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Code_Aster
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Version
5.0
Titrate:
Dualisation of the boundary conditions
Date:
09/02/01
Author (S):
J. PELLET
Key:
R3.03.01-B
Page:
1/30
Manual of Reference
R3.03 booklet: Boundary conditions and loadings
HI-75/01/001/A
Organization (S):
EDF/MTI/MN
Manual of Reference
R3.03 booklet: Boundary conditions and loadings
Document: R3.03.01
Dualisation of the boundary conditions
Summary:
One explains the principle of the multipliers of Lagrange to solve the linear systems under
stresses closely connected resulting from the imposition of the boundary conditions of the kinematic type. The matrix of rigidity
obtained not being more definite positive, certain algorithms of resolution thus become unusable. One
thus seek a technique to be able to continue to use the algorithm of factorization LDL
T
without
permutation and without elimination. The technique suggested is that of the “Lagrange doubles” (used in
Code Castem 2000). It is shown that this technique is effective. One gives some indications on
conditioning of the matrices obtained by this technique.
The problem of the search of the clean modes of the constrained systems is then examined. One shows
that a possible solution is to add the boundary conditions dualized to the matrix of “rigidity” and of not
to touch with the matrix of “mass”.
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Code_Aster
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Version
5.0
Titrate:
Dualisation of the boundary conditions
Date:
09/02/01
Author (S):
J. PELLET
Key:
R3.03.01-B
Page:
2/30
Manual of Reference
R3.03 booklet: Boundary conditions and loadings
HI-75/01/001/A
Contents
1 Introduction ............................................................................................................................................ 3
2 Dualisation of the boundary conditions kinematics, principle of the multipliers of Lagrange .......... 3
3 Disadvantages of this dualisation ......................................................................................................... 7
4 Principle of the “doubles Lagrange” ......................................................................................................... 10
5 additional Advantage .................................................................................................................... 13
6 Remark on the conditioning of the system .................................................................................... 14
7 clean Modes and parameters of Lagrange ......................................................................................... 16
7.1 Introduction .................................................................................................................................... 16
7.2 Mechanical problem to solve .................................................................................................. 16
7.3 System reduces ............................................................................................................................... 16
7.4 System Dualisé ............................................................................................................................ 19
7.5 ......................................................................................................................................... Example 21
7.6 Conclusions ................................................................................................................................... 23
Appendix 1 ................................................................................................................................................. 24
Appendix 2 ................................................................................................................................................. 26
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Code_Aster
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Version
5.0
Titrate:
Dualisation of the boundary conditions
Date:
09/02/01
Author (S):
J. PELLET
Key:
R3.03.01-B
Page:
3/30
Manual of Reference
R3.03 booklet: Boundary conditions and loadings
HI-75/01/001/A
1 Introduction
We are interested in this document in the dualisation of the boundary conditions (known as
kinematics). Two problems distinct from linear algebra are examined:
·
the resolution of the linear systems: paragraphs 2, 3, 4, 5, 6,
·
the search of the clean modes: paragraph 7.
2 Dualisation of the boundary conditions kinematics,
principle of the multipliers of Lagrange
In Code_Aster (as in the other codes of finite elements), one is brought to solve
many linear systems.
Often such a system can be regarded as the algebraical expression of a problem of
minimization of a positive quadratic functional calculus
()
J U
where
U
belongs to
R
N
or N is the number
nodal unknown factors while being constrained by a certain number of relations closely connected
()
C U
I
- =
D
I
0
(boundary conditions of the Dirichlet type).
It is with this problem of minimization under stresses closely connected that one is interested here. In the continuation of
document, one will take as example the case (and the vocabulary) of linear static mechanics. One
will speak about matrix of rigidity, vector displacement,… but the technique suggested remains valid
for the problems of thermal evolution, into linear or not-linear.
That is to say the discretized problem:
()
Pb1
:
min
U V
U


R
N
J
where:
·
()
J U
is a quadratic form (total potential Energy)
()
(
) ()
J U
With, U
B, U
=
-
1
2
With
is a positive symmetrical matrix
(
)
(
)
With, U
U
0
but not inevitably definite
(
With
=
0
is possible for
U
0
)
·
V
is the space of displacements kinematically acceptable
(it is under space closely connected of
R
N
).
This discrete problem is solved numerically while expressing that the “derivative” of
()
J U
in
V
is
null. One is then brought back to solve a linear system of equations.
The problem is “to derive”
()
J U
in
V
R
N
.
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Code_Aster
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Titrate:
Dualisation of the boundary conditions
Date:
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Author (S):
J. PELLET
Key:
R3.03.01-B
Page:
4/30
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HI-75/01/001/A
In general, for practical reasons, the expression of
()
J U
is calculated in the base of all them
nodal displacements (without taking account of the stresses):
U
belongs then to
R
N
where N is the number
total of nodal unknown factors.
If
V
is under vector space of
R
N
generated by
(
)
N p
-
basic nodal displacements,
derivation of
()
J U
in
V
is done very simply: it is enough “to forget” in the matrices
With
and
B
lines and columns agents with the removed ddl
(
)
U
I
=
0
.
If the constrained ddl are not put at zero but not assigned to a given value:
U
D
I
I
=
, it is necessary to modify
B
.
Finally if the stresses “mix” the ddl between them (linear relations between unknown factors) it is necessary to modify
With
and
B
.
The principle of the multipliers of Lagrange makes it possible to solve the problem without touching with
matrices
With
and
B
. The price to be paid is an increase in the number of unknown factors in the system with
to solve.
Instead of solving the problem in space
V
of dimension N - p, one solves it in space
R
n+p
,
noted additional unknown factors
I
being called multiplying of Lagrange.
Principle and justification
Let us take again the preceding problem by clarifying space
V
Problem 1:
()
()
(
) ()
()
{
}
min
,
,
U V
U
U
With, U
B, U
V
U
U
=
-
=
=
=


J
J
C
D
I
p
N
I
I
1
2
1
R
C
I
are linear forms on
R
N
, them
D
I
are constant data. One supposes
more than the p forms
C
I
are independent between them: the dimension of the space generated by
C
I
is
p
.
One can show (cf [Appendix 1]) that this problem is equivalent to the following:
Problem 2:
()
{
}
(
) ()
()
{
}
to find
where
such as where
such as
U V
V
U
C U
With B v
v V
V
v
C v
I
0
0
I
=
=
=
-
=
=
=
=


R
R
N
I
N
D
I
p
I
p
,
,
,
1
0
0
1
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Code_Aster
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Version
5.0
Titrate:
Dualisation of the boundary conditions
Date:
09/02/01
Author (S):
J. PELLET
Key:
R3.03.01-B
Page:
5/30
Manual of Reference
R3.03 booklet: Boundary conditions and loadings
HI-75/01/001/A
Let us rewrite problem 2 differently:
Problem 3:
to find
such as
U
R
N
-
=
I C U D
I
I
0
éq 2-1
=
v
V
C v
0
I
,
0
éq 2-2
(
)
-
=
v V
With B v
0
,
0
éq 2-3
The equations [éq 2-2] and [éq 2-3] show (while identifying
R
N
and its dual) that:
()
I C
I
is orthogonal with
V
0
(
)
With B
-
is orthogonal with
V
0
V
0
is under vector space of
R
N
orthogonal with
(
)
{
}
C
I
I
p
=
1,
(
V
0
is of dimension (N - p)
because the p conditions
C
I
are supposed to be independent).
Since decomposition of
R
N
all in all direct of 2 pennies orthogonal spaces is single, one
in deduced that
(
)
With B
-
belongs to the vector space generated by
C
I
.
There is thus a family of scalars
I
called multipliers of Lagrange such as:
(
)
With B
C
I
- +
=
I
I
0
This equality is true in
R
N
.
Problem 3 becomes then:
Problem 4:
(
)
(
)
(
)
To find
U
C U D
With B
C
I
I
I
=
=
-
=
- +
=




R
N
I
I
I
,
I
p
I
p
,
,
,
R
1
1
0
0
The reciprocal one (Pb4 => Pb3) is obvious: if there exists
I
such as:
(
)
(
)
With B
C
v
With B v
v
I
- +
=
-
= -
=
I
I
I I
I
V
C
0
0
0
,
.
then
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Code_Aster
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Version
5.0
Titrate:
Dualisation of the boundary conditions
Date:
09/02/01
Author (S):
J. PELLET
Key:
R3.03.01-B
Page:
6/30
Manual of Reference
R3.03 booklet: Boundary conditions and loadings
HI-75/01/001/A
Problem 4 is the sought problem. It will be said that it is the problem with conditions
dualized kinematics. Matriciellement one can write it:
()
()
()
KX
F
WITH C
C
0
U
B
D
K
X
F
T
=








=



éq 2-4
where:
U
X
+
R
R
R
N
p
N p
;
;
WITH A
A.C.
K WITH
+
+
N N
p N
N p N p
,
,
,
;
;
One realizes that this system can be obtained while seeking to make extreme the functional calculus:
()
(
) () (
)
L
U
With, U
B, U
Cu D
,
=
-
+
-
1
2
éq 2-5
This functional calculus is called Lagrangian initial problem. Main interest of this
method is to free itself from the stresses:
U
and
are sought in
R
N
and
R
p
(
X
in
R
n+p
).
Coefficients
I
coefficients of Lagrange of the problem are called (one will say sometimes them
“Lagrange”).
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Code_Aster
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Titrate:
Dualisation of the boundary conditions
Date:
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Author (S):
J. PELLET
Key:
R3.03.01-B
Page:
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R3.03 booklet: Boundary conditions and loadings
HI-75/01/001/A
3
Disadvantages of this dualisation
One sees according to the expression of Lagrangian that the matrix
K
is not positive any more (what was the case of
With
). Indeed:
(
)
(
)
=
+
<
U Cu
U
With U
Cu
0
0
0
0
0
0
0
0
0
0
1
2
0
L
,
,
!
The loss of the positivity of the matrix
K
involve that the resolution of the system
KX F
=
cannot any more
to be done in general by the conventional algorithms of gradient or, by the factorization of Cholesky.
The algorithm of factorization LDL
T
without permutation of the lines and columns is not guaranteed any more either:
it is the latter algorithm which one wants to be able to continue to use.
Let us illustrate the problem on the following example:
Example 1:
a spring of stiffness
K
connect 2 nodes NR
1
and NR
2
NR
1
NR
2
X
K
2 unknown factors: U
1
, U
2
; 2 modal forces: F
1
, F
2
(N = 2)
With
K
K
K
K
=
-
-




1 stress:
U
U
1
2
+
=
(p = 1)
The dualized problem is written:
KX F
=
with:
K
X
F
=
-
-




=




=




K
K
K
K
U
U
F
F
0
1
2
1
2
Recall of the condition necessary and sufficient so that algorithm LDL
T
- SP (without permutation)
function:
Let us note
K
I
under matrix of
K
formed of I first lines and columns of
K
.
(If
K
is of command
N
,
K
N
=
K
,
K
1
= [K
11
]).
There will be no null pivot in algorithm LDL
T
- SP if and only if (
K
I
is invertible for
all
I
included/understood enters
1
and
N
).
This condition will be noted: cond1
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Code_Aster
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Dualisation of the boundary conditions
Date:
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Key:
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Page:
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HI-75/01/001/A
The matrix
K
above is written with like classification of the unknown factors, the command of the components
of
X
:
·
(
)
X
=
U U
1
2
,
,
K
=
-
-




K
K
K
K
0
K
2
=
-
-




K
K
K
K
do not check the condition cond1.
On example 1 let us test new classifications:
·
(
)
X
=
,
U U
1
2
K
=
-
-




0
K
K
K
K
K
1
do not check the condition cond1.
·
(
)
X
=
U
U
1
2
,
(
)
K
K
K
K
=
-
-




=
=
= -
+
K
K
K
K
K
K
0
1
2
2
3
2
det
det
det
-
K
> 0 are supposed (stiffness of the spring)
-
K
3
is invertible only if
+
0
. The case
+ =
0
corresponds indeed to one
“bad” physical blocking: the condition:
U
U
1
2
-
= cte does not lock the “movements
rigid body " for
With
(without energy).
So that the total problem has a single solution, it is necessary indeed that the conditions
C U D
I
I
=
generate a space of acceptable displacements which does not contain any
movements of rigid body of
With
.
With the notations of [§1] one will write:
{}
ker A
V
=
0
0
One will suppose in the continuation of the document that this condition is checked. I.e.
stresses
C
I
at least the movements of rigid body lock of
With
(they
can be more numerous of course). When this condition is checked and that them
conditions
C
I
are independent between them, one will say that the problem is physically
posed well.
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Dualisation of the boundary conditions
Date:
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Author (S):
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Key:
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Page:
9/30
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HI-75/01/001/A
-
K
2
is invertible only if
0
It is thus seen that classification
(
)
U
U
1
2
,
check the condition cond1 if
0
.
If blocking
U
U
1
2
+
=
is reduced to:
()
()
()
()
U
U
1
2
1 0
0 1
=
=
=
=
,
,
,
,
it goes
that does not go
The symmetry of the problem shows that to be able to deal with the problem
()
()
,
,
=
0 1
it is necessary to number
(
)
X
U
U
=
2
1
,
.
From this very simple example, one can draw some general conclusions (all negative):
·
if one numbers all them
I
after
U
I
, if
With
is singular, the condition cond1 will not be
checked for
K
With
I
=
,
·
that is to say a condition
C U D
=
and
the multiplier of associated Lagrange. The equation
C U D
=
in general does not utilize all the unknown factors
U
I
: the equation constrained some ddl. If
is numbered before the ddl that it constrained the condition cond1 will not be checked. Indeed
let us look at under matrix
K
J
where
J
is the number of the equation giving
.
K
K
0
0
0
J
J
=


-
1
·
That is to say a physical structure which, by bad luck, “is retained by its physical last ddl”
i.e. such as if this ddl is not locked, the matrix is singular, and such as if one it
lock the matrix is invertible. The use of a ddl of Lagrange
for this blocking is
impossible. Indeed, if one numbers
before the physical last ddl, there will be a null pivot with
level of
, and if one numbers it after (thus in any last ddl), the matrix
K
N
-
1
will not be
not invertible since blocking is not taken yet into account. One will see with [§4] that
technique of the “double lagrange” makes it possible to solve this problem.
To finish this paragraph we can make the following remark: If
K
is invertible, one
knows that there is a classification of the unknown factors making it possible to factorize
K
by LDL
T.
This
classification is for example that resulting from algorithm LDL
T
with permutation (pivot
maximum for example). But this renumerotation relates to only the lines of the matrix; it
y thus has loss of the symmetry of
K
. It is enough to consider the following example:
Example 2:
[]
[]
With
C
K
X
U
=
=


=
=
0
1
0 1
1 0
K
is invertible, but there is not any common permutation of the lines and columns of
K
allowing a resolution by LDL
T
.
All these remarks show that the dualisation proposed in this paragraph does not allow
not to use LDL
T
_SP.
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Code_Aster
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Titrate:
Dualisation of the boundary conditions
Date:
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Author (S):
J. PELLET
Key:
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Page:
10/30
Manual of Reference
R3.03 booklet: Boundary conditions and loadings
HI-75/01/001/A
4
Principle of the “Lagrange doubles”
The method suggested here is that implemented in code CASTEM 2000 (communication
personal of Th. CHARAS and P. VERPEAUX). An intuitive presentation could be of it
following:
It is seen that the dualized problem [éq 2-4] has null terms on the diagonal: those corresponding to
ddl of Lagrange. This property is also noticed on the Lagrangian one [éq 2-5]: there are no terms
quadratic in
.
This nullity of the diagonal terms prevents certain permutations of lines and columns: one cannot
to place Lagrange before the physical ddl that it constrained.
The idea is then to break up each coefficient of Lagrange
in 2 equal parts
1
and
2
.
The equation
C U D
=
is then replaced by:
(
)
(
)
C U
D
C U
D
-
-
=
+
-
=

1
2
1
2
where
is a nonnull constant.
Let us show the equivalence of the old problem and the new one:
Problem 1: “simple Lagrange”
to find:
such as (S):
U
With C
B
Cu D
T




+
=
=




R
R
N
p
(S)
=
=
+
+
=
=


+
+
=
-
+
=
+
-
=





1
2
1
2
1
2
1
2
1
2
With C
B
Cu D
With C
C
B
Cu
D
Cu
D
T
T
T



is a constant
0
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Code_Aster
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Dualisation of the boundary conditions
Date:
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J. PELLET
Key:
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Page:
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HI-75/01/001/A
From where the problem are equivalent to the precedent:
Problem 2: “double Lagrange”
to find:
such as ():
U
With C
C
B
Cu
D
Cu
D
T
T
×




+
=
-
+
=
+
-
=




R
R
R
N
p
p
,

1
2
1
2
1
2
1
2
The new problem can be written:
K' X'
F'
=
with:
(
)
(
)
X
U
F'
B, D, D
'
,
=
=




1
2
K'
With
C
C
C
L
L
C
L
L
T
T
=
-
-




The problem corresponds to make extreme the functional calculus:
(
)
(
) ()
(
)
(
) (
)
L
',
,
,
,
U
With, U
B, U
Cu D
Cu D
1
2
1
2
1
2
1
2
1
2
2
=
-
+
-
+
- -
-
-
One can show (cf Appendix 2) that if one observes a certain rule of classification of the unknown factors, and
by choosing the constant
>
0
, the matrix
K'
check the condition cond1.
This rule is as follows:
That is to say a relation of blocking
Cu D
- =
0
, it corresponds to him 2 multipliers of Lagrange
1
and
2
.
This relation utilizes a certain number of ddl physical.
Regulate R
0:
For each relation of blocking, it is necessary to place
1
before the constrained first ddl physical and
2
after the constrained last ddl physical.
To decrease the occupation memory of the matrix
K
, it is necessary to seek to minimize the bandwidth.
It is what one does in Code_Aster in “framing” the relations “with nearest”:
1
is placed just
before the constrained first ddl,
2
is placed just after the constrained last ddl.
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Dualisation of the boundary conditions
Date:
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Author (S):
J. PELLET
Key:
R3.03.01-B
Page:
12/30
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R3.03 booklet: Boundary conditions and loadings
HI-75/01/001/A
Illustration:
That is to say a problem with 4 ddls physical:
U, U, U, U
1 2 3 4
.
This system is subjected to 2 conditions:
U has
U has
B
U has
U has
B
11 1
13 3
1
22 2
24 4
2
+
=
+
=


let us call
11
12
,
2 ddls of Lagrange associated with the 1
era
condition and
2
1
22
,
those associated
2
ème
condition.
By supposing that the physical ddls were numbered in the order:
U, U, U, U
1 2 3 4
, classification
total of the ddls retained by Aster is then:
11
1
2
1
2
3
12
4
22
,
,
U
U U
U
11
12
and
frame “with more close” the constrained ddls (
U
U
1
3
and
)
2
1
12
and
frame “with more close” the constrained ddls (
U
U
2
4
and
)
Technique of the “Lagrange doubles” associated with the rule R
0
thus allows to solve any system
linear posed physically well with the algorithm of LDL
T
without permutation. The demonstration
suppose nevertheless that the matrix
With
that is to say symmetrical and positive (not inevitably definite).
Note:
Assumptions:
With
symmetrical and
With
positive are necessary to use LDL
T
(or LU)
without permutation as the two following counterexamples show it:
·
With
0 1
1 0
1
=
is symmetrical but nonpositive,
·
With
0
1
1 1
2
= -
is positive but nonsymmetrical.
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Dualisation of the boundary conditions
Date:
09/02/01
Author (S):
J. PELLET
Key:
R3.03.01-B
Page:
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HI-75/01/001/A
5 Advantage
additional
One will show in this paragraph that the technique of the “Lagrange doubles” can allow
economically to solve a series of problems which would differ only by their conditions with
limits kinematics (for example a variable area of contact).
Note:
This possibility is not currently used in the code.
That is to say a system with stresses
KX F
=
:
Let us write this system while emphasizing a particular stress (this calculation remains obviously
valid when there are several stresses)
Cu D
- =
0
. To simplify the writing, one chooses
=
1
.
That is to say
1
the first ddl of Lagrange associated with the stress
2
the second ddl of Lagrange associated with the stress
{}
U X
= -
1
2
,
~
K
= matrix
K
projected on
U
;
B
= vector
F
projected on
U
The system is written with these variables:
~
K C
C
C
C
U
B
D
D
T
T
-
-








=




1
1
1
1
1
2
Let us change the coefficient
(
)
2
2
,
:
-
1
3
and let us write the new system:
(
)
(
)
()
~
~
S
K U
C
B
C U
D
C U
D
C U
D
K U B
T
+
+
=
+
=
-
+
=
-
+
=
+
+
=
=



-
-
-
1
2
1
2
1
2
1
2
1
2
0
3
éq 5 1
éq 5 2
éq 5 3
This last system is uncoupled: one can solve [éq 5-3] to obtain
U
then to calculate
1
and
2
.
It is noticed that the resolution of [éq 5-3] corresponds to the initial problem without the stress
Cu D
- =
0
. Values of
1
and
2
then do not have any more the same physical significance. In others
terms, the system
()
S
with the same solution in
U
that the subsystem
~
K U B
=
; 2 unknown factors
additional
1
and
2
do not disturb the solution in
U
. The total system can appear of size
higher (+2) with what is necessary, but by means of computer, that can be very convenient.
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Dualisation of the boundary conditions
Date:
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Author (S):
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Key:
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Page:
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HI-75/01/001/A
Indeed, let us imagine now that we know in advance that certain relations kinematics
are likely to be slackened. Let us number them
2
associated these relations at the end of the system. Us
let us can then triangulate partially once and for all the system while stopping before these ddl.
left the triangulated matrix is most important in volume: all physical ddl and all them
1
.
When a concrete problem arises, i.e. when one knows the list of the linear relations
active, it is enough to update the last lines of the matrix (
-
if the relation is active,
+
3
if
it is not it). One can then finish the triangulation and solve the problem economically.
6
Notice on the conditioning of the system
When one looks at the form of the matrix which one finally will factorize
K'
(Cf [§3]), one sees
that its various submatrices
With
,
C
,
I
can be of order of magnitude very different. One knows
that in general this situation is not favorable numerically (limited precision of the computers).
It should be noticed that the equations of connection
Cu D
- =
0
can be multiplied by a constant
arbitrary
()
without changing the problem. Moreover, we saw that the matrices
I
were
also arbitrary
(
)
>
0
. We thus have two parameters allowing “to regulate” it
conditioning of the matrix.
We will not make a general demonstration but we are satisfied to examine the case more
commonplace which is: a spring, a ddl, a connection.
The matrix
K'
is written if
K
is the rigidity of the spring:
K'
=
-
+
-




K
The conditioning of this matrix is related to the dispersion of its eigenvalues
µ
I
:
Let us calculate the polynomial characteristic of
K'
:
()
(
)
(
)
P
K
µ
µ
µ
µ
=
+
-
+
+
2
2
2
2
= -
<
= + +
+
>
= + -
+
<
µ
µ
µ
1
2
2
2
3
2
2
2
0
8
2
0
8
2
0
K
K
K
K
K
is the eigenvalue of the nonconstrained system. This eigenvalue is the required order of magnitude
for
µ µ µ
1
2
3
,
,
.
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It is noticed that
µ
µ
1
3
0
0
<
<
,
and
µ
2
0
,
>
i.e. 2 eigenvalues added by
coefficients of Lagrange is < 0 (it is besides because of that that LDL
T
_SP is not guaranteed without
precautions).
One seeks to obtain eigenvalues of the same order of magnitude:
µ
µ
µ
1
2
3
µ µ
µ
2
3
1
2
2
2
2
4
éq 6-1
If
µ
µ
<<
K
K
,
then
3
2
0
: it is not the sought result.
If
µ
µ
µ
>>
K,
then
2
3
1
2
The three eigenvalues are then in absolute value about
who is an arbitrary constant
very large in front
K
. This solution is not that which one will retain because the value
K
is in the case
General (with great number of ddl) of an order of magnitude comparable with the other eigenvalues of
system.
One will choose rather:
µ
µ
µ
=
-
-
K
K
K
K
1
2
3
2
2

Practically in Code_Aster, one chooses a value of
single for all the system. This
value is the average of the extreme values of the diagonal terms associated the physical ddl:
()
()
(
)
min
max
/
has
has
II
II
+
2
. Moreover, one takes
=
.
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7
Clean modes and parameters of Lagrange
7.1 Introduction
This paragraph wants to answer the two following questions:
Q1:
Which is the system of values (and vectors) clean tiny room to be solved when one
mechanical model is subjected to stresses linear kinematics
homogeneous?
Q2:
Which is the model dualized (with parameters of Lagrange) equivalent or precedent?
7.2
Mechanical problem to solve
One supposes a mechanical system already discretized by finite elements.
The nodal unknown factors are noted
{} (
)
U
=
=
U
I
N
I
1,
.
Nodal displacements are not all independent: there exists
()
p
N
<
linear relations
homogeneous between these displacements:
()
(
)
B
J
p
J
U
=
=
0
1
,
.
These linear relations are independent between them, i.e. the row of the matrix
B
containing the coefficients of
p
linear relations is
p
.
That is to say
K
the matrix of rigidity of the mechanical system without stresses.
That is to say
M
the matrix of mass of the mechanical system without stresses.
Which is the system with the eigenvalues to solve to find the modes clean of the structure
stress?
7.3 System
reduced
Let us notice that if one writes the linear relations kinematics in the form:
B U
=
0
éq 7.3-1
where:
B
is a matrix
p N
×
U
is the vector of the nodal unknown factors
R
N
then:
B U
=
0
éq 7.3-2
and the relation is also valid for speeds.
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Moreover, if
B
is of row
p
, then there is a square submatrix of
B
of row
p
. Let us note
B
1
this submatrix.
Then let us make a partition of the unknown factors of
U
in
U
1
and
U
2
such as:
[
]
(
)
B U
B
B
U
U
U
U
B
B
=


=
=
×
=
× -
-
0
0
1
2
1
2
1
2
1
2

R
R
p
N p
p p
p
N p
stamp
stamp
The linear relations can then be written:
B U
B U
1
1
2
2
0
+
=
what makes it possible to express the unknown factors
U
1
according to
U
2
since
B
1
is invertible.
U
B
B U
1
1 1
2
2
= -
-
éq 7.3-3
Stamp reduced rigidity:
The elastic deformation energy of the not forced discretized structure is
W
def
=
1
2 THE U.K.U
T
.
If one partitionne the matrix
K
same manner as one partitionné
U
, one obtains:
K
T
=


K
K
K
K
1
12
12
2
then:
2
1
1
1
2
2
2
2
12
1
1
12
2
W
def
T
T
T
T
T
=
+
+
+
THE U.K.U
THE U.K.U
The U.K.
U
The U.K.
U
Let us introduce the linear stresses then [éq 7.3-3]:
2
2
2
1
1
1 1
2
2
2
2
2
2
12
1 1
2
2
2
2
1
12
2
2
2
2
W
def
T
T
T
T
T
T
T
T
T
T
=
+
-
-
=
-
-
-
-
U B B
K B B U
THE U.K.U
The U.K.
B B U
U B B K
U
THE U.K.U
~
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with:
~
K
K
B B
K B B
K
B B
B B K
2
2
2
1
1
1 1
2
12
1 1
2
2
1
12
=
+
-
-
-
-
-
-
T
T
T
T
T
éq 7.3-4
It is thus seen that one expressed the deformation energy of the reduced structure like a form
bilinear of
U
2
. Nodal unknown factors of
U
1
were eliminated. Nodal unknown factors
U
2
are forced.
Stamp of reduced mass:
Let us adopt the same partition for the matrix of mass
M
. We can write the relation [éq 7.3-2]:
B U
B U
1
1
2
2
0
+
=
Same calculation as previously then leads us to:
2
2
2
2
W
cin
T
=
“~”
U MR. U
with:
~
M
M
B B
MR. B
B
MR. B
B
B B M
2
2
2
1
1
1 1
2
12
1 1
2
2
1
12
=
+
-
-
-
-
-
-
T
T
T
T
T
éq 7.3-5
Conclusion:
The system to solve to find the modes (and the frequencies) clean of the structure forced
is:
To find them
(
)
(
)
(
)
N p
I
I
N p
I
I
-
×
-
-
X
K
MR. X
,
~
~
=
=
=
=
2
2
2
2
0
R
R
such as
:
with
~
K
2
and
~
M
2
defined by [éq 7.3-4] and [éq 7.3-5].
Application to the locked ddl:
In this case:
B
I
B
1
2
0
=
=


from where:
~
K
K
2
2
=
and
~
M
M
2
2
=
i.e. it is enough “to forget” in
K
and
M
lines and columns corresponding to the locked ddl.
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7.4 System
Dualisé
We saw with [§5] that the taking into account of the coefficients of Lagrange (double) in a matrix
With
led to the matrix
A'
:
With
With
B
B
B
I
I
B
I
I
T
T
'
=
-
+
-




where:
·
B
is the matrix of the conditions kinematics:
B U
=
0
,
·
+
R
arbitrary
0
,
·
R
arbitrary
0
,
Let us apply the dualisation of the C.L to the matrices
K
and
M
, by partitionnant the ddl in
X
X
1
2
,
like
with [§7.3]. We obtain the problem with the eigenvalues according to:
()
S
K
T
K
T
K
T
K
T
K
K
K
K
K
K
K
K
m
T
m
T
m
T
m
T
m
m
m
m
m
m
m
m
K
K
B
B
K
K
B
B
B
B
I
I
B
B
I
I
M
M
B
B
M
M
B
B
B
B
I
I
B
B
I
I
1
12
1
1
12
2
2
2
1
2
1
2
2
1
12
1
1
12
2
2
2
1
2
1
2
-
+
-








-
-
-












=
X
0
for an own pulsation
and a clean vector
X
this system, one can write:
(
)
K X
K
X
B
1
1
12
2
1
1
2
0
+
+
+
=
T
éq 7.4-1
(
)
K
X
K X
B
12
1
2
2
2
1
2
0
+
+
+
=
T
éq 7.4-2
(
) (
)
B X
B X
1
1
2
2
1
2
0
+
-
-
=
éq 7.4-3
(
) (
)
B X
B X
1
1
2
2
1
2
0
+
+
-
=
éq 7.4-4
with:
K
K
M
I
I
I
K
m
K
m
=
-
=
-
=
-
2
2
2
;
;
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The system
()
S
is of command
(
)
N
p
+
2
if
N
p
= a number of ddls physical
= a number of relations kinematics


.
The characteristic polynomial in
2
is degree a priori
N
p
+
2
. Its term of higher degree is worth:
() (
)
-
+
=
m
I
I
N
m
p
1
2
if
m
I
is I
ème
diagonal term of
M
.
·
it is thus seen that if
m
0
, the term of higher degree is
0 (because them
m
I
are > 0) and thus
the dualized system
()
S
has more eigenvalues than the reduced system:
(
)
N p
-
. Both
systems are thus not equivalent. It is what one notes on the example of [§7.5],
·
let us choose
N
N
=
=
0
:
[éq 7.4-3] and [éq 7.4-4]
1
2
1
1 1
2
2
=
= -




-
X
B B X
[éq 7.4-1]
(
)
1
2
1
1
1 1
2
12
2
1
2
=
= -
-
+
-
-
B
K B B
K
X
T
[éq 7.4-2]
(
)
-
+
-
-
+
=
-
-
-
K
B
B X
K X
B B
K B
B
K
X
12
1 1
2
2
2
2
2
1
1
1 1
2
12
2
0
T
T
(
)
-
=
~
~
K
M
X
2
2
2
2
0
with:
~
~
K
K
B
B
K
B B
K B
B
B B
K
M
MR. B B
M
B B
MR. B B
B B
M
2
12
1 1
2
2
2
1
1
1 1
2
2
1
12
2
12
11
2
2
2
1
1
11
2
2
1
12
= -
+
+
-
= -
+
+
-




-
-
-
-
-
-
-
-
T
T
T
T
T
T
T
T
T
T
It is noted that the definitions of
~
K
2
and
~
M
2
are identical to those of the equations
[éq7.3-4] and [éq 7.3-5].
It is thus seen that any clean vector
X
dualized system is also clean vector of
reduced system (with the same own pulsation) if one projects it on space
U
2
.
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Reciprocally, any clean vector
X
2
reduced problem can be prolonged in a clean vector
dualized system
[
]
X
X X
T
T
T
T
T
=
1
2
1
2
,
,
,



.
with:
(
)
(
)
X
B B X
B
K B B
K
X
B
K B B
K
X
1
1 1
2
2
1
1
1
11
2
12
2
2
1
1
11
2
12
2
1
2
1
2
= -
= -
-
+
= -
-
+




-
-
-
-
-
T
T
The two systems are thus equivalent, they have the same clean modes and the same values
clean.
The dualized system, although of size higher than the reduced system, does not have more values
clean that the reduced system (the dimension of clean space is the same one).
Conclusion:
The dualized system is equivalent to the reduced system as soon as one chooses
m
m
=
=
0
,
i.e. if one takes the matrix of dualized rigidity but which one does not modify the matrix of
mass. It is what is made in Aster.
7.5 Example
That is to say the system:
U
1
U
2
K
m
m
K
M
=
-
-




=


K
K
K
K
m
m
0
0
the stress is added to him
(
)
U
U
1
+
=
2
0
0
; That is to say
=
.
The reduced system is then:
·
K
K
K
M
M
M
1
2
12
1
2
12
0
=
=
= -
=
=
=
K
K
K
m
m
;
;
;
;
;
·
B
B
1
2
=
=
;
()
(
)
=
+
=
+
~
;
~
K
M
2
2
2
2
1
1
K
m
()
=
+
+
=

2
2
2
2
1
1
1
K
m
; X
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It is noted that the eigenvalue
2
depends on the report/ratio
=
.
If
=
=
0
2
,
K
m
If
=
=
1
2
2
K
m
If
2
K
m
Let us choose
(
)
= =
1
to simplify and let us write the dualized system:
K
K
K
K
m
m
X
X
K
K
K
K
K
K
K
K
K
K
K
K
m
m
m
m
m
m
m
m
m
m
m
m
-
-
-
-






-
-
-














=
0
0
0
1
2
1
2
the eigenvalues of this system are:
2
2
=


K
m
K
m
K
m
K
m
,
,
,
It is noted that one finds the real eigenvalue (the 4°), but that one finds 3 eigenvalues
parasites due to nonthe nullity of
m
and
m
.
If one chooses
m
m
=
=
0
, calculation shows that the characteristic polynomial is degree 1 and that its
only solution is:
{
}
2
2
1
1
=
=
- +



K
m
X
,
who is the sought solution.
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7.6 Conclusions
·
If
K
and
M
are the matrices of rigidity and mass of a not-constrained system.
·
If the linear stresses can be written in the form:
[
]
B
B
U
U
B
1
2
1
2
1
0


=
with
invertible
·
Then the clean modes of the forced structure are those of the reduced system:
(
)
~
~
K
M
X
2
2
2
2
0
-
=
with:
~
~
K
K
B
B
K
B B
K B
B
B B
K
M
MR. B B
M
B B
MR. B B
B B
M
2
12
1 1
2
2
2
1
1
1 1
2
2
1
12
2
12
11
2
2
2
1
1
11
2
2
1
12
= -
+
+
-
= -
+
+
-




-
-
-
-
-
-
-
-
T
T
T
T
T
T
T
T
T
T
·
The dualized system (double Lagrange) which is written:
(
)
~
~
~
~ ~~
K
MR. X
-
=
2
0
with:
[
]
~
~
~
~
~
~
X
X
X
K
K
K
B
B
K
K
B
B
B
B
I
I
B
B
I
I
M
M
M
M
M
T
T
T
T
T
T
T
=
=
-
-








=






1
12
1
2
1
12
1
1
12
2
2
2
1
2
1
2
1
12
12
2
0 0
0 0
0
0
0 0
0
0
0 0
has the same solutions then as the reduced system.
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Titrate:
Dualisation of the boundary conditions
Date:
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Author (S):
J. PELLET
Key:
R3.03.01-B
Page:
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Manual of Reference
R3.03 booklet: Boundary conditions and loadings
HI-75/01/001/A
Appendix 1
That is to say problem 1
()
min
U V
U
J
()
(
) ()
J U
With, U
B, U
=
-
1
2
V
under space refines
{
}
R
R
N
N
I
p
=
-
=
=
U
C U D
I
I
such as
0
1
,
With
and
B
are defined on
R
N
With
positive symmetrical matrix of command N.
Problem 2
To find
U V
such as:
(
) ()
(
)
With, v
B, v
v
V
0
0
0
0
0
-
=
{
}
V
U
C U D
I
I
=
-
=
=
R
N
I
p
such as
0
1
,
,
{
}
V
v
C U
I
0
0
0
1
=
=
=
R
N
I
p
such as
,
,
One will show that the two preceding problems are equivalent.
Let us notice first of all that problem 2 is equivalent to problem 2 '.
Problem 2 '
To find
U V
such as:
(
) (
)
(
)
With, v U
B, v U
v V
- -
- =
0
{
}
V
U
C U D
I
I
=
-
=
=
R
N
I
p
such as
0
1
,
,
There is indeed bijection between the whole of
{
}
v U, U V, v V
-
and the unit
V
0
.
Let us show that problem 2 ' is equivalent to problem 1:
That is to say
U
solution of 2 '
Then,
v V
()
()
(
) () (
) ()
(
) (
) (
)
(
) (
) (
)
(
)
(
)
(
)
J
J
v
U
Front, v
B, v
With, U
B, U
Front, v
With, v U
With, U
Front, v
With, v
With, U
With U v U v
-
=
-
-
+
=
-
- -
=
-
+
=
-
-
1
2
1
2
1
2
1
2
1
2
2
1
2
0
,
S
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Key:
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Page:
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Manual of Reference
R3.03 booklet: Boundary conditions and loadings
HI-75/01/001/A
Let us calculate the derivative first of all of
()
J U
:
U V
v
V
,
0
0
()
(
)
()
(
) (
) ()
(
)
J
J
J
'
lim
lim
,
,
,
U v
U
v
U
To the v
Front v
B v
With B v
=
+
-
=
+
-




=
-
0
0
0
0
0
0
0
0
0
2
That is to say
U
the solution of Pb1
v V
, let us pose
v
v U
v
V
0
0
0
= -
;
.
(
)
()
()
(
)
+
-
J
J
J
U
v
U
U v
v
V
0
0
0
0
0
0
'
,
It is seen that
()
I U
who is a linear form on
V
0
must be systematically positive. This is not
possible that if this form is identically null.
One concludes from it that:
() (
)
(
) (
)
I
,
U
v U
With B v U
v V
-
=
-
-
=
0
S
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Key:
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R3.03 booklet: Boundary conditions and loadings
HI-75/01/001/A
Appendix 2
Definitions, notations
With
is the matrix of unconstrained rigidity
(
)
N N
×
(symmetrical and positive)
C
is the matrix of blocking:
Cu D
- =
0
(
C
stamp
(
)
N p
p N
×
<
)
U
is the vector of the physical ddl
R
N



1
is the vector of the first ddl of Lagrange
R
p



2
is the vector of the second ddl of Lagrange
R
p
(
)
X
U
=
×
×
,
,



1
2
R
R
R
N
p
p
One notes:
·
U
the whole of the physical ddl,
·
1
the whole of the first ddl of Lagrange,
·
2
the whole of the second ddl of Lagrange.
+
R
K
stamp of a symmetrical nature 2p + N
K
With
C
C
C
L
L
C
L
L
T
T
=
-
-




The matrix
K
written above corresponds to a certain classification of the unknown factors:
(
)
X
U
=
,
,



1
2
The genuine matrix
K
which one seeks to show that it is factorisable by LDL
T
without permutation
is not written with this classification. The only rule of classification taken into account is
following:
Regulate R0:
Both ddl of Lagrange associated with an equation with connection
C U D
I
I
- =
0
the ddl frame
physiques constrained by this equation.
Thereafter, to simplify the writing, one will take
=
1
.
One seeks to show that very under matrix
K
I
of
K
is invertible.
That is to say under matrix
K
I
data. It corresponds to a sharing of the ddl: those of row
I, those of
row > I.
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Key:
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HI-75/01/001/A
We will note:
~
U
the subset of U corresponding to the ddl of row
I.
~
~
U
the subset of U corresponding to the ddl of row > I.
L
1
is the whole of the couples
(
)
()
()









1
1
12
1
1
12
, row
row
such as
<
I
{}
{}
L
L
1
1
1
2
=
=






1
1
12
;
L
3
is the whole of the couples
(
)
()
()









3
1
3
2
3
1
3
2
, row
row
such as
I
<
<
{}
{}
L
L
3
3
1
2
=
=






3
1
3
2
;
L
2
is the whole of the couples
(
)
()
()












2
1
2
2
21
2
2
, row
row
such as
<
I
{}
{}
L
L
2
1
2
2
=
=






21
2
2
;
One has
L
L
ij
ij
=
==
1 3
1 2
,
#
.
The matrix
C
can cut out in 3 parts corresponding to cutting (L
1
, L
2
, L
3
)
C
1
C
2
C
3
Each matrix C
I
can cut out in 2 parts corresponding to cutting
()
~, ~~
U U
~
C
I
~
~
C
I
Matrix A can cut out in 4 parts corresponding to cutting
()
~, ~~
U U
With
With
With
With
With
=


~
~
~
T
Using these notations, the problem to be solved is to show that the matrix K
I
is invertible.
K
I
I
C
I
I
C
I C
C
C
C
With
I
T
T
T
=
-
-
-








0
0
0
0
1
1
2
1
1
2
~
~
~
~
~
~
~
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Code_Aster
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Key:
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Page:
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Manual of Reference
R3.03 booklet: Boundary conditions and loadings
HI-75/01/001/A
This matrix corresponds to the vector
X
U
I
=


















1
1
12
2
1
~
It should be shown that:
K X
X
I
I
I
.
=
=
0
0
The problem is equivalent to:
Problem 1:
()
(
)
S
T
T
=
-
+
+
=
-
+
=
+
=
+
+
+
=




=
=
=
=















-
-
-
-


















1
1
12
1
1
1
12
1
2
1
2
1
1
1
12
2
2
1
1
1
12
2
1
0
0
0
0
0
0
0
~ ~
~ ~
~ ~
~
~
~.~
~
C U
C U
C U
C
C
With U
U
General case:
It is supposed that
~
U
;
L
1
;
L
2
()
S






1
1
12
=
éq An2-1
~ ~
C U
1
0
=
éq An2-2



21
2
=
~ ~
C U
éq An2-3
(
)
2
0
1 11
2
2
~
~ ~
~ ~
C
C C
With U
T
T



+
+
=
éq An2-4
From [éq An2-4], one deduces:
(
)
2
0
1 11
2
2
~ ~
~ ~ ~
~ ~
u.a.
u.a. C
With U
T T
T
T



+
+
=
From [éq An2-2], one obtains:
(
)
~ ~
~
~ ~
~ ~
~ ~ ~ ~
~ ~~
u.a.
u.a. C
With U
u.a. C U
U With
T
T
T
T
T
T
T
1
2
2
2
2
0
0
0
=
+
=
+
=
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Code_Aster
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Dualisation of the boundary conditions
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Key:
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Page:
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Manual of Reference
R3.03 booklet: Boundary conditions and loadings
HI-75/01/001/A
However
~
With
is symmetrical positive (under matrix of a positive matrix) and
~ ~
C C
2
2
T
is also a matrix
symmetrical positive, therefore this sum can be null only if the two terms are null.
=
=




~ ~.~ ~
~ ~~
u.a. C U
U With
T
T
T
2
2
0
0
=
=
~ ~
C U
2
2
1
0
0
éq An2-5
~
With
is a positive matrix, one wants to show that:
~ ~~
~
U With
U
T
= =
0
0
éq An2-6
It remains us to be shown that:
~
U
=
=
0
0
11
and
·
~
U
=
0
Let us prolong
~
U
on
R
N
by
()
~
~
~, ~~
U
U
U U
=
=
0
:
U With
U With
U
With
Cu
C U
C U
C U
C U
C U
C U
T
T
=
=
=




=




=




~ ~ ~
ker
~ ~
~ ~
~ ~
0
0
0
0
1
2
3
1
2
3
Indeed
~
C
3
0
=
bus if not, there would exist ddl of
~
U
constrained by equations not yet
taken into account (of row > I) what is contrary with R
0
.
Prolongation
U
of
~
U
is thus in the cores of
With
and
C
. One will show that it is then null.
Let us take again the problem with “simple Lagrange”.
()
S
2
=
+
=
=




With C
B
Cu D
T



If
U
0
0
is such as
With
0
0
=
and
Cu
0
0
=
.
If
U
1
is solution of S
2
, it is seen that then
U
U
1
0
+ µ
is also solution. What is impossible bus
we suppose our problem posed physically well.
One concludes from it that
U
U
=
=
0
0
~
.
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Page:
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Manual of Reference
R3.03 booklet: Boundary conditions and loadings
HI-75/01/001/A
·
11
0
=
[éq An2-4] gives:
~
C
1 11
0
T



=
éq An2-7
In the same way that the rule R
0
imposing
~
C
3
0
=
, it is seen that
~
~
C
1
0
=
.
[éq An2-6] gives:
C
C
C
1
1
1
1
1
1
1
1
1
0
0
0
T
T
T










=


=


=
~
~
~
éq An2-8
Let us reason by the absurdity: if
11
0
is such as
C
1 11
0
T
=
, it is that there is a combination
linear of the lines of
C
1
who is null, which is contradictory with the fact that the lines of
C
1
from/to each other are independent (physical problem good posed).
Thus
11
0
=
.
Particular case:
When one (or more) of the assemblies
~
U
, L
1
, L
2
is empty, the system (S) is simplified. One can
to check that the reasoning which one made in the general case, makes it possible to show
similar results.