/smash/get/diva2:612216/FULLTEXT01.pdf
Kungliga Tekniska Högskolan
Sound And Vibration department
MWL
Marcus Wallenberg Laboratory for Sound and Vibration Research
Master Thesis Topic
:
Evaluation of COMSOL Multiphyscis for rotordynamical analysis
Presented by: Hessameddin Moradi Motlagh
Supervisors:
Professor Ulf
Dr
Carlsson
Martin Karlsson
Examiner: Professor Hans Bodén
Acknowledgment
This report is the result of my master thesis in Sound And Vibration department of
KTH university. It has been done in ÅF Group, KTH with corporation of COMSOL
Multiphysics between November-March 2012. However, The report is delivered in
December 2012.
During my thesis working time, different people in different periods of time helped me.
At KTH university, I want to thank professor Ulf Carlsson my supervisor at KTH who supported me through the procedure, the extension permission to deliver my report and his great guidelines.
On Åf Group, I would like to thank Örjan Johansson that gave me the opportunity to do my thesis in the company and Martin Karlsson for introducing me good information sources such as book references and his bearing parameters.
On COMSOL, Remi Magnard with his vital technical supports and Per Backlund who gave me the permission from COMSOL to use COMSOL’s license during my thesis project time.
Also, I should thank to professor Hans Bodén at KTH to being my mentor and examinor.
Hessam Moradi
December 2012 ii
ABSTRACT
Rotary machinery systems are widely used in different industries. As they have rotational movements, the resonances will be the function of rotational speed. Nowadays, there are different CAE (Computer Aided Engineering ) softwares that use finite element method to solve complex problems by simplifying the model to a limited DOF (Degree Of
Freedom) system. Some of the CAE software have built-in module for rotary machinery
to solve eigenvalue problems, unbalance response, stability, etc. COMSOL multiphysics
is a CAE software that can deal with structural mechanics. There is not any specific module for the rotor dynamics analysis in COMSOL multiphysics version 4.2a; however,
It is possible to make a model based on 1D beam elements or even structural mechanics elements and solve the problem by using equivalent gyroscopic moment and unbalance force terms. The scope of this thesis is based on 1D beam element solution. Moreover, the result of the COMSOL solution has been compared with ANSYS based on 3D element
In this thesis, the main purpose is implementing rotor dynamics concepts in COMSOL
multiphysics based on beam element
. Second chapter is focused on the theory part of the rotordynamics where the equations in the stationary cases are derived which measn the rotational speed of the rotor does not change by time; moreover, a brief overview of finite element method has been mentioned. The third chapter covers the implementation of the rotor dynamics in COMSOL where a very brief overview of different COMSOL’s sections that are used in this thesis are described; Moreover, load implementation , 3D mapping based on extrusion integration and comparison a result data with ANSYS are mentioned. In the fourth chapter rotordynamics analysis is described based on Campbell
, stability analysis and harmonic response from a model that is made in COMSOL.
Final chapter is the conclusion part about this thesis based on the results that have been shown.
1
In this thesis COMSOL version 4.2a has been used
2
ANSYS version 13 has been used for comparison the results with COMSOL
3
It is possible to use shell or solid elements to have a better result in dynamic behavior of the rotor which is out of the scope of this master thesis.
4
Generating Campbell plot based on eigenvalue solutions , requires a post processing code that is done by a Matlab script available in the appendix part.
iii
Contents
2
4
2.1 Unbalance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
2.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
. . . . . . . . . . . . . . . . . . . . . . . . .
5
2.2 Gyroscopic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
2.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
2.2.2 Derivation Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.3 Finite Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
16
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.2 Beam Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.2.1 Bearing load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.2.2 Shaft load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.2.3 Disk load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.2.4 3D Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.2.5 Solvers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.3 COMSOL modeling verification . . . . . . . . . . . . . . . . . . . . . . . . 21
3.3.1 Verification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.3.1.1 Lateral Mode . . . . . . . . . . . . . . . . . . . . . . . . . 21
25
4.1 Campbell Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
4.2 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
4.3 Harmonic response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
32
33
A.1 Matlab Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
iv
Contents
A.2 ANSYS batch file [5] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
A.3 Bearing plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
v
Table of symbols
5
x, y, z
,r
θ
(θ
x
, θ y
, θ z
)
G r
G c
e
I p
I
ω
Ω
j
F
b
F
c
J p
J
δ
µ
Special coordinates
Rotational coordinates [rad]
Gravity center
Geometrical center
Eccentricity [m]
Bearing forces (Stiffness and Damping) [N]
Centrifugal force [N]
Polar mass moment of inertia [kg.m
2
]
Mass moment of inertia around x or y [kg.m
2
Polar area moment of inertia [m
4
]
Area moment of inertia around x or y [m
4
]
Rotational speed [rad/s]
Natural Frequency [rad/s]
Imaginary symbol
Spring constant for inclination [N.m/rad]
Mass of the shaft per unit length [Kg/m]
]
Tab. 0.1: Table of symbols
5
In this thesis bold symbols are vector or matrix notation, otherwise are scalar.
1
1 Introduction
Typically, combination of shafts, bladed disks that are supported by bearings make a rotor dynamic system. The parts that can rotate (Shafts and disks) are called rotor and the static parts as stator (like supports). Turbo-machinaries such as gas turbines, compressors, Jet engines are some example of the rotor dynamics systems.
The rotating parts can rotate with a constant or varying rotational speed. The constant rotational speed is called stationary and non-stationary otherwise. In this thesis all of the derivation formula are considered in stationary case; however, in the industries rotors usually work with varying rotational speed but if the change of the rotational speed rate respect to time is fairly small then it is possible to consider a quasi-stationary rotordynamic case.
.
One of the simplest models for rotordynamics problems is based on Jeffcot rotor where the rotor assumed as a rigid disk that is located on a massless elastic shaft. This simple model can be used for some kinds of analysis such as unbalance effect. Another assumption is
rotor where the dynamic study of the rotor becomes more simplified [4]
When the rotor parts are rotating in some specific rotational frequencies that are called critical speeds the rotor can have high vibration amplitudes. In this case it can damage different part of rotor, stator and any structures that are connected to them. Critical speeds are indeed resonances of the rotor where the rotor eigenfrequencies match rotor
rotational frequencies. The first and second critical speeds are more dominant and it is possible to detected them by using Campbell plot. Before producing a product it is important for designers to detect these frequencies in a computer modeling stage.
The results help to produce more reliable and effective products. In rotor dynamics usually there are some supports as bearings. These supports have different types like fluid bearings, magnet bearings, etc. Always, a bearing exerts a force to shaft to hold it in a right position in order to prevent any contact with stator. This force can be calculated based on the bearing type. For example in fluid bearings, the bearing parameters which
are stiffness and damping can be calculated by using CFD
1
Symmetry in shaft stiffness and moment of inertia [6]
2
The first one is called major critical speed [6]
3
Computational Fluid Dynamics
2
1 Introduction of forces like unbalance force that exerts to the system. In this case, if the gravity and the geometry centerline of rotor does not sit on each other this force shows up.
It could be typically to have unbalance effect for the beam or shaft due to installation of
the components, thermal effect or inhomogeneity of the shafts or disks. [6]
Gyroscopic moment is a typical moment that exerts to the rotating part of a rotor.
This moment shows up when an instantaneous inclination angel from bearing centerline observes and it is a function of rotor rotational speed, mass moment of inertia and inclination angel time derivative in stationary case. It is typical to observe the effect of this moment when a disk is placed in a location of a rotating shaft where the lateral displacement of the shaft inclines the disk.
In this thesis, the major work is setting the rotor dynamics equations inside COMSOL’s framework in a correct manner without focusing on finite element approach that is handled by COMSOL engine codes. However, Matlab code is required for post processing the results to plot the data for some king of solutions such as eigenvalue problems. To
implement rotor dynamics codes that are mentioned in chapter 2, rotors are considered
as flexible elements and disks as rigid. It means that the equation of rotor dynamics for a disk can be set as a point load and for a beam as an edge load (load per unit length).
3
2 Theory
In this chapter the main theory of rotordynamic is derived based on newton second law.
The deviation formula in this section is based on massless shaft with a rigid disk and it
is assumed that no torsional moment and no displacement in z direction happen.[6]
2.1 Unbalance
2.1.1 Introduction
In dynamics when a mass rotating around an axis in a rotating frame, it produces a centrifugal force with proportionality to ω
2 in the outward of the radial direction. It has
been shown in figure 2.1 as the vector F
c
.
Figure. 2.1:
Centrifugal force
Based on Newton’s second law the total force in the radial direction should be equal to zero as we do not have any movement in the radial direction. This means that
F
c
= −F =
d dt
(mv) = mω
2
r
The same situation can happen in rotor dynamic system when the centerline geometry of the bearing and the center of mass of the rotor do not coincide.
4
2 Theory
Figure. 2.2:
Unbalance situation
2.1.2 Derivation Formula
Based on figure 2.3 A, This restoring force is due to the location change of the geometry
centerline G position.
c
that is assumed to be placed on the bearing centerline before deformation.
So F = −kr − c˙r is the restoring force that wants to push back the beam to the original
By writing Newton second law at the point G
r
we have
m
¨r
Gr
+ c˙r + kr = 0 (2.1)
The transformation r
Gr
coordinate can be written as r
Gr
= r + e. So ¨
r
and e are the displacement and eccentricity respectively.
= ¨r + ¨e. Where
The equation 2.1 can be decomposed into x and y components. Where
r
= rcos(ωt)i + rsin(ωt)j and e = ecos(ωt)i + esin(ωt)j regarding to x = rcos(ωt) and
y
= rsin(ωt)
So rewriting 2.1 as two equations in x and y components. As in this thesis we are looking
for the stationary case solution, it means that all time derivation of ω are zero so
m
(¨x + ¨ecos(ωt) − eω
2
m
(¨y + ¨esin(ωt) − eω
2
cos
(ωt)) + c ˙x + kx = 0
sin
(ωt)) + c ˙y + ky = 0
(2.2)
5
2 Theory
As eccentricity is constant so equation 2.2 can be written as :
m
(¨x − eω
m
(¨y − eω
2
2
cos
(ωt)) + c ˙x + kx = 0
sin
(ωt)) + c ˙y + ky = 0
(2.3)
Both terms F
x
= eω
2
cos
(ωt) and F
B shows the existing forces diagram.
y
= eω
2
sin
(ωt) are unbalance forces that exerts to
the disk due to eccentricity . Figure 2.3 A is the geometry of the problem and figure 2.1
Equation 2.3 can be written as
m
¨x + c ˙x + kx = meω
2
m
¨y + c ˙y + ky = meω
2
cos
(ωt)
sin
(ωt)
(2.4)
6
2 Theory
Figure. 2.3:
Unbalance effect derivation formula.
F
= F
k
+ F
d
is the restoring force from the beam and F unbalance force and F
k
and F
d c
is are spring and damping forces belong to the structure respectively
The solution of equation 2.4 can be obtained by substituting x = rcos(ωt + θ) and
y
= rsin(ωt + θ) into 2.4 and equating the coefficient of cosine and sine terms we get :
r
θ
=
√
(ω
2
= tan
eω
2
−ω
−
1
(
2
n
)
2
+(
cω/m
ω
2 −ω 2
n cω m
)
)
2
7
2 Theory
2.2 Gyroscopic
2.2.1 Introduction
Gyroscopic effect is a common phenomenon in a rotary machinery system especially in high rotational speed. Gyroscopic effect is a moment that can be generated in a rotary system when there are instantaneous inclinations changing in time around the rotation axis.
Figure. 2.4:
Inclination of the disk around y axis due to the beam mode shape
In this section to use Newton formula it is important to deal with vector derivation. In vector calculus the derivative of a vector with respect to time can have a direction based on the absolute value time derivations and the direction change of the vector. As it is
shown in 2.5 A, It is possible to decompose W
t
2
− W
t
1 into two vectors. One of them shows the vector direction change without changing the magnitude (W one shows the magnitude change without direction change (W q
⊥
) and the other
) . If the absolute value of the vector does not change, this vector movement called stationary. In this thesis we deal
with stationary rotations that we can see the results in 2.5 B and 2.5 C. It is possible to
8
show that W
⊥ is always perpendicular to W
t
1 if ∆θ → 0.
2 Theory
Figure. 2.5:
The derivation of a vector with respect to time.
W
t2
= W p
+ W q where W p
= W
t1 and W
⊥ is a vector which is produced because of the direction change of W
t1
. W q is a vector which is produced because of changing in absolute value of W
t1
.
A: General form whenW
t1 goes to W
t2 during ∆t = t
2
− t
1 time.
B:W
t1 goes to W
t2 when the absolute value is constant (W
t1
= W
t2
) during ∆t = t
2
− t
1 time
C:W
t1 goes to W
t2 when the absolute value is constant (W
t1
= W
t2
) during ∆t = t
2
− t
1
∆t → 0 time when
9
2.2.2 Derivation Formula
2 Theory
Figure. 2.6:
Gyroscopic term extraction
The restoring moments are M
y
Gyroscopic moment is J
p
θ
1 and M
x
in x and y axis direction respectively.
Figure 2.6 can be used to show the effect of gyroscopic moment. Here, we have inclination
without any lateral displacement. It is assumed that the inclination angel θ is small .
The projected inclination angel on the plane xz and yz are θ
x
θ x
is in the y direction and θ
y
and θ
y
respectively where is in the −x direction based on right hand screw rule.
Line OB is the tangential direction at point O. It could be possible, the centerline of the disk does not sit on the inlination angel of the beam. In this case OA which is the centerline of the disk does not coincide on OB . This difference is shown by angle τ that is called dynamic unbalance.
When disk is rotating with rotational speed ω it can produce a rotational momentum
J p
ω
in the direction of rotational speed. With the same logic in section 2.2.1 this
angular momentum direction changes by time and in the stationary case the net angular
10
2 Theory momentum is perpendicular to it. The net of angular momentum is J
p
a torque J
p
ω ˙
1 during 4T −→ 0.
ωθ
1 and produce
By using Newton second law for moments which expresses that the derivative of angular
momentum is equal to the applied moment, equation 2.5 can be written. By decomposing
moments into x and y coordinates based on the important result from section 2.2.1
equation 2.6 can be written.[6]
J ¨
1
+ J
p
θ
1
+ δθ = 0 (2.5)
(
J ¨
+ J
p
ω ˙
1y
+ δθ
x
− J
p
θ
1x
+ δθ
y
= 0
= 0
(2.6)
Where δ [N.m/rad] is torsional spring constant, J [kg.m
2 around x or y axis and J
p
[kg.m
2
] is mass moment of inertia
] is polar mass moment of inertia
Based on figure 2.6 the inclination of the disk with respect to the beam can make the
relation between θ
1
= θ + τ
. So in x and y components equation 2.7
can be derived.
and θ,where θ
1
(
θ
1x
θ
1y
= θ
x
= θ
y
+ τcos(ωt)
+ τsin(ωt)
By inserting equation 2.7 into equation 2.6 equation 2.12 can be obtained
(2.7)
(
θ x
+ J
p
θ y
θ y
− J
p
θ x
+ δθ
x
+ δθ
y
= (J − J
p
)τω
2
cos
(ωt)
= (J − J
p
)τω
2
sin
(ωt)
(2.8)
From the equation 2.8 for eigenvalue solution the only terms that is depends on theta
are −J
p
ω ˙
y
and J
p
ω ˙
x
. These terms are gyroscopic components in x and y direction respectively that makes the omega dependent eigenfrequency solution. The eigenvalue
solution of 2.8 gives forward whirling and backward whirling frequencies.When the rotor
is rotating it has orbital path or circular path depends on the forces and moments that exerts to it. In the orbital movement the motion would be composed of two frequencies that rotates in the same or opposite direction of the shaft’s rotation. The one that is in the direction of the rotational speed is called forward whirling and the other opposite direction is called backward whirling.
1
τ is dynamic unbalance
2
Note that we are delivering stationary case where ω is constant
11
2 Theory
In section 2.2.2 gyroscopic is considered when there was not any lateral displacement. But
in overall we have six degree of freedom for solid. For rotating disk or beam with constant
ω the deflections in the direction of the bearing centerline is negligible in comparison
with lateral movements. Based on Newton second law we can write these equations.[6]
F x
= −αx − γθ
x
F y
M xz
= −αy − γθ
y
= −γx − δθ
x
M yz
= −γy − δθ
y
(2.9)
Where α, γ, δ are spring constants and for a beam depends on boundary conditions based on beam theory. The response due to transition can be obtained by writing Newton second law at point G
r
(
m x
¨ = F
x m y
¨ = F
y
= −αx − γθ
x
= −αy − γθ
y
(2.10)
But x
Gr
= x + e cos(ωt) and y equation of motion becomes:
Gr
= y + e sin(ωt) and by inserting them into 2.10 the
(
m
¨x + αx + γθ
x m
¨y + αy + γθ
y
= m e ω
2
= m e ω
2
cos
(ωt)
sin
(ωt)
(2.11)
The same principle in section 2.2.2 can be used to drive the gyroscopic in this case. The
added terms to equation 2.8 in no damping case, are restoring moments because of the
inclination. The result would be
(
J ¨
x
+ J
p
ω ˙
y
+ γx + δθ
x
θ y
− J
p
θ x
+ γy + δθ
y
= (J − J
p
)τω
2
cos
(ωt)
= (J − J
p
)τω
2
sin
(ωt)
Equation 2.11 and 2.12 can be written as matrix form.
(2.12)
M¨
+ G ˙q + Kq = F
3
M is the mass matrix, G is the gyroscopic matrix, K is the stiffness matrix, q is the displacement fields and F is unbalance force.
12
2 Theory
m
0 0 0
0
m
0 0
0 0
J
0
0 0 0
J
θ
¨x
¨y
¨
¨
x
θ y
+ω
0 0
0 0
0 0 0
0 0 −J
0
0
p
0
0
J p
0
θ
˙
˙x
˙y
˙
x
θ y
+
α
0
γ
0
0
α
0 γ
γ
0
δ
0
0
γ
0
δ
x y
θ x
θ y
= meω 2
cos
(ωt)
sin
(ωt)
(2.13)
0
0
As gyroscopic terms depends on ω so the eigenfrequencies are dependent on rotational
speed of the rotor. Equation 2.13 can be written when damping is included too[6]. The
matrix form can be written as:
M
¨q + (G + C) ˙q + Kq = F
(2.14)
M
is the inertia matrix , G is the gyroscopic matrix, C is the damping matrix, K is the stiffness matrix and q and F are displacement and force vectors respectively. For an element the matrices are:
M
=
G
=
C
=
K
=
F
=
0 0
0 0
0 0
0 0 −ωJ
p c
11
0
0
c
12
0
c
11
0
0
c
21
0
c
0
0
ωJ
0
0
p
0
c
12
22
0
0
0
0
m
0 0 0
0
m
0 0
0
0
J
0
0
J
c
21
0
c
22
α
0
γ
0
0
α
0 γ
γ
0
δ
0
0
γ
0
δ
meω
2
meω
2
cos
(ωt)
sin
(ωt)
0
0
It is obvious that in the bearing positions equation 2.16 hold.
(
K total
C total
= K
structure
= C
structure
+ K
bearing
+ C
bearing
(2.15)
(2.16)
13
2 Theory
2.3 Finite Element
Finite element is widely use for complex structures and approximate the solution by limiting the number of DOF of the structure by controlling the mesh size. This method is used in COMSOL or any other finite element software to solve the problem.
In this method the solution for each element is linear combination of the shape functions.
The shape functions are special coordinate functions that are used for interpolation data between nodes. The shape functions have some properties such as:
• Each shape function only should be 1 in one node and 0 in the rest nodes
• The sum of the shape functions are 1
• It should be continuous and it should be derivative up to the required order
The element matrices can be calculated in different ways. For example in structural mechanics, the stiffness is obtained from the potential energy of an element.
dU
=
1
2
EI
∂
2
u
∂
2
x dx
Based on the finite element method u the displacement vector components can be written as:
u
=ϕq where ϕ is the shape function matrix and q is the generalized coordinate. By substituting it to potential energy equation and integrating over the beam’s element length l, the stiffness element matrix obtains as :
k e
=
´
0
l
EI
¨
T
By using kinetic energy the mass matrix, gyroscopic matrix and force matrix can be
The element displacement can be written as linear combination of shape functions. The element equation can be written as
M
¨q + (G + C) ˙q + Kq = F
where for a rigid disk it has been shown in 2.15 where K = 0. For a rotor element without
damping consideration it can be shown that the symmetric element matrices are [6]:
14
2 Theory
M =
µl
420
µr
2
120l
156
0
0
22l
54
0
0
−13l
156
−22l
4l
2
0
0
0
0
54
−13l
13l
−3l
2
0 0
4l
2
13l
0
0
−3l
2
36
0
0
36
−3l
4l
2
S
3l 0
−36
0
0
0
4l
2
−3l
0
−36 −13l
0
0
−3l −3l
2
0
3l 0 0 −l
2
36
0
0
−3l
2
S
Y
M
156
0
0
−22l
156
22l 4l
2
0 0 4l
2
Y
36
0
M
3l 4l
0
2
4l
2
+
G
=
µl
420
0
−
36 0
3l
0
0
−
36 3l
36
3l
0
0 0
3l
−
4l
0
0
3l
2
l
0
0
−l
2
0
S
Y
0
0 0
3l −36 0
2
M
−
3l 0 0
0
−
3l −4l
2
0
K
=
µl
420
12
0 12
0
−
6l 4l
2
−
6l
12
6l
0
0
0
−
12 6l 0
0
−
6l 2l
0
S
0 4l
2
0 −6l 12
2
0
0 2l
2
Y
M
0 12
0 6l 4l
2
−
6l 0 0 4l
2
These element matrices are calculated based on calculating the kinetic and potential energy of an element (both translation and rotation) . Moreover, the shape function is considered for two nodes beam element with eight degree of freedom.
The assembled matrix for the whole system (Disk-Shaft system) can be found in [6] where
the matrix elements have 12 × 12 dimensions.
15
3 COMSOL
3.1 Introduction
COMSOL multiphysics is a finite element software that supports different kinds of physics such as structural mechanics, acoustics, etc. They can interact with each other in order to simulate coupling physic problems.
The module that is used in this thesis is “Beam” physic where the solution is based on
Euler beam equations are valid when the shape of each cross section remains plane after
deformation and the curvature radios is much longer than beam’s thickness. [3]
Beam module is available in both 2D and 3D workspace. 3D workspace has been used as the solution on neutral axis can be extruded to have 3D visualization of different modes by using coupling integration that will be described later.
COMSOL has object structural form with scope definition functionality. It means that, it is possible to define different variables to the specific part of the physics. The advantage is saving memory and reducing calculation time. There is a special variable in COMSOL that is called “Probes”. A probe has the ability to be called from parametric sweep event handler over and over during problem solving stage. Parametric sweep is a recursive function that runs itself over given parameters. This ability helps to solve a physic one time with different number of parameters that should be change during the calculations.
For example in this thesis bearing data are dependent on rotational frequency of the rotor. It means that bearings are a function of rotational speed. So it is possible to define a parameter in parametric sweep that plays the rotational speed role for the bearing data.
For coupled physics problems field parameters have the scope through the connections.
It means that in the connection that the material is belonging to several physics, the field properties of any of the physics has global scope. In the case that, there is not any connection between different physics from geometrical point of view, it is possible to use integration operator in COMSOL to make the scope of a particular physic element shared globally.
16
3 COMSOL
Another interesting feature in COMSOL is “coupling integration”. The task of coupling part is mapping the source data to the destination data. This feature can be used for extrusion purpose and make a 3D modeling.
3.2 Beam Element
In COMSOL beam element should design as an edge element respect to geometrical view perspective. The cross section of any edge can be set by adding “Cross section data” feature . There are different types of standard cross sections for a beam element that can be chosen by user.
Boundary conditions in the beam can be set on points and edges. As it is mentioned in the theoretical part, the axial displacement and torsional modes are much smaller than lateral modes in constant rotational speed and when we have plan motion, so it is
By adding “prescribed displacement/rotation” feature in the beam it is possible to fix the displacement and rotation around axial direction. In “Beam” physic, loads could be point loads or edge loads. Bearing loads can directly being implemented as point load in a beam simulation based on neutral axis. Gyroscopic moment for a disk, can be assumed as a point load moment as the disk assumed as a point mass, however; for the beam edge load has been used as the beam’s neutral axis geometry is available.
3.2.1 Bearing load
it is possible directly enter the bearing forces as the function of our spatial variable fields components.
"
F bx
F by
#
= −
"
k xx k yx k xy k yy
# "
u v
#
−
"
c xx c yx c xy c yy
# "
˙v
˙u
# or as the equation form in COMSOL :
F x
= −k
xx
u − k
xy
v−c
xx u t
− c
xy v t
F y
= −k
yy
v − k
yx
u−c
yy v t
− c
yx u t
1
However; it is possible to use spring load directly when there is not cross terms in the spring force equation.
17
3 COMSOL where k
ij
and c
ij
are spring constant and damping parameter respectively are diagonal of the bearing .
3.2.2 Shaft load
Gyroscopic moment can act to the beam the same as disks. To calculate mass moment inertia of the disk or a cylinder around the axial axis can be written as J
J p
=
1
2
m
(r
2
i
+ r
2
o
) for a pipe where m is the mass r
i
and r
o p
=
1
2
mr
2 and are inner and outer radius of
J p
=
1
2
ρV
(r
2
i
+ r
2
o
) → J
p
=
1
2
ρAL
(r
i
2
+ r
2
o
) (3.1) where
A
= π(r
2
o
− r
2
i
)
By substituting 3.2 in 3.1 the equation can be obtained.
(3.2)
J p
=
1
2
ρπ
(r
2
o
− r
2
i
)(r
2
i
+ r
2
o
)L = −
1
2
ρπ
(r
4
o
− r
4
i
)L (3.3)
Equation 3.3 can be written per unit length as an edge load. In this case the gyroscopic
moments in both x and y directions per unit length based on gyroscopic terms in equation
(
M x
M
= −(
y
= (
1
2
1
2
ρπ
(r
2
o
ρπ
(r
2
o
− r
− r
2
i
2
i
)(r
)(r
2
i i
2
+ r
2
o
))ω ˙θ
y
+ r
2
o
))ω ˙θ
x
= −
=
1
2
1
2
ρπ
ρπ
(r
(r
4
o
4
o
− r
4
i
− r
4
i
)ω ˙θ
)ω ˙θ
x y
(3.4) where M
x
and M
y
are the moments per unit length. Equation 3.4 can be used directly
in COMSOL as an edge load for any pipe beam cross section form.
It is obvious that by inserting r
i
section.
= 0 the equation 3.4 would be valid for a circular cross
2
i = j
3
i 6= j
4
The default variables in COMSOL are ρ = beam.rho,r
o
= 0.5 ∗ beam.do_pipe, r
i
= 0.5 ∗
beam.di_pipe, ˙
y
θ x
= thxt
5
If the cross section selected circular instead of pipe, beam.do_pipe and beam.di_pipe are not available
in COMSOL anymore. instead they should be written based on beam.area. In this case 3.4 when
( the cross section is selected as circular it can be written as
−.5beam.rho ∗
.5beam.rho ∗
beam.area
2
pi beam.area
2
pi
;however, this formula can be used for pipe sections directly in COMSOL too.
18
3 COMSOL
3.2.3 Disk load
Gyroscopic moment can act on disk the same way as on a shaft but as a point load.
Moreover, as in the simulation the disk is considered as a point mass, mass contribution effect should be considered in order to compensate the lack of disk in simulation.
In this case the gyroscopic moment can be entered directly as a point load moment.
(
M x
= −(
M y
= (
1
2
1
2
ρπr
ρπr
4
4
t
)ω ˙θ
t
)ω ˙θ
x y
Where t is the thickness of the disk. For the mass contribution by adding a point mass load and substituting the moment of inertia of the disk in x, y, z directions, the lack of the disk can be compensated.
F
= −mω
2
M x
u
= −J
x
ω
2
θ x
M y
= −J
y
M z
ω
2
θ y
= −J
z
ω
2
θ z
3.2.4 3D Modeling
In order to make a 3D model version based on neutral axis solution of beam element,
COMSOL has different solutions. The one that has been used in this thesis is General
Extrusion in coupling model option. The task of this extrusion is mapping source displacement field to destination geometry displacement. This transformation matrix can be written as
"
"
"
x y new y new z z new new new x new
#
#
#
=
=
=
"
"
"
cos
(thz) −sin(thz)
sin
(thz)
cos
(thz)
cos
(thx) −sin(thx)
sin
(thx)
cos
(thx)
cos
(thy) −sin(thy)
sin
(thy)
cos
(thy)
# "
# "
# "
x y y z z x
#
#
#
(3.5)
u
= x
new v
= y
new w
= z
new
− x
− y
− z
(3.6)
6
I should have special thank from Remi Magnard in COMSOL who helped me to implement 3D mapping
19
3 COMSOL
θ x
= genext(thx)
θ y
= genext(thy)
θ z
= genext(thz)
(3.7)
Where u, v, w, thx, thy, thz are COMSOL’s displacement and rotation field components.
By inserting equation 3.6 and 3.7 in 3.5 and use super position of the displacements , the
result becomes :
x new
= genext(u) + x ∗ (cos(θ
y
) + cos(θ
z
) − 2) − y ∗ sin(θ
z
) + z ∗ sin(θ
y
)
y new
= genext(v) + x ∗ sin(θ
x
) + y ∗ (cos(θ
z
) + cos(θ
z
) − 2) − z ∗ sin(θ
x
)
z new
= genext(w) − x ∗ sin(θ
y
) + y ∗ sin(θ
x
) + z ∗ (cos(θ
x
) + cos(θ
y
) − 2)
(3.8) where “genext” is the name of the coupling function for the displacement field mapping.
To use equation 3.8 it is important that coupling function should be defined beforehand
in COMSOL and the neutral axis of the beam should be selected in the source term.
Figure 3.3 has been created by 3D extrusion mapping based modeling.
Figure. 3.1:
3D mapping in COMSOL based on extrusion from the model in figure 3.3
20
3 COMSOL
3.2.5 Solvers
There are different kinds of solvers to deal with different study types in COMSOL multiphysics. In this thesis, all of the analyzing parts has been done in Eigenfrequency and Frequency domain solvers. In all of the solvers it is possible to use Parametric Sweep that is an interesting feature in COMSOL. This feature can make an event dispatcher to the solver each time when its value being called by the solver. To generate campbell plots and root locus diagrams, this feature has been used to call the appropriate bearing
in the equations as well as gyroscopic omega dependency.
3.3 COMSOL modeling verification
3.3.1 Verification
This section is verifying the results of simple models in COMSOL multiphysics. Lateral mode verification has been shown analytically in the first step. Afterward a model from
COMSOL multiphysics and ANSYS 13 commercial software have been compared in eigenvalue solution for campbell plot to investigate gyroscopic whirling modes. The model in ANSYS 13 is based on 3D modeling finite element solution. The ANSYS model
is based on [5] where small changes are considered. The code is available in appendix.
3.3.1.1 Lateral Mode
To evaluate the lateral mod, a simple model is designed in COMSOL based on these parameters:
L
= 1000 mm
d
= 100 mm
E
= 211 ∗ 10
9
P a
ρ
= 7850
kg m
3
(3.9)
For a beam with simply supported boundary condition, the first bending eigenfrequency can be obtained theoretically as
f
=
π
r
2
EI mL
3
(3.10)
Where “E” is Young Modulus, “I” is area moment of inertia around center of the cross section, “m” is mass and “L” is the length of beam.
7
In rotating machinery, bearing parameters are ω dependent
21
3 COMSOL
By substituting the parameter values from 3.9 in 3.10 the first bending eigenfrequency
can be obtained analytically.
f
=
π
2 r
211∗10
9
7850∗
π
4
∗
π
∗.
05
4
4
∗
0.1
4
∗
1
4
= 203.59 Hz
The corresponded model in COMSOL is shown in figure 3.2 that corresponded with
analytical solution.
Figure. 3.2:
COMSOL first lateral mode at 203.59 Hz
3.3.1.2 A COMSOL model comparison with ANSYS 13 CAE software
Figure 3.3 shows the geometry part of the problem. The model is a simple beam that is
rigidly supported at one end and free in the other end. The campbell plot can be seen
from both COMSOL and ANSYS 13 CAE software in Figure 3.4.
Figure. 3.3:
The comparison model. Dimensions are in millimeter
22
3 COMSOL
E
[
N m
2
] (young’s modulus) ρ [
211E9
kg m
3
] Density ν [−] Poisson’s ratio
7850 0.3
Tab. 3.1: Model properties
Figure. 3.4:
Campbell plot. COMSOL versus ANSYS13
Dashed lines: ANSYS13 solution data
Solid lines : COMSOL multiphysics data
The first and second backward and forward whirling modes due to gyroscopic effect are
that the first whirling mode time derivative inclinations between θ than the first one as the whirling modes are better separated.
x
and θ
y
is not strong as the second whirling modes. The gyroscopic term in the second whirling is stronger
23
3 COMSOL
At (ω = 345 RP M, f = 5.721 Hz) one lateral or torsional mode can be seen in the model and as the beam and disk are homogenous, the corresponded spring matrix is constant and so it can be seen as a constant line.
The ANSYS and COMSOL model solutions are fairly following each other. The error increases for higher natural frequencies where the mesh shape and density of the mesh is dominant to get a closer result to the exact solution. In ANSYS everything is calculated based on 3D modeling finite element and therefore it is more reliable and closer to the exact solution; however, both results still based on numerical solutions. In 3D modeling ANSYS, moment of inertia is totally dependent on mesh quality; however in
COMSOL based on 1D Euler beam solution, moment of inertia has been calculated based on geometry model and cross section of the beam mathematically . So it is totally independent of mesh.
24
4 Analyzing rotordynamics
In this chapter, three cases rotor dynamics test are discussed . The first case is an eigenvalue solution and plot campbell diagram which is eigenfrequency versus shaft rotational speed in order to detect critical speeds. The second case is stability analyzing by using root locus diagram and the third one is harmonic response to track the displacement field magnitude and phase difference with unbalance force excitation. The model that has been used here to show the results, have supports that are not rigid at the both sides of the shaft . Instead, the supports are flexible with cross term bearing parameters. The plots of the bearing parameters are available in the Appendix. The model geometry is
shown in figure 4.1 with the properties in the below table.
E
[
N m
2
] (young’s modulus) ρ [
kg m
3
209E9
] Density ν [−] Poisson’s ratio
7850
Tab. 4.1: Model properties
0.28
Figure. 4.1: COMSOL model geometry
4.1 Campbell Diagram
Campbell diagram is one of the most important diagrams in the very first step of analyzing rotor dynamic system. It shows natural frequency behavior of the rotor when rotor
25
4 Analyzing rotordynamics speed changes and mainly is used to find specific eigenfrequencies of the system that are equal with the rotational speed of the rotor. In this diagram horizontal axis is rotational frequency of the rotor and vertical axis is the eigenfrequency. It is possible to visualize the
critical speeds if we find the intersection of line Y = ω
with Y
i
= f(Ω
i
)
for any i where
Y i
is the eigenfrequency of the system solution for ith mode for different rotational speed of the rotor. These are the frequencies that should be avoided as we have resonances.
The main reason is because of unbalance force that could exist in a rotary system. This unbalance force is proportional to squared of rotational speed of the rotor and when it reach any of the natural frequencies frequency of the rotor, it could reach stator and damage the rotor because of high vibration amplitude.
In COMSOL eigenvalue solution is based on the assumption of Y = A(x, y, z)e
λt
where
λ
is eigenvalue. By using this definition the characteristic equation, gives the natural frequency of the system as the imaginary part of the eigenvalue solution. So λ = ±jΩ
d
where Ω
d
is the damped natural frequency.
1
ω: Shaft rotational speed
2
Ω
i
: Eigenfrequency for any i
26
4 Analyzing rotordynamics
Figure. 4.2:
Campbell plot example Model 1
figure 4.1 with bearing parameters
It is assumed that the rotor speed is between 0 to 9000 CPM
black line with first and second eigenfrequencies which are the first backward and forward whirling modes are 887 RPM and 893 RPM. The third critical speed whirling modes happens at 3942 RPM. The critical speed belongs to the second forward whirling modes are far away from the rotor working frequency range, so there is not any concern about it.
Another example of a campbell plot is shown in figure 4.3. The geometry and boundary
condition belongs to figure 3.3 and the rotor working frequency is between 0 to 140 Hz.
In this case, the whirling modes belong to gyroscopic terms are in the range of the rotor speed.
3
Cycle Per Minute
27
4 Analyzing rotordynamics
Figure. 4.3: Campbell plot example
The first critical speed is f
f c
2 existence.
c
1
= 14.5 Hz as the lateral mode and the second one is
= 65.82 Hz which is the first backward whirling mode because of the gyroscopic term
4.2 Stability
Generally, stability means the amplitude of the eigenvalue solution should not growth in time when there is no external excitation. Otherwise, it is unstable system. As it is discussed, in rotor dynamics the solution of eigenvalue depends on rotational speed due to gyroscopic moment. To detect the stability in rotor dynamics it is common to use
Root-Locus plot where the modal damping ration is in X direction and eigenfrequencies of the rotor is in Y direction. When modal damping ratio is negative it means that the system is self-excited and bearing forces feeding energy to the system.
Figure 4.4 shows root-locus plot for model in figure 4.1 when the supports are not rigid.
28
4 Analyzing rotordynamics
Figure. 4.4:
Stability example
As it has been shown in figure 4.4 the solution is stable for most of the frequency
range;however, for close to whirling mode motions the solution has more unstable tendency as they have negative modal damping ratio.
4.3 Harmonic response
Harmonic response is an important analyzing process in rotor dynamics. The idea is that rotor should not have large displacement in any points due to any external forces in order
to not reach stator. As it is mentioned already in section 2.1 the unbalance force has
harmonic nature and this harmonic force can make high displacement amplitude. It is possible to cancel large displacement locations due to unbalance force by using external mass in an appropriate position which is not belong to this thesis scope.
In harmonic balance in the stationary case , if u is the solution of the displacement, the amplitude and phase can be written as :
29
4 Analyzing rotordynamics
θ
= arctan(
u img u real
)
u amp
= q
u
2
img
+ u
2
real
The result of harmonic response for the model based on figure 4.1 are shown blew.
Figure. 4.5: Displacement of the disk
.
Eccentricity unbalance is considered 0.01 m. As it has been shown in the figure 4.5 the
maximum displacement happens in the same frequency of the critical speeds in figure 4.2
The phase difference of the same model has been shown in figure 4.6. As it has been
shown in both critical speed frequencies the phase difference is −
π
2 or 270
o
.
30
4 Analyzing rotordynamics
Figure. 4.6: Phase difference between excitation and displacement of the disk
31
5 Conclusion
Using 1D element is simple method but it does not possible to use it for some certain problems e.g. disks contains blades. In COMSOL,moreover; 3D mapping gives an intuition about mode shapes of the beam in a tangible way but all disks are considered as solid elements in point load and mass contribution load implementation. It means that no deformation on disks can be demonstrated. Besides, 3D mapping is just extruding the solution of neutral axis of the beam to another geometry which does not show the real solution of the other locations of the beam in space. A closer solution to the reality can be achieved if 3D element approach being implemented like the ANSYS model that has been shown for comparison. In 3D modeling it is possible to make a complex geometry for disks that contain different type of blades and work on meshing part freely. Moreover; it is possible to have complex boundary conditions contain surface or volume loads.
The implementation of gyroscopic in COMSOL is easy even for beam where edge load should be used. The reason belongs to the load implementation in COMSOL as it will automatically multiply to the shape function and integrated along the domain to get the force vector.
That would be the next working approach in COMSOL that solve everything in structural mechanics instead of 1D beam. Fortunately, COMSOL made this project very easy, as no finite element code is written particularly for this project . It means that all discretization and assembling of matrices has been done in COMSOL in the background stage and that is the purpose of using any CAE software. Loads are implemented based on analytical solutions that are available in any rotordynamics textbooks and the most important task in this project was dealing with COMSOL to implement rotordynamics analytical formula based on Unbalance and Gyroscopic terms in the appropriate fields and solving progress and post processing of data.
One restriction in the loading section that has been used in this thesis, is modeling only a disk as in COMSOL it is not possible to define one node as a beam element. Maybe one idea is to resolve this problem by making two elements (three points) and make their length very small and use point load approach and mass contribution to solve the gyroscopic implementation; however, implementing boundary conditions for the disk is not an easy task.
32
A Appendix
A.1 Matlab Code
function
out = correctionTable(s , referenceColumn , scale ,sort ,sortType)
% Function to adapt output tables from COMSOL with Ansys model export for better %visualization
lineIndex=1; fid=fopen(s) ; repeatIndex=1; objectIndex=1;
TYPES={’ac ’ , ’dc ’ }; stringValues={’Model: ’ , ’Version : ’ , ’Date: ’ , ’Table: ’ };
i f
(nargin==2) scale=1;
sort
=0;
elseif
(nargin==3)
[temp1 temp2]=size( scale ) ;
i f
(temp1~=1 && temp2~=1)
disp
( ’ scale␣argument␣should␣be␣a␣vector ’ ) ;
return elseif
(temp2==1) scale=scale ’ ;
end for
i=1:length( scale )
i f
(~isnumeric( scale ( i )))
disp
( ’The␣scale␣argument␣should␣be␣a␣vector␣of␣numbers’ ) ;
return end end sort
=0;
else i f
(nargin==4) sortSize=size(sort) ;
i f
(~isnumeric(sort))
disp
( ’Error :␣The␣data␣selection␣in␣the␣sorting␣vector␣should␣contains␣numerical␣values
! ’ )
return end i f
(nargin==5)
i f
(~strcmpi( sortSize(1) , ’1 ’ ) && ~strcmpi( sortSize (2) , ’1 ’ ))
disp
( ’Error :␣The␣fourth␣argument␣should␣be␣a␣row␣or␣column␣vector␣that␣identifies␣ the␣column␣which␣sorting␣should␣be␣done. ’ )
return else i f
(strcmpi( sortSize (2) , ’1 ’ ))
sort
=sort ’ ;
end
33
A Appendix
end
sortType=’ac ’ % Ascending
else i f
(~strcmpi(sortType ,TYPES(1)) && ~strcmpi(sortType ,TYPES(2)) ) sortType=TYPES(1) ;
end end end end
headerNames={};
for
i=length(s):−1:1
i f
(strcmpi(s( i ) , ’ . ’ )) endIndex=i −1;
for
j=i −1:−1:1
i f
(strcmpi(s( j ) , ’\ ’ ) | | strcmpi(s( j ) , ’/ ’ )) name=s( j+1:endIndex) ;
break end end end break end
temp=textscan( fid , ’%s ’ , ’whitespace ’ , ’\n’ ) ;
for
i=1:length(temp{1}) header=1; temp2=temp{1}( i ) ;
i f
(~strcmpi(char(temp2{1}(1)) , ’%’ )) %Remove comments currentLine=char(temp{1}( i )) ; colLine(lineIndex)=textscan(currentLine , ’%s ’ ) ;
%Errors
i f
(length(colLine{lineIndex})<3)
return elseif
(referenceColumn>length(colLine{lineIndex})) display ( ’Error ,␣the␣reference␣Column␣number␣is␣bigger␣than␣the␣size␣of␣array ’ )
return elseif
(length( scale )~=length(colLine{lineIndex})−1)
i f
(nargin==2) scale=ones(1 ,length(colLine{lineIndex})−1);
else
display ( ’Error ,␣length␣of␣the␣scale␣argument␣should␣be␣the␣same␣of␣the␣input␣ table ’ ’ s␣column␣−␣1 ’ )
return end else for
j=1:length(sort)
i f
(sort( j )>length(colLine{lineIndex}))
disp
( ’Error :␣one␣of␣the␣elements␣in␣the␣sort␣argument␣is␣bigger␣than␣the␣ number␣of␣available␣columns ’ )
return elseif
(nargin>=4 && sort( j )<=0)
disp
( ’Error :␣One␣of␣the␣sorting␣vector␣has␣negative␣or␣zero␣value . ’ )
return end end end
% Separation and scaling
i f
(lineIndex>1)
34
A Appendix
i f
(strcmpi(colLine{lineIndex−1}(referenceColumn) ,colLine{lineIndex}( referenceColumn)))
for
j=1:length(colLine{lineIndex})
i f
(j<referenceColumn) scaleIndex=j ;
else
scaleIndex=j−1;
end i f
( j~=referenceColumn) colLine{lineIndex}{j}= num2str(str2double(colLine{lineIndex}{j })∗scale
(scaleIndex)) ;
i f
(~sort) out{objectIndex}{repeatIndex}=[out{objectIndex}{repeatIndex} char
(9) colLine{lineIndex}{j }];
else
% Sorting
for
k=1:length(sort)
i f
(sort(k)==objectIndex) tempNum=str2num(out{objectIndex}{repeatIndex}) ;
i f
(strcmpi(sortType ,Type(1)))
i f
(tempNum(length(tempNum))−colLine{lineIndex}{j}>0) out{objectIndex}{repeatIndex}=[out{objectIndex}{ repeatIndex} char(9) colLine{lineIndex}{j }];
else else end
out{objectIndex}{repeatIndex}=[out{objectIndex}{ repeatIndex} char(9) colLine{lineIndex}{j }];
end end end
%end Sorting
end
objectIndex=objectIndex+1;
end end
objectIndex=1;
else
repeatIndex=repeatIndex+1;
for
j=1:length(colLine{lineIndex})
i f
(j<referenceColumn) scaleIndex=j ;
else
scaleIndex=j−1;
end i f
( j~=referenceColumn) colLine{lineIndex}{j}=num2str(str2double(colLine{lineIndex}{j })∗scale ( scaleIndex)) ; out{objectIndex}{repeatIndex}=[colLine{lineIndex}{referenceColumn} char(9) colLine{lineIndex}{j }]; objectIndex=objectIndex+1;
end end
objectIndex=1;
end
35
A Appendix
else for
j=1:length(colLine{lineIndex})
i f
(j<referenceColumn) scaleIndex=j ;
else
scaleIndex=j−1;
end i f
( j~=referenceColumn) colLine{lineIndex}{j}=num2str(str2double(colLine{lineIndex}{j })∗scale ( scaleIndex)) ; out{objectIndex}{repeatIndex}=[colLine{lineIndex}{referenceColumn} char(9) colLine{lineIndex}{j }]; objectIndex=objectIndex+1;
end end
objectIndex=1;
end
lineIndex=lineIndex+1;
else
% Header Making
currentHeaderLine=char(temp{1}( i )) ; colLineHeader(lineIndex)=textscan(currentHeaderLine , ’%s ’ ) ;
for
j=1:length(stringValues)
i f
(strcmpi(char(colLineHeader{lineIndex}(2)) ,char(stringValues{j }))) header=0;
break end end i f
(header) spaceData=isspace(currentHeaderLine) ; tempName=’ ’ ;
for
j=3:length(currentHeaderLine)
i f
(spaceData( j )==1 && j~=length(currentHeaderLine) && (spaceData( j+1)==1 | | strcmpi(currentHeaderLine( j ) , ’\t ’ ))) indexJ=j ;
break end end
j=indexJ ;
while
j<=length(currentHeaderLine)
i f
(spaceData( j )==0) tempName=[tempName currentHeaderLine( j ) ] ;
elseif
( j~=length(currentHeaderLine) && spaceData( j+1)==0) tempName=[tempName currentHeaderLine( j ) currentHeaderLine( j+1)] ; j=j+1;
else i f
(~isempty(tempName)) headerNames{length(headerNames)+1}=tempName;
end
tempName=’ ’ ;
end i f
( j==length(currentHeaderLine) && ~isempty(tempName)) headerNames{length(headerNames)+1}=tempName;
36
A Appendix
end
j=j+1;
end end end end
% Header correction
for
i=1:length(headerNames) tt=char(headerNames( i )) ; indexes=regexp(tt , ’ [A−Za−z0−9␣ . ] ’ ) ;
for
j=length(indexes):−1:2
i f
(indexes( j )−indexes(j−1)~=1)
i f
(strcmpi( tt (indexes( j )) , ’␣ ’ )) headerNames( i )={tt (indexes( j+1): indexes(length(indexes))) };
else
headerNames( i )={tt (indexes( j ) : indexes(length(indexes))) };
end break end end end
% Saving
a=’ ’ ; k=1;
for
i=1:length(headerNames)
save
([name ’_’ char(eval([ ’headerNames( ’ num2str( i ) ’ ) ; ’ ]) ) ’ . txt ’ ] , ’−ascii ’ , ’a ’ ) a=’ ’ ; f i l e=fopen([name ’_’ char(eval([ ’headerNames( ’ num2str( i ) ’ ) ; ’ ]) ) ’ . txt ’ ] , ’w’ ) ;
for
m=1:length(out{i })
i f
(~isempty(a))
eval
( ’a=[eval ([ ’ ’a ’ ’ ; ] ) ␣char(10)␣out{i}{m}]; ’ ) ;
else eval
( ’a=[out{i}{m}]; ’ ) ;
end end fprintf
( file ,a) ;
fclose
( f i l e ) ;
end
37
A.2 ANSYS batch file [5]
/PREP7
! Defines the elements
ET,1 ,MESH200
ET,2 ,SOLID186
KEYOPT,1 ,1 ,7
KEYOPT,1 ,2 ,0
! Defines the material data
MAT,1 ,
MP,Ex,1,211e9
MP,dens,1,7850
MP,prxy,1 ,0.3
! Defines the 2D model
BLC4,0 ,0 ,1 ,0.015
BLC4,1 ,0 ,0.05 ,0.3
aadd, all lsel , , loc ,x,0 lesize , all , , ,1 allsel lsel , , loc ,y, ,0 lsel ,a, loc ,y,0.015
lsel ,u, loc ,x,1 ,1.05
lesize , all , , ,25 allsel lsel , , loc ,x,1 ,1.05
lsel ,u, loc ,x,1 lsel ,u, loc ,x,1.05
lesize , all , , ,2 allsel lsel , , loc ,x,1 lsel ,a, loc ,x,1.05
lesize , all ,0.03
allsel
type
,1 amesh, all
! Rotate the 2D model 360 degrees
type
,2
EXTOPT,ESIZE,3 ,0 ,
EXTOPT,ACLEAR,1
EXTOPT,ATTR,1 ,1 ,2
VROTAT, all , , , , , ,1 ,2 ,360, ,
ACLEAR, all cm,e_model,elem
FINISH
/SOLU asel , , loc ,x,0
DA, all ,ALL, allsel eplo coriolis ,on, , ,on pi=acos(−1) antype,modal modopt,qrdamp,5 ,0.1 , ,on,
! Solving the modal analysis and writes out the result to utdata2 . t x t
∗do, i ,1 ,51 spn=(i−1)∗14∗2∗pi/60
A Appendix
38
cmomega,e_model,spn, , ,0 ,0 ,0 ,1 ,0 ,0 mxpand,5 solve
∗enddo finish
/post1 neigv=5 nint=51 speeddel=14 pi=acos(−1)
∗dim,rew, table , nint , neigv+1
∗dim,imw, table , nint , neigv+2
∗do, i ,1 , nint nrpm=(i−1)∗speeddel
∗set ,rew( i ,1) ,nrpm
∗set ,imw( i ,1) ,nrpm
∗set ,imw( i ,2) ,nrpm/60
∗do, j ,1 ,neigv set , i , j ,0 ,0
∗get ,rew( i , j+1), active , , set , freq set , i , j ,0 ,1
∗get ,imw( i , j+2), active , , set , freq
∗enddo
∗enddo
∗cfopen , utdata2 , txt i i=1
∗vwrite ,imw( ii ,1) ,imw( ii ,2) ,imw( ii ,3) ,imw( ii ,4) ,imw( ii ,5) ,imw( ii ,6) ,imw( ii ,7)
(f7 .0 , f6 .1 , f8 .2 , f8 .2 , f8 .2 , f8 .2 , f8 .2)
∗cfclose
A Appendix
39
A.3 Bearing plots
A Appendix
Figure. A.1: Stiffness data
40
A Appendix
Figure. A.2: Damping data
41
Bibliography
[1] COMSOL Manual Help, COMSOL Multiphysics.
[2] www.livephysics.com.
[3] A.C, N., VibroAcoustics part1, KTH, 2008.
[4] Genta, G., Vibration Dynamics and Control, Wiley Series in Nonlinear Science, 2009.
[5] Samuelsson, J., Rotor dynamic analysis of 3d-modeled gas turbine rotor in ansys,
Master’s thesis, Linköping University of Technology, 2009.
[6] Toshio Yamamoto, Y. I., Linear and Nonlinear Rotordynamics: A Modern Treatment
with Applications
, Wiley Series in Nonlinear Science, 2002.
42
List of Figures
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
Unbalance effect derivation formula.
. . . . . . . . . . . . . . . . . . . . . .
7
Inclination of the disk around y axis due to the beam mode shape
. . . . . . . . . .
8
The derivation of a vector with respect to time.
. . . . . . . . . . . . . . . . .
9
. . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3D mapping in COMSOL based on extrusion from the model in figure 3.3
. . . 20
COMSOL first lateral mode at 203.59 Hz
. . . . . . . . . . . . . . . . . . . . 22
The comparison model. Dimensions are in millimeter
. . . . . . . . . . . . . . 22
Campbell plot. COMSOL versus ANSYS13
. . . . . . . . . . . . . . . . . . . 23
4.1 COMSOL model geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
Campbell plot example Model 1 figure 4.1 with bearing parameters . . . . . 27
4.3 Campbell plot example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.5 Displacement of the disk . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
4.6 Phase difference between excitation and displacement of the disk . . . . . 31
A.1 Stiffness data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
A.2 Damping data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
43
List of Tables
0.1 Table of symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
3.1 Model properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
4.1 Model properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
44
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