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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 Ruhl’s 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
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, ([email protected])
189
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. C0 class 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.
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, ([email protected])
190
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 Hamilton’s 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 is A, mass
density is ρ and it is acted upon by an external force q(x, t) acting in the direction of y axis. The beam
is thus subjected to lateral vibrations in y direction. 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-z will remain same. According to the Euler-Bernoulli beam theory a plane
cross section at a distance x remains plane even after bending and has a rotation about z axis given by
the slope dv dxφ = of the elastic curve as shown in Figure 4.2. (x, y) are the co-ordinates of point P
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
xx yvε ′′= − ; 0yyε = ; 0zε = ; 0xy yz zxε ε ε= = = (4.2)
xx Eyvσ ′′= − ; 0yyσ = ; 0zzσ = ; 0xy yz zxτ τ τ= = = (4.3)
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, ([email protected])
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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 1
2 2
l
xx xxV
A
U dV Ey v dAdxσ ε ′′= =∫ ∫ ∫ 2
0
1
2
l
zzEI v dx′′= ∫ (4.4)
where zzI is second moment of area of cross section of the beam about an axis parallel to z-axis. The
kinetic energy is
v
φ=′v
x x
y
y y
z
q(x,t)
P (y,z)
x
x
o
P′
Figure 4.1 An Euler-Bernoulli beam before deformation.
y
v ′
v
P v′
vyu x′−=
x
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2
0
1
2
l
T Av dxρ= ∫ � (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 Hamilton’s principle,
which states that,
( )2
1
0
t
t
T U W dtδ δ− + = ∫ (4.7)
Differential equation of motion is
),( txqvAvEI zz =+′′′′ ��ρ (4.8)
Boundary conditions are
00
l
zzEI v vδ′′ ′ = and 0
0l
zzEI v vδ′′′ = (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 in x-y plane. Let v be the
nodal linear displacement and φ 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 ��ρ (4.10)
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, ([email protected])
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Figure 4.4 Beam element in x-y plane
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 vφ φ= is the nodal
displacement vector. The linear and angular displacements at the nodes 1 and 2 of the element
( 1 1 2, ,v vφ and 2φ ) 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 freedom 1 1 2, ,v vφ and 2φ . With the help of boundary conditions of the element at the
two nodes
1 4(0)v v a= = 1 3(0)v aφ′ = =
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 aφ′ = = + + (4.13)
1
2
x
y
1v
dx x
2v
z
o
2φ1φ
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Solving for the four constants 1a , 2a , 3a , 4a and substituting in equation (4.12), it gives
( ) ( ) ( ) ( ) 2
1 2 1 2 1 2 1 2 1 13 2 2
2 1 3 1( ) 2v x v v x v v x x v
l l l lφ φ φ φ φ = − + + − − + + + +
(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
( ) 3323
1 32 lllxxN +−= ( ) 33223
2 2 lxllxlxN +−=
( ) 323
3 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 ��ρ+= ′′′′ (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 Qρ ′′ ′′+ = ∫ ∫�� (4.18)
which can be written as
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195
[ ]{ } [ ]{ } { }( ) ( ) ( )ne ne neM q K q Q+ =�� (4.19)
with
[ ] { } ∫=l
dxNANM0
ρ = 2 2
2
156 22 54 13
4 13 3
156 22420
sym 4
l l
l l lAl
l
l
ρ
− − − −
[ ] { }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 2
ne TQ v vϕ ϕ=
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 SM q 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
M q K q Q + = �� (4.21)
Assume a solution of the form, { } { } j
0
tq q e λ= . Substituting this in equation (4.21) and setting the
determinant equal to zero, the associated eigen value problem becomes
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, ([email protected])
196
{ } { }2
0 0
S S
K q M qλ = (4.22)
equation (4.22) can be expressed in the form of
( ){ } { }0 0S S
M p K q − = , with 21p λ= (4.23)
Pre-multiplying equation (4.23) by inverse of S
K the eigen value problem becomes
[ ] [ ]( ){ } { }00D p I q− = (4.24)
where [ ] ( ) 1S 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 a frequency equation in the form of a polynomial of degree n in p, where n is number
of degrees of freedom. Equation (4.25) possesses n real and positive roots and pr related to the system
natural frequencies by 21r rp λ= with r = 1, 2, …, n. If { }rq , represents the eigen vector (mode
shape) corresponding to the eigen value rp , 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 4.1 Obtain 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 ρ = 7800 Kg/m3,
Young’s modulus of shaft material E = 2.1x1011 N/m2 and length of the shaft l = 3 m.
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197
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 m2 and I = 4.91 × 10-10 m4
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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198
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 0
8 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 are 1 4 1 40, 0, 0 and 0w w M M= = = = for simply supported ends. Above
global equation after eliminating 1st and 7
th rows 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 { } { }2
nw wω= −�� , hence the above equation takes the following
form
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199
{ }
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
ω−
′ − − − − ′ − −
− × = −
′−
The above equation has the following form
( )2[ ] [ ] { } {0}nM K wω− =
Therefore, eigen values can be obtained as
det|A| = 0 where A = [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.
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, ([email protected])
200
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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201
2 2
4n
EIn
mlω π= rad/sec
where Am ρ= 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 nω π π−
−
× × ×= = =
× ×
Corresponding to different modes (i.e. n) the natural frequency is tabulated in Table 4.1. MATLAB computer 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 material
L = 3.0; % m Shaft span
N_element =10; % Number of elements
N_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_node
fprintf(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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202
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
end nat_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 vector
fprintf(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 vector
if(isequal(i,1))
y(1, 1:N_mode_plot) = 0; %zeros(1, 1:N_mode_plot); % sorting linear displ. vetors
elseif(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. vetors else
y(jj, 1:N_mode_plot) = eigen_vectors((i-1), 1:N_mode_plot); % sorting linear displ. vetors
end
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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203
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 command
line 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 matrix m_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 matrix
m_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
m_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)
% Error Mode
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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-1.5
-1
-0.5
0
0.5
1
1.5
1 3 5 7 9 11
Node Number
Norm
alised Eigen-V
ectors
Mode I-94.08 rad/sec
Mode II-376.3604
rad/secMode III-847.1608
rad/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-m2, 01.0=d m, 1=L m and
E = 2.1×1011 N/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 3 1
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
3
2 2 2
2 2
2 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
θ θ
θ θ
−− − + = − − − −
��
��
��
��
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
θ
θ
θ
θ
θ
��
��
��
��
��
��
��
��
��
��
+
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
θ
θ
θ
θ
θ
− −
=
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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 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
m y
I
θ
θ
θ
θθ
��
��
��
��
��
��
��
��
3 2
1 1
2
2 2 22
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
Mh 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
θ
θ
θ
θθ
− − − − −
− − − + = − − − 5
0
0
M
For free vibration, on substituting xx 2ω−=�� , in the above equation we get an eigen value problem of
the following form
[ ] [ ] { }02 =+− KMnω
which can be solve to give the following natural frequency of the system
1 1
2 867.18 29.44n nω ω= ⇒ = rad/s and 2 2
2 48.3652 10 289.23n nω ω= × ⇒ = rad/s.
Example 4.4 A 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,Young’s 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
Bearing Bearing
0.1 m Φ
Bearing Bearing
Rigid discs
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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
Norm
alised Eigen-V
ectors 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 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. Neglect the gyroscopic effects.
a b
A B
Figure E4.1 An overhang rotor system
Exercise 4.2 Find critical speeds of the rotor system shown in Figure E4.2. Take EI = 2 MN.m2 for
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 4.3 Find the fundamental bending critical speed of the rotor system shown in Figure E3.7.
B1 and B2 are simply supported bearings and D1 and D2 are rigid discs. The shaft is made of steel with
modulus of rigidity E = 2.1 (10)11 N/m
2 and uniform diameter d = 10 mm. The various shaft lengths
are as follows: B1D1 = 50 mm, D1D2 = 75 mm, and D2B2 = 50 mm. The mass of discs are: md1 = 4 kg
and md2 = 6 kg. Consider the shaft as massless and neglect the diametral mass moment of inertia of
both discs.
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)11 N/m2 of Young’s modulus, E, and
7800 kg/m3 of mass density, ρ. 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 B2
D1 D2
Figure E4.3 Example 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): For F(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 at 0x x= , it can be
written as
)(),( 0
*
0 xxFtxF −= δ (4.28)
The consistent force vector can be written as
( ) *
0 0 0 0
0
{ } ( ){ } { ( )}
h
neP F x x N dx F N xδ= − =∫ (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)
h 1 2
F(x, t) = F(t)
1
2
1 2
( ) 12
1
0 2
1 2
12
( )
( ){ } ( ){ }
( )
( )
h
ne
F t h
F t hP F t N dx
F t h
F t h
= = −
∫
F0 F0 F0
h/2
h h h
(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 and
F Fa F b
h
−= =
Figure 4.16 A linearly varying loading
The assumed form of the force becomes
1( ) ( )2 11
2
( , ) 1 { }e ne
f
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
( , )
21 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, θumb = 300 phase. The phase is measured with
respect to some physical reference on the shaft. Let us assume that reference mark is designed with y-
axis. The angular speed, ω, 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, ω, is in
counter-clockwise direction as seen in Figure 4.18. FE equation for element 1 in the vertical plane x-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
yv 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 0
156 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
w
v
vx
w
w
v
v
−
− − ′− − − − ′− − + − ′ − ′
��
��
��
��
��
��
��
��
1 1
1 1
1 1
1 1
2 2
2 2
2 2
2 2
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
− ′ − − ′ − = + − ′ − ′
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215
In FE equation for element 2, since rows 5th & 7th are EOM corresponding to displacements 3w 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
w
v
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 0
12 6 0 0 12 6
4 0 0 6 2
12 6 0 0
4 0 0
12 6
sym 4
w
w
v
v
w
w
v
v
− ′− − ′− − ′ − ′
2
2
2
j 2
6
3
3
6
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
ω−
−
− − − −
= + − × +
× +
The unbalance force is obtained by j
b bm r e θ in the horizontal direction F4, and the unbalance force in
the vertical direction F3 can be obtained by using Figure 4.19.
(a) (b)
Figure 4.19 Unbalance forces in the horizontal and vertical directions for different direction of rotor
rotations (a) Counter clockwise rotor rotation direction (b) Clockwise rotor rotation direction
FE equation for element 3 is given as
yz jFF =
tjj
bby eermF ωθ=y
z
yz FjF )(−=
θ
ω
θ
y
z ω
tjj
bby eermF ωθ=
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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 0
156) 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 0
4 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
0
0
0
0
0
0
0
0
2.5 10 [0.
6 2
(4 4) 0 0 6 2
12 6 0 0
4 0 0
12 6
sym 4
w
w
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
S
M
S
M
e
S
M
S
M
ω
−
− − − −
+ −
× +
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= = = = .
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, ([email protected])
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, 3
rd, 13
th and 15
th 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 0
1339.4 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
′− ′− − ′− −
′− + −
′− −
′ − ′ ′
j
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
ω−
−
=
× −
× +
Since excitation is a simple harmonic i.e.,{ } { } j tw W e ω= , hence { } { }2w wω= −�� , 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 eω ω ωω− + =
which can be solved as
( ) }{][][}{12 FMKW
−= ω
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, ([email protected])
218
where ω 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 4.5 For 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.
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219
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