orcaflex線理論

1、第 8 頁 共 8 頁1 overviewwe use a finite element model for a line as shown in the figure below.figure1: line modelthe line is divided into a series of line segments which are then modelled by straight massless model segments with a node at each end.the model segments only model the axial and torsional p

2、roperties of the line. the other properties (mass, weight, buoyancy etc.) are all lumped to the nodes, as indicated by the arrows in the figure above.nodes and segments are numbered 1,2,3,. sequentially from end a of the line to end b. so segment n joins nodes n and (n+1).nodeseach node is effective

3、ly a short straight rod that represents the two half-segments either side of the node. the exception to this is end nodes, which have only one half-segment next to them, and so represent just one halfsegment.each line segment is divided into two halves and the properties (mass, weight, buoyancy, dra

4、g etc.) of each halfsegment are lumped and assigned to the node at that end of the segment.forces and moments are applied at the nodes - with the exception that weight can be applied at an offset. where a segment pierces the sea surface, all the fluid related forces (e.g. buoyancy, added mass, drag)

5、 are calculated allowing for the varying wetted length up to the instantaneous water surface level.segmentseach model segment is a straight massless element that models just the axial and torsional properties of the line. a segment can be thought of as being made up of two co-axial telescoping rods

6、that are connected by axial and torsional spring+ dampers.the bending properties of the line are represented by rotational springs+ dampers at each end of the segment, between the segment and the node. the line does not have to have axial symmetry, since different bend stiffness values can be specif

7、ied for two orthogonal planes of bending.this section has given only an overview of the line model. see structural model for full details.2 structural model detailsthe following figure gives greater detail of the line model, showing a single mid-line node and the segments either side of it. the figu

8、re includes the various spring+ dampers that model the structural properties of the line, and also shows the xyz-frames of reference and the angles that are used in the theory below.figure: detailed representation of line modelas shown in the diagram, there are 3 types of spring+ dampers in the mode

9、l:the axial stiffness and damping of the line are modelled by the axial spring+ damper at the centre of each segment, which applies an equal and opposite effective tension force to the nodes at each end of the segment. the bending properties are represented by rotational spring+ dampers either side

10、of the node, spanning between the nodes axial direction nz and the segments axial direction sz. if torsion is included (this is optional) then the lines torsional stiffness and damping are modelled by the torsional spring+ damper at the centre of each segment, which applies equal and opposite torque

11、 moments to the nodes at each end of the segment. if torsion is not included then this torsional spring+ damper is missing and the two halves of the segment are then free to twist relative to each other.3 calculation stagesthe program calculates the forces and moments on a mid-node in 5 stages:1. te

12、nsion forces.2. bend moments. 3. shear forces.4. torsion moments.5. total load.4 tension forcesfirstly the tensions in the segments are calculated. to do this, program calculates the distance (and its rate of change) between the nodes at the ends of the segment, and also calculates the segment axial

13、 direction sz, which is the unit vector in the direction joining the two nodes.linear axial stiffnessin the case of linear axial stiffness the tension in the axial spring+damper at the centre of each segment iscalculated as follows. it is the vector in direction sz and whose magnitude is given by:te

14、 = ea.e + (1 -2u).(po.ao - pi.ai) + ea.e.(dl/dt)/l0wherete = effective tensionea = axial stiffness of line, as specified on the line types form (= effective youngs modulus x cross-section area)e = mean axial strain = (l - l0) / (l0)l = instantaneous length of segmentl = expansion factor of segmentl0

15、 = unstretched length of segmentu = poisson ratiopi, po = internal pressure and external pressure respectively (see line pressure effects)ai, ao = internal and external cross section areas respectively (see line pressure effects)e = damping coefficient of the line, in seconds (this is defined below)

16、dl/dt = rate of increase of length.note: the effective tension te can be negative, indicating effective compression. for the relationship between effective tension and pipe wall tension see line pressure effects.-this effective tension force vector is then applied (with opposite signs) to the nodes

17、at each end of the segment. each mid-node therefore receives two tension forces, one each from the segments on each side of it.non-linear axial stiffnesswhen the axial stiffness is non-linear then the tension calculation is as follows. it is the vector in direction sz and whose magnitude is given by

18、:te = var tw() + (1 -2).(po.ao - pi.ai) + eanom.e.(dl/dt)/l0wherevar tw is the function relating strain to wall tension, as specified by the variable data source defining axial stiffness.eanom is the nominal axial stiffness which is defined to be the axial stiffness at zero strain.as in the linear c

19、ase the effective tension force vector is then applied (with opposite signs) to the nodes at each end of the segment. each mid-node therefore receives two effective tension forces, one each from the segments on each side of it.damping coefficient ethe damping coefficient e represents the structural

20、damping in the line. it is calculated automatically based on the axial target damping value specified on the general data form.e = e(critical) . (target axial damping) / 100wheree (critical) =(2.segmentmass.l0 / ea) is the critical damping value for a segment and segment mass includes the mass of an

21、y contents but not the mass of any attachments.note: if the axial stiffness is non-linear then we use the nominal axial stiffness eanom in the formula for e.5 bend momentsthe bend moments are then calculated. there are bending spring + dampers at each side of the node, spanning between the nodes axi

22、al direction nz and the segments axial direction sz. each of these spring + dampers applies to the node a bend moment that depends on the angle between the segment axial direction sz and the nodes axial direction nz.these axial directions are associated with the frames of reference of the node and s

23、egment. the nodes frame of reference nxyz is a cartesian set of axes that is fixed to (and so rotates with) the node. nz is in the axial direction and nx and ny are normal to the line axis and correspond to the end x- and y-directions that are specified by the gamma angle on the line data form (see

24、end orientation).the segment has two frames of reference - sx1 y1 z at the end nearest end a, and sx2 y2 z at the other end. these two frames have the same sz-direction, which was calculated in step 1 above, so the bend angle 2 between nz and sz can now be calculated. the effective curvature vector

25、c is then calculated - it is the vector whose direction is the binormal direction, which is the direction that is orthogonal to sz and nz, and whose magnitude is 2 / ( l0), where l0 is the unstretched length of segment.linear, isotropic bending stiffnessin the case of linear, isotropic bending stiff

26、ness the bend moment, m2, generated by the bending spring + damper is calculated. isotropic bend stiffness means that the bend stiffnesses for the x and y-directions are equal. the bend moment m2 is the vector in the binormal direction whose magnitude is given by:m2 = ei.|c| + d.d|c|/dtwhereei = ben

27、ding stiffness, as specified on the line types formd = (/100).dcdc = the bending critical damping value for a segment = l0.(segmentmass.ei.l0) = target bending damping, as specified on the general data form.the bend angle (1) and bend moment vector (m1) on the other side of the node are calculated s

28、imilarly, so the node experiences two bend moments, one each from the segments on each side of it.linear, non-isotropic bending stiffnessif the bend stiffnesses for bending about the x and y-directions are different, then the above equation is separated into its components in the sx2 and sy2 directi

29、ons, giving:component of m2 is the sx2 direction = eix.cx + dx.dcx/dtcomponent of m2 is the sy2 direction = eiy.cy + dy.dcy/dtwhereeix, eiy = bending stiffnesses of segment, as specified on the line types formcx, cy = components of the curvature vector c in the sx2 and sy2 directionsdx = (/100).l0.(

30、segmentmass.eix.l0)dy = (/100).l0.(segmentmass.eiy.l0).the curvature used in the calculation of bending moments is the value in the true plane of bending, taking full account of the 3d motions of the adjacent nodes. the bending damping term d represents the effect on bending of the structural dampin

31、g in the line; its level is set by the target bending damping data item on the general data form.the bend angle () and bend moment vector (m1) on the other side of the node are calculated similarly.non-linear, isotropic bending stiffnessin this case the bend moment m2 is given by:m2 = var bm( / l0)

32、+ d.d|c|/dtwherevar bm is the function relating curvature to bend moment as specified by the variable data source defining bending stiffnessd = (/100).dcdc = the bending critical damping value for a segment = l0.(segmentmass.einom.l0)einom is the nominal bending stiffness which is defined to be the

33、bending stiffness at zero curvature.the bend angle (1) and bend moment vector (m1) on the other side of the node are calculated similarly.6 shear forceshaving calculated the bend moments at each end of the segment, the shear force in the segment can be calculated. each model segment is a straight st

34、iff rod in which the bend moment vector varies from m1 at one end (the end nearest end a of the line) to m 2 at the other end, where these bend moments are calculated as described in step 2 : bend moments above. because the model segment is stiff in bending, the bend moment varies linearly along the

35、 segment and the shear force in the segment is the constant vector equal to the rate of change of bend moment along the length. the shear force is therefore given by:shear force vector = (m 2 - m1) / lwhere l is the instantaneous length of the segment. note that m1 and m2 are vectors, so this is a v

36、ector formula that defines both the magnitude and direction of the shear force.this shear force vector is applied (with opposite signs) to the nodes at each end of the segment.7 torsion momentsthe torsion is then calculated (providing torsion has been included). to do this, the directions sx1, sy1,

37、sx2 and sy2 must first be calculated, since so far only the segment axial direction sz has been found.the directions sx2 and sy2 at the end of the segment are determined from the orientation nxyz of the adjacent node, by rotating nxyz until its z-direction is aligned with sz. this rotation is theref

38、ore through angle 2 and it is a rotation about the binormal direction (i.e. the direction that is orthogonal to both nz and sz). (note that rotations about the binormal direction are bending rotations only - i.e. they involve no twisting.) the directions sx1 and sy1 at the other end of the segment a

39、re derived in the same way, but starting from the orientation of the node at that other end of the segment.the twist angle in the segment can then be calculated - it is the angle between the directions sx1 and sx2. note that this twist angle is also the angle between sy1 and sy2 - in other words the

40、 orientations sx1 y1 z and sx2 y2 z, at the two ends of the segment, differ by just a twist through angle .linear torsional stiffnessin the case of linear torsional stiffness the torque generated by the torsion spring+damper can then be calculated it is a moment vector whose direction is the segment

41、 axial direction sz and whose magnitude is given by:torque = k. / l0 + c.(d/dt) + a.tewherek = torsional stiffness, as specified on the line types form = segment twist angle (in radians), between directions sx1 and sx2l0 = unstretched length of segmentd/dt = rate of twist (in radians per second)c =

42、torsional damping coefficient of the line (this is defined below)a = torque per unit tension, as specified on the line types formte = effective tension in the segmentthis torque moment vector is then applied (with opposite signs) to the nodes at each end of the segment.non-linear torsional stiffness

43、if the torsional stiffness is non-linear then the calculation of torque is as follows. it is a moment vector whose direction is the segment axial direction sz and whose magnitude is given by:torque = var torque( / l0) + c.(d/dt) + a.tewhere var torque is the function relating twist per unit length t

44、o torque as specified by the variable data source defining torsional stiffness.damping coefficient cthe damping coefficient c represents the torsional effect of structural damping in the line. it is calculated automatically based on the target torsional damping value specified on the general data fo

45、rm using the formulac = c(critical).(target torsional damping) / 100where c(critical) is the critical damping value for a segment, given byc(critical) = (2.iz.k/l0).here, iz is the rotational moment of inertia of the segment about its axis, allowing only for the structural mass of the line, not the

46、mass of any contents (since the contents are assumed to not twist with the pipe).note: if the torsional stiffness is non-linear then we use the nominal torsional stiffness knom in the formula for c. in this case we define knom to be the torsional stiffness at zero twist per unit length.8 total loada

47、s described above, each mid-node experiences two tension forces, two bend moments, two shear forces and two torque moments (one each from the segments either side of the node). these loads are then combined with the other non-structural loads (weight, drag, added mass etc.) to give the total force a

48、nd moment on the node. program then calculates the resulting translational and rotational acceleration of the node, and then integrates to obtain the nodes velocity and position at the next time step. 9 hydrodynamic and aerodynamic loadsdragboth hydrodynamic and aerodynamic drag forces are applied t

49、o the line. the same drag formulation is used for hydrodynamic and aerodynamic drag forces.note: aerodynamic drag is only included if the include wind loads on lines option is enabled in the environment data.the drag forces applied to a line are calculated using the cross flow principle. that is, th

50、e fluid velocity relative to the line is split into its components vn normal to the line axis, and vz parallel to the line axis. the components of drag force normal to the line axis are then based on vn, and its x and y-components vx, vy. the component of drag force parallel to the line axis is base

51、d on vz.the drag force formulae use drag coefficients, cdx, cdy and cdz (specified on the line type data form) and the drag areas appropriate to each direction.for the directions normal to the line axis (x and y) the drag area is taken to be the projected area d.l where d is the outside diameter and

52、 l is the length of line represented by the node. for the axial direction (z) the drag area is taken to be the skin surface area .d.l.if cdx or cdy vary with reynolds number then orcaflex calculates the reynolds number using the normal component of relative velocity, vn. this reynolds number is then

53、 used to find cdx or cdy (or both) using the specified variable data table.there is a choice of the following three possible drag formulations for the drag force components (fx, fy, fz) in the local line directions. the formulations differ in how the drag force components vary with the incidence ang

54、le between the flow and the line axial direction. the formulations are reviewed in casarella and parsons.in the formulae below, is the fluid density and p is the proportion wet or proportion dry as appropriate.1）standard formulationfx = p.( .(d.l).cdx.vx.|vn| )fy = p.( .(d.l).cdy.vy.|vn| )fz = p.( .

55、(d.l).cdz.vz.|vz| )this formulation is the most commonly-used and was the formulation used by versions of orcaflex before a choice of formulation was introduced. it has been proposed or used by various authors, including richtmyer, reber and wilson. it is appropriate for general flow conditions.2）po

56、de formulationfx = same as standard formula for fx, abovefy = same as standard formula for fy, abovefz = +/- w.( .(d.l).cdz.|v| )where the sign of fz, i.e. whether it is towards end a of the line or towards end b, is the same as the axial component of the relative flow vector.this formulation is pre

57、ferred by some analysts for systems with near-tangential flow.warning: the pode formula for fz is discontinuous at= 900, since then the axial component of the flow vector is zero and so the direction of fz is undefined. in this case orcaflex sets fz to zero.3）eames formulation for bare cablesfx = w.

58、( .(d.l).cdx.vx.|vn| + .(d.l).cdz.vx.(|v| - |vn|) )fy = w.( .(d.l).cdy.vy.|vn| + .(d.l).cdz.vy.(|v| - |vn|) )fz = w.( .(d.l).cdz.vz.|v| )4）drag force variation with incidence anglethe above formulae for the drag force components can be re-written in a form that highlights how the drag force varies with incidence angle.consider the case where the line is axial