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Titrate:
A finite element of cable-pulley
Date:
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Author (S):
Mr. AUFAURE
Key:
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Page:
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Organization (S): EDF/IMA/MN
Handbook of reference
R3.08 booklet: Machine elements with average fiber
Document: R3.08.05
A finite element of cable-pulley
Summary:
The pulleys play an important part in the structures of cables such as the air powerlines. It
is thus useful to know to model them in one at the same time realistic and powerful way. The formulation is presented
finite element of a length of cable passing by a pulley: expressions of the internal forces and the matrix of
rigidity. The pulley can fixed or be supported by a flexible structure. Its position on the cable with balance is not
not known a priori: this position is that for which on both sides the tension of the cable is the same one.
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Titrate:
A finite element of cable-pulley
Date:
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Author (S):
Mr. AUFAURE
Key:
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Contents
1 Notations ................................................................................................................................................ 3
2 Introduction ............................................................................................................................................ 4
3 Assumptions and definition of a finite element of cable-pulley .................................................................... 5
4 Forces intern of a finite element of cable-pulley .................................................................................. 7
5 Matrix of rigidity ................................................................................................................................... 8
6 Matrix of mass ................................................................................................................................... 9
7 Introduction of the element of cable-pulley into Code_Aster .............................................................. 9
8 Example of application .......................................................................................................................... 10
9 Conclusion ........................................................................................................................................... 11
10 Bibliography ...................................................................................................................................... 11
Appendix 1 Calcul of the matrix of rigidity ................................................................................................ 12
Appendix 2 Figure of balance of an inextensible cable weighing subjected to a tension given in extrémité13
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A finite element of cable-pulley
Date:
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1 Notations
With
surface of the cross-section of the cable.
E
Young modulus.
F
vector of the forces intern element.
F
force intern with node N.
N
H
horizontal component of the tension [§An1].
I
stamp unit of command 3.
3
K
stamp rigidity of the element.
L
current length of the element.
L
length at rest.
0
L
initial length.
I
L
1
NR NR
3
1
l1
euclidian norm of l1
L
2
NR NR
3
2
l2
euclidian norm of l2.
NR
current tension of the cable constituting the element.
NR
initial tension (prevoltage).
I
S
arrow of a range of cable [§An1].
S
length of a range [§An1].
T
current temperature.
T
initial temperature.
I
U
vector-displacement of the nodes compared to the initial position.
U
displacement of node N compared to its initial position.
N
W
weight per unit of length.
X
vector-position of node N in initial configuration.
N
thermal dilation coefficient.
current deformation compared to the initial configuration.
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A finite element of cable-pulley
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2 Introduction
One uses pulleys, during the construction of the air powerlines, for the operation of pose
cables. The cable in the course of pose [Figure 2-a] is fixed at the one of the supports of stop of the canton, it
rest on pulleys placed at the bottom of the insulators of the supports of alignment and it is retained by one
force on the level of the second support of stop. While exploiting this force - or by moving its point
of application - one adjusts the arrow of the one of the ranges, that which is subjected to constraints
of environment. Then one removes the pulleys and one fixes the cable at the insulators. The length of the cable
in the various ranges is then fixed and it determines the later behavior of the line under
statical stresses (wind, overload of white frost) and in dynamic mode (movement due to
forces of Laplace created by the currents of short-circuit).
Appear 2-a: Pose of a cable in a canton in two ranges
The finite element of cable-pulley presented here makes it possible to model the operation of pose and thus to calculate,
in a natural way, the length of cable in the various ranges.
The idea of this finite element came to us some time after the conversation [bib1] and we have
presented its formulation in [bib2].
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A finite element of cable-pulley
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3
Assumptions and definition of a finite element of cable-pulley
N3
N1
N2
Appear 3-a: Tronçon of cable passing by a pulley - reality
Let us take a length of cable NR NR
1
2 passer by by the pulley N3 [Figure 3-a]. This pulley is not
inevitably fixes and can, for example and as it is the case in the example of [§8], being gone up to
the end of a cable.
N3
N1
N2
Appear 3-b: Tronçon of cable passing by a pulley - modelled
The pulley is supposed to be specific [Figure 3-b]. One makes moreover following assumptions:
· The position of balance of N3 is not known, but it is necessarily on
section N1 N2 deformed starting from its initial position.
In modeling of the lines, the horizontal movements are of low amplitude and this
assumption is generally not restrictive;
· The two strands N3 N1 and N3 N2 are always rectilinear, like elements of cable
1st command.
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It rises from this assumption that:
L
= X + U - X - U
1
1
1
3
3
éq 3-1
L
= X + U - X - U
2
2
2
3
3
L
T
1
=
1
L 1l
éq 3-2
L
T
2
=
l2 l2
The overall length of the two strands is:
-
in the current position, where the temperature is T:
L = l1 + l2;
-
in the initial position, indicated by index I, where the tension is Ni and
temperature Ti:
L
= (l1) + (L
I
;
I
2) I
-
in a not forced position, indicated by index 0, where the temperature is T0:
L
= (L) + (L
0
1
.
0
2) 0
· The pulley is without friction and thus the tension is the same one in the two strands.
It rises from this assumption that the deformation is also the same one and one takes for
value of this one the measurement of the relative lengthening of the section compared to the initial state:
L - L
=
I
éq 3-3
l0
must remain small, so that section A is regarded as constant.
It will be noted that, within the framework of the finite element method, the linear loads
do not prevent the tension from being constant of N1 with N2. These forces are indeed
concentrated with the N1 nodes and N2 and on the axis of the N3 pulley.
· The relation of behavior is elastic:
NR = E [
WITH - (T - T)]+ NR
I
I
éq 3-4
One calls finite element of cable-pulley, a length of cable N1 N2 satisfactory N3
with the preceding assumptions.
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4
Forces intern of a finite element of cable-pulley
Let us point out the following definition: one calls internal forces of a finite element of structure the forces
that it is necessary to exert in its nodes to maintain it in its current deformed configuration.
In the case of a finite element of cable-pulley, the internal forces result immediately from
statics. One has indeed [Figure 4-a]:
NR
F
=
L
1
1
éq 4-1
1
L
NR
F
=
L
2
2
2
L
and, to ensure balance:
F
= - (F + F
3
1
2).
F1 and F2 having even module, F3 is directed according to the bisectrix of the angle (NR NR NR
1
3
2).
F3 = - (F1 + F2)
N3
N1
N2
F1
F1 = F2
F2
Appear 4-a: Forces internal of an element of cable-pulley
F3 is applied to the axis of the pulley.
The system F of the internal forces of the element is thus:
F
1
F = F2.
F
3
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5
Stamp rigidity
The matrix of rigidity K of the element is derived from Fréchet of F in the direction of displacement
U of the nodes:
F = K U.
K is calculated by the following traditional formula, used intensively in [bib3], [bib4] and [bib5]:
D
F = lim
(Fu + U).
éq 5-1
0 D
The detail of calculations is given in [§An1] and the final expression of K is as follows:
NR
NR
K11 +
I
K
3
12
- K11 - K12 -
I
3
1
L
1
L
NR
NR
K =
KT
K
T
12
22 +
I3
- K22 - K12 -
I3
2
L
2
L
NR
NR
1
1
T
T
- K11 - K12 -
I3 - K22 - K12 -
I
K
3
11 + K22 + K12 + K12 +
+ NI3
1
L
2
L
1l
2
L
éq 5-2
NR is given by [éq 3-4] and [éq 3-3];
EA NR 1
K
=
-
L lT
11
;
2 1 1
0l
1
L 1l
EA
K
=
L lT
12
1 2;
0
L 1l 2l
EA NR 1
K
=
-
L lT
22
.
2 2 2
0l
2
L 2l
K is symmetrical, because of the symmetry of K11 and K22 and of total symmetry per blocks.
But K depends on displacements on NR, NR and NR
1
2
3 via L, L
1
2 and NR: the element
finished cable-pulley is thus a nonlinear element.
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6
Stamp of mass
This matrix intervenes obviously only in the dynamic problems. The element of cable-pulley
is not used in Code_Aster that for the quasi-static problems of pose of cables [§7].
Note:
One presents nevertheless in [bib2] the example of a dynamic problem comprising a cable
pulley.
The matrix of mass of the element N1 N2 N3 is obtained by assembling the matrices of mass
“coherent” of the elements of cable with two nodes N3 N1 and N3 N2 [bib6] and while adding
specific mass of the pulley.
It should be noted that, during a dynamic analysis, this matrix of mass must be updated
because the lengths l1 and l2 vary.
7 Introduction of the element of cable-pulley into
Code_Aster
The element of cable-pulley is supported by a mesh SEG3.
In command AFFE_MODELE, under the key word factor AFFE, one must define them as follows
arguments of the key words:
key word
GROUP_MA
PHENOMENE
MODELISATION
argument
group meshs
“MECANIQUE”
“CABLE-POULIE”
of cable-pulley
The constitutive material must be elastic.
In command AFFE_CARA_ELEM, the cable-pulleys are treated like cables.
As the element of cable-pulley is nonlinear [§5] and that, for the moment, it is used only in statics
[§6], it is accessible only by operator STAT_NON_LINE. Under the key word factor COMP_ELAS, them
arguments of the key words are as follows:
key word
GROUP_MA
RELATION
DEFORMATION
argument
group meshs
“ELAS”
“GREEN”
of cable-pulley
Lastly, the force of gravity acting on the nodes NR and NR
1
2d' an element of cable-pulley [3-b]
is following because it depends on the length of the strands NR NR and NR NR
3
1
3
2. For a structure
comprising at least an element of cable-pulley, one must specify it in STAT_NON_LINE under
key word factor EXCIT:
EXCIT:
(CHARGE: charge of gravity TYPE_CHARGE: “SUIV”)
One finds an example of application in test SSNL100 A [V6.02.100].
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8 Example
of application
This example is that of the pose of a cable with two ranges and imposed tension of adjustment.
On [Figure 7-a], O is the anchoring of the cable on the support of stop of left. P1 is a first pulley
placed at the foot of the lifting chain PC
1
, fixed out of C at the support of alignment. P2 is one
second pulley placed on the support of stop of straight line. O, P1 and P2 are, to simplify, located on
horizontal.
Z
C
X
Q
P
NR
R
1
R
Q
p
1
2 2
2
O
'P
Q 1
'
'
R'
1
Q
R
2
2
1
Appear 7-a: Equilibre of a cable in two ranges, subjected to a tension of adjustment given.
OP1 = P1 P2 = 100 m; W = 30 NR/m; E × A = 5 X 107 NR; NR p = 5.000 NR
In initial position, the cable at rest, supposed in weightlessness, is right: line in feature of axis of
[Figure 7-a]. For modeling in finite elements, this line is cut out in:
· ten elements of cable with two nodes between O and Q1;
· an element of cable-pulley Q1 R1 P1;
· nine elements of cable between R1 and Q2;
· an element of cable-pulley Q2 R2 P2.
One simultaneously subjects the cable to gravity and the tension of adjustment NR p exerted in R2.
position of balance (line in feature full with [Figure 7-a]) is reached in 11 iterations by the operator
STAT_NON_LINE of Code_Aster. The arrows are 7.955 m and 7.867 m, respectively in
range of left and that of straight line.
For an inextensible cable of a range of 100 m, weight linear 30 NR/m and subjected to a tension
in end of 5.000 NR, the theoretical arrow is 7.941 m [§An2].
While exploiting the tension of adjustment, one can adjust one of the two arrows to a value fixed at
advance.
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9 Conclusion
The finite element of cable-pulley presented in this note, of a very simple mechanical formulation, has
performances comparable with those of an element of ordinary cable. It is very convenient and even
essential for a realistic modeling of the air powerlines. It should find
other applications, in particular in Robotique.
10 Bibliography
[1]
J.L. LILIEN: Private communication.
[2]
Mr. AUFAURE: With finite element off cable passing through has pulley. Computers & Structures 46,
807-812 (1993).
[3]
J.C. SIMO, L. VU-QUOC: With three-dimensional finite-strain rod model. Leaves II: computational
aspects. Comput. Meth. appl. Mech. Engng 58, 79-116 (1986).
[4]
A. CARDONA, Mr. GERADIN: With beam finite element nonlinear theory with finite rotations. Int.
J. Numer. Meth. Engng. 26, 2403-2438 (1988).
[5]
A. CARDONA: Year integrated approach to mechanism analysis. Thesis, University off Liege,
Belgium (1989).
[6]
Mr. AUFAURE: Modeling of the cables in Code_Aster, in the course of drafting.
[7]
H. MAX IRVINE: Cable Structures. MIT Close (1981).
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Appendix 1 Calcul of the matrix of rigidity
One shows here how applies the formula [éq 5-1] to the calculation of the first three lines of the matrix
K [éq 5-2], those which relate to the F1 force. The other lines are obtained is by permutation
indices, is by summation of two preceding lines.
The first lines are thus obtained by calculating the derivative:
D
lim
1
F (U + U)
éq An1-1
0 D
and by putting in factor the vector U.
According to the relations [éq 4-1], [éq 3-4] and [éq 3-3], one a:
L U + U + L U + U
- L
I
L U + U
F
1
2
1
1 (U + U
)
(
) (
)
= EA
- (T - Ti)
(
)
+ Ni
0
L
1
L (U + U)
with, according to [éq 3-1]:
L (U + U
) = X + U + U
- X - U - U
1
1
1
1
3
3
3
and, according to [éq 3-2]:
L (U + U
)
T
1
=
1
L (U + U
) 1l (U + U
).
Consequently:
D
lim
1
L (U + U) = 1
U - U
3
0 D
D
1
lim
T
1
L (U + U) =
1
L U 1
U - U
3
0 D
1
L (U)
() (
)
and by permutation of indices 1 and 2:
D
1
lim
T
2
L (U + U) =
l2 U u2 - U
3
0 D
2
L (U)
() (
).
Finally:
D
1
1
lim
= -
T
L U
U - U
0
3
1
1
3
D
1
L (U +
U)
() (
)
1
L (U)
.
While carrying the preceding expressions in [éq An1-1] and by putting in factor the vector U, one
easily obtains the first three lines of K.
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Appendix
2 Figure of balance of an inextensible cable weighing
subjected to a tension given in end
Let us take a cable [Figure An2-a] whose end is fixed at the point O and of which the other end P
is subjected to the tension NR p. O and P is on same horizontal the and distant ones of S. the weight
linear is W. The arrow S. is sought.
Z
O (0, 0)
X
P (S, 0)
NR p
S
Appear An2-a: heavy Câble in balance
One finds in [bib7], p 6, the following well-known formulas:
· appear of balance of the cable:
H
W S
W S
Z (X) =
cosh
- X
cosh
;
éq An2-1
W
H
2
-
2 H
· tension
:
W S
NR (X) = H cosh
- X.
éq An2-2
H
2
H is the horizontal, constant tension along the cable since the external force distributed - the weight -
is vertical.
H is calculated by [éq An2-2] written out of P:
W S
2 NR p W S
cosh
-
=
.
0
2 H
W S 2 H
W S is thus root of the transcendent equation:
2 H
2 NR
cosh X
p
=
X.
W S
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This equation has two roots [Figure An2-b] if:
2 NR p > 0p,
W S
with:
p
= sinh X
0
0
and:
X
= cotanh X
X
0
0
0 > 0.
2 Np
W S
1
cosh X
W S
2 H
X
W S
Appear An2-b: Calcul of 2 A.m.
The smallest root, which corresponds to the greatest tension of the cable, is only useful. The other root
corresponds to a sag of the considerable cable, of about size of its range.
W S being calculated, the arrow results from [éq An2-1]:
2 H
S 2 H
W S
S =
cosh
-
1.
2 W S
2 H
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