Research Article The Static WKB Solution to Catenary ...downloads.hindawi.com/journals/mpe/2014/231726.pdf · The Static WKB Solution to Catenary Problems with Large Sag and Bending

Research ArticleThe Static WKB Solution to Catenary Problems withLarge Sag and Bending Stiffness

Yuhung Hsu and Chanping Pan

Department of Construction Engineering National Taiwan University of Science and Technology Taipei City 10607 Taiwan

Correspondence should be addressed to Yuhung Hsu jeromeehsugmailcom

Received 30 June 2014 Accepted 11 September 2014 Published 28 September 2014

Academic Editor Bruno Briseghella

Copyright copy 2014 Y Hsu and C Pan This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Large sag with a bending stiffness catenary is a subject that draws attention in the realm of fatigue analysis estimation ofsuspension cable sag for bridge cable hoisting and ocean engineering of the employment ofmooring systemsHowever the bendingstiffness is the cause of boundary layers at the anchorage of cables thereby finding a solution of the differential equation can beextremely difficult Previous studies have tackled this problem with the perturbation method yet due to the complexity of thematching process and solution finding the method might not be an ideal solution for engineering applications Moreover thefinite difference method and the finite element method in numerical analysis can often be ineffective because of inappropriateparameter configuration and the drastic variation of functions in the boundary layers Therefore this study proposed a novelbending moment expression of a large sag catenary The expression was derived from the sag identified using bending momentequations and a solution was found by applying theWKBmethod (Wentzel-Kramers-Brillouin method) to overcome the complexproblem of boundary layers Consequently a simple solution of various mechanical properties in a cable with bending stiffness andlarge sag could be obtained

1 Introduction

Slim tension members have been comprehensibly applied tocivil and ocean engineering to create cables of cable-stayedand suspension bridges that satisfy traffic demands andmooring systems that satisfy the demands of mining deep-sea petrol and natural gas Such structural problems could beunderstood as general cable problems where the mechanicalbehavior of cables varies significantly with the magnitude oftension In the study of cables of bridges researchers oftenregard the ratio of midpoint sag to span as an appropriatestandard for simplifying the parabola theory When the ratioof midpoint sag to span is 18 the horizontal force happensto be the total cable weight When the ratio of midpoint sagto span is smaller than 18 (horizontal force greater than thetotal cable weight) the resulting parabola is appropriate andcorrect However when the ratio of midpoint sag to span isgreater than 18 (horizontal force smaller than the total cableweight) a greater error can be found in the parabola [1] Therelatively high tension applied on the cables of a cable-stayedbridge can result in a deflected shape that can be simplified

as a parabola because of its relatively small sag and theirbending stiffness have only minor effects and are negligiblehowever deflected shapes of cables with relatively low tension(main cables of a suspension bridge or marine risers inocean engineering) can be regarded as a catenary becausethe large sag increases the influence of the bending stiffnessand cannot be neglected In addition the bending momentand stress of the anchor points are important parameters infatigue analysis Damages to cables caused by fatigue resultin significant losses thus the understanding of mechanicalbehavior in cables with bending stiffness demands immediateattention

The problem however of incorporating the bendingstiffness effect in cables results from the complex differentialequations Bending stiffness is the cause of boundary layersin anchorages and rapid variations of bending momentsoccur near the differential regions of the cable anchorageBurgess [2] utilized the finite difference method to analyzecables with bending stiffness and low tension and indicatedthe importance of parameter configuration to avoid the lossof effectiveness when employing general finite difference

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 231726 11 pageshttpdxdoiorg1011552014231726

2 Mathematical Problems in Engineering

or finite element methods Reasonable precision should beobtained by adopting a smaller step size and finer gridMoreover an approximation algorithm (eg perturbationmethod) should be adopted to find the analytical expressionwhich cannot be found by an analytical approach because ofthe complexity of differential equations

In dealing with the boundary layer problems the per-turbation method consists of two main techniques [3ndash5]matched asymptotic expansions and multiple scales methodMost scholars adopted the matched asymptotic expansionsto analyze boundary layers in cables with bending stiffnessas M S Triantafyllou and G S Triantafyllou [6] analyzedthe hanging string Wolfe [7] analyzed the inextensible cableIrvine [8] analyzed the stay-cables Stump and Fraser [9]analyzed the high-speed transport of thin-sheet materialsand Stump and van der Heijden [10] analyzed bending andtwisting rods In later research Denoel and Detournay [11]employed multiple scales method Furthermore Denoel andCanor [12] embraced a new perturbation method patchingasymptotics for analysis Nonetheless methods of perturba-tion are not convenient choices for engineering applicationsAll perturbation methods require the matching of solutionsof cable segments outside the boundary layers and beamsegments inside the boundary layers and thematching proce-dures as well as the solutions are extraordinarily complicated

To correct the weaknesses of a conventional perturba-tion method this study proposed a novel catenary bendingmoment equation The core idea originated from the com-parison between two catenary models which consisted ofanchorages with the same horizontal distance and verticalelevation The self-weight and horizontal force between thetwo models were also identical The only difference is thatone model contains bending stiffness but the other does notThe bending moment solution of the model with a bendingstiffness catenary can be expressed by that of the modelwithout a bending stiffness catenary By winding aroundthe difficulty of a conventional perturbation method whichrequires a fourth order differential equation for finding thesolution the proposed equation could directly identify cablesag with the help of bending moment equations

To satisfy the demands of different cable forces and toverify the correctness of the proposed bending momentequation two respective second order differential equationsof parabola and catenary incorporating bending stiffnesswere established and discussed The parabola was expressedby a small sag linear moment-curvature relation and thesolution of the parabola was used to verify the catenary resultThe catenary was expressed by a large sag nonlinearmoment-curvature relation moreover the WKB method (Wentzel-Kramers-Brillouinmethod is named after physicistsWentzelKramers and Brillouin who all developed it in 1926) inthe perturbation method [3ndash5] was adopted to solve thesag differential equations The solution could resolve thecomplexity of boundary layers and does not require thematching of cable segments and beam segments as does ageneral perturbation method With only one equation theproposed method could describe the drastic variations infunctions inside and outside the boundary layer and find thesag and bendingmoment effortlessly Furthermore this study

established the influence of bending stiffness on cablesrsquo staticbehavior and the standard where the parabola could replacethe catenary

2 Static Theories of Catenaries and Parabolaswithout Bending Stiffness

21 The Catenary Solution Figure 1 illustrates a catenaryproblem that considers only tensile forces without the bend-ing moment Self-weight effects are given and set and thecatenary is located in the first quadrantThe equation of staticequilibrium parallel to the 119909-axis is

120597

120597119909

(119879

120597119909

120597119904

) =

120597

120597119909

(119867) = 0 (1)

119879 is the tension 119867 is the horizontal tension and 119867 is aconstantThe equation of static equilibrium perpendicular tothe 119910-axis is

120597

120597119909

(119879

120597119910

120597119904

) =

120597

120597119909

(119867

120597119910

120597119909

) = 119867

1205972119910

1205971199092

= 119898119892

120597119904

120597119909

(2)

119898 is the per-unity-length mass of the cable 119892 is accelerationof gravity and 119904 is the length of the catenary curve 119871

is the horizontal distance between the anchorages 119898119892 isuniformly distributed along the curved cable and 120597119904120597119909 =

radic1 + (119889119910119889119909)2

= radic1 + 11991010158402 Dimensionless parameters are

introduced as follows 119909 = 119909119871 119910 = 119910119871 and 119898119892119871119867 =

119902 where 119902 implies the effects of cable self-weight on theequation of static equilibrium Equation (2) is modified andbecomes

11991010158401015840

=

119898119892119871

119867

radic1 + 11991010158402

= 119902radic1 + 11991010158402 (3)

119910 is the dimensionless sag of the catenary where the primedenote differentiation with respect to 119909 It can be substitutedinto the boundary conditions of the hinges on the two ends(119910(0) = 1 and 119910(1) = tan 120579) and the solution of (3) is thesolution of the catenary Consider

119910 =

1

119902

cosh (119902119909 + 119886 minus

119902

2

) minus cosh (119886 minus

119902

2

) (4)

Consider 119886 = sinhminus1(119902 tan 120579(2 sinh(1199022))) The twohinges are at the same elevation level (120579 = 0

∘ 119886 = 0) (4)can be simplified as

119910 =

1

119902

cosh [119902 (119909 minus

1

2

)] minus cosh (

119902

2

) (5)

22 The Parabola Solution As shown in Figure 2 given thatthe self-weight distributes uniformly along the chordwisedirection 120597119904120597119909 = sec120579 is a constant and 120579 is the chord-wise inclination of the suspension cable The suspensioncable length indicates the shortest distance between the twoanchorages at the two ends Equation (3) is modified as

11991010158401015840

= 119902 sec120579 (6)

Mathematical Problems in Engineering 3

mgds

ds dy

dx

H

L

x

H

A

B

mg

y

RA

L

RB

Ltan120579T120597y

120597s

T120597x

120597s

T120597y

120597s+ d(T120597y

120597s)

T120597x

120597s+ d(T120597x

120597s)

Figure 1 Free-body of static equilibrium of catenaries without bending stiffness

H

mgds

ds dy

dxH

T

H

L

x

H

A

B

mg

y

T120597y

120597s RBLtan120579

120579

L

RA

T120597y

120597s+ d(T120597y

120597s)

Figure 2 Free-body of static equilibrium of a suspended cable with its self-weight distributed uniformly without bending stiffness

The boundary condition is set to the hinges at the twoends and the solution can be expressed by a parabolaConsider

119910 =

119902 sec1205792

(1199092minus 119909) + 119909 tan 120579 (7)

The two hinges are at the same elevation level (120579 = 0∘)

and (7) is simplified as

119910 =

119902

2

(1199092minus 119909) (8)

Equation (6) must accept the small sag cable assumptionMost previous studies regarded a ratio ofmidpoint sag to spanbelow 18 as small sag cable (119910(05119871)119871 lt 18119867 gt 119898119892119871) Onthe contrary a ratio of midpoint sag to span greater than 18is considered as large sag cable (119910(05119871)119871 gt 18 119867 lt 119898119892119871)[1]

3 Conventional Solution of the Equationof Static Equilibrium of Catenaries andParabolas with Bending Stiffness

31 Differential Equation of Catenaries with Bending StiffnessAs shown in Figure 3 the bending moment and shearare incorporated into the catenary The equation of staticequilibrium parallel to the 119909-axis is

120597

120597119904

(119879

120597119909

120597119904

+ 119881

120597119910

120597119904

) = 0 (9)

The horizontal tensile force 119867 = 119879(120597119909120597119904) + 119881(120597119910120597119904) isa constant The equation of static equilibrium perpendicularto the 119910-axis is

120597

120597119904

(119879

120597119910

120597119904

minus 119881

120597119909

120597119904

) = 119898119892 (10)

4 Mathematical Problems in Engineering

H x

H

A

B

mg

ymgds

ds dy

dx

M

L

RA

Ltan120579

RB

T120597y

120597sminus V

120597x

120597s+ d(T120597y

120597sminus V

120597x

120597s)

M+ dM

T120597x

120597s+ V

120597y

120597s+ d(T120597x

120597s+ V

120597y

120597s)

T120597y

120597sminus V

120597x

120597s

T120597x

120597s+ V

120597y

120597s

Figure 3 Free-body of static equilibrium of catenaries with bending stiffness

The equilibrium of cable segments yields

119881 =

120597119872

120597119904

(11)

Compilation yields

119867

1205972119910

1205971199092minus

1205972119872

1205971199092

= 119898119892

120597119904

120597119909

(12)

The moment-curvature relation of a large-sag cable is

119872 = 119864119868120581 =

11986411986811991010158401015840

(1 + 11991010158402)32

(13)

119864119868 is flexural rigidity 119864 the elastic modulus and 119868 is thesecond moment of area In (13) bending moments appliedwith compression on the upper part of the cable is regardedas positive By substituting (13) into (12)

119867

1205972119910

1205971199092minus

1205972

1205971199092[119864119868

11991010158401015840

(1 + 11991010158402)32

] = 119898119892radic1 + 11991010158402 (14)

32 Solution of Parabolas with Bending Stiffness Due to thedifficulty in finding a solution to (14) previousmethods couldnot solve catenaries consisting of bending stiffness and largesag In the textbook [1] the most common solution is tosimplify (12) it is a classical small sag simplified solutionusing linear curvature Assuming a tense cable caused by anenormous tensile force neglect the nonlinear terms in (13)and simplify it as119872 = 119864119868119910

10158401015840 Assuming (12) being a parabolasimplify it as

11986711991010158401015840minus 119864119868119910

1015840101584010158401015840= 119898119892 sec120579 (15)

Dimensionless parameter 1205852

= 1198671198712119864119868 is introduced

to show the influence of bending stiffness Equation (15) ismodified as

1199101015840101584010158401015840

minus 120585211991010158401015840

= minus1205852119902 sec120579 (16)

Boundary conditions of hinges (119910(0) = 0 119910(1) = tan 120579and 119910

10158401015840(0) = 119910

10158401015840(1) = 0) are substituted into the solution to

(16) yielding a sag equation that includes hinges and excludesbending moments at the two ends Consider

119910 =

119902 sec1205792

(1199092minus 119909) + tan 120579119909

+

119902 sec1205791205852

[1 minus cosh (120585119909) + tanh(

120585

2

) sinh (120585119909)]

(17)

The two hinges are at the same elevation level (120579 = 0∘)

and (17) is simplified as

119910 =

119902

2

(1199092minus 119909) +

119902

1205852[1 minus cosh (120585119909) + tanh(

120585

2

) sinh (120585119909)]

(18)

Dimensionless bending moment equation of hinges on ahorizontal cable is

11991010158401015840

1205852

=

119872

(119867119871)

= 119872

=

119902

1205852[1 minus cosh (120585119909) + tanh(

120585

2

) sinh (120585119909)]

(19)

Bending moments applied with compression on theupper part of the cable are regarded as positive The substitu-tion of the fixed-end boundary condition shows that the slopeof the hinge pivot is not equal to zero but to the slope of thecable (119910(0) = 0 119910(1) = tan 120579 1199101015840(0) = tan 120579 1199101015840(1) = tan 120579)The obtained dimensionless fixed-end bending moment is

119872119860

=

119872119860

(119867119871)

= 119902 sec120579 [minus

1

1205852+

coth (1205852)

(2120585)

]

119872119861=

119872119861

(119867119871)

= 119902 sec120579 [

1

1205852minus

coth (1205852)

(2120585)

]

(20)

Mathematical Problems in Engineering 5

H H

A B

L

x

y

y1

y2MA

RA RB

s1

s2120591 = y2 minus y1

m1gs1 = m2gs2

MB

Figure 4 The comparison model between number 1 and number 2 catenaries

119872119860and 119872

119861are bending moments at the anchorage having

positive values when rotating counterclockwise The sagequation is

119910 =

119902 sec1205792

(1199092minus 119909) + tan 120579119909 +

119902 sec1205792120585

coth(

120585

2

)

times [1 minus cosh (120585119909) + tanh(

120585

2

) sinh (120585119909)]

(21)

For a horizontal parabola (120579 = 0∘) (21) can be simplified

as

119910 =

119902

2

(1199092minus 119909) +

119902

2120585

coth(

120585

2

)

times [1 minus cosh (120585119909) + tanh(

120585

2

) sinh (120585119909)]

(22)

The dimensionless bending moment equation for thefixed-end on a horizontal cable is

11991010158401015840

1205852

=

119872

(119867119871)

= 119872

= 119902

1

1205852+

1

2120585

[sinh (120585119909) minus coth(

120585

2

) cosh (120585119909)]

(23)

Equations (17) and (21) are solutions based on parabolasThis is different from the large sag cable approximation in [7ndash11]

4 Novel Catenary Model of Static Equilibriumwith Bending Stiffness

41 Novel Catenary Bending Moment Equation To overcomethe difficulty of finding a solution using (14) this studyadopted a novel concept to modify (14) Figure 4 is a com-parison model between number 1 and number 2 catenariesBased on (5) number 1 catenary was set to include the tensileforce but not the bending moment Consider

119867

12059721199101

1205971199092

= 1198981119892

1205971199041

120597119909

(24)

1198981is the per-unity-length mass of number 1 catenary 119904

1is

the length of number 1 catenary curve and 1199101is the sag of

number 1 catenary

Based on (12) number 2 catenary was introduced toinclude the shear and bending moment Consider

119867

12059721199102

1205971199092

minus

1205972119872

1205971199092

= 1198982119892

1205971199042

120597119909

(25)

1198982is the per-unity-length mass of number 2 catenary 119904

2is

the length of number 2 catenary curve and 1199102is the sag of

number 2 catenaryThe mechanical relation existing between number 1 and

number 2 catenaries was independent and indirect The twocatenaries had two definitely different shapes 119904

1= 1199042 and

had different masses along the chord 1198981

= 1198982 However

they shared the following in common they had the samehorizontal distance and vertical elevation between the twoanchorages the same identical total cable self-weight and thesame horizontal tensile force 119867

Mathematically (24) can be subtracted by (25) as follows

1205972119872

1205971199092

= 119867(

12059721199102

1205971199092

minus

12059721199101

1205971199092) minus 119892(119898

2

1205971199042

120597119909

minus 1198981

1205971199041

120597119909

) (26)

By integrating (26)

120597119872

120597119909

= 119867(

1205971199102

120597119909

minus

1205971199101

120597119909

) minus 119892 (11989821199042minus 11989811199041) + 1198621 (27)

11989821198921199042= 11989811198921199041and (27) is modified as

120597119872

120597119909

= 119867(

1205971199102

120597119909

minus

1205971199101

120597119909

) + 1198621 (28)

Another integration could obtain the bending momentfunction 119872 Consider

119872 = 119867(1199102minus 1199101) + 1198621119909 + 119862

2 (29)

Equation (29) is a new expression of the cable bendingmoment equation The first characteristic allows it to avoidthe difficulty of (14) which requires a fourth order differentialequation to find the solution instead (29) directly findsthe sag from the bending moment equation The secondcharacteristic is that a relation is not needed between sag 119910

2

and 1199101of the number 1 and number 2 catenaries Simply put

(29) is a mathematical equation that expresses the bendingmoment of number 2 catenary from the sag of number1 catenary The two catenaries happen to have the sameself-weight and horizontal tensile force In other words any

6 Mathematical Problems in Engineering

form of 1199101can be selected to calculate the 119910

2with bending

stiffnessConstants of integration 119862

1and 119862

2are obtained based

on the boundary conditions The two catenaries shared thesame anchorage elevation 119910

2(0) = 119910

1(0) and 119910

2(119871) = 119910

1(119871)

119872119860and 119872

119861are the bending moments at the anchorage that

showpositive valuewhen rotating counterclockwise Bendingmoments when applied with compression on the upper partof the cable are regarded as positive The substitution of119872(0) = minus119872

119860and 119872(119871) = 119872

119861results in 119862

1= (119872119860

+ 119872119861)119871

and 1198622= minus119872

119860 Equation (29) is modified as

119872 = 119867(1199102minus 1199101) + 119872

119860(

119909

119871

minus 1) +

119872119861119909

119871

(30)

The dimensionless bending moment equation is

119872

(119867119871)

= 119872 = 1199102minus 1199101+ 119872119860(119909 minus 1) + 119872

119861119909 (31)

Equation (30) can also be easily derived by examining thestatic equilibrium of the free-body diagram of number 1 andnumber 2 catenaries

42 Verification of the Correctness of the Novel BendingMoment Equation Using the Parabola This study used theparabola to verify (31) Let (7) be 119910

1 11991010158401015840

1= 119902 sec120579 By

substituting 11991010158401015840

1= 119902 sec120579 and 119872 = 119864119868119910

10158401015840

2into (31)

11991010158401015840

2= 1205852(1199102minus 1199101) + 1205852119872119860(119909 minus 1) + 120585

2119872119861119909 (32)

The correctness of (29) can be verified with 1199102(which

is (21)) a solution obtained by substituting the fixed-endboundary condition

43 Novel Catenary Differential Equation The moment-curvature relation of the large sag catenary is119872 = 119864119868119910

10158401015840(1 +

11991010158402)32 and nonlinear terms are not neglected Parameter 120591 is

set to

120591 = 1199102minus 1199101 (33)

120591 represents difference of sag between number 1 and number2 catenaries caused by bending moments 119910

2= 1199101+ 120591 and

11991010158401015840

2= 11991010158401015840

1+ 12059110158401015840 By substituting the results into (31) to obtain

the differential equation of large sag catenary

11991010158401015840

2

(1 + 11991010158402

2)

32=

11991010158401015840

1+ 12059110158401015840

(1 + 11991010158402

2)

32

= 1205852(1199102minus 1199101) + 1205852119872119860(119909 minus 1) + 120585

2119872119861119909

(34)

Given that (4) is 1199101 11991010158401

= sinh(119902119909 + 119886 minus 1199022) and 11991010158401015840

1=

119902 cosh(119902119909+119886minus1199022) In a cable-stayed bridge 95of the cableshave 120585 ge 50 [13] (1+119910

10158402

2)32

asymp (1+11991010158402

1)32 and (1+119910

10158402

2)32 can

be replaced by (1 + 11991010158402

1)32

= [1 + sinh2(119902119909 + 119886 minus 1199022)]

32

=

cosh3(119902119909+119886minus1199022) Subsequently (34) becomes the differen-tial equation with a bending moment sag of 120591 Consider

12059110158401015840

1205852

minus cosh3 (119902119909 minus

119902

2

) 120591

= minus

119902

1205852cosh (119902119909 minus

119902

2

)

+ cosh3 (119902119909 minus

119902

2

) [119872119860(119909 minus 1) + 119872

119861119909]

(35)

The large horizontal tensile force and small bendingstiffness make 1120585

2 much smaller as compared to otherparameters The 120591

10158401015840 in (35) is multiplied by small parameter11205852 to form a paradigmatic boundary layer problem that

involves the multiplication of a highest order derivativewith small parameter 1120585

2 This implies that an enormousfunction variation exists in the differential region of thecatenary anchorage Leaving the differential region functionvariations quickly come to a mild plane and maintain at asteady state The division that signifies the rapid and drasticfunction variation is known as the boundary layer and thedifferential region at the catenary anchorage is known as thethickness of the boundary layer Since (35) did not have ananalytical solution this study adopted the WKB method inthe perturbation method to find the approximation solution

5 WKB Catenary Solution withBending Stiffness

First the homogeneous solution of (35) was found as follows

12059110158401015840

1205852

minus cosh3 (119902119909 + 119886 minus

119902

2

) 120591 = 0 (36)

TheWKB approximation method developed by WentzelKramers and Brillouin [2ndash4] was adopted to find the solutionto (36)

In the classical Sturm-Liouville equation [5] 120591101584010158401205852

+

1199021(119909)120591 = 0 When 119902

1(119909) gt 0 the first order approxima-

tion is 120591 = [1198621sin(120585 intradic119902

1119889119909) + 119862

2cos(120585 intradic119902

1119889119909)] 4radic119902

1

When 1199021(119909) lt 0 the first order approximation is 120591 =

[1198621sinh(120585 intradic119902

1119889119909) + 119862

2cosh(120585 intradic119902

1119889119909)] 4radic119902

1

Equation (36) is 1199021= minuscosh3(119902119909 + 119886 minus 1199022) lt 0 and the

WKB approximation is

120591ℎ=

1198621sinh [120585 int cosh32 (119902119909 + 119886 minus 1199022) 119889119909] + 119862

2cosh [120585 int cosh32 (119902119909 + 119886 minus 1199022) 119889119909]

cosh34 (119902119909 + 119886 minus 1199022)

(37)

Mathematical Problems in Engineering 7

The result of 120585 int cosh32(119902119909 + 119886 minus 1199022)119889119909 in (37) is

120575 = 120585int cosh32 (119902119909 + 119886 minus

119902

2

) 119889119909

=

2120585

3119902

radiccosh (119902119909 + 119886 minus

119902

2

) sinh(119902119909 + 119886 minus

119902

2

)

minus 119894119865 (120601 | 2)

(38)

119865(120601 | 2) is the elliptic integral of the first kind 120601 = 119894(119902119909 + 119886 minus

1199022)2 and 119894 is the imaginary unit Consider

119865 (120601 | 2) = int

120601

0

1

radic1 minus 2sin2120601119889120601

= int

119894119902

2radic1 + 2sinh2 [(119902119909 + 119886 minus 1199022) 2]

119889119909

(39)

11205852 is too small and therefore the nonhomogeneous solution

could be obtained by neglecting 120591101584010158401205852 in (36) Consider

minuscosh3 (119902119909 minus

119902

2

) 120591119901= minus

119902

1205852cosh (119902119909 minus

119902

2

)

+ cosh3 (119902119909 minus

119902

2

)

times [119872119860(119909 minus 1) + 119872

119861119909]

120591119901=

119902

1205852sech2 (119902119909 minus

119902

2

) minus 119872119860(119909 minus 1) minus 119872

119861119909

(40)

The final result is 120591 = 120591ℎ+ 120591119901 Define

120591 =

1198621sinh 120575

cosh34 (119902119909 + 119886 minus 1199022)

+

1198622cosh 120575

cosh34 (119902119909 + 119886 minus 1199022)

+

119902

1205852sech2 (119902119909 + 119886 minus

119902

2

) minus 119872119860(119909 minus 1) minus 119872

119861119909

(41)

The total sag is 119910 = 1199102= 1199101+ 120591 Consider

119910 =

1

119902

cosh (119902119909 + 119886 minus

119902

2

) minus cosh (119886 minus

119902

2

)

+

1198621sinh 120575

cosh34 (119902119909 + 119886 minus 1199022)

+

1198622cosh 120575

cosh34 (119902119909 + 119886 minus 1199022)

+

119902

1205852sech2 (119902119909 + 119886 minus

119902

2

) minus 119872119860(119909 minus 1) minus 119872

119861119909

(42)

For a horizontal catenary (120579 = 0∘ 119886 = 0) (41) can be

simplified as

120591 =

1198623sinh 120575

cosh34 (119902119909 minus 1199022)

+

1198624cosh 120575

cosh34 (119902119909 minus 1199022)

+

119902

1205852sech2 (119902119909 minus

119902

2

) minus 119872119860(119909 minus 1) minus 119872

119861119909

(43)

120575 = (21205853119902)[radiccosh(119902119909 minus 1199022) sinh(119902119909 minus 1199022) minus 119894119865(120601 | 2)]The total sag is

119910 =

1

119902

cosh [119902 (119909 minus

1

2

)] minus cosh (

119902

2

)

+

1198623sinh 120575

cosh34 (119902119909 minus 1199022)

+

1198624cosh 120575

cosh34 (119902119909 minus 1199022)

+

119902

1205852sech2 (119902119909 minus

119902

2

) minus 119872119860(119909 minus 1) minus 119872

119861119909

(44)

A solution could be found by substituting (42) and(44) into the boundary conditions The advantage of thehorizontal catenaries in (44) is their arbitrary abilities to besubstituted into any desired boundary conditions When thecatenaries are hinges and do not possess bending moments119872119861

= 119872119860

= 0 Considering that 119910(0) = 0 and 119910(1) = 0let 1205751

= (21205853119902)[radiccosh(1199022) sinh(1199022) minus 119894119865(1198941199024 | 2)] andsimplify the solution to yield

1198623= 0 119862

4= minus

119902 sech1205751

1205852cosh54 (1199022)

(45)

The substitution of (45) into (43) generates the bendingmoment sag equation for horizontal hinged catenaries asfollows

120591 =

119902

1205852[sech2 (119902119909 minus

119902

2

)

minus

sech 1205751cosh 120575

cosh54 (1199022) cosh34 (119902119909 minus 1199022)

]

(46)

The substitution of (45) into (44) generates the total sagequation for horizontal hinged catenaries as follows

119910 =

1

119902

cosh [119902 (119909 minus

1

2

)] minus cosh (

119902

2

)

+

119902

1205852[sech2 (119902119909 minus

119902

2

)

minus

sech 1205751cosh 120575

cosh54 (1199022) cosh34 (119902119909 minus 1199022)

]

(47)

Boundary conditions of fixed-ends (119910(0) = 0 119910(1) = 01199101015840(0) = 0 1199101015840(1) = 0) are substituted into the solution to (47)

and the obtained constants of integration are

1198623= 0

1198624

=minus

4 sinh (1199022) sech1205751[1205852cosh3 (1199022) minus 2119902

2]

1205852cosh54(1199022) [4120585cosh52(1199022) tanh 120575

1minus3119902 sinh(1199022)]

(48)

119872119861= minus119872

119860 and the bending moment 119872

119860 of the left fixed-

end is

8 Mathematical Problems in Engineering

119872119860

=

sech2 (1199022) sinh (1199022) [41205852cosh3 (1199022) minus 5119902

2] minus 4119902120585cosh52 (1199022) tanh (120575

1)

1205852[4120585cosh52 (1199022) tanh (120575

1) minus 3119902 sinh (1199022)]

(49)

Substitute (48) and (49) into (43) to obtain the bendingmoment sag equation for horizontal fixed-end catenaries asfollows

120591 =

1198624cosh 120575

cosh34 (119902119909 minus 1199022)

+

119902

1205852sech2 (119902119909 minus

119902

2

) + 119872119860 (50)

Substitute (48) and (49) into (44) to obtain the total sagequation for horizontal fixed-end catenaries as follows

119910 =

1

119902

cosh [119902 (119909 minus

1

2

)] minus cosh (

119902

2

)

+

1198624cosh 120575

cosh34 (119902119909 minus 1199022)

+

119902

1205852sech2 (119902119909 minus

119902

2

) + 119872119860

(51)

Substitute (43) into (31) to obtain the bending momentequation of horizontal fixed-end catenaries as follows

119872 =

1198624cosh 120575

cosh34 (119902119909 minus 1199022)

+

119902

1205852sech2 (119902119909 minus

119902

2

) (52)

1198624is illustrated in (45) and (48) Without the necessity of a

general perturbation method [10 11] that requires the match-ing of cable and beam segments the proposed equations (44)and (52) can describe the drastic function variations withinand without the boundary layer with a single equation

6 Results and Discussion

Figure 5 shows a comparison between the total sag of parabo-las and catenaries The figure indicates that under any 119902 and120585 the WKB solution of catenaries derived from (51) wasbetween (5) and (22) At the same 119902 (10) the three solutionsdemonstrated greater differences as 120585 reduced As 119902 = 10

and 120585 = 20 the total sag retrieved by (51) of the cablespan midpoint was minus0105 This corresponds to [12] At thesame 120585 (120585 = 40) the three solutions demonstrated greaterdifferences as 119902 increased When 119902 = 05 (22) shows a goodapproximation of (51)

Fundamental mechanics show that the differencebetween (5) and (8) increased with 119902 Therefore whendiscussing the influence of 120585 on sag the sag of (5) and (8)should be divided by (51) and (22) to compare with sag 120591

caused by the bending momentThe bending moment sag 120591corresponds to the solution of (22) that contains a bendingstiffness parabola Consider

120591 =

119902

2120585

coth(

120585

2

) [1 minus cosh (120585119909) + tanh(

120585

2

) sinh (120585119909)]

(53)

First the influence of bending moment sag 120591 on catenar-ies is evaluated

Define the ratio of spanmidpoint bendingmoment sag tohinged catenaries corresponding to division of (5) and (46)Equation (5) is the sag of the catenary (119909 = 05) and (46)is the bending moment sag equation for horizontal hingedcatenaries (119909 = 05) Consider

1199031= (

119902

120585

)

21 minus sech120575

1cosh54 (1199022)

cosh (1199022) minus 1

(54)

1199031is the ratio of span midpoint bending moment sag to

hinged catenariesUnder four different 119902rsquos (05 1 2 and 3) Figure 6

views the ratio of span midpoint bending moment sag 120591

to catenaries supported by a hinge The ratio was found todrastically fluctuate between 10 le 120585 le 30 At all the 119902 values120591 occupied a higher ratio when 120585 was smaller a ratio thatamounted to 786 120591 occupied a smaller ratio when 120585 waslarger When 120585 = 20 the ratio was only 17 and when120585 = 50 the ratio was only 03 Bending stiffness had effectiveinfluences when 120585 lt 20 in the hinged catenary

Define the ratio of span midpoint bending moment sagto fixed-end catenaries corresponding to division of (5) and(50) Equation (50) is the bending moment sag equation forhorizontal fixed-end catenaries (119909 = 05) Consider

1199032=

119902 (1198624+ 1199021205852+ 119872119860)

cosh (1199022) minus 1

(55)

1199032is the ratio of span midpoint bending moment sag to

fixed-end catenaries1198624is illustrated in (48)119872

119860is illustrated

in (49)Figure 7 demonstrates the fixed-end catenary sag ratio

occupied by spanmidpoint bendingmoment sag 120591Thefigureshowed a variation trend identical to Figure 6 but with agreatly enhanced value When 119902 = 10 120585 = 10 had a 359ratio 120585 = 50 had a 69 ratio and 120585 = 100 had a 34 ratioObviously whether the boundary condition is hinge or fixed-end results in significant differences Fixed-end cables shouldtake bending stiffness influences into account

Define the ratio of spanmidpoint bendingmoment sag ofparabolas to catenaries corresponding to division of (50) and(53) Equation (53) is the bending moment sag equation forhorizontal fixed-end parabolas (119909 = 05) Consider

1199033=

(119902120585) csch (1205852) sinh2 (1205854)1198624+ 1199021205852+ 119872119860

(56)

Mathematical Problems in Engineering 9

0 01 02 03 04 05 06 07 08 09 1

000

Eq (51)Eq (22)Eq (5)

minus012minus010minus008minus006minus004minus002

y

x

(a) 119902 = 10 120585 = 40

0 01 02 03 04 05 06 07 08 09 1

000

minus012minus010minus008minus006minus004minus002

y

x

Eq (51)Eq (22)Eq (5)

(b) 119902 = 10 120585 = 100

0 01 02 03 04 05 06 07 08 09 1

y

x

000

minus025

minus020

minus015

minus010

minus005

Eq (51)Eq (22)Eq (5)

(c) 119902 = 20 120585 = 40

0 01 02 03 04 05 06 07 08 09 1

000

minus005minus006

minus004minus003minus002minus001

y

x

Eq (51)Eq (22)Eq (5)

(d) 119902 = 05 120585 = 40

Figure 5 Comparisons between the total sag of parabolas and catenaries

10 20 30 40 50 60 70 80 90 100000

002

004

006

008

r1

120585

q = 05

q = 10q = 20

q = 30

Figure 6The ratio of spanmidpoint bendingmoment sag to hingedcatenaries (119909 = 05)

1199033is the ratio of span midpoint bending moment sag of

parabolas to catenaries 1198624is illustrated in (48) 119872

119860is

illustrated in (49)The ratio differences between bending stiffness and solu-

tions of parabolas and catenaries were compared using (50)and (53) Figure 8 signifies that the ratio varied with 120585 underthe four different 119902 values (05 1 2 and 3) The parabolasolution was not applicable because a large 119902 led to increaseddifferences on the contrary a smaller 119902 led to reduceddifferences and a parabola solution could replace a catenary

10 20 30 40 50 60 70 80 90 100

r2

00

01

02

03

04

120585

q = 05

q = 10q = 20

q = 30

Figure 7The ratio of span midpoint bending moment sag to fixed-end catenaries (119909 = 05)

solution When 119902 = 05 the ratio varied between 1019and 1035 Equation (22) was a good approximation of (51)When 119902 = 10 the ratio varied between 1076 and 114When 119902 = 30 the ratio could be as high as 2427 and (22)was an inappropriate selection

Figure 9 shows the comparison between the parabola andcatenary fixed-end bending moments As (20) was based ona small sag linear hypothesis 119872

119860varied linearly with 119902

However (49) varied nonlinearly

10 Mathematical Problems in Engineering

10 20 30 40 50 60 70 80 90 100

r3

q = 05

q = 10q = 20

q = 30

120585

10

15

20

25

Figure 8 The ratio differences between bending stiffness andsolutions of parabolas and catenaries

Eq (20)Eq (49)

00 05 10 15 20 25 30000

001

002

003

004

q

MA

Figure 9The comparison between the parabola and catenary fixed-end bending moments (120585 = 30)

Define the ratio of fixed-end bending moments ofparabolas to catenaries corresponding to division of (20) and(49) Consider

1199034=

119902 sec120579 [minus11205852+ coth (1205852) (2120585)]

119872119860

(57)

1199034is the ratio of fixed-end bending moments of parabolas to

catenaries 119872119860is illustrated in (49)

Figure 10 adopts (20) and (49) to evaluate the ratiosof influences of bending stiffness on fixed-end bendingmoments The figure showed a variation trend extremelysimilar to Figure 8 When 119902 = 05 the ratio varied between1025 and 1035 and (20) was a good approximation of(49) When 119902 le 05 the parabola solution demonstrated thatratio differences increased with 119902 and when 119902 = 10 the ratiofluctuated between 1101 and 1143When 119902 = 30 the ratiocould be as high as 250 and the selection of (20) was aninappropriate choice

10 20 30 40 50 60 70 80 90 100

r4

120585

10

15

20

25

q = 05

q = 10q = 20

q = 30

Figure 10The ratios of influences of bending stiffness on fixed-endbending moments

Figure 11 shows the distribution of parabola and catenarybending moments The figure indicates that 120585 changed theboundary layer thickness and 119872

119860 At the same 119902 (119902 = 10)

increased 120585 and reduced 119872119860 This resulted in a reduced

thickness of boundary layers and a more drastic variationin bending moment functions 119902 increased the differencebetween (23) and (52) When 119902 = 05 the catenary solutioncould be replaced by the parabola solution In addition thetwo solutions of 119902 = 30 showed significant difference invalues and shapes and in such a circumstance the parabolawas not applicable

7 Conclusions

(1) This study proposed a novel large sag catenary bend-ing moment expression that allows finding the sagdirectly from the bending moment equation andfinding the WKB catenary solution using the WKBmethod The matching of cable and beam segmentsis no longer necessary thereby overcoming the com-plexity of boundary layers Moreover with only asingle equation the proposed method could simulta-neously describe the drastic function variations insideand outside the boundary layerThemethod providesa simple calculation of cables with bending stiffnessand large sag and fulfills the engineering needs fortensile cable fatigue stress analysis and estimation ofsuspension cable sag for bridge hoisting

(2) Statistical analysis revealed a significantly differentinfluence on the behavior of cables with bending stiff-ness given that the boundary condition was hinges orfixed-ends Bending stiffness was regarded as effectivewhen 120585 lt 20 for the hinged catenary Howeverthe fixed-end catenary should always consider theinfluence of bending stiffnessWhen catenary 119902 le 05the horizontal tensile force was massive (119867 ge 2119898119892119871)The parabola shows a very good approximation ofcatenary It could replace theWKB catenary solutionIn addition 120585 had the ability to change the boundarylayer thickness where bending moments were dis-

Mathematical Problems in Engineering 11

02 04 06 08 10

minus0012

minus0010

minus0008

minus0006

minus0004

minus0002

M

x

Eq (52)Eq (23)

(a) 119902 = 10 120585 = 40

02 04 06 08

minus0005

minus0004

minus0003

minus0002

minus0001

M

x

Eq (52)Eq (23)

10

(b) 119902 = 10 120585 = 100

02 04 06 08

minus0020

minus0015

minus0010

minus0005

M

x

10

Eq (52)Eq (23)

(c) 119902 = 20 120585 = 40

Eq (52)Eq (23)

02 04 06 08

minus0005

minus0006

minus0004

minus0003

minus0002

minus0001

M

x

10

(d) 119902 = 05 120585 = 40

Figure 11 The distribution of parabola and catenary bending moments

tributed and fixed-end bending moment Higher 120585

lowered the fixed-end bending moment reduced theboundary layer and increased the drastic variation ofthe bending moment functions

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] H M Irvine Cable Structures Dover Publications 1992[2] J J Burgess ldquoBending stiffness in a simulation of undersea

cable deploymentrdquo International Journal of Offshore and PolarEngineering vol 3 no 3 pp 197ndash204 1993

[3] E J Hinch Perturbation Methods Cambridge University PressCambridge UK 1991

[4] J Kevorkian and J D ColeMultiple Scale and Singular Pertur-bation Methods Springer Berlin Germany 1996

[5] A H Nayfeh Introduction to Perturbation Techniques JohnWiley amp Sons New York NY USA 2011

[6] M S Triantafyllou and G S Triantafyllou ldquoThe paradox of thehanging string an explanation using singular perturbationsrdquoJournal of Sound and Vibration vol 148 no 2 pp 343ndash351 1991

[7] PWolfe ldquoThe effect of bending stiffness on inextensible cablesrdquoInternational Journal of Engineering Science vol 30 no 9 pp1187ndash1192 1992

[8] H M Irvine ldquoLocal bending stresses in cablesrdquo InternationalJournal of Offshore and Polar Engineering vol 3 no 3 pp 172ndash175 1993

[9] D M Stump and W B Fraser ldquoBending boundary layers ina moving striprdquo Nonlinear Dynamics vol 21 no 1 pp 55ndash702000

[10] D M Stump and G H M van der Heijden ldquoMatched asymp-totic expansions for bent and twisted rods applications for cableand pipeline layingrdquo Journal of Engineering Mathematics vol38 no 1 pp 13ndash31 2000

[11] V Denoel and E Detournay ldquoMultiple scales solution for abeam with a small bending stiffnessrdquo Journal of EngineeringMechanics vol 136 no 1 Article ID 006001QEM pp 69ndash772010

[12] V Denoel and T Canor ldquoPatching asymptotics solution ofa cable with a small bending stiffnessrdquo Journal of StructuralEngineering vol 139 no 2 pp 180ndash187 2013

[13] A B Mehrabi and H Tabatabai ldquoUnified finite differenceformulation for free vibration of cablesrdquo Journal of StructuralEngineering vol 124 no 11 pp 1313ndash1322 1998

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