Finite Element Analysis of Simple Rotors IIT

7/24/2019 Finite Element Analysis of Simple Rotors IIT

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

CHAPTER 4

FINITE ELEMENT ANALYSIS OF SIMPLE ROTOR SYSTEMS

In the past several methods have been successfully developed to analyze the dynamic behaviour of

rotor bearing systems. Of the several methods (e.g. the transfer matrix, influence coefficient,

mechanical impedance and finite element methods) amongst the most popular approaches Finite

Element Method (FEM) is one, which is particularly well suited for modeling large scale and

complicated rotor systems. Several works showed that the use of finite elements for the modeling of

rotor bearing systems makes it possible to formulate increasingly complicated problems.

Euler-Bernoulli beam accounts for the major effects of bending in beams, which is due to pure

bending. In this theory, any plane cross-section of the beam before bending is assumed to remain

plane after bending and remain normal to elasic axis. Therefore, a beam cross section has not only

translation but also rotation. Rayleigh accounts for the energy arising out of this cross-sectional

rotation, which he called rotary inertia. Subsequently, Timoshenko accounted for the shear strain

energy in the beam due to bending caused by shear force. Thus, Timoshenko beam usually refers to a

beam in which both the rotary inertia and shear deformation effects are taken into account. The effects

of rotary inertia and shear deformation are predominant in transverse vibration of beam having large

cross-section (i.e. thick beam). If the beam is rotating, then gyroscopic effects also perform an

important role along with the rotary inertia and shear effects.

4.1 Literature Review

Historically Ruhl (1970) and, Ruhl and Booker (1972) were amongst the first people to utilize the

finite element method to study the stability and unbalance response of turborotor systems. In their

finite element formulations, only elastic bending energy and translational kinetic energy are included.

However many effects such as rotary inertia, gyroscopic moments, shear deformations, internal and

external damping, which can be very important for some configurations as discussed in the book by

Dimentberg (1961) were all neglected in their finite element analysis. McVaugh and Nelson (1976)

generalized Ruhls work by utilizing a Rayleigh beam model to devise a finite element formulation

including the effects of rotary inertia, gyroscopic moments and axial load to simulate a flexible rotor

system supported on linear stiffness and viscous damping bearings. In order to facilitate the

computations of natural whirl speeds and unbalance response, the element equations were transformed

into a rotating frame of reference for the case of isotropic bearings. Also to save the computational

time Guyan reduction procedure (1965) was adopted to reduce the size of the system matrices. Zorzi

and Nelson (1977) extended the work of McVaugh and Nelson and by the inclusion of both internal


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viscous and hysteretic damping in the same finite element model. At about same time, Rouch and Kao

(1979) and Nelson (1980) utilized Timoshenko beam theory for establishing shape functions and

based on these shape functions the system finite element matrices of governing equations were

derived. In these system finite element matrices, it is found that a shear parameter is included in the

shape functions to take into account the effect of transverse deformations. Comparison is made of the

finite element analysis with the classical closed form Timoshenko beam theory analysis for non-

rotating and rotating shafts. Ozguven and Ozkan (1984) and Edney et. al. (1990) presented the

combined effects of shear deformations and internal damping to analyze natural whirl speeds and

unbalance responses of rotor bearing systems. By using the homogeneous solutions of the governing

equations for static deflections as the shape functions, Ueghorn and Tabarrok (1992) developed a

finite element model for free lateral vibration analysis of linearly tapered Timoshenko beams.

However the mass matrix they derived is only approximate although the stiffness matrix is exact.

Tseng and Ling (1996) developed a new finite element model of a Timoshenko beam to analyse the

small amplitude, free vibrations of non uniform beams on variable two parameter foundations. An

important characteristic of this model is that the cross sectional area, the second moment of area, and

shear foundation modulus are all assumed to vary in polynomial forms, implying that the beam

element can deal with commonly seen non-uniform beams having different cross sections such as

rectangular, circular, tubular and even complex, thin walled sections as well as the foundations of

beams which vary in general way. This new beam element model enables user to handle vibration

analysis of more general beam likes structures. Chen and Peng (1997) studied the stability of the

rotating shaft with dissimilar stiffness and discussed the influences of the stiffness ratio and axial

compressive loads. A finite element model of a Timoshenko beam is adopted to approximate the

shaft, and the effects gyroscopic moments and torsional rigidities are taken into account. Results

showed that with the existence of the dissimilar stiffness unstable zones would occur. Critical speeds

will decrease and instability regions will enlarge if the stiffness ratio is increased. The increase of the

stiffness ratio consequently makes the rotating shaft unstable. When the axial compressive load

increases, the critical speed decrease and zones of instability enlarges. Ku (1998) developed an

alternative finite element shaft model (i.e. C0class Timoshenko beam finite element model) to study

the combined effects of shear deformations and internal damping on forward and backward whirl

speeds and the onset speeds of instability threshold of a flexible rotor systems supported on linear

stiffness and viscous damping bearings. Mohiuddin and Kulief (1999) presented a finite element

formulation of the dynamic model of a rotor bearing system. The elastodynamic model coupled

bending and torsional motion of the rotating shaft as derived using Lagrangian approach. The model

accounts for the gyroscopic effects as well as inertia coupling between bending and torsional

vibrations. A reduced order model was obtained using model truncation. Model transformations

involved the complete mode shapes of general rotor system with gyroscopic effects and anisotropic

bearings.


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Vibrating beams are most frequently modeled using Euler-Bernoulli model of beam both because of

its simplicity and because it is well established an accurate approximation to real motion in case of

thin beams. In the present chapter Euler-Bernoulli beam theory is considered. Equations of motion are

derived using Hamiltons principle. By using Galerkin method, finite element formulation with

consistent mass matrix and stiffness matrix is obtained. For the present case whirling motion of the

beam in vertical and horizontal directions are uncoupled. The analysis in the vertical plane (i.e.x-y

plane) is presented and in the horizontal plane (i.e. x-z plane) it will be identical. For obtaining

bending natural whirl frequencies the eigen value problem formulation is also presented. Numerical

examples are presented for system natural whirl frequencies, mode shapes and unbalance responses.

4.2 Euler-Bernoulli Beam Theory

Consider a beam of length l as shown in Figure 4.1. The cross sectional area of the beam isA, mass

density is and it is acted upon by an external force q(x, t) acting in the direction ofyaxis. The beam

is thus subjected to lateral vibrations inydirection. For Euler-Bernoulli beam the motion in vertical

plane and horizontal plane are uncoupled. So the analysis in the two perpendicular planes can be done

separately and it will be identical. The analysis is presented here is for vertical plane x-y. The analysis

in the horizontal plane x-zwill remain same. According to the Euler-Bernoulli beam theory a plane

cross section at a distancexremains plane even after bending and has a rotation aboutzaxis given by

the slope dv dx= of the elastic curve as shown in Figure 4.2. (x, y) are the co-ordinates of pointP

under consideration. Thus the displacement field of this beam can be defined as (see Figure 4.3)

xu yv= ; ( , )yu v x t = ; 0zu = (4.1)

The corresponding strain and stress field are

xxyv = ; 0yy = ; 0z = ; 0xy yz zx = = = (4.2)

xxEyv = ; 0

yy = ; 0zz = ; 0xy yz zx = = = (4.3)


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Figure 4.2 The Euler-Bernoulli beam after deformation.

Figure 4.3 The Euler-Bernoulli beam displacement field

The strain energy is

2 2

0

1 12 2

l

xx xxV

A

U dV Ey v dAdx = = 2

0

12

l

zzEI v dx= (4.4)

wherezz

I is second moment of area of cross section of the beam about an axis parallel to z-axis. The

kinetic energy is

v

=v

x

y

z

q(x,t)

P (y,z)

x

x

o

P

Figure 4.1 An Euler-Bernoulli beam before deformation.

v

v

Pv

vyux =


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2

0

1

2

l

T Av dx= (4.5)

The work done by the external forces is

0

l

w qvdx= (4.6)

The element equation of motion and boundary conditions can be obtained from Hamiltons principle,

which states that,

( )2

1

0

t

t

T U W dt + = (4.7)

Differential equation of motion is

),( txqvAvEIzz =+ (4.8)

Boundary conditions are

00

l

zzEI v v = and 0 0l

zzEI v v = (4.9)

4.3 Finite Element Formulation

Discretise the given beam into several finite elements and consider one element in co-ordinate system

x-y-z as shown in Figure 4.4. Deformations of the element are considered inx-yplane. Letv be the

nodal linear displacement and be the nodal angular displacement of the shaft element (positive sign

convention for linear and angular displacements at nodal points are shown in Figure 4.4). The element

has two nodes 1 and 2. For free vibrations the element equation (4.8) can be written as

0)()(

=+

ee

zz vAvEI

(4.10)


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Figure 4.4 Beam element inx-yplane

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

descretised generalized displacements of the element nodes. In the present finite element model, each

element has two nodes and each nodes have two generalized displacements (one linear and other

rotational), therefore displacement can be obtained within the element by using appropriate shape

functions.

{ }( )( ) ( , ) ( ) ( ) neev x t N x q t = (4.11)

where ( )N x is the shape function matrix, and { }( )

1 1 2 2( )Tne

q t v v = is the nodal

displacement vector. The linear and angular displacements at the nodes 1 and 2 of the element

( 1 1 2, ,v v and 2 ) assumed to be known. Thus element has four boundary conditions and four

constants in the shape function can be determined. Therefore assume the transverse displacement v(x)

to be a cubic polynomial within the element as

3 2

1 2 3 4( )v x a x a x a x a= + + + (4.12)

where 1a , 2a , 3a , and 4a are constants to be determined from boundary conditions. The above

equation (4.12) satisfies the governing differential equation of a beam equation (4.10). In addition the

cubic displacement shape function satisfies the continuity condition of both the linear and angular

displacements at the nodes. Expressing the transverse displacement of the element as a function of

nodal degrees of freedom1 1 2, ,v v and

2 . With the help of boundary conditions of the element at the

two nodes

1 4(0)v v a= = 1 3(0)v a = =

3 2

2 1 2 3 4( )v l v a l a l a l a= = + + + 2

2 1 2 3( ) 3 2v l a l a l a = = + + (4.13)

1

2

1v

dxx

2v

z

o

21


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Solving for the four constants 1a , 2a , 3a , 4a and substituting in equation (4.12), it gives

( ) ( ) ( ) ( ) 21 2 1 2 1 2 1 2 1 13 2 22 1 3 1

( ) 2v x v v x v v x x v

l l l l

= + + + + + +

(4.14)

Collecting the terms of the nodal degrees of freedom and writing in the matrix form equation (4.11)

can be obtained

{ } )()( )()(),( nee tqxNtxv = with 4321 NNNNN = where

( ) 33231 32 lllxxN += ( ) 332232 2 lxllxlxN +=

( ) 3233 32 llxxN += ( )3223

4 llxlxN = (4.15)

4.3.1 Weak Form

Substituting approximated shape functions of equation (4.11) in elemental equation of motion (4.10),

residue is given by

)()()( eezz

e vAvEIR += (4.16)

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

function, residue can be minimized as

{ } =l

e dxRN

0

)( 0 (4.17)

using equations (4.11), (4.16) and (4.17), the weak form can be obtained as

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

l lne ne ne

zzN A N q dx N EI N q dx Q + = (4.18)

which can be written as


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[ ]{ } [ ]{ } { }( ) ( ) ( )ne ne neM q K q Q+ = (4.19)

with

[ ] { } =l

dxNANM

0

=2 2

2

156 22 54 13

4 13 3

156 22420sym 4

l l

l l lAl

ll

[ ] { }0

l

zzK N EI N dx = =2 2

3

2

12 6 12 6

4 6 2

12 6

sym 4

zz

l l

l l lEI

ll

l

and { } { }( ) 1 1 2 2ne T

Q v v =

where [M] is the mass matrix, [K] is the stiffness matrix and {Q} is the generalized force vector.

4.4 System Equations of Motion

Obtain elemental mass matrix and stiffness matrix for each element as in equation (4.19). Then by

considering connectivity add corresponding elemental mass and stiffness matrix to get global mass

and stiffness matrix. So equation of motion for whole system becomes

[ ] { } [ ] { } { }S S Sq K q Q+ = (4.20)

4.5 Eigen Value Problem

For obtaining natural frequencies, reduced system of equations is obtained by applying boundary

conditions to equation (4.20). For this rows and columns are eliminated corresponding to applied

boundary conditions. This set the corresponding natural frequencies equal to zero which are

associated with the rigid body translation and rotation. Remaining natural frequencies are obtained by

setting the determinant of reduced system of equations equal to zero. Reduced system of equations of

motion after applying boundary conditions becomes

{ } { } { }SS S

q K q Q + = (4.21)

Assume a solution of the form, { } { } j0 tq q e = . Substituting this in equation (4.21) and setting the

determinant equal to zero, the associated eigen value problem becomes


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{ } { }20 0S S

K q M q = (4.22)

equation (4.22) can be expressed in the form of

( ){ } { }0 0S S

M p K q = , with21p = (4.23)

Pre-multiplying equation (4.23) by inverse ofS

K the eigen value problem becomes

[ ] [ ]( ){ } { }0 0D p I q = (4.24)

where [ ] ( )1

S S

D K M

= is known as the dynamic matrix. Equation (4.24) possesses a nontrivial

solution if and only if the determinant of the coefficients { }0q vanishes. Thus the characteristic

equation can be written as

[ ] [ ]( ) { }( ) det 0p D p I = = (4.25)

where ( )p is afrequency equationin the form of a polynomial of degree ninp, where nis numberof degrees of freedom. Equation (4.25) possesses nreal and positive roots andprrelated to the system

natural frequencies by21

r rp = with r = 1, 2, , n. If { }rq , represents the eigen vector (mode

shape) corresponding to the eigen valuerp , then n solutions of the eigen value problem (4.24) can be

written as

[ ] [ ]( ){ } { }0r rD p I q = with r= 1, 2, , n (4.26)

Example 4.1Obtain natural frequencies of a continuous rotor system as shown in Figure 4.5. The

following data are given: diameter of shaft d= 10 mm, density of shaft material = 7800 Kg/m3,

Youngs modulus of shaft materialE= 2.1x1011N/m2and length of the shaft l= 3 m.


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Figure 4.5 A simply supported shaft

Solution:The shaft is divided into three elements as shown in Figure 4.5. From the data given, we

have

A= 7.854 10-4 m2andI= 4.91 10-10m4

Finite element (FE) equation for element 1 can be written as

1 1 1

1 1 13

2 2 2

2 2 2

156 22 54 13 12 6 12 6

4 13 3 4 6 21.46 10 103.11

156 22 12 6

sym 4 sym 4

w w S

w w M

w w S

w w M

+ =

FE equation for element 2 can be written as

2 22

2 2 23

3 3 3

33 3

156 22 54 13 12 6 12 6

4 13 3 4 6 21.46 10 103.11

156 22 12 6

sym 4 sym 4

w w S

w w M

w w S

Mw w

+ =

FE equation for element 3 can be written as

3 23

3 3 33

4 4 4

44 4

156 22 54 13 12 6 12 6

4 13 3 4 6 21.46 10 103.11

156 22 12 6

sym 4 sym 4

w w S

w w M

w w S

Mw w

+ =


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Assembled FE equation of the whole system can be obtained as

1

1

2

23

3

3

4

4

1.46 10

156 22 54 13 0 0 0 0

4 13 3 0 0 0 0

312 0 54 13 0 08 13 3 0 0

312 0 54 13

8 13 3

156 22

sym 4

w

w

w

w

w

w

w

w

1

1

11

2

2

3

3

44

4

4

0

0103.11

0

0

12 6 12 6 0 0 0 0

4 6 2 0 0 0 0

24 0 12 6 0 0

8 6 2 0 0

24 0 12 6

8 6 2

12 6

sym 4

wS

w M

w

w

w

w

Sw

Mw

+ =

Boundary conditions are1 4 1 40, 0, 0 and 0w w M M = = = = for simply supported ends. Above

global equation after eliminating 1stand 7

throws and columns reduces to

1 1

2 2

2 23

3 3

3 3

4 4

1.46 10 103.11

4 13 3 0 0 0 4 6 2 0 0 0

312 0 54 13 0 24 0 12 6 0

8 13 3 0 8 6 2 0

312 0 13 24 0 6

8 3 8 2

sym 4 sym 4

w w

w w

w w

w w

w w

w w

+

0

0

0

0

0

0

=

For simple harmonic motion, we have { } { }2nw w= , hence the above equation takes the following

form


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{ }

1

2

23 2

3

3

4

4 6 2 0 0 0 4 13 3 0 0 0

24 0 12 6 0 312 0 54 13 0

8 6 2 0 8 13 3 0103.11 1.46 10 0

24 0 6 312 0 13

8 2 8 3

sym 4 sym 4

n

w

w

w

w

w

w

=

The above equation has the following form

( )2[ ] [ ] { } {0}nM K w =

Therefore, eigen values can be obtained as

det|A| = 0 whereA = [K]-1[M]

Natural frequencies can be obtained from above conditions. Natural frequencies of the simply

supported beam are given in Table 4.1. Exact natural frequencies from analytical closed-form

equations and approximate natural frequencies with the finite element method for different number of

elements have been tabulated in Table 1 for study of convergence. It is clear that with 10 elements

itself the convergence has occurred with quite accuracy.

Table 4.1 Convergence study of natural frequency (rad/sec)Natural frequency by FEM (number of elements)

Mode no.Natural frequency by

analytical method (3) (6) (10) (100)

I 14.18 14.19 14.18 14.18 14.18

II 56.12 57.39 56.77 56.73 56.72

III 127.6 141.6 128.1 127.7 127.6

Corresponding to each natural frequency there will be a corresponding mode shape. Table 4.2 shows a

typical eigen vector corresponding to the first natural frequency, corresponding to the eigen value

problem formulated above. Hence, the linear displacement corresponding to simply supported beam

for the first natural frequency can be taken out as shown in Table 4.3. Figure 4.6 shows the different

mode shapes.


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Table 4.2 Eigen vector corresponding to the first natural frequency.

Table 4.3 Displacement at various nodes corresponding to the first natural frequency

DOF values

1w 0

2w -0.42

3w -0.42

4w 0

Figure 4.6 Mode shapes

Analytical solution: The natural frequency for the continuous simply supported beam using analytical

method (i.e., closed form expression) is given as

DOF First column

1w -0.51

2w -0.42

2w -0.25

3w -0.42

3w 0.25

4w 0.51


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2 24n

EIn

ml = rad/sec

where Am = is mass per unit length, l is the length of the beam and n is mode number. For the

present case, we have

11 101

2 2 2 2 225 4

2.1 10 4.91 10(2.078) 14.18

7800 7.85 10 (3)n

n n n

= = =

Corresponding to different modes (i.e. n) the natural frequency is tabulated in Table 4.1. MATLABcomputer codes (input and main files) are given below:

% input_file.m Geometrical and physical data of the rotor-bearing system

d = 0.01; % m Diameter of shaft element

E = 2.1e11; % N/m^2 Young's modulus of the shaft materialL = 3.0; % m Shaft span

N_element =10; % Number of elementsN_mode_plot =5; % Number of mode shapes to be plotted (Fixed at present)

Ndof_node = 2; % Number of Dofs at each elemental node

rho = 7850; % kg/m^3 Mass density of the shaft element

% "simpl_supprt.m" Free Vibration of simple beams

% Reading rotor bearing geometrical and physical parameters

clear; % clear all previous variables

euler_in;

A=pi*(d^2)/4.0;

h = L/N_element;

I=pi*(d^4)/64.0;

fido=fopen('euler_out.m', 'w');

fprintf(fido,'\n Geometrical and physical parameter of rotor-bearing system \n');

fprintf(fido,' h= %8.4g m\n E= %8.4g N/m^2\n d= %8.4g m\n',h, E, d);

fprintf(fido,' A= %8.4g m^2\n I= %8.4g m^4\n rho= %8.4g kg/m^3\n',A, I, rho);fprintf(fido,' N_element= %4f\n Ndof_node= %4f\n', N_element, Ndof_node);

% Calling elemetal stiffness and mass matrices function and printing

[k_element, m_element] = stiff_mass_mat(A, E, I, h, Ndof_node, rho);

fprintf(fido,'\n Elemetal stiffness matrix \n');

for i=1:2*Ndof_nodefprintf(fido, '%8.4g ',k_element(i,:));

fprintf(fido,'\n');end

fprintf(fido,'\n Elemetal mass matrix \n');

for i=1:2*Ndof_node

fprintf(fido, '%8.4g ',m_element(i,:));fprintf(fido,'\n');

end

% Assembling stiffness and mass matrices and printing

[k_global, m_global] = assm_k_m_global(k_element, m_element, N_element, Ndof_node);


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fprintf(fido,'\n Global stiffness matrix \n');

for i=1:(2*Ndof_node+2*(N_element-1))fprintf(fido, '%8.4g ',k_global(i,:));

fprintf(fido,'\n');

end

fprintf(fido,'\n Global mass matrix \n');for i=1:(2*Ndof_node+2*(N_element-1))

fprintf(fido, '%8.4g ',m_global(i,:));

fprintf(fido,'\n');

end

% Applying boundary conditions (simply support)

k_global((2*Ndof_node+2*(N_element-1)-1), :) = [];k_global(:, (2*Ndof_node+2*(N_element-1)-1)) = [];

k_global(1, :) = [];k_global(:, 1) = [];

m_global((2*Ndof_node+2*(N_element-1)-1), :) = [];

m_global(:, (2*Ndof_node+2*(N_element-1)-1)) = [];

m_global(1, :) = [];

m_global(:, 1) = [];

% Eigen value of simply supported beam

dynm_mat = m_global\k_global; % using back division operator, Stiffness matrix is singular for free-free beam

case.eigen_vectors = zeros(size(dynm_mat));

[eigen_vectors eigen_values] = eig(dynm_mat);

fprintf(fido, '\nNatural Frequencies of free-free beam \n');for i = 1:(2*Ndof_node+2*(N_element-1)-2)

nat_freq(i) = sqrt(eigen_values(i, i)); % taking sqrt of diagonal terms

endnat_freq = sort(nat_freq); % sorting nfs in ascending order

fprintf(fido, '%8.4g \n', nat_freq);

exact_nf(1:5) = [1 4 9 16 25]*(pi^2)*sqrt(E*I/(rho*A*L^4));

fprintf(fido, '\nExact first natural frequencies of simply supported beam \n');

fprintf(fido, '\n%8.4g \n', exact_nf);

%Eigen Vectors of simply supported beam

eigen_vectors = fliplr(eigen_vectors); %fliping eigen vector L to R so as to match corresponding eigen vectorfprintf(fido, '\nEigen vectors (linear displ.)\n');

jj = 0;for i = 1:2:(2*Ndof_node+2*(N_element-1))

jj = (i) - jj;

x(jj) = jj; % making node number vectorif(isequal(i,1))

y(1, 1:N_mode_plot) = 0; %zeros(1, 1:N_mode_plot); % sorting linear displ. vetorselseif(isequal(i,(2*Ndof_node+2*(N_element-1)-1)))

y(jj, 1:N_mode_plot) = 0; %zeros(1, 1:N_mode_plot); % sorting linear displ. vetorselse

y(jj, 1:N_mode_plot) = eigen_vectors((i-1), 1:N_mode_plot); % sorting linear displ. vetorsend

fprintf(fido, '%8.4g ', y(jj, 1:N_mode_plot));

fprintf(fido,'\n');end;

% Plotting eigen vectors first five

plot(x, y(:, 1), 'k-o', x, y(:, 2), 'k--*', x, y(:, 3), 'k:x', x, y(:, 4), 'k-.>', x, y(:, 5), 'k-^');

legend('I mode', 'II mode','III mode','IV mode','V mode');

title('Mode Shapes for Free-Free beam');


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xlabel('Node number --->');ylabel('Relative linear displ.--->');

% Closing all input and output files

status = fclose('all');

function [k_element, m_element]= stiff_mass_mat(A, E, I, h, Ndof_node, rho) % stiff_mass_mat.m

% STIFF_MASS_MAT elemental stiffness and mass matrices % H1 line can be seen from matlab commandline by lookfor stiff_mass_mat

% STIFF_MASS_MAT(A, E, I, h, Ndof_node, rho), where

% A is area of cross-section of beam,

% EI modulus of rigidity of beam,% h shaft element length,

% Ndof_node is the number of degree of freedom of each node% rho is the mass density of the shaft material

% Elemental Stiffness Matrix

k_element = zeros(2*Ndof_node,2*Ndof_node);k_element = (E*I/h^3)*[12 6*h -12 6*h;

6*h 4*h^2 -6*h 2*h^2;

-12 -6*h 12 -6*h;

6*h 2*h^2 -6*h 4*h^2];

% Elemetal mass matrixm_element = zeros(2*Ndof_node,2*Ndof_node);m_element = (rho*A*h/420)*[156 22*h 54 -13*h;

22*h 4*h^2 13*h -3*h^2;54 13*h 156 -22*h;

-13*h -3*h^2 -22*h 4*h^2];

function [k_global, m_global] = assm_k_m_global(k_element, m_element, N_element, Ndof_node)

% file name assm_k_m_global.m

% ASSM_K_M_GLOBAL Assembles elemental mass and stiffness matrices% assm_k_m_global(k_element, m_element, N_element, Ndof_node), where

% k_element is elemental stiffness matrix of shaft element

% m_element is elemental mass matrix of shaft element% N_element number of shaft element% Ndof_node is the number of degree of freedom of each node

% "assm_k_global.m" Assembling stiffness matrix

k_global = zeros((2*Ndof_node+2*(N_element-1)),(2*Ndof_node+2*(N_element-1)));

for i = 1 : N_element

j=(2*i-1); % makes sure that subsequent elemental matrix is having proper shift

k_global1 = zeros((2*Ndof_node+2*(N_element-1)),(2*Ndof_node+2*(N_element-1)));% Assigning elemental stiffnes matrix for 1st element to the global matrix

k_global1((j:j+3), (j:j+3)) = k_element;

k_global = k_global1 + k_global;

end

% "assm_m_global.m" Assembling mass matrixm_global = zeros((2*Ndof_node+2*(N_element-1)),(2*Ndof_node+2*(N_element-1)));for i = 1 : N_element

j=(2*i-1); % makes sure that subsequent elemental matrix is having proper shiftm_global1 = zeros((2*Ndof_node+2*(N_element-1)),(2*Ndof_node+2*(N_element-1)));

% Assigning elemental stiffnes matrix for 1st element to the global matrix

m_global1((j:j+3), (j:j+3)) = m_element;m_global = m_global1 + m_global;

end


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204

Example 4.2 To demonstrate the application and accuracy of the finite element model of Euler-

Bernoulli beam, consider a simply supported continuous shaft. The analysis is performed with five

elements as shown in Figure 4.7. For the present analysis shaft is considered as uniform in cross

section. Following data is considered for the analysis: diameter of the shaft = 0.1m, length of the shaft

= 3.75m, number of elements=5, length of the each element = 0.75m, elastic modulus of the shaft

material = 2.11E11 N/m2, mass density of the shaft material = 7339.449 kg/m3. Comparisons of the

FEM results are made with the classical closed form solutions and are given in Table 4.4. Results are

also given for 10-element model. For obtaining mode shapes eigen vectors are normalized by the

highest eigen vector of corresponding eigen values for linear displacements and are plotted in Figure

4.8.For five element model first four lowest natural whirl frequencies are in good agreement with

classical closed form solutions. Results obtained by FEM are greater than classical closed form

solutions. This is because shape functions used are obtained by approximating displacement within

the element by cubic polynomial neglecting higher order terms. So displacement is less and hence

effective stiffness obtained from FEM formulation is more which increases the natural frequencies

than classical closed form solutions. Error is within 3% for all first four natural frequencies. For ten-

element model error is less than 1% for all first four natural frequencies.

Figure 4.7 A five element simply supported shaft model (length of each element is 0.75m)

Table 4.4 Comparison of natural whirl frequencies by FEM with classical closed form solutions for

simply supported uniform shaft

FEM natural whirl frequency

(rad/sec)

% ErrorMode

No.,

n

Classical closed form

natural whirl frequency

(rad/sec) (5) (10) (5) (10)

1 94.0777 94.08403 94.08176 0.0072 0.0050

2 376.3108 376.9310 376.3604 0.1648 0.0132

3 846.6994 853.4207 847.1608 0.7938 0.0545

4 1505.2433 1539.9174 1507.7459 2.3035 0.1662

Number in bracket indicates number of elements

Element No.

Node No.

Support Support


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205

-1.5

-1

-0.5

0

0.5

1

1.5

1 3 5 7 9 11

Node Number

NormalisedEig

en-Vectors

Mode I-94.08 rad/sec

Mode II-376.3604

rad/sec

Mode III-847.1608rad/secMode IV-1507. 7459

rad/sec

Figure 4.8 Mode shapes for simply supported beam (10-element model)

Example 4.3 Find the bearing critical speed of rotor system shown in Figure 4.9 using finite element

analysis. The following rotor data are given: 10=m kg, 02.0=dI kg-m, 01.0=d m, 1=L m and

E= 2.11011N/m2.

massless shaft

10 cm Figure 4.9

0.6 m 0.4 m

Solution:

Figure 4.10 A finite element discretisation of the rotor system

On dividing the rotor in four element (i.e. h= 0.25 m) as shown in Figure 4.10, so that

10

34.91 10

EI

h

= N/m

h h h h

(1) (2) (3) (4)

2 31

4

5


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206

The elemental equation for element (1) can be written as

1 1 1

2 2

1 1 1

32 2 2

2 22 2 2

0 0 0 0 12 6 12 6

0 0 0 0 6 4 6 2

0 0 0 0 12 6 12 6

0 0 0 0 6 2 6 4

y y Sh h

Mh h h hEI

y y Sh hh

Mh h h h

+ =

Similarly for other elements, the elemental equations can be written by changing the corresponding

nodal variables. If disc is assumed to be in element (3), the mass matrix for element (3) becomes

=

d

d

I

mM

000

000

0000

0000

][

On assembling elemental equations for all the elements, we have

1

1

2

2

3

3

4

4

5

5

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 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

d

d

y

y

y

ym

I

y

+

3 2

2 2 2

2 2 23

2 2 2

2 2

12 6 12 6 0 0 0 0 0 0

6 4 6 2 0 0 0 0 0 0

12 6 24 0 12 6 0 0 0 0

6 2 0 8 6 2 0 0 0 0

0 0 12 6 24 0 12 6 0 0

0 0 6 2 0 8 6 2 0 0

0 0 0 0 12 6 24 0 12 6

0 0 0 0 6 2 0 8 6 2

0 0 0 0 0 0 12 6 12 6

0 0 0 0 0 0 6 2 6 4

h h

h h h h

h h

h h h h h

h hEI

h h h h hh

h h

h h h h h

h h

h h h h

+

1 1

1 1

2

2

3

3

4

4

5 5

5 5

0

0

0

0

0

0

y S

M

y

y

y

y S

M

=


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207

The boundary conditions are :5151 0 yyyy ==== and 051 ==MM . Finally, we get the

assembled equation in which the boundary conditions have been applied in the following form

1

2

2

3

3

4

4

5

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 00 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

d

d

y

y

m y

I

3 21 1

22 2 2

2

3

3 2 2 23

4

2 2 24

2 25

4 6 2 0 0 0 0 0

06 24 0 12 6 0 0 0

02 0 8 6 2 0 0 0

00 12 6 24 0 12 6 0

00 6 2 0 8 6 2 0

0 0 0 12 6 24 0 6

0 0 0 6 2 0 8 2

0 0 0 0 0 6 2 4

h h h

yh h

h h h h

yh hEI

h h h h h h

yh h

h h h h

h h h

+ = 5

0

0

M

For free vibration, on substituting xx2

= , in the above equation we get an eigen value problem of

the following form

[ ] [ ] { }02 =+ KMn

which can be solve to give the following natural frequency of the system

1 1

2 867.18 29.44n n = = rad/s and 2 22 48.3652 10 289.23n n = = rad/s.

Example 4.4A typical simply supported rotor disc system as shown in Figure 4.11 is analyzed to

show the application of the present finite element model. The physical properties of the rotor bearing

system are given as: diameter of shaft = 0.1 m, length of shaft = 3.5 m,Youngs modulus of material

of shaft = 4.08E11 N/m2, mass density of the shaft material= 7830 kg/m

3, number of rigid disks = 4

and mass of each rigid disk = 60.3 kg. The rotor is modeled as seven element, Figure 4.11 and

fourteen-element member Figure 4.12. In case of seven element member, two identical rigid bearings

are located at node number two and seven, and four rigid disks are located at node numbers three,


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208

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

number three and thirteen, and rigid disks are located at node numbers five, seven, nine and eleven.

The shaft is considered as uniform in cross section.

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

corresponding to the locations of rigid disks. Also assembled mass and stiffness matrices are obtained,

boundary conditions are applied and dynamic matrix is calculated. The natural whirl frequencies are

obtained by solving eigen value problem and mode shapes can be drawn by using MATLAB package.

Natural whirl frequencies are obtained for 7 elements and 14 elements model and are given in Table

4.5. Results show that convergence has already occurred with 7 elements model. Mode shapes are

shown in Figure 4.13.

Figure 4.11 Rotor Bearing System with Rigid Disks (7-elements of 0.5 m each)

Figure 4.12 Rotor Bearing System with Rigid Disks (14-elements of 0.25 m each)

Rigid discs

BearingBearing

0.1 m

BearingBearing

Rigid discs


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209

Table 4.5 Natural whirl frequencies of rotor bearing system with rigid disks.

Natural whirl frequencies (rad/sec)Mode No.

For 7 elements For 14 elements

1 116.1756 115.8136

2 438.4482 438.2795

3 861.5609 860.7477

4 1209.4922 1207.0327

-1.5

-1

-0.5

0

0.5

1

1.5

1 2 3 4 5 6 7 8

Node Number

NormalisedEig

en-Vectors Mode I-116.1756

rad/sec

Mode II- 438.4482

rad/sec

Mode III-861.5609

rad/sec

Mode IV-1209.4922

rad/sec

Figure 4.13 Mode shapes for rotor bearing system with rigid disks (7-element model)

Exercise 4.1 Obtain the bending critical speed of the rotor system as shown in Figure E4.1. Take themass of the disc, m= 5 kg and the diametral mass moment of inertia, Id= 0.02 kg-m

2. Take shaft

length a= 0.3 m and b= 0.7 m. The diameter of the shaft is 10 mm. Neglect the gyroscopic effects.

a b

A B

Figure E4.1 An overhang rotor system

Exercise 4.2Find critical speeds of the rotor system shown in Figure E4.2. TakeEI= 2 MN.m2for

the shaft and mass moment of inertia of disc is negligible.


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210

3 m

1.5 m

Fixed end 80 kg 100 kg

Figure E4.2 An overhang rotor system

Exercise 4.3Find the fundamental bending critical speed of the rotor system shown in Figure E3.7.

B1and B2are simply supported bearings and D1and D2are rigid discs. The shaft is made of steel with

modulus of rigidityE= 2.1 (10)11

N/m2and uniform diameter d= 10 mm. The various shaft lengths

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

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

both discs.

Exercise 4.4 Find the transverse natural frequency of a rotor system as shown in Figure E4.4.

Consider shaft as massless and is made of steel with 2.1 (10)11N/m2 of Youngs modulus, E, and

7800 kg/m3 of mass density, . The disc has 10 kg of mass. The shaft is simply supported at ends (In

Figure 3.4 all dimensions are in cm).

60 40

10

30

Figure E4.4 Example 4.4

B1 B2D1 D2

Figure E4.3Example 4.3


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211

4.6 Forced Vibration Analysis (The consistence load matrix)

In the present section elemental force vector has been obtained for various types of loading over the

element.

Case (a): ForF(x, t) to be uniform distributed load.

Figure 4.14 A beam element under uniform loading

The consistent force vector is given as

(4.27)

Case (b): For the concentrated load acting at0x x= , it can be

written as

)(),( 0*

0 xxFtxF = (4.28)

The consistent force vector can be written as

( ) *

0 0 0 0

0

{ } ( ){ } { ( )}

h

neP F x x N dx F N x= = (4.29)

Figure 4.15 A beam element under concentrated loads at different positions

For concentrated load as shown in Figure 4.15, the following elemental force vector can be obtained

(a)

h1 2

F(x, t) =F(t)

1

2

1 2

( ) 121

0 21 2

12

( )

( ){ } ( ){ }

( )

( )

h

ne

F t h

F t hP F t N dx

F t h

F t h

= =

F0F0 F0

h/2

hhh

(b) (c)


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212

{ }

{ }

( )

0 0

( )

0 0

( )0 0

(i) For 0, { } 1 0 0 0

(ii) For , { } 0 0 0 1

1 1 1 1(iii) For , { }2 2 8 2 8

Tne

Tne

T

ne

x P F

x h P F

hx P F h h

= =

= =

= =

(4.30)

Case(c): For the load varying linearly over the element as shown in Figure 4.16, let us assume that

( ) bxatxF e +=),( (4.31)

From Figure 4.16, we have

( )

( )

1 1

2 2

at 0 ( , ) ( )

and ( , ) ( )

e

e

x F x t F t F a

x h F x t F t F a bh

= = => =

= = => = +

which gives 2 11 andF F

a F bh

= =

Figure 4.16 A linearly varying loading

The assumed form of the force becomes

1( ) ( )2 11

2

( , ) 1 { }e nef

FF F x xF x t F x N F

Fh h h

= + = = (4.32)

Hence the consistent load vector will be


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213

( )

( )

( )

( )

7 3

1 27 3 20 20

20 201 1 21 12 2

1 220 301( ) ( ) 20 30

3 7 3 720 20 20 1 220 20

1 12

1 1 230 201 230 20

{ } { } { }h

ne ne

f

F F hh h

F F hFh hP N N dx F

Fh h F F h

h hF F h

+ +

= = = +

(4.33)

Case(d) For load varying quadratically as shown in Figure 4.17 at least three nodes are required

and corresponding force values must be specified; the element force can be written as

{ }1

( )( ) 2

1 2 3 2

3

( , )

2

1 1

2where 1

41

nee

f f f f

f f

F

F x t a bx cx N N N F N F

F

x x

h h

x xN N

h h

x x

h h

= + + = =

=

(4.34)

Hence, the load vector can be obtained as

=h

ne

f

neFdxNNP

0

)()(}{}{}{ (4.35)

Obtained the same as an exercise.

Example 4.5 Perform the forced vibration analysis of the rotor system as shown in Figure 4.17. The

shaft is having diameter of d = 10 mm and disc has mass of mdisc = 1.5 kg. Unbalance mass in

formation is given as: mumb= 0.005 kg, rumb= 0.05 m, umb= 300phase. The phase is measured with

respect to some physical reference on the shaft. Let us assume that reference mark is designed withy-

axis. The angular speed, , is in counter clockwise direction as seen from left. Let us assume disc in

element 2. The shaft element is taken such that the disc is always at the node point.


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214

Figure 4.17 A single disc rotor system Figure 4.18 Unbalance on the disc

Solution: Let us assume the disc is on element 2. The shaft element is taken such that the disc is

always at the node point. The phase is measured with respect to some physical reference on the shaft.

Let us assume that reference mark is aligned with y-axis at time t = 0. The angular speed, ,is in

counter-clockwise direction as seen in Figure 4.18. FE equation for element 1 in the vertical planex-z

is given as

1 1 1

1 1 13

2 2 2

2 2 2

156 22 54 13 12 6 12 6 0

4 13 3 4 6 2 01.46 10 103.11

156 22 12 6 0

sym 4 sym 4 0

zw w S

w w M

w w S

w w M

+ = +

EOM in the horizontal plane (x-y) is given as

1 1 1

1 1 13

2 2 2

2 2 2

156 22 54 13 12 6 12 6 0

4 13 3 4 6 2 01.46 10 103.11

156 22 12 6 0

sym 4 sym 4 0

v v S

v v Mx

v v S

v v M

+ = +

On combining elemental equations and for vertical & horizontal plane, we get

1

1

1

13

2

2

2

2

156 22 0 0 54 13 0 0 12 6 0 0 12 6 0 0

4 0 0 13 3 0 0 4 0 0 6 2 0 0156 22 0 0 54 13 12 6 0 0 12 6

4 0 0 13 3 4 0 0 61.40 10 103.11

156 22 0 0

4 0 0

156 22

sym 4

w

wv

vx

w

w

v

v

+

1 1

1 1

1 1

1 1

2 2

2 2

2 2

2 2

0

00

2 0

12 6 0 0 0

4 0 0 0

12 6 0

sym 4 0

z

z

y

y

z

z

y

y

w S

w Mv S

v M

w S

w M

v S

v M

= +


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215

In FE equation for element 2, since rows 5th& 7thare EOM corresponding to displacements3w and

3v , the lumped mass at node 3 of the disc will contribute mdisc 3w and mdisc 3v .

( )

( )

2

2

2

233

3

3

33

3

156 22 0 0 54 13 0 012 6

4 0 0 13 3 0 0

156 22 0 0 54 13

4 0 0 13 3

1.46 10 103.11156 1.5 /1.46 10 22 0 0

4 0 0

156 1.5 /1.46 10 22

sym 4

w

wv

v

w

w

v

v

+ +

+

2

2

2

2

3

3

3

3

0 0 12 6 0 0

4 0 0 6 2 0 012 6 0 0 12 6

4 0 0 6 2

12 6 0 0

4 0 0

12 6

sym 4

w

wv

v

w

w

v

v

2

2

2

j 2

6

3

36

3

3

0

0

0

0

( j)2.5 10 [0.866 j0.5]

0

2.5 10 [0.866 j0.5]

0

z

z

y

y

t

z

z

y

y

S

M

S

Me

S

M

S

M

= +

+

+

The unbalance force is obtained byj

b bm r e in the horizontal directionF4, and the unbalance force in

the vertical directionF3can be obtained by using Figure 4.19.

(a) (b)

Figure 4.19 Unbalance forces in the horizontal and vertical directions for different direction of rotorrotations (a) Counter clockwise rotor rotation direction (b) Clockwise rotor rotation direction

FE equation for element 3 is given as

yz jFF =

tjj

bbyeermF =y

z

yzFjF )(=

y

z

tjj

bby eermF =


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216

3

3

3

33

4

4

4

4

156 22 0 0 54 13 0 0 12 6 0 0 12 6 0 0

4 0 0 13 3 0 0 4 0 0 6 2 0 0

156 22 0 0 54 13 12 6 0 0 12 6

4 0 0 13 3 4 0 0 61.46 10 103.11

156 22 0 0

4 0 0

156 22

sym 4

w

w

v

vx

w

w

v

v

+

3 3

3 3

3 3

3 3

4 4

4 4

4 4

4 4

0

0

0

2 0

12 6 0 0 0

4 0 0 0

12 6 0

sym 4 0

z

z

y

y

z

z

y

y

w S

w M

v S

v M

w S

w M

v S

v M

= +

Global FE equation : On assembling all elemental equations, we get

3

156 22 0 0 54 13 0 0 0 0 0 0 0 0 0 0

4 0 0 13 3 0 0 0 0 0 0 0 0 0 0

156 22 0 0 54 13 0 0 0 0 0 0 0 0

4 0 0 13 3 0 0 0 0 0 0 0 0

(156 ( 220 0 54 13 0 0 0 0 0 0

156) 22)

(4 4) 0 0 13 3 0 0 0 0 0 0

(156 ( 220 0 54 13 0 0 0 0156) 22)

(4 4) 0 0 13 3 0 0 0 01.46 10

(1183.4 ( 220 0 54 13 0 0

156) 22)

x

+ +

+

+ +

+

+

+

1

1

1

1

2

2

2

2

3

3

3

3

4

4

(4 4) 0 0 13 3 0 0

(1183.4 ( 220 0 54 13

156) 22)

(4 4) 0 0 13 3

156 22 0 0

4 0 0

156 22

sym 4

w

w

v

v

w

w

v

v

w

w

v

v

w

w

+

+ + +

4

4

v

v

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 04 0 0 6 2 0 0 0 0 0 0 0 0

(12 ( 60 0 12 6 0 0 0 0 0 0

12) 6)

(4 4) 0 0 6 2 0 0 0 0 0 0

(12( 6 6) 0 0 12 6 0 0 0 0

12)

(4 4) 0 0 6 2 0 0 0 0103.11

(12( 6 6) 0 0 12 6 0 0

12)

(4 4) 0 0 6 2 0 0

(12( 6 6) 0 0

12)

+

+

+

+ +

+ +

+ +

+

+ +

1

1

1

1

2

2

2

2

3

3

3

3

4

4

4

4

00

0

0

0

0

0

0

2.5 10 [0.

6 2

(4 4) 0 0 6 2

12 6 0 0

4 0 012 6

sym 4

ww

v

v

w

w

v

v

w

w

v

v

w

w

v

v

=

+

1

1

1

1

j

6

4

4

4

4

0

0

0

0

5 j0.866] 0

0 0

2.5 10 [0.866 j0.5] 0

0 0

0

0

0

0

z

z

y

y

t

z

z

y

y

SM

S

M

e

S

M

S

M

+

+

Boundary conditions :1 1 4 40, 0, 0, 0w v w v= = = =

and 1 1 4 40, 0, 0, 0z y z yM M M M= = = = .


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217

Correspondingly we will be having : 2, 4, 5, 6, , 12, 14, 16 (total 12 equation) with RHS known.

Rest of 4 equation can be removed since it contain shear forces as additional unknown. From

displacement information we will be having 1st, 3rd, 13th and 15th column multiplied by zero (i.e.

corresponding displacement from B.Cs.) so these columns can also be eliminated since it will not

contribute any term in totality. Shear forces which are nothing but forces being transmitted through

the bearing can be obtained once displacements are known.

FE equation after applying B.Cs. reduces to

1

1

2

2

2

23

3

4 0 13 3 0 0 0 0 0 0 0 0

4 0 0 13 3 0 0 0 0 0 0

312 0 0 0 54 13 0 0 0 0

8 0 0 13 3 0 0 0 0

312 0 0 0 54 13 0 0

8 0 0 13 3 0 01.46 10

1339.4 0 0 0 13 0

8 0 0 13 01339.4 0 0 13

8 0 13

4 0

sym 4

w

v

w

w

v

vx

w

w

3

3

3

4

4

v

v

w

v

+

1

1

2

2

2

2

3

3

3

3

4

4

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

4 0 0 13 3 0 0 0 0 0 0

312 0 0 0 54 13 0 0 0 0

8 0 0 13 3 0 0 0 0

312 0 0 0 54 13 0 0

8 0 0 13 3 0 0103.11

1339.4 0 0 0 13 0

8 0 0 3 0

1339.4 0 0 13

8 0 3

4 0

sym 4

w

v

w

w

v

v

w

w

v

v

w

v

+

6

6

0

0

0

0

0

0

2.5 10 [0.5 j0.866]

0

2.5 10 [0.866 j0.5]

0

0

0

te

=

+

Since excitation is a simple harmonic i.e.,{ } { } tw W e = , hence { } { }2w w= , where {W} is a

vector contains complex quantity i.e., magnitude and phase information of various displacement

components. Hence EOM can be written as

2 j j j[ ]{ } [ ]{ } { }t t tM W e K W e F e + =

which can be solved as

( ) }{][][}{ 12 FMKW =


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218

where is the rotor spin speed. It should be noted that matrices [M] and [K] contain real quantities,

whereas force vector {F} and corresponding displacement vector {W} will contain complex terms.

The response (amplitudes and phases) can be plotted with respect to shaft speeds.

(a) Response amplitudes versus speeds(b) Response phases versus speeds

Figure 4.20 Forced response variations with respect to spin speed of the shaft

Exercise 4.5For exercises 4.1 to 4.4 plot linear and angular displacements (with both amplitude and

phase) of the discs with respect to the rotational speed of the rotor (take the rotational frequency of the

rotor minimum of 0.1 rad/s and maximum at least 5 rad/s above the second critical speed). Assume

imbalances of 20 gm-mm at one of disc with 30-degree phase with some shaft reference point. Check

whether critical speeds are in agreement with the obtained by free vibration analysis.

References

Bickford, W.B., Greenhill, L.M., and Nelson, H.D., 1985, A Conical Beam Finite Element for Rotor

Dynamics Analysis,Transactions of the ASME, Journal of Vibration, Stress and Reliability in

Design, vol. 107, pp. 421-430.

Booker, J.F., and Ruhl, R.L., 1986, A Finite Element Model for Distributed Parameter Turborotor

Systems , ASME, Journal of Vibration, Stress and Reliability in Design, vol. 108, pp. 177-181,paper C432/142.

Booker, J.F., and Ruhl, R.L., 1972, A Finite Element Model for Distributed Parameter TurborotorSystems,ASME, Journal of Engineering for Industry, vol. 94, p. 126.

Chen, L.W., and Ku, D.M., 1991, Finite Element Analysis of Natural Whirl Speeds of RotatingShafts,Computers & Structures, vol. 40, pp. 741-747.

Chen, W.J., A Note on Computational Rotor Dynamics, Transactions of the ASME journal,Journal of Vibration and Acoustics, vol.120, pp. 228-233.

Childs, D.W., and Graviss, K., 1982, A Note on Critical-Speed Solutions for Finite Element Based

Rotor Models.ASME, Journal of Mechanical Design, vol. 104, pp. 412-416.

Childs, D., Turbomachinery Rotordynamics, Phenomena, Modeling, and Analysis, John Wiley &

Sons, Inc., 1993.

Chong-Won Lee Vibration Analysis of Rotors, Kluwer Academic Publishers, 1993.Crandall, H., S.,Rotardynamics software, Rotating Machinery Dynamics (Proc.), 1992, pp. 3-21.


32/33


219

Cook, D., Malkus, D.W., and Plesha, M.E., Concepts and Applications of Finite Element Analysis,John Wiley & Sons, Third Edition, 1989.

Den Hartog, J.,P.,Mechanical Vibration,Dover Publications, Inc.,1985, pp. 262 263.Diana, G., Massa, E., and Pizzigoni, 1975, Finite Element Method for Computing Oil Whirl

Instability of A Rotating Shaft Supported by Elastic Bearings, I. Mech. E., pp. 659-663.

Dimarganas, A.D., 1975, A General Method for Stability Analysis of Rotating Shafts, Inggenieur-Archiv, vol.44, p. 126.

Dimentberg, F.M., 1961,Flexural Vibrations of Rotating Shafts,Butterworths, London.

Dutt, J.K., and Nakara, B.C., 1992, Stability of Rotor Systems with Viscoelastic Supports,Journal

of Sound and Vibration, vol.153, No.1, pp. 89-96.Edney, S.L., Fox, C.H.J., and Williams, E.J., 1990, Tapered Timoshenko Finite Elements for Rotor

Dynamics Analysis, Journal of Sound & Vibration, vol.137, pp. 463-481.

Frisiwell, M., I., Penny, J., E., T., Gravery, S., D., Smart, M., G., 1998 Computing Critical speeds

for Rotating Machines with speed dependent Bearing Properties , Journal of Sound & Vibration,

vol.213, No. 1, pp. 139-158.

Genta, G., and Gugliotta, A., 1988, A Conical Element for Finite Element Rotor Dynamics,

Journal of Sound & Vibration, vol. 120, pp. 175-182.Genta,G., and Tonoli, A., 1996, A Harmonic Finite Element for The Analysis of Flexural,

Torsional, and Axial Rotor Dynamic Behavior of Discs,, Journal of Sound & Vibration, vol. 196,

pp. 19-43.Genta, G., 1988, Whirling of Unsymmetrical Rotors, A Finite Element Approach Based on Complex

Coordinates,Journal of Sound & Vibration, vol. 124, pp. 27-53.Genta, G., 1985, A Consistent Matrix Approach for Finite Element Method ; Dynamic Modeling of

Rotors, Current Advances in Mechanical Design & Production, 3rd

Cairo University MDPConference, Cairo, pp 115-122.

Genta, G., Vibration of Structures and Mechanics, Practical Aspects, Springer-Verlong, 1993.

Goodwin, M. J., 1989,Dynamics of Rotor Bearing Systems, Unwin Hyman Inc.

Gunter, E.J., 1966, Dynamic Stability of Rotor Bearing System, NASA, SP-113.Guyan, R.J., 1965, Reduction of Stiffness and Mass Matrices, AIAA, Journal 3, pp. 380.

Shiau, T.N., and Hwang, J.L., 1989, A New Approach to the Dynamic Characteristics of UndampedRotor-Bearing Systems, ASME, Journal of Vibration, Acoustics, Stress, and Reliability in

Design, vol.111, pp. 379-385.

Kahraman, A., Nevzat Ozguven, H., Houser, D.R., and Zakrajsek, J.J., 1992, Dynamic Analysis ofGeared Rotors by Finite Elements,Trans ASME, Journal of Mechanical Design, vol. 114, p. 507

Kao, J.S., and Rouch, K.E., 1979, A Tapered Beam Finite Element For Rotor Dynamics Analysis,

Journal of Sound & Vibration, vol. 66, pp. 119-140.

Kim, Y.D., and Lee, C.W., 1986, Finite Element Analysis of Rotor Bearing Systems Using A ModalTransformation Matrix,Journal of Sound & Vibration, vol. 111, pp. 441-456.

Khulief, Y.A., and Mohiuddin, M.A., 1994, Modal Characteristics of Rotors Using A Conical Shaft

Finite Element,Computer Methods in Applied Mechanics & Engineering, vol. 115, pp. 125-144.

Khulief, Y.A., and Mohiuddin, M.A., 1999, Coupled Bending Torsional Vibration of Rotors using

Finite Elements, Journal of Sound & Vibration, Vol. 223(2), pp. 297-316.

Ku, D.M, 1998, Finite Element Analysis of Whirl Speeds of Rotor-Bearing Systems with Internal

Dampimg, Journal Mechanical Design, Vol. 102, pp. 599-619.Lund, J.W., 1974, Stability of Damped Critical Speeds of a Flexible Rotor on Fluid Film Bearings,

Transactions of the ASME, Journal of Engineering for Industry, Vol. 96, pp.509-517.McVaugh, J.M., and Nelson, H.D., 1976, The Dynamics of Rotor Bearing Systems Using Finite

Elements,Transactions of the ASME, Journal of Engineering for Industry, vol. 98, pp. 593-600.

Meiorvitch, L., 1967,Analytical Methods in Vibrations, MacMillan Book Co.Nelson, H.D., and Zorzi, E.S., 1977, Finite Element Simulation of Rotor Bearing Systems With

Internal Damping,Transactions of the ASME, Journal of Engineering for Power, vol. 99, pp.71-76.

Nelson, H.D., 1980, A Finite Rotating Shaft Element Using Timoshenko Beam Theory,,

Transactions of the ASME, Journal of Mechanical Design, vol.102, pp.793-803.


33/33


Nelson, H.D., Sept. 1977, A Finite Rotating Shaft Element Using Timoshenko Beam Theory,Engineering Research Center Report ERC-R-77023, Arizona State University, p. 61.

Nelson, H.D., and Zorzi, E.S., 1980, The Dynamics of Rotor Bearing Systems With Axial Torque - AFinite Element Approach,Journal of Mechanical Engineering Design, vol.102, 158-161.

Ozguven, H.N., and Ozkan, Z.L., 1984, Whirl Speeds & Unbalance Response of Multibearing

Rotors Using Finite Elements, Transactions of the ASME, Journal of Vibration, Acoustics,Stress, and Reliability in Design, vol.106, pp. 72-79.

Polk, S.R., May1974, Finite Element Formulation & Solution of Flexible Rotor Rigid DiscSystems for Natural Frequencies & Critical Whirl Speeds, MSE Engineering Report, Arizona

State University.Rao, J. S.,Rotor Dynamics,New Age International Publishers, Third Edition, 1996, pp. 209 - 211.

Ruhl, R.L., 1970, Dynamics of Distributed Parameter Rotor Systems: Transfer Matrix & Finite

Element Techniques,Ph.D. Dissertation, Cornell University, Ithaca, N.Y.

Ruhl, R.L., and Booker, J.F., 1972, A Finite Element Model for Distributed Parameter Turborotor

Systems,Transactions of the ASME, Journal of Engineering for Industry, vol. 94, pp. 128-132.

Tondl, A., 1965, Some Problems of Rotor Dynamics, Champan and Hall Ltd., England, pp. 100 102

Thorkildsen, T., June 1979, Solution of a Distributed mass and Unbalanced Rotor Systems using aConsistent Mass Matrix Approach, MSE Engineering Report, Arizona state University.

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