Numerical solutions of various cables and their applications by menez

Architectural Engineering February 2014, Volume 2, Issue 1, PP.1-9

Numerical Solutions of Various Cables and Their Applications Yuemin Yang The First Highway Survey and Design Institute of China, 710075, China Email: yuemin_yang@163.com

Abstract This article deeply analyzes the curve form, deflection and deformation of the length given cable under various loads by the method of discretization, solution of nodal equilibrium, and iterative solution of the whole system balance. The curves of all other kinds of cables and their mechanical properties are also discussed. The solution of the cable is adopted to analyze the effect of the cable's sag on the non-linearity of cable-stayed bridge. Besides, the solutions are applied to determine the ideal curves of arch bridges. Keywords: Analysis of Main Cables; Cable's Sag; Ideal Curves of Arch Bridges

1 INTRODUCTION The most important step in cable supported structure design, especially the suspension bridge, cable stayed bridge and cableway, is the nonlinear analysis of the structure under various loads. Then how to exactly present the mechanical properties of its cable, tower and beam becomes essential. Because the beams and towers are stiff elements whose relations between displacement and applied load are well known, the objective of this paper is to find an exact formulation of the flexible cable's mechanical properties. Besides, their applications are also discussed. For example, since the solution of the cable is no bending result for flexible structures, hence, it can be adopted to determine the ideal curves of arch bridges, and the structures alike.

2 MAIN CABLE'S MECHANICAL PROPERTIES AND ITS DEFORMATION We divide the cable into n segments along x direction. Each segment has the length: dxi  cos i  dsi

dy tgi  i dxi

(1) (2)

The weight, q(xi) dsi, of each segment can be equivalently represented by two neighbouring nodal joint loads. When n is large enough, the discretized curve is identical to the original smooth one. Each segment is assumed as straight line. Starting from point A, its equilibrium equations are: N1 cos 1  H dx1 1 N1 sin 1  VA  q( x1 )  2 cos 1

tg1 

VA 1 q1dx1  H 2 H cos 1

(3) (4) (5)

If we define the internal axial force of the cable as Ni, to a generic point i, we have: Ni cos i  Ni 1 cosi 1  H Ni sin i  Ni 1 sin i 1  Pi  Qi dxi dxi 1 1 1 Qi  q( xi )  q( xi 1 ) 2 cos i 2 cos i 1 -1http://www.ivypub.org/AE

(6) (7) (8)

Ni-1 Q 1

k p

i+1

i-1

p2+Q 2 pi +Q

H A

pk +Q k Pi+Qi

FIG.1 DISCRETE MODELING OF THE CABLE

where Pi is the applied concentrated load at point i. From Eqs.(6) (7) (8), we obtain: tgi  tgi 1 

Pi 1 q( xi 1 )dxi 1 1 q( xi )dxi   H 2 H cos i 1 2 H cos i

(9)

If Eq.(9) is divided by dx  0 ,we get the differential equation of the cable [1] [3]: Pi q ( x)  ( x  xi )  1  y 2 H H Eq.(9) is a discrete form of Eq.(10).  ( x) is Dirac function. Eqs.(5) and (9) can be simplified as: y  

sin i  ai cos i  bi

(10)

(11)

in which ai  tgi 1 

Pi 1 q( xi 1 )dxi 1  H 2 H cos i 1

(12)

1 q( xi )dxi 2 H When i  1 , we have P1=‐VA, 0  0 , q(x0)=0. The solution of (11) gives: bi  

cos i 

(13)

ai  bi  ai 2  bi 2  (ai 2  1)(bi 2  1)

(14)

(ai 2  1)

i  arcsin(ai cos i  bi )

(15)

From Eqs.(5) and (9), it is clear that when VA and H are given all the unknown quantities in Eq.(9) can be obtained in recursion. The coordinate yi of each point is: yi 1  yi  dxi tgi i  1 n  1

(16)

We begin with starting estimates for VA and the horizontal force H by assuming the cable in the shape of a triangle, where the cable is straight, as shown in Fig.2. The length of cable AC is L0/2. 1 1 f  ( L0 )2  ( a)2 2 2

(17)

VA a

H x

B f

y  C

FIG. 2 ESTIMATING OF THE STARTING VALUES -2http://www.ivypub.org/AE

According to the balance of the whole cable in Fig.2, here all the loads on the cable are left out, we get:

VA 

1 q dx ( Pi  i i )(a  xi )  a cos  H A  VA  ctg  H 2f tg  a L  Lk  L0

(18) (19) (20) (21)

’

where L0 is the given length of the cable, and Lk is the cable's length in k th iteration: Lk   ( yi  yi 1 )2  ( xi  xi 1 )2

(22)

If we define the angle of the resultant of H and VA as φ with respect to horizontal line, the iterative sequences for φk and Hk are given as follows [1] [5]:

k  k 1  arctg ( H k  H k 1 (1 

yn ) a

L ) L0

(23) (24)

Substituting into Eqs.(5) and (9) for the starting values VA and H, we get the coordinate yn of point B and the length Lk of the cable. When yn and Lk are put into (21), (23), (24), we get new values VA and H. The iteration is carried on until the changes between two following iterations are small enough or less than the given tolerance values. The iterations (23) and (24) converge so fast that after 4~5 repeats the error between two following iterations is less than 10-5, whereas the iteration is started with any initial values larger than zero. To any ratio f/a, the numerical solution of Eq.(9) is almost the same as the analytical solution of Eq.(10). Elastic extension of the cable is given by

L  

Ni dxi EAi cos  i

(25)

If the given length of free cable is L0, and assuming that the cable can't extend under load, we define this state of the cable as state I. Then we define the state of the cable with extension as state II, and its length is L. Since we only consider geometric non-linearity of the cable, the cable's extension is only related to its internal force in state II. But, so far the cable's length in state II is an unknown, the problem can't be solved for. Therefore, we need first to take the calculated extension of Eq.(25) in state I, △L+L0, as an approximate length of state II. Namely, the cable's length in state II is taken as △L+L0. With the length assumption of △L+L0 as L0 of Eq.(24) in state II, we can carry on the analysis of the problem obtaining a new extension △L of the cable in state II. If the new extension △L is equal to the assumed one, that is the answer we are looking for. Otherwise, we still need to take the recent extension △L+L0 as cable's length L0 of Eq.(24) in state II, and repeat the calculation until the requirement is satisfied. From context we know that this second iteration is to satisfy the cable's deformation compatibility, and it converges even faster.

3 DETERMINATION OF VARIOUS CABLE FORMS 3.1 Curve Form of the Cable When its Span and Vertical Height Are Given The main cable of a suspension bridge can be considered as a combination of two symmetric half cables. The form of the half cable is then the same as a tilted cable. To a tilted cable, the derivation of its equilibrium is exactly the same as above. The setting up of equilibrium equations starts from point A. The iterative variable is the horizontal force H at point A. The iterative sequence of H goes as[5]: H k  H k 1 (1 

df ) f0

df  yn  f0

where yn is the vertical coordinate of point B during the iterative process, and f0 is the designed height of point B. -3http://www.ivypub.org/AE

(26) (27)

N1 α1

B +Q Pn n f P k +Q

Q1 A

Pk-1+Q k-1

x P2 +Q 2

P +Q 1

a FIG.3 TILTED CABLE MODELLING

3.2 Curve Form of the Cable When H Is Given In order to keep balance of the tower, side span cable must offer the same horizontal force as the middle span, i.e., the horizontal force in the side span cable is given. We may take the vertical load P at point A as variable in the iteration. The balance of the whole cable requires that

M BP  P  a  f  H  0

(28)

where MBP is the moment about B produced by all the loads on the cable. It is easy to verify the increment of MBP is a smaller amount during the iteration. The differentiation of (28) gives out:

dP 

H df a

(29)

The iterative relation of variable P develops in the way[5]:

Pk  Pk 1 

H df a

(30)

The setting up of equilibrium is the same as above, which is started from point A. Under the same conditions, we obtain exactly the same result by Eqs. (26) and (30).

3.3 Curve Form of A Length Given Tilted Cable When the length of a cable is given, the equilibrium equations for the cable are set up in the same manner as above. The iterative variables and their iterative relations are [1] [5]: H k  H k 1 (1 

Pk  Pk 1 

L ) L0

Hk df a

(31) (32)

in which L  Lk  L0 .

3.4 Iterative Solution of the Cable With a Horizontal Restriction At its Middle In order to increase the stiffness of the cable, a horizontal restriction is often imposed on a middle point k of the cable [8]. Let the cable's length of point k's left side be LL0, and k's right side be LR0.The governing equations of point k are: Nk cos k  H k  Nk 1 cosk 1

Nk sin  k  Nk 1 sin  k 1  Pk  Qk Let:

H R  N K cos  K

H L  Nk 1 cos k 1

(33) (34) (35)

Here point k could be any generic point. The constant horizontal force in the cable of k's left is HL, and that of k's right -4http://www.ivypub.org/AE

is HR.HK is the horizontal restriction force on point k. From (33) and (34) we obtain: tg k 

Nk 1 sin  k 1 Pk Q   k HR HR HR

(36)

Here Eq.(36) can also be written as the form of Eq.(11). Thus ak 

Nk 1 sin  k 1 Pk 1 q( xk 1 )dxk 1   HR H R 2 H R cos  k 1

(37)

1 q( xk )dxk 2 HR

(38)

bk  

The starting value of HL is the same as Eqs.(18) (19). Then the starting value of HR can be achieved, as in Fig.2, by the consideration of whole cable's balance: VB 

q( x )dx 1  ( Pi  i i )xi a cos 

(39)

H B  VBctg  H R The iterative sequences are

(40)

[1] [5]

: yn ) a L  H L,k 1 (1  L ) LL0

k  k 1  arctg (

(41)

H Lk

(42)

H Rk  H R,k 1 (1 

LR ) LR0

(43)

LL  LLk  LL0 LR  LRk  LR 0 ;

where

(44)

LLk, LL0; LRk, LR0 are the calculated lengths and assumed lengths of the cables of k's left and right sides, respectively. The iteration has a great accuracy and converges fast as well. The horizontal restriction force Hk is the difference between HL and HR. According to the discussion above, we may find that the influence on the vertical displacement produced by horizontal restriction of the hanging members of the suspension bridge in the example of reference [1] is about 0.4661%, which can be ignored thoroughly. This can also be achieved by an observation that the horizontal force H in the main cable with a general ratio f/a is usually bigger than the dead load including the main cable and the deck. If we compare the horizontal restriction of the hanging members with the dead load of entire bridge, it is clear that the horizontal restriction is meaningless and negligible, so as stated by the digital result above. All the solutions of various cables above are typical applications of the shooting method in Numerical Analysis.

4 EFFECT OF CABLE'S SAG ON THE NON-LINEARITY OF CABLE STAYED BRIDGE B α β

A u

C a

f A a

P0 (a)

(b)

FIG. 4 MODELS OF STRAIGHT AND SAGGED CABLES

To discuss the effect of cable's sag, we consider two cable structures with same properties as shown in Fig.4 (a) and -5http://www.ivypub.org/AE

Fig.4 (b). Two cables have the same free length. Member AC in both structures only acts as a horizontal restriction [8]. In Fig.4 (a), we assume the cable to be straight. In the analysis of the cable in Fig.4 (a), the exact formulation of out-offbalance force is adopted, i.e., the deformation compatibility requirement. The extending amount of the cable is:

a a  sin  sin 

(45)

a a tg   f u f

(46)

EA L L

(47)

L 

tg 

Where: The internal force of the cable is:

N

The out-off-balance force at joint A writes[2]: R  P0  N cos 

(48)

The solution to Fig.4 (a) is carried out by N-R method where the remainder is Eq.(48). The numerical method given in the paper, i.e., the iteration sequence Eq.(31), is adopted to find out the solution to Fig.4 (b). We assume: the weight density of the cable γ=80kN/m3, E=2.05x108kPa [8], and the axial force of the cable at joint A when A is unmoved: A  0.011  1860 (49) NA   n n where A is the area of the cable, [б]=1860MPa is the allowable stress of the cable, and n is safety factor. Then the applied external load at A is: P0  N A cos 

(50)

In order to describe the deformation of the straight cable exactly, we equilibrate the weight qL of the cable by two nodal joint loads at points A and B, i.e., the load at A is: 1 (51) P  P0  qL 2 where q=Aγ. This formulation is to take the mean value of the cable's internal forces as the axial force of the weightless straight cable. Although the mean value can predict the deformation very well, there are some errors in its axial force at joints A and B. If we assume f=150.0m, a=300.0m, and n=3, the solutions for the cables in Fig.4 are given in Table 1. TABLE 1 EFFECT OF THE CABLE'S SAG

Displace. u at A Expansion △L Hor. Restric. H Axial Force NA

Straight cable

Sagged Cable

2.3416544m 1.0537385m 6296.85404kN 7062.2143kN

2.2991289m 1.0540650m 6298.6041kN 6998.2064kN

Errors 1.85% 0.03099% 0.0278% 0.915%

Straight cable by N-R method 2.327599m 1.047375m 6259.0640kN 7019.6983kN

It is clear from Table 1 that the expansions of two cables are almost the same. However, there are some errors in the displacement u. Here u is measured under the reference f=150.0m.When P0 is small enough, the height f of the sagged cable is f≤150.0m.This means a negative displacement with respect to f=150.0m. Thus, some amount of the load P0 is transferred to overcome the weight sagging. The rest load produces the positive displacement u only. The total displacement of the sagged cable, including the negative part, is much larger than that of the straight member. The straight cable in Fig.4 (a) is the limit stage of the sagged cable. In engineering, however, we often take the straight member as reference. Thus, the negative part has no meaning in analysis. The errors occurred to H and NA of the straight cable arose out of the analysis model. These errors can't occur in practice. These errors should only be taken as reference. The cable in Table 1 is relative long and sagged heavy in engineering. Then the cable commonly used in engineering -6http://www.ivypub.org/AE

will have less sag effect. If we set points A and B of the cable in Table 1 at same horizontal level, then the effect of the sag on cable's length is about 7.8044x10-3% [7] with respect to the straight line AB. Hence we can think the lengths are the same. Then, for the cables with f≠0, their lengths have less difference from straight lines. Thus, their expansions are almost the same. It is clear that the effect of cable's sag can't be estimated by the expansion. Only the difference between the vertical positive displacements can show the effect. To imitate the sag effect by a straight member then, and when the load increment is large, i.e., the loading is started from zero, the modified elastic modulus of the straight member may be larger than its original value. If a small load increment after some loading is concerned, according to more detail analysis, we may use the tangent stiffness of the sagged cable, which is smaller than that of the straight member. We should pay more attention to the variation of the cable's stiffness for different loading during analysis. The last column in Table 1 is the results where the out-off-balance force at joint A reads[2] [7]: R  P  K (u)  u

(52)

[2]

where K(u) is the vertical stiffness of point A : EA cos2  L By the well known formulations of cable element, the horizontal length is[3] [6]: K (u ) 

a

HL0 HL V V W (sh1  sh1 )+ 0 W H H EA

(53)

(54)

the vertical height

f 

HL0  V 2 V  W 2  WL0 V 1 )  (  )  1 ( )  1 ( W  H H  EA W 2

(55)

where

VB  P0  W  P0  qL0  V ; H A  H B  H

(56)

If we let W  0 in Eqs.(54),(55), we get the solution of a straight cable: HL0 (57)  L0 cos  EA VL (58) f  0  L0 sin  EA V where (59) tg  H Based on the data in Table 1, and letting a=300 in (54) and (57), we obtain the solutions for H. The substitution of H into (55), (58) yields the height f: a

Straight cable: H=6297.15172kN f=152.33445m Sagged cable: H=6298.0001kN f=152.291842m The results above are almost the same as in Table 1. The minute differences between them arose out of that the different cable lengths are used in the calculation. The author believes that the extended length should be used in the calculation of the cable's expansion because we usually use the extended length of the member in its stiffness. Indeed, the difference between extended length and the original length of the cable is rather small, at most no more than





1860

 0.9073% (60) 2.05  105 When a safety factor is considered, the extension  of the cable in engineering is even smaller. Hence, in practice the difference often is ignored. The author proposes that the straight model could be used to represent the flexible cable's properties in the structure analysis. The effect of the cable's sag should be evaluated by the model in Fig.4 (b). Here we keep the same P0 in the global model of the structure as the one in the separated cable model in Fig.4 (b). The differences of the element nodal forces between the global and the separated cable models are taken as extra loads to E



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be put on the same joints of the global structure in the iterative process. To a cable with large span, we should adopt different elastic modulus according to loading increment. When the loading increment is small, we may adopt the tangent elastic modulus, as Ernst’s formula expressed. The results show that when the model in Fig.4 (b) is used to modify the corresponding straight member in the entire model, the axial force, as in a cable stayed bridge deck, is a little larger than unmodified. The modification has no effect on convergence of the iteration.

5 DETERMINATION OF IDEAL CURVES FOR ARCH BRIDGE yi

B H

FIG. 5 MODELLING OF ARCH BRIDGE

FIG. 6 SIDE SPAN

If we take the discrete elements with axial pressure as the analysis model of the arch bridge shown in Fig.5, the model will behave as an unstable structure when the bending moment of each section vanishes[4]. There is no solution to such a model in mathematics. If we turn upside down the arch bridge, and the pressing elements are changed into stretching, then it is stable. There must exist a solution to such a problem. When the pressing model is studied, we have to consider the flexural stiffness of the member. This implies that there must exist some bending moment at each section. The moment can't vanish at all. If we think the arch bridge as an upside down cable, we can easily find its exact solution by the ways given in the article. It can be proved that if we apply finite element method to the arch bridge found by the method in the paper, we may find the bending moment at each section almost is zero. The minute moment is thoroughly induced by angle restriction and axial deformation. There exists a huge horizontal pushing force in the side span as shown in Fig.5. The side span can also be considered as an upside down cable, as shown in Fig.6. The solution for Fig.6 can also be obtained by the way presented in the paper. Here we only need to turn round the cable about line A-B from downside to upside. This is the ideal curve for arch bridge we are looking for. If the iteration (30) can't converge, this means the horizontal pushing force H is too small. Then we should take the iteration (26) to find out an appropriate H.

6 CONCLUSIONS Through the solution of nodal equilibrium, and iterative solution of the whole system balance, the cable's solution satisfies its balance equation, all boundary conditions and the attached length restriction. So, its solution is exact in mathematics. From the opinion of error analysis, the iterative solution for cable is in fact to distribute the final accumulated errors to the discrete points on the cable by adjusting the initial values. The solution is very stable. According to the discussion in the paper, we know the cable is a hardening member, and the straight member is its limit stage. When the load increment is large, or the load is put on the structure from zero, the modified elastic modulus of the imitating straight member may be larger than its original value, because the positive displacement is smaller than the straight member. If a small load increment is concerned, according to more detail analysis, we may use the tangent stiffness of the sagged cable in the analysis, which is smaller than that of the straight member. If we set up the discrete model in terms of pressing elements to look for the ideal curve of an arch bridge, we can't get an exact solution in mathematics. If we consider the arch bridge as an upside down cable, we can get its exact solution.

REFERENCES [1]

Yue-min Yang. “Geometrically Nonlinear Analysis of Suspension Bridge and Properties of its Main Cable”. WOLD SCI-TECH R&D. 31(4) (2009): 711-717

[2]

Yue-min Yang. “Comparison of Nonlinear Equations’ Solutions in Mechanics and Their Modification”. WOLD SCI-TECH R&D. -8http://www.ivypub.org/AE

33(3) (2011): 376-379 (in Chinese) [3]

Ahmadi-Kashani K, Bell A J. “The Analysis of Cables Subject to Uniformly Distributed Loads”. Engineering Structures,10 (1988): 174-184

[4]

Bathe, Klaus-Jürgen. Numerical Methods in Finite Element Analysis. New Jersey: Prentice-Hall, Inc. Englewood Cliffs, 1976

[5]

David Kincaid, Ward Cheney. Numerical Analysis-Mathematics of Scientific Computing. Thomson Learning, 2002

[6]

Irvine H M. Cable Structures. The MIT Press, 1981

[7]

Stricklin J A, etc. Static Geometric and Material Nonlinear Analysis-Advances in Computational Methods and Design. Huntsville, Alabama: UAH Press, The Univ. of Alabama, 1972

[8]

Niels J•Gimsing. Cable Supported Bridges—Concept & Design, (2nd Edition). John Wiley & Sons Ltd, 1997

[9]

HUANG Yan, LAN Wei-ren. “STATIC ANALYSIS OF CABLE STRUCTURE”. Applied Mathematics and Mechanics (English Edition). 2006, 27(10):1425-1430

[10] Huu-Tai Thai, Seung-Eock Kim. “Nonlinear Static and Dynamic Analysis of Cable Structures”. Finite Elements in Analysis and Design. Volume 47, Issue 3, March 2011, 237-246 [11] Wang Chunjiang, etc. “A New Catenary Cable Element. International Journal of Space Structures”. Volume 18, Number 4/ December 2003, 269-275 [12] Hiroyuki Sugiyama, etc. “A Non-Incremental Nonlinear Finite Element Solution for Cable Problems”. ASME 2003 International Design Engineering Technical Conferences and Computers and Information in Engineering Conference. DETC2003/VIB-48321, 171-181

AUTHORS Yuemin Yang, born in 1957, earned his bachelor degree of solid mechanics in Xi’an Institute of Metallurgy and Architecture Engineering, Xi’an, China, in 1981. He was awarded master degree of structure engineering in Chong-Qing Institute of Architecture Engineering in 1985. Finally, he got his doctor degree of bridge engineering in China Academy of Railway Sciences, Beijing, China, in 1995. His research area covers applied mechanics, dynamics of structures, especially the vibration of bridges. He is now a member of bridge designing in The First Highway Survey and Design Institute of China. His favorite jog is the designing of arch bridges. The largest stone arch bridge of a span 146m in the world, Danhe Bridge, is his work. Besides, During the study for his doctor degree in Beijing, he finished an experiment of coupled vibration between vehicle and bridge, the first experiment in China. By now, he has published the papers: Geometrically Nonlinear Analysis of Suspension Bridge and Properties of its Main Cable；Comparison of Nonlinear Equations’ Solutions in Mechanics and Their Modification, and Linear Multistep Numerical Solutions for Vibration Equation, etc, in the journal WORLD SCI-TECH R&D. Other articles were also published in other journals. Dr. Yang is a leading member of Shannxi Mechanics Academy. He used to be the leader of Bridge Designing Office in the institute. The article, Linear Multistep Numerical Solutions for Vibration Equation, is awarded the hot paper by the journal. Now he is interested in numerical analysis of engineering problems.

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