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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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Organization (S): EDF/MTI/MN
Handbook of Référence
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
constraints 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 LDLT 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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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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Dualisation of the boundary conditions
Date:
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Author (S):
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Key:
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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 RN or N is the number
nodal unknown factors while being constrained by a certain number of relations closely connected C (U
I
) - di = 0
(boundary conditions of the Dirichlet type).
It is with this problem of minimization under constraints 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
J (U)
U V
RN
where:
·
J (U) is a quadratic form (Energie potential total)
1
J (U) =
(With, U) - (B, U)
2
A is a positive symmetrical matrix (With, U) 0 U
) but not inevitably definite
(To = the 0 is possible for U 0)
·
V is the space of displacements kinematically acceptable
(it is under space closely connected of RN).
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 RN.
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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 constraints): U belongs then to RN where N is the number
total of nodal unknown factors.
If V is under vector space of RN generated by (N - p) of basic nodal displacements,
derivation of J (U) in V is done very simply: it is enough “to forget” in the matrices A and B them
lines and columns correspondents with the removed ddl (ui =)
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 constraints “mix” the ddl between them (linear relations between unknown factors) it A should be modified
and B.
The principle of the multipliers of Lagrange makes it possible to solve the problem without touching with
matrices A 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 Rn+p space,
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 J (U)
U V
1
J (U) =
(With, U) - (B, U)
2
V =
N
{uR C (U) =d, I =, 1
I
I
} p
Ci are linear forms on RN, the di are constant data. One supposes
more than the p forms Ci are independent between them: the dimension of the space generated by
Ci are p.
One can show (Cf. [Annexe 1]) that this problem is equivalent to the following:
Problem 2:
to find U V where
V = {U RN Ci (U) = I
D, I =,
1}
p
such
(With - b) (v)
that where
= 0 v
V
0
V
N
0
=
{v such
R
that Ci (v) =
0 I =
,
1}
p
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Let us rewrite problem 2 differently:
Problem 3:
to find U R N such as
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 (by identifying RN and its dual) that:
(I) Ci
is orthogonal in V0
(With - b)
is orthogonal in V0
V
{Ci I =1, p}
0 are under vector space of RN orthogonal with
(
) (V0 is of dimension (N - p)
because the p conditions Ci are supposed to be independent).
Since the decomposition of RN all in all direct of 2 pennies orthogonal spaces is single, one
in deduced that (With - b) belongs to the vector space generated by Ci.
There is thus a family of scalars I called multipliers of Lagrange such as:
(With - b) + C
I
I
= 0
I
This equality is true in RN.
Problem 3 becomes then:
Problem 4:
To find U RN, I R, (I =, 1 p)
(
I =,
1 p) C U - D
I
I
= 0
(
With - B
) + C
I I = 0
I
The reciprocal one (Pb4 => Pb3) is obvious: if there is I such as:
(With - b) + C
I
= 0, then v
V (With - b) v = - C. v
I
I I
= 0
0
I
I
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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 CT U
B
=
C
0
éq 2-4
D
(K) (X) (F)
where:
U N
p
X n+
R
R
R p
;
;
WITH A
; C
N, N
With
; K
p, N
An+ p, n+ p
One realizes that this system can be obtained while seeking to make extreme the functional calculus:
1
L (U,) =
(With, U) - (B, U) + (Cu -) D
éq 2-5
2
This functional calculus is called Lagrangien of the initial problem. Principal interest of this
method is to free itself from the constraints: U and are sought in RN and RP (X in
Rn+p).
Coefficients I are called coefficients of Lagrange of the problem (one will say sometimes them
“Lagrange”).
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3
Disadvantages of this dualisation
One sees according to the expression of Lagrangien that the matrix K is not positive any more (what was the case of
With). Indeed:
1
U Cu 0 L (U,)! =
(With, U) + Cu
0
0
0
0
0
0
0
0
0 < 0
2
The loss of the positivity of the matrix K involves that the resolution of the system KX = F cannot any more
to be done in general by the traditional algorithms of gradient or, by the factorization of Cholesky.
The algorithm of factorization LDLT 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 connects 2 nodes N1 and N2
X
N1
K
N2
2 unknown factors: u1, u2; 2 modal forces: f1, f2 (N = 2)
K
- K
With =
- K
K
1 constraint: U + U
1
2
= (p = 1)
The dualized problem is written: KX = F
with:
K
- K
U
F
1
1
K =
- K
K
X =
U F
2
= f2
0
Recall of the condition necessary and sufficient so that algorithm LDLT-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, K1 = [k11]).
There will be no null pivot in algorithm LDLT-SP if and only if (Ki is invertible for
any I ranging between 1 and N).
This condition will be noted: cond1
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The matrix K above is written with like classification of the unknown factors, the command of the components
X:
·
X = (U, U
1
2,)
K
- K
K =
-
K
K
0
K
- K
K2 =
do not check the condition cond1.
- K
K
On example 1 let us test new classifications:
·
X = (, U, U
1
2)
0
K =
K
-
K
- K K
K1 does not check the condition cond1.
·
X = (U, U
1
2)
K
- K
det K1 = K
K =
0
det K
2
2
=
2
- K K
det K3 = - K (+)
-
K is supposed > 0 (stiffness of the spring)
-
K3 is invertible only if + 0. The case + = 0 corresponds indeed to one
“bad” physical blocking: the condition: U - U
1
2 = cte does not block the “movements
rigid body " for A (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 A.
With the notations of [§1] one will write:
ker A V0 = {}
0
One will suppose in the continuation of the document that this condition is checked. I.e.
constraints Ci block at least the movements of rigid body of A (they
can be more numerous of course). When this condition is checked and that them
conditions Ci are independent between them, one will say that the problem is physically
posed well.
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-
K2 is invertible only if 0
It is thus seen that classification (U, U
1
2) checks the condition cond1 if 0.
If blocking U + U
1
2 = is reduced to:
u1 =
(,) = (1,)
0 it goes
u2 =
(,) = (0,)
1 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 I after the ui, if A 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 ui: 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
0
- 1
K
J
J
= 0 0
· 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 blocked, the matrix is singular, and such as if one it
block 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), matrix kN-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 LDLT. This
classification is for example that resulting from algorithm LDLT 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 = []
0
0
1
U
K =
X
C = []
=
1
1
0
K is invertible, but there is not any common permutation of the lines and columns of K
allowing a resolution by LDLT.
All these remarks show that the dualisation proposed in this paragraph does not allow
not to use LDLT_SP.
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4
Principle of the “Lagrange doubles”
The method suggested here is that implemented in code CASTEM 2000 (communication
personal of Th. CHARRAS 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 Lagrangien [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 into 2 equal parts 1 and 2.
The equation C U = D is then replaced by:
C U - (1 - 2) = D
C U + (1 - 2) = D
where is a nonnull constant.
Let us show the equivalence of the old problem and the new one:
Problem 1: “simple Lagrange”
U
RN
With + CT
= B
to find:
such as (S):
p
Cu = D
R
1 = 2
With + CT1 + CT2 = B
= 1 + 2
(S)
Cu - 1 + 2 = D
With + CT = B
Cu + 1 - 2 = D
Cu =
D
is a constant 0
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From where the problem are equivalent to the precedent:
Problem 2: “double Lagrange”
With + CT1 CT2 = B
U
N
R
to find:
such as (): Cu
- 1 + 2 = D
1 2 p
p
,
R × R
Cu
+ 1 - 2 = D
The new problem can be written:
K' X' = F'
with:
X'
= (U, 1 2)
F' =
(B, D,) D
WITH CT
CT
K' =
C
- L L
C L - L
The problem corresponds to make extreme the functional calculus:
1
(U, 1, 2) =
(With, U) - (B, U) + (1, Cu -) D
2
2
+ (, Cu -)
D - (1 - 2, 1 - 2)
2
One can show (cf Annexe 2) that if one observes a certain rule of classification of the unknown factors, and
by choosing the constant > 0, the K' matrix 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 R0:
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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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 = B has
11 1
13 3
1
U + has U = B has
22 2
24 4
2
let us call 1
2
1
2
1, 1 2 ddls of Lagrange associated with the 1ère condition and 2, 2 those associated
2nd 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:
1
1
2
2
1, 1
U, 2, 2
U, 3
U, 1, 4
U, 2
1
2
1 and 1 frames “with more close” the constrained ddls (U and U
1
3)
1
2
2 and 1 frame “with more close” the constrained ddls (U and U
2
4)
The technique of the “Lagrange doubles” associated with the R0 rule thus makes it possible to solve any system
linear posed physically well with the algorithm of LDLT without permutation. The demonstration
suppose nevertheless that matrix A is symmetrical and positive (not inevitably definite).
Note:
Assumptions: With symmetrical and A positive are necessary to use LDLT (or LU)
without permutation as the two following counterexamples show it:
0
1
·
A1 = 1 0 is symmetrical but nonpositive,
0 1
·
A2 =
is positive but nonsymmetrical.
- 1 1
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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 zone of contact).
Note:
This possibility is not currently used in the code.
That is to say a system with constraints KX = F:
Let us write this system while emphasizing a particular constraint (this calculation remains obviously
valid when there are several constraints) Cu - D = 0. To simplify the writing, = 1 is chosen.
That is to say
1 the first ddl of Lagrange associated with the constraint
2 the second ddl of Lagrange associated with the constraint
U = X - {1 2
,
}
~
K = matrix K projected on U; B = vector F projected on U
The system is written with these variables:
~
K
CT CT U
B
1
C
- 1
1
=
D
C 1 - 1 2
D
Let us change the coefficient (2 2
,): - 1 3 and let us write the new system:
~
K
U (1 2) CT
+
+
=
B
(1 +2) = 0
éq 5 - 1
(S) C
U - 1 + 2 = D
C U - 1 + 2 = D
éq 5 - 2
1
2
~
C U + +
3
= D
K U = B
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 constraint
Cu - D = 0. The 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 out of U as the subsystem K U = B; 2 unknown factors
additional 1 and 2 does not disturb the solution out of 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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Indeed, let us imagine now that we know in advance that certain relations kinematics
are likely to be slackened. Let us number the 2 associated ones with 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 the 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 A, 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 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 K' matrix 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 µi:
Let us calculate the polynomial characteristic of K':
P (µ) = (µ + 2) (- µ2 + kµ + 22)
µ1 = -
2
< 0
+ K +
2
K + 82
µ2 =
> 0
2
+ K -
2
K + 82
µ3 =
< 0
2
K is the eigenvalue of the nonconstrained system. This eigenvalue is the required order of magnitude
for µ1, µ2, µ3.
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It is noticed that µ1 < 0, µ3 < 0 and µ2, > 0 i.e. the 2 eigenvalues added by
coefficients of Lagrange is < 0 (it is besides because of that that LDLT_SP is not guaranteed without
precautions).
One seeks to obtain eigenvalues of the same order of magnitude:
µ1 µ2 µ3
µ
2
2
2
2 µ3
µ1
2
4
éq 6-1
If << K, then µ3 0 µ2 K: it is not the sought result.
If >> K, then µ µ 2 µ
2
3
1
The three eigenvalues are then in absolute value about which is an arbitrary constant
very large in front of 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:
=
K µ1 - 2k
µ2 2k
µ3 - K
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 (a)+max (has
II
II)/2. Moreover, one takes =.
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Date:
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Key:
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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 constraints 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 is p (< N) linear relations
homogeneous between these displacements: B ()
0 (J 1, p
J U =
=
).
These linear relations are independent between them, i.e. the row of the matrix B
containing the coefficients of the p linear relations is p.
That is to say K the matrix of rigidity of the mechanical system without constraints.
That is to say M the matrix of mass of the mechanical system without constraints.
Which is the system with the eigenvalues to solve to find the modes clean of the structure
constraint?
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 nodal unknown factors RN
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. Notons B1
this submatrix.
Then let us make a partition of the unknown factors of U in U1 and U2 such as:
U1
B U = 0
[
B
B]
0
1
2
U
=
2
U
p
1
R
U
N p
2
-
R
B1 = matrix p × p
B2 = matrix p × (N - p)
The linear relations can then be written:
B U + B U
1
1
2
2
= 0
what makes it possible to express the U1 unknown factors according to U2 since B1 is invertible.
U
= - B 1
- B U
1
1
2
2
éq 7.3-3
Stamp reduced rigidity:
1
The elastic deformation energy of the not forced discretized structure is W
T
def =
THE U.K.U.
2
If one partitionne the matrix K in the same way that one partitionné U, one obtains:
K
K
1
12
K = T
K
K
12
2
then:
2 W
T
T
T
T
T
def
= U1 K1 U1 + U2 K2 U2 + U2 K12 U1 + U1 K12 U2
Let us introduce the linear constraints then [éq 7.3-3]:
2
T
T
- T
1
W
-
T
def
= U2 B2 B1 K1 B1 B2 U2 + U2 K2 U2
T
1
-
T
T
- T T
- U2 K12 B1 B2 U2 - U2 B2 B1 K12 U2
T
=
~
U2 K2 U2
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with:
~
K
= K + BT BT K B 1
- B - K B 1
- B - BT B-TKT
2
2
2
1
1
1
2
12
1
2
2
1
12
éq 7.3-4
It is thus seen that one expressed the deformation energy of the reduced structure like a form
bilinear of U2. The nodal unknown factors of U1 were eliminated. The nodal unknown factors U2
are forced.
Stamp of reduced mass:
Let us adopt the same partition for the matrix of mass Mr. Nous 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 W
T
cin
=
“~
U2 m2 “U2
with:
~
M
= M + BT BT MR. B 1
- B - MR. B-1 B - BT B-TMT
2
2
2
1
1
1
2
12
1
2
2
1
12
éq 7.3-5
Conclusion:
The system to solve to find the modes (and the frequencies) clean of the structure forced
is:
2
~
2 ~
To find them (N - p) (
N p
Xi, I) R
× R such as: (K2 - I m2) I
X
= 0
~
~
with K 2 and m2 defined by [éq 7.3-4] and [éq 7.3-5].
Application to the blocked ddl:
In this case:
B
= I
1
B
2 = 0
~
~
from where: K = K
2
2 and M = M
2
2
i.e. it is enough “to forget” in K and M the lines and columns corresponding to the blocked 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
A led to the A' matrix:
AT BT BT
A' =
B
- I + I
B
I
- I
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:
T
T
T
T
K
K
1
12
K
B1
K
B
M
M
1
1
12
m
B1
m
B1
T
T
T
T
(K
K
B
B
M
M
B
B
S)
12
2
K
2
K
2
2
12
2
m
2
m
2
-
X = 0
K
B1 K
B2 - K
I + K
I
m
B1 m
B2 - m I m I
K
B1 K
B2
K
I - K
I
m
B1 m
B2
m
I - m
I
for an own pulsation and a clean vector X of this system, one can write:
K X + K X + BT
1
1
12
2
1 (1
+ 2
) = 0
éq 7.4-1
K
X + K X + BT
12
1
2
2
2 (1
+ 2
) = 0
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 - 2 M
2
2
I
I
I; = K
- m
;
= K -
m
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N
= a number of ddls physical
The system (S) is of command (N + 2 p) if
.
p = a number of relations kinematics
The characteristic polynomial in 2 is a priori degree N + 2 p. Son term of higher degree is worth:
N
(- m
2 p
I) (+ m)
if semi is the ième diagonal term of Mr.
i=1
· it is thus seen that if m 0, the term of higher degree is 0 (because the semi ones 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
X
= - B1 B2 X2
[éq 7.4-1]
- 1
- T
1
-
1
= 2 =
B1 (- K1 B1 B2 + K12) X
2
2
[éq 7.4-2]
-
- 1
T
T
1
2
2 +
2
2 -
-
2
1
(-
-
K
B
B X
K X
B B
K B
B
1
1
2 + K 12) X
12
1
2
= 0
(~ - 2 ~
K
M2) X
2
2
= 0
with: ~K
= - KT B 1
- B + K + BT BT K B-1 B - BT BT K
2
12
1
2
2
2
1
1
1
2
2
1
12
~
M
= - MT B 1
- B + M + BT BT MR. B 1
- B - BT BT M
2
12
1
2
2
2
1
1
1
2
2
1
12
~
~
It is noted that the definitions of K 2 and m2 are identical to those of the equations
[éq7.3-4] and [éq 7.3-5].
It is thus seen that any clean vector X of the dualized system is also clean vector of
reduced system (with the same own pulsation) if one projects it on U2 space.
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Reciprocally, any clean vector X2 of the reduced problem can be prolonged in a clean vector
system dualized XT
[XT, XT T T
= 1 2, 1, 2].
with:
X = - B 1 - B X
1
1
2
2
1
= -
BT
1
-
1
1 (- K B
B + K
1
1
2
12) X2
2
1
= -
BT
1
-
2
1 (- K B
B + K
1
1
2
12) X2
2
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:
K
m
m
u1
u2
K
- K
m 0
K =
M
=
- K
K
0
m
the constraint U is added to him
U
1 +
2 = 0
()
0; That is to say =.
The reduced system is then:
·
K = K; K = K; K = - K; M = m; M = m; M
1
2
12
1
2
12 = 0
·
B =; B
1
2 =
~
= K (1+) 2
~
K
; M
2
2
2
=
(m1+)
2
K 1
2
(+)
=
; X = 1
m 1+ 2
2
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It is noted that the eigenvalue 2 depends on the report/ratio =.
K
If = 0
2
,
= m
2
If = 1
2
K
= m
If
2
K
m
Let us choose (= =)
1 to simplify and let us write the dualized system:
K
- K
m
0
X
K
K
m
m
1
- K
K
0
m
X
K
K
m
m
2
-
= 0
K
K
- K
K
m
m
- m
m
1
K
K
K
- K
m
m
m
- m2
the eigenvalues of this system are:
K
2
K K 2
=
K
,
,
,
m
m
m
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 Mr.
If one chooses m = m = 0, calculation shows that the characteristic polynomial is degree 1 and that its
only solution is:
K
2
2
=
m
X =
{- 1, +}
1
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 constraints can be written in the form:
[
U
B
B] 1
0
B
1
2
invertible
U =
with 1
2
· Then the clean modes of the forced structure are those of the reduced system:
(~
2 ~
K - M2) X
2
2
= 0
with:
~
K
= - KT B 1
- B + K + BT BT K B 1
- B - BT BT K
2
12
1
2
2
2
1
1
1
2
2
1
12
~
M
= - MT B 1
- B + M + BT BT MR. B 1
- B - BT BT M
2
12
1
2
2
2
1
1
1
2
2
1
12
· The dualized system (double Lagrange) which is written:
(~~
~
~
- 2
) ~~
K
MR. X = 0
with:
~
~
XT
= [X X
1
12
1
2
]
K
K
BT BT
1
12
1
1
~
~
KT
K
BT BT
K = 12
2
2
2
B
B
1
2
- I
I
B
B
I
1
2
- I
M
M
1
12
0
0
T
~
~
M
M
0 0
M = 12
2
0
0
0
0
0
0
0
0
has the same solutions then as the reduced system.
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Appendix 1
That is to say problem 1
min J (U)
U V
1
J (U) =
(With, U) - (B, U)
2
V under space refines RN = {U RN such
C
that
U - D
I
I = 0 I = 1
,}
p
With and B are defined on RN
With positive symmetrical matrix of command N.
Problem 2
To find U V such as: (With, v) - (B, v) = 0 v
V
0
0
0
0)
V = {U RN such as C U - D
I
I = 0, I = 1,}
p
V0 = {v RN such
C
that
U
0
I
= 0, I = 1,}
p
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) = 0 v V)
V = {U RN such as C U - D
I
I = 0, I = 1,}
p
There is indeed bijection between the whole of the {v - U, U V, v}
V and the V0 unit.
Let us show that problem 2 ' is equivalent to problem 1:
That is to say U solution of 2 '
Then, v V
1
1
J (v) - J (U) =
(AV, v) - (B, v) - (With, U) + (B, U)
2
2
1
1
=
(AV, v) - (With, v - U) - (With, U)
2
2
1
1
=
(
S
AV, v) - 2 (With, v) + (With, U) =
(A (UV, UV) 0
2
2
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Let us calculate first of all the derivative of J (U):
U V, v V
,
0
0
J (U + v
0) - J (U)
I (U) v0 = lim
0
=
lim (With, v0) + (AV, v
0
0) - (B, v0)
(With the b) v
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 (U + v0) - J (U)
0 (I (U) v0 0, v0 V0)
It is seen that I (U) who am a linear form on V0 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
= 0
) (
)
(
) (
)
,
v
V
S
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Appendix 2
Definitions, notations
A is the matrix of unconstrained rigidity (N × N) (symmetrical and positive)
C is the matrix of blocking: Cu - D = 0 (C stamps N × p (p <)
N)
U is the vector of physical ddl RN
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) RN × R p × R 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 stamps of a symmetrical nature 2p + N
WITH CT
CT
K =
C
- L L
C L - L
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 LDLT 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 frame the ddl
physiques constrained by this equation.
Thereafter, to simplify the writing, = 1 will be taken.
One seeks to show that very under Ki matrix of K is invertible.
That is to say under Ki matrix given. It corresponds to a sharing of the ddl: those of row I, those of
row > I.
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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
2
1
2
1 is the whole of the couples (1, 1)
ruffle
such as
(1) <
(
ng
ruffle
1)
ng
I
L1 = {1
2
1}; L2
1
1 = {1}
L
1
2
1
2
3 is the whole of the couples (3, 3)
ruffle
such as I <
(3) <
(3
ng
ruffle
)
ng
L1 = {1
2
3}; L2
3
3 = {3}
L
1
2
1
2
2 is the whole of the couples (2, 2)
ruffle
such as
(2)
ng
I < ruffle (2)
ng
L1 = {1
2
2}; L2
2
2 = {2}
J
There are L =
Li
#.
i=1,3
j=1,2
The matrix C can cut out in 3 parts corresponding to cutting (L1, L2, L3)
C1
C2
C3
~ ~~
Each matrix C
(U, U)
I can cut out in 2 parts corresponding to cutting
~
C
~
~
I
Ci
~ ~~
Matrix A can cut out in 4 parts corresponding to cutting (U, U)
~A
With
With =
~
~
T
With
With
Using these notations, the problem to be solved is to show that the Ki matrix is invertible.
-
~
I
I
0
C
1
~
I
- I
0
C
K
1
I
=
~
0
0
- I C
2
~
~
~
~
CT CT CT
With
1
1
2
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1
12
1
This matrix corresponds to the vector Xi = 1
2
~
U
It should be shown that: K .X = 0
X
I
I
I = 0
The problem is equivalent to:
Problem 1:
- 1
2
~ ~
1 + 1 +
1
C U = 0
~
~
U = 0
1
2
~
1 - 1 +
1
C U = 0
(
S) =
~
1
2
0
1
1 = 1 =
-
~
2 + C2 U =
0
1
0
~
1
2
~
1
~
2 =
T
1
C (1 + 1)
T
+ C2 2 + A.~u = 0
General case:
~
It is supposed that U; L1; L2
(S)
1
2
1
= 1
éq An2-1
~ ~
C U
1
= 0
éq An2-2
12 = ~ ~
2
C U
éq An2-3
~T
~T ~
~
2
1
+ (
~
C
2
C C2 + A) U = 0
1 1
éq An2-4
From [éq An2-4], one deduces:
T T
T ~ T ~
~
2 ~ ~
1
~
+
(
~
u.a.
U
2
C
2
C + A) U = 0
1 1
From [éq An2-2], one obtains:
~T ~ T
T ~ T ~
~
u.a.
= 0 ~u
1
(C C + A
2
2
) ~u = 0
~T ~ T ~ ~
~T ~~
u.a. C U + U With
2
2
= 0
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~
~T ~
However A is symmetrical positive (under matrix of a positive matrix) and C C
2
2 is also a matrix
symmetrical positive, therefore this sum can be null only if the two terms are null.
T T ~
~ ~
u.a.
~
C U
2
2
= 0
~T ~~
U With =
0
~ ~
C U
1
2
= 0
2
= 0
éq An2-5
~
A is a positive matrix, one wants to show that:
~T ~~
~
U To = 0 U = 0
éq An2-6
It remains us to be shown that: ~
U = 0
1
and
= 0
1
·
~
U = 0
Let us prolong ~
U on RN by ~~ = 0 = (~, ~~
U
U
U U):
~
C With
U
~T
=
Au~ = 0
U ker A
~
C U
C u~
1
0
1
~
~
Cu = C U
2
= C U
2
=
0
C U
3
C~ u~
3
0
~
Indeed C3 = 0 bus if not, there would not yet exist ddl of ~u constrained by equations
taken into account (of row > I) what is contrary in R0.
The prolongation U of ~
U is thus in the cores of A and C. One will show that it is then null.
Let us take again the problem with “simple Lagrange”.
(
With + CT = B
S2) = Cu =
D
If u0 0 is such as Au0 = 0 and Cu0 = 0.
If u1 is solution of S2, it is seen that then U + U
1
µ 0 is also solution. What is impossible bus
we suppose our problem posed physically well.
One concludes from it that U = 0 U
~ = 0.
Handbook of Référence
R3.03 booklet: Boundary conditions and loadings
HI-75/01/001/A
Code_Aster ®
Version
5.0
Titrate:
Dualisation of the boundary conditions
Date:
09/02/01
Author (S):
J. PELLET
Key:
R3.03.01-B
Page:
30/30
·
11 = 0
[éq An2-4] gives:
~
CT 1
1 1
= 0
éq An2-7
~
~
~
In the same way that the rule imposing R0 C3 = 0, one sees that C1 = 0.
[éq An2-6] gives:
~CT 1
0
CT 1
1
1
1 1
= ~~
=
0
=
éq An2-8
CT 1
0
1
1
Let us reason by the absurdity: if 1
T1
1 0 is such as C1 1
= 0, it is that there is a combination
linear of the lines of C1 which is null, which is contradictory with the fact that the lines of C1
from/to each other are independent (physical problem good posed).
Thus 11 = 0.
Particular case:
When one (or more) of the ~ sets
U, L1, L2 are 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.
Handbook of Référence
R3.03 booklet: Boundary conditions and loadings
HI-75/01/001/A
Outline document