ENSMP 2ème 3ème année Cours d'éléments finis novembre

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1 ENSMP 2ème/3ème année, Cours d'éléments finis, 22–26 novembre 2004 Disque et anneaux en rotation z r r r p p p On étudie trois solides en rotation comme indiqué ci-contre, un disque et deux anneaux. Ils sont chargés en pression imposée à leur périphérie, ce qui est caractéristique d'un chargement de disque de turbine. Dans un disque en rotation autour de son axe, la force centrifuge donne naissance à des champs de contrainte biaxiaux (?rr–???), la contrainte dominante étant la contrainte circonférencielle, qui est susceptible de provoquer l'éclatement. Dans un premier temps, on se pose le problème du disque, qui a une solution analytique, et on cherche à caractériser le profil qu'il faudrait donner à celui-ci pour que les champs de contraintes soient uniformes dans toute la structure. On trouve la solution par résolution du problème inverse au sein d'une boucle d'optimisation. On considère ensuite successivement le cas d'un anneau homognène, puis celui d'un anneau contenant un renfort en matériau composite de module plus important, ce qui permet de faire évoluer la forme de la pièce vers des dessins plus économiques en masse.

  • solution sig rranalytical

  • material ?

  • solution sig

  • geometry visualization

  • file

  • disk

  • option


Publié le : lundi 1 novembre 2004
Lecture(s) : 77
Tags :
Source : mms2.ensmp.fr
Nombre de pages : 7
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1
ENSMP 2ème/3ème année, Cours d’éléments finis, 22–26 novembre 2004
Disque et anneaux en rotation
z disque, qui a une solution analytique, et on cherche à
caractériser le profil qu’il faudrait donner à celui ci pour
p que les champs de contraintes soient uniformes dans
toute la structure. On trouve la solution par résolution
r
du problème inverse au sein d’une boucle d’optimisation.
On considère ensuite successivement le cas d’un anneaup
homognène, puis celui d’un anneau contenant un renfort
r en matériau composite de module plus important, ce qui
permet de faire évoluer la forme de la pièce vers des
p
dessins plus économiques en masse.
Le projet fait donc appel à des connaissances der
mécanique des milieux continus pour bien comprendre
les efforts en présence. Il fournit l’occasion de mener une
On étudie trois solides en rotation comme indiqué
procédure d’optimisation de forme élémentaire.
ci contre, un disque et deux anneaux. Ils sont chargés
Code utilisé : ZéBuLoNen pression imposée à leur périphérie, ce qui est
Mots clés :foces centrifuges, optimisation, matériauxcaractéristique d’un chargement de disque de turbine.
compositesDans un disque en rotation autour de son axe, la
force centrifuge donne naissance à des champs de
contrainte biaxiaux ( – ), la contrainte dominanterr
étant la contrainte circonférencielle, qui est susceptible
de provoquer l’éclatement.
Dans un premier temps, on se pose le problème du








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s
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2
2 2Theory pR R0
= 1+
2 2 2Rotating disk R − R0 r
1. The equations for the stress distribution provoked by
rotation in the case of solid circular disk : From these two solutions the equations for the stress
3+ distributions for the full problem (for solid disk and disk2 2 2= R − rr
with a circular hole respectively) can be easily found.8
3+ 1+ 3 Solution. Solid disk2 2 2 2= R − r
8 8 Curves of stress distribution along radius are presented
2. The equations for the stress distribution provoked by on the fugure below.
rotation in the case of a disk with a circular hole :
2 23+ R R2 2 2 0 2= R + R − − rr 0 28 r
Rectangular cross section2 23+ R R 1+ 3 14002 2 2 0 2 sig rr= R + R + − r sig 0 28 r 3+ sig r
1200 Analytical solution sig rr
Analytical solution sig
Disk submitted to pressure on the outer surface
1000
1. Stress distribution in the case of solid circular disk :
800For the disk without a hole at the origin and with
no body forces, only the case of stress distribution 600
symmetrical with respect to the axis may exist,
400
namely = = constr
2002. Stress distribution in the case of a disk with a
circular hole : 0

2 2pR R0
-200= 1−r 0 50 100 150 2002 2 2R − R0 r radius
qrwrwnssnqnnrwqnqqssssqqrwqqnrwnss
Sigma3
From condition = = const an equation discribing Finite element model and materials properties :r
the optimal shape of the disk can be obtained : Because of the axisymmetry (axe of rotation y) and
symmetry in respect to axe x only 1/4 of central cross2 2 2R r
h= h exp 1− section is taken into account. Material of the disk is0 22p R
supposed to be elastic isotropic with the Young modulus
y
E = 194 MPa and Poisson coeficient = 0.3. Angular
velocity of the disk = 2000 rad/c, mass per unitShape of optimazed cross section
20 −9 3
Analytical solution volume of material = 8· 10 t/mm , outer radius of
Optimization
the disk R= 225 mm, minimal thickness h = 5 mm.0
15
z x
Available files
10
1. disk_old.mast – Zmaster geometry visualization
file. Contains information about points, lines and
5
domains.
2. disk.mast.tmpl – parametrical Zmaster geometry
0
0 50 100 150 200 visualization file (based on disk_old.mast).
radius
3. disk.inp – calculation command file. ContainsMinimizing £(A), the optimisation procedure will
calculation and post processing data.look for the best set of geometrical parameters A =
(z ,z ,z ,z ,z ,z ) which correspond to the uniform and1 2 3 4 5 6 4. optim.inp – optimization command file. Contains
equal stress fields and in the disk.r parameters of optimization.Z h i
2 2£(A)= ( − p) +( − p) dvr 5. elast.mat – material constants file.
V
ssqswqrsnqrwss
Thickness4
6. SIMU2 – small shell which performs some opera (a) Command ***calcul which marks the beginning
tions to generate a data (current value of criteria in of a FEM calculation definition folowed by the
the defined integration point of each element) from the calculation type (mechanical for the static
a file of results to be compared with required value mechanical problems).
(zero.txt in our case).
(b) Keyword ***resolution is used to declare
7. DES, SOL – gnuplot files and data needed the solution procedure including the loading
for visualisation of results of calculations before sequences.
(sigma_non_opt) and after (sigma_opt, res.opt)
(c) Command ***bc specify the both the geometri
optimisation process.
cal and force boundary conditions. Note that this
Proposed work example has two types of boudary conditions :
Files examination
i. **centrifugal – is used to apply body forces
1. Edit the file disk_old.mast and study the syntax to to a defined set of elements (in our case
define points, lines, and domains. Use a command ALL_ELEMENT - which permits to apply
Zmaster disk_old.mast which launches a graphical the boundary conditions to all the elements
environment and shows geometry written in *.mast of a mesh) due to rotational acceleration.
file. Corresponding mesh file (disk_old.geof) can Verify the syntax of this command. To apply
be built using a command Zrun B disk_old.mast. a centrifugal force, it is required to give
To visualise this mesh the command Zmaster the volumetric mass for the material in
disk_old.geof has to be taped. the material file. This is specified with the
***coefficient command (see below).2. Edit the file disk.mast.tmpl. Check that some
geometrical parameters (coordinates of the points) ii. **impose_nodal_dof – this boundary condi
are changed with parametrical values (optimisation tion imposes degrees of freedom located
parameters). at nodes to defined values in time. The
3. Edit the calculation data file disk.inp. condition is general and therefore applies to
r5
all types of DOF for all types of problem. evaluation of the function to be optimized. In
Find an aplication of symmetry condition our case it evokes the executive code SIMU2
and pressure on the outer surface (nset ext) which calculates values of criteria in the points
of a disk. defined earlier and writes them to the file
disk.OUT.(d) Option *gauss_var in ***output – permits to
obtain the results ( and ) in the specified (d) Option ***values defines the starts values andr
gauss points. This information will be used for limits of variations of variables of optimisation.
the criteria of optimization. (e) Option ***compare i_file_file allows to com
(e) Keyword ***material is used to define material pare two files (fname1 and fname2) which
properties (look elast.mat file). contain the columns of doubles. Here we
compare the current value of criteria with the(f) Keyword ****return ends the file.
required value (zero.txt).
4. Edit the file elast.mat. Verify the parameters and
6. Open the gnuplot visualization files DES and SOLcoefficients of the model.
and study their syntaxis.
5. Edit the optimization command file optim.inp.
Calculations
(a) Command ****optimize marks a beginning
1. Use the single type of optimization which allowsof optimization definition. Different optimizer
to verify if all the components of optimizationtypes are available, here we’ll use type single
mechanism are written and work properly and makesand type levenberg_marquardt.
the calculations using the start values of parameters
(b) The ***files option is used to declare files where of optimization. To start this process use the
the optimization variables are to be replaced command Zrun o optim. Check that geometry and
during the trial evaluations of the error function. stress distributions correspond to the non optimized
(c) The ***shell option allows to declare external state. Compare obtained stress distributions with the
(system) commands to be executed before the analytical solutions.
ssq6
2. Change the type of optimization to leven Disk with a circular hole. Curves of stress distribution
berg_marquardt. Start again the optimization along radius are presented on the fugure below.
process. After a certain number of iterations the
process will give the optimized values of parameters
of optimization. Using them plot the curve which Rectangular cross section
3000corresponds to the optimized shape of a surface of a sig rr
sig
sig rdisk. Compare with an analytical solution. For this Analytical solution sig rr2500
Analytical solution sig
purpose use the gnuplot files.
2000
3. Try to change the minimal thickness of a disk and
see what will happen and explain the observations. 1500
1000
Optimized cross section 500
700
sig rr
sig
sig r 0600
500 -500
100 120 140 160 180 200 220
radius
400
Finite element model and materials properties : Use
300 the same parameters of material. Dimentions of a model
are : inner radius of a disk R = 100 mm, outer radius0200
R= 220 mm ;
100
Proposed work
0 Files examination Edit and study the available files.
Names and meanings of the files are keeped from the
-100
0 50 100 150 200 previous problem.
radius
qqqqqqqq
Sigma
Sigma7
Calculations Perform the same calculations and changed. Explain the new formulation of criteria and the
analysis as in the previous problem. Note that criteria is results obtained.

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