On Longitudinal Vibrations in the Hoist Cables of the Pieter Schelte

Transcription

1 Delft University of Technology Faculty Electrical Engineering, Mathematics and Computer Science Delft Institute of Applied Mathematics On Longitudinal Vibrations in the Hoist Cables of the Pieter Schelte Thesis submitted to the Delft Institute of Applied Mathematics in partial fulfillment of the requirements for the degree ACHELOR OF SCIENCE in Applied Mathematics by Vera van den Dool Delft, the Netherlands January 25 Copyright c 25 by Vera van den Dool. All rights reserved.

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3 Sc thesis APPLIED MATHEMATICS On Longitudinal Vibrations in the Hoist Cables of the Pieter Schelte VERA VAN DEN DOOL Delft University of Technology Supervisor Dr.ir. W.T. van Horssen Other members of the Committee Ir. J. van Leeuwen Dr. J.G. Spandaw Dr.ir. F.J. Vermolen January, 25 Delft

4 Abstract In this thesis the hoisting of a jacket from the sea bottom is modeled as a forced, longitudinally vibrating string, with nonhomogeneous boundary conditions. The problem is studied by using perturbation theory, where approximations of the solutions of the first and second order problem will be constructed. 3

5 Contents Introduction 6 2 Problem derivation 8 2. System Description Jacket Lifting System Main Hoist System Model derivation Overview of the system Assumptions Equation of motion for the system Initial and boundary conditions Problem formulation Coordinate transformation Solving the problem Introducing orders Introducing operators Solving the seperate problems Solving the O problem Solving the Oε problem Resonance Resonance in the O problem Resonance in the Oε problem Results Summary of results Conclusions and recommendations 44 Appendix: A Orthogonality of the eigenfunctions 46 4

6 List of Figures. The Pieter Schelte Schematic overview of the Tilting Lift eams Jacket lifting procedure Composition of the main hoist system System overview Coordinate transformation List of Tables 2. Variables Constants Simplifications

7 Chapter Introduction Allseas is a Swiss-based company which specializes in offshore pipeline installation and sub-sea construction. Only a few of pipelay and support vessels are operational worldwide. Oil and gas platforms consist of large jackets carrying the topside, both of which have to be moved or deconstructed after use. Following the usual procedure, construction removal is performed by cutting down the construction to be removed into smaller sections which are removed separately. A new concept for a vessel allowing the removal of constructions in a single lift was put forward by Allseas in 987. As a result, the deconstruction of support structure could from now on be performed onshore, with the following advantages: A safer work environment for the people involved; A more cost effective solution for the companies involved; Fewer environmental risks, and; A large reduction of the required offshore man-hours. The concept vessel allowing for single lift platform installation and decommissioning was named the Pieter Schelte, named after marine engineer Pieter Schelte Heerema, father of Allseas owner Edward Heerema. Construction of the vessel started in 2 and it is expected to be operational early 25. When finished, the Pieter Schelte, at 382 meters long and 24 meters wide, will be able to remove jackets up to a weight of 25, tons in a single lift and install platforms weighing up to 48, tons in one piece. During the jacket lifting procedure, the Pieter Schelte is connected with this jacket through a lifting cable. At this stage, motion of the sea influences the vertical position of the Pieter Schelte and thus also the hoisting cables 6

8 Figure.: The Pieter Schelte. and jacket. In case the jacket is close to the bottom of the sea, this motion may cause the jacket to rebounce on its foundation. In such a case, rapid release and subsequent increase of the tension in the hoisting cable will occur. Possible consequences of this are snapping of the hoisting cable or loop formation. oth of these can result in extra costs and delays. Also the fatigue caused by large stress fluctuations decreases the lifespan of the hoisting cable. In 24, H.R. van Rijn wrote a master thesis researching the hoist cable behavior in case of jacket rebouncing, called Response of Pieter Schelte main hoist cables due to longitudinal shock loads 3]. Following from analysis of the work of H.T. van Rijn, it was found that more research with respect to the hoist cable behavior during jacket lifting was required. This thesis will therefore aim to derive a more detailed mathematical analysis of the hoist cable behavior during the inital phases of the jacket lifting. The research question this thesis aims to answer from a mathematical perspective is: What are the vibrations occurring in the hoist cable during the initial phase of the jacket lifting? In order to answer this research question, first a description of the relevant mechanical features of the Pieter Schelte is given in Chapter 2. In this chapter, also an analytical model describing the behavior of the hoist cables will be derived. Thereafter, the solutions to this model will be calculated in Chapter 3. The results following from this are presented and discussed in Chapter 4. 7

9 Chapter 2 Problem derivation 2. System Description 2.. Jacket Lifting System At the stern of the Pieter Schelte, a Jacket Lifting System JLS is installed. This jacket lifting system consists of two 5 meter long Tilting Lift eams TLs, which can incline up to a 5 angle with respect to the vessel. These beams are kept in place by the the hoist system and an inner upender. Figure 2.: Schematic overview of the Tilting Lift eams. 8

10 A jacket is lifted from its foundation as follows:. First, the Pieter Schelte is positioned with the stern above the jacket, see figure 2.3a. 2. Hereafter, the tilting lift beams are positioned in the correct angle, see figure 2.3b-c. 3. Then, the hoist cables are attached to the jacket and pulled in. This first causes the cables to pretension, followed by the lift off of the jacket, see figure 2.3d-e. 4. The jacket is now tilted until it has the same angle as the tilting lift beams. The jacket is further lifted until its center of gravity is high enough in relation to the tilting lift beams, see figure 2.3f. 5. The TLs are tilted a few degrees inwards, to ensure the jacket lands onto support sledges on the beams, see figure 2.3g. 6. The TLs are tilted further until they are completely horizontal and the jacket is on board, see figure 2.3h-i. a b c d 9

11 e f g h i Figure 2.2: Jacket lifting procedure Main Hoist System The main hoist system MHS consists of two independent lifting arrangements for each TL. These lifting arrangements consist of an upper and lower block, with two sets of axially aligned sheaves. The lower blocks are attached to the jacket. The MHS winches pull on two sets of steel cables, which have 36 falls over the sheaves. An overview of this is given in figure 2.3.

12 Figure 2.3: Composition of the main hoist system. 2.2 Model derivation The main hoist system is now modeled as a single vertical string, with the jacket modeled as a rigid mass M jacket attached to it Overview of the system The ship and thus the system moves with the motions of the sea z t. As the length of the hoist cable is not directly dependent on the movement of the sea, for the origin of the system the tip of the TL is chosen. This way, the motion z t can be looked at as an external motion working on the system. The position of a piece of the cable in pretensioned equilibrium is given by xt. When longitudinal vibrations occur, this piece of cable moves up or down. This displacement is given by uxt, t. The point where the hoist cable is attached to the jacket is taken as x = Lt. In order to lift the jacket, the cable is hoisted up. The velocity with which this is done is called v h, where hoisting upwards corresponds to a negative v h. Note that v h is the first time derivative of the position xt. For the motion of the sea z t, the position of a piece of cable xt, the hoisting velocity v h and the displacement uxt, t, downwards is taken to be the positive direction. An overview of the system is given in figure 2.4, an overview of variables is given in table 2.. For some purposes, it is required to know the location of the cable or the jacket with respect to the surrounding water or the seabed. The absolute

13 position of a point will therefore be given by R: R = z t + xt + ux, t z t xt ux, t v h Lt M Figure 2.4: System overview. Name Lt xt uxt, t z t v h axt, t Rxt, t Description Length of the hoist cable Position in the cable Displacement from pretensioned equilibrium Motion of the origin of the system TL tip Hoisting speed Acceleration of hoist cable within frame of reference Absolute position, outside frame of reference Table 2.: Variables Assumptions In order to make a mathematical model, certain assumptions have to be made. The main hoist system is modeled as a string with a constant axial stiffness. This cable has a constant axial stiffness if there is tension on the cable, the axial stiffness is otherwise. 2

14 The hoist cable is considered to be pretensioned at the beginning of the lifting. The hoist cable has a constant mass distribution m c, given in kg/m. The movement of the TL tip z t is modeled with an independent function which models wave movement. This means the movement of the TL tip is prescribed and will not be influenced by other factors like for example, the possible tilting of the ship due to the weight of the jacket. Many factors, like the dynamics of the sea, the possibly uneven shape of the jacket, etc. could influence lateral and rotational forces. In this thesis those factors are neglected and only longitudinal forces and vibrations are considered. When the jacket is pulled through the water, this causes drag. This is translated into a damping term. During the lifting, parts of the construction will rise above water. As this thesis focuses on the initial phase of the lifting, the assumption is made that the jacket is fully submerged at all times, and that the hoist cable does not experience any drag from the water or air. Friction in the hoist cable occurring when the cable extends is considered in the form of damping. This damping depends on the hoist cable velocity. The foundation is modeled as a partial spring, which will only be active if the jacket hits the bottom, i.e. if RLt, t > RL,. This force is proportional to how far the spring is pushed in: RLt, t RL,. This leads to a force { k RLt, t RL, if RLt, t > RL, F k t = otherwise The hoisting velocity v h is assumed to be small and constant, and no initial accelerations will be considered. Note that the hoisting velocity is the rate with which the cable length changes: v h = dx dt, where hoisting upwards corresponds to a negative hoisting velocity Equation of motion for the system The equation of motion for this problem can be derived using Newton s second law of motion: F = m a,

15 i.e., the acceleration of a piece of the hoist cable times its mass should equal the force working on it. The acceleration of a piece of the hoist cable within the frame of reference is called ax, t. As the origin of the system, the TL tip, is constantly moving along with the sea z t, there is a second acceleration to be taken into account, being the second time derivative of the movement of the TL tip z t. Newton s second equation applied to a small part x of the cable gives thus F = x m c ax, t + z t There are two kinds of forces working on a small part of the hoist cable. The first is the gravitational force, equal to the mass of a small piece of the hoist cable times the gravitational acceleration. The second force comes from the tension in the cable. The tension T x on one side of the small piece of rope may be slightly smaller or bigger than on the other side, which yields a net force. Together this gives: Hence follows F = x m c g + T x + x T x x m c ax, t + z t = x m c g + T x + x T x, m c ax, t + z t = m c g + According to Hooke s law, T x + x T x. x f = kx, the tension on the cable is proportional to the extension of the cable. Hence it follows: T x = u x, lim x T x + x T x x u u = lim x x + x x x = 2 u x x x The above mentioned acceleration within the frame of reference, ax, t, is the second time derivative of the displacement üx, t, with Du Dt = Dux, t Dt ax, t = D2 ux, t Dt 2 = u t + ẋt u x, = 2 u t 2 + 2ẋt 2 u x t + ẋt2 2 u x 2 + ẍt u x Here ẋt is the hoisting speed v h. Since this velocity is assumed to be constant, the hoisting acceleration ẍt =. This gives ax, t = 2 u t 2 + 2v 2 u h x t + 2 u v2 h x

16 Name Description Unit Longitudinal stiffness of rope N] m c Mass density of the cable kg g Gravitational acceleration m ] s 2 M jacket, M Mass jacket kg] β Viscous damping constant ] F k t Force exerted by the foundation N] F u uoyant force N] M added Added mass term kg] C A Added mass constant ] Table 2.2: Constants. Inserting 2.2. and into gives 2 u t 2 + 2v 2 u h x t + vh 2 ] 2 u m c x 2 = g z t In order to take into account the damping caused by friction in the cable, an additional term is added to the equation. This damping is assumed to be proportional to the displacement speed Du Dt = u t + v h u x. Adding this term into the equation gives: 2 u t 2 + 2v 2 u h x t + vh 2 ] 2 u u m c x 2 + β t + v u h = g z t x An overview of used constants is given in table Initial and boundary conditions Initial conditions For the problem to be solved, the initial displacement and velocity have to be known. This corresponds to the initial conditions: The initial displacement ux, = gx The initial velocity du x, = hx dt 5

17 oundary conditions Since the frame of reference was chosen to have the TL-tip xt = as its origin, the boundary condition in xt = is trivial: u, t = The boundary conditions at x = Lt, the point where the jacket is attached to the hoist cable, can also be found using Newton s second law of motion. The forces working on the jacket are the following: A gravitational force M jacket g. The tension in the cable, which on the boundary only works upwards, u x. The force F k t exerted by the foundation if the jacket hits the bottom, i.e. if RLt, t > RL,. A damping force dependent on the displacement speed of the jacket:. This force is assumed to be linear. C damp DR Dt An upward force the buoyant force F u proportional to the weight of the displaced water. The acceleration working on the jacket consists of two parts: The acceleration a = D2 ux,t Dt 2. The acceleration of the entire system, outside the reference frame, z t. When hoisting the jacket, not only the weight of the jacket M jacket has to be considered, a significant volume of water has to be displaced. This is translated into an added mass term M added. This added mass is assumed to be proportional to the jacket mass. This leads to renaming M jacket = M and M added = C A M. Combined this yields: M g u x F DR kt + C damp Dt F u = D 2 ux, t M + C A Dt 2 + z t Using DR Dt = ż t + dx dt + u x dx dt + u t = ż t + v h + u + u x t,

18 this gives M g u x F kt F u + C damp ż t + v h + u + u = x t 2 u M + C A t 2 + 2v 2 u h x t + 2 u v2 h x 2 + z t Taking the terms containing u to the left hand side of the equation yields u x C u damp v h x + u 2 u + M + C A t t 2 + 2v 2 u h x t + 2 u v2 h x 2 = and thus M g F k t F u + C damp ż + v h + M + C A z t C damp v h u x C u damp t + M + C A Simplification by introducing = Lt: 2 u t 2 + 2v 2 u h x t + 2 u v2 h x 2 = M g F k t F u + C damp ż + v h + M + C A z t M+C A, α = C damp M+C A gives at xt = αv h u x α u t + 2 u t 2 + 2v 2 u h x t + 2 u v2 h x 2 g = F kt + F u + C A M + C A + α ż + v h + z t Symbol α Simplifies M+C A C damp M+C A Table 2.3: Simplifications. 7

19 2.3 Problem formulation The equation of motion for the system and boundary conditions have been obtained. This leads to the following system: ] 2 u + 2v 2 u t 2 h x t + vh 2 2 u m c + β u x 2 t + v h u x = g z t, ux, t x= =, αv h u x α u ft = t + 2 u t 2 + 2v 2 u h x t + v2 h 2 u = ft, x 2 g +C A F kt+f u M+C A + α ż + v h + z t In order to simplify the problem, the coordinate system is reversed, as shown in figure 2.5. Lt z xt ux, t ux, t xt v h v h z Lt M M Figure 2.5: Coordinate transformation. The equation of motion remains the same, the boundary conditions are reversed: The gravitational force now works downwards: M jacket g. The tension in the cable works upwards + u x. The force F k t exerted by the foundation if Lt > L, works upwards. 8

20 The damping force dependent on the displacement speed of the jacket works downwards. C damp DR Dt The buoyant force F u proportional to the weight of the displaced water works upwards. This leads to the reversed problem ] 2 u + 2v 2 u t 2 h x t + vh 2 2 u m c + β u x 2 t + v h u x = g z t, ux, t x=lt =, + αv h u x + α u t 2 u 2v 2 u t 2 h x t v2 h 2 u x 2 x= = ft, ft = g +C A F kt+f u M+C A α ż + v h z t Coordinate transformation y applying a coordinate transformation, the variable boundary at x = Lt is dealt with. Taking uxt, t = uxt, t, xt = Ltxt, gives the following equalities for derivatives: u t = u x x t + u xl t = t u Lt x + u t, 2 u L t 2 = t x 2 2 u Lt x 2 2 L tx 2 u Lt x t + 2 L t 2 x Lt 2 L tx u Lt x + 2 u t 2, u x = u x x x = u Lt x, 2 u x 2 = 2 u Lt 2 x 2, 2 u t x = t x 2 u L Lt 2 x 2 L t u Lt 2 x + Lt 2 u x t Using L t = v h and thus L t = further simplifies these equation: 9

21 u t = x v h u Lt x + u t, 2 u t 2 = vh x 2 2 u Lt x 2 2 vh x 2 u Lt x t + 2 v2 h x u Lt 2 x + 2 u t 2, u x = u Lt x, 2 u x 2 = 2 u Lt 2 x 2, 2 u t x = v h x 2 u Lt 2 x 2 v h u Lt 2 x + Lt can now be rewritten to 2 u x t vh x Lt + ux, t x= =, ft = 2 2 u + αv h And further to vh ft = 2 vh x 2 u x 2 Lt x t vh 2 2 u m c Lt 2 u Lt x 2v h v h x ] + α 2 u Lt 2 x v2 h x x 2 ] x vh Lt ] u Lt 2 x + 2 u t 2 + β + 2v h x v h u Lt x + u t ] u x + u t v h u Lt 2 x + Lt 2 u x t g +C A F kt+f u M+C A α ż + v h z t. Lt x 2 mc Lt 2 ] vh x Lt v h x 2 u Lt 2 x 2 ] + v h 2 2 u u Lt x x 2 2 vh x Lt v h u Lt 2 x + Lt ] = g z t, 2 u x t ] ] vh 2 2 u Lt 2 x 2 x= = ft, ] 2 2 u + 2 vh x 2 Lt x + β v h Lt x u x ] 2 v h Lt x 2 u x t + 2 u + β u t 2 t = g z t, ux, t x= =, ] 2 Lt α v h Lt x 2 vh Lt x + 2 u x vh Lt x 2] 2 u x 2 2 v h Lt x ] 2 u x t + α u t 2 u t 2 x= = ft, g +C A F kt+f u M+C A α ż + v h z t. Introducing the operator L, with ] 2 vh m L = x c Lt Lt 2 2 v h x Lt ] 2 2 x ] 2 vh x + β v h x Lt Lt x x t + 2 t 2 + β t, u x t ] ] v2 + 2 h x u Lt 2 x + 2 u t 2 2

22 and the following operator for the boundary conditions u] = ] 2 Lt + α v h Lt + 2 vh Lt ux, t x= 2 v h 2 u Lt x t + α u 2 u x vh 2 u Lt x 2 t 2 u t 2 x= leads to the following system to be solved: Lu] = g z ] t, ft u] =, ft = g +C A F kt+f u M+C A α ż + v h z t., For convenience, u and x will from now on be referred to as u and x

23 Chapter 3 Solving the problem 3. Introducing orders In order to solve the problem as given in 2.3., a perturbation method is applied. ux, t is assumed to be of the form u u + εu + ε 2 u 2, u = u + εu + Oε Assumptions now have to be made on the order of the variables, to rewrite the problem with orders: t = O F k t, F u, g, z = O v h = ṽ h ε = Oε β = βε = Oε α = αε = Oε Lt = + εt = O Terms containing Lt can be rewritten using a series expansion for v h Lt = ṽhε + εt = ṽhε + εt Terms of Oε 3 will be neglected. Note that = ṽhε εt + ε2 t 2 L 2 = ṽhε ṽhε 2 t L 2 + ṽhε 3 t 2 L 3. +εt : vh ṽh ε = ṽhε 2 2 t Lt L 2 = ṽh 2 ε 2 L 2 + Oε 3,

24 and m c Lt 2 = m c + εt 2 = m c L 2 + εt L 2 2 = m c L 2 2 εt L 2 = m c L ε2 t 2 L 4 m 2 c εt m L c ε 2 t 2 L 6, 3..4 and Lt = + εt = + εt = εt + ε2 t 2 L 2 = εt L 2 + ε2 t 2 L These previous results allow rewriting of the problem to 2 ] vh ε x 2 mc 2 mc εt + 3 mc ε2 t 2 2 u L 2 L 4 L 6 x vh ε x + β vh ε 2 v hε 3 t x L 2 2 vh ε v hε 2 t L 2 ux, t x= =, ] εt + ε2 t 2 + α vh ε 2 L 2 L 3 L v hε 3 2 t + 2 vh ε u L 2 L x ] 2 ft = vh ε 2 vh ε v hε 2 t L 2 v hε 2 t L 2 ] u x x ] 2 u x t + 2 u u + βε t 2 t = g z t, 2 u x 2 ] 2 u u x t + αε t 2 u t 2 x= = ft, g +C A F kt+f u M+C A αε ż + v h z t. 3.2 Introducing operators 3..6 The linear operator L can be rewritten with orders: ] ] m L = c 2 L 2 x t 2 + ε 2 ṽh x 2 x t + β t 2 m c t 2 L 4 x 2 + ε 2 ṽh 2 L 2 x 2 m 3 c t 2 2 ] L 6 + x 2 ṽ2 h ṽh x2 L 2 + β x + 2ṽht 2 L 2. x t 3.2. This can be rewritten as L =L + εl + ε 2 L 2,

25 with L = m c L 2 2 x t 2, L = 2 ṽh x 2 x t + β t 2 m c t 2 L 4 x 2, ṽ 2 h L 2 = L 2 x 2 m 3 c t 2 2 L 6 + x 2 ṽ2 h x2 L 2 The boundary conditions can be split this way as well: ] εt + ε2 t 2 + α v hε 2 2 L 2 L 3 L + 2 vh ε u x u] = ] vh ε 2 2 u L 2 vh ε x 2 v hε 2 t L 2 ux, t x= This can be written as with u] = u] = + ṽh β x + 2ṽht 2 L 2 x t ] 2 u u x t + αε t 2 u t 2 x= u] = u] + ε u] + ε 2 2 u], u x 2 u t 2 x= 2 u] = u x= t u L 2 u x= t 2 L 3 u x= x 2 v h + α v h + 2 ] 2 u x t, u + α t x= ] ] 2 vh u x ṽ 2 2 u Finally the forcing function ft can be rewritten as with From this follows: f t =, x v ht 2 u L 2 x t x= ft = f t + εf t, g + C A F kt + F u M + C A z t, f t = α ż + v h. Lu] = L u] + εl u] + ε 2 L 2 u] = L u + εu + ε 2 u 2 ] + εl u + εu + ε 2 u 2 ] + ε 2 L 2 u + εu + ε 2 u 2 ] = L u ] + ε L u ] + L u ] + ε 2 L u 2 ] + L u ] + L 2 u ] + Oε

26 The following problems will now be solved: The problem to solve u x, t: the O problem L u ] = g z t, ] f t u ] =. The problem to solve u x, t: the Oε problem L u ] = L u ], u ] = u ] + ] f t The solutions to the Oε 2 problem will not be computed in this thesis. 3.3 Solving the seperate problems 3.3. Solving the O problem The O problem boils down to: Take m c L 2 u x, t x= =, f t = 2 u + 2 u x 2 t 2 u x 2 u t 2 = g z t, x= = f t, g +C A F kt+f u M+C A z t u x, t = w x, t + v x, t, where w satisfies the nonhomogeneous boundary conditions. As u and w then satisfy the same nonhomogeneous boundary conditions, then satisfies homogeneous boundary conditions. v = u w Solving w w satisfies the nonhomogeneous boundary conditions, i.e. w satisfies { w x 2 w t 2 w x= =. x= = f t,

27 Take This implies w x = d 2t, 2 w t 2 = d t + xd 2t. Substituting in leads to w = d t + xd 2 t w x 2 w t 2 x= = d 2 t d t = f t, w x= = d t + d 2 t = Combining this yields: d t = d 2 t, d2 d 2 dt 2 + d 2 t = f t This is a nonhomogeneous ordinary differential equation for d 2 t, for which the solution is given by d 2 t = A tv t + A 2 tv 2 t, where v t and v 2 t are linearly independent solutions to the homogeneous problem d 2 d 2 dt 2 + d 2 t =, 3.3. A t and A 2 t are given by t A t = W v 2tf tdt, A 2 t = t W v tf tdt, 3.3. and where W is the Wronskian of v and v 2. The functions v t and v 2 t are obtained by solving 3.3., yielding the linearly independent solutions v t = sin t, v 2 t = cos t

28 The Wronskian of v t and v 2 t thus equals sin L W = t cos cos t sin = sin 2 t + cos 2 d 2 is now given by =. d 2 = A t sin d = d 2. w is therefore given by t t t t + A 2 t cos t, with w =d t + xd 2 t = d 2 t + xd 2 t =x d 2 t =x A t sin t + A 2 t cos t, A t = A 2 t = t L t cos t f tdt, t f tdt. L sin Solving v The equality u x, t = w x, t + v x, t, can be substituted in the differential equation from 3.3., leading to m c L 2 2 u x u t 2 2 v m c L 2 x 2 2 w m c L 2 x v t w t 2 = g z t = g z t Using 2 w =, x2 27

29 leads to the following nonhomogeneous problem with homogeneous boundary conditions for v mc L 2 2 v x v t 2 = g z t 2 w t 2, v L x 2 v t 2 x= =, v x= = In order to solve this nonhomogeneous problem, first the eigenvalues and eigenfunctions of the related homogeneous problem, mc L 2 2 y + 2 y =, x 2 t 2 y x 2 y t 2 x= =, y x= =, are calculated. This is done by applying the method of separation of variables, writing yx, t = XxT t This leads to Solving for X From follows m c L 2 T T = X X = µ X + µ 2 X = Xx = c cos µx + c 2 sin µx From also follows: T = µ 2 m c L 2 T Combining these results gives the boundary conditions for X: X T X T x= = X T + µ 2 m c L 2 XT x= = X + µ 2 m c L 2 X x= =, and X x= =

30 Applied to X this yields: µc 2 + µ 2 m c L 2 c =, c cos µ + c 2 sin µ = From this can be derived: c 2 m c µ m c µ c = c 2 m c µ, c = c 2 tan µ = c 2 tan µ = tan µ This yields an infinite amount of solutions for µ, where these µ n can be calculated numerically. All these solutions for µ n correspond to the a solution for X, and therefore X can be written as X n x = c cos µ n x µ n sin µ n x m c Solving for T From follows T + µ2 n m c L 2 T = T n t = c 3 n cos µ n t + c 4 n sin µ n t. m c m c Combining the solutions for Xx and T t then leads to yx, t = i n cosµ n t m c cos µ n x µ n sin µ n x m c + j n sinµ n t cos µ n x µ n sin µ n x. L m c m c Now that the eigenfuctions of are found, v can be written as v x, t = a n tφ n x,

31 where φ n corresponds to the eigenfunctions solved for X, φ n x = cos µ n x The derivatives of v can be written as µ n m c sin µ n x v x 2 2 v t 2 = = a n t d2 φ n x dx 2, d 2 a n t dt 2 φ n x Note that d 2 φ n dx 2 = d2 dx 2 = µ 2 n = µ 2 nφ n. cos µ n x µ n sin µ n x m c cos µ n x µ n m c sin µ n x As v satisfies the differential equation this can now be written as m c 2 v L 2 x v t 2 = g z t 2 w t 2, m c L 2 m c L 2 a n t d2 φ n x dx 2 + a n tµ 2 nφ n x + m a c n L 2 µ 2 n + d2 a n dt 2 d 2 a n dt 2 φ nx = g z t 2 w t 2, d 2 a n dt 2 φ nx = g z t 2 w t 2, φ n x = g z t 2 w t The eigenfunctions are orthogonal with weight + δx see appendix A, therefore both sides of can be multiplied by + δx and φ m, then integrated from to : 3

32 Call a n t = a m t L 2 = m c L 2 m c L 2 µ 2 n + d2 a n dt 2 φ n xφ m + g z t 2 w t 2 φ m µ 2 m + d2 a m dt 2 m a m t c µ 2 m + d2 a m dt 2 = k m t = φ 2 m + δx dx + L δx dx δx dx + L δx dx g z t 2 w t 2 φ m g z t 2 w φ t 2 m + δx dx. + δx dx g z t 2 w φ2 m t 2 φ2 m φ m + δx + δx dx dx, then the solutions for a m satisfy the nonhomogeneous ordinary differential equation m c L 2 These solutions are given by with D t and D 2 t given by µ 2 ma m t + d2 a m dt 2 = k m t a m t = D tv t + D 2 tv 2 t t D t = W v 2tk m tdt, D 2 t = t W v tk m tdt, with v t and v 2 t the independent solution to the homogeneous problem related to 3.3.4, µ m v t = sin t, m c µ m v 2 t = cos t, m c 3

33 and W the Wronskian of v and v 2, W = µ m m c Consequently D t = t t D 2 t = mc µ m µ m cos mc sin µ m µ m t k m tdt, m c t k m tdt, m c hence a m t = D t sin Summary The following u has been calculated, µ n µ n t + D 2 t cos t m c m c u x, t = w x, t + v x, t, with w x, t = x A t sin t + A 2 t cos t, and with and a n t satisfying a n t = D t sin A t = A 2 t = t L t v x, t = cos t f tdt, t f tdt, L sin a n tφ n x, φ n x = cos µ n x µ n m c sin µ n x, µ n µ n t + D 2 t cos t, m c m c 32

34 with D t = t t D 2 t = mc µ n µ n cos mc sin µ n µ n t k n tdt, m c t k n tdt, m c and k n t = g z t 2 w φ2 n t 2 φ n + δx + δx dx dx Solving the Oε problem The Oε problem must satisfy L u ] = L u ], u ] = u ] + The problem thus boils down to: mc L 2 2 u + 2 u x 2 t 2 u x 2 u t 2 u x= =. = 2 ṽh x 2 u x t β u t x= = t L 2 ] f t. + 2 t 2 u, m cl 4 x 2 u x + 2 ṽh 2 u x t α u t + f t x=, In order to solve this nonhomogeneous PDE with nonhomogeneous boundary conditions, u x, t is split up into w x, t, a part satisfying the nonhomogeneous boundary conditions, and v x, t. u x, t = w x, t + v x, t Solving w Call then w satisfies Taking rt = t u L 2 x + 2 ṽh 2 u x t α u t + f t, x= w x 2 w t 2 x= = rt, w x= = w = d 3 t + xd 4 t,

35 leads to w x = d 4t, 2 w t 2 = d 3t + xd 4t. Substituting this into yields w x 2 w t 2 x= = d 4 t d 3t = rt, w x= = d 3 t + d 4 t = This gives the following equation from which d 4 t can be calculated: d 2 d 4 dt 2 + d 4 = rt This is a nonhomogeneous ordinary differential equation for d 4 t, to which the solution is given by d 4 t = tv t + 2 tv 2 t, where v t and v 2 t are linearly independent solutions to the related homogeneous problem, where W is the Wronskian of v and v 2, and t and 2 t are given by t t = 2 t = t W v 2trtdt, W v trtdt The homogeneous problem corresponding to is the same as Therefore v, v 2 and the Wronskian are known and given in and Hence d 4 t is given by d 4 t = t sin t + 2 t cos t, and w x, t is thus given by w x, t = x t sin t + 2 t cos t, with t = 2 t = t t cos t rtdt, sin t rtdt

36 Solving v As v x, t = u x, t w x, t, v now satisfies the following nonhomogeneous problem with homogeneous boundary conditions, mc L 2 2 v + 2 v x 2 t 2 = 2 ṽh x 2 u x t β u t + 2 t 2 u 2 w, m cl 4 x 2 t 2 v L x 2 v t 2 x= =, v x= = Note that the eigenvalues and eigenfunctions of the related homogeneous problem, mc L 2 2 y + 2 y =, x 2 t 2 y x 2 y t 2 x= =, y x= =, are known, since this problem is the same as 3.3.2, which was solved earlier. The eigenfunctions φ n x are given by φ n x = cos µ n x µ n m c sin µ n x, and v x, t can thus be written as the sum of eigenfunctions: v x, t = The derivatives of v are given by 2 v t 2 2 v x 2 b n tφ n x = = d 2 b n dt 2 φ nx, b n 2 φ n x dx This can be substituted into the Oε problem from 3.3.7, which leads to d 2 b n dt 2 φ nx m c L 2 b n t d2 φ n x dx 2 u can also be rewritten in the equation, = 2 ṽh x 2 u u =v x, t + w x, t = a n tφ n x + w x, t, 35 x t β u t t 2 u w +2 m c L 4 x 2 2 t

37 which leads to the following: d 2 b n dt 2 φ nx =2x ṽh m c L 2 da n dt dφ n x dx b n t d2 φ n x dx 2 β da n dt φ nx + 2 t m c L 4 a n d 2 φ n x dx 2 + 2x ṽh 2 w x t β w 2 w t t Since the eigenfunctions are orthogonal with weight + δx, both sides of can be multiplied by + δx and φ m, then integrated from to. = + d 2 b n dt 2 φ nφ m + 2 ṽh x β + δx δx + δx φ m dx φ m da n dt da n dt φ ndx + 2x ṽh 2 w x t β w 2 w t t 2 m c L 2 dφ n dx dx + L δx + 2 t m c L 4 φ m + δx φ m δx dx. b n d 2 φ n dx 2 dx φ m Recall d 2 φ n dx 2 = µ2 nφ n Using and the orthogonality of the eigenfunctions, the equation can be further simplified: d 2 b m dt 2 φ 2 m + L δx dx + µ 2 m m c L 2 b m φ 2 m + L δx dx = β da m φ 2 m + L dt δx dx + 2 ṽh x + L δx da n dφ n φ m dt dx dx 2µ 2 t ma m m c L 4 + L δx φ 2 mdx + 2x ṽh 2 w x t β w 2 w t t 2 φ m + L δx dx a n d 2 φ n dx 2 dx 36

38 Dividing both sides of by + δx φ 2 mdx, leads to d 2 b m dt 2 Rename + µ 2 m m c L 2 b m = β da m 2µ 2 t dt ma m m c L 4 ṽh 2x + L δx φ m dan dt + φ2 m + δx dx 2x ṽh 2 w x t β w t 2 w and rename Define J m t = I n = + + φ2 m t 2 + δx = β da m 2µ 2 t m dt m c L 4 a m ṽh 2x L + + δx φ2 m + δx dx 2x ṽh 2 w x t β w t 2 w φ2 m ṽh 2x + δx + δx dx φ2 m 2x ṽh q m t = β da m dt 2 w x t β w φ2 m t 2 w t 2 + δx 2µ 2 t m m c L 4 a m + The equality in can be written as dφ n dx dx φ m + δx dx dφ φ n m dx dx t 2 + δx dφ φ n m dx dx da n dt φ m + δx dx , φ m + δx dx dx I n da n dt + J mt dx dx. d 2 b m dt 2 + µ2 m m c L 2 b m t = q m t This is, once again, a nonhomogeneous ordinary differential equation for b m t, to which the solution is given by 37

39 b m t = C tv t + C 2 tv 2 t, where v t and v 2 t are linearly independent solutions to the homogeneous problem, as given in , C t and C 2 t are given by t C t = W v 2tq m tdt, C 2 t = t W v tq m tdt, and where W is the Wronskian of v and v 2, as given in solution for b m t is thus given by The b m = C t sin µ m t + C 2 t cos µ m t, m c m c with C t = C 2 t = t mc µ m t µ m Summary This yields the final formula for u : cos µ m t q m tdt, m c mc sin µ m t q m tdt. m c u x, t = w x, t + v x, t, where w is now given by w = x t sin t + 2 t cos t, with v x, t is given by with t = 2 t = φ n x = t t v x, t = cos t rtdt, sin t rtdt b n tφ n x, cos µ n x µ n sin µ n x, m c 38

40 and with b n = C t sin µ n t + C 2 t cos µ n t, m c m c C t = C 2 t = t t mc µ n µ n cos µ n t q n tdt, m c mc sin µ n t q n tdt. m c Resonance When dealing with differential equation, special attention needs to be paid to the possibility of resonance. Resonance occurs when the natural frequencies the frequencies of the solutions to the homogeneous problem coincide with the forcing frequencies the frequency of the nonhomogeneous part, in this case the right hand side of the problem. Recall the O and Oε problem, { L u ] = g z t, C., and { L u ] = L u ], +.C oth of these problems are solved with the operator L, therefore have the same natural frequencies µn m c. Furthermore, the nonhomogeneous part of the Oε problem contains solutions to the O problem. Resonance is thus suspected in the Oε problem Resonance in the O problem Resonance in the O problem will occur, if the frequencies of the solutions to the homogeneous O problem, coincide with the forcing frequencies of the O problem, the frequencies of the right hand side of the differential equation in The O problem with homogeneous boundary conditions was reformulated to mc L 2 2 v x v t 2 = g z t 2 w t 2, v L x 2 v t 2 x= =, v x= =,

41 and further to m a c n L 2 µ 2 n + d2 a n dt 2 φ n x = g z t 2 w t 2 g z t 2 w m a m t c L 2 µ 2 m + d2 a m dt 2 = φ2 m The natural frequencies to are given by f m = µ m t 2 φ m + δx dx dx + δx m c If for any m, the frequencies of z or 2 w x correspond to this frequency, resonance will occur. If this is the case, special research is required. For now the assumption will be made that no resonance occurs in the O problem. t Resonance in the Oε problem Resonance occurs in the Oε problem, if the forcing frequency of the Oε problem coincides with the natural frequency of the Oε problem. The Oε problem was given by L u ] = L u ], ] f t u ] = u ] +, and this boiled down to mc L 2 2 v + 2 v x 2 t 2 v L x 2 v t 2 x= =, v x= =. This problem was reformulated as d 2 b m dt 2 + µ 2 m m c L 2 b m t m c L 4 a m + = β da m 2µ 2 m dt 2x ṽh + 2 w = 2 ṽh x 2 u x t β u t + 2 t 2 u 2 w, m cl 4 x 2 t 2 x t β w φ2 m t 2 w t 2 + δx 4 ṽh 2x + δx φ2 m + δx dx φ m + δx dx. dx dφ φ n m dx dx da n dt

42 The natural frequencies are again given by f m = µ m m c Under the assumption that no resonance occurs in the O problem, a m t can be rewritten as: µ m µ m a m t = c sin t + c 2 cos t + p m t m c m c where the first two terms are the solutions to the homogeneous problem for a m t, and p m t is a particular solution to the problem in This equality is substituted in 3.4.8, which can thus be rewritten as d 2 b m dt 2 + µ 2 m m c L 2 b m ] = β mc c µ m cos µ m mc t c 2 m c µ m sin µ m t + dp m m c dt ] 2µ 2 t µ m µ m m m c L 4 c sin t + c 2 cos t + p m t m c m c ṽh 2x L + + dφ δx φ n m dx dx φ2 m + δx dx ] mc c µ n cos µ n mc t c 2 m c µ n sin µ n t + dp n m c dt 2x ṽh 2 w x t β w t 2 w φ t + 2 m + δx dx. + δx dx φ2 m 3.4. As there are multiple terms on the right hand side of 3.4. containing sines and cosines with frequency µn m c, resonance occurs. If resonance occurs in the Oε problem, u t will grow indefinitely for increasing t. This makes the results invalid on large timescale. If t is of O however, the approximation is valid, with an error of Oε 2. 4

43 Chapter 4 Results 4. Summary of results The problem to be solved was the following: vh Lt x 2 mc Lt 2 ] ] 2 2 u + 2 vh x 2 Lt x + β v h Lt x u x ] 2 v h Lt x 2 u x t + 2 u + β u t 2 t = g z t ux, t x= = ] 2 Lt + α v h Lt + 2 vh Lt 2 v h Lt ] 2 u x vh 2 u Lt x ] 2 2 u x t + α u t 2 u t 2 x= = ft ft = g +C A F kt+f u M+C A α ż + v h z t. 4.. y applying a perturbation method, the problem was split up into the more manageable problems of O and Oε, leading to ux, t = u x, t + εu x, t + Oε This has resulted in the following equalities: u x, t = w x, t + v x, t, w x, t = x A t = t A 2 t = t A t sin t + A 2 t cos cos t f tdt sin t f tdt f t = g +C A F kt+f u M+C A z t v x, t = a ntφ n x 42 t

44 a m t = D t sin µm + D 2 t cos µm D = t mc µ m cos µm L m c t k m tdt D 2 = t mc µ m sin µm L m c t k m tdt k m t = L m c t g z t 2 w t 2 φ m + δx dx m φ2 + δx dx φ n x = cos µ n x µn m c sin µ n x u x, t = w x, t + v x, t, w x, t = x t = t 2 t = t t sin t + 2 t cos cos t rtdt sin t rtdt rt = t u L 2 x + 2 ṽh 2 u L x t α u t + f t v x, t = b ntφ n x b m t = C t sin µm + C 2 t cos µm L m c t C t = t mc µ m cos µm C 2 t = t mc µ m sin µm q m t = I n = J m t = β dam dt 2µ 2 m t L m c t L m c t a m + L m c t t x= L m c t q m tdt q m tdt m cl 4 I n dan dt ṽh 2x + δx φ dφn m dx dx m φ2 + δx dx 2x ṽh 2 w x t β w t 2 w t 2 m φ2 + δx dx + J m t φ m + δx dx 43

45 Chapter 5 Conclusions and recommendations In this thesis, the hoisting of a jacket from the sea is modeled as a forced, longitudinally vibrating string. This has resulted in a nonhomogeneous partial differential equation for ux, t with nonhomogeneous boundary conditions. A regular perturbation method has been applied, splitting ux, t up into ux, t = u x, t + εu x, t + Oε 2. This allowed for an approximation of ux, t to be constructed through two more manageable problems, an O problem and an Oε problem. Due to this perturbation method, the separate problems have been solved analytically, the results are found in chapter 4. The O problem is solved easily. However in specific cases, resonance may occur. The problem then requires additional research. The forcing frequency with the Oε problem coincides with the natural frequency of the Oε problem, and thus resonance occurs. Consequently, for increasing t, u x, t will grow indefinitely. Nonetheless, the perturbation expansion ux, t u + εu is a valid approximation for ux, t on a small timescale, i.e. for times t of O. In this case, the approximation for ux, t is accurate with an error of Oε 2. The analysis of the vibrations in the hoist cable is especially important to research the probability of jacket rebouncing. Jacket rebouncing, if it occurs, will happen during the initial phase of lifting, thus for small t. The approximation found in this thesis is therefore suitable for this purpose. The approximation is also valid for describing the vibrations in the hoist cable after potential jacket rebouncing. Due to the occuring resonance, the solutions for ux, t in this thesis are not valid on a large time scale. If an approximation for the vibrations in the hoist cable is required on a large time scale, the use of a multiple time scale method is recommended. 44

46 ibliography ] Richard Haberman. Applied Partial Differential Equations: With Fourier Series and oundary Value Problems. Pearson, 22. 2] D.F. Parker. Fields, Flows and Waves: An introduction to Continuum Mechanics. Springer, 23. 3] H.R. van Rijn. Response of Pieter Schelte main hoist cables due to longitudinal shock loads. Master of Science Thesis, Delft University of Technology,

47 Appendix A Orthogonality of the eigenfunctions The eigenfunctions satisfy the following problem: mc 2 φ n µ 2 x nφ 2 n = L 2 φ n L x + µ2 nφ n x= = φ n x= = A.. Consider two eigenfunctions φ n and φ m. Introduce Lφ = M c L 2 2 φ x 2. A..2 Then: Lφ n + µ 2 nφ n =, Lφ m + µ 2 mφ m =. A..3 Consider This can be rewritten using A..3 φ m Lφ n φ n Lφ m dx = φ m Lφ n φ n Lφ m dx. = = φ m µ 2 nφ n φ n µ 2 mφ m dx µ 2 mφ m φ n µ 2 nφ m φ n dx µ 2 m µ 2 nφ m φ n dx A..4 A..5 46

48 y applying Green s formula ], A..4 can also be rewritten to φ m Lφ n φ n Lφ m dx dφ n = φ m dx φ dφ m n dx = φ m dφ n dx φ n dφ ] m dx φ m dφ n dx φ n dφ ] m dx The boundary conditions can be used to calculate this integral: φ x = x= µ2 φ This gives = φ m Lφ n φ n Lφ m dx dφ ] n dx dφ m dx φ x= =. φ m = µ2 nφ m φ n µ2 mφ m φ n x= A..6 A..7 L µ2 nφ n φ n L ] µ2 mφ m =µ 2 n µ 2 m φ mφ n A..8 Note that this can be rewritten to an integral form using the Dirac-delta function: = µ 2 n µ 2 m φ mφ n The following equality now holds: And thus µ 2 m µ 2 nφ m φ n dx = + µ 2 m µ 2 n µ 2 n µ 2 m δxφ mxφ n xdx δx + + L δx µ 2 n µ 2 m δxφ mxφ n xdx µ 2 m µ 2 nφ m xφ n x = δx φ m xφ n xdx = φ m xφ n xdx = A..9 A.. A.. 47

49 This means that the eigenfunctions are orthogonal with weight + δx. 48