CN109408899B - Deepwater suspension cable nonlinear motion response calculation method - Google Patents
Deepwater suspension cable nonlinear motion response calculation method Download PDFInfo
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- CN109408899B CN109408899B CN201811145405.9A CN201811145405A CN109408899B CN 109408899 B CN109408899 B CN 109408899B CN 201811145405 A CN201811145405 A CN 201811145405A CN 109408899 B CN109408899 B CN 109408899B
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Abstract
The invention discloses a deepwater suspension cable nonlinear motion response calculation method, which is based on an elastic wave theory, establishes a dynamic model of a suspension cable according to a Hamilton principle from the angle of energy, deduces three-dimensional nonlinear motion equations of the suspension cable in three directions of a normal direction, a tangential direction and an auxiliary normal direction, considers plane motion, obtains respective nonlinear expressions through Taylor expansion, adopts a finite difference method to solve according to the characteristics of the equations, verifies the accuracy and reliability of the solution method, and simultaneously analyzes the reasons of errors; the method is provided for analyzing the nonlinear response law of the deepwater hoisting cable and the hoisting load under external disturbance, and can also be used for analyzing the transverse and longitudinal movement of the hoisting cable.
Description
Technical Field
The invention relates to the field of deepwater hoisting cables, in particular to a deepwater hoisting cable nonlinear motion response calculation method.
Background
The nonlinear motion response of the deep water hoisting cable is a new problem in the process of development of ocean engineering technology to deep sea, is a complex nonlinear time-varying process, and is difficult to solve due to nonlinearity caused by elastic deformation of the hoisting cable, nonlinearity of external excitation and nonlinearity of boundary conditions, and the engineering application field is limited to an underwater hoisting system, so that domestic and foreign research results are relatively few.
Disclosure of Invention
The invention aims to provide a method for calculating nonlinear motion response of a deepwater hoisting cable, so as to solve the problems in the background technology.
In order to achieve the purpose, the invention provides the following technical scheme:
a nonlinear motion response calculation method for a deepwater lifting cable comprises the following specific steps:
s1, performing deepwater cable dynamics modeling to obtain a deepwater cable nonlinear motion equation: based on the elastic wave theory, the structural characteristics and the elastic mechanical properties of the suspension cable are considered, the bending, shearing and torsional rigidity of the suspension cable are neglected, and S is used 0 Representing the geometry of the hoist cable when it is not stretched, S i Indicates the static equilibrium position, S f Representing hoist cable dynamic geometry, such as the aerial hoist cable dynamic geometry diagrams shown in FIGS. 1-2; establishing an arc coordinate s, connecting one end of a hoist cable with the mother ship, connecting the other end with a hoisting load, and setting R in the figure i (s) and R f (s, t) represent the position vectors of a point on the hoist cable at the static equilibrium position and on the dynamic curve, respectively, the three-dimensional displacement of the hoist cable relative to the equilibrium position can be represented as:
R(s,t)＝R f (s,t)-R i (s) (1-1)
respectively arranging R (s, t) along the normal direction
Tangential direction
And minor normal direction
Divided into three components R 1 (s,t)、R 2 (s,t)、R 3 (s, t) can be:
in consideration of the continuity and nonlinearity of the movement of the hoisting cable, the classical mechanics theory is not suitable for analyzing the continuous nonlinear movement of the hoisting cable any more, so from the energy angle, according to the Hamilton principle, the total energy of the hoisting cable is considered to be composed of the strain energy, the kinetic energy, the gravitational potential energy and the work done by the external force, and then the three-dimensional nonlinear movement equation of the deepwater hoisting cable is deduced.
The expression describing the transient strain energy of the hoist cable is:
wherein e is f Strain energy, s, representing cable transients f Arc length coordinate, s, representing transient configuration of cable after elongation 0 Indicating the arc length coordinate when the hoist cable is not extended.
The suspension cable is in the transient configuration x f The strain energy in time was:
wherein
Is a balance position i Strain energy of time hoist cable, L i Indicating the length of the hoist cable in the equilibrium position, L i Is a balance position x i The cross-sectional area of the hoist cable, E being the modulus of elasticity, P, of the hoist cable i (s i T) is the static tension of the suspension cable during balance, epsilon is the dynamic component of the Lagrange strain after the central line is stretched, and the expression is shown as follows;
when fluid around the hoist cable is not considered, its gravitational potential energy can be expressed as:
wherein
Expressed in equilibrium position χ i The gravitational potential energy of the hoisting cable, rho is the density of the hoisting cable, l τ And l n Individual watchShowing the direction cosines of the tangent and the normal.
At this time, for the suspension cable in the water, due to the effect of the buoyancy, the direction of the suspension cable is opposite to the direction of the gravity, and the potential energy generated by the buoyancy can be expressed as follows:
wherein
Expressed in equilibrium position χ i Potential energy produced by time buoyancy, ρ w The density of water is expressed and the other quantities have the same meaning as the formula.
Then the hoist cable is in the transient configuration χ f The kinetic energy of (c) can be expressed as:
wherein V f The absolute velocity of a mass point on a dynamic suspension cable structure is expressed as follows:
the work done by external force F acting on the hoist cable can be expressed as:
in which the external force F is divided into three directional components F along the direction of displacement 1 、F 2 And F 3 。
Then according to Hamilton principle there are:
expressions (1-4), (1-7), (1-8), (1-9), (1-10) and (1-119) are substituted into the expression (1-12), and a three-dimensional nonlinear motion equation of the hoist cable in three directions can be obtained.
Equation of tangential motion:
normal equation of motion:
secondary normal motion equation:
-ρA i U 3，tt ＝[(P i +EA i ε)U 3，s ] ，s +F 3 (1-15)
assuming that the constitutive relation of the suspension cable is linear, only the in-plane motion is considered, the terms related to the minor normal direction are ignored, and the nonlinear motion equation of the suspension cable is obtained by combining the basic equation of the elastic wave:
s2, solving the nonlinear motion response numerical value of the deep water suspension cable:
deriving a more applicable differential format according to Taylor expansion, wherein the matrix of the obtained differential format is as follows:
the differential format of displacement and velocity in relation to time takes the student's Mohammad's findings:
wherein U, v, a respectively represent displacement, velocity and acceleration, alpha 1 、α 2 、β 1 、β 2 The integration parameters are respectively 0.5, 1.0, 0.5 and 1.0.
Normal and tangential accelerations at different nodes in equations (3-49) and hoist cable nonlinear equations of motion (2-16)
And (2-17), the relationship between the external force F, can be obtained by defining discrete kinetic equations:
MA i+1 +C|V i |V i +KU i ＝(F excit ) i (3-50)
where M is the mass of the hoist cable per unit length, including the additional mass, A represents acceleration, V represents velocity, U represents displacement, F excit Representing an external stimulus.
As a further scheme of the invention: the propagation speed of the nonlinear elastic wave in the step S1 is a function related to parameters such as position coordinates, horizontal tension, and hoist cable weight ratio λ.
As a still further scheme of the invention: the essence of the finite difference method in step S2 for solving the partial differential equation is to discretize the continuous problem and convert it into a finite form linear equation set for solution, and the main solution steps include: carrying out grid division on the solution domain, and replacing a continuous function with the numerical value of a grid intersection point; secondly, constructing a proper differential format, discretizing a differential equation and deriving a linear equation set; and thirdly, performing interpolation approximation on the approximate values on the discrete points to obtain an approximate solution of the solution domain.
Compared with the prior art, the invention has the beneficial effects that: based on an elastic wave theory, from the perspective of energy, a dynamic model of the hoist cable is established according to a Hamilton principle, three-dimensional nonlinear motion equations of the hoist cable in three directions of a normal direction, a tangential direction and a sub-normal direction are deduced, plane motion is considered, respective nonlinear expressions are obtained through Taylor expansion, a finite difference method is adopted to solve according to the characteristics of the equations, the accuracy and the reliability of the solving method are verified, and meanwhile, the reason of error occurrence is analyzed; the method is provided for analyzing the nonlinear response law of the deepwater hoisting cable and the hoisting load under external disturbance, and can also be used for analyzing the transverse and longitudinal movement of the hoisting cable.
Drawings
FIG. 1 is a schematic view of the geometry of the dynamics of an aerial hoist cable.
FIG. 2 is a schematic view of the geometry of the dynamics of an aerial hoist cable.
Fig. 3 is a schematic diagram of the stress of the underwater hoisting cable in the deepwater hoisting cable nonlinear motion response calculation method.
Detailed Description
The technical solutions in the embodiments of the present invention will be clearly and completely described below with reference to the drawings in the embodiments of the present invention, and it is obvious that the described embodiments are only a part of the embodiments of the present invention, and not all of the embodiments. All other embodiments, which can be derived by a person skilled in the art from the embodiments given herein without making any creative effort, shall fall within the protection scope of the present invention.
Referring to fig. 1 to 3, in an embodiment of the present invention, a method for calculating a nonlinear motion response of a deep water hoisting cable includes the following specific steps:
s1, performing deepwater cable dynamics modeling to obtain a deepwater cable nonlinear motion equation: based on the elastic wave theory, the structural characteristics and the elastic mechanical properties of the suspension cable are considered, the bending, shearing and torsional rigidity of the suspension cable are ignored, and S is used 0 Representing the geometry of the hoist cable when it is not stretched, S i Indicates the static equilibrium position, S f A schematic diagram showing the geometry of the hoist cable dynamics, such as the geometry of the hoist cable dynamics in air shown in FIGS. 1-2; establishing an arc coordinate s, connecting one end of a hoist cable with the mother ship, connecting the other end with a hoisting load, and setting R in the figure i (s) and R f (s, t) respectively represent the position vector of a certain point on the hoisting cable on the static equilibrium position and the dynamic curve, and then the hoisting cable phaseThe three-dimensional displacement for the equilibrium position can be expressed as:
R(s,t)＝R f (s,t)-R i (s) (1-1)
respectively arranging R (s, t) along the normal direction
Tangential direction
And minor normal direction
Divided into three components R 1 (s,t)、R 2 (s,t)、R 3 (s, t) can be:
in consideration of the continuity and nonlinearity of the movement of the hoisting cable, the classical mechanics theory is not suitable for analyzing the continuous nonlinear movement of the hoisting cable any more, so from the energy angle, according to the Hamilton principle, the total energy of the hoisting cable is considered to be composed of the strain energy, the kinetic energy, the gravitational potential energy and the work done by the external force, and then the three-dimensional nonlinear movement equation of the deepwater hoisting cable is deduced.
The expression describing the transient strain energy of the hoist cable is:
wherein e is f Strain energy, s, representing cable transients f Arc length coordinate, s, representing transient configuration of cable after elongation 0 Indicating the arc length coordinate when the hoist cable is not extended.
The suspension cable is in the transient configuration x f The strain energy in time was:
wherein
Is a balance position x i Strain energy of time hoist cable, L i Indicating the length of the hoist cable in the equilibrium position, L i Is a balance position i The cross-sectional area of the hoist cable, E being the modulus of elasticity, P, of the hoist cable i (s i And t) is the static tension of the suspension cable at equilibrium, and the expression is as follows:
P i (s i ,t)＝EA i e i (1-5)
epsilon is the dynamic component of the Lagrange strain after the central line is stretched, and the expression is as follows:
when fluid around the hoist cable is not considered, its gravitational potential energy can be expressed as:
wherein
Expressed in equilibrium position χ i The gravitational potential energy of the hoisting cable, rho is the density of the hoisting cable, l τ And l n Representing the tangential and normal direction cosines, respectively.
In this case, as shown in fig. 3, when the suspension cable in water is subjected to buoyancy and the direction of the suspension cable is opposite to the direction of gravity, the potential energy generated by the buoyancy can be expressed as follows:
wherein
Expressed in equilibrium position χ i Potential energy produced by time buoyancy, ρ w The density of water is expressed and the other quantities have the same meaning as the formula.
Then the hoist cable is in the transient configuration χ f The kinetic energy of (c) can be expressed as:
wherein V f The absolute velocity of a mass point on a dynamic suspension cable structure is expressed as follows:
the work done by an external force F acting on the hoist cable can be expressed as:
in which the external force F is divided into three directional components F along the direction of displacement 1 、F 2 And F 3 。
Then according to Hamilton principle there are:
expressions (1-4), (1-7), (1-8), (1-9), (1-10) and (1-119) are substituted into the expression (1-12), and a three-dimensional nonlinear motion equation of the hoist cable in three directions can be obtained.
Equation of tangential motion:
normal equation of motion:
secondary normal motion equation:
-ρA i U 3，tt ＝[(P i +EA i ε)U 3，s ] ，s +F 3 (1-15)
it can be seen from the above derived equations that each equation contains two unknowns, tension P and curvature κ, for the hoist cable in the equilibrium state of force. Since the tension has an important influence on the nonlinear motion of the hoist cable, and the curvature is directly related to the configuration of the hoist cable and closely related to the overall stress and motion of the hoist cable, the two quantities need to be analyzed when the hoist cable is in a balanced state. Discussing the configuration of the hoisting cable in the balanced state, neglecting the action of displacement and external force, considering that the balanced state of the hoisting cable is instantaneous, and all time-related parameters can be made to be zero, and then the equation for calculating the tension and curvature of the balanced state of the hoisting cable can be obtained by the equation as follows:
P i κ i ＝(ρ-ρ w )A i gl n (1-17)
the above two equations give the balance configuration of the hoist cable, when the influence of the flow field on the hoist cable is taken into account, namely rho w When the suspension cable is not equal to 0, the equation considers the effect of buoyancy and represents the balance configuration of the suspension cable in water. Introduction of phi i Is shown by
The angle from the vertical direction, the curvature and the direction cosine can be respectively expressed as:
κ i ＝φ i ,s (1-18)
l τ ＝sinφ i (1-19)
l n ＝cosφ i (1-20)
integral transformation is performed on equations (1-16) and (1-17), and expressions of tension and curvature of the hoist cable can be obtained by substituting (1-18), (1-19) and (1-20):
wherein P is 0 The above two equations apply to a hoist cable in a relaxed state for its horizontal tension.
According to the two formulas, the tension and curvature of the suspension cable and the arc length coordinate s are in a nonlinear relation, the equation cannot be resolved, and in order to facilitate calculation and consider the nonlinearity of the suspension cable at the same time, Taylor series expansion is carried out on the two equations to four orders, so that the following can be obtained:
for convenience of expression, let λ be equal to P 0 And/[ rho ] Ag, which represents the ratio of horizontal tension to gravity per unit length of the hoist cable, and has a dimension of 1/m. When s is equal to L, the first group of the compound,
indicating the ratio of horizontal tension to hoist cable weight. The significance of the introduction of lambda is that, when neglecting horizontal tension or the horizontal tension in the hoist cable is small,
can be kept small, then equations (1-23) and (1-24) can maintain convergence of the series whether the hoist cable is in a slack or tight condition.
Neglecting the small quantity of the fourth order and above, writing out (1-23) and (1-24) again, and obtaining:
κ(s,t)＝λ(1-λ 2 s 2 ) (1-26)
as can be seen from the formulas (1-25) and (1-26), λ is one of the important factors affecting the tension and curvature (configuration) of the hoist cable, and its magnitude reflects the configuration and tightness of the hoist cable. When λ is large, the high-order term in the formula is not negligible, which means that the hoist cable is in a slack state, and vice versa, the hoist cable is in a tensioned state.
S2, simplifying a nonlinear motion equation of the deep water suspension cable: according to the derived three-dimensional motion equation in the balance state of the suspension cable, rewriting the motion equation as follows:
-ρA i R 1,tt ＝[(P+EAε)(1+R 1,s -κR 2 )] ,s -κ(P+EAε)(R 2,s -κR 1 )+F 1 (2-1)
-ρAR 2,tt ＝[(P+EAε)(R 2,s -κR 1 )] ,s -κ(P+EAε)(1+R 1,s -κR 2 )+F 2 (2-2)
-ρAR 3,tt ＝[(P+EAε)R 3,s ] ,s +F 3 (2-3)
wherein F 1 ,F 2 ,F 3 Representing external forces in three directions, P (s, t) and k (s, t) represent the tension and curvature of the hoist cable, respectively, and epsilon is dynamic strain expressed as:
assuming that the constitutive relation of the hoist cable is linear and only in-plane motion is considered, ignoring the terms related to the minor normal, the dynamic strain can be expressed as:
ε＝R 1,s -κR 2 (2-5)
considering only the in-plane motion by substituting P (s, t), κ (s, t) and equations (2-1) and (2-2), the hoist cable nonlinear motion equation can be derived as:
after the above equations (2-6) and (2-7) are collated, it is rewritten:
from the foregoing analysis, it can be seen that the expressions for the tension and curvature at the equilibrium state of the hoist cable for the parameter λ ignore the terms of the 4 th order and above 4 th order thereof, and therefore, the terms of the parameter λ 4 th order and above 4 th order in equations (2-8) and (2-9) are truncated to obtain:
combining the non-linear terms in equations (2-10) and (2-11) yields:
the above two equations (2-12) and (2-13) are very complex, and the solution becomes extremely difficult and cannot be resolved due to the plurality of nonlinear terms included in the equations. Further simplification of the two equations is required. By observing the characteristics of the equation, the basic equation of the elastic wave is combined:
as can be seen from the above equation, the parameter affecting the entire elastic wave propagation characteristic is the propagation velocity, and the visco-elastic constitutive relation also affects, that is, the terms contributing to the elastic wave propagation velocity are mainly concerned with equations (2-12) and (2-13), and other irrelevant terms are negligible, so that the simplification of equations (2-12) and (2-13) continues to be obtained:
let C 1 2 ＝a 1 +a 2 U 1,s +a 3 U 2 ，C 2 2 ＝b 1 +b 2 U 1,s +b 3 U 2 Respectively representing the tangential and normal propagation speeds of the nonlinear elastic wave, and the coefficients are as follows:
thus, the differential equations (2-16) and (2-17) of the nonlinear plane motion of the deep water suspension cable are obtained. It can be seen that the propagation velocity of the nonlinear elastic wave is a function related to parameters such as position coordinates, horizontal tension, and hoist cable weight ratio λ.
S3, solving the nonlinear motion response numerical value of the deep water suspension cable: the finite difference method is adopted for solving, the essence of the finite difference method for solving the partial differential equation is to discretize a continuous problem and convert the discretization into a linear equation set in a finite form for solving, and the main solving steps comprise: carrying out grid division on the solution domain, and replacing a continuous function with the numerical value of a grid intersection point; secondly, constructing a proper differential format, discretizing a differential equation and deriving a linear equation set; and thirdly, performing interpolation approximation on the approximate values on the discrete points to obtain an approximate solution of the solution domain.
For the established equations (2-16) and (2-17), a suitable differential format is first constructed to perform discrete approximation on the partial differential terms. Introducing a variable, and performing Taylor expansion on the variable to obtain:
different points x can be selected according to different equations and solving precision i And an order number. The related research results show that if the precision is limited in the equation solving process by adopting the three-point and two-order differential format, the calculation precision is much better when the five-point and four-order differential format is adopted.
Firstly, a fourth-order format of a first-order partial derivative is constructed, and 5 points are found to be x respectively i-2 、x i-1 、x i 、x i+1 And x i+2 Then write out the division x i The Taylor series of the other four points except the point to the strain quantity is as follows:
equations (3-2), (3-3), (3-4) and (3-5) are modified and multiplied by constants a, b, c and d to obtain:
summing equations (3-2), (3-3), (3-4) and (3-5), preserving the first derivative term du (x) i ) The other high-order terms are omitted, and the solving condition of-2 a-b + c +2d is introduced as 1 (3-10)
The first derivative term is reserved for the result after the superposition of (3-2), (3-3), (3-4) and (3-5) for determining the four values of the undetermined coefficients a, b, c and d, which is denoted by "1" in equation (3-10). In the same way, to eliminate the second derivative term, let: 4a + b + c +4d ═ 0 (3-11)
"0" in the above equation indicates that determining the four values of a, b, c, d eliminates the second derivative term. Similarly, to eliminate the third and fourth derivative terms, one can obtain:
-8a-b+c+8d＝0 (3-12)
16a+b+c+16d＝0 (3-13)
thus, four linear equations (3-10), (3-11), (3-12) and (3-31) are obtained for solving a, b, c and d, and a ═ 2/4! B-16/4! C 16/4! D-2/4! . Then a, b, c and d are brought back to the equations (3-6), (3-7), (3-8) and (3-9) to obtain the first derivative du (x) i ) The expression of/dx:
when u (x) is related to u (x) i ) Are symmetrically distributed, note that equations (3-14) are a central difference form of the fourth order, to obtain du (x) 1 ) An approximate solution of/dx, which can be given by u (x) 2 )、u(x 3 )、u(x 4 ) And u (x) 5 ) Is expressed in terms of taylor series expansion:
also, to obtain the first derivative du (x) 1 ) Dx, introduction of the condition a +2b +3c +4d ═ 1 (3-19)
And simultaneously eliminating second-order and above high-order terms, and enabling:
a+4b+9c+16d＝0 (3-20)
a+8b+27c+64d＝0 (3-21)
a+16b+81c+256d＝0 (3-22)
by solving the simultaneous equations (3-19), (3-20), (3-21) and (3-22), a can be obtained as 96/4! B-72/4! C 32/4! D-6/4! Bringing back equations (3-15), (3-16), (3-17) and (3-18), du (x) can be obtained 1 ) The expression of/dx is:
same method, du (x) 2 ) Approximate solution of/dx with u (x) 1 )、u(x 3 )、u(x 4 ) And u (x) 5 ) Is expressed in terms of taylor series, such that the system of undetermined coefficients equations is:
-a+b+2c+3d＝1 (3-24)
a+b+4c+9d＝0 (3-25)
-a+b+8c+27d＝0 (3-26)
a+b+16c+81d＝0 (3-27)
then du (x) is obtained 2 ) The expression of/dx is:
du (x) N-1 ) Dx and du (x) N ) The expression of/dx is:
thus from equations (3-14), (3-23), (3-28), (3-29) and (3-30) the matrix for the difference format can be derived as:
continuing to derive the fourth order format of the second partial differential as described above, the second derivative term is retained for equations (3-2), (3-3), (3-4) and (3-5) with 4a + b + c +4d ═ 2 (3-32)
Omitting the third order and the higher order terms above the third order, the coefficient equation can be obtained as follows:
-8a-b+c+8d＝0 (3-33)
16a+b+c+16d＝0 (3-34)
-32a-b+c+32d＝0 (3-35)
solving equations (3-32), (3-33), (3-34), and (3-35) may yield a ═ -2/4! B is 32/4! C 32/4! D-2/4! And the four coefficients are brought back to the original equation to obtain du 2 (x i )/dx 2 Expression (c):
for du 2 (x 2 )/dx 2 When x is equal to x, the approximate expression of (a) can be adopted 1 ,x 3 ,x 4 ,x 5 ,x 6 Linear combination of time u (x) represents:
au(x 1 )+bu(x 3 )+cu(x 4 )+du(x 5 )+eu(x 6 ) (3-37)
to omit du (x) 2 ) Dx, let-a + b +2c +3d +4e become 0 (3-38)
Similarly, to omit the third order and the above items, let:
-a+b+8c+27d+64e＝0 (3-39)
a+b+16c+81d+256e＝0 (3-40)
-a+b+32c+243d+1024e＝0 (3-41)
retention d 2 u(x 2 )/dx 2 Introduction of conditionsa+b+4c+9d+16e＝2 (3-42)
Solving equations (3-38), (3-39), (3-40), (3-41), and (3-42) yields a ═ 2/4! B is 32/4! C 32/4! D-2/4! Substituting the equation to obtain du 2 (x 2 )/dx 2 The expression of (c):
then d is obtained in the same way 2 u(x N-1 )/dx 2 Expression (c):
finally, d is obtained 2 u(x 1 )/dx 2 And d 2 u(x N )/dx 2 Due to d 2 u(x 1 )/dx 2 And d 2 u(x N )/dx 2 Including boundary conditions, so choose when x ═ x 2 ,x 3 ,x 4 ,x 5 Time u (x) and du (x) 1 ) Linear combination of/dx, i.e.:
solving the equation set (3-45), obtaining the coefficient, and then inputting the coefficient into the original equation to obtain d 2 u(x 1 )/dx 2 Expression (c):
then d is obtained in the same way 2 u(x N )/dx 2 The expression of (c):
equations (3-31) and (3-48) are the difference format established for solving the spatial differentials of partial differential equations (2-16) and (2-17).
The differential format of displacement and velocity in relation to time takes the student's Mohammad's findings:
wherein U, v, a respectively represent displacement, velocity and acceleration, alpha 1 、α 2 、β 1 、β 2 The integration parameters are respectively 0.5, 1.0, 0.5 and 1.0.
Normal and tangential accelerations at different nodes in equations (3-49) and hoist cable nonlinear equations of motion (2-16)
And (2-17), the relationship between the external force F, can be obtained by defining discrete kinetic equations:
MA i+1 +C|V i |V i +KU i ＝(F excit ) i (3-50)
where M is the mass of the hoist cable per unit length, including the additional mass, A represents acceleration, V represents velocity, U represents displacement, F excit Representing an external stimulus.
Although the present invention has been described in detail with reference to the foregoing embodiments, it will be apparent to those skilled in the art that various changes in the embodiments and/or modifications of the invention can be made, and equivalents and modifications of some features of the invention can be made without departing from the spirit and scope of the invention.
Claims (6)
1. A method for calculating nonlinear motion response of a deep water suspension cable is characterized in thatThe method mainly comprises the following steps of carrying out deepwater hoisting cable dynamics modeling to obtain a deepwater hoisting cable nonlinear motion equation: based on the elastic wave theory, the structural characteristics and the elastic mechanical properties of the suspension cable are considered, the bending, shearing and torsional rigidity of the suspension cable are ignored, and S is used 0 Representing the geometry of the hoist cable when it is not stretched, S i Indicates the static equilibrium position, S f Representing hoist cable dynamic geometry; establishing an arc coordinate s, connecting one end of a hoist cable with the mother ship, connecting the other end of the hoist cable with a hoisting load, and R i (s) and R f (s, t) represent the position vectors of a point on the hoist cable at the static equilibrium position and on the dynamic curve, respectively, the three-dimensional displacement of the hoist cable relative to the equilibrium position can be represented as:
R(s,t)＝R f (s,t)-R i (s) (1-1)
respectively making R (s, t) be tangential
Normal direction
And minor normal direction
Divided into three components R 1 (s,t)、R 2 (s,t)、R 3 (s, t) can be:
from the perspective of energy, according to the Hamilton principle, the total energy of the suspension cable is considered to be composed of the strain energy, the kinetic energy, the gravitational potential energy and the work done by the external force of the suspension cable;
the expression describing the transient strain energy of the hoist cable is:
wherein f represents a transient variable of the hoist cable, e f Strain energy, s, representing cable transients f Arc length coordinate, s, representing transient configuration of cable after elongation 0 An arc length coordinate representing when the hoist cable is not extended;
then the hoist cable is in the transient configuration χ f The strain energy in time was:
where i represents a physical quantity at the position of the hoist cable equilibrium,
is a balance position i Strain energy of time hoist cable, L i Indicating the length of the hoist cable in the equilibrium position, A i Is a balance position i The cross-sectional area of the hoist cable, E being the modulus of elasticity, P, of the hoist cable i (s i And t) is the static tension of the suspension cable at equilibrium, and the expression is as follows:
P i (s i ,t)＝EA i e i (1-5)
epsilon is the dynamic component of the Lagrange strain after the central line is stretched, and the expression is as follows:
R 1 is a tangential displacement component R of the hoist cable 1 Abbreviated form of (s, t), R 2 Is a tangential displacement component R of the hoist cable 2 Abbreviated form of (s, t), R 1,s For tangential displacement R of the hoist cable 1 Partial derivative of the arc length s, R 2,s For normal displacement R of cable 2 Partial derivative of the arc length s, R 3,s For normal displacement R of cable-hanging pair 3 The partial derivative of the arc length coordinate s, and kappa is the curvature of the geometrical configuration of the suspension cable;
when fluid around the hoist cable is not considered, its gravitational potential energy can be expressed as:
wherein
Expressed in equilibrium position χ i The gravitational potential energy of the hoisting cable, rho is the density of the hoisting cable, l τ And l n Denotes the cosine of the tangential and normal directions, omega, respectively i Expressed in equilibrium position χ i The water depth of the hoisting cable;
at this time, for the suspension cable in the water, due to the effect of the buoyancy, the direction of the suspension cable is opposite to the direction of the gravity, and the potential energy generated by the buoyancy can be expressed as follows:
wherein
Expressed in equilibrium position χ i Potential energy produced by time buoyancy, ρ w The other quantities have the same meanings as in the formulae (1-7), U 1 、U 2 Representing the velocity components of the water flow in the tangential direction and the normal direction, respectively;
then the hoist cable is in the transient configuration χ f The kinetic energy of (c) can be expressed as:
wherein V f The absolute velocity of a mass point on a dynamic suspension cable structure is expressed as follows:
wherein R(s) i T) represents the displacement of the hoist cable in the state of static equilibrium, R 1,t (s, t) denotes the tangential displacement R 1 Partial derivative of (s, t) with respect to time t, R 2,t (s, t) denotes the normal displacement R 2 Partial derivative of (s, t) with respect to time t, R 3,t (s, t) represents the secondary normal displacement R 3 Partial derivative of (s, t) with respect to time t, e 1 、e 2 、e 3 Respectively representing the direction vectors of the tangential direction, the normal direction and the auxiliary normal direction;
the work done by external force F acting on the hoist cable can be expressed as:
in which the external force F is divided into three directional components F along the direction of displacement 1 、F 2 And F 3 U is the speed of the water flow, R 3 Is a normal displacement component R of a suspension cable pair 3 Abbreviated form of (s, t);
then according to Hamilton principle there are:
where δ is the variation sign, t 1 、t 2 Respectively representing upper and lower integral limits;
expressions (1-4), (1-7), (1-8), (1-9), (1-10) and (1-11) are substituted into the expression (1-12), and a three-dimensional nonlinear motion equation of the suspension cable in three directions can be obtained;
tangential motion equation:
wherein R is 1,tt Indicating tangential displacement R of hoist cable 1 Solving a second-order partial derivative of the time t, namely the acceleration of the suspension cable in the tangential direction, wherein a lower corner mark tt represents that a physical quantity solves the second-order partial derivative of the time;
normal equation of motion:
wherein R is 2,tt Indicating normal displacement R of hoist cable 2 Calculating the second-order partial derivative of time t, i.e. the normal acceleration, U, of the cable 2,s Indicating normal velocity U of water flow 2 A first partial derivative of s;
secondary normal motion equation:
-ρA i R 3,tt ＝[(P i +EA i ε)U 3,s ] ,s +F 3 (1-15)
wherein R is 3,tt Indicating normal speed R of cable-hanging pair 3 Calculating the second-order partial derivative of time t, i.e. the acceleration of the cable in the auxiliary normal direction, U 3,s Indicating the normal velocity U of the water flow 3 A first partial derivative of s;
it can be seen from the equations obtained by the above derivation that each equation contains two unknowns of tension P and curvature kappa when the suspension cable is in a stress balance state; discussing the configuration of the hoisting cable in the balanced state, neglecting the action of displacement and external force, considering that the balanced state of the hoisting cable is instantaneous, and making all time-related parameters zero, the equation for calculating the tension and curvature of the balanced state of the hoisting cable can be obtained from the above equation as follows:
P i κ i ＝(ρ-ρ w )A i gl n (1-17)
the above two equations give the balance configuration of the hoist cable, when the influence of the flow field on the hoist cable is taken into account, namely rho w When not equal to 0, the equation considers the effect of buoyancy and represents the balance configuration of the underwater suspension cable; introduction of phi i To represent tangent vector
The angle from the vertical direction, the curvature and the direction cosine can be respectively expressed as:
κ i ＝φ i ,s (1-18)
l τ ＝sinφ i (1-19)
l n ＝cosφ i (1-20)
integral transformation is performed on equations (1-16) and (1-17), and the expressions of the tension and curvature of the hoist cable can be obtained by substituting (1-18), (1-19) and (1-20):
wherein P is 0 The above two equations are applicable to the hoist cable in a slack state for the horizontal tension of the hoist cable;
according to the two formulas, the tension and curvature of the suspension cable and the arc length coordinate s are in a nonlinear relation, the equation cannot be resolved, and in order to facilitate calculation and consider the nonlinearity of the suspension cable at the same time, Taylor series expansion is carried out on the two equations to four orders, so that the following can be obtained:
for convenience of expression, let λ be equal to P 0 The/[ rho ] Ag represents the ratio of horizontal tension to gravity of the unit length of the suspension cable, and the dimension of the ratio is 1/m; when s is equal to L, the first group of the compound,
the ratio of the horizontal tension to the self weight of the suspension cable is shown, wherein L is the length of the suspension cable;
neglecting the small amount of the fourth order and above, rewriting (1-23) and (1-24) to obtain:
κ(s,t)＝λ(1-λ 2 s 2 ) (1-26)。
2. the method for calculating the nonlinear motion response of the deep water hoist cable according to claim 1, characterized in that the equation of the nonlinear motion of the deep water hoist cable is subsequently simplified: according to the derived three-dimensional motion equation in the balance state of the suspension cable, rewriting the motion equation as follows:
-ρA i R 1,tt ＝[(P+EAε)(1+R 1,s -κR 2 )] ,s -κ(P+EAε)(R 2,s -κR 1 )+F 1 (2-1)
-ρAR 2,tt ＝[(P+EAε)(R 2,s -κR 1 )] ,s -κ(P+EAε)(1+R 1,s -κR 2 )+F 2 (2-2)
-ρAR 3,tt ＝[(P+EAε)R 3,s ] ,s +F 3 (2-3)
wherein F 1 ,F 2 ,F 3 Representing external forces in three directions, P (s, t) and k (s, t) represent the tension and curvature of the hoist cable, respectively, and epsilon is dynamic strain expressed as:
assuming that the constitutive relation of the hoist cable is linear and only in-plane motion is considered, ignoring the terms related to the minor normal, the dynamic strain can be expressed as:
ε＝R 1,s -κR 2 (2-5)
substituting P (s, t), κ (s, t) -into equations (2-1) and (2-2), considering only the in-plane motion, one can derive the hoist cable nonlinear motion equation as:
after the above equations (2-6) and (2-7) are collated, it is rewritten:
from the foregoing analysis, it can be seen that the expressions for the tension and curvature at the equilibrium state of the hoist cable for the parameter λ ignore the terms of 4 th order and 4 th order or more, and therefore, the terms of 4 th order and 4 th order or more for the parameter λ in equations (2-8) and (2-9) are truncated to obtain:
combining the non-linear terms in equations (2-10) and (2-11) yields:
the two equations (2-12) and (2-13) are very complex, and a plurality of nonlinear terms contained in the equations make the solution extremely difficult and an analytic solution cannot be obtained; therefore, the two equations need to be further simplified, and the characteristics of the equations are observed, and the basic equations of the elastic waves are combined:
wherein u is the displacement of the elastic wave, C represents the propagation speed of the elastic wave, tau is the propagation period of the elastic wave, and X represents the position coordinate of the elastic wave on the X axis;
as can be seen from the above equation, the parameter affecting the entire elastic wave propagation characteristic is the propagation velocity, and the visco-elastic constitutive relation also affects, that is, the terms contributing to the elastic wave propagation velocity are mainly concerned with equations (2-12) and (2-13), and other irrelevant terms are negligible, so that the simplification of equations (2-12) and (2-13) continues to be obtained:
let C 1 2 ＝a 1 +a 2 U 1,s +a 3 U 2 ，C 2 2 ＝b 1 +b 2 U 1,s +b 3 U 2 Respectively representing the tangential and normal propagation speeds of the nonlinear elastic wave, and the coefficients are as follows:
3. the method for calculating the nonlinear motion response of the deepwater hoisting cable according to claim 2, wherein the equation of the nonlinear motion of the deepwater hoisting cable is solved after subsequent simplification, namely the numerical solution of the nonlinear motion response of the deepwater hoisting cable is as follows:
for the established equations (2-16) and (2-17), firstly constructing a difference format and performing discrete approximation on partial differential terms; introducing a variable, and performing Taylor expansion on the variable to obtain:
different points x can be selected according to different equations and solving precision i And the order;
firstly, a fourth-order format of a first-order partial derivative is constructed, and 5 points are found to be x respectively i-2 、x i-1 、x i 、x i+1 And x i+2 Then write out the division x i The Taylor series of the other four points except the point to the strain quantity is as follows:
equations (3-2), (3-3), (3-4) and (3-5) are modified and multiplied by constants a, b, c and d to obtain:
summing equations (3-2), (3-3), (3-4) and (3-5), preserving the first derivative term du (x) i ) The other high-order terms are omitted, and the solving condition of-2 a-b + c +2d is introduced as 1 (3-10)
The first derivative term is reserved for the result after the (3-2), (3-3), (3-4) and (3-5) are superposed for determining four values of the undetermined coefficients a, b, c and d, which is expressed by '1' in the equation (3-10); in the same way, to eliminate the second derivative term, let: 4a + b + c +4d ═ 0 (3-11)
"0" in the above equation indicates that determining the four values of a, b, c, d eliminates the second derivative term; similarly, to eliminate the third and fourth derivative terms, one can obtain:
-8a-b+c+8d＝0 (3-12)
16a+b+c+16d＝0 (3-13)
thus, four linear equations (3-10), (3-11), (3-12) and (3-13) are obtained for solving a, b, c and d, and a ═ 2/4! B-16/4! C 16/4! D-2/4! (ii) a Then a, b, c and d are brought back to the equations (3-6), (3-7), (3-8) and (3-9) to obtain the first derivative du (x) i ) The expression of/dx:
when u (x) is related to u (x) i ) Are symmetrically distributed, note that equations (3-14) are a central difference form of the fourth order, to obtain du (x) 1 ) An approximate solution of/dx, which can be given by u (x) 2 )、u(x 3 )、u(x 4 ) And u (x) 5 ) Is expressed in terms of taylor series expansion:
also, to obtain the first derivative du (x) 1 ) Dx, introduction of the condition a +2b +3c +4d ═ 1 (3-19)
And simultaneously eliminating second-order and above high-order terms, and enabling:
a+4b+9c+16d＝0 (3-20)
a+8b+27c+64d＝0 (3-21)
a+16b+81c+256d＝0 (3-22)
by solving the simultaneous equations (3-19), (3-20), (3-21) and (3-22), a can be obtained as 96/4! Where b is-72/4! C 32/4! D-6/4! Bringing back equations (3-15), (3-16), (3-17) and (3-18), du (x) can be obtained 1 ) The expression of/dx is:
same method, du (x) 2 ) Approximate solution of/dx with u (x) 1 )、u(x 3 )、u(x 4 ) And u (x) 5 ) Is expressed in terms of taylor series, such that the system of undetermined coefficients equations is:
-a+b+2c+3d＝1 (3-24)
a+b+4c+9d＝0 (3-25)
-a+b+8c+27d＝0 (3-26)
a+b+16c+81d＝0 (3-27)
then du (x) is obtained 2 ) The expression of/dx is:
du (x) N-1 ) Dx and du (x) N ) The expression of/dx is:
thus from equations (3-14), (3-23), (3-28), (3-29) and (3-30) the matrix for the difference format can be derived as:
continuing to derive the fourth order format of the second partial differential as described above, the second derivative term is retained for equations (3-2), (3-3), (3-4) and (3-5) with 4a + b + c +4d ═ 2 (3-32)
Omitting the third order and the higher order terms above the third order, the coefficient equation can be obtained as follows:
-8a-b+c+8d＝0 (3-33)
16a+b+c+16d＝0 (3-34)
-32a-b+c+32d＝0 (3-35)
solving equations (3-32), (3-33), (3-34), and (3-35) may yield a ═ -2/4! B is 32/4! C 32/4! D-2/4! And the four coefficients are brought back to the original equation to obtain du 2 (x i )/dx 2 Expression (c):
for du 2 (x 2 )/dx 2 When x is equal to x, the approximate expression of (a) can be adopted 1 ,x 3 ,x 4 ,x 5 ,x 6 Linear combination of time u (x) represents:
au(x 1 )+bu(x 3 )+cu(x 4 )+du(x 5 )+eu(x 6 ) (3-37)
to omit du (x) 2 ) (ii)/dx, let-a + b +2c +3d +4e equal 0 (3-38)
Similarly, to omit the third order and the above items, let:
-a+b+8c+27d+64e＝0 (3-39)
a+b+16c+81d+256e＝0 (3-40)
-a+b+32c+243d+1024e＝0 (3-41)
retention d 2 u(x 2 )/dx 2 The introduction condition a + b +4c +9d +16e is 2 (3-42)
Solving equations (3-38), (3-39), (3-40), (3-41), and (3-42) yields a ═ 2/4! B is 32/4! C 32/4! D-2/4! Substituting the equation into the equation to obtain du 2 (x 2 )/dx 2 The expression of (c):
then d is obtained in the same way 2 u(x N-1 )/dx 2 Expression (c):
finally, d is obtained 2 u(x 1 )/dx 2 And d 2 u(x N )/dx 2 Due to d 2 u(x 1 )/dx 2 And d 2 u(x N )/dx 2 Including boundary conditions, so choose when x ═ x 2 ,x 3 ,x 4 ,x 5 Time u (x) and du (x) 1 ) Linear combination of/dx, i.e.:
solving the equation set (3-45), obtaining the coefficient, and then inputting the coefficient into the original equation to obtain d 2 u(x 1 )/dx 2 Expression (c):
then d is obtained in the same way 2 u(x N )/dx 2 Expression (c):
equations (3-31) and (3-48) are the difference format established for solving the spatial differentials of partial differential equations (2-16) and (2-17);
the differential format of displacement and velocity in relation to time takes the student's Mohammad's findings:
wherein U is 1 、U 2 Respectively representing the displacement of the hoist cable in different directions, v 1 、v 2 Respectively representing the speeds of the hoist cable in different directions, a 1 、a 2 Respectively representing accelerations of the hoist cable in different directions, alpha 1 、α 2 、β 1 、β 2 Respectively taking the integral parameters of 0.5, 1.0, 0.5 and 1.0;
the relationship between normal and tangential acceleration at different nodes in equations (3-49) and external force F in the hoist cable nonlinear motion equations (2-16) and (2-17) can be obtained by defining discrete kinetic equations:
MA i+1 +C|V i |V i +KU i ＝(F excit ) i (3-50)
where M is the mass of the hoist cable per unit length, including the additional mass, A represents acceleration, V represents velocity, U represents displacement, F excit Denotes an external stimulus, C denotes a viscosity coefficient, and K denotes an elastic coefficient.
4. The method of claim 2, wherein the propagation velocity of the nonlinear elastic wave is a function related to the position coordinates, the horizontal tension, and the hoist cable weight ratio λ parameter.
5. The method for calculating the nonlinear motion response of the deepwater hoisting cable according to claim 3, wherein the essence of solving the partial differential equation by the finite difference method is to discretize a continuous problem and convert the discretization into a finite form of a linear equation system for solving.
6. The method for calculating the nonlinear motion response of the deepwater hoisting cable according to claim 5, wherein the main solving step of the finite difference method comprises the following steps: carrying out grid division on the solution domain, and replacing a continuous function with the numerical value of a grid intersection point; secondly, constructing a proper differential format, discretizing a differential equation and deriving a linear equation set; and thirdly, performing interpolation approximation on the approximate values on the discrete points to obtain an approximate solution of the solution domain.
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