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Optimization of Cable Tensioning in Cable-Stayed Bridges
Le Hong Lam1Nguyen Huy Cung2Vu Hong Nghiep2
1 Hohai UniversityNanjing210098
2 Ho Chi Minh City University of TransportVietnam.
[email protected]@yahoo.com
AbstractDuring the structural analysis of cable-stayed bridges, some specific problems arise that are notcommon in other types of bridges. One of these problems is the derivation of an optimal sequence forthe tensioning of the stay cables. This paper describes a novel solution to this problem. Finite elementsare used for modeling the bridge structure. Numerical examples are presented, which illustrate the highobtainable accuracy of the method.
Keywordscable-stayed bridgesoptimizationfinite elementcatenary cable element
1. Introduction
In the design cable-stayed bridge, an important step is determining the tensioning forces of stay
cables to achieve a desired geometry of the bridge after construction, especially under the reaction of
dead load. Many structural analysis techniques were proposed to solve this problem. The different
models of cable have been investigated.
The elastic cable is assumed to be perfectly flexible and possesses only tension stiffness; it is
incapable of resisting compression, shear and bending forces. When the weight of the cable is neglected,
the cable element can be considered as a straight member. But under action of its own dead load and
axial tensile force, a cable supported at its end will sag into a catenary shape. The axial stiffness of a
cable will change with changing sag. When a straight cable element for a whole inclined cable stay is
used in the analysis, the sag effect has to be taken into account. On the consideration of the sag
nonlinearity in the inclined cable stays, it is convenient to use an equivalent straight cable element with
an equivalent modulus of elasticity, which can well describe the catenary action of the cable. The
concept of a cable equivalent modulus of elasticity was first introduced by Ernst [1],[2],[5],[6]. This
cable element has been used popularly in cable stayed bridge design.
More accurate than cable element with an equivalent modulus of elasticity, some cable elements
have been proposed such as nonlinear geometry cable element [4] proposed by the last of authors,
nonlinear cable element [13] proposed by R.Karoumi, elastic catenary cable element with an unknown
initial length [9].
In this paper, the other model of cable is used for modeling cable in cable stayed bridge. When
adjusting the cable, the length of cable changes, lead to the changing of sag. The horizontal component
of cable tension is determined by using the built equations of the length of cable. From then on,determining the tensioning forces of stay cables under dead load to achieve a desired geometry of the
bridge after construction.
2. Proposed cable
Consider the inclined cable AB of span l subjected to its own weight, having projection equal to w
(Figure 1). Now, we build the equations of the length of cable. For the below expressions it is assumed
that the cable is perfectly flexible and Hookes law is applicable to the cable material.
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Figure 1: Geometry of the inclined cable
The length of cable [1], [2]:
3
0 2
0
cos
cos 2
l DL
H
= + (1)
Where
2 32
0
012
lw l
D Q dz= = , Q is the simple beam shear force for given span l and H0 the
horizontal component of cable tension.
Let 0 be the elastic strain, we have [1]:
( )20 000 0
1cos .
l lH Hds tg ds
EF EF
= = +
Where
E : Youngs modulus of cable.
F : Cross section area.
: The inclination of the cable to the horizontal at any point.
On the other hand,
0
Qtg tg
H = +
Hence
( )
( )
2
20 00
00 0
220
2
0 00
2 20 0 0
2
0 00 0 0
1 1
1 2
1 1. 1 . .2
= + = + +
= + + +
= + + +
l l
l
l l l
H H Qtg ds tg ds
EF EF H
H Q Qtg tg ds
EF H H
H H Htg ds tg Qds Q ds
EF EF H EF H
(2)
Because of
0
0l
Qds= , (2) can be express:
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0 00 2
0
.
cos
H l D
EF EFH
= + (3)
If we skip the elastic strain of cable, the length of cable is:
3' 0 0
0 0 0 2 20 0
.cos
cos 2 cos
H l Dl DL L
H EF EFH
= = + (4)
When adjusting the cable, the position A changes to A due to the displacement of pylon Figure 2.
Also, the original sag f0increase to f, the horizontal component of cable tension H0change to H. The
expression of the length of cable can be written:
'
1 1 1L L = (5)
Where
L1: Length of cable.
1 : The elastic strain.
Figure 2: The inclined cable in working state.
From the Fig.2-2, we see:
3 31 1
1 2 2
1 0 0
cos cos'
cos 2 2
D Dl uL A B
H H
= + = + (6)
( ) ( )2 2 2 2
1
2
2 2
' 2 2 2 2
1 2 1
sin coscos
= + = + =
+ + = =
=
A B d v l u d l vd ul AB vd ul
vd ul vd ul vd ulAB AB AB
AB AB AB AB
lv u
(7)
And
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1
1 2 2
22
cos 1 1 1 1'
cos 1 cos 1
cos
cos . 1 sin .cos cos .cos
cos . 1 sin .co
+ +
= = +
+ + + = +
= + +
= +
l u u vd ul l u vd ull AB
A B l AB AB l AB
vd ul u vd ul u
AB l ll
v u u
l l l
v
l
( )2
2
s cos 1
cos . 1 sin .cos sin
+
= +
u
l
v u
l l
(8)
By substituting (7) and (9) into (6), we have:
( )3 3
1 2 2
cos . 3 cos .sin cos sin cos sin
cos 2 2= + +
l D DL v u v u
H H l
(10)
The elastic strain:
( ) 20 12
1
2
2
2
2
2
2
.1 cos
cos
1 cos . 1 sin .cos sin
2 21 1 sin .cos sin
cos
2 21 sin .cos sin
cos
= + = +
= + +
+ +
+
H l u D H u Dl
EF EFH EF l EFH
H u v u Dl
EF l l l EFH
Hl u v u D
EF l l l EFH
Hl v u u
EF l l l
( )2
2
1 sin 2 1 cos 2cos
1 sin 2 cos 2cos
+
= + +
= +
D
EFH
Hl v u u D
EF l l l EFH
Hl v u D
EF l l EFH
(11)
By substituting (10) and (11) into (5), we have:
( )3 3
'
1 2 2
2
cos . 3 cos .sin cos sin cos sin
cos 2 2
1 sin 2 cos 2cos
= + +
l D DL v u v u
H H l
Hl v u D
EF l l EFH
(12)
Suppose that the cable doesnt elongate, then
' 1
1 0L L= (13)
By substituting (4) and (12) into (13), then transforming and reducing, we have:
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( )5 2
3 2 3 20 002
0 0
22 5
. . .cos .cos .1 sin 2 cos 2 .sin .cos .cos
2. . .
.cos .
1 3 sin .cos 3 sin cos 02
+ + + +
+ + =
E F D Dv u E F H H v u H
l l l H l H l
D E F v u
H Dl l l l
(14)
By solving eq.(14) we can determine the horizontal component of cable tension H.
3. Optimization cable tension under dead load
The purpose is determined the tension of cable that satisfy two following conditions:
i) The displacement of beam is zero.
ii) The displacement of pylon is minimum.
To analyze the cable stayed bridge, the pylon and girder are modeled using beam column
element. The stay cables are replaced by nodal forces as shown in Figure 3.
Figure 3: The stay cables are replaced by nodal forces.
The nodal forces of stay cable are determined by the following formulas:
1 3= =P P H (15)
2
2 tan2 8
= +wl wl
Pf
(16)
2
4 tan2 8
= wl wl
Pf
(17)
P1
P3
P4
P2
Y
Figure 4: The nodal force of cable.
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The general form of the equilibrium equations are:
[ ]{ } { }K u F= (18)
Where
[K]: The stiffness matrix of system.
{u}: The nodal displacement vector.
{F}: The force vector.
`
Figure 5: Procedure to optimize the tension of cables force:
Begin
Input the properties of cable stayed bridge
Calculate [K], {F} with the
beginning sag f0
Solve the eq (18) to calculatevector {u}
Solve the eq (14) to calculate H, then
calculate vector {F} from (15),(16),(17)
Solve the eq (18) to calculatevector {u}
Check the condition:
Max {u} [u]
NO
YES
Calculate the initial cables force
Export results.END
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4. Numerical example
In order to estimating the accuracy of this method, let us analyze the same problem which
introduced in [7] and [9].
4.1 Problem
Consider a cable stayed bridge which the detailed geometry and the finite element model are
shown in Figure 6. The material and sectional properties of stay cables, deck and pylons are shown in
Table 1.
Figure 6: Geometry of the cable-stayed bridge studied in example.
Table 1. The material and sectional properties of stay cables, deck and pylons
Member E(GPa) A(m2) I(m4) w(KN/m)
Exterior cable 207 0.042 - 3.2
Interior cable 207 0.016 - 1.2
Girder 207 0.32 1.131 87.5
Pylon (0.0-20.3m) 207 0.269 0.432 -
Pylon (20.3-40.6m) 207 0.228 0.345 -
Pylon (40.6-61.0m) 207 0.203 0.211 -
The display of cable stayed bridge model using Matlab is shown in Figure 7.
Figure 7: Calculating modeling of the system.
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4.2 Numerical results
4.2.1. Displacement of nodes
Node Ux Uy Rz
1 0.033200 0.000000 -0.000877
2 0.025973 -0.000371 0.0002383 0.014000 0.000225 -0.000060
4 0.000000 0.000000 0.000025
5 -0.014000 0.000180 -0.000042
6 -0.025867 -0.000322 0.000123
7 -0.033842 0.000093 -0.000457
8 -0.033842 0.000093 0.000457
9 -0.041818 -0.000322 -0.000123
10 -0.053685 0.000180 0.000042
11 -0.067685 0.000000 -0.000025
12 -0.081685 0.000225 0.000060
13 -0.093658 -0.000371 -0.000238
14 -0.100885 0.000000 0.000877
15 0.000000 0.000000 0.000000
16 0.000000 -0.022732 0.000000
17 0.000000 -0.034466 0.000000
18 0.000000 -0.047544 0.000000
19 0.000000 0.000000 0.000000
20 0.000000 -0.022732 0.000000
21 0.000000 -0.034466 0.000000
22 0.000000 -0.047544 0.000000
4.2.2. Initial force components of elements
Element Pi Pj Mi Mij Mj
1 -1046.4336 -1046.4336 0.0000 570.2548 -
1911.8732
2 -1733.5221 -1733.5221 -1911.8732 -152.1121 -
1444.7338
3 -2026.9961 -2026.9961 -1444.7338 43.1139 -
1521.42124 -2026.9961 -2026.9961 -1521.4212 -34.2901 -
1599.5417
5 -1718.1429 -1718.1429 -1599.5417 84.7407 -
1283.3597
6 -1154.8083 -1154.8083 -1283.3597 -296.8176 -
2362.6583
7 -0.0000 -0.0000 -2362.6583 350.5709 -
2362.6583
8 -1154.8083 -1154.8083 -2362.6583 -296.8176 -
1283.3597
9 -1718.1429 -1718.1429 -1283.3597 84.7407 -
1599.5417
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10 -2026.9961 -2026.9961 -1599.5417 -34.2901 -
1521.4212
11 -2026.9961 -2026.9961 -1521.4212 43.1139 -
1444.7338
12 -1733.5221 -1733.5221 -1444.7338 -152.1121 -
1911.873213 -1046.4336 -1046.4336 -1911.8732 570.2548
0.0000
14 -2842.0370 -2746.3793 0.0000 0.0000
0.0000
15 -2746.3793 -2710.0464 0.0000 0.0000
0.0000
16 -2710.0464 -2677.5379 0.0000 0.0000
0.0000
17 -2842.0370 -2746.3793 0.0000 0.0000
0.000018 -2746.3793 -2710.0464 0.0000 0.0000
0.0000
19 -2710.0464 -2677.5379 0.0000 0.0000
0.0000
20 1136.1296 1154.5049 0.0000 0.0000
0.0000
21 822.6040 828.9788 0.0000 0.0000
0.0000
22 486.8279 491.4248 0.0000 0.0000
0.0000
23 512.4597 517.0566 0.0000 0.0000
0.0000
24 673.8735 680.2481 0.0000 0.0000
0.0000
25 1254.7082 1273.0840 0.0000 0.0000
0.0000
26 1254.7082 1273.0840 0.0000 0.0000
0.0000
27 673.8735 680.2481 0.0000 0.0000
0.0000
28 512.4597 517.0566 0.0000 0.00000.0000
29 486.8279 491.4248 0.0000 0.0000
0.0000
30 822.6040 828.9788 0.0000 0.0000
0.0000
31 1136.1296 1154.5049 0.0000 0.0000
0.0000
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4.2.3. Diagrams
Figure 8: Displacement diagram.
Figure 9: Axial force diagram.
Figure10: Moment diagram.
4.2.4. Remarks
Displacement of deck is very small. The maximum displacement is 0.000371m (at node2 and node 13). Therefore, under dead load, the designed geometry of deck seems
invariable.
Due to the equilibrium condition of forces, displacements of pylons are almost equal zero. The present study and the study of Ki - Hae [9] exactly reproduce the target profile of the
deck, while Wangs study [7] yields an inaccurate profile. The tensions of the stay cables
in present study and the study of Kim-Lee are not considerably different but considerably
different in Wangs study. This comparison is shown in Table 2 and Figure 11.
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Table 2 The comparison
Cable no.Present
study
Ki Haes
study
Wangs
study
Relative error
between Present &
Ki Hae
Relative error
between Present &
Wang
Relative error
between Ki -
Hae & Wang
20 11.45 11.38 10.73 0.61% 5.71% 6.29%
21 8.25 8.26 8.26 0.12% 0.00% 0.12%
22 4.89 4.85 4.79 0.82% 1.24% 2.04%
23 5.14 5.11 4.55 0.58% 10.96% 11.48%
24 6.77 6.79 7.81 0.30% 15.02% 15.36%
25 12.63 12.55 11.3 0.63% 9.96% 10.53%
Force Unit: MN
Figure 11: The comparison of three methods.
5. Conclusion
This paper presents the cable element to model the stay cables and procedure to solve an important
problem in cable stayed bridge analysis. The numerical example is presented for demonstrating the
validity and effectiveness of the proposed method in comparison to other methods in [7], [9]. Although
this paper only presents the formulations and numerical example in the two-dimensional problem, the
proposed method can easily extended to the three-dimensional problems.Its believed that this method
provides the powerful tool for engineers in design of cable stayed bridge.
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Reference
[1]. MS Troitsky. Cable-stayed bridge theory and design. BSP Professional Books-1998.[2]. Le Dinh Tam, Pham Duy Hoa. Cable-stayed bridge. Science & engineering Press, 2001. (in Vietnamese).[3].Nguyen Viet Trung. Cable supported bridge. Construction Press, Hanoi 2004. (in Vietnamese).[4].Nguyen Huy Cung. Cable stayed bridge design. Engineering graduate thesis. University of Transportation2007. (in Vietnamese).
[5]. Cable-Stayed Bridges-Developments Recent and Their Future. Developments in Civil Engineering-Vol.40.Elsevier[6]. Wang PH, Tseng TC, Yang CG.Initial shape of cable-stayed bridges. Computers & Structures Vols. 46. No.6. pp. 1095-I 106. 1993[7]. Pao-Hsii Wang, Hung-Ta Lin, Tzu-Yang Tang. Study on nonlinear analysis of a highly redundant cable-stayed bridge. Computers & Structures 80 (2002) 165182[8]. H. M. Alit and A. M. Abdel-Ghaffarf.Modeling the nonlinear seismic behavior of cable-stayed bridges withpassive control bearings.Computers & Structures Vol. 54. No. 3. pp. 461492, 1995[9]. Ki Seok Kim, Hae Sung Lee. Analysis of target configuration under dead load for cable stayed bridges.Computers and Structures 79 (2001).[10].D.W.Chen, F.T.K.Au, L.G.Tham, P.K.K.Lee. Determination of initial cable forces in prestressed concretecable stayed bridges for given design deck profiles using the force equilibrium method. Computers andStructures74. 2000.[11].J.H.O. Negrao, L.M.C Simoes. Optimization of cable stayed bridges with three dimensional modeling.Computers and Structures 64 (1997).[12].J.F. Debongnie. Fundamentals of finite elements.Les Editions de lUniversit de Lige, Lige, 2003, ISBN 2-93032254-3[13].Nguyen Hoai Son. FEM with Matlab. Ho Chi Minh National university publishing house, 2001. (inVietnamese).[14].Raid Karoumi.Response of Cable-Stayed and Suspension Bridges to Moving Vehicles-Analysis methods andpractical modeling techniques. Ph.D Thesis.[15].Zienkiewicz, O. C. and Taylor, R. L., The Finite Element Method, 5th edition, Butterworth-Heinemann, 2000.