F
F
V
M
,
0
0
K
K
I
(19)
где
V
− center radius vector;
0
I
− body inertia tensor.
Since the system's axes
OABC
are directed along the main axes of inertia, the inertia
tensor
0
I
has a diagonal view and its components can be easily calculated.
.
2
,
4
3
4
2
02
2
2
03
01
MR
I
M
H
К
I
I
Now let's consider the ways of orientation of the moving trihedron
OABC
relatively to fixed
z
y
x
O
. There is a well known method of determining the orientation of the trihedron
OABC
with
respect to
z
y
x
O
using Euler angles
,
,
through
and
the polar angles of the axis
OC
,
and through
-the angle between
OC
z
plane and
COA
plane. Let us indicate through
3
2
1
,
,
the components of the angular velocity vector respectively in the axes
,
OA
,
OB OC
:
;
cos
sin
sin
1
;
sin
sin
cos
2
(20)
.
cos
3
If at some point in time the trihedron
OABC
coincides with the trihedron
z
y
x
O
so that
,
0
from (20) it follows that
.
,
,
0
3
2
1
The component turns out to be
equal to zero, no matter what the values
,
,
,
are, which in general is wrong. That is why it is
inconvenient to use Euler's corners in those cases where the trihedron
OABC
coincides with the
trihedron
z
y
x
O
at some point in time, except for those cases when the vector
lies in a plane
z
y
O
at this moment.
Let's consider another way to determine the trihedron orientation [9, 10]. As it was stated at
the beginning, the trihedron
OABC
coincides with
z
y
x
O
. Transition of the trihedron to the final
position is carried out by performing the following three consecutive operations: turning it by an
angle
1
around the axis
x
O
OA
, then turning it by an angle
2
around the axis
OB
in the new
position and, finally, by an angle
3
around the axis
OC
in the new position. Let's define through
G the matrix of guiding cosines of axes
,
OA
,
OB OC
: in relation to fixed axes
z
O
y
O
x
O
,
,
. If
3
2
1
,
,
x
x
x
X
- the coordinates of the vector in the system
OABC
, and
3
2
1
,
,
y
y
y
Y
- its
coordinates in the system
z
y
x
O
, then
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Семей қаласының Шәкәрім атындағы мемлекеттік университетінің хабаршысы № 4(92)2020
122
.
Y
G
X
(21)
Let's result without the proof the following lemma: if the trihedron
OABC
rotates on an angle
near an axis
OA
, the matrix of guide cosines in new position
1
G
is set by the formula
BG
G
1
,
where
cos
sin
0
sin
cos
0
0
0
1
1
B
. (22)
Similarly, when turning by angle
around an axis
OB
, we get
G
B
G
2
2
,
where
cos
0
sin
0
1
0
sin
0
cos
2
B
, (23)
and when rotates by angle
around the axis
OC
.
G
B
G
3
3
,
where
1
0
0
0
cos
sin
0
sin
cos
2
B
. (24)
3
2
1
,
,
B
B
B
- orthogonal matrices, besides,
.
3
,
2
,
1
,
1
i
B
B
i
i
If the starting position
OABC
is the same
z
y
x
O
and the end position is reached by turning
by angles
,
,
,
3
2
1
, then
,
1
1
2
2
3
3
B
B
B
G
(25)
or
2
1
2
1
2
3
2
1
3
1
3
2
1
3
1
3
2
3
2
1
3
1
3
2
1
3
1
3
2
c
c
c
s
s
s
s
c
c
s
s
s
s
c
c
c
c
c
s
c
s
s
c
s
s
s
c
c
c
G
. (26)
Here,
i
c
and
,
3
,
2
,
1
,
i
s
i
definition for
i
cos
и
i
sin
.
The infinitesimal rotation associated with
, should be considered as a set of three
consecutive infinitesimal rotations with angular velocities
3
2
1
,
. Then, according to the known
vector property of infinitely small rotations, we can consider
the sum of three separate angular
velocity vectors
,
0
,
,
0
,
0
,
0
,
2
2
1
1
3
3
,
0
,
0
recorded in different coordinate
systems. However, the components of these vectors with respect to any coordinate system can be
obtained using orthogonal transformations
3
2
1
,
,
B
B
B
. Let's write out components of a vector in the
system connected with a moving body:
3
3
3
2
2
2
3
3
1
1
1
2
2
3
3
B
B
B
B
B
B
(27)
and detailed:
;
2
3
1
3
2
1
s
c
c
;
2
3
1
3
2
2
c
s
c
(28)
.
3
1
2
3
s
Let's resolve the system (28) regarding
3
2
1
,
,
:
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123
;
2
2
3
1
3
1
c
s
c
;
2
3
1
3
2
c
s
(29)
.
3
2
2
3
1
3
2
3
c
s
c
s
Therefore, there is a one-to-one correspondence between
and
vectors for all
3
2
1
,
,
, except
0
cos
2
. Thus, in the case of small fluctuations of mass near the equilibrium position, this
method of trihedron orientation
OABC
relatively to
z
y
x
O
excludes those "undesirable
paradoxes" that were encountered when using Euler angles.
Taking out the second order members in (26) and (28), we get
1
1
1
1
2
1
3
2
3
G
,
.
(30)
Let us define
as the contact surface of the shell with the mass, through
rs
ss
s
,
,
- the components of the stress tensor. Then, the reaction of the shell is given by
the following formulas:
;
1
1
1
2
0
d
B
B
F
(31)
,
1
1
1
2
3
1
1
1
2
0
0
d
B
B
B
B
r
K
where
,
cos
,
sin
2
,
sin
0
R
h
R
r
1
,
0
,
0
3
and
-angle between the positive
axes
z
and
r
and the points of junction of the mass with the shell.
Through
s
ss
s
Q
N
N
N
,
,
and
0
,
,
ss
s
M
M
M
signify the components of the force and
moment tensors on
, taking into account the known expressions:
;
2
2
dr
N
h
h
;
2
2
dr
r
M
h
h
(32)
and ratio
Rdrd
d
of formula (31) transformed to:
;
2
0
1
1
1
2
0
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