

S.N

.

Boundary conditions

Angular

displacement

Linear displacement at any location of the

shaft

Maximum linear

displacement (δmax)

1

A cantilever shaft with a concentrated load P at

free end

2

2x l

Pl

EIϕ

== ( )

2

36

Pxy l x

EI= −

3

3x l

Ply

EI==

2

A cantilever shaft with a concentrated load P at

x = a

2

2x l

Pa

EIϕ

== ( )

2

36

Pxy l x

EI= − for 0 x a< <

( )2

36

Pay x a

EI= − for a x l< <

( )2

36x l

Pay l a

EI== −

3

A cantilever shaft with a uniformly distributed

load q(x) = q0 over the whole span

3

0

2x l

q l

EIϕ

== ( )

22 20 6 4

24

q xy x l lx

EI= + −

4

0

8x l

q ly

EI==

492

4

A cantilever shaft with a linearly varying

distributed load q(x) over the whole span

3

0

24x l

q l

EIϕ

== ( )

22 2 2 30 10 10 5

120

q xy l l x lx x

lEI= − + −

4

0

30x l

q ly

EI==

5

A cantilever shaft with a concentrated moment

M at the free-end

x l

Ml

EIϕ

==

2

2

Mxy

EI=

2

2x l

Mly

EI==

6

A simply supported shaft with a concentrated

load P at the mid-span

2

1 216

Pl

EIϕ ϕ= =

223

12 4

Px ly x

EI

= −

for 0

2

lx< <

3

0.5 48x l

Ply

EI==

493

7

A simply supported shaft with a concentrated

load P at any point

( )2 2

16

Pb l b

lEIϕ

−=

( )1

2

6

Pab l b

lEIϕ

−=

( ) [ ]2 2 2 for 0

6

Pbxy l x b x a

lEI= − − < <

( ) ( )

[ ]

3 2 2 3

6

for

Pb ly x a l b x x

lEI b

a x l

= − + − −

< <

( )2 2

3/22 2

3 9 3l b

x

Pb l by

lEI−

=

−=

and at center, if a b>

( )2 2

0.53 4

48x l

Pby l b

EI== −

8

A simply supported shaft with a uniformly

distributed load q(x) = q0 over the whole span

3

1 224

ql

EIϕ ϕ= = ( )3 2 3

224

qxy l lx x

EI= − +

4

0.5

5

384x l

qly

EI==

9

A simply supported shaft with a concentrated

moment M at one end point

16

Ml

EIϕ =

23

Ml

EIϕ =

2

21

6

Mlx xy

EI l

= −

2

/ 3

5

9 3x l

Mly

EI=

=

and at centre

2

0.5 16x l

Mly

EI==

494

10

3

01

7

360

q l

EIϕ =

3

02

45

q l

EIϕ =

( )4 2 2 40 7 10 3360

q xy l l x x

lEI= − +

4

0

0.5190.00652

x l

q ly

EI==

and at center 4

0

0.50.00651

x l

q ly

EI==

495

Exercise Problems

Use both the influence coefficient and transfer matrix methods for the entire problem unless otherwise

stated in the problem. The Durkerley’s formula could be used to rough estimate of the fundamental

natural frequency.

Exercise 8.1 Obtain the transverse natural frequencies of a rotor as shown in Figure E8.1. The rotor is

assumed to be fixed supported at one end and free at the other. Take mass of the disc m = 2 kg and its

diametral mass moment of inertia kg-m2. The shaft is assumed to be massless, and its

length and diameter are 0.2 m and 0.01 m, respectively. Take the Young’s modulus

N/m2 for the shaft material.

Figure E8.1

Exercise 8.2 Find the transverse natural frequencies and mode shapes of the rotor system shown in

Figure E8.2 by the influence coefficient method. Take EI = 2 MNm2 for the shaft and the diametral

mass moment of inertia of the disc is negligible.

Figure E8.2

Exercise 8.3 For Exercise 8.2 when the left and right discs have respectively diametral mass moment

of inertias as kg-m2 and kg-m

2, obtain the transverse natural frequencies and

mode shapes of the rotor system. [Hint: Use the eigen value formulation for the present case]

Exercise 8.4 Obtain the transverse natural frequency of a rotor system as shown in Figure E8.4. The

mass of the disc m, is 5 kg and the diametral mass moment of inertia, Id, is 0.02 kg-m2. Shaft lengths

are a = 0.3 m and b = 0.7 m. The diameter of the shaft is 10 mm.

0.05d

I =

112.1 10E = ×

.dI =1

0 05 .dI =2

0 06

496

Figure E8.4 An overhang rotor system

Exercise 8.5 Obtain transverse natural frequencies of a rotor system as shown in Figure E8.5. The

mass of the disc is kg and the diametral mass moment of inertia is Id = 0.02 kg-m2. Shaft

lengths are a = 0.3 m and b = 0.7 m, the diameter of the shaft is 10 mm and the modulus of elasticity

of the shaft is E = 2.1 × 1011

N/m2. Consider two different cases, i.e. when bearing A is (i) a simple

support and (ii) a flexible support, which provides a bending stiffness equal two times the bending

stiffness of a cantilevered shaft having length a. Bearing B is a fixed bearing.

Figure E8.5 An overhung rotor system

Exercise 8.6 Find the transverse natural frequencies and mode shapes of the rotor system shown in

Figure E8.6. B1 and B2 are bearings, which provide simply supported end condition, and D1 and D2

are rigid discs. The shaft is made of steel with the Young’s modulus E = 2.1 (10)11

N/m2 and uniform

diameter d = 10 mm. Various shaft lengths are as follows: B1D1 = 50 mm, D1D2 = 75 mm, and D2B2 =

50 mm. The mass of discs are: m1 = 4 kg and m2 = 6 kg. Consider the shaft as massless. Consider the

following cases (i) neglect the diametral mass moment of inertia of both discs and (ii) take

kg-m2 and kg-m

2. [Hint: Use the eigen value formulation for the second case]

Figure E8.6

5m =

.dI =1

0 05

.dI =2

0 06

497

Exercise 8.7 Find transverse natural frequencies and mode shapes of the rotor system shown in Figure

E8.7. B1 and B2 are bearings, which provide simply supported end condition and D1, D2, D3, D4 and

D5 are rigid discs. The shaft is made of the steel with the Young’s modulus E = 2.1 (10)11

N/m2 and

uniform diameter d = 20 mm. Various shaft lengths are as follows: B1D1 = 150 mm, D1D2 = 50 mm,

D2D3 = 50 mm, D3D4 = 50 mm, D4D5 = 50 mm, and D5B2 = 150 mm. The mass of discs are: m1 = 4

kg, m2 = 5 kg, m3 = 6 kg, m4 = 7 kg, and m5 = 8 kg. Consider the shaft as massless. Consider the

following cases (i) consider the disc as point masses, i.e., neglect the diametral and polar mass

moment of inertia of all discs, and (ii) consider discs as thin and take diameter of discs as d =1 8 cm,

d =2 10 cm, d =3 12 cm, d =4 14 cm, and d =5 16 however, neglect the gyroscopic effects.

Figure E8.7 A multi-disc rotor system with simply supported end conditions

For a Jeffcott rotor with an off-set disc, the following influence coefficients are valid:

11 12

21 22

x

y zx

x f

M

α α

ϕ α α

=

with

2 3 22 2

11 12

2 221 22

3 2

3 3

3 3( )

3 3

a l a ala b

EIl EIl

al a lab b a

EIl EIl

α α

α α

− − −

=

− − −−

where l is the span length, and a and b are the distance of the disc from left and right supports,

respectively.

Exercise 8.8 Find all the transverse natural frequencies and draw corresponding mode shapes of the

rotor system shown in Figure E8.8. B1 is fixed support (i.e., with the transverse linear and angular

(slope) displacements equal to zero), and B2 and B3 are bearings with simply support condition (i.e.,

with the transverse linear displacement equal to zero). Shaft segments have the following dimensions:

B1D1 = 50 mm, D1B2 = 50 mm, B2D2 = 25 mm, D2B3 = 25 mm, and B3D3 = 30 mm. The shaft is made

498

of steel with Young’s modulus E = 2.1×1011

N/m2. The mass of discs are: m1 = 1 kg, m2 = 1.5 kg and

m3 = 0.75 kg. Compare the order of magnitude of the torsional and the bending natural frequencies so

obtained for the same system. Consider two cases (i) the shaft as massless and discs as rigid lumped

masses. (ii) consider discs as thin and take diameter of discs as d =1 12 cm, d =2 6 cm, and

d =3 12

cm, however, neglect the gyroscopic effects.

Figure E8.8

Exercise 8.9 Find the unbalance response and critical speeds of a rotor system shown in Figure E8.9.

B is a bearing with fixed end conditions, and D1 and D2 are rigid discs. The shaft is made of steel with

the Young’s modulus E = 2.1(10)11

N/m2 and a uniform diameter d =10 mm. Various shaft lengths are

as follows: BD2 = 50 mm, and D2D1 = 75 mm. The diametral mass moments of inertia of discs are:

= 0.04 kg-m2 and = 0.1 kg-m

2. Take an unbalance of 2 gm at a radius of 5 cm at a convenient

location and orientation. Neglect the mass of the shaft.

Figure E8.9 A two-disc overhang rotor

Exercise 8.10 Obtain the transverse natural frequencies of an overhung rotor system as shown in

Figure E8.10. The end B1 of the shaft is having fixed end conditions and the other end is free. The

length of the shaft is 0.4 m and the diameter is 0.1 m. The disc is thin and has 1 kg of mass, 0.04 kg-

m2 of the polar mass moment of inertia, and 0.02 kg-m

2 of diametral mass moment of inertia. Neglect

the mass of the shaft and consider the gyroscopic effects. Take the shaft speed of 10,000 rpm. Obtain

the Campbell diagram and show critical on that.

Figure E8.10 A overhung rotor

1dI

2dI

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Exercise 8.11 Obtain transverse natural frequencies of the rotor system shown in Figure E8.11 for

following parameters: (i) discs are point masees with m1 = 5 kg, m2 = 8 kg,, l = 1 m, d = 0.02 m, ρ =

7800 kg/m3 and E = 2.1×10

11 N/m

2, and (ii) for the case disc is thin with radii of 10 cm and 15 cm

with other parameters as of case (i).

Figure E8.11 Two disc rotor system

Exercise 8.12 Find transverse natural frequencies and mode shapes of the rotor system a shown in

Figure E8.12. B1 and B2 are fixed supports, and D1 and D2 are rigid discs. The shaft is made of the

steel with the Young’s modulus of E = 2.1 (10)11

N/m2, and has uniform diameter of d = 10 mm.

Different shaft lengths are as follows: B1D1 = 50 mm, D1D2 = 75 mm, and D2B2 = 50 mm. Mass of

thin discs are: m1 = 2 kg and m2 = 3 kg and radius are: r1 = 5 cm and r2 = 8 cm. Consider the shaft as

massless.

Figure E8.12

Exercise 8.13 Find the fundamental transverse natural frequency of the rotor system shown in Figure

E8.13 by using the Durkerley’s formula. Take EI = 2 MN-m2 for the shaft, and the mass moment of

inertia of the disc is negligible. [Answer: 1nf

ω = 49.01 rad/sec]

Figure E8.13 An overhang rotor system

B1 B2

D1 D2

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Exercise 8.14 Obtain transverse natural frequencies of a rotor system as shown in Figure E8.14. The

mass of discs are m1 = 5 kg, and m2 = 8 kg. Shaft length is such that l = 0.3 m, the diameter of the

shaft is 15 mm, and the modulus of elasticity of the shaft is E = 2.1 × 1011

N/m2. Both the bearings are

roller supports. [Hint: 3

118

l

EIα = ,

3

2248

l

EIα = and

3

12 2132

l

EIα α= = ].

Figure E8.14 An overhung rotor system

Exercise 8.15 Obtain transverse natural frequencies of a rotor system as shown in Figure E8.15. The

mass of discs are m1 = 5 kg, and m2 = 8 kg, and radius are r1 = 5 cm and r2 = 8 cm. Shaft length is

such that l = 0.3 m, the diameter of the shaft is 15 mm, and the modulus of elasticity of the shaft is E =

2.1 × 1011

N/m2. Both the bearings are roller supports. [Hint: Need to consider the diametral mass

moment of inertia].

Figure E8.15 An overhung rotor system

Exercise 8.16 Find transverse natural frequencies and mode shapes of the rotor system shown in

Figure E8.16. B is a fixed bearing, which provide fixed support end condition; and D1, D2, D3, D4 and

D5 are rigid discs. The shaft is made of the steel with the Young’s modulus E = 2.1 (10)11

N/m2 and

the uniform diameter d = 20 mm. Various shaft lengths are as follows: D1D2 = 50 mm, D2D3 = 50

mm, D3D4 = 50 mm, D4D5 = 50 mm, and D5B2 = 50 mm. The mass of discs are: m1 = 4 kg, m2 = 5 kg,

m3 = 6 kg, m4 = 7 kg, and m4 = 8 kg. Consider the shaft as massless. Two cases to be considered (i)

Consider the disc as point masses, i.e., neglect the diametral and polar mass moment of inertia of all

501

discs; (ii) consider discs as thin and take diameter of discs as d =1 12 cm, d =2 6 cm, and d =3 12 cm,

d =4 14 cm, and d =5 16 , however, neglect the gyroscopic effects.

Figure E8.16 A multi-disc overhung rotor

Exercise 8.17 Consider a rotor system as shown in Figure E8.17 for the transverse natural frequency.

Two flexible massless shafts are connected by a coupling (i.e., a pin joined). A thin disc of mass 3 kg

is attached to one of the shaft (let us take toward the left side shaft) and it is not interfering the relative

motion between the two shafts. Other ends of shafts have fixed conditions. Take the length of each of

the shaft as 0.5 m and the diameter as 0.05m. Young’s modulus E = 2.1 (10)11

N/m2. Use TMM.

Figure E8.17

Exercise 8.18 Consider a rotor system as shown in Figure E8.18 for the transverse natural frequency.

Two flexible massless shafts are connected by a coupling (i.e., a pin joined). A thin disc of mass 3 kg

is attached to one of the shaft (let us take toward the left side shaft) and it is not interfering the relative

motion between the two shafts. Other ends of shafts have fixed conditions. Take the length of the

shaft as 0.6 m (left side), 0.4 m (right side), and the diameter as 0.05m. Young’s modulus E = 2.1

(10)11

N/m2. Use the influence coefficient method and the TMM.

Figure E8.18

502

Exercise 8.19 Obtain the transverse natural frequency of co-axial shafts rotor system which is

modelled as shown in Fig. E8.19. Discs have the mass of 3 kg and 2 kg, respectively, on shaft A and

shaft B. The shaft A and B are respectively 2 cm and 1.5 cm diameters with a length of 40 cm each.

Neglect the inertia of shafts. The bearing between two co-axial shafts provides an effective transverse

stiffness of 100 MN/m. Take E = 2.1×1011

N/m2. Use TMM.

Figure E8.19

Exercise 8.20 Objective questions with multiple choice answer. Select a single answer only.

(i) The Durkerley’s formula gives lower bound of the fundamental frequency.

(a) True (b) False

(ii) There is an approximation involved in the formulation of the influence coefficient method for

multi-DOF rotor system.

(a) True (b) False

(iii) There is an approximation involved in the solution procedure by the influence coefficient method

for multi-DOF rotor system.

(a) True (b) False

(iv) There is an approximation involved in the solution procedure by the transfer matrix method for

multi-DOF rotor system.

(a) True (b) False

(v) There is an approximation involved in the formulation of the transfer matrix method for multi-

DOF rotor system.

(a) True (b) False

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(vi) For a N-disc rotor with negligible moments of inertia the number of transverse natural frequency

would be

(a) N (b) 2N (c) 4N (d) N2

(vii) For a N-disc rotor with appreciable diametral moment of inertia, however, without gyroscopic

effects, the number of transverse natural frequency would be

(a) N (b) 2N (c) 4N (d) N2

(viii) For a N-disc rotor with appreciable diametral moment of inertia and with gyroscopic effects, the

number of transverse natural frequency would be

(a) N (b) 2N (c) 4N (d) N2

(ix) A point matrix relates

(a) State vectors at either sides of a disc (b) State vectors at either sides of a shaft

(c) State vectors from a shaft to another shaft (d) State vectors at a disc to another disc

(x) A field matrix relates

(b) State vectors at either sides of a disc (b) State vectors at either sides of a shaft

(c) State vectors from a shaft to another shaft (d) State vectors at a disc to another disc

(x) A transfer matrix relates

(a) State vectors from a shaft to another shaft (b) State vectors at a disc to another disc

(c) State vectors from one end of the rotor system to other end (d) either (a) or (b)

(xi) An overall transfer matrix relates

(a) State vectors from a shaft to another shaft (b) State vectors at a disc to another disc

(c) State vectors from one end of the rotor system to other end (d) either (a) or (b)

504

References:

Meirovitch, L., 1986, Elements of Vibration Analysis, McGraw Hill Book Co., NY.

Scarborough, J.B., 1966, Numerical Mathematical Analysis, 6th ed., Baltimore, MD: Johns Hopkins

Press.

Thomson, W.T. and Dahleh, M.D., 1998, Theory of Vibration with Applications, Fifth Edition,

Pearson Education Inc., New Delhi.

Timoshenko, S.P. and Young, D.H., 1968, Elements of Strength of Materials, An east-west edition,

Fifth edition: New Delhi.

Young, W.C., and Budynas, R.G., 2002, Roark’s Formulas for Stress and Strain, McGraw Hill

International Edition, General Engineering Series, New York.

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