Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
1/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
Organization (S):
EDF/IMA/MN, IAT St CYR
Manual of Reference
R3.08 booklet: Machine elements with average fiber
Document: R3.08.01
“Exact” elements of beams (right and curved)
Summary:
This document presents the elements of beam of Code_Aster based on an exact resolution of the equations of
continuous model carried out for each element of the mesh.
The beams can be right (Elements
POU_D_T
and
POU_D_E
)
or curves (Elements
POU_C_T
).
section, constant or variable over the length, can be of an unspecified form. The material is homogeneous,
isotropic, elastic linear.
The assumptions selected are as follows:
·
Assumption of Euler: transverse shearing is neglected, as well as the inertia of rotation.
This assumption is checked for strong twinges (element
POU_D_E
).
·
Assumption of Timoshenko: transverse shearing and all the terms of inertia are taken into account.
This assumption is to be used for weak twinges (elements
POU_D_T
and
POU_C_T
).
·
Assumption of Saint-Coming: torsion is free.
The processing of the various loadings and the sizes awaited in result (forced - efforts) is
also presented.
Code_Aster
®
Version
3
Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
2/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
Contents
1 equations of the movement ................................................................................................................ 6
1.1 The traction and compression .................................................................................................................. 6
1.1.1 Local equilibrium equation ........................................................................................................ 6
1.1.2 Method from Lagrangian ......................................................................................................... the 7
1.2 Pure torsion (torsion of Saint-Coming) ........................................................................................ 8
1.2.1 Local equilibrium equation ........................................................................................................ 8
1.2.1.1 Circular beam of section ........................................................................................ 8
1.2.1.2 Unspecified beam of section .................................................................................... 9
1.2.2 Method from Lagrangian ......................................................................................................... the 9
1.3 The bending pure ............................................................................................................................ 10
1.3.1 Local equilibrium equation ...................................................................................................... 10
1.3.2 Method from Lagrangian ....................................................................................................... the 13
2 Element of right beam ...................................................................................................................... 15
2.1 Longitudinal movement of traction and compression ....................................................................... 15
2.1.1 Determination of the matrix of rigidity ................................................................................. 15
2.1.2 Determination of the second member ....................................................................................... 16
2.1.3 Calculation of the efforts to the nodes of the beam .......................................................................... 17
2.1.4 Determination of the matrix of mass ................................................................................. 18
2.2 Free movement of torsion around the longitudinal axis ............................................................... 19
2.3 Movement of bending ................................................................................................................... 19
2.3.1 Bending in the plan (x0z) .................................................................................................... 20
2.3.2 Bending in the plan (xOy) .................................................................................................... 22
2.3.3 Determination of the matrix of coherent mass with the matrix of rigidity ...................... 23
2.3.3.1 Bending in the plan (xoz) ........................................................................................ 23
2.3.3.2 Movement of bending around the axis (O Z) ............................................................. 24
2.4 Stamp of mass reduced by the technique of the concentrated masses .......................................... 25
3 particular right Beams ................................................................................................................ 27
3.1 Eccentricity of the axis of torsion compared to the neutral axis ............................................................. 27
3.2 Variable sections ......................................................................................................................... 29
3.2.1 Calculation of the matrix of rigidity ............................................................................................. 30
3.2.1.1 Determination of the equivalent section (Seq) ........................................................ 30
3.2.1.2 Determination of a constant of equivalent torsion (Ceq) .................................. 33
3.2.1.3 Determination of the equivalent geometrical moments .......................................... 34
3.2.2 Calculation of the matrix of mass ............................................................................................. 37
3.2.2.1 By the method of the equivalent masses ............................................................... 37
3.2.2.2 By the method of the masses concentrated (diagonal matrix) .................................. 38
4 geometrical Rigidity - prestressed Structure .................................................................................... 40
5 Beam curves ....................................................................................................................................... 46
Code_Aster
®
Version
3
Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
3/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
5.1 Stamp flexibility for the bending in the plan of the beam [C1] ................................................ 51
5.2 Stamp flexibility for the bending out of the plan of the beam [C2] ................................................ 52
6 Loadings ........................................................................................................................................ 55
6.1 Loading by deformation ......................................................................................................... 55
6.1.1 For the right beam of Euler and the right beam of Timoshenko ........................................... 55
6.1.2 For the beam curves of Timoshenko .................................................................................. 56
6.2 Loading due to gravity ....................................................................................................... 57
6.3 Loadings distributed .................................................................................................................... 61
6.3.1 Right beam with constant section ......................................................................................... 61
6.3.2 Right beams with variable section ......................................................................................... 62
6.3.3 Beam curves ....................................................................................................................... 62
6.4 Thermal loading .................................................................................................................. 63
6.5 Electric loading ................................................................................................................... 63
6.5.1 Secondary conductor right finished or infinite .............................................................................. 64
6.5.2 Secondary conductor describes by part of mesh ASTER ....................................... 65
7 Torque of the efforts - Torque of the stresses (or efforts generalized) - nodal Forces and réactions.66
7.1 The torque of the efforts ..................................................................................................................... 66
7.2 The tensor of the stresses ............................................................................................................. 67
7.3 Calculation of the nodal forces and the reactions ................................................................................... 70
8 Element of bar .................................................................................................................................. 71
9 Bibliography ......................................................................................................................................... 71
Foreword
This reference material of the elements of beam was carried out starting from a work carried out
by m.t. Bourdeix, P. Hemon, O. Wilk of the Institute Aerotechnics of the National Academy of Arts and
Trades, within the framework of an External Contract of Research and Development with this laboratory.
The volume of this document is due at the same time to the required precision and the didactic character of
the talk, which is voluntarily preserved.
Code_Aster
®
Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
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R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
Notations
The notations used here are not all identical to those used in [U1.04] and [U4.24.01],
for reasons of compactness and homogeneity with [R3.08.03].
One gives the correspondence between this notation and that of the documentation of use.
DX, DY, DZ and DRX, DRY, DRZ are in fact the names of the degrees of freedom associated with the components
displacement
U v W
X
y
Z
,
,
,
.
C
constant of torsion
JX
E E
y
Z
,
eccentricity of the center of torsion/shearing
EY EZ
,
E
Young modulus
E
Poisson's ratio
NAKED
G
modulate of Coulomb =
(
)
E
2 1
+
I I
y
Z
,
geometrical moments of bending compared to the axes
y Z
,
IY IZ
,
I
p
polar geometrical moment
I
X
polar moment of inertia around the longitudinal axis
X
K K
y
Z
,
coefficients of shearing
1
1
AY
AZ
K
stamp rigidity
M
stamp of mass
M
M
M
X
y
Z
,
,
moments around the axes
X y Z
,
MT MFY MFZ
,
,
NR
normal effort with the section
NR
S
surface of the section
With
U v W
,
translations on the axes
X y Z
,
DX DY DZ
V V
y
Z
,
sharp efforts along the axes
y Z
,
VY VZ
,
density
RHO
C
T
transverse shear stress
X
y
Z
,
,
rotations around the axes
X y Z
,
DRX DRY DRZ
Code_Aster
®
Version
3
Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
5/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
Introduction
A beam is a solid generated by a surface of surface
S
of which the geometrical center of inertia
G
followed a curve
C
called the average fiber or neutral fiber. The surface
S
(section is the cross-section
transversal) or profile, and it is supposed that if it is evolutionary, its evolutions (size, form) are
continuous and progressive when
G
described the average line.
For the study of the beams in general, one makes the following assumptions:
·
the cross-section of the beam is indeformable,
·
transverse displacement is uniform on the cross-section.
These assumptions make it possible to express displacements of an unspecified point of the section, in
function of displacements of the point corresponding located on the average line, and according to one
increase in displacement due to the rotation of the section around the transverse axes. This
last can be neglected (
POU_D_E
) or to be the subject of a modeling (
POU_D_T
and
POU_C_T
).
The discretization in “exact” elements of beam is carried out on a linear element with two nodes and six
degrees of freedom by nodes. These degrees of freedom are the three translations
U v W
,
and three
rotations
X
y
Z
,
,
.
Z
y
X
1
2
U
X
v
y
W
Z
U
X
v
y
W
Z
Waited until the deformations are local, it is built in each node of the mesh a base
local depending on the element on which one works. The continuity of the fields of displacements is
ensured by a basic change, bringing back the data in the total base.
In the case of the right beams, one traditionally places the average line on axis X of the base
local, transverse displacements being thus carried out in the plan
()
y Z
,
.
Finally when we arrange sizes related to the degrees of freedom of an element in a vector
or an elementary matrix (thus of dimension 12 or 12
2
), one arranges initially the variables for
node 1 then those of node 2. For each node, one stores initially the sizes related to the three
translations, then those related to three rotations. For example, a vector displacement will be structured
in the following way:
U, v, W
,
U, v, W
,
1
1
1
X
y
Z
2
2
2
X
y
Z
1
1
1
2
2
2
,
,
node 1
node 2
!
“
# # # #
$
# # # #
!
“
# # # #
$
# # # #
Code_Aster
®
Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
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R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
1
Equations of the movement
In this chapter one presents the equations of the movement of the beams in traction and compression, in
torsion and in bending in the elastic range. In each case, these equations are deduced by
application of the equations of Lagrange, resulting from the principle of Hamilton, or by writing balance
room of a segment of beam. We chose to point out the two methods, the reader will be able
to refer to that which is most familiar for him. One limits oneself here to the cases where the only loadings are
loadings distributed (not of concentrated forces).
1.1
traction and compression
The traction and compression is the translatory movement on the longitudinal axis of the beam.
1.1.1 Local equilibrium equation
A segment length is considered
dx
subjected to an axial load NR [1.1.1-a] intern and a force
external
F
ext.
by unit of length.
NR (X)
F
ext.
X
O
dx
X + dx
X
NR (X + dx)
NR (X)
Appear 1.1.1-a: Segment of beam charged axially
The beam has a section
()
S X
and consists of a material of density
()
X
and of module
of Young
()
E X
. The fundamental principle of mechanics makes it possible to write:
()
() ()
()
-
=
+
+
NR X
NR X dx
F
S ds
S S S
U.S.
T
ds
U
X
ext.
X
X dx
X
X dx
() + (+
) +
where is displacement on the axis of the segment.
2
2
Thus:
()
NR X dx
NR X
dx
dx
F
S ds
dx
S
U
T ds
ext.
X
X dx
X
X dx
(+
)
() +
-
=
+
+
1
1
2
2
While passing in extreme cases when
dx
0
, one obtains:
()
D NR X
dx
F
S
U
T
X
extx
() +
()
=
2
2
éq 1.1.1-1
Code_Aster
®
Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
7/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
Only the first order terms are preserved and one replaces in [éq 1.1.1-1], then one uses the law
of Hooke and the assumption that the beam consists of longitudinal fibers working only in
traction and compression to express the axial load by:
NR X
ES ux
() =
éq 1.1.1.- 2
One obtains thus after simplification by
dx
:
X ES
U
X
F
S
U
T
ext.
+
=
2
2
éq 1.1.1-3
who represents local balance with the first command of a beam, for a movement of
traction and compression.
1.1.2 Method of the Lagrangian one
Taking again the segment of beam of the figure [Figure 1.1.1-a] total kinetic energy of the beam of
length
L
is written:
E
S
U
T
dx
C
O
L
=
1
2
2
.
One will note for the continuation
E
S
U
T
C
E
=
1
2
2
elementary kinetic energy.
The internal energy of deformation, thanks to the law of Hooke is written:
E
ES
U
X
dx
p
O
L
int
.
=
1
2
2
One will note in the same way
E
ES
U
X
p
E
int
=
1
2
2
.
There is also the work of the external force given by:
E
F
U dx
E
F
U
p
ext.
O
L
p
ext.
ext.
exte
=
and at the elementary level
=
.
The Lagrangian one is given by:
L
=
E
E
E
C
p
p
ext.
-
-
int
and Lagrangian density:
L
E
E
E
C
p
p
E
E
exte
=
-
-
int
.
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
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R3.08 booklet: Machine elements with average fiber
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For the monodimensional continuous system, the equation of Lagrange is written in this case:
L
U
X
L
U
T
L
U
-
-
'
=
%
0
éq 1.1.2-1
where
U
and
% U
indicate the derivative respectively compared to
X
and compared to time. Its
application brings back for us obviously to the equation of the movement of a beam in traction and compression
[éq 1.1.1-3].
1.2
Pure torsion (torsion of Saint-Coming)
Torsion is the rotational movement around the longitudinal axis of the beam. It is supposed here that it
center of gravity is confused with the center of rotation (of torsion) [R3.03.03], and it is neglected
roll of the section. The case of the eccentricity of the center of torsion compared to the center of
gravity is treated with [§3.1].
1.2.1 Local equilibrium equation
A segment length is considered
dx
put in rotation under the action of one moment
M
X
[Figure 1.2.1-a] intern and of an external couple
X
by unit of length.
NR (X)
X
O
dx
X + dx
X
M (X + dx)
M (X)
X
positive
Appear 1.2.1-a: Segment of beam in rotation around (OX)
The segment is turned of an angle
X
compared to the not deformed position. We have as follows:
1.2.1.1 Circular beam of section
(
)
()
-
+
+
() +
+
+
=
M X
M X dx
S ds
I
T ds
X
X
X
X dx
X
X
X dx
X
X
2
2
with
I
R ds
X
S
=
2
is the plane moment of inertia of the section
S
around the axis of rotation
()
0, X
.
As for traction, one obtains after division by
dx
and passage in extreme cases:
DM
dx
I
T
X
X
X
X
+
=
2
2
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
9/72
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R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
The law of behavior is introduced:
M
G I
X
X
p
X
=
where
G
is the module of Coulomb (or modulus of rigidity) and
I
p
polar geometrical moment by
report/ratio in the center of gravity of the section. (One has besides:
I
I
X
p
=
for a material of mass
voluminal homogeneous).
We obtain the expression then:
X G I
X
I
T
p
X
X
X
X
+
=
2
2
éq 1.2.1.1-1
who represents local balance with the first command of a segment of beam for a movement of torsion.
1.2.1.2 Unspecified beam of section
To take account of the roll while remaining on the free assumption of torsion, in the case of them
noncircular sections one is led to replace the moment
I
p
by a constant of torsion
C
(lower than
I
p
) in the equation of torsion ([R3.03.03] for the calculation of
C
).
By definition,
M
G C
X
X
X
=
. One obtains then:
X G C
X
C
T
X
X
X
+
=
2
2
éq 1.2.1.2-1
When the center of gravity of the section is not the center of rotation, this expression is not
valid and the movements of torsion and bending are coupled.
1.2.2 Method of the Lagrangian one
We have same manner as with [§1.1.2] the kinetic energy (for example for a beam of
circular section):
int
E
I
T
dx
E
G I
T
dx
E
dx
C
O
L
X
p
O
L
p
X
p
X
O
L
X
ext.
=
,
=
,
=
.
internal potential energy
and the work of the external couple
X
1
2
1
2
2
2
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
10/72
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R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
By applying the equation of Lagrange [éq 1.1.2-1] to the variable
X
, one leads naturally to
[éq 1.2.1-1] giving the movement of a beam in pure torsion.
1.3
The pure bending
The bending is the rotation and translatory movement around an axis perpendicular to the axis
longitudinal of the beam. One speaks here about pure bending (around OY or OZ). One limits oneself to the case of
right beams. The curved beams are treated with [§5].
One describes the equation of bending in plan (O, X, Z), the extension to the plan (O, X, y) is immediate
[Figure 1.3-a].
y
X
O
y
Z
Appear 1.3-a: Bending of a beam in plan (O, X, Z)
The translation along the axis (O, Z) is noted
W
and rotation around (O, y) is noted
y
.
1.3.1 Local equilibrium equation
A segment length is considered
dx
subjected to the sharp effort
V
Z
, bending moment
M
y
,
an external effort
T
Z
ext.
distributed uniformly by unit of length, and an external couple
m
y
ext.
distributed
uniformly by unit of length [Figure 1.3.1-a].
tz
ext.
V (x+dx)
my
ext.
Z
y
X
V (X)
M (X)
M (x+dx)
X
x+dx
y
Moments
positive
dx
Appear 1.3.1-a: Segment of beam in bending in plan (O, X, Z)
Code_Aster
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Titrate:
“Exact” elements of beams (right and curved)
Date:
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J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
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The local balance of the forces and the moments (on the section of X-coordinate
X dx
+
) gives for the forces:
(
)
(
)
()
-
-
-
+
+
+
+
+
V X V X dx
T
ds
S ds
W
T
M X
M X dx
m
ds
V X ds
I ds
T
Z
Z
X
X dx
Z
X
X dx
y
y
y
X dx
Z
X
X dx
y
X
X dx
y
ext.
ext.
() +
+
+
=
() +
+
+
=
and for the moment:
X
2
2
2
2
One neglects the terms in
dx
2
. While passing in extreme cases when
dx
tends towards 0, one obtains:
V
X
T
S
W
T
My
X
V
m
I
T
Z
Z
Z
y
y
y
ext.
ext.
+
=
+
= +
2
2
2
2
-
.
It is noted that the effort uniformly distributed
T
Z
ext.
product a term which is of the second command in balance
moments and is thus neglected. One introduces then the relations of behavior of resistance
materials.
M
I.E.(internal excitation)
X
y
y
y
= +
V
K SG
W
X
Z
Z
y
=
+
éq 1.3.1-1
the expression [éq 1.3.1-1] of
V
Z
is due to Timoshenko [bib4] where
K
Z
is the coefficient of shearing
in direction Z. It characterizes the model of beam of Timoshenko; it will be seen thereafter that it
model of beam of Euler corresponds to a simplification of the model of Timoshenko.
I
Z
is the moment
geometrical of the section compared to the axis (O, y).
Consequently, one leads to the two equations coupled in
W
and
y
for the bending in the plan
(O, X, Z).
X K SG
W
X
T
S
W
T
Z
y
Z
ext.
+
+
=
2
2
éq 1.3.1-2
X I.E.(internal excitation)
X
K SG
W
X
m
I
T
y
y
Z
y
y
y
y
ext.
-
+
+
= +
2
2
éq 1.3.1-3
When the beam is uniform, i.e. the section and the material are constant on the axis
longitudinal, the equations [éq 1.3.1-2] and [éq 1.3.1-3] are reduced to only one equation in
W
. For
that, one only once derives compared to X-coordinate X the equilibrium equation from the moments [éq 1.3.1-3].
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
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R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
I.E.(internal excitation)
X
K SG
W
X
X
I
X T
y
y
Z
y
y
y
3
3
2
2
3
2
-
+
= +
.
It will be noted that this handling eliminates the presence from the term resulting from an external couple
uniformly distributed. Then, the equation [éq 1.3.1-2] can be put in the form:
y
Z
Z
Z
X
W
X
1
K SG T
K G
W
T
ext.
=
+
2
2
2
2
-
.
I.E.(internal excitation)
W
X
S
W
T
I 1
E
K G
W
X
T
I
K G
W
T
T
y
y
Z
y
Z
Z
ext.
4
4
2
2
4
2
2
2
4
4
0
+
+
+
=
-
-
éq 1.3.1-4
There remains useful for this type of equation to point out the physical significance of the various terms, so at the time
simplifications to be aware of the neglected effects.
I.E.(internal excitation)
W
X
y
4
4
balance the charging density in the direction of the translation due to the moment of
bending.
S
W
T
2
2
is the term of inertia of translation.
I
W
X
T
y
4
2
2
represent the inertia of rotation of bending.
I
E
K G
W
X
T
y
Z
4
2
2
is an additional term of the inertia of rotation due to the taking into account of
transverse shearing (assumption of Timoshenko).
2
4
4
I
K G
W
T
y
Z
result from the coupling between the inertia of rotation and the inertia of translation coming from
sharp effort.
The model of beam of Timoshenko (
POU_D_T
or
POU_C_T
), takes into account the whole of these
terms, in particular those which relate to the sharp effort. One can thus modelize beams
of weak twinge.
The model of beam of Euler (
POU_D_E
) is a simplification since the deformations in effort
edge are neglected as well as the inertia of rotation (what is justified because it does not intervene in
dynamic studies that for the high modes). These assumptions are justified in the case of one
beam of sufficiently large twinge. So for the model of Euler, the equation of the movement of
bending, in the general case of the beams with variable section is written:
X
I.E.(internal excitation)
W
X
S
W
T
T
y
Z
ext.
2
2
2
2
2
0
-
+ =
.
éq 1.3.1-5
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In addition, it is indeed the shearing action which causes the rotation of the cross-sections by
report/ratio with the neutral axis. To neglect this effect thus amounts writing that
V
Z
=
0
what brings to [éq 1.3.1-1].
y
W
X
=
-
.
éq 1.3.1-6
who is the translation of the assumption of Euler.
Concerning the bending in the plan (O, X, y), the same step leads to [éq 1.3.1-7] for
beam of Timoshenko with:
X K SG
v
X
T
S
v
T
X I.E.(internal excitation)
X
K SG
v
X
m
I
T
y
Z
y
Z
Z
y
Z
Z
Z
Z
ext.
ext.
+
=
+
=
-
-
-
2
2
2
2
éq 1.3.1-7
and when the section is constant:
I.E.(internal excitation)
v
T
S
v
T
I
E
K G
v
X
T
I
K G
v
T
T
Z
Z
y
Z
y
y
ext.
4
4
2
2
4
2
2
2
4
4
1
0
-
-
=
+
+
+
.
éq 1.3.1-8
The use of the assumption of Euler (i.e. in the plan (O, X, y)
Z
v
X
=
) allows to lead to
the equation of the movement of bending for a beam of Euler according to [éq 1.3.1-9].
X
I.E.(internal excitation)
v
X
S
v
T
T
Z
y
ext.
2
2
2
2
2
0
-
+
=
.
éq 1.3.1-9
1.3.2 Method of the Lagrangian one
The kinetic energy is expressed by:
E
I
T
dx
S
W
T
dx
C
O
L
y
y
O
L
=
+
according to displacements in rotation and translation
1
2
1
2
2
2
.
The potential energy intern is worth:
E
I.E.(internal excitation)
X
dx
W
X
dS dx
p
O
L
y
O
L
C
S
y
T
int
=
+
+
1
2
1
2
2
where
C
T
is transverse shear stress and the term
W
X
y
+
deformation of
shearing. The model of beam of Euler neglects this term while the model of Timoshenko emits
an assumption on the distribution of the stresses
C
T
in the section, compatible with the expression
[éq 1.3.1-1]. In the general case of the model of Timoshenko, the potential energy interns is written:
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E
I.E.(internal excitation)
X
dx
K SG
W
X
dx
p
O
L
y
O
L
Z
y
int
.
=
+
+
1
2
1
2
2
2
The potential of the external loads is expressed as for him by:
E
T
dx
m
dx
p
Z
O
L
y
O
L
y
ext.
ext.
ext.
=
-
-
.
The use of the equation of Lagrange [éq 1.1.2-1] applied once to the variable
W
then with the variable
y
us brings back to the two equations [éq 1.3.1-2] and [éq 1.3.1-3] describing the movement in bending of one
segment of beam.
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2
Element of right beam
One describes in this chapter obtaining the elementary matrices of rigidity and mass for the element
of right beam, according to the model of Euler (
POU_D_E
) or of Timoshenko (
POU_D_T
). Matrices of
rigidity are calculated with the option
“RIGI_MECA”
, and matrices of mass with the option
“MASS_MECA”
for the coherent matrix, and the option '
MASS_MECA_DIAG'
for the matrix of mass
diagonalized.
2.1
Longitudinal movement of traction and compression
A difficulty to write the variational formulation comes owing to the fact that there can be in the structures
composed of beams of the concentrated loadings (assimilable to of the Dirac functions). The equilibrium equation
[éq 1.1.1-1] must be replaced by:
()
()
()
dN
dx X
F
X
F
X
ext.
ic
I
NR
I
+
+
=
=
1
0
One omitted by simplicity the inertias which would undergo the same processing as the forces
external
F
ext.
.
I
represent the function of Dirac function located at the point
I
, them
F
ic
are the concentrated forces applied
with the beam.
For the application of the finite element method, the equilibrium equation must be written in the form
principle of virtual work which is in this case:
()
NR FD
dx dx
F
v dx
F
v
ext.
ic
I
NR
I
=
+
=
1
éq 2.1-1
Any confusion being excluded,
I
indicate the measurement of Dirac function associated with the point
I
,
v
is a field of
longitudinal displacement kinematically acceptable unspecified.
In practice, it is supposed that there is no force concentrated inside the elements of beam, but
only with the nodes ends.
2.1.1 Determination of the matrix of rigidity
It corresponds to the expression of the virtual work of the interior forces according to a displacement
given. I.e.:
NR FD
dx dx
L
0
for an element length L.
The elastic relation of behavior is introduced:
()
NR X
ES of
dx
=
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While choosing by function test:
()
()
()
()
v X
X
X
L
v X
X
X
L
=
= -
=
=
1
2
1
and
one obtains directly:
() ()
[
]
() ()
[
]
NR ddx dx
ES
L
dx dx
ES
L U L
U
NR ddx dx
ES
L
dx dx
ES
L U L
U
O
L
O
L
O
L
O
L
=
-
= -
-
=
=
-
1
2
0
0
and
The matrix of rigidity of the element is thus:
K
=
-
-
ES
L
1
1
1
1
Note:
In the expression of the virtual work of the interior efforts,
U
intervenes only for
()
U 0
and
()
U L
:
U
was not discretized inside the element. This is why the element is qualified
of “exact”: one obtains the exact solution with the nodes, but only with the nodes.
2.1.2 Determination of the second member
The second member is the expression of the virtual work of the efforts applied.
The second associate member with the loading distributed and the functions tests previously introduced
is:
()
()
F
F
F
F
X
X
L dx
F
F
X xL dx
ext.
ext.
1
2
1
0
1
2
0
1
1
=
-
=
with
Note:
In
AFFE_CHAR_MECA_F
, one can introduce
F
ext.
like an unspecified function of
X
.
On the level of the calculation of
F
1
and
F
2
, on the other hand, integration is made by supposing that
F
ext.
vary linearly between the values taken with the nodes ends. If one must modelize one
radial force distributed nonlinear, it is then necessary to discretize more finely.
But let us insist on the fact that whatever the form of
()
F
X
ext.
(polynomial or different), if one
can calculate the integrals exactly
F
1
and
F
2
, the solution of the static problem will be exact
with the nodes of the problem.
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The virtual work of the concentrated forces (given by assumption to the nodes of the elements) does not intervene
not directly on the level of the element.
One introduces these forces concentrated in the form of nodal forces, directly in the vector
assembled of the second member.
2.1.3 Calculation of the efforts to the nodes of the beam
Complete P.T.V [éq 2.2-1] is written indeed on the assembled system.
In addition, by writing the formula of integration per part on all the structure (beam
[
]
X X
O
,
1
):
() () () ()
[
]
[]
()
NR v dx
NR X v X
NR X v X
NR
v
NR v dx
X
O
O
I
I
NR
I
X
J
M
J
=
-
+
-
=
=
,
,
1
1
1
1
éq 2.1.1-2
J
representing all the intervals without discontinuity of normal effort, therefore without concentrated force, and
[]
NR
I
jumps of
NR
between these intervals.
Indeed, by bringing this expression closer to the PTV, one finds, for each loading concentrated (in
choosing functions test
v
adapted):
[]
I
NR
NR
F
I
ic
=
=
1,
Each finite element of beam is by assumption an interval without discontinuity. There can thus be
discontinuity of the internal efforts
NR
from one element to another if there is a force concentrated on the node
connecting the two elements.
The internal efforts for an element are determined in the following way:
The equilibrium equation inside an element is:
NR
F
X
item
,
+
=
0
The formula of integration by parts [éq 2.1.1-2] on the element gives:
()
() ()
() ()
[
]
() ()
NR X v dx
NR L v L
NR
v
F
X v X dx
L
X
ext.
O
L
0
0
0
=
-
+
,
While considering
()
NR L
and
()
NR O
like data, one could have obtained this formula directly
PTV [éq 2.1-1].
By still taking the functions test:
()
()
()
()
v X
X
X
L
v X
X
X
L
=
= -
=
=
1
2
1
and
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One obtains:
() ()
[
]
()
() ()
[
]
()
()
()
[]
()
()
-
-
= -
+
-
=
+
-
=
-
ES
L U L
U O
NR O
F
ES
L U L
U O
NR L
F
NR O
NR L
K U O
U L
F
F
1
2
1
2
that is to say
I.e. the internal efforts are obtained while cutting off with the product
K U
nodal forces
equivalent to the distributed loads
F
ext.
.
It is also observed that they are of opposite sign. So that the sign is the same one from one element to another, it
is thus necessary to change the sign of
()
NR O
calculated by this method. It is what is made by the calculation of
the option
EFGE_ELNO_DEPL
.
2.1.4 Determination of the matrix of mass
The matrix of mass to be coherent with the matrix of rigidity is given from same
functions test. However, it is not possible any more to calculate exactly the associated nodal forces
without making assumption on the form of the solution. The calculation of the matrix of mass will involve one
error of discretization.
A dynamic calculation will thus require a discretization of the structure of beam in small elements,
what is not the case for a static calculation. It goes without saying that in the case of a dynamic calculation, it
calculation of the efforts which one will lead as to [§2.1.2] by cutting off the nodal forces from inertia is
also approximate. The solution
U
is selected in the space generated by the functions tests (it be-with
to say the polynomials of degree to most equal to 1):
() () () ()
U
U
X
U L
X
=
+
0
1
2
The matrix of mass appears in the expression of virtual work due to the inertias:
W =
=
U
U
T
V MR. U
U
% %,
.
1
2
Work is also written:
()
()
()
() ()
() ()
()
()
() ()
(
)
()
()
() ()
(
)
W =
LV X U X, T dx
= dS = S
U X, T
=
X U T +
X U T
W = L S xx X X
dx
W =
S L
X
X X X
dx
O
m
m
S
O
T
T
O
% %
% %
% %.
with:
in the case of a homogeneous material.
While taking
, one a:
1
1
2
2
1
2
1
2
1
2
1
2
V
U
V
U
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The matrix of mass is thus written:
M
M
=
S
L dx
L
dx
L
dx
L dx
=
SSL
O
O
O
O
12
1 2
1 2
22
6
2 1
1 2
and made calculations:
2.2
Free movement of torsion around the longitudinal axis
The problem is similar to that of traction compression. For a beam
, charged by
torques distributed
()
X
X
and of the concentrated moments
ic
, the principle of virtual work
is written:
()
M ddx dx
dx
X
X
ic
I
NR
I
=
+
=
1
,
The law of behavior is:
()
M X
G C ddx
X
X
=
Except for the variables, this equation with the same form as that of the movement of traction and compression.
By using the same reasoning, one obtains the same expressions for the matrices of mass and
of stiffness elementary is:
K
M
= G C
L
=
C L
1
1
1
1
6
2
1
1
2
-
-
and
.
The calculation of the matrix of mass like having previously required to discretize the field
solution.
The second member, due to the couple
X
distributed, is in the same way obtained that for the movement of
traction and compression:
()
()
()
()
1
2
1
2
1
X
dx
X
dx
X =
xL
X = xL
O
L
X
O
L
X
-
2.3
Movement of bending
We place ourselves here within the framework of a right beam at constant section of Timoshenko type. Us
let us take account of the effects of transverse shearing. The beam of Euler-Bernoulli will be then treated
by simplification of the equations of Timoshenko.
The description of the bending is more complex than the preceding movements, but a judicious choice
functions tests will enable us to obtain of the same results forms.
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2.3.1 Bending in the plan (x0z)
With obvious notations and while not being interested initially, as in the cases
precedents, with the efforts of inertia, the principle of virtual work is written for the movement of bending
in the plan (x0z):
(
)
(
)
()
()
V
M
T
m
T
m
Z
y
Z
y
ic
I
NR
I
ic I
ext.
ext.
+ +
=
+
+
+
=
'
'
,
1
éq 2.3.1-1
for all
()
,
kinematically acceptable.
The matrix of rigidity results from the expression of the virtual work of the interior forces which one goes
to clarify by using the relation of behavior then while integrating by parts:
(
)
()
(
)
() ()
()
(
)
() ()
()
(
)
[
]
(
)
(
)
() ()
() ()
[
]
V
M
K SG W
E I
K SG W L
L
L
W
K SG W
K SG
I.E.(internal excitation)
L
L
I.E.(internal excitation)
Z
y
Z
y
y y
Z
Z
Z
y
y
y
y
y y
+ +
=
+
+ +
=
+
-
+
-
+
+
+
+
-
-
'
'
'
'
'
'
'
''
'
'
'
'
“
'
0
0
0
0
0
The functions tests which one will choose “will check the equilibrium equations without second member”,
i.e. [éq 1.3.1-2] and [éq 1.3.1-3]:
(
)
“
'
“
+
=
-
+
=
0
0
I.E.(internal excitation)
K SG
y
Z
éq 2.3.1-2
Under these conditions, nodal forces, expression of the work of the interior forces in these
virtual displacements given are expressed exactly, without assumption on the form of the solution, in
function of displacements in end of beam as in the preceding cases:
(
)
() ()
()
(
)
() ()
()
(
)
[
]
() ()
() ()
[
]
V
M
K SG W L
L
L
W
I.E.(internal excitation)
L
L
Z
y
Z
y
y
y
+ +
=
+
-
+
+
-
'
'
'
'
'
'
0
0
0
0
0
éq 2.3.1-3
Note:
It is clear that the condition [éq 2.3.1-2] led to functions tests depending explicitly
geometrical and material characteristics of the beam, but that A do not laugh at awkward.
The couples of functions selected tests are:
()
(
)
,
,
,…,
=
=
+
I
I
I
4
1
4
Code_Aster
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Titrate:
“Exact” elements of beams (right and curved)
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where, while having noted
y
y
Z
I.E.(internal excitation)
K SGL
=
12
2
, functions
I
are defined by:
()
()
()
1
3
2
5
2
3
2
6
2
1
1
2
3
1
6
1
1
1
4
2
2
2
1
1
3
4
(X) =
+
X
L
X
L
X
L +
(X) = L + xL
X
L
(X) =
L
+
X
L
+
X
L
+
X
L
(X) =
+
X
L
X
y
y
y
y
y
y
y
y
y
- -
+
-
-
+
-
- +
()
()
L + +
(X) =
+
X
L
+
X
L
+
X
L
(X) = L + xL
X
L
(X) =
L
+
X
L
+
X
L
+
X
L
(X) =
y
y
y
y
y
y
y
-
-
-
-
-
1
1
1
2
3
6
1
1
1
2
2
2
3
3
2
7
4
3
2
8
(
)
1
1
3
2
2
+
X
L
+
+
X
L
y
y
-
éq 2.3.1-4
One checks without difficulty that the couples
(
)
I
I
,
+
4
check well [éq 2.3.1-2]. Moreover:
(
)
()
(
)
()
()
(
)
()
(
)
()
()
(
)
()
(
)
()
()
(
)
()
(
)
()
()
1
5 0
1
5
2
6 0
2
6
3
7 0
3
7
4
8 0
4
8
1 0
0 0
0 1
0 0
0 0
1 0
0 0
0 1
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
()
()
()
()
=
=
=
=
=
=
=
=
L
L
L
L
éq 2.3.1-5
The matrix of rigidity results easily from [éq 2.3.1-3] (by ordering the columns according to
()
() ()
()
(
)
W O
O
W L
L
y
y
,
.
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
()
()
()
()
K =
I.E.(internal excitation)
L
+
L
L
+
L L
L
L
+
L
y
y
y
y
y
12
1
1
2
1
2
4
12
2
2
12
1
2
4
12
3
2
2
2
-
-
-
-
Sym
It is clear that the calculation of the efforts acts in the same way that to [§2.1.3].
2.3.2 Bending in the plan (xOy)
The matrix of rigidity for a movement of bending in the plan (xOy) is obtained in the same way that
in the preceding case. The functions tests which lead to an exact expression of the nodal forces
must this time check (equation similar to [éq 2.3.1-2]):
(
)
“
'
“
'
-
=
-
-
=
0
0
I.E.(internal excitation)
K SG
Z
Z
éq 2.3.2-1
The couples of functions selected tests are:
(
) (
) (
) (
)
1
5
2
6
3
7
4
8
,
;
,
;
,
;
,
-
-
-
-
I
being given by [éq 2.3.1-4] while having replaced
y
by
Z
Z
y
I.E.(internal excitation)
K
SGL
=
12
2
. The matrix of rigidity
obtained is:
(
)
(
)
(
)
(
)
K =
I.E.(internal excitation)
L
+
L
L
+
L
L
L
L
+
L
Z
Z
Z
Z
Z
12
1
1
2
1
2
4
12
2
2
12
1
2
4
12
3
2
2
2
-
-
-
-
sym
Code_Aster
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Titrate:
“Exact” elements of beams (right and curved)
Date:
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Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
2.3.3 Determination of the matrix of coherent mass with the matrix of rigidity
Option
“MASS_MECA”
of the operator
CALC_MATR_ELEM
.
2.3.3.1 Bending in the plan (xoz)
Let us consider the movement of bending in the plan (X O Z), the work of the inertias is written:
(
)
W =
W w+
dx
=
dS
=
y dS
m
y
I
y
O
L
m
S
I
S
Z
Z
% %
% %
.
with
and
2
In the case of a homogeneous material, we have:
m
I
Z
= S
= I
Z
.
and
()
W X, T
and
()
y
X, T
are discretized on the basis of function tests introduced for the calculation of
stamp rigidity, that is to say:
() ()
()
() ()
()
() ()
()
() ()
()
W X T
X W T
X
T
X W T
X
T
X T
X W T
X
T
X W T
X
T
W
W
W
y
y
y
y
y
T
W
y
y
W
(,)
()
()
(,)
()
()
~
~
=
+ + +
=
+
+
+
=
=
=
in other words:
with
and
1
1
2
3
2
4
5
1
6
7
2
8
1
2
1
2
3
4
1
2
1
2
1
2
W
W
y
T
y
y
=
=
with
W
~
5
6
7
8
By integrating these notations in the expression of the work of the inertias, one a:
W
dx
W
dx
m
O
L
T
W
wt
T
T
m
O
L
T
W
wt
T
T
y
y
y
y
=
+
=
+
I
I
Z
Z
or:
W
W
W
W
W
W
W
W
~
~
~
~
~
~
~
~
% %
% %
% %
% %
.
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
24/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
One deduces the form from it of the matrix of mass:
()
() () ()
()
()
()
(
)
M
M
= m m =
S X X
X + I X
X
X dx
=
S
+
ij
ij
O
L
I
J
Z
i+
j+
Z
L + L + L
11L
11L Z
L
L + L + L
L + L + L24
L + L Z + L
L
3L
I
J
Z
Z
Z
Z
Z
Z
Z
Z
-
-
-
-
-
4
4
2
13
35
7
10
3
210
120
24
9
70
3
10
6
13
420
3
40
105
60
120
13
420
1
2
2
2
2
2
2
2
2
2
2
3
3
3
2
2
2
for 1 to 4 and from 1 to 4.
That is to say:
(
)
Z
40
L
24
L
L
L
120
L + L + L
L + L + L
L + L + L
Z
Z
Z
Z
Z
Z
Z
Z
2
Z
Z
Z
Z
Z
Z
Z
Z
+
I
+
L
+
L
+
L + L + L
-
-
-
-
-
-
-
-
-
-
2
3
3
3
2
2
2
2
2
2
3
3
3
2
140
60
13
35
7
10
3
11
210
11
120
24
105
60
120
2
2
1
6
5
1
10
2
6
5
1
10
2
2
15
6
3
1
10
2
L
L + L
L
L + L + L
Z
Z
Z
Z
Z
30
6
6
6
5
1
10
2
2
15
6
3
2
2
-
-
It should well be noted, as in [§2.1.4], that in the dynamic case, one is not ensured to have one
exact solution with the nodes, as it is the case in statics.
2.3.3.2 Movement of bending around the axis (O Z)
In the same way, for the movement of bending around the axis (O Z), in the plan (X O y), the work of the forces
of inertia is written:
(
)
v
v
dx
Z dS
I
Z
I
Z
O
L
I
S
y
y
y
+
=
=
m
with:
% %
% %
.
2
This time
()
v X, T
and
()
Z
X, T
are discretized in accordance with [§2.3.2] by:
()
() ()
() ()
() ()
()
()
()
() ()
() ()
() ()
()
()
=
+
=
+
+
v X, T
X v T
X
T
X v T
X
T
X, T
X v T
X
T
X v T
X
T
Z
Z
Z
Z
Z
1
1
2
3
2
4
5
1
6
7
2
8
1
2
1
2
-
-
-
-
Code_Aster
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Titrate:
“Exact” elements of beams (right and curved)
Date:
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Key:
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Page:
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R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
We obtain the matrix of following mass then:
(
)
M =
S
+
Z
L + L + L
L
L Z
L
L + L + L
L
L
L
L + L Z + L
L
L Z
L
L
L
L
L + L + L
L
L
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
1
2
13
35
7
10
3
11
210
11
120
24
9
70
3
10
6
13
420
3
40
24
105
60
120
13
420
3
40
24
140
60
3
120
13
35
7
10
3
11
210
11
2
2
2
2
2
2
2
2
2
2
3
3
3
2
2
2
2
2
3
3
2
2
2
+
+
-
-
-
+
+
-
-
-
-
-
(
)
2
2
2
3
3
2
120
24
105
60
3
120
2
2
2
2
1
6
5
1
10
2
6
5
1
10
2
2
15
6
3
1
10
2
30
6
6
6
5
1
10
2
2
15
6
3
Z
Z
Z
Z
L
L + L + L
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
+
I
+
L
L
L + L + L
L
L + L
L
L + L + L
-
-
-
-
- +
- -
- +
In the model of beam of Euler-Bernoulli, the effects of transverse shearing are neglected. It
is thus enough, to obtain the matrices of mass and rigidity associated with this model, to cancel them
variables
y
and
Z
contained in the matrices of mass and rigidity of the model of Timoshenko.
(
y
and
Z
the coefficients of form utilize
K
y
and
K
Z
, opposite of the coefficients of shearing).
It will be noted that in the Euler-Bernoulli model programmed in Aster, the inertia of rotation is
also neglected. It is thus necessary, for this model, to cancel the terms in “
I
Z
“and”
I
y
“in
stamp of mass of the model of Timoshenko.
2.4
Stamp of mass reduced by the technique of the concentrated masses
The matrix of mass is thus reduced to a diagonal matrix and is obtained by the option
“MASS_MEGA_DIAG”
of the operator
CALC_MATR_ELEM
.
The element beam is considered with constant section
S
and with constant density
.
The technique known as of “Lumping” consists in summoning on the diagonal all the terms of the line of
coherent matrix and to cancel all the extra-diagonal terms.
Code_Aster
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
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J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
With regard to the diagonal component related to the movement of traction and compression
()
M
11
and
that related to the movement of torsion
()
M
44
, we have:
(
)
M
S L
M
I + I
L
I I
y
Z
y
Z
11
44
2
2
=
=
,
: geometrical moments.
One can consider that these components were obtained by sharing the element of beam into two
equal shares length
L
2
then by associating the mass and inertia obtained the node of
half-element. For
M
44
, the preceding expression corresponds to a choice: one could also
to write:
M
C L
44
2
=
.
Note: Comparison with the methods of numerical integration.
One can note that if one carries out a in the following way approached integration:
()
(
)
F
=
(E)
N
F has
has
I
E, N
E
i=1, N
IE
IE
my
: nodes of the element
: node of the element numbers
one obtains an identical result (for a beam:
()
my E
L
=
and
N
=
2
).
The diagonal components related to the movements of bending which are programmed are:
M
S L,
M
S L,
M
S
L, S L
I
L,
M
S
L, S L
I
L.
y
Z
22
33
55
3
2
66
3
2
2
2
105
48
2
15
105
48
2
15
=
=
=
+
=
+
Min
Min
The components well are found
M
22
and
M
33
dependant on the translations of the movements of bending by
technique of the masses concentrated with the nodes. On the other hand, the origin of the formulas used for
components
M
55
and
M
66
dependant on rotations, is unknown. One can simply notice that one
find the values:
S L + I
L,
S L + I
L
Z
y
3
3
105
2
15
105
2
15.
for the diagonal components of the matrix of equivalent mass [§2.3]. But this matrix is not
not diagonal. Nevertheless, the results obtained by this method remain correct.
Code_Aster
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Titrate:
“Exact” elements of beams (right and curved)
Date:
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J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
3
Particular right beams
It is a question in this chapter of taking into account right beams whose section has properties which
were ignored until now, in particular the beams having a center of torsion excentré by
report/ratio with the neutral axis (the section does not have 2 axes of symmetry), and those whose section evolves/moves
continuously on their axis.
3.1
Eccentricity of the axis of torsion compared to the neutral axis
The center of torsion is the point which remains fixed when the section is subjected to the only moment of
torsion. It is also called center of shearing because an effort applied in this point does not produce
rotation
X
.
O
F
C
O
M
C
(not of rotation)
(not of displacement in C)
At the point C, the effects of bending and torsion are uncoupled, one can thus use the established results
in the preceding chapter. One finds the components of displacement as in point 0 while considering
rigid relation of body:
U (O) = U (C) + OC
=
Q
0
0
OC =
0
E
E
X
y
Z
with
vector rotation
and
O
Z
y
E
y
C
E
Z
Code_Aster
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Titrate:
“Exact” elements of beams (right and curved)
Date:
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Key:
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R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
In fact, one obtains:
U = U
U = U + E
U = U
E.
X
C
C
C
X
y
y
Z
X
Z
Z
y
X
-
éq 3.1-1
The change of variables given by [éq 3.1-1] is written matriciellement:
U
U
U
U
U
U
E
E
E
E
X
y
Z
X
y
Z
X
y
Z
X
y
Z
Z
y
Z
y
C
C
C
C
C
C
C
C
C
C
C
C
1
1
1
1
1
1
2
2
2
2
2
2
1 0 0
0
0 0
0 1 0
0
0 0 1
0 0
0
0 0 0
1
0 0
0 0 0
0
1 0
0 0 0
0
0 1
1 0 0
0
0 0
0 1 0
0 0
0 0 1
0 0
0
0 0 0
1
-
-
=
0
0 0
0 0 0
0
1 0
0 0 0
0
0 1
1
1
1
1
1
1
2
2
2
2
2
2
P
!
“
# # # # # # # # #
$
# # # # # # # # #
U
U
U
U
U
U
X
y
Z
X
y
Z
X
y
Z
X
y
Z
It is thus enough to determine the elementary matrices of mass
()
M
C
and of stiffness
()
K
C
in
identify (C, X, y, Z) where the movements of bending and torsion are uncoupled then to be transported
in the reference mark related to the neutral axis (O, X, y, Z) by the following transformations:
K
P K P
K
P K P
=
=
T
C
T
C
and
.
Values of
E
y
and
E
Z
are to be provided to Code_Aster via the operand
SECTION: “GENERAL”
of the operator
AFFE_CARA_ELEM
, default values being obviously
zero values.
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
3.2 Sections
variables
It is possible to take into account evolutionary sections in a continuous way for the beams
straight lines of Timoshenko and Euler (
POU_D_E
and
POU_D_T
only). One distinguishes two types from
variation of section:
·
linear or refines,
·
quadratic or homothetic.
The distinction between the two types is conceived easily by taking the example of a beam
rectangular:
·
if only one of side dimensions varies, one supposes in a linear way, then the surface of
cross-section varies linearly, and is given by:
S (X) = S
+ S
xL
1
2
1
1
1
-
·
when two side dimensions vary (in a linear way), the surface of section will evolve/move
in a quadratic way.
S (X) = S
+
S
S
X
L
1
2
1
2
1
1
-
Code_Aster makes it possible to treat sections '
RING
“,”
RECTANGLE
“and”
GENERAL
', but for
obvious reasons of geometry, all these types of section cannot admit the two types of
variation. The following table summarizes the existing possibilities.
Section
Constant
Linear
Quadratic
ring
yes
not
yes
rectangle
yes
yes
according to y
yes
general
yes
not
yes
For the section '
RECTANGLE
', it is the user who chooses the type of variation, while specifying
“REFINES”
or
“HOMOTHETIC”
in
AFFE_CARA_ELEM
. It should well be noted that in the case
“REFINES”
, them
dimensions can vary only according to Y.
We consider generally that the section varies according to the formula [éq 3.2-1]:
S (X) = S + C xL
m
1
1
éq 3.2-1
S
1
is the initial section in
X
=
0
C
is fixed by the knowledge of the final section
S
2
in
X
L
=
.
m
give the degree of variation:
m
=
1
linear variation,
m
=
2
quadratic variation.
The section varying, it goes from there in the same way inertias
()
I X
y
,
()
I X
Z
and
()
I X
p
.
Code_Aster
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Titrate:
“Exact” elements of beams (right and curved)
Date:
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Key:
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Page:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
We will thus have:
I (X) = I
+ C xL
y
y
m+
1
1
2
éq 3.2-2
I (X) = I
+ C xL
Z
Z
m+
1
1
2
éq 3.2-3
I (X) = I
+ C xL
p
p
m+
1
1
2
éq 3.2-4
C
is given for each formula starting from the value for
X = L: I, I, I
y
Z
p
2
2
2
.
The Coulomb and Young moduli (E) (G) are supposed to be constant.
The principle adopted by Code_Aster consists in calculating equivalent characteristics of section,
constants on the beam, starting from the real characteristics data at the two ends. These
equivalent characteristics thus depend on the phenomenon to which they contribute, in particular,
are distinct for the effects of rigidity or inertia.
3.2.1 Calculation of the matrix of rigidity
3.2.1.1 Determination of the equivalent section (S
eq
)
The determination of the equivalent section does not use nor the method taken with [§2.1.1] to obtain
stamp exact rigidity nor an approximation of the solution by a polynomial function like
described with [§2.1.4]. In fact, the method employed deviates from the finite element method and even of
the method of Galerkin, it consists in carrying out a resolution of the problem of the beam with section
variable without efforts distributed imposed, which makes it possible to clarify the efforts at the ends in function
displacements. This method is “coherent” with that of [§2.1.1] because the definite functions tests
in 2.1.1 1 or 0 is worth on the ends of the beam, therefore [éq 2.1-1] the nodal forces can be
“comparable” with efforts.
In addition, this method makes it possible to obtain exact results for the static problem without force
distributed and led as we will see it with a value
S
eq
included/understood enters
S
1
and
S
2
who, in the case
General, guarantees the convergence of the solution approximated towards the exact solution (without however
to know the command of convergence).
The section of the beam being variable, the equation of traction and compression in statics without effort distributed
imposed is written:
X E S (X)
U
X
=
X
L
NR (X) = E S (X) ux
0
0
with
éq 3.2.1-1
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
31/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
We determine the matrix of rigidity in the general case [éq 3.2-1], we deduce some thereafter
values of the equivalent sections for the cases
m
=
1
(linear progression) and
m
=
2
(progression
quadratic).
While integrating [éq 3.2.1-1], we have:
E S (X) ux = C
1
or, by taking account of the expression of
()
S X
:
E S + C xL
U
X = C
m
1
1
1
The constant of integration is given starting from the values of thrust loads to the nodes.
We integrate once again in order to obtain the efforts with the nodes according to
displacements
()
U
U
0
1
=
and
()
U L
U
=
2
:
(
)
U
X =
C
E S + C
X
L
U (X) =
C
E S
L
C + C
X
L + C if m =
U (X) =
C
E S
L
C L m +c xL +C if m
m
m
1
1
1
1
2
1
1
1
2
1
1
1
1
1
2
-
-
-
from where
and
ln
It is noted that the expression of
()
U X
is far from being polynomial.
By taking account of the fact that:
(
)
(
)
C
=
NR
X =
C
= + NR
X = L
C
=
+ C
U () U (L)
+ C
for
for
m
m
1
1
1
2
2
1
1
0
1
0
1
1
-
-
-
- +
- +
and that:
Code_Aster
®
Version
3
Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
32/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
we obtain:
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
for
and for
m
NR E C M
L C
U
U
NR E C M
L C
U
U
m
NR E C M
m
L
U
C
C
+ U
C
C
NR E C M
m
L
U
m
m
m
m
=
=
+
-
=
+
-
=
=
-
-
+
- +
+
- +
=
-
-
-
-
-
1
1
1
2
1
1
1 1
1
1 1
1
1
1
1
2
2
1
2
1
1
1
1
1
1
2
1
1
2
1
1
ln
ln
(
)
(
)
(
)
(
)
1
1 1
1
1 1
1
1
2
1
1
+
- +
-
+
- +
-
-
-
-
C
C
U
C
C
.
m
m
m
m
By replacing C by its value, that is to say:
(
) ()
(
) ()
(
)
(
)
· if
· if
m
C
S
S
,
NR
E
L
S
S
S
S
U
U
NR
E
L
S
S
S
S
U
U
m
C
S
S
NR
E
L S S U
U
NR
E
L S S U
U
=
=
-
=
-
-
-
=
-
-
-
=
=
-
=
-
=
-
1
1
2
1
2
1
1
2
1
2
1
1
2
2
2
1
2
1
2
1
2
1
1
1 2
1
2
2
1 2
2
1
ln
ln
ln
ln
We note that the matrices of rigidity, in the two treated cases, will have the same form as for
a constant section if one takes as equivalent section:
(
)
S
S
S
S
S
S
S S
eq
eq
=
=
for a section varying linearly
for a section varying in a quadratic way
2
1
2
1
1 2
-
-
ln ln
Code_Aster
®
Version
3
Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
33/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
3.2.1.2 Determination of a constant of equivalent torsion (C
eq
)
The equation of pure torsion of a beam with variable section, is written:
()
X G C X
Q
X
C (X)
C 1 C xL
(m or)
X
m2
=
=
+
=
+
0
1
2
1
with:
éq 3.2.1-2
The method is the same one as for the calculation of the equivalent section: it is a question of integrating the equation
the preceding one in order to obtain the efforts (torques
X
X
,
1
2
) according to displacements
with the nodes
(
)
X
X
1
2
,
and to deduce some, by comparison with the formulas with constant section,
expression the one geometrical moment polar are equivalent.
By integration of [éq 3.2.1-2], we have:
G C
C xL
X
D
m
X
1
2
1
1
+
=
+
,
D
1
the constant of integration is determined by the torques applied to the nodes.
(
)
()
(
)
(
)
Of:
we deduce:
=
=
X
m
X
m
X
D
G C
C xL
X
D
G C
L
C m+
C xL
D
1
1
2
1
1
1
2
1
1
1
+
-
+
+
- +
- +
,
.
We determine
D
2
starting from the system:
(
)
(
) (
)
(
)
(
)
(
)
(
)
(
)
X
X
m
X
m
X
m
D L
G C C m+ D
D L
G C C m+
C
D
D
C
C
1
2
1
1
1
2
1
1
1
2
2
1
2
1
1
1 1
1
1
1
=
=
=
that is to say:
-
+
-
+
+
+
+
+
-
- +
- +
- +
.
By taking account of the fact that:
D X
D X
L
1
1
1
2
0
= -
=
=
=
for
for
,
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
34/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
we have finally:
(
)
(
)
(
)
(
)
(
)
(
)
for
m
G
L
C C
C C
C
C
G
L
C C
C I
C C
X
X
p
X
X
=
=
-
-
-
=
-
-
1
2
2
1
2
1
1
2
2
1
2
2
1
1
2
1
2
3
2
3
2
3
2
3
1
2
2
3
2
2
3
2
3
2
3
2
1
We thus take in the case of a section varying linearly, a polar geometrical moment
equivalent
C
eq
following form:
(
)
(
)
C
C C
C C
C
C
eq
=
-
-
2
2
1
1
2
2
1
2
3
2
3
2
3
2
3
linear variation
(
)
(
)
(
)
(
)
(
)
(
)
for
m
G
L
C C
C C
C
C
G
L
C C
C C
C C
X
X
X
X
=
=
-
-
-
=
-
-
2
3
3
1
2
1
1
2
2
1
2
2
1
1
2
2
1
3
4
3
4
3
4
3
4
1
2
3
4
3
4
3
4
3
4
2
1
In the case of a section varying in a quadratic way, the polar geometrical moment is written:
(
)
(
)
C
C C
C C
C
C
eq
=
-
-
3
2
1
1
2
2
1
3
4
3
4
3
4
3
4
quadratic variation
3.2.1.3 Determination of the equivalent geometrical moments
In fact, it does not seem possible to find, as we did for the section or the moment
geometrical polar, of equivalent geometrical moments
(
)
I
I
y
Z
eq
eq
and
who would come to substitute themselves
at the geometrical moments
(
)
I
I
y
Z
and
in the expression of the terms of the matrix of rigidity.
We expose here the method suggested by J.R. BANERJEE and F.W. WILLIAMS [bib3] which clarifies
the matrix of rigidity in the case of a movement of bending of a Euler-Bernoulli beam to section
variable (linear or quadratic).
The results correspond to those programmed in Code_Aster for a beam of the type
Euler-Bernoulli with variable section (linear or quadratic). By extension, the same step is there
applied for the beams of the Timoshenko type. The results are not here detailed.
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
35/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
Let us consider the bending in the plan (X O y).
On the basis of the static equation of the movement of bending of a beam of the Euler-Bernoulli type:
2
2
2
2
0
X
I.E.(internal excitation) (X)
v
X
v
X
Z
Z
=
=
,
and
()
v X
is expressed according to four constants of integration
(
)
C
C
C
C
1
2
3
4
,
. These constants
are determined by the values of displacements to the nodes:
v (O)
v
v (L)
v
(O)
(L)
,
v
v
B
C
C
C
C
B
V (X)
X I.E.(internal excitation) (X)
v
X
M (X)
I.E.(internal excitation) (X)
v
Z
Z
Z
Z
Z
Z
y
Z
Z
Z
=
=
=
=
1
2
1
2
1
2
3
4
2
2
2
1
2
1
2
that is to say:
stamp (4 X 4).
Efforts:
and moments:
=
,
=
=
X,
2
also express themselves according to these constants of integration, and one can write:
T
M
T
M
D
C
C
C
C
D
y
Z
y
Z
1
1
2
2
1
2
3
4
=
,
stamp (4 X 4).
The matrix of rigidity corresponds to the product
D B
-
1
. The terms of this matrix are clarified in
following tables.
Let us recall that
or
Let us pose
=
,
I X
I 1 C xL
m
,
W
I.E.(internal excitation) Cl, W
I.E.(internal excitation)
C
L
, W
I.E.(internal excitation)
C
L.
Z
Z
m
Z
Z
Z
1
1
()
+
=
=
=
=
+
2
1
2
2
3
3
1
2
1
1
Code_Aster
®
Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
36/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
(
)
(
)
(
)
(
) (
)
(
) (
)
(
)
(
)
(
) (
)
(
)
Variation closely connected
Quadratic variation
=
=
=
=
m
C
I
I
m
C
I
I
C
C + c+
C
2 C c+
C C
2 C c+ c+
C c+ c+
C
C
C c+
C
C c+
c+ ln c+
2 C
C
c+
3
2
1
1
2
1
2
4
3
3
3
1
1 2
3
2
1 2
3
4
2
1
4
1
2
1
1
2
1
2
1
1
2
2
3
4
2
5
4
2
6
3
5
2
2
3
1
1
4
-
-
+
+
=
-
=
-
-
The matrix
K
is written then:
K
+
=
1
3
1
2
2
3
1
2
3
1
4
2
2
1
5
3
1
2
3
1
6
W
W
W
W
W
W
W
W
W
W
-
-
-
Sym
Now let us consider the bending in the plan (X O Z).
For the sections with quadratic variation, the step is identical. But it differs for the sections
with linear variation (according to y only).
One calculates the terms of the matrix of rigidity corresponding to the bending in plan (0, X, Z) by
values given in the following table.
(
)
(
)
(
) (
)
Variation closely connected: bending in the plan (0xz)
C
I
I
ln DC
C
C
C
C
C
C
C
=
-
+
-
+
-
+ -
=
-
=
-
+
+ -
2
1
1
1
1
1
1
1
1
5
2
4
6
3
5
1
2
1
2
1
2
2
2
1
2
ln
Code_Aster
®
Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
37/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
In the case of the beams of Timoshenko, for the coefficients of shearing, one applies to the section
reduced
K S
relations used for the section, namely:
()
(
)
(
) ()
()
(
)
() ()
()
()
K S
K S
K S
ln K S
ln K S
K S
K S
K S
ln K S
ln K S
K S
S S K K
K S
S S K K
y
eq
y
y
y
y
Z
eq
Z
Z
Z
Z
y
eq
y
y
Z
eq
Z
Z
2
2
=
-
-
=
-
-
=
=
2
1
1
1
2
1
1
2
1
2
2
1
2
1
2
1
2
1
1 2
1 2
if the variation is closely connected
if the variation is quadratic
and one introduces the additional terms into
K
in the same way that for a constant section.
Calculations are not here detailed. A matrix is obtained
K
of the same form than previously
with for main amendment the value of
:
·
variation
closely connected
(
) (
)
(
)
= +
+ - +
+
C
ln C
C
C C
2
1
2
12
2
2 3
·
variation
quadratic
(
)
= + +
+ +
C
C
C C
C
3
2 3 2
1
3
3
3
3.2.2 Calculation of the matrix of mass
3.2.2.1 By the method of the equivalent masses
“Average” values are calculated for the section, the reduced section, and the moments, namely:
()
S
S + S
S
L
S X dx = S + S + S S
I
I + I
I
I + I
I
I + I + I + I
O
L
y
y
y
Z
Z
Z
y
y
Z
Z
X
=
=
=
=
=
1
2
1
2
1 2
2
1
3
2
2
2
1
2
1
2
1
2
1
2
if the variation is closely connected
if the variation is quadratic
whatever the variation
The matrix of mass is then calculated like that of a beam having these characteristics.
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
3.2.2.2 By the method of the masses concentrated (diagonal matrix)
·
If the section varies in a way closely connected, the programmed matrices correspond, with regard to
movements of traction and compression and torsion, with those of the beams prismatic,
by using sections and equivalent inertias of torsion:
-
for traction compression:
L
S
S
S
S
3
8
0
0
3
8
1
2
2
1
+
+
-
for torsion:
(
)
.
L I
I
I
I
y
Z
y
Z
X
X
1
1
2
2
1
4
1 0
0 1
2
+
+
+
-
for the movements of bending:
()
(
)
()
()
()
()
(
)
S L
MR. M
S L
0
M
M
v W
v W
M
M
S L, S L
L I I
M
M
S L,
,
,
,
Z
y
Z
y
eq
eq
y
y
eq
2
1
5 5
6.6
2
11 11
12 12
1
1
2
2
5 5
11 11
3
2
6.6
12 12
3
2
0
2
105
48
15
105
1
1
2
1
1
,
,
,
,
,
min
min
=
=
+
+
=
=
with
(
)
S L
L I I
S
S
S
eq
Z
Z
eq
2
1
1
2
48
15
2
2
+
+
=
+
with
·
If the section varies in a homothetic way, the matrices are programmed, for the different ones
movements, in the following way:
-
for the traction and compression:
(
)
(
)
5
12
0
0
5
12
1
2
1
2
1
2
S + S L
S + S L
U
U
Code_Aster
®
Version
3
Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
39/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
-
for torsion:
(
)
L
I
I
I
I
y
y
Z
Z
X
X
1
2
1
2
1
4
1 0
0 1
2
+
+
+
-
for the bending in each of the two plans:
(
)
()
(
)
(
)
()
()
()
()
(
)
(
)
(
)
=
=
,
+
=
with
and:
5
12
0
5
12
2
105
2
48
15
1
2
5 5
6.6
1
2
11 11
12 12
1
1
2
2
5 5
11 11
1
2
3
1
2
3
6.6
12
1
1
2
1
2
S + S L
M
M
S + S L
M
M
v W
v W
M
M
S
S
L
S
S
L
I
I
L
M
M
Z
y
Z
y
y
y
2
,
,
,
,
,
,
,
,
min
+
+
+
(
)
(
)
(
)
12
1
2
3
1
2
3
2
105
2
48
15
1
2
=
,
+
min
.
S
S
L
S
S
L
I
I
L
Z
Z
+
+
+
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
4
Geometrical rigidity - prestressed Structure
Option: “
RIGI_MECA_GE
“
In the case of a prestressed structure, therefore subjected to initial efforts (known and independent
time), one cannot neglect in the equilibrium equation the terms introduced by the change
of geometry of the virgin state of stress in a prestressed state [bib2].
V
O
V
*
V
ij
O
ij
virgin state
of stress
state
prestressed
state
deformed
Appear 4-a: the various states
This change of geometry does not modify the equilibrium equation, within the framework of the assumption of small
disturbances (HP) around
V
O
(and of
V
*
), that by the addition of a linear term in displacements
whose associated matrix is called geometrical matrix of rigidity and who expresses himself by:
W
U
X
v
X FD
G
K D
I
ijo
V
K D
J
O
=
3
3
where
U
3D
is displacement (resp.
v
3D
virtual displacement kinematically acceptable) taken with
to leave
V
*
(but compared to
V
O
within the framework of the HP) and
O
the prestressing (of Cauchy if one
wants) since one is within the framework of the HP.
W
G
being a symmetrical bilinear form in
U
3D
and
v
3D
, it can be interpreted like the variation
of a potientiel
U
G
.
W
U
G
G
=
One a:
2
3
3
U
X
U
X
G
kD
I
V
ijo
K D
J
O
=
U
For a model of beam 3D, the tensor of stresses is reduced in the local axes of the beam to
components
xx
xy
xx
,
and
, from where:
2
2
2
3
3
3
3
3
3
U
U
X
U
X
U
X
U
y
U
X
U
Z
G
xx
O
V
I D
I D
xy
O
I D
I D
xz
O
I D
I D
O
=
+
+
Code_Aster
®
Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
41/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
Terms
U
X
U
X
U
y
U
X
U
Z
X D
X D
X D
X D
X D
3
2
3
3
3
3
,
and
are neglected [bib5]. Moreover, in the reference mark
room of the center of torsion of the beam:
(
)
()
()
()
(
)
()
()
(
)
()
()
,
,
,
'
'
'
'
'
'
'
U
X y Z
U X
Z
X
y
X
U
X y Z
v X
Z
X
U
X y Z
W X
Z
X
U
U Z
y
v Z
W y
X D
y
Z
yD
y
Z D
X
y
Z
Z
y
X
X
X
X
3
3
3
0
0
=
+
-
=
-
=
+
=
+
-
-
-
-
+
and
from where one fires, according to the preceding assumption:
(
) (
)
(
)
(
)
()
2
2
2
2
2
U
v Z
W y
W y
v Z
G
xx
O
V
X
X
xy
O
X
X
xz
O
X
X
O
=
-
+
+
+
+
+
-
-
'
'
'
'
'
'
'
'
However, the generalized efforts are connected to the stresses by the expressions:
NR
V
V
M
Z
M
y
xx
O
S
y
xy
O
S
Z
xz
O
S
y
xx
S
Z
xx
S
°
=
=
=
=
= -
One deduces some:
() ()
(
)
(
) ()
(
)
2
2
2
2
2
2
2
2
2
2
2
U
NR
v
W
M v
M W
V W
V v
y
Z
y
Z
G
O
L
yo
X
zo
X
yo
X
zo
X
xx
O
V
X
xy
O
xz
O
X X
O
=
+
-
-
+
-
+
+
+
+
°
'
'
'
'
'
'
'
'
'
'
By supposing, moreover, that
xx
O
is constant in the discretized element (what is inaccurate for example
for a vertical beam subjected to its actual weight) and that
X
vary linearly compared to
X
X
L
X
L
L
L
X
X
= -
+
= -
+
1
1
1
1
2
1
2
,
'
from where
, it comes:
(
) ()
(
)
xx
O
V
X
O
y
Z
O
y
Z
O
y
Z
O
y
Z
O
y
Z
NR
L
I
I
S
NR
L
I
I
S
NR
L
I
I
S
NL I I
S
+
=
+
-
+
-
+
+
2
2
2
1
2
1
2
'
,
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
By neglecting in particular the terms which had with the influence of the shearing action on the mode of buckling
or of vibration, and by supposing that the distributed loads are null on an element, one a:
(
)
(
)
NR (X)
V (X)
V (X)
M
M
M
xL + M
M
M
M
xL + M
y
Z
y
y
y
y
Z
Z
Z
Z
=
,
=
=
=
=
constant
constant,
constant,
2
1
1
2
1
1
-
-
Under this assumption and for the model of Euler-Bernoulli (for the model of Timoshenko, one uses
even matrix), one obtains the following matrix:
With
With
With
With
=
1
2
3
0
Higher triangular part of the geometrical matrix of rigidity with:
(
)
()
()
()
With
1
1
1
1
1
1
1
1
2
1
2
1
2
1
2
1
2
3
4
5
6
1
2
1 2
2
2
2
10
3
1 2
2
2
2
10
4
12
12
12
12
5
2
15
6
2
15
=
-
-
+
-
-
-
-
+
-
+
-
+
-
-
U
v
W
NR
L
M
L
M
L
V
NR
NR
L
M
L
M
L
V
NR
I
I
NR
S L
M
M
L V
M
M
L V
L NR
L NR
X
y
Z
O
y
O
y
O
zo
O
O
zo
zo
yo
O
y
Z
O
zo
zo
yo
yo
y
O
zo
O
O
Code_Aster
®
Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
(
)
()
(
)
()
With
2
2
2
2
2
2
2
1
2
1
2
1
2
1
2
1
2
1
7
8
9
10
11
12
1
2
1 2
2
2
2
10
3
1 2
2
2
2
10
4
2
2
2
2
2
2
12
12
=
-
+
+
-
+
-
-
+
-
+
+
-
+
-
-
-
+
U
v
W
NR
L
M
L
M
L
V
NR
NR
L
M
L
M
L
V
NR
M
L
M
L
V
M
L
M
L
V
I
I
NR
S L
M
M
L V
M
X
y
Z
O
y
O
y
O
zo
O
O
zo
zo
yo
O
y
O
yo
zo
zo
zo
yo
y
Z
O
zo
zo
yo
y
O
(
)
()
(
)
()
()
(
)
()
()
M
L V
NR
M
M
L V
L NR
NR
M
M
L V
L NR
yo
zo
O
zo
zo
yo
O
O
y
O
y
O
yo
O
2
1
2
1
2
12
12
5
10
12
12
30
6
10
12
12
30
-
-
-
+
-
-
-
+
+
-
Code_Aster
®
Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
44/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
(
)
()
(
)
()
(
)
()
With
3
2
2
2
2
2
2
1
2
1
2
1
2
1
2
7
8
9
10
11
12
7
8
1 2
2
2
2
10
9
1 2
2
2
2
10
10
12
12
12
12
11
2
15
12
2
15
=
-
-
-
-
-
-
+
+
+
-
+
-
+
-
-
U
v
W
NR
L
M
L
M
L
V
NR
NR
L
M
L
M
L
V
NR
I
I
NR
S L
M
M
L V
M
M
L V
L NR
L NR
X
y
Z
O
yo
yo
zo
O
O
zo
zo
yo
O
y
Z
O
zo
zo
yo
yo
yo
zo
O
O
By using the equalities
M
X
V =
y
Z
-
0
and
M
X
V =
Z
y
+
0
, the programmed matrix is found.
Moreover, to be able to deal with the problems of discharge of thin beams, requested
primarily by moments bending and efforts normal, it is necessary to add the assumption of
rotations moderated in torsion [bib 4], [bib 5].
This results in the following shape of the field of displacements:
(
)
()
()
() ()
(
)
()
() ()
(
)
U X y Z
U X
Z
X
X
X
y
X
X
X
y
X
Z
Z
X
y
,
=
+
+
-
-
In addition, if the center of torsion
C
is not confused with the center of gravity, it is necessary to write:
(
)
()
(
)
(
)
()
(
)
V X y Z
v X C
Z Z
W X y Z
W X C
y y
C
X
C
X
,
,
,
,
=
- -
=
+ -
These two amendments bring additional terms in the geometrical matrix of rigidity:
The assumption of moderate rotations results in adding with
2U
G
the term:
()
()
2
1
U
M
M
V
V
G
zo
O
L
X y
X
yo
X Z
X
yo X y
zo X y
=
-
+
+
+
,
,
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
45/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
Terms of the matrix
With
to add are:
(
)
(
)
(
)
(
)
4 5
2
10 11
2
4 6
2
10 12
2
1
2
1
2
-
+
-
-
-
-
-
+
:
:
:
:
M
M
M
M
zo
zo
yo
yo
With regard to the eccentricity of the center of torsion, it is necessary to add the terms corresponding to:
(
)
(
)
()
U
NR Z v
NR y W
y V
Z V
y M
Z M
G
O C oL X
O
C O
L
X
C
yo
C zo
X
O
L
X
C
zo
C
yo
X
O
L
2
2
=
-
-
+
+
-
'
'
'
'
'
'
Moreover, it is necessary to carry out a change of reference mark as with [§3.1].
Code_Aster
®
Version
3
Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
46/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
5 Beam
curve
To calculate the matrix of rigidity for a curved element of beam, we make calculation while passing
by various stages.
We leave the equilibrium equations which integrated will give us a matrix (noted
J
)
allowing to determine the efforts in a point of the beam knowing the efforts in another point.
This matrix will take into account the local basic change.
Then by writing the potential energy of the element and by noticing the decoupling of the bending in
the plan of the element of the bending out of this plan, one determines the two matrices of flexibility.
Finally the matrices of flexibility being calculated, one obtains the matrix of local rigidity by using it
principle of Castigliano, which must be recomputed in the total base to be assembled.
Z
0
y
2
y
Z
Z
2
X
2
arc
NR
I
NR
J
X
O
0
y
0
M
C
X
1
Appear 5-a
Cf: Instruction manual of Code_Aster (booklet [U4.2]: modeling index C p26/30).
To attach the efforts applied in a point P of the structure to the efforts obtained in another point
Q of the structure, one integrates the equilibrium equations static of a curved beam (without effort distributed).
We here will limit we to study the curved beam with constant section (with taking into account of
transverse shearing) and with constant radius of curvature.
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
C
NR
I
NR
J
&& '&
X, '&
y, '&
Z
(
)
&& '&
X
&& '&y
&& '&z
base local curved beam
Appear 5-b: Identifies average (local reference mark)
The equilibrium equations static are:
NR
V
=
V + NR
R =
V
=
M
M
R
=
M + MR. V =
M
+V
=
S
S
2 S
T S
S
T
2
S
,
,
,
,
,
,
-
-
-
1
1
1
1
2
1
0
0
0
0
0
0
this for:
Q
Y
Z
M
P
&& '&
N
1
&& '&
S
X
Appear 5-c: Curvilinear reference mark
Code_Aster
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
To integrate, the conditions out of P are used:
NR = F
V
=
F
V
= F
M
= M
M
=
M
M
= M
y
X
Z
T
y
X
Z
1
2
1
2
-
-
While integrating and while passing in the system of following axis:
P
&& '&
N
1
&& '&
S
Q
X
Z
y
Appear 5-d: System of axis chosen by integration
One obtains:
(
)
NR
V
V
M
M
M
R
R
R R
T
1
2
1
2
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
1
0
0
0
1
-
-
-
-
-
=
(
cos
sin
sin
cos
cos
cos
sin
sin
sin
cos
cos
)
sin
J
!
“
# # # # # # # # # # # #
$
# # # # # # # # # # # #
Fx
Fy
Fz
MX
My
Mz
We now will take into account the mechanical characteristics by using energy
potential:
Ep
NR
ES
V
K SG
V
K SG
M
I.E.(internal excitation)
M
I.E.(internal excitation)
M
I.E.(internal excitation)
ds
T
2
S
S
1
2
=
1
2
2
12
1
22
2
2
12
1
2
2
~
~
~
~
~
~
+
+
+
+
+
(the sign “
~
“means that we use the efforts intern) (note:
~
F
F
= -
by the principle
of action-reaction).
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
Recall of the relations of behavior
within the framework of the model of Timoshenko:
NR
ES custom
V
K SG
W
S
V
K SG
W
S
M
I.E.(internal excitation) S
M
I.E.(internal excitation)
S
M
I.E.(internal excitation)
S
T
=
=
-
=
+
=
= -
= -
~
~
~
~
~
1
1
1
2
2
2
2
1
1
1
2
2
2
{}
{}
(
)
(
)
Torque of the interior efforts:
Kinematic torque:
from where:
+
+
+
+
=
+
T
C
S T
S M
S
N
N
S
N
N
NR
M
U
W
W
Ep
NR V V
ES
K SG
K SG
NR
V
V
ds
M
MR. M
T
S
S
T
S
S
~
~
~
~
,
,
,
+
+
1
1
2
2
1
1
2
2
1
2
1
2
1
2
1
2
1
2
1
0
0
0
1
0
0
0
1
1
2
1
2
1
2
1
0
0
0
1
0
0
0
1
1
2
1
2
I.E.(internal excitation)
I.E.(internal excitation)
I.E.(internal excitation)
M
M
M
ds
T
Code_Aster
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
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Key:
R3.08.01-A
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
or:
Ep
Fx
Fy
Fz
MX
My
Mz
ES
K SG
K SG
GC
I.E.(internal excitation)
I.E.(internal excitation)
Fx
Fy
Fz
MX
My
Mz
Rd
P
T
O
B
T
Q
P
=
being the matrix obtained previously.
1
2
1
1
1
0
1
0
1
1
1
2
1
2
J
J
J
One can thus calculate the matrix of flexibility
[]
C
:
[] =
C
J
J
T
ES
K SG
K SG
GC
I.E.(internal excitation)
I.E.(internal excitation)
Rd
0
1
2
1
2
1
1
1
0
1
0
1
1
We can notice that the matrix
J
can break up into two pennies matrices
independent, a part concerning the bending in the plan of the element, the other concerning the bending
out of the plan of the element.
Note:
This decomposition will make it possible to reverse the matrices more easily than we go
to obtain a little further.
J
J
J
,
1
2
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
Bending in the plan of the element:
Bending out of the plan of the element:
=
(
)
=
(
)
NR
V
M
R
R
Fx
Fy
Mz
M
M
V
R
R
T
1
2
1
1
2
0
0
1
1
1
0
0
-
-
-
-
-
cos
sin
sin
cos
cos
sin
cos
sin
cos
sin
cos
sin
J
!
“
# # # # #
#
$
# # # # # #
1
2
J
!
“
# # # # # #
$
# # # # # #
MX
My
Fz
5.1
Stamp flexibility for the bending in the plan of the beam [C
1
]
[]
(
)
(
)
C
R
R
ES
K SG
I.E.(internal excitation)
Rd
O
O
ES
K SG
R
I.E.(internal excitation)
ES
K SG
R
I.E.(internal excitation)
R
1
1
2
1
1
1
1
1
0
0
1
1
1
1
2
2
1
2
2
2
1
2
2
=
(
)
=
+
+
+
(
cos
sin
cos
sin
cos
sin
cos
sin
cos
cos sin
cos sin
cos
sin
cos
)
-
-
-
-
-
-
-
-
J
I.E.(internal excitation)
ES
K SG
R
I.E.(internal excitation)
R
I.E.(internal excitation)
I.E.(internal excitation)
Rd
2
2
2
1
2
2
2
2
2
1
sin
cos
sin
sin
sym.
+ +
-
Appendix:
(
)
(
)
(
)
(
)
cos
sin
cos
sin
cos
sin (
)
sin
sin
cos
sin (
)
sin cos
sin
2
2
2
2
2
1
2
1
4 2
2
1
2
1
4 2
2
2
O
O
O
O
O
D
D
D
D
D
-
-
+
=
=
+
=
=
+
=
=
from where:
(
)
(
)
()
(
)
()
(
)
(
)
()
(
)
C
R
ES
R
K SG
R
I.E.(internal excitation)
C
R
ES
R
K SG
R
I.E.(internal excitation)
11
1
1
3
2
22
1
1
3
2
4
2
2
4
2
2
4
6
8
2
4
2
2
4
2
2
4
2
2
=
+
(
) +
(
) +
+
,
=
+
+
(
) +
,
sin
sin
sin
sin
sin
sin
sin
-
-
-
-
Code_Aster
®
Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
52/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
[
]
(
)
(
)
C
R
I.E.(internal excitation)
C
R
ES
R
K SG
R
I.E.(internal excitation)
R
I.E.(internal excitation)
C
R
I.E.(internal excitation)
C
R
I.E.(internal excitation)
33
1
2
12
1
2
3
1
2
3
2
2
3
2
13
1
2
2
23
1
2
2
2
2
2
2
2
2
1
=
,
=
+
,
=
,
=
.
-
-
-
-
-
-
sin
sin
sin
cos
sin
cos
5.2
Stamp flexibility for the bending out of the plan of the beam [C
2
]
[]
(
)
(
)
(
)
C
R
R
I.E.(internal excitation)
I.E.(internal excitation)
K SG
Rd
I.E.(internal excitation)
I.E.(internal excitation)
I.E.(internal excitation)
I.E.(internal excitation)
R
I.E.(internal excitation)
R
I.E.(internal excitation)
I.E.(internal excitation)
I.E.(internal excitation)
R
I.E.(internal excitation)
2
1
2
2
1
1
0
0
1
1
1
0
0
0
1
0
0
0
1
2
2
1
1
2
1
2
2
1
=
=
O
O
+
+
+
+
cos
sin
sin
cos
cos
sin
cos
sin
cos sin
cos sin
cos
cos
sin
sin
cos
cos
sin
-
-
-
-
-
-
J
(
)
+
.
+
+
R
I.E.(internal excitation)
R
I.E.(internal excitation)
R
I.E.(internal excitation)
K SG
Rd
cos sin
sym
cos
sin
1
2
2
2
2
1
2
1
1
-
()
(
)
()
(
)
()
(
)
()
(
)
()
[
]
()
(
)
()
(
)
()
(
)
C
R
I.E.(internal excitation)
R
I.E.(internal excitation)
C
R
I.E.(internal excitation)
R
I.E.(internal excitation)
C
R
I.E.(internal excitation)
R
I.E.(internal excitation)
Rb
K SG
C
R
I.E.(internal excitation)
R
I.E.(internal excitation)
C
R
I.E.(internal excitation)
R
I.E.(internal excitation)
C
R
I.E.(internal excitation)
11
2
1
22
2
1
33
2
3
3
1
2
12
2
2
1
2
13
2
2
2
1
23
2
2
4
2
2
4
2
2
4
2
2
4
2
2
2
6
8
2
4
2
2
2
2
4
2
2
4
4
2
2
2
=
+
+
+
=
+
+
=
+
+
+
+
=
+
=
+
+
=
-
-
-
-
-
-
-
-
sin
sin
sin
sin
sin
sin
sin
sin
sin
sin
sin
sin
(
)
sin
cos
sin
2
2
1
2
2
2
2
+
-
+
R
I.E.(internal excitation)
Code_Aster
®
Version
3
Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
53/72
Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
Having determined the matrix of flexibility, we will be able to calculate the matrix of rigidity.
One thus has:
[]
E
E
p
E
E
E
P
T
P
P
=
)
,
(
this
: for outside
1
2 T
C T
T
,
Like one a:
T
J T
Q
P
E
E
=
(Recall:
T
T
P
Q
E
E
and
are not described in the same base).
One also has:
[]
E
p
E
E
E
Q
T
T
Q
Q
=
,
1
1
2
1
T
J
C J T
T
-
-
Using the theorem of Castigliano, one can obtain displacements associated with the external efforts
T
E
.
(
)
[]
(
)
U
C T
S N N
U
J
C J T
S N N
T
T
p
P
E
Q
E
E
E
Q
T
Q
P
Q
P
Q
= []
=
P
1
1
in the local base of the point
and
in the local base of the point
and
,
,
,
,
.
1
2
1
2
-
-
By breaking up the problem into two subproblems, and using the principle of superposition, one
can write:
(
)
T
T
T
T
T
T
T
T
T
T
Ptotal
=
+
=
=
+
=
for intern
and
E
effort
coming
Po
P
E
effort
coming
Po
Q
E
I
I
E
effort
coming
Po
P
E
effort
coming
Po
Q
QQ
E
I
I
PP
PQ
PP
PQ
Qtotal
QP
QP
QQ
I
int
int
int
int
:
,
-
-
-
-
(one applies two torques of effort (independent one of the other) to the point
P
and at the point
Q
).
[]
(
)
[
]
Action coming on a side:
Action coming on other side:
=
=
=
= +
=
=
1
1
T
T
K U
K
C
T
J
T
J
J K J
U
K J U
PP
P
PQ
Q
I
E
I
E
T
Q
T
Q
p
-
-
-
-
-
1
Code_Aster
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Version
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
We obtain as follows:
T
K U
K J
U
T
J K U
J K J U
Ptotal
Qtotal
E
P
T
Q
E
P
T
Q
=
=
+
and the same:
-
-
The matrix of rigidity for “displacements”
U
P
(given in its local base) and for
“displacements”
U
Q
(given in its local base) is:
K
K
K J
J
K
J K J
=
-
-
T
T
.
Moreover, in the case of beams with hollow circular section (elbows of pipings), one divides
I
I
y
Z
and
by coefficients of flexibility given by the user, to take into account the variation of rigidity
had with ovalization (cf [§7.2]).
For the matrix of mass, one considers only the reduced matrix (concentrated masses), and one makes
the simplifying assumption that the expression given for the right beams [§2.4] remains valid in
considering a right element length
R.
.
One obtains:
(
)
()
()
()
()
M
M
M
M
M
M
S R
M
M
I
I R
M
M
I R
S R
S R
M
M
I R
S R
S R
y
Z
y
Z
11
22
33
77
88
99
44
10 10
55
11 11
3
2
66
12 12
3
2
2
2
2
15
105
48
2
15
105
48
=
=
=
=
=
=
=
=
+
=
=
+
=
=
+
.
.
.
min
,
.
min
,
This assumption is restrictive and does not allow to take well into account the inertia of bending or of
torsion which had with the curvature. It is thus in this case to better modelize a beam curves by several
elements
POU_C_T
.
Code_Aster
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
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J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
6 Loadings
Various types of loading developed for the elements of beam
“
MECA_POU_D_E
“
,
“
MECA_POU_D_T
“
,
“
MECA_POU_C_T
“
in linear elasticity are:
Types or options
MECA_POU_D_E
MECA_POU_D_T
MECA_POU_C_T
*
CHAR_MECA_EPSI_R
: loading by
imposition of a deformation (of type
thermal stratification)
developed
developed
developed
*
CHAR_MECA_PESA_R
: loading due to
gravity
developed
developed
developed
*
CHAR_MECA_FR1D1D
: loading distributed
by actual values
developed
developed
developed
*
CHAR_MECA_FF1D1D
: loading distributed
by function
developed
developed
developed
*
CHAR_MECA_TEMP_R
: loading
“thermal”
developed
developed
developed
*
CHAR_MECA_FRELEC
: loading “forces
electric " created by a secondary conductor
right
developed
developed
not
*
CHAR_MECA_FRLAPL
: loading “forces
electric " created by a secondary conductor
unspecified
developed
developed
not
6.1
Loading by deformation
OPTION:
“CHAR_MECA_EPSI_R
“
One calculates the loading starting from a state of deformation (this option was developed to take
in account the thermal stratification in pipings). The deformation is given by the user to
the aid of the key word
EPSI_INIT
in
AFFE_CHAR_MECA
.
6.1.1 For the right beam of Euler and the right beam of Timoshenko
The model takes into account only one work in traction and compression and pure bending (not of effort
edge, not of torque).
The following relations of behavior are used:
NR
ES
U
X
M
E I
X
M
E I
X
y
y
y
Z
Z
Z
=
=
=
,
,
.
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Titrate:
“Exact” elements of beams (right and curved)
Date:
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Key:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
Thus, one obtains the second elementary member associated with this loading:
at item 1:
at item 2:
=
,
=
,
=
,
=
,
=
,
=
,
F
E S
U
X
M
E I
X
M
E I
X
F
E S
U
X
M
E I
X
M
E I
X
X
y
y
y
Z
Z
Z
X
y
y
y
Z
Z
Z
1
1
1
1
1
2
2
2
2
2
1
2
While being given “
U
X
X
X
y
Z
,
and
“for a beam, one can affect a loading to him.
6.1.2 For the curved beam of Timoshenko
y
X
The method used can be compared with that previously presented. One uses to obtain it
loading with the nodes, the matrix of local rigidity
K
who multiplied by displacements with the nodes
U
give the efforts
F
applied to the nodes:
F
K
=
U.
The method takes into account only the deformation related to the length of the beam. But one does not take
in account that the length projected on
X
, i.e. the shortest distance connecting the two points
extremes of the curved beam:
2R sin
(
R
: radius of curvature). This length is multiplied by one
rate of deformation, which gives the state of deformation of the beam. Then one projects on
X
and on
y
.
Code_Aster
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Titrate:
“Exact” elements of beams (right and curved)
Date:
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Key:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
One has thus for
U
:
U
R
EPX
R
EPX
R
EPX
R
EPX
with
EPX
U
X
=
=
point 1
point 2
-
2
2
0
0
0
0
2
2
sin
cos
sin
sin
sin
cos
sin
sin
6.2
Loading due to gravity
OPTION: “
CHAR_MECA_PESA_R
“
The force of gravity is given by the module of acceleration
G
and a normalized vector indicating
direction
N
.
The principle to distribute the loading on the two nodes of the beam is to take account of
functions of form
()
X
associated each degree of freedom of the element [§2]. We thus have for
a loading in gravity an equivalent nodal force
'
Q
:
'
'
Q
X
S G N dx
O
L
=
()
Note:
The functions of form used are (simplifying assumption) those of the model
Euler-Bernoulli.
It is necessary, of course, to be placed in the local reference mark of the beam to make this calculation.
Code_Aster
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Titrate:
“Exact” elements of beams (right and curved)
Date:
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J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
Axial load (in
X
):
(
)
F
S
dx
X
L
X
L
F
L S
S
F
L S
S
S S
S
S xL
F
L
S
S S
S
X
I
O
L
X
X
X
I
=
=
,
=
=
+
=
+
=
+
=
+
+
2
from where:
at item 1,
at item 2,
for a section varying linearly
and
-
-
G X
G X
G X
G X
1
1
2
1
2
1
2
1
1
1 2
1
3
6
6
3
3
2
1
2
1
,
(
)
2
2
1 2
2
1
2
1
2
12
2
3
12
2
-
,
=
+
+
,
+
for a section varying homothétiquement
F
L S
S S
S
S
S
S
S
X
L
X
G X
Torque:
Without taking into account of the roll, it is non-existent.
·
In the plan (X O Z):
F
S dx
M
S dx
F
S dx
M
S dx
Z
O
L
y
O
L
Z
O
L
y
O
L
1
1
2
2
1
2
3
4
=
=
=
=
G Z
G Z
G Z
G Z
Code_Aster
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Titrate:
“Exact” elements of beams (right and curved)
Date:
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J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
1
3
2
3
2
3
3
2
3
2
1
2
3
1
2
2
3
7
1
=
+,
=
+
,
=
,
=
+
=
2
4
from where:
X
L
X
L
L
X
L
X
L
X
L
X
L
X
L
L
X
L
X
L
F
L
S
Z
-
-
-
- +
-
G Z
20
3
20
20 30
3
20
7
20
30 20
8
5
2
30
2
2
1
2
1
2
2
1
2
1
1 2
2
1
2
2
1
+
,
=
+
,
=
+,
=
+
,
=
+
+
,
for a section varying linearly,
and:
S
M
L S
S
F
L
S
S
M
L S
S
F
L
S
S S
S
y
Z
y
Z
-
G Z
G Z
G Z
G Z
M
L
S
S S
S
F
L
S
S S
S
M
L S
S S
S
y
Z
y
1
2
2
2
1
1 2
2
1
1 2
2
2
1
1 2
2
2
2
60
2
5
8
30
2
2
60
=
,
=
+
+
,
=
+
+
,
for a section varying homothétiquement.
G Z
G Z
G Z
-
-
-
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Titrate:
“Exact” elements of beams (right and curved)
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Key:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
·
In the plan (X O Z):
F
S dx
M
S dx
F
S dx
M
S dx
y
O
L
Z
O
L
y
O
L
Z
O
L
1
1
2
2
1
2
3
4
=
=
=
=
G y
G y
G y
G y
-
-
from where:
for a section varying linearly,
and:
=
+,
=
+
,
=
+,
=
+
,
=
+
F
L
S
S
M
L S
S
F
L
S
S
M
L S
S
F
L
S
S S
y
Z
y
Z
y
1
1
2
2
1
7
20
3
20
20 30
3
20
7
20
30 20
8
5
1
2
2
1
2
1
2
2
1
2
1
1 2
G y
G y
G y
G y
G y
-
+
,
=
,
=
+
+
,
=
-
-
,
for a section varying homothétiquement.
2
30
2
2
60
2
5
8
30
2
2
60
2
2
1
1 2
2
1
1 2
2
2
1
1 2
2
1
2
2
S
M
L
S
S S
S
F
L
S
S S
S
M
L
S
S S
S
Z
y
Z
+
+
-
G y
G y
G y
Code_Aster
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Titrate:
“Exact” elements of beams (right and curved)
Date:
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Key:
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Manual of Reference
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
6.3 Loadings
distributed
OPTIONS:
“CHAR_MECA_FR1D1D”
,
“CHAR_MECA_FF1D1D”
,
The loadings are given under the key word
FORCE_POUTRE
, that is to say by actual values in
AFFE_CHAR_MECA
(option
CHAR_MECA_FR1D1D
), that is to say by functions in
AFFE_CHAR_MECA
(option
CHAR_MECA_FF1D1D
).
The various possibilities are:
constant loading
variable loading
linearly
right beam with constant section
developed
developed
right beam with section varying linearly
developed
not
right beam with section varying
homothétiquement
developed
not
curved beam
developed
not
The loading is given only by forces distributed, not by moments distributed.
The method used to calculate the loading to be imposed on the nodes is that presented to [§2.1.2].
6.3.1 Right beam with constant section
For a loading constant or varying linearly, one obtains:
F
L N
N
F
L N
N
X
X
1
2
1
2
1
2
3
6
6
3
=
+
,
=
+
.
N
N
1
2
and are the components of the axial loading as in points 1 and 2 coming from the data of
the user replaced in the local reference mark.
Code_Aster
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Titrate:
“Exact” elements of beams (right and curved)
Date:
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Author (S):
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T
T
T
T
y
y
Z
Z
1
2
1
2
,
and
are those of the sharp effort. And one a:
F
L
T
T
M
L
T
T
F
L
T
T
M
L
T
T
F
L
T
T
M
L
T
T
F
y
y
y
Z
y
y
y
y
y
Z
y
y
Z
Z
Z
y
Z
Z
Z
1
1
2
1
1
2
2
1
2
2
1
2
1
1
2
1
1
2
2
7
20
3
20
20
30
3
20
7
20
30
20
7
20
3
20
20 30
2
2
2
=
+
,
=
+
,
=
+
,
=
+
,
=
+
,
=
+
,
-
-
=
+
,
=
+
.
L
T
T
M
L
T
T
Z
Z
y
Z
Z
3
20
7
20
30 20
1
2
2
1
2
2
6.3.2 Right beams with variable section
The provided loading must be constant. One uses a method similar to that used by the weight
clean [§6.2].
6.3.3 Beam
curve
The provided loading must be constant along the element. The loading reduced to the nodes is
equivalent with that which one can obtain by taking again the results of [§6.3.1]. With a load
constant, that becomes:
F
L N
F
L N
F
L T
F
L T
M
L T
M
L T
F
L T
F
L T
M
L T
M
L T
X
X
y
y
y
y
Z
y
Z
y
Z
Z
Z
Z
y
Z
y
Z
1
2
1
2
1
2
1
2
1
2
2
2
2
2
12
12
2
2
12
12
2
2
2
2
=
=
=
=
=
=
=
=
=
=
,
,
,
,
,
,
,
,
,
.
-
-
Code_Aster
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“Exact” elements of beams (right and curved)
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6.4 Loading
thermics
OPTION:
“
CHAR_MECA_TEMP_R
“
To obtain this loading, it is necessary to calculate the deformation induced by the difference in temperature
T T
reference
-
:
(
)
(
)
(
)
U
L
T T
U
L
T T
reference
reference
1
2
=
=
thermal expansion factor
-
-
-
:
.
Then, the induced forces are calculated:
F
K U
=
like
K
is the matrix of local rigidity to the element, one must then carry out a change of
base to obtain the values of the components of the loading in the total reference mark.
6.5 Loading
electric
OPTION:
“CHAR_MECA_FRELEC”,
“CHAR_MECA_FRLAPL”.
This type of loading makes it possible to take into account the force of Laplace acting on a conductor
the main thing, due to the presence of a secondary conductor.
The secondary conductor is right and it is not based on part of the Aster mesh if one uses
the option
“CHAR_MECA_FRELEC”
.
The secondary conductor is not necessarily right and it can be based on part of the mesh
Aster if the option is used
“CHAR_MECA_FRLAPL”
.
Linear density of the force of Laplace exerted in a point
M
main conductor by
secondary conductor is written:
F
E
E
R
R
() =
M
ds
1
2
1
2
3
2
Note:
To obtain the force of Laplace, the vector turned over following the calculation carried out by one of
these two options must be multiplied by the temporal function of intensity specified by
the operator
“DEFI_FONC_ELEC”
.
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“Exact” elements of beams (right and curved)
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6.5.1 Secondary conductor finished or infinite right
OPTION:
“CHAR_MECA_FRELEC”
For a finished secondary conductor, we have:
(
)
'
'
'
F M
E
N
D
(
)
sin
sin
=
2
1
1
2
-
1
M
P
1
P
2
1
2
with
&&
N
=
D
D
=
E
2
D
D
D
E
1
E
2
For an infinite secondary conductor, we have:
F
E
N
()
M
D
=
1
Three types of loading are possible:
·
conductor infinite parallel right,
·
multiple infinite parallel conductors right,
·
right conductor in unspecified position finished or infinite.
In the case of a conductor infinite parallel right, its position can be given in two manners:
·
maybe by a vector translation of the main conductor to the secondary conductor,
·
maybe by the distance enters the two conductors and by a point of the secondary conductor.
One thus obtains a constant loading on all the main element. One can thus use the technique
presented to [§6.3] [§6.4] to calculate the loading with the nodes.
F
F
F
F
M
E
F
M
E
F
F
1
2
1
1
2
2
1
2
2
2
12
12
=
()
=
()
=
()
=
()
, () =
.
constant
MR. L
MR. L
MR. L
MR. L
M
M
-
For the case of the parallel conductors right infinite multiple, one must give it directly
vector “forces of normalized Laplace”. The aforementioned being usually given in the total reference mark, it is necessary
to determine the vector “forces of Laplace” in the local reference mark of the main element. Thus of same
manner that previously, one calculates the loading with the nodes.
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In the case of a right conductor in unspecified position finished or infinite, its position is given by
two points
P
1
and
P
2
such as the current circulates of
P
1
towards
P
2
. The loading is not
obligatorily constant along the main element.
The selected method to calculate the loading reduced to the nodes is obviously the same one
that before. But here, integration is done numerically by discretizing the element in some
numbers (in practice: 100 points enters
P
1
and
P
2
).
One integrates as follows:
F
F
F
F
1
3
2
2
3
2
2
3
1
2
3
=
()
+
=
()
+
M
X
L
X
L
dx
M
X
L
X
L
dx
O
L
O
L
-
-
'
'
'
'
'
'
M
E
F M
L
X
L
X
L
X
L
dx
M
E
F M
L
X
L
X
L
dx
O
L
O
L
1
1
3
2
2
1
3
2
2
=
()
+
=
()
+
-
-
-
-
-
Note:
(
)
(
)
Like
and
()
F
X
E
F
X
X
E
M
M
=
()
= =
0
0
1
1
one uses only the functions of form associated with the problem with bending.
6.5.2 Secondary conductor describes by part of mesh ASTER
OPTION:
“CHAR_MECA_FRLAPL”
The secondary conductor as a whole is not necessarily right. But it is described
only by right elements. Its length is obligatorily finished.
Its position can be supplemented by the use of a vector translation or a symmetry plan (by one
not and a normalized normal vector of the symmetry plane) compared to the main conductor.
Except the fact that it is necessary to summon the interaction of the various elements of the secondary conductor on
main conductor, the method is the same one as previously (case (III)).
Note:
For numerical integration, one uses only 5 points included/understood enters
P
1
and
P
2
.
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7
Torque of the efforts - Torque of the stresses (or efforts
generalized) - nodal Forces and reactions
OPTIONS:
“EFGE_ELNO_DEPL”,
“SIEF_ELGA_DEPL”,
“SIGM_ELNO_DEPL”,
“SIPO_ELNO_DEPL”,
“FORC_NODA”,
“REAC_NODA”.
The option “
EFGE_ELNO_DEPL
“allows to calculate the torque of the efforts to the 2 nodes of each element
of “beam”.
Options “
SIGM_ELNO_DEPL
“and”
SIPO_ELNO_DEPL
“allow to calculate the maximum values
components of the tensor of the stresses intervening in a model of “beam”.
The option
“SIEF_ELGA_DEPL”
the calculation of the nodal forces allows (option
“FORC_NODA”
and of
reactions
“REAC_NODA”
).
Note:
When one of these options end in “
_C
“, that means that the values of
displacements (and thus of the efforts) are complex.
7.1
The torque of the efforts
OPTION:
“EFGE_ELNO_DEPL”
One seeks to calculate with the two nodes of each element “beam” constituting the mesh of
studied structure, efforts exerted on the element “beam” by the remainder of the structure. The values are
data in the local base of each element.
By integrating the equilibrium equations, one obtains [§2.1.3] the efforts in the local reference mark of the element:
(
)
R
K
U
M
U
F
R
LOC
LOC
LOC
LOC
LOC
LOC
LOC
Y
T
Y
Z
Y
Z
T
Y
Z
E
E
E
Z
NR
V
V
M
M
M
NR
V
V
M
M
M
=
+
=,
,
where:
% %
,
,
,
,
,
,
,
,
,
-
-
-
-
-
-
-
1
1
1
1
1
1
2
2
2
2
2
2
K
LOC
E
elementary matrix of rigidity of the “exact” element of beam,
M
LOC
E
elementary matrix of mass of the element beam,
F
LOC
E
vector of the efforts “distributed” on the element beam,
U
LOC
vector “degree of freedom” limited to the element beam,
% %
U
LOC
vector “acceleration” limited to the element beam.
One changes then the signs of the efforts to node 1 [§2.1.3].
OPTION:
“SIEF_ELGA_DEPL”
The option
“SIEF_ELGA_DEPL”
is established for reasons of compatibility with other options.
It is used only for calculation of the nodal forces and the reactions.
It is calculated by:
R
K
U
LOC
LOC
LOC
E
=
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7.2
The tensor of the stresses
OPTIONS:
“SIGM_ELNO_DEPL”,
“SIPO_ELNO_DEPL”.
One seeks to calculate the maximum values of the components
xx
xy
xz
,
and
who are connected to
efforts by:
(
)
NR
dS
V
dS
V
dS
M
y
Z
dS
M
Z
dS
M
y
dS
xx
S
y
xy
S
Z
xz
S
T
xz
xy
S
y
S
xx
Z
S
xx
=
,
=
,
=
,
=
,
=
,
=
.
-
-
With the option
“SIGM_ELNO_DEPL”
, one calculates the maximum effect of the whole of the efforts on
xx
xy
xz
,
and
.
·
For
xy
, one a:
xz
I
Max
Yi
Yi
T I
X I
Ti
y
y
V
With
M
J
R
I I
With K
With
K
J X
RT
=
+
,
(=,)
:
for the node
with
surface of the section,
constant of shearing,
: constant of torsion,
: radius of torsion.
1 2
:
:
·
For
xz
, one a:
xz
I
Max
Zi
Zi
Ti
X I
Ti
V
With
M
J
R
=
+
,
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“Exact” elements of beams (right and curved)
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·
For
xx
, one a:
-
For a rectangular section:
xz
I
Max
I
I
Yi
Y I
Zi
Zi
Zi
Zi
I
I
Yi
Yi
Zi
Zi
Zi
Zi
Y
Z
Y
Z
NR
With
M
I
H
M
I
H
NR
NR
With
M
I
H
M
I
H
NR
I
I
H
H
= + +
+
,
+
+
<
,
,
,
if
if
with
: geometrical moments,
: sides of the rectangle.
2
2
0
2
2
0
-
-
For a circular section:
() ()
() ()
xx
I
Max
I
I
Y I
Z I
I
Yi
I
I
Y I
Zi
I
Yi
NR
With
M
M
R
I
NR
NR
With
M
M
R
I
NR
R
= +
+
+
,
+
+
<
,
if
if
with
: radius of the section.
2
2
2
2
0
0
-
-
For a general section:
xx
I
Max
I
I
Y I
Yi
Z I
Z I
Z I
Yi
NR
With
M
I
R
M
I
R
=
+
-
.
In this last case, one makes calculation at the point
(
)
R
R
Yi
Zi
,
.
R
R
Y I
Yi
and
and
R
Z
I
are them
distances to neutral fiber.
To find these formulas, one uses the relations between the stresses and the deformations then those
between the internal efforts and
(
)
U X
xy
xz
X X
y X
Z X
,
;
;
;
;
;
,
,
,
:
(
)
(
)
xx
xx
xy
xy
xz
xz
xx
y X
Z X
xy
xy
X X
xy
Z
xy
xz
X X
xz
E
G
G
U X Z
y
Z
V X
y
W X
y
=
=
=
,
=
, +
=
=,
=
+
=, +
,
,
,
,
,
2
2
2
2
-
-
-
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NR
V
V
M
M
M
EA
K GA
K GA
GJ
I.E.(internal excitation)
I.E.(internal excitation)
U X
Y
Z
T
Y
Z
Y
Z
Y
Z
xy
xz
X X
y X
Z X
=
0
0
,
,
,
,
(The various types of movement are uncoupled when one works with the main axes (by
definition)).
One obtains:
(
)
(
)
(
)
xx
Y
Y
Z
Z
xy
Y
Y
T
X
xy
Y
Y
T
X
X y Z
E
NR
EA Z
M
I.E.(internal excitation)
y M
I.E.(internal excitation)
X y Z
G
V
K GA Z
M
G J
X y Z
G
V
K GA Z
M
G J
,
=
+
,
,
=
,
,
=
,
-
-
-
With the option “
SIPO_ELNO_DEPL
“, calculation is somewhat different. The effects are sought
maxima of each effort
(
)
NR V
V
M
M
M
Y
Z
T
X
Y
,
,
,
,
,
on the components
xx
,
xy
and
xz
.
One finds the preceding results in broken up form. The vector result is written:
(
)
(
)
11
21
31
41
51
61
12
22
32
42
52
62
1
1
3
4
5
6
1 2
2
2
,
,
,
,
,
,
,
=
=
=
,
=
,
=
,
=
,
with:
for the node
for the rectangular sections:
I
I
I
I
I
I
I
Zi
Zi
I
Ti
X I
Ti
I
Y I
Y I
Zi
I
Zi
Zi
Yi
NR
With
I I
NR
With
V
With
M
J
R
M
I
H
M
I
H
=
-
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Titrate:
“Exact” elements of beams (right and curved)
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(
)
5
6
5
6
I
Y I
Yi
I
I
Zi
Y I
I
Y I
Zi
I
Yi
Yi
Zi
I
Zi
Zi
Yi
M
I
R
M
I
R
R
R
M
I
R
M
I
R
=
,
=
,
=
,
=
.
for the unspecified sections at the point
:
-
-
,
In the case of beam with hollow circular section (pipes), the flexibility of the thin hulls not being
not represented well, certain sizes should be corrected. Two coefficients are used for that:
·
a coefficient of flexibility “flex” (also used by rigidity [§5]),
·
a coefficient of intensification of the stresses “isigm”.
In particular, one can take as a starting point the the rules
RCC_M
.
(
)
(
)
I
I
M
I
R
M
I
R
Y
Y
Z
Z
Z
Y
=
=
E
,
=
.
conventional
conventional
cflex
cflex
only in bending.
corrig
isigm
cflex
corrected
isigm
cflex
isigm
1
1
5
6
7.3
Calculation of the nodal forces and the reactions
OPTION:
“
FORC_NODA
“
This option calculates a vector of nodal forces on all the structure. It produces a field with
nodes in the control
CALC_NO
by assembly of the elementary terms.
For this calculation, one usually uses in 3D [R5.03.01] for example the principle of virtual work
and one writes:
F Q
=
T
where
Q
T
represent the matrix symbolically associated with the operator divergence. For an element, one
writing agricultural work of virtual deformations:
()
()
()
Q
U
T
H
W H
=
For the elements of beam, one calculates simply the nodal forces by assembly of the forces
nodal elementary calculated by the option
SIEF_ELGA_DEPL
, which is expressed by:
[] [
] []
F
K
U
LOC
LOC
LOC
=
[
]
§7.1
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“Exact” elements of beams (right and curved)
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OPTION:
“
REAC_NODA
“
This option, called by
CALC_NO
, allows to obtain the reactions
R
with the supports starting from the forces
nodal
F
by:
R F F
F
= -
+
tank
iner
F
F
tank
iner
and
being nodal forces associated the loadings given (specific and distributed) and
with the efforts of inertia.
8
Element of bar
Key word
“BAR”
A bar is a right beam of constant section comprising only the degrees of freedom of
traction and compression only. The equation of the movement, matrices of mass and rigidity, and
the efforts are thus those of the beams (right of constant section) relating to the traction and compression.
Concerning the characteristics of the section, the surface of the cross section constitutes the only data
useful [U4.24.01 §11].
9 Bibliography
[1]
J.S. PRZEMIENIECKI. Theory off Matrix Structural Analysis, Mc Graw Hill, New York. 1968.
[2]
Mr. GERADIN, D. RIXEN. Theory of the vibrations, application to the dynamics of the structures,
Masson, Paris. 1993.
[3]
J.R. BANERJEE, F.W. WILLIAMS. Exact Bernoulli-Euler static stiffness matrix for A arranges off
tappered beam-columns. Interm. J. for Numerical Methods in Engineering, vol. 23,
p 1615-1628. 1986.
[4]
J.L. BATOZ, G. DHATT. Modeling of the structures by finite elements - HERMES.
[5]
V. OF TOWN OF GOYET. Nonlinear static analysis by the finite element method
formed space structures of beams with nonsymmetrical sections. Thesis of the University
from Liege. 1989.
[6]
C. MASSONNET, S. CESCOTTO. Mechanics of materials. Cd. Boeck-Wesmael S.A. 1994.
[7]
P. DESTUYNDER. Introduction to the calculation of the structures. Run C. Scientific computation. 1990.
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“Exact” elements of beams (right and curved)
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