Finite Element
Analysis of a
Rotating Disk
Eliannah Hunderfund
MANE 4240 – Introduction to Finite Elements
Professor Ernesto Gutierrez-Miravete
Rensselaer Polytechnic Institute, Hartford, CT
May 17, 2015
Table of Contents
Abstract ........................................................................................................................................... 3
Introduction and Background ......................................................................................................... 3
System Overview ............................................................................................................................ 4
Figure 1 .................................................................................................................................... 4
Figure 2 .................................................................................................................................... 4
Figure 3 .................................................................................................................................... 4
Figure 4 .................................................................................................................................... 5
Theory & Methodology ................................................................................................................... 5
Analytical Solution .......................................................................................................................... 6
Figure 5 ........................................................................................................................................ 7
Figure 6 ........................................................................................................................................ 8
Figure 7 ........................................................................................................................................ 8
Figure 8 ........................................................................................................................................ 8
Comsol Multiphysics ....................................................................................................................... 9
Figure 9 ........................................................................................................................................ 9
Figure 10 ...................................................................................................................................... 9
Figure 11 ...................................................................................................................................... 9
Figure 12 .................................................................................................................................... 10
Figure 13 .................................................................................................................................... 10
Figure 13 .................................................................................................................................... 11
Figure 14 .................................................................................................................................... 11
Figure 15 .................................................................................................................................... 12
Figure 16 .................................................................................................................................... 12
Figure 17 .................................................................................................................................... 13
Figure 18 .................................................................................................................................... 13
Figure 19 .................................................................................................................................... 14
Results ........................................................................................................................................... 14
Figure 20 .................................................................................................................................... 15
Figure 21 .................................................................................................................................... 16
Figure 22 .................................................................................................................................... 16
1
Figure 23 .................................................................................................................................... 17
Conclusion ..................................................................................................................................... 17
Appendix A .................................................................................................................................... 19
Appendix B .................................................................................................................................... 21
References .................................................................................................................................... 22
2
Abstract
This project evaluates the finite element analysis of stress and displacement induced through
centripetal force in a thin rotating disk of uniform thickness and constant angular velocity with
a 1m outer diameter and a .1m inner diameter made of Ti 6-4 Titanium. The stress and
displacement were determined using Maple software to obtain an exact solution for the 1D
case, and then compared to the FEM solution obtained through Comsol Multiphysics software
for the 1D and 2D case.
Introduction and Background
This project evaluates a thin rotating disk of uniform thickness and constant angular velocity of
3000 rev/m (314.16 rad/s). The disk will be made of Ti 6-4 Titanium with outer diameter of 1m
and inner diameter of 0.1m. It is meant to represent a simplified case of a rotor as used in
aerospace jet engines, which Titanium is commonly used in. Stress and displacement in rotors is
critical to characterize because the stress must not exceed ultimate tensile stress of the
material in order to avoid the failure mode of burst, in addition the displacement is critical to
maintain tip clearance on the outer diameter of the blade.
When designing a thin rotating disk (or rotor) the speed at which the disk is rotating must be
limited to that which does not induce a centripetal force so great it exceeds the tensile strength
of the material (“burst”). In addition, the displacement at that velocity, induced by the
centripetal force, must not be greater than the outer boundary surrounding the outer
diameter, or tip clearance.
This project will evaluate the accuracy of the characterization of displacement and stress using
analytical and FEM methods in Maple and Comsol Multiphysics software, respectively, in 1D as
well as a look at the 2D case evaluated in Comsol Multiphysics.
3
System Overview
The first case this paper will evaluate is a 1D rotating disk made of Ti 6-4 Titanium, with inner
radius (a) of 0.1m and outer radius (b) of 1m, illustrated in Figure 1.
r(a)=0.1m
r(b)=1.0m
Figure 1
The material properties of Ti 6-4 Titanium are listed below in Figure 2.
Material Properties Titanium Ti 6-4
Density (𝜌)
4430 kg/m3
Poisson’s Ratio (𝑣)
0.33
Modulus of Elasticity (E)
1.14 GPa
Yield Tensile Strength (𝜎𝑡𝑒𝑛𝑠𝑖𝑙𝑒,𝑦𝑖𝑒𝑙𝑑 )
880 MPa
Figure 2
The solid mechanics principle to consider in this problem of centripetal acceleration and
centripetal force as described in Figure 3, Equation 1 and 2, below.
𝑣
𝜔
a = -𝑟 ∙ 𝜔2 ,
𝐹 = 𝑚𝑎 = 𝑚 ∙ (-𝑟 ∙ 𝜔2 )
r
Figure 3
4
a = -𝑟 ∙ 𝜔2
Eq. (1)
𝐹 = 𝑚𝑎 = 𝑚 ∙ (-𝑟 ∙ 𝜔2 )
Eq. (2)
In this project the angular speed of the disk will be evaluated at a constant angular velocity of 𝜔
= 3000 rotations/minute or 314.16 rad/s.
The second case evaluated in this project will be a thin 2D rotating disk with the same material
properties and angular speed described earlier with a thickness of 0.1m applied, as illustrated
below in Figure 4.
t=0.1m
Figure 4
Theory & Methodology
The governing equation to determine the displacement and stress for the 1D rotating disk will
be the second order differential equation described in Equation 3.
−
𝜕2 𝑢
𝜕𝑟 2
−
1 𝜕𝑢
𝑟 𝜕𝑟
+
𝑢
𝑟2
=
(1−𝑣 2 )𝜌𝜔2 𝑟
𝐸
Eq. (3)
The values for 𝑟, 𝐸, 𝜌, 𝑎𝑛𝑑 𝜔 have been outlined previously and the variable 𝑢 will represent
displacement in meters. This ODE governing equation assumes plane stress. To solve, the
boundary conditions need to be assumed that the displacement at the inner diameter of the
disk, r(a), is 0 and that the displacement at the outer diameter, r(b0, is constant. In addition,
this governing equation assumes that the disk is free of interference on the outer diameter. The
boundary conditions are listed below in Equation 4.
5
𝑑𝑢
Boundary conditions: 𝑢(𝑎) = 0 and 𝑑𝑟 (𝑏) = 0
Eq. (4)
When this boundary value ODE is solved for the 1D case, the displacement will be known and a
number of other values can be determined. Radial and tangential strain can be derived using
Equations 5 and 6, below.
Radial Strain: 𝜀𝑟 =
𝑑𝑢
Eq. (5)
𝑑𝑟
Tangential Strain: 𝜀𝑡 =
𝑢
Eq. (6)
𝑟
From radial and tangential strain, the radial and tangential (hoop) components of stress can be
determined using Equations 7 and 8, below.
𝐸
Eq. (7)
Radial Stress: 𝜎𝑟 = (1−𝑣2 )·(𝜀𝑟 + 𝑣 · 𝜀𝑡 )
𝐸
Eq. (8)
Tangential Stress: 𝜎𝑡 = (1−𝑣2 )·(𝜀𝑡 + 𝑣 · 𝜀𝑟 )
Finally, from radial and tangential (hoop) stress, the Von Mises stress can be derived which is
critical to understand because the value of Von Mises stress is compared to the Tensile Yield
Stress to predict if the material will fail. The Von Mises stress is described in Equation 9 for the
1D case (no axial stress). The relationship of Von Mises Stress and Ultimate Tensile Yield
Strength is described in Equation 10.
(𝜎𝑟 −𝜎𝑦 )
Von Mises Stress: 𝜎𝑣𝑚,𝑥,𝑦 = √
2
2
2
+(𝜎𝑦 −0) +(0−𝜎𝑦 ) )
2
𝜎𝑣𝑚 ≤ 𝜎𝑡𝑒𝑛𝑠𝑖𝑙𝑒,𝑦𝑖𝑒𝑙𝑑
Eq. (9)
Eq. (10)
Analytical Solution
To solve the boundary value ODE analytically described in Equation 3, Maple software was
used. The code input for the Maple Software is included as Appendix A. The exact solution was
obtained by solving the 2nd Order ODE using boundary conditions and appropriate material
properties and angular velocity, as described in the System overview. The radial displacement
was solved to be the following, exact solution, descried in Equation 11.
u(r)
Eq. (11)
6
The graph for u(r) is in Figure 5 with radial position from r(a) o r(b) in meters on the x axis and
displacement in meters on the y axis. The displacement increases from r=a to b. The maximum
displacement at r(b) is equal to 0.0008291136940 m or 8.29114e-4 m.
Figure 5
The graph obtained for radial and tangential strain with radial position from r=a to b on the xaxis and strain on the y-axis is in Figure 6.
7
Figure 6
The graph obtained for radial and tangential (hoop) stress with radial position from r=a to b on
the x-axis and stress in Pa is in Figure 7.
Figure 7
Finally, the graph for Von Mises stress with radial position from r=a to b on the x-axis and Von
Mises Stress in Pa is in Figure 8. The stress is greatest at r=a and decreases as r approaches b.
The maximum Von Mises Stress at r=a is 2.846194981e8 Pa (284.6 MPa).
Figure 8
8
It is interesting to note that at 𝜔= 5575 rot/min or 552.4 rad/s, the Von Mises stress at r(a) is
equal to the Tensile Yield Strength of the material. So, this reveals that the disk should be
limited to less than 5575 rot/min with some safety factor to prevent material failure. This
relationship is described in Equation 12.
At 𝜔= 5575 rot/min, 𝜎𝑣𝑚 = 880 𝑀𝑃𝑎 = 𝜎𝑡𝑒𝑛𝑠𝑖𝑙𝑒,𝑦𝑖𝑒𝑙𝑑
Eq. (12)
Comsol Multiphysics
The 1D case was evaluated using Comsol Multiphysics using the Coefficient PDE, initial values
equal to the boundary conditions described in Equation 4 and a Dirichlet Condition applied at
r=a. The Comsol files are included in Appendix B. The mesh sizes shown in Figures 9, 10 and 11
show what a coarse, normal and extremely fine mesh look like, respectively, applied to the 1D
geometry. All cases would be considered sufficient for the 1D case, with negligible difference in
the results for displacement.
Figure 9
Figure 10
Figure 11
The displacement line graph for u(r) for a normal mesh size is in Figure 12. In Figure 13 is the
maximum displacement obtained at u(r=b) for coarse, normal and extremely fine mesh sizes.
9
Figure 12
Mesh Size
Extremely Coarse
Normal
Extremely Fine
Displacement at r=b (meters)
0.000829316326517072
0.000829113996679808
0.000829113694000417
Figure 13
In Figure 14, the graph describes the displacement of the entire 1D disk in a surface plot. Note,
this is just an expansion of the axisymmetric u(r) function shown in Figure 12.
10
Figure 13
The 2D case, described in Figure 4, was also evaluated using Comsol Multiphysics using the
Solid Mechanics input, material properties of Ti64, initial values equal to the boundary
conditions described in Equation 4, a prescribed displacement of 0 at r(a) at one corner and a
prescribed acceleration described in Equation 1. The geometry of the disk is shown in Figure 14.
The blue node represents the prescribed displacement of 0.
Figure 14
An extremely fine mesh was applied to the 2D geometry, which is shown in Figure 15.
11
Figure 15
The 2D displacement was obtained using Comsol Multiphysics and shows very different results
than the 1D, because there is now a thickness which must be considered. The line graph shows
the displacement of the bottom surface at y=0 from Figure 15. The maximum displacement
now occurs at r=0.92 and is 0.000633680287828534 m. The line graph for the u(r) displacement
is shown in Figure 16.
Figure 16
12
The Von Mises Stress was also obtained from Comsol Multiphysics and was graphed in a line
graph and surface graph in Figures 17 and 18, respectively. In the 2D surface plot it is evident
that the disk is stretching to become thinner due to the centripetal force. The maximum Von
Mises Stress is still at r=a and is 237196407.729246 or 237.2 MPa.
Figure 17
Figure 18
13
Finally, the 3D surface plot of the Von Mises plot is shown in Figure 19 and simply an expansion
of the 2D Surface Plot, applied around 360 degrees.
Figure 19
Results
In the results section of this project the values obtained analytically, through Maple, and using
FEM, using Comsol Multiphysics, will be compared. The first thing to be compared will be the
displacement u(r) for the analytical 1D solution, the extremely coarse, normal and extremely
fine FEM 1D meshes and the 2D extremely fine mesh. The graph for u(r) for each of these cases
is shown in Figure 20.
14
Displacement u(r)
0.0009
0.0008
0.0007
u (meters)
0.0006
0.0005
0.0004
0.0003
0.0002
0.0001
0
0
0.2
0.4
0.6
0.8
1
1.2
r(meters)
u_exact_1D
2D_displacement_comsol
1D_displacement_comsol_normal
1D_displacement_comsol_extremely_coarse
1D_displacement_comsol_extremely_fine
Figure 20
Analyzing the graph in Figure 20, it is clear that there is a difference between the 1D and 2D
displacements. However, there is not a large difference between the analytical and FEM 1D, nor
between the mesh sizes. There is one noticeable discrepancy between r=0.2m and r=0.4m for
the 1D FEM extremely coarse mesh, as would be expected for a large mesh size. In addition,
although the 1D and 2D results differ, the 1D results are conservative showing greater
displacement. This shows that FEM using Comsol Multiphysics is a good way to obtain the
approximate solution for the case of a simple 1D rotating disk. The error between the exact 1D
solution and the 1D solution for FEM is shown in Figure 21.
15
Method
Exact_Maple
Comsol_Extremely_Coarse
Comsol_Normal
Comsol_Extremely_Fine
1D u(r) at r=b
% Error
0.000829114
0
0.000829316
0.024363357
0.000829114
4.00451E-07
0.000829114
3.69068E-05
Figure 21
This is a primary example of when a smaller mesh size does not provide more accurate results.
Although it is miniscule, the percent error for the extremely fine mesh size is actually greater
than the normal mesh size. The extremely fine mesh was not necessary to obtain a more
accurate solution for this simple case, and sometimes increasing mesh size can cause a less
favorable divide of the area of interest.
In Figure 22 is the comparison between the exact 1D solution and the 2D FEM solution for
displacement. This shows a 30% difference between the 1D max displacement and the FEM 2D
max displacement, but again this is conservative error.
Method
1D_Maple
2D_Comsol_Extremely_Fine
% error of 1D_Maple compared to 2D_Comsol
Max Displacement u(r)
0.000829114
0.00063368
30.84105912
Figure 22
Finally, the results for exact 1D Von Mises is compared to the FEM 2D Von Mises using an
extremely fine mesh in Comsol Multiphysics. The graph for the two Von Mises values over r=a
to r=b is graphed in Figure 23. The difference here can be attributed the thickness which was
not included in the 1D evaluation, however the error in this case is again conservative. So, the
1D case is appropriate to find a conservative solution with higher Von Mises Stresses than
expected. The shape of the graph and the point of max stress, at r=a, is the same between the
1D and 2D cases.
16
Von Mises Stress
3.00E+08
Von Mises Stress (Pa)
2.50E+08
2.00E+08
1.50E+08
1.00E+08
5.00E+07
0.1
0.127
0.154
0.181
0.208
0.235
0.262
0.289
0.316
0.343
0.37
0.397
0.424
0.451
0.478
0.505
0.532
0.559
0.586
0.613
0.64
0.667
0.694
0.721
0.748
0.775
0.802
0.829
0.856
0.883
0.91
0.937
0.964
0.991
0.00E+00
r (meters)
2D_Von_Mises_Comsol
1D_Von_Mises_Maple
Figure 23
In Figure 24 is the comparison between the exact 1D solution and the 2D FEM solution for Von
Mises Stress. This shows a 20% difference between the 1D max stress and the FEM 2D max
stress, but this error is conservative and would not put the calculation at risk for providing a
false positive margin to yield strength of safety factor.
Method
2D_Comsol_Extremely_Fine
1D_Maple_Exact
% error of 1D_Maple compared to 2D_Comsol
Von Mises Stress at r = a (Pa)
2.37E+08
2.85E+08
19.99
Conclusion
For the problem of evaluating a simplified rotating disk to find displacement and stress, the results
obtained through this project show that FEM using Comsol Multiphysics can obtain an extremely
accurate solution in comparison. In addition, for simple geometry, an extremely fine mesh size is not
17
necessary. It is possible to model the 1D rotating disk in less time than it takes to obtain the exact
solution analytically, and when this is expanded to a problem with complex geometry and transient
conditions the required work input to obtain an accurate solution will be much less in FEM software
than using analytical methods. In the case of a rotating disk, it was also shown that the 1D case, which
will generally be quicker and simpler to evaluate, can show conservative and somewhat accurate results.
For the connection to an aerospace rotor, the simplified case would be useful as a first pass to
understand where stresses and displacements are in a preliminary design before investing more time
into a 2D or 3D analysis. In the case of complex 3D geometry in rotors, the only method for predicting
displacement and stress without actually testing is FEM, so its accuracy is imperative.
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Appendix A
Maple Code Input
>
>
>
>
>
>
>
>
>
>
>
19
>
>
>
>
20
Appendix B
1D_DISK_COMSOL.mph
2D_DISK_COMSOL.mph
21
References
http://solidmechanics.org/text/Chapter4_1/Chapter4_1.htm
http://www.ewp.rpi.edu/hartford/~sarric/SMS/Readings/32669_04.pdf
http://asm.matweb.com/search/SpecificMaterial.asp?bassnum=MTP641
http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/Part_II/04_ElasticityPola
r/ElasticityPolars_04_BodyForcesRotatingDiscs.pdf
http://www.doiserbia.nb.rs/img/doi/1450-5584/2006/1450-55840601065A.pdf
http://courses.washington.edu/me354a/Thick%20Walled%20Cylinders.pdf
http://www.hpmsl.neu.edu/content/2012Papers/Haghpanah-ASME-2012.pdf
http://www.solid.iei.liu.se/Education/TMHL61/formula_tables/formulas_table_solid_mechanics.pdf
https://indigo.uic.edu/bitstream/handle/10027/10185/Chianese_Stefano.pdf?sequence=1
22