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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.

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