CHAPTER 10

TRANSVERSE VIBRATIONS-VI:

FINITE ELEMENT ANALYSIS OF ROTORS WITH GYROSCOPIC EFFECTS

Previously, gyroscopic effects on a rotor with a single disc were discussed in great detail by using the

quasi-static and dynamic analyses with the help of the analytical approach. For multi-DOF systems

with gyroscopic effect of thin discs was described while discussing the transfer matrix method. There

we have not considered gyroscopic effects due to thick flexible shafts (i.e., the shaft with distributed

mass and stiffness properties). In previous chapter, we dealt with analysis of simple rotor with the

analytical (i.e., the continuous) and numerical (finite element) approaches without considering the

important effects like shear deformation and gyroscopic effects. In the present chapter, the analysis

would be extended to initially a single disc of a simple rotor. Then it will be extended for more

general rotor systems with the help of a powerful analysis tool of the finite element method. The

Timoshenko beam theory would be used for the development of governing equation of the continuous

system analysis. The finite element formulation will be developed for the spinning Timoshenko beam

which includes higher effects like the rotary inertia, shear deformation and gyroscopic effects. The

eigen value problem would be developed through the state space form of the governing equation.

With the help of examples the extraction of modal parameters from the special eigen value problem

will be explained. The standard Campbell diagram for various cases are obtained, which shows the

natural whirl frequency variation with the spin speed of the shaft for asynchronous whirl. This

diagram can be used to obtain the critical speeds of such rotor systems.

10.1 Rotor Systems with a Single Disc

In the present section a finite element analysis for an overhang rotor with gyroscopic effects as shown

in Figure 10.1 would be illustrated. However, the procedure then can be extended to other boundary

conditions (Jeffocott rotor, rotor with intermediate support, etc.). The shaft has been modelled with

consistent mass and stiffness matrix. However, gyroscopic couple due to disc is only considered and

for the shaft it is neglected. The shaft is treated as flexible and massless.

Figure 10.1 A cantilever rotor system

593

Now since we would be considering the gyroscopic couple, hence we need to consider both plane

motion simultaneously. Let us model the rotor as single element with two nodes. The elemental

equations of motion in z-x plane as shown in Figure 10.2(a) (without gyroscopic effects) can be

written as

1

1 1 1

2

2 2 2

2 2 1 1

2 2

3

2 2

22

156 22 54 13

12 6 12 64 13 3

4 6 2156 22

12 6

sym 4sym 4

x

y y xz

x

d y y xz

l lSu ul ll l l

Ml l lEImm lu l u Slm

I l Ml

m

ϕ ϕ

ϕ ϕ

− − − − − −− + = + − − +

��

��

��

��

(10.1)

with 420

Alm

ρ=

The elemental equations of motion in y-z plane as shown in Figure 10.2(b) (without gyroscopic effect)

can be written as

1

1 1 1

2

2 2 2

2 2 1 1

2 2

3

2 2

22

156 22 54 13

12 6 12 64 13 3

4 6 2156 22

12 6

sym 4sym 4

y

x x yz

y

d x x yz

l lSv vl ll l l

Ml l lEImm lv l v Slm

I l Ml

m

ϕ ϕ

ϕ ϕ

− − − − − −− + = + − − +

��

��

��

��

(10.2)

(a) Element in z-x plane (b) Element in y-z plane

Figure 10.2 A rotor element

In chapter 5 gyroscopic couple effect of disc alone was described and it is given as (please note that

the disc is at node 2)

594

2

2

2

2

0 0 0 0

0 0 0

0 0 0 0

0 0 0

yp

p x

u

I

v

I

ϕω

ϕ

− −

�

�

�

�

(10.3)

It can be expanded in the following form to accommodate both the nodes of the shaft element

1

1

2

2

1

1

2

2

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0

y

x

p y

p x

u

v

u

I

v

I

ϕ

ϕω

ϕ

ϕ

− −

�

�

�

�

�

�

�

�

(10.4)

It could be seen that the gyroscopic effect leads to coupling of motions in y-z and z-x planes. In

equations (10.1) and (10.2) now gyroscopic couple effect (equation (10.4)) would be added and

equations of motion of the element, in both planes, can be written as

1

1

2

2

2 2

1

2 2

1

22

2

2

156 22 0 0 54 13 0 0

4 0 0 13 3 0 0

156 22 0 0 54 13

4 0 0 13 3

156 22 0 0

4 0 0

156 22

sym 4

y

x

d

y

x

d

l l

l l lul l

l l l

vml

mm

uIl

m

m vl

m

Il

m

ϕ

ϕ

ϕ

ϕ

− −

−

−

+ −

+

+ −

+

��

��

��

��

��

��

��

��

595

1

1

2

2

1

2 2

1

2

3

2

2

0 0 0 0 0 0 0 0 12 6 0 0 12 6 0 0

0 0 0 0 0 0 0 0 4 0 0 6 2 0 0

0 0 0 0 0 0 0 0 12 6 0 0 12 6

0 0 0 0 0 0 0 0 4 0 0 6 2

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0

y

x

p y

p x

u l l

l l l

v l l

l lEI

u l

I

v

I

ϕ

ϕω

ϕ

ϕ

−

− − − − + −

�

�

�

�

�

�

�

�

1

1

2

2

1

1

2

2

2

22

12 6 0 0

4 0 0

12 6

sym 4

y

x

y

x

u

v

l

l u

l

l v

l

ϕ

ϕ

ϕ

ϕ

− −

1

1

1

1

2

2

2

2

0_

0

0

0

0

0

0

0

x

xz

y

yz

x

xz

y

yz

S

M

S

M

S

M

S

M

− − −

= +

(10.5)

The above formulation could alternatively be obtained by a more formal approach by using the

additional kinetic energy term of the disc developed in Chapter 5 due to gyroscopic moments in finite

element formulations of Euler-Bernoulli beams as described in Chapter 9, which will lead to the

similar results. This will be illustrated while developing the finite element formulation with the

Timoshenko beam model (in which the rotary and gyroscopic effects of shafts will also be considered)

in subsequent sections. For the present illustration of an overhang rotor, boundary conditions are

given as

1 1 2 2 2 21 1 0 and 0y x xz x yz yu v M S M Sϕ ϕ= = = = = = = = (10.6)

After application of boundary conditions (10.6), on elimination of he first four rows and columns

equations of motion (10.5) reduces to

2 2 2

2 2

2 2 2

2

3

2 2

2

156 22 0 0

0 0 0 0 12 6 0 04 0 0

0 0 0 4 0 0

0 0 0 0 12 6156 22

0 0 0 sym 4

sym 4

d

y y yp

px x

d

ml

mu u ulI

I lEImm

m v v l vll

m I l

I

m

ϕ ϕ ϕω

ϕ ϕ

+ −

− + − + − + − −

�� �

�� �

�� �

�� �2

2

0

0

0

0xϕ

=

(10.7)

which is equation of motions of the following standard form

596

[ ]{ } [ ]{ } [ ]{ } { }0M G Kη ω η η− + =�� � (10.8)

For the formulation of standard form of the eigen value problem, equation (10.8) need to be

transformed to the state space form as discussed in Chapter 5. The following example will illustrate

the procedure of obtaining whirl natural frequencies for a typical case.

For the asynchronous whirl the eigen value problem of the form as equation (10.8) is not a standard

one. For obtaining the standard form of the eigen value, equation (10.8) has to expressed in the state

space form. To illustrate the transformation, let us consider the following simple example.

Example 10.1: Obtain the state space form of the following standard governing equation for

vibrations of a spring-mass-damper system.

0=++ kxxcxm ��� (a)

Solution: Let us express the velocity as

vx =� (b)

From equation (a), we get

c kv v x

m m= − −� (c)

Equations (b) and (c) which can be combined as

0 1

/ /

x x

v k m c m v

= − −

�

� (d)

which can be written in the standard state space form, as

{ } [ ]{ }h D h=� (e)

with

597

{ } [ ] { }0 1

; ; / /

x xh D h

v k m c m v

= = = − −

��

�

Answer

Now coming back to conversion of equation (10.8) to the state space form, let

{ } { }x v=� (10.9)

So that equation (10.8) takes the following form

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

v M G v M K xω− −

= −� (10.10)

Combining above two equations, we have

{ } [ ]{ }hDh =� (10.11)

with

[ ][ ] [ ] [ ] [ ]

1 1

0 1D

M K M Gω− −

=

− ; { }

{ }{ }x

hv

=

(10.12)

Let us assume the solution of equation (10.11) of the following form

{ } { }0

th h e

υ= (10.13)

where ν is the whirl frequency. On substituting equation (10.13) into equation (10.11), we get

{ } [ ]{ }0 0h D hν = (10.14)

which is a standard eigen value problem. Eigen values of equation (10.14) appear as pure imaginary

conjugate pairs with magnitudes equal to natural whirl frequencies. For the case when gyroscopic

effect is not present, the eigen value will be of the form jα± and jα± . However, with gyroscopic

effect it takes the form jα +± and jα −± where α α α+ −> > and they correspond to forward and

backward whirls, respectively. The detailed illustration of the solution of eigen value problem will be

598

made in next example, when the method of finite element will be applied to rotors with gyroscopic

effects.

Example 10.2 Obtain the forward and backward synchronous transverse critical speeds for a general

motion of a rotor as shown in Figure 10.3. The rotor is assumed to be fixed supported at one end.

Take mass of the thin disc m = 1 kg with the radius of 3.0 cm. The shaft is assumed to be massless and

its length, L, and diameter, d, are 0.2 m and 0.01 m, respectively. Take shaft Young’s modulus E = 2.1

× 1011

N/m2. Using the finite element method and considering the mass of the shaft with material

density ρ = 7800 kg/m3 obtain first two forward and backward synchronous bending critical speeds by

drawing the Campbell diagram.

Figure 10.3

Solution: On taking single element for the present problem, we have the following data:

m = 1 kg, 2 41

24.50 10

pI mr

−= = × kg-m2, 2 41

42.25 10

pI mr

−= = × kg-m2,

E = 2.1 × 1011

N/m2, ρ = 7800 kg/m

3,

l = L = 0.2 m, d = 0.15 m, 2 2 5

4 40.01 7.854 10A d

π π −= = = × , 4 4 10

64 640.01 4.9087 10I d

π π −= = = × m4,

547800 7.854 10 0.2

2.9172 10420 420

Alρ −−× × ×

= = × kg, 11 10

4

3 3

2.1 10 4.9087 101.2885 10

0.2

EI

l

−× × ×= = × N/m

From equation (10.7), we have

2 2

2 2

2 2

2 3

2 2

104.5508 0.1284 0 0 0 0 0 0

0.0272 0 0 0 0 0 0.450010 10

104.5508 0.1284 0 0 0 0

sym 0.0272 0 0.4500 0 0

y y

x x

u u

v v

ϕ ϕω

ϕ ϕ

− −

− − ×

− −

�� �

�� �

�� �

�� �

599

2

2

2

5

2

1.5463 0.1546 0 0 0

0.0206 0 0 010

1.5463 0.1546 0

sym 0.0206 0

y

x

u

v

ϕ

ϕ

− + =

−

(a)

Hence, equation (a) has the same form as that of equation (10.8). The state space form can be written

as

{ } [ ]{ }hDh =� (b)

with

[ ][ ] [ ]

[ ] [ ] [ ] [ ]1 1

0 ID

M K M Gω− −

=

− ; { }

{ }{ }

xh

x

= �

; { } 2

2

2

2

y

x

u

xv

ϕ

ϕ

=

; { } 2

2

2

2

y

x

u

xv

ϕ

ϕ

=

�

�

��

�

(c)

2

104.5508 0.1284 0 0

0.0272 0 0[ ] 10

104.5508 0.1284

sym 0.0272

M−

− = −

, (d)

3

0 0 0 0

0 0 0 0.4500[ ] 10

0 0 0 0

0 0.4500 0 0

G−

=

−

, 5

1.5463 0.1546 0 0

0.0206 0 0[ ] 10

1.5463 0.1546

sym 0.0206

K

− = −

(e)

To compare eigen values and eigen vectors without and with gyroscopic effects the results from both

the analysis are provided. Table 10.1 lists eigen values and eigen vectors corresponding to one

element eigen analysis with 1 rad/s ( ~ 10 rpm, i.e. very slow speed) of rotor speed (i.e., negligible

small gyroscopic effects). In eigen values the real part is related with the damping and the imaginary

part is related with the natural whirl frequency of the system. It can be observed that the imaginary

part of eigen value in serial numbers 1 to 4 are all the same. 1 and 2 are complex conjugate. Thus, the

whirl frequency from serial number1 is positive and the whirl natural frequency from serial number 2

is exactly same with a negative sign. The negative frequency has no physical significance, hence, can

be omitted. Without gyroscopic effect eigen value from serial numbers 1 and 3 are same, and eigen

values from serial number 4 is complex conjugate of 3. Hence, actually there is only one natural whirl

frequency (i.e., 192.68 rad/s see Table 10.1) from the first four eigen values. Similarly, from serial

numbers 5 to 8 there is another natural whirl frequency (i.e., 1257.55 rad/s see Table 10.1). Hence,

without gyroscopic effect case with one element we could be able to get only two natural frequencies

600

(refer Table 10.3). Eigen vectors are also appearing in complex conjugate (see Table 10.1). It should

be noted that with zero speed the dynamic matrix [D] becomes singular, since all the diagonal

elements are zero with 0ω = . Hence, it will be worthwhile to write the eigen value problem as

[ ] [ ] [ ]( ){ } { }1

0M K I xλ−

− = (f)

which is a familiar form as discussed in detail in Chapters 7 and 9 for respectively torsional and

transverse vibrations.

The eigen value and eigen vector for matrix [D], for example, at 1500 rad/s, are provided in Table

10.2. It can be observed that these eigen values are also purely imaginary and occurring in complex

conjugate (e.g., serial numbers: 1 and 2, 3 and 4, 5 and 6, and 7 and 8). Serial numbers 1 to 4 belong

to the first natural mode (as in the case of without gyroscopic effects these are all same, however, now

1 and 3 are not same). Eigen values 2 and 4 are complex conjugate of eigen values 1 and 3,

respectively; hence only eigen values 1 and 3 need to be considered (for the same reason as given for

the case of without gyroscopic effects). Serial numbers 1 and 3 belong to the backward and forward

whirls, respectively (which are below and above the first natural frequency without gyroscopic

couple, i.e. 192.68 rad/s). Similarly serial numbers 5 and 7 belong to the backward and forward

whirls, respectively (which are below and above the second natural frequency without gyroscopic

couple , i.e. 1257.55 rad/s); and serial numbers 6 and 8 are exactly same as 5 and 7 with a negative

sign (since real part in all eigen values are anyway zero).

Eigen vectors corresponding to these eigen values are also listed in Table 10.2. It can be observed that

they are appearing as complex number. However, on close observation, it can be seen that either real

part is zero or imaginary part is zero (e.g., for the eigen value 1, the real part of 1st and 2

nd eigen

vector components are zero, whereas, for the imaginary part of 3rd

and 4th eigen vector components

are zero. When the real part is zero the phase is 900 and when the real part is zero the phase is 0

0.

Table 10.1 Eigen value and eigen vectors without gyroscopic couple (i.e., at speed of 1 rad/s)

S.N. Eigen value Eigen vectors

1 -0.0000 + 0.1927×103j 4.8421×10

-4+0.0000j

0.0036 +0.0000j

0.0000-4.8421×10-4

j

0.0000 -0.0036j

0.0000+0.0933j

0.0000- 0.7009j

0.0933+0.0000j

0.7009+0.0000j

601

2 -0.0000 - 0.1927×103j 4.8421×10

-4+0.0000j

0.0036 +0.0000j

0.0000+4.8421×10-4

j

0.0000+0.0036j

0.0000-0.0933j

0.0000-+0.7009j

0.0933+0.0000j

0.7009+0.0000j

3 0.0000 + 0.1926×103j -4.8423×10

-4+0.0000j

-0.0036 +0.0000j

0.0000-4.8423×10-4

j

0.0000 -0.0036j

0.0000-0.0933j

0.0000- 0.7009j

0.0933+0.0000j

0.7009+0.0000j

4 0.0000 - 0.1926×103j -4.8423×10

-4+0.0000j

-0.0036 +0.0000j

0.0000+4.8423×10-4

j

0.0000 +0.0036j

0.0000+0.0933j

0.0000+0.7009j

0.0933+0.0000j

0.7009+0.0000j

5 -0.0000 + 2.7567×103j 0.0000 +1.8793×10

-7j

0.0000 -2.5650×10-4

j

1.8793×10-7

+0.0000j

-2.5650×10-4

–0.0000j

-5.1807×10-4

-0.0000j

0.7071 + 0.0000j

-0.0000 +5.1807×10-4

j

0.0000 - - 0.7071j

6 -0.0000 - 2.7567×103j 0.0000 -1.8793×10

-7j

0.0000 +2.5650×10-4

j

1.8793×10-7

+0.0000j

-2.5650×10-4

–0.0000j

-5.1807×10-4

-0.0000j

0.7071 + 0.0000j

-0.0000 -5.1807×10-4

j

0.0000 + 0.7071j

7 0.0000 + 2.7584×103j 0.0000 +1.8721×10

-7j

0.0000 -2.5635×10-4

j

1.8721×10-7

+0.0000j

-2.5635×10-4

–0.0000j

-5.1639×10-4

-0.0000j

0.7071 + 0.0000j

-0.0000 +5.1639×10-4

j

0.0000 - - 0.7071j

8 0.0000 - 2.7584×103j 0.0000 -1.8721×10

-7j

0.0000 +2.5635×10-4

j

1.8721×10-7

+0.0000j

-2.5635×10-4

–0.0000j

-5.1639×10-4

-0.0000j

0.7071 + 0.0000j

-0.0000 -5.1639×10-4

j

0.0000 + 0.7071j

602

Table 10.2 Eigen value and eigen vectors with gyroscopic effects at 1500 rad/s

S.N. Eigen value Eigen vectors

7 0.0000 + 4.2662×103j 6.9367×10

-8 +0.0000j

0.0002 + 0.0000j

0.0000 -6.9367×10-8

j

-0.0000 - 0.0002j

-0.0000 + 0.0003j

-0.0000 + 0.7071j

0.0003 + 0.0000j

0.7071 + 0.0000j

8 0.0000 - 4.2662×103j 6.9367×10

-8 -0.0000j

0.0002 - 0.0000j

0.0000 +6.9367×10-8

j

-0.0000 + 0.0002j

-0.0000 - 0.0003j

-0.0000 - 0.7071j

0.0003 - 0.0000j

0.7071+ 0.0000j

5 -0.0000 + 1.8034×103j 0.0000 +1.3635×10

-6j

-0.0000 - 0.0004j

-1.363510-6

+0.0000j

0.0004 - 0.0000j

-0.0025 - 0.0000j

0.7071 + 0.0000j

-0.0000 - 0.0025j

0.0000 + 0.7071j

6 -0.0000 - 1.8034×103j -0.0000 -1.3635×10

-6j

-0.0000 + 0.0004j

-1.3635×10-6

-0.0000j

0.0004 + 0.0000j

-0.0025 + 0.0000j

0.7071+ 0.0000j

-0.0000 + 0.0025j

0.0000 - 0.7071j

3 0.0000 + 0.2105×103j -0.0005 + 0.0000j

-0.0033 + 0.0000j

0.0000 + 0.0005j

0.0000 + 0.0033j

-0.0000 - 0.0996j

-0.0000 - 0.7000j

-0.0996 + 0.0000j

-0.7000 + 0.0000j

4 0.0000 - 0.2105×103j -0.0005 - 0.0000j

-0.0033 - 0.0000j

0.0000 - 0.0005j

0.0000 - 0.0033j

-0.0000 + 0.0996j

-0.0000 + 0.7000j

-0.0996 - 0.0000j

-0.7000 + 0.0000j

1 -0.0000 + 0.1742×103j -0.0000 - 0.0005j

-0.0000 - 0.0040j

0.0005 - 0.0000j

0.0040 - 0.0000j

0.0881 + 0.0000j

0.7016 + 0.0000j

603

0.0000 + 0.0881j

0.0000 + 0.7016j

2 -0.0000 – 0.1742×103j -0.0000 + 0.0005j

-0.0000 + 0.0040j

0.0005 + 0.0000j

0.0040 + 0.0000j

0.0881 - 0.0000j

0.7016 + 0.0000j

0.0000 - 0.0881j

0.0000 - 0.7016j

Figs. 10.4 (a), and (b) show Campbell diagrams for the present problem with 1 and 20 elements,

respectively. Table 10.3 lists the critical speeds up to the third mode with and without gyroscopic

couple. It should be noted that the critical speed without gyroscopic couple is always in between the

corresponding mode forward and backward whirl critical speeds, which indicate the splitting of

natural frequency in the case of gyroscopic effect. It can be observed that the split in the whirl natural

frequency is distinct at higher speeds and at higher mode numbers. Fig 10.5(a) and (b) show mode

shapes for linear and angular displacements, respectively.

Table 10.3 Critical speeds with different number of elements

Mode no.

Critical speed without

gyroscopic effects (rad/s)

Critical speed with gyroscopic effects (rad/s)

for 1 element

for 20 elements

1

element

20

elements

Backward

whirl

Forward

whirl

Backward

whirl

Forward

whirl

1

192.65 192.65 190.32 195.02 190.32 195.02

2

1257.55 2724.22 1709.47 †

1706.29 6773.36

3

* 7993.37 * * 7543.02 19858.0

† Critical speed could not be obtained because estimated natural whirl frequency with one element

never intersected the nf

ω ω= line. This is due to the fact that the FEM always gives over estimation

of the natural frequencies and error in second natural is expected to be more with single element,

hence variation of the natural whirl frequency was diverging nature.

* With a single element, it is possible to get critical speeds only up to the second mode.

604

Shaft spin speed, ω (rad/s)

Figure 10.4(a) Campbell diagram (with 1 element)

Shaft spin speed, ω (rad/s)

Figure 10.4(b) Campbell diagram (with 20 elements)

0 500 1000 1500 20000

500

1000

1500

2000

2500

3000

0 1000 2000 3000 4000 5000 6000 7000 80000

2000

4000

6000

8000

10000

1B 1F

2B 2F

Nat

ura

l w

hir

l fr

equen

cy, ω

nf

(rad

/s)

ω = ωnf

2B

2F 3B

3F

ω = ωnf

Nat

ura

l w

hir

l fr

equen

cy, ω

nf

(rad

/s)

605

Shaft length

Figure 10.5(a) Normalized linear displacement mode shapes at ω = 100 rad/s (forward);

(1) - first mode shape (1

191.43F

nfω = rad/s), (2) - second mode shape(1

2647.20F

nfω = rad/s)

Shaft length

Figure10.5(b): Normalized angular displacement mode shapes at ω = 100 rad/s (forward);

(1) - first mode shape(1

191.43F

nfω = rad/s), (2) - second mode shape (1

2647.20F

nfω = rad/s)

0 0.05 0.1 0.15 0.2-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0 0.05 0.1 0.15 0.2-1

-0.5

0

0.5

1

1

2

Rel

ativ

e li

nea

r dis

pla

cem

ents

1

2

Rel

ativ

e an

gula

r dis

pla

cem

ents

606

10.2 TIMOSHENKO BEAM THEORY

In previous section we discussed the gyroscopic effect on a disc alone in a slender flexible beam. The

cross-sectional dimensions of the beam were considered to be small in comparison with the length. In

this case the rotation of a beam element is equal to slope of the elastic line of the shaft. When we have

to investigate higher frequency modes an infinitesimal element would have appreciable amount of

rotary inertia. When the beam is thick in that case the cross-sectional dimensions of the beam are

considered to be comparable in comparison with the length and the shear effect becomes predominant.

The Euler-Bernoulli beam theory is based on the assumption that plane cross-sections remain plane

and perpendicular to the longitudinal axis after bending. This assumption implies that all shear strains

are zero. When the normality assumption is not used, i.e., plane sections remain plane but not

necessarily to the longitudinal axis after deformation, the transverse shear strain yzε is not zero.

Therefore, the rotation of a transverse normal plane about x-axis is not equal to - dv/dz. Beam theory

based on these relaxed assumptions is called a shear deformation beam theory, most commonly

known as the Timoshenko beam theory.

Consider a short stubby beam as shown in Figure 10.6(a) before deformation. Let L be the length of

the shaft, A be the cross sectional area, E be the Young’s modulus, and ρ be the mass density of the

shaft material. Assume that the deformation of the beam is purely due to the shear and that a vertical

element before deformation remains vertical after deformation and moves by distance vs (subscript s

to represents the pure shear) in the transverse y direction as shown in Figure 10.6(b). For the pure

shear case also there will be no coupling in deformations in the two transverse directions (i.e., in the x

and y directions).

Since there is no coupling of motion in two transverse directions and on neglecting displacement in

the axial direction, the displacement field is given by

0xu = ( , )y su v z t= 0zu = (10.15)

Line elements tangential to the elastic line of the beam undergo a rotation β(z,t) due to the shear as

shown in Figure 10.6(c). Engineering strains from displacement fields of equation (10.15) are

2yz svε ′= (10.16)

and

0=xxε , 0=yyε , 0=zzε , 0=yzε , 0=zxε (10.17)

where prime ( ′ ) represents derivative with respect to z.

607

Figure 10.6(a) A beam before deflection

Figure 10.6(b) The beam after the shear deflection

Figure 10.6(c) Rotation due to shear

608

The stress field consists of only one component

yz sGvτ ′= (10.18)

where, G is the shear modulus of the beam material. Equation (10.18) shows that the shear stress

across the cross section of the beam is uniform and independent of the coordinate y. However, if we

introduce explicit dependency of stress field on another independent variable, y, then analysis

becomes unnecessarily complicated. In actual case this is not true and therefore a shear correction

factor ksc is applied to equation (10.18), rather than making the theory more complicated by having

one more independent parameter y in the analysis.

Figure 10.7 Typical cross sections of the shaft

Several definitions of the shear correction factors are found in literatures. Present investigation uses

the one given by Cowper (1966) (mentioned also by Shames and Dym, 2005). These are given for

some cases as (Figure 10.7)

( )6 1

7 6sck

ν

ν

+=

+ for a circular cylinder (10.11)

and

( )( )

( )( ) ( )

2

2 2

6 1 1

7 6 1 20 12sc

rk

r r

ν

ν ν

+ +=

+ + + +

for a hollow cylinder (10.20)

with

0ir r r=

where, ν is the Poisson’s ratio, and ri and ro are the inner and outer radii of the cross section of shaft

as shown in Figure 10.7. Thus considering the shear correction factor, equation (10.18) can be written

as

609

yz sc sk Gvτ ′= − (10.21)

In general, the beam is not subjected to the pure shear only. The significant component of the

deformation arises due to the bending moment as in the case of Euler and Rayleigh (as compared to

Timoshenko beam model it considers only the rotary inertia effect without shear deformation effects)

beams. Combining the effect of shear considered above with the Euler beam theory (Chapter 9), the

following displacement field for a point P(x, y, z) as shown in Figure 10.8 is assumed

0xu = ; ( , )yu v z t= and { }( , ) ( , )z xu y z t y v z tϕ β′= = − (10.22)

with

s bv v v= + ,

s b xv v v β ϕ′ ′ ′= + = + (10.23)

(a) Shear (b) Bending (Euler)

Figure 10.8 A displacement field of the Timoshenko beam

The total slope of the beam, v′ , consists of two parts, one due to the bending, which is ϕx(z,t) and

other due to only the shear, which is β(z,t). The axial displacement of a point at a distance y from

center line is only due to the bending slope and that the shear force has no contribution to this. So

there are two independent variables in this problem. One independent variable is the total

displacement of the point, v(z,t) and other is the slope due to the bending ϕx(z,t). The engineering

strain field from equation (10.22) is

zzz x

uy

zε ϕ

∂′= =

∂; ( )1 1

2 2yz x

yzuu

vy z

ε ϕ +

∂∂ ′= = − −∂ ∂

(10.24)

610

and

0xxε = ; 0yyε = ; 0xyε = ; 0zxε = (10.25)

The corresponding stress field with the shear correction factor is given as

zz xEyσ ϕ ′= ; ( )yz sc xk AG vτ ϕ′= − − (10.26)

and

0yyσ = ; 0xxσ = ; 0xy zxτ τ= = (10.27)

The strain (conservative) energy is

( ){ }22

0

1

2

L

xx x sc xU EI k AG v dzϕ ϕ′ ′= + −∫ (10.28)

The kinetic energy is

{ }2 2

0

1

2

L

xx xT Av I dzρ ρ ϕ= +∫ �� (10.29)

If f(z,t) is the distributed force for transverse loads then the work done (non-conservative) by external

forces can be written as

0

( , )

L

ncW f z t vdz= ∫ (10.30)

The elemental equation of motion and boundary conditions can be obtained from Hamilton’s

principle, as follows

( )2

1

0

t

nc

t

T U W dtδ δ − + = ∫ (10.31)

Substituting equation (10.28), (10.29) and (10.30), into equation (10.31), we get

{ } ( ){ }2

1

22 2 2

0 0 0

1 1

2 2( , ) 0

t L L L

xx x xx x sc x

t

Av I dz EI k AG v dz f z t vdz dtδ ρ ρ ϕ ϕ ϕ δ

′ ′+ − + − + = ∫ ∫ ∫ ∫�� (10.32)

611

On operating the variation operator, from equation (10.32) we get

{ } ( ) ( ){ }2

1 0 0 0

( ) ( ) ( ) ( , ) 0

t L L L

xx x x xx x sc x x

t

Av v I dz EI k AG v v dz f z t vdz dtρ δ ρ ϕ δ ϕ ϕ δ ϕ ϕ δ ϕ δ

′ ′ ′ ′+ − + − − + = ∫ ∫ ∫ ∫� �� �

(10.33)

On changing the order of variation and differentiation in equation (10.33), we get

2

1 0

( ) ( )

t L

xx x x

t

Av v I dz dtt t

ρ δ ρ ϕ δϕ ∂ ∂

+ + ∂ ∂

∫ ∫ ��

( )2 2

1 10 0

( ) ( , ) 0

t tL L

xx x x sc x x

t t

EI k AG v v dz dt f z t vdz dtx x

ϕ δϕ ϕ δ δϕ δ ∂ ∂ ′ ′− + − − + =

∂ ∂ ∫ ∫ ∫ ∫ (10.34)

On performing integration by parts of terms, which has both the differential and variational operators

(i.e., first four terms), in equation (10.34), we get

{ } { } { } { }2 2

2 2

1 11 10 0 0 0

( ) ( ) ( ) ( )

t tt tL L L L

xx x x xx x x

t tt t

Av v dz Av v dz dt I dz I dz dtρ δ ρ δ ρ ϕ δϕ ρ ϕ δϕ

− + − +

∫ ∫ ∫ ∫ ∫ ∫� ��� ��

[ ] { } ( )( ){ } ( )( )2 2 2 2

1 1 1 10 00 0

( ) ( )

L Lt t t tL L

xx x x xx x x sc x sc x

t t t t

EI dt EI dz dt k AG v v dt k AG v v dz dtϕ δϕ ϕ δϕ ϕ δ ϕ δ

′ ′′ ′ ′′ ′− + − − + −

∫ ∫ ∫ ∫ ∫ ∫

( )( ){ }2 2

1 10 0

( , ) 0

t tL L

sc x x

t t

k AG v dz dt f z t vdz dtϕ δϕ δ

′+ − + = ∫ ∫ ∫ ∫ (10.35)

The first and third terms of equation (10.35) will vanish, since variations are not defined in temporal

domain. Remaining terms can be rearranged in the following form

( ){ }2

1 0

( , )

t L

sc x

t

Av k AG v f z t vdz dtρ ϕ δ

′′ ′− + − − ∫ ∫ �� ( ){ }

2

1 0

t L

xx x xx x sc x x

t

I EI k AG v dz dtρ ϕ ϕ ϕ δϕ

′′ ′− − − − ∫ ∫ ��

[ ] ( )( ){ }2 2

1 10 0

( ) 0

L Lt t

xx x x sc x

t t

EI dt k AG v v dtϕ δϕ ϕ δ ′ ′− − − = ∫ ∫ (10.36)

612

Variations vδ and xδϕ in spatial domain are arbitrary, this yields the differential equations of motion

as

( )sc xAv k AG v fρ ϕ′′ ′− − =�� (10.37)

and

( ) 0xx x xx x sc xI EI k AG vρ ϕ ϕ ϕ′′ ′− − − =�� (10.38)

and boundary conditions as

( )0

0L

sc xk AG v vϕ δ′ − = and 0

0L

xx x xEI ϕ δϕ′ = (10.39)

Equations of motions (10.37) and (10.38) can be combined to a single equation. On rearranging

equation (10.37) for free vibrations, we get

x

sc

v vk G

ρϕ ′ ′′= − �� (10.40)

which gives

x

sc

v vk G

ρϕ′′ ′′′ ′= − �� ; x

sc

v vk G

ρϕ′ ′′= −�� �� ���� and x

sc

v vk G

ρϕ ′′′ ′′′′ ′′= − �� (10.41)

On differentiating equation (10.38) with respect to z, we get

( ) 0xx x xx x sc xEI I k AG vϕ ρ ϕ ϕ′′′ ′ ′′ ′− + − =�� (10.42)

On substituting equations (10.40) and (10.41) into equation (10.42), we get

0xx xx sc

sc sc sc

EI v v I v v k AG v v vk G k G k G

ρ ρ ρρ

′′′′ ′′ ′′ ′′ ′′− − − + − − =

�� �� ���� ��

which gives, the Timoshenko beam equation for free vibrations as

24 2 4 4

4 2 2 2 41 0xx

xx xx

sc sc

Iv v E v vEI A I

z t k G t z k G t

ρρ ρ

∂ ∂ ∂ ∂+ − + + =

∂ ∂ ∂ ∂ ∂ (10.43)

613

It should be noted that due to the rotary inertia and the shear deformation in the Timoshenko beam,

two extra terms (the third and fourth terms in the left hand side of the above equation) are appearing

in the equation of motion as compared to the Euler-Bernoulli beam, in which only the first two terms

in the left hand side appears. The third and fourth terms containing ksc are related to the shear effect,

whereas without it is related to the rotary inertia (i.e., xxI vρ ′′�� ). It should be noted that similar equation

of motion could be developed for z-x plane of the following form

24 2 4 4

4 2 2 2 41 0

yy

yy yy

sc sc

Iu u E u uEI A I

z t k G t z k G t

ρρ ρ

∂ ∂ ∂ ∂+ − + + =

∂ ∂ ∂ ∂ ∂ (10.44)

where u is the linear transverse displacement in the x-axis direction, and Iyy is the moment of inertia of

the cross section about the y-axis.

Exact Solution:

Equation (10.43) is solved for a simply supported end condition of a shaft. For obtaining the closed

form solution of equation (10.43) for the specified boundary conditions, it can be simplified by using

a general solution of the following form

( , ) sin sinn nf

nzv z t A t

L

πω

=

(10.45)

On substituting equation (10.45) into equation (10.43), we get

( ) ( )2 4 4 2

2 44 21 0nf nf

sc xx sc

L AL L En n

Ek G EI E k G

ρ ρ ρω π ω π

− + + + − =

(10.46)

On solving equation (10.46) for nfω , we get

22 4

2nf

b b ac

aω

− + −= (10.47)

with

( )4 2

21

xx sc

AL l Eb n

EI E k G

ρ ρπ

= + +

; 2 4

sc

La

Ek G

ρ= − ; ( )4πnc −= (10.48)

614

The above exact solution will be used for comparing natural frequencies for a simply supported end

condition of a shaft with the finite element method as a benchmark solution for the convergence

study. On the similar lines for other boundary conditions expressions of natural frequencies could be

attempted.

10.3 Finite Element Formulations of the Timoshenko Beam

For the finite element analysis, we need to discretise the beam into number of elements as shown in

Fig. 10.9. Consider a finite element of the shaft of length l in an elemental co-ordinate system x-y-z.

Since the two orthogonal transverse plane motions are uncoupled, the deformation of the element is

initially considered in the y-z plane with two nodes 1 and 2 as shown in Figure 10.10. The motion in

z-x plane could be analysed in the similar lines. Let v be the nodal linear transverse displacement of

the shaft element. Let v dv dz′ = be the total slope of the beam, which consists of two parts, one due

to the bending, which is ϕx and other due to the shear, which is β and is given by ( )xv ϕ′ − . From the

displacement field of equation (10.22), we know that the axial displacement of a point P(x, y, z) at a

distance y from the centre line is only due to the bending slope and that the shear force has no

contribution to this. In previous section we observed that there are two independent variables in this

problem (i.e., v and xϕ ).

Figure 10.9 Discretization of a beam into number of elements

Figure 10.10 A typical beam element in y-z plane

615

From equations (10.37) and (10.38) without considering the work done by the external forces,

equations of motion are

( ) 0sc x

Av k AG vρ ϕ′

′− − =�� (10.49)

and

( ) 0xx x xx x sc xEI I k AG vϕ ρ ϕ ϕ′′ ′− + − =�� (10.50)

In the finite element model, the continuous displacement field can be approximated in terms of

descretised generalized-displacements of element nodes. In the present study, each element in single

plane (e.g., y-z) has two nodes and each node has two generalised displacements (one linear and the

other rotational). Therefore, displacements could be obtained within the element by using appropriate

shape functions to be derived in the subsequent subsection.

10.3.1 Weak Formulations of the Timoshenko Beam Element for the Static Case

Let the static case be considered first by dropping time derivative terms. Equations of motion (10.49)

and (10.50) could be written as below while boundary conditions remains the same

( ) 0sc x

k AG v ϕ′

′ − = (10.51)

and

( ) 0xx x sc xEI k AG vϕ ϕ′′ ′+ − = (10.52)

On assuming approximate solution of the following form, we have

{ }( )( ) ( ) ( )neev z S z u= (10.53)

and

{ }( )( ) ( ) ( )nee

x z T z uϕ = (10.54)

On substituting approximate solutions of equations (10.53) and (10.54) in equations of motion (10.51)

and (10.52), the residue of each equation is given by

( )( ) ( ) ( )

1

e e e

sc xR k AG v ϕ′ ′= − (10.55)

and

( )( ) ( ) ( ) ( )

2

e e e e

x sc xR EI k AG vϕ ϕ′′ ′ ′= + − (10.56)

616

Employing the Galerkin principle, one has

{ } ( )

1

0

0

l

eS R dz =∫ (10.57)

and

{ } ( )

2

0

0

l

eT R dz =∫ (10.58)

Using equations (10.55) and (10.56) into equations (10.57) and (10.58), we get

{ } ( )( ) ( )

0

0

l

e e

sc xS k AG v dzϕ′ ′

− =∫ (10.59)

and

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

0

0

l

e e e

x sc xT EI k AG v dzϕ ϕ′′ ′ ′+ − =∫ (10.60)

On performing integration by parts, it gives

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

00

0

ll

e e e e

sc x sc xS k AG v S k AG v dzϕ ϕ′ ′′− − − =∫ (10.61)

and

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

00 0 0

0

l l lle e e e

x x sc sc xT EI T EI dz T k AGv dz T k AG dzϕ ϕ ϕ′ ′ ′ ′′− + − =∫ ∫ ∫ (10.62)

On looking into the completeness and compatibility requirements, we need the compatibility

requirement up to ( )ev ′ and

( )e

xϕ ′ (i.e., up to the first derivative) and the completeness requirement up

to ( )ev and

( )e

xϕ (i.e., up to the value of variables itself). Hence, according to these requirements a

linear interpolation function would serve the purpose. The variables ( )ev and

( )e

xϕ do not have same

the same physical units; they can be interpolated, in general, with different degrees of interpolation.

As we know for thin beams we have ( ) ( )e e

xvϕ ′= − , hence a linear in ( )e

v implies a constant in ( )e

xϕ .

This will make the bending energy ( )2

1 ( )

2

0

l

e

xEI dzϕ ′∫ to zero. This numerical problem is known as

shear locking (since it originates from the shear effect). To overcome this, several alternative methods

have been developed in the literature (Reddy, 2003).

617

10.3.2 Derivation of Shape Functions

Now on assuming that ( )ev is a cubic polynomial, then

( )e

xϕ should be of the same order as ( )e

v ′ that is

therefore a quadratic. These are exact shape functions in the static analysis. The inter-element

compatibility requires that 1v ,

1xϕ ,

2v and 2xϕ must be continuous. Therefore, ( )e

v and ( )e

xϕ can be

assumed as

( )( ) 2 2

1 2( , )ev v v a bξ η ξ η ξη ξ η ξη= + + + − (10.63)

and

1 2

( ) ( , )e

x x x cϕ ξ η ϕ ξ ϕ η ξη= + + (10.64)

with

1z

lξ = − and

z

lη = (10.65)

where a, b and c are unknown interpolation coefficients, and ξ and η are called natural coordinates.

All four boundary conditions of the element in Fig. 10.10 are satisfied as

( )

10

e

zv v

== ; ( )

2

e

z lv v

== ;

1

( )

0

e

x xz

ϕ ϕ=

= and 2

( )e

x xz l

ϕ ϕ=

= (10.66)

because ξη and 2 2( )ξ η ξη− vanish at boundaries of the element. Therefore a, b, c are assumed to be

internal degrees of freedom (DOFs), which will be eliminated subsequently. Because of these DOFs

we could be able to make the shape function of the required degree of polynomial (i.e. a cubic for ( )ev

and a quadratic for ( )e

xϕ ) without violating desired boundary conditions. It could be observed that

without internal DOFs, the assumed shape function would be linear for both ( )ev and

( )e

xϕ . The

presence of ξη and 2 2( )ξ η ξη− is necessary in order to complete the polynomial of the required

degree to avoid the shear locking. Equation (10.66) can be written as

( ) 2 2( , ) 0 0 ( ) 0ev ξ η ξ η ξη ξ η ξη = −

1 21 2

T

x xv v a b cϕ ϕ

{ }( )ne

S u= (10.67)

and

( ) ( , ) 0 0 0 0eϕ ξ η ξ η ξη= 1 21 2

T

x xv v a b cϕ ϕ

{ }( )ne

T u= (10.68)

with

618

0

1 0 0 0 0 0 0z

S=

= and 0 0 1 0 0 0 0z l

S=

= (10.69)

and

0

0 1 0 0 0 0 0z

T=

= and 0 0 0 1 0 0 0z l

T=

= (10.70)

where, { }( )ne

u is the nodal displacement vector. On substituting equations (10.67)-(10.70) in

equations (10.61) and (10.62), we get

{ } { } { } { }

0

( ) ( )

0 0

( )

0

( )

0

0

0

0

sc x z

sc xl l z lne ne

sc sc

k AG v

k AG v

k AG S S dz u k AG S T dz u

ϕ

ϕ

=

=

′ − − ′ −

′ ′ ′− =

∫ ∫ (10.71)

and

{ } { } { } { } { } { }

0

( ) ( ) ( )

0 0 0

0

0

0

0

0

xx x z

l l lne ne ne

xx sc sc xx x z l

EI

EI T T dz u k AG T S dz u k AG T T dz u EI

ϕ

ϕ

=

=

′−

′ ′ ′− + =

∫ ∫ ∫ (10.72)

On combining equations (10.71) and (10.72), the stiffness relationship could be obtained as

[ ] { }{ }

{ }

( )( )

7 17 7

7 10

ne

ne fK u

××

×

=

(10.73)

with

[ ] [ ] [ ] [ ] [ ] [ ]1 2 3 4 5K K K K K K= − + − + (10.74)

[ ] { }1

0

l

scK k AG S S dz′ ′= ∫

1 0 1 0 0 0 0

0 0 0 0 0 0 0

1 0 1 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 1 (3 ) 0 0

0 0 0 0 0 1 (5 ) 0

0 0 0 0 0 0 0

sc

l l

l l

k AG

l

l

− −

=

(10.75)

619

[ ] { }2

0

l

scK k AG S T dz′= ∫

0 1 2 0 1 2 0 0 1 6

0 0 0 0 0 0 0

0 1 2 0 1 2 0 0 1 6

0 0 0 0 0 0 0

0 1 6 0 1 6 0 0 0

0 0 0 0 0 0 1 30

0 0 0 0 0 0 0

sck AG

− − −

= −

−

(10.76)

[ ] { }3

0

l

K T EI T dz′ ′= ∫

0 0 0 0 0 0 0

0 1 3 0 1 6 0 0 1 12

0 0 0 0 0 0 0

0 1 6 0 1 3 0 0 1 12

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 1 12 0 1 12 0 0 1 30

xxEI

=

(10.77)

[ ] { }4

0

l

scK k AG T S dz= ∫

0 0 0 0 0 0 0

1 2 0 1 2 0 1 6 0 0

0 0 0 0 0 0 0

1 2 0 1 2 0 1 6 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

1 6 0 1 6 0 0 1 30 0

sck AG

−

= − − − −

(10.78)

[ ] { }5

0

l

scK k AG T T dz= ∫

0 0 0 0 0 0 0

0 1 3 0 1 6 0 0 1 12

0 0 0 0 0 0 0

0 1 6 0 1 3 0 0 1 12

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 1 12 0 1 12 0 0 1 30

sck AG

=

(10.79)

and

[ ]

0 02 2 6

02 3 2 6 6 12

0 02 2 6

02 6 3 6 12

0 0 0 0 06 6

0 0 0 0 05 30

06 12 6 12 30 30 3

k k k k k

l l

k kl EI k kl EI k kl

l l

k k k k k

l l

k kl EI k kl EI k klK

l l l

k k

k k

l

k kl k kl k kl EI

l

−

− + −

− − − −

− = − +

−

+

(10.80)

where

620

sck k AG= (10.81)

Equation (10.73) has a vector { }( )ne

u , which contain three internal DOFs that need to be eliminated at

this stage with the help of the static condensation scheme. Defining the external DOFs (four in

numbers) of the element as masters and internal DOFs (three in numbers) of the element as slaves, the

stiffness matrix in equation (10.73) is subdivided as

[ ] [ ][ ] [ ]

{ }{ }

{ }

( )11 12 ( )4 4 4 3

21 223 4 3 3 7 70

ne

nex x

x x x

k k fu

k k

=

(10.82)

with

[ ]( )

( )

2

11 3

2 2

6 sym

3 0.5 22;

6 3 6

3 ( 6 1) 3 0.5 2

xx

sc

l lEIk

lk l

l l l l l

Φ + = − −

− + − Φ +

[ ]22

2

10 0 0

0 6 130

50 1

6

sck AGk

l

l l

=

Φ +

[ ] [ ]12 21

0 2 0 2

0 0 0 0 ;12

2 2

T sck AGk k

l l

− = = −

{ }

{ }( )

( )

ne

ne

d

au

b

c

=

2

12 xx

sc

EI

k GAlΦ = ; { }

1 2

( )

1 2

Tne

x xd v vϕ ϕ = and { }1 1 2 2

( ) Tne

y yz y yzf S M S M = − −

1 0( )y sc x z

S k AG v ϕ=

′= − − ; 1 0yz xx x z

M EI ϕ=

′= ;

2( )y sc x z l

S k AG v ϕ=

′= − − ; 2yz xx x z l

M EI ϕ=

′= .

where Φ is called the shear parameter, Sy is the shear force, and Myz is the bending moment. Equation

(10.82) can be written as

[ ]{ } [ ] { }( ) ( )

11 12

ne ne

a

k d k b f

c

+ =

and [ ]{ } [ ] { }( )

21 22 0ne

a

k d k b

c

+ =

(10.83)

The second set of equation (10.83) gives

621

[ ] [ ]1

22 21

a

b k k

c

−

= −

1 21 2

T

x xv vϕ ϕ 1

2

1

2

0 02 2

2 2

6 63 3

x

x

l lv

l l

v

l l

ϕµ µµ µ

µ µ ϕµ µ

−

= − − − −

(10.84)

with

)1(1 Φ+=µ (10.85)

On substituting equation (10.84) into the first set of equation (10.83), we get

[ ]{ } [ ] [ ] [ ]{ }( ) { }1( ) ( ) ( )

11 12 22 21

ne ne nek d k k k d f

−+ − = (10.86)

which can be can be written as

[ ]{ } { } )()( nenefdk = (10.87)

with

[ ] [ ] [ ][ ] [ ]1

11 12 22 21k k k k k−

= − (10.88)

where [ ]k is the condensed stiffness matrix. Now internal DOFs could be eliminated from the shape

assumed functions also. On substituting a, b, and c obtained from equation (10.84) into equation

(10.67), we get

1 2

1

2

1

2

( ) 2 2

1 2

1

2 2

2

1

2 2

2

( , ) 0 0 ( ) 0

0 0 ( ) 0

0 02 2

0 0 ( ) 02 2

6 63 3

Te

x x

x

x

x

x

v v v a b c

va

bv

c

l lv

l l

v

l l

ξ η ξ η ξη ξ η ξη ϕ ϕ

ϕξ η ξη ξ η ξη

ϕ

ϕ µ µξ η ξη ξ η ξη µ µ

µ µϕ µ µ

= −

= + −

−

= + − − − − −

1

2

1

2

x

x

v

v

ϕ

ϕ

which could be simplified as

622

1 2

( )

1 1 2 3 2 4

e

x xv N v N N v Nϕ ϕ= + + + { } )(nedN= (10.89)

with

[ ]1

( ) ( ) ( )1

i i iN z z zα β= + Φ

+ Φ;

2

12 xx

sc

EI

k GAlΦ = ; i = 1, 2, 3, 4.

2 3

1 1 3 2α η η= − +; 1 1β η= − ;

2 3

2 ( 2 )lα η η η= − +; 2

2 ( ) / 2lβ η η= − ;

2 3

3 3 2α η η= −;

3β η= ;

2 3

4 ( )lα η η= − +; 2

4 ( ) / 2lβ η η= − + ;

z

lη = and { }

1 2

( )

1 2

Tne

x xd v vϕ ϕ =

where { }( )ne

d is the nodal displacement vector and N is the translational shape function vector.

The ( )i zα functions are associated with the bending deformation and the ( )i zβ functions are due to

the shear deformation of a Timoshenko beam.

Similarly, on substituting a, b and c obtained from equation (10.84) in equation (10.68), we get

{ }1 2

( )( )

1 1 2 3 2 4

nee

x x xM v M M v M M dϕ ϕ ϕ= + + + = (10.90)

with

[ ]1

( ) ( ) ( )1

i i iM z z zε δ= + Φ

+ Φ ;

2

12 xx

sc

EI

k GAlΦ = ; i =1, 2, 3, 4

2

1 (6 6 ) / ;lε η η= − 1 1δ = ;

2

2 1 4 3 ;ε η η= − + 2 1 ;δ η= −

2

3 (6 6 ) / ;lε η η= − 3 0δ = ;

2

4 3 2 ;ε η η= − 4δ η= ;

z

lη =

623

where M is the rotational shape function vector. Functions, ( )i zε , are associated with the bending

deformation and functions, ( )i zδ , are due to the shear deformation of a Timoshenko beam.

10.3.3 Weak Formulation of the Timoshenko Beam Element for the Dynamic Case

For dynamic case the shape functions derived in the previous section will still be valid and could take

the following form

1 2

( )

1 1 2 3 2 4( , ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )e

x xv z t N z v t N z t N z v t N z tϕ ϕ= + + + { }( )

( ) ( )ne

N z d t= (10.91)

and

{ }1 2

( )( )

1 1 2 3 2 4( , ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )nee

x x xz t M z v t M z t M z v t M z t M z d tϕ ϕ ϕ= + + + = (10.92)

On substituting equations (10.91) and (10.92) in equations of motion (10.49) and (10.50), residues of

each equations of motion are given by

( )( ) ( ) ( ) ( )

1

e e e e

sc xR Av k AG vρ ϕ′′ ′= − −�� (10.93)

and

( )( ) ( ) ( ) ( ) ( )

2

e e e e e

xx x xx x sc xR EI I k AG vϕ ρ ϕ ϕ″ ′= − + −�� (10.94)

Using the Galerkin method to minimize the residue, one has

{ }1

( )

0

0

l

eN R dz =∫ (10.95)

and

{ } ( )

2

0

0

l

eM R dz =∫ (10.96)

On substituting of equations (10.93) and (10.94) in equations (10.95) and (10.96) the weak

formulation can be obtained as

{ } { } { } { } { } { }

0

( ) ( ) ( )

0 0 0

( )

0

( )

0

sc x z

l l lne ne ne

sc sc

sc x z l

k AG v

A N N d dz k AG N N d dz k AG N M d dzk AG v

ϕ

ρϕ

=

=

′ − −

′ ′ ′+ − = ′ −

∫ ∫ ∫��

(10.97)

624

and

{ } { } { } { } { } { } { } { }( ) ( ) ( ) ( ) 0

0 0 0 0

0

0

l l l lne xx xne ne ne z

xx xx sc sc

xx x z l

EII M M d dz EI M M d dz k AG M N d dz k AG M M d dz

EI

ϕρ

ϕ

=

=

′−

′ ′ ′+ − + = ′

∫ ∫ ∫ ∫��

(10.98)

Equations (10.97) and (10.98) can be written as

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

0

( ) ( ) ( )

1 1 2

( )

0

( )

0

sc x z

ne ne ne

sc x z l

k AG v

M d K d K dk AG v

ϕ

ϕ

=

=

′ −

+ − = ′ −

�� (10.99)

and

[ ]{ } [ ]{ } [ ]{ } [ ]{ }( ) ( ) ( ) ( ) 0

2 3 4 5

0

0

ne xne ne ne z

x z l

EIM d K d K d K d

EI

ϕ

ϕ

=

=

′

+ − + = ′

�� (10.100)

with

[ ] { }1

0

l

M A N N dzρ= ∫

2

2 2 2

2 2 2 2

2 2 2 2 2 2

4(70 147 78) sym.

(35 77 44) (7 14 8)

840(1 ) 4(35 63 27) (35 63 26) (70 147 78)

(35 63 26) (7 14 6) (35 77 44) (70 14 8)

l lAl

l

l l l l

ρ

Φ + Φ+

Φ + Φ+ Φ + Φ+ = +Φ Φ + Φ+ Φ + Φ+ Φ + Φ+ − Φ + Φ+ − Φ + Φ+ − Φ + Φ+ Φ + Φ+

[ ] { }2

0

l

xxM I M M dzρ= ∫

2 2

2

2 2 2 2

36 sym

3(5 1) (10 5 4)

36 3(5 1) 3630(1 )

3(5 1) (5 5 1) 3(5 1) (10 5 4)

xxl lI

ll

l l l l

ρ

− Φ − Φ + Φ + = − Φ −+ Φ − Φ − Φ − Φ − Φ − Φ + Φ +

[ ] { }1

0

l

scK k AG N N dz′ ′= ∫

2

2 2

2 2 2

2 2 2 2

12(5 10 6) sym

6 (5 10 8)

60(1 ) 12(5 10 6) 6 12(5 10 6)

6 (5 10 2) 6 (5 10 8)

scl lk AG

l l

l l l l

Φ + Φ +

Φ + Φ + = + Φ − Φ + Φ + − Φ + Φ +

− Φ + Φ + − Φ + Φ +

[ ] { }2

0

l

scK k AG N M dz′= ∫

2 2

22

2 2 2 2

12(5 6) sym

6 (5 10 8)

12(5 6) 6(5 5 1) 12(5 6)60(1 )

6 (5 10 2) 6 (5 10 8)

scl lk AG

ll

l l l l

Φ + Φ + Φ + = − Φ + Φ + Φ − Φ ++ Φ

− Φ + Φ + − Φ + Φ +

[ ] { }3

0

l

xxK M EI M dz′ ′= ∫

2 2

2 3

2 2 2 2

12 sym

6 ( 2 4)

12 6 12(1 )

6 ( 2 2) 6 ( 2 4)

xxl lEI

ll

l l l l

Φ + Φ + = − −+ Φ

− Φ + Φ − − Φ + Φ +

625

[ ] { }4

0

l

scK M k AG N dz′= ∫ 2 2 2

2

2 2 2 2 2 2

12(5 6) sym

6(5 5 1) (5 10 8)

12(5 6) 6 12(5 6)60(1 )

6(5 5 1) (5 10 2) 6(5 5 1) (5 10 8)

scl lk AG

ll

l l l l

Φ + − Φ + Φ − Φ + Φ + = − Φ + − Φ ++ Φ − Φ + Φ − − Φ + Φ + Φ + Φ − Φ + Φ +

[ ] { }5

0

l

scK M k AG M dz= ∫

2 2

2

2 2 2 2

36 sym

3(5 1) (10 5 4)

36 3(5 1) 3630(1 )

3(5 1) (5 5 1) 3(5 1) (10 5 4)

scl lk AG

ll

l l l l

− Φ − Φ + Φ + = − Φ −+ Φ − Φ − Φ − Φ − Φ − Φ + Φ +

with 2

12 xx

sc

EI

k GAlΦ =

On combining equations (10.99) and (10.100), we get

[ ]{ } [ ]{ } { }( ) ( ) ( )ne ne ne

M d K d f+ =�� (10.101)

with

[ ] [ ] [ ] [ ] [ ]1 2 T RM M M M M= + = +

[ ] [ ] [ ] [ ]2

2

10 TTTT MMMM Φ+Φ+=

[ ] [ ] [ ] [ ]2

2

10 RRRR MMMM Φ+Φ+=

and

[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]1 2 3 4 5 0 1K K K K K K K K= − + − + = + Φ

where { }( )ne

f is generalized force vector, the mass matrix [ ]M consists of the translational mass

matrix [ ]TM and the rotational mass matrix [ ]RM . Details of the mass matrix and stiffness matrices

are given in Appendix 10.1. Now, through examples the effect of rotary inertia and shear deformation

would be illustrated.

Example 10.2 Obtain natural frequency parameters defined by a non-dimensional form as nω

( )EIAL nn /244 ωρω = of a uniform, non-rotating simply supported Timoshenko beam for the first

four modes. The slenderness parameter R (=r/2L) is to be varied from 0.02 (thin beam) to 0.10 (thick

beam). Show the comparison for different number of elements and with the exact analytical formula.

626

Solution: Since in previous chapters the usual finite element procedures have been dealt in detail (i.e.,

regarding the elemental equations, assembly procedures, application of boundary conditions, eigen

value problem formulations, etc.), hence here those details for the present problem is omitted. The

uniform, non-rotating simply supported Timoshenko beam for obtaining natural frequencies, is first

discretised into five numbers of elements. From the eigen value formulation of the present problem

first four lowest natural frequencies obtained and are compared with the published work (Ku 1998)

and by the classical closed form solution given by Shames and Dym, 2005. Ku (1998) has considered

C0 continuity (i.e., compatibility up to linear displacements) for the Timoshenko beam model whereas

model represented here considers C1 continuity (i.e., the compatibility up to the linear and angular

displacements). Comparisons of non-dimensional natural frequencies for first four modes are

presented in tabular form in Tables 10.4-10.7, respectively. In published work of Ku, 1998, only first

two natural frequencies are available. The present study extracts natural frequencies in first four

modes. These have been compared with classical closed form solutions. These studies are also

conducted by discretising the beam into the three and seven number of elements in order to make

convergence comparisons and are summaries in Tables 10.4-10.7.

Table 10.4 Natural frequency parameters nω of a uniform, non-rotating simply supported

Timoshenko beam for the first mode.

R

Present work Ku (1998) Shames and

Dym (2005) *(3) (5) (7) (3) (5) (7)

0.02 3.1316 3.1300 3.1294 3.13460 3.1297 3.1293 3.1325

0.04 3.0992 3.0964 3.0956 3.11039 3.0942 3.0938 3.1061

0.06 3.0497 3.0449 3.0439 3.07206 3.0403 3.0399 3.0644

0.08 2.9881 2.9810 2.9792 3.01909 2.9736 2.9733 3.0103

0.10 2.9190 2.9098 2.9073 2.96344 2.8995 2.8992 2.9472

*Number in the bracket indicates number of elements considered.

Table 10.5 Natural frequency parameters nω of a uniform, non- rotating simply supported

Timoshenko beam for the second mode.

R

Present work Ku (1998) Shames and

Dym (2005) (3) (5) (7) (3) (5) (7)

0.02 6.2467 6.2033 6.1957 6.2875 6.2018 6.1913 6.2122

0.04 6.0555 5.9864 5.9702 6.0335 5.9589 5.9497 6.0210

0.06 5.8041 5.7082 5.6831 5.7170 5.6540 5.6462 5.7573

0.08 5.5373 5.4198 5.3873 5.3973 5.3439 5.3374 5.4671

0.10 5.2779 5.1454 5.1076 5.0995 5.0481 5.0481 5.1792

627

Table 10.6 Natural frequency parameters nω of a uniform, non-rotating simply supported

Timoshenko beam for the third mode.

R

Present work Shames and Dym

(2005) (3) (5) (7) (10) (15)

0.02 9.8418 9.2159 9.1661 9.1444 9.1344 9.1934

0.04 9.6009 8.6629 8.5740 8.5285 8.5049 8.6364

0.06 9.2564 8.0541 7.9361 7.8726 7.8389 7.9839

0.08 8.8619 7.4939 7.3590 7.2849 7.2450 7.3659

0.10 8.1258 7.0046 6.8613 6.7813 6.7381 6.8137

Table 10.7 Natural frequency parameters nω of a uniform, non-rotating simply supported Timoshenko

beam for the fourth mode.

R

Present work Shames and Dym

(2005) (3) (5) (7) (10) (15)

0.02 13.2640 12.2161 12.0504 11.9727 11.9361 12.0421

0.04 12.5987 11.2187 10.9781 10.8396 10.7659 10.9357

0.06 11.6571 10.2279 9.9587 9.78718 9.6931 9.8244

0.08 10.3753 9.3770 9.1074 8.9223 8.8190 8.8755

0.10 8.4569 7.8377 7.7423 7.6888 7.6595 8.0456

Due to the shear effect, deflection of the beam increases which reduces the effective stiffness. So the

non-dimensionalised natural frequency decreases as the diameter of the beam is increased (i.e. with

the slenderness ratio). This trend could be observed in all four modes. The accuracy of the present

solution is well within 5% for all cases. By FEM we over-predict with coarser discretisation and it

could be seen that as the number of elements are increased from 3 to 7 for first two modes and from 3

to 15 for the third and fourth modes the natural frequency parameter is decreasing. Especially

converge (not much improvement) could be observed in Tabled 10.6 and 10.7 between the number of

elements 10 to 15. The difference in the closed form solution of Shames and Dym (2005) and the

FEM results could be due to the approximate nature of the eigen function ( sinn z

L

πchosen in the

closed-form solution).

Example 10.3 A typical simply supported rotor disc system as shown in Figure 10.11 is to be

analyzed for obtaining the whirl natural frequencies to show the application of the finite element

method. Physical properties of the rotor system are given as: the diameter of shaft is 0.1m, the length

of shaft is 3.5 m, the Young’s modulus of material of shaft is 2.08×1011

N/m2, the mass density of the

628

shaft material is 7830 kg/m3, the Poisson’s ratio is 0.3, the number of rigid discs are 4, and the mass of

each rigid disc is 60.3 kg.

Solution: The rotor is discretised into the seven and fourteen elements, respectively, as shown in

Figures 10.11 and 10.12. In the case of seven elements, two identical rigid bearings are located at

node numbers two and seven; and four rigid discs are located at node numbers three, four, five and

six. In case of fourteen element member, two identical rigid bearings are located at node numbers

three and thirteen, and rigid discs are located at node numbers five, seven, nine and eleven. The shaft

is assumed uniform along the span.

Rigid discs are considered as point masses and these point masses are added to mass matrix

corresponding to locations of rigid discs linear accelerations. The assembled mass and stiffness

matrices are obtained by usual method and boundary conditions are applied to get the reduced form

dynamic matrix. Natural whirl frequencies are obtained by solving eigen value problem. Natural whirl

frequencies are obtained for 7 elements and 14 elements model and are given in Table 10.8. Results

show that good convergence has already occurred with 7 elements model.

Figure 10.11 Rotor bearing system with rigid disks (7-elements of 0.50 m length each)

Figure 10.12 Rotor bearing system with rigid disks (14-elements of 0.25 m length each)

629

For the present case R is less than 0.02 so effect of shear is less but as rotary inertia is also considered,

natural whirl frequencies of the rotor bearing system decreases as compared to Euler-Bernoulli case.

This decrease is 0.3%, 1.8%, 2% and 3% for the first, second third and fourth modes, respectively.

The difference is higher at higher modes since the rotary inertia becomes predominant at higher

frequencies. It should be noted that in the present problem, the rotor is assumed to be non-rotating and

because of this gyroscopic couple is absent. The case of rotating shaft with the rotary inertia, shear

deformation and gyroscopic couple would be considered in the next section.

Table 10.8 Natural whirl frequencies of rotor bearing system with rigid disks (For non-rotating

Timoshenko beam case)

Mode No.

Natural whirl frequencies without

considering shear and rotary inertia

effects (rad/sec)

Natural whirl frequencies with

shear and rotary inertia effects

(rad/sec)

For 7 elements For 14 elements For 7 elements For 14 elements

1 116.1756 115.8136 115.6111 115.3816

2 438.4482 438.2795 431.0719 430.8983

3 861.5609 860.7477 839.1099 838.1357

4 1209.4922 1207.0327 1168.2006 1165.2664

10.4 Whirling of Timoshenko Shafts

In the present section the analysis of whirling of spinning Timoshenko shaft is presented which

includes higher effects like the rotary inertia, the shear effect and the gyroscopic effect. A finite

element formulation including these effects is presented by using the consistent mass matrix

approach. Because of the gyroscopic effect, two perpendicular transverse motions are now coupled

and most importantly natural frequencies of the rotor system depend upon the angular speed of the

shaft. It leads to the forward and backward whirls phenomena, to which we have already described in

detail for single mass rotor system in Chapter 5. The finite element formulation in accordance with

real co-ordinate system with gyroscopic effect yields the gyroscopic matrix as skew symmetric and all

other matrices as symmetric. The eigen value problem (without damping) gives eigen values as pure

imaginary and eigen vectors as complex. The finite element analysis of the Timoshenko rotating beam

are compared with results obtained by Weaver et al. (1990) and Eshleman and Eubanks (1969), who

determined critical speeds from approximate relations.

10.4.1 Equations of Motion of a Spinning Timoshenko Shaft

The analysis of the previous section could be extended for the present case with inclusion of

gyroscopic effects. However, this would lead to coupling of motion in two orthogonal planes. The

630

translation of the cross section centreline neglecting axial motion is given by two displacements u(z ,t)

and v(x ,t) in the x and y direction respectively, which consists of contribution due to both the bending

deformation and contribution due to the shear deformation.

( , ) ( , ) ( , )b su z t u z t u z t= + and ( , ) ( , ) ( , )b sv z t v z t v z t= + (10.102)

The rotation of the cross section is described by the rotation angles ( , )x z tϕ and ( , )y z tϕ about x and y

axes, respectively; which are associated with bending deformation of the element as

( , )

( , ) b

x

v z tz t

zϕ

∂= −

∂ and

( , )( , ) b

y

u z tz t

zϕ

∂=

∂ (10.103)

On differentiating equations (10.102), we get

( , )

( , ) ( , )b

s

u z tu z t u z t

z

∂′ ′= +

∂ and

( , )( , ) ( , )b

s

v z tv z t v z t

z

∂′ ′= +

∂ (10.104)

On noting equations (10.103), equations (10.104) take the form

( , ) ( , ) ( , )y su z t z t u z tϕ′ ′= + and ( , ) ( , ) ( , )x sv z t z t v z tϕ′ ′= − +

or

( , ) ( , ) ( , )s yu z t u z t z tϕ′ ′= − and ( , ) ( , ) ( , )s xv z t v z t z tϕ′ ′= + (10.105)

where su′ and

sv′ are shear strains. For the differential shaft element located at z, the potential energy

of an element of length, l, can be expressed as

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

0

1

2

l

nc y x sc s sU EI k AG u v dzϕ ϕ ′ ′ ′ ′= + + +

∫

Where the first term is the elastic bending and shear deformation energy. On noting equation (10.105)

, we get

( ) ( ){ } ( ) ( ){ }2 22 2

0

1

2

l

nc y x sc y xU EI k AG u v dzϕ ϕ ϕ ϕ ′ ′ ′ ′= + + − + +

∫ (10.106)

631

The kinetic energy of a shaft element rotating at a constant speed, ω, including the translational and

rotational forms is given by

( ) ( ) ( ){ }1 12 2 2 2 2

2 2

0 0

l l

y x P x y y x PT A u v I I dz I dzρ ρ ϕ ϕ ω ϕ ϕ ϕ ϕ ω= + + + + − +∫ ∫� � � �� � (10.107)

In equation (10.107) various kinetic energy terms are contributed as follows: the first term is due to

linear motions, the second term is due to tilting motions, the third term is due to gyroscopic couples

and the last term is due to spinning of the rotor. In Chapter 5 derivation of terms related to gyroscopic

moments have been described in detail, while discussing the energy method. Angular momentum in y-

z plane and z-x plane are opposite in direction and have same magnitude. The quantity x yϕ ϕ is a

constant quantity in the conservative system, so it gives

( )0

x yd

dt

ϕ ϕ= ; so that x y x yϕ ϕ ϕ ϕ= −� � (10.108)

Noting equation (10.108), equation (10.107) becomes

( ) ( ){ }1 12 2 2 2 2

2 2

0 0

2

l l

y x P y x PT A u v I I dz I dzρ ρ ϕ ϕ ωϕ ϕ ω= + + + − +∫ ∫� � �� � (10.109)

If ( , )xf z t and ( , )yf z t are distributed transverse forces in x and y directions, respectively; then the

work done by external forces is

( )0

l

nc x yW f u f v dz= +∫ (10.110)

from the Hamilton’s principle, we have

{ }2

1

( ) 0

t

nc

t

T U W dtδ δ− + =∫ (10.111)

On substituting equation (10.106), (10.109), and (10.110) into equation (10.111), we get

( ) ( ){ }2

1

2 2 2 2 2

0

1

22

t l

y x P y x P

t

A u v I I I dz dtδ ρ ρ ϕ ϕ ωϕ ϕ ω

+ + + − + ∫ ∫ � � �� �

632

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

1

2 22 2

0

1

2

t l

y x sc y x

t

EI k AG u v dz dtδ ϕ ϕ ϕ ϕ ′ ′ ′ ′− + + − + + ∫ ∫

( )2

1 0

0

t l

x y

t

f u f v dz dtδ

+ + =

∫ ∫ (10.112)

On operating the variation operator in equation (10.112), we get

{ }2

1 0

( ) ( ) ( ) ( ) 2 ( ) 2 ( )

t l

x x y y P y x P y x

t

Au u Av v I I I I dz dtρ δ ρ δ ρ ϕ δ ϕ ρ ϕ δ ϕ ωδ ϕ ϕ ωϕ δ ϕ

+ + + − − ∫ ∫ � � � � � �� � � �

( ) ( ) ( ) ( ){ }2

1 0

( ) ( )

t l

y y x x sc y y sc x x

t

EI EI k AG u u k AG v v dz dtϕ δ ϕ ϕ δ ϕ ϕ δ ϕ ϕ δ ϕ

′ ′ ′ ′ ′ ′ ′ ′− + + − − + + + ∫ ∫

{ }2

1 0

0

t l

x y

t

f u f v dzdtδ δ+ + =∫ ∫ (10.113)

On changing the order of variation and differentiation in equation (10.113), we get

2

1 0

( ) ( ) ( ) ( ) ( ) ( )

t l

y y x x P y x P y x

t

Au u Av v I I I I dz dtt t t t t

ρ δ ρ δ ρ ϕ δϕ ρ ϕ δϕ ω δϕ ϕ ωϕ δ ϕ ∂ ∂ ∂ ∂ ∂

+ + + − − + ∂ ∂ ∂ ∂ ∂

∫ ∫ � � �� �

( ) ( )2

1 0

( ) ( ) ( ) ( )

t l

y y x x sc y y sc x x

t

EI EI k AG u u k AG v v dz dtz z z z

ϕ δϕ ϕ δϕ ϕ δ δϕ ϕ δ δϕ ∂ ∂ ∂ ∂ ′ ′ ′ ′− + + − − + + +

∂ ∂ ∂ ∂ ∫ ∫

{ }2

1 0

0

t l

x y

t

f u f v dzdtδ δ+ + =∫ ∫ (10.114)

On performing integration by parts of terms, which has both the differential and variational operators,

in equation (10.114), we get

{ } { } { } { }2 2

2 2

1 11 10 0 0 0

( ) ( ) ( ) ( )

t tt tl l l l

t tt t

Avu u dz Au u dz dt Av v dz Av v dz dtρ δ ρ δ ρ δ ρ δ

− + −

∫ ∫ ∫ ∫ ∫ ∫� �� � �� +

{ } { } { } { }2 2

2 2

1 11 10 0 0 0

( ) ( ) ( ) ( )

t tt tl l l l

y y y y xx x x xx x x

t tt t

I dz I dz dt I dz I dz dtρ ϕ δϕ ρ ϕ δϕ ρ ϕ δϕ ρ ϕ δϕ

+ − + − +

∫ ∫ ∫ ∫ ∫ ∫� �� � ��

633

{ }2

2

110 0

( ) ( ) ( )

t tl l

P x y P x y P y x

tt

I dz I I dz dtωϕ δ ϕ ωϕ δ ϕ ωϕ δ ϕ

− + − +

∫ ∫ ∫ � �

{ } { } { }2 2 2 2

1 1 1 10 00 0

( ) ( ) ( ) ( )

l lt t t tl L

y y y y x x x x

t t t t

EI dt EI dz dt EI dt EI dz dtϕ δϕ ϕ δϕ ϕ δϕ ϕ δϕ

′ ′′ ′ ′′ − + − +

∫ ∫ ∫ ∫ ∫ ∫

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

1 1 10 00

lt t tl l

sc y sc y sc y y

t t t

k AG u u dt k AG u u dz dt k AG u dz dtϕ δ ϕ δ ϕ δϕ

′ ′′ ′ ′− − + − + −

∫ ∫ ∫ ∫ ∫

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

1 1 10 00

lt t tl l

sc x sc x sc x x

t t t

k AG v v dt k AG v v dz dt k AG v dz dtϕ δ ϕ δ ϕ δϕ

′ ′′ ′ ′− + + + − +

∫ ∫ ∫ ∫ ∫

2

1 0

( ) 0

t l

x y

t

f u f v dz dtδ δ

+ + = ∫ ∫ (10.115)

The first, third, fifth, seventh and ninth terms of equation (10.115) will vanish, since variations are

not defined in temporal domain. Remaining terms can be rearranged in the following form

( ){ }2

1 0

t l

sc y x

t

Au k AG u f udz dtρ ϕ δ

′′ ′− + + − ∫ ∫ �� ( ){ }

2

1 0

)

t l

sc x y

t

Av k AG v f vdz dtρ ϕ δ

′′ ′− + − − ∫ ∫ ��

( ){ }2

1 0

t l

y y sc y P x y

t

I EI k AG u I dz dtρ ϕ ϕ ϕ ωϕ δϕ

′′ ′− − − + − ∫ ∫ �� �

( ){ }2

1 0

t l

x x sc x P y x

t

I EI k AG v I dz dtρ ϕ ϕ ϕ ωϕ δϕ

′′ ′− − − − + ∫ ∫ �� �

[ ] ( )( ){ } ( )( ){ }2 2 2 2

1 1 1 10 0 0 0

( ) ( ) 0

l l l lt t t t

y y x x sc y sc x

t t t t

EI dt EI dt k AG u u dt k AG v v dtϕ δϕ ϕ δϕ ϕ δ ϕ δ ′ ′ ′ ′ − − − + − − = ∫ ∫ ∫ ∫

(10.116)

634

Variations uδ , yδϕ , vδ and xδϕ in spatial domain are arbitrary, this yields the differential equations

of motion as

( ) ; ( ) 0;

( ) ; ( ) 0.

sc y x y sc y y P x

sc x y x sc x x P y

Au k AG u f EI k AG u I I

Av k AG v f EI k AG v I I

ρ ϕ ϕ ϕ ρ ϕ ωϕ

ρ ϕ ϕ ϕ ρ ϕ ωϕ

′′ ′ ′′ ′− + = + + − + =

′′ ′ ′′ ′− − = + − − − =

�� ���

�� ���

(10.117)

and boundary (geometrical and natural) conditions are

0

( ) 0l

sc yk AG u uϕ δ′ − = , 0

0l

x xEIϕ δϕ′ = ,

and

0( ) 0

l

sc xk AG v vϕ δ′ + = , 0

0l

y yEIϕ δϕ′ = (10.118)

It should be noted that now all four equations of motion (10.117) are coupled due to gyroscopic

couple terms and need to be solved simultaneously. In the next subsection the finite element

formulation of governing equations are presented.

10.4.2 Finite Element Formulation

Now the finite element formulation for governing equations (10.117) will be developed by using the

Galerkin’s method. As compared to previous section finite element formulations, here two plane

governing equations need to be considered simultaneously. In the finite element model, the

continuous displacement field can be approximated in terms of generalised displacements of the

element nodes. In the present finite element model (see Figures 10.13 and 10.14), each element has

two nodes and each node have four generalized displacements (two linear and other two rotational).

The linear displacement within the element can be obtained by using appropriate shape functions and

are defined as

{ }( )

( , ) ( ) ( )ne

uu z t N z tη= and { }

( )

( , ) ( ) ( )ne

vv z t N z tη= (10.119)

{ }1 1 2 2

( )

1 1 2 2( )

Tne

x y x yt u v u vη ϕ ϕ ϕ ϕ =

635

and

1 2 3 4

1 2 3 4

0 0 0 0

0 0 0 0

u

v

N N N N N

N N N N N

=

= − −

Figure 10.13 A typical beam element in y-z plane

Figure 10.14 A typical rotor element in z-x plane

In the matrix form, it could be combined as

[ ]{ }( )( , )

( ) ( )( , )

ne

t

u z tN z t

v z tη

=

(10.120)

with,

[ ] 1 2 3 4

1 2 3 4

0 0 0 0( )

0 0 0 0

u

t

v

N N N N NN z

N N N N N

= = − −

where Ni ,(i =1, 2, 3, 4) is called the translational shape function matrix and are same as derived in

previous section, which are given in equation (10.89), { }( )

( )ne

tη is nodal displacement vector and

636

angular displacement (it should be carefully noted that in previous section one plane motion was

considered and accordingly in equation (10.89) the order of stacking of displacements are different,

which can be chosen as per convenience, however, the mass and stiffness matrices might take

different form that should be taken care of) within the element can be obtained by

{ }( )

( , ) ( ) ( )x

ne

xz t M z tϕϕ η = and { }

( )

( , ) ( ) ( )y

ne

yz t M z tϕϕ η = (10.121)

and

1 2 3 4

1 2 3 4

0 0 0 0

0 0 0 0

y

x

N M M M M

N M M M M

ϕ

ϕ

= − −

=

In the matrix form

[ ]{ }( )( , )

( )( , )

ney

t

x

z tM z

z t

ϕη

ϕ

=

where,

[ ] 1 2 3 4

1 2 3 4

0 0 0 0( )

0 0 0 0

y

x

t

M M M M MM x

M M M MM

ϕ

ϕ

− − = =

??

is called the rotational shape function matrix. Mi , i = 1, 2, 3, 4 are given in equation (10.90).

10.4.3 The Weak Form Finite Element Formulations

Since in the previous section, already the weak form finite element formulation without gyroscopic

couple effect has been performed in great detail, hence, for the present case the same is described

briefly. Neglecting the work done by external forces and substituting equations (10.119) and (10.121)

in equations of motion (10.117), residues can be given as

( ) ( ) ( ) ( )

1

( ) ( ) ( ) ( )

2

( )

( )

e e e e

sc y

e e e e

sc x

R Au k AG u

R Av k AG v

ρ ϕ

ρ ϕ

′

′

′= − +

′= − −

��

��

637

( )

( ) ( ) ( ) ( ) ( ) ( )

3

( ) ( ) ( ) ( ) ( ) ( )

4

( )e e e e e e

x sc x x P y

e e e e e e

y sc y y P x

R EI k AG v I I

R EI k AG u I I

ϕ ϕ ρ ϕ ωϕ

ϕ ϕ ρ ϕ ωϕ

′′ ′= + − − −

′′ ′= + + − +

�� �

�� �

(10.122)

The Galerkin method is used to minimize the residue. So applying the weight function equivalent to

shape functions, residue can be minimized as

{ } ( )

1

0

0

l

e

uN R dz =∫ ; { }

2

( )

0

0

l

e

vN R dz =∫ ;

{ } ( )

3

0

0x

l

eM R dzϕ =∫ ; { } ( )

4

0

0y

l

eM R dzϕ =∫ (10.123)

using equation (10.122) and (10.123) weak formulations can be obtained and equation of motion for

the finite element takes the form

[ ]{ } [ ]{ } [ ]{ } { }( ) ( ) ( ) ( )

( )ne ne ne ne

M q G q K q f tω− + =�� � (10.124)

with

[ ] [ ] [ ]RT MMM += ; [ ] [ ] [ ]10 KKK Φ+= ; [ ] [ ] [ ] [ ]GGGG 2

10 Φ+Φ+=

where [M] is the elemental mass matrix, [K] is the elemental stiffness matrix and [G] is the elemental

gyroscopic matrix. The mass matrix [M] consists of the translational mass matrix [ ]TM and the

rotational mass matrix[ ]RM which are given as

[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]2 2

0 1 2 0 1 2; T T T T R R R RM M M M M M M M= + Φ + Φ = + Φ + Φ (10.125)

Details of the mass, stiffness and gyroscopic matrices are given in Appendix 10.2.

10.4.4 Rigid Disc Element

Rigid disc equations of motion have already been described in detail in Chapter 5. The rigid disc

equation of motion can be written as,

[ ] { } [ ] { } [ ]{ } { }0 ( )d dd d d

M G f tη ω η η− + =�� � (10.126)

638

where [ ]d

M is the disc mass matrix, [ ]d

G is the gyroscopic matrix and { }( )d

f t is the vector of

generalized external forces Appendix 10.2.

10.4.5 System Equations of Motion

Equations of motion of the complete system can be obtained by assembling the contribution of each

elemental equation of motion, i.e. equation (10.126). The system equation of motion becomes

[ ]{ } [ ]{ } [ ]{ } { }( )M G K f tη ω η η− + =�� � (10.127)

where [ ]M is assembled mass matrix, [ ]G is assembled gyroscopic matrix, [ ]K is assembled

stiffness matrix, { }( )f t is the assembled external force vector and { }( )tη is the assembled vector of

nodal co-ordinates of the whole system.

10.4.6 Eigen Value Problem

For the sake of computation of eigen values and eigen vectors the homogeneous part of the equation

(10.127) written in the first order state vector form as

[ ] [ ]

[ ] [ ]{ }

[ ] [ ][ ] [ ]

{ } { }0 0

00

M Mh h

M G Kω

−+ =

−

� (10.128)

in compact form

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

0M h K h+ =� (10.129)

where

{ }{ }{ }

hη

η

=

�

; [ ][ ] [ ]

[ ] [ ]* 0 M

MM Gω

=

− ; [ ]

[ ] [ ][ ] [ ]

* 0

0

MK

K

−=

Assumed solution of the form,

{ } { }0

th h e

λ= (10.130)

therefore

{ } { }0

th h eλλ=� (10.131)

On substituting equations (10.130) and (10.131) into equation (10.129), we get

639

[ ] { } [ ] { } 00

*

0

*=+ tt ehKehM λλλ (10.132)

Therefore associated eigen value problem becomes

[ ] [ ]( ){ } { }0 0D I hλ+ = with [ ] [ ] [ ]1

* *D M K

−

= (10.133)

Eigen values of equation (10.133) appear as pure imaginary conjugate pairs with magnitude equal to

the natural whirl frequencies. When gyroscopic effect is dropped, eigen values appear as repeated

pairs of pure imaginary roots. When gyroscopic effect is considered, it leads to splitting of eigen

values into forward and backward whirls (i.e. with increase in spin speed, increasing eigen values

represent backward whirls whereas decreasing eigen value represented forward whirls). Because of

the presence of gyroscopic terms, the whirl speeds are a function of the rotational or spin speed of the

shaft. To determine the critical speeds, a Campbell diagram can be drawn to find the intersection

points of the whirl speeds with one per rev excitation i.e. for synchronous whirl λ ω= and higher

multiples of per revolution excitations.

Example 10.4 Obtain the forward and backward critical speeds of a uniform cross section, a simply

supported spinning Timoshenko shaft. Consider gyroscopic effects also. Show the results up to second

modes with different number of elements and for different value of the slenderness ratio.

Solution: In order to evaluate natural whirl frequencies of a uniform cross section, simply supported

rotating (spinning) Timoshenko beam model in which gyroscopic effects also play predominant role,

is discretised into five different elements. The comparison is carried out for the forward and backward

critical speeds , λ=±1, and the results are compared to the values published by Eshleman and Eubanks

(1969). Critical speeds are non-dimensionalised by using natural frequency parameter, nω

( )2 4 2 /n n

AL EIω ρ ω= . The results are also obtained for ten number of elements. These results are

listed in Table 10.9 and Table 10.10 for first and second critical speed respectively. The effect of

shear goes on increasing as slenderness parameter R (R=r/2L, where r is radius of beam), is changed

from 0.02 to 0.10. The results are in agreement with the value published by Eshleman & Eubanks

(1969).

640

Table 10.9 First non-dimensionalised critical speeds of a simply supported rotating Timoshenko

beam.

R Backward whirl (λ = -1) Forward whirl (λ = 1)

Present work Eshleman &

Eubanks

(1969)

Present work Eshleman &

Eubanks

(1969) (5) (10) (5) (10)

0.02 3.1240 3.1234 3.1251 3.1360 3.1354 3.1373

0.04 3.0745 3.0739 3.0780 3.1187 3.1180 3.1245

0.06 3.0028 3.0018 3.0067 3.0899 3.0887 3.1037

0.08 2.9183 2.9166 2.9193 3.0494 3.0473 3.0757

0.10 2.8291 2.8269 2.8234 2.9985 2.9954 3.0416

Table 10.10 Second non-dimensionalised critical speeds of a simply supported rotating Timoshenko

beam.

R Backward Whirl (λ=-1) Forward Whirl (λ=1)

Present work Eshleman &

Eubanks

(1969)

Present Work Eshleman &

Eubanks

(1969) (5) (10) (5) (10)

0.02 6.1597 6.1522 6.1560 6.2487 6.2408 6.2489

0.04 5.8591 5.8440 5.8387 6.1278 6.1084 6.1515

0.06 5.5132 5.4914 5.4493 5.9280 5.8945 6.0045

0.08 5.1858 5.1582 5.0670 5.6734 5.6279 5.8250

0.10 4.8937 4.8621 4.7231 5.3964 5.3258 5.6287

Example 10.5 Obtain the forward and backward critical speeds of the rotor system of Example 10.3

by considering the rotary and gyroscopic effects of shaft along with the discs. In present example the

effects of spinning of the rotor has been considered, and hence obtain the Campbell diagram.

Solution: A typical rotor disc system as shown in Figure 10.12 is analyzed to show the application of

the present finite element model. A typical rotor bearing system as shown in Figures 10.12 and 10.13

is analysed to show the application of the finite element model for present case (i.e. rotary inertia,

shear and gyroscopic effects). The analysis is performed with different number of elements (7 & 14)

in order to establish the convergence. The physical properties of the shaft are same as in Problem 4.4.

The natural whirl frequencies as a function of spin speed are obtained by solving eigen value problem

at different spin speeds. The critical speed map called Campbell Diagram is plotted for first four

natural whirl frequencies as a function of spin speed. Critical speeds are obtained as intersection

points of natural whirl frequencies with synchronous whirl. Variation of natural whirl frequencies of

the rotor system as a function of spin speed including rotary inertia, shear effect and gyroscopic effect

641

0 500 1000

1500

2000

2500

3000

0

500

1000

1500

Spin Speed (rad/sec)

Na

tura

l W

hirl F

requ

ency

(rad

/sec)

Synchronous W hirl

4B

4F

3F

3B

2B

2F

1F 1B

is given in Table 10.11 for seven element model and in Table 10.12 for fourteen element model.

Forward and backward whirl natural frequencies for first four mode for each spin speed is given in the

Tables 10.11 and 10.12. Figure 10.15 shows the Campbell diagram for seven elements model and

Figure 10.16 shows Campbell diagram for fourteen-element model. Critical speeds obtained from

Campbell diagram are given in Table 10.13 for seven and fourteen element model.

Figure 10.15 Campbell diagram for rotor disc system (7-elements model)

Figure 10.16 Campbell diagram for rotor bearing system (14-element model)

116.31

439.74

864.74

1208.62

0 5 0 0 10 00

1 500

2 000

2 5 00

3 0 00

0

50 0

10 00

15 00

S p in S pe e d (ra d /se c )

Na

tura

l W

hirl F

requ

en

cy (

rad/s

ec)

S yn ch ro n ou s W h ir l

2 F

2 B

3 B

4 B

1 B 1 F

3 F

4 F

116.08

863.61

1205.21

439.55

642

Table 10.12 Variation of natural whirl frequencies of the rotor system as a function of spin speed

including rotary inertia and shear effect (fourteen element model with rigid discs)

Spin speed

(rad/sec)

Natural Whirl Frequencies

(rad/sec)

Spin speed

(rad/sec)

Natural Whirl Frequencies

(rad/sec)

Backward Forward Backward Forward

0 115.3816 115.3816 1256.64 108.0046 123.0490

430.8983 430.8983 405.7617 455.3722

838.1357 838.1357 798.4256 874.7258

1165.2664 1165.2664 1117.5498 1206.8125

418.88 112.8870 117.9086 1675.52 105.6212 125.6569

422.5738 439.1482 397.3076 463.3240

825.2409 850.6818 784.5246 886.2290

1150.0536 1179.7865 1100.2929 1219.3679

837.76 110.4272 120.4653 2094.4 103.2791 128.2859

414.1895 447.3102 388.8451 471.1565

812.0016 862.8781 770.3143 897.3932

1134.1449 1193.6279 1082.4155 1231.3258

1047.2 109.2111 121.7540 2513.27 100.9801 130.9332

409.9800 451.3544 380.3932 478.8624

805.2550 868.8453 755.8155 908.2256

1125.9320 1200.3007 1063.9768 1242.7197

Table 10.13 Critical speeds of rotor bearing system with rigid discs including shear and

rotary inertia

Mode

No.

Critical Speeds

without gyroscopic

effects (rad/sec)

Natural Whirl Frequencies (rad/sec)

For 7 elements For 14 elements

Backward Forward Backward Forward

1 115.3816 114.9266 116.3128 114.6986 116.0819

2 430.8983 422.6644 439.7359 422.5013 439.5510

3 838.1357 813.6305 864.7369 812.7908 863.6147

4 1165.2664 1125.3705 1208.6196 1122.9023 1205.2135

643

APPENDIX 10.1: TIMOSHENKO BEAM MODEL

10.1.1 Translational mass matrix

[ ] [ ] [ ] [ ]2

2

10 TTTT MMMM Φ+Φ+=

where

[ ]2

20

2 2

156 sym

22 4

54 13 156420(1 )

13 3 22 4

T

l lAlM

l

l l l l

ρ

= + Φ − − −

[ ]2

20

2 2

294 sym

38.5 7

126 31.5 294420(1 )

31.5 7 38.5 7

T

l lAlM

l

l l l l

ρ

= + Φ − − −

[ ]2

20

2 2

140 sym

17.5 3.5

70 17.5 140420(1 )

17.5 3.5 17.5 3.5

T

l lAlM

l l l

ρ

= + Φ − − −

10.1.2 Rotational mass matrix

[ ] [ ] [ ] [ ]2

2

10 RRRR MMMM Φ+Φ+=

where

[ ] 20

6sym

5

1 2

10 15

6 1 6(1 )

5 10 5

1 1

10 30 10 15

DR

l

l

IM

l l

l l

ρ

= − −+ Φ − −

[ ] 21

0 sym

1

2 6

1(1 ) 0 0

2

1 1

2 6 2 6

DR

l

IM

l l

ρ

− =

+ Φ − −

644

[ ] 22

0 sym

03

0 0 0(1 )

10

6 2 6

DR

l

IM

l l

ρ

= + Φ

10.1.3 Stiffness Matrix

[ ] [ ] [ ]10 KKK Φ+=

where

[ ]2

30

2 2

12 sym

6 4

12 6 12(1 )

6 2 6 4

l lEIK

ll

l l l l

= − −+ Φ

−

[ ]2

31

2 2

0 sym

0

0 0 0(1 )

0 0

lEIK

l

l l

= + Φ

−

645

APPENDIX 10.2 ROTATING TIMOSHENKO BEAM MODEL

10.2.1 Translational Mass Matrix

[ ] [ ] [ ] [ ]2

0 1 2M M M M= + Φ + Φ

[ ]

2

2

20

2 2

2 2

156 sym

0 156

0 22 4

22 0 0 4

54 0 0 13 156420(1 )

0 54 13 0 0 156

0 13 3 0 0 22 4

13 0 0 3 22 0 0 4

l l

l lAlM

l

l

l l l l

l l l l

ρ

− = + Φ

− − − − −

[ ]

2

2

21

2 2

2 2

294 sym

0 294

0 38.5 7

38.5 0 0 7

126 0 0 31.5 294420(1 )

0 126 31.5 0 0 294

0 31.5 7 0 0 38.5 7

31.5 0 0 7 38.5 0 0 7

l l

l lAlM

l

l

l l l l

l l l l

ρ

− = + Φ

− − − − −

[ ]

2

2

22

2 2

2 2

140 sym

0 140

0 17.5 3.5

17.5 0 0 3.5

70 0 0 17.5 140420(1 )

0 70 17.5 0 0 140

0 17.5 3.5 0 0 17.5 3.5

17.5 0 0 3.5 17.5 0 0 3.5

l l

l lAlM

l

l

l l l l

l l l l

ρ

− = + Φ

− − − − −

10.2.2 Rotational Mass Matrix

[ ] [ ] [ ] [ ]2

2

10 RRRR MMMM Φ+Φ+=

646

[ ] 20

6sym

5

60

5

1 20

10 15

1 20 0

10 15

6 1 6(1 )0 0

5 10 5

6 10 0 0

5 10

1 1 20 0 0

10 30 10 15

1 1 20 0 0 0

10 30 10 15

R

l

l

l

l

IM

l l

l

l l

l l

ρ

− = + Φ− −

−

− − − −

;

[ ] 21

0 sym

0 0

10

2 6

10 0

2 6

10 0 0 0

(1 ) 2

10 0 0 0 0

2

1 10 0 0

2 6 2 6

1 10 0 0 0

2 6 2 6

R

l

l

IM

l l

l l

ρ

− =

+ Φ − − − − −

[ ] 22

0 sym

0 0

0 03

0 0 03

0 0 0 0 0(1 )

0 0 0 0 0 0

0 0 0 0 06 3

0 0 0 0 0 06 3

R

l

l

IM

l l

l l

ρ

= + Φ

10.2.3 Stiffness Matrix

[ ] [ ] [ ]10 KKK Φ+=

[ ]

2

2

30

2 2

2 2

12 sym

0 12

0 6 4

6 0 0 4

12 0 0 6 12(1 )

0 12 6 0 0 12

0 6 2 0 0 6 4

6 0 0 2 6 0 0 4

l l

l lEIK

ll

l

l l l l

l l l l

− = − −+ Φ

− −

−

;

[ ]

2

2

31

2 2

2 2

0 sym

0 0

0 0

0 0 0

0 0 0 0 0(1 )

0 0 0 0 0 0

0 0 0 0 0

0 0 0 0 0 0

l

lEIK

l

l l

l l

= + Φ −

−

647

10.2.4 Gyroscopic Matrix

[ ] [ ] [ ] [ ]2

2

10 GGGG Φ+Φ+=

with

[ ]22

20

2

2 2

0 skew-sym

36 0

3 0 0

0 3 4 0

0 36 3 0 060(1 )

36 0 0 3 36 0

3 0 0 3 0 0

0 3 0 0 3 4 0

l

l lArG

ll

l

l l l

l l l l

ρ

−

− = −+ Φ − − −

− −

;

[ ]22

21

2

2 2 2

0 skew-sym

0 0

15 0 0

0 15 5 0

0 0 15 0 060(1 )

0 0 0 15 0 0

15 0 0 5 0 0 0

0 15 5 0 15 5 5 0

l

l lArG

ll

l

l l

l l l l l

ρ

= + Φ

− −

;

[ ]22

22

2

2 2

0 skew-sym

0 0

0 0 0

0 0 10 0

0 0 0 0 060(1 )

0 0 0 0 0 0

0 0 0 5 0 0 0

0 0 5 0 0 0 10 0

lArG

l

l

l l

ρ

= + Φ −

10.2.5 Rigid disc element matrices:

[ ]

=

d

d

d

d

d

I

I

m

m

M

000

000

000

000

; [ ]

−=

000

000

0000

0000

p

p

d

I

IG ; { } { }

Td

x y y xx u u ϕ ϕ=

648

Concluding Remarks

In the present chapter, finite element analyses of simple rotor system including the gyroscopic effect

are considered. Initially, elemental equations are derived for a rigid disc by considering the effect of

gyroscopic couple. The gyroscopic couple introduces complexity by four folds, firstly both the

orthogonal motions need to be considered simultaneously for the analysis; secondly, the gyroscopic

effect makes the natural whirl frequency dependent upon the spin speed of the rotor; thirdly, the

gyroscopic effect introduces velocity term in the equation of motion; fourthly, the eigen value

problem has to be formulated in the state space form. The finite element method is applied to solution

of governing partial differential equation of the Timoshenko beam with the rotary and gyroscopic

effects. Elemental matrices are presented for the mass, stffness, and gyroscopic effect. With the

numerical illustration standard analysis by using the Campbell diagram is presented to obtain the

variations of whirl natural frequencies with spin speed of the shaft. Crtitical speeds are obtained from

such Campbell diagrams.

Exercise 10.1 Obtain the forward and backward synchronous bending critical speeds for a general

motion of a rotor as shown in Figure E10.1. The rotor is assumed to be fixed supported at one end.

Take mass of the disc m = 2 kg, polar mass moment of inertia Ip = 0.01 kg-m2 and diametral moment

of inertia Id = 0.005 kg-m2. The shaft is assumed to be massless and its length and diameter are 0.2 m

and 0.1 m, respectively. Take shaft Young’s modulus E = 2.1 × 1011

N/m2. Using the finite element

method and considering the mass of the shaft with material density ρ = 7800 kg/m3 obtain first two

forward and backward synchronous bending critical speeds by drawing the Campbell diagram.

Figure E10.1

Exercise 10.2 Obtain the transverse forward and backward synchronous critical speeds for a general

motion of a rotor system as shown in Figure E10.2. Take the mass of the disc, m = 10 kg, the

diametral mass moment of inertia, Id = 0.02 kg-m2 and the polar mass moment of inertia, Ip = 0.04 kg-

m2. The disc is placed at 0.25 m from the right support. The shaft has the diameter of 10 mm, the total

span length of 1 m and the Young’s modulus of 2.1×1011

N/m2. Consider the mass of the shaft with

the material density ρ = 7800 kg/m3 and gyroscopic effects of the disc.

649

Figure E10.2 A rotor system

Exercise 10.5 Formulate the standard eigen value problem for the following rotor system equations of

motion

[ ]{ } [ ]{ } [ ]{ } { }M x G x K x fω− + =�� �

where [M], [G] and [K] are the mass, gyroscopic and stiffness matrices respectively, {x} and {f} are

the response and force vectors respectively and ω is the rotational frequency of the rotor. Discuss

characteristics of the eigen values and interpret them physically for the above case.

Exercise 10.6 Find bending critical speeds of a rotor system shown in Figure E10.6 by using the finite

element analysis. Consider the shaft as massless and E = 2.1×1011

N/m2. The mass moment of inertias

of the disc about its principal axes are Ip = 0.02 kg-m2 and Id = 0.01 kg-m

2 and the tilting of the disc

from the vertical axis is 20 due to the mounting error. Discuss whether the gyroscopic effect will be

present or not? Obtain critical speeds accordingly. Consider the mass of the shaft with the material

density ρ = 7800 kg/m3.

Figure E10.6

Exercise 10.7 A shaft EI of total length l is supported “freely” at its ends. At a quarter length between

the end bearings the shaft carries a disc of mass m and of diameter moment of inertia Id. Find

a. The non-rotational natural frequency.

b. The whirl frequency.

For the circular cylinder following relations for the polar moment of inertia, Ip, and the diametral

moment of inertia, Id, are given

650

1 2

2pI mr= , and ( )1 2 2

123

dI m r l= +

where m is the mass of the cylinder, r is the radius of the cylinder and l is the length of the cylinder.

Exercise 10.8 The rotor of a turbine 13.6 kg in mass is supported at the mid span of a shaft with

bearings 0.4064 m. apart. The rotor is known to have an unbalance of 0.2879 kg-cm. Determine the

forces exerted on the bearings at speed of 6000 rpm if the diameter of the steel shaft is 2.54 cm.

Assume the shaft to be simply supported at the bearings. Take E=200 GN-m-2

. (Answer: RA = RB =679.2

N).

Exercise 10.9 A shaft of total length l on end bearings carries two disks at the quarter-length points.

The disks have mass m and inertia Id, the shaft stiffness is EI.

(a) Set up the equations for the whirling shaft, where the whirling frequency is equal to spin speed.

(b) Make the frequency equation dimensionless in terms of (i) the critical speed function

2 22 ml ω

EIω = and (ii) the disk effect 2

dI

mlµ = .

(c) Find the critical speed for the following three cases: 0µ = , µ → ∞ and 1/12µ = .

Exercise 10.10 Derive equations of motion and boundary conditions of a rotor system for transverse

(bending) vibrations by using the Hamilton’s principle for the following strain energy, U, and kinetic

energy, T, terms.

{ }1 2 2

2

0

l

x yU EI dzϕ ϕ′ ′= +∫

and

{ } { }1 12 2 2 2 2

2 2

0 0 0 0

l l l l

x y d x y p p x yT A u u dz I dz I dz I dzρ ϕ ϕ ω ω ϕ ϕ= + + + + −∫ ∫ ∫ ∫� � � � � �

with

; y x

x y

du du

dz dzϕ ϕ= − =

where ux and uy are the rotor translational displacements in the horizontal and vertical directions

respectively, Id and Ip are the diametral and polar mass moment of inertias of the shaft respectively, EI

is the modulus of rigidity of the shaft, A is the cross-sectional area of the shaft, l is the length of the

shaft, ρ is the mass density of the shaft material, ω is the shaft rotational frequency and dot, and

651

prime, represent the time, t, and spatial (in the shaft axial direction, z) derivatives respectively. Give

the physical significance of contributions of the each term in the strain and kinetic energy expressions.

Exercise 10.11 Develop the finite element formulation for the following equations of motion of a

rotor by using the Galerkin method (there is no need to obtain elements of the mass matrix, stiffness

matrix etc. instead keep them in the integral form)

34 4 22

4 2 2 2 22 0

yx x xuu u u

EI mr mz z t z t t

ω ∂∂ ∂ ∂

− + + = ∂ ∂ ∂ ∂ ∂ ∂

and

4 4 232

4 2 2 2 22 0

y y yxu u uu

EI mr mz z t z t t

ω ∂ ∂ ∂∂

− − + = ∂ ∂ ∂ ∂ ∂ ∂

where m (= ρA) is the mass per unit length of the shaft material, r is the radius of gyration of the shaft,

ux and uy are the rotor translational displacements in the horizontal and vertical directions respectively,

ω is the shaft rotational frequency, EI is the modulus of rigidity of the shaft and A is the cross-

sectional area of the shaft. Comment about the shape function, whether there would be any changes as

compared to the Euler-Bernoulli beam model.

Exercise 10.12 Consider a rotor system shown in Figure E10.12. The shaft has the distributed mass

and stiffness properties. Bearings B1 and B2 are simply supported bearings. The span of the shaft is 3

m, the diameter of the shaft is 10 mm, the mass density of the shaft material is 7800 kg/m3 and

Young’s modulus of the shaft material is 2.1(10)11

N/m2. Proceed with dividing the beam in three

elements and writing the elemental and global equations and application of the boundary conditions.

Determine bending critical speeds of the rotor system by using the developed finite element system

equations.

Figure E10.12

Exercise 10.13 Obtain the bending critical speed of the rotor system as shown in Figure E10.13. Take

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

length a = 0.3 m and b = 0.7 m. The diameter of the shaft is 10 mm. Take Timoshenko beam and

consider the gyroscopic effects.

652

Figure E10.13 An overhang rotor system

Exercise 10.14 Find the bearing critical speed of rotor system shown in Figure E10.14 using finite

element analysis. The following rotor data are given: 10=m kg, 02.0=dI kg-m2, 01.0=d m,

1=L m and E = 2.1×1011

N/m2. Take Timoshenko beam and consider the gyroscopic effects.

Figure E10.14

Exercise 10.15 Choose a single answer from the multiple choice questions

(i) For the eigen value problem of the rotor system with gyroscopic couple and without damping, the

eigen values are expected to be

(A) Complex (B) Pure real (C) Pure imaginary (D) Zero

(ii) Using the finite element method with fewer numbers of elements, the natural frequency of the

rotor system as compared to the exact one is expected to be

(A) Always more (B) Always less (C) Equal (D) Either less or more

References:

Couper, G.R., 1966, ASME Journal of Applied Mechanics, 33(2), 335-340. The shear coefficient in

Timoshenko’s beam theory.

Eshleman, R.L., and Eubanks, R.A., 1969, ASME Journal of Engineering for Industry, 91(4B), 1180-

1188. On the critical speeds of a continuous rotor.

Ku (1998), D.-M., 1998, Mechanical Systems and Signal Processing, 12(5), 599-610. Finite element

analysis of whirl speeds for rotor-bearing systems with internal damping.

Shames, I. H., and Dym, C. L., 2005, Energy and Finite Element Methods in Structural Mechanics,

New Age International (P) Ltd., New Delhi.

Weaver, W., Timoshenko, S.P., and Young, D.H., 1990, Vibration Problems in Engineering,

John Wiley & Sons, New York.

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