Torsional Vibrations of Rotor

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


INTRODUCTIONTorsional vibrations is predominant whenever there is large discs on relatively 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 IC engines)

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

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

� Torques related to gear mesh frequencies, turbine blade passing frequencies, etc.

For machines having massive rotors and flexible shafts (where the system 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 load fluctuations should not coincide with the torsional natural frequencies.

Hence, the determination of torsional natural frequencies of the system is very important.


Consider a rotor system as shown Fig.1The 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 and allow 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 ω and executing torsional vibrations, θ(t), due to this it has actual speed of {ω + θ(t)}. It should be noted that the spinning speed remains same however angular velocity 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


From the theory of torsion of shaft, we have

where,Kt is the torsional stiffness of shaft,Ip is the rotor polar moment of inertia, kg-m2,J is the shaft polar second moment of area, l is the length of the shaft and θ is the angular displacement of the rotor. From the free 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 torsionalnatural frequency. On substituting Eqs. (3) and (4) into Eq. (2), we get

t

T GJK

lθ= =

External torque of disc p t pI K Iθ θ θ= � − =� ��

(1)

(2)

ˆ( ) sin nf� t � � t=2 2ˆ sinnf nf nf� � � � t � �= − = −��

(3)(4)

/nf t pK Iω = (5)

θ̂ nfω


A two-disc torsional system is shown in Fig. 3. In this case whole of the rotor is free to rotate as the shaft being mounted on frictionless bearings.From the free body diagram in Fig. 4(a)

For free vibration, we have SHM, so the 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 Iθ θ= =� ��

11 1 2 0p t tI K Kθ θ θ+ − =��

2 2 2 1 0p t tI K Kθ θ θ+ − =��(6)(7)

21 1nf� � �= −�� 2

2 1nf� � �= −��and (8)Substituting equation (4) into equations (6) & (7), it gives

1

21 1 2 0p nf t tI K Kω θ θ θ− + − =

2

22 2 1 0p nf t tI K Kω θ θ θ− + − =

(9)(10)


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 θ1 = θ2

1

2

21

22

00

t p nf t

t t p nf

K I K

K K I

ω θθω

� �− − � � � �=� �

− − � ��

[ ]{ } { }0K θ =or (11)

( )1 21 2

4 2 0p p nf p p t nfI I I I Kω ω− + = (12)

( ) ( )1 2 1 2 1 2

0.50 andnf nf p p t p pI I K I Iω ω � �= = + �

(13)

1nfω

Fig .5 First mode shape

(14)


From Eq. (14) it can be concluded that, The first root of equation (12) represents the case when both discs simply rolls together in phase with each other as shown in Fig. 5. It is the rigid body mode, which is of a little practical significance. This mode it generally occurs whenever the system has free-free end conditions (for example aeroplane during flying).

From equation (11), for , we get

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

Fig. 6 shows mode shape of two-rotor system, showing two discs vibrating in opposite directions.

2nf nfω ω=

(15)( )( )1 1 2 1 2 1 2

ˆ ˆ 0t p p p p p t tK I I I I I K Kθ θ� �− + − = �

2 11 2ˆ ˆ

p pI Iθ θ = −or

Fig.6 Second mode shape


From mode shapes, we have

Since both the masses are always vibrating in opposite direction, there must be a point on the shaft where torsional vibration is not taking place i.e. a torsional node. The location of the node may be established by treating each end of the real 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 lθ θ θ

θ= � =

2 1 1 2 2

2nf t p t pK I K Iω = =

1 2 1 2 2 2

2 2 and t nf p t nf pK I K Iω ω= =

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

(16)

(17)

(18)

(19)


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

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

Substituting equation (1), we have

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

System with a Stepped Shaft

1 2 3

1 1 1 1

et t t tK 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)


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

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

( )1 2 1

2e e nf pl a GJ Iω+ = ( )3 2 2

2e e n pl b GJ Iω+ =

( )2

1/2

1 2

1 1 2 2 3 31 2

1and

p p ten te

p p

I I KK

I I l GJ l GJ l GJω

� �+� � == � � + +� � �

and

where

(22)

(21)

From equation (13), we have

4 42 3 2 22

2

, ,64 4

ee e

Jl l J d J d

Jπ π= = =

Since above equation is for shaft segment 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 asa ab b

′=

′ (23)


When there are several number of discs in the rotor system it becomes is multi-DOF system. When the mass of the shaft itself may be significantthen the analysis described in previous sections (i.e. single or two-discs rotor 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 impedance o 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. At particular station in the system, we have two state variables: the angular twist θ and Torque T.

MODF Systems


� Governing equations of motion of the whole system.� Point Matrix:

The equation of motion for the disc 2 is given by

For free vibrations, angular oscillations of the disc is given by

Substituting back into equation (24), we get

Angular displacements on the either side of the rotor are equal, hence

Equations (26) and (27) can be combined as

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

2 22 2 pR LT T I θ− = ��

2ˆsin tθ θ ω=

2 22 2

ˆ sinnf nftθ ω θ ω ω θ= − = −��

2

222 2 nf pR LT T Iω θ− = −

2 2R Lθ θ=

Fig. 8 A multi-disc rotor system

(b) Free body diagram of rotor section 2

2

2 2

1 01nf pR L

IT T

θ θω

� �� = � �−� � � � �

{ } [ ] { }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)


� Field Matrix:For shaft element 2 as shown in Fig. 9(a), the angle of twist is related to its torsional 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 12

TK

θ θ− =

2 1L RT T=

{ } [ ] { }2 2 1L R

S F S=

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

(29)

(30)

(31)

(32)

22 2

1 10 1

RL

k

T T

θ θ� � � � � �= � �

� � � � �or


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 ωn so for each value of ωn different value of Rθ4 is obtained and using Eq. (34) relative displacements of all other stations can be obtained, 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

θ θ� �� = � �

� � � � �

0 0R n RT T= =

( )21 0nft ω = 0 0Rθ ≠since

( )110 01 we get nfR R tθ θ ω= =

(33)

(34)

(35)

(36)


� Example 1.Obtain the torsional natural frequency of the system shown in Fig.10 using the transfer matrix method. Check results with closed formsolution available. Take G = 0.8×1011 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 given as

Mode shapes are given as

Fig.10 4;

11 2 4

-60.8 10 N/m 0.6 m; (0.1) 9.82 10 m32

G l Jπ= × = = = ×

-611

6

0.8 109.82 10 1.31 10 Nm/rad

0.6t

GJk

l×= = × = ×

( )1 2

2 21 2

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

22.6 5.66p p t

n np p

I I k

I Iω ω

+ + ×= = = =

×

10nω = { } { }2 0

Rθ θ= 2537.77 rad/snω = { } { } { }1

2

2 0 04.0R p

p

I

Iθ θ θ= − = −ForFor


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 22 0 0

2

2 2 2 20

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

kkI I II I kT T T

I k k

TI I I k I k

θ θ θω ω ωω ω

ω θω ω ω ω

� �� = = � � � � � � � �� − − −− −� � � � � � ��

� �− � �� = � �− − − − � � �

( )( )

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

ωθ θω ω ω ω

− −

− −

� �− × ×� � � �� = � �− + × − × +� � � � �


• Since ends of the rotor are free, the following boundary conditions will apply

• 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 frequencies one 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 ω ω θ−= − + × =

{ }00θ ≠

2 5 2[9.77 10 28.26] 0n nω ω−× − =

1 20 and 537.77 rad/secn nω ω= =

10nω = { } { }2 0

Rθ θ=

2537.77 rad/snω = { } { }2 0

4.0Rθ θ= −which are also exactly the same as obtained by closed form solutions


� Example 2.Find torsional natural frequencies and mode shapes of the rotor system shown in Figure 1. B is a fixed end and D1 and D2 are rigid discs. The shaft is made of steel with modulus of rigidity G = 0.8 (10)11 N/m2 and uniform 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 ofdiscs are: Ip1 = 0.08 kg-m2 and Ip2 = 0.2 kg-m2. Consider the shaft as massless and use (i) the analytical method and (ii) the transfer matrix method.

Fig. 11

Solution:Analytical method:• From free body diagrams of discs as shown 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

θ θ θ θ

θ θ θ

+ + =

+ =

��

��

• The above equations for free vibrations and they are homogeneous second order differential equations.

• In free vibrations discs will execute simple harmonic motions.


• 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

21 2 2 1

222 2

- 00

p n

p n

k k I k

k k I

ω θθω

� �+ − � � � �=� �

− − � ��

1 2 1 2 2

4 22 1 2 1 2( ) 0p p n p p p nI I I k I k I k k kω ω− + + + =

( )1 2 2 1 2 2 1 2

1 2

2

2 1 2 2 1 2 1 224

2p p p p p p p p

np p

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

I Iω

+ + ± + + −=

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 GJk k

l l

I I

= = = =

= =

Example 2 contd...


• 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 nω ω= =

2

1 2

221

2 2

-0.2336 for and -10.700 forp n

n n

k I

k

ωθ ω ωθ

= =

Fig.13 Mode shapes

Example 2 contd...


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 Fθ θ=

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

ω ω

θ θω

� �� − +� �� = � �

� �� − +� ��

2

2

01 2

0 0 1n pR

Ipp T

k k

ω� �= × + − +� ��

� �0 0RT ≠since

Example 2 contd...


• 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 22 2

1 2 2

-1- 1 1 0n p n p

n p n p

I II I

k k k

ω ωω ω

� �� − + − + =� ��

2 1 2 1 1 2

2 2 1 2 1 1 22 21 14

2 4np p p p p p

k k k k k kI I I I I I

ω� � � �

= + ± + −� � � ��


In some machine the shaft may not be continuous from one end of the machine to the other, but may have a gearbox installed at one or more locations. So shafts will be having different angular velocities as shown in Fig.15(a). For the purpose of analysis the gear system must be reduced to system with a continuous shaft so that they may be treated as described in the preceding section.

Geared Systems

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


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 ke and the disc moment of inertia Ipe .

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

By imagining the rotor Ip2 to be held rigidly whilst shaft 1 is rotated through some angle θ1 at the gearbox. The shaft 2 is rotated through an angle θ1/n at gearbox, where n is the gear ratio.

The strain energy stored in shaft 2

While applying the same input at the gear location to the equivalent system results in the stain energy stored in the equivalent shaft and can be expressed as

( )1 122 2 2 12 2

2E k k nθ θ= =

1 212 eE k θ=

(37)

(38)


Equating equations (37) and (38) give(39)

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

where θ2 and θe are the angle of twist of actual shaft 2 and the shaft 2 in equivalent system respectively. It can be seen from Figure 15(b) that θe = θ1 and ω1 and ω2 are the angular frequencies of the shafts 1 and 2, respectively. Noting that

Equation (40) can be written as

which simplifies to

22 /ek k n=

( ) ( )1

1 12 2 12 22

2 2

p p eeI Iω θ ω θ+ = +� � (40)

( ) 11 2 12 1and enT k T keθ θ θ= = =

2

22

112

21

12 p e

e

d nT d TI I

n dt k dt kω ω

� �� + = +� � � �

� ��

21

2

2 22 2

12

2

p

e

I d n T d n TI

dt k dt knω ω� � � �� + = + � � � ��

2

2/peI I n= (41)


• The general rule, for forming the equivalent system for the purpose of analysis, is to divide all shaft stiffness and rotor polar mass moment of inertias of the geared system by n2 (where n is the gear ratio n = ω2/ω1= ωgeared shaft/ωreference). When analysis is completed, it should be remembered that the elastic line of the real system is modified (as compared to with that of the equivalent system) by dividing the displacement amplitudes for equivalent shaft by gear ratio n as shown in Fig.16.

Fig.16 The elastic line in the original system


� Example 3• For geared system as shown in Fig.16 find the natural frequency and

mode shapes. Find also the location of nodal point on the shaft (i.e. the location of the point where the angular twist due to torsional vibration is zero). The shaft ‘A’ has 5 cm diameter and 0.75 m length and the shaft ‘B’ has 4 cm diameter and 1.0 m length. Take modulus of rigidity of the shaft G equals to 0.8 × 1011 N/m2, polar mass moment of inertia of discs are IA = 24 Nm2 and IB = 10 Nm2. Neglect the inertia of gears.

20 cm diameter

10 cm diameter

Gea

r Pai

r B

A

Fig.17

Solution:On taking shaft B has input shaft (or reference 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 shaftA A B B

B B A A

D T Nn

D T Nωω

� �= = = = = = = = =� �

� �


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

4 7 4�6.136 10 m ;

32A 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 equivalent system, the system will look as shown in Figure 4.19. The equivalent system of the shaft system A has the following torsional stiffness and mass moment of inertia properties

4 7 4� 2.51 10 m

32B BJ d −= = ×

44 2

2 2 2 2

6.545 10 241.6362 10 N m/rad and 6 Nm

2 2A PA

A PAe

K IK I

n n×= = = × = = =

which gives the equivalent length as 11 7

4

0.8 10 2.513 101.2288m

1.6362 10B

AeAe

GJl

K

−× × ×= = =×

Example 3 contd..


• 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 BeK 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 /9.81PA PB ee

nPA PBe

I I K

I Iω

� �+ � �+ ×� �= = =� �×� � � �

Example 3 contd..


• The node location can be

obtained from Fig. 20 as

• The negative sign indicates that both discs are either ends of node location

• 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

101.667

6An e PB

n B PAe

l Il I

θθ

= = − = − = −

1 21.667 n nl l=

1 22.2288n nl l+ =

20.8358 mnl =

Let 1rad then 1.667 radB Aeθ θ= = −

Hence, 0.8333radA

Ae

n

θθ = = −

Example 3 contd..


• 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 equivalent two mass rotor (Fig.19) can be considered as two single DOF systems (Fig. 22).

• The stiffness and mass moment of inertia properties of the system is given as

Fig. 21 Mode shape and nodal point location in the actual system

Fig. 22 A single DOF system

114 2

22

0.8 10 2.513 102.435 10 N/m and kgm

0.8358 9.81B

l PBn

e

GJK I

l× ×= = = × =

Example 3 contd..


• It gives the natural frequency as

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

42

2

2.435 10 9154.62 rad/sec

10

l

PBn

eK

Iω × ×= = =

Fig.23 Actual and equivalent geared systems

Example 3 contd..


• 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 to be input shaft and according the gear ratio will be obtained).

• It is assumed that equivalent shaft (i.e. B) has same diameter as the reference shaft (i.e. A). The equivalent mass moment of inertia and stiffness can be written as

• The total equivalent length and the torsional stiffness would be

100.5

20B B

A A

Dn

Dωω

= = = =

42 4 2

2 2 2

2.011 1040N m and 8.044 10 N/m

(0.5)PB PB

PB Be e

I II K

n n×= = = = = ×11 7

44

0.8 10 6.136 108.044 10 0.610 m

8.044 10A

B BB

ee

e

GJK l

l

−× × ×= = × ∴ = =×

A B

11 74 2

ee 0.61 0.75 1.36 m

and

0.8 10 6.136 103.61 10 N/m

1.360A

e

l l l

GJKe l

−

= + = + =

× × ×= = = ×

Example 3 contd..


• The natural frequencies of two mass rotor system is given by

1 0nω =

1 12 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

ω� � � �+ +� � � �= =� � � � � �

and

2

124(24 40)

9.81 3.61 10 153.65 rad/s24 40nω +� �= × × × =� �× �

Fig.24 Equivalent two mass rotor system

• The node location can be obtained as

1

2

101.667

6PBn

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..


• The stiffness of will be (equivalent stiffness corresponding to shaft A speed)

• The shaft stiffness corresponding to shaft B speed can be defined in two ways i.e

• On equating above equations the location of the node in the actual system can be obtained as

• which is same as by previous method.

22

ABe

n

GJK

l=

22

BB

GJK

l=

2 2

2 22

A

nB Be

GJK n K n

l= =

2

22 0.84B

Ae

Jl n l

J= =

Example 3 contd..


For the marine vessel power transmission shafts or machine tool drives, there may be many rotor inertias in the system and gear box may be a branch point where more than two shafts are attached. In such cases where there are more than two shafts attached as shown in Fig. 25 to the gearbox the system is said to be branchedFor Fig.25 having branched system, state vectors for different branch can be written as

For branch A, Taking θ0A=1 as reference valuefor angular displacementsince left hand end of branch A is free endhence for free vibrations we have T0A = 0. Equation for branch A takes the form

Branched Systems

Fig.25 Branched system

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

; nA nCA nB B C

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

11 12

21 22

10

nA

a a

T a a

θ � �� = � ��

(42)

(43)


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 11andnA nAT a aθ= =

nAoB

ABnθθ =

( )11 12

21 220

nA AB

nB oB

b b n

b b T

θθ � �� = � �

� � � ��

( ) 1211 nA ABnB oBb n b Tθ θ= +

( ) ( )( )22 21 22210 nA AB nA ABoB oBb n b T T b b nθ θ= + � = −

(44)

(45)

(46)

(47)

and

(48)


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

• Another conditions at branch to be satisfied regarding torque transmitted 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 ACoc

an n

θθ = =

1 1 12 2 2nA nA oB nB oC nCT T Tθ θ θ= +

oB oC oB oC

nA nB nA nC AB ACnA nA

T T T TT T

n nθ θ θ θ� = + � = +

oBoC

ABAC nA

TT n T

n� �

= −� � �

21 112

2221oC

ABAC

b aT n a

b n� �

= +� � �

(49)

(50)

(51)

(52)

/1111 12

21 22 21 21 11 2220

AC

AC AC ABnc

a nc c

c c n a b a n b n

θ � �� = � �

� � � �� +� � � � �


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 into transfer matrices for each station considered, where upon the state vector at each station may be evaluated. The plot of angular displacement against shaft position then indicates the system mode shapes.

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

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

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

(53)22 21 1122 21

22

21 11

20AC

AB

ACAC

c b a nc an c a

n b n+ + =


� Example 4Obtain the torsional critical speeds of the branched system as shown in Fig. 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. Take gear 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 asFig. 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 4.97 100.01 1 0.01 10 1

n

n n

ωω ω

− −− � �� − × ×× � �= = � �� − − � � �

with


• 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 CEBn n

U P Fω ω

−

−

� �×= = � �− − × + �

with

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

[ ]6

2 9 2

1 1.27 10[ ] [ ]

0.006 7.64 10 1B BFCn n

U P Fω ω

−

−

� �×= = � �− − × + �

with

22 21 1122

22

21 11

21 20BD

BC

BDBD

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

09 (1 3.97 10 )

n nn n

n n n

n

ω ω ω ω

ω ω ωω

−−

− −

−

− − × + − × −

− × − − ×+ =− ×

Example 4 contd..


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 gear train is free.

2 10 4 19 60.04372 3.3921 10 1.21 10 0n n nω ω ω− −− + × − × =

1,2,3

2 60; 135.48 10nω = × and 91064.2 ×

01 =nω 2 11640nω = 3 51387nω = rad/s

Example 4 contd..


Damping may come from o The shaft material o Torsional vibration dampers.

The torsional vibration damper is a device which may be used to join together two-shaft section as shown in Fig. 27. It transmits a torque, which is dependent upon of the angular velocity on one shaft relative to the other.Torsional dampers can be used as a means of attenuating system vibration and to tune system resonant frequencies to suit particular operating conditions. The damping in the system introducesphase lag angles to the system displacement and torque. The displacement and torque parameters must now be represented mathematically both in-phase and quadraturecomponents.Fig. 28 shows a general arrangement of torsional MDOF rotor system with damping. EOM of the nth rotor from free body diagram (Fig. 29) can be written as

Damping in Torsional Systems

Fig. 27 Torsional vibration damper


And

Torques RTn and LTn may 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 θ− − �

L n R nθ θ=

(54)

(55)

Fig. 28 General arrangement of MDOF system with damping.

Fig. 29 Free body diagram of rth rotor.

1 2sin cosn n nT T t T tω ω= +

1 2sin cosn n nt tθ θ ω θ ω= +

(56)

(57)

1 1

2 2

1 1

2 2

2

2

1 0 0 00 1 0 0

1 0

0 1n L nn

T TI C

T TC I

θ θθ θ

ω ω

ω ω

� �� = � �− −� � � ��

� � � �� − �

(58)


The previous equation can be simplified as

The characteristics of the shaft elementat station n are 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 each ends i.e.

Substituting T, θ 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 nth shaft segment

( ) ( ) ( )1 1 1R n L n n n n n n nT T K Bθ θ θ θ− − −= = − + −� �

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 Bθ θ ω θ ω θ− − −= = − − −

( )( 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 Bθ θ ω θ ω θ− − −= = − + −

(59)

(60)

(61)

(62)


Substituting T, θ, 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 2sin cos sin cosL n L n R n R nT t T t T t T tω ω ω ω− −+ = +

( 1) ( 1)1 1 2 2andL n R n L n R nT 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 00 0 0 1 0 0 0 1

n nL n R n

K B K B

B K B KT T

T T

θ θω ωθ θω ω

−

− � � � �� =

� � � ��

1[ ] { } [ ] { }n L n n R nL S M S −=

11 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)


Equations of motion of continuous system can be derived by (i) force balance of a differential element (ii) Hamilton’s principle (iii) Lagrangesmethod. While the force balance is a convenient approach for most of the problems, Hamiltons principle and Lagranges method are applied for a complex system. These two approaches need the consideration of the energy of the system.

�Hamilton’s PrincipleHamilton’s principle is stated as an integral equation in which the energy is integrated over an interval of time. Mathematically, the principle can be stated as

where L is the Lagrangian, T and V are kinetic and potential energy of the 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 during an interval of time from t1 to t2 , the actual path will be that for which the integral has a stationary value. It can be shown that in fact stationary value will be, in fact, the minimum value of the integral. The Hamilton’s principle can yield governing differential equations as well as boundary conditions

Torsional Vibration for Continuous Systems

[ ]2 2

1 1

( ) 0t t

t t

T V W dt Ldtδ δ δΠ = − + = =� � (67)


�Lagrange’s EquationHamilton’s principle is stated as an integral equation where total energy is integrated over an time interval.

On the other hand, Lagrange’s equation is differential equations, in which one considers energies of the system instantaneously in time.

Hamilton’s principle can be used to derive the Lagrange’s equation in a set of generalized coordinates. Lagrange’s 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 much simpler in torsional vibration of shafts and it is identical to the axial vibration of rods

0=∂∂−��

�

��

�

∂∂

ii qL

qL

dtd (68)


Governing Differential Equations

To obtain the governing differential equations the variational 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


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


The displacement field at a point P is

where (x, y, z) is the Cartesian coordinate of the point P, (r, β, z) is the cylindrical coordinate of the point and θ (x, t) is the angular displacement and 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 θβθθβθ ==−=−==

( )( ) ( )

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

ε ε ε ε

ε θ ε θ

= = = = = = = + =

= + =− = + =

xu

u xxx ∂

∂=,

(69)

(70)


(71)

The strain energy is given by

(72)

The kinetic energy is given by

(73)

; ;

;0 ;0 ;0 yz

xGzG

xGyG

GE

xyzxxyzx

yzzzyyxxxx

∂∂−==

∂∂==

======

θγτθγτ

γτσσεσ

1 1 1 1 1 12 2 2 2 2 2

2 2 21 12 22 2

0 0

(

12

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

l l

A

U dV

Gz Gy dAdx GJ dxx x x

σ ε σ ε σ ε τ γ τ γ τ γ

θ θ θ

� �= + + + + + �

� �∂ ∂ ∂� � � � � �= + =� �� ∂ ∂ ∂� � � � � ��

�

��

( )2 2with A

J y z dA= +�

{ } 12 2 2 22

0 0

12

l l

x y zA

T u u u dAdx J dxρ ρ θ� �= + + =� � ��


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 be obtained

(76)From the Hamilton’s principle, δ∏ = 0 , Eq.(76) becomes

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

dtdu

u xx =�

),( );,( ;0 txyutxzuu zyx θθ �� =−==

1 *0 02

0

( )l

W M M x x dxδ θ� �= + − ��

( )2 2

1 1

1 12 2 *, 0 02 20

( )t t l

xt tT U W dt J GJ M M x x dxdtρ θ θ θ δ θ� �Π = − − = − − − − ��

2

1

*, , 0 00

0 ( )t l

x xtJ GJ M M x x dxdtδ ρ θδθ θ δθ δθ δ δθ� �Π = = − − − − ��

Π


(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 and they are arbitrary, so

and (80,81)

2 2 2

1 1 1

*, 0 00 0 0

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

xt t tJ dtdx GJ dxdt M M x x dxdt

t xρ θ δθ θ δθ δθ δ δθ∂ ∂� � � � � �− − + − = �� ∂ ∂ � �

� � � � � ��

2 2 2 2

1 1 11

2

1

, ,0 0 00

*0 00

( ) ( ) ( ) ( )

( ) 0

ltl l t t t l

x xxt t tt

t l

t

J dx J dtdx GJ dt GJ dxdtx

M M x x dxdt

ρ θ δθ ρ θ δθ θ δθ θ δθ

δθ δ δθ

∂� � � �− − + � � ∂

� �− + − = �

� � � � � �

� �

� ��

{ }2

1

*, 0 0 , 00

( ) ( ) ( ) 0l t l

xx xtJ GJ M M x x dtdx GJρ θ δθ θ δ δθ θ δθ� �− − + − − = ��

lδθδθδθ and , ,

0

*, 0 0( ) ( ) 0xxJ GJ M M x xρ θ δθ θ δ− − − − =�� 0)(

0, =l

xGJ δθθ

0


Equations (80) and (81) represent equations of motion and boundary conditions respectively.

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

(82)where , i = 1, 2, …, r are called shape functions

� Galerkin MethodOn substituting equation (82) into equation (80), the residue as,

(83)

{ }1

( )2( ) 21 2

( )( )

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

( )

neer

r

t

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

t

θθ

θ θ

θ

� �� = + + + = =� � � � � ��

� ��

)(xNi

( )eR

( ) ( ) ( ) *, 0 0( , ) ( ) ( )e e exxR J GJ M x t M t x xρ θ θ δ= − − − −��


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

(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 θ is the dependent variable a and b are constants to be determined by element end conditions as shown in Fig. 33. (C0 of continuity and C1 of compatibility conditions).

ridxRNl

ei , ,2 ,1 ;0

0

)(�==�

( ) ( ) *, 0 0

0

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

e ei xxN J GJ M x t M t x x dx i rρ θ θ δ� �− − − − = = ��

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

0 0 0

( , ) ( ) ( ) 0

1, 2, ,

l l lhe e e

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

i r

ρ θ θ θ δ� �− + − + − = �

=

� � ��

�

bxae +=)(θ


At any general element (e) the following end condition existsat ith node x = 0 θ (e) = ; at jth node x = l θ (e) = θj (88)

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

(89)

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

(90)

iθ

( ) lba iji / ; θθθ −==

�{ } )()( )()( nee txN θθ =

iθ

Fig.33 A beam element with end conditions


with (91)

where Ni and Nj are interpolation functions corresponding to ith and jth nodes, respectively.

� Checking of Compatibility Requirement:

In Figure 34 the node j is the common node of element (e) and (e+1). For element (e), we can write equation (90), as

(92)

{ } ( )( ); ; 1 / ; /

ne ii j i j

j

N N N N x l N x lθ

θθ� �

� �= = = − =� � � ��

Fig. 34 Two neighbouring elements

��

��

−=j

ie lxlxθθ

θ )/()/1()(


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

(93)

We will check the compatibility requirements at common node j.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 of higher order, on taking first derivative of the equation (92) with respect to x, we get

(96)

��

��

−=+

k

je lxlxθθ

θ )/()/1()1(

jhx

e θθ ==

)(

jx

e θθ ==

+

0

)1(

�{ } )(,

)(, )( ne

xe

x tN θθ =


with

(97)

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

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

(99)

For x = l in 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 x

j

N N N N l N lθ

θθ� �

� �� = = = − = � ��

��

��

−=j

iex ll

θθ

θ )/1()/1()(,

��

��

−=+

k

jex ll

θθ

θ )/1()/1()1(,

( ) lijlx

ex /)(

, θθθ −==

( ) ljkx

ex /

0

)1(, θθθ −=

=

+


The completeness condition ensures the convergence as the elementsize 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 00

0 0 0

( , ) ( ) ( ) l l l

he e ex x xJ N dx GJ N dx GJ N N M x t M t x x dxρ θ θ θ δ� �+ = + + − ��

{ } { } { } { }

{ }

( ) ( ), ,

0 0

( ) *, 0 0

00

( , ) ( ) ( )

l lne ne

x x

hl

i ex

j

J N N dx GJ N N dx

NGJ N M x t M t x x dx

N

ρ θ θ

θ δ

� �+ =� � � �

� �� + + − �

� �

� �

�

��


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

x x ex x l

GJJ N N dx GJ N N dx N M x t M t x x dx

GJ

θρ θ θ δ

θ=

=

� �−� �� + = + + −� � � � �−� ��

� � ��

[ ] { } [ ] { } { } { }ERneenee TTKM +=+ )()()()( θθ��

[ ] { }2

( )

20 0

2 11 26

l le i i j

j i j

N N N JlM J N N dx J dx

N N Nρρ ρ

� � � �= = =� � � � � � �

� � � ��


(107)

(108)

where is the element mass matrix, is the element stiffness matrix, is the reaction element torque vector, and is the external

element torque vector. Matrices and are called consistent mass and stiffness matrices.

From strength of materials pure torsion theory we have

{ }ET

[ ] { }2

( ) , , ,, , 2

, , ,0 0

1 11 1

l le i x i x j x

x xj x i x j x

N N N GJK GJ N N dx GJ dx

N N N l

� � −� �� = = =� � � � � −� � � �

� �

{ }��

��

��

�� −

=��

�� −

==

=

lx

ex

x

ex

R

RR GJ

GJ

T

TT

)(,

0

)(,

2

1

θθ

{ } { } *0 0

0

( , ) ( ) ( ) l

ET N M x t M t x x dxδ� �= + − ��

[ ]( )eK

TGJ l

θ=

[ ]( )eM

[ ]( )eM [ ]( )e

K{ }RT


Rayleigh-Ritz Method

Rayleight-Ritz method can be also used to develop finite element formulation from the functional i.e. the potential energy, equation (76), of the system. On substituting equation (82) into equation (76), get functional 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 equation (110), we get

2

1

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

( )t le e e e e

xtJ GJ M M x x dxdtρ θ θ θ δ θ� �Π = − − − − ��

( )eΠ

0)(

=∂Π∂

i

e

θri ,,2,1 �=

( )2

1

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

, 0 000 ( )

ee e et l xe ext

i i i i

J GJ M M x x dxdtθθ θρ θ θ δ

θ θ θ θ� �∂∂Π ∂ ∂= = − − + −� �∂ ∂ ∂ ∂ �

� ��

�


0

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

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 ( ) 0

t ee e el t l xe e ext

i i i it

J dx J GJ M M x x dxdtθθ θ θρ θ ρ θ θ δ

θ θ θ θ� �∂∂ ∂ ∂− + + + − =� �∂ ∂ ∂ ∂ �

� � ��

{ } { } ( )2

1

( ) ( ) *0 00

( ) 0net l ne i

i it

dNdNJ N N GJ M M x x N dxdt

dx dxρ θ θ δ� �� + + + − =� � ��

� � ��

ri ,,2,1 �=

{ } { } { } ( ){ }2

1

( ) ( ) *0 00 0 0

( ) 0net l l lne

t

dN dNJ N N dx GJ dx M M x x N dx dt

dx dxρ θ θ δ� �� + + + − =� � ��

� � � ��


(111)

where is the element mass matrix, is the element stiffness matrix is the elemental reaction torque vector and is the elemental

external torque vector.

Assembled System EquationsOnce the elemental equation of the form as equation (105) has been derived next step is to obtain the system equation by assembling all such elements of the system as shown in Figure 35,

[ ] { } [ ] { } { } { }( )( ) ( ) ( ) ( ) ( )nee e ne ne neR EM K T Tθ θ+ = +��

[ ]( )eM [ ]( )e

K

{ }( )neET

�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

{ }( )neRT


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

(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 element finite 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

T

TT

T

lGJJl

θθ

θθρ��

��

1RT2RT

)1(1ET )1(

2ET

2 2

3 3

(2)22

(2)33

2 1 1 11 2 1 16

R E

R E

T TJl GJTl T

θθρθθ

− � �� − � �� + = + � � � �−� � � � � � � ��

��

��

1 1

2 2 2 2

3 3

(1)

1 1(1) (2)

2 2

(2)3 3

2 1 0 1 11 2 2 1 1 1 1 1

60 1 2 0 0 1

R E

R R E E

R E

T TJl GJ

T T T Tl

T T

θ θρ θ θ

θ θ

� � � �−� � −� � � �� + + − + − = − + + � � � �

� � � � � � � ��

��

��

��


Application of Boundary Conditions:For the present illustration since node 1 is fixed and node 3 is free, hence the following conditions are specified by the problem

(115,116)On applying the boundary conditions, i.e. equations (115) and (116) into equation (114), we get

(117)

On taking last two equations, we get

(118)

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

1 1 0 andθ θ= =��3

0RT =

��

�

�

��

�

�

++��

�

�

��

�

�−=

��

�

�

��

�

�

��

�

�

��

�

−+−−

+��

�

�

��

�

�

��

�

�

��

�

+)2(

)2()1(

)1(

3

2

3

2

3

22

11

00

0

1001111

0110

2101221012

6E

EE

ER

T

TT

TT

lGJJl

θθ

θθρ

��

��

��

��

��

�� +

=��

��

��

�

−−

+��

��

��

�)2(

)2()1(

3

2

3

2

3

22

1112

2114

6 E

EE

T

TT

lGJJl

θθ

θθρ��

��


Free Torsional VibrationFor free vibration, the solution can be assumed of the following form

so that (119,120)

where Θ is the amplitude of vibration, is the natural frequency and .On substituting equations (119) and (120) into homogeneous

part of equation (118), we get

(121)

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

(122)

tj net

t ω

θθ

��

��

ΘΘ

=��

��

3

2

3

2

)()( 222

33

( )( )

nj tn

te

tωθ ω

θΘ� � � �

= − Θ� ��

��

��

nω1j = −

��

��

=��

��

ΘΘ

��

��

��

��

�

−−

+��

��

�−

00

1112

2114

6 3

22

lGJJl

n

ρω

0

33

3322

22

22

=��

��

� −��

��

� −−

��

��

� −−��

��

� −

Jll

GJJll

GJ

Jll

GJJllGJ

nn

nn

ρωρω

ρωρω


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

(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), since otherwise the natural frequency will be complex in nature.

� Example 5Determine the natural frequencies and mode shapes for the rotor system as shown in Figure 5. Neglect the mass of the shaft and assume discs are having lumped masses.

( ) 077

102

2222 =+−

ba

ba

nn ωω / ; / 6with a GJ l b Jlρ= =

2nω

��

��

��

��

�+��

��

�+=2

222

7710

47

107

10b

aba

ba

ba

nω


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 element equation can be written, noting the equation (38), as

(126)

lIp1

G J

Ip2

mass less shaft

��

��

��

��−

=��

��

��

�

−−

+��

��

��

�

2,

1,

2

1

2

1

1111

00

2

1

x

x

p

p

GJ

GJ

lGJ

I

I

θθ

θθ

θθ��

��

Solution:

Example 5 contd..


Figure 36 Discretised single element Two disc rotor system.

Since for the present problem both the nodes are free, the following end conditions 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

Ip1 Ip2

�1 �2

(1)

1 2

021 == TT1 2

. 0i e GJ GJθ θ′ ′= =

��

��

−=��

��

2

12

2

1

θθ

ωθθ

n��

��

��

��

=��

��

��

�

�

��

�

−−

−−

00

2

1

2

2

1

1

θθ

ω

ω

pn

pn

Il

GJl

GJl

GJI

lGJ

Example 5 contd..


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

(130);

which gives a frequency equation, it gives two torsional natural frequencies and 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 from equation (129).

0

1

1

2

2

=−−

−−

pn

pn

Il

GJl

GJl

GJI

lGJ

ω

ω ( ) 02121

22 =��

�

� +− ppppnn IIl

GJIIωω

( )1 2

1 2

0p p

p p

k I Ip and p

I I

+= =

Example 5 contd..


(rad/s)


Geared Element for Branched Systems

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 1

Gear 2

�g1

�g2

�1

�2 = 1/n × �1

1

2

Gear ratioωω

=2 1

1g gn

θ θ= −

2 1 2Since is defined in terms of we can eliminate g g gθ θ θ

Gear 1

Gear 2

Shaft for gear 2


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

(134)

The stiffness matrix of shaft element connected to gear 2 will have to be modified, 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)

2

1 12g

g g

II

nθ� �� + � �

� ��

1

1gn

θ−

-1/n× �g1 = �1 �2

(2)(1)K1

Figure 38 Equivalent Gear Element

( )

1

21 1 2 1

2

1 2

12

1.

2g

U K

Kn

θ θ

θθ

= −

� �= +� �

� �


The external work done is give as(136)

On applying Lagrangian equation,

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

{ }1

1 1

1

1 11 122

1 2111 2

2

0

.

0

g

g g

g

U K K Kn n n n F

KU KK nn

θθ

θ θθθ

θθ

�� ∂ � �= + = �� ∂ � ��

� =� � � �� ∂ � ��= + =� � � �� ∂ � � �

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-m2, and IPB = IPC = IPD = 0. Take gear 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.

1

1 1 1 2 2 1 2 2gW f f f fn

θθ θ θ= − − = −


A

C

(1)

0 .4m

D

B

(3)0.

1mF

E

0.2m

(2)

Figure 38

Branched power system.kg-m2

kg-m2

kg-m2

kg-m2

kg-m2

kg-m2

01.0=PAI005.0=PEI

006.0=PFI

005.0=PBI006.0=PCIFigure 8

006.0=PDI

Gear ratio is defined as

Similarly,

/ ; /BC B C C B BCn nω ω θ θ= = −

/D B BDnθ θ= −

Example 6 contd..


Element (1)

(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

θθ

θθ

11

11

00

��

��

C C(2)

22

222 )/(2/1)(2/1 nkkU BECE θθθθ +=−=

2/ BCBC nIPEI

Figure. 39 Element (1)


Example 6 contd..


22, 2 2 2, 2 2

2(2) 2 22

2 2

222 2

2 2

( / ) ( / ) ; ( / )

/ //

/ // 0/0

B EBC B BC E Bc B E

BC BC

BC

B C BCBC BCC BC B

E EBCE E

U k n k n U k n k

k n k nk

k n k

T nk n k nI nTk n kI

θ θθ θ θ θ

θθθθ

= + = +

� �∴ = � �

�

� � −� �� ∴ + = � ��

� � � � � ��

��

��

Now,

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 /

///

00/

33

22

32

θθ

θθ��

��


(B)

(C)

Example 6 contd..


1 12 22 2

1 1 2 3 2 3

2 2

3 3

0 00 0 0/ / / /0 ( / / ) 0 0 0

0 / 00 0 00 / 00 0 0

A A AA

BC BD BC BDB C BC D BD BB

BCE E EE

BDE F FF

k kI T

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

k n kI T

k n kI T

θθθθθθθθ

−� � � �� − + ++ + � � � � � �� ∴ + =

� ��

��

��

��

��

The assembled equations of motion can be written as

Now to have eigen value problem (EVP),

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

02 =− MK ω

61 10062.201 ×=k 6

2 10566.12 ×=k 33 10398.785 ×=k

610

785398.0019635.000566.1218867.40

19635.018867.4507.202062.20100062.201062.201

×

��

�

�

��

�

−−

=∴k��

�

�

��

�

=

006.00000005.00000007167.0000001.0

M

(D)

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

/ / 0B c BC D BDT T n T n− − =(E)

(F)

and

Example 6 contd..


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

rad/s; rad/s;

rad/s rad/s54 101472.2 ×=nω5

3 105085.0 ×=nω

52 101156.0 ×=nω01 =nω

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

Example 6 contd..


A multi-cylinder reciprocating machine contains many reciprocating and 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 unpredictable effects like variable inertia, internal dampings, misalignments in the transmission units, uneven firing order etc.

The analysis can be best carried out, by lumping the inertias of rotating and reciprocating parts at discrete points on the main shaft. The problem then reduces to the forced torsional vibration study of an N disc system subjected to varying torques at different cylinder points.

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

Modelling of reciprocating machine systems


On the other hand, parts mounted on the shafting, like the damper, flywheel, generator etc. are rigid and will have very high polar mass moments of inertia.

The system containing the crankshaft, coupling, generator/ auxiliary drive shaft/ other driven shaft like pumps, and mounted parts can then be reduced to a simple system with a series of rigid rotor (representing the inertias) 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 of reciprocating parts.

� Consider the piston shown in two different positions in Figure 4.43, and imagine 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 of the 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 same acceleration 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 crank throw or radius. Hence, the total polar mass moment of inertia varies from Irot to Irot+ mrec r2, when the crankshaft is rotating.

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

� We consider as an approximation the system to have an average inertia given by

� where r is crank radius.

12

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.

� Considering an idealisation of a crank throw into an ordinary shaft having the 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 crank pin to twisting, if the main shaft is subjected to twisting.

� Moreover the beam formulae, if used will not very accurate, because of short 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 more rectangular blocks and application of the theory becomes extremely difficult.


� Because of this uncertainties in analytical calculation to estimate the torsional stiffness of crank throws, several experiments have been carried 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. The torsional 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 the shaft from the coupling up to the point of rigidity.

� In case where one part of the system is connected to the other part through gears, or other transmission units, it is convenient to reduce all the inertias and stiffness to one reference speed.

Fig 44 Equivalent length of a crank


� 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.

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

� Torsional oscillation in the crankshaft and in the shafting of driven machinery is vibration phenomenon of practical importance in the design of 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 firing period. Even though the torque is periodic the fact that it fluctuates so violently within the period, constitutes one of the inherent disadvantages of a reciprocating machine, from the dynamics point of view, as compared with a turbine where the torque is practically uniform.

� It is possible to express the torque by a reciprocating engine into its harmonic components of several orders of the engine speed, and these harmonic components can be excite the engine driven installations into forced torsional vibrations.


�It is a commonly known fact that failures can occur in reciprocating machine installations, when the running speed of the engine is at or near a dominant torsional critical speed of the system.

�High dynamic stresses can occur in the main shafting of such engine installations and to avoid these conditions, it is essential that the torsional vibration characteristics of the entire installation be analysed before the unit is put into operation.

�Any analysis of torsional vibration characteristics of reciprocating machinery 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 the permissible values, to check the safety of installation.


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Torsional Vibrations of Rotor

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