Code_Aster ®
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Titrate:
Conditions of solid connection of body
Date:
12/02/01
Author (S):
J. PELLET
Key:
R3.03.02-A
Page:
1/12
Organization (S): EDF/MTI/MN
Handbook of Référence
R3.03 booklet: Boundary conditions and loadings
Document: R3.03.02
Conditions of solid connection of body
Summary
One presents in this documentation a manner of modelling indeformable parts of structure, in
small displacements and rotations, thanks to key word LIAISON_SOLIDE of AFFE_CHAR_MECA.
Handbook of Référence
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Code_Aster ®
Version
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Titrate:
Conditions of solid connection of body
Date:
12/02/01
Author (S):
J. PELLET
Key:
R3.03.02-A
Page:
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Contents
1 Introduction ............................................................................................................................................ 3
2 Principle of the use of the key word .......................................................................................................... 3
3 Which are the treated cases of figure?..................................................................................................... 4
4 Processing of the cases 2DA and 3DA ............................................................................................................ 5
4.1 Case 2DA .......................................................................................................................................... 5
4.2 Case 3DA .......................................................................................................................................... 5
5 Processing of the case 2DB .......................................................................................................................... 6
5.1 General case ..................................................................................................................................... 6
5.2 Particular cases ................................................................................................................................ 6
6 Processing of the case 3DB .......................................................................................................................... 7
6.1 General case ..................................................................................................................................... 7
6.1.1 Processing of the points A, B, C and determination of the vector rotation .................................... 7
6.1.2 Relations concerning a point M (A, B, C) .......................................................................... 8
6.1.3 Summary of the equation to write ............................................................................................... 9
6.2 Particular cases ................................................................................................................................ 9
7 How to detect the particular cases?.............................................................................................. 11
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Code_Aster ®
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Titrate:
Conditions of solid connection of body
Date:
12/02/01
Author (S):
J. PELLET
Key:
R3.03.02-A
Page:
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1 Introduction
Key word LIAISON_SOLIDE of commands AFFE_CHAR_MECA and (AFFE_CHAR_MECA_F) allows
to model an indeformable part of a structure.
The principle selected is to write linear relations between the ddls “solid” part; these relations
expressing the fact that the distances between the nodes are invariable.
Important remark:
The relations expressing the indeformability of a solid are in general not linear.
linearization of the problem supposes that the problem can be solved in theory of “small
displacements ". To be convinced some, let us take the example of a segment AB in 2D:
B
With
The indeformability of AB is written:
([
2
2
2
2
X + dx
With
With) - (X + dx
B
B)] + (
[y +dy
With
With) - (y + Dy
B
B)] = (X - X
With
B) + (y - y
With
B)
(
2
2
dx - dx
B
With) + (
2 X - X
B
With) (dx - dx
B
With) + (Dy - Dy
B
With) + (
2 y - y
B
With) (Dy - Dy
B
To) = 0
X, y, X, y
With
With
B
B co-ordinates of A and B
by noting dx, Dy, dx, Dy
With
With
B
B displacements of A and B
it is seen that the expression is quadratic in dx, dx, Dy and Dy
With
B
With
B. To be able to linearize it,
it is necessary to eliminate the quadratic terms and for this reason, one is obliged to suppose that them
elements dx, dx, Dy and Dy
With
B
With
B are small compared to the length of AB.
This remark wants to say that one cannot use this key word when the structure becomes deformed (or
turn) too much. In such situations, “to rigidify” a solid part, one is obliged to use one
“hard” material (compared to the remainder of the structure).
2
Principle of the use of the key word
Key word LIAISON_SOLIDE is a key word factor répétable at will. With each occurrence of the word
key, the user defines a “piece of model” which it wishes to rigidify.
This “piece of model” defined by key words GROUP_MA, GROUP_NO, MAILLE and NOEUD, one
deduced the list from the nodes to be rigidified.
Once this drawn up list, one writes the linear relations necessary to express that the “piece
rigid " has nothing any more but the degrees of freedom of a solid (3 in 2D or 6 in 3D).
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Titrate:
Conditions of solid connection of body
Date:
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Author (S):
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Key:
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3
Which are the treated cases of figure?
According to ddls carried by the nodes of the list of the nodes to rigidify, one places oneself in one of the four
following cases of figure. If one does not find oneself in one of these cases of figure, the code stops in
fatal error
The cases 2DA and 2DB correspond to plane” or axisymmetric problems “.
The cases 3DA and 3DB correspond to 3D problems.
Case 2DA:
All the nodes of the list of the nodes to be rigidified carry the ddls DX, DY (and possibly
DRZ) but they do not carry DRX, DRY and DZ and there is at least a node of the list of
nodes to be rigidified which carries DRZ.
Case 2DB:
All the nodes of the list of the nodes to be rigidified carry DX, DY but they do not carry DRX,
DRY and DZ.
Case 3DA:
All the nodes of the list of the nodes to be rigidified carry DX, DY, DZ (and possibly DRX,
DRY, DRZ) and there exist a node of the list of the nodes to be rigidified which carries DRX, DRY, DRZ.
Case 3DB:
All the nodes of the list of the nodes to be rigidified carry DX, DY, DZ and there is not node
list of the nodes to be rigidified carrying at the same time DRX, DRY, DRZ.
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Titrate:
Conditions of solid connection of body
Date:
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Author (S):
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Key:
R3.03.02-A
Page:
5/12
4
Processing of the cases 2DA and 3DA
In these 2 cases of figure, one could find a node of the list of the nodes to be rigidified which carried all
degrees of freedom of the solid. That is to say this node.
in 2D: DX, DY, DRZ
in 3D: DX, DY, DZ, DRX, DRY, DRZ
That is to say a node M of the list of the nodes to be rigidified unspecified.
In theory of small displacements, the movement of a solid body is expressed by:
U is of displacement of A
U
= U + AM
With
M
With

where the vector rotation of the solid
4.1 Case
2DA
The linear relations are written:

X
MR. a: DX (M) - DX ()
With + y DRZ ()
To = 0
with AM =

y
DY (M)

- DY ()
With - X DRZ ()
To = 0
+ if M
por DRZ
you
: DRZ (M) - DRZ ()
To = 0
4.2 Case
3DA
DX (M) - DX ()
WITH - DRY ()
A.Z + DRZ ()
A. y = 0
MR. a: DY (M) - DY () A DRZ () A.X + DRX (A).z = 0
DZ (M)

- DZ ()
WITH - DRX ()
A. y + DRY ()
A.X = 0
DRX (M) - DRX ()
To = 0

+ if M
por DRX
you, DRY, DRZ: DRY (M) - DRY ()
To = 0
DRZ (M) - DRZ ()
With =

0
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Code_Aster ®
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Titrate:
Conditions of solid connection of body
Date:
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Author (S):
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Key:
R3.03.02-A
Page:
6/12
5
Processing of the case 2DB
5.1 Case
General
With and B in the list of the nodes to be rigidified/D (A, B) 0
· determination
of
:
nx
That is to say N = AB K =
(K unit vector according to O
N
Z).
y
(U

- U
B
With). AB = 0
U - U - K AB
B
With
(U - U

B
With) .n - (K AB) .n = 0
· since
AB 0, one can determine:
1
= (
DX B .N
DX A.N
DY B .N
DY B .N
K AB) (
() X -
() X + () y - () y)
.n
'
N
1
That is to say =
X
(
;
N

K AB)
=
=
.n

'
ny
· equations to be written:
-
(U - U) .AB
B
With
= 0
(1 equation for 2 points A
and B)
X
-
M (A, B):
AM =
y
DX (M) - DX ()
With + y
'
'
'
'

(DX (B).n - DX
X
(a).n + DY
X
(B).n - DY
y
()
With ny) = 0
DY (M) - DY () A X
'
'
'
'


(DX (B).n - DX
X
()
A.N + DY
X
(B).n - DY
y
()
With ny) = 0
5.2 Case
private individuals
· list nodes to be rigidified = {}
With there is nothing to write A tear,
· list nodes to be rigidified = {Have} where all Have them have the same co-ordinates.
That is to say Ao the first node of the list of the nodes to be rigidified
DX

(Have) - DX (A0) = 0
WITH A
I
O it is necessary to write DY


(Have) - DY (A0) = 0
Note:
is unspecified
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Titrate:
Conditions of solid connection of body
Date:
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Key:
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Page:
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6
Processing of the case 3DB
6.1 Case
General
With, B, C in the list of the nodes to be rigidified such as ABC is a triangle of nonnull surface
6.1.1 Processing of the points A, B, C and determination of the vector rotation
B
B = AB; C = AC; m = AM
B
That is to say N = B C

b'= bn; c'= C N

With
C
C
M, UM - UA = M
éq 6.1-1
U

- U = B

·
B
With

for the points B and C: U - U = C
C
With

U - U .N =.
B
B
With
'
éq 6.1-2
U - U .C =.
N
B
With
éq 6.1-3
U - U .B
B
With
= 0
éq 6.1-4
U - U .N =.
C
C
With
'
éq 6.1-5
U - U .B = -.
N
C
With
éq 6.1-6
U - U .C
C
With
= 0
éq 6.1-7
U - U .C + U - U .B
B
With
C
With
= 0
éq 6.1-8
Of the 6 equations concerning the points B and C,
-
3 are to be written: [éq 6.1-4], [éq 6.1-7] and [éq 6.1-8] (they do not utilize)
-
3 are used to determine:
.B'= (UB - U A) .N




.c'= (CPU - U A) .n éq
6.1-9


.n'=

(UB - U A) .C
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Conditions of solid connection of body
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6.1.2 Relations concerning a point M (A, B, C)

12
That is to say U
the vector
ABCM
: (U A V
, A, WA, U B V
, B, ......., C
W, U MR. V
, M, WM)

R
the vector
3



(
X, y, Z) R
The equation [éq 6.1-9] can be written: Mr. = MR. U
1
2
ABCM
b' b' b'
X
y
Z
with M = it
it
it
X
y
Z M is invertible bus ABC is of nonnull surface
1
1


N
N
N
X
y
Z
- nx - ny - N N N N 0 0 0 0 0 0

Z
X
y
Z

and
M2 = - nx - ny - N
0
0
0
N
N
N
0 0 0
Z
X
y
Z

- cx - cy - C C C C 0 0 0 0 0 0

Z
X
y
Z

= -
MR. M
1
U
1
2.
ABCM
éq 6.1-10
The equation [éq 6.1-1] U
- U - M
M
With
= 0 can be written:
Mr. U
M
4
ABCM +
3. = 0
éq 6.1-11
0
- M
M

Z
y
with
M = M
0

Z
- M
where
M
3
X
= (M, M, M
X
y
Z)
- M
M
0

y
X

- 1
1



and
M4 =
- 1
0
0
1




- 1
0
1
Mr. U
+ Mr. Mr. M
1
.U
= 0 Mr. U
4
ABCM
3
1
2
ABCM
5
= 0
ABCM
M (A, B, C), the 3 equations should be written corresponding to the 3 lines of the matrix
M = M + M - Mr. M
1
5
4
3
1
2
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6.1.3 Summary of the equation to be written
· calculation
of
B, C, N, b', it
· calculation
of
M-1 M
1
2
U - U B
B
A. = 0

· for the points B and C: U - U .C
C
With
= 0
U - U .c + U - U B
B
With
C
A. = 0
·
M (A, B, C):
- calculation
of
M = M + MR. M 1
- M
5
4
3
1
2
-
writing of the 3 equations corresponding to M
6.2 Case
private individuals
· list nodes to be rigidified = {}
With there is nothing to write A tear,
· list nodes to be rigidified = {Have} where all Have them have the same co-ordinates.
That is to say Ao the first point of the list of the nodes to be rigidified
DX (Have) - DX (A0) = 0


WITH A DY (Have) - DY (A0)
I
O
= 0

DZ

(Have) - DZ (A0) = 0
is unspecified, which does not pose a problem.
· list nodes to be rigidified = {Have} where all Have them are aligned (right).
The solid {Have} does not have more whereas 5 movements of possible rigid bodies.
It misses rotation around.
That is to say:
- two
points
With and B/
AB 0
-
B = AB
-
n1 a vector not no orthogonal with B (thus with)
-
N = bn
2
1
B
B
n1
With
N2
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Conditions of solid connection of body
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· not b: U - U = B
B
With
(U - U) .B
B
With
= 0
éq 6.2-1
(U - U) .N1 = (N1) .B
B
With
éq 6.2-2
(U - U) .N2 = (N2) .B
B
With
éq 6.2-3
-
the equation [éq 6.2-1] is to be written
-
the equations [éq 6.2-2] and [éq 6.2-3] are used to calculate
The component of on B is unspecified, one does not hold account of it:
= 1n1 + 2n2
(n1) = 2n1 N
2
that is to say K = n1n2.b
K 0

(


K 1 = - U B - U .n
With
2
N
2)
(
)
= 1n2 N
1





K 2 = - (U B - U A) .n1
K =
K 1n1 +
K 2n2
has



That is to say N
(
B,
B
W, U MR. V
, M, WM)
1 = B
; N

2 =
;

U

ABM = U
V
,
, W, U V
With
With
With
B,
R9

C

K = M1 U ABM with: M1 = [M,
2 - M,
2
] 0

0
has
- B
has
- C
M2 =

B - has
0
B
- C


C
- has

C - B

0


M (A, B)
U M - U A - m
= 0
m = AM = (M, M, M
X
y
Z)
MR. U
4
ABM + m3 = 0
0
- M
M
Z
y
- 1 0
0
1





·
with m3 = Mz
0
- MX
M4 = 0 -

1
0
0
1

-

M
M
y
X
0

0
0
- 1


1
1
MR. U
5
ABM = 0
with
M5 = M4 + Mr. M
K
3
1
for each point M, it is necessary to write the 3 equations corresponding to the 3 lines of the matrix
M5.
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Conditions of solid connection of body
Date:
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· summary of the equations to be written:
- calculation
of
B, N, N,
1
2 K
- calculation
of
M1
-
for point b: U - U B
B
A. = 0
-
M (A, B)
1
- calculation
of
M = M + Mr. M
5
4
3
1
K
-
writing of the 3 equations corresponding to M5
7
How to detect the particular cases?
In the paragraphs [§6] and [§7], we saw that it could arrive of the particular cases when
certain nodes geometrically were confused or aligned on the same line.
The numerical criteria selected to detect these particular cases are:
· 2 points A and B are confused if:
AB
-
10 6. DMIN
· 3 points A, B, C are aligned if:
(AB AC

) 1/2
- 6
10 .DMIN
where: DMIN notes the length of smallest stops meshs of the grid.
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Code_Aster ®
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Titrate:
Conditions of solid connection of body
Date:
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Author (S):
J. PELLET
Key:
R3.03.02-A
Page:
12/12
Intentionally white left page.
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