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:
1/72
Organization (S): EDF/IMA/MN, IAT St CYR
Handbook of Référence
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 grid.
The beams can be right (Eléments POU_D_T and POU_D_E) or curves (Eléments 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-Venant: torsion is free.
The processing of the various loadings and the sizes awaited in result (forced - efforts) is
also presented.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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
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 of Lagrangien ......................................................................................................... 7
1.2 Pure torsion (torsion of Saint-Venant) ........................................................................................ 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 of Lagrangien ......................................................................................................... 9
1.3 The bending pure ............................................................................................................................ 10
1.3.1 Local equilibrium equation ...................................................................................................... 10
1.3.2 Method of Lagrangien ....................................................................................................... 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 inflection ................................................................................................................... 19
2.3.1 Inflection in the plan (x0z) .................................................................................................... 20
2.3.2 Inflection 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 Inflection in the plan (xoz) ........................................................................................ 23
2.3.3.2 Movement of inflection 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 - Structure prestressed .................................................................................... 40
5 Beam curves ....................................................................................................................................... 46
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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
5.1 Stamp flexibility for the inflection in the plan of the beam [C1] ................................................ 51
5.2 Stamp flexibility for the inflection 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 grid ASTER ....................................... 65
7 Torque of the efforts - Torseur of the constraints (or efforts generalized) - Forces nodal and réactions.66
7.1 The torque of the efforts ..................................................................................................................... 66
7.2 The tensor of the constraints ............................................................................................................. 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 Institut Aérotechnique of Conservatoire National of Arts and
Trades, within the framework of Contrat Externe de Recherche and Développement 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.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
4/72
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
eccentricity of the center of torsion/shearing
EY, EZ
y, Z
E
Young modulus
E
Poisson's ratio
NAKED
G
E
modulate of Coulomb = (21+)
I, I
geometrical moments of inflection compared to the axes y, Z
IY, IZ
y
Z
I
polar geometrical moment
p
I X
polar moment of inertia around the longitudinal axis X
K, K
coefficients of shearing
y
Z
1
1
AY
AZ
K
stamp rigidity
M
stamp of mass
M, M, M moments around axes X, y, Z
MT, MFY, MFZ
X
y
Z
NR
normal effort with the section
NR
S
surface of the section
With
U, v, W
translations on axes X, y, Z
DX DY DZ
V, V
sharp efforts along axes y, Z
VY, VZ
y
Z
density
RHO
transverse shear stress
cT
rotations around axes X, y, Z
DRX DRY DRZ
X, y, Z
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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
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 is the cross-section (section
transversal) or profile, and it is supposed that if it is evolutionary, its evolutions (size, form) are
continuous and progressive when G describes 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 the three
rotations X, y, Z.
Z
y
1
2
X
U X
U
X
v y
v y
W Z
W Z
Waited until the deformations are local, it is built in each node of the grid 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 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 122), 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
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
6/72
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 inflection 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
One considers a segment length dx subjected to an axial load NR [1.1.1-a] intern and a force
external fext per unit of length.
X
X + dx
NR (X)
NR (X)
NR (X + dx)
O
X
fext
dx
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 module
of Young E (X). The fundamental principle of mechanics makes it possible to write:
2
x+dx
X dx
U.S.
NR (X) + NR (X + dx) +
fext (S)
+
ds
(S) S (S)
()
-
=
X
ds
X
t2
where U is displacement on
X
center segment.
Thus:
NR (X + dx) - NR (X)
1 x+dx
1
2
x+dx
U
+
F
ext. (S) ds
=
S
ds
dx
dx X
dx X
t2
While passing in extreme cases when dx 0, one obtains:
D NR (X)
2 U
+ F (X) =
S
(X
ext.
)
dx
t2
éq 1.1.1-1
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
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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
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:
U
NR (X) = ES
éq 1.1.1.- 2
X
One obtains thus after simplification by dx:
U
2 U
ES
fext
S
X
X +
=
t2
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 Lagrangien
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:
1
2
L
U
E
=
S
dx
C
2 O
T
.
1
U 2
One will note for the continuation E
=
S
it
2
T elementary kinetic energy.
The internal energy of deformation, thanks to the law of Hooke is written:
1
2
L
U
E
=
ES
dx
p
.
int
2 O
X
1
U 2
Of the same E will be noted
=
E
S
p
inte
2
X.
There is also the work of the external force given by:
L
E
=
F
U dx
p
ext.
ext.
O
and at the elementary level E
= F
U
p
ext.
.
exte
The Lagrangian one is given by:
L = E - E
- E
C
p
p
int
ext.
and Lagrangian density:
L = E - E
- E
C
p
p
.
E
inte
exte
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
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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:
8/72
For the monodimensional continuous system, the equation of Lagrange is written in this case:
L
L
L
-
=
0
éq 1.1.2-1
U
X
U -
T
U
'
%
where U and % U respectively indicate the derivative 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-Venant)
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
warping 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
One considers a segment length dx put in rotation under the action of one moment M X
[Figure 1.2.1-a] intern and of an external couple X per unit of length.
X
X + dx
M (X)
X
M (X + dx)
NR (X)
O
X
positive
dx
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
2
x+dx
x+
-
dx
M (X) + M (X + dx) +
X
X
(S)
ds =
I
ds
X
X
X
X
X
t2
with I
=
R ds
2 is the plane moment of inertia of the section S around the axis of rotation (0, X).
X
S
As for traction, one obtains after division by dx and passage in extreme cases:
DM
2
X
+
= I
X
dx
X
X
t2
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
9/72
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 the polar geometrical moment by
report/ratio in the center of gravity of the section. (One has besides: I
= I for a material of mass
X
p
voluminal homogeneous).
We obtain the expression then:
2
X
X
+
=
G I p
I
X
X
X
X
t2
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 warping while remaining on the free assumption of torsion, in the case of them
noncircular sections one is led to replace 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
. One obtains then:
X
2
X
G C
X
+ X =
C
X
X
t2
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 inflection are coupled.
1.2.2 Method of Lagrangien
We have same manner as with [§1.1.2] the kinetic energy (for example for a beam of
circular section):
2
L 1
E
=
I
X
dx
C
,
O X
2
T
internal potential energy
2
L 1
E
=
G I
X
dx
,
int
p
O
p
2
T
and the work of the external couple
L
E
=
dx
p
X
.
ext.
O
X
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
10/72
By applying the equation of Lagrange [éq 1.1.2-1] to 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 inflection 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 inflection in plan (O, X, Z), the extension to the plan (O, X, y) is immediate
[Figure 1.3-a].
Z
y
y
O
X
Appear 1.3-a: Flection 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
One considers a segment length dx subjected to the shearing action Vz, the moment bending M y,
an external effort tz distributed uniformly per unit of length, and an external couple m
distributed
ext.
yext
uniformly by unit of length [Figure 1.3.1-a].
V (x+dx)
tzext
V (X)
Z
M (x+dx)
y
Moments
y
positive
dx
X
x+dx
M (X)
X
myext
Appear 1.3.1-a: Segment of beam in inflection in plan (O, X, Z)
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
11/72
The local balance of the forces and the moments (on the section of X-coordinate X + dx) gives for the forces:
2
x+dx
x+dx
W
- V (X) +V
Z
Z (X + dx) +
T
ds =
X
zext
S ds
X
t2
and for the moment:
2
x+dx
x+dx
x+dx
- M (X) + M
y
y
y (X + dx) +
m
ds
y
-
Vz (X
)
=
ext.
X
ds
X
I ds
y
X
t2
The terms in dx2 are neglected. While passing in extreme cases when dx tends towards 0, one obtains:
V
2
Z
W
+ tz
=
S
X
ext.
t2
My
2
- V + m
y
Z
y
= +
I
.
X
y
ext.
t2
It is noted that the effort uniformly distributed tz produces a term which is of the second command in balance
ext.
moments and is thus neglected. One introduces then the relations of behavior of resistance
materials.
M
= + I.E.(internal excitation)
y
y
y X
W
V
= K SG
Z
Z
+
X
y
éq 1.3.1-1
the expression [éq 1.3.1-1] of Vz is due to Timoshenko [bib4] where kz 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. Iz is the moment
geometrical of the section compared to the axis (O, y).
Consequently, one leads to the two equations coupled out of W and there for the inflection in the plan
(O, X, Z).
W
2 W
K SG
Z
+
+ T
y
Z
=
S
éq 1.3.1-2
X
X
ext.
T 2
2
y
W
I.E.(internal excitation)
-
K SG
y
y
Z
+ + m
y
y
= +
I
X
X
X
y
ext.
t2
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 out of W. For
that, one only once derives compared to X-coordinate X the equilibrium equation from the moments [éq 1.3.1-3].
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
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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:
12/72
3
2
3
y
W
I.E.(internal excitation)
- K SG
y
+
= + I
y
y
.
x3
Z
x2
X
y
X t2
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:
2
2
y
W
1
W
=
-
+
T
.
X
2
Z
X
K
ext.
2
Z SG
kz G T
4 W
2 W
E 4 W
2I
4
y W
I.E.(internal excitation)
+ S
- I 1+
+
- T
y
= 0 éq 1.3.1-4
x4
t2
y
K G
2
2
4
zext
Z
X T
K G
Z
T
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.
4 W
I.E.(internal excitation) y x4 balance charging density in the direction of the translation due to the moment of
inflection.
2 W
S t2 is the term of inertia of translation.
4 W
I
y x2 t2 represents the inertia of rotation of inflection.
E
4 W
Iy K G 2 2 is an additional term of the inertia of rotation due to the taking into account of
Z
X T
transverse shearing (assumption of Timoshenko).
2Iy 4 W
result from the coupling between the inertia of rotation and the inertia of translation coming from
K G
Z
4
T
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 model 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
inflection, in the general case of the beams with variable section is written:
2 W
2 W
I.E.(internal excitation)
+
-
= 0.
2
y
S
T
éq 1.3.1-5
X
x2
t2
zext
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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 Vz = 0 what brings to [éq 1.3.1-1].
W
y = -.
éq 1.3.1-6
X
who is the translation of the assumption of Euler.
Concerning the inflection in the plan (O, X, y), the same step leads to [éq 1.3.1-7] for
beam of Timoshenko with:
v
2 v
K SG
y -
+ T
Z
y
=
S
X
X
ext.
t2
éq 1.3.1-7
2
Z
v
I.E.(internal excitation)
K SG
Z
Z
y
+ m
Z
Z
=
I
X
X -
-
X
Z
ext.
t2
and when the section is constant:
4 v
2 v
E
4 v
2 I 4 v
I.E.(internal excitation)
- S
- I
Z
1+
+
+ T
Z
0.
éq 1.3.1-8
T 4
T 2
Z
K G
=
2
2
4
yext
y
X T
K G
y
T
v
The use of the assumption of Euler (i.e. in the plan (O, X, y) Z =) makes it possible to lead to
X
the equation of the movement of inflection for a beam of Euler according to [éq 1.3.1-9].
2 v
2 v
I.E.(internal excitation)
-
+
= 0.
2
Z
S
T
éq 1.3.1-9
X
x2
t2
yext
1.3.2 Method of Lagrangien
The kinetic energy is expressed by:
2
2
L 1
y
L 1
W
E
=
I
dx +
S
dx
C
O
y
2
T
O
2
T
according to displacements in rotation and translation.
The potential energy intern is worth:
2
L 1
y
L 1
W
E
=
I.E.(internal excitation)
dx +
+ dS dx
p
int
O2
X
O
C
2
S
X
y
T
W
where C is transverse shear stress and the term
+ deformation of
T
X
y
shearing. The model of beam of Euler neglects this term while the model of Timoshenko emits
an assumption on the distribution of the constraints C in the section, compatible with the expression
T
[éq 1.3.1-1]. In the general case of the model of Timoshenko, the potential energy interns is written:
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“Exact” elements of beams (right and curved)
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2
2
L 1
y
L 1
W
E
=
I.E.(internal excitation)
dx +
K SG
+
dx
p
.
O2
X
O
Z
2
X
y
int
The potential of the external loads is expressed as for him by:
L
L
E
= - T
dx
- m dx
p
Z
.
O
y
O
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 brings back for us to the two equations [éq 1.3.1-2] and [éq 1.3.1-3] describing the movement in inflection 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 Timoshenko (POU_D_T). Matrices of
rigidity are calculated with option “RIGI_MECA”, and the matrices of mass with the option
“MASS_MECA” for the coherent matrix, and 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
NR
(X) + F (X) + F C I (X
ext.
I
) = 0
dx
i=1
One omitted by simplicity the inertias which would undergo the same processing as the forces
external fext.
C
I represents the function of Dirac function located as in point I, the fi 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:
FD
NR
NR
dx =
F
v dx +
F C
I (v
ext.
I
)
éq 2.1-1
dx
i=1
Any confusion being excluded, I indicates the measurement of Dirac function associated with item 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.:
L FD
NR
dx
for an element length L.
0
dx
The elastic relation of behavior is introduced:
NR (X) = ES dx
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While choosing by function test:
X
X
v (X) = 1 (X) = 1
and
v (X) = 2 (X) =
L
L
one obtains directly:
L D
L
ES of
ES
NR
1 dx =
-
dx
= -
[(uL) - (U) 0]
O
dx
O
L dx
L
and
L D
L ES of
ES
NR
2 dx =
dx
=
[(uL) - U () 0]
O
dx
O
L dx
L
The matrix of rigidity of the element is thus:
ES 1
-
1
K =
L - 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:
f1
1
X
with
F
=
F
1
ext. (X) 1 - dx
F
0
L
2
1
X
F
=
F
2
ext. (X) dx
0
L
Note:
Into AFFE_CHAR_MECA_F, one can introduce fext like an unspecified function of X.
On the level of the calculation of f1 and f2, on the other hand, integration is made by supposing that fext
vary linearly between the values taken with the nodes ends. If one must model one
radial force distributed nonlinear, it is then necessary to discretize more finely.
But let us insist on the fact that whatever the shape of F
(X
ext.
) (polynomial or different), if one
can calculate exactly the integrals f1 and f2, 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
NR v dx
, X
= [NR (x1) v (x1) - NR (xo) v (xo)]+ [NR] I (v)
I
i=1
éq 2.1.1-2
M
- NR v dx
, X
J
J
=1
J representing all the intervals without discontinuity of normal effort, therefore without concentrated force, and
[NR] jumps of NR between these intervals.
I
Indeed, by bringing this expression closer to the PTV, one finds, for each loading concentrated (in
choosing suitable functions test v):
I 1, NR
[NR]
F C
=
=
I
I
Each finite element of beam is by assumption an interval without discontinuity. There can thus be
discontinuity of the efforts intern 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 reference mark
, X
= 0
The formula of integration by parts [éq 2.1.1-2] on the element gives:
L
L
NR (X) v dx
, X
= [NR (L) v (L) - NR ()
0 v ()
0
] + F
ext. (X) v (X) dx
0
O
By regarding NR (L) and NR (O) as data, one could have obtained this formula directly
PTV [éq 2.1-1].
By still taking the functions test:
X
X
v (X) = 1 (X) = 1
and
v (X) = 2 (X) =
L
L
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One obtains:
ES
-
[(uL) - (uo)] = - NR (O) + F
L
1
ES [(uL) - (uo)] = NR (L) + F
L
2
- NR (O)
(
U O) f1
that is to say
K
NR (L) = [] (
U L) -
f2
I.e. the internal efforts are obtained by cutting off with the product K U the nodal forces
equivalent to the distributed loads fext.
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
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)
0 1
(X) + (
U L) 2
(X)
The matrix of mass appears in the expression of virtual work due to the inertias:
U
W =
T
V
MR. U
%, U =
1
.
u2
Work is also written:
L
W =
V (X) U
% (X, T) dx
O
m
with:
=
dS = S
m
in the case of a homogeneous material.
S
While taking
U (X, T) = (X) U (T) + (X) U (T), one a:
1
1
2
2
L
1 (X)
W = S
T
V
(X
1
X
2
) % U
O
2 (
dx
X)
() ()
L 1 (X)
W = T
V S
(X
1
X
2
) % U.
O
2 (
dx
X)
() ()
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The matrix of mass is thus written:
L
L
2
dx
1
1 dx
2
M =
S
O
O
L
L
2
1 dx
2
dx
O
O2
and made calculations:
SSL
2
1
M =
6 1
2
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
C
X (X) and of the concentrated moments I, the principle of virtual work
is written:
D
NR
M
dx =
dx
C
X
X
+
I
I (),
dx
i=1
The law of behavior is:
D
M
X
X (X)
= G C dx
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:
G C 1 - 1
K =
L
- 1 1
C L
2 1
and
M =
.
6
1 2
The calculation of the matrix of mass like having previously required to discretize the field
solution.
The second member, due to couple X distributed, is in the same way obtained that for the movement of
traction and compression:
L
X
1 (X)
dx
X
1 (X) = 1
O
L
L
X = X
2 (X)
dx
X
2 ()
L
O
2.3
Movement of inflection
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 inflection 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 Inflection 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 inflection
in the plan (x0z):
NR
V
(“+) + Me = (T + m + +, éq 2.3.1-1
ext.
ext.
) Tc
C
I ()
m
Z
y
Z
y
I
I
I ()
i=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
'
Z (“+) + Me
=
K SG
y
Z
'+ '+ +
'
(W y) () E Iy y
= K S
Z
[G (wL) (“(L) + (L))- (W) 0 (” () 0+ () 0)]
-
K SG
Z
(W '' + ') + K SG
Z y ('+)
+ EIy [y (L) “(L) - y () 0” () 0] - I.E.(internal excitation)
y y “
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
éq 2.3.1-2
I.E.(internal excitation) “-
K SG
y
Z
(+) = 0
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
Z (“+) + Me = K SG
y
Z
'
+
- 0 '0 + 0
[(wL) ((L) (L)) (W) (() ())] éq 2.3.1-3
+ EIy [y (L) “(L) - y () 0” () 0]
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:
(,) = (, ii+) I
4
=,
1…, 4
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12 I.E.(internal excitation) y
where, while having noted y =
, functions
K
2
I are defined by:
Z SGL
3
2
1
X
X
X
1 (X) =
2
3
+ 1
1+
L -
L
y
-
+
L (
y)
y
6
X
X
5 (X) =
-
L (
1
1+
y) L
L
3
L
X
4 +
2
y X
2 + there X
2 (X) =
-
+
1+
L
2
L -
2
L
y
2
1
X
X
6 (X) =
3
4 +
+ 1+
L (
y)
1+
L - (
y)
y
1
X 3
X 2
X
(X) =
3
-
2
+ 3
+
1+
L
L
y
L
y
- 6
X
X
(X) =
1
7
-
L (1+
y) L
L
L
X 3
2 -
2
y X
y X
(X) =
4
-
+
+
1+
L
2
L
2
L
y
éq 2.3.1-4
1
2
X
X
(X) =
3
+ -
2 +
1+
L
(
y)
8
L
y
One checks without difficulty that the couples (I, i+4) check well [éq 2.3.1-2]. Moreover:
(1, 5) = (1,) 0,
= 0, 0
(0)
(1 5)
()
(L)
(2, 6) = (0,) 1 (,) = (0,) 0
(0
2
6
)
(L)
(
éq 2.3.1-5
3, 7) = (0,)
0
(,) = (1,)
0
(0
3
7
)
(L)
(4, 8) = (0,) 0 (,) = (0,) 1
(0
4
8
)
(L)
The matrix of rigidity results easily from [éq 2.3.1-3] (by ordering the columns according to
((wo), (O)
,
(
W L)
, (L
y
y
)
).
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L
L
1
-
- 1
-
(2
2
4+
2
2
y
) L
L
(2 - y) L
12 I.E.(internal excitation)
K =
y
12
2
12
L3 (+
1
y
)
L
Sym
1
(2
4+
2
y) L
12
It is clear that the calculation of the efforts acts in the same way that to [§2.1.3].
2.3.2 Inflection in the plan (xOy)
The matrix of rigidity for a movement of inflection 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
éq 2.3.2-1
I.E.(internal excitation) “-
K SG
Z
Z
('-) = 0
The couples of functions selected tests are:
(1, - 5); (- 2, 6); (3, - 7); (- 4, 8)
12 I.E.(internal excitation)
Z
I being given by [éq 2.3.1-4] while having replaced y by Z =
. The matrix of rigidity
K
2
y SGL
obtained is:
L
L
1
- 1
(2
2
4+
2
2
Z) L
L (2 - Z
) L
12 I.E.(internal excitation)
-
K =
Z
12
2
12
L3 (+
1
Z
)
L
1
-
(2
4+
2
Z) L
sym
12
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2.3.3 Determination of the matrix of coherent mass with the matrix of rigidity
Option “MASS_MECA” of operator CALC_MATR_ELEM.
2.3.3.1 Inflection in the plan (xoz)
Let us consider the movement of inflection in the plan (X O Z), the work of the inertias is written:
L
W = (W W % + %
m
y
I y
O
Z
) dx
with
= dS
and =
y
2 dS
m
.
S
Iz
S
In the case of a homogeneous material, we have:
m = S
and
I = Iz.
Z
(
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) = 1 (X) w1 (T)
+ 2 (X)
(T
y
) + 3 (X)
w2 (T)
+4 (X) (T
y
)
1
2
(X, T
y
) = 5 (X) w1 (T) + 6 (X) (T
y
) +7 (X) w2 (T) +8 (X) (T
y
)
1
2
w1
1
y
1
2
in other words:
W
T
= W
with
W
W =
and
=
~
~
W
W
2
3
y
2
4
5
T
6
y = W
front
EC.
=
y
y
~
7
8
By integrating these notations in the expression of the work of the inertias, one a:
L
W
T
T
T
T
=
W W +
% W
W
% dx
m
O
W
W
Iz
y
y
~
~
~
~
L
or:
W
T
T
T
T
=
W W +
% W
W
% dx
m
.
O
W
W
Iz
y
y
~
~
~
~
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02/12/96
Author (S):
J.M. PROIX, P. MIALON, m.t. BOURDEIX
Key:
R3.08.01-A
Page:
24/72
One deduces the form from it of the matrix of mass:
L
M = (mij) m =
S
ij
(X)
I
(X) J (X) + I
Z (X) i+4 (X)
j+4 (X) dx
O
for I from 1 to 4 and J from 1 to 4.
That is to say:
2
2
2
2
13L 7L
L
- 11L2
11L2
2
9
3
13 2
3 2
2
Z
Z
Z
L
L
L
L
L
L
L
Z
Z
Z
Z
Z
+
+
-
-
+
+
+
+
35
10
3
210
120
24
70
10
6
420
40
24
2
2
L3
L3
3
2
2
2
3
3
3
2
Z
L
- 13L
3L
L
- L
L
L
Z
Z
Z
Z
Z
S
+
+
-
-
-
-
140
60
M =
105
60
120
420
40
24
120
(
2
2
2
2
2
2
+
1
13L
7 L
L
- 11L
11L
L
Z
)
Z
Z
Z
Z
+
+
+
+
35
10
3
210
120
24
L3
L
3
L
3
2
Z
Z
+
+
105
60
120
6
- 1 Z
- 6
- 1 Z
+
+
5L
10
2
5L
10
2
2L
L
2
2
Z
Lz
1
Z - L L
L
Z
Z
I
+
+
-
-
+
Z
+
15
6
3
10
2
30
6
6
(
2
+
1
6
1
Z
Z)
-
5L
10
2
2L
L
L 2
+
Z + Z
15
6
3
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 inflection around the axis (O Z)
In the same way, for the movement of inflection around the axis (O Z), in the plan (X O y), the work of the forces
of inertia is written:
L
(v v+
%
%
m
Z
I
Z
O
y
) dx
with:
=
Z
dS =
I
I
.
2
y
S
y
This time v (X, T) and Z (X, T) are discretized in accordance with [§2.3.2] by:
v (X, T) = 1 (X) v T -
1 ()
X
2 ()
T
Z () + 3 (X) v2 (T) - X
4 () Z
(T)
1
2
Z
(X, T) = - 5 (X) v1 (T) +6 (X) Z
(
T) - 7 (X) v2 (T) +8 (X) Z
(
T)
1
2
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
25/72
We obtain the matrix of following mass then:
2
2
2
2
13L 7L
L
11L2
11L2
2
9
3
- 13 2
3 2
2
Z
Z
Z
L
L
L
L
L
L
L
Z
Z
Z
Z
Z
+
+
+
+
+
+
-
-
35
10
3
210
120
24
70
10
6
420
40
24
3
3
3
2
2
2
2
2
3
3
3 2
L
L Z
L
13L
3L
-
Z
Z
L
L
L
L
Z
Z
Z
S
+
+
+
+
-
-
M =
105
60
120
420
40
24
140
60
120
(
2
2
2
2
2
2
+
13L
7 L
L
- 11L
11L
1
L
Z
)
Z
Z
Z
Z
+
+
-
-
35
10
3
210
120
24
3
3
3 2
L
L
L
Z
Z
+
+
105
60
120
6
1
- 6
1
Z
Z
-
-
5L
10
2
5L
10
2
2
2
2L
L
Z
L
- 1
Z
Z - L Lz
L Z
I
+
+
+
-
+
Z
15
6
3
10
2
30
6
6
+ (
-
1+
Z
Z) 2
6
1
+
5L
10
2
2
2L
L
Z
L Z
+
+
15
6
3
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 utilize the coefficients of form ky and kz, 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 “Iz” 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 operator CALC_MATR_ELEM.
The element beam is considered with section constant S and 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.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
26/72
With regard to the diagonal component related to the movement of traction and compression (M11) and
that related to the movement of torsion (M44), we have:
L
M
= S
11
2
L
M
=
44
(
I + I)
I, I
y
Z
y
Z: geometrical moments.
2
One can consider that these components were obtained by sharing the element of beam into two
L
equal shares length
then by associating the mass and inertia obtained the node of
2
half-element. For M44, the preceding expression corresponds to a choice: one could also
L
to write: M
= C
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:
(E)
F
=
F
my (aei)
E
N
i=1, N
(ae: nodes I of the élém
ent
E, N
I
: 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 inflection which are programmed are:
L
M
= S
,
22
2
L
M
= S
,
33
2
L3
L2
2L
M
= Min
S, S
+
I,
55
y
105
48
15
L3
L2
2L
M
= Min
S, S
+
I.
66
Z
105
48
15
One finds well the components M22 and M33 related to the translations of the movements of inflection by
technique of the masses concentrated with the nodes. On the other hand, the origin of the formulas used for
components M55 and M66 related to rotations, is unknown. One can simply notice that one
find the values:
L3
2L
S
+ I
,
Z
105
15
L3
2L
S
+ I
y
.
105
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.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
27/72
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.
F
M
O
C
C
O
(not of rotation)
(not of displacement in C)
At the point C, the effects of inflection 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
X
with
= 0 vector rotation
0
0
and
OC = ey
E Z
Z
E
C
Z
y
O
ey
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
28/72
In fact, one obtains:
U
= U
X
C
X
U
= U + E
y
éq 3.1-1
C
y
Z
X
U
= U - E.
Z
C
Z
y
X
The change of variables given by [éq 3.1-1] is written matriciellement:
ux
ux
c1
1 0 0
0
0 0
1
U
y
uy
C
1
0 1 0 - E
0 0
1
U
Z
Z
U
c1
0 0 1 E
Z
y
0 0
0
1
X
X
c1
0 0 0
1
0 0
1
y
c1
0 0 0 0 1 0
y1
Z
Z
c1
0 0 0
0
0 1
1
U
X
=
U
C
2
1 0 0
0
0 0
x2
U
y
U
c2
0 1 0 - E
0 0
y2
U
Z
Z
U
C
Z
2
0 0 1
ey
0 0
2
X
c2
0
0 0 0
1
0 0
x2
y
C
y
2
0 0 0
0
1 0
2
Z
C
2
0 0 0
0
0 1
Z
2
!# # # # # # # # “
# # #
$
# # # # # # #
P
It is thus enough to determine the elementary matrices of mass (Mc) and stiffness (Kc) in
locate (C, X, y, Z) where the movements of inflection 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 = PT K P
C
and
K = PT K P
C
.
The values of ey and ez are to be provided to Code_Aster via the operand
SECTION: “GENERALE” of operator AFFE_CARA_ELEM, default values being obviously
zero values.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
29/72
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 (X) = S + 2
1
- 1
1
S
L
1
· when two side dimensions vary (in a linear way), the surface of section will evolve/move
in a quadratic way.
2
S
X
S (X) = S +
2
1 1
- 1
S
L
1
Code_Aster makes it possible to treat sections “CERCLE”, “RECTANGLE” and “GENERALE”, 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
yes
according to y
general
yes
not
yes
For section “RECTANGLE”, it is the user who chooses the type of variation, by specifying “AFFINE”
or “HOMOTHÉTIQUE” in AFFE_CARA_ELEM. It should well be noted that in case “AFFINE”, them
dimensions can vary only according to Y.
We consider generally that the section varies according to the formula [éq 3.2-1]:
X m
S (X) = S + C
1 1
L
éq 3.2-1
S1 is the initial section in X = 0
C is fixed by the knowledge of the final section S2 in X = L.
m gives the degree of variation: m = 1 variation linear, m = 2 variation quadratic.
The section varying, it goes from there in the same way inertias I (X
y
), I (X
Z
) and I (X
p
).
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
30/72
We will thus have:
X m+2
I (X) = I
1+ C
y
y1
L
éq 3.2-2
X m+2
I (X) = I
1+ C
Z
z1
L
éq 3.2-3
X m+2
I (X) = I
1+ C
p
p1
L
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 (Seq)
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 Seq value lain between S1 and S2 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:
U
E S (X)
= 0 0
X
L
X
X
éq 3.2.1-1
U
with
NR (X) = E S (X) X
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
31/72
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:
U
E S (X)
= C
X
1
or, by taking account of the expression of S (X):
X m U
E S 1+ C
= C
1
L X
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)
0 = u1 and (
U L) = u2:
U
C
X - m
1
=
+
1 C
X
E S
L
1
C
L
X
from where U (X) =
1
ln +
1 C
+ C if m = 1
E S
C
L
2
1
C
L
1
and
U (X)
=
1
+ C if m2
E S
C
m 1
-
2
1
(
X
L -)
m 1+ C
L
It is noted that the expression of (
U X) is far from being polynomial.
By taking account of the fact that:
C
= - NR
Po
ur X =
1
1
0
C
= + NR for X = L
1
2
and that:
(+
1 c)-m 1
+ U (0) - U (L)
C
=
2
(+
1 c)-m 1
+ - 1
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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
we obtain:
E C M
NR
1
1 =
ln (
u1 - U
L 1+ c) (
2)
for m =
1
E C M
NR
1
2 =
ln (
u2 - U
L 1+ c) (
1)
E C M
1
1
1
(1 -)
m
(1+ c) m
(1+ c) m
NR
1 =
- U
+u
1
2
L
1 (1+c) M-1
1 (1+c) M-1
and for m = 2
E C M
(1+ c) m 1 -
(1+ c) m 1
-
1
(1 -)
m
NR
U
2 =
U
.
L
1
1 - (1+ c) m 1
- - 2
1 (1+c) m 1
By replacing C by its value, that is to say:
· if m = 1
S
C =
2 -,
1
S1
E
(S2 - S1)
N1 =
(u1 - u2)
L
ln S2 - ln S1
E
(S2 - S1)
N2 =
(u2 - u1)
L
ln S2 - ln S1
· if m = 2
S
C =
2 - 1
S1
E
N1 =
S S
1 2 (
u1 - u2)
L
E
N2 =
S S
1 2 (
u2 - u1)
L
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
2
1)
Seq =
for a section varying linearly
S -
ln ln S
2
1
S
=
S S
eq
for a section varying in a quadratic way
1 2
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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
3.2.1.2 Determination of a constant of equivalent torsion (Ceq)
The equation of pure torsion of a beam with variable section, is written:
Q
G C (X) X
0
X
X =
éq 3.2.1-2
X m+2
with: C (X) = C 1
1
+ C
(m or
1
2)
L
=
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) according to displacements
1
2
with the nodes (X,
X and to deduce some, by comparison with the formulas with constant section,
1
2)
expression the one geometrical moment polar are equivalent.
By integration of [éq 3.2.1-2], we have:
X m+2
G C 1+ C
X
D
1
L
=
,
X
1
D1 the constant of integration is determined by the torques applied to the nodes.
- (m+2)
D
X
Of:
X
=
1
+
1 C
,
X
G C
L
1
we deduce:
- (m+)
D
- L
X
1
1
X (X)
=
1+ C
D.
G C
2
+
1
(
C m+)
1
L
We determine D2 starting from the system:
- D L
1
X =
+ D
1
G C C
2
1
(m+) 1
- D L
1
- m 1
+
X
=
1+ C
+ D
2
G C C
2
1
(m+) (
) ()
1
- m 1
+
X (1+ c) (
) +x
that is to say: D
1
2
2 =
(
.
1+ c) (m+) 1 - 1
By taking account of the fact that:
D
= - for X
=
1
1
0
D
= for X
=
L
1
2
,
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
34/72
we have finally:
for m = 1
2
2
G
(2C C213 - C C123)
1
=
2
2
-
L
1
2
(
X
X
C2 3 - C13)
(
)
2
2
G
(2C C213 - C I1p 32)
2
=
2
2
-
L
2
1
(
X
X
C 3 C
2
13)
(
)
We thus take in the case of a section varying linearly, a polar geometrical moment
Ceq equivalent of the following form:
(2
2
2 C C
2
13 - C C
1
2 3)
Ceq =
(
linear variation
2
2
C2 3 - C13)
for m = 2
3
3
G
(C C214 - C C124)
1
=
3
3
3
-
L
1
2
(
X
X
C24 - C14) (
)
3
3
G
(C C214 - C C124)
2
=
3
3
3
-
L
2
1
(
X
X
C 4 C
2
14)
(
)
In the case of a section varying in a quadratic way, the polar geometrical moment is written:
(3
3
C C
2
14 - C C
1
2 4)
Ceq = 3
(
quadratic variation
3
3
C24 - C14)
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 and Z
who would come to substitute themselves
eq
eq)
at the geometrical moments (I
and I
y
Z) 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 inflection 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.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
35/72
Let us consider the inflection in the plan (X O y).
On the basis of the static equation of the movement of inflection of a beam of the Euler-Bernoulli type:
2
2 v
v
I.E.(internal excitation) (X)
= 0
Z
, and
=
2
X
2
X
Z
X
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)
1
= v2
(O) = (L) =,
Z
Z
Z
Z
1
2
that is to say:
v1
C1
Z
C
1
= B 2, B
v
stamp (4 X 4).
2
C
3
Z
2
C4
2
v
Efforts: V (X) =
I.E.(internal excitation) (X)
y
2
X
Z
X
2
v
and moments: M (X) =
I.E.(internal excitation) (X)
Z
Z
,
x2
also express themselves according to these constants of integration, and one can write:
Ty
C
1
1
M
Z
C
1
= D 2
, D
T
stamp (4 X 4).
y
C
2
3
M
Z
C
2
4
The matrix of rigidity corresponds to the product D B-1. The terms of this matrix are clarified in
following tables.
X m+2
Let us recall that I (X) =
I 1 + C
, m
1 or 2,
Z
z1
L
=
C
C 2
C 3
Let us pose
W1 = I.E.(internal excitation)
, W2 =
I.E.(internal excitation)
, W
I.E.(internal excitation)
.
z1 L
z1
L
3
Z
=
1
L
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
36/72
Variation closely connected
Quadratic variation
m = 1
m = 2
1
1
I
3
I 4
C =
2
- 1
C =
2
- 1
I
I
1
1
C + 2
4
2
1
(C +3 c+) 3
C
2 C
2
(c+)
3
3
(
C C +)
1
2 C (c+)
1 (2 c+)
3
2 C
2
4
(c+)
1 (2 c+)
3
4 C
= C -
2 2
C
5
2
4
(c+) 1
= C -
4 2
C
2
6
3
5
(c+) 1
3
(
C
c+ 2) ln (c+)
1 - 2 C (c+) 1
The matrix K is written then:
W W
- W
W
3
1
2
2
3
1
2 3
1
W
- W
+W
K =
1
4
2
2
1
5
Sym
W
- W
3
1
2 3
W
1 6
Now let us consider the inflection 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 inflection in plan (0, X, Z) by
values given in the following table.
Variation closely connected: inflection in the plan (0xz)
I
C
= 2 - 1
I1
(
ln C +)
1
2
1
C
1
2 - 1
1
(C +) 1 1
- 2
1
1
+ C - 2
5 = C 2 - 4
6 = C 3 - 5
(C + 2) ln (C +) 1 - 2c
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
37/72
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 S2 - K S
y
y
1
K S
2
1
y
)
(
)
=
eq
(
ln K S
y
2 -
1
1
2
) L (N K S
y
)
(
if the variation is closely connected
K S2 - K S
Z
Z
1
2
1
)
(K S
Z
) =
eq
ln (K S
z2 2) - ln (K S
z1 1)
(K S
y
) = S S K K
eq
1 2
y
y
1
2
(
if the variation is quadratic
K S
Z
) = S S K K
eq
1 2 Z
Z
1
2
and one in the same way introduces the additional terms into K that for a constant section.
Calculations are not here detailed. One obtains a matrix K of the same forms than previously
with for principal modification the value of:
· variation
closely connected
2
= (C + 2) ln (C +)
1 - 2c +
c3 (C + 2)
12
· variation
quadratic
c3
2
=
+
c3 (c2 + C
3 +)
3
C +1
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 =
1
2 if the variation is closely connected
2
1
L
S + S + S S
S =
S (X) dx =
1
2
1 2 if the variation is quadratic
L O
3
I + I
y
y
I
1
2
y
=
2
I + I
Z
Z
1
2
Iz =
whatever the variation
2
I + I + I + I
y
y
Z
Z
I
=
1
2
1
2
X
2
The matrix of mass is then calculated like that of a beam having these characteristics.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
38/72
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:
S
3 + S
1
2
0
8
-
for traction compression: L
S
3
+ S
0
2
1
8
(L I +I +I +I
y
Z
y
Z
1 0
1
1
2
2)
X
-
for torsion:
1
4
0 1.
x2
-
for the movements of inflection:
S L
1
v1
(w1)
2
M
5,5 (M6,6)
0
z1 (y1)
S L
2
2
v2 (w2)
0
M
11,
11
(M12,12) Z
2 (y2)
S L3
2
eq
S L
eq
L
with M
= M
= min
,
5 5
,
11 11
,
105
48
+
+
15 (I
I
y
y
1
1)
S L3
2
eq
S L
eq
L
M
= M
= min
,
48
+
+
15 (I
I
z1
z2)
6,6
12 12
,
105
S + S
with S
1
2
eq =
2
· 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
1
S + S2)
L
0
12
1
U
(
S +5
U
1
S2)
2
0
L
12
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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
-
for torsion:
(I +I +I +I
y
y
Z
Z
1 0
1
2
1
2)
X
L
1
4
0 1 x2
-
for the inflection in each of the two plans:
(5
v W
1
S + S2)
1 (1)
L
12
M
Z
y
5 5
, (M6,6)
0
1
(1)
(
S +5
1
S2)
L
v
2 (2
W)
12
11
M 11
,
(M1212,) z2 (2y)
(
3
3
1
S + S2) L
(1
S + S2) L
L
with M
= M
= min
,
+
5 5
,
11 11
,
I + I
2
105
2
48
(1y 2y) 15
(
3
3
1
S + S2) L
(1
S + 2
S) L
L
and: M
=
12 = min
,
+
I + I
.
2
105
2
48
(1z 2z)
6,6
M12,
15
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
40/72
4
Geometrical rigidity - Structure prestressed
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].
Vo
V *
V
oij
ij
virgin state
state
state
of constraint
prestressed
deformed
Appear 4-a: Les various states
This change of geometry does not modify the equilibrium equation, within the framework of the assumption of small
disturbances (HPP) around Vo (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:
U 3D
v 3D
W
K
O
K
=
FD
G
X ij X
I
V
J
O
where U 3D is displacement (resp. v 3D virtual displacement kinematically acceptable) taken with
to start from V * (but compared to Vo within the framework of the HPP) and O the prestressing (of Cauchy if one
wants) since one is within the framework of the HPP.
WG being the symmetrical bilinear shape out of U 3D and v 3D, it can be interpreted like the variation
of a potientiel UG.
W
= U
G
G
One a:
3D
3
U
U D
2 U
K
O
K
G
=
X
ij X
I
V
J
O
For a model of beam 3D, the tensor of constraints is reduced in the local axes of the beam to
components xx, xy and xx, from where:
3D
3D
3D
3D
3D
3
U
U
U
U
U
U D
2 U
O
I
I
=
+ 2 O
I
I
+ 2 O
I
I
G
xx
X
X
xy X
y
xz X
Z
Vo
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
41/72
2
U 3D
3D
3D
3D
3D
X
ux ux
ux U
Terms
X
,
and
are neglected [bib5]. Moreover, in the reference mark
X
X y
X
Z
room of the center of torsion of the beam:
U 3D
X (X, y, Z) = (ux) + zy (X) - y Z (X)
U 3D
y (X, y, Z) = v (X) - Z y (X)
U 3D
Z (X, y, Z) =
(
W X) + Z X
(X)
u'
'
'
+z - y
y
Z
- Z
y
and U =
v'
'
- Z X
0
- X
w'
'
+ there X
X
0
from where one draws, according to the preceding assumption:
2
2
2 U
O
=
(v'-z '
'
O
'
O
'
X) + (w'+ there X)
2 xy (w' there X)
2
X
xz (v' Z
G
xx
X) (
X)
+
+
+
-
-
Vo
However, the generalized efforts are connected to the constraints by the expressions:
NR °
O
= V
O
=
V
O
xx
y
xy
Z = xz
S
S
S
M = Z M = - y
y
xx
Z
xx
S
S
One deduces some:
L
2 U
=
NR °
O
'
O
'
O
O
G
(“+” -
'-
'+
'-
'
O
(v) 2 (W) 2) 2 M v 2 M W 2V W 2V v
y
X
Z
X
y
X
Z
X
O
'
O
O
+
'
xx (2 2
y + Z) (X
) 2 + (2 y + Z
xy
xz) X
X
Vo
By supposing, moreover, that oxx is constant in the discretized element (what is inaccurate for example
for a vertical beam subjected to its actual weight) and that X varies linearly compared to
X
X
1
1
'
1
2
X
X
= 1
1
2, from where
, it comes:
L +
= -
+
L
X
L
L
O
O
NR
I y + Iz
NR
I y + Iz
-
1
O2
2
2
'
L
S
L
S
,
xx (y + Z) (X)
= (1 2)
O
O
NR
I + I
NR
I + I
V
y
Z
y
Z
O
-
2
L
S
L
S
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
42/72
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)
X
= const
,
handle
M y = (M - M
y2
y1) + M
L
y1
V (X)
y
= constant,
X
V (X)
M
= constant,
Z = (M
- M
z2
z1)
+ M
Z
L
z1
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 =
1
2
0
A3
Higher triangular part of the geometrical matrix of rigidity with:
1
2
3
4
5
6
1
U
1
v
1
W
X
1
1y
1z
1
M o1
y
O
-
NR
2L
NR O
2
12 L
O
O
10
M Y2 V
-
+ Z
2L
2
M o1
Z
O
-
NR
2L
NR O
3
12
-
With
O
1
=
L
O
V
10
Mz2
-
- y
2L
2
O
O
O
O
(
- Mz + Mz
+ My - My
I
O
y + Iz)
1
2
1
2
NR
12
12
4
(S L)
(LVo
O
y)
(LVz)
-
-
12
12
2 L NR O
5
15
2 L NR O
6
15
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
43/72
7
8
9
10
11
12
2
U
v2
2
W
X
2
y2
Z
2
1
M o1
y
NR O
2L
NR O
2
- 12 L
O
O
10
M Y2 V
+
+ Z
2L
2
M o1
Z
NR O
2L
NR O
3
- 12
-
L
O
Original version
10
Mz2
+
- y
2L
2
M O
O
O
O
O
(
M oy2)
y
M
Mz - Mz
- My +
With
1
1
1
2
Z
1
2
=
O
2L
2L
(Iy +Iz)
(
)
NR
12
4
-
12
M O
O
O
Original version
S L
O
O
y2
V
M
(LVz)
Z
z2
y
()
(LVy)
+
-
+
+
2L
2
2L
2
-
-
12
(
12
M O - M O
z1
z2)
NR O
12
(L No)
5
-
-
10
(LVo
30
y)
+ (12
- M O + M O
y1
y2)
NR O
12
(L No)
6
-
-
10
(LVo
30
y)
+ 12
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
44/72
7
8
9
10
11
12
2
U
v2
2
W
X
2
y2
Z
2
7
M oy1
O
-
NR
2L
NR O
8
12
-
L
O
O
10
M Y2 V
-
- Z
2L
2
M o1
Z
O
-
NR
2L
NR O
9
12
With
O
O
3
=
L
V
10
Mz2
-
+ y
2L
2
O
O
O
O
(
- Mz + Mz
+ My - My
I
O
y + Iz)
(1
2)
(1 2)
NR
12
10
12
+
(S L)
(LVo
O
y)
(LVz)
-
-
12
12
2 L NR O
11
15
2 L NR O
12
15
My
M
By using the equalities
-
Z
V = 0 and
+V = 0, one finds the programmed matrix.
X
Z
X
y
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 - zC) X
W
(X, y, Z) =
(
W X, C)+ (y - teststemyç) X
These two modifications bring additional terms in the geometrical matrix of rigidity:
The assumption of moderate rotations results in adding with 2UG the term:
L
2 1
U
=
- M O
O
O
O
, +
, +
+
O
(X y) M
X
y (X Z)
V
V
G
Z
X
y
X y
Z
X y
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
45/72
The terms of matrix A to be added are:
O
O
(
M
M
4 -)
5
1
: +
Z (10 -)
11
z2
: -
2
2
O
O
(
M
M
4 -)
6
y1
: -
(10 - 12)
y2
: +
2
2
With regard to the eccentricity of the center of torsion, it is necessary to add the terms corresponding to:
L
L
L
U 2
= NR O Z v' '
- NR O y W '
O
O
'
G
C
'-
+
O
X
C O
X
(y V Z V
C
y
C Z)
X
O
X
+ (
L
2
y M O - Z M O
'
C
Z
C
y) (X
)
O
Moreover, it is necessary to carry out a change of reference mark as with [§3.1].
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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
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 inflection in
the plan of the element of the inflection 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.
z2
Z
y
z0
x2
arc
y2
Nj
M
NR
y0
I
C
0
x1
xo
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.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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
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'&
&&y
'&
'&
Ni
&&z
&&x
Nj
C
“&” & '&
&& (X, y, Z) bases local curved beam
Appear 5-b: average Repère (local reference mark)
The equilibrium equations static are:
M1
NR - V =
M
-
= 0
, S
1
0
T, S
R
NR
M
V
+
= 0
1
M + T
1
- V
= 0
, S
R
, S
R
2
V
= 0
M
+V
=
2, S
2, S
1
0
this for:
Q
'&
&&s
Y
M
'&
&&n 1
P
X
Z
Appear 5-c: Curvilinear reference mark
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
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To integrate, the conditions out of P are used:
NR = F
M
= M
y
T
y
V
= - F M
= - M
1
X
1
X
V
= F
M = M
2
Z
2
Z
While integrating and while passing in the system of following axis:
Q
X
'&
&&s
'&
&&n 1
y
Z
P
Appear 5-d: System of axis chosen by integration
One obtains:
NR
cos
- sin
0
0
0
0 Fx
V
1
sin
cos
0
0
0
0
Fy
V
2
0
0
1
0
0
0 Fz
=
M
0
0
R
T
(
cos -)
1
cos - sin 0
MX
M
0
0
R sin
sin
cos
1
0
My
M
R R
2
(cos - 1
) -
sin
0
0
0
1
Mz
!# # # # # # # # # # # “
# # # # # # # # # #
$
# # #
J
We now will take into account the mechanical characteristics by using energy
potential:
~
~
~
~
~
~
1
2
2
2
2
2
S
2
2
NR
V
V
M
M
M
Ep
1
2
T
1
2
=
+
+
+
+
+
ds
2 s1 ES K SG K SG
I.E.(internal excitation)
I.E.(internal excitation)
I.E.(internal excitation)
1
2
1
2
~
(the sign “~” means that we use the efforts intern) (note: F = - F by the principle
of action-reaction).
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
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Recall of the relations of behavior
within the framework of the model of Timoshenko:
U
NR = ES S
~
W
V
1
1
= K SG
1
-
S
2
~
W
V
2
2
= K SG
2
+
S
1
~
M
= I.E.(internal excitation)
T
S
~
M
1
1
= - I.E.(internal excitation)
S
~
M
2
2
= - I.E.(internal excitation)
2
S
Torque of the interior efforts:
Kinematic torque:
~
~
S
NR + T
s+ N +
1
1 N
{}
2
T
2
~
~
{}
C
M S + M
U.S. + W N + W N
T
1
1
2
2
from where:
1
0
0
ES
NR
1 s2
1
Ep =
0
0
2 (
NR, V, V
1
2)
V1
ds
s1
K SG
1
V
1
2
0
0
K SG
2
1
0
0
I.E.(internal excitation)
M
1
T
S
1
+ 2
(M, M, M
T
1
2) 0
0 M ds
2
S
1
I.E.(internal excitation)
1
1
1 m2
0
0
EI2
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
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or:
1
ES
1
Fx T
Fx
K SG
1
Fy
Fy
1
0
1
B Fz
K SG
Fz
Ep
T
=
J
2
J
Rd
2 O MX
Q
1
MX
My
GC
My
1
Mz
0
Mz
P
I.E.(internal excitation)
P
1
1
I.E.(internal excitation)
2
J being the matrix obtained previously.
One can thus calculate the matrix of flexibility []
C:
1
ES
1
K SG
1
1
0
K SG
[]
C
=
JT
2
J
1
Rd
0
GC
1
0
I.E.(internal excitation)
1
1
I.E.(internal excitation)
2
We can notice that the matrix J can break up into two pennies matrices
independent, a part concerning the inflection in the plan of the element, the other concerning the inflection
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
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
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Inflection in the plan of the element:
NR
cos
- sin
0 Fx
V
=
sin
cos
0 Fy
1
M
R (cos -) 1 - R sin 1 Mz
2
!# # # # # “
# # # # # # $
#
1
J
Inflection out of the plan of the element:
M
cos
-
sin
R (cos
-) MX
T
1
M
= sin
cos
R sin
1
My
V
Fz
2
0
0
1
!# # # # # “
# # #
$
# # # #
J2
5.1
Stamp flexibility for the inflection in the plan of the beam [C1]
1
cos
sin R
(cos - 1
) ES
[
1
C1] =
- sin cos
- R sin
J Rd
O
K SG
1
1
0
0
1
1
I.E.(internal excitation)
2
2
2
cos sin R2 (cos -) 12
cos
sin cos
sin R2 (cos-)
1 sin
R (cos 1
-)
+
+
-
+
-
ES K SG
I.E.(internal excitation)
ES
K SG
I.E.(internal excitation)
1
2
1
2
EI2
2
2
2
2
=
sin cos R sin
R sin
+
+
-
Rd
O
ES
K SG
I.E.(internal excitation)
I.E.(internal excitation)
1
2
2
1
sym.
I.E.(internal excitation)
2
Appendix:
cos 2D +
S
in 2D =
O
O
1
1
cos 2D
=
(+sin cos
) =
(2 - sin (2))
O
2
4
1
1
sin 2D
=
(+sin C bone) =
(2 - sin (2))
O
2
4
sin2
sin cos D =
O
2
from where:
R
R
R3
C111 =
(2 +sin (2))+
(2 - S (
in 2)
) +
(6 - 8 sin +s (in2),
4ES
4 K SG
4 I.E.(internal excitation)
1
2
R
R
R3
C122 =
(2+ (
sin 2) +
(2 - sin (2))+
(2 - (
sin 2),
4ES
4 K SG
4 I.E.(internal excitation)
1
2
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
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R
C133 =
,
EI2
R
R3
R3
R3
C1
2
2
2
12
= -
sin +
sin -
sin -
cos -
[2
2],
2ES
2k SG
2EI
2EI
1
2
2
R2
C113 =
(sin -),
EI2
R2
C123 =
(cos -)
1.
EI2
5.2
Stamp flexibility for the inflection out of the plan of the beam [C2]
1
0
0
cos
sin
0 I.E.(internal excitation)
[
1
C2] =
- sin
cos
0 0
0 J Rd
O
I.E.(internal excitation)
2
R (cos
-)
1
R
1
sin 1
1
0
0
K SG
2
2
2
cos sin
cos
sin cos
sin
R (cos -)
1 cos R
2
sin
+
-
+
+
I.E.(internal excitation)
I.E.(internal excitation)
I.E.(internal excitation)
I.E.(internal excitation)
I.E.(internal excitation)
I.E.(internal excitation)
1
1
1
2
2
sin cos
R (cos-)
1
=
sin
R cos
sin
+
-
+
Rd
O
I.E.(internal excitation)
I.E.(internal excitation)
I.E.(internal excitation)
1
EI1
R2 (cos-) 2
1
R2 sin2
1
sy.
m
+
+
I.E.(internal excitation)
I.E.(internal excitation)
K SG
1
2
R
R
C211 = -
(2+ (
sin 2) +
(2+s (in2)
4EI
4EI1
R
R
C222 =
(2
-
(
sin 2) +
(2
+ S (
in 2)
4EI
4EI1
R3
R3
Rb
C233 = -
[6 - 8 sin +s (
in 2)]+
(2+ (
sin 2) +
2EI
4EI
K SG
1
2
R
R
C2
2
2
12
= -
sin +
sin
2EI
2EI1
R2
R2
C213 =
(2+ (
sin 2) - 4sin) +
(2 - (
sin 2)
4EI
4EI1
R2
2
R
C2
(sin2 +2cos - 2)
2
23
= -
+
sin
2EI
2EI1
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
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Having determined the matrix of flexibility, we will be able to calculate the matrix of rigidity.
One thus has:
1
T
E
E
=
T
C
T, this
T, (
: for outside)
P [
] E
E
E
p
P
P
2
Like one a:
Te
J Te
=
Q
P
(Recall: Te and Te
are not described in the same base).
P
Q
One also has:
T
1
T
E
E
1
-
1
-
E
E
p
=
T
J
C J T
,
T
Q
[] Q
Q
2
Using the theorem of Castigliano, one can obtain displacements associated with the external efforts
Te.
U = [C] Te
p
in the local base of the point P
,
P
(S N, N
1
2) P
T
and U
= J 1
-
- 1 in the local base of Po Q
int
,
Q
[]
C J Te
Q
(S N, N
1
2) Q
Te
and
Te.
P
Q
By breaking up the problem into two subproblems, and using the principle of superposition, one
can write:
Te
=
E
+
E
=
I
I
-
-
I: for intern,
P
T
T
T
T
total
PP
PQ
PP
PQ
(
)
effort
effort
coming
coming
point
point
P
Q
and Te
=
Te
+ Te
=
I
I
-
-
Q
QQ
T
T
total
QP
QP
QQ
effort
effort
coming
coming
point
point
P
Q
(one applies two torques of effort (independent one of the other) to the point P and the point Q).
Action coming on a side: Ti = - Te = - K Up
- 1
PP
P
(K = [] C)
Action coming on other side: Ti = +
1
J Te =
1
J
T
T
PQ
Q
[J
K
J
] U = K
J
U
Q
Q
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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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We obtain as follows:
Te
=
-
K U
K JT U
Ptotal
P
Q
and the same: Te
= - J K U + J K JT U
Qtotal
P
Q
The matrix of rigidity for “displacements” U P (given in its local base) and for
“displacements” UQ (given in its local base) is:
K - K J
T
K =
- J K J K JT.
Moreover, in the case of beams with hollow circular section (elbows of pipings), one divides I and I
y
Z
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:
S R.
M = M = M = M = M = M
11
22
33
77
88
99 =
2
(I + I
y
Z) R.
M
= M
44
10 10 =
2
2 I R
3
2
.
y
S (R) S (R)
M = M
55
11 11 =
+ min
,
15
105
48
3
2
2 I R
.
S
Z
(R)
S (R)
M
= M
66
12 12 =
+ min
,
15
105
48
This assumption is restrictive and does not allow to take well into account the inertia of inflection or of
torsion which had with the curvature. It is thus in this case to better model a beam curves by several
elements POU_C_T.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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
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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
developed
developed
developed
imposition of a deformation (of type
thermal stratification)
* CHAR_MECA_PESA_R: loading due to developed
developed
developed
gravity
* CHAR_MECA_FR1D1D: loading distributed
developed
developed
developed
by actual values
* CHAR_MECA_FF1D1D: loading distributed
developed
developed
developed
by function
* CHAR_MECA_TEMP_R: loading
developed
developed
developed
“thermal”
* CHAR_MECA_FRELEC: loading “forces
developed
developed
not
electric " created by a secondary conductor
right
* CHAR_MECA_FRLAPL: loading “forces
developed
developed
not
electric " created by a secondary conductor
unspecified
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
assistance of 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 inflection (not of effort
edge, not of torque).
The following relations of behavior are used:
U
NR = ES, X
M
= E I
y
y
y,
X
M
= E I
Z
Z
Z.
X
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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
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Thus, one obtains the second elementary member associated with this loading:
U
at item 1: F
=
E S
,
x1
1 X
M
=
E I
y
,
y
y
1
1 X
M
=
E I
Z
,
Z
Z
1
1 X
U
at item 2: F
=
E S
,
x2
2 X
M
=
E I
y
,
y
y
2
2 X
M
=
E I
Z
,
Z
Z
2
2 X
“U
y
“
While being given
Z
,
and
for a beam, one can affect a loading to him.
X
X
X
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 which multiplied by displacements with the nodes
U gives 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 Y.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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
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One has thus for U:
- 2R sin EPX cos
2R sin EPX
sin
point 1
0
0
U =
0
0
point 2
2R sin EPX
cos
2R sin EPX sin
U
with EPX = X
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:
'
L
'
Q =
(X) S G N dx
O
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.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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
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Axial load (in X):
L
F
=
G X
S
dx
X
I
I
O
X
X
=
1
1
, =
L
2
L
S
S
from where: F
= G
X L 1
2
X
+
at item 1,
1
3
6
S
S
F
= G
X L 1
2
X
+
at item 2,
2
6
3
X
for a section varying S linearly = S +
1
(S2 - S1)
,
L
3 S + 2
S S + S
1
1 2
2
and F
= gx L
X
,
1
12
S +
S S + S
2
2
1 2
3 2
F
=
X
X
G L
,
2
12
2
X
for a section varying homothétiquement
S
S +
S
S
1
(2 - 1)
L
Torque:
Without taking into account of warping, it is non-existent.
· In the plan (X O Z):
L
F
=
G Z S dx
Z
1
O 1
L
M
=
G Z S dx
y
1
O2
L
F
= G Z
S dx
Z
2
O 3
L
M
=
G Z S dx
y
2
O 4
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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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3
2
3
2
X
X
X
X
X
1 = 2
L
3
+1, =
+ 2
,
2
L -
L
-
L
L - L
3
2
3
2
X
X
X
X
3 = -
2
3
, = L
+
4
L +
L
-
L
L
7 S
3 S
from where: F
=
2
Z
1
G
Z L
+
,
1
20
20
2
1
S
2
S
M
= - G
Z
L
+
,
1
y
20 30
3 S
7
1
S2
F
=
G
Z L
+
,
2
Z
20
20
2
1
S
2
S
M
=
G
Z L
+
,
2
y
30 20
for a section varying linearly,
8 S +5 S S
+ 2
1
1 2
S2
and: F
=
G zL
,
1
Z
30
- 2 S - 2 S S - S
M
=
G zL2
1
1 2
2
y
,
1
60
2 S +5 S S
+8 S
F
=
G zL
1
1 2
2
Z
,
2
30
S +2 S S + 2 S
M
=
G zL2 1
1 2
2
y
,
2
60
for a section varying homothétiquement.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
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· In the plan (X O Z):
L
F
=
G y S
D
X
y
1
O 1
L
M
=
G y - S D
X
Z
1
O
2
L
F
=
G y S
D
X
y
2
O 3
L
M
=
G y - S
D
X
Z
2
O
4
7 S
3 S
from where: F
=
G
y L
1 +
2,
y1
20
20
S
S
M
=
G
y L2
1 + 2,
z1
20 30
3 S
7 S
F
=
G
y L
1 +
2,
y2
20
20
S
S
M
= - G
y L2
1 + 2,
z2
30 20
for a section varying linearly,
8 S +5 S
1
S
1 2 + 2 S2
and: F
=
G y L
y
,
1
30
2 S + 2 S S + S
2
1
1 2
2
M
=
gy L
,
1
Z
60
2 S +5 S S
8
+
1
1 2
S2
F
=
gy L
,
2
y
30
- S - 2 S S - 2 S
2
1
1 2
2
M
=
gy L
,
2
Z
60
for a section varying homothétiquement.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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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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6.3 Loadings
distributed
OPTIONS:
“CHAR_MECA_FR1D1D”,
“CHAR_MECA_FF1D1D”,
The loadings are given under key word FORCE_POUTRE, that is to say by actual values in
AFFE_CHAR_MECA (option CHAR_MECA_FR1D1D), is 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
developed
not
homothétiquement
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:
N
N
F
= L
1
2
X
+
,
1
3
6
N
N
F
= L
1
2
X
+
.
2
6
3
N and N
1
2 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.
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Titrate:
“Exact” elements of beams (right and curved)
Date:
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Author (S):
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Key:
R3.08.01-A
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T, T
, T
and T
y
y
Z
Z are those of the sharp effort. And one a:
1
2
1
2
7 T
3 T
y
y
F
= L
1
2
y
+
,
1
20
20
T
T
y
y
M
= L2
1
2
Z
+
,
1
20 30
3 T
7 T
y
y
F
= L
1
2
y
+
,
2
20
20
T
T
y
y
M
= - L2
1
2
Z
+
,
2
30 20
7 T
3 T
Z
Z
F
= L
1
2
Z
+
,
1
20
20
T
T
Z
Z
M
= - L2
1
2
y
+
,
1
20 30
3 T
7 T
Z
Z
F
1
2
Z
=
L
+
,
2
20
20
T
T
Z
Z
M
= L2
1
+ 2.
y2
30 20
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:
N
N
F
= L
, F
= L
X
X
,
1
2
2
2
T
T
F
= L y
, F
= L y
y
y
,
1
2
2
2
T
T
M
= L
y
2
, M
= - L
y
2
Z
Z
,
1
2
12
12
T
T
F
= L Z
, F
= L Z
Z
Z
,
1
2
2
2
T
T
M
= - L2 Z
, M
= L2 Z
y
y
.
1
2
12
12
Handbook of Référence
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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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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 - Tréférence:
U
= -
L
1
(T - Tréférence)
U
=
L
2
(T - Tréférence)
(: thermal dilation coefficient).
Then, the induced forces are calculated:
F =
K U
as 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 grid if one uses
option “CHAR_MECA_FRELEC”.
The secondary conductor is not necessarily right and it can be based on part of the grid
Aster if one uses option “CHAR_MECA_FRLAPL”.
Linear density of the force of Laplace exerted in a point M of the principal conductor by
secondary conductor is written:
1
E R
F (M) =
E
2
ds
2 1
R 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
operator “DEFI_FONC_ELEC”.
Handbook of Référence
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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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6.5.1 Secondary conductor finished or infinite right
OPTION: “CHAR_MECA_FRELEC”
For a finished secondary conductor, we have:
'
'
'
E
N
F (M) =
1 (sin - sin
1
2
)
2
D
P
P
1
2
D
E 2
E 2D
with
N =
D
1
2
D = D
E 1
&&
M
1
For an infinite secondary conductor, we have:
N
F (M) = e1 D
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 principal 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 principal element. One can thus use the technique
presented to [§6.3] [§6.4] to calculate the loading with the nodes.
L
F
= F
1
(M) 2
L
F
= F
2
(M) 2
2
L
M
=
E
F
1
1
(M) 12
2
L
M
= - E F (M)
, F
2
1
(M) = constant M
.
12
For the case of the parallel conductors right infinite multiple, one must give it directly
vector “forces of normalized Laplace”. This one 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 principal element. Thus of same
manner that previously, one calculates the loading with the nodes.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
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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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In the case of a right conductor in unspecified position finished or infinite, its position is given by
two points P1 and P2 such as the current circulates of P1 towards P2. The loading is not
obligatorily constant along the principal 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 between P1 and P2).
One integrates as follows:
3
2
L
X
X
F
=
F
1
M
() 2
3
+1 dx
O
L -
L
3
2
L
X
X
F
=
F
2
M
() -
2
+ 3
dx
O
L
L
'
3
2
L
'
'
X
X
X
M
=
- E
F (M) L
-
+ 2
dx
1
O
1
L
L - L
'
3
2
L
'
'
X
X
M
=
- E
F (M) L
-
+
dx
2
O
1
L
L
Note:
Like F (M) X = 0
and
(e1 F (M)
) X = 0 (X = E
1)
one uses only the functions of form associated with the problem with inflection.
6.5.2 Secondary conductor describes by part of grid 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 principal conductor.
Except the fact that it is necessary to summon the interaction of the various elements of the secondary conductor on
principal conductor, the method is the same one as previously (case (III)).
Note:
For numerical integration, one uses only 5 points ranging between P1 and P2.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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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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7
Torque of the efforts - Torseur of the constraints (or efforts
generalized) - Forces nodal and reactions
OPTIONS:
“EFGE_ELNO_DEPL”,
“SIEF_ELGA_DEPL”,
“SIGM_ELNO_DEPL”,
“SIPO_ELNO_DEPL”,
“FORC_NODA”,
“REAC_NODA”.
Option “EFGE_ELNO_DEPL” makes it possible to calculate the torque of the efforts to the 2 nodes of each element
of “beam”.
Options “SIGM_ELNO_DEPL” and “SIPO_ELNO_DEPL” make it possible to calculate the maximum values
components of the tensor of the constraints intervening in a model of “beam”.
Option “SIEF_ELGA_DEPL” allows the calculation of the nodal forces (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 grid 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
= Ke U
+ Me U %
- F E
LOC
LOC
LOC
LOC
LOC
LOC
where: R
= - NR
,
1 - V1, - V1, - M1, - M1, - M1, NR
2, V
2, V
2, M
2, M
2, M
2
LOC
(
Y
Z
T
Y
Z
Y
Z
T
Y
Z)
Ke
elementary matrix of rigidity of the “exact” element of beam,
LOC
Me
elementary matrix of mass of the element beam,
LOC
F E
vector of the efforts “distributed” on the element beam,
LOC
U
vector “degree of freedom” limited to the element beam,
LOC
%
U
vector “acceleration” limited to the element beam.
LOC
One changes then the signs of the efforts to node 1 [§2.1.3].
OPTION: “SIEF_ELGA_DEPL”
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
Ke
=
U
LOC
LOC
LOC
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
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Key:
R3.08.01-A
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7.2
The tensor of the constraints
OPTIONS:
“SIGM_ELNO_DEPL”,
“SIPO_ELNO_DEPL”.
One seeks to calculate the maximum values of components xx, xy
and xz which is connected to
efforts by:
NR = dS
xx,
S
V = dS
y
xy,
S
V = dS
Z
xz,
S
MT = (y - Z
xz
xy,
S
) dS
M = Z dS
y
,
S xx
M = - y D
S
Z
.
S
xx
With option “SIGM_ELNO_DEPL”, one calculates the maximum effect of the whole of the efforts on
xx, xy and xz.
· For
xy, one a:
I
I
V
M
I Max
Y
T
I
xz
=
+
R, for
node I I
(=,
1 2)
I
I
T
Y
With
J X
with: With K there A air
: E of the section,
ky: constant of shearing,
J X: constant of torsion,
RT: radius of torsion.
· For
xz, one a:
I
I
V
M
I Max
Z
T
I
xz
=
+
R,
I
I
T
Z
With
J X
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
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Titrate:
“Exact” elements of beams (right and curved)
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Key:
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· For
xx, one a:
-
For a rectangular section:
I
I
I
I
I
NR
M
H
M
H
I Max
Y
Z
Z
Z
xz
= + +
+
if NR
0
,
I
I
I
With
I
Y
2
IZ
2
I
I
I
I
I
NR
M
Y HZ
MZ HZ
-
+
+
if
0
NR <
,
I
I
I
With
I
Y
2
IZ
2
with IY, IZ: geometrical moments,
Y
H
HZ
,
: sides of the rectangle.
-
For a circular section:
I
I
NR
2
2 R
I Max
I
I
xx
= +
+
M
+ M
if NR
0
,
I
(Y) (Z) I
With
I
Y
I
I
NR
2
2 R
I
I
-
+
M
+ M
if
0
NR <
,
I
(Y) (Z) I
With
I
Y
with R: radius of the section.
-
For a general section:
I
I
I
NR
M
M
I Max
Y
I
Z
I
xx
=
+
R
-
R.
I
I
Z
I
Y
With
IY
IZ
In this last case, one makes calculation at the point (R I, R I
I
I
I
Y
Z). R
R
Y
Y
and
and
RZ are them
distances to neutral fiber.
To find these formulas, one uses the relations between the constraints and the deformations then those
between the internal efforts and (U, X; xy; xz; X
X; y X;
,
,
Z
, X):
xx = E xx
xy =
G 2 xy
xz =
G 2 xz,
xx = U, x+ Z y, X
- y Z, X
2
xy = xy - Z X, X (xy =V, X - Z)
2
xy = xz + there X, X
(xz =W, x+y),
Handbook of Référence
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
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Key:
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NR
EA
U, X
V
K GA
Y
Y
0
xy
V
K GA
Z
Z
xz
=
M
GJ
T
X, X
M
0
I.E.(internal excitation)
Y
Y
y, X
M
I.E.(internal excitation)
Z
Z Z, X
(The various types of movement are uncoupled when one works with the principal axes (by
definition)).
One obtains:
NR
M
M
Y
Z
xx (X, y, Z) =
E
+ Z
- y
,
EA
I.E.(internal excitation)
Y
EIZ
V
M
Y
T
xy (X, y, Z)
= G
- Z
,
K
Y GA
G J
X
V
M
Y
T
xy (X, y, Z) =
G
- Z
,
K
Y GA
G J
X
With 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 components xx, xy and xz.
One finds the preceding results in broken up form. The vector result is written:
(1 1 1 1 1 1
1, 2
3
, 4
, 5
, 6
2
2
2
2
2
2
1, 2
3
, 4
5
, 6
)
I
NR
with: I
1
=
for node I I =
I
(1, 2)
With
I
NR
i1 =
I
With
I
V
I
Z
3
=
,
I
Z
With
I
M
I
T
I
4
=
R,
I
T
J X
for the rectangular sections:
I
I
M
H
I
Y
Z
5
=
,
I
I
2
Y
I
I
M
H
I
Z
Y
6
= -
,
I
I
2
Z
Handbook of Référence
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Titrate:
“Exact” elements of beams (right and curved)
Date:
02/12/96
Author (S):
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Key:
R3.08.01-A
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I
M
I
Y
I
5
=
R,
I
IY
I
M
I
Z
I
6
= -
R,
I
IY
for the unspecified sections at the point (I
I
Y
R, Z
R):
I
M
I
Y
I
5
=
R,
I
Z
IY
I
M
I
Z
I
6
= -
R.
I
Y
IZ
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 constraints “isigm”.
In particular, one can take as a starting point the rules RCC_M.
I traditional
I traditional =
(cflex)
1 only in inflection.
cflex
MY
isigm
=
R
,
5
I
Z
Y corr E
ig
cflex
MZ
isigm
=
R
6
Y
(isigm)
1.
Corrected IZ
cflex
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 command 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 = QT
where QT represents the matrix symbolically associated with the operator divergence. For an element, one
writing agricultural work of virtual deformations:
(QT) H = (U)
(wh)
For the elements of beam, one calculates simply the nodal forces by assembly of the forces
nodal elementary calculated by option SIEF_ELGA_DEPL, which is expressed by:
[F] = [K] [U
[
]
§7.1
LOC
LOC
LOC]
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
71/72
OPTION: “REAC_NODA”
This option, called by CALC_NO, makes it possible to obtain the reactions R with the supports starting from the forces
nodal F by:
R = F - Fchar + Finer
Fchar
Finer
and
being nodal forces associated the loadings given (specific and distributed) and
with the efforts of inertia.
8
Element of bar
Key word “BARRE”
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 Université
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. Calcul scientific. 1990.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
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:
72/72
Intentionally white left page.
Handbook of Référence
R3.08 booklet: Machine elements with average fiber
HI-75/96/060/A
Outline document