2019-RG-math-Vinogradov-translation
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Chapter 7. The simplest method for solving boundary value problems with stiff ordinary differential equations without orthonormalization - the method of "conjugation of sections of the integration interval", which are expressed by matrix exponents.
The idea of overcoming the difficulties of computation by dividing the interval of integration into conjugate areas belongs to Dr.Sc. Professor Yu.I.Vinogradov (his doctoral thesis was defended including on this idea), and the simplest realization of this idea through the formulas of the theory of matrices belongs to the Ph.D. A.Yu.Vinogradov.
We divide the interval of integration of the boundary value problem, for example, into 3 sections. We will have points (nodes), including edges:
x |
, x |
, x |
, |
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We have boundary conditions in the form: |
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UY (x |
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3 |
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x3 .
u, v.
We can write the matrix equations of conjugation of sections:
Y (x |
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x )Y (x ) Y |
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Y (x ) K(x |
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1 |
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Y (x |
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)Y (x |
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3 |
3 |
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We can rewrite it in a form more convenient for us further:
x1 ) ,x2 ) ,
x3 ) .
EY (x |
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x )Y (x ) Y |
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(x |
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0 |
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0 |
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EY (x ) K (x x |
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2 |
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EY (x |
) K (x |
2 |
x |
)Y (x |
) Y (x |
2 |
2 |
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3 |
3 |
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where E is the identity matrix.
x1
x2
x3
) ,
) ,
) .
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Then in a combined matrix form we obtain a system of linear algebraic equations in the following form:
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This system is solved by Gauss’ method with the separation of the main element.
At points located between nodes, the solution is to be solved by solving Cauchy’s problems with the initial conditions in the i-th node:
Y (x)
K (x
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It is not necessary to apply orthonormalization for boundary value problems for stiff ordinary differential equations, since on each section of the integration interval the calculation of each matrix exponent is fulfilled independently and from the initial orthonormal identity matrix, which makes it unnecessary to use orthonormalization, unlike S.K.Godunov’s method, which greatly simplifies programming in comparison with S.K.Godunov's method.
It is possible to calculate Cauchy’s matrices not in the form of matrix exponents, but with Runge-Kutta’s methods from the starting identity matrix, and the vector of the particular solution of the inhomogeneous system of differential equations can be calculated on each site by RungeKutta’s methods from the starting zero vector. In the case of Runge-Kutta’s methods, error estimates are well known, which means that calculations can be performed with a known accuracy.
43
Chapter 8. Calculation of shells of composite and with frames by the simplest method of "conjugation of sections of the integration interval".
8.1. The variant of recording of the method for solving stiff boundary value problems without orthonormalization - the method of "conjugation of sections, expressed by matrix exponents "- with positive directions of matrix formulas of integration of differential equations.
We divide the interval of integration of the boundary value problem, for example, into 3 sections. We will have points (nodes), including edges:
x |
, x |
, x |
, |
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0 |
1 |
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2 |
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We have boundary conditions in the form: |
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UY (x |
) |
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0 |
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VY (x |
) |
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3 |
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x3 .
u, v.
We can write the matrix equations of conjugation of sections:
Y (x ) K (x |
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x |
)Y (x |
) Y |
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(x |
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1 |
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0 |
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Y (x |
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x )Y (x ) Y |
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Y (x |
) K (x |
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)Y (x |
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2 |
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We can rewrite it in a form more convenient for us further:
x0
x1
x2
) , ) ,
) .
EY (x ) K (x |
x |
)Y (x |
0 |
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1 |
1 |
0 |
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EY (x2 ) K(x2 x1 )Y (x1
EY (x3 ) K (x3 x2 )Y (x2 where E is the identity matrix.
As a result, we obtain a system of linear algebraic
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) Y |
0 |
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1 |
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) Y (x2 x1
) Y (x3 x2
equations:
) ,
) ,
) .
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.
This system is solved by Gauss’ method with the separation of the main element.
It turns out that it is not necessary to apply orthonormalization, since sections of the integration interval are chosen so long that the computation on them is stable.
At points near the nodes, the solution is found by solving the corresponding Cauchy’s problems with the origin at the i-th node:
Y (x)
8.2. Composite shells of rotation.
K (x
x |
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i |
i |
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x |
i |
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Let us consider the conjugation of segments of the composite shell of rotation.
Suppose we have 3 sections, where each section can be expressed by its differential equations and the physical parameters can be expressed differently - different formulas on different sections:
In the general case (for the example of section 12), the physical parameters of the section
(vector P12 (x) ) are expressed in terms of the required parameters of the system of ordinary differential equations of this section (through the vector Y12 (x) ) as follows:
P12 (x) M12Y12 (x) ,
where the matrix M12 is a square non-degenerate matrix.
45
With the transition of the conjugation point, we can write in a general form (but using the
conjugation point
x1
):
P |
(x |
) P |
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01 |
1 |
01 12 |
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L |
P |
(x ) |
01 12 |
12 |
1 |
,
where
P01 12
is the discrete increment of physical parameters (forces, moments) during the
transition from the "01" section to the "12" section, and the square nondegenerate matrix
L01 12
is
diagonal and consists of units and minus ones on the main diagonal to establish the correct correspondence between the positive directions of forces, angular momenta, displacements and angles when going from "01" to "12", which may be different (in different differential equations of different conjugate regions) in the equations to the left of the conjugation point and in the equations to the right of the conjugation point.
The last two equations combine to form the equation:
M |
Y |
(x ) P |
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01 01 |
1 |
01 12 |
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M |
Y |
(x ) |
01 12 |
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12 12 |
1 |
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At the conjugation point
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M |
Y |
(x |
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12 12 |
2 |
12 23 |
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L |
M |
Y |
(x |
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12 23 |
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23 23 |
2 |
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.
If the shell consisted of identical parts, then we could write in a combined matrix form a system of linear algebraic equations in the following form:
U |
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But in our case the shell consists of 3 sections, where the middle section can be considered, for example, a frame expressed in terms of its differential equations.
Then instead of vectors Y (x0 ) , Y (x1 ) , Y (x2 ) , Y (x3 ) we should consider vectors:
Y01 (x0 ),Y01 (x1 ),Y12 (x1 ),Y12 (x2 ),Y23 (x2 ),Y23 (x3 ) .
46
Then the matrix equations
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will take the form: |
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VY |
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EY01 (x1 ) K01 (x1 |
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L12 23M 23Y23 (x2 ) |
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After rearranging the summands, we get: |
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(x1 |
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x0 )Y01 (x0 ) EY01 (x1 ) Y01 |
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M 01Y01 (x1 ) L01 12 M12Y12 (x1 ) P01 12 |
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K |
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x )Y (x ) EY (x |
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M12Y12 (x2 ) L12 23M 23Y23 (x2 ) P12 23 ,
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x |
2 |
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As a result, we can write down the final system of linear algebraic equations:
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U |
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Y01 |
(x0 ) |
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Y * |
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Y01 |
(x1 ) |
01 |
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0 |
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M 01 |
L01 12 M12 |
0 |
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P01 12 |
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Y12 |
(x1 ) |
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K |
12 |
(x |
2 |
x ) |
E |
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Y * |
(x |
2 |
x ) |
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Y12 |
(x2 ) |
12 |
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M12 |
L12 23 M 23 |
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P12 23 |
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Y23 |
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K |
23 |
(x |
3 |
x |
2 |
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Y * |
(x |
3 |
x |
2 |
) |
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Y23 |
(x3 ) |
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This system is solved by the Gauss method with the separation of the main element.
At points located between nodes, the solution is to be solved by solving Cauchy’s problems with the initial conditions in the i-th node:
Y (x) K (x x |
)Y (x |
) Y |
i |
i |
|
It is not necessary to apply orthonormalization for
ordinary differential equations.
|
) . |
(x xi |
boundary value problems for stiff
8.3. Frame, expressed not by differential, but algebraic equations.
Let us consider the case when the frame (at a point
differential equations, but in terms of algebraic equations.
x1
)
is expressed not in terms of
Above we wrote down that:
P |
(x |
) P |
L |
P |
(x |
) |
01 |
1 |
01 12 |
01 12 |
12 |
1 |
|
We can represent the vector
P01(x1 ) |
of force factors and displacements in the form: |
P01(x1 ) R01(x1 ) ,
S01(x1 )
48
where
R01(x1 ) |
is the displacement vector, |
Algebraic equation for the frame:
S01(x1 ) |
is the vector of forces and moments. |
GR S ,
where G is the matrix of the rigidity of the frame, R is the vector of the frame movements, the vector of force factors that act on the frame.
At the point of the frame we have:
S
is
R 0, S GR ,
that is, there is no discontinuity in the movements R 0 , but there is a resultant vector of force factors S GR, which consists of forces and moments on the left plus forces and moments to the right of the point of the frame.
P |
(x |
) |
R |
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01 |
1 |
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S |
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L |
P |
(x ) |
01 12 |
12 |
1 |
,
P |
(x |
) |
0 |
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01 |
1 |
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GR |
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R |
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(x |
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0 |
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01 |
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1 |
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S |
01 |
(x ) |
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GR |
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1 |
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M |
Y |
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(x ) |
0 |
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01 |
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01 |
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1 |
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GR |
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L01 12
L01 12
L01 12
P |
(x |
) |
12 |
1 |
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R |
(x |
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12 |
1 |
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S |
12 |
(x |
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1 |
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M |
Y |
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12 |
12 |
,
) |
, |
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) |
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(x |
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1 |
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,
M |
Y |
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(x ) g |
* |
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01 |
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01 |
1 |
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L |
M |
Y |
(x ) |
01 12 |
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12 12 |
1 |
, где
g |
* |
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0 |
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GR |
||
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,
which is true if we do not forget that in this case we have:
P |
(x |
) |
01 |
1 |
|
R |
(x |
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01 |
1 |
S |
01 |
(x |
|
1 |
) )
,
that is, the vector of displacements and force factors is first compiled from displacements
(above) R01(x1 ) , and then from force factors (below) S01(x1 ) .
49
Here it is necessary to remember that the displacement vector
R |
(x ) |
01 |
1 |
is expressed in
terms of the required state vector |
Y01(x1 ) : |
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g |
* |
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0 |
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0 |
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, |
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GR |
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GR |
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(x ) |
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01 |
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1 |
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R |
(x ) |
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p |
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M |
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P |
(x ) |
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01 |
1 |
M |
Y (x ) M |
Y |
(x ) |
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01 |
1 |
S |
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(x ) |
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01 |
01 |
1 |
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01 |
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M |
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01 |
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where for convenience was introduced re-designation |
M 01 |
M |
p |
. |
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Then we can write: |
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R01 (x1 ) |
M |
p |
M |
p |
Y01 (x1 ) , |
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11 |
12 |
p 11
p 21
M |
p |
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12 |
Y |
(x |
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M |
p |
01 |
1 |
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22 |
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g |
* |
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0 |
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0 |
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0...0 |
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Y |
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(x |
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0 |
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Y |
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(x |
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GR |
(x ) |
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G M |
p |
M |
p |
Y |
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(x ) |
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G M |
p |
M |
p |
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01 |
1 |
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G M |
p |
M |
p |
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01 |
1 |
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11 |
12 |
01 |
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11 |
12 |
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11 |
12 |
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01 |
1 |
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We write the matrix equations for this case:
|
UY |
01 |
(x |
) u, |
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0 |
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VY |
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(x |
2 |
) v. |
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12 |
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EY01 (x1 ) K01 (x1 |
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* |
(x1 |
x0 ) , |
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x0 )Y01 (x0 ) Y01 |
||||||||||
M Y |
(x ) g* L |
M Y (x ) , |
||||||||
01 01 |
1 |
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01 12 |
12 12 |
1 |
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EY12 (x2 ) K12 (x2 |
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* |
(x2 |
x1 ) . |
|
x1 )Y12 (x1 ) Y12 |
Let us write the vector g* in the equation:
M 01Y01 (x1 ) |
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0 |
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p Y01 (x1 ) L01 12 M12Y12 (x1 ) , |
||
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p |
M |
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G M11 |
12 |
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(M 01 |
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0 |
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p ) Y01 (x1 ) L01 12 M12Y12 (x1 ) . |
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G M |
p |
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11 |
M12 |
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50
To ensure the non-cumbersomeness, we introduce the notation:
(M |
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0 |
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) |
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01 |
G M |
p |
M |
p |
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11 |
12 |
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M
*
.
Then equation
M |
Y |
(x ) g |
* |
L |
M |
Y |
(x ) |
|
|||||||
|
01 01 |
1 |
|
01 12 |
|
12 12 |
1 |
will take the form:
* |
|
(x ) |
M Y |
01 |
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1 |
L |
M |
Y |
(x ) |
01 12 |
|
12 12 |
1 |
.
For convenience, we rearrange the terms in the matrix equations system of linear algebraic equations is written clearly:
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UY |
01 |
(x |
) u, |
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0 |
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VY |
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(x |
) v. |
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12 |
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2 |
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K |
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(x |
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x |
)Y |
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(x |
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) EY |
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(x ) Y |
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* |
(x |
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x |
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01 |
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01 |
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01 |
01 |
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0 |
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1 |
0 |
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0 |
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1 |
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1 |
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* |
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(x1 ) L01 12 M12Y12 (x1 ) 0 , |
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M Y01 |
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K |
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(x |
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x )Y |
(x ) EY |
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(x |
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) Y |
* |
(x |
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x |
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12 |
2 |
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2 |
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2 |
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1 |
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12 |
1 |
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12 |
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12 |
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1 |
Thus, we obtain the resulting system of linear algebraic equations:
so that the resulting
) ,
) .
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U |
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0 |
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0 |
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0 |
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Y |
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(x |
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u |
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01 |
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* |
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K |
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(x |
x |
) |
E |
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0 |
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0 |
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0 |
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Y |
(x |
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x |
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01 |
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01 |
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1 |
0 |
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Y |
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(x |
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1 |
0 |
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0 |
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M |
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L |
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M |
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0 |
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0 |
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* |
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01 |
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01 12 |
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12 |
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Y |
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(x |
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* |
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0 |
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0 |
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K |
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(x |
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x ) |
E |
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12 |
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1 |
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Y |
(x |
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x ) |
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12 |
2 |
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(x |
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12 |
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1 |
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2 |
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0 |
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0 |
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0 |
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V |
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12 |
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v |
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.
If an external force-moment action g p is applied to the frame, then
should be rewritten in the form g* |
0 |
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|
0 |
, then: |
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GR g p |
GR (x ) g p |
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01 |
1 |
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g |
* |
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0 |
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GR |
g* |
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0 |
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0...0 |
Y |
|
(x ) |
0 . |
G M11p M12p |
Y01 (x1 ) g p |
G M11p M12p |
|
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01 |
1 |
g p |