Two-node Catenary Cable Element With Rigid End Effect and

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TWO-NODE CATENARY CABLE ELEMENTWITH RIGID-END EFFECT AND CABLE

SHAPE ANALYSIS

Y. B. YANG * , and JIUNN-YIN TSAY y

*Department of Construction National Yunlin University of Science and Technology

Yunlin 64002, Taiwan yDepartment of Civil Engineering

National Taiwan University, Taipei 10617, Taiwan [email protected]

Received 24 June 2010Accepted 27 July 2010

The eect of rigid ends is considered in the formulation of a general two-node cable element forthe analysis of cable-supported structures. The stiness matrix of the catenary cable elementwas derived as the inverse of the exibility matrix, with allowances for self-weight and pre-tension eects. In modeling the cables of suspension bridge, distinction is made between single cables (e.g., stay cables and hangers) and multi segment cables (e.g., main cables). Theunstressed length of each cable element in terms of the pretension force is determined by a trial-and-error procedure prior to structural analysis. Cable shape analysis was conducted todetermine the conguration of main cables for cable-supported bridges under the dead loads. Itwas demonstrated that the eect of rigid ends cannot be ignored for taut cables, that is, cableswith large pretensions. Further, the cable element derived can be reliably used in the analysis of cable-supported bridges, regardless of the sag magnitudes.

Keywords : Generalized cable element; cable-supported bridge; cable shape analysis; geometricnonlinear analysis.

1. Introduction

Cable-supported bridges with an elegant outlook and ecient structural perform-ance are often constructed as long-span bridges for connecting two places separatedby valleys, rivers, or seas, or as landmark bridges in scenic or metropolitan areas.Most cable elements adopted in the analysis of cable-supported bridges do not take

into account the eect of rigid ends. It was noted by Gimsing1

that there may exist alocal bending eect for cables that are connected to the other components of thebridge by sockets or rigid joints. Typical examples of rigid ends can be found at the

Corresponding author, on leave from National Taiwan University.

International Journal of Structural Stability and DynamicsVol. 11, No. 3 (2011) 563 580#.c World Scientic Publishing CompanyDOI: 10.1142/S021945541100421X

563

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reliability and eciency of the algorithm are demonstrated in the study of severalexamples.

2. Two-Node Cable Element Considering the Rigid-End Eect

Conventionally, truss elements have been adopted to simulate the cables installed oncable-supported structures with Ernst' equivalent modulus 14 :

E eq E

1 wc lAE

12T 31

where E denotes the elastic modulus, A the cross-sectional area, l the horizontalprojected length, wc the weight per unit length, and T the tension force of the cable.

For cables under their self-weight alone, the sag can be quite large. For cablessubjected to large pretensions, in addition to the self-weight, the sag may be quitesmall. Such elements are not suitable for modeling cables under the condition of zeropretension or in natural shapes, since the equivalent modulus reduces to zero for zeropretension, which is not physically justied.

The two-node catenary cable element presented by Yang and Tsay 3 is capable of simulating both the self-weight and pretension eects. It is good for modeling notonly cables with large pretensions (i.e., small sags), but also cables with small pre-tensions (i.e., large sags). In addition, the temperature eect on the cable can be

easily included in updating the uniform loads acting on the cable. All these areimprovements over the equivalent truss elements conventionally used.

In this study, the two-node catenary element is generalized to include the eect of rigid ends, as shown in Fig. 1. The projection length of each rigid end of the cable isassumed to be proportional to the same directional component of the resultanttension force. Let S R and S L denote the lengths of the two end segments, as shown inFig. 1. By assuming the cable to be of the catenary shape, one can modify thehorizontal and vertical projection lengths, that is, l and h , of the cable with no rigid

H

V

T

W

x

y

H

V-W

SR

SL

l

hl 0

Fig. 1. Cable element with two rigid ends.

Two-Node Catenary Cable Element with Rigid-End Eect 565

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ends to include the contribution of rigid ends as follows 2

l Hl 0EA

H q

sinh 1 V

H

sinh 1

V ql0H

S R H ffiffiffiffiffiffiffiffiffiffiH 2 V 2p S L

H

ffiffiffiffiffiffiffiffiH 2 V ql02p 2

h l0EA

V ql0

2 1q ffiffiffiffiffiffiffiffiffiH 2 V 2p ffiffiffiffiffiffiffiH 2 V ql0 2q S R

V

ffiffiffiffiffiffiffiffiffiffiffiH 2 V 2p S L

V ql0

ffiffiffiffiffiffiffiffiH 2 V ql02p 3

where V and H respectively denote the vertical and horizontal components of the

cable force, q is the dead load per unit length of the cable, l0 is the unstressed lengthof the cable excluding the end segments, and L0 the unstressed length of the cableincluding the contributions of the two end segments, that is,

L 0 l0 S R S L 4By letting W denote the total dead load acting on the cable, we have q W =L 0, or

q W

L 01 T 5

by taking into account the eect of thermal expansion. Here, is the thermalcoecient of expansion of the cable and T is the temperature rise from the designtemperature.

As can be seen from Eqs. ( 2) and ( 3), both the projection lengths ( l; h ) of the cableare functions of the horizontal and vertical tension components H and V of the cableforce, that is, l f H ; V , h g H ; V . The total dierentials of the projectionlengths ( l; h ) in Eqs. (2) and ( 3) can be expressed as

dldh

@ f @ H

@ f @ V

@ g @ H

@ g @ V 0BB@ 1CCA

dH dV F

dH dV 6

The coecients f ij in the exibility matrix [F ] can be obtained directly as thedierentials of l and h. By letting

V V ql0 7we have

f 11 @ f

@ H l0EA

1q sinh

1 V

H sinh1 V

H 1q V ffiffiffiffiffiffiffiffiffiffiH 2 V 2p V

ffiffiffiffiffiffiffiffiH 2 V 2p " #

566 Y. B. Yang & J.-Y. Tsay

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S R1

ffiffiffiffiffiffiffiffiffiffiH 2 V 2p

H 2

ffiffiffiffiffiffiffiffiH 2

V 2

3

q 0B@

1CA

S L1

ffiffiffiffiffiffiffiffiffiffiH 2 V 2p H 2

ffiffiffiffiffiffiffiffiH 2 V 2 3r

0BB@1CCA

8a

f 12 @ f @ V

1q

H

ffiffiffiffiffiffiffiffiffiH 2 V 2p H

ffiffiffiffiffiffiffiH 2 V 2p " #S R

HV

ffiffiffiffiffiffiffiffiffiffiH 2 V 2 3q S

LH V

ffiffiffiffiffiffiffiH 2 V 2 3r 8b

f 21 @ g @ H f 12 8c

f 22 @ g @ V

l0EA

1q

V

ffiffiffiffiffiffiffiffiH 2 V 2p V

ffiffiffiffiffiffiH 2 V 2p " #S R 1 ffiffiffiffiffiffiffiffiffiffiH 2 V 2p

V 2

ffiffiffiffiffiffiffiffiH 2 V 2 3q 0B@ 1CA

S L1

ffiffiffiffiffiffiffiffiffiffiH 2 V 2p V 2

ffiffiffiffiffiffiffiffiH 2 V 2 3r

0BB@1CCA

8d

For a 3D catenary cable element, there are three force components acting at each of the two ends. The 3D exibility matrix F 3 3 in the local coordinates can beobtained as a direct extension from the 2D in-plane matrix F 2 2 by including theout-of-plane coecient f 33 as follows2,15

f 33 lH

L0EA

1q

sinh 1 V

H sinh 1 V H 8e

S R

ffiffiffiffiffiffiffiffiffiffiH 2 V 2p S L

ffiffiffiffiffiffiffiffiH 2 V 2p 8f

and letting the other out-of-plane coecients equal to zerof 13

f 31

f 23

f 32

0

8g

The stiness matrix [K ] of the cable element can be obtained simply as the inverse of the exibility matrix, namely

K 6 6 k kk k 9


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where

k

F 13 3

f 11 f 12 0f 21 f 22 00 0 f 3324 35

1

10

Obviously, the stiness matrix k of the cable element is symmetric.

For the case of vertical hangers with l 0, the horizontal component H vanishes(H 0), and Eq. ( 3) reduces to

h l0EA

V ql0

2 1q jV j jV j S R S L 11For V > ql0, which is the case for taut cables , the vertical reaction force V can be

given asV

ql02

EAl0 h S R S L l0 12

On the other hand, for V < ql0, which is the case for loose cables , the vertical reactionforce V can be given as

V ql0

2 q h S R S L

2 ql0EA

13

The vertical reaction forces given in Eqs. ( 12) and ( 13) can be applied in calculatingthe forces of hangers in the erection and completion stages, whose major function isto connect the main cables to the deck of suspension bridges.

3. Unstressed Length of Single Cables

Cables used are hangers of a suspension bridge or stay cables of cable-stayed bridges asregarded as single cables, since they have no intermediate loading points. As can beseen from Eq. ( 8), the unstressed length L 0 of the cable with rigid ends (or the length l0

with no rigid ends), instead of the pretension force, is a key parameter in determiningthe exibility coecients and therefore the stiness coecients of the catenary cableelement. Prior to execution of structural analysis, the unstressed length L0 of eachcable with rigid ends can be determined from Eq. ( 2) as the horizontal projectionlength of the cable by a trial-and-error procedure, as described below:

(1) As for the horizontal projection length in Eq. ( 2), the modulus of elasticity E ,cross-sectional area A, self-weight q , and lengths of the two rigid ends, S R , S L ,(if there are any) of the cable are all assumed to be known.

(2) The pretension force T of each cable is rst given. The reaction forces ( H ; V )existing at the ends of each cable are obtained as the components of the pre-tension force T .

(3) Given the nodal coordinates of the two ends of the cable, both the horizontalprojection l and the distance between the two end points can be computed.


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(4) The distance between the two end points after deduction of the end segmentlengths ( S R S L ) will be used as the rst trial value for the unstressed length l0.

(5) By substituting this trial value into Eq. ( 2), an improved horizontal projection

length l can be computed.(6) Compute the dierence between the previous and computed projection lengths:

l l l 14(7) If the dierence l is less than a preset tolerance, then the value l0 used in the

computation is the solution desired, and total unstressed length L0 (i.e.,L 0 l0 S R S L ) can be determined accordingly.

(8) Otherwise, depending on whether l is positive or negative, a new trial value l0larger or smaller than the previous trial value should be used, and the above

procedure can be repeated.As for the unstressed length of hangers ( l 0), a similar procedure can be con-

ducted based on the vertical projection length in Eq. ( 11). The vertical projection his computed as the distance between the two ends of the hanger. Also, the distancebetween the two ends of the hanger excluding the rigid ends is used as a trial value forthe unstressed length l0. By substituting this value into Eq. ( 11), an improved ver-tical projection length h can be computed. Let h denote the dierence between theprevious and computed projection lengths, that is,

h h h : 15If the dierence h is less than a preset tolerance, then the unstressed length l0 usedin the computation is the solution desired. Otherwise, the procedure should berepeated for a new trial value l0. Using such a procedure, the total unstressed lengthL 0 (i.e., L0 l0 S R S L ) of the hanger can be obtained.

Alternatively, a simplied formula can be applied for calculation of the unstressedlength for cables with a sag-to-span ratio n ( f =l) less than 1:8. By assuming thecable to be of a parabolic shape excluding the rigid ends, the unstressed length l0 canbe computed as follows

l0 Z l

0 ffiffiffiffiffiffiffiffiffiffiffi1 dydx 2s dx Z

l

0 ffiffiffiffiffiffiffiffi1 4f l 2xl 1 2s dx 16

where dy=dx indicates the slope at any point of the cable, and l is the horizontalprojection length between the two supports of the cable excluding the rigid ends.Accordingly, the unstressed length of the cable without rigid ends can be approxi-mately obtained as follows 4

l0

ffiL m 1

8

3n 2

32

5 n 4

256

7 n 6

Main span

17a

l0 ffiL s sin

83

L sL m

2 f L s" #

21

sin 3 Side span 17b


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where n is the sag-to-span ratio (as dened above), Lm is the length of the main spanlength, dened as the horizontal distance between the two towers, LS is the length of the side span, f is the sag of the main span, and is the angle of sag to the side span,

that is, tan 1f =L s.

4. Unstressed Length of Multi-Segment Cables

For suspension bridges, the main cables normally consist of a number of intermediateloading points, as shown in Fig. 2. Before the cable element can be used in simulatingthe cables in structural analysis, the unstressed length of each segment of the maincables should rst be determined. Given the geometric conguration of the entirebridge, the geometric shape of each main cable is established from loading point toloading point without considering the saddle eect induced on the parts of the maincable near the pylon tops and anchorages. A trial-and-error procedure for deter-mining the unstressed length and nodal coordinates will be conducted from the left tothe right end of the cable-only system. The coordinate system in such a procedure isdened as positive for the rightward x -axis and positive for the downward y-axis, asindicated in Fig. 2. Each hanger is assumed to carry the part of deck loads associatedwith it, say, based on the lumped mass concept.

The material and section properties of each segment are assumed to be known:

elastic modulus E

, cross-sectional area A

, unit weight W

, and rigid-end length S

Rand S L (if there are any). For each segment, there exist three unknowns H ; V ; l0 inEq. (2). The trial-and-error procedure for nding the unstressed length of eachsegment of the cable is as follows:

(1) Since the location of each hanger on the deck is given, the horizontal projectionlength l of each segment of the main cable can be determined.

1

2

34

56

7 89

1

23

4

5 6 7 8

y

x

V

H

P1 P2 P3 P4 P5 P6 P7 P8

Fig. 2. Multi segment cable with intermediate nodal loads.


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(2) To start with, we shall assume the main cable to be of a parabolic shape, therst trial values of the coordinates of each nodal point can be computed.Accordingly, the vertical projection length h of each segment can be computed.

(3) The trial value of the horizontal tension force H of the segment under uniformloads can be assumed as follows 4

H qL 2m

8f 18where Lm is the length of the main span length, and f is the sag of the mainspan.

(4) By dening the parameter as

ql

2H 19the trial values for l0 and V can be computed as follows 5

l0 ffiffiffiffiffiffiffiffiffiffil 2 sinh 2

h 2s S R S L 20aor approximately

l0

ffiffiffiffiffiffiffiffiffil 2 h 2p S R S L 20b

for small values of (or for large cable tension), andV

q 2 jhj

coshsinh l0 21

(5) By substituting the above trial values for H , V , and l0 into Eq. ( 2), an improvedhorizontal projection length l of the rst segment can be computed.

(6) Compute the dierence l between the given (xed) and computed projectionlengths:

l l l 22(7) If the dierence l is less than a preset tolerance, then the value l0 used in the

calculation is the solution desired.(8) Otherwise, depending on whether l is positive or negative, a new trial value of

l0 larger or smaller than the previous trial value should be used, and the pro-cedure utilizing Eq. ( 2) should be repeated.

(9) After obtaining a convergent value for the unstressed length l0, the verticalprojection length h of the segment can be determined from Eq. ( 3).

(10) As the left-end coordinates of the rst segment are already known, the right-end coordinates of this segment with the associated unstressed length can bedetermined accordingly.

(11) The procedure presented above can be repeated for the second segment, thenthe third segment, and so on. Finally, the coordinates of all nodal points of themain cable of the bridge can be determined.


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(12) As the left-end reaction forces of the rst segment are already known, the left-end reaction forces of the second segment can be obtained by the followingequilibrium conditions

H k H k 1 23V k V k 1 q l0k 1 S R S L P k 24

where P k denotes the vertical force of the kth hanger. Such a procedure can berepeated to obtain the force components for each segment.

In the rst trial for nding the unstressed length and nodal coordinates of allsegments of the main cable, errors of the projection lengths ( dl, dh ) may be accu-

mulated for each node. To remove the errors, a new set of trial values of the reactionforces (H ; V ) should be used for the next iteration. The following is the relationshipbetween the accumulated errors and the adjusted values ( dH , dV ) for the forcecomponents:

dldh X

n

k1F ij k

dH

dV 8

Documents

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