Code_Aster ®
Version
5.0
Titrate:
Elements of Fourier for the axisymmetric structures
Date:
21/12/00
Author (S):
X. DESROCHES
Key:
R3.06.04-A
Page:
1/12
Organization (S): EDF/MTI/MN
Handbook of Référence
R3.06 booklet: Machine elements and thermal for the continuous mediums
Document: R3.06.04
Elements of Fourier for the structures
axisymmetric
Summary
The elements of Fourier are intended to calculate the response of structure for axisymmetric geometry solicited
by nonaxisymmetric loadings broken up into Fourier series.
One exposes in this document a general theory of Analyze of Fourier with coupling of the symmetrical modes
and antisymmetric in the anisotropic case. The case of isotropic, or orthotropic materials of axis OZ, where
modes are uncoupled, is studied separately.
The elements of Fourier are usable in Code_Aster starting from modeling AXIS_FOURIER.
meshs supports of these elements are triangles and quadrangles of degree 1 and 2.
Handbook of Référence
R3.06 booklet: Machine elements and thermal for the continuous mediums
HI-75/00/006/A

Code_Aster ®
Version
5.0
Titrate:
Elements of Fourier for the axisymmetric structures
Date:
21/12/00
Author (S):
X. DESROCHES
Key:
R3.06.04-A
Page:
2/12
1 Introduction
The analysis of Fourier is intended to calculate the response of structures for axisymmetric geometries
subjected to nonaxisymmetric loadings. In this case, it is necessary to develop them
loadings in Fourier series. Generally convergence is reached for 4 or 5 harmonics,
but the speed of this convergence depends on the nature of the loading: more the loading is
regular and more the corresponding series converges quickly. The most unfavourable case is that of one
force concentrated for which the practice shows that it is necessary to go to beyond (at least 7 harmonics).
In Code_Aster, the decomposition of the loading in Fourier series is supposed to be made
au préalable by the user. Code_Aster makes it possible to calculate the answers to this loading,
harmonic by harmonic (modeling AXIS_FOURIER), and overall after recombination of
harmonics between them (operators COMB_CHAM_NO and COMB_CHAM_ELEM).
One will expose in a first chapter the general framework of the anisotropy, while insisting on decoupling
modes in the orthotropic case. The second chapter clarifies the calculation of the matrix of rigidity
in the isotropic case.
For the use of the elements of Fourier in Code_Aster, one returns to the note of use of
modeling Fourier [U1.01.07].
2
Analyze of anisotropic Fourier
2.1 Theory
general
All the fields considered (forces, displacements, strains, stresses) are expressed in
cylindrical co-ordinates with following convention on the command of the components:
1 radial component according to R
2 axial component according to Z
3 tangential component according to
Example: (ur, uz, U), (Fr, fz, F)
Z
R

uz
M
U
U

R
The grid is localized in plan (R, Z), the symmetry of revolution being done around axis OZ.
trihedron (R, Z,) is directed in the direct direction.
Handbook of Référence
R3.06 booklet: Machine elements and thermal for the continuous mediums
HI-75/00/006/A

Code_Aster ®
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Titrate:
Elements of Fourier for the axisymmetric structures
Date:
21/12/00
Author (S):
X. DESROCHES
Key:
R3.06.04-A
Page:
3/12
Z
rez
R

er
C
One breaks up displacement U (or the loading F) according to U = custom + ua
custom
where
(resp. ua)
indicate the symmetrical part (resp. antisymmetric) of the development in Fourier series of U by
report/ratio with the variable.
One obtains:


custom = custom
R
L (R, Z) cosl

l=0




custom =
v S
S
Z
L (R, Z) cosl left symmetrical U
l=0



custom =
ws,
sin
L (R Z) (- L)
l=0



ua = ua
R
L (R, Z) sin L
l=0




ua = goes
has
Z
L (R, Z) sin L left antisymétriqu U
E
l=0



ua = wa, cos
L (R Z)
L
l=0

To note the choice of the sign ­ for custom, which makes it possible to simplify later calculations. If one notes
Custom = (S S S
has
L
U, lv, L
W) (resp.U
L
L) the L - ième component symmetrical (resp. antisymmetric) of
development in Fourier series of U, one obtains:

cosl
0

sinl
0






S
has
U =

cosl
Ul +
sin L
Ul
éq 2.1-1
l=0

0
- sinl

0
cosl



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R3.06 booklet: Machine elements and thermal for the continuous mediums
HI-75/00/006/A

Code_Aster ®
Version
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Titrate:
Elements of Fourier for the axisymmetric structures
Date:
21/12/00
Author (S):
X. DESROCHES
Key:
R3.06.04-A
Page:
4/12
If one indicates by the vector deformation linearized, one realizes that can be broken up into
following Fourier series:
cosl I
0

sinl
4
4 2
I
0

=
,


4
4,
S
2
has
L +

éq 2.1-2
0
- sinl
2 4
I
0
cosl
2
2 4
I
L
l=
,
,
2
0

with = {R, Z, rz, R, Z}
S
S
S
has
has
has
L =
L
B Ul
L = L
B Ul
with (see [bib1]):



0
0
R




0
0
Z



1
L

0
-



Bs
R
R
L =



0
Z R


L

1
0
-
R
R R

L

0


R
Z

There is Ba
Bs
=
L
L
L
(this is due to the choice of the symmetrical development of U in (cos, cos, ­ sin) to the place
of (cos, cos, sin)). One will omit starting from now the indices has and S and one will note Bl the operator
allowing to calculate the deformations corresponding to the harmonic L.
2.2
Coupling and decoupling of the symmetrical and antisymmetric modes
By taking again the preceding notations, one a:
cosl I
0

sin
2
2,
L
1
I2
02,1
U =
custom +
ua
0
- sinl
L
1 2

0
cos
L
,
,
L
1 2

L
L

Handbook of Référence
R3.06 booklet: Machine elements and thermal for the continuous mediums
HI-75/00/006/A

Code_Aster ®
Version
5.0
Titrate:
Elements of Fourier for the axisymmetric structures
Date:
21/12/00
Author (S):
X. DESROCHES
Key:
R3.06.04-A
Page:
5/12
what is written, by introducing matrices M S and M has
L
L:
U = (M sUs
has
has
L
L + MR. U
L
L)
L
U
S
S
has
has
L = MR. U
L
L + MR. U
L
L
One deduces from it that:
S S
has has
L = Me L L + Me L L
cosl I4
04,2

with
M L =

02 4
- sinl I
,
2
sinl I4
04 2
M has
'
,
L =

0
cosl
2 4
I
,
2
Calculation of the deformation energy
2
W
T
=
D dsd
with ds = rdrdz
L
L L
0 S
2
2
= D
T S T
M DM S S
'ds + D
T has T
M a' DM has has
'ds
L
L
L L
L
L
L
L
0
S
0
S
2
2
+ D
T has T
M a' DM S S
'ds + D
T S T
M DM is had has
'ds
L
L
L L
L
L
L
L
0
S
0
S
sinl I4
0
D
D
1
3 cosl I4
0

Since M has
'DM S
L
'L =
T



0
cosl I 2D
D
3
2
0
- sinl I2

2
has
S
D sin L cosl
1

- D3 (sinl)
Me DM
L
'L =


T D
2
3 (cos L)
- D sinl cosl
2

2
and that
sin L cosl D =
, if D

=

0
0
T has
S
T S
has
3
there is thus no term (L, L) or (L, L) in W.
0
There is then no coupling (U has U.S.) or (U.S.U has
,
,
).
Handbook of Référence
R3.06 booklet: Machine elements and thermal for the continuous mediums
HI-75/00/006/A

Code_Aster ®
Version
5.0
Titrate:
Elements of Fourier for the axisymmetric structures
Date:
21/12/00
Author (S):
X. DESROCHES
Key:
R3.06.04-A
Page:
6/12
2.3
Calculation of the constraints
Just as, can be broken up into following Fourier series:
= (M S
has
has
L + M
L
'L L)
L
Law of Hooke = D, one deduces:

L D1 -
L D3
sinl D
cosl
1
D3
= cos
sin

S
has
T
L +
T

cosl D3 - sinl D2
sinl D
cosl
3
D2 L
L
Maybe, while revealing the matrices M S and M has
'L
'L:

D1 04,2
0
D
4,4
3

= M S
S
has
'L


L +
T
L
0
0
L


D
2,4
2
- D3
2,2




04,4 - D3
D1 04,2

+ M a'
S
has
L
T

L +
L

D3
02,2
0
D
2,4
2



D
0
0
D
1
4,2
4,4
3
By posing Ds
and Da =
, one deduces the parts from them symmetrical and
0
D
- Dt
2,4
2
3
02,2
antisymmetric of the constraint relating to the harmonic L:

S
S S
has has
S
S
has
has
L = D L + D L = D
L
B ul + D L
B ul

éq 2.3-1

has
S has
S.A.
has
S
S
has
L = - D L + D L = - D
L
B ul + D L
B ul
Note:
In the case of the orthotropism compared to OZ, there is Da = 0 and [éq 2.1-1] is reduced to:

S
S
S
L = D
L
B ul


has
S
has
L = D
L
B ul
I.e. if displacements are symmetrical (or antisymmetric), the constraints are it
too.
Handbook of Référence
R3.06 booklet: Machine elements and thermal for the continuous mediums
HI-75/00/006/A

Code_Aster ®
Version
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Titrate:
Elements of Fourier for the axisymmetric structures
Date:
21/12/00
Author (S):
X. DESROCHES
Key:
R3.06.04-A
Page:
7/12
3
Calculation of the matrix of rigidity
3.1 Case
General
Are U and two fields kinematically acceptable unspecified. By applying the principle of
virtual work with the element of volume v, one can write:
(T.) FD = (T U
. F) FD
v
v
After decomposition in Fourier series and integration compared to, one obtains, for fields
S.A.S.A.
L, L, L
U, L
u.a. .A. unspecified and for any harmonic L:
(T S S T has has
T
S
S T
has
has
L
.l + L .l) L
ds = (read. fl + read. fl) L
ds
S
S
L
L
Maybe, while using [éq 2.3-1] and while posing:
K S
T
=
B Ds B ds
L
L
L
L
SSL
K has
T
=
B Ds B ds = K S = K
L
L
L
L
L
L
SSL
K have
T
=
B Da B ds
L
L
L
L
SSL
One obtains the system of equations according to:

K custom + Kas ua = F S
L L
L
L
L

éq 3-1
T Kasus


+ K ua = F has
L
L
L L
L
where T
have
have
Kl = - Kl one sees that if Da 0, the decoupling of the modes in symmetrical harmonics and
antisymmetric is not possible any more. On the other hand, if Da = 0 (orthotropism compared to OZ) then
Kas
L
= 0 and [éq 3-1] are reduced to:

K custom = F S
L
L
L
K ua = F has

L L
L
While taking for vectors displacement (resp. force) corresponding to the harmonic L the vectors:
U
S
S
S
has
has
has
L = {R
U, Z
U, U,
R
U, Z
U, U} L
F
S
S
S
has
has
has
L = {Fr, fz, F, F

R, F Z, F} L
Handbook of Référence
R3.06 booklet: Machine elements and thermal for the continuous mediums
HI-75/00/006/A

Code_Aster ®
Version
5.0
Titrate:
Elements of Fourier for the axisymmetric structures
Date:
21/12/00
Author (S):
X. DESROCHES
Key:
R3.06.04-A
Page:
8/12
One has then:
have


G
G
K
K
K U = F
with K
L
L


L
L
L
L = T
Kas
K
L
L
3.2 Calculation
of
K gl in the isotropic case
In this case there are thus K have
T
S
L
= 0. One details in the continuation the calculation of K =
B D B ds
L

S
L
L
L
L
In the isotropic case, one a:
D1 D 2d2
0

D 2D1 D


2
0


0
D 2D 2d1
0

D Ds
=
=

0
0
0
D

3

D

3
0

0
0
D



3
E (1 -)
with
D1 = (1+) (1 - 2)
E
D2 = (1+) (1 - 2)
E
D3 = (21+)
One can write:

R



Z
S


U


R
T


S
'
ur uz
U
ur uz
U
ur uz
U


= L
B U
Z = L
B
,
,
,
,
,
,
,
,

rz
R R R R R R Z Z Z




S

U




R





Z
fcts of form
derived from the fcts of form
0
0
0
1 0
0
0 0
0
0 0 0 0 0 0 0 1 0


1
0
- L 0 0 0 0 0 0
with B' L =

0
0
0
0 1
0
1 0
0


L
0
- 1 0 0 1 0 0 0


0
L
0
0 0
0
0 0
1


Handbook of Référence
R3.06 booklet: Machine elements and thermal for the continuous mediums
HI-75/00/006/A

Code_Aster ®
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Titrate:
Elements of Fourier for the axisymmetric structures
Date:
21/12/00
Author (S):
X. DESROCHES
Key:
R3.06.04-A
Page:
9/12
While indicating by {WJ}
functions of form of the element considered, one a:
J =1 with N
ur
node J



R
W
“# $
J


!

0
0
!
U
R
Z

W
R

J



!
0
0

!
U
R

W

J
R
!

0
0


R
!




U



R
W


J




0
0
R


!

R

!

U
R (J)


U

W

J


U
Z
=
=!

0
0
!uz (J)


R
R


W




U
J

U (J)


0
0





!

R
!



R



W
J


U

0
0

R
!

Z
!

Z

W



J



U

!
0
0
!
Z


Z

Z

WJ


!
0
0


U
% & '
!
Z

block P





J
Z

One notes (P) = (P
P
1,
,
! NR) where NR is the number of nodes of the element.
Then K
T
=
Pt B D B P ds
L

'
'
L
L
L
SSL
I, J
Kl symmetrical and is made of blocks (Kl) 3× 3:
(I, J
K)
T P you
'
= I B D B P ds
L
L
L
J
L
SSL
Handbook of Référence
R3.06 booklet: Machine elements and thermal for the continuous mediums
HI-75/00/006/A

Code_Aster ®
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Titrate:
Elements of Fourier for the axisymmetric structures
Date:
21/12/00
Author (S):
X. DESROCHES
Key:
R3.06.04-A
Page:
10/12
I, J
The calculation of the blocks (Kl) is clarified below:
D1+12 D3
0
- L (D1+ D)
3
D2
0
lD3
0
D2
0


0
L 2d3
0
0
0
0
0
0
lD

3
- L (D1+ D) 3
0
L 2d1+ D3
- lD2
0
- D3 0
- lD2
0



D2
0
- lD2
D1
0
0
0
D2
0
tB' D B'


L
L =
0
0
0
0
D3
0
D3
0
0



lD3
0
- D3
0
0
D3
0
0
0

0
0
0
0
D3
0
D3
0
0



D2
0
- lD2
D2
0
0
0
D1
0


0
lD3
0
0
0
0
0
0
D

3
I, J
I, J
I, J
K
K
K

I, J
11
12
13
'
'
I, J
I, J
I, J
T
T
L
I
P
L
B D L
B
J
P = (Kij)
=
with
3 ×
K
K
K
21
22
23
3 I, J
I, J
I, J
K
K
K
31
32
33


2
I, J
D1 + L D3
W

W
W W

D2
W

W
J
I
K11 =

I
J
I
J +
WI
+ W
2
I
W
J
W + D1
+ D3


R

R

R
Z Z
R
R
J

R

2
I, J
L D3
W W

W W
K22 =
W W
I
J
I
J

2
I
J + D3
+ D1
R
R

R

Z
Z



2
I, J
L D1 + D3
W W

W

W
I
J
I J
D3
W
W
J
I

K33 =
W W
2
I
J + D3
+
W
W


R


R R
Z

Z -
I
+
R
R
R
J




W

W

W

W
I, J
D2
W

K
= D
I
J
2
+ D
I
J
3
+
W
J
12

R Z
Z R
R
I
Z


W

W
W

W
I, J
D2
W
K
= D
I
J
3
+ D
I
J
2
+
W
I
21
R
Z

Z R
R
J Z

I, J
L
L
W
L
W

K
13 = -
(D1+ D)
3 W W - D2W
I + D W
J
3
R 2
I
J
R
J
R

R
I
R


I, J
L
L
W
L
J
W
K
I
31 = -
(D1+ D)
3 W W - D2W
+ D W

3
R 2
I
J
R
I
R
R
J R

I, J
L
W
L
I
W
K23 = - D2
W + D W
J

3
R
Z
J
R
I
Z
I, J
L
W
L
J
W
K32 = - D2W
+ D
I
3
W

R
I
Z
R
Z
J



The blocks K I, J are not symmetrical except for I = J (on the diagonal of K). One notices in
fact that the blocks K I, J can be written for any harmonic (L = 0 included/understood).
Handbook of Référence
R3.06 booklet: Machine elements and thermal for the continuous mediums
HI-75/00/006/A

Code_Aster ®
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Titrate:
Elements of Fourier for the axisymmetric structures
Date:
21/12/00
Author (S):
X. DESROCHES
Key:
R3.06.04-A
Page:
11/12

D3
K I, J = K I, J
0
+ l2
W W
11
11
r2
I
J

D3
KI, J = K I, J
0
+ l2
W W
22
22

r2
I
J

D1
KI, J = K I, J
0
+ l2
W W
33
33

R2 I J
KI, J

= K I, J
12
12
0

K I, J = K I, J
21
021
KI, J

= - L K I, J
13
13
0
KI, J = - L K I, J
31
031
KI, J = - L K I, J
23
023

KI, J = - L K I, J
32
032




where the blocks K I J
0, are independent of the harmonic L.
4 Loadings
It is supposed that the loading was broken up according to the same base which displacements, that is to say:

cosl
0

sinl
0






S
has
F =

cosl
Fl +
sin L
Fl
l=0

0
- sinl

0
cosl



There is not coupling for the same harmonic between the parts symmetrical and antisymmetric of
loading because of orthogonality of the goniometrical functions sin L and cosl, this for all
types of loading. This wants to say in particular that the equivalent nodal forces are them
same for the harmonics symmetrical and antisymmetric if the amplitudes F S and F has
L
L are them
same.
For the nature of the acceptable loadings with Fourier modeling, one returns to the note
of use [U1.01.07].
Handbook of Référence
R3.06 booklet: Machine elements and thermal for the continuous mediums
HI-75/00/006/A

Code_Aster ®
Version
5.0
Titrate:
Elements of Fourier for the axisymmetric structures
Date:
21/12/00
Author (S):
X. DESROCHES
Key:
R3.06.04-A
Page:
12/12
5
Conclusion and Perspectives
Currently, it is supposed that the decomposition of the loading was made as a preliminary by the user,
i.e. {F S, F
L
L}
is known. This decomposition could be carried out by an operator
l0
of Code_Aster which would project the loading on the modes of Fourier.
For the moment, only the nonanisotropic case is established, i.e. it y forever coupling of
modes. The extension to the anisotropy can constitute a later development.
6 Bibliography
[1]
DUVAUT G.: “Mechanical of the continuous mediums” p282
[2]
ASKA HS. : “Axisymmetric Structures in Fourier series”, May 1982, ISD
Handbook of Référence
R3.06 booklet: Machine elements and thermal for the continuous mediums
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