Американский Научный Журнал ON THE TENSION OF A CYLINDRICAL ROD OF VARIABLE CROSS-SECTION (41-47)

The theory of small elastoplastic deformations is widespread in the field of structural analysis. In this paper consider the stretching of an in_nitely long cylindrical rod of variable cross-section. The results of solving the linearized equations of the theory of small elastic-plastic deformations [1-7] in the case of an axisymmetric problem are used. It is assumed that a simple stretch occurs in the initial state. In the first approximation, the relations for the components of displacements, deformations, and stresses are obtained. Solutions are expressed in terms of zero -and first-order Bessel functions. Скачать в формате PDF
American Scientific Journal № ( 36 ) / 2020 41

of professional development and evaluation of
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CONCLUSION
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Pedagogical in novation is a special area of
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Шишарина. - М.:
PalmariumAcademicPublishin g, 2018. - 208 c.
7 Кравченко, А.И. История менеджмента.
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образованию / А.И. Кравченко. - М.:
Академический проект, 2018. - 623 c
8 http://iac.kz/ru/project/pisa
9 Ирсалиев С .А., Камзолдаев М.Б.,
Ташибаева Д.Н., Копеева А.Т., «УЧИТЕЛЯ
КАЗАХСТАНА:ПОЧЕМУ МОЛОДЫЕ ЛЮДИ
ВЫБИРАЮТ ЭТУ ПРОФЕССИЮ И ЧТО ИХ
МОТИВИРУЕТ ОСТАВАТЬСЯ В НЕЙ?», Астана:
Общественное объединение «Центранализа и
стратегии «Белес», 2019. – 126 с.

UDC 51 -74
GRNTI 27.35.31

ON THE TENSION OF A CYLINDRICAL ROD OF V ARIABLE CROSS -SECTIO N

Petrov N.I.
Department of general physics,
Chuvash State University,
45 Moskovsky Prospekt, Cheboksary, Russia
Abstract . The theory of small elastoplastic deformations is widespread in the field of structural analysis. In
this paper consider the stretching of an in_nitely long cylindrical rod of variable cross -section. The results of
solving the linearized equations of the theory of small elastic -plastic deformations [1 -7] in the case of an
axisymmetric problem are used. It is assumed that a simple stretch occurs in the initial state. In the first
approximation, the relations for the components of displacements, deformations, and stresses are obtained.
Solutions are expressed in terms of zero -and first -order Bessel functions.
Keywords: stretching, displacement, deformation, stress, boundary, conditions, linearization, Bessel
function.

In the case of axisymmetric deformation, the
relations of the theory of small elastic -plastic
deformations have the form [1 -2]
−= 2
3
�� ,
−= 2
3
�� ,
−= 2
3
�� ,

42 American Scientific Journal № ( 36 ) / 2020
= 2
3
�� ,
�= Ф(��), �+�+�= 0 ,
�= 1
√2[(−)2+(−)2+(−)2+62]12 ,
��= √2
3[(�−�)2+(�−�)2+(�−�)2+6�2]12 . (1)
Denote Ф(��)= 3Ψ (��)��. Relations (1) will take
the form
−= 2(��)� , −= 2(��)� ,
−= 2(��)� , = 2(��)�.
(2)
We will look for the solution in the form of series
by degrees of the parameter �
��= ∑ ��
∞
�=0
��� , ��= ∑ ��
∞
�=0
��� .
Assume that a simple stretch occurs in the initial
state
For the function Ψ(��) decomposition takes place
(��)= (0)(��0)+��(0)
��� ��′+�2[1
2
�2(0)
���2 (��′)2+�(0)
��� ��′′]+⋯
where are the function values Ψ and its derivatives
are taken in the original state when ��= ��0.
The linearized relations (2) in the first
approximation have the form
′= ′+2��′+(�+�)�′ ,
′= ′+2��′+(�+�)�′ , (3)
′= ′+2��′ ,
′ = 2��′ ,
where ��0= �0, ��′= �′ ,
�= (��0),�−�= � (�0)
�� �0.
We satisfy the continuity equation by assuming
�′= 1
�
′
� , �′= −1
�
′
,
(4)
For components of the strain tensor , we obtain
expressions
0= ����� , 0= 0= 0 = 0 ,
�0= �0= −1
2�0 , �0=0 .

American Scientific Journal № ( 36 ) / 2020 43

�′= �′
� = 1
�2
′
−1
� 2′
� ,�′= �′
� = − 1
�2
′
,
�′= �′
= 1
� 2′
� ,
�′= 1
2(�′
+�′
� )= 1
2(−1
� 2′
2− 1
�2
′
� +1
�
2′
�2).(5)
Relations (3) will take the form
′= ′+2�(1
�2
′
−1
� 2′
� )+(�−�)1
� 2′
� ,
′= ′+2�(− 1
�2
′
)+(�−�)1
r ∂2′
∂z∂r ,
′= ′+2�1
�
∂2′
∂z∂r ,
′ = 2�1
2 (−1
� 2′
2− 1
�2
′
� +1
�
2′
�2).(6)
The equilibrium equations have the form
′
� +′
+′−′
� = 0 ,



(7)
′
� +′
+′
� = 0 .


Using the relations (6) and the equilibrium
equations (7), we obtain an equation for determining
the function ′
4′
�4−2
�
3′
�3+ 3
�2
2′
�2− 3
�3
′
� +
+(3�−�)
�
4′
�22−(3�−�)
�
1
�
3′
2� +4′
4= 0.(8)
In [5, 6], the solution of equation (8) is considered
in polynomials. In this paper, we consider the solution
of equation (8) as
′(�,)= (�)∙sin � . (9)
Then from (8) and (9), we get
�4
��4−2
�
�3
��3+(3
�2−3�−�
� �2)�2
��2+(− 3
�3+3�−�
�
�2
�)�
�� +�4= 0 (10 )
Equation (10) can be written as follows
(�2
�� 2−1
�
�
�� +�2)(�2
�� 2−1
�
�
�� +�̅2)= 0 ,(11 )

44 American Scientific Journal № ( 36 ) / 2020
where
�2= �2
2�[(�−3�)+�√3(�−�)(�+3�) ],
�̅2= �2
2�[(�−3�)−�√3(�−�)(�+3�) ].(12 )
The solution of equation (11) is the sum of the
solutions of the equations
�21
��2−1
�
�1
�� +�21= 0,�22
��2−1
�
�2
�� +�̅22= 0.(13 )
Substitutions
1(�)= ��1(�� ),2= ��2(�̅�) (14 )
from (11) we get the equations
�2�2�1′′+�� �1′+(�2�2−1)�1= 0,
�̅2�2�2′′+�̅��2′+(�̅2�2−1)�2= 0, (15)
The integrals of equations (15) are linear
combinations of first -order Bessel and Neumann
functions. Neumann functions turn to infinity when �=
0 and they should be absent from the solution . In order
to ′(�,) did not contain the alleged members take in
her expression , an arbitrary constant paired, that is,
believe
′(�,)= �[�̅�1(�� )+��̅1(�̅�)]sin � ,(16 )
where 1(�� ) – the Bessel function of the first
order .
According to (4), (5), (6) and (9) we get
�′= −�
�∙��� � ,�′= 1
�
�
�� ��� � ,
�′= �
�[1
�− �
�� ]��� � ,�′= − �
�2��� � ,
�′= �
�
�
�� ��� �,�′= 1
2�[�2
��2−1
�
�
�� +�2]��� � ,
′= [� 1
��
�3
��3−� 1
��2
�2
��2+(� 1
��3−3��
�)�
�� +2� �
�2]��� � ,
′= {� 1
��
�3
��3−� 1
��2
�2
��2+[(2�−3�)�
�+� 1
��3]�
�� −2� �
�2}��� � ,
′= [� 1
��
�3
��3−� 1
��2
�2
��2+�(�
�+ 1
��3)�
�� ]��� � ,
′ = �1
�(�2
��2−1
�
�
�� +�2)sin � .(17 )
Add up the equations (13)
(�21
��2+�22
��2)−1
�(�1
�� +�2
�� )= −�21−�̅22 ,
�2
��2−1
�
�
�� = − �2
2�[(�−3�)1+�√3(�−�)(�+3�)1]+
+(�−3�)2− �√3(�−�)(�+3�)2 , (18)

American Scientific Journal № ( 36 ) / 2020 45

where = 1+2.
Equation (18) is written as
�2
��2−1
�
�
�� = −�2(��∙1+�−�∙2) ,(19 )
Where
�= ����� √3(�−�)(�+3�)
(�−3�) .
Differentiating equality (13) and adding, we get


�3
��3−1
�
�2
��2+ 1
�2
�
�� = −�2(���1
�� +�−��2
�� ).(20 )

In view of (19), (20), relations (16) take the form
�′= −�
�(1+2)��� � ,�′= 1
�(�1
�� +�2
�� )��� � ,
�′= �
�[1
�(1+2)−(�1
�� +�2
�� )]��� � ,�′= −�(1+2)��� � ,
�′= �
�(�1
�� +�2
�� )��� � ,
�′= �2
2�[(1−��)1+(1−�−�)2]��� � ,
′= �
�[2�1
�(1+2)−(3�+���)�1
�� −(3�+��−�)�2
�� ]��� � ,
′= �
�{[(2�−3�)−���]�1
�� +[(2�−3�)−��−�]�2
�� −
−2�1
�(1+2)}��� � ,
′= ��
�[(1−��)�1
�� +(1−�−�)�2
�� ]��� � ,
′ = ��2
�((1−��)1+(1−��)2)sin � ,(21 )

where 1= �̅��1(�� ),2= ��̅�1(�̅�).
�1
�� = �̅�2�0(�� ),�2
�� = ��̅2�0(�̅�)
where 0(�� ) – Bessel function of zero order.
Expressions (21) take the form
�′= −� [�̅1(�� )+�1(�̅�)]��� � ,
�′= [�̅�0(�� )+��̅0(�̅�)]��� � ,
�′= �
�{[�̅1(�� )+�1(�̅�)−�[�̅�0(�� )+��̅0(�̅�)]]}��� � ,
�′= −�� [�̅1(�� )+�1(�̅�)]��� � ,
�′= �[�̅�0(�� )+��̅0(�̅�)]��� � ,
�′= �2
2[(1−��)�̅�1(�� )+(1−�−�)�1(�̅�)]��� � ,

46 American Scientific Journal № ( 36 ) / 2020
′= �
�[2�[�̅1(�� )+�1(�̅�)]−�(3�+���)�̅�0(�� )+(3�+��−�)��̅0(�̅�)]��� � ,
′= �
�[−2�[�̅1(�� )+�1(�̅�)]+[(2�−3�)−���]�̅�0(�� )+
+[(2�−3�)−��−�]��̅0(�̅�)]��� � ,
′= �� [(1−��)�̅�0(�� )+(1−��)��̅0(�̅�)]��� � ,
′ = �2�((1−��)�̅1(�� )+(1−�−�)�1(�̅�))sin � ,(22 )

Consider the tension of an infinitely long
cylindrical rod of variable cross section. The equation
of the surface of the rod is presented in the form

�= �+�� (), (23)

where �= ����� ,�− small parameter ( �≪ 1). The cylinder stretches along the axis , the side
surface is free of stress. We write the boundary
conditions on the surface in the form
��� (�� )+��� (� )= 0,
cos (�� )+cos (� )= 0, (24)

where �− normal to the surface,
,, − stress components in a cylindrical
coordinate system � .
The displacement of the points of the cylinder
occurs in the meridional planes, we put

�= �(�,),�= 0,�= �(�,), (25)

where �,�,�−are the components of
displacement along the axes � .
The linearized boundary condition s (24) have the
form

′= 0,′−0��
� = 0,�= � (26)

We satisfy expressions (22) by conditions (26)
2�[�̅1(�� )+�1(�̅�)]−�[(3�+���)�̅�0(�� )+(3�+��−�)��̅0(�̅�)]= 0,
2��̅1(�� )+2�� 1(�̅�)−(3�+���)�̅�� 0(�� )+(3�+��−�)��̅�0(�̅�)= 0.(27 )

From the first condition we find

�̅= (3�+��−�)�̅�0(�̅�)−2�1(�̅�)
2�1(�� )−(3�+���)�� 0(�� )� (28 )

Then from the second condition we find
�= �������
0 {(1−��)[(3�+��−�)�̅�0(�̅�)−2�1(��̅̅̅̅)]1(�� )+
+(1−�−�)[2�1(�� )−(3�+���)�� 0(�� )]1(�̅�)
2�1(�� )−(3�+���)��0(�� ) } (29 )

Putting �= � ��� ,�= ����� , we determine the
desired solution from expressions (22), where the value
of the coefficient � is found from relation (29) and the
coefficient �̅ from relation (28).

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in the theory of elastic -plastic body. Moscow: Nauka,
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[3] Ishlinsky A. Y., Ivl ev D. D. Mathematical

American Scientific Journal № ( 36 ) / 2020 47

theory of plasticity. Moscow: Fizmatlit, 2001. 704 p.
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[4] Ivlev D. D., Mikhailova M. V.,Petrov N. I. On
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