# The Snap Tension Analysis of Taut-Slack Mooring Line with tanh Method

Authors: Su-xia Zhang; Rui-zhi Shi; Shengli Chen; Xi-jun Liu
Journal: Mathematical Problems in Engineering
 (2017)

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## Abstract

The continuous model is introduced, and the nonlinear partial differential equations of taut-slack mooring line system are transferred to nonlinear algebraic equations through tanh method, and four solitary solutions are obtained further. At the same time, to express the results clearly, the curve surfaces of strains, displacements, and tension are plotted. The results show that there are four different solutions in the system. With the pretension increasing, the tension changes from one solitary solution to snap tension, and when the pretension is increased further, the curve converts to continuous line, until straight line, which is corresponding to the taut mooring line. In the process of increasing of pretension, the mooring line transfers from slack to taut, accompanied with tension skipping, which is reduced by the system parameters, and different combination of parameters may introduce different tension in line, and the uncertainty may cause the breakage of mooring system. The results have an agreement with experiment, which shows that the calculating method in this paper may be believable and feasible. This work may provide reference for design of mooring system.


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## Body

## 1. Introduction

With the development of exploitation of deep-water resources, TLP and SPAR platforms are designed and used in some areas and are considered as the excellent platforms in deep water, in which mooring line is one of the most important parts of platforms.Mooring lines are flexible structures and have unidirectional stiffness; that is, they can only bear tension and cannot bear compressive loading. When the platform moves with large amplitude, the mooring line will transfer from taut to slack, accompanied with snap tension. The research showed that the snap tension was usually several times to dozen times more than the mean tension [1], which may be the real reason causing breakage of mooring line.Many works have been done by the scholars to reveal the mechanism of breakage of mooring line. Niedzwecki and Thampi [2] investigated the snap load behavior of marine cable systems in regular seas. Huang and Vassalos studied the impulse loads under alternative taut-slack conditions by using the model of bilinear string [3, 4]. They neglected the effect of floating structure, the cable was simplified into the model subjected to periodic sinusoidal excitation at the top end of cable, and the snap tension of the mooring cable was calculated [5]. The results show that the effect of tension magnification is due to the shock. Experiments show that, under sinusoidal excitation and quasi-static condition, the result calculated by the quasi-static method is near the experiment result. But under the impact condition, the impact tension is much larger than the result calculated by the quasi-static method [6]. The equivalent force model is used to analyze the effect of mooring systems on the horizontal motions [7]. Zhang et al. studied the analytical solution of taut-slack mooring line with tanh method and obtained snap tension curves in four cases [8]. Until now, most of the solutions to the equations are numerical simulation, and the analytical solution is not fully developed.At the same time, different methods solving the nonlinear evolution equations have been developed. The tanh method, developed for years, is one of the most direct and effective algebraic methods for finding exact solutions of nonlinear diffusion equations. Recently, much work has been concentrated on the various extension and applications of the method [9]. Huibin and Kelin [10] introduced a power series in tanh as a possible solution and substituted this expansion directly into a higher-order KdV equation. Wazwaz [11] employed the tanh method for traveling wave solutions of nonlinear equations and the work was extended to equations that do not have tanh polynomial solutions. Abdou and Soliman [12] presented and implemented the solutions of nonlinear physical equations for constructing multiple traveling wave in a computer algebraic system by means of computerized symbolic computation and a modified extended tanh-function method. Zheng et al. [13] presented the generalized extended tanh-function method for constructing the exact solutions of nonlinear partial differential equations (NPDEs) in a unified way making use of a new generalized analysis. As a result, the solitary wave solutions and other new and more general solutions were obtained. Based on the symbolic computation, Zhi and Zhang [14] combined the tanh-function method with the symmetry method to construct new type of solutions of the nonlinear evolution equations for the first time. With the combined method, some new types of solutions of the coupled (2 + 1)-dimensional nonlinear system of Schrodinger equations were obtained.In this paper, to reveal the mechanism of snap tension in mooring line while transforming from taut to slack or from slack to taut, the continuous model is introduced and tanh method is used to resolve the nonlinear equations of taut-slack mooring line. Four types of analytical solutions of taut-slack mooring line are obtained, and the displacement and tension curve surfaces in different cases are presented, and the effects of pretension will be discussed in Section4. The snap tension is obtained, which may be the reason causing the breakage of mooring line.


## 2. The tanh Method

Consider a given evolution, say in two variables [15],(1)Hu,ut,ux,uxx,…=0,where u=ux,t and H is a polynomial about u and its derivatives.The fact that the solutions of many nonlinear equations can be expressed as a finite series of tanh function motivates us to seek for the solutions of (1) in the form(2)ux,t=uξ=∑i=0maiTi=∑i=0maitanhξi,where ξ=κx-ct+ξ0. Notice that the highest order of dpu/dξp is(3)odpudξp=m+p,p=1,2,3,…,ouqdpudξp=q+1m+p,q=1,2,3,…;p=1,2,3,….So m can be obtained by balancing the derivative term of the highest order with the nonlinear term in (1). k,c,a0,…,am are parameters to be determined. Substituting (2) into (1) will yield a set of algebraic equations for k,c,a0,…,am because all coefficients of tanh(ξ)i have to vanish. Using these relations, k,c,a0,…,am can be obtained. Therefore, the traveling solitary wave solutions are obtained. Usually m is a positive integer; however, once in a while, the value of m is a negative or a fraction. In these cases, we can introduce a transformation uξ=vξsign(m)/d, where d is the denominator of m, and transform (1) into another equation for v, whose balancing number will be a positive integer. Then it can be dealt with by the above method. Therefore the final solution u may be a rational or radical function about tanh⁡(ξ).


## 3. Equations of Motion

The model of mooring line is shown in Figure1. The horizontal uniform mooring line is considered, and the shear deformation is neglected. The left end of the line is fixed, and the right end is attached to the buoy; here the condition is simplified by assuming harmonic excitation Absin⁡ωt. e1 and e2 are the unit vector in tangential and normal direction, respectively, and e3 is the binormal unit vector. χo and χf represent the static and dynamic configuration, and U(U1,U2,U3) is the displacement relative to the static configuration.Figure 1
The model of mooring line.
(a)
(b)Only the in-plane motions of mooring line are considered, and the equations are as follows [16]: (4)ρAU1,tt=P+EAε1+U1,s-ψU2,s-ψP+EAεU2,S+ψU1-ρ-ρwAglt+F1(5)ρAU2,tt=P+EAεU2,S+ψU1,s+ψP+EAε1+U1,s-ψU2-ρ-ρwAgln+F2,where ε is the dynamic strain, and the expression is as follows:(6)ε=U1,s-ψU2+12U1,s-ψU22+U2,s+ψU12.EA represents the cable section modulus, ρA represents the cable mass/length, and F1,F2 are components of an arbitrary external excitation, including the hydrodynamic loading and exciting from the buoy attached at the end, mainly the end exciting. In other words, the transforming from taut to slack of mooring line is mainly caused by the large amplitude motion of attached buoy. The hydrodynamic loading is calculated with Morison’s equation as follows:(7)F1=Catρwπd24U1,tt+Cdtρwd2U1,t-vtU1,t-vtF2=Canρwπd24U2,tt+Cdnρwd2U2,t-vnU2,t-vn.Can and Cat are the added mass coefficients in tangential and normal direction, respectively, Cdt and Cdn are fluid drag coefficients in tangential and normal direction, respectively, v is the velocity of fluid particle, and the subscripts t and n denote the projection in tangential and normal directions, respectively. U1 and U2 are the displacements in tangential and normal directions, subscript s denotes partial differentiation with respect to arc lengths, and t denotes partial differentiation with respect to time. Moreover, Ps and ψs are the equilibrium tension and curvature distributions describing a catenary as given by (8)P=P02+ρAgs2,ψ=ρAgP0P02+ρAgs2,where P0 denotes the horizontal component of the equilibrium cable tension.Now considering the effects of curvature and tension in line, substitute (6) and (8) into (4) and (5), and the equations can be reduced as(9)g32λ3s2-12λ7s6-λU1+gλ2s+2gλ4s3+2Eρλ3sU2+gλsU1,s+2Eρλ6s3-λ4sU12+gλ2s2+Eρλ3s2+gλ4s4-Eρλ-2gU2,s+6Eρλ6s3-λ4sU22+Eρλ3s2-λU2U2,sU2,ss+Eρ2λ6s2-3λ8s4+2λ10s6-12λ12s8-12λ4U13+3Eρλ5s-2λ7s3+λ9s5U23+Eρ+P0ρA+12gλs2U1,ss-1+Catρwπd24ρAU1,tt+Cdtρwd2ρAU1,t-vtU1,t-vt+gλs+3EρU1,sU1,ss+EρU2,sU2,ss+32EρU1,s2U1,ss+4Eρλ2-2λ4s2+λ6s4U2U2,s+Eρλ3s2-λU+2Eρ3λ5s2-3λ7s4+λ9s6-λ3U22U2,s+2Eρλ6s3-λ4sU12U1,s+Eρλ3sU2,s2U22,s3+Eρ12λ6s4+12λ2-λ4s2U1,s2U1+Eρ12λ2-λ4s2+12λ6s4U12U1,ss=0g1-λ4s4-λ2s2+g-λ2s-2λ4s3U1+gλsU2,s+52Eρλ-λ3s2U1,s2-1+Canρwπd24ρAU2,tt+Cdnρwd2ρAU2,t-vnU2,t-vn+2Eρλ4s2-Eρλ6s4-gλ-12gλ7s6+32gλ3s2-Eρλ2U2+3Eρ-λ5s+2λ7s3-λ9s5U13+Eρ2λ6s2-3λ8s4+2λ10s6-12λ12s8-12λ4U23+2g-Eρλ3s2-gλ4s4-gλ2s2+EρλU1,s+12Eρ3λ7s4-λ9s6-3λ5s2+λ3U12+32Eρ3λ7s4-λ9s6-3λ5s2+λ3U22+12gλs2+P0ρAU2,ss+12Eρλ3s2-λU2,s2+Eρλ-λ3s2U1,s3+2Eρλ3sU2U2,s+2Eρ3λ7s4-3λ5s2-λ9s6+λ3U1,sU12+2Eρλ6s3-λ4sU2,sU22-Eρλ3sU1U1,s2+Eρλ3s2-λU2U2,ss+Eρλ-λ3s2U1U1,sU1,ss+Eρ12λ6s4-λ4s2+12λ2U2U2,s2+32EρU2,s2U2,ss+Eρ12λ2-λ4s2+12λ6s4U22U2,ss+Eρλ-λ3s2U1U1,ss-2Eρλ3sU1U1,s=0,where λ=ρAg/P0. The dynamic tension T is that (10)T=P0+EAε.In the quasi-static fluid, vt=0 and vn=0. Then the initial equationsmay be simplified and rewritten as follows:(11)A1U1,ss-A2U1,tt+A3+A4U1+A5U2+A6U1,s+A7U2,s+A8U12+A9U22+A10U13+A11U23+A12U1,sU1,ss+A13U2,sU2,ss+A14U1,s2U1,ss+A15U2U2,s+A16U22U2,s+A17U12U1,s+A18U1,s2U1+A19U2,s2U2+A20U2,s3+A21U2U2,sU2,ss+A22U12U1,ss+A23sgnU1,tU1,t2=0(12)B1U2,ss-B2U2,tt+B3+B4U1+B5U2+B6U1,s+B7U2,s+B8U12+B9U22+B10U13+B11U23+B12U1,s2+B13U2,s2+B14U1,s3+B15U2U2,s-B16U1U1,s+B17U1,sU12+B18U2,sU22-B19U1U1,s2+B20U2U2,s2+B21U1U1,ss+B22U2U2,ss+B23U1U1,sU1,ss+B24U2,s2U2,ss+B25U22U2,ss+B26sgnU2,tU2,t2=0,where Ai and Bi are coefficients, shown in the Appendix.


## 4. Results and Discussions

Assuming thatU1, U2 are the polynomials of T with m and n ranks, respectively, the nonlinear term A14κ4U1′2U1′′ in (11) is 3m+4-order polynomials of T, and A21κ3U2U2′U2′′ is 3n+3-order polynomials of T, while the highest order linear term is A1κ2U1′′, which is m+2 polynomials of T. Similarly, in (12), the nonlinear terms B23κ3U1U1′U1′′ and B24κ4U2′2U2′′ are 3m+3 and 3n+4 polynomials of T, respectively, and the highest order linear term B1κ2U2′′n+2. To look for the traveling wave solution of (11) and (12), the parameters m and n are first determined by balancing the linear term of highest order with the nonlinear term:(13)max⁡3m+4,3n+3=m+2max⁡3n+4,3m+3=n+2.It is solved that(14)1m=-1n=-12m=-1n=-23m=-2n=-14m=-12n=-12.For cases (2), (3), and (4), there is no nontrivial solution in terms of tanh function. For case (1), thefollowing transformation is introduced:(15)U1ξ=1b0+b1tanh⁡ξ,U2ξ=1a0+a1tanh⁡ξ,ξ=κx-c1t,where a0,a1,b0,b1,κ, and c1 are undetermined coefficients.Submitting the transformation into (9), collecting the coefficients of T with the same orders, the nonlinear algebraic equations can be obtained:(16)P1=A5b0+A3+A9b02+A4a0+A15κb1b0+A16κb1b02+A8a02+A6κa1+A10a03+A11b03+A18κ2a0a12+A17κa1a02+A19κ2b0b12+A7κb1+A20κ3b13=0P2=2A8a1a0+2A9b1b0+3A10a1a02+A4a1+A5b1+3A11b1b02+A15κb12+A18κ2a13+A19κ2b13+2A16κb0b12+2A17κa0a12=0P3=3A10a0a12+A8a12+A9b12+3A11b0b12+A16κb13+A17κa13=0P4=A11b13+A10a13=0P5=B7κb1+B12κ2a12-B19κ2a0a12+B6κa1+B4a0+B8a02+B9b02+B10a03+B3+B11b03+B5b0+B13κ2b12+B15κb1b0+B18κb1b02+B14κ3a13-B16κa0a1+B17κa1a02+B20κ2b12b0=0P6=2B8a1a0+2B9b1b0+3B10a1a02+B4a1+B5b1+3B11b1b02+B15κb12+2B18κb12b0+B20κ2b13-B19κ2a13=0P7=3B10a0a12+B8a12+B9b12+3B11b0b12+B18κb13+B17κa13=0P8=B11b13+B10a13=0.Solving them with RATH package [17, 18], the four solutions are obtained.Case 1.
The parameters are that(17)c1=-B1B2B2κ=-B4A9+A4B9-4A4B3A92+4A3B4A92-A9B4A52+B9A4A52B92-B4A9+A4B9B7A9a0=-A52A9a1=-B4A9+A4B9-4A4B3A92+4A3B4A92-A9B4A52+B9A4A522-B4A9+A4B9A9b0=-B4A9+A4B92-B3A9+A3B9b1=-A6-B4A9+A4B9-4A4B3A92+4A3B4A92-A9B4A52+B9A4A522-B3A9+A3B9B7A4A9B9.The constraint parameters are that(18)B8=-B4-4A62B92A92A4B3+4A62B92A92A3B4-A62B92A52B4A9+A62B93A52A4+A42A93B72B4-A43A92B72B94A42A92B72B9A3-A9B3A7=A9B7B9,B5=A5B9A9,A8=--4A62B92A92A4B3+4A62B92A92A3B4-A62B92A52B4A9+A62B93A52A4+A42A93B72B4-A43A92B72B94A4A92B72B9A3-A9B3,B6=A6B4A4,A1=A2B1B2.And other parameters are arbitrary.Case 2.
The parameters are that(19)c1=B1B2B2,a0=-A52A9,a1=--B4A9+A4B9-4A4B3A92+4A3B4A92-A9B4A52+B9A4A522-B4A9+A4B9A9b1=A6-B4A9+A4B9-4A4B3A92+4A3B4A92-A9B4A52+B9A4A522-B3A9+A3B9B7A4A9B9,b0=--B4A9+A4B92-B3A9+A3B9κ=--B4A9+A4B9-4A4B3A92+4A3B4A92-A9B4A52+B9A4A52B92-B4A9+A4B9B7A9.The constraint parameters are that(20)B8=-B4-4A62B92A92A4B3+4A62B92A92A3B4-A62B92A52B4A9+A62B93A52A4+A42A93B72B4-A43A92B72B94A42A92B72B9A3-A9B3,A7=A9B7B9,B5=A5B9A9A8=--4A62B92A92A4B3+4A62B92A92A3B4-A62B92A52B4A9+A62B93A52A4+A42A93B72B4-A43A92B72B94A4A92B72B9A3-A9B3,B6=A6B4A4,A1=A2B1B2.And other parameters are arbitrary.Case 3.
The parameters are that(21)a1=--B4A9+A4B9-4A4B3A92+4A3B4A92-A9B4A52+B9A4A522-B4A9+A4B9A9,c1=-B1B2B2b1=A6-B4A9+A4B9-4A4B3A92+4A3B4A92-A9B4A52+B9A4A522-B3A9+A3B9B7A4A9B9,a0=-A52A9b0=--B4A9+A4B92-B3A9+A3B9,κ=--B4A9+A4B9-4A4B3A92+4A3B4A92-A9B4A52+B9A4A52B92-B4A9+A4B9B7A9.The constraint parameters are that(22)B8=-B4-4A62B92A92A4B3+4A62B92A92A3B4-A62B92A52B4A9+A62B93A52A4+A42A93B72B4-A43A92B72B94A42A92B72B9A3-A9B3,B5=A5B9A9,B6=A6B4A4A8=--4A62B92A92A4B3+4A62B92A92A3B4-A62B92A52B4A9+A62B93A52A4+A42A93B72B4-A43A92B72B94A4A92B72B9A3-A9B3,A7=A9B7B9,A1=A2B1B2.And other parameters are arbitrary.Case 4.
The parameters are that(23)c1=B1B2B2,κ=-B4A9+A4B9-4A4B3A92+4A3B4A92-A9B4A52+B9A4A52B92-B4A9+A4B9B7A9b0=--B4A9+A4B92-B3A9+A3B9,a1=-B4A9+A4B9-4A4B3A92+4A3B4A92-A9B4A52+B9A4A522-B4A9+A4B9A9a0=-A52A9,b1=-A6-B4A9+A4B9-4A4B3A92+4A3B4A92-A9B4A52+B9A4A522-B3A9+A3B9B7A4A9B9.The constraint parameters are that(24)B8=-B4-4A62B92A92A4B3+4A62B92A92A3B4-A62B92A52B4A9+A62B93A52A4+A42A93B72B4-A43A92B72B94A42A92B72B9A3-A9B3,B5=A5B9A9,B6=A6B4A4A8=--4A62B92A92A4B3+4A62B92A92A3B4-A62B92A52B4A9+A62B93A52A4+A42A93B72B4-A43A92B72B94A4A92B72B9A3-A9B3,A7=A9B7B9,A1=A2B1B2.And other parameters are arbitrary.To express the solutions clearly, the plots of the solutions are drawn, as shown in Figures2–5, and the parameters used are shown in Table 1. The four solitary solutions are 2 left traveling waves and 2 right traveling waves according to the expressions of c. From the figures, it can be seen that, in the four solitary solutions, two of them introduce solitary tension in the system, and the tension does not change with the time delay. The other solitary solutions in U1 and U2induce solitary wave tension, but with the time, the tension attenuates and disappears, and the energy is absorbed by the line and depleted by the damping from structure and external medium. The solitary tension wave in line may induce the breakage of line structure.Table 1
Parameters used in the calculation.
Quantity
Symbol
Value
Modulus of elasticity
E
1.0 × 1011 Pa
Density of cable
ρ
c
3750 kg/m3
Cross-sectional area
A
314.2 mm2
Drag coefficient in normal direction
C
D
n
1.05
Inertia coefficient
C
m
1
Exciting frequency
ω
5 rad/s
Static pretension
P
0
27 kN
Length
L
2000 m
Water density
ρ
w
1025 kg/m3
Drag coefficient in tangent direction
C
D
t
0.01
Added mass coefficients
C
a
t
C
a
n
1
Exciting amplitude
A
b
10 mFigure 2
The displacements and tension surfaces in Case1.
(a)
Displacement ofU1
(b)
Displacement ofU2
(c)
Tension surfaceFigure 3
The displacements and tension surfaces in Case2.
(a)
Displacement ofU1
(b)
Displacement ofU2
(c)
Tension surfaceFigure 4
The displacements and tension surfaces in Case3.
(a)
Displacement ofU1
(b)
Displacement ofU2
(c)
Tension surfaceFigure 5
The displacements and tension surfaces in Case4.
(a)
Displacement ofU1
(b)
Displacement ofU2
(c)
Tension surfaceIn the problem of taut-slack mooring line, the pretension is an important factor which relates to the line station. At the same exciting loading, when the pretension is altered, the tension curves are different, as shown in Figure6. It can be seen that when the pretension is 10000 N, the solution is 2-solitary solution, and with the pretension increasing, the tension changes from solitary solution to snap tension, and with the continuous increasing, the curve converts to continuous line, until straight line, which is corresponding to the taut mooring line. In the process of increasing the pretension, the mooring line transfers from slack to taut, accompanied with tension skip.Figure 6
Tension curve with different pretension.
(a)
P
0
=
10000 N
(b)
P
0
=
15000 N
(c)
P
0
=
20000 N
(d)
P
0
=
25000 N
(e)
P
0
=
30000 N
(f)
P
0
=
40000 NGivenP0 as 26000 N, the tension at s=1900 m is calculated as shown in Figure 7. The results show that there is a tension skip at about t=7 s, and the tension decreases. When the parameters are given the same as those in [19], where the length of cable is 30 m, excitation frequency is 1.2 Hz, and amplitude is 8 cm, the calculated result is shown in Figure 8, and the experimental result is shown in Figure 9. The calculated maximum tension is 945.5 N, and it is 790.8 N in the experiment. This is because the cable is elastic, and the pretension is not so accurate in the experiment. Excluding the effect, the theoretical result is consistent with the experimental one. It indicates that the calculating method in this paper may be believable and feasible.Figure 7
Tension history curve in mooring line ats=1900 m with the method in this paper.Figure 8
Tension history curves in mooring line simulated with the same parameters in [19].Figure 9
Tension history curves in mooring line end in middle stage at exciting end [19].The results show that the skip tension is reduced by the system parameters, and different combination of parameters may introduce different tension in line, and the uncertainty may cause the breakage of mooring system.


## 5. Conclusions

In this paper, the continuous model is introduced, and the nonlinear partial differential equations of taut-slack mooring line system are transferred to nonlinear algebraic equations through tanh method, and four solitary solutions are obtained further. To express the results clearly, the curve surfaces of displacementU1, U2 and tension T are plotted. The results show that there are four different solutions in the system. With the pretension increasing, the tension changes from solitary solution to snap tension, and when the pretension increased further, the curve converts to continuous line, until straight line, which is corresponding to the taut mooring line. In the process of increasing the pretension, the mooring line transfers from slack to taut, accompanied with tension skip, which is reduced by the system parameters, and different combination of parameters may introduce different tension in line, and the uncertainty may cause the breakage of mooring system. The results have an agreement with experiment, which shows that the calculating method in this paper may be believable and feasible.



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