Dynamics of an Elastic Cable Carrying a Moving Mass Particle

Nonlinear Dynamics 33: 11–32, 2003.© 2003 Kluwer Academic Publishers. Printed in the Netherlands.

Dynamics of an Elastic Cable Carrying a Moving Mass Particle

M. AL-QASSABMechanical Engineering Department, University of Bahrain, P.O. Box 32038, Manama, Kingdom of Bahrain;E-mail: [email protected]

S. NAIRMechanical, Materials and Aerospace Engineering Department, Illinois Institute of Technology,10 West 32nd St., Chicago, IL 60616, U.S.A.

J. O’LEARYCivil and Architectural Engineering Department, Illinois Institute of Technology, 3300 South Federal Street,Chicago, IL 60616, U.S.A.

(Received: 10 March 2003; accepted: 17 June 2003)

Abstract. The dynamic behavior of an elastic catenary cable due to a moving mass along its length is investigated.The equations of motions are derived using the Hamilton’s principle for general supports that include the horizontaland inclined cables with small and large sags and for variable velocity of the moving mass. Those equations ofmotions are in general nonlinear partial differential equations due to the initial curvature of the cable. The equationsare also complex due to the presence of three different types of accelerations of the moving mass. Those are thenormal, Coriolis and centrifugal accelerations. Therefore, we used the Galerkin procedure with sine function(Fourier representation) and anti-derivative functions of the compactly supported wavelets as trial basis and useddirect integration methods to integrate the discretized equations of motions. Newton–Raphson method is used foriterations. Several examples are studied and the results as obtained by Fourier and wavelet representations arecompared. Because of the localization feature, wavelets are proven to minimize the spurious oscillations speciallythose appearing in the cable tension.

Keywords: Elastic cable, cable dynamics, moving mass, wavelets.

1. Introduction

Cables are used in ski lifts and tramways as a means of transportation. Stationary cables thatare capable of carrying rocket propelled trolleys have been used by the Sandia Laboratories,Albuquerque, to simulate aircraft flight for evaluation of airborne equipment and to preciselytarget items for impact testing [1]. A similar idea of aircraft flight simulation will be conductedat the White Sands Missile Range. A Kevlar cable of 4572 m in length will be supportedbetween two mountain tops and a helicopter will be used to travel along the cable length forflight testings.

Smith [2] considered a stretched stationary string that vibrates in the transverse directiononly due to a mass particle moving along its span with a constant horizontal velocity com-ponent. The dynamic tension in the string is neglected because of the large initial strain. Thegeneral solution of the vibrating system is given by the superposition solutions of the staticdeflection, free vibration, and the forced vibration due to the applied forces. The author re-duced the problem to an integral equation with the interaction force between the string and themass particle as an unknown. An approximate solution is found based on constant interaction

12 M. Al-Qassab et al.

force equals to the weight of the moving mass. This solution is valid for small velocity ratios1

much lower than 1/3. Another approximate solution is found based on a massless string; i.e.the string inertia is neglected. The exact solution of the displacement of the moving masshave been obtained for values of the velocity ratio greater than 1/3, which means that themass is sufficiently fast to exit the string before the initial disturbance reaches it after twicetraversing the length of the string. Kanninen and Florence [3] considered two forces of thesame magnitude traveling in opposite directions on a stretched infinite string. The forces arethe result of the use of explosives located over the string. The Laplace transform is usedfor the solution. The velocity and displacement distributions for the supersonic and subsonicvelocities are given. Flaherty [4] studied the effect of a moving force with varying speed onthe transverse velocity of an infinite string. The solution of the wave propagation equationunder the moving force was obtained by using the Laplace and Fourier transformations. Thetwo cases of decreasing and increasing velocities of the moving force are considered. In bothcases the velocities pass through the characteristic speed of the string. The results of the bothcases showed that the singularity occurs at the moment the velocity of the moving force passesthrough the characteristic speed. Yen and Tang [5] used the perturbation method to find thenonlinear vibration of an infinite elastic string supported by an elastic foundation due to a loadmoving in the vicinity of the critical speed of the linear string. Both the axial and the transversedisplacements are expanded in power series of the dimensionless load intensity. Although,the solutions have been strictly given for the subcritical and supercritical velocities only, theauthors have shown that each solution can be valid at the critical velocity and extended tothe region of the other solution. An example is given which showed that solutions at thelinear critical speed exist and for velocities greater than the critical speed the modes havetwo different amplitudes. Fryba [6] considered the vibration of a massless string subjected toa moving mass. The deflection of the string was represented by the Green’s function. Thenthe problem was reduced to a nonhomogeneous differential equation for the deflection ofthe mass. The nonhomogeneous term contains a parameter that relates the string tension andlength and the moving mass and its speed. The solution is given for different cases of thatparameter. Sagartz and Forrestal [7] studied the response of an infinitely long stretched stringloaded by an accelerating force. The solution of the wave propagation equation that governsthe motion of the string due to the applied force was obtained by employing the Laplaceand Fourier transforms. An expression for the kink angle was provided and its location wasdiscussed for an example of constant acceleration of magnitude equal to the wave speed ofthe string. Forrestal et al. [8] considered a slightly curved stretched cable carrying a massmoving with constant acceleration. The maximum speed sought for the mass was less thanthe propagation speed. They assumed that there are two kinds of moving forces acting on thecable. The first force is the weight of the mass and the second force is the reaction from thenormal acceleration of the mass. The solution due to the latter force is given and added to thesolution given by Sagartz and Forrestal [7] due to a moving force. The authors have given anexpression to the kink angle and plots illustrating the deflection of the cable at several timevalues. A comparison of the theoretical and experimental results of the deflection of a selectedpoint on the cable showed an accurate prediction. Rodeman et al. [1] obtained the solution ofan infinitely long stretched cable subjected to a mass particle moving at constant horizontalacceleration taken into consideration the unknown reaction force between the mass and thecable. The cable deflection is given by an integral form in term of the unknown reaction from

1 Velocity ratio is the velocity of the moving mass to the characteristic speed of the wave in the string.

Dynamics of an Elastic Cable Carrying a Moving Mass Particle 13

which the delay differential equation is found for the deflection of the moving mass. Thedelay differential equation is solved numerically and asymptotically. The small parameterused in the asymptotic solution is the ratio of the moving mass inertia to the cable tension.These solutions have been used to obtain the cable response history. The graphical resultsshowed that as the mass velocity passes through the wave speed to supersonic speed the cabledeflection experiences two opposite jumps which could not be predicted by the moving forcemodel. D’Acunto [9] studied the free boundary problem of an infinitely long string subjectedto a moving load. He considered that both the string vibration and the motion of the movingload are influenced by each other and they are unknowns. The author has reduced the problemto the one that describe the motion of the moving mass and he discussed the existence anduniqueness of the solutions. Wu and Chen [10] used the finite element method to study thedynamics of a suspended cable due to a moving load. They considered horizontal cables withsag to span ratios less than 1/8. The updated Lagrangian formulation is used to derive theequations of motions of the cable structure. Then the finite element is used to discretize thedisplacements and obtained the property matrices which they solved by the Newmark directintegration scheme. For the spacial case of the free vibration of a bare cable, they found thenatural frequencies and a comparison is made with results obtained by Henghold et al. [11]given in a form of table. They found a good agreement. The authors showed by mean of graphsthe response of the dynamic shape to changing the cable stiffness ratio, the moving mass ratioand the nondimensional speed of the moving mass. Pierucci [12] studied the response of aninfinite string due to a harmonic force moving at constant speed. The equation of motion issolved by Fourier transform and discussed the wave scattering in the string. He mentionedthat there are waves on the string that are traveling to the left and to the right. When the forceis moving at subsonic speed there is a wave traveling ahead of the force and a wave travelingto the left away from the force. At supersonic speed there are no traveling waves ahead of theforce, instead all the waves move to the left. Tadjbakhsh and Wang [13] studied the vibrationsof an inclined taut cable with a riding, accelerating mass. Their analysis is for the case ofinextensible cables. Although, they used a catenary cable for the static analysis, their model isgood for cables that have no minimum sag, i.e. the lower point of the cable dose not go belowthe horizontal line that passes through the lower support. They used the Galerkin method toapproximate the dynamic displacements of the cable in the normal and tangential directions ofcable configuration. The basis functions used for the Galerkin procedure are the sine functionfor the normal displacement and the integral of the curvature of the cable and sine for thetangential displacement, while the coefficients of the series, which are dependent on time, arethe same for the both displacements.

In this paper, we aim to study a wider case of elastic cable carrying a moving mass. Wewill allow horizontal and inclined cables with arbitrary sag to span ratios. The moving masscould be small or large with constant velocity, constant acceleration or variable velocity andvariable acceleration. The dynamic equations will be derived from Hamilton’s principle whichare in general nonlinear. These equations will be solved using the Galerkin procedure withFourier and wavelets analysis. The dynamic configuration will be measured from the catenaryequilibrium of the cable. The location of the mass, the response of the cable and the cabletension will be shown graphically for different cases. Discussion on the convergence of themethod and a comparison between the classical Galerkin procedure and the wavelet-Galerkinmethod will be given.


Figure 1. The static and dynamic configurations.

2. General Formulation

The equations of motions of a vibrating cable due to a moving mass will be derived fromthe Hamilton’s principle. Consider a cable suspended between two points at any arbitrarylevels. All dependent variables will be referenced to the stretched static shape described bythe Cartesian coordinates x(s) and y(s) where s is the spatial variable measured along thecable length of the stretched configuration. The dynamic displacements in the longitudinaland transverse directions are described by u(s, t) and v(s, t), respectively, where t is time(Figure 1). Denoting by so(t) the distance the moving mass has traveled along the cable length,its position vector po(t) can be represented as follows:

po(t) = [x (so)+ u (so, t)] ı +[y (so)+ v (so, t)

] ,

where ı and are the fixed global unit vectors in the horizontal and vertical directions, re-spectively. For the general case of the characteristic motion of the moving mass, we assumethat the mass has a local velocity dso (t) /dt and a local acceleration d2so (t) /dt2 along the arclength of the cable with a direction tangent to the cable at the location of the moving mass.Then the velocity vector of the moving mass will be

vo (t) = dpodt

=(∂u (so, t)

∂t+ dso

dt

(dx (so)

dso+ ∂u (so, t)

∂so

))ı

+(∂v (so, t)

∂t+ dso

dt

(dy (so)

dso+ ∂v (so, t)

∂so

)) . (1)


Equation (1) and the velocity vector of the cable at any arbitrary point on the cable, which is∂u(s, t)/∂t ı+∂v(s, t)/∂t , can be used to obtain the kinetic energy of the system as follows:

K = 1

2

l∫0

ρ(u2 + v2)ds

+ 1

2

l∫0

M δ(s − so(t))((u+ so(x′ + u′))2 + (v + so(y′ + v′))2

)ds, (2)

where dot means partial differentiation with respect to time, t , and the prime means partialdifferentiation with respect to the space variable s, l is the cable length, ρ is the mass of thecable per unit length, M is the mass of the moving particle and δ(s) is the Dirac delta. If dSis the current arc length of the cable, the Lagrangian strain ε (s, t) becomes

ε (s, t) = dS − ds

ds=

√(x′ + u′)2 + (y′ + v′)2 − 1. (3)

The potential energy can be written as

� =l∫

0

(T ε + 1

2EAε2 − ρ g v −M g δ (s − so (t)) v

)ds, (4)

where T is the cable static tension and it is a function of s, g is the acceleration of gravity, Eis the modulus of elasticity and A is the cross sectional area of the cable. Using Equation (2)and (4) in the Hamilton’s principle

δ

t2∫t1

(K −�) dt = 0 (5)

and performing the variation on u and v and integrating by parts lead tot2∫t1

l∫0

{(ρ u+M δ(s − so(t))[u+ 2 so u

′ + s2o (x

′′ + u′′)+ so (x′ + u′)])δu

+ (ρ v +M δ(s − so(t))[v + 2 so v′ + s2

o (y′′ + v′′)+ so (y′ + v′)])δv

+ T + EA εε + 1

[(x′ + u′)δu′ + (y′ + v′)δv′] − [ρ +M δ(s − so(t))]g δv}

ds dt = 0. (6)

Note that the boundary terms of integration have been set equal to zero, which arel∫

0

{[ρ u+M δ(s − so)(u+ so(x′ + u′))]δu

+ [ρ v +M δ(s − so)(v + so(y′ + v′))]δv}∣∣t2t1

ds = 0

t2∫t1

M δ(s − so){[u+ so(x′ + u′)]δu+ [v + so(y′ + v′)]δv}∣∣l0 dt = 0


and the following relation has been used

∂δ (s − so)∂t

+ so ∂δ (s − so)∂s

= 0.

The boundary conditions of a fixed cable at its two ends are given by

u(0, t) = v(0, t) = 0,

u(l, t) = v(l, t) = 0, (7)

and the initial conditions are

u (s, 0) = ∂u (s, 0)

∂t= 0,

v (s, 0) = ∂v (s, 0)

∂t= 0. (8)

3. Static Configuration

The governing equations of the static shape of the bare cable are given as follows:{T x′}′ = 0, (9)

{T y′}′ + ρ g = 0. (10)

By referring to Figure 1, the boundary conditions for x and y are as follows:

x (0) = 0, x (l) = b, (11)

y (0) = 0, y (l) = −h. (12)

Solving (9–12) leads to the catenary shape of the cable.

4. Discretization

The Galerkin method is used to obtain approximate dynamic displacements of the cable. Inthe procedure of the Galerkin method, basis functions that satisfy the boundary conditions areused. Thus we assume series solutions for u (s, t) and v (s, t)

u (s, t) =∞∑m=1

Um (t) φm (s) , (13)

v (s, t) =∞∑m=1

Vm (t) φm (s) , (14)

where φm (s) is the basis function satisfying the boundary conditions given in Equations(7). Um (t) and Vm (t) are the time dependant functions for the longitudinal and transversedisplacements, respectively. They are the only functions that admit variations, hence δu =φmδUm and δv = φmδVm. Therefore, when substituting Equations (13) and (14) in Equation


(6), sets of coupled ordinary differential equations can be obtained for arbitrary functions δUmand δVm. In the following equations, the summation symbol has been omitted for brevity andthe repeated indices indicate summation.

Longitudinal direction (for arbitrary δUm)ρ

l∫0

φmφn ds +M φm(so)φn(so)

Un(t)

+ 2M so φm(so)φ′n(so) Un(t)

+M φm(so)(so φ′n(so)+ s2

o φ′′n(so)) Un(t)

+l∫

0

T + EA εε + 1

(x′ + Unφ′n)φ

′m ds

= −M φm(so)(so x′(so)+ s2

o x′′(so)). (15)

Transverse direction (for arbitrary δVm)ρ

l∫0

φmφn ds +M φm(so)φn(so)

Vn(t)

+ 2M so φm(so)φ′n(so) Vn(t)

+M φm(so)(so φ′n(so)+ s2

o φ′′n(so)) Vn(t)

+l∫

0

{T + EA εε + 1

(y′ + Vnφ′n)φ

′m − ρ g φm

}ds

= −M φm(so)(so y′(so)+ s2

o y′′(so)− g), (16)

where n,m = 1, 2, . . . , N . N being the maximum number in the series. The right handsides in the above equations represent the external forces due to the presence of the movingmass. Those forces are of two kinds, namely the weight of the moving mass which appears inEquation (16) only and the forces due to the motion of the moving mass as centrifugal forces.

In order to nondimensionalize the above equations we define the following dimensionlessparameters

s∗ = s/ l, x∗ = x/l, y∗ = y/l, U ∗m = Um/l, V ∗

m = Vm/l,s∗o = so/ l, t∗ = t

√To/ρl2, µ = M/ρl

T ∗ = T /To, α = EA/To, σ = ρgl/To, β = ρgl/EA.Introducing the above parameters in Equations (15) and (16) to get in the longitudinal direction

1∫0

φmφn ds∗ + µ φm(s∗o )φn(s∗o ) U ∗

n (t∗)


+ 2µ s∗o φm(s∗o )φ

′n(s

∗o ) U

∗n (t

∗)

+ µ φm(s∗o )(s∗o φ′n(s

∗o )+ s∗

2

o φ′′n(s

∗o )) U

∗n (t

∗)

+1∫

0

T ∗ + α εε + 1

(x∗′ + U ∗nφ

′n)φ

′m ds∗

= −µ φm(s∗o )(s∗o x∗′(s∗o )+ s∗

2

o x∗′′(s∗o )) (17)

and in the transverse direction 1∫

0

φmφn ds∗ + µ φm(s∗o )φn(s∗o ) V ∗

n (t∗)

+ 2µ s∗o φm(s∗o )φ

′n(s

∗o ) V

∗n (t

∗)

+ µ φm(s∗o )(s∗o φ′n(s

∗o )+ s∗

2

o φ′′n(s

∗o )) V

∗n (t

∗)

+1∫

0

{T ∗ + α εε + 1

(y∗′ + V ∗n φ

′n)φ

′m − σ φm

}ds∗

= −µ φm(s∗o )(s∗o y∗′(s∗o )+ s∗

2

o y∗′′(s∗o )− σ ). (18)

In order to find the unknown functions U ∗(t∗) and V ∗(t∗), those equations have to be integ-rated over time starting at t∗ = 0 to the moment the mass exit the cable at s∗o = 1. Beforemaking any attempt to do the time integration, it is more convenient to put Equations (17) and(18) in a matrix form as shown below:

M γ + C γ + K γ + R(γ ) = F, (19)

where

γ ={

UV

}; U = [U ∗

1 U ∗2 · · · U ∗

N ]T ; V = [V ∗1 V ∗

2 · · · V ∗N ]T ; (20)

M =[

M 00 M

]; Mmn =

1∫0

φmφn ds∗ + µ φm(s∗o )φn(s∗o ); (21)

C =[

C 00 C

]; Cmn = 2µ s∗o φm(s

∗o )φ

′n(s

∗o ); (22)

K =[

K 00 K

]; Kmn = µ φm(s∗o )(s∗o φ′

n(s∗o )+ s∗2

o φ′′n(s

∗o )); (23)

R(γ ) ={

Rx(U,V)Ry(U,V)

}; (24)


Rx m =1∫

0

T ∗ + α εε + 1

(x∗′ + U ∗nφ

′n)φ

′m ds∗; (25)

Ry m =1∫

0

{T ∗ + α εε + 1

(y∗′ + V ∗n φ

′n)φ

′m − σφm

}ds∗; (26)

F ={

Fx

Fy

}; Fx m = −µ φm(s∗o )(s∗o x∗′

(s∗o )+ s∗2

o x∗′′(s∗o ));

Fy m = −µ φm(s∗o )(s∗o y∗′(s∗o )+ s∗

2

o y∗′′(s∗o )− σ ). (27)

Equation (19) is the nonlinear form of the cable vibrations due to the moving mass representedin the standard matrix form. It is clear that the matrices (21–23) and the force vectors (27) aredependent on the mass location, s∗o , consequently they are dependant on time, t∗. It means thatthose matrices and vectors have to be evaluated at each time step of the numerical integration.The size of the matrices in Equation (19) is 2N × 2N and the size of the submatrices inEquations (21–23) is N × N . The size of the vectors in Equations (24–27) is 2N × 1 and thesize of their subvectors is N × 1, where N is as defined before.

Equations (19–27) will be integrated numerically using trapezoidal rule with the Newton–Raphson iterative method, where the Jacobian matrix needed in the iteration scheme is givenbelow

J(γ in+1) = ∂ R(γ in+1)

∂ γ in+1

(28)

=[

J1 J2

JT2 J3

]; (29)

and

J1 nm =1∫

0

{T∗ + α εε + 1

+ α − T ∗

(ε + 1)3(x∗′ + U ∗

k φ′k)

2}φ′nφ

′m ds∗, (30)

J2 nm =1∫

0

α − T ∗

(ε + 1)3(x∗′ + U ∗

k φ′k)(y

∗′ + V ∗k φ

′k)φ

′nφ

′m ds∗, (31)

J3 nm =1∫

0

{T∗ + α εε + 1

+ α − T ∗

(ε + 1)3(y∗′ + V ∗

k φ′k)

2}φ′nφ

′mds∗. (32)


5. Fourier Representation

In this section, we use φn = sin(nπs/ l) as the basis function. Therefore, Equations (13) and(14) become

u(s, t) =∞∑m=1

Um(t) sin(mπsl

), (33)

v(s, t) =∞∑m=1

Vm(t) sin(mπsl

). (34)

The boundary conditions on the dynamic displacements u and v are satisfied by the aboverelations. The normalized form of (33) and (34) will be used to obtain the matrices.

6. Wavelet Representation

A wavelet is a localized function on the real line R [14–16]. It occupies only an interval, say[a, b], and outside this interval it is zero. This interval is determined by translation and dilationof the wavelet along the real line. If ψ ∈ L2[R] is a wavelet then its translation and dilation isrepresented by

ψjk(x) = 2j/2ψ(2j x − k),where j ,k ∈ Z and Z is the set of all integers. The integer k is the parameter that is responsibleof translation and the integer j is the parameter that is responsible of dilation. By changing kthe wavelet will move along the real line R and by changing j the width of the wavelet willchange. If j is greater than zero then the wavelet will be narrow and of high frequency. If j isless than zero the wavelet will be wide and of low frequency. In comparison with the Fouriertransform we see that the wavelets transform is characterized by two parameters which are jand k, while Fourier transform is characterized by one parameter say n, in einx . Fourier trans-form is restricted to the use of sine and cosine functions only, whereas in wavelets transformthere are infinitely many functions. Both transforms have the similarity of being powerfulin representing functions that might be a signal, an image, a solution of partial differentialequation, etc. But, the Fourier transform has a disadvantage in representing functions that areof singular nature and the Gibbs phenomenon is expected. The main advantage of waveletsis that they are local functions and the ability to overcome the Gibbs phenomenon is highlypossible.

The basis functions we adapt here are the anti-derivative wavelets that satisfy the boundaryconditions of our problem. The anti-derivative wavelets are derived by Xu and Shann [17]to smooth out the wavelets and to easily satisfy the boundary conditions. They consideredthe anti-derivative of Daubechies compactly supported wavelets to represent functions in theSobolev frames. For Dirichlet boundary conditions the anti-derivative wavelet is defined as

2jk(x) =x∫

0

ψjkds − x ψjk, for 0 ≤ x ≤ R, (35)


where

ψjk = 1

R

R∫0

ψjk ds,

R = 2p − 1 and p is the wavelet order. For the evaluation of the integral term in (35) we canuse the algorithm given by Chen et al. [18], so that

2jk(x) = 2−j/2[θ(2j x − k)− θ(−k)− x

R(θ(2jR − k)− θ(−k))

],

for 0 ≤ x ≤ R, (36)

where

θ(x) =x∫

0

ψ ds, j ≥ −1

and

k ∈ Dj ⇐⇒{

1 − R ≤ k ≤ p − 1 if j = −1,

p − R ≤ k ≤ 2jR − p if j ≥ 0.(37)

The approximate solutions of the dynamic displacements are represented by

uJ (s, t) =J−1∑j=−1

∑k∈Dj

Ujk(t)2jk(s), (38)

vJ (s, t) =J−1∑j=−1

∑k∈Dj

Vjk(t)2jk(s), (39)

where2jk(s) is the family of the anti-derivative wavelets given in Equation (36), J ≥ 0 andDjis given in Equation (37). Equations (38) and (39) are referred to as multilevel representationof displacement components with total number of terms determined fromN = 2J+1R+R−p.

Although the integration scheme of the trapezoidal rule is unconditionally stable, it requiressmall time steps for better accuracy. We found that is the case when using the trapezoidal rulefor the wavelets representation. Since we are using Daubechies wavelets, the time step has tobe one of the dyadic points (k/2j ) and a smaller time step means a larger computer memoryto store the wavelet. Therefore, we are going to use the Newmark method that has artificialdamping to overcome the numerical noise resulting from the integration.

7. Results and Discussions

7.1. VIBRATION OF A BARE CABLE

Consider a cable of length l = 2,002.37 m that has level supports, h = 0, a span b = 2,000 m,elastic parameter β = ρgl/EA = 1/5000 and α = 844. The linear theory of vibrationof parabolic cables as developed by Irvine and Caughey [19] predicts that the first natural


Figure 2. The first natural frequency vs the number of terms, N , using Fourier and wavelets representations.

frequency is ω(1) = 5.34979. Using the anti-derivative wavelet of Daubechies’ compactlysupported wavelets of the order of p = 4, Figure 2 shows the first natural frequency for bothFourier and wavelets representations plotted against the number of terms, N , and comparedwith the value by the linear theory. Both methods are showing fast convergence with a smallnumber of terms.

7.2. VIBRATION OF A HORIZONTAL CABLE CARRYING A MOVING MASS

Now we consider that the mass is moving at constant speed, dso/dt = Vo starting from theleft support of the cable. Thus, at t = 0, so = 0. In dimensionless form, the mass velocitybecomes

ds∗odt∗

= V ∗o

√σ ,

where V ∗o = Vo/√gl. Now we compare our solution with the finite element solution given by

Wu et al. [10] who used the updated Lagrangian formulation and Newmark direct integrationscheme to obtain the dynamic configuration of cable carrying a moving mass. For µ = 1.0and V ∗

o = 1.0, Figure 3 shows the path of the moving mass along the span of the cable usingN = 30. Shown in the figure also the results from [10] for the same velocity ratio and forµ = 1.0 and µ = 2.0. We notice that our solution, which is for µ = 1.0, depicts the onefor µ = 2.0 of reference [10]. Although Wu and Chen [10] reproduce, as a limiting case,the solution of the example given by Fryba [6], which is a force moving at constant speedon a string, but this is different than a moving mass because of the fact that a moving forcedoes not have a coupling acceleration with the string. Besides, our solution reproduced theexact natural frequencies of a bare cable. We notice in Figure 3 that there is a local maximain our solution at so/ l = 0.655 while their solution for µ = 2.0 gives a local maxima atso/ l = 0.8. This shift of the local maxima might be attributed to the fact that we have usedthe full nonlinear terms, while they used only the linear and quadratic terms. Therefore, wethink that their mass ratio has a factor of two missing.


Figure 3. The path of the moving mass along the span of the cable obtained by the Fourier method for µ = 1.0,and from [10] for µ = 1.0 and 2.0. The velocity ratio is V ∗

o = 1.0.

The dynamic configuration and the cable tension at selected time values are shown inFigure 4 as obtained by Fourier and wavelet methods. As the mass enters the cable from theleft support, a forward wave travels from left to right. It hits the right support and reflectsback. It meets with the moving mass when the mass is half way of the cable span and someof it transmits through the mass and some of it get reflected by the mass. The wave that isreflected by the mass travels to the right of the mass and when it reaches the right supportit gets reflected back again. The wave that gets transmitted through the mass travels to theleft of the mass. This pattern of wave scattering repeats itself until the mass exits the cableat the right support. The differences between Fourier and wavelets results are not apparentin the cable dynamic response. However, the differences are clear when comparing the cabletension. The Fourier solution shows oscillations at all times and along the entire cable length,while the wavelets solution shows horizontal straight lines with very small local oscillations atthe position of the moving mass. The wavelet prediction is in agreement with the assumptionmade in the linear theory of free vibration of cables which states that the dynamic tensionof the cable is a function of time alone. We expect to notice similar behavior, since, that theexample we are considering here has a small sag-to-span ratio, about 1 : 47. Therefore, thedynamic tension given by the wavelets solution is more accurate. The oscillations found in theFourier solution can be attributed to the fact that the Fourier representation does not convergeuniformly to such functions. It is noticed that a high tension(about 2.6 times the static tension)is found when the mass is before reaching mid-span, while it stays around a constant valueafter the mass has passed the mid-span.

In practice, it might be of interest to have the moving mass traveling along the cable withvariable velocity and variable acceleration. There are many cases of this form, but we willconsider the case in which the moving mass start from rest at the left support and moves witha constant acceleration to a desired maximum velocity when the mass reaches one third ofthe cable. Then the moving mass shall travel at that velocity until it reaches two thirds of thecable. Finally, the moving mass shall be brought to zero at the right support with constant


Figure 4. (a) Dynamic configuration and (b) tension at different time instants. µ = 1.0 and V ∗o = 1.0. – – – static

shape and tension; - - - Fourier representation; — wavelet representation.


deceleration. If we denote by Vmax, the maximum velocity attained by the moving mass, itsdimensionless form is achieved by dividing by

√To/ρ, hence

V ∗max = Vmax

√σ ,

where Vmax = Vmax/√gl. The variable acceleration in dimensionless form is given by

a∗o(t

∗) =

V ∗maxt∗1

; 0 ≤ t∗ ≤ t∗1 ,0; t∗1 < t

∗ ≤ t∗2 ,− V ∗

maxt∗3 −t∗2 ; t∗2 < t∗ ≤ t∗3 ,

(40)

where t∗1 is the time required by the moving mass to reach one third of the cable, t∗2 is the timerequired by the moving mass to reach two thirds of the cable and t∗3 is the total time requiredby the moving mass to reach the right support of the cable. The variable velocity, V ∗

o and theposition of the moving mass, s∗o are achieved by integrating Equation (40) once and twice,respectively.

As an example we will chose Vmax = 1.0 and same cable configuration as the previousexample, i.e. σ = 0.1688832, that leads to V ∗

max = 0.41095. Figure 5 shows the graphicalrepresentation of the moving mass motion. Under this motion the path of the moving massalong the cable is shown in Figure 6. We observe that the maximum deflection is at the middleof the cable length which is in contrary with the previous example of constant velocity andsame mass ratio of this example (Figure 3), in which the maximum deflection occurs at aroundone-third of the cable length. We also observe that the maximum deflections in both examplesare almost the same. The dynamic shape of the cable is shown in Figure 7 at different timeinstants.

7.3. VIBRATION OF AN INCLINED CABLE CARRYING A MOVING MASS

As a last example we consider an inclined cable carrying a moving mass particle that has aconstant acceleration. The location of the mass in its dimensionless form will be

s∗o = a∗o σ

2t∗

2, (41)

where a∗o = ao/g and ao is the constant acceleration of the moving mass. The length and

the span of the cable will be taken same as the previous examples but supported at differentelevations, h = 90 m. Using Fourier representation for a∗

o = 1 the path of the moving massalong the cable is given in Figure 8. The minimum point occurs at around so/ l = 0.8. Thedynamic shape and the total tension of the cable at selected time values are shown in Figure 9using Fourier and wavelets representations. Both solutions are obtained using N = 10. Wenotice that the cable static tension is shown constant at this scale, because in this problemthe equilibrium configuration was chosen to be very close to the chord. Regarding the cabledisplacement, we see that the discrepancies between Fourier and wavelets solutions are small.While, regarding the total cable tension, the Fourier solution has many oscillations more thanthe wavelets solution. In fact, as the moving mass travels along the cable length, the cabletension is discontinuous at the mass location. Away from the mass location the cable tensionis slowly changing with the spatial variable, s and mostly changes with time, t . Therefore,the oscillations in the Fourier solution are due to the nonuniform convergence of the Fourierseries to such functions and a larger value of N produces more oscillations.


Figure 5. The motion of the moving mass with variable velocity and variable acceleration with V ∗max = 0.41095.


Figure 6. Path of the moving mass along the cable for a variable velocity and variable acceleration with maximumvelocity of Vmax = 1.0 and for µ = 1.0. – – – static shape; — dynamic shape.

The wavelets solution can be improved by taking larger value of N , but then, we have touse a smaller time step because it is related to the inverse of the maximum natural frequencyof the system. Since s∗o has to be a dyadic point we should select a time step such that whensubstituted in Equation (41) we get

s∗o = R ( k2ν)2, (42)

where ν is a positive integer and k = 1, 2, . . . , 2ν . Therefore, 2, 2 ′ and 2 ′′ need to beconstructed at the smallest dyadic point appears in the numerical integration scheme whichis 1/22ν+1 here. Then in the computer program, those functions have to be stored in a singlecolumn arrays of size of R 22ν+1 each. The results shown in Figure 9 are obtained at ν = 9.For ν ≥ 10 we exceed the computer stack limit and the computer program does not work.

To overcome this problem, we write the system of equations in terms of s∗o using therelation in equation (41). The time derivatives are transformed to

∂( )

∂t∗= s∗o

∂( )

∂s∗o, (43)

∂2( )

∂t∗2 = s∗2

o

∂2()

∂s∗2o

+ s∗o∂( )

∂s∗o, (44)

where s∗o and s∗o are function of s∗o . Using Equations (43) and (44) in Equation (19) we obtainthe following system of equations

s∗2

o M∂2γ

∂s∗2o

+ [s∗oM + s∗oC] ∂γ∂s∗o

+ K γ + R(γ ) = F, (45)

where γ = γ (s∗o ) and the vectors and matrices are as given before. By denoting the followingmatrices

M∗ = s∗2

o M


Figure 7. The dynamic shape of the cable at different time instants for a moving mass with variable velocityand variable acceleration with maximum dimensionless velocity Vmax = 1.0 and µ = 1.0. – – – static shape;— dynamic shape.

and

C∗ = s∗oM + s∗oC,

Equation (45) becomes analogous to Equation (19). The initial conditions are

γ (0) = 0, (46)

∂γ (0)

∂s∗o= 0, (47)

∂2γ (0)

∂s∗2o

= 1

3s∗oM−1(0)

dF(0)ds∗o

, (48)

where Equation (48) is obtained from Equation (45) at s∗o = 0.


Figure 8. The path of the moving mass along the cable length for µ = 1.0 and acceleration ratio a∗o = 1.0.– – – static shape; — dynamic shape.

The previous solution procedure can be used to solve Equation (45) subjected to the initialconditions (46–48) and replacing M and C by M∗ and C∗ respectively. The increment on theposition of the moving mass is now achieved by the linear relation

s∗o = k :s∗oand the array size to store the wavelet becomes R 2ν+1, which is very much smaller than whatwas required in the previous formulations. ForN = 17 and using an increment of:s∗o = 2−12,Figure 10 shows that some oscillations are diminishing from the wavelet solution but not fromFourier solution. This support our argument that the cable tension behaves like a piecewisecontinuous function and that the Fourier solution does not converge to such functions.

8. Conclusions and Recommendations

The vibration of a catenary elastic cable that is fixed at its supports and carrying a movingmass particle have been investigated. The general formulation have been derived using theHamilton’s principle. The cable configuration was not restricted to small sags and the movingmass particle was assumed to travel along the cable with general motion. The solution wasobtained using the Galerkin procedure with two methods of representations. The first methodis the Fourier representation with sine functions as the trial basis to remove the spatial depend-ence. The second method is the multilevel wavelets representation with anti-derivative of thecompactly supported wavelets as the trial basis to remove the spatial dependence.

Using Fourier and wavelets models to compare with the finite element solution done byWu et al. [10] for a horizontal cable carrying a mass moving at constant speed, we noticedthat our solutions of µ = 1.0 captured their solution of µ = 2.0. We believe that theirsolution has a factor of two missing. We also believe that our solution is more accurate becausewe have used the full nonlinear terms, while they used linear and quadratic terms only. Thecomparison between the Fourier and wavelets representations showed that both methods arein good agreement when obtaining the path of the moving mass along the cable and when


Figure 9. (a) The dynamic shape and (b) the total tension of the cable at three time instants for µ = 1.0 and anacceleration ratio a∗o = 1.0. N = 10. – – – static shape; — - — Fourier; — wavelets.

obtaining the cable displacements at different time instants. However, when obtaining thecable tension at different time instants, the wavelets solution was different than the Fouriersolution. The Fourier solution showed many oscillations while the wavelets solution showedstraight lines with localized oscillations at the location of the moving mass. We know that asthe mass travels along the cable length of small sag-to-span ratio the cable tension changesand it behaves as a piecewise continuous function. We also know that the Fourier solution doesnot converge uniformly when representing such functions and the Gibbs phenomenon occurs.In the wavelet solution, as we move to the higher levels with more terms the oscillationsdiminish, as we have seen in Figures 9 and 10.

The moving mass considered in this paper was assumed to be a lumped mass acting on thecable at a single point and its aerodynamic characteristics are ignored and the friction betweenthe mass and the cable is also ignored. The cable bending moment has been ignored too. Thiswork can be extended by considering that the moving mass is a trolley acting on the cable attwo points and has known moment of inertia and powered by a rocket with a thrust applied


Figure 10. (a) The dynamic shape and (b) the total tension of the cable at three time instants for µ = 1.0 and anacceleration ratio a∗o = 1.0. N = 17. – – – static shape; — - — Fourier; — wavelets.

at a desired inclination. Brakes can be also included so that the trolley can be brought to restbefore it reaches the second cable support.

References

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Dynamics of an Elastic Cable Carrying a Moving Mass Particle