torsional vibrations of rotor systems

8/13/2019 Torsional Vibrations of Rotor Systems

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Dr. R. Tiwari ([email protected]) 1

ByDr. Rajiv Tiwari

Department of Mechanical EngineeringIndian Institute of Technology Guwahati 781039

Under AICTE Sponsored QIP Short Term Course on

Theory & Practice of Rotor Dynamics(15-19 Dec 2008)

IIT Guwahati

TORSIONAL VIBRATIONS OF ROTOR

SYSTEMS


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INTRODUCTION

Torsional vibrations is predominant whenever there is large discs onrelatively thin shafts (e.g. flywheel of a punch press).

Torsional vibrations may originate from the following forcing

Inertia forces of reciprocating mechanisms (such as pistons in ICengines)

Impulsive loads occurring during a normal machine cycle (e.g. duringoperations of a punch press)

Shock loads applied to electrical machinery (such as a generator linefault followed by fault removal and automatic closure)

Torques related to gear mesh frequencies, turbine blade passingfrequencies, etc.

For machines having massive rotors and flexible shafts (where thesystem natural frequencies of torsional vibrations may be close to, or

within, the excitation torque frequency range during normal operation)torsional vibrations constitute a potential design problem area.

In such cases designers should ensure the accurate prediction ofmachine torsional frequencies and frequencies of any torsional loadfluctuations should not coincide with the torsional natural frequencies.

Hence, the determination of torsional natural frequencies of the system isvery important.


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Consider a rotor system as shown Fig.1

The shaft is considered as massless and it provides torsional stiffness only.

The disc is considered as rigid and has no flexibility.

If an initial disturbance is given to the disc in the torsional mode andallow it to oscillate its own, it will execute the free vibrations as shown in

Fig. 2.It shows that rotor is spinning with a nominal speed of and executingtorsional vibrations, (t), due to this it has actual speed of {+ (t)}.

It should be noted that the spinning speed remains same however angularvelocity due to torsion have varying direction over a period.

The oscillation will be simple harmonic motion with a unique frequency,which is called the torsional natural frequency of the rotor system.

Simple System with Single Rotor Mass

Fig.1a A single-mass cantilever rotor system Fig.1(b) Free body diagram of disc

Fig. 2 Torsional vibrations of a rotor


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From the theory of torsion of shaft, we have

where,Kt is the torsional stiffness of shaft,Ipis the rotorpolar moment ofinertia, kg-m2,Jis the shaft polar second moment of area, lis the length

of the shaft and is the angular displacement of the rotor. From thefree body diagram of the disc as shown in Fig. 1(b)

Equation (2) is the equation of motion of the disc due to free torsionalvibrations. The free (or natural) vibration has the simple harmonic

motion (SHM). For SHM of the disc, we have

where is the amplitude of the torsional vibration and is the torsional

natural frequency. On substituting Eqs. (3) and (4) into Eq. (2), we get

t

T GJK

l= =

External torque of disc p t pI K I = =

(1)

(2)

( ) sin nf t t=2 2

sinnf nf nf t = =

(3)

(4)

/nf t pK I = (5)

nf


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A two-disc torsional system is shownin Fig. 3. In this case whole of therotor is free to rotate as the shaftbeing mounted on frictionless

bearings.From the free body diagram in Fig.

4(a)

For free vibration, we have SHM, sothe solution will take the form

Two-Disc Torsional System

Fig.3 A two-disc torsional system

Fig. 4 Free body diagram of discs

1 21 2External torque and External torquep pI I = =

11 1 2 0p t tI K K + =

2 2 2 10p t tI K K + =

(6)(7)

2

1 1nf = 2

2 1nf = and (8)

Substituting equation (4) into equations (6) & (7), it gives

1

2

1 1 2 0p nf t tI K K + =

2

22 2 1 0p nf t tI K K + =

(9)

(10)


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In matrix form

The non-trial solution of equation (11) is obtained by taking determinant

of the matrix [K] which gives the frequency equation as

The roots of equation (12) are given as

From equation (11) corresponding to first natural frequency for = 0,

we get 1 = 2

1

2

21

22

0

0

t p nf t

t t p nf

K I K

K K I

=

[ ]{ {0K =or (11)

( )1 21 2

4 2 0p p nf p p t nf

I I I I K + = (12)

( ) ( )1 2 1 2 1 20.5

0 andnf nf p p t p pI I K I I = = + (13)

1nf

Fig .5 First mode shape

(14)


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From Eq. (14) it can be concluded that,

The first root of equation (12) represents the case when both discssimply rolls together in phase with each other as shown in Fig. 5. Itis the rigid body mode, which is of a little practical significance. Thismode it generally occurs whenever the system has free-free endconditions (for example aeroplane during flying).

From equation (11), for , we get

The second mode shape (Eq. 15) represents the case when bothmasses vibrate in anti-phase with one another.

Fig. 6 shows mode shape of two

-rotor system, showing two discs

vibrating in opposite directions.

2nf nf =

(15)( )( )

1 1 2 1 2 1 2 0t p p p p p t t K I I I I I K K

+ =

2 11 2

p pI I = or

Fig.6 Second mode shape


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From mode shapes, we have

Since both the masses are always vibrating in opposite direction, theremust be a point on the shaft where torsional vibration is not taking placei.e. a torsional node.

The location of the node may be established by treating each end of thereal system as a separate single-disc cantilever system as shown in Fig.6.

Since value of natural frequency is known (the frequency of oscillation of

each of the single-disc system must be same), hence we write

where is defined by equation (13), and are torsional stiffness of two(equivalent) single-rotor system, which can be obtained from equation

(17), as

The length l1 and l2 then can be obtained by (from equation 1)

1 2 1 1

1 2 2 2

l

l l l

= =

2 1 1 2 2

2

nf t p t pK I K I = =

1 2 1 2 2 2

2 2 andt nf p t nf p

K I K I = =

1 21 2andt tl GJ K l GJ K = = with1 2l l l+ =

(16)

(17)

(18)

(19)


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Fig. 7(a) shows a two-disc stepped shaft. In such cases the actual shaftshould be replaced by an unstepped equivalent shaft for the purpose ofthe analysis as shown in Fig. 7(b).

The equivalent torsional spring of shafts connected in series, can bewritten as

Substituting equation (1), we have

where are equivalent lengths of shaft segments having equivalent shaftdiameter d3 and le is the total equivalent length of unstepped shaft havingdiameter d3 as shown in Fig. 7(b).

System with a Stepped Shaft

1 2 3

1 1 1 1

et t t t K K K K

= + +

1 2 31 2 3 31 2 e e e ee e el l J J l J J l J J l l l= + + = + +

1 1 11 2 2 3 31 / , / , /e e e e e el l J J l l J J l l J J = = =

with

(20)

(21)


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From Fig. 7(b) and noting equations (17) and (19), in equivalent shaft thenode location can be obtained as

Fig. 7(a) Two discs with stepped shaft (b) Equivalent uniform shaft

( )1 2 1

2

e e nf pl a GJ I + = ( )3 2 22

e e n pl b GJ I + =

( )2

1/2

1 2

1 1 2 2 3 31 2

1and

p p te

n te

p p

I I KK

I I l GJ l GJ l GJ

+ =

= + +

and

where

(22)

(21)

From equation (13), we have

4 4

2 3 2 22 2

, ,64 4

ee e

Jl l J d J d

J

= = =

Since above equation is for shaftsegment in which node is assumedto be present, we can write

2 2ande ea a J J b b J J = =

Above equations can be combined as

a a

b b

=

(23)


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When there are several number of discs in the rotor system it becomes ismulti-DOF system. When the mass of the shaft itself may be significantthen the analysis described in previous sections (i.e. single or two-discsrotor systems) is inadequate to model such system, however, they could

be extended to allow for more number of lumped masses (i.e. rigid discs)but resulting mathematics becomes cumbersome.

Alternative methods are:

o Transfer matrix methods

o Methods of mechanical impedanceo Finite element methods

Transfer matrix method: A multi-disc rotor system, supported on frictionless supports, is shown in

Fig. 7. Fig. 8 shows the free diagram of a shaft and a disc, separately. Atparticular station in the system, we have two state variables: the angular

twist and Torque T.

MODF Systems


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Governing equations of motion of the whole system.

Point Matrix:

The equation of motion for the disc 2is given by

For free vibrations, angularoscillations of the disc is given by

Substituting back into equation (24),

we get

Angular displacements on the eitherside of the rotor are equal, hence

Equations (26) and (27) can becombined as

Fig. 9(a) Free body diameter of shaft section 2

2 22 2 pR LT T I =

2sin t =

2 2

2 2 sinnf nf t = =

2

2

22 2 nf pR LT T I =

2 2R L =

Fig. 8 A multi-disc rotor system

(b) Free body diagram of rotor section 2

2

2 2

1 0

1nf pR L

IT T

=

{ } [ ] { }R 2 22

SL

P S=or

where {S}2 is the state vector at station 2 and [P]2 is the point matrix for station .

(24)

(25)

(26)

(27)

(28)


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Field Matrix:

For shaft element 2 as shown in Fig. 9(a), the angle of twist is related to itstorsional stiffness and to the torque, which is transmitted through it, as

Since the torque transmitted is same at either end of the shaft, hence

Combining (29) and (30), we get

where [F]2 is the field matrix for the shaft element 2.

Now we have

where [U]2 is the transfer matrix, which relates the state vector at right of station

2 to the state vector at right of station 1.

2 1

2

T

K =

2 1L RT T=

{ } [ ] { }2 2 1L RS F S=

{ [ ] { } [ ][ ] { } [ ] {22 2 1 12 2 2R RR S P S P F S U S = = ==

(29)

(30)

(31)

(32)

22 2

1 10 1

RL

kT T =

or


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On the same lines, we can write

where [T] is the overall system transfer matrix

The overall transformation can be written as

For free-free boundary conditions,

On using equation (35) into equation(34), the second set of equation gives

On taking

In Eq. (36), t11 contains n so for each value of n different value of R4 is

obtained and using Eq. (34) relative displacements of all other stations can beobtained, by which mode shapes can be plotted.

{ } [ ] { }

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

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

1 01

2 1 2 1 02

3 2 3 2 1 03

1 0n n n n n

RR

R R

R R

R R

S U S

S U S U U S

S U S U U U S

S U S U U U S T S =

=

= =

= =

= =

11 12

21 22 0nR R

t t

t tT T

=

00

R n RT T= =

( )21 0nft = 0 0R since

( )110 01 we get nfR R t = =

(33)

(34)

(35)

(36)


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Example 1.

Obtain the torsional natural frequency of the system shown in Fig.10using the transfer matrix method. Check results with closed form

solution available. Take G= 0.81011 N/m2

Solution:

We have following properties of the rotor

The torsional stiffness is given as

Analytical method: The natural frequencies in the closed form are givenas

Mode shapes are given as

Fig.104

;

11 2 4

-60.8 10 N/m 0.6 m; (0.1) 9.82 10 m32G l J

= = = =

-611

6

0.8 109.82 10 1.31 10 Nm/rad

0.6t

GJk

l

= = =

( )1 2

2 2

1 2

6( ) 22.6 5.66 1.31 100; and 537.97 rad/sec

22.6 5.66

p p t

n n

p p

I I k

I I

+ + = = = =

1

0n

= { } { }2 0

R

=

2

537.77 rad/sn = { } { } { }

1

2

2 0 04.0

R p

p

I

I

= =

ForFor


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Example 1 contd...

Transfer matrix method:State vectors can be related between stations 0 & 1 and 1 & 2, as

The overall transformation of state vectors between 2 & 0 is given as

On substituting values of various rotor parameters, it gives

1 1 0

2 2 2 1 2 2 1 0

{ } [ ] { }

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

R

R R

S P S

S P F S P F P S

=

= =

( )

( ) ( )

2 1 12 2

1

2 1 2 2

2 2 22 2

2 0 0

2

2 2 2 2

0

1 11 0 1 0 1 01 1

1 1 110 1

1 1

1 1

Rtt

n p n p n pn p n p t

n p t t

n p n p p t n p t

kk

I I II I kT T T

I k k

TI I I k I k

= =

=

( )

( )

5 2 7

2 5 4 2 7 22 0

1 1.73 10 7.64 10

5.66 9.77 10 22.6 9.77 10 1

Rn

n n n nT T

= + +


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Since ends of the rotor are free, the following boundary conditions willapply

On application of boundary conditions, we get the following condition

Since , we have

which gives the natural frequency as

which are exactly the same as obtained by the closed form solution.

Mode shapes can be obtained by substituting these natural frequenciesone at a time into equation (A), as

Example 1 contd...

0 2 0RT T= =

2 5 421 0[ 28.26 9.77 10 ]{ } 0n nt

= + =

{ }0

0

2 5 2[9.77 10 28.26] 0n n =

1 20 and 537.77 rad/secn n = =

10n = { } { }2 0

R =

2

537.77 rad/sn

= { } { }2 04.0

R

=

which are also exactly the same as obtained by closed form solutions


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Example 2.Find torsional natural frequencies and mode shapes of the rotor systemshown in Figure 1. B is a fixed end and D1 and D2 are rigid discs. Theshaft is made of steel with modulus of rigidity G= 0.8 (10)11 N/m2 anduniform diameter d= 10 mm. The various shaft lengths are as follows:BD-1 = 50 mm, and D1D2 = 75 mm. The polar mass moment of inertia of

discs are: Ip1 = 0.08 kg-m2 and Ip2 = 0.2 kg-m2. Consider the shaft asmassless and use (i) the analytical method and (ii) the transfer matrixmethod.

Fig. 11

Solution:Analytical method: From free body diagrams of discs asshown in Fig.12, equations of motion can be written as

1

2

1 1 1 2 1 2

2 2 2 1

( - ) 0

( - ) 0

p

p

I k k

I k

+ + =

+ =

The above equations for free vibrations and they are homogeneous secondorder differential equations.

In free vibrations discs will execute simple harmonic motions.


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For the simple harmonic motion, hence equations of motion take the form

On taking determinant of the above matrix, it gives the frequency

equation as

which can be solved for , as

For the present problem the following properties are gives

1

2

2

1 2 2 1

222 2

- 0

0

p n

p n

k k I k

k k I

+ =

1 2 1 2 2

4 2

2 1 2 1 2

( ) 0p p n p p p n

I I I k I k I k k k + + + =

( )1 2 2 1 2 2 1 2

1 2

2

2 1 2 2 1 2 1 224

2

p p p p p p p p

n

p p

I k I k I k I k I k I k k k I I

I I

+ + + + =

Fig.12 Free body diagram of discs

1 2

1 21 2

1 2

2 2

1878 N/m and 523.598 N/m

0.08 kgm and 0.2 kgmp p

GJ GJ k k

l l

I I

= = = =

= =

Example 2 contd...


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Natural frequencies are obtained as

The relative amplitude ratio can be obtained as (Fig.13)1 2

44.792 rad/s and 175.02 rad/sn n

= =

2

1 2

2

21

2 2

-0.2336 for and -10.700 forp n n n

k I

k

= =

Fig.13 Mode shapes

Example 2 contd...


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Transfer matrix method

For Figure 4.14 state vectors can be related as

The above state vector at various stations can be related as

On application of boundary conditions the second equation of equation(A), we get

Fig.14 Two-discs rotor system with station numbers

{ } [ ] [ ] [ ] [ ]{ }2 2 2 1 1 0

RP F P F =

1 1

2

2 2

2 1 2 2

22 0

1 2

1 11- 1

1

n p n p

R n p

I I

k k k k

T TIpp

k k

+

=

+

2

2

0

1 2

0 0 1n p

R

Ipp T

k k

= + +

0 0RT since

Example 2 contd...


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On substituting for p, we get

which can be solved to give

It should be noted that it is same as obtained by the analytical method

Example 2 contd...

2 2

2 1

2 2

2 2

1 2 2

-1- 1 1 0

n p n p

n p n p

I II I

k k k

+ + =

2 1 2 1 1 2

2 2 1 2 1 1 22 21 1 42 4

n

p p p p p p

k k k k k k

I I I I I I

= + +


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In some machine the shaft may not be continuous from one end of themachine to the other, but may have a gearbox installed at one or morelocations. So shafts will be having different angular velocities asshown in Fig.15(a).

For the purpose of analysis the gear system must be reduced tosystem with a continuous shaft so that they may be treated asdescribed in the preceding section.

Geared Systems

Fig. 15(a) Actual system (b) Equivalent system without gearbox


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In actual system as shown in Fig. 15(a) the shaft torsional stiffness k2and rotor moment of inertia Ip2.

Let the equivalent system as shown in Fig.15(b) has the shaft torsionalstiffness keand the disc moment of inertia Ipe .

The strain and kinetic energies must be the same in both the real andequivalent systems for theoretical model to be valid.

By imagining the rotor Ip2 to be held rigidly whilst shaft 1 is rotatedthrough some angle 1 at the gearbox. The shaft 2 is rotated throughan angle

1/nat gearbox, where nis the gear ratio.

The strain energy stored in shaft 2

While applying the same input at the gear location to the equivalentsystem results in the stain energy stored in the equivalent shaft andcan be expressed as

( )1 122 2 2 12 22

E k k n = =

1 2

12 eE k=

(37)

(38)


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Equating equations (37) and (38) give

(39)

If consideration is now given to the kinetic energies of both the realand equivalent systems, which must also be equated

where 2 and e are the angle of twist of actual shaft 2 and the shaft 2in equivalent system respectively. It can be seen from Figure 15(b)that e = 1 and 1 and 2 are the angular frequencies of the shafts 1and 2, respectively. Noting that

Equation (40) can be written as

which simplifies to

2

2/ek k n=

( ) ( )1

1 1

2 2 12 22

2 2

p p ee

I I + = + (40)

( )11 2 12 1and enT k T k e = = =

2

22

11

2

2

1

1

2

p e

e

d nT d T I I

n dt k dt k

+ = +

21

2

2 22 2

12

2

p

e

I d n T d n T I

dt k dt k n

+ = +

2

2

/peI I n= (41)


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The general rule, for forming the equivalent system for the purpose ofanalysis, is to divide all shaft stiffness and rotor polar mass moment of

inertias of the geared system by n2 (where nis the gear ratio n= 2/1= geared shaft/reference). When analysis is completed, it should beremembered that the elastic line of the real system is modified (ascompared to with that of the equivalent system) by dividing the

displacement amplitudes for equivalent shaft by gear ratio nas shownin Fig.16.

Fig.16 The elastic line in the original system


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Example 3

For geared system as shown in Fig.16 find the natural frequency andmode shapes. Find also the location of nodal point on the shaft (i.e. thelocation of the point where the angular twist due to torsional vibration iszero). The shaft A has 5 cm diameter and 0.75 m length and the shaftB has 4 cm diameter and 1.0 m length. Take modulus of rigidity of the

shaft Gequals to 0.8 1011 N/m2, polar mass moment of inertia ofdiscs are IA = 24 Nm2 and IB= 10 Nm2. Neglect the inertia of gears.

20 cm diameter

10 cm diameter

GearP

air B

A

Fig.17

Solution:On taking shaft B has input shaft (orreference shaft) as shown in Fig.18 the gear

ratio can be defined as

input speed 20 rpm of reference shaftGear ratio 2

output speed 10 rpm of driven shaft

A A B B

B B A A

D T Nn

D T N

= = = = = = = = =


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The area moment of inertia and the torsional stiffness can be obtained

as4 7 4

6.136 10 m ;32

A AJ d = =

4

4

6.545 10 Nm/rad

2.011 10 Nm/rad

AA

A

B

GJK

l

K

= =

=

On replacing shaft A with reference to the shaft B by an equivalentsystem, the system will look as shown in Figure 4.19. The equivalentsystem of the shaft system A has the following torsional stiffness andmass moment of inertia properties

4 7 4 2.51 10 m

32B B

J d = =

44 2

2 2 2 2

6.545 10 241.6362 10 N m/rad and 6 Nm

2 2

A PAA PA

e

K IK I

n n

= = = = = =

which gives the equivalent length as11 7

4

0.8 10 2.513 101.2288m

1.6362 10

B

AeA

e

GJl

K

= = =

Example 3 contd..


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The equivalent stiffness of the full shaft is given as

The equivalent shaft length is given as

The natural frequency of the equivalent two mass rotor system asshown in Fig.19 is given as

Fig.19 Equivalent single shaft system

4

4 4

1 1 1 1 11.1085 10 m/N

2.011 10 1.6362 10e A Be

K K K

= + = + =

which gives 9021.2 N/meK =

42

e A B A B4e1.2288 1 2.2288 mB

A

dl l l n l l

d= + = + = + =

( )

( )

1/ 2 1/ 2

2

( ) 6 10 9021.2/9.81153.62 rad/sec

( ) 6 10 /9.81

PA PB ee

n

PA PBe

I I K

I I

+ + = = =

Example 3 contd..


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The node location can be

obtained from Fig. 20 as

The negative sign indicates that both discs are either ends of nodelocation

The absolute location of the node position is given as

Also from Fig. 20 we have which gives

Fig. 20 Mode shape and nodal point location in the equivalent system

1

2

10

1.6676

An e PB

n B PAe

l I

l I

= = = =

1 21.667n nl l=

1 22.2288

n nl l+ =

20.8358 mnl =

Let 1rad then 1.667 radB Ae

= =

Hence, 0.8333radA

A

e

n

= =

Example 3 contd..


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The mode shape and node location in the actual system is shown in

Fig. 21

Alternative way to obtain natural frequency is to use the equivalenttwo mass rotor (Fig.19) can be considered as two single DOFsystems (Fig. 22).

The stiffness and mass moment of inertia properties of the system isgiven as

Fig. 21 Mode shape and nodal pointlocation in the actual system

Fig. 22 A single DOF system

114 2

22

0.8 10 2.513 10

2.435 10 N/m and kgm0.8358 9.81

B

l PBn

e

GJ

K Il

= = = =

Example 3 contd..

E l 3 d


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It gives the natural frequency as

which is same as obtained earlier. The whole analysis can be done byreplacing shaft B with reference to shaft A speed by an equivalentsystem. For more clarity some of the basic steps are given as follows.

4

2

2

2.435 10 9154.62 rad/sec

10

l

PB

n

eK

I

= = =

Fig.23

Actual and equivalent geared systems

Example 3 contd..

E l 3 td


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It is assumed here that we are choosing reference shaft as input shaft(i.e. for present case shaft A is reference shaft hence it is assumed tobe input shaft and according the gear ratio will be obtained).

It is assumed that equivalent shaft (i.e. B) has same diameter as thereference shaft (i.e. A). The equivalent mass moment of inertia andstiffness can be written as

The total equivalent length and the torsional stiffness would be

100.5

20

B B

A A

Dn

D

= = = =

42 4 2

2 2 2

2.011 1040N m and 8.044 10 N/m(0.5)

PB PBPB Be e

I II Kn n

= = = = =

11 74

4

0.8 10 6.136 108.044 10 0.610 m

8.044 10

AB B

Be

e

e

GJK l

l

= = = =

A B

11 74 2

ee0.61 0.75 1.36 m

and

0.8 10 6.136 103.61 10 N/m

1.360

A

e

l l l

GJKe

l

= + = + =

= = =

Example 3 contd..

E l 3 td


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The natural frequencies of two mass rotor system is given by

1 0n =

1 1

2 2

2

( ) ( ) 9.81

PA PB PA PB

PA PB PA PB

e ee e

e e

I I I IK Kn

I I I I

+ + = =

and

2

1

24(24 40)9.81 3.61 10 153.65 rad/s

24 40n

+

= =

Fig.24 Equivalent two mass rotor system

The node location can be obtained as

1

2

101.667

6

PBn

n PA

eIl

l I= = =

we have

1 2 1 21.36e en nl l l l+ = + =

1 2 1 2(1.667 ) 1.36 0.85m and 0.51mn n n nl l l l+ = = =

which gives

Example 3 contd..

Example 3 contd


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The stiffness of will be (equivalent stiffness corresponding to shaft Aspeed)

The shaft stiffness corresponding to shaft B speed can be defined intwo ways i.e

On equating above equations the location of the node in the actualsystem can be obtained as

which is same as by previous method.

22

AB

en

GJK

l=

22

BB

GJK

l=

2 2

22

2

A

n

B Be

GJK n K n

l= =

2

2

2 0.84B

A

e

Jl n l

J= =

Example 3 contd..

B h d S


36/104


For the marine vessel power transmission shafts or machine tool

drives, there may be many rotor inertias in the system and gear boxmay be a branch point where more than two shafts are attached. Insuch cases where there are more than two shafts attached as shown inFig. 25 to the gearbox the system is said to be branched

For Fig.25 having branched system, state vectors for different branchcan be written as

For branch A,Taking 0A=1 as reference value

for angular displacement

since left hand end of branch A is free end

hence for free vibrations we have T0A = 0.

Equation for branch A takes the form

Branched Systems

Fig.25 Branched system

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

;nA nC A nB B C

S A S S B S S C S = = =

11 12

21 22

1

0nA

a a

T a a

=

(42)

(43)


37/104


which can be expanded as

At branch point, between shaft A and B, we have

where nAB is the gear ratio between shaft A and B. For branch B, TnB= 0,since end of branch is free.

For branch B from equation (42), and noting the condition above, we have

Above equation can be expanded as

21 11

andnA nA

T a a= =

nAoB

ABn

=

( )11 12

21 220

nA AB

nB oB

b b n

b b T

=

( ) 1211

nA ABnBoB

b n b T = +

( ) ( )

( )22 21 22210 nA AB nA ABoB oBb n b T T b b n = + =

(44)

(45)

(46)

(47)

and

(48)


38/104


At branch C, we have the following condition (nothing equation 45)

Another conditions at branch to be satisfied regarding torquetransmitted at various branches is

Equation (50) can be written as

On substituting equation (48) and (43), we get

Substituting (49) and (51) into equation (42), we get:

11nA

AC AC

oc

a

n n

= =

1 1 1

2 2 2nA nA oB nB oC nC

T T T = +

oB oC oB oC

nA nB n A nC AB AC

nA nA

T T T T T T

n n = + = +

oBoC

AB

AC nA TT n Tn

=

21 11

2

22

21oC

AB

AC

b aT n a

b n

= +

(49)

(50)

(51)

(52)

/1111 12

21 22 21 21 11 22

2

0

AC

AC AC ABnc

a nc c

c c n a b a n b n

=

+


39/104


From equation (52) second equation will give the frequency equation as

The roots of the above equation are system natural frequencies.

As before, these frequencies may then be substituted back intotransfer matrices for each station considered, where upon the statevector at each station may be evaluated. The plot of angular

displacement against shaft position then indicates the system modeshapes.

Using this method, there will not be any change in elastic line due togear ratio, since these have now already been allowed for in theanalysis.

Moreover, for the present case we have not gone for equivalentsystem at all.

For the case when the system can be converted to a single shaft theequivalent system approach has the advantage.

(53)22 21 11

22 21

22

21 11

20AC

AB

AC

AC

c b a nc an c a

n b n+ + =


40/104


Example 4Obtain the torsional critical speeds of the branched system as shown inFig. 26. Take polar mass moment of inertia of rotors as: IPA = 0.01 kg-m2, IPE= 0.005 kg-m2,IPF = 0.006 kg-m2, and IPB = IPC = IPD = 0. Takegear ratio as: nBC = 3 and nBD = 4. The shaft lengths are: lAB = lCE = lDF =1 m and diameters are dAB = 0.4 m, dCE = 0.2 m and dDF = 0.1 m. Take

shaft modulus of rigidity G= 0.8 1011 N/m2

Solution:

The branched system has the following

mass moment of inertias

For branch A the state vector at stations are related as

Fig. 26 A branched rotor system

01.0=PAI0.005PEI =

0.006PFI =

OAAnA SUS }{][}{ =

[ ] AABA PFU ][][=11 2 99

2 2

1 0 1 4.97 10 4.97 101 4.97 10

0.01 1 0.01 10 1

n

n n

= =

with

Example 4 contd


41/104


For branch B the state vector at stations are related as

Similarly, for branch C, we have

From equation (53), the frequency equation can be written as

On substitution, we get

{ } [ ] { }nB B OBS U S=

[ ]8

2 10 2

1 7.95 10[ ] [ ]

0.005 3.97 10 1B CEB

n n

U P F

= =

+ with

{ } [ ] { }nC C OC S U S=

[ ]6

2 9 2

1 1.27 10[ ] [ ]

0.006 7.64 10 1B BFC

n n

U P F

= =

+ with

22 21 1122

22

21 11

21 20BD

BC

BD

BD

c a c b a nn c a

n b n+ + =

( )2 11 2

9 2 2

9 2 2 11 2

10 2

0.006 (1 4.97 10 )4 1 7.64 10 ( 0.01 )

4

(1 7.64 10 )( 0.005 )(1 4.97 10 )4 0

9 (1 3.97 10 )

n nn n

n n n

n

+

+ =

Example 4 contd..

Example 4 contd..


42/104


which can be simplified to

The roots of the polynomial are

Natural frequencies are given as

It can be seen that the rigid body mode exist since ends of the geartrain is free.

2 10 4 19 60.04372 3.3921 10 1.21 10 0n n n

+ =

1,2,3

2 60; 135.48 10n = and91064.2

01 =n 2 11640n = 351387n = rad/s

Example 4 contd..

Damping in Torsional Systems


43/104


Damping may come from

o The shaft materialo Torsional vibration dampers.

The torsional vibration damper is a device which may be used to jointogether two-shaft section as shown in Fig. 27. It transmits a torque,

which is dependent upon of the angular velocity on one shaft relative tothe other.

Torsional dampers can be used as

a means of attenuating system vibration

and to tune system resonant frequenciesto suit particular operating conditions.

The damping in the system introduces

phase lag angles to the system

displacement and torque. The displacement and torque parameters mustnow be represented mathematically both in-phase and quadraturecomponents.

Fig. 28 shows a general arrangement of torsional MDOF rotor system

with damping. EOM of the nthrotor from free body diagram (Fig. 29) canbe written as

Damping in Torsional Systems

Fig. 27 Torsional vibration damper


44/104


And

Torques RTnand LTnmay be written in

whilst the angular displacements is

Differentiating equations (56) and (57) with

respect to time and substituting

in equations (54) and (55) leads to

n nR n L nT T C

L n R n =

(54)

(55)

Fig. 28 General arrangement of MDOF system with damping.

Fig. 29 Free body diagram of rthrotor.

1 2sin cos

n n nT T t T t = +

1 2sin cosn n nt t = +

(56)

(57)

1 1

2 2

1 1

2 2

2

2

1 0 0 0

0 1 0 0

1 0

0 1n L nn

T TI C

T TC I

=

(58)


45/104


The previous equation can be simplified as

The characteristics of the shaft element

at station nare represented in the

equation describing the torque appliedto the shaft at the location of rotor n, as

While the torque transmitted through the shaft is the same at eachends i.e.

Substituting T, and in equation (60) and on separating the in-

phase and quadrature components , we get

{ [ ] {L nR n nS P S=

Fig. 30 Free body diagram of nthshaft segment

( ) ( ) ( )1 1 1R n L n n n n n n nT T K B = = +

1L n R nT T =

( )( 1) ( 1) ( 1)1 1 1 1 2 2R n L n L n n R n n L n n R nnT T K K B B = =

( )( 1) ( 1) ( 1)2 2 2 2 1 1R n L n n L n n R n n L n n R nT T K K B B = = +

(59)

(60)

(61)

(62)

Substituting T and in equation (61) we get


46/104


Substituting T, , and in equation (61), we get

On separating in-phase and quadrature components, we get

Combining (62) and (63) we get:

which can be written as

which can be simplified as

From equations (59) and (65), we get

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

sin cos sin cosL n L n R n R n

T t T t T t T t + = +

( 1) ( 1)1 1 2 2

andL n R n L n R n

T T T T = =

1 1

2 2

1 1

2 2 1

0 0 1 00 0 0 1

0 0 1 0 0 0 1 0

0 0 0 1 0 0 0 1n nL n R n

K B K BB K B K

T T

T T

=

1[ ] { } [ ] { }n L n n R nL S M S =

1 1 1{ } [ ] [ ] { } [ ] { }L n n n R n n R nS L M S F S = =

1 1{ } [ ] { } [ ] [ ] { } [ ] { }R n n L n n n R n n R nS P S P F S U S = = =

Remaining analysis will remain same for obtaining natural frequency

& made shapes.

(63)

(64)

(65)

(66)

Torsional Vibration for Continuous Systems


47/104


Equations of motion of continuous system can be derived by (i) force

balance of a differential element (ii) Hamiltons principle (iii) Lagrangesmethod. While the force balance is a convenient approach for most of theproblems, Hamiltons principle and Lagranges method are applied for acomplex system. These two approaches need the consideration of theenergy of the system.

Hamiltons PrincipleHamiltons principle is stated as an integral equation in which the energyis integrated over an interval of time. Mathematically, the principle can bestated as

where L is the Lagrangian, Tand Vare kinetic and potential energy ofthe system and W is the work done. The physical interpretation of

equation (67) is that out of all possible paths of motion of a system duringan interval of time from t1 to t2 , the actual path will be that for which theintegral has a stationary value. It can be shown that in fact stationaryvalue will be, in fact, the minimum value of the integral. The Hamiltonsprinciple can yield governing differential equations as well as boundary

conditions

Torsional Vibration for Continuous Systems

[ ]2 2

1 1

( ) 0

t t

t t

T V W dt Ldt = + = = (67)


48/104


Lagranges Equation

Hamiltons principle is stated as an integral equation where totalenergy is integrated over an time interval.

On the other hand, Lagranges equation is differential equations, inwhich one considers energies of the system instantaneously in time.

Hamiltons principle can be used to derive the Lagranges equation ina set of generalized coordinates. Lagranges equation can be written

as

The study of torsional vibration of rotors is very important especially in

applications where high power transmission and high speed are there.As compared to transverse vibrations governing equations are muchsimpler in torsional vibration of shafts and it is identical to the axialvibration of rods

0=

ii qL

qL

dtd (68)


49/104


Governing Differential Equations

To obtain the governing differential equations thevariational principle is used that requires the system

potential energy functional.

P(y, z)

(x, t) (x, t)M(x, t)

x0

x y

zl

0

x

y

z

Figure 31 A rod under time dependent torque


50/104


The potential energy functional is obtained by considering thedisplacement field of any point P, as shown in Figure 31 and 32.

P (y, z)

P(y, z)

y

z

uz

uy

o

(x, t)

P (y, z)

P(y, z)

r

y

z

uz

uy

Q

r z

y

Figure 2 A point under torsional displacement

Th di l t fi ld t i t P i


51/104


The displacement field at a point P is

where (x, y, z) is the Cartesian coordinate of the point P, (r, , z) is thecylindrical coordinate of the point and (x, t) is the angular displacementand u(x, y, z) is the linear displacement of the point.

The strain field is given by

where etc.

The stress field is given by

),(cos);,(sin;0 txyrutxzruu zyx =====

( )( ) ( )

1

, , , , ,2

1 1 1 1

, , , , , ,2 2 2 2

0; 0; 0; 0;

; ;

xx x x yy y y zz z z yz y z z y

xy y z z y x zx z x x z x

u u u u u

u u z u u y

= = = = = = = + =

= + = = + =

x

uu xxx

=,

(69)

(70)

;0;0;0 yz GE yzzzyyxxxx ======


52/104


(71)

The strain energy is given by

(72)

The kinetic energy is given by

(73)

;;

yz

xGzGxGyG xyzxxyzx

yzzzyyxxxx

==

==

1 1 1 1 1 1

2 2 2 2 2 2

2 2 2

1 12 2

2 2

0 0

(

1

2

xx xx yy yy zz zz xy xy yz yz zx zx

V

l l

A

U dV

Gz Gy dAdx GJ dxx x x

= + + + + +

= + =

( )2 2withA

J y z dA= +

{ } 12 2 2 220 0

1

2

l l

x y z

A

T u u u dAdx J dx

= + + =

where etc and velocity components noting the equation (69) candu

u x=


53/104


where , etc. and velocity components, noting the equation (69) can

be written as (74)

The external work done is given by

(75)

The Lagrangian is = T (U+W). The following integration can beobtained

(76)

From the Hamiltons principle, = 0 , Eq.(76) becomes

which can be written, by performing the integration by parts, as

dtux

),();,(;0 txyutxzuu zyx ===

1 *

0 020

( )

l

W M M x x dx = +

( )2 2

1 1

1 12 2 *, 0 02 20

( )t t l

xt t

T U W dt J GJ M M x x dxdt = =

2

1

*, , 0 000 ( )

t l

x xt J GJ M M x x dxdt = =

2 2 2 *l t t l t l


54/104


(77)which can be written, again by performing integration by parts, as

(78)

which can be further simplified as

(79)

Since are variations of angular displacement andthey are arbitrary, so

and (80,81)

2 2 2

1 1 1

*

, 0 00 0 0

( ) ( ) ( ) 0l t t l t l

xt t t

J dtdx GJ dxdt M M x x dxdt

t x

+ =

22 2 2

1 1 11

2

1

, ,0 0 0

0

*

0 00

( ) ( ) ( ) ( )

( ) 0

ltl l t t t l

x xxt t tt

t l

t

J dx J dtdx GJ dt GJ dxd

x

M M x x dxdt

+

+ =

{2

1

*

, 0 0 , 00( ) ( ) ( ) 0

l t l

xx xt

J GJ M M x x dtdx GJ + =

l and,, 0

*

, 0 0( ) ( ) 0xxJ GJ M M x x = 0)(

0, =

l

xGJ

0

Equations (80) and (81) represent equations of motion and boundary


55/104


conditions respectively.

Finite Element FormulationFor the finite element solution let the approximate solution for an elementbe, in the form of a general polynomial, defined as

(82)where , i= 1, 2, , rare called shape functions

Galerkin Method

On substituting equation (82) into equation (80), the residue as,

(83)

{ }

1

( )2( ) 2

1 2

( )( )

( , ) ( ) ( ) ( ) ( ) ( )

( )

nee

r

r

tt

x t a b cx N x N x N x N x t

t

= + + + = =

)(xNi

( )e

R

( ) ( ) ( ) *

, 0 0( , ) ( ) ( )e e e

xxR J GJ M x t M t x x =

On minimising the residue by using the Galerkin method, we havel


56/104


(84)On substituting equation (83) into equation (84), we get

(85)

by integrating by parts of second term in equation (85), so as to give

(86)

The interpolation function of linear form is given by

(87)

where is the dependent variableaand bare constants to be determinedby element end conditions as shown in Fig. 33. (C0 of continuity and C1 of

compatibility conditions).

ridxRN

l

e

i ,,2,1;00

)(==

( ) ( ) *

, 0 0

0

( , ) ( ) ( ) 0 1, 2, ,

l

e e

i xxN J GJ M x t M t x x dx i r = =

( ) ( ) ( ) *

, , , 0 00

0 0 0

( , ) ( ) ( ) 0

1, 2, ,

l l lh

e e e

i i x i x x iJN dx GJN GJN dx N M x t M t x x dx

i r

+ + =

=

bxae +=)(


57/104


At any general element (e) the following end condition exists

at ithnode x= 0 (e) = ; atjthnode x= l (e) = j (88)

On application of equations (88) into equation (87) constants can beobtained as

(89)

Noting equation (82) the shape function can be written as

(90)

i

lba iji /; ==

{ })()( )()(

neetxN =

i

Fig.33 A beam element with end conditions

with (91)( )


58/104


( )

where Niand Nj are interpolation functions corresponding to ith andjthnodes,

respectively.

Checking of Compatibility Requirement:

In Figure 34 the nodejis the common node of element (e) and (e+1). For

element (e), we can write equation (90), as

(92)

{ } ( )( )

; ; 1 / ; / ne i

i j i j

j

N N N N x l N x l

= = = =

Fig. 34 Two neighbouring elements

=j

ielxlx

)/()/1()(

Similarly for the element (e+1), we can write


59/104


(93)

We will check the compatibility requirements at common nodej.

For x= l in element (e), equation (92), we get

(94)

For x= 0 in element (e+1), equation (93), we get

(95)

To verify whether the present interpolation function gives compatibility ofhigher order, on taking first derivative of the equation (92) with respect tox, we get

(96)

=+

k

je

lxlx

)/()/1()1(

jhx

e ==

)(

jx

e ==

+

0

)1(

{ })(

,

)(

, )( ne

x

e

x tN =

with


60/104


(97)

For element (e), we can write equation (96), as

(98)

Similarly for the element (e+1), we can write

(99)

For x= lin element (e), equation (98), we get

(100)

For x= 0 in element (e+1), equation (98), we get

(101)

{ } ( )( ), , , , ,; ; 1/ ; 1/ ne ix i x j x i x j xj

N N N N l N l

= = = =

=

j

iex ll

)/1()/1()(,

=

+

k

je

x ll

)/1()/1()1(

,

( ) lijlxe

x /

)(

, ==

( ) ljkx

e

x /0

)1(

, ==

+

Th l t diti th th l t


61/104


The completeness condition ensures the convergence as the element

size is decreased and in the limit it should converge to the exact valuefor infinitesimal size element.

Equation (86) can be written in vector notation as

(102)

On substituting equation (90) into equation (102), we get

(103)

{ } { } { } { }( ) ( ) ( ) *, , , 0 000 0 0

( , ) ( ) ( )

l l lh

e e e

x x xJ N dx GJ N dx GJ N N M x t M t x x dx + = + +

{ } { } { } { }

{ }

( ) ( )

, ,

0 0

( ) *, 0 0

00

( , ) ( ) ( )

l l

ne ne

x x

hl

i ex

j

J N N dx GJ N N dx

NGJ N M x t M t x x dxN

+ =

+ +

Noting from equation (91) that Ni is having 1 and 0 value at node 1 and 2

respectively and N is having 0 and 1 value at node 1 and 2 respectively


62/104


respectively, and Nj is having 0 and 1 value at node 1 and 2 respectively.

On substituting these values in equation first term of right hand side of

equation (103), we get

(104)

which can be simplified to standard finite element formulation of

equations of motion as

(105)

with

(106)

{ } { } { } { } { }

( )

,0( ) ( ) *

, , 0 0( )

0 0 0,

0

( , ) ( ) ( )0

el l l

xxne ne

x xe

xx l

GJ

J N N dx GJ N N dx N M x t M t x x dxGJ

=

=

+ = + +

[ ] { } [ ] { } { } { }ERneenee

TTKM +=+ )()()()(

[ ] { }2

( )

2

0 0

2 1

1 26

l le i i j

j i j

N N N JlM J N N dx J dx

N N N

= = =

[ ] { }2

( ) , , , 1 1l l

e i x i x j xN N N GJK GJ N N dx GJ dx

= = =


63/104


(107)

(108)

where is the element mass matrix, is the element stiffness matrix,

is the reaction element torque vector, and is the externalelement torque vector. Matrices and are called consistent mass

and stiffness matrices.

From strength of materials pure torsion theory we have

{ ET

[ ] { }, , 2, , ,0 0

1 1

j

x x

j x i x j x

K GJ N N dx GJ dxN N N l

= = =

{ }

=

=

=

=

lx

e

x

x

e

x

R

R

RGJ

GJ

T

TT

)(

,

0

)(

,

2

1

{ } { } *0 00

( , ) ( ) ( )

l

ET N M x t M t x x d x = +

[ ]( )e

K

T

GJ l

=

[ ]( )e

M

[ ]( )e

M [ ]( )e

K

{ }RT

Rayleigh-Ritz Method


64/104


Rayleight-Ritz method can be also used to develop finite elementformulation from the functional i.e. the potential energy, equation (76), ofthe system. On substituting equation (82) into equation (76), getfunctional for the element

(109)

minimising the functional for element, , is equivalent to

(110)

Hence, for a typical node i, on substituting the equation (109) into the


2

1

1 1( ) ( )2 ( )2 ( ) * ( ), 0 02 20

( )t le e e e ext

J GJ M M x x dxdt = ( )e

0)(

=

i

e

ri ,,2,1 =

( )2

1

( )( ) ( ) ( ),( ) ( ) *

, 0 00

0 ( )

ee e et l

xe e

xt

i i i i

J GJ M M x x dxdt

= = +

0

On performing integration of parts of first term in left hand side of theabove equation, we get


65/104


0

On substituting equation (14) into the above equation, we get

The above equations can be written in the matrix form as

which gives the standard finite element formulation of equations of

motion, as

( )2

2

1

1

( )( ) ( ) ( ),( ) ( ) ( ) *

, 0 00 0

( ) 0t ee e e

l t lxe e e

xt

i i i it

J dx J GJ M M x x dxdt

+ + + =

{ } { } ( )2

1

( ) ( ) *

0 00

( ) 0net l ne i

i it

dNdNJ N N GJ M M x x N dxdt

dx dx

+ + + =

ri ,,2,1 =

{ } { } { } ( ){ }2

1

( ) ( ) *

0 0

0 0 0

( ) 0net l l lne

t

dN dN J N N dx GJ dx M M x x N dx dt

dx dx

+ + + =

(111)[ ] [ ] { } { } { }( )( ) ( ) ( ) ( ) ( )nee e ne ne ne

R EM K T T + = +


66/104


where is the element mass matrix, is the element stiffness matrix

is the elemental reaction torque vector and is the elementalexternal torque vector.

Assembled System EquationsOnce the elemental equation of the form as equation (105) has beenderived next step is to obtain the system equation by assembling all such

elements of the system as shown in Figure 35,

[ ] [ ] { } { } { }R E

[ ]( )e

M [ ]( )e

K

{ }( )ne

ET

1 2

h

1 2

x

1 2

1 2 3

(2)(1)

(b) Beam is discretised into two elements

Figure 35. (a) A cantilever beam

{ }( )ne

RT

The element finite element equation for the first element can be written, bynoting equations (105) to (108) and for first element i= 1 andj=2, as


67/104


(112)

where and are reaction forces at node 1 and at node 2 respectively.

and represent equivalent external force at node 1 and 2 of element(1). Similarly, for the second element i= 2 and j= 3, hence the elementfinite element equation can be written as by considering the equation (105)

(113)

Now equations (112) and (113) can be assembled in the following form

(114)

+

=

+

)1(

)1(

2

1

2

1

2

1

2

1

1111

2112

6 E

E

R

R

TT

TT

lGJJl

1RT

2RT

)1(

1ET )1(

2ET

2 2

3 3

(2)

22

(2)33

2 1 1 1

1 2 1 16

R E

R E

T TJl GJ

Tl T

+ = +

1 1

2 2 2 2

3 3

(1)

1 1

(1) (2)2 2

(2)3 3

2 1 0 1 1

1 2 2 1 1 1 1 16

0 1 2 0 0 1

R E

R R E E

R E

T T

Jl GJ T T T T l

T T

+ + + = + +

Application of Boundary Conditions:

For the present illustration since node 1 is fixed and node 3 is free, hence


68/104


the following conditions are specified by the problem

(115,116)

On applying the boundary conditions, i.e. equations (115) and (116) into


(117)

On taking last two equations, we get

(118)

The equation (118) can be solved if external forces are specified

1 1 0 a n d = =

30RT =

++

=

+

+

+)2(

)2()1(

)1(

3

2

3

2

3

22

11

0

0

0

100

1111

0110

210

1221

012

6E

EE

ER

T

TT

TT

l

GJJl

+

=

+

)2(

)2()1(

3

2

3

2

3

22

11

12

21

14

6 E

EE

T

TT

l

GJJl

Free Torsional Vibration

For free vibration, the solution can be assumed of the following form


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so that (119,120)

where is the amplitude of vibration, is the natural frequency and

.On substituting equations (119) and (120) into homogeneouspart of equation (118), we get

(121)

Equation (121) is the standard eigen value problem. For non-trivial solutionthe following determinant should be zero, i.e.

(122)

tj net

t

=

3

2

3

2

)(

)( 222

33

( )

( )nj t

n

te

t

=

n

1j =

=

+

0

0

11

12

21

14

6 3

22

l

GJJln

0

33

33

22

22

22

=

Jl

l

GJJl

l

GJ

Jl

l

GJJl

l

GJ

nn

nn

On taking determinate of equation (122), we get a polynomial and it iscalled characteristics equation or frequency equation of the following form


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(123,124)

Equation (123) is a quadratic in and can be solved as

(125)

Note that only the positive sign is considered in equation (125), sinceotherwise the natural frequency will be complex in nature.

Example 5Determine the natural frequencies and mode shapes for the rotor systemas shown in Figure 5. Neglect the mass of the shaft and assume discs arehaving lumped masses.

( ) 077

102

2222 =+

ba

ba

nn / ; / 6with a GJ l b Jl= =

2

n

+

+=

2

22

2

77

104

7

10

7

10

b

a

b

a

b

a

b

an

mass less shaft

Example 5 contd..


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Figure 36 Two disc rotor system

Since the present problem contains only lumped masses, let us discretisethe shaft into one element as shown in Figure 6. The finite elementequation can be written, noting the equation (38), as

(126)

lIp1

G J

Ip2

mass less shaft

=

+

2,

1,

2

1

2

1

11

110

0

2

1

x

x

p

p

GJ

GJ

l

GJI

I

Solution:

Ip1 Ip2Example 5 contd..


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Figure 36 Discretised single element Two disc rotor system.

Since for the present problem both the nodes are free, the following endconditions apply

(127)

For simple harmonic motion, we have the following relation

(128)On substituting equations (127) and (128) into the equation (126), we get

(129)

l

1 2

(1)

1 2

021 == TT 1 2. 0i e GJ GJ = =

=

2

12

2

1

n

=

0

0

2

1

2

2

1

1

pn

pn

Il

GJ

l

GJl

GJI

l

GJ

For non trivial solution of the equation (129) the following determinate

Example 5 contd..


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For non-trivial solution of the equation (129), the following determinateshould be zero

(130)

;

which gives a frequency equation, it gives two torsional natural frequenciesand are given as

(131)

which are exactly the same as which we obtained by the closed form

solution. Mode shapes can be obtained for these natural frequencies fromequation (129).

0

1

1

2

2

=

pn

pn

Il

GJ

l

GJl

GJI

l

GJ

( ) 02121

22 =

+ ppppnn II

l

GJII

( )1 21 2

0 p p

p p

k I Ip and pI I

+= =


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76/104



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78/104



79/104



80/104



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82/104



83/104



84/104



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(rad/s)


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Geared Element for Branched Systems

Gear 1

g1

1 Gear 1


88/104


Figure 37 Gear Element

Figure 7 shows the example of branched systems, Gear element.

for no slip condition (132,133)

from the state vector of the system.

Gear 2

g1

g22= 1/n 1

1

2

Gear ratio

=

2 1

1g g

n

=

2 1 2Since is defined in terms of we can eliminateg g g

Gear 2

Shaft for gear 2

The equivalent inertia force of the gear pair w. r. t shaft reference 1 is

(134)21 1

2g

g g

II

+


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The stiffness matrix of shaft element connected to gear 2will have to bemodified, since the angular deflection of the left hand side of that element is

now (see Figure 38.)

The P.E of the shaft element is

(135)

n

1

1

gn

-1/ng1= 1 2

(2)(1)K1

Figure 38 Equivalent Gear Element

( )

1

2

1 1 2 1

2

1 2

12

1.

2

g

U K

Kn

=

= +

The external work done is give as

(136)1

1 1 1 2 2 1 2 2

g

W f f f f


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On applying Lagrangian equation,

(137)

which is equations of motion (stiffness matrix) of the gear element.

{ }

1

1 1

1

1 11 1

22

1 211

1 2

2

0

.

0

g

g g

g

U K K K

n n n nF

KU KKnn

= + = =

= + =

Example 6.Obtain the torsional critical speeds of the branched system as shown in

Figure 8. Take polar mass moment of inertia of rotors as: IPA = 0.01 kg-m2, IPE = 0.005 kg-m2,IPF = 0.006 kg-m

2, and IPB = IPC = IPD = 0. Takegear ratio as: nBC= 3 and nBD= 4. The shaft lengths are: lAB= lCE= lDF= 1m and diameters are dAB = 0.4 m, dCE = 0.2 m and dDF = 0.1 m. Take

shaft modulus of rigidity G = 0.8 1011 N/m2.

1 1 1 2 2 1 2 2W f f f f n = =

C

E

0.2

m

(2)

Branched power system.

kg m2

0060=PDI

Example 6 contd..


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A

(1)

0.4

m

D

B

(3)

0.1m

F

E(2)

Figure 38

kg-m2kg-m2

kg-m2

kg-m2

kg-m2

kg-m2

01.0=PAI

005.0=PEI

006.0=PFI

005.0=PB

I

006.0=PCIFigure 8

006.0=PDI

Gear ratio is defined as

Similarly,

/ ; /BC B C C B BCn n = =

/D B BDn =

Element (1)

Example 6 contd..


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(A)

Element (2)

To find stiffness matrix we will consider P.E., from equation (A)

=

+

B

A

B

A

B

A

B

A

T

T

kk

kk

I

I

11

11

0

0

C C(2)

2

2

2

22 )/(2/1)(2/1 nkkU BECE +==

2/ BCBC nI PEI

Figure. 39 Element (1)


2

2, 2 2 2, 2 2( / ) ( / ) ; ( / )

B EBC B BC E Bc B EU k n k n U k n k = + = +

Now,Example 6 contd..


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2

(2) 2 2

2

2 2

22

2 2

2 2

/ /

/

/ // 0

/0

BC BC

BC

B C BCBC BCC BC B

E EBCE E

k n k nk

k n k

T nk n k nI n

Tk n kI

=

+ =

Since

Element (3)

BCCBC nTT /=

D F(3)

2/ BCPD nI PFI

Similar to element (2) we can write

=

+

F

BDD

F

B

BD

BDBD

F

B

F

BdD

T

nT

knk

nknk

I

nI /

/

//

0

0/

33

2

2

3

2


(B)

(C)

1 12 22 2

0 00 0 0A A AA

k kI T

k k k k k k

The assembled equations of motion can be written as

(D)

Example 6 contd..


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2 22 2

1 1 2 3 2 3

2 2

3 3

/ / / /0 ( / / ) 0 0 0

0 / 00 0 0

0 / 00 0 0

BC BD BC BDB C BC D BD BB

BCE E EE

BDE F FF

k k k n k n k n k nI I n I n

k n kI T

k n kI T

+ ++ + + =

Now to have eigen value problem (EVP),

we haveN/m; N/m; N/m

02 = MK

6

1 10062.201 =k6

2 10566.12 =k3

3 10398.785 =k

610

785398.0019635.00

0566.1218867.40

19635.018867.4507.202062.201

00062.201062.201

= k

=

006.0000

0005.000

00007167.00

00001.0

M

(D)

It should be noted that in the assembled form second row in the torquecolumn is zero, since we have the following condition

/ / 0B c BC D BDT T n T n =

(E)

(F)

and

Using equations (F), the natural frequency can be obtained as

rad/s; rad/s;51011560 0

Example 6 contd..


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rad/s; rad/s;

rad/s rad/s54 101472.2 =n

5

3 105085.0 =n

5

2 101156.0 =n01 =n

Relative amplitude of angular displacements (i.e. mode shapes)can beobtained by substituting these natural frequencies one at a time in theequation (D). It should be noted that angular displacements obtainedfrom equation (D) are actual displacements and no use of gear ratio isrequired.

A multi-cylinder reciprocating machine contains many reciprocating androtating parts such as pistons connecting rods crankshafts flywheels

Modelling of reciprocating machine systems


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rotating parts such as pistons, connecting rods, crankshafts, flywheels,dampers and cracks.

The system is so complicated that it is difficult, if not impossible, to

undertake an exact analysis of its vibrational characteristics.

The actual system is characterised by the presence of unpredictableeffects like variable inertia, internal dampings, misalignments in thetransmission units, uneven firing order etc.

The analysis can be best carried out, by lumping the inertias of rotatingand reciprocating parts at discrete points on the main shaft. The problemthen reduces to the forced torsional vibration study of an Ndisc system

subjected to varying torques at different cylinder points.

The crankshaft and the other drive or driven shafts are generally flexiblein torsion, but have low polar moments of inertia, unlike in the case ofsome large turbines or compressors.

On the other hand, parts mounted on the shafting, like the damper,

flywheel, generator etc. are rigid and will have very high polar massmoments of inertia


97/104


flywheel, generator etc. are rigid and will have very high polar massmoments of inertia.

The system containing the crankshaft, coupling, generator/ auxiliary driveshaft/ other driven shaft like pumps, and mounted parts can then be

reduced to a simple system with a series of rigid rotor (representing theinertias) connected by the massless flexible shafts as shown in Figure 4.42

Figure 4.42

Polar mass moments of inertia

Determination of polar mass moments of inertia is a straightforward

matter for rotating parts, however, it is not quite so simple in the case ofi ti t


98/104


atte o otat g pa ts, o e e , t s ot qu te so s p e t e case oreciprocating parts.

Consider the piston shown in two different positions in Figure 4.43, andimagine the crankshaft with a polar mass moment of inertia, Irot, to be

non-rotating but executing small torsional oscillations.

In first case there is no motion for the piston, with small oscillations ofthe crank and hence the equivalent inertia of the piston is zero.

Figure4.43 Equivalent mass moment of inertia of piston

Whereas in second figure, the piston has practically the sameacceleration as that of the crank pin and the equivalent inertia is mrecr2, where mrec is the mass of the reciprocating parts and r is the


99/104


2, rec p g pcrank throw or radius. Hence, the total polar mass moment of inertiavaries from Irot to Irot+ mrecr

2, when the crankshaft is rotating.

The inertia of connecting rod can be taken care of by consideringdynamic equivalent two masses one at piston & other at crank pin. Itwill contribute to both Irot and mrec by small amount.

We consider as an approximation the system to have an averageinertia given by

where ris crank radius.

1

2

2rot recI I m r= +

Torsional stiffness of shafts connecting the rotors

In determining the torsional stiffness of the shafts connecting the rotors,

the main difficulty arises from the crank webs.


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y

Considering an idealisation of a crank throw into an ordinary shaft havingthe same flexibility as shown in Figure 4.44. Through this idealisation is

physically possible, but the calculations involved are extremely difficult.

This is because the crank webs are subjected to bending and the crankpin to twisting, if the main shaft is subjected to twisting.

Moreover the beam formulae, if used will not very accurate, because ofshort stubs involved rather than long beams usually considered.

Further torques applied at the free end also give rise to sidewise

displacement, which is prevented in the machine.

For high-speed lightweight engines, the crank webs are no morerectangular blocks and application of the theory becomes extremelydifficult.

Fig 44 Equivalent length of a crank


101/104


Because of this uncertainties in analytical calculation to estimate thetorsional stiffness of crank throws, several experiments have beencarried out on a number of crank shafts of large slow speed engines,

In general the procedure that is applied to reduce the reciprocating

machine system to a mathematical model, is to use a basic diameter,which corresponding to the journal diameter of the crankshaft. Thetorsional stiffness are all calculated based on the basic diameter,irrespective of their actual diameter.

For the end rotors (i.e. the generator rotor) compute the stiffness of theshaft from the coupling up to the point of rigidity.

In case where one part of the system is connected to the other partthrough gears, or other transmission units, it is convenient to reduce all

the inertias and stiffness to one reference speed.

Torsional vibration in reciprocating machinery

Finite element methods, which is described in subsequent sections,

could be used for more accurate analysis of such complex system.


102/104


Once the mathematical model is developed, it can be used in illustratingthe critical speed calculations, forced vibration response, and thecoefficient of cyclic irregularity

Torsional oscillation in the crankshaft and in the shafting of drivenmachinery is vibration phenomenon of practical importance in the designof reciprocating engines.

The average torque delivered by a cylinder in a reciprocating machine,is a small fraction of the maximum torque, which occurs during the firingperiod. Even though the torque is periodic the fact that it fluctuates soviolently within the period, constitutes one of the inherent disadvantagesof a reciprocating machine, from the dynamics point of view, ascompared with a turbine where the torque is practically uniform.

It is possible to express the torque by a reciprocating engine into itsharmonic components of several orders of the engine speed, and theseharmonic components can be excite the engine driven installations intoforced torsional vibrations.

It is a commonly known fact that failures can occur in reciprocatingmachine installations, when the running speed of the engine is at or


103/104


, g p gnear a dominant torsional critical speed of the system.

High dynamic stresses can occur in the main shafting of such engineinstallations and to avoid these conditions, it is essential that thetorsional vibration characteristics of the entire installation be analysedbefore the unit is put into operation.

Any analysis of torsional vibration characteristics of reciprocatingmachinery should finally predict the maximum dynamic stresses ortorque developed in the shafting & couplings of the system, as

accurately as possible, so that they can be compared with thepermissible values, to check the safety of installation.


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torsional vibrations of rotor systems

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