Rtiwari Rd Book 03c

Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, ([email protected])

159

3.3 Mechanical Impedance (and Receptance) Method

By this method behavior of complete system is obtained from the behavior of individual

components of the system. It is particularly convenient to use when the characteristics of shaft and

those of bearings are determined from independent experimental or theoretical investigations

carried over a range of frequencies. The simple addition of shaft impedance and bearing

impedance gives the impedance of complete system, which may be used to find system critical

speeds and forced response. General principles are discussed first with reference to a simple

spring-mass system as shown in Figure 3.18(a). Mechanical impedance Z: is defined as the force

required to produce unit displacement. It is generally a complex quantity because of the phase lag

of displacement behind force due to damping.

Figure 3.18 A simple spring-mass system

The impedance of spring alone is given as (Figure 3.18(b))

ss

f kxZ k

x x= = = (49)

where fs is the force on the spring, x is the corresponding displacement and k is the stiffness of the

spring. The impedance of the mass alone is given as (Figure 3.18(c))

22m

m

f m xmxZ m

x x x

ωω

−= = = = −

�� (50)

m

m

f

fm

x

(a) A simple spring-mass

system

m

f

(d) Equivalent system for analysis

purpose (since mass & the spring

have same amount of displacement

they can be thought of connected

in parallel)

k x

fs

(b) Impedance of spring

alone

(c) Impedance of spring

alone


160

where fm is the force on the mass, x is the corresponding displacement, m is the mass and ω is the

simple harmonic forcing frequency to the system. The spring and mass are effectively connected in

parallel (Figure 3.18(d)) as far as force is concerned. For subsystems connected in parallel the net

system impedance is given by the sum of individual subsystem impedances (for example the

equivalent stiffness of two spring connected in parallel is given by the sum of the individual spring

stiffness). So for the system shown in Figure 3.18(d) the impedance at the forcing point is

2

s mZ Z Z k mω= + = − (51)

At the resonance (i.e. forcing frequency is equal to the system natural frequency) system impedance

will be zero (since any force produces an infinite amplitude, exception being when the forcing point is

at a node whereupon impedance tends towards infinitely) so that

2 0nk mω− = that is n

k

mω = (52)

where nω is the natural frequency of the system. It is noteworthy that for subsystems connected in

series, the net system ‘receptance’ is sum of the individual subsystem receptances (receptance being

the inverse of impedance). The above approach may be applied to machines whose shafts carry many

rotor inertias when the shaft free-free impedance has been determined independently of those of the

bearing, pedestals and foundations. Shaft free-free impedance means the impedance when the shaft is

not constrained at either support point. This can be determined experimentally by suspending the shaft

so that it is supported only in the vertical direction as shown in Figure 3.19, then determining the

impedance horizontal response to a known horizontal forcing. The individual impedances component

impedances may be combined according to the rules described above to determine the impedance of

the complete system.

Figure 3.19 Suspended shaft in vertical direction: Free-free end conditions

F(t)


161

Figure 3.20 A flexible rotor under forced excitation

The impedance of a light flexible shaft carrying a number of disc masses may be determined

theoretically as follows. The shaft is considered to be forced, in the first instance, at locations A and B

(see Figure 3.20). The forcing causes the reaction forces and moments …2211 ,,, MFMF etc. to be

set up as a consequence of disc mass inertias and these deform the shaft according to the relationship

11 12 13 14 1,21

21 22 23 24 2,22

1 2 3 4 ,2

1,1 1,2 1,3 1,4 1,21

2,1 2,2 2,3 2,4 2,22

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

nl

nl

n n n n n nl n

n n n n n na

n n n n n na

n n n na n

α α α α αδα α α α αδ

α α α α αδα α α α αδα α α α αδ

α α α α αδ

+ + + + +

+ + + + +

=

…

…

� � � � � ��

…

…

…

� � � � � ��

…

1

2

1

2

,2

n

n n n

F

F

F

M

M

M

�

�

(53)

where ijα will be determined from the beam theory and back subscripts : l refers to the linear and a

refers to the angular shaft displacement. In general the loading applied to any rotor mass i to cause its

acceleration is

( ) ( )xmxmF liliim +−=+= δωδ 2�� (54)

and

( ) ( )2

i im i d a d aM I Iδ φ ω δ φ= + = − +�� (55)

Light flexible shaft

Zero displacement, “datum position”

x A

B δl φ

δα

Tangent to shaft Line parallel to AB


162

where mi and id

I are the disc mass and the disc diametral mass moment of inertia respectively, x and

φ are shaft linear and angular displacements respectively, caused by movements at shaft locations A

and B and not caused by the inertia forces. The reaction loading on the shaft due to inertia will be

equal and opposite and is given by

( )2

s i i lF m xω δ= + (56)

and

( )2

is i d aM Iω δ φ= + (57)

where ω is the excitation frequency. Substituting equations (56) and (57), written for each disc, into

equation (53), we get

11 12 13 14 1,21

21 22 23 24 2,22

1 2 3 4 ,2

21,1 1,2 1,3 1,4 1,21

2,1 2,2 2,3 2,4 2,22

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

1

nl

nl

n n n n n nl n

n n n n n na

n n n n n na

n n n na n

α α α α αδα α α α αδ

α α α α αδα α α α αδωα α α α αδ

α α α αδ

+ + + + +

+ + + + +

=

…

…

� � � � � ��

…

…

…

� � � � � ��

( )( )

( )( )( )

( )

1

2

1 1 1

2 2 2

1 1

2 2

2 ,2n

l

l

n l n n

d a

d a

n n d a n n

m x

m x

m x

I

I

I

δ

δ

δ

δ φ

δ φ

α δ φ

+ + + + + +

�

�

…

(58)

which can be rearranged as

}]{[}]{[ xRA =δ (59)

with

[ ]

( )( )

( )

2

1 11 2 12 1,2

2

1 21 2 22 2,2

2

1 2 ,1 2 2 ,2 2 ,2

1/

1/

1/

n

n

n

d n

d n

n n d n n

m a m a I a

m a m a I aA

m a m a I a

ω

ω

ω

− − − − − −

=

− − −

…

…

� � � �

…


163

[ ] { } { }

11 1 12 2 1,2 1 1

21 1 22 2 2,2 2 2

2 ,1 1 2 ,2 2 2 ,2

; ;

n

n

n

n d l

n d l

a n nn n n n d

a m a m a I x

a m a m a I xR x

a m a m a I

δδ

δ

δ φ

= = =

…

…

� ��

…

(60)

which gives

{ } [ ]{ }C xδ = with 1[ ] [ ] [ ]C A R−= (61)

In general the application of inertia loads iF and iM to the shaft at some point causes proportional

reaction forces AF and BF at points A and B, where

=

i

i

BB

AA

B

A

M

F

bb

bb

F

F

21

21 (62)

the generalized form of equation (62), allowing for all inertias, is

1 2 ,2

1 2 1 2

1 2 ,2

TA A A nA

n n

B B B nB

b b bFF F F M M M

b b bF

=

��

� (63)

where ijb may be determined from static equilibrium considerations. Substituting for disc inertia

forces and moments from equations (56) and (57) into equation (63), we get

( )( )

( )( )

( )

1

2

1 1 1

2

2 2 2

21 2 ,2

21 2 ,2

1 1

2

n

l

l

A A A nAn l n n

B B B nBd a

d a n n

m x

m x

b b bFm x

b b bFI

I

ω δω δ

ω δω δ φ

ω δ φ

+

+ +=

+

+

�…

…

�

(64)

On substituting for s'δ in equation (64) from equation (61) (i.e. 1 11 1 12 2 1,2l n nc x c x cδ φ= + + +… etc.)

then gives


164

[ ]{ }A

B

FD x

F

=

(65)

with

[ ]

( )( )

( )

{ } { }

1 11 1 12 1 13 1 1,2

1 2 ,2 2 21 2 22 2 23 2 2,22

1 2 ,2

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

1 2

1

1

1n n n n

n

A A A n n

B B B n

d n d n d n d n n

T

n

m c m c m c m c

b b b m c m c m c m cD

b b b

I c I c I c I c

x x x

ω

φ

+ + =

+

=

…

… …

… � � � � �

…

�

Figure 3.21 Rigid body linear and angular displacements of the shaft

However, displacements 2211 ,,, φφ xx etc. are related to displacements at forcing points, Ax and

Bx ,

by a relationship of the form

=B

A

x

xGx ][}{ (66)

with

xB

xA a

l

x

al

xxxx BAA

−+=


165

{ } [ ]

1

11 12

2

1 2

1,1 1,2

1

2 ,1 2 ,2

; n n

n

n n

n n

n

xg g

x

g gx x G

g g

g g

φ

φ

+ +

= =

� ��

� ��

Element of [G] can be obtained by simple consideration of geometry. Substituting equation (66) into

equation (65), it gives

[ ]A A AA AB A

B B BA BB B

F x Z Z xZ

F x Z Z x

= =

(67)

with

[ ] [ ][ ]Z D G=

where [Z] is the impedance matrix for the shaft and rotor assembly, relating forcing and

displacements at points A and B. In the matrix [Z] now all quantities, one can known by theoretical

analysis. The more general form of equation (67) that allows also for the motion in y-direction is

0 0

0 0

0 0

0 0

xA AA AB A

xB BA BB B

yA AA AB A

yB BA BB B

F Z Z x

F Z Z x

F Z Z y

F Z Z y

=

(68)

The response at any other location along the shaft can be obtained by pre-multiplying inverse of

impedance matrix in equation (68) to get displacements at A and B, then substituting these in

equations (66) and (61). If BF is chosen to be zero, the point A can be chosen as any convenient

location along the shaft and corresponding (direct) impedance so evaluated. Equation (68) can be

expanded for any number of forcing points, for example for third forcing point C


166

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

xA AA AB AC A

xB BA BB BC B

xC CA CB CC C

yA AA AB AC A

yB BA BB BC B

yC CA CB CC C

F Z Z Z x

F Z Z Z x

F Z Z Z x

F Z Z Z y

F Z Z Z y

F Z Z Z y

=

�

�

�

… … … … … … … … …

�

�

�

(69)

If forcing is caused by imbalance and AF and BF are reaction forces at pinned bearing supports, the

forced response of the system can be determined by assigning zero values to , , and A B A Bx x y y . From

equation (69) the third and sixth equations gives

xc cc cF Z x= and yc cc cF Z y=

which gives displacements at C as

/ and /c xc cc c yc ccx F Z y F Z= = (70)

From equation (69), 1st, 2nd, 4th, and 5th equations gives

; ; ; xA AC c xB BC c yA AC c yB BC cF Z x F Z x F Z y F Z y= = = = (71)

Noting equation (70), equation (71) can be combined as

0 0

/0 0=

/0 0

0 0

xA AC AC

xB xc ccBC c BC

yA yc ccAC c AC

yB BC BC

F Z Z

F F ZZ x Z

F F ZZ y Z

F Z Z

=

(72)

Equation (72) gives reaction forces at A and B due to forcing at C. These reaction forces may be

substituted back into equation (69), which may be written more generally, to give response at any

other locations as


167

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

xAAA AB ACA

xBBA BB BCB

xDDA DB DCD

yAAA AB ACA

yBBA BB BCB

yDDA DB DCD

FR R Rx

FR R Rx

FR R Rx

FR R Ry

FR R Ry

FR R Ry

=

�

�

�

…… … … … … … ……

�

�

�

(73)

with [ ] 1[ ]R Z −=

where R are component of receptance matrix. In equation (73), [R] is already known to us

corresponding to new sets of chosen points, {F} matrix is now completely known. So the new Dx and

Dy can be obtained as

D DA xA DB xB DD xC DA xA DB xB DD xDx Z F Z F Z F R F R F R F′ ′ ′= + + = + + (74)

and

D DA yA DB yB DD yC DA yA DB yB DD yDy Z F Z F Z F R F R F R F′ ′ ′= + + = + + (75)

Example 3.5 Obtain transverse synchronous critical speeds of a rotor system as shown in Figure 3.3.

Take the mass of the disc, m = 10 kg, the diametral mass moment of inertia, Id = 0.02 kg-m2, the polar

mass moment of inertia, Ip = 0.04 kg-m2. The disc is placed at 0.25 m from the right support. The

shaft is having diameter of 10 mm and total span length of 1 m. The shaft is assumed to be massless.

Take shaft Young’s modulus E = 2.1 × 1011 N/m2. Neglect gyroscopic effects. Take one plane motion

only.

a b

l = a + b

Figure 3.22 A rotor system

Influence coefficients are defined as:

11 12

21 22

y F

M

α α

α αθ

=

with

( )( )

2 2 2 3 2

11 12

2 2

21 22

/ 3 ; 3 2 / 3

( ) / 3 ; 3 3 / 3

a b EIl a l a al EIl

ab b a EIl al a l EIl

α α

α α

= = − − −

= − = − − −

Solution: Figure 3.23 shows a schematic of the rotor system.


168

Figure 3.23 Schematic diagram of a rotor system

From equation (53), we have

1 11 12 1

1 21 22 1

l

a

F

M

δ α αδ α α

=

with

4 4 10 4(0.01) 4.909 10 m64 64

I d xπ π −= = = ; 0.75a = m; 0.25b = m; 1.0l = m.

( ) { }

2 2 2 24

11 11 10

2 3 2 2 3 2

4

12 11 10

21

(0.75) (0.25)1.137 10 m/N

3 3 2.1 10 4.909 10 1

3 2 3 (0.75) 1 2 0.75 0.75 13.3031 10 m/N

3 3 2.1 10 4.909 10 1

( ) 0.75 0.25 (0.75 0.25)

3 3 2.1 1

a b

EIl

a l a al

EIl

ab b a

EIl

α

α

α

−−

−−

×= = = ×

× × × × ×

− − × × − × − ×= − = − = − ×

× × × × ×

− × × −= =

× ×

( )

4

1211 10

2 22 23

22 11 10

3.0314 10 m/N0 4.909 10 1

3 0.75 1 3 0.75 1(3 3 )1.4146 10 m/N

3 3 2.1 10 4.909 10 1

al a l

EIl

α

α

−

−−

= − × =× × ×

× × − × −− −= − = − = ×

× × × × ×


1

1

2 2

1 11 12

2 2

1 21 22

1[ ]

1

d

d

m IA

m I

ω α ω α

ω α ω α

− −=

− − and [ ] 1

1

11 1 122

21 1 22

d

d

m IR

m I

α αω

α α

=

and

[ ] 1 1

12 21 22 12

2 2

1 21 1 11

11 1since

1

d dI I a b d bA

c d c aA ad bcm m

ω α ω α

ω α ω α

−− − − = = −−−

with

0.75m 0.25m

A B


169

1 1

4 2

11 22 12 21 1 1 11 22( ) ( ) 1d dA m I m Iω α α α α ω α α= − − + +


11 1

1

1 1 1 1 1 1

1

2 211 1 1222 121

2 221 1 221 12 1 11

2 2 2 2

22 11 12 21 1 22 12 12 22

2 2 2

1 12 11 1 1 11 21 1 1 12 12

(1 )1[ ] [ ] [ ]

(1 )

(1 ) (1 )1

(1 )

dd d

d

d d d d d d

d

m II IC A R

m IA m m

I m I m I I I I

A m m m m m I

α αω α ω α

α αω α ω α

ω α α ω α α ω α α ω α α

ω α α ω α α ω α α

− −

= = −

− + − +=

+ − +1

2

1 11 22(1 ) dm Iω α α

−

(A)

Figure 3.24 Displacement of the shaft

From free body diagram of shaft, we have

( )

1

1 1 1 1

1 1 1

0

10 0

so that

11

A B

A B B

A B

F F F F

aM F l M Fa F F Ml l

aF F F F Ml l

= ⇒ = +

= ⇒ − − = ⇒ = +

= − + = − −

∑

∑

Combining equations (B) & (C), we get

1

1

( / ) (1/ )

( / ) (1/ )

A

B

F Fb l l

F Ma l l

− =

(D)

(B)

(C)

F1

FA

FB

M1

aδ

l

a

Tangent to shaft

Parallel

Rigid body displacements

of the shaft


170

Noting equation (D), from equation (65), we have

1

1

[ ]A

B

F xD

F φ

=

(E)

with

[ ]

( ) ( )

1 1

1 1

1 1

1 11 1 12

21 22

1 11 21 1 12 22

1 11 21 1 12 22

(1 )( / ) (1/ )

(1 )( / ) (1/ )

/ (1 ) 1/ ( / ) (1/ ) (1 )

( / ) (1 ) (1/ ) ( / ) (1/ ) (1 )

d d

d d

d d

m c m cb l lD

I c I ca l l

b l m c l I c b l m c l I c

a l m c l I c a l m c l I c

+ − = +

+ − − +=

+ + + +

Figure 2.25 Rigid body displacement of the shaft

From Figure 2.25, we have

( )1 1

b A

A A B A B

x x a a b ax x a x x x x

l l l l l

− = + = − + = +

(F)

and

( ) ( ) BAAB xlxl

l

xx/1/11 +−=

−=φ (G)

On combining equations (F) and (G), we get

1

1

[ ]A

B

x xG

xφ

=

with ( / ) ( / )

[ ](1/ ) (1/ )

b l a lG

l l

= −

(H)

Hence, from equations (E) and (H), we have

xA xB x1

1φ

a b


171

( ) ( )

−

++++

+−−+=

B

A

dd

dd

B

A

x

x

ll

lalb

cIlcmlacIlcmla

cIlcmlbcIlcmlb

F

F

)/1()/1(

)/()/(

)1()/1()/()/1()1()/(

)1()/1()/(/1)1(/

2212121111

2212121111

11

11

which can be simplified as

A AA AB A

B BA BB B

F Z Z x

F Z Z x

=

(I)

with

( ) ( ){ }{ }( )

( ) ( ){ }( )

{ }( )

{ }( )

{ }( )

1 1

1 1

1

1

1 11 21 1 11 21

1 12 22 1 12 22

1 11 21 1

1 12 22

/ (1 ) 1/ ( 1/ ) / (1 ) 1/ /

( / ) (1/ ) (1 ) 1/ ( / ) (1/ ) (1 ) 1/

( / ) (1 ) (1/ ) / ( / )

( / ) (1/ ) (1 ) 1/

d d

d d

AA AB

BA BB

d

d

b l m c l I c l b l m c l I c a l

b l m c l I c l b l m c l I c lZ Z

Z Za l m c l I c b l a l m

a l m c l I c l

+ − − + + − +

− + − − +

=

+ + +

+ + −

{ }( )

{ }( )1

1

11 21

1 12 22

(1 ) (1/ ) /

( / ) (1/ ) (1 ) 1/

d

d

c l I c a l

a l m c l I c l

+ + + + +

Since A & B are pinned support 0A Bx x= = even at critical speeds hence AA ABZ Z= = ∞ i.e.

AAB BA

B

FZ Z

x= = ∞ = . Hence denominator of any of impedance can be put equal to zero to get the

frequency equation. Noting the equation (A), the common denominator of the [Z] matrix (or its

components) are determinant of matrix [A] i.e.

1 1

4 2

11 22 12 21 1 1 11 22( ) ( ) 1 0n d n dm I m Iω α α α α ω α α− − + + =

or

1

1 1

1 11 224 2

11 22 12 21 1 1 11 22 12 21

( ) 10

( ) ( )

d

n n

d d

m I

m I m I

α αω ω

α α α α α α α α

+− + =

− −

From the present problem data, we have the frequency equation of the following form

42 2

2 8 2 8

(10 1.137 0.02 14.146) 10 10

(10 0.02) (1.137 14.146 3.03 ) 10 (10 0.02) (1.137 14.146 3.03 ) 10n nω ω

−

− −

× + × ×− + =

× × × − × × × × − ×

or


172

4 2 4 8(8.44 10 ) 0.7243 10 0n nω ω− × + × =

which can be solved as

4 4 2 4

2 48.44 10 (8.44 10 ) 4 0.7243 10 8.44 8.26710

2 2nω

+ × ± × − × × ±= = ×

Natural frequencies of the system is given as

129.45nω = rad/sec and

2289.23nω = rad/sec.

3.4 Dynamic Stiffness Matrix Method

It is similar to the transfer matrix method, in that it involves a division of the shaft into a number of

smaller elements for the purpose of analysis. The difference is there in system equation formulation

and the response at all stations in the system is determined simultaneously. The method has the

advantage that once the system matrix has been assembled it can be used to calculate system stability

thresholds as well as response and critical speeds. The major disadvantage is that it requires the

storing and manipulation of large matrices and so is more demanding of computer power than is the

transfer matrix method.

(a) A disc (b) A shaft segment

Figure 3.26 Free body diagram

If a beam element alone as shown in Figure 3.26(b), without concentrated mass, is considered, the

relationships between applied forces and moments and resulting deflections and slopes, are given as

yA yB

θA

θB

QA QB

MB MA MA

MC

QA

QC

Uv


173

=

b

B

A

A

B

B

A

A

y

y

kkkk

kkkk

kkkk

kkkk

Q

M

Q

M

θ

θ

44434241

34333231

24232221

14131211

(76)

Elements of equation (76) can be obtained by rearranging equations (24-26) of the transfer matrix

method. From the transfer matrix method, a field matrix relates state vectors at two stations as follows

BAQ

M

y

lEI

lEI

lEI

lEI

ll

Q

M

y

−

−=

−

θθ

1000

1002

10

621

2

32

(77)

which can be expanded as

( ) ( )

( ) ( )

2 3

2

/ 2 / 6

/ / 2

A B B B B

A b B B

A B B

A B

y y l l EI M l EI Q

l EI M l EI Q

M M lQ

Q Q

θ

θ θ

− = − + + +

= + +

= −

=

(78)

Equation (78) can be regrouped as

2 3

2

1 2 6

0 12

A B B

l ly l y MEI EI

Ql lEI EI

θ θ

− − = +

(79)

and

0 0 1

0 0 0 1A B B

M y l M

Q Qθ− −

= +

(80)

On rearranging equations (79-80), we have


174

2 3

2

1 0 1 0 0 / 2 / 6

0 1 0 1 0 0 / / 2A B A B

y l y M Ml EI l EI

Q Ql EI l EIθ θ− −

− = +

(81)

and

0 0 0 0 1 0 1

0 0 0 0 0 1 0 1A B A B

y y M l M

Q Qθ θ− − −

+ = − +

(82)

On combining equations (81-82), we get

2 3

2

1 0 1 0 0 / 2 / 6

0 1 0 1 0 0 / / 2

0 0 0 0 1 0 1

0 0 0 0 0 1 0 1

A A

B B

y Ml l EI l EI

Ql EI l EI

ly M

Q

θ

θ

− − − = − −− −

(83)

which takes the form of equation (76), with

12 3

11 12 13 14

221 22 23 24

31 32 33 34

41 42 43 44

1 0 10 0 / 2 / 6

0 1 0 10 0 / / 2

0 0 0 01 0 1

0 0 0 00 1 0 1

k k k k ll EI l EI

k k k k l EI l EI

k k k k l

k k k k

−−

− = − −

−

(84)

For the most common type of forcing (i.e. the unbalance) the applied moments and shear forces takes

the form

j j j j; ; ; t t t t

A A A A B B B BM M e Q Q e M M e Q Q eω ω ω ω= = = = (85)

where AM , … are complex in general. Deflections and slopes, similarly can be written as

j j j j; ; ; t t t t

A A A A B B B By Y e e y Y e eω ω ω ωθ θ= = Θ = = Θ (86)

where AY , … are complex in general. Substituting equations (85) and (86) into equation (76), it gives


175

11 12 13 14

21 22 23 24

31 32 33 34

41 42 43 44

AA

AA

BB

bB

k k k k YM

k k k kQ

k k k k YM

k k k kQ

Θ = Θ

(87)

which can be written in more compact form as

=

}{

}{

][][

][][

}{

}{

2221

1211

B

A

B

A

d

d

uu

uu

F

F (88)

with

{ }M

FQ

=

; { }Y

d

= Θ

; [ ] 13 14

12

23 24

k ku

k k

=

; …

Equation (88) can be expanded to allow for the shear forces, moments, slopes and displacements in

the horizontal direction, as

11 12

11 12

21 22

21 22

{ } { }[ ] 0 [ ] 0

{ } { }0 [ ] 0 [ ]

{ } { }[ ] 0 [ ] 0

{ } { }0 [ ] 0 [ ]

A A

Ah Ah

B B

Bh Bh

F du u

F du u

F du u

F du u

ν ν

ν ν

=

(89)

where subscripts: v and h refer to the vertical and horizontal directions. Equation (89) relates to the

forces & moments at ends A and B to displacements at ends A and B. Now considering the forces and

moments acting on the concentrated mass (i.e. disc as shown in Figure 3.26(a)) at the end of the

element, equations of motion for disc mass are

;

;

A v cv v Ah h ch h

c A P Ah d A ch Ah P A d Ah

Q U Q B my Q U Q B mx

M M I I M M I I

ν

ν ν ν νωθ θ ωθ θ

+ − − = + − − =

− − = − + =

��

� ��

(90)

where p AI ωθ� is the gyroscopic moment, where B is the bearing force (equal to zero if the station

considered is not a baring location), U is a known imbalance, M is the magnitude of concentrated

mass, PI is the polar mass moment of inertia ( 0PI = for gyroscopic effects are to ignored) and Id is

the diametral moment of inertia (Id is related to the rotary inertia). Note that the slopes and


176

displacements on each side of the concentrated mass are the same. The bearing reaction force may be

expressed in the form

j j; and t t

v v h hB B e B B eω ω= = (91)

where vB and

hB are complex in general. On substituting equation (91) into equation (90) and it

gives: (shear force & bending moment at end A are related with at end C).

2

2

2 2

2 2

;

j ;

j

A c

Ah h ch h

A c p Ah d Av

Ah ch p A d Ah

Q B m Y Q U

Q B m X Q U

M M I I

M M I I

ν ν ν ν

ν ν

ν

ω

ω

ω ω

ω ω

= − + −

= − + −

= − Θ + Θ

= + Θ + Θ

(92)

On substituting equation (92) into equation (89), it gives

2 2

11 12 13 14

2

21 22 23 24

2 2

11 12 13 14

2

21 22 23 24

31 32 33 34

41 42 43 44

31 32

0 j 0 0

0 0 0 0

j 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

v

v

h

h

v

v

h

h

C

d p

C

Cp d

C h h

B

B

B

B

Mk k I I k k

Q U Bk m k k k

M I k k I k k

Q U B k m k k k

M k k k k

k k k kQ

k kM

Q

ν ν

ω ωωω ω

ω

−

− + +

− − − + +

=

33 34

41 42 43 440 0 0 0

C

C

C

C

B

B

B

B

Y

Z

Y

Zk k

k k k k

Θ

Φ Θ Φ

(93)

which can be written as

{ }{ }

0 00 01 0

10 11 11

[ ] [ ] { }

[ ] [ ] { }

F M M d

M M dF

=

(94)


177

where subscripts refer to node numbers and the matrix ][ 0M is the dynamic stiffness matrix for

element 0. Similarly for element 1, it will take the form:

{ }{ }

[ ][ ] [ ]

0 11 12 1

21 21 22

{ }

{ }

F M M d

dF M M

=

(95)

where 1{ }F is similar to }{ 1F but also contains imbalance forcing terms and bearing force terms (as

does 0{ }F ). Equations (94) and (95) may be combined to eliminate the internal force and moment

terms of the matrix }{ 1F and it will give overall equation for element 0 and 1

{ }{ }{ }

000 01 0

*

1 10 11 11 12 1

2 21 22 2

[ ] [ ] [0] { }

[ ] [ ] [ ] [ ] { }

[0] [ ] [ ] { }

FM M d

F M M M M d

F M M d

= +

(96)

where { }*1F contains only the imbalance and bearing force terms of matrix { }1F . Equation (96) may

be extended for any number of system elements to give an overall system matrix equation of the form

=

−

}{

}{

}{

}{

}{

*1

*2

*1

0

n

n

F

F

F

F

F

…

}]{[}{

}{

}{

}{

}{

}{

*

1

2

1

0

sZPor

d

d

d

d

d

n

n

=

−

…

(97)

The left hand side of equation (97) contains only known imbalance forcing terms, known applied

forces and moments at the shaft ends (usually zero), and unknown bearing reaction forces. Bearing

reaction forces can be found in the following steps

}{][}{ 1 PZs −= (98)

Equation (98) will give an expression for displacement at each bearing location. In the case of

bearings, which behave as pinned supports, these expressions can be equated to zero and solved


178

simultaneously to give values for the bearing reaction forces. The back substitution of these forces

into equation (98) then enables the shaft displacements at all other locations to be evaluated. It is

noteworthy that the system dynamic stiffness matrix [Z] is banded about the leading diagonal; this can

be made use of when storing the matrix in the computer, since only non-zero elements values need to

be stored.

Example 3.6 Obtain transverse synchronous critical speeds of a rotor system as shown in Figure 3.27.

Take the mass of the disc, m = 10 kg, the diametral mass moment of inertia, Id = 0.02 kg-m2, the polar

mass moment of inertia, Ip = 0.04 kg-m2. The disc is placed at 0.25 m from the right support. The

shaft is having diameter of 10 mm and total span length of 1 m. The shaft is assumed to be massless.

Take shaft Young’s modulus E = 2.1 × 1011 N/m2. Neglect gyroscopic effects. Take one plane motion

only.

a b

l = a + b

Figure 3.27 A rotor system

Solution: Figure 3.28 shows the free body diagram of shaft elements and disc without gyroscopic

effects.

(a) (b)

(c) (d) (e)

Figure 3.28 A shaft-disc element free-body diagram

QA, yA

A

MA,

θA

MB,

θB

B

QB, yB

A B

C MC

QC

MA

QA

C A

C A

y y

θ θ

=

=

D C A B D C

MD QD QC MC


179

For the shaft element DC as shown in Figure 3.28(b), we have

2 2

1

2

1

12 6 12 6

4 6 2

12 6

sym 4

D D

D D

C C

C C

Q yl l

M l l lk

Q yl

M l

θ

θ

− − = − −− − −

(A)

with

1 3

1

EIk

l=

For the beam element AB as shown in Figure 3.28(c), we have

2 2

2

2

2

12 6 12 6

4 6 2

12 6

sym 4

A A

A A

B B

B B

Q yl l

M l l lk

Q yl

M l

θ

θ

− − = − −− − −

(B)

with

2 3

2

EIk

l=

For a disc as shown in Figure 2.38(e), we have the following relationship

andA C C C A d CQ Q my M M I θ− = − = ��

which can be rearranged as

andA C C A C d CQ Q my M M I θ= + = − �� (C)

Following conditions hold at disc

CA yy = and AC θθ = (D)


180

Substituting equations (C) and (D) into equation (B), we get

2 2

2

2

2

12 6 12 6

4 6 2

12 6

sym 4

C C C

C d C C

B B

B B

Q my yl l

M I l l lk

Q yl

M l

θ θ

θ

+ − − − = − −− − −

��

��

(E)

For SHM, we have 2

C n Cy yω= −�� and 2

C n Cθ ω θ= −�� , from equation (E), we have

( )( )

2

2

2 2 2

22

2

2

12 / 6 12 6

4 / 6 2

12 6

sym 4

nC C

C Cd n

B B

B B

m k l lQ y

M l I k l lk

Q yl

Ml

ω

θω

θ

+ − − −= − − − − −

(F)

On combining equations (A) and (F), we get

1 1 1 1 1 1

2 2

1 1 1 1 1 1 1

1 2

1 1 1 1 1 2 2 2 2 22

2 2

1 1 2 22

1 1 1 1 1 1 2 2 2 22

12 6 12 6 0 0

6 4 6 2 0 0

12 1212 6 6 6 12 6

4 46 2 6 6 6 2

D

D

nC C

C C

Bd n

B

k l k k l k

l k l k l k l kQ

k kMk l k l k l k k l k

mQ Q

M M l k l kl k l k l k l k l k

Q I

M

ω

ω

−

− − + − − + −

+ − + =

− + − + + − − − −

2

2 2

2 2 2 2 2 2

2 2

2 2 2 2 2 2 2 2

0 0 12 6 12 6

0 0 6 2 6 4

D

D

C

C

B

B

y

y

l ky

k l k k l k

l k l k l k l k

θ

θ

θ

− − −

−

(G)

Boundary conditions are

0,0 ==== BDBD MMyy (H)

On application of boundary conditions, equation (G) can be written as


181

( )

( )

2 2

1 1 1 1 1 1

1 2

1 1 1 1 2 2 2 22

2 2

1 1 2 22 2

1 1 1 1 2 2 2 22

2 2

2 2 2 2 2 2

4 6 2 0

12 1206 6 6 6

0

0 4 42 6 6 2

0

0 6 2 4

D

n C

C

Bd n

l k l k l k

k kl k l k l k l k

m y

l k l kl k l k l k l k

I

l k l k l k

θω

θθω

−

− + − + − + = − + + − −

−

(I)

Equation (I) represents an eigen value problem. The non-trivial solution can be obtained by equating

determinant equal to zero, which will give natural frequencies of the system.

( ) ( ){ } ( )24 2 2 2 2 2 2 2 2 2 2 2

1 2 1 2 1 2 1 2 1 1 2 2 2 1 1 2 1 2 1 2(16 ) 48 144 0n d n dmI k k l l k k l l m k l k l I k k k k l l l lω ω + − + − + + =

After substituting values k1 = 2.443×102 N/m , k2 = 6.597×10

2 N/m, l1 = 0.75 m, l2 = 0.25 m, m = 10

kg and 0.02dI = kgm2, we get

5 4 9 2 131.814 10 7.132 10 1.316 10 0n nω ω× − × + × =



2193.36nω = rad/sec

Example 3.7 Obtain the bending critical speed of the rotor system as shown in Figure 3.29. Take the

mass of the disc, m = 5 kg and the diametral mass moment of inertia, Id = 0.02 kg-m2. Take shaft

length a = 0.3 m and b = 0.7 m. The diameter of the shaft is 10 mm. Use the dynamic stiffness matrix

method. Neglect the gyroscopic effects.

a b

A B

Figure 3.29 An overhang rotor system

For a beam element as shown in Figure 3.30, following relations holds


182

1 1

2 2

1 1

3

2 2

2

2 2

12 6 12 6

4 6 2

12 6

sym 4

F yl l

M l l lEI

F yll

M l

θ

θ

− − = −

M1, θ1 F1, y1 F2, y2 M2, θ2

Figure 3.30 Positive sense of beam displacements and forces

Solution: Figure 3.31 shows the free body diagram of the rotor elements with neglecting the

gyroscopic effects.

(b) (c)

lCM

QC

2Q DM

D

C D

Figure 3.31 Free body diagram of rotor elements

For shaft segment AC as shown in Figure 3.31(b), we have following relations

1 1

2

1 1 1

3

11

2

1

12 6 12 6

4 6 2

12 6

sym 4

A A

A A

C C

C C

Q Ql l

M Ml l lEI

Q Qll

M Ml

− − = − −− − −

(A)

For the disc as shown in Figure 3.31 (c) for SHM, we have

ba

C DB A

(l )1 (l )2

AQ

lMA QCMC

1 MB

QB

QAMA

Shaft

Disc

(a)

(d)


183

2

2

A b A B n

B A d B A B d n

Q Q my Q Q y

M M I M M I

ω

θ θω

− = ⇒ = −

− = ⇒ = +

��

��

(B)

On substituting and rearranging equation (B) into equation (A), we get

21 2 1

221 1 1

3

11

2

1

12 6 6

4 6 2

12 6

sym 4

BB n

Bb d n

CC

CC

yl l lQ m y

l l lM I EI

yllQ

lM

ωθω θ

θ

− − −+ = −−− −−

(C)

On rearranging equation (C), we get

2

1 1 1 1 1 1

2 2 2

1 1 1 1 1 2

1 1 1

2

1 1

(12 ) 6 12 6

(4 ) 6 2

12 6

sym 4

B Bn

B Bd n

C C

C C

Q yk m l k k l k

M l k I l k l k

Q yk l k

M l k

ωθω

θ

+ − − − = − −− − −

(D)

where 311

EIkl

=

For shaft C-D as shown in Figure 3.31(d), we have

−

−

−

−

−

=

−

−

D

D

C

C

D

D

C

C

y

y

kl

klk

klklkl

klkklk

M

Q

M

Q

θ

θ

2

2

2

222

2

2

2222

2

2

222222

4

612

1264

612612

(E)

where 3

22 / lEIk =

By assembling equations (D) and (E), we get


184

−

−−−

−++−

−+−+−−

−−

−+

=

−

−

+−

+−

D

D

C

C

b

B

d

D

D

CC

CC

B

B

y

y

y

klklklkl

klkklk

klklklklklklklkl

klkklklkkklk

klklIklkl

klkklmk

M

Q

MM

QQ

M

Q

θ

θ

θωω

222222

2222

222222

22

22222

221

2122111

2111

222221121111

12111

21

2111

111112

1

4612600

61261200

126)44()66(26

612)66()1212(612

0026)4(6

006126)12(

The boundary conditions are 0B B DQ M M= = = and 0C Dy y= = . On right hand side already third

and fourth columns elements are zero. So fifth row and fifth column has eliminated from equation (3)

and it is given as

( ) ( )( )

2

1 1 1 1 1 1

2 2

1 1 1 1 1 1 1 1

1 1 1 1 2 1 1 2 2 2 2

2 2 2 2

1 1 1 1 1 1 2 2 1 1 2 2 2 2

2 2

2 2 2 2 2 2

(12 ) 6 12 6 00

6 (4 ) 6 2 00

12 6 12 12 6 6 6

6 2 6 6 (4 4 ) 12

0 0 6 12 4

B

d

B

c

D

k m l k k l ky

l k l k I l k l k

k l k k k l k l k l k

l k l k l k l k l k l k l k

l k l k l k

ωω

θ

θθ

+ −

− − =− − + − + − − + +

0

0

[ ]{ } {0}K d =

The non-trivial solution can be obtained by equating determinant equal to zero, which will give

natural frequencies of the system, as

( ) ( ){ } ( )4 2 2 2 2 2 4 2 2 2 2 2 2 4 4

2 2 2 2 1 1 1 2 2 1 1 2 1 2 2 2 1 1 1 2 1 24 3 4 48 3 144 0n d n dmI k l k l k l k k l m k l k l l I k l k l k k l lω ω − + − + − − =

Given data: mass of the disc, m = 5 kg, diametral mass moment inertia 02.0=dI kg-m2, shaft

lengths 0.3 m and 0.7 m. Diameter of the shaft d = 0.01 m. After substituting values k1 = 3.818×103

N/m, k2 = 3.005×102 N/m, l1 = 0.3, l2 = 0.7, m = 5 kg and Id = 0.02kgm

2, we get

4 4 9 2 125.494 10 2.4375 10 4.0967 10 0n nω ω− × + × − × =



2206.44nω = rad/sec

Exercise 3.6. Obtain the bending critical speeds of an overhang rotor system as shown in Figure E3.6.

The end B1 of the shaft is having fixed end conditions. Length of the shaft is 0.4 m and diameter is 0.1


185

m. The disc is thin and has 1 kg of mass, 0.04 kg-m2 of polar mass moment of inertia and 0.02 kg-m2

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. Use the dynamic stiffness method.

3.5 Dunkerley’s Formula

It can be used to calculate the machine natural frequencies without recourse to the numerical

methods. This method gives very crude estimation of natural frequency. From influence coefficient

method the natural frequency of the system is obtained by the following conditions

0=′A (99)

For three degrees of freedom system; above equation will give

…+

+++

−2

2333222111

3

2

1)(

1

ωωmamama (100)

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

( ) ( ) ( )2 2 2

1 2 3 11 1 22 2 33 31/ 1/ 1/ a m a m a mω ω ω+ + = + + (101)

2 2 2

1 1 2 2 3 3 11 22 33/ / / 1/ 1/ 1/m k m k m k ω ω ω= + + = + + (102)

where 2211 ,ωω and 33ω are the natural frequencies of the system (i.e. roots of the polynomial in

equation (100)) when only mass 1 2 3, or m m m is present. In most cases the fundamental frequency

1ω will be much lower than the other natural frequencies, so equation (4) may be approximated, in

general, to

B1

D1

Figure E3.6


186

2 2 2 2

1 11 22 33

1 1 1 1

ω ω ω ω= + + (103)

Dunkerley first suggested this. Equation (103) always gives a value for fundamental frequency, which

is slightly lower than the true value, by virtue of the approximation involved.

Example 3.8 Find the fundamental critical speed of the rotor system shown in Figure3.32. Take EI =

2 MN.m2 for the shaft and mass moment of inertia of disc is negligible.

3 m

1.5 m

Fixed end 80 kg 100 kg

Figure 3.32 An overhang rotor system

Solution:

F1 F2

l l/2 l/2

(a) Case 1 (b) Case 2

Figure 3.33 Overhang rotor systems with a single disc

For Figure 3.33, we have

11 223 3

1 2

3 3;

EI EIk k

l l= = with 1l l= and 2 / 2l l=


187

2 2

11 223 3

1 1 2 2

3 3;

EI EIω ω

m l m l= =

The system critical speed is given as

3 3

1 2

2 2 2

11 22

41 21 1 14.1625 10

3 3

m l m l

EI EIω ω ω−= + = + = ×

or 49.014ω = rad/sec (Fundamental frequency)

Exercise 3.7 Find the fundamental bending critical speed of the rotor system shown in Figure E3.7.

B1 and B2 are simply supported bearings and D1 and D2 are rigid discs. The shaft is made of steel with

modulus of rigidity E = 2.1 (10)11 N/m2 and uniform diameter d = 10 mm. The various shaft lengths

are as follows: B1D1 = 50 mm, D1D2 = 75 mm, and D2B2 = 50 mm. The mass of discs are: md1 = 4 kg

and md2 = 6 kg. Consider the shaft as massless and neglect the diametral mass moment of inertia of

both discs.

Finite element method is now very popular among all other methods for continuous and complex

rotor-bearing systems. In subsequent chapters we will discuss this method in detail.

B1 B2

D1 D2

Figure E3.7

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