An Asymptotic Linear Thin-Walled Rod Model Coupling Twist and Bending
A linear one-dimensional model for thin-walled rods with open strongly curved cross-section, obtained by asymptotic methods is presented. A dimensional analysis of the linear three-dimensional equilibrium equations lets appear dimensionless numbers which reflect the geometry of the structure an...
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| Date: | 2010 |
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| Language: | English |
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Інститут механіки ім. С.П. Тимошенка НАН України
2010
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| Cite this: | An Asymptotic Linear Thin-Walled Rod Model Coupling Twist and Bending / A. Hamdouni, O. Millet // Прикладная механика. — 2010. — Т. 46, № 9. — С. 123-143. — Бібліогр.: 48 назв. — анг. |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine| _version_ | 1859596170358161408 |
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| author | Hamdouni, A. Millet, O. |
| author_facet | Hamdouni, A. Millet, O. |
| citation_txt | An Asymptotic Linear Thin-Walled Rod Model Coupling Twist and Bending / A. Hamdouni, O. Millet // Прикладная механика. — 2010. — Т. 46, № 9. — С. 123-143. — Бібліогр.: 48 назв. — анг. |
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| description | A linear one-dimensional model for thin-walled rods with open strongly
curved cross-section, obtained by asymptotic methods is presented. A dimensional analysis of the
linear three-dimensional equilibrium equations lets appear dimensionless numbers which reflect the
geometry of the structure and the level of applied forces. For a given force level, the order of
magnitude of the displacements and the corresponding one-dimensional model are deduced by
asymptotic expansions. In the case of low force levels, we obtain a one dimensional model whose
kinematics, traction and twist equations correspond to Vlassov ones. However this model couples
twist and bending effects in the bending equations, at the difference from Vlassov model where the
twist angle and the bending
displacement are uncoupled.
Запропоновано отриману асимптотичним методом одновимірну модель для тонкостінного стержня з відкритим сильно скривленим поперечним перерізом, яка враховує
взаємо- зв’язок між скручуванням та згином. За допомогою аналізу розмірностей в лінійних
тривимірних
рівняннях рівноваги знайдено безрозмірні величини, які характеризують геометрію стержня та рівень
прикладених сил. Для заданого рівня сил методом асимптотичного розкладу отримані порядок змі-
щень та відповідна одновимірна модель. У випадку низького рівня сил отримано одновимірну модель,
кінематичні рівняння, рівняння кручення та згину відповідають моделі Власова. Однак ця модель враховує в рівняннях згину взаємодію між згином і крученням на відміну від моделі Власова,
яка таку взаємодію не враховує.
Запропоновано отриману асимптотичним методом одновимірну модель для тонкостінного стержня з відкритим сильно скривленим поперечним перерізом, яка враховує
взаємо- зв’язок між скручуванням та згином. За допомогою аналізу розмірностей в лінійних
тривимірних
рівняннях рівноваги знайдено безрозмірні величини, які характеризують геометрію стержня та рівень
прикладених сил. Для заданого рівня сил методом асимптотичного розкладу отримані порядок зміщень та відповідна одновимірна модель. У випадку низького рівня сил отримано одновимірну модель,
кінематичні рівняння, рівняння кручення та згину відповідають моделі Власова. Однак ця модель враховує в рівняннях згину взаємодію між згином і крученням на відміну від моделі Власова,
яка таку взаємодію не враховує.
|
| first_indexed | 2025-11-27T21:13:09Z |
| format | Article |
| fulltext |
2010 П РИКЛАДНАЯ МЕХАНИКА Том 46, № 9
ISSN0032–8243. Прикл. механика, 2010, 46, №9 123
A . H a m d o u n i , O . M i l l e t
AN ASYMPTOTIC LINEAR THIN-WALLED ROD MODEL COUPLING
TWIST AND BENDING
LEPTIAB, Université de La Rochelle, France
e-mail oliver.millet@univ-lr.fr
Abstract. A linear one-dimensional model for thin-walled rods with open strongly
curved cross-section, obtained by asymptotic methods is presented. A dimensional analysis
of the linear three-dimensional equilibrium equations lets appear dimensionless numbers
which reflect the geometry of the structure and the level of applied forces. For a given force
level, the order of magnitude of the displacements and the corresponding one-dimensional
model are deduced by asymptotic expansions. In the case of low force levels, we obtain a
one dimensional model whose kinematics, traction and twist equations correspond to
Vlassov ones. However this model couples twist and bending effects in the bending
equations, at the difference from Vlassov model where the twist angle and the bending
displacement are uncoupled.
Key words: thin-walled rod model, linear elasticity, asymptotic methods.
1. Introduction.
Thin and thin-walled structures (plates, shells, rods and thin-walled rods) are widely
used in industry because they provide a maximum of stiffness with a minimum of weight.
However there exists many different models in the literature. Therefore engineers must
know a priori their respective domain of validity and what model to use in function of the
given data of the problem (geometry of the structure, applied loads, boundary conditions).
Classical models (Kirchhoff – Love, Koiter, Bernouilli, Vlassov...) are generally
obtained from three-dimensional equilibrium equations by making a priori (kinematic and
static) assumptions on the unknowns of the problem. Therefore, the domain of validity of these
classical models with respect to the given data of the problem is difficult to specify rigorously.
Asymptotic methods enable to deduce rigorously plate, shell and rod models from the
three-dimensional equations without making any priori assumption. In linear plate and shell
theory, since the pioneering work of Goldenveizer [11], there exists a large literature on the
subject [2, 7, 38 – 41].
In linear theory of rods, the first works on the subject are due to Rigolot [33]. More
recently, other justifications of linear and non linear rod models by asymptotic expansion
have been developed in [3, 20 – 22, 42]. Let us also cite the synthesis [46] of previous works
[44, 45] which recall the different possible approaches in linear theory of elastic rods
(displacement formulation, mixed formulation in stress-displacements).
These results then have been extended to thin-walled rods. The approach used is based
on the asymptotic behavior of Poisson equation in a thin domain when the thickness tends
towards zero [34, 35, 46]. This way, Rodriguez and Viaño [36] have justified a linear elastic
model of Vlassov for thin-walled rod by asymptotic method similar to Vlassov one.
However their approach uses "a priori" scaling assumptions on the displacement field which
is an unknown of the problem. Moreover, it is based on an expansion at the second order of
the equations with respect to the diameter ε and then the relative thickness η is assumed to
tend towards zero. These two operations do not a priori commute and the result depends on
the choice made (see fig. 1).
124
Fig. 1
Existing asymptotic approaches.
This is a classical result well known for multi-scales asymptotic approaches. It is
encountered in shell theory (with the relative thickness and the shallowness as small
parameters), in homogenization of composite or periodic structures [1, 9, 19].
We propose in this paper to use the constructive approach based on asymptotic
expansions, already developed by the authors for plates [24 – 31], shells [8, 16, 17] and thin-
walled rods [12, 13, 18], to deduce a linear model for thin-walled rod from three-
dimensional equations. The approach used is based on a decomposition of the three-
dimensional equations on Frenet basis of the initial configuration. Then a dimensional
analysis of equilibrium equations lets appear pertinent dimensionless numbers
characterizing the geometry and the applied loads. These numbers are measurable and
enable to define the domain of validity of the obtained model. Thus the order of magnitude
of the displacements and the corresponding asymptotic model are directly deduced from the
level of applied forces. This constitutes the constructive character of our approach.
In this paper we limit our analysis to thin-walled rods with strongly curved profile
subjected to low force levels. In lemma 1, we begin with deducing the order of magnitude of
the displacements from the level of applied forces. Then the asymptotic expansion of
equations leads to the kinematics and to the one-dimensional equilibrium equations of
results 1 to 4. The kinematics and the one-dimensional traction and twist equations
correspond exactly to Vlassov ones [47]. However, whereas Vlassov theory relies on a priori
physical assumptions, in the approach developed here the unknowns of the problem are
directly deduced from the three-dimensional equations.
On the other hand, the one-dimensional bending equations obtained in result 4 differ
from Vlassov ones. They involve a supplementary term coupling bending and torsion
effects, whereas they are uncoupled in Vlassov model. (Such a limitation of Vlassov theory
already have been noticed by other authors [5, 6, 23, 43]).
We recall that in linear elastic theory, the thin-walled rods possess the following
particular property: an external bending loading whose resultant induces a torque, will
generally induce not only a bending displacement but also a twist. In contrary, a torque will
induce only a twist, but no bending, at the difference from the model obtained in this paper.
That is why we call it model "with coupling between twist and bending". However, let us
notice that such a coupling between twist and bending effects exists in the models used for
flexural-torsional buckling or in dynamics models for flexural and torsional vibration
analysis (see for example [14, 15, 32, 37, 48]), but not for classical linear elastic analysis.
2. The three-dimensional problem.
We assume once and for all that an origin O and an orthonormal basis 1 2 3( , , )e e e have
been chosen in 3
R . We index by a star *( ) all dimensional variables and the variables
without a star will denote dimensionless variables. Let *
ω be an open cylindrical surface of
3
R , 3( )Oe its axis, whose length is L and diameter d . We note
*
g
γ and *
d
γ its lateral
boundary, * *
1 = {0}γ ω × and * *
2 = { }Lγ ω × its extremities.
125
Fig. 2
Scheme of the rod and of the profile in the plane of a section
Let us consider now a thin-walled rod with open cross-section and 2h thickness, whose
middle surface is *
ω . The thin-walled rod occupies the set
* *
= [ , ]h hωΩ × − of 3
R in its
reference configuration. We call * *
1 1= ] , [h hγΓ × − and * *
2 2= ] , [h hγΓ × − the extreme faces,
* *
= ] , [
g g
h hγΓ × − and * *= ] , [
d d
h hγΓ × − the lateral faces, * *= { }hω
±
Γ × ± the upper and
lower faces. Let *
M be a generic point of the beam. We decompose the vector
*
OM
uuuuur
as
follows:
* * * * * * *
3 3=OM x e G C C m r n+ + +
uuuuur uuuuuur uuuuuur
, (1)
where *
3x is the coordinate of the current cross-section containing *
M on the axis *
3(0 )x ,
*
G the point of intersection between the axis *
3(0 )x and the current cross-section, *
C an
arbitrary chosen point in the plane of the cross-section (see Fig. 2) located by its cartesian
coordinates * *
1 2( , )c c
x x , and *
r the thickness variable. We call *
C the intersection curve
between *
ω and the cross-section. The orthogonal projection *
m of *
M on the middle
surface is located by its cartesian coordinates * * *
1 2= ( , )x x x or by its curvilinear abscissa *
s
along *
C . The origin *
0s of the curvilinear abscissa is an arbitrary chosen point of *
C . We
note n the unit normal and t the unit tangent vector of *
C . Moreover, we call *
l and *
q
the coordinates of the vector
* *
C m
uuuuuur
in the basis ( , )t n . Finally, we call *
α the angle 1( , )e t
and *
c the curvature of the curve *
C (see fig. 2).
In what follows, we consider only thin-walled rods such as << 1
d
L
, << 1
h
d
and
*|| || << 1h c
∞
. We assume that the rod is subjected to the applied body forces
* * * * * 3
3 3= :
t n
f f t f n f e+ + Ω → R and to the applied surface forces
* * * * * 3
3 3 0= :
t n
g g t g n g e
± ± ± ±
±
+ + Γ → R . Moreover, the rod is assumed to be clamped on its
extremities *
1Γ and *
2Γ , and free on its lateral faces
*
g
Γ and *
d
Γ . The unknown of the
problem is then the displacement * * 3:U Ω → R . Within the framework of linear elasticity,
the displacement U* and the Cauchy stress tensor *
σ satisfy the linear equilibrium
equations:
126
* * * *
* *
0,1
* * *
* *
,
= in ;
= 0 on ;
. = on ;
. = 0 on ,
g d
Div f
U
N g
T
σ
σ
σ
±
±
− Ω
Γ
Γ
Γ
(2)
where N and T denote the unit outward normal vector to the upper and lower faces and to
the lateral extremities respectively. Within the framework of linear elasticity, the
constitutive law of the Hookean material considered writes * * *= ( ) 2Tr e I eσ λ µ+ , where
* *
*
* *
1
=
2
U U
e
M M
∂ ∂
+
∂ ∂
denotes the linear strain tensor, λ and µ denote the Lamé constants
of the material, and the overbar the transposition operator. Finally the boundary conditions
on
* *
g d
Γ ∪ Γ are considered on average upon the thickness, in order the twist to be of the
same order as the bending in the asymptotic model obtained.
3. Dimensional analysis of equilibrium equations and reduction to a one-scale
problem.
First, we decompose the equations such as to separate the axial components from the
components in the plane of the cross-section. To do this, let us decompose *
U on Frenet
basis 3( , , )t n e of the initial configuration as follows:
* * * *
3 3=
t n
U u t u n u e+ + . (3)
Then the gradient of the vector *
U can be decomposed in the basis 3( , , )t n e on the
following form:
* * *
* * *
* * *
3
* * *
* * ** * * *
* * ** * *
3* * * *
3
* * *
* 3 3 3
* * *
3
= =
t t t
n
n n n
t
u u u
k c u
s r x
U U U
k c UU u u u
k c us r x
M s r x
u u u
k
s r x
∂ ∂ ∂
−
∂ ∂ ∂
∂ ∂ ∂
+ Λ ∂ ∂ ∂ ∂
+∂ ∂ ∂
∂ ∂ ∂ ∂
∂ ∂ ∂
∂ ∂ ∂
,
where *
* *
1
=
1
k
r c−
and where
0 1
= 1 0
−
Λ
denotes the two-dimensional matrix of the
wedge product. In the same way , the three-dimensional equilibrium equations can be
decomposed in the basis 3( , , )t n e and writes in *
Ω :
* * *
* * * *3
* * *
3
2 = ;tn tt t
tn t
k c f
r s x
σ σ σ
σ
∂ ∂ ∂
+ − + −
∂ ∂ ∂
* * *
* * * * * *3
* * *
3
= ;nn tn n
tt nn n
k c c f
r s x
σ σ σ
σ σ
∂ ∂ ∂
+ + − + −
∂ ∂ ∂
* * *
* * * *3 3 33
3 3* * *
3
= .n t
n
k c f
r s x
σ σ σ
σ
∂ ∂ ∂
+ − + −
∂ ∂ ∂
127
The detailed expression of the components of *
σ will be given directly in their
dimensionless form (4). The associated boundary conditions on the upper and lower faces
*
±
Γ are given by:
* * * * * *
3 3= , = , =
tn t nn n n
g g gσ σ σ
± ± ±
and the boundary conditions on the lateral extremities
*
,g d
Γ reduce to:
* * *
3= 0, = 0, = 0.
tt tn t
σ σ σ
It is important to notice that a boundary layer with respect to the shear stress *
3tσ
appears on the free lateral extremities. This is a classical phenomenon in plate and shell
theory. In order to avoid this boundary layer which is not the subject of this paper, we relax
the boundary conditions on
*
,g d
Γ as follows: *
3 = 0
h
t
h
drσ
−
∫ .
3.1. Dimensional analysis of equations. Let us define the following dimensionless
physical data and dimensionless unknowns of the problem:
*
= ;t
t
rt
u
u
u
*
= ;n
n
rn
u
u
u
*
3
3
3
= ;
r
u
u
u
*
3
3 = ;
x
x
L
*
= ;
s
s
d
*
= ;
r
r
h
*
= ;
r
c
c
c
t
f =
*
t
rt
f
f
;
n
f =
*
n
rn
f
f
;
3f =
*
3
3r
f
f
;
t
g =
*
t
rt
g
g
;
n
g =
*
n
rn
g
g
; 3g =
*
3
3r
g
g
,
where the variables indexed by ( )
r
are the reference ones. The new variables which appear
(without a star) are dimensionless. To avoid any assumption on the order of magnitude of
the displacement components, the reference scales
tr
u ,
nr
u and 3r
u are firstly assumed to
be equal to h . Thus we a priori allow small displacements in the framework of the theory of
linear elasticity.
In a natural way we introduce *=|| ||
r
c c
∞
which denotes the maximum of curvature
(the smaller radius of curvature) of the middle surface *
ω . As in shell theory, the order of
magnitude of the curvature is a fundamental data in the asymptotic expansion of equations.
Therefore we will have to distinguish the rods with shallow cross profile from the rods with
strongly curved profile.
First the dimensional analysis of the coefficient *
k leads to
1
=
1
r
k
hc rc−
. Setting
=
r
hcν , the assumption of thin walled-rod ensures that 1ν � . We then have the following
expansion 2= 1 ( ) ...k rc rcν ν+ + + .
On the other hand, the dimensional analysis of the stress tensor leads to:
3
3
3 3 33
=
tt tn t
tn nn n
t n
σ σ σ
σ σ σ σ
σ σ σ
with
tt
σ = n
u
r
β
∂
∂
2 3
3
( 2)[1 ( ) ...] t
n
u u
rc rc cu
s x
β ν ν η ν βηε
∂ ∂
+ + + + + − +
∂ ∂
;
nn
σ =
2 3
3
( 2) [1 ( ) ...]n t
n
u u u
rc rc cu
r s x
β β ν ν η ν βηε
∂ ∂ ∂
+ + + + + − +
∂ ∂ ∂
;
33σ =
2 3
3
[1 ( ) ...] ( 2)n t
n
u u u
rc rc cu
r s x
β β ν ν η ν β ηε
∂ ∂ ∂
+ + + + − + +
∂ ∂ ∂
;
128
tn
σ =
2
[1 ( ) ...]t n
t
u u
rc rc cu
r s
ν ν η ν
∂ ∂
+ + + + +
∂ ∂
;
3tσ =
2 3 3
3
3 3
[1 ( ) ...] , =t n
n
u u u u
rc rc
s x r x
ν ν η ηε σ ην
∂ ∂ ∂ ∂
+ + + + +
∂ ∂ ∂ ∂
, (4)
where we set * =σ µσ , =
d
L
ε , =
h
d
η and =
λ
β
µ
. Now let us denote ω the dimensionless
middle surface obtained from *
ω , whose current point will be noted m . Its associated
curvature C is obtained by dimensional analysis of *
C . Then the dimensional analysis of
the three-dimensional linear equilibrium equations leads in = ] 1,1[ωΩ × − to:
tn
r
σ∂
∂
+
2
(1 ( ) ...) 2tt
tn
rc rc c
s
σ
ν ν η ν σ
∂
+ + + −
∂
+
3
3
t
x
σ
ηε
∂
∂
=
t t
F f− ;
nn
r
σ∂
∂
+
2
(1 ( ) ...) tn
tt nn
rc rc c c
s
σ
ν ν η ν σ ν σ
∂
+ + + + −
∂
+
3
3
n
x
σ
ηε
∂
∂
=
n n
F f− ;
3n
r
σ∂
∂
+
2 3
3(1 ( ) ...) t
n
rc rc c
s
σ
ν ν η ν σ
∂
+ + + −
∂
+
33
3x
σ
ην
∂
∂
= 3t
F f− . (5)
The associated boundary conditions on the upper and lower faces
±
Γ become:
3 3 3= , = , =
tn t t nn n n n
G g G g G gσ σ σ
± ± ± . (6)
Therefore, this dimensional analysis naturally reveals the following dimensional
numbers characterizing the thin-walled rod problems in linear elasticity (they are
measurable data of the problem and must be considered as given data):
ε =
d
L
; η =
h
d
; ν =
r
hc ;
t
F = tr
hf
µ
;
n
F = nr
hf
µ
;
3F = 3r
hf
µ
;
t
G = tr
g
µ
;
n
G = nr
g
µ
; 3G = 3r
g
µ
;
i) The shape ratio ε characterizes the inverse of the shooting-pain of the rod. This is a
known parameter of the problem which satisfies 1ε� .
ii) The dimensional number η denotes the ratio between the thickness h of the rod to
the length of its profile. This number is also a data of the problem which satisfies 1η� .
iii) The shape ratio =
r
hcν is the ratio between the thickness to the smaller radius of
curvature of the middle surface ω of the rod. Its is a given geometrical data of the problem.
iv) The force ratios , ( { , ,3})
i i
F G i t n∈ represent respectively the ratio of the resultant
on the thickness of the body forces (respectively of the surface forces) to µ considered as a
reference stress. These numbers only depend on known physical quantities and must be
considered as known data of the problem.
3.2. One-scale assumption. To reduce the problem to a one-scale problem, ε is chosen
as the small reference parameter of the problem. (If not we have multi-scale problems which are
much more complicated. It is not the subject of this paper).
The other dimensional numbers are then linked to ε , or more precisely to the powers
of ε . In a natural way, as in shell theory, we have to distinguish thin-walled rods:
with strongly curved profile where =ν ε ;
with shallow profile where 2=ν ε .
This distinction is fundamental because these two families of thin-walled rods do not
have the same asymptotic behavior.
On the other hand, three cases can be distinguished and studied:
the thick rods where = 1η . This is not the subject of this paper;
129
the thin-walled rods where =η ε . It is the case studied here;
the very thin-walled rods where = p
η ε , > 1p . This case is not studied in this paper.
Finally, the applied loads are an essential given data of the problem. In the framework
of a one-scale asymptotic expansion, the force ratios must be linked also to ε . This is
equivalent to fix the order of magnitude of the applied forces which are given data. In the
case of thin-walled rods with strongly curved profile, we will consider applied forces such
as: 6 6 5
3 3= = , = = , = =
t t n n
F G F G F Gε ε ε .
These force ratios, which characterize the level of applied forces, are chosen in order all
kinds of loading to be involved at the same order in the asymptotic one-dimensional
equilibrium equations.
In the sequel, we shall consider a thin-walled rod with a strongly curved profile
corresponding to = =η ν ε , submitted to force levels such as 6= =
t n
F F ε , 6= =
t n
G G ε
and 5
3 3= =F G ε . The problem then reduces to a dimensionless one-scale problem, which
can be easily written from (5) and (6), using the expressions (4) of the stresses.
4. Asymptotic expansion of equations.
The standard asymptotic technique then proceeds as follows. First we postulate that the
solution 3= ( , , )
t n
U u u u of the problem admits a formal expansion with respect to the
powers of ε :
0 0 0 1 1 1 2 2 2 2
3 3 3 3( , , ) = ( , , ) ( , , ) ( , , ) ...
t n t n t n t n
u u u u u u u u u u u uε ε+ + + . (7)
The expansion of U with respect to ε implies an expansion of the components of the
stresses σ with respect to ε as well. Then we replace
t
u ,
n
u , 3u by their expansions in
equilibrium equations and we equate to zero the factor of the successive powers of ε . This
way we obtain a succession of coupled problems 0P , 1P , 2P … . Its resolution leads to the
search asymptotic one-dimensional model corresponding to the force level considered.
It is important to notice that with the approach developed here, the order of magnitude
of the displacements (which are unknowns of the problem) are directly deduced from the
level of applied forces. In particular, for the force levels considered here, the axial displacement
is one order smaller then the other ones. This is the result of the following lemma:
Lemma 1.
For force levels such as
6= =
t n
F F ε ,
6= =
t n
G G ε and
5
3 3= =F G ε , we have
0
3 = 0u .
The proof of this lemma is rather long and technical. The demarche is similar to the
proof of result 1 and is not detailed here. Hence, for the level forces considered, the
reference scales of the axial displacement 3 =
r
u h is not properly chosen. In order for 3u to
be of the order of one unit, the reference scales of the displacement must satisfy 3 =
r
u hε .
Therefore the new reference scale for the axial displacement *
3u that we have to consider is
3 =
r
u hε . The other reference scales for the tangential and normal displacements
= =
tr nr
u u h stay unchanged.
Remark 1. It is important to notice that this lemma only leads to the right scalings for
the displacements corresponding to the level of applied forces considered. However, it
would have been possible to start directly from these right scalings or reference scales for
the displacements, as it is often made in the literature.
5. The one-dimensional model.
In the last section, we have determined the right reference scales (or equivalently the
order of magnitude) of the displacements corresponding to the force levels considered. In
this section, we perform the asymptotic expansion of equations which leads to the search
one-dimensional model.
130
According to the force levels considered 6= =
t n
F F ε , 6= =
t n
G G ε and 5
3 3= =F G ε ,
the dimensionless equilibrium equations must be written again with 3 =
r
u hε and
= =
tr nr
u u h as reference scales. The dimensionless components of the displacement will
still be noted with ,
t n
u u and 3u . Thus for the level forces considered here, the new
dimensionless equilibrium equations are the same as the previous ones (4) – (6). Only 3u
must be changed into 3uε in the new expressions of the components of the stresses. Then
we assume again that there exists a formal expansion with respect to ε , similar to (7), of the
new dimensionless solution 3( , , )
t n
u u u .
5.1. A Vlassov kinematics. Result 1: For applied force levels such as
6= =
t n
F F ε ,
6= =
t n
G G ε and
5
3 3= =F G ε , the leading term
0 0 0
3( , , )
t n
u u u is a displacement of Vlassov
type which satisfies:
0
t
u% = 0
1 2cos( ) sin( ) ( )c c
u u q sα α+ − Θ ; 0
n
u% = 0
1 2sin( ) cos( ) ( )c c
u u l sα α− + + Θ ;
0
3u% =
0
1 2
3 1 2
3 3 3
c c
du du d
u x x
dx dx dx
ω
Θ
− − − ,
where 3u denotes the axial or traction displacement; 1
c
u and 2
c
u denote the tangential
displacements of the point C ; 0
Θ denotes the angle of rotation around the axis 3( , )C e ; ω
is called the sectorial area defined as follows; =d ds qω − .
Proof: The asymptotic expansion of the new dimensionless equations leads again to
problems 0P , 1P , 2P .…
Problem 0P :
The cancellation of the factor of 0
ε leads to 0P which can be written:
0
0
0 0
= 0
= 0;
= 1
= 0.
= 0
tn
tn
nn nn
r
in for r
r
σ
σ
σ σ
∂
∂
Ω ±
∂
∂
Therefore, we get 0 0=
tn nn
σ σ in Ω which implies that all the components of 0
σ are
equal to zero. Then writing the components of the stresses in terms of displacements, we
obtain
0
= 0t
u
r
∂
∂
and
0
= 0n
u
r
∂
∂
in Ω , or in an equivalent way (in the next, for the simplicity of
the notations, we will adopt the following ones: a function u which depends only of 3( , )s x will be
noted u% ; a function u which depends only on 3( )x will be noted u ):
0 0 0 0
3 3= ( , ), = ( , )
t t n n
u u s x u u s x% % . (8)
Let us now prove that 0 0
3 3 3= ( )u u x .
Problem 1P :
The cancellation of the factor of ε leads to problem 1P which easily implies that
1 1 1
3= = = 0
tn nn n
σ σ σ . Writing the stresses in terms of displacements, we obtain in Ω :
1 0 1=
t t t
u r uψ− +% % ; (9)
1 0 1
=
2
n n n
u r u
β
ψ
β
− +
+
% % ; (10)
131
0 0
3 3=u u% (11)
with
0
t
ψ%
0
0
= ,n
t
u
cu
s
∂
+
∂
%
%
0
0 0
= t
n n
u
cu
s
ψ
∂
−
∂
%
% % . (12)
From the last expressions, the components of the stresses at order 1 reduce to:
1
1 0 01
= (2 ) = 4
2
n
tt n n
u
r
β
σ β β ψ ψ
β
∂ +
+ +
∂ +
% % ; (13)
1
1 0 0
33 = = 2
2
n
n n
u
r
β
σ β βψ ψ
β
∂
+
∂ +
% % ; (14)
0
1 3
3 =
t
u
s
σ
∂
∂
%
. (15)
The boundary conditions on the lateral surfaces at order one for =s s
+
and =
g
s s write
1 = 0
tt
σ and
1 1
31
= 0
t
drσ
−
∫ . (16)
Problem 2P .
The cancellation of the factor of 2
ε leads to problem 2P which reduces in Ω to
2 1
= 0tn tt
r s
σ σ∂ ∂
+
∂ ∂
;
2
1
= 0nn
tt
c
r
σ
σ
∂
+
∂
;
2
3 = 0n
r
σ∂
∂
(17)
with the associated boundary conditions for = 1r ±
2 2 2
3= 0, = 0, = 0
tn nn n
σ σ σ . (18)
Let us integrate equation (17) upon the thickness. With the boundary condition (18), we
obtain
1 1
1
= 0
tt
drσ
−
∫ . Replacing 1
tt
σ with its expression (13) in terms of displacement, we get
0 = 0
n
ψ% . Then, from (9) – (14) we deduce that 1 1 1 1
33= = 0 and =
tt n n
u uσ σ % .
Then problem 2P leads, according to the boundary conditions, to 2 2 2
3= = = 0
tn nn n
σ σ σ .
The last equations are equivalent in terms of displacements to:
0 2 0
2 1 2 2 1 2 1 1
3 3
3
= , = , =
2 2 2
t n
t t t n n n
r u
u r u u r u u r u
s x
β ψ β
ψ ψ
β β
∂ ∂
− + − + − +
+ ∂ + ∂
% %
% %% % % (19)
with
1 1
1 1 1 1
= and =n t
t t n n
u u
cu cu
s s
ψ ψ
∂ ∂
+ −
∂ ∂
% %
% %% % .
From the last expressions of the displacements, we obtain the following expressions of
the components of the stresses 2
σ at order two:
0
2 11 1
= 4 4
( 2) 2
t
tt n
r
s
β ψ β
σ ψ
β β
+ ∂ +
− +
+ ∂ +
%
% ; (20)
0
2 1
33 = 2 2
( 2) 2
t
n
r
s
β ψ β
σ ψ
β β
∂
− +
+ ∂ +
%
% ; (21)
0 0
2 3
3
3
= t
t
u u
s x
σ
∂ ∂
+
∂ ∂
% %
. (22)
The associated boundary conditions on the lateral surface =s s
−
and =s s
+
write
2 = 0
tt
σ and
132
1 2
31
= 0
t
drσ
−
∫ . (23)
That leads, in terms of displacements, to 1 1( ,0) = ( ,0) = 0
n n
s sψ ψ
− +
% % and
0 0
( ) = ( ) = 0t t
s s
s s
ψ ψ
− +
∂ ∂
∂ ∂
% %
;
0 0 0 0
3 3
3 3
( ) = ( ) = 0t t
u u u u
s s
s x s x
− +
∂ ∂ ∂ ∂
+ +
∂ ∂ ∂ ∂
% % % %
. (24)
Problem 3P .
The cancellation of the factor of 3
ε leads to problem 3P which reduces in Ω to:
3 2
= 0tn tt
r s
σ σ∂ ∂
+
∂ ∂
; (25)
3
2
= 0nn
tt
c
r
σ
σ
∂
+
∂
; (26)
3 2
3 3 = 0n t
r s
σ σ∂ ∂
+
∂ ∂
(27)
with the associated boundary conditions for = 1r ±
3 = 0
tn
σ ; 3 = 0
nn
σ ; (28)
3
3 = 0
n
σ . (29)
As previously, let us integrate equation (26) upon the thickness. With the boundary
condition (28), we obtain
1 2
1
= 0
tt
drσ
−
∫ . According the the expression (20) of 2
tt
σ , we get
1 = 0
n
ψ% . Thus expressions (20) and (21) of 2
tt
σ and 2
33σ reduce to:
0 0
2 2
33
1
= 4 ; = 2
( 2) ( 2)
t t
tt
r r
s s
β ψ β ψ
σ σ
β β
+ ∂ ∂
− −
+ ∂ + ∂
% %
. (30)
In the same way, we shall now integrate (27) upon the thickness. Using (29), and then
(23) and (22), we obtain:
0 0
3
3
= 0t
u u
s x
∂ ∂
+
∂ ∂
% %
, (31)
which is nothing else than the non-distorsion Vlassov assumption obtained for the leading
term of the expansion of the displacement. Using the previous results obtained, the
expressions of the stresses at order three reduce to
2 0 2 0 2
3 3 3
32
1 1 1 1
= 4 ; = 4 ; = 0
2 2 2 2
t t
tn nn n
r r
c
ss
β ψ β ψ
σ σ σ
β β
+ ∂ − + ∂ −
+ + ∂∂
% %
. (32)
On the other hand, according to (32), the boundary conditions at order three
3 3
3 3( , ) = ( , ) = 0
tn tn
s x s xσ σ
− +
, leads in terms of displacements to:
2 0 2 0
3 32 2
( , ) = ( , ) = 0t t
s x s x
s s
ψ ψ
− +
∂ ∂
∂ ∂
% %
. (33)
Problem 4P .
The cancellation of the factor of 4
ε leads to problem 4P which reduces in Ω to:
4 3 2
3
2 = 0tn tt tt
tn
c rc
r s s
σ σ σ
σ
∂ ∂ ∂
+ − +
∂ ∂ ∂
; (34)
4 3
3 3 2 2
= 0nn tn
tt nn tt
c c rc
r s
σ σ
σ σ σ
∂ ∂
+ + − +
∂ ∂
;
4 3 2
3 3 33
3
= 0n t
r s x
σ σ σ∂ ∂ ∂
+ +
∂ ∂ ∂
, (35)
with the boundary conditions for = 1r ±
133
4 = 0
tn
σ ; (36)
4 = 0
nn
σ ; (37)
4
3 = 0
n
σ . (38)
Using the boundary conditions (36) and (37), an integration of equations (34) and (35)
upon the thickness lead to:
3 2
1 3
1
2 = 0tt tt
tn
c rc dr
s s
σ σ
σ
−
∂ ∂
− +
∂ ∂
∫ ; (39)
3
1 3 3 2 2
1
= 0tn
tt nn tt
c c rc dr dr
s
σ
σ σ σ
−
∂
+ − +
∂
∫ . (40)
In the same way, after multiplying equations (25) and (26) with rc , the integration upon
the thickness leads to
2
1 13
1 1
= tt
tn
c dr rc dr
s
σ
σ
− −
∂
∂
∫ ∫ ; (41)
1 13 2 2
1 1
=
nn tt
c dr rc drσ σ
− −
∫ ∫ . (42)
We then use (41) [respectively (42)] to simplify (39)[respectively (40)] which reduce to
3
1 3
1
= 0tt
tn
c dr
s
σ
σ
−
∂
−
∂
∫ ; (43)
3
1 3
1
= 0tn
tt
c dr
s
σ
σ
−
∂
+
∂
∫ . (44)
On the other hand, let us derive (44) with respect to s . We have:
2 3 3
1 3
21
= 0tn tt
tt
dc
c dr
ds ss
σ σ
σ
−
∂ ∂
+ +
∂∂
∫ . (45)
Now using (43) and (44) to eliminate 3
tt
σ in (45), we obtain according to (32):
4 0 3 0 2 0
2
4 3 2
1
= 0t t t
dc
c
c dss s s
ψ ψ ψ∂ ∂ ∂
− +
∂ ∂ ∂
% % %
, whose general solution is given by
2 0
2
= ( ) ( )t
Acos Bsin
s
ψ
α α
∂
+
∂
%
, with ( ) =
d
c s
ds
α
. Using the boundary conditions (33) and (24),
we obtain
0
= 0t
s
ψ∂
∂
%
or equivalently 0
3= ( )
t
xψ Θ% . Therefore the tangential displacements are
solution of the following differential system:
0
0 0
0
=
n
t
ψ
ψ
=
Θ
%
%
= 0 ⇒
0
0
0
0 0
= 0;
.
t
n
n
t
u
cu
s
u
cu
s
∂
−
∂
∂
+ = Θ
∂
%
%
%
%
In a Cartesian basis, we get after a few calculations
0 0 0 0
1 1 2 2 2 2 1 1= ( ) = ( )c c c c
u u x x and u u x x− − Θ + − Θ% % ,
where 1
c
u and 2
c
u represents at the leading order the displacements of the arbitrary point C
in the directions 1e and 2e . The angle 0
Θ characterizes the rotation of the section around
the axis 3( , )C e . The point C is generally identified to the shear center of the sections. In
the basis (t,n), we then have:
134
0 0
1 2
0 0
1 2
= cos( ) sin( ) ( ) ;
= sin( ) cos( ) ( ) ;
c c
t
c c
n
u u u q s
u u u l s
α α
α α
+ − Θ
− + + Θ
%
%
with
1 1 2 2
1 1 2 2
( ) = ( )cos ( )sin ;
( ) = ( )sin ( )cos .
c c
c c
l s x x x x
q s x x x x
α α
α α
− + −
− − + −
This last expression characterizes a rigid displacement in the plane of the sections and is
similar to Vlassov kinematics. (Excepted for the sign of q in the expression of
0
t
u% . This is due to
an orientation of the normal n opposite to Vlassov one). Moreover, the axial displacement 0
3u
can be determined from (31). We obtain the expresion of 0
3u% of result 1.
5.2. Traction equation. Result 2: For applied level forces such as 6= =
t n
F F ε ,
6= =
t n
G G ε and 5
3 3= =F G ε , the leading terms of the displacements 3u , 0
Θ , 1
c
u and 2
c
u
satisfy the following traction equation:
2
3
2
3
d u
ES
dx
3
1
1 3
3
c
d u
ES
dx
−
3
2
2 3
3
c
d u
ES
dx
− –
3 0
3
3
d
ES
dx
ω
Θ
− 3= Pµ− , where E and µ are respectively Young modulus and Lamé coefficient
of the material, and where:
S =
1
1
s
s
drds
+
−
−
∫ ∫ ; S
ω
=
1
1
s
s
drdsω
+
−
−
∫ ∫ ; 1S
1
11
=
s
s
x drds
+
−
−
∫ ∫ ;
2S =
1
21
s
s
x drds
+
−
−
∫ ∫ ; 3P =
1
3 3 31
[ ]
s s
s s
f drds g g ds
+ −+ +
−
− −
+ −∫ ∫ ∫ .
Proof: We just proved that 0 0=
t
ψ Θ% . So we have 2 2
33= = 0
tt
σ σ and 3 3= = 0
tn nn
σ σ , that
leads to the following expressions of the stresses at order three:
1 0
3 2 3
3
1 1
= 4 4 2
( 2) 2 2
t
tt n
u
r
s x
β ψ β β
σ ψ
β β β
+ ∂ + ∂
− + +
+ ∂ + + ∂
% %
% ; (46)
1 0
3 2 3
33
3
1
= 2 2 4
( 2) 2 2
t
n
u
r
s x
β ψ β β
σ ψ
β β β
∂ + ∂
− + +
+ ∂ + + ∂
% %
% ;
1 1 0
3 3
3
3 3
= 2t
t
u u d
r
s x dx
σ
∂ ∂ Θ
+ −
∂ ∂
% %
. (47)
Problem 4P then reduces in Ω to:
4 3
= 0tn tt
r s
σ σ∂ ∂
+
∂ ∂
;
4
3
= 0nn
tt
c
r
σ
σ
∂
+
∂
; (48)
4 3
3 3 = 0n t
r s
σ σ∂ ∂
+
∂ ∂
. (49)
Using (37), the integration of equation (48) upon the thickness leads to
1 3
1
= 0
tt
drσ
−
∫ .
Then replacing 3
tt
σ with its expression (46), we get
0
2 3
3
1
4 2 = 0
2 2
n
u
x
β β
ψ
β β
+ ∂
+
+ + ∂
%
% . On the
other hand, using (38) the integration of equation (49) upon the thickness leads to
1 3
31
= 0
t
drσ
−
∫ . According to (47), we have equivalently in terms of displacements
1 1
3 3( ) ( ) = 0
t
u s u x∂ ∂ + ∂ ∂% % , and the expressions of the stresses reduce to:
135
1 1 0 0
3 3 33
33 3
3 3
1 3 2
= 4 ; = 2 ; = 2
( 2) ( 2) 1
t t
tt t
u d
r r r
s s x dx
β ψ β ψ β
σ σ σ
β β β
+ ∂ ∂ + ∂ Θ
− − + −
+ ∂ + ∂ + ∂
% % %
. (50)
This last equation leads to 4
3 = 0
n
σ according to (38) and (49).
Problem 5P :
The cancellation of the factor of 5
ε leads to problem 5P which reduces in Ω to
5 4 3 3
4 3
3
2 = 0tn tt tt t
tn
c rc
r s s x
σ σ σ σ
σ
∂ ∂ ∂ ∂
+ − + +
∂ ∂ ∂ ∂
; (51)
5 4
4 4 2 3
= 0nn tn
tt nn tt
c c rc
r s
σ σ
σ σ σ
∂ ∂
+ + − +
∂ ∂
; (52)
5 4 3
3 3 33
3
3
=n t
f
r s x
σ σ σ∂ ∂ ∂
+ + −
∂ ∂ ∂
, (53)
with the boundary conditions for = 1r ±
5 5 5
3 3= 0, = 0, =
tn nn n
gσ σ σ
± . (54)
Using the boundary condition (54), an integration of (53) upon the thickness leads to:
4 3
1 13 33
3 3 31 1
3
= [ ]
s s s
t
s s s
drds f drds g g ds
s x
σ σ
+ −+ + +
− −
− − −
∂ ∂
+ − − −
∂ ∂
∫ ∫ ∫ ∫ ∫ .
Using the boundary condition
1 4
31
= 0
t
drσ
−
∫ on the free lateral surface for =s s
−
et
=s s
+
, we obtain:
3
1 133
3 3 31 1
3
= [ ]
s s s
s s s
drds f drds g g ds
x
σ
+ −+ + +
− −
− − −
∂
− − −
∂
∫ ∫ ∫ ∫ ∫ .
Finally replacing 2
33σ with its expressions (50), we obtain the traction equation of result 2.
5.3. Twist equation. Result 3: For applied forces such as
6= =
t n
F F ε ,
6= =
t n
G G ε
and
5
3 3= =F G ε , the leading terms of the displacement 3u ,
0
Θ , 1
c
u and 2
c
u satisfy the
following twist equation:
3 4 4 4 0 2 0
3 1 2 3
1 23 4 4 4 2
33 3 3 3 3
=
c c
d t
E d u E d u E d u E d d dM
S J J J J M
dxdx dx dx dx dx
ω ω ω ωω ω
µ µ µ µ
Θ Θ
− − − + − − ,
where:
S
ω
=
1
1
s
s
drdsω
+
−
−
∫ ∫ ; J
ωω
=
1 2
1
s
s
drdsω
+
−
−
∫ ∫ ; 1J
ω
=
1
11
s
s
x drdsω
+
−
−
∫ ∫ ; 2J
ω
=
1
21
s
s
x drdsω
+
−
−
∫ ∫ ;
d
J
ω
=
1 2
1
2 (1 )
s
s
r cq drds
+
−
−
−∫ ∫ ; 3M =
1
3 3 31
[ ]
s s
s s
f drds g g dsω ω
+ −+ +
−
− −
+ −∫ ∫ ∫ ,
1 1
1 1
= [ ] [ ] .
s s s s
t n n n t t t
s s s s
M lf drds l g g ds qf drds q g g ds
+ − + −+ + + +
− −
− − − −
+ − − − −∫ ∫ ∫ ∫ ∫ ∫
Proof: Let us follow step by step for equation (51) and (52) the same demarche as for
problem 4P . We can prove in the same way that 1
t
ψ% does not depend on 3x and we set
1 1
3= ( )
t
xψ Θ% . Thus the displacement at order 1 has the same form as the displacement at the
leading order. On the other hand, according to the previous result, problem 5P reduces in Ω to:
5 4 3
3
3
= 0tn tt t
r s x
σ σ σ∂ ∂ ∂
+ +
∂ ∂ ∂
; (55)
5
4
= 0nn
tt
c
r
σ
σ
∂
+
∂
; (56)
136
5 4 3
3 3 33
3
3
=n t
f
r s x
σ σ σ∂ ∂ ∂
+ + −
∂ ∂ ∂
, (57)
with the following expressions of the stresses at order three: 3 = 0
tt
σ and
0
3 3
33
3
3 2
=
1
u
x
β
σ
β
+ ∂
+ ∂
%
; (58)
0
3
3
3
= 2
t
d
r
dx
σ
Θ
− . (59)
Problem 6P :
The cancellation of the factor of 6
ε leads to the following tangential and normal
equations of problem 6P which write in Ω :
6 5 4 4
5 3
3
2 =tn tt tt t
tn t
c rc f
r s s x
σ σ σ σ
σ
∂ ∂ ∂ ∂
+ − + + −
∂ ∂ ∂ ∂
; (60)
6 5
5 5 2 4
=nn tn
tt nn tt n
c c rc f
r s
σ σ
σ σ σ
∂ ∂
+ + − + −
∂ ∂
, (61)
with the boundary conditions for = 1r ±
6 =
tn t
gσ
± ; (62)
6 =
nn n
gσ
± . (63)
Let us integrate equations (60) and (61) upon the thickness. Using the boundary
conditions (62) and (63), we obtain the system:
5 4 4
1 15 3
1 1
3
2 = [ ]tt tt t
tn t t t
c rc dr f dr g g
s s x
σ σ σ
σ
+ −
− −
∂ ∂ ∂
− + + − − −
∂ ∂ ∂
∫ ∫ ; (64)
5
1 15 5 2 4
1 1
= [ ]tn
tt nn tt n n n
c c rc dr f dr g g
s
σ
σ σ σ
+ −
− −
∂
+ − + − − −
∂
∫ ∫ . (65)
Let us now use equations of problem 5P . First multiplying equations (55) and (56) with
rc , we obtain:
5 4 3 5
1 1 2 43
1 1
3
= 0; = 0tn tt t nn
tt
rc rc rc dr rc rc dr
r s x r
σ σ σ σ
σ
− −
∂ ∂ ∂ ∂
+ + +
∂ ∂ ∂ ∂
∫ ∫ .
An integration by parts of the previous equations leads to:
4 3
1 5 3
1
3
= 0tt t
tn
c rc rc dr
s x
σ σ
σ
−
∂ ∂
− + +
∂ ∂
∫ ; (66)
( )
1 5 2 4
1
= 0
nn tt
c rc drσ σ
−
− +∫ . (67)
Then replacing (66) and (67) in (64) and (65) respectively, we get:
5 3 4
1 15 3 3
1 1
3 3
= [ ]tt t t
tn t t t
c rc dr f dr g g
s x x
σ σ σ
σ
+ −
− −
∂ ∂ ∂
− − + − − −
∂ ∂ ∂
∫ ∫ ; (68)
5
1 15
1 1
= [ ]tn
tt n n n
c dr f dr g g
s
σ
σ
+ −
− −
∂
+ − − −
∂
∫ ∫ . (69)
Now we shall multiply (68) with ( )q s and (69) with ( )l s . We obtain:
5 3 4
1 15 3 3
1 1
3 3
= [ ]tt t t
tn t t t
q qc qrc q dr qf dr q g g
s x x
σ σ σ
σ
+ −
− −
∂ ∂ ∂
− − + − − −
∂ ∂ ∂
∫ ∫ ; (70)
137
5
1 15
1 1
= [ ]tn
tt n n n
l lc dr lf dr l g g
s
σ
σ
+ −
− −
∂
+ − − −
∂
∫ ∫ . (71)
Using the following equalities:
5
tt
q
s
σ∂
∂
=
5
5( )
;tt
tt
q q
s s
σ
σ
∂ ∂
−
∂ ∂
5
tn
l
s
σ∂
∂
=
5
5( )
tn
tn
l l
s s
σ
σ
∂ ∂
−
∂ ∂
equations (70) and (71) reduce to
5 3 4
1 15 5 3 3
1 1
3 3
( )
= [ ]tt t t
tt tn t t t
q q
qc qrc q dr qf dr q g g
s s x x
σ σ σ
σ σ
+ −
− −
∂ ∂ ∂ ∂
− − − + − − −
∂ ∂ ∂ ∂
∫ ∫ ;
5
1 15 5
1 1
( )
= [ ]tn
tn tt n n n
l l
lc dr lf dr h g g
s s
σ
σ σ
+ −
− −
∂ ∂
− + − − −
∂ ∂
∫ ∫ .
Now let us integrate the previous equations with respect to s after subtraction. We
obtain
5 5 3 4
1 5 5 3 3
1
3 3
( )
=
s
tt tn t t
tt tn t
s
q l q l
cl cq rcq q drds M
s s s s x x
σ σ σ σ
σ σ
+
−
−
∂ ∂ ∂ ∂ ∂ ∂
− − + + − − +
∂ ∂ ∂ ∂ ∂ ∂
∫ ∫ (72)
where
t
M , whose expression is given in result 3, denotes the twist torque calculated at point
C . To simplify the previous equations, we use on one hand the geometrical properties
= 0
q
cl
s
∂
+
∂
and = 1
l
cq
s
∂
−
∂
, and on the other hand the boundary conditions 5 = 0
tt
σ and
5 = 0
tn
σ on =s s
−
and =s s
+
. Then equation (72) reduces to
3 4
1 5 3 3
1
3 3
=
s
t t
tn t
s
rcq q drds M
x x
σ σ
σ
+
−
−
∂ ∂
− +
∂ ∂
∫ ∫ .
Now we multiply equation (55) by r and integrate it upon a section. We get:
5 4 3
1 3
1
3
= 0
s
tn tt t
s
r r r drds
r s x
σ σ σ
+
−
−
∂ ∂ ∂
+ +
∂ ∂ ∂
∫ ∫ .
Using the boundary condition 4 = 0
tt
σ on =s s
−
and =s s
+
, an integration by part of
the first term leads to:
3
1 15 3
1 1
3
=
s s
t
tn
s s
drds r drds
x
σ
σ
+ +
− −
− −
∂
∂
∫ ∫ ∫ ∫ . (73)
On the other hand, we shall multiply equation (57) with the sectorial area ω and
integrate the result upon a section. We get:
5 4 3
1 13 3 33
31 1
3
=
s s
n t
s s
drds f drds
r s x
σ σ σ
ω ω ω ω
+ +
− −
− −
∂ ∂ ∂
+ + −
∂ ∂ ∂
∫ ∫ ∫ ∫ . (74)
Using the property
4 4 4
4 43 3 3
3 3
( ) ( )
= =t t t
t t
d
q
s s ds s
σ ωσ ω ωσ
ω σ σ
∂ ∂ ∂
− +
∂ ∂ ∂
and the boundary condition (54), we obtain
4 3
1 143 33
3 3 3 31 1
3
( )
= [ ]
s s s
t
t
s s s
q drds f drds g g ds
s x
ωσ σ
σ ω ω ω
+ −+ + +
− −
− − −
∂ ∂
+ + − − −
∂ ∂
∫ ∫ ∫ ∫ ∫ .
With the boundary condition
1 4
31
= 0
t
drσ
−
∫ on =s s
−
and =s s
+
, the last equation
reduces to:
3
1 14 33
3 3 3 31 1
3
= [ ]
s s s
t
s s s
q drds f drds g g ds
x
σ
σ ω ω ω
+ −+ + +
− −
− − −
∂
+ − − −
∂
∫ ∫ ∫ ∫ ∫ .
138
Now let us derive the last equation with respect to 3x . We obtain the relation:
4 2 3
1 13 33 3
21 1
3 33
=
s s
t
s s
dM
q drds drds
x dxx
σ σ
ω
+ +
− −
− −
∂ ∂
− −
∂ ∂
∫ ∫ ∫ ∫ . (75)
where the expression of 3M is given in result 3. To finish let us replace 5
tn
σ and 4
3tσ with
their expressions (73) and (75) in equation (72). We get:
3 2 3
1 3 33 3
21
3 33
(1 ) =
s
t
t
s
dM
cq r drds M
x dxx
σ σ
ω
+
−
−
∂ ∂
− − +
∂ ∂
∫ ∫ .
Finally, replacing 3
3tσ and 3
33σ with their expressions (58) – (59), we obtain the twist
equilibrium equation of result 3.
5.4. Bending equations. Result 4: For force levels such as
6= =
t n
F F ε ,
6= =
t n
G G ε
and
5
3 3= =F G ε , the leading terms of the displacements 3u ,
0
Θ , 1
c
u and 2
c
u are solutions
of the following bending equations:
3 4 4 4 0 2 0
3 1 2 31
1 11 12 1 1 13 4 4 4 2
33 3 3 3 3
=
c c
d
E d u E d u E d u E d d dM
S J J J J P
dxdx dx dx dx dx
ω
µ µ µ µ
Θ Θ
− − − + − − ;
3 4 4 4 0 2 0
3 1 2 32
2 12 22 2 2 23 4 4 4 2
33 3 3 3 3
=
c c
d
E d u E d u E d u E d d dM
S J J J J P
dxdx dx dx dx dx
ω
µ µ µ µ
Θ Θ
− − − + − − ,
where
1S =
1
11
s
s
x drds
+
−
−
∫ ∫ ; 2S =
1
21
s
s
x drds
+
−
−
∫ ∫ ; 11J =
1 2
11
s
s
x drds
+
−
−
∫ ∫ ;
22J =
1 2
21
s
s
x drds
+
−
−
∫ ∫ ; 12J =
1
1 21
s
s
x x drds
+
−
−
∫ ∫ ; 1J
ω
=
1
11
s
s
x drdsω
+
−
−
∫ ∫ ;
2J
ω
=
1
21
s
s
x drdsω
+
−
−
∫ ∫ ; 1d
J =
1 2
1
2 cos
s
s
r c drdsα
+
−
−
∫ ∫ ; 2d
J =
1 2
1
2 sin
s
s
r c drdsα
+
−
−
∫ ∫
and
1P =
1 1
1 1
cos cos [ ] sin sin [ ]
s s s s
t t t n n n
s s s s
f drds g g ds f drds g g dsα α α α
+ − + −+ + + +
− −
− − − −
+ − − − −∫ ∫ ∫ ∫ ∫ ∫ ;
2P =
1 1
1 1
sin sin [ ] cos cos [ ]
s s s s
t t t n n n
s s s s
f drds g g ds f drds g g dsα α α α
+ − + −+ + + +
− −
− − − −
+ − + + −∫ ∫ ∫ ∫ ∫ ∫ ;
31M =
1 1
1 3 1 3 3 32 2 3 2 3 31 1
[ ] ; = [ ] .
s s s s
s s s s
x f drds x g g ds M x f drds x g g ds
+ − + −+ + + +
− −
− − − −
+ − + −∫ ∫ ∫ ∫ ∫ ∫
Proof: Let us start again from equations (68) – (69). We have:
5 3 4
1 15 3 3
1 1
3 3
= [ ]tt t t
tn t t t
c rc dr f dr g g
s x x
σ σ σ
σ
+ −
− −
∂ ∂ ∂
− − + − − −
∂ ∂ ∂
∫ ∫ ; (76)
5
1 15
1 1
= [ ]tn
tt n n n
c dr f dr g g
s
σ
σ
+ −
− −
∂
+ − − −
∂
∫ ∫ . (77)
We shall multiply them respectively with cosα and sinα . Then an integration upon a
section leads to:
5 3 4
1 5 3 3
1
3 3
cos cos cos cos = cos
s s
tt t t
tn t
s s
c rc drds p ds
s x x
σ σ σ
α ασ α α α
+ +
−
− −
∂ ∂ ∂
− − + −
∂ ∂ ∂
∫ ∫ ∫ ;
5
1 5
1
sin sin = sin
s s
tn
tt n
s s
c drds p ds
s
σ
α ασ α
+ +
−
− −
∂
+ −
∂
∫ ∫ ∫ ,
139
with
1
1
= [ ]
t t t t
p f dr g g
+ −
−
+ −∫ and
1
1
= [ ]
n n n n
p f dr g g
+ −
−
+ −∫ . Using the following
properties:
5
cos tt
s
σ
α
∂
∂
=
5
5(cos )
sintt
tt
c
s
ασ
ασ
∂
+
∂
;
5
sin tn
s
σ
α
∂
∂
=
5
5(sin )
costn
tn
c
s
ασ
ασ
∂
−
∂
,
we get:
5 3 4
1 5 5 3 3
1
3 3
(cos )
sin cos cos cos =
s
tt t t
tt tn
s
c c rc drds
s x x
ασ σ σ
ασ ασ α α
+
−
−
∂ ∂ ∂
+ − − +
∂ ∂ ∂
∫ ∫
= cos
s
t
s
p dsα
+
−
−∫
5
1 5 5
1
(sin )
cos sin = sin
s s
tn
tn tt n
s s
c c drds p ds
s
ασ
ασ ασ α
+ +
−
− −
∂
− + −
∂
∫ ∫ ∫ .
By substraction, we obtain finally:
3 4
1 3 3
11
3 3
cos cos =
s
t t
s
rc drds P
x x
σ σ
α α
+
−
−
∂ ∂
− + −
∂ ∂
∫ ∫ . (78)
On the other hand, let us multiply equation (57) with 1x and integrate the result upon a
section. We get
5 4 3
1 13 3 33
1 1 1 1 31 1
3
=
s s
n t
s s
x x x drds x f drds
r s x
σ σ σ
+ +
− −
− −
∂ ∂ ∂
+ + −
∂ ∂ ∂
∫ ∫ ∫ ∫ .
Then using the equality
4 4
43 1 3
1 3
( )
= cost t
t
x
x
s s
σ σ
ασ
∂ ∂
−
∂ ∂
, the boundary condition (54) and
4
3 = 0
t
σ for =s s
±
, we obtain:
3
1 14 33
3 1 1 3 1 3 31 1
3
cos = [ ]
s s s
t
s s s
x drds x f drds x g g ds
x
σ
ασ
+ −+ + +
− −
− − −
∂
− + − − −
∂
∫ ∫ ∫ ∫ ∫ . (79)
Now let us derive equation (79) with respect to 3x . We get
4 2 3
1 3 33 31
1 21
3 33
cos =
s
t
s
dM
x drds
x dxx
σ σ
α
+
−
−
∂ ∂
− + −
∂ ∂
∫ ∫ . (80)
Adding equations (78) and (80), we have
3 2 3
1 3 33 31
1 121
3 33
cos =
s
t
s
dM
rc x drds P
x dxx
σ σ
α
+
−
−
∂ ∂
− + − −
∂ ∂
∫ ∫ .
Finally, replacing 3
3tσ and 3
33σ by their respective expressions, we obtain the bending
equation in the direction 1e of result 4. The bending equation in the direction 2e is obtained
in the same way, by permutation of the indices.
6. Comparison with Vlassov model.
To compare the one-dimensional thin-walled beam model obtained at results 1 to 4 to
Vlassov model, we shall first go back to the initial dimensional domain *
Ω and to the
dimensional variables *
t
u , *
n
u , *
3u , *
f and *
g . To do this, let us define
*0 0 0 *0 0 0 *0 0 0
3 3 3= = ; = = ; = =
t tr t t n nr n n r t
u u u hu u u u hu u u u huε . (81)
We then have the following result:
Result 5: For force levels such as
6= =
t n
F F ε ,
6= =
t n
G G ε and
5
3 3= =F G ε , the
displacement
*0 *0 *0
3( , , )
t n
u u u is of Vlassov type:
*0
t
u% = * * * *0
1 2cos( ) sin( ) ( )c c
u u q sα α+ − Θ ; *0
n
u% = * * * *0
1 2sin( ) cos( ) ( )c c
u u l sα α− + + Θ ;
*0
3u% =
* * *0
* * * *1 2
3 1 2* * *
3 3 3
c c
du du d
u x x
dx dx dx
ω
Θ
− − − . (82)
140
Starting from result 1 and from the dimensional analysis on the geometric parameters
performed, the proof of this result does not constitute any difficulty and is left to the reader.
We just need to set * * * * 2 *0 0
1 1 2 2 3 3= , = , = , = , =c c c c
u hu u hu u hu dε ω ω εΘ Θ and to use the
relations 2/ =h L ε and 2=hL d between the small parameters.
In the same way, we shall go back to dimensional variables in traction, twist and
bending equations of results 2 to 4. However, we will not give here the complete
dimensional equations, but only the reduced ones which are sufficient for a comparison with
Vlassov model. We recall that the one-dimensional equations reduce to a much more simple
form if they are written in a particular base, called "reduced basis". In this reduced basis, the
directions 1e and 2e correspond to the principal inertial axis of the profile. Moreover the
origin of the frame coincides with the center of gravity of the profile and the origin of the
sectorial area with the shear center. We then have the following result whose proof does not
constitute any difficulty and is left to the reader:
Result 6: For force levels such as
6= =
t n
F F ε ,
6= =
t n
G G ε and
5
3 3= =F G ε , the
leading terms of the displacements
*
3u ,
*0
Θ ,
*
1
c
u and
*
2
c
u are solution of the following
reduced one-dimensional equilibrium equations:
*02
3* *
3*2
3
=
d u
ES P
dx
− ; (83)
*0 *04 2 *
* * * 3
* * **4 *2 *
3 3 3
=
t
d
d d dM
EJ J M
dx dx dxω ω ω
µ
Θ Θ
− + ; (84)
* *04 2 *
1* * * 31
11 1 1*4 *2 *
3 3 3
=
c
d
d u d dM
EJ J P
dx dx dx
µ
Θ
− + ; (85)
* *04 2 *
2* * * 32
22 2 2*4 *2 *
3 3 3
=
c
d
d u d dM
EJ J P
dx dx dx
µ
Θ
− + . (86)
The dimensional expressions of the forces and of the geometric constants involved in
result 6 may be obtained easily fom results 2 to 4. We shall quote that the kinematics, the
one-dimensional reduced traction and twist equilibrium equations of results 5 and 6
correspond exactly to Vlassov ones [47]. However the one-dimensional bending equations
(85) – (86) differ from Vlassov ones which write (in the reduced basis):
*4 *
1* * 31
11 1*4 *
3 3
=
c
d u dM
EJ P
dx dx
+ ; (87)
*4 *
2* * 32
22 2*4 *
3 3
=
c
d u dM
EJ P
dx dx
+ . (88)
Therefore, at the difference from Vlassov model, the bending equations (85) – (86) contain a
supplementary term coupling twist and bending effects. This coupling term is linked to the new
geometrical constants *
1d
J and *
2d
J and does not seem to have any equivalent in the literature. It
corresponds most probably to a correction at the second order of Vlassov model. Thus the model
obtained by asymptotic expansion in this paper should improve Vlassov one where the twist
angle and the bending displacements are uncoupled. (We recall that from Vlassov model, an
external bending loading whose resultant induces a torque, will induce not only a bending dis-
placement but also a twist. In contrary, a torque will induce only a twist, but no bending, unlike
the model obtained in this paper where these two effects are coupled).
Let us quote that such a limitation of Vlassov theory (lack of coupling) already have been
noticed by other authors [5, 6, 23, 43]. To improve Vlassov model, the authors proposed to
add directly supplementary terms characterizing coupling effects in equilibrium equations.
141
7. Conclusion.
In this paper we deduced by asymptotic expansion a one-dimensional linear model for
thin-walled rods obtained for a strongly curved profile subjected to low force levels. The
obtained kinematics, the one-dimensional traction and twist equilibrium equations of results
1 to 3 correspond exactly to Vlassov ones [47]. However, whereas Vlassov approach relies
on a priori physical assumptions, with our approach the kinematics and equilibrium
equations are directly deduced from the three-dimensional equilibrium equations for the
level of applied forces considered. Thus the domain of validity of the obtained model can be
specified precisely thanks to the dimensionless numbers introduced.
Another major result is that that this asymptotic approach leads to an explicit analytical
expression of the geometrical constants involved in the one-dimensional equilibrium
equations. In particular, we obtain a general analytical expression of the twist rigidity *
d
J
ω
,
whereas in the literature only an approximate expression depending on an empiric
coefficient is given [47].
Finally, it is important to notice that the one-dimensional bending equations of result 4
differ from Vlassov ones. At the difference from Vlassov model, we obtain a supplementary
term coupling twist and bending effects. This coupling is due to the new geometrical
constants *
1d
J and *
2d
J and does not seem to have any equivalent in the literature.
Р Е З ЮМ Е . Запропоновано отриману асимптотичним методом одновимірну модель для тон-
костінного стержня з відкритим сильно скривленим поперечним перерізом, яка враховує взаємо-
зв’язок між скручуванням та згином. За допомогою аналізу розмірностей в лінійних тривимірних
рівняннях рівноваги знайдено безрозмірні величини, які характеризують геометрію стержня та рівень
прикладених сил. Для заданого рівня сил методом асимптотичного розкладу отримані порядок змі-
щень та відповідна одновимірна модель. У випадку низького рівня сил отримано одновимірну мо-
дель, кінематичні рівняння, рівняння кручення та згину відповідають моделі Власова. Однак ця мо-
дель враховує в рівняннях згину взаємодію між згином і крученням на відміну від моделі Власова,
яка таку взаємодію не враховує.
1. Caillerie D. Thin elastic and periodic plates // Math. Meth. in the Appl. Sci., 6, 159 – 191 (1984).
2. Ciarlet P.G., Destuynder P. // Justification of the two-dimensional linear plate model, J. Mécanique, 18,
315 – 344 (1979).
3. Cimetière A., Geymonat G., Dret H. Le, Raoult A., Tutek Z. Asymptotic theory and analysis for
displacements and stress distribution in nonlinear elastic straight slender roads // J. Elasticity, 19, 111 –
161 (1998).
4. Cimetière A., Hamdouni A., Millet O. Le modèle linéaire usuel de plaque déduit de l'élasticité non linéaire
tridimensionnelle // C. R. Acad. Sci. Paris, série IIb, 326, 159 – 162 (1998).
5. Davies J.M., Leach P. First-order generalised beam theory // J. Construct Steel Research, 31, 187 – 220
(1994).
6. Davies J.M., Leach P. Second-order generalised beam theory // J. Construct Steel Research, 31, 221 – 241
(1994).
7. Destuynder P. A classification of thin shell theories // Acta Applicandae Mathematicae, 4, 15 – 63 (1985).
8. Elamri K., Hamdouni A., Millet O. Le modèle linéaire de Novozhilov-Donnell déduit de l'élasticité non
linéaire tridimensionnelle // C. R. Acad. Sci. Paris, Série II b, 327, 1285 – 1290 (1999).
9. Geymonat G. Sur la commutativité des passages à la limite en théorie asymptotique des poutres
composites // C. R. Acad. Sci., Série II, 305, 225 – 228 (1987).
10. Gjelsvik A. // The theory of thin-walled bars, John Wiley & Sons, N.Y. (1981).
11. Goldenveizer A.L. Derivation of an approximate theory of shells by means of asymptotic integration of
the equations of the theory of elasticity // Prikl. Math. Mech., 27, 593 – 608 (1963).
12. Grillet L., Hamdouni A., Millet O. An asymptotic non-linear model for thin-walled rods // C. R. Acad.
Sci., Paris, série Mécanique, 332, 123 – 128 (2004).
13. Grillet L., Hamdouni A., Millet O. Justification of the kinematic assumptions for thin-walled rods with
shallow profile // C. R. Mécanique, 333, 493 – 489 (2005).
142
14. Gulyaev V.I., Lugovoi P.Z., Gaidaichuk V.V., Solov'ev I.L., Gorbunovich I.V. Effect of the Length of a
Rotating Drillstring on the Stability of its Quasistatic Equilibrium // Int. Appl. Mech. – 2007. – 43, N 9.
– P. 1017 – 1023.
15. Gulyaev V.I., Lugovoi P.Z., Khudolii S.N., Glovach L.V. Theoretical Identification of Forces Resisting
Longitudinal Movement of Drillstring in Curved Wells // Int. Appl. Mech. – 2007. – 43, N 11. –
P. 1248 – 1255.
16. Hamdouni A., Millet O. Classification of thin shell models deduced from the nonlinear three-dimensional
elasticity. Part I: the shallow shells // Arch. Mech., 55(2), 135 – 175 (2003).
17. Hamdouni A., Millet O. Classification of thin shell models deduced from the nonlinear three-dimensional
elasticity. Part II: the strongly curved shells // Arch. Mech., 55(2), 177 – 219 (2003).
18. Hamdouni A., Millet O. An asymptotic non-linear model for thin-walled rods with strongly curved open
cross-section // Int. J. of Non-Linear Mech., 41, 396 – 416 (2006).
19. Lewinski T. Effective models of composite periodic plates. I-asymptotic solution // Int. J. Solids
Structures, 27(9), 1155 – 1172 (1991).
20. Madani K. Étude des structures élastiques élancées à rayon de courbure faiblement variable: une
classification de modèles asymptotiques // C.R. Acad. Sci., Série IIb, 326, 605 – 608 (1998).
21. Marigo JJ., Ghidouche H., Sedkaoui Z. Des poutres flexibles aux fils extensibles: une hierarchie de
modèles asymptotiques // C.R. Acad. Sci., Série IIb, 326, 79 – 84 (1998).
22. Marigo JJ., Madani K. Quelques modèles d'anneaux élastiques suivant les types de chargement // C.R.
Acad. Sci., Série IIb, 326, 805 – 810 (1998).
23. Massonnet Ch. A new approach (including shear lag) to elementary mechanics of material // Int. J. Solids
Structures, 19(1), 33 – 54 (1983).
24. Millet O., Hamdouni A., Cimetière A. Justification du modèle bidimensionnel linéaire de plaque par
développement asymptotique de l'équation de Navier // C. R. Acad. Sci., Paris, série II b, 324, 289 –
292 (1997).
25. Millet O., Hamdouni A., Cimetière A. Justification du modèle bidimensionnel non linéaire de plaque par
développement asymptotique des équations d'équilibre // C. R. Acad. Sci., Paris, série II b, 324, 349 –
354 (1997).
26. Millet O., Hamdouni A., Cimetière A. Construction d'un modèle eulérien de plaques en grands
déplacements par méthode asymptotique // C. R. Acad. Sci., Paris, série II b, 325, 257 – 261 (1997).
27. Millet O., HamdouniA., Cimetière A. Dimensional analysis and asymptotic expansions of the equilibrium
equations in nonlinear elasticity. Part I: The membrane model // Arch. Mech., 50(6), 853 – 873 (1998).
28. Millet O., Hamdouni A., Cimetière A. Dimensional analysis and asymptotic expansions of the
equilibrium equations in nonlinear elasticity. Part II: The "Von Karman Model" // Arch. Mech., 50(6),
975 – 1001 (1998).
29. Millet O., Hamdouni A., Cimetière A. A classification of thin plate models by asymptotic expansion of
nonlinear three-dimensional equilibrium equations // Int. J. of Non-Linear Mechanics, 36, 165 – 186
(2001).
30. Millet O., Hamdouni A., Cimetière A. An eulerian approach of membrane theory for large displacements
// Int. J. of Non-Linear Mechanics, 38, 1403 – 1420 (2003).
31. Millet O., Cimetière A., Hamdouni A. An asymptotic elastic-plastic plate model for moderate
displacements and strong strain hardening // Eur. J. of Mech. A/ Solids, 22, 369 – 384 (2003).
32. Prokic A., Lukic D. Dynamic behavior of braced thin-walled beams // Int. Appl. Mech., 43(11), 1290 –
1303 (2007).
33. Rigolot A. Déplacements finis et petites déformations des poutres droites: Analyse asymptotique de la
solution à grande distance des bases // J. Mécanique appliquée, 1(2), 175 – 206 (1977).
34. Rodriguez J.M., J.M. Viaño J.M. Asymptotic analysis of Poisson's equation in a thin domain. Application
to thin-walled elastic beams, in: P.G. Ciarlet, L. trabucho and J.M. Viano eds., Asymptotic Methods for
Elastic Structures, Walter de Gruyer, Berlin, 181 – 193 (1995).
35. Rodriguez J.M., J.M. Viaño J.M. // Asymptotic general bending and torsion models for thin-walled
elastic beams, in: P.G. Ciarlet, L. trabucho and J.M. Viano eds., Asymptotic Methods for Elastic
Structures,Walter de Gruyer, Berlin, 255 – 274 (1995).
36. Rodriguez J.M., J.M. Viaño J.M. Asymptotic derivation of a general linear model for thin-walled elastic
rods // Comput. Methods Appl. Mech. Engrg, 147, 287 – 321 (1997).
37. Rudavskii Y.K., Vikovich I.A. Forced flexural-and-torsional vibrations of a cantilever beam of constant
cross section // Int. Appl. Mech., 43(8), 129 – 1303 (2007).
143
38. Sanchez-Palencia E. Statique et dynamique des coques minces, I - Cas de flexion pure non inhibée”, C.
R. Acad. Sci., Paris, série I, 309, 411 – 417 (1989).
39. Sanchez-Palencia E. Statique et dynamique des coques minces, I - Cas de flexion pure inhibée -
Approximation membranaire // C. R. Acad. Sci., Paris, série I, 309, 531 – 537 (1989).
40. Sanchez-Palencia E. Passage à la limite de l'élasticité tridimensionnelle à la théorie asymptotique des
coques minces // C. R. Acad. Sci., Paris, série II b, 311, 909 – 916 (1990).
41. Sanchez-Hubert J. et Sanchez-Palencia E. Coques élastiques minces. Propriétés asymptotiques, Masson,
Paris (1997).
42. Sanchez-Hubert J., Sanchez Palencia E. Statics of curved rods on account of torsion and flexion // Eur. J.
Mech A/Solids, 18, 365 – 390 (1999).
43. Saucha J., Rados J. A critical review of Vlassov's general theory of stability of in-plane bending of thin-
walled elastic beams // Meccanica, 6, 177 – 190 (2001).
44. Trabucho L., Viaño J.M. Existence and characterization of higher-order terms in an asymptotic
expansion method for linearized elastic beams // Asymptotic Analysis, 2, 223 – 255 (1989).
45. Trabucho L., Viaño J.M. A new approach of Timoshenko's beam theory by asymptotic expansion
method // Math. Mod. and Numer. Anal., 5, 651 – 680 (1990).
46. Trabucho L., Viaño J.M. Mathematical modelling of rods // P.G. Ciarlet and J.L. Lions eds., Handbook
of Numerical Analysis, Vol. IV, North Holland, Amsterdam, 487 – 974 (1996).
47. Vlassov V.Z. Pièces longues en voiles minces, Eyrolles, Paris (1962).
48. Zhang L., Tong G.S. Elastic flexural-torsional buckling of thin-walled cantilevers // Thin-Walled
Strucures, 46, 27 – 37 (2008).
Поступила 30.01.2009 Утверждена в печать 15.04.2010
*From the Editorial Board: The article corresponds completely to submitted manuscript.
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| id | nasplib_isofts_kiev_ua-123456789-95414 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 0032-8243 |
| language | English |
| last_indexed | 2025-11-27T21:13:09Z |
| publishDate | 2010 |
| publisher | Інститут механіки ім. С.П. Тимошенка НАН України |
| record_format | dspace |
| spelling | Hamdouni, A. Millet, O. 2016-02-25T21:04:47Z 2016-02-25T21:04:47Z 2010 An Asymptotic Linear Thin-Walled Rod Model Coupling Twist and Bending / A. Hamdouni, O. Millet // Прикладная механика. — 2010. — Т. 46, № 9. — С. 123-143. — Бібліогр.: 48 назв. — анг. 0032-8243 https://nasplib.isofts.kiev.ua/handle/123456789/95414 A linear one-dimensional model for thin-walled rods with open strongly curved cross-section, obtained by asymptotic methods is presented. A dimensional analysis of the linear three-dimensional equilibrium equations lets appear dimensionless numbers which reflect the geometry of the structure and the level of applied forces. For a given force level, the order of magnitude of the displacements and the corresponding one-dimensional model are deduced by asymptotic expansions. In the case of low force levels, we obtain a one dimensional model whose kinematics, traction and twist equations correspond to Vlassov ones. However this model couples twist and bending effects in the bending equations, at the difference from Vlassov model where the twist angle and the bending displacement are uncoupled. Запропоновано отриману асимптотичним методом одновимірну модель для тонкостінного стержня з відкритим сильно скривленим поперечним перерізом, яка враховує взаємо- зв’язок між скручуванням та згином. За допомогою аналізу розмірностей в лінійних тривимірних рівняннях рівноваги знайдено безрозмірні величини, які характеризують геометрію стержня та рівень прикладених сил. Для заданого рівня сил методом асимптотичного розкладу отримані порядок змі- щень та відповідна одновимірна модель. У випадку низького рівня сил отримано одновимірну модель, кінематичні рівняння, рівняння кручення та згину відповідають моделі Власова. Однак ця модель враховує в рівняннях згину взаємодію між згином і крученням на відміну від моделі Власова, яка таку взаємодію не враховує. Запропоновано отриману асимптотичним методом одновимірну модель для тонкостінного стержня з відкритим сильно скривленим поперечним перерізом, яка враховує взаємо- зв’язок між скручуванням та згином. За допомогою аналізу розмірностей в лінійних тривимірних рівняннях рівноваги знайдено безрозмірні величини, які характеризують геометрію стержня та рівень прикладених сил. Для заданого рівня сил методом асимптотичного розкладу отримані порядок зміщень та відповідна одновимірна модель. У випадку низького рівня сил отримано одновимірну модель, кінематичні рівняння, рівняння кручення та згину відповідають моделі Власова. Однак ця модель враховує в рівняннях згину взаємодію між згином і крученням на відміну від моделі Власова, яка таку взаємодію не враховує. en Інститут механіки ім. С.П. Тимошенка НАН України Прикладная механика An Asymptotic Linear Thin-Walled Rod Model Coupling Twist and Bending Об асимптотической линейной модели для тонкостенных стержней с взаимосвязью между скручиванием и изгибом Article published earlier |
| spellingShingle | An Asymptotic Linear Thin-Walled Rod Model Coupling Twist and Bending Hamdouni, A. Millet, O. |
| title | An Asymptotic Linear Thin-Walled Rod Model Coupling Twist and Bending |
| title_alt | Об асимптотической линейной модели для тонкостенных стержней с взаимосвязью между скручиванием и изгибом |
| title_full | An Asymptotic Linear Thin-Walled Rod Model Coupling Twist and Bending |
| title_fullStr | An Asymptotic Linear Thin-Walled Rod Model Coupling Twist and Bending |
| title_full_unstemmed | An Asymptotic Linear Thin-Walled Rod Model Coupling Twist and Bending |
| title_short | An Asymptotic Linear Thin-Walled Rod Model Coupling Twist and Bending |
| title_sort | asymptotic linear thin-walled rod model coupling twist and bending |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/95414 |
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