Добавил:
vinogradov-design.narod.ru к.ф.-м.н. http://nauchkor.ru/users/572f1d7d5f1be72184501535 Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз: Предмет: Файл:

2019-RG-math-Vinogradov-translation

.pdf
Скачиваний:
4
Добавлен:
11.05.2019
Размер:
1.54 Mб
Скачать

41

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

,

0

1

 

2

 

We have boundary conditions in the form:

 

 

 

UY (x

)

 

 

0

 

 

VY (x

)

 

 

3

 

 

x3 .

u, v.

We can write the matrix equations of conjugation of sections:

Y (x

) K (x

 

x )Y (x ) Y

 

(x

 

0

 

0

0

 

 

1

1

 

 

 

Y (x ) K(x

 

 

x

)Y (x

) Y

 

(x

 

 

 

 

 

1

1

2

2

 

 

1

Y (x

) K (x

2

x

)Y (x

) Y (x

2

2

 

3

3

 

 

 

We can rewrite it in a form more convenient for us further:

x1 ) ,x2 ) ,

x3 ) .

EY (x

) K (x

 

x )Y (x ) Y

 

(x

 

0

 

0

0

 

1

1

 

 

EY (x ) K (x x

)Y (x

) Y (x

 

1

 

1

2

2

1

EY (x

) K (x

2

x

)Y (x

) Y (x

2

2

 

3

3

 

where E is the identity matrix.

x1

x2

x3

) ,

) ,

) .

42

Then in a combined matrix form we obtain a system of linear algebraic equations in the following form:

U

 

0

 

0

 

 

 

0

 

 

 

Y (x

)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E

K (x

 

x )

0

 

 

 

0

 

 

 

0

 

 

0

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

Y (x

)

 

0

 

E

K (x

 

x

)

 

0

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

2

 

 

 

 

 

 

Y (x

)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

0

 

E

 

 

K (x

 

x

)

 

2

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

Y (x

)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

0

 

0

 

 

V

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

 

 

u

 

 

(x

x

)

Y

 

0

1

 

 

(x

x

)

Y

 

1

2

 

 

(x

x

)

Y

 

2

3

 

 

v

 

.

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

x

)Y (x

) Y

 

(x

 

i

i

 

 

 

x

i

)

 

 

.

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

,

0

1

 

2

 

We have boundary conditions in the form:

 

 

 

UY (x

)

 

 

0

 

 

VY (x

)

 

 

3

 

 

x3 .

u, v.

We can write the matrix equations of conjugation of sections:

Y (x ) K (x

 

x

)Y (x

) Y

 

(x

 

 

 

 

1

 

1

0

0

 

 

 

1

Y (x

) K(x

 

x )Y (x ) Y

 

(x

 

2

 

2

2

 

 

1

1

 

 

 

Y (x

) K (x

3

x

)Y (x

) Y (x

3

3

 

2

2

 

 

 

We can rewrite it in a form more convenient for us further:

x0

x1

x2

) , ) ,

) .

EY (x ) K (x

x

)Y (x

0

1

1

0

 

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

 

(x

x

 

) Y

0

 

1

 

) Y (x2 x1

) Y (x3 x2

equations:

) ,

) ,

) .

44

U

 

 

 

0

 

0

 

 

0

 

Y (x

)

 

 

 

u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

K (x

x

)

 

E

 

0

 

 

0

 

0

 

Y

(x

 

x

)

 

 

 

 

 

 

 

 

 

1

0

 

 

 

 

 

 

 

 

 

Y (x

)

 

 

 

1

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

K (x

 

x )

E

 

0

 

1

 

Y

(x

 

x )

 

 

2

 

 

 

 

2

 

 

 

 

1

 

 

 

 

 

Y (x

)

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

0

K (x

 

x

)

E

 

2

 

Y

(x

 

x

)

 

 

 

3

 

 

 

 

3

 

 

 

 

 

 

2

 

 

 

Y (x

)

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

0

 

0

 

 

V

 

3

 

 

 

 

v

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

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

)Y (x

) Y

 

(x

 

i

i

 

 

 

x

i

)

 

 

.

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

 

01

1

01 12

 

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

 

 

01 01

1

01 12

 

L

M

Y

(x )

01 12

 

12 12

1

.

At the conjugation point

x

2

 

, we similarly obtain the equation:

M

Y

(x

) P

 

 

12 12

2

12 23

 

L

M

Y

(x

)

12 23

 

23 23

2

 

.

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

 

 

 

0

 

0

 

 

0

 

Y (x

)

 

 

 

u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

K (x

x

)

 

E

 

0

 

 

0

 

0

 

Y

(x

 

x

)

 

 

 

 

 

 

 

 

 

1

0

 

 

 

 

 

 

 

 

 

Y (x

)

 

 

 

1

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

K (x

 

x )

E

 

0

 

1

 

Y

(x

 

x )

 

 

2

 

 

 

 

2

 

 

 

 

1

 

 

 

 

 

Y (x

)

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

0

K (x

 

x

)

E

 

2

 

Y

(x

 

x

)

 

 

 

3

 

 

 

 

3

 

 

 

 

 

 

2

 

 

 

Y (x

)

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

0

 

0

 

 

V

 

3

 

 

 

 

v

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

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

 

 

 

 

 

 

 

 

UY (x

0

) u,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

VY (x

 

 

) v.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

EY (x1 ) K (x1

x0 )Y (x0 ) Y

 

(x1

 

x0 ) ,

 

 

 

 

EY (x

) K(x

2

x )Y (x ) Y

(x

2

 

x ) ,

 

 

 

 

 

2

 

 

 

 

 

1

 

 

 

1

 

 

 

 

 

 

 

 

 

1

 

 

EY (x

 

) K (x

 

x

)Y (x

) Y

 

(x

 

 

x

 

)

 

3

3

 

 

3

2

 

 

 

 

 

 

 

 

 

 

2

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

will take the form:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

UY

01

(x

0

) u,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

VY

23

(x

3

) v.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

EY01 (x1 ) K01 (x1

 

 

 

 

 

 

 

 

 

 

 

 

 

*

(x1

x0 ) ,

 

x0 )Y01 (x0 ) Y01

 

 

M 01Y01 (x1 ) P01 12

 

L01 12 M12Y12 (x1 ) ,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

*

(x2

 

x1 ) ,

 

EY12 (x2 ) K12 (x2 x1 )Y12 (x1 ) Y12

 

 

 

M12Y12 (x2 ) P12 23

 

 

L12 23M 23Y23 (x2 )

,

 

 

EY23 (x3 ) K 23 (x3

 

 

 

 

 

 

 

 

 

 

 

 

 

*

 

(x3

 

x2 ) .

x2 )Y23 (x2 ) Y23

 

 

After rearranging the summands, we get:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

UY01 (x0 ) u,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

VY23 (x3 ) v.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

K01 (x1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

*

(x1

x0 )

,

x0 )Y01 (x0 ) EY01 (x1 ) Y01

 

M 01Y01 (x1 ) L01 12 M12Y12 (x1 ) P01 12

,

 

 

K

12

(x

2

x )Y (x ) EY (x

2

) Y *

(x

2

x )

,

 

 

 

1

12

1

 

 

 

 

12

 

 

 

 

12

 

 

 

 

1

 

M12Y12 (x2 ) L12 23M 23Y23 (x2 ) P12 23 ,

47

K

 

(x

 

x

 

)Y

 

(x

 

) EY

 

(x

) Y

*

(x

 

 

23

3

2

23

2

23

23

3

 

 

 

 

 

 

3

 

 

 

x

2

)

 

 

.

As a result, we can write down the final system of linear algebraic equations:

 

 

U

 

 

0

 

 

0

 

0

 

 

0

 

 

0

Y01

(x0 )

 

 

u

 

 

K

01

(x x

)

E

 

 

0

 

0

 

 

0

 

 

0

Y *

(x x

 

)

 

1

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Y01

(x1 )

01

1

0

 

 

 

0

 

 

M 01

L01 12 M12

0

 

 

0

 

 

0

P01 12

 

 

 

 

 

 

 

 

 

 

Y12

(x1 )

 

 

 

 

0

 

 

0

K

12

(x

2

x )

E

 

 

0

 

 

0

Y *

(x

2

x )

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

Y12

(x2 )

12

 

 

1

 

 

 

0

 

 

0

 

 

0

 

M12

L12 23 M 23

 

0

P12 23

 

 

 

 

 

 

 

 

 

 

Y23

(x2 )

 

 

 

 

0

 

 

0

 

 

0

 

0

K

23

(x

3

x

2

) E

Y *

(x

3

x

2

)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Y23

(x3 )

23

 

 

 

 

 

0

 

 

0

 

 

0

 

0

 

 

0

 

 

V

 

 

v

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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

 

 

01

1

 

S

 

 

 

 

 

L

P

(x )

01 12

12

1

,

P

(x

)

0

 

 

01

1

 

GR

 

 

 

 

 

 

R

 

(x

)

 

0

 

 

 

01

 

1

 

 

 

 

 

 

 

 

 

 

S

01

(x )

 

GR

 

 

 

 

1

 

 

 

 

M

Y

 

 

(x )

0

 

01

 

 

01

 

 

1

 

GR

 

 

 

 

 

 

 

 

 

 

L01 12

L01 12

L01 12

P

(x

)

12

1

 

 

R

(x

 

 

12

1

 

S

12

(x

 

 

1

M

Y

 

 

12

12

,

)

,

 

)

 

 

 

(x

)

 

1

 

,

M

Y

 

(x ) g

*

 

01

 

 

01

1

 

 

L

M

Y

(x )

01 12

 

12 12

1

, где

g

*

 

 

0

GR

 

,

which is true if we do not forget that in this case we have:

P

(x

)

01

1

 

R

(x

 

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 ) :

 

 

 

 

 

g

*

 

0

 

 

0

 

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

GR

 

GR

 

(x )

 

 

 

 

 

 

 

 

 

 

 

 

 

 

01

 

1

 

 

 

 

 

R

(x )

 

 

 

 

 

 

 

 

p

 

 

 

M

P

(x )

 

01

1

M

Y (x ) M

Y

(x )

 

 

 

 

 

 

01

1

S

 

(x )

 

 

 

01

01

1

 

 

 

01

1

 

M

 

 

01

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

where for convenience was introduced re-designation

M 01

M

p

.

 

Then we can write:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

R01 (x1 )

M

p

M

p

Y01 (x1 ) ,

 

 

 

 

 

11

12

p 11

p 21

M

p

 

 

 

12

Y

(x

)

 

 

 

M

p

01

1

 

 

 

 

22

 

 

 

 

 

 

 

,

g

*

 

0

 

 

 

 

0

 

 

 

 

0...0

 

 

Y

 

(x

)

 

0

 

 

Y

 

(x

)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

GR

(x )

 

G M

p

M

p

Y

 

(x )

 

G M

p

M

p

 

01

1

 

G M

p

M

p

 

01

1

 

 

 

 

 

11

12

01

 

11

12

 

 

 

 

11

12

 

 

 

 

 

 

 

01

1

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

We write the matrix equations for this case:

 

UY

01

(x

) u,

 

 

 

 

 

 

0

 

 

 

 

 

VY

 

(x

2

) v.

 

 

 

 

12

 

 

 

 

 

 

EY01 (x1 ) K01 (x1

 

 

 

 

 

 

*

(x1

x0 ) ,

x0 )Y01 (x0 ) Y01

M Y

(x ) g* L

M Y (x ) ,

01 01

1

 

 

 

 

 

01 12

12 12

1

 

EY12 (x2 ) K12 (x2

 

 

 

 

 

 

*

(x2

x1 ) .

x1 )Y12 (x1 ) Y12

Let us write the vector g* in the equation:

M 01Y01 (x1 )

 

 

0

 

p Y01 (x1 ) L01 12 M12Y12 (x1 ) ,

 

 

p

M

 

 

 

G M11

12

(M 01

 

 

 

0

 

p ) Y01 (x1 ) L01 12 M12Y12 (x1 ) .

G M

p

 

 

 

11

M12

 

50

To ensure the non-cumbersomeness, we introduce the notation:

(M

 

 

 

0

 

 

)

01

G M

p

M

p

 

 

 

 

 

 

11

12

 

 

 

 

 

 

 

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

 

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:

 

 

 

 

 

 

 

UY

01

(x

) u,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

VY

 

 

(x

) v.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

12

 

2

 

 

 

 

 

 

 

 

 

 

 

K

 

(x

 

x

)Y

 

(x

 

) EY

 

(x ) Y

 

*

(x

 

x

 

01

 

01

 

01

01

 

0

 

1

0

 

 

0

 

 

 

 

 

1

 

1

 

 

 

 

 

*

 

(x1 ) L01 12 M12Y12 (x1 ) 0 ,

 

 

 

 

 

M Y01

 

 

K

 

(x

 

x )Y

(x ) EY

 

(x

 

) Y

*

(x

 

x

12

2

 

2

 

2

 

 

1

 

 

12

1

 

 

 

12

 

12

 

 

1

Thus, we obtain the resulting system of linear algebraic equations:

so that the resulting

) ,

) .

 

 

U

 

 

0

 

 

 

0

 

 

0

 

Y

 

(x

)

 

 

 

u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

01

 

*

 

 

 

 

K

 

(x

x

)

E

 

 

 

0

 

 

0

 

 

 

0

 

Y

(x

 

x

)

01

 

 

 

 

 

 

 

 

 

 

 

01

 

 

1

0

 

 

 

 

 

 

 

 

 

 

 

Y

 

(x

)

 

1

0

 

 

 

0

 

 

M

 

L

 

 

M

 

0

 

 

 

 

 

0

 

 

 

 

 

*

 

 

 

 

01

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

01 12

 

12

 

 

Y

 

(x

)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

*

 

 

 

 

 

 

0

 

 

0

 

K

 

(x

 

x )

E

 

 

12

 

1

 

Y

(x

 

x )

 

 

 

 

 

12

2

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

1

 

 

Y

 

(x

 

)

12

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

0

 

 

0

 

 

 

0

 

 

V

 

12

 

 

 

 

 

v

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

If an external force-moment action g p is applied to the frame, then

should be rewritten in the form g*

0

 

 

0

, then:

GR g p

GR (x ) g p

 

 

 

 

 

 

01

1

 

g

*

 

0

 

 

 

 

 

 

 

GR

g*

 

0

 

0...0

Y

 

(x )

0 .

G M11p M12p

Y01 (x1 ) g p

G M11p M12p

 

 

 

 

01

1

g p

Соседние файлы в предмете Численные методы