Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2013, Article ID 760939, 17 pages http://dx.doi.org/10.1155/2013/760939 Research Article Vibrations due to Flow-Driven Repeated Impacts Sumin Jeong and Natalie Baddour Department of Mechanical Engineering, University of Ottawa, 161 Louis Pasteur, Ottawa, ON, Canada K1N 6N5 Correspondence should be addressed to Natalie Baddour; [email protected] Received 15 March 2013; Accepted 21 May 2013 Academic Editor: Jia-Jang Wu Copyright © 2013 S. Jeong and N. Baddour. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We consider a two-degree-of-freedom model where the focus is on analyzing the vibrations of a fixed but flexible structure that is struck repeatedly by a second object. e repetitive impacts due to the second mass are driven by a flowing fluid. Morison’s equation is used to model the effect of the fluid on the properties of the structure. e model is developed based on both linearized and quadratic fluid drag forces, both of which are analyzed analytically and simulated numerically. Conservation of linear momentum and the coefficient of restitution are used to characterize the nature of the impacts between the two masses. A resonance condition of the model is analyzed with a Fourier transform. is model is proposed to explain the nature of ice-induced vibrations, without the need for a model of the ice-failure mechanism. e predictions of the model are compared to ice-induced vibrations data that are available in the open literature and found to be in good agreement. erefore, the use of a repetitive impact model that does not require modeling the ice-failure mechanism can be used to explain some of the observed behavior of ice-induced vibrations. 1. Introduction In this paper, we consider the analysis of the vibrations of a structure that is repeatedly struck by a second object. e repetitive nature of the impacts of the second, free object on the primary (assumed fixed but flexible) structure is due to a flowing fluid that repeatedly drives the free object onto the primary structure. e analysis of this model is based on two equations from mechanics: Morison’s equation and the conservation of momentum. Morison’s equation is adopted to add the influence of fluid flow on the properties of the structure that is assumed to be immersed in the fluid. e conservation of momentum is combined with the coefficient of restitution to derive the model in which the force on the structure is considered to be a series of impacts due to the free object being driven by the fluid flow. Numerical simulations are conducted to obtain insight into the dynamics predicted by the model. is model is proposed and investigated in order to attempt to explain some of the observed nature of ice-induced vibrations. 2. Motivation Research analyzing ice mechanics has been conducted by the National Research Council of Canada, the Canadian Coastguard, U.S. Navy, various universities, the offshore industry, and several oil companies [1]. In the early stages, field and laboratory experiments were used to investigate the mechanism for ice-induced vibrations (IIV). Blenkarn [2] presented ice force data which he recorded in drilling platforms of Cook Inlet from the winter of 1963 to 1969. His paper became the basis of the negative damping or self- excited model. Toyama et al. [3] suggested phase division of an ice forcing function based on small scale tests and this approach was also used by other researchers. Ranta and R¨ aty [4] also proposed a similar idea of phase division. A more recent experimental publication was presented by Barker et al. [5], in which extensive small scale tests of wind turbines in Danish waters were investigated. e results provided information on ice forces which vary with the shapes of structures. Recent publications of Sodhi’s experiments [6–8] also contributed to the understanding of the dependency of ice forces on ice velocities. Along with the experimental progress, theoretical models of IIVs were also proposed by attempting to define the origin of the vibrations. Aſter Peyton’s early publication [9], Neill [10] suggested that crushed ice tends to break into a certain size. He proposed that the fracture size and velocity of the accompanying ice sheet determine a characteristic failure
Vibrations due to Flow-Driven Repeated Impacts
Mar 21, 2016
We consider a two-degree-of-freedom model where the focus is on analyzing the vibrations of a fixed but flexible structure that is struck repeatedly by a second object. The repetitive impacts due to the second mass are driven by a flowing fluid. . Conservation of linear momentum and the coefficient of restitution are used to characterize the nature of the impacts between the two masses. This model is proposed to explain the nature of ice-induced vibrations, without the need for a model of the ice-failure mechanism. The predictions of the model are compared to ice-induced vibrations data that are available in the open literature and found to be in good agreement.
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 760939 17 pageshttpdxdoiorg1011552013760939
Research ArticleVibrations due to Flow-Driven Repeated Impacts
Sumin Jeong and Natalie Baddour
Department of Mechanical Engineering University of Ottawa 161 Louis Pasteur Ottawa ON Canada K1N 6N5
Correspondence should be addressed to Natalie Baddour nbaddouruottawaca
Received 15 March 2013 Accepted 21 May 2013
Academic Editor Jia-Jang Wu
Copyright copy 2013 S Jeong and N Baddour This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
We consider a two-degree-of-freedom model where the focus is on analyzing the vibrations of a fixed but flexible structure that isstruck repeatedly by a second objectThe repetitive impacts due to the secondmass are driven by a flowing fluidMorisonrsquos equationis used to model the effect of the fluid on the properties of the structure The model is developed based on both linearized andquadratic fluid drag forces both of which are analyzed analytically and simulated numerically Conservation of linear momentumand the coefficient of restitution are used to characterize the nature of the impacts between the two masses A resonance conditionof the model is analyzed with a Fourier transformThis model is proposed to explain the nature of ice-induced vibrations withoutthe need for a model of the ice-failure mechanism The predictions of the model are compared to ice-induced vibrations data thatare available in the open literature and found to be in good agreementTherefore the use of a repetitive impact model that does notrequire modeling the ice-failure mechanism can be used to explain some of the observed behavior of ice-induced vibrations
1 Introduction
In this paper we consider the analysis of the vibrations ofa structure that is repeatedly struck by a second object Therepetitive nature of the impacts of the second free objecton the primary (assumed fixed but flexible) structure is dueto a flowing fluid that repeatedly drives the free object ontothe primary structure The analysis of this model is based ontwo equations from mechanics Morisonrsquos equation and theconservation of momentum Morisonrsquos equation is adoptedto add the influence of fluid flow on the properties of thestructure that is assumed to be immersed in the fluid Theconservation of momentum is combined with the coefficientof restitution to derive the model in which the force on thestructure is considered to be a series of impacts due to the freeobject being driven by the fluid flow Numerical simulationsare conducted to obtain insight into the dynamics predictedby the model This model is proposed and investigated inorder to attempt to explain some of the observed nature ofice-induced vibrations
2 Motivation
Research analyzing ice mechanics has been conducted bythe National Research Council of Canada the Canadian
Coastguard US Navy various universities the offshoreindustry and several oil companies [1] In the early stagesfield and laboratory experiments were used to investigatethe mechanism for ice-induced vibrations (IIV) Blenkarn[2] presented ice force data which he recorded in drillingplatforms of Cook Inlet from the winter of 1963 to 1969His paper became the basis of the negative damping or self-excited model Toyama et al [3] suggested phase division ofan ice forcing function based on small scale tests and thisapproach was also used by other researchers Ranta and Raty[4] also proposed a similar idea of phase division A morerecent experimental publication was presented by Barkeret al [5] in which extensive small scale tests of wind turbinesin Danish waters were investigated The results providedinformation on ice forces which vary with the shapes ofstructures Recent publications of Sodhirsquos experiments [6ndash8]also contributed to the understanding of the dependency ofice forces on ice velocities
Along with the experimental progress theoretical modelsof IIVs were also proposed by attempting to define the originof the vibrations After Peytonrsquos early publication [9] Neill[10] suggested that crushed ice tends to break into a certainsize He proposed that the fracture size and velocity of theaccompanying ice sheet determine a characteristic failure
2 Mathematical Problems in Engineering
frequency which in turn decides the forcing frequency [6]This point of view defined the origin of IIVs as being viaa characteristic failure mechanism a point of view whichwas supported by Matlock et al [11] In contrast to thecharacteristic failure mechanism Blenkarn [2] explainedIIVs as a self-excited vibration due to negative dampingAccording to this theory IIVs originate from the interactionbetween a flexible structure and decreasing ice crushingforces with increasing stress rate Maattanen one of the bigcontributors of the self-excited model verified his modelthrough field and laboratory experiments [7 12 13] SimilarlyXu andWang [14] also proposed an ice force oscillator modelbased on the self-excited model
As research in the field progressed different physicalprocesses to explain the origin of the ice force were proposedIn particular attention was turned to whether the ice wasconsidered to fail in crushing (compression) or in bendingIt is thought that compressive (crushing) failures occur as aresult of ice interacting with a narrow vertical (cylindrical)structure [15] Tsinker defines crushing as the completefailure of granularization of the solid ice sheet into particlesof grains or crystal dimension no cracking flaking or anyother failure mode occurs during pure crushing Specificmodels that correspond to ice in crushing failure are givenin [3 16] Even though these two articles model the samephysical phenomena they are different in their modellingapproaches
The concept of adding ice-breaking cones to cylindricalstructures was proposed in the late 1970s changing theeffective shape of the structure from cylindrical to conicalThe ice force on a conical structure is smaller than the forceon a cylindrical structure of similar size [5 8 17 18] It isthought that the main reason for the reduction in ice forceis that a well-designed cone can change the ice-failure modefrom crushing to bending It is considered that the primaryfailure mechanism for ice interacting with a conical structureis that of bending failure In this failure mode the ice sheetsimpacting on a cone fail by bending and typically the icebreaks almost simultaneously in each event of ice failureWith conical structures and bending ice failure analyticalmodels are more likely to be characteristic failure frequencymodels that were initially proposed in the 70srsquo since the iceforce can effectively be modelled as only depending on theproperties of the ice With conical structures the local iceforces were experimentally found to drop to almost zeroafter each event of ice failure [19 20] and are consequentlymodelled analytically with a periodic function For examplethe forcing function suggested by Qu et al [21] is a simpleone-degree-of-freedom model corresponding to a periodicice force This model is based on their own previous work[19 20] aswell as that ofHirayama andObara [18]Thismodelattempts to represent an ice sheet failing through the bendingmode and simply models the forcing function as a saw-toothshaped periodic force From a modelling point of view thebending failure of ice is easier to model since the ice-forcingfunction is periodic and is decoupled from the motion of thestructure This type of model thus becomes a regular forcedvibration problem and once the form of the external ice forceis chosen it is a relatively straightforward problem to solve
Despite extensive studies no theoretical model com-pletely explains the IIV mechanismThe variety of modellingapproaches and dependence of the ice load on the structuralproperties have caused modelling and design difficulties Forinstance there is nomethod for choosing the proper dynamicice case for the design and optimization of an ice-resistantjacket platform [22 23] In addition most IIV models donot consider the influence of fluid flow even though moststructures subjected to IIVs are offshore structures Sincefluid flow is one of the main driving forces of IIVs it cancontribute to the dynamics of ice forces as well as that of thestructure but it has not been considered in past research
Since the first modeling approaches were introduced nomajor advances in IIV modeling have been proposed andthere is no general consensus on one correct approach Allexisting modeling approaches analyze the process of IIVsfrom a microscopic point of view with a heavy emphasis onthe mode in which the ice fails For example in [5] numeroussmall scale tests were conducted to attempt to distinguish icefailures into four different modes by the sizes of failed icepieces and the structural responsesTheprocedure of definingice-failure modes is thus neither clear nor generally agreedupon
In this paper we propose that amacroscopic point of viewmay give a different insight For instance a detailed analysisof all possible deformations is not generally considered whena collision of two particles is analyzed In that case the wholecollision process can be efficiently analyzed with a singleparameter the coefficient of restitutionThemain idea of thispaper is that if the collision of the particles occurs repetitivelythe movement of the two particles may resemble those ofIIVs
This simple concept provides the inspiration for the newmodeling approach proposed in this paper The vibrationcharacteristics of IIVs are proposed as being the result of arepetitive forcing function with the moving fluid repeatedlypushing the ice towards the structure If all the microscopicice-failure processes can be condensed into a single macro-scopic parameter the coefficient of restitution the remainingvibration characteristics are similar to those of repetitivedriven collisions of two particles The ice is modeled asone of the particles in the collision and the structure is theother IIVs can then be modeled and simulated by usingthe conservation of momentum along with the coefficient ofrestitution which is a completely different perspective fromexisting IIV models that take a microscopic view and focuson the precise mechanism of ice failure
3 Modeling of Flow-Driven Repeated Impactswith the Conservation of Momentum
We consider a two degree-of-freedom system that consists oftwo masses Figure 1 is a schematic diagram of two masseswhich collide repetitively Mass 119872 represents the free massand mass119898 represents the flexible structure
The force which drives the free mass needs to be modeledso that collisions occur repetitivelyThe freemass is driven byvarious external forces such as wind and thermal expansion
Mathematical Problems in Engineering 3
M
u u
x x x0
z z
m
k
c
Figure 1 Mechanical model of the system
but the main driving force is fluid flow Depending on theinitial displacement and velocity of the flow the velocity ofthe free mass is determined just prior to the time of impactand the maximum velocity of the free mass is taken to be thevelocity of the flowThe free mass moves until it collides withthe structure After the impact the free mass either stopsmoves forward or bounces back in the opposite directiondepending on the coefficient of restitution and conservationof total system (free mass and structure) momentum Sincethe free mass and fluid flow now have a relative velocitywith respect to each other the velocity of the free mass ischanged by the effect of fluid drag until the free mass reachesthe velocity of the flow and then moves again towards thestructure for another impact The free mass moves towardsthe structure and transfers its momentum to the structureagain when the relative displacement between the mass andstructure becomes 0 (ie when a collision occurs) In somecases the structure can transfermomentum to themasswhenthemomentumof the structure is larger then that of themassThis whole process keeps repeating but due to structuraldamping and fluid drag the momenta which are transferredto each other gradually decrease
Since mass 119872 is floating in the flow it is free to moveThe flow however moves with its own velocity which inthis case is always assumed to be positive (to the right inthe figure) The displacement of the free mass is given by thevariable 119911 and that of the structure is given by the variable119909 The mass is driven by the drag force 119891
119863 which is present
when there is a relative velocity between the flow and themass There are four conditions depending on the relativevelocities
(i) When ge 0 and || ge || 119891119863le 0
(ii) When ge 0 and || lt || 119891119863gt 0
(iii) When lt 0 and || ge || 119891119863ge 0
(iv) When lt 0 and || lt || 119891119863gt 0
Outside of a collision with the structure no force is trans-mitted to the free mass except the drag force due to the fluidThe structure imparts a force to the freemass when a collisionbetween them occursThe drag force from the fluid is definedas
119891119863= minus
1
2120588119862119863119860( minus )
2 (1)
where 120588 119862119863 and 119860 are the density of the fluid drag
coefficient and area related to the drag coefficient respec-tively [24] The drag force which influences the systemof Figure 1 changes the forcing direction according to therelative velocities as explained above The drag force for thefree mass of Figure 1 therefore is expressed as
119891119863= minus
1
2120588119862119863119860 | minus | ( minus ) (2)
The equation ofmotion of the freemass can be obtained from(2) since no other forces are involved and is thus given as
119872 = minus1
2120588119862119863119860 | minus | ( minus ) (3)
Since (3) is a nonlinear equation it should be linearizedor simulated numerically The linearized version of (3) isdeveloped by assuming a small relative velocity Equation (3)is linearized as
119872 = minus119887 ( minus ) (4)
where
119887 =1
2120588119862119863119860 | minus | (5)
| minus | is assumed to be small and constant The solution of(4) consists of homogeneous and particular solutions and canbe obtained as
119911 (119905) = 1198881+ 119905 + 119888
2119890minus(119887119872)119905
(6)
The velocity of the free mass is then given by the derivative of(6) as
(119905) = minus 1198882
119887
119872119890minus(119887119872)119905
(7)
Equation (6) is a closed form solution of linearized equation(3) The solution of nonlinear equation (3) will be developedlater since it requires a numerical approach
The structure is modeled as a single-degree-of-freedomsystem consisting of a mass damper and spring withoutexternal forces (other than impacts with the free mass) asin Figure 1 Since the source of structural vibrations is thetransfer of momenta during collisions outside the time ofthe collisions the structure vibrates as a free simple harmonicoscillator The equation of structural motion is expressed as
119898 + 119888 + 119896119909 = 0 (8)
For an underdamped case (8) has the following solution
119909 (119905) = 1198883119890minus120577120596119899119905 cos120596
119889119905 + 1198884119890minus120577120596119899119905 sin120596
119889119905 (9)
with 120577 120596119889 and 120596
119899having the usual interpretation and
definition as damping coefficient damped frequency andnatural frequency The velocity of the structure at time 119905 canbe calculated as
(119905) = minus 1205961198891198883119890minus120577120596119899119905 sin120596
119889119905 minus 120577120596
1198991198883119890minus120577120596119899119905 cos120596
119889119905
+ 1205961198891198884119890minus120577120596119899119905 cos120596
119889119905 minus 120577120596
1198991198884119890minus120577120596119899119905 sin120596
119889119905
(10)
4 Mathematical Problems in Engineering
From (6) (7) (9) and (10) the displacements and veloci-ties of the mass and structure at any given time can be foundSubstituting these expressions into the equations for theconservation of total systemmomentum and the equation forthe coefficient of restitution the displacements and velocitiesimmediately after impact can be evaluated Each of the twomasses however has two unknowns after impact Thus theunknowns after impact are the displacement and velocityof both masses Since only two equations can be developedfrom the conservation of momentum and the coefficientof restitution equation the number of unknowns exceedsthe number of equations by two Salapaka et al [25] alsoinvestigated a similar modeling approach They developedthe equation of a mass-spring system driven by a vibratingtable which impacts themass at a certain position repetitivelyFaced with the same problem of excessive unknowns theyassumed that the displacements of each mass before andafter impact remain the same thus reducing the number ofunknowns after impact to two [25]
The instants just before and just after impact will bedenoted by 119905
119896and 119905119891 Since the impact occurs when the dis-
placements of the twomasses are the same the displacementsat time 119905
119896are given as
119909 (119905119896) = 119911 (119905
119896) = 119909119896 (11)
The same notation is used for velocities From the conserva-tion of momentum the total momentum of the two masses119898 and119872 is conserved so that
119898119891+119872
119891= 119898119896+119872
119896 (12)
The second relevant impact equation is obtained from thecoefficient of restitution [26] as
119891minus 119891= 119890 (
119896minus 119896) (13)
Equations (12) and (13) can be rewritten in matrix form as
[119898 119872
1 minus1] [
119891
119891
] = [119898119896+119872
119896
119890119896minus 119890119896
] (14)
or
[119891
119891
] =minus1
119898 +119872[minus1 minus119872
minus1 119898][
119898119896+119872
119896
119890119896minus 119890119896
] (15)
From (15) the velocities after impact 119891and 119891 are
119891=
1
119898 +119872(119898119896+119872
119896+ 119890119872
119896minus 119890119872
119896) (16)
119891=
1
119898 +119872(119898119896+119872
119896minus 119890119898
119896+ 119890119898
119896) (17)
After impact the displacements are assumed to be thesame as before impact From (11)
119909 (119905119891) = 119909 (119905
119896) = 119909119896
119911 (119905119891) = 119911 (119905
119896) = 119909119896
(18)
Once the impact occurs the momentum of the free mass istransmitted to the structure or vice versa 119888
1 1198882 1198883 and 119888
4of
(6) (7) (9) and (10) need to be repeatedly calculated aftereach impact in order to obtain the new responses Since thecharacteristics of the system responses are reset after eachimpact the time right after the impact 119905
119891 is also reset to 0
to calculate 1198881 1198882 1198883 and 119888
4 Substituting 119905
119891= 0 into (6) and
(7)
119911 (119905119891) = 119911 (0) = 119888
1+ 1198882
(119905119891) = (0) = minus
119887
1198721198882
(19)
The displacements before and after impact are assumed to bethe same and the velocity of the free mass after the impactis calculated from (17) Equation (19) is equal to 119909
119896and
119891
respectively
119911 (0) = 1198881+ 1198882= 119909119896
(0) = minus119887
1198721198882= 119891
(20)
From (20) 1198881and 1198882are calculated as
1198881= 119909119896minus119872
119887( minus
119891)
1198882=119872
119887( minus
119891)
(21)
Applying the same assumption and (16) to (9) and (10)
119909 (119905119891) = 119909 (0) = 119888
3= 119909119896
(119905119891) = (0) = minus120577120596
1198991198883+ 1205961198891198884= 119891
(22)
It then follows that 1198883and 1198884are obtained from (22) as
1198883= 119909119896
1198884=
1
120596119889
(119891+ 120577120596119899119909119896)
(23)
Substituting the calculated 1198881 1198882 1198883 and 119888
4into (6) and (9)
the responses of the two masses after impact are expressed as
119911 (119905) = [119909119896minus119872
119887( minus
119891)] + 119905 +
119872
119887( minus
119891) 119890minus(119887119872)119905
119909 (119905) = 119909119896119890minus120577120596119899119905 cos120596
119889119905 +
1
120596119889
(119891+ 120577120596119899119909119896) 119890minus120577120596119899119905 sin120596
119889119905
(24)
All the equations are developed based on the linearized dragforce Equation (24) holds until the next impact After thenext impact 119888
1 1198882 1198883 and 119888
4should be recalculated according
to new values of 119909119896 119896 119911119896 and
119896obtained from the impact
equations
Mathematical Problems in Engineering 5
4 Numerical Simulations
41 Linearized Drag Force For the numerical simulationsthere were two primary concerns how to identify the instantof an impact and how to define the impact condition Sincethe basis of this repeated impact model is the exchangeof momenta through impacts the time and condition ofthe impact need to be defined as precisely as possibleThe solution chosen to address this was to use a timestepping algorithm In the beginning 119888
1 1198882 1198883 and 119888
4are
calculated with the initial conditions By substituting 1198881 1198882
1198883 and 119888
4into (24) the responses of the free mass and
structure are obtained Based on the calculated response therelative displacement between the free mass and structure iscompared to the impact condition at every time step Theimpact condition is theoretically given by
|119909 (119905) minus 119911 (119905)| = 0 (25)
The actual numerical simulation however cannot achieveenough precision to capture the exact impact conditionTherefore a tolerance for the impact condition is requiredto allow for numerical round-off errors of the numericalsimulation
The tolerance should be considered carefully because toosmall tolerance has no meaning and a large tolerance makesimpacts occur more often than they should In the algorithmthe tolerance for the impact condition is set flexibly by thefollowing equation
Δ =1
10min (|120575119909| |120575119911|) (26)
where Δ is the impact condition such that |119909(119905) minus 119911(119905)| le Δ
and 120575119909 = (119905)Δ119905 and 120575119911 = (119905)Δ119905 Furthermore Δ119905 is thetime stepThe flexible impact condition hasmerit over a fixedone When the responses of the two masses are very smallthe fixed impact condition cannot capture some impactsTheflexible impact condition however is determined accordingto the size of the responses therefore the impact conditionis able to capture impacts whether the responses are slow orfast
After the comparison of the relative displacement andthe impact condition if the relative displacement is largerthan the impact condition the same values of 119888
1 1198882 1198883 and
1198884will be forwarded to the next time step Otherwise the
relative displacement is equal to or less than the impactcondition and thus an impact takes place Once the algorithmcatches the impact 119888
1 1198882 1198883 and 119888
4are calculated based on
the values from the previous time step and new conditionscorresponding to the impact as explained in the previoussection In addition the algorithm is performed on twodifferent time scales one is global time and the other is localtime The global time is the overall time starting with thebeginning of the algorithm and the local time counts timeduration between impacts thus after each impact the localtime is reset to 0 while the global time increases continuouslyWith 0 local time new responses of the system based on thenew 1198881 1198882 1198883 and 119888
4begin and evolve until next impact This
process keeps repeating until the system reaches the steadystate
The steady-state condition is also implemented in thealgorithm to avoid infinitesimal impacts which occur at theend of the responses The steady-state condition involvesdisplacements velocities and accelerations The algorithmverifies at every step whether or not both responses of the iceand structure satisfy the steady-state conditions The steady-state conditions are given as
(i) max(|119911(119905)| |119909(119905)|) lt 00001m(ii) max(|(119905)| |(119905)|) lt 00001ms(iii) max(|(119905)| |(119905)|) lt 00001ms2
The value of 00001 was chosen through trial and error for thesteady-state condition When the system reaches the steadystate the structure is constantly forced by the free mass andso the equation of motion in the steady state is expressed as
119898 + 119888 + 119896119909 = minus119887 ( minus ) (27)
Since the system is in the steady state
119896119909 = 119887 (28)
In the steady state condition the twomasses collide andmovetogether therefore the responses are given by
119909 (119905) = 119911 (119905) =119887
119896 (29)
which is a constant (steady state) displacement since isthe (assumed) constant velocity of the flow The algorithmreplaces (24) with (28) when the steady-state condition issatisfied
Figure 2 is plotted using the same structural propertiesin Table 1 which are obtained from [27] The two free massproperties mass 119872 and area 119860 are arbitrarily chosen as1600 kg which is the same as the structural mass and 1m2respectively The reason for this is as follows If the free massis much larger than the structural mass then its displacementand velocity would be hardly affected by the collisionwith thestructure and in fact repeated impacts would not occur Onthe other hand if the free mass is much smaller than that ofthe structure then the structure itself would hardly be affectedby the impacts Thus the most meaningful dynamics willoccur when the two masses are relatively the same order ofmagnitude For simplicity we begin our analysis by choosingthem to be the same and later in the paper we investigate theeffect of different choices of for the free mass
The linearization factor | minus | is set as 01ms becauseit is the maximum difference between the free mass velocityand flow velocity which is set as 01ms Therefore the valueof 119887 is
119887 =1
2120588119862119863119860 | minus | = 4995 kgs (30)
where 120588 is the density of water and119862119863 the drag coefficient is
1The coefficient of restitution another important parameteris assumed to be 1 which implies no energy dissipationthrough impact In addition the initial displacement and
6 Mathematical Problems in Engineering
0
minus02
minus04
minus06
minus08
minus10 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall Responses
Disp
lace
men
t (m
)
Time (s)
0
minus0002
minus0004
minus0006
186 188 190 192 194 196 198 200
StructureIce
(b) Responses from 185 to 200 seconds
Disp
lace
men
t (m
)
Time (s)
StructureIce
0
380 385 390 395 400
times10minus4
minus1
minus2
minus3
(c) Responses from 380 to 400 seconds
Figure 2 Responses of the ice and structure with a linearized drag force
Table 1 Parameters for simulations
Parameter Model structureℎ (mm) 41119863 (mm) 76119898 (kg) 1600119888 (kNsm) 02119896 (MNm) 1ℓ (m) 4493120590119888(MPa) 14
119909max (mm) 20
velocity of the free mass are minus001m and 001ms respec-tively and 0m and 0ms for the structure Figure 2 shows theresponse based on these parameters
Since the motivation for this model was to attempt toexplain IIV without analyzing specific ice-failure mecha-nisms we attempt to compare our results to some measuredIIV data There is very little IIV data in the open literatureespecially for displacement measurements Only laboratorymeasurements are available from [28 29] for displacementmeasurements Since the structural data for their tests arenot presented it is inadequate to compare their experimentaldata to Figure 2 directly but general vibration characteristicscan be compared Structural vibrations of IIVs becomequasi-static vibrations at low ice velocities and steady-statevibrations at medium ice velocities [30] although there isno quantitative definition of low and high ice velocities Aquasi-static vibration is defined by Karna [30] as a transientresponse which is followed by its maximum response atthe peak ice force and is not amplified by the dynamicsof the structure Figure 3(b) represents typical quasi-static
Mathematical Problems in Engineering 7
(a) [28]
1
0
2 4 6 8 10
Stru
ctur
aldi
spla
cem
ent(
cm)
Ice velocity = 76mms Time (s)
(b) [29]
2 4 6 8 10
1
0Stru
ctur
aldi
spla
cem
ent(
cm)
Time (s)Ice velocity = 395mms
(c) [29]
Figure 3 Measured structural responses from the literature
vibrations at ice velocity 76mms and Figure 3(c) shows thecharacteristic of a steady-state vibration at free mass velocity395mms Figure 3(a) however presents the displacementresponse although it is plotted at a lower free mass velocitythan Figure 3(b) This demonstrates that the vibration char-acteristics of IIVs are not determined by the velocity of theice alone but by involving all parameters
As shown in Figure 3(a) actual IIVs display multiplevibration characteristics from quasi-static to steady-statevibrations even at the same ice velocity contrary to thecharacteristic failure frequency model which predicts onetype of characteristic vibration at one ice velocity In thisrespect the proposed impact model possesses an advantageover the characteristic failure frequency model In an impactmodel the velocity of the ice varies from time to time whichis more realistic because part of the ice (free mass) contactingthe structure is deformed as the vibrations proceed Figure 2clearly shows multiple vibration characteristics Figure 2(b)displays quasi-static vibrations similar to Figure 3(b) From380 to 400 seconds the vibrations become stable and steadystate shown in Figure 2(c) which is close to the vibrationcharacteristic of Figure 3(c) The vibration characteristics ofthe impact model however cannot be clearly distinguishedfor the entire duration rather they show a mixture of thecharacteristics Therefore it is reasonable to conclude thatthe vibration characteristics of the impact IIV model aregoverned by the momenta of the ice and structure at the timeof impact
The weakness of modelling IIVs as repeated impacts asproposed in this paper model is how to model the relevantproperties of the ice The definitions of the ice mass and areain the impact IIVmodel imply that themass and area directlyaffect the impact or interactionwith the structure Among thetwo ice properties the influence of the ice area 119860 is directlyrelated to the velocity of the flowbecause of the relationwith 119887given by 119887 = (12)120588119862
119863119860|minus | It is inappropriate to simulate
the linearized model with large flow velocities since then thelinearized model would not be applicable
The mass of the free mass plays an important role in thismodel In order to understand its role it will be investigatedby conducting several numerical simulations Figures 4 5and 6 are plotted with the same parameters of Figure 2 exceptwith differentmasses which are proportional to the structuralmass 50 and 150 percent of 1600 kg so that the structuralmass can be used as a benchmark Including Figure 2 whichis plotted with the free mass of 100 percent of the structuralmass the responses with three different free masses arecompared At the beginning of the responses the heaviermass transfers higher momentum thus impacts happen lessfrequently but with largermagnitudes On the other hand thelighter mass allows the structure to approach the steady-statevibration in relatively short time In Figure 6 the magnitudesof the structural responses are not too different but theheavier mass induces more quasi-static vibrations for longertime Therefore it can be concluded that the free mass hasa greater effect on determining the vibration characteristicsthan on the magnitude of the structural response
Another important factor in this model is the coefficientof restitution The coefficient of restitution also requiresexperimental research to be defined quantitatively Theconducted numerical simulations can only show how thecoefficient of restitution influences the responses of thesystem Figure 2 is plotted with the coefficient of restitution119890 = 1 which is an unrealistic case Figures 7 8 and 9are simulated based on the same parameters of Figure 2 butwith different values of 119890 It becomes obvious that the factorthat most influences the magnitudes of the vibrations is thecoefficient of restitution rather than the velocities of the freemass The system with 119890 = 01 approaches the steady stateafter the first impact The system with 119890 = 05 not onlyapproaches the steady-state vibration faster but also showslower structural magnitudes than the system with 119890 = 1Figure 2 Figure 8 indicates that the system with 119890 = 05
loses the majority of its momentum after approximately54 seconds This suggests that vibrations due to repeatedimpacts can be controlled by manipulating the coefficient
8 Mathematical Problems in Engineering
002
0
minus002
minus004
minus006
minus008
minus01
minus0120 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with119872 = 800
0 50 100 150 200 250 300 350 400
02
0
minus02
minus04
minus06
minus08
minus1
minus12
minus14
Time (s)
Disp
lace
men
t (m
)
StructureIce
(b) Overall responses with119872= 2400
Figure 4 Responses with different ice masses 1
186 188 190 192 194 196 198 200
Disp
lace
men
t (m
)
StructureIce
Time (s)
0
minus2
minus4
minus6
times10minus4
(a) Responses with119872 = 800 from 185 to 200 seconds
0
minus0002
minus0004
minus0006
minus0008
186 188 190 192 194 196 198 200
StructureIce
Time (s)
Disp
lace
men
t (m
)
(b) Responses with119872= 2400 from 185 to 200 seconds
Figure 5 Responses with different ice masses 2
of restitution via the properties of the contacting area Forinstance the magnitudes of vibrations can be significantlyreduced by increasing the surface roughness of the structure
42 Quadratic Drag Force In the previous section themodelwas developed based on the linearized drag force whichassumes small relative velocity between the velocities of thefree mass and structure The linearized model howeveris definitely vulnerable to two factors namely when theinterval between impacts exceeds several seconds and whenthe velocity of the flow is no longer small Figure 10 clearlyshows the error between the models based on the linearized
drag and quadratic drag as time passes The displacement oflinearized equations (3) and (6) is compared with numer-ically solved equation (3) by using the built-in ordinarydifferential equation solver ofMathCadThe difference growsas time goes by but at high flow velocity the differencedecreases in Figure 10(b) This is because of the linearizationfactor | minus | which calibrates the difference between thelinear and quadratic models The solution however to thequadratic drag force needs to be developed for a more precisesimulation
A numerical solution is needed to solve (3) but insteadof using the built-in ordinary differential equation solver
Mathematical Problems in Engineering 9
StructureIce
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
Time (s)
1
0
minus1
minus2
minus3
minus4
times10minus4
(a) Responses with119872 = 800 from 385 to 400 seconds
StructureIce
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
0
minus2
minus4
minus6
times10minus4
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 6 Responses with different ice masses 3
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Overall responses with 119890 = 01
0
minus005
minus01
minus015
minus02
minus025
minus03
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400Time (s)
StructureIce
(b) Overall responses with 119890 = 05
Figure 7 Responses with different coefficients of restitution 1
of MathCad a simple numerical solution is developed for afaster algorithm From (3)
119872 = minus1
2120588119862119863119860 ( minus ) | minus | = minus119861 ( minus ) | minus | (31)
or
=119889
119889119905= minus
119861
119872( minus ) | minus | (32)
where
119861 =1
2120588119862119863119860 (33)
Using 119894 as a counter for the a number of time steps thedisplacement of (34) can be expressed as
119911119894= 119911119894minus1
+119889119911119894minus1
119889119905Δ119905 = 119911
119894minus1+ 119894minus1Δ119905 (34)
in which Δ119905 is previously defined as a time step z is given by
119894= 119894minus1
+119889119894minus1
119889119905Δ119905 =
119894minus1+ 119894minus1Δ119905 (35)
By substituting (32) into (35)
119894= 119894minus1
+ [minus119861
119872(119894minus1
minus )1003816100381610038161003816119894minus1 minus
1003816100381610038161003816] Δ119905(36)
10 Mathematical Problems in Engineering
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 2 4 6 8 10 12 14
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Responses with 119890 = 01 from 0 to 15 seconds
Disp
lace
men
t (m
)
Time (s)
StructureIce
46 48 50 52 54 56 58 60
2
0
minus2
minus4
minus6
minus8
times10minus4
(b) Responses with 119890 = 05 from 45 to 60 seconds
Figure 8 Responses with different coefficients of restitution 2
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus001386 388 390 392 394 396 398 400
Time (s)
Disp
lace
men
t (m
)
(a) Responses with 119890 = 01 from 385 to 400 seconds
StructureIce
Time (s)
Disp
lace
men
t (m
)
398 3985 399 3995 400
times10minus6
6
5
4
3
(b) Responses with 119890 = 05 from 398 to 400 seconds
Figure 9 Responses with different coefficients of restitution 3
Given the initial displacement and velocity of the free massthe displacement can be numerically solved by using (34)and (36) as simultaneous equationsThe simulation results of(34) and (36) are also plotted in Figure 10 At the high flowvelocity Figure 10(b) (34) and (36) predict the displacementclose enough to the one evaluated with the built-in ordinarydifferential equation solver In the algorithm of the numericalsimulation (34) and (36) replace (6) and (7)
Using the same parameters as those used in Figure 2but with a different flow velocity the responses of themodel with the quadratic drag force are compared withthe one with the linearized drag force in Figure 11 At high
flow velocities over 1ms the quadratic drag force modelapproaches steady-state vibration more quickly than thelinearized drag force model because at high flow velocity thequadratic drag induces more damping while it transfers moreforces than the linearized drag In Figure 12 the quadraticdrag force model exhibits the characteristic of the steady-state vibration while the linearized drag force model stilldisplays quasi-static vibration As velocities increase whichcorresponds to flowvelocities increasing vibrations approachthe steady state The linearized drag force model howeverdoes not show a steady-state vibration until 400 secondsin Figure 12 The linearization factor allows the linearized
Mathematical Problems in Engineering 11
0 100 200 300 400 500 600
100
80
60
40
20
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(a) Flow velocity 01ms
0 100 200 300 400 500 600
1000
800
600
400
200
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(b) Flow velocity 1ms
Figure 10 Comparison of the linearized and quadratic Drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
Disp
lace
men
t (m
)
(a) Overall responses with the quadratic drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
0
minus02
minus04
minus06
minus08
minus1
Disp
lace
men
t (m
)
(b) Overall responses with the linear drag
Figure 11 Responses with the quadratic and linear drags 1
drag force model to follow the quadratic drag force modelin the beginning but when the system is in a series of lowvelocity impacts the linearization factor causes a differenceThe linearized drag force model therefore is inappropriateto simulate the repeated impact model at high flow veloci-ties
Since the free mass area 119860 plays a similar role as theflow velocity in the model the model with wide area is alsosimulated with the quadratic drag force The influence of thedrag force on the response of the free mass increases as thesize of the area increases because the area is related to the dragforce by (1) 119891
119863= minus(12)120588119862
119863119860( minus )
2 Figure 13 is plottedwith the area 119860 = 2m2 the flow velocity = 01ms and
the rest of parameters are the same as the parameters usedin Figure 11 The flow velocity of Figure 13 is identical to thatof Figure 2 but the responses quickly approach the steady-state vibrations as if they are driven by high flow velocityFigure 13(b) also shows a similar vibration characteristic toFigure 12(a) because both flow velocity and area influence thedrag force in the impact model Therefore the area can beanother controlling factor of the vibrations Since the areais related to the size of the contacting area of the structurereducing the contacting area has a similar effect to that ofreducing the flow velocity
In the previous section the roles of the free mass andthe coefficient of restitution have been defined by changing
12 Mathematical Problems in EngineeringD
ispla
cem
ent (
m)
395 396 397 398 399 400Time (s)
StructureIce
8
6
4
2
0
times10minus4
(a) Responses with the quadratic drag from 395 to 400 secondsD
ispla
cem
ent (
m)
386 388 390 392 394 396 398 400Time (s)
StructureIce
0
minus0005
(b) Responses with the linear drag from 385 to 400 seconds
Figure 12 Responses with the quadratic and linear drags 2
005
0
minus005
minus01
minus015
minus02
minus025
minus03
minus035
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
(a) Overall responses with 119860 = 2
Disp
lace
men
t (m
)
395 396 397 398 399 400Time (s)
StructureIce
1
0
minus1
times10minus5
(b) Responses with 119860 = 2 from 395 to 400 seconds
Figure 13 Responses with 119860 = 2m2
relevant parameters To verify if those definitions still holdwith the quadratic drag force the impact model with thequadratic drag is plotted with change in free mass and thecoefficient of restitution Figures 14 15 and 16 are plottedon the same parameters used in Figures 2 4(a) and 4(b)respectively except that they are based on the quadratic dragforce The responses of Figures 14 15 and 16 prove thatthe defined role of the free mass holds with the quadraticdrag force The responses of the system with heavier freemass Figure 16(b) is still unstable while the system with thefree mass 119872 = 800 kg Figure 15(b) is in the steady-statevibration The quadratic drag force however induces more
drag force thus the system responses tend to be stabilizedfaster than those with the linearized drag forces
Figures 17 and 18 are plotted on the same conditions withFigures 7(a) and 7(b) respectively but with the quadraticdrag forceThe coefficient of restitution also governs themag-nitudes of responses in the impact model with the quadraticdrag force however the system with the quadratic drag forceis more sensitive to change in the coefficient of restitutionthan the one with the linearized drag force The system with119890 = 05 Figure 18(b) loses the momentum at approximately43 seconds under the influence of the quadratic drag forcewhile the system with the linearized drag force and 119890 = 05
Mathematical Problems in Engineering 13
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
01
0
minus01
minus02
minus03
minus04
minus05
minus06
minus07
(a) Overall responses with119872 = 1600
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
StructureIce
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 1600 from 385 to 400 seconds
Figure 14 Responses with119872 = 1600
0
minus001
minus002
minus003
minus004
minus005
minus006
minus0070 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall responses with119872 = 800
395 396 397 398 399 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
0
5
minus5
times10minus6
(b) Responses with119872 = 800 from 395 to 400 seconds
Figure 15 Responses with119872 = 800
Figure 8(b) loses the majority of its momentum at approxi-mately 54 seconds This is again because the quadratic dragforce dissipates energy more quickly than the linearized dragforceHowever it is demonstrated in Figures 17 and 18 that thecoefficient of restitution plays the same role in both quadraticand linearized drag forces
5 Resonance Conditions
One of the main objectives of the analysis aims to find aresonance condition (which is often referred to as lock-in
when investigating IIVs) The structural vibration frequencyincreases as the free mass velocity increases but when lock-in occurs the structural frequency remains at the naturalfrequency even with increasing free mass velocity [6] In theimpact model however the velocities of the free mass cannotbe the primary lock-in condition because the velocities of thefree mass are not fixed
One possible interpretation of modelling IIVs with theimpact model is that it considers the forces to be a seriesof discrete impulses as shown in Figure 19(a) while otherIIV models consider them as continuous periodic forces
14 Mathematical Problems in Engineering
0 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
(a) Overall responses with119872= 2400
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 16 Responses with119872 = 2400
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
(a) Overall responses with 119890 = 01
StructureIce
0 2 4 6 8 10 12 14
0002
0
minus0002
minus0004
minus0006
minus0008
Disp
lace
men
t (m
)
Time (s)
(b) Responses with 119890 = 01 from 0 to 15 seconds
Figure 17 Responses with 119890 = 01
In fact the impact IIV model proposes a way to calculate thestrength of each impulse and the times between impulses andthese depend on the momenta of the ice and structure aswell as the coefficient of restitution for the collision Undercertain conditions it is possible for the discrete impulsesof Figure 19(a) to approach the regular periodic series ofimpulses of Figure 19(b) which is suspected to cause the lock-in condition
This view of the impact IIV model can be analyzed byusing a Fourier transform [31] There are several definitions
of the Fourier transform (FT) that may be used and in thispaper the chosen version of the FT is given by
119865 (120596) = int
infin
minusinfin
119891 (119905) 119890minus119894120596119905
119889119905 (37)
The corresponding inverse Fourier transform is then given by
119891 (119905) =1
2120587int
infin
minusinfin
119865 (120596) 119890119894120596119905119889120596 (38)
Mathematical Problems in Engineering 15
005
0
minus005
minus01
minus015
minus02
minus0250 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with 119890 = 05
Time (s)
Disp
lace
men
t (m
)
StructureIce
36 38 40 42 44 46 48 50
4
2
0
minus2
minus4
times10minus4
(b) Responses with 119890 = 05 from 35 to 50 seconds
Figure 18 Responses with 119890 = 05
f(t)
fj
tj t
(a) Random impulses
f(t)
f
T 2T 3Tt
pT
(b) Uniform impulses
Figure 19 Ice forces as a series of impulses
The discrete ice forces of Figure 19(a) can be expressed asa series of impulses as
119891 (119905) =
infin
sum
119895=1
119891119895120575119863(119905 minus 119905119895) (39)
where 120575119863indicates the Dirac delta function and 119891
119895indicates
the strength of each impact If the sizes and intervals ofthe impulses are uniform as shown in Figure 19(b) (39) isrewritten as
119891 (119905) =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (40)
In reality the repeated impacts will not all have the samestrength or occur at a fixed interval however it is sufficient forthe situation of Figure 19(a) to approach that of Figure 19(b)for this analysis to be relevant
Equation (40) is often called a comb functionTheFouriertransform of the comb function is given [32] by
119865 (120596) = 2120587
infin
sum
119873=1
119891120575119863(120596 minus
2120587119873
119879) (41)
For a typical single-degree of freedom forced vibrationsproblem
+ 2120577120596119899 + 1205962
119899119909 =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (42)
Taking Fourier transforms of both sides and rearranging give
119883 (120596) = 2120587suminfin
119873=1119891120575119863 (120596 minus 2120587119873119879)
(minus1205962 + 2120577120596119899120596119894 + 1205962
119899)
(43)
Taking the inverse Fourier transforms to find the structureresponse in time gives
119909 (119905) = 119891
infin
sum
119873=1
119890119894(2120587119873119879)119905
(minus4120587211987321198792 + 119894120577120596119899 (4120587119873119879) + 1205962
119899) (44)
Equation (44) experiences a resonance when the denom-inator of any of the terms in the sum goes to zero that is when120596119899= 2120587119873119879 where 119873 is an integer larger than 0 This says
that the structure will experience a resonance when the natu-ral frequency is an integral multiple of the forcing frequency
16 Mathematical Problems in Engineering
as would be expected In the previous numerical simulationsthis condition has been observed in several cases Figures12(a) and 13(b) For instance the structural frequency ofFigure 13(b) is 2120587119879 = 212058705 = 1257 rads The naturalfrequency of the structure is radic119896119898 = radic10000001600 =
25 rads With119873 = 2 the structural frequency is close to thenatural frequency The result is what is seen in Figure 13(b)The two masses vibrate at the same frequency of 2120587119879 whichis half of the natural frequency The structural frequencyremains at that level which is lock-in and so does thefrequency of the free mass response
Thus the impact model successfully predicts the exis-tence of a lock-in condition and also predicts that it can occurwhen the frequency of the structure and freemass impacts aresynchronized together at a multiple of the natural frequencyof the structure The question of the precise conditionsnecessary for the frequency of the impacts to occur at amultiple of the natural frequency still remains open
6 Summary and Conclusions
This paper analyzes the vibrations caused by repeated impactsof a flow-driven freemass onto a second structure Conserva-tion of linearmomentum and the coefficient of restitution areused to characterize the nature of the impacts between thetwo masses This model is referred to as the impact modelTheproperties of the impactmodel were investigated throughnumerical simulations The mass of the free mass is one ofthe factors determining the vibration characteristics sinceit is related to the momentum of each impact This modelis proposed to explain the nature of ice-induced vibrationswithout requiring a microscopic model of the mechanism ofice failure The magnitudes of the vibrations are governedin part by the coefficient of restitution which when usedto explain IIVs encapsulates all ice properties mentionedin other IIV models In addition to the ice properties thecoefficient of restitution also governs the interaction betweenthe structure and the free mass The free mass is used torepresent the ice when modelling IIVs The last propertyof the impact model is that of the free mass area whichinfluences the dissipation of the system momenta throughthe drag force A free mass with a small area shows the samebehavior of a mass driven by lower flow velocity
Therefore through the impact model two factors whichcontrol the characteristics of IIVs have been proposed theice area and the coefficient of restitution By changing thesize of the contact area between the ice and structure thedriving force of the ice can be reduced as if the ice weredriven by lower flow velocities Furthermore by changingfactors related to the coefficient of restitution such as thesurface roughness or shape of the structure IIVs can bereduced to avoid the lock-in conditionwhich occurswhen thefrequencies of the structure and ice impacts are synchronizedat a multiple of the natural frequency of the structure Inaddition to the inclusion of the effect of flow the newapproach to ice forces through the impact model permitsthe answering of questions which originate from the weaktheoretical points of existing IIV models
Particular strengths of the model are that it qualitativelymatches observed results and in particular that a mechanismof ice failure is not required in order to perform simulationsand predictions It also proposes a simple mechanism toexplain the potentially destructive resonance lock-in condi-tions
References
[1] J P Dempsey ldquoResearch trends in ice mechanicsrdquo InternationalJournal of Solids and Structures vol 37 no 1-2 pp 131ndash153 2000
[2] K A Blenkarn ldquoMeasurement and analysis of ice forces oncook inlet structuresrdquo in Proceedings of the Offshore TechnologyConference (OTC rsquo70) pp 365ndash378 1970
[3] Y Toyama T Sensu M Minami and N Yashima ldquoModel testson ice-induced self-excited vibration of cylindrical structuresrdquoin Proceedings of the 7th International Conference on Port andOcean Engineering under Arctic Conditions vol 2 pp 834ndash8441983
[4] M Ranta and R Raty ldquoOn the analytic solution of ice-inducedvibrations in a marine pile structurerdquo in Proceedings of the 7thInternational Conference on Port and Ocean Engineering underArctic Conditions vol 2 pp 901ndash908 1983
[5] A Barker G Timco H Gravesen and P Voslashlund ldquoIce loadingon Danish wind turbinesmdashpart 1 dynamic model testsrdquo ColdRegions Science and Technology vol 41 no 1 pp 1ndash23 2005
[6] D S Sodhi ldquoIce-induced vibrations of structuresrdquo SpecialReport 89-5 International Association for Hydraulic ResearchWorking Group on Ice Forces 1989
[7] D S Sodhi and C E Morris ldquoCharacteristic frequency of forcevariations in continuous crushing of sheet ice against rigidcylindrical structuresrdquo Cold Regions Science and Technologyvol 12 no 1 pp 1ndash12 1986
[8] D S Sodhi C E Morris and G F N Cox ldquoDynamic analysisof failure modes on ice sheets encountering sloping structuresrdquoin Proceedings of the 6th International Conference on OffshoreMechanics and Arctic Engineering vol 4 pp 281ndash284 1987
[9] H R Peyton ldquoSea ice forcesrdquo in Proceedings of the Conferenceon Ice Pressure against Structures 1966
[10] C R Neill ldquoDynamic ice forces on piers and piles an assess-ment of design guidelines in the light of recent researchrdquoCanadian Journal of Civil Engineering vol 3 no 2 pp 305ndash3411976
[11] G Matlock W Dawkins and J Panak ldquoAnalytical model forice-structure interactionrdquo Journal of the Engineering MechanicsDivision pp 1083ndash11092 1971
[12] M M Maattanen ldquoIce-induced vibrations of structures self-excitationrdquo Special Report 89-5 International Association forHydraulic Research Working Group on Ice Forces 1989
[13] M M Maattanen ldquoNumerical simulation of ice-induced vibra-tions in offshore structuresrdquo in Proceedings of the 14th NordicSeminar on Computational Mechanics 2001 Keynote lecture
[14] J Xu and L Wang ldquoIce force oscillator model and its numericalsolutionsrdquo in Proceedings of the 7th International Conference onOffshore Machanics and Arctic Engineering pp 171ndash176 1988
[15] G TsinkerMarine Structures Engineering Chapman-Hall NewYork NY USA 1st edition 1995
[16] G Huang and P Liu ldquoA dynamic model for ice-inducedvibration of structuresrdquo Journal of OffshoreMechanics andArcticEngineering vol 131 no 1 Article ID 011501 6 pages 2009
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
2 Mathematical Problems in Engineering
frequency which in turn decides the forcing frequency [6]This point of view defined the origin of IIVs as being viaa characteristic failure mechanism a point of view whichwas supported by Matlock et al [11] In contrast to thecharacteristic failure mechanism Blenkarn [2] explainedIIVs as a self-excited vibration due to negative dampingAccording to this theory IIVs originate from the interactionbetween a flexible structure and decreasing ice crushingforces with increasing stress rate Maattanen one of the bigcontributors of the self-excited model verified his modelthrough field and laboratory experiments [7 12 13] SimilarlyXu andWang [14] also proposed an ice force oscillator modelbased on the self-excited model
As research in the field progressed different physicalprocesses to explain the origin of the ice force were proposedIn particular attention was turned to whether the ice wasconsidered to fail in crushing (compression) or in bendingIt is thought that compressive (crushing) failures occur as aresult of ice interacting with a narrow vertical (cylindrical)structure [15] Tsinker defines crushing as the completefailure of granularization of the solid ice sheet into particlesof grains or crystal dimension no cracking flaking or anyother failure mode occurs during pure crushing Specificmodels that correspond to ice in crushing failure are givenin [3 16] Even though these two articles model the samephysical phenomena they are different in their modellingapproaches
The concept of adding ice-breaking cones to cylindricalstructures was proposed in the late 1970s changing theeffective shape of the structure from cylindrical to conicalThe ice force on a conical structure is smaller than the forceon a cylindrical structure of similar size [5 8 17 18] It isthought that the main reason for the reduction in ice forceis that a well-designed cone can change the ice-failure modefrom crushing to bending It is considered that the primaryfailure mechanism for ice interacting with a conical structureis that of bending failure In this failure mode the ice sheetsimpacting on a cone fail by bending and typically the icebreaks almost simultaneously in each event of ice failureWith conical structures and bending ice failure analyticalmodels are more likely to be characteristic failure frequencymodels that were initially proposed in the 70srsquo since the iceforce can effectively be modelled as only depending on theproperties of the ice With conical structures the local iceforces were experimentally found to drop to almost zeroafter each event of ice failure [19 20] and are consequentlymodelled analytically with a periodic function For examplethe forcing function suggested by Qu et al [21] is a simpleone-degree-of-freedom model corresponding to a periodicice force This model is based on their own previous work[19 20] aswell as that ofHirayama andObara [18]Thismodelattempts to represent an ice sheet failing through the bendingmode and simply models the forcing function as a saw-toothshaped periodic force From a modelling point of view thebending failure of ice is easier to model since the ice-forcingfunction is periodic and is decoupled from the motion of thestructure This type of model thus becomes a regular forcedvibration problem and once the form of the external ice forceis chosen it is a relatively straightforward problem to solve
Despite extensive studies no theoretical model com-pletely explains the IIV mechanismThe variety of modellingapproaches and dependence of the ice load on the structuralproperties have caused modelling and design difficulties Forinstance there is nomethod for choosing the proper dynamicice case for the design and optimization of an ice-resistantjacket platform [22 23] In addition most IIV models donot consider the influence of fluid flow even though moststructures subjected to IIVs are offshore structures Sincefluid flow is one of the main driving forces of IIVs it cancontribute to the dynamics of ice forces as well as that of thestructure but it has not been considered in past research
Since the first modeling approaches were introduced nomajor advances in IIV modeling have been proposed andthere is no general consensus on one correct approach Allexisting modeling approaches analyze the process of IIVsfrom a microscopic point of view with a heavy emphasis onthe mode in which the ice fails For example in [5] numeroussmall scale tests were conducted to attempt to distinguish icefailures into four different modes by the sizes of failed icepieces and the structural responsesTheprocedure of definingice-failure modes is thus neither clear nor generally agreedupon
In this paper we propose that amacroscopic point of viewmay give a different insight For instance a detailed analysisof all possible deformations is not generally considered whena collision of two particles is analyzed In that case the wholecollision process can be efficiently analyzed with a singleparameter the coefficient of restitutionThemain idea of thispaper is that if the collision of the particles occurs repetitivelythe movement of the two particles may resemble those ofIIVs
This simple concept provides the inspiration for the newmodeling approach proposed in this paper The vibrationcharacteristics of IIVs are proposed as being the result of arepetitive forcing function with the moving fluid repeatedlypushing the ice towards the structure If all the microscopicice-failure processes can be condensed into a single macro-scopic parameter the coefficient of restitution the remainingvibration characteristics are similar to those of repetitivedriven collisions of two particles The ice is modeled asone of the particles in the collision and the structure is theother IIVs can then be modeled and simulated by usingthe conservation of momentum along with the coefficient ofrestitution which is a completely different perspective fromexisting IIV models that take a microscopic view and focuson the precise mechanism of ice failure
3 Modeling of Flow-Driven Repeated Impactswith the Conservation of Momentum
We consider a two degree-of-freedom system that consists oftwo masses Figure 1 is a schematic diagram of two masseswhich collide repetitively Mass 119872 represents the free massand mass119898 represents the flexible structure
The force which drives the free mass needs to be modeledso that collisions occur repetitivelyThe freemass is driven byvarious external forces such as wind and thermal expansion
Mathematical Problems in Engineering 3
M
u u
x x x0
z z
m
k
c
Figure 1 Mechanical model of the system
but the main driving force is fluid flow Depending on theinitial displacement and velocity of the flow the velocity ofthe free mass is determined just prior to the time of impactand the maximum velocity of the free mass is taken to be thevelocity of the flowThe free mass moves until it collides withthe structure After the impact the free mass either stopsmoves forward or bounces back in the opposite directiondepending on the coefficient of restitution and conservationof total system (free mass and structure) momentum Sincethe free mass and fluid flow now have a relative velocitywith respect to each other the velocity of the free mass ischanged by the effect of fluid drag until the free mass reachesthe velocity of the flow and then moves again towards thestructure for another impact The free mass moves towardsthe structure and transfers its momentum to the structureagain when the relative displacement between the mass andstructure becomes 0 (ie when a collision occurs) In somecases the structure can transfermomentum to themasswhenthemomentumof the structure is larger then that of themassThis whole process keeps repeating but due to structuraldamping and fluid drag the momenta which are transferredto each other gradually decrease
Since mass 119872 is floating in the flow it is free to moveThe flow however moves with its own velocity which inthis case is always assumed to be positive (to the right inthe figure) The displacement of the free mass is given by thevariable 119911 and that of the structure is given by the variable119909 The mass is driven by the drag force 119891
119863 which is present
when there is a relative velocity between the flow and themass There are four conditions depending on the relativevelocities
(i) When ge 0 and || ge || 119891119863le 0
(ii) When ge 0 and || lt || 119891119863gt 0
(iii) When lt 0 and || ge || 119891119863ge 0
(iv) When lt 0 and || lt || 119891119863gt 0
Outside of a collision with the structure no force is trans-mitted to the free mass except the drag force due to the fluidThe structure imparts a force to the freemass when a collisionbetween them occursThe drag force from the fluid is definedas
119891119863= minus
1
2120588119862119863119860( minus )
2 (1)
where 120588 119862119863 and 119860 are the density of the fluid drag
coefficient and area related to the drag coefficient respec-tively [24] The drag force which influences the systemof Figure 1 changes the forcing direction according to therelative velocities as explained above The drag force for thefree mass of Figure 1 therefore is expressed as
119891119863= minus
1
2120588119862119863119860 | minus | ( minus ) (2)
The equation ofmotion of the freemass can be obtained from(2) since no other forces are involved and is thus given as
119872 = minus1
2120588119862119863119860 | minus | ( minus ) (3)
Since (3) is a nonlinear equation it should be linearizedor simulated numerically The linearized version of (3) isdeveloped by assuming a small relative velocity Equation (3)is linearized as
119872 = minus119887 ( minus ) (4)
where
119887 =1
2120588119862119863119860 | minus | (5)
| minus | is assumed to be small and constant The solution of(4) consists of homogeneous and particular solutions and canbe obtained as
119911 (119905) = 1198881+ 119905 + 119888
2119890minus(119887119872)119905
(6)
The velocity of the free mass is then given by the derivative of(6) as
(119905) = minus 1198882
119887
119872119890minus(119887119872)119905
(7)
Equation (6) is a closed form solution of linearized equation(3) The solution of nonlinear equation (3) will be developedlater since it requires a numerical approach
The structure is modeled as a single-degree-of-freedomsystem consisting of a mass damper and spring withoutexternal forces (other than impacts with the free mass) asin Figure 1 Since the source of structural vibrations is thetransfer of momenta during collisions outside the time ofthe collisions the structure vibrates as a free simple harmonicoscillator The equation of structural motion is expressed as
119898 + 119888 + 119896119909 = 0 (8)
For an underdamped case (8) has the following solution
119909 (119905) = 1198883119890minus120577120596119899119905 cos120596
119889119905 + 1198884119890minus120577120596119899119905 sin120596
119889119905 (9)
with 120577 120596119889 and 120596
119899having the usual interpretation and
definition as damping coefficient damped frequency andnatural frequency The velocity of the structure at time 119905 canbe calculated as
(119905) = minus 1205961198891198883119890minus120577120596119899119905 sin120596
119889119905 minus 120577120596
1198991198883119890minus120577120596119899119905 cos120596
119889119905
+ 1205961198891198884119890minus120577120596119899119905 cos120596
119889119905 minus 120577120596
1198991198884119890minus120577120596119899119905 sin120596
119889119905
(10)
4 Mathematical Problems in Engineering
From (6) (7) (9) and (10) the displacements and veloci-ties of the mass and structure at any given time can be foundSubstituting these expressions into the equations for theconservation of total systemmomentum and the equation forthe coefficient of restitution the displacements and velocitiesimmediately after impact can be evaluated Each of the twomasses however has two unknowns after impact Thus theunknowns after impact are the displacement and velocityof both masses Since only two equations can be developedfrom the conservation of momentum and the coefficientof restitution equation the number of unknowns exceedsthe number of equations by two Salapaka et al [25] alsoinvestigated a similar modeling approach They developedthe equation of a mass-spring system driven by a vibratingtable which impacts themass at a certain position repetitivelyFaced with the same problem of excessive unknowns theyassumed that the displacements of each mass before andafter impact remain the same thus reducing the number ofunknowns after impact to two [25]
The instants just before and just after impact will bedenoted by 119905
119896and 119905119891 Since the impact occurs when the dis-
placements of the twomasses are the same the displacementsat time 119905
119896are given as
119909 (119905119896) = 119911 (119905
119896) = 119909119896 (11)
The same notation is used for velocities From the conserva-tion of momentum the total momentum of the two masses119898 and119872 is conserved so that
119898119891+119872
119891= 119898119896+119872
119896 (12)
The second relevant impact equation is obtained from thecoefficient of restitution [26] as
119891minus 119891= 119890 (
119896minus 119896) (13)
Equations (12) and (13) can be rewritten in matrix form as
[119898 119872
1 minus1] [
119891
119891
] = [119898119896+119872
119896
119890119896minus 119890119896
] (14)
or
[119891
119891
] =minus1
119898 +119872[minus1 minus119872
minus1 119898][
119898119896+119872
119896
119890119896minus 119890119896
] (15)
From (15) the velocities after impact 119891and 119891 are
119891=
1
119898 +119872(119898119896+119872
119896+ 119890119872
119896minus 119890119872
119896) (16)
119891=
1
119898 +119872(119898119896+119872
119896minus 119890119898
119896+ 119890119898
119896) (17)
After impact the displacements are assumed to be thesame as before impact From (11)
119909 (119905119891) = 119909 (119905
119896) = 119909119896
119911 (119905119891) = 119911 (119905
119896) = 119909119896
(18)
Once the impact occurs the momentum of the free mass istransmitted to the structure or vice versa 119888
1 1198882 1198883 and 119888
4of
(6) (7) (9) and (10) need to be repeatedly calculated aftereach impact in order to obtain the new responses Since thecharacteristics of the system responses are reset after eachimpact the time right after the impact 119905
119891 is also reset to 0
to calculate 1198881 1198882 1198883 and 119888
4 Substituting 119905
119891= 0 into (6) and
(7)
119911 (119905119891) = 119911 (0) = 119888
1+ 1198882
(119905119891) = (0) = minus
119887
1198721198882
(19)
The displacements before and after impact are assumed to bethe same and the velocity of the free mass after the impactis calculated from (17) Equation (19) is equal to 119909
119896and
119891
respectively
119911 (0) = 1198881+ 1198882= 119909119896
(0) = minus119887
1198721198882= 119891
(20)
From (20) 1198881and 1198882are calculated as
1198881= 119909119896minus119872
119887( minus
119891)
1198882=119872
119887( minus
119891)
(21)
Applying the same assumption and (16) to (9) and (10)
119909 (119905119891) = 119909 (0) = 119888
3= 119909119896
(119905119891) = (0) = minus120577120596
1198991198883+ 1205961198891198884= 119891
(22)
It then follows that 1198883and 1198884are obtained from (22) as
1198883= 119909119896
1198884=
1
120596119889
(119891+ 120577120596119899119909119896)
(23)
Substituting the calculated 1198881 1198882 1198883 and 119888
4into (6) and (9)
the responses of the two masses after impact are expressed as
119911 (119905) = [119909119896minus119872
119887( minus
119891)] + 119905 +
119872
119887( minus
119891) 119890minus(119887119872)119905
119909 (119905) = 119909119896119890minus120577120596119899119905 cos120596
119889119905 +
1
120596119889
(119891+ 120577120596119899119909119896) 119890minus120577120596119899119905 sin120596
119889119905
(24)
All the equations are developed based on the linearized dragforce Equation (24) holds until the next impact After thenext impact 119888
1 1198882 1198883 and 119888
4should be recalculated according
to new values of 119909119896 119896 119911119896 and
119896obtained from the impact
equations
Mathematical Problems in Engineering 5
4 Numerical Simulations
41 Linearized Drag Force For the numerical simulationsthere were two primary concerns how to identify the instantof an impact and how to define the impact condition Sincethe basis of this repeated impact model is the exchangeof momenta through impacts the time and condition ofthe impact need to be defined as precisely as possibleThe solution chosen to address this was to use a timestepping algorithm In the beginning 119888
1 1198882 1198883 and 119888
4are
calculated with the initial conditions By substituting 1198881 1198882
1198883 and 119888
4into (24) the responses of the free mass and
structure are obtained Based on the calculated response therelative displacement between the free mass and structure iscompared to the impact condition at every time step Theimpact condition is theoretically given by
|119909 (119905) minus 119911 (119905)| = 0 (25)
The actual numerical simulation however cannot achieveenough precision to capture the exact impact conditionTherefore a tolerance for the impact condition is requiredto allow for numerical round-off errors of the numericalsimulation
The tolerance should be considered carefully because toosmall tolerance has no meaning and a large tolerance makesimpacts occur more often than they should In the algorithmthe tolerance for the impact condition is set flexibly by thefollowing equation
Δ =1
10min (|120575119909| |120575119911|) (26)
where Δ is the impact condition such that |119909(119905) minus 119911(119905)| le Δ
and 120575119909 = (119905)Δ119905 and 120575119911 = (119905)Δ119905 Furthermore Δ119905 is thetime stepThe flexible impact condition hasmerit over a fixedone When the responses of the two masses are very smallthe fixed impact condition cannot capture some impactsTheflexible impact condition however is determined accordingto the size of the responses therefore the impact conditionis able to capture impacts whether the responses are slow orfast
After the comparison of the relative displacement andthe impact condition if the relative displacement is largerthan the impact condition the same values of 119888
1 1198882 1198883 and
1198884will be forwarded to the next time step Otherwise the
relative displacement is equal to or less than the impactcondition and thus an impact takes place Once the algorithmcatches the impact 119888
1 1198882 1198883 and 119888
4are calculated based on
the values from the previous time step and new conditionscorresponding to the impact as explained in the previoussection In addition the algorithm is performed on twodifferent time scales one is global time and the other is localtime The global time is the overall time starting with thebeginning of the algorithm and the local time counts timeduration between impacts thus after each impact the localtime is reset to 0 while the global time increases continuouslyWith 0 local time new responses of the system based on thenew 1198881 1198882 1198883 and 119888
4begin and evolve until next impact This
process keeps repeating until the system reaches the steadystate
The steady-state condition is also implemented in thealgorithm to avoid infinitesimal impacts which occur at theend of the responses The steady-state condition involvesdisplacements velocities and accelerations The algorithmverifies at every step whether or not both responses of the iceand structure satisfy the steady-state conditions The steady-state conditions are given as
(i) max(|119911(119905)| |119909(119905)|) lt 00001m(ii) max(|(119905)| |(119905)|) lt 00001ms(iii) max(|(119905)| |(119905)|) lt 00001ms2
The value of 00001 was chosen through trial and error for thesteady-state condition When the system reaches the steadystate the structure is constantly forced by the free mass andso the equation of motion in the steady state is expressed as
119898 + 119888 + 119896119909 = minus119887 ( minus ) (27)
Since the system is in the steady state
119896119909 = 119887 (28)
In the steady state condition the twomasses collide andmovetogether therefore the responses are given by
119909 (119905) = 119911 (119905) =119887
119896 (29)
which is a constant (steady state) displacement since isthe (assumed) constant velocity of the flow The algorithmreplaces (24) with (28) when the steady-state condition issatisfied
Figure 2 is plotted using the same structural propertiesin Table 1 which are obtained from [27] The two free massproperties mass 119872 and area 119860 are arbitrarily chosen as1600 kg which is the same as the structural mass and 1m2respectively The reason for this is as follows If the free massis much larger than the structural mass then its displacementand velocity would be hardly affected by the collisionwith thestructure and in fact repeated impacts would not occur Onthe other hand if the free mass is much smaller than that ofthe structure then the structure itself would hardly be affectedby the impacts Thus the most meaningful dynamics willoccur when the two masses are relatively the same order ofmagnitude For simplicity we begin our analysis by choosingthem to be the same and later in the paper we investigate theeffect of different choices of for the free mass
The linearization factor | minus | is set as 01ms becauseit is the maximum difference between the free mass velocityand flow velocity which is set as 01ms Therefore the valueof 119887 is
119887 =1
2120588119862119863119860 | minus | = 4995 kgs (30)
where 120588 is the density of water and119862119863 the drag coefficient is
1The coefficient of restitution another important parameteris assumed to be 1 which implies no energy dissipationthrough impact In addition the initial displacement and
6 Mathematical Problems in Engineering
0
minus02
minus04
minus06
minus08
minus10 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall Responses
Disp
lace
men
t (m
)
Time (s)
0
minus0002
minus0004
minus0006
186 188 190 192 194 196 198 200
StructureIce
(b) Responses from 185 to 200 seconds
Disp
lace
men
t (m
)
Time (s)
StructureIce
0
380 385 390 395 400
times10minus4
minus1
minus2
minus3
(c) Responses from 380 to 400 seconds
Figure 2 Responses of the ice and structure with a linearized drag force
Table 1 Parameters for simulations
Parameter Model structureℎ (mm) 41119863 (mm) 76119898 (kg) 1600119888 (kNsm) 02119896 (MNm) 1ℓ (m) 4493120590119888(MPa) 14
119909max (mm) 20
velocity of the free mass are minus001m and 001ms respec-tively and 0m and 0ms for the structure Figure 2 shows theresponse based on these parameters
Since the motivation for this model was to attempt toexplain IIV without analyzing specific ice-failure mecha-nisms we attempt to compare our results to some measuredIIV data There is very little IIV data in the open literatureespecially for displacement measurements Only laboratorymeasurements are available from [28 29] for displacementmeasurements Since the structural data for their tests arenot presented it is inadequate to compare their experimentaldata to Figure 2 directly but general vibration characteristicscan be compared Structural vibrations of IIVs becomequasi-static vibrations at low ice velocities and steady-statevibrations at medium ice velocities [30] although there isno quantitative definition of low and high ice velocities Aquasi-static vibration is defined by Karna [30] as a transientresponse which is followed by its maximum response atthe peak ice force and is not amplified by the dynamicsof the structure Figure 3(b) represents typical quasi-static
Mathematical Problems in Engineering 7
(a) [28]
1
0
2 4 6 8 10
Stru
ctur
aldi
spla
cem
ent(
cm)
Ice velocity = 76mms Time (s)
(b) [29]
2 4 6 8 10
1
0Stru
ctur
aldi
spla
cem
ent(
cm)
Time (s)Ice velocity = 395mms
(c) [29]
Figure 3 Measured structural responses from the literature
vibrations at ice velocity 76mms and Figure 3(c) shows thecharacteristic of a steady-state vibration at free mass velocity395mms Figure 3(a) however presents the displacementresponse although it is plotted at a lower free mass velocitythan Figure 3(b) This demonstrates that the vibration char-acteristics of IIVs are not determined by the velocity of theice alone but by involving all parameters
As shown in Figure 3(a) actual IIVs display multiplevibration characteristics from quasi-static to steady-statevibrations even at the same ice velocity contrary to thecharacteristic failure frequency model which predicts onetype of characteristic vibration at one ice velocity In thisrespect the proposed impact model possesses an advantageover the characteristic failure frequency model In an impactmodel the velocity of the ice varies from time to time whichis more realistic because part of the ice (free mass) contactingthe structure is deformed as the vibrations proceed Figure 2clearly shows multiple vibration characteristics Figure 2(b)displays quasi-static vibrations similar to Figure 3(b) From380 to 400 seconds the vibrations become stable and steadystate shown in Figure 2(c) which is close to the vibrationcharacteristic of Figure 3(c) The vibration characteristics ofthe impact model however cannot be clearly distinguishedfor the entire duration rather they show a mixture of thecharacteristics Therefore it is reasonable to conclude thatthe vibration characteristics of the impact IIV model aregoverned by the momenta of the ice and structure at the timeof impact
The weakness of modelling IIVs as repeated impacts asproposed in this paper model is how to model the relevantproperties of the ice The definitions of the ice mass and areain the impact IIVmodel imply that themass and area directlyaffect the impact or interactionwith the structure Among thetwo ice properties the influence of the ice area 119860 is directlyrelated to the velocity of the flowbecause of the relationwith 119887given by 119887 = (12)120588119862
119863119860|minus | It is inappropriate to simulate
the linearized model with large flow velocities since then thelinearized model would not be applicable
The mass of the free mass plays an important role in thismodel In order to understand its role it will be investigatedby conducting several numerical simulations Figures 4 5and 6 are plotted with the same parameters of Figure 2 exceptwith differentmasses which are proportional to the structuralmass 50 and 150 percent of 1600 kg so that the structuralmass can be used as a benchmark Including Figure 2 whichis plotted with the free mass of 100 percent of the structuralmass the responses with three different free masses arecompared At the beginning of the responses the heaviermass transfers higher momentum thus impacts happen lessfrequently but with largermagnitudes On the other hand thelighter mass allows the structure to approach the steady-statevibration in relatively short time In Figure 6 the magnitudesof the structural responses are not too different but theheavier mass induces more quasi-static vibrations for longertime Therefore it can be concluded that the free mass hasa greater effect on determining the vibration characteristicsthan on the magnitude of the structural response
Another important factor in this model is the coefficientof restitution The coefficient of restitution also requiresexperimental research to be defined quantitatively Theconducted numerical simulations can only show how thecoefficient of restitution influences the responses of thesystem Figure 2 is plotted with the coefficient of restitution119890 = 1 which is an unrealistic case Figures 7 8 and 9are simulated based on the same parameters of Figure 2 butwith different values of 119890 It becomes obvious that the factorthat most influences the magnitudes of the vibrations is thecoefficient of restitution rather than the velocities of the freemass The system with 119890 = 01 approaches the steady stateafter the first impact The system with 119890 = 05 not onlyapproaches the steady-state vibration faster but also showslower structural magnitudes than the system with 119890 = 1Figure 2 Figure 8 indicates that the system with 119890 = 05
loses the majority of its momentum after approximately54 seconds This suggests that vibrations due to repeatedimpacts can be controlled by manipulating the coefficient
8 Mathematical Problems in Engineering
002
0
minus002
minus004
minus006
minus008
minus01
minus0120 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with119872 = 800
0 50 100 150 200 250 300 350 400
02
0
minus02
minus04
minus06
minus08
minus1
minus12
minus14
Time (s)
Disp
lace
men
t (m
)
StructureIce
(b) Overall responses with119872= 2400
Figure 4 Responses with different ice masses 1
186 188 190 192 194 196 198 200
Disp
lace
men
t (m
)
StructureIce
Time (s)
0
minus2
minus4
minus6
times10minus4
(a) Responses with119872 = 800 from 185 to 200 seconds
0
minus0002
minus0004
minus0006
minus0008
186 188 190 192 194 196 198 200
StructureIce
Time (s)
Disp
lace
men
t (m
)
(b) Responses with119872= 2400 from 185 to 200 seconds
Figure 5 Responses with different ice masses 2
of restitution via the properties of the contacting area Forinstance the magnitudes of vibrations can be significantlyreduced by increasing the surface roughness of the structure
42 Quadratic Drag Force In the previous section themodelwas developed based on the linearized drag force whichassumes small relative velocity between the velocities of thefree mass and structure The linearized model howeveris definitely vulnerable to two factors namely when theinterval between impacts exceeds several seconds and whenthe velocity of the flow is no longer small Figure 10 clearlyshows the error between the models based on the linearized
drag and quadratic drag as time passes The displacement oflinearized equations (3) and (6) is compared with numer-ically solved equation (3) by using the built-in ordinarydifferential equation solver ofMathCadThe difference growsas time goes by but at high flow velocity the differencedecreases in Figure 10(b) This is because of the linearizationfactor | minus | which calibrates the difference between thelinear and quadratic models The solution however to thequadratic drag force needs to be developed for a more precisesimulation
A numerical solution is needed to solve (3) but insteadof using the built-in ordinary differential equation solver
Mathematical Problems in Engineering 9
StructureIce
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
Time (s)
1
0
minus1
minus2
minus3
minus4
times10minus4
(a) Responses with119872 = 800 from 385 to 400 seconds
StructureIce
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
0
minus2
minus4
minus6
times10minus4
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 6 Responses with different ice masses 3
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Overall responses with 119890 = 01
0
minus005
minus01
minus015
minus02
minus025
minus03
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400Time (s)
StructureIce
(b) Overall responses with 119890 = 05
Figure 7 Responses with different coefficients of restitution 1
of MathCad a simple numerical solution is developed for afaster algorithm From (3)
119872 = minus1
2120588119862119863119860 ( minus ) | minus | = minus119861 ( minus ) | minus | (31)
or
=119889
119889119905= minus
119861
119872( minus ) | minus | (32)
where
119861 =1
2120588119862119863119860 (33)
Using 119894 as a counter for the a number of time steps thedisplacement of (34) can be expressed as
119911119894= 119911119894minus1
+119889119911119894minus1
119889119905Δ119905 = 119911
119894minus1+ 119894minus1Δ119905 (34)
in which Δ119905 is previously defined as a time step z is given by
119894= 119894minus1
+119889119894minus1
119889119905Δ119905 =
119894minus1+ 119894minus1Δ119905 (35)
By substituting (32) into (35)
119894= 119894minus1
+ [minus119861
119872(119894minus1
minus )1003816100381610038161003816119894minus1 minus
1003816100381610038161003816] Δ119905(36)
10 Mathematical Problems in Engineering
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 2 4 6 8 10 12 14
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Responses with 119890 = 01 from 0 to 15 seconds
Disp
lace
men
t (m
)
Time (s)
StructureIce
46 48 50 52 54 56 58 60
2
0
minus2
minus4
minus6
minus8
times10minus4
(b) Responses with 119890 = 05 from 45 to 60 seconds
Figure 8 Responses with different coefficients of restitution 2
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus001386 388 390 392 394 396 398 400
Time (s)
Disp
lace
men
t (m
)
(a) Responses with 119890 = 01 from 385 to 400 seconds
StructureIce
Time (s)
Disp
lace
men
t (m
)
398 3985 399 3995 400
times10minus6
6
5
4
3
(b) Responses with 119890 = 05 from 398 to 400 seconds
Figure 9 Responses with different coefficients of restitution 3
Given the initial displacement and velocity of the free massthe displacement can be numerically solved by using (34)and (36) as simultaneous equationsThe simulation results of(34) and (36) are also plotted in Figure 10 At the high flowvelocity Figure 10(b) (34) and (36) predict the displacementclose enough to the one evaluated with the built-in ordinarydifferential equation solver In the algorithm of the numericalsimulation (34) and (36) replace (6) and (7)
Using the same parameters as those used in Figure 2but with a different flow velocity the responses of themodel with the quadratic drag force are compared withthe one with the linearized drag force in Figure 11 At high
flow velocities over 1ms the quadratic drag force modelapproaches steady-state vibration more quickly than thelinearized drag force model because at high flow velocity thequadratic drag induces more damping while it transfers moreforces than the linearized drag In Figure 12 the quadraticdrag force model exhibits the characteristic of the steady-state vibration while the linearized drag force model stilldisplays quasi-static vibration As velocities increase whichcorresponds to flowvelocities increasing vibrations approachthe steady state The linearized drag force model howeverdoes not show a steady-state vibration until 400 secondsin Figure 12 The linearization factor allows the linearized
Mathematical Problems in Engineering 11
0 100 200 300 400 500 600
100
80
60
40
20
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(a) Flow velocity 01ms
0 100 200 300 400 500 600
1000
800
600
400
200
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(b) Flow velocity 1ms
Figure 10 Comparison of the linearized and quadratic Drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
Disp
lace
men
t (m
)
(a) Overall responses with the quadratic drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
0
minus02
minus04
minus06
minus08
minus1
Disp
lace
men
t (m
)
(b) Overall responses with the linear drag
Figure 11 Responses with the quadratic and linear drags 1
drag force model to follow the quadratic drag force modelin the beginning but when the system is in a series of lowvelocity impacts the linearization factor causes a differenceThe linearized drag force model therefore is inappropriateto simulate the repeated impact model at high flow veloci-ties
Since the free mass area 119860 plays a similar role as theflow velocity in the model the model with wide area is alsosimulated with the quadratic drag force The influence of thedrag force on the response of the free mass increases as thesize of the area increases because the area is related to the dragforce by (1) 119891
119863= minus(12)120588119862
119863119860( minus )
2 Figure 13 is plottedwith the area 119860 = 2m2 the flow velocity = 01ms and
the rest of parameters are the same as the parameters usedin Figure 11 The flow velocity of Figure 13 is identical to thatof Figure 2 but the responses quickly approach the steady-state vibrations as if they are driven by high flow velocityFigure 13(b) also shows a similar vibration characteristic toFigure 12(a) because both flow velocity and area influence thedrag force in the impact model Therefore the area can beanother controlling factor of the vibrations Since the areais related to the size of the contacting area of the structurereducing the contacting area has a similar effect to that ofreducing the flow velocity
In the previous section the roles of the free mass andthe coefficient of restitution have been defined by changing
12 Mathematical Problems in EngineeringD
ispla
cem
ent (
m)
395 396 397 398 399 400Time (s)
StructureIce
8
6
4
2
0
times10minus4
(a) Responses with the quadratic drag from 395 to 400 secondsD
ispla
cem
ent (
m)
386 388 390 392 394 396 398 400Time (s)
StructureIce
0
minus0005
(b) Responses with the linear drag from 385 to 400 seconds
Figure 12 Responses with the quadratic and linear drags 2
005
0
minus005
minus01
minus015
minus02
minus025
minus03
minus035
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
(a) Overall responses with 119860 = 2
Disp
lace
men
t (m
)
395 396 397 398 399 400Time (s)
StructureIce
1
0
minus1
times10minus5
(b) Responses with 119860 = 2 from 395 to 400 seconds
Figure 13 Responses with 119860 = 2m2
relevant parameters To verify if those definitions still holdwith the quadratic drag force the impact model with thequadratic drag is plotted with change in free mass and thecoefficient of restitution Figures 14 15 and 16 are plottedon the same parameters used in Figures 2 4(a) and 4(b)respectively except that they are based on the quadratic dragforce The responses of Figures 14 15 and 16 prove thatthe defined role of the free mass holds with the quadraticdrag force The responses of the system with heavier freemass Figure 16(b) is still unstable while the system with thefree mass 119872 = 800 kg Figure 15(b) is in the steady-statevibration The quadratic drag force however induces more
drag force thus the system responses tend to be stabilizedfaster than those with the linearized drag forces
Figures 17 and 18 are plotted on the same conditions withFigures 7(a) and 7(b) respectively but with the quadraticdrag forceThe coefficient of restitution also governs themag-nitudes of responses in the impact model with the quadraticdrag force however the system with the quadratic drag forceis more sensitive to change in the coefficient of restitutionthan the one with the linearized drag force The system with119890 = 05 Figure 18(b) loses the momentum at approximately43 seconds under the influence of the quadratic drag forcewhile the system with the linearized drag force and 119890 = 05
Mathematical Problems in Engineering 13
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
01
0
minus01
minus02
minus03
minus04
minus05
minus06
minus07
(a) Overall responses with119872 = 1600
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
StructureIce
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 1600 from 385 to 400 seconds
Figure 14 Responses with119872 = 1600
0
minus001
minus002
minus003
minus004
minus005
minus006
minus0070 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall responses with119872 = 800
395 396 397 398 399 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
0
5
minus5
times10minus6
(b) Responses with119872 = 800 from 395 to 400 seconds
Figure 15 Responses with119872 = 800
Figure 8(b) loses the majority of its momentum at approxi-mately 54 seconds This is again because the quadratic dragforce dissipates energy more quickly than the linearized dragforceHowever it is demonstrated in Figures 17 and 18 that thecoefficient of restitution plays the same role in both quadraticand linearized drag forces
5 Resonance Conditions
One of the main objectives of the analysis aims to find aresonance condition (which is often referred to as lock-in
when investigating IIVs) The structural vibration frequencyincreases as the free mass velocity increases but when lock-in occurs the structural frequency remains at the naturalfrequency even with increasing free mass velocity [6] In theimpact model however the velocities of the free mass cannotbe the primary lock-in condition because the velocities of thefree mass are not fixed
One possible interpretation of modelling IIVs with theimpact model is that it considers the forces to be a seriesof discrete impulses as shown in Figure 19(a) while otherIIV models consider them as continuous periodic forces
14 Mathematical Problems in Engineering
0 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
(a) Overall responses with119872= 2400
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 16 Responses with119872 = 2400
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
(a) Overall responses with 119890 = 01
StructureIce
0 2 4 6 8 10 12 14
0002
0
minus0002
minus0004
minus0006
minus0008
Disp
lace
men
t (m
)
Time (s)
(b) Responses with 119890 = 01 from 0 to 15 seconds
Figure 17 Responses with 119890 = 01
In fact the impact IIV model proposes a way to calculate thestrength of each impulse and the times between impulses andthese depend on the momenta of the ice and structure aswell as the coefficient of restitution for the collision Undercertain conditions it is possible for the discrete impulsesof Figure 19(a) to approach the regular periodic series ofimpulses of Figure 19(b) which is suspected to cause the lock-in condition
This view of the impact IIV model can be analyzed byusing a Fourier transform [31] There are several definitions
of the Fourier transform (FT) that may be used and in thispaper the chosen version of the FT is given by
119865 (120596) = int
infin
minusinfin
119891 (119905) 119890minus119894120596119905
119889119905 (37)
The corresponding inverse Fourier transform is then given by
119891 (119905) =1
2120587int
infin
minusinfin
119865 (120596) 119890119894120596119905119889120596 (38)
Mathematical Problems in Engineering 15
005
0
minus005
minus01
minus015
minus02
minus0250 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with 119890 = 05
Time (s)
Disp
lace
men
t (m
)
StructureIce
36 38 40 42 44 46 48 50
4
2
0
minus2
minus4
times10minus4
(b) Responses with 119890 = 05 from 35 to 50 seconds
Figure 18 Responses with 119890 = 05
f(t)
fj
tj t
(a) Random impulses
f(t)
f
T 2T 3Tt
pT
(b) Uniform impulses
Figure 19 Ice forces as a series of impulses
The discrete ice forces of Figure 19(a) can be expressed asa series of impulses as
119891 (119905) =
infin
sum
119895=1
119891119895120575119863(119905 minus 119905119895) (39)
where 120575119863indicates the Dirac delta function and 119891
119895indicates
the strength of each impact If the sizes and intervals ofthe impulses are uniform as shown in Figure 19(b) (39) isrewritten as
119891 (119905) =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (40)
In reality the repeated impacts will not all have the samestrength or occur at a fixed interval however it is sufficient forthe situation of Figure 19(a) to approach that of Figure 19(b)for this analysis to be relevant
Equation (40) is often called a comb functionTheFouriertransform of the comb function is given [32] by
119865 (120596) = 2120587
infin
sum
119873=1
119891120575119863(120596 minus
2120587119873
119879) (41)
For a typical single-degree of freedom forced vibrationsproblem
+ 2120577120596119899 + 1205962
119899119909 =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (42)
Taking Fourier transforms of both sides and rearranging give
119883 (120596) = 2120587suminfin
119873=1119891120575119863 (120596 minus 2120587119873119879)
(minus1205962 + 2120577120596119899120596119894 + 1205962
119899)
(43)
Taking the inverse Fourier transforms to find the structureresponse in time gives
119909 (119905) = 119891
infin
sum
119873=1
119890119894(2120587119873119879)119905
(minus4120587211987321198792 + 119894120577120596119899 (4120587119873119879) + 1205962
119899) (44)
Equation (44) experiences a resonance when the denom-inator of any of the terms in the sum goes to zero that is when120596119899= 2120587119873119879 where 119873 is an integer larger than 0 This says
that the structure will experience a resonance when the natu-ral frequency is an integral multiple of the forcing frequency
16 Mathematical Problems in Engineering
as would be expected In the previous numerical simulationsthis condition has been observed in several cases Figures12(a) and 13(b) For instance the structural frequency ofFigure 13(b) is 2120587119879 = 212058705 = 1257 rads The naturalfrequency of the structure is radic119896119898 = radic10000001600 =
25 rads With119873 = 2 the structural frequency is close to thenatural frequency The result is what is seen in Figure 13(b)The two masses vibrate at the same frequency of 2120587119879 whichis half of the natural frequency The structural frequencyremains at that level which is lock-in and so does thefrequency of the free mass response
Thus the impact model successfully predicts the exis-tence of a lock-in condition and also predicts that it can occurwhen the frequency of the structure and freemass impacts aresynchronized together at a multiple of the natural frequencyof the structure The question of the precise conditionsnecessary for the frequency of the impacts to occur at amultiple of the natural frequency still remains open
6 Summary and Conclusions
This paper analyzes the vibrations caused by repeated impactsof a flow-driven freemass onto a second structure Conserva-tion of linearmomentum and the coefficient of restitution areused to characterize the nature of the impacts between thetwo masses This model is referred to as the impact modelTheproperties of the impactmodel were investigated throughnumerical simulations The mass of the free mass is one ofthe factors determining the vibration characteristics sinceit is related to the momentum of each impact This modelis proposed to explain the nature of ice-induced vibrationswithout requiring a microscopic model of the mechanism ofice failure The magnitudes of the vibrations are governedin part by the coefficient of restitution which when usedto explain IIVs encapsulates all ice properties mentionedin other IIV models In addition to the ice properties thecoefficient of restitution also governs the interaction betweenthe structure and the free mass The free mass is used torepresent the ice when modelling IIVs The last propertyof the impact model is that of the free mass area whichinfluences the dissipation of the system momenta throughthe drag force A free mass with a small area shows the samebehavior of a mass driven by lower flow velocity
Therefore through the impact model two factors whichcontrol the characteristics of IIVs have been proposed theice area and the coefficient of restitution By changing thesize of the contact area between the ice and structure thedriving force of the ice can be reduced as if the ice weredriven by lower flow velocities Furthermore by changingfactors related to the coefficient of restitution such as thesurface roughness or shape of the structure IIVs can bereduced to avoid the lock-in conditionwhich occurswhen thefrequencies of the structure and ice impacts are synchronizedat a multiple of the natural frequency of the structure Inaddition to the inclusion of the effect of flow the newapproach to ice forces through the impact model permitsthe answering of questions which originate from the weaktheoretical points of existing IIV models
Particular strengths of the model are that it qualitativelymatches observed results and in particular that a mechanismof ice failure is not required in order to perform simulationsand predictions It also proposes a simple mechanism toexplain the potentially destructive resonance lock-in condi-tions
References
[1] J P Dempsey ldquoResearch trends in ice mechanicsrdquo InternationalJournal of Solids and Structures vol 37 no 1-2 pp 131ndash153 2000
[2] K A Blenkarn ldquoMeasurement and analysis of ice forces oncook inlet structuresrdquo in Proceedings of the Offshore TechnologyConference (OTC rsquo70) pp 365ndash378 1970
[3] Y Toyama T Sensu M Minami and N Yashima ldquoModel testson ice-induced self-excited vibration of cylindrical structuresrdquoin Proceedings of the 7th International Conference on Port andOcean Engineering under Arctic Conditions vol 2 pp 834ndash8441983
[4] M Ranta and R Raty ldquoOn the analytic solution of ice-inducedvibrations in a marine pile structurerdquo in Proceedings of the 7thInternational Conference on Port and Ocean Engineering underArctic Conditions vol 2 pp 901ndash908 1983
[5] A Barker G Timco H Gravesen and P Voslashlund ldquoIce loadingon Danish wind turbinesmdashpart 1 dynamic model testsrdquo ColdRegions Science and Technology vol 41 no 1 pp 1ndash23 2005
[6] D S Sodhi ldquoIce-induced vibrations of structuresrdquo SpecialReport 89-5 International Association for Hydraulic ResearchWorking Group on Ice Forces 1989
[7] D S Sodhi and C E Morris ldquoCharacteristic frequency of forcevariations in continuous crushing of sheet ice against rigidcylindrical structuresrdquo Cold Regions Science and Technologyvol 12 no 1 pp 1ndash12 1986
[8] D S Sodhi C E Morris and G F N Cox ldquoDynamic analysisof failure modes on ice sheets encountering sloping structuresrdquoin Proceedings of the 6th International Conference on OffshoreMechanics and Arctic Engineering vol 4 pp 281ndash284 1987
[9] H R Peyton ldquoSea ice forcesrdquo in Proceedings of the Conferenceon Ice Pressure against Structures 1966
[10] C R Neill ldquoDynamic ice forces on piers and piles an assess-ment of design guidelines in the light of recent researchrdquoCanadian Journal of Civil Engineering vol 3 no 2 pp 305ndash3411976
[11] G Matlock W Dawkins and J Panak ldquoAnalytical model forice-structure interactionrdquo Journal of the Engineering MechanicsDivision pp 1083ndash11092 1971
[12] M M Maattanen ldquoIce-induced vibrations of structures self-excitationrdquo Special Report 89-5 International Association forHydraulic Research Working Group on Ice Forces 1989
[13] M M Maattanen ldquoNumerical simulation of ice-induced vibra-tions in offshore structuresrdquo in Proceedings of the 14th NordicSeminar on Computational Mechanics 2001 Keynote lecture
[14] J Xu and L Wang ldquoIce force oscillator model and its numericalsolutionsrdquo in Proceedings of the 7th International Conference onOffshore Machanics and Arctic Engineering pp 171ndash176 1988
[15] G TsinkerMarine Structures Engineering Chapman-Hall NewYork NY USA 1st edition 1995
[16] G Huang and P Liu ldquoA dynamic model for ice-inducedvibration of structuresrdquo Journal of OffshoreMechanics andArcticEngineering vol 131 no 1 Article ID 011501 6 pages 2009
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
Mathematical Problems in Engineering 3
M
u u
x x x0
z z
m
k
c
Figure 1 Mechanical model of the system
but the main driving force is fluid flow Depending on theinitial displacement and velocity of the flow the velocity ofthe free mass is determined just prior to the time of impactand the maximum velocity of the free mass is taken to be thevelocity of the flowThe free mass moves until it collides withthe structure After the impact the free mass either stopsmoves forward or bounces back in the opposite directiondepending on the coefficient of restitution and conservationof total system (free mass and structure) momentum Sincethe free mass and fluid flow now have a relative velocitywith respect to each other the velocity of the free mass ischanged by the effect of fluid drag until the free mass reachesthe velocity of the flow and then moves again towards thestructure for another impact The free mass moves towardsthe structure and transfers its momentum to the structureagain when the relative displacement between the mass andstructure becomes 0 (ie when a collision occurs) In somecases the structure can transfermomentum to themasswhenthemomentumof the structure is larger then that of themassThis whole process keeps repeating but due to structuraldamping and fluid drag the momenta which are transferredto each other gradually decrease
Since mass 119872 is floating in the flow it is free to moveThe flow however moves with its own velocity which inthis case is always assumed to be positive (to the right inthe figure) The displacement of the free mass is given by thevariable 119911 and that of the structure is given by the variable119909 The mass is driven by the drag force 119891
119863 which is present
when there is a relative velocity between the flow and themass There are four conditions depending on the relativevelocities
(i) When ge 0 and || ge || 119891119863le 0
(ii) When ge 0 and || lt || 119891119863gt 0
(iii) When lt 0 and || ge || 119891119863ge 0
(iv) When lt 0 and || lt || 119891119863gt 0
Outside of a collision with the structure no force is trans-mitted to the free mass except the drag force due to the fluidThe structure imparts a force to the freemass when a collisionbetween them occursThe drag force from the fluid is definedas
119891119863= minus
1
2120588119862119863119860( minus )
2 (1)
where 120588 119862119863 and 119860 are the density of the fluid drag
coefficient and area related to the drag coefficient respec-tively [24] The drag force which influences the systemof Figure 1 changes the forcing direction according to therelative velocities as explained above The drag force for thefree mass of Figure 1 therefore is expressed as
119891119863= minus
1
2120588119862119863119860 | minus | ( minus ) (2)
The equation ofmotion of the freemass can be obtained from(2) since no other forces are involved and is thus given as
119872 = minus1
2120588119862119863119860 | minus | ( minus ) (3)
Since (3) is a nonlinear equation it should be linearizedor simulated numerically The linearized version of (3) isdeveloped by assuming a small relative velocity Equation (3)is linearized as
119872 = minus119887 ( minus ) (4)
where
119887 =1
2120588119862119863119860 | minus | (5)
| minus | is assumed to be small and constant The solution of(4) consists of homogeneous and particular solutions and canbe obtained as
119911 (119905) = 1198881+ 119905 + 119888
2119890minus(119887119872)119905
(6)
The velocity of the free mass is then given by the derivative of(6) as
(119905) = minus 1198882
119887
119872119890minus(119887119872)119905
(7)
Equation (6) is a closed form solution of linearized equation(3) The solution of nonlinear equation (3) will be developedlater since it requires a numerical approach
The structure is modeled as a single-degree-of-freedomsystem consisting of a mass damper and spring withoutexternal forces (other than impacts with the free mass) asin Figure 1 Since the source of structural vibrations is thetransfer of momenta during collisions outside the time ofthe collisions the structure vibrates as a free simple harmonicoscillator The equation of structural motion is expressed as
119898 + 119888 + 119896119909 = 0 (8)
For an underdamped case (8) has the following solution
119909 (119905) = 1198883119890minus120577120596119899119905 cos120596
119889119905 + 1198884119890minus120577120596119899119905 sin120596
119889119905 (9)
with 120577 120596119889 and 120596
119899having the usual interpretation and
definition as damping coefficient damped frequency andnatural frequency The velocity of the structure at time 119905 canbe calculated as
(119905) = minus 1205961198891198883119890minus120577120596119899119905 sin120596
119889119905 minus 120577120596
1198991198883119890minus120577120596119899119905 cos120596
119889119905
+ 1205961198891198884119890minus120577120596119899119905 cos120596
119889119905 minus 120577120596
1198991198884119890minus120577120596119899119905 sin120596
119889119905
(10)
4 Mathematical Problems in Engineering
From (6) (7) (9) and (10) the displacements and veloci-ties of the mass and structure at any given time can be foundSubstituting these expressions into the equations for theconservation of total systemmomentum and the equation forthe coefficient of restitution the displacements and velocitiesimmediately after impact can be evaluated Each of the twomasses however has two unknowns after impact Thus theunknowns after impact are the displacement and velocityof both masses Since only two equations can be developedfrom the conservation of momentum and the coefficientof restitution equation the number of unknowns exceedsthe number of equations by two Salapaka et al [25] alsoinvestigated a similar modeling approach They developedthe equation of a mass-spring system driven by a vibratingtable which impacts themass at a certain position repetitivelyFaced with the same problem of excessive unknowns theyassumed that the displacements of each mass before andafter impact remain the same thus reducing the number ofunknowns after impact to two [25]
The instants just before and just after impact will bedenoted by 119905
119896and 119905119891 Since the impact occurs when the dis-
placements of the twomasses are the same the displacementsat time 119905
119896are given as
119909 (119905119896) = 119911 (119905
119896) = 119909119896 (11)
The same notation is used for velocities From the conserva-tion of momentum the total momentum of the two masses119898 and119872 is conserved so that
119898119891+119872
119891= 119898119896+119872
119896 (12)
The second relevant impact equation is obtained from thecoefficient of restitution [26] as
119891minus 119891= 119890 (
119896minus 119896) (13)
Equations (12) and (13) can be rewritten in matrix form as
[119898 119872
1 minus1] [
119891
119891
] = [119898119896+119872
119896
119890119896minus 119890119896
] (14)
or
[119891
119891
] =minus1
119898 +119872[minus1 minus119872
minus1 119898][
119898119896+119872
119896
119890119896minus 119890119896
] (15)
From (15) the velocities after impact 119891and 119891 are
119891=
1
119898 +119872(119898119896+119872
119896+ 119890119872
119896minus 119890119872
119896) (16)
119891=
1
119898 +119872(119898119896+119872
119896minus 119890119898
119896+ 119890119898
119896) (17)
After impact the displacements are assumed to be thesame as before impact From (11)
119909 (119905119891) = 119909 (119905
119896) = 119909119896
119911 (119905119891) = 119911 (119905
119896) = 119909119896
(18)
Once the impact occurs the momentum of the free mass istransmitted to the structure or vice versa 119888
1 1198882 1198883 and 119888
4of
(6) (7) (9) and (10) need to be repeatedly calculated aftereach impact in order to obtain the new responses Since thecharacteristics of the system responses are reset after eachimpact the time right after the impact 119905
119891 is also reset to 0
to calculate 1198881 1198882 1198883 and 119888
4 Substituting 119905
119891= 0 into (6) and
(7)
119911 (119905119891) = 119911 (0) = 119888
1+ 1198882
(119905119891) = (0) = minus
119887
1198721198882
(19)
The displacements before and after impact are assumed to bethe same and the velocity of the free mass after the impactis calculated from (17) Equation (19) is equal to 119909
119896and
119891
respectively
119911 (0) = 1198881+ 1198882= 119909119896
(0) = minus119887
1198721198882= 119891
(20)
From (20) 1198881and 1198882are calculated as
1198881= 119909119896minus119872
119887( minus
119891)
1198882=119872
119887( minus
119891)
(21)
Applying the same assumption and (16) to (9) and (10)
119909 (119905119891) = 119909 (0) = 119888
3= 119909119896
(119905119891) = (0) = minus120577120596
1198991198883+ 1205961198891198884= 119891
(22)
It then follows that 1198883and 1198884are obtained from (22) as
1198883= 119909119896
1198884=
1
120596119889
(119891+ 120577120596119899119909119896)
(23)
Substituting the calculated 1198881 1198882 1198883 and 119888
4into (6) and (9)
the responses of the two masses after impact are expressed as
119911 (119905) = [119909119896minus119872
119887( minus
119891)] + 119905 +
119872
119887( minus
119891) 119890minus(119887119872)119905
119909 (119905) = 119909119896119890minus120577120596119899119905 cos120596
119889119905 +
1
120596119889
(119891+ 120577120596119899119909119896) 119890minus120577120596119899119905 sin120596
119889119905
(24)
All the equations are developed based on the linearized dragforce Equation (24) holds until the next impact After thenext impact 119888
1 1198882 1198883 and 119888
4should be recalculated according
to new values of 119909119896 119896 119911119896 and
119896obtained from the impact
equations
Mathematical Problems in Engineering 5
4 Numerical Simulations
41 Linearized Drag Force For the numerical simulationsthere were two primary concerns how to identify the instantof an impact and how to define the impact condition Sincethe basis of this repeated impact model is the exchangeof momenta through impacts the time and condition ofthe impact need to be defined as precisely as possibleThe solution chosen to address this was to use a timestepping algorithm In the beginning 119888
1 1198882 1198883 and 119888
4are
calculated with the initial conditions By substituting 1198881 1198882
1198883 and 119888
4into (24) the responses of the free mass and
structure are obtained Based on the calculated response therelative displacement between the free mass and structure iscompared to the impact condition at every time step Theimpact condition is theoretically given by
|119909 (119905) minus 119911 (119905)| = 0 (25)
The actual numerical simulation however cannot achieveenough precision to capture the exact impact conditionTherefore a tolerance for the impact condition is requiredto allow for numerical round-off errors of the numericalsimulation
The tolerance should be considered carefully because toosmall tolerance has no meaning and a large tolerance makesimpacts occur more often than they should In the algorithmthe tolerance for the impact condition is set flexibly by thefollowing equation
Δ =1
10min (|120575119909| |120575119911|) (26)
where Δ is the impact condition such that |119909(119905) minus 119911(119905)| le Δ
and 120575119909 = (119905)Δ119905 and 120575119911 = (119905)Δ119905 Furthermore Δ119905 is thetime stepThe flexible impact condition hasmerit over a fixedone When the responses of the two masses are very smallthe fixed impact condition cannot capture some impactsTheflexible impact condition however is determined accordingto the size of the responses therefore the impact conditionis able to capture impacts whether the responses are slow orfast
After the comparison of the relative displacement andthe impact condition if the relative displacement is largerthan the impact condition the same values of 119888
1 1198882 1198883 and
1198884will be forwarded to the next time step Otherwise the
relative displacement is equal to or less than the impactcondition and thus an impact takes place Once the algorithmcatches the impact 119888
1 1198882 1198883 and 119888
4are calculated based on
the values from the previous time step and new conditionscorresponding to the impact as explained in the previoussection In addition the algorithm is performed on twodifferent time scales one is global time and the other is localtime The global time is the overall time starting with thebeginning of the algorithm and the local time counts timeduration between impacts thus after each impact the localtime is reset to 0 while the global time increases continuouslyWith 0 local time new responses of the system based on thenew 1198881 1198882 1198883 and 119888
4begin and evolve until next impact This
process keeps repeating until the system reaches the steadystate
The steady-state condition is also implemented in thealgorithm to avoid infinitesimal impacts which occur at theend of the responses The steady-state condition involvesdisplacements velocities and accelerations The algorithmverifies at every step whether or not both responses of the iceand structure satisfy the steady-state conditions The steady-state conditions are given as
(i) max(|119911(119905)| |119909(119905)|) lt 00001m(ii) max(|(119905)| |(119905)|) lt 00001ms(iii) max(|(119905)| |(119905)|) lt 00001ms2
The value of 00001 was chosen through trial and error for thesteady-state condition When the system reaches the steadystate the structure is constantly forced by the free mass andso the equation of motion in the steady state is expressed as
119898 + 119888 + 119896119909 = minus119887 ( minus ) (27)
Since the system is in the steady state
119896119909 = 119887 (28)
In the steady state condition the twomasses collide andmovetogether therefore the responses are given by
119909 (119905) = 119911 (119905) =119887
119896 (29)
which is a constant (steady state) displacement since isthe (assumed) constant velocity of the flow The algorithmreplaces (24) with (28) when the steady-state condition issatisfied
Figure 2 is plotted using the same structural propertiesin Table 1 which are obtained from [27] The two free massproperties mass 119872 and area 119860 are arbitrarily chosen as1600 kg which is the same as the structural mass and 1m2respectively The reason for this is as follows If the free massis much larger than the structural mass then its displacementand velocity would be hardly affected by the collisionwith thestructure and in fact repeated impacts would not occur Onthe other hand if the free mass is much smaller than that ofthe structure then the structure itself would hardly be affectedby the impacts Thus the most meaningful dynamics willoccur when the two masses are relatively the same order ofmagnitude For simplicity we begin our analysis by choosingthem to be the same and later in the paper we investigate theeffect of different choices of for the free mass
The linearization factor | minus | is set as 01ms becauseit is the maximum difference between the free mass velocityand flow velocity which is set as 01ms Therefore the valueof 119887 is
119887 =1
2120588119862119863119860 | minus | = 4995 kgs (30)
where 120588 is the density of water and119862119863 the drag coefficient is
1The coefficient of restitution another important parameteris assumed to be 1 which implies no energy dissipationthrough impact In addition the initial displacement and
6 Mathematical Problems in Engineering
0
minus02
minus04
minus06
minus08
minus10 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall Responses
Disp
lace
men
t (m
)
Time (s)
0
minus0002
minus0004
minus0006
186 188 190 192 194 196 198 200
StructureIce
(b) Responses from 185 to 200 seconds
Disp
lace
men
t (m
)
Time (s)
StructureIce
0
380 385 390 395 400
times10minus4
minus1
minus2
minus3
(c) Responses from 380 to 400 seconds
Figure 2 Responses of the ice and structure with a linearized drag force
Table 1 Parameters for simulations
Parameter Model structureℎ (mm) 41119863 (mm) 76119898 (kg) 1600119888 (kNsm) 02119896 (MNm) 1ℓ (m) 4493120590119888(MPa) 14
119909max (mm) 20
velocity of the free mass are minus001m and 001ms respec-tively and 0m and 0ms for the structure Figure 2 shows theresponse based on these parameters
Since the motivation for this model was to attempt toexplain IIV without analyzing specific ice-failure mecha-nisms we attempt to compare our results to some measuredIIV data There is very little IIV data in the open literatureespecially for displacement measurements Only laboratorymeasurements are available from [28 29] for displacementmeasurements Since the structural data for their tests arenot presented it is inadequate to compare their experimentaldata to Figure 2 directly but general vibration characteristicscan be compared Structural vibrations of IIVs becomequasi-static vibrations at low ice velocities and steady-statevibrations at medium ice velocities [30] although there isno quantitative definition of low and high ice velocities Aquasi-static vibration is defined by Karna [30] as a transientresponse which is followed by its maximum response atthe peak ice force and is not amplified by the dynamicsof the structure Figure 3(b) represents typical quasi-static
Mathematical Problems in Engineering 7
(a) [28]
1
0
2 4 6 8 10
Stru
ctur
aldi
spla
cem
ent(
cm)
Ice velocity = 76mms Time (s)
(b) [29]
2 4 6 8 10
1
0Stru
ctur
aldi
spla
cem
ent(
cm)
Time (s)Ice velocity = 395mms
(c) [29]
Figure 3 Measured structural responses from the literature
vibrations at ice velocity 76mms and Figure 3(c) shows thecharacteristic of a steady-state vibration at free mass velocity395mms Figure 3(a) however presents the displacementresponse although it is plotted at a lower free mass velocitythan Figure 3(b) This demonstrates that the vibration char-acteristics of IIVs are not determined by the velocity of theice alone but by involving all parameters
As shown in Figure 3(a) actual IIVs display multiplevibration characteristics from quasi-static to steady-statevibrations even at the same ice velocity contrary to thecharacteristic failure frequency model which predicts onetype of characteristic vibration at one ice velocity In thisrespect the proposed impact model possesses an advantageover the characteristic failure frequency model In an impactmodel the velocity of the ice varies from time to time whichis more realistic because part of the ice (free mass) contactingthe structure is deformed as the vibrations proceed Figure 2clearly shows multiple vibration characteristics Figure 2(b)displays quasi-static vibrations similar to Figure 3(b) From380 to 400 seconds the vibrations become stable and steadystate shown in Figure 2(c) which is close to the vibrationcharacteristic of Figure 3(c) The vibration characteristics ofthe impact model however cannot be clearly distinguishedfor the entire duration rather they show a mixture of thecharacteristics Therefore it is reasonable to conclude thatthe vibration characteristics of the impact IIV model aregoverned by the momenta of the ice and structure at the timeof impact
The weakness of modelling IIVs as repeated impacts asproposed in this paper model is how to model the relevantproperties of the ice The definitions of the ice mass and areain the impact IIVmodel imply that themass and area directlyaffect the impact or interactionwith the structure Among thetwo ice properties the influence of the ice area 119860 is directlyrelated to the velocity of the flowbecause of the relationwith 119887given by 119887 = (12)120588119862
119863119860|minus | It is inappropriate to simulate
the linearized model with large flow velocities since then thelinearized model would not be applicable
The mass of the free mass plays an important role in thismodel In order to understand its role it will be investigatedby conducting several numerical simulations Figures 4 5and 6 are plotted with the same parameters of Figure 2 exceptwith differentmasses which are proportional to the structuralmass 50 and 150 percent of 1600 kg so that the structuralmass can be used as a benchmark Including Figure 2 whichis plotted with the free mass of 100 percent of the structuralmass the responses with three different free masses arecompared At the beginning of the responses the heaviermass transfers higher momentum thus impacts happen lessfrequently but with largermagnitudes On the other hand thelighter mass allows the structure to approach the steady-statevibration in relatively short time In Figure 6 the magnitudesof the structural responses are not too different but theheavier mass induces more quasi-static vibrations for longertime Therefore it can be concluded that the free mass hasa greater effect on determining the vibration characteristicsthan on the magnitude of the structural response
Another important factor in this model is the coefficientof restitution The coefficient of restitution also requiresexperimental research to be defined quantitatively Theconducted numerical simulations can only show how thecoefficient of restitution influences the responses of thesystem Figure 2 is plotted with the coefficient of restitution119890 = 1 which is an unrealistic case Figures 7 8 and 9are simulated based on the same parameters of Figure 2 butwith different values of 119890 It becomes obvious that the factorthat most influences the magnitudes of the vibrations is thecoefficient of restitution rather than the velocities of the freemass The system with 119890 = 01 approaches the steady stateafter the first impact The system with 119890 = 05 not onlyapproaches the steady-state vibration faster but also showslower structural magnitudes than the system with 119890 = 1Figure 2 Figure 8 indicates that the system with 119890 = 05
loses the majority of its momentum after approximately54 seconds This suggests that vibrations due to repeatedimpacts can be controlled by manipulating the coefficient
8 Mathematical Problems in Engineering
002
0
minus002
minus004
minus006
minus008
minus01
minus0120 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with119872 = 800
0 50 100 150 200 250 300 350 400
02
0
minus02
minus04
minus06
minus08
minus1
minus12
minus14
Time (s)
Disp
lace
men
t (m
)
StructureIce
(b) Overall responses with119872= 2400
Figure 4 Responses with different ice masses 1
186 188 190 192 194 196 198 200
Disp
lace
men
t (m
)
StructureIce
Time (s)
0
minus2
minus4
minus6
times10minus4
(a) Responses with119872 = 800 from 185 to 200 seconds
0
minus0002
minus0004
minus0006
minus0008
186 188 190 192 194 196 198 200
StructureIce
Time (s)
Disp
lace
men
t (m
)
(b) Responses with119872= 2400 from 185 to 200 seconds
Figure 5 Responses with different ice masses 2
of restitution via the properties of the contacting area Forinstance the magnitudes of vibrations can be significantlyreduced by increasing the surface roughness of the structure
42 Quadratic Drag Force In the previous section themodelwas developed based on the linearized drag force whichassumes small relative velocity between the velocities of thefree mass and structure The linearized model howeveris definitely vulnerable to two factors namely when theinterval between impacts exceeds several seconds and whenthe velocity of the flow is no longer small Figure 10 clearlyshows the error between the models based on the linearized
drag and quadratic drag as time passes The displacement oflinearized equations (3) and (6) is compared with numer-ically solved equation (3) by using the built-in ordinarydifferential equation solver ofMathCadThe difference growsas time goes by but at high flow velocity the differencedecreases in Figure 10(b) This is because of the linearizationfactor | minus | which calibrates the difference between thelinear and quadratic models The solution however to thequadratic drag force needs to be developed for a more precisesimulation
A numerical solution is needed to solve (3) but insteadof using the built-in ordinary differential equation solver
Mathematical Problems in Engineering 9
StructureIce
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
Time (s)
1
0
minus1
minus2
minus3
minus4
times10minus4
(a) Responses with119872 = 800 from 385 to 400 seconds
StructureIce
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
0
minus2
minus4
minus6
times10minus4
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 6 Responses with different ice masses 3
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Overall responses with 119890 = 01
0
minus005
minus01
minus015
minus02
minus025
minus03
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400Time (s)
StructureIce
(b) Overall responses with 119890 = 05
Figure 7 Responses with different coefficients of restitution 1
of MathCad a simple numerical solution is developed for afaster algorithm From (3)
119872 = minus1
2120588119862119863119860 ( minus ) | minus | = minus119861 ( minus ) | minus | (31)
or
=119889
119889119905= minus
119861
119872( minus ) | minus | (32)
where
119861 =1
2120588119862119863119860 (33)
Using 119894 as a counter for the a number of time steps thedisplacement of (34) can be expressed as
119911119894= 119911119894minus1
+119889119911119894minus1
119889119905Δ119905 = 119911
119894minus1+ 119894minus1Δ119905 (34)
in which Δ119905 is previously defined as a time step z is given by
119894= 119894minus1
+119889119894minus1
119889119905Δ119905 =
119894minus1+ 119894minus1Δ119905 (35)
By substituting (32) into (35)
119894= 119894minus1
+ [minus119861
119872(119894minus1
minus )1003816100381610038161003816119894minus1 minus
1003816100381610038161003816] Δ119905(36)
10 Mathematical Problems in Engineering
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 2 4 6 8 10 12 14
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Responses with 119890 = 01 from 0 to 15 seconds
Disp
lace
men
t (m
)
Time (s)
StructureIce
46 48 50 52 54 56 58 60
2
0
minus2
minus4
minus6
minus8
times10minus4
(b) Responses with 119890 = 05 from 45 to 60 seconds
Figure 8 Responses with different coefficients of restitution 2
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus001386 388 390 392 394 396 398 400
Time (s)
Disp
lace
men
t (m
)
(a) Responses with 119890 = 01 from 385 to 400 seconds
StructureIce
Time (s)
Disp
lace
men
t (m
)
398 3985 399 3995 400
times10minus6
6
5
4
3
(b) Responses with 119890 = 05 from 398 to 400 seconds
Figure 9 Responses with different coefficients of restitution 3
Given the initial displacement and velocity of the free massthe displacement can be numerically solved by using (34)and (36) as simultaneous equationsThe simulation results of(34) and (36) are also plotted in Figure 10 At the high flowvelocity Figure 10(b) (34) and (36) predict the displacementclose enough to the one evaluated with the built-in ordinarydifferential equation solver In the algorithm of the numericalsimulation (34) and (36) replace (6) and (7)
Using the same parameters as those used in Figure 2but with a different flow velocity the responses of themodel with the quadratic drag force are compared withthe one with the linearized drag force in Figure 11 At high
flow velocities over 1ms the quadratic drag force modelapproaches steady-state vibration more quickly than thelinearized drag force model because at high flow velocity thequadratic drag induces more damping while it transfers moreforces than the linearized drag In Figure 12 the quadraticdrag force model exhibits the characteristic of the steady-state vibration while the linearized drag force model stilldisplays quasi-static vibration As velocities increase whichcorresponds to flowvelocities increasing vibrations approachthe steady state The linearized drag force model howeverdoes not show a steady-state vibration until 400 secondsin Figure 12 The linearization factor allows the linearized
Mathematical Problems in Engineering 11
0 100 200 300 400 500 600
100
80
60
40
20
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(a) Flow velocity 01ms
0 100 200 300 400 500 600
1000
800
600
400
200
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(b) Flow velocity 1ms
Figure 10 Comparison of the linearized and quadratic Drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
Disp
lace
men
t (m
)
(a) Overall responses with the quadratic drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
0
minus02
minus04
minus06
minus08
minus1
Disp
lace
men
t (m
)
(b) Overall responses with the linear drag
Figure 11 Responses with the quadratic and linear drags 1
drag force model to follow the quadratic drag force modelin the beginning but when the system is in a series of lowvelocity impacts the linearization factor causes a differenceThe linearized drag force model therefore is inappropriateto simulate the repeated impact model at high flow veloci-ties
Since the free mass area 119860 plays a similar role as theflow velocity in the model the model with wide area is alsosimulated with the quadratic drag force The influence of thedrag force on the response of the free mass increases as thesize of the area increases because the area is related to the dragforce by (1) 119891
119863= minus(12)120588119862
119863119860( minus )
2 Figure 13 is plottedwith the area 119860 = 2m2 the flow velocity = 01ms and
the rest of parameters are the same as the parameters usedin Figure 11 The flow velocity of Figure 13 is identical to thatof Figure 2 but the responses quickly approach the steady-state vibrations as if they are driven by high flow velocityFigure 13(b) also shows a similar vibration characteristic toFigure 12(a) because both flow velocity and area influence thedrag force in the impact model Therefore the area can beanother controlling factor of the vibrations Since the areais related to the size of the contacting area of the structurereducing the contacting area has a similar effect to that ofreducing the flow velocity
In the previous section the roles of the free mass andthe coefficient of restitution have been defined by changing
12 Mathematical Problems in EngineeringD
ispla
cem
ent (
m)
395 396 397 398 399 400Time (s)
StructureIce
8
6
4
2
0
times10minus4
(a) Responses with the quadratic drag from 395 to 400 secondsD
ispla
cem
ent (
m)
386 388 390 392 394 396 398 400Time (s)
StructureIce
0
minus0005
(b) Responses with the linear drag from 385 to 400 seconds
Figure 12 Responses with the quadratic and linear drags 2
005
0
minus005
minus01
minus015
minus02
minus025
minus03
minus035
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
(a) Overall responses with 119860 = 2
Disp
lace
men
t (m
)
395 396 397 398 399 400Time (s)
StructureIce
1
0
minus1
times10minus5
(b) Responses with 119860 = 2 from 395 to 400 seconds
Figure 13 Responses with 119860 = 2m2
relevant parameters To verify if those definitions still holdwith the quadratic drag force the impact model with thequadratic drag is plotted with change in free mass and thecoefficient of restitution Figures 14 15 and 16 are plottedon the same parameters used in Figures 2 4(a) and 4(b)respectively except that they are based on the quadratic dragforce The responses of Figures 14 15 and 16 prove thatthe defined role of the free mass holds with the quadraticdrag force The responses of the system with heavier freemass Figure 16(b) is still unstable while the system with thefree mass 119872 = 800 kg Figure 15(b) is in the steady-statevibration The quadratic drag force however induces more
drag force thus the system responses tend to be stabilizedfaster than those with the linearized drag forces
Figures 17 and 18 are plotted on the same conditions withFigures 7(a) and 7(b) respectively but with the quadraticdrag forceThe coefficient of restitution also governs themag-nitudes of responses in the impact model with the quadraticdrag force however the system with the quadratic drag forceis more sensitive to change in the coefficient of restitutionthan the one with the linearized drag force The system with119890 = 05 Figure 18(b) loses the momentum at approximately43 seconds under the influence of the quadratic drag forcewhile the system with the linearized drag force and 119890 = 05
Mathematical Problems in Engineering 13
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
01
0
minus01
minus02
minus03
minus04
minus05
minus06
minus07
(a) Overall responses with119872 = 1600
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
StructureIce
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 1600 from 385 to 400 seconds
Figure 14 Responses with119872 = 1600
0
minus001
minus002
minus003
minus004
minus005
minus006
minus0070 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall responses with119872 = 800
395 396 397 398 399 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
0
5
minus5
times10minus6
(b) Responses with119872 = 800 from 395 to 400 seconds
Figure 15 Responses with119872 = 800
Figure 8(b) loses the majority of its momentum at approxi-mately 54 seconds This is again because the quadratic dragforce dissipates energy more quickly than the linearized dragforceHowever it is demonstrated in Figures 17 and 18 that thecoefficient of restitution plays the same role in both quadraticand linearized drag forces
5 Resonance Conditions
One of the main objectives of the analysis aims to find aresonance condition (which is often referred to as lock-in
when investigating IIVs) The structural vibration frequencyincreases as the free mass velocity increases but when lock-in occurs the structural frequency remains at the naturalfrequency even with increasing free mass velocity [6] In theimpact model however the velocities of the free mass cannotbe the primary lock-in condition because the velocities of thefree mass are not fixed
One possible interpretation of modelling IIVs with theimpact model is that it considers the forces to be a seriesof discrete impulses as shown in Figure 19(a) while otherIIV models consider them as continuous periodic forces
14 Mathematical Problems in Engineering
0 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
(a) Overall responses with119872= 2400
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 16 Responses with119872 = 2400
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
(a) Overall responses with 119890 = 01
StructureIce
0 2 4 6 8 10 12 14
0002
0
minus0002
minus0004
minus0006
minus0008
Disp
lace
men
t (m
)
Time (s)
(b) Responses with 119890 = 01 from 0 to 15 seconds
Figure 17 Responses with 119890 = 01
In fact the impact IIV model proposes a way to calculate thestrength of each impulse and the times between impulses andthese depend on the momenta of the ice and structure aswell as the coefficient of restitution for the collision Undercertain conditions it is possible for the discrete impulsesof Figure 19(a) to approach the regular periodic series ofimpulses of Figure 19(b) which is suspected to cause the lock-in condition
This view of the impact IIV model can be analyzed byusing a Fourier transform [31] There are several definitions
of the Fourier transform (FT) that may be used and in thispaper the chosen version of the FT is given by
119865 (120596) = int
infin
minusinfin
119891 (119905) 119890minus119894120596119905
119889119905 (37)
The corresponding inverse Fourier transform is then given by
119891 (119905) =1
2120587int
infin
minusinfin
119865 (120596) 119890119894120596119905119889120596 (38)
Mathematical Problems in Engineering 15
005
0
minus005
minus01
minus015
minus02
minus0250 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with 119890 = 05
Time (s)
Disp
lace
men
t (m
)
StructureIce
36 38 40 42 44 46 48 50
4
2
0
minus2
minus4
times10minus4
(b) Responses with 119890 = 05 from 35 to 50 seconds
Figure 18 Responses with 119890 = 05
f(t)
fj
tj t
(a) Random impulses
f(t)
f
T 2T 3Tt
pT
(b) Uniform impulses
Figure 19 Ice forces as a series of impulses
The discrete ice forces of Figure 19(a) can be expressed asa series of impulses as
119891 (119905) =
infin
sum
119895=1
119891119895120575119863(119905 minus 119905119895) (39)
where 120575119863indicates the Dirac delta function and 119891
119895indicates
the strength of each impact If the sizes and intervals ofthe impulses are uniform as shown in Figure 19(b) (39) isrewritten as
119891 (119905) =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (40)
In reality the repeated impacts will not all have the samestrength or occur at a fixed interval however it is sufficient forthe situation of Figure 19(a) to approach that of Figure 19(b)for this analysis to be relevant
Equation (40) is often called a comb functionTheFouriertransform of the comb function is given [32] by
119865 (120596) = 2120587
infin
sum
119873=1
119891120575119863(120596 minus
2120587119873
119879) (41)
For a typical single-degree of freedom forced vibrationsproblem
+ 2120577120596119899 + 1205962
119899119909 =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (42)
Taking Fourier transforms of both sides and rearranging give
119883 (120596) = 2120587suminfin
119873=1119891120575119863 (120596 minus 2120587119873119879)
(minus1205962 + 2120577120596119899120596119894 + 1205962
119899)
(43)
Taking the inverse Fourier transforms to find the structureresponse in time gives
119909 (119905) = 119891
infin
sum
119873=1
119890119894(2120587119873119879)119905
(minus4120587211987321198792 + 119894120577120596119899 (4120587119873119879) + 1205962
119899) (44)
Equation (44) experiences a resonance when the denom-inator of any of the terms in the sum goes to zero that is when120596119899= 2120587119873119879 where 119873 is an integer larger than 0 This says
that the structure will experience a resonance when the natu-ral frequency is an integral multiple of the forcing frequency
16 Mathematical Problems in Engineering
as would be expected In the previous numerical simulationsthis condition has been observed in several cases Figures12(a) and 13(b) For instance the structural frequency ofFigure 13(b) is 2120587119879 = 212058705 = 1257 rads The naturalfrequency of the structure is radic119896119898 = radic10000001600 =
25 rads With119873 = 2 the structural frequency is close to thenatural frequency The result is what is seen in Figure 13(b)The two masses vibrate at the same frequency of 2120587119879 whichis half of the natural frequency The structural frequencyremains at that level which is lock-in and so does thefrequency of the free mass response
Thus the impact model successfully predicts the exis-tence of a lock-in condition and also predicts that it can occurwhen the frequency of the structure and freemass impacts aresynchronized together at a multiple of the natural frequencyof the structure The question of the precise conditionsnecessary for the frequency of the impacts to occur at amultiple of the natural frequency still remains open
6 Summary and Conclusions
This paper analyzes the vibrations caused by repeated impactsof a flow-driven freemass onto a second structure Conserva-tion of linearmomentum and the coefficient of restitution areused to characterize the nature of the impacts between thetwo masses This model is referred to as the impact modelTheproperties of the impactmodel were investigated throughnumerical simulations The mass of the free mass is one ofthe factors determining the vibration characteristics sinceit is related to the momentum of each impact This modelis proposed to explain the nature of ice-induced vibrationswithout requiring a microscopic model of the mechanism ofice failure The magnitudes of the vibrations are governedin part by the coefficient of restitution which when usedto explain IIVs encapsulates all ice properties mentionedin other IIV models In addition to the ice properties thecoefficient of restitution also governs the interaction betweenthe structure and the free mass The free mass is used torepresent the ice when modelling IIVs The last propertyof the impact model is that of the free mass area whichinfluences the dissipation of the system momenta throughthe drag force A free mass with a small area shows the samebehavior of a mass driven by lower flow velocity
Therefore through the impact model two factors whichcontrol the characteristics of IIVs have been proposed theice area and the coefficient of restitution By changing thesize of the contact area between the ice and structure thedriving force of the ice can be reduced as if the ice weredriven by lower flow velocities Furthermore by changingfactors related to the coefficient of restitution such as thesurface roughness or shape of the structure IIVs can bereduced to avoid the lock-in conditionwhich occurswhen thefrequencies of the structure and ice impacts are synchronizedat a multiple of the natural frequency of the structure Inaddition to the inclusion of the effect of flow the newapproach to ice forces through the impact model permitsthe answering of questions which originate from the weaktheoretical points of existing IIV models
Particular strengths of the model are that it qualitativelymatches observed results and in particular that a mechanismof ice failure is not required in order to perform simulationsand predictions It also proposes a simple mechanism toexplain the potentially destructive resonance lock-in condi-tions
References
[1] J P Dempsey ldquoResearch trends in ice mechanicsrdquo InternationalJournal of Solids and Structures vol 37 no 1-2 pp 131ndash153 2000
[2] K A Blenkarn ldquoMeasurement and analysis of ice forces oncook inlet structuresrdquo in Proceedings of the Offshore TechnologyConference (OTC rsquo70) pp 365ndash378 1970
[3] Y Toyama T Sensu M Minami and N Yashima ldquoModel testson ice-induced self-excited vibration of cylindrical structuresrdquoin Proceedings of the 7th International Conference on Port andOcean Engineering under Arctic Conditions vol 2 pp 834ndash8441983
[4] M Ranta and R Raty ldquoOn the analytic solution of ice-inducedvibrations in a marine pile structurerdquo in Proceedings of the 7thInternational Conference on Port and Ocean Engineering underArctic Conditions vol 2 pp 901ndash908 1983
[5] A Barker G Timco H Gravesen and P Voslashlund ldquoIce loadingon Danish wind turbinesmdashpart 1 dynamic model testsrdquo ColdRegions Science and Technology vol 41 no 1 pp 1ndash23 2005
[6] D S Sodhi ldquoIce-induced vibrations of structuresrdquo SpecialReport 89-5 International Association for Hydraulic ResearchWorking Group on Ice Forces 1989
[7] D S Sodhi and C E Morris ldquoCharacteristic frequency of forcevariations in continuous crushing of sheet ice against rigidcylindrical structuresrdquo Cold Regions Science and Technologyvol 12 no 1 pp 1ndash12 1986
[8] D S Sodhi C E Morris and G F N Cox ldquoDynamic analysisof failure modes on ice sheets encountering sloping structuresrdquoin Proceedings of the 6th International Conference on OffshoreMechanics and Arctic Engineering vol 4 pp 281ndash284 1987
[9] H R Peyton ldquoSea ice forcesrdquo in Proceedings of the Conferenceon Ice Pressure against Structures 1966
[10] C R Neill ldquoDynamic ice forces on piers and piles an assess-ment of design guidelines in the light of recent researchrdquoCanadian Journal of Civil Engineering vol 3 no 2 pp 305ndash3411976
[11] G Matlock W Dawkins and J Panak ldquoAnalytical model forice-structure interactionrdquo Journal of the Engineering MechanicsDivision pp 1083ndash11092 1971
[12] M M Maattanen ldquoIce-induced vibrations of structures self-excitationrdquo Special Report 89-5 International Association forHydraulic Research Working Group on Ice Forces 1989
[13] M M Maattanen ldquoNumerical simulation of ice-induced vibra-tions in offshore structuresrdquo in Proceedings of the 14th NordicSeminar on Computational Mechanics 2001 Keynote lecture
[14] J Xu and L Wang ldquoIce force oscillator model and its numericalsolutionsrdquo in Proceedings of the 7th International Conference onOffshore Machanics and Arctic Engineering pp 171ndash176 1988
[15] G TsinkerMarine Structures Engineering Chapman-Hall NewYork NY USA 1st edition 1995
[16] G Huang and P Liu ldquoA dynamic model for ice-inducedvibration of structuresrdquo Journal of OffshoreMechanics andArcticEngineering vol 131 no 1 Article ID 011501 6 pages 2009
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
4 Mathematical Problems in Engineering
From (6) (7) (9) and (10) the displacements and veloci-ties of the mass and structure at any given time can be foundSubstituting these expressions into the equations for theconservation of total systemmomentum and the equation forthe coefficient of restitution the displacements and velocitiesimmediately after impact can be evaluated Each of the twomasses however has two unknowns after impact Thus theunknowns after impact are the displacement and velocityof both masses Since only two equations can be developedfrom the conservation of momentum and the coefficientof restitution equation the number of unknowns exceedsthe number of equations by two Salapaka et al [25] alsoinvestigated a similar modeling approach They developedthe equation of a mass-spring system driven by a vibratingtable which impacts themass at a certain position repetitivelyFaced with the same problem of excessive unknowns theyassumed that the displacements of each mass before andafter impact remain the same thus reducing the number ofunknowns after impact to two [25]
The instants just before and just after impact will bedenoted by 119905
119896and 119905119891 Since the impact occurs when the dis-
placements of the twomasses are the same the displacementsat time 119905
119896are given as
119909 (119905119896) = 119911 (119905
119896) = 119909119896 (11)
The same notation is used for velocities From the conserva-tion of momentum the total momentum of the two masses119898 and119872 is conserved so that
119898119891+119872
119891= 119898119896+119872
119896 (12)
The second relevant impact equation is obtained from thecoefficient of restitution [26] as
119891minus 119891= 119890 (
119896minus 119896) (13)
Equations (12) and (13) can be rewritten in matrix form as
[119898 119872
1 minus1] [
119891
119891
] = [119898119896+119872
119896
119890119896minus 119890119896
] (14)
or
[119891
119891
] =minus1
119898 +119872[minus1 minus119872
minus1 119898][
119898119896+119872
119896
119890119896minus 119890119896
] (15)
From (15) the velocities after impact 119891and 119891 are
119891=
1
119898 +119872(119898119896+119872
119896+ 119890119872
119896minus 119890119872
119896) (16)
119891=
1
119898 +119872(119898119896+119872
119896minus 119890119898
119896+ 119890119898
119896) (17)
After impact the displacements are assumed to be thesame as before impact From (11)
119909 (119905119891) = 119909 (119905
119896) = 119909119896
119911 (119905119891) = 119911 (119905
119896) = 119909119896
(18)
Once the impact occurs the momentum of the free mass istransmitted to the structure or vice versa 119888
1 1198882 1198883 and 119888
4of
(6) (7) (9) and (10) need to be repeatedly calculated aftereach impact in order to obtain the new responses Since thecharacteristics of the system responses are reset after eachimpact the time right after the impact 119905
119891 is also reset to 0
to calculate 1198881 1198882 1198883 and 119888
4 Substituting 119905
119891= 0 into (6) and
(7)
119911 (119905119891) = 119911 (0) = 119888
1+ 1198882
(119905119891) = (0) = minus
119887
1198721198882
(19)
The displacements before and after impact are assumed to bethe same and the velocity of the free mass after the impactis calculated from (17) Equation (19) is equal to 119909
119896and
119891
respectively
119911 (0) = 1198881+ 1198882= 119909119896
(0) = minus119887
1198721198882= 119891
(20)
From (20) 1198881and 1198882are calculated as
1198881= 119909119896minus119872
119887( minus
119891)
1198882=119872
119887( minus
119891)
(21)
Applying the same assumption and (16) to (9) and (10)
119909 (119905119891) = 119909 (0) = 119888
3= 119909119896
(119905119891) = (0) = minus120577120596
1198991198883+ 1205961198891198884= 119891
(22)
It then follows that 1198883and 1198884are obtained from (22) as
1198883= 119909119896
1198884=
1
120596119889
(119891+ 120577120596119899119909119896)
(23)
Substituting the calculated 1198881 1198882 1198883 and 119888
4into (6) and (9)
the responses of the two masses after impact are expressed as
119911 (119905) = [119909119896minus119872
119887( minus
119891)] + 119905 +
119872
119887( minus
119891) 119890minus(119887119872)119905
119909 (119905) = 119909119896119890minus120577120596119899119905 cos120596
119889119905 +
1
120596119889
(119891+ 120577120596119899119909119896) 119890minus120577120596119899119905 sin120596
119889119905
(24)
All the equations are developed based on the linearized dragforce Equation (24) holds until the next impact After thenext impact 119888
1 1198882 1198883 and 119888
4should be recalculated according
to new values of 119909119896 119896 119911119896 and
119896obtained from the impact
equations
Mathematical Problems in Engineering 5
4 Numerical Simulations
41 Linearized Drag Force For the numerical simulationsthere were two primary concerns how to identify the instantof an impact and how to define the impact condition Sincethe basis of this repeated impact model is the exchangeof momenta through impacts the time and condition ofthe impact need to be defined as precisely as possibleThe solution chosen to address this was to use a timestepping algorithm In the beginning 119888
1 1198882 1198883 and 119888
4are
calculated with the initial conditions By substituting 1198881 1198882
1198883 and 119888
4into (24) the responses of the free mass and
structure are obtained Based on the calculated response therelative displacement between the free mass and structure iscompared to the impact condition at every time step Theimpact condition is theoretically given by
|119909 (119905) minus 119911 (119905)| = 0 (25)
The actual numerical simulation however cannot achieveenough precision to capture the exact impact conditionTherefore a tolerance for the impact condition is requiredto allow for numerical round-off errors of the numericalsimulation
The tolerance should be considered carefully because toosmall tolerance has no meaning and a large tolerance makesimpacts occur more often than they should In the algorithmthe tolerance for the impact condition is set flexibly by thefollowing equation
Δ =1
10min (|120575119909| |120575119911|) (26)
where Δ is the impact condition such that |119909(119905) minus 119911(119905)| le Δ
and 120575119909 = (119905)Δ119905 and 120575119911 = (119905)Δ119905 Furthermore Δ119905 is thetime stepThe flexible impact condition hasmerit over a fixedone When the responses of the two masses are very smallthe fixed impact condition cannot capture some impactsTheflexible impact condition however is determined accordingto the size of the responses therefore the impact conditionis able to capture impacts whether the responses are slow orfast
After the comparison of the relative displacement andthe impact condition if the relative displacement is largerthan the impact condition the same values of 119888
1 1198882 1198883 and
1198884will be forwarded to the next time step Otherwise the
relative displacement is equal to or less than the impactcondition and thus an impact takes place Once the algorithmcatches the impact 119888
1 1198882 1198883 and 119888
4are calculated based on
the values from the previous time step and new conditionscorresponding to the impact as explained in the previoussection In addition the algorithm is performed on twodifferent time scales one is global time and the other is localtime The global time is the overall time starting with thebeginning of the algorithm and the local time counts timeduration between impacts thus after each impact the localtime is reset to 0 while the global time increases continuouslyWith 0 local time new responses of the system based on thenew 1198881 1198882 1198883 and 119888
4begin and evolve until next impact This
process keeps repeating until the system reaches the steadystate
The steady-state condition is also implemented in thealgorithm to avoid infinitesimal impacts which occur at theend of the responses The steady-state condition involvesdisplacements velocities and accelerations The algorithmverifies at every step whether or not both responses of the iceand structure satisfy the steady-state conditions The steady-state conditions are given as
(i) max(|119911(119905)| |119909(119905)|) lt 00001m(ii) max(|(119905)| |(119905)|) lt 00001ms(iii) max(|(119905)| |(119905)|) lt 00001ms2
The value of 00001 was chosen through trial and error for thesteady-state condition When the system reaches the steadystate the structure is constantly forced by the free mass andso the equation of motion in the steady state is expressed as
119898 + 119888 + 119896119909 = minus119887 ( minus ) (27)
Since the system is in the steady state
119896119909 = 119887 (28)
In the steady state condition the twomasses collide andmovetogether therefore the responses are given by
119909 (119905) = 119911 (119905) =119887
119896 (29)
which is a constant (steady state) displacement since isthe (assumed) constant velocity of the flow The algorithmreplaces (24) with (28) when the steady-state condition issatisfied
Figure 2 is plotted using the same structural propertiesin Table 1 which are obtained from [27] The two free massproperties mass 119872 and area 119860 are arbitrarily chosen as1600 kg which is the same as the structural mass and 1m2respectively The reason for this is as follows If the free massis much larger than the structural mass then its displacementand velocity would be hardly affected by the collisionwith thestructure and in fact repeated impacts would not occur Onthe other hand if the free mass is much smaller than that ofthe structure then the structure itself would hardly be affectedby the impacts Thus the most meaningful dynamics willoccur when the two masses are relatively the same order ofmagnitude For simplicity we begin our analysis by choosingthem to be the same and later in the paper we investigate theeffect of different choices of for the free mass
The linearization factor | minus | is set as 01ms becauseit is the maximum difference between the free mass velocityand flow velocity which is set as 01ms Therefore the valueof 119887 is
119887 =1
2120588119862119863119860 | minus | = 4995 kgs (30)
where 120588 is the density of water and119862119863 the drag coefficient is
1The coefficient of restitution another important parameteris assumed to be 1 which implies no energy dissipationthrough impact In addition the initial displacement and
6 Mathematical Problems in Engineering
0
minus02
minus04
minus06
minus08
minus10 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall Responses
Disp
lace
men
t (m
)
Time (s)
0
minus0002
minus0004
minus0006
186 188 190 192 194 196 198 200
StructureIce
(b) Responses from 185 to 200 seconds
Disp
lace
men
t (m
)
Time (s)
StructureIce
0
380 385 390 395 400
times10minus4
minus1
minus2
minus3
(c) Responses from 380 to 400 seconds
Figure 2 Responses of the ice and structure with a linearized drag force
Table 1 Parameters for simulations
Parameter Model structureℎ (mm) 41119863 (mm) 76119898 (kg) 1600119888 (kNsm) 02119896 (MNm) 1ℓ (m) 4493120590119888(MPa) 14
119909max (mm) 20
velocity of the free mass are minus001m and 001ms respec-tively and 0m and 0ms for the structure Figure 2 shows theresponse based on these parameters
Since the motivation for this model was to attempt toexplain IIV without analyzing specific ice-failure mecha-nisms we attempt to compare our results to some measuredIIV data There is very little IIV data in the open literatureespecially for displacement measurements Only laboratorymeasurements are available from [28 29] for displacementmeasurements Since the structural data for their tests arenot presented it is inadequate to compare their experimentaldata to Figure 2 directly but general vibration characteristicscan be compared Structural vibrations of IIVs becomequasi-static vibrations at low ice velocities and steady-statevibrations at medium ice velocities [30] although there isno quantitative definition of low and high ice velocities Aquasi-static vibration is defined by Karna [30] as a transientresponse which is followed by its maximum response atthe peak ice force and is not amplified by the dynamicsof the structure Figure 3(b) represents typical quasi-static
Mathematical Problems in Engineering 7
(a) [28]
1
0
2 4 6 8 10
Stru
ctur
aldi
spla
cem
ent(
cm)
Ice velocity = 76mms Time (s)
(b) [29]
2 4 6 8 10
1
0Stru
ctur
aldi
spla
cem
ent(
cm)
Time (s)Ice velocity = 395mms
(c) [29]
Figure 3 Measured structural responses from the literature
vibrations at ice velocity 76mms and Figure 3(c) shows thecharacteristic of a steady-state vibration at free mass velocity395mms Figure 3(a) however presents the displacementresponse although it is plotted at a lower free mass velocitythan Figure 3(b) This demonstrates that the vibration char-acteristics of IIVs are not determined by the velocity of theice alone but by involving all parameters
As shown in Figure 3(a) actual IIVs display multiplevibration characteristics from quasi-static to steady-statevibrations even at the same ice velocity contrary to thecharacteristic failure frequency model which predicts onetype of characteristic vibration at one ice velocity In thisrespect the proposed impact model possesses an advantageover the characteristic failure frequency model In an impactmodel the velocity of the ice varies from time to time whichis more realistic because part of the ice (free mass) contactingthe structure is deformed as the vibrations proceed Figure 2clearly shows multiple vibration characteristics Figure 2(b)displays quasi-static vibrations similar to Figure 3(b) From380 to 400 seconds the vibrations become stable and steadystate shown in Figure 2(c) which is close to the vibrationcharacteristic of Figure 3(c) The vibration characteristics ofthe impact model however cannot be clearly distinguishedfor the entire duration rather they show a mixture of thecharacteristics Therefore it is reasonable to conclude thatthe vibration characteristics of the impact IIV model aregoverned by the momenta of the ice and structure at the timeof impact
The weakness of modelling IIVs as repeated impacts asproposed in this paper model is how to model the relevantproperties of the ice The definitions of the ice mass and areain the impact IIVmodel imply that themass and area directlyaffect the impact or interactionwith the structure Among thetwo ice properties the influence of the ice area 119860 is directlyrelated to the velocity of the flowbecause of the relationwith 119887given by 119887 = (12)120588119862
119863119860|minus | It is inappropriate to simulate
the linearized model with large flow velocities since then thelinearized model would not be applicable
The mass of the free mass plays an important role in thismodel In order to understand its role it will be investigatedby conducting several numerical simulations Figures 4 5and 6 are plotted with the same parameters of Figure 2 exceptwith differentmasses which are proportional to the structuralmass 50 and 150 percent of 1600 kg so that the structuralmass can be used as a benchmark Including Figure 2 whichis plotted with the free mass of 100 percent of the structuralmass the responses with three different free masses arecompared At the beginning of the responses the heaviermass transfers higher momentum thus impacts happen lessfrequently but with largermagnitudes On the other hand thelighter mass allows the structure to approach the steady-statevibration in relatively short time In Figure 6 the magnitudesof the structural responses are not too different but theheavier mass induces more quasi-static vibrations for longertime Therefore it can be concluded that the free mass hasa greater effect on determining the vibration characteristicsthan on the magnitude of the structural response
Another important factor in this model is the coefficientof restitution The coefficient of restitution also requiresexperimental research to be defined quantitatively Theconducted numerical simulations can only show how thecoefficient of restitution influences the responses of thesystem Figure 2 is plotted with the coefficient of restitution119890 = 1 which is an unrealistic case Figures 7 8 and 9are simulated based on the same parameters of Figure 2 butwith different values of 119890 It becomes obvious that the factorthat most influences the magnitudes of the vibrations is thecoefficient of restitution rather than the velocities of the freemass The system with 119890 = 01 approaches the steady stateafter the first impact The system with 119890 = 05 not onlyapproaches the steady-state vibration faster but also showslower structural magnitudes than the system with 119890 = 1Figure 2 Figure 8 indicates that the system with 119890 = 05
loses the majority of its momentum after approximately54 seconds This suggests that vibrations due to repeatedimpacts can be controlled by manipulating the coefficient
8 Mathematical Problems in Engineering
002
0
minus002
minus004
minus006
minus008
minus01
minus0120 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with119872 = 800
0 50 100 150 200 250 300 350 400
02
0
minus02
minus04
minus06
minus08
minus1
minus12
minus14
Time (s)
Disp
lace
men
t (m
)
StructureIce
(b) Overall responses with119872= 2400
Figure 4 Responses with different ice masses 1
186 188 190 192 194 196 198 200
Disp
lace
men
t (m
)
StructureIce
Time (s)
0
minus2
minus4
minus6
times10minus4
(a) Responses with119872 = 800 from 185 to 200 seconds
0
minus0002
minus0004
minus0006
minus0008
186 188 190 192 194 196 198 200
StructureIce
Time (s)
Disp
lace
men
t (m
)
(b) Responses with119872= 2400 from 185 to 200 seconds
Figure 5 Responses with different ice masses 2
of restitution via the properties of the contacting area Forinstance the magnitudes of vibrations can be significantlyreduced by increasing the surface roughness of the structure
42 Quadratic Drag Force In the previous section themodelwas developed based on the linearized drag force whichassumes small relative velocity between the velocities of thefree mass and structure The linearized model howeveris definitely vulnerable to two factors namely when theinterval between impacts exceeds several seconds and whenthe velocity of the flow is no longer small Figure 10 clearlyshows the error between the models based on the linearized
drag and quadratic drag as time passes The displacement oflinearized equations (3) and (6) is compared with numer-ically solved equation (3) by using the built-in ordinarydifferential equation solver ofMathCadThe difference growsas time goes by but at high flow velocity the differencedecreases in Figure 10(b) This is because of the linearizationfactor | minus | which calibrates the difference between thelinear and quadratic models The solution however to thequadratic drag force needs to be developed for a more precisesimulation
A numerical solution is needed to solve (3) but insteadof using the built-in ordinary differential equation solver
Mathematical Problems in Engineering 9
StructureIce
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
Time (s)
1
0
minus1
minus2
minus3
minus4
times10minus4
(a) Responses with119872 = 800 from 385 to 400 seconds
StructureIce
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
0
minus2
minus4
minus6
times10minus4
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 6 Responses with different ice masses 3
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Overall responses with 119890 = 01
0
minus005
minus01
minus015
minus02
minus025
minus03
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400Time (s)
StructureIce
(b) Overall responses with 119890 = 05
Figure 7 Responses with different coefficients of restitution 1
of MathCad a simple numerical solution is developed for afaster algorithm From (3)
119872 = minus1
2120588119862119863119860 ( minus ) | minus | = minus119861 ( minus ) | minus | (31)
or
=119889
119889119905= minus
119861
119872( minus ) | minus | (32)
where
119861 =1
2120588119862119863119860 (33)
Using 119894 as a counter for the a number of time steps thedisplacement of (34) can be expressed as
119911119894= 119911119894minus1
+119889119911119894minus1
119889119905Δ119905 = 119911
119894minus1+ 119894minus1Δ119905 (34)
in which Δ119905 is previously defined as a time step z is given by
119894= 119894minus1
+119889119894minus1
119889119905Δ119905 =
119894minus1+ 119894minus1Δ119905 (35)
By substituting (32) into (35)
119894= 119894minus1
+ [minus119861
119872(119894minus1
minus )1003816100381610038161003816119894minus1 minus
1003816100381610038161003816] Δ119905(36)
10 Mathematical Problems in Engineering
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 2 4 6 8 10 12 14
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Responses with 119890 = 01 from 0 to 15 seconds
Disp
lace
men
t (m
)
Time (s)
StructureIce
46 48 50 52 54 56 58 60
2
0
minus2
minus4
minus6
minus8
times10minus4
(b) Responses with 119890 = 05 from 45 to 60 seconds
Figure 8 Responses with different coefficients of restitution 2
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus001386 388 390 392 394 396 398 400
Time (s)
Disp
lace
men
t (m
)
(a) Responses with 119890 = 01 from 385 to 400 seconds
StructureIce
Time (s)
Disp
lace
men
t (m
)
398 3985 399 3995 400
times10minus6
6
5
4
3
(b) Responses with 119890 = 05 from 398 to 400 seconds
Figure 9 Responses with different coefficients of restitution 3
Given the initial displacement and velocity of the free massthe displacement can be numerically solved by using (34)and (36) as simultaneous equationsThe simulation results of(34) and (36) are also plotted in Figure 10 At the high flowvelocity Figure 10(b) (34) and (36) predict the displacementclose enough to the one evaluated with the built-in ordinarydifferential equation solver In the algorithm of the numericalsimulation (34) and (36) replace (6) and (7)
Using the same parameters as those used in Figure 2but with a different flow velocity the responses of themodel with the quadratic drag force are compared withthe one with the linearized drag force in Figure 11 At high
flow velocities over 1ms the quadratic drag force modelapproaches steady-state vibration more quickly than thelinearized drag force model because at high flow velocity thequadratic drag induces more damping while it transfers moreforces than the linearized drag In Figure 12 the quadraticdrag force model exhibits the characteristic of the steady-state vibration while the linearized drag force model stilldisplays quasi-static vibration As velocities increase whichcorresponds to flowvelocities increasing vibrations approachthe steady state The linearized drag force model howeverdoes not show a steady-state vibration until 400 secondsin Figure 12 The linearization factor allows the linearized
Mathematical Problems in Engineering 11
0 100 200 300 400 500 600
100
80
60
40
20
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(a) Flow velocity 01ms
0 100 200 300 400 500 600
1000
800
600
400
200
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(b) Flow velocity 1ms
Figure 10 Comparison of the linearized and quadratic Drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
Disp
lace
men
t (m
)
(a) Overall responses with the quadratic drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
0
minus02
minus04
minus06
minus08
minus1
Disp
lace
men
t (m
)
(b) Overall responses with the linear drag
Figure 11 Responses with the quadratic and linear drags 1
drag force model to follow the quadratic drag force modelin the beginning but when the system is in a series of lowvelocity impacts the linearization factor causes a differenceThe linearized drag force model therefore is inappropriateto simulate the repeated impact model at high flow veloci-ties
Since the free mass area 119860 plays a similar role as theflow velocity in the model the model with wide area is alsosimulated with the quadratic drag force The influence of thedrag force on the response of the free mass increases as thesize of the area increases because the area is related to the dragforce by (1) 119891
119863= minus(12)120588119862
119863119860( minus )
2 Figure 13 is plottedwith the area 119860 = 2m2 the flow velocity = 01ms and
the rest of parameters are the same as the parameters usedin Figure 11 The flow velocity of Figure 13 is identical to thatof Figure 2 but the responses quickly approach the steady-state vibrations as if they are driven by high flow velocityFigure 13(b) also shows a similar vibration characteristic toFigure 12(a) because both flow velocity and area influence thedrag force in the impact model Therefore the area can beanother controlling factor of the vibrations Since the areais related to the size of the contacting area of the structurereducing the contacting area has a similar effect to that ofreducing the flow velocity
In the previous section the roles of the free mass andthe coefficient of restitution have been defined by changing
12 Mathematical Problems in EngineeringD
ispla
cem
ent (
m)
395 396 397 398 399 400Time (s)
StructureIce
8
6
4
2
0
times10minus4
(a) Responses with the quadratic drag from 395 to 400 secondsD
ispla
cem
ent (
m)
386 388 390 392 394 396 398 400Time (s)
StructureIce
0
minus0005
(b) Responses with the linear drag from 385 to 400 seconds
Figure 12 Responses with the quadratic and linear drags 2
005
0
minus005
minus01
minus015
minus02
minus025
minus03
minus035
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
(a) Overall responses with 119860 = 2
Disp
lace
men
t (m
)
395 396 397 398 399 400Time (s)
StructureIce
1
0
minus1
times10minus5
(b) Responses with 119860 = 2 from 395 to 400 seconds
Figure 13 Responses with 119860 = 2m2
relevant parameters To verify if those definitions still holdwith the quadratic drag force the impact model with thequadratic drag is plotted with change in free mass and thecoefficient of restitution Figures 14 15 and 16 are plottedon the same parameters used in Figures 2 4(a) and 4(b)respectively except that they are based on the quadratic dragforce The responses of Figures 14 15 and 16 prove thatthe defined role of the free mass holds with the quadraticdrag force The responses of the system with heavier freemass Figure 16(b) is still unstable while the system with thefree mass 119872 = 800 kg Figure 15(b) is in the steady-statevibration The quadratic drag force however induces more
drag force thus the system responses tend to be stabilizedfaster than those with the linearized drag forces
Figures 17 and 18 are plotted on the same conditions withFigures 7(a) and 7(b) respectively but with the quadraticdrag forceThe coefficient of restitution also governs themag-nitudes of responses in the impact model with the quadraticdrag force however the system with the quadratic drag forceis more sensitive to change in the coefficient of restitutionthan the one with the linearized drag force The system with119890 = 05 Figure 18(b) loses the momentum at approximately43 seconds under the influence of the quadratic drag forcewhile the system with the linearized drag force and 119890 = 05
Mathematical Problems in Engineering 13
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
01
0
minus01
minus02
minus03
minus04
minus05
minus06
minus07
(a) Overall responses with119872 = 1600
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
StructureIce
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 1600 from 385 to 400 seconds
Figure 14 Responses with119872 = 1600
0
minus001
minus002
minus003
minus004
minus005
minus006
minus0070 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall responses with119872 = 800
395 396 397 398 399 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
0
5
minus5
times10minus6
(b) Responses with119872 = 800 from 395 to 400 seconds
Figure 15 Responses with119872 = 800
Figure 8(b) loses the majority of its momentum at approxi-mately 54 seconds This is again because the quadratic dragforce dissipates energy more quickly than the linearized dragforceHowever it is demonstrated in Figures 17 and 18 that thecoefficient of restitution plays the same role in both quadraticand linearized drag forces
5 Resonance Conditions
One of the main objectives of the analysis aims to find aresonance condition (which is often referred to as lock-in
when investigating IIVs) The structural vibration frequencyincreases as the free mass velocity increases but when lock-in occurs the structural frequency remains at the naturalfrequency even with increasing free mass velocity [6] In theimpact model however the velocities of the free mass cannotbe the primary lock-in condition because the velocities of thefree mass are not fixed
One possible interpretation of modelling IIVs with theimpact model is that it considers the forces to be a seriesof discrete impulses as shown in Figure 19(a) while otherIIV models consider them as continuous periodic forces
14 Mathematical Problems in Engineering
0 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
(a) Overall responses with119872= 2400
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 16 Responses with119872 = 2400
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
(a) Overall responses with 119890 = 01
StructureIce
0 2 4 6 8 10 12 14
0002
0
minus0002
minus0004
minus0006
minus0008
Disp
lace
men
t (m
)
Time (s)
(b) Responses with 119890 = 01 from 0 to 15 seconds
Figure 17 Responses with 119890 = 01
In fact the impact IIV model proposes a way to calculate thestrength of each impulse and the times between impulses andthese depend on the momenta of the ice and structure aswell as the coefficient of restitution for the collision Undercertain conditions it is possible for the discrete impulsesof Figure 19(a) to approach the regular periodic series ofimpulses of Figure 19(b) which is suspected to cause the lock-in condition
This view of the impact IIV model can be analyzed byusing a Fourier transform [31] There are several definitions
of the Fourier transform (FT) that may be used and in thispaper the chosen version of the FT is given by
119865 (120596) = int
infin
minusinfin
119891 (119905) 119890minus119894120596119905
119889119905 (37)
The corresponding inverse Fourier transform is then given by
119891 (119905) =1
2120587int
infin
minusinfin
119865 (120596) 119890119894120596119905119889120596 (38)
Mathematical Problems in Engineering 15
005
0
minus005
minus01
minus015
minus02
minus0250 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with 119890 = 05
Time (s)
Disp
lace
men
t (m
)
StructureIce
36 38 40 42 44 46 48 50
4
2
0
minus2
minus4
times10minus4
(b) Responses with 119890 = 05 from 35 to 50 seconds
Figure 18 Responses with 119890 = 05
f(t)
fj
tj t
(a) Random impulses
f(t)
f
T 2T 3Tt
pT
(b) Uniform impulses
Figure 19 Ice forces as a series of impulses
The discrete ice forces of Figure 19(a) can be expressed asa series of impulses as
119891 (119905) =
infin
sum
119895=1
119891119895120575119863(119905 minus 119905119895) (39)
where 120575119863indicates the Dirac delta function and 119891
119895indicates
the strength of each impact If the sizes and intervals ofthe impulses are uniform as shown in Figure 19(b) (39) isrewritten as
119891 (119905) =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (40)
In reality the repeated impacts will not all have the samestrength or occur at a fixed interval however it is sufficient forthe situation of Figure 19(a) to approach that of Figure 19(b)for this analysis to be relevant
Equation (40) is often called a comb functionTheFouriertransform of the comb function is given [32] by
119865 (120596) = 2120587
infin
sum
119873=1
119891120575119863(120596 minus
2120587119873
119879) (41)
For a typical single-degree of freedom forced vibrationsproblem
+ 2120577120596119899 + 1205962
119899119909 =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (42)
Taking Fourier transforms of both sides and rearranging give
119883 (120596) = 2120587suminfin
119873=1119891120575119863 (120596 minus 2120587119873119879)
(minus1205962 + 2120577120596119899120596119894 + 1205962
119899)
(43)
Taking the inverse Fourier transforms to find the structureresponse in time gives
119909 (119905) = 119891
infin
sum
119873=1
119890119894(2120587119873119879)119905
(minus4120587211987321198792 + 119894120577120596119899 (4120587119873119879) + 1205962
119899) (44)
Equation (44) experiences a resonance when the denom-inator of any of the terms in the sum goes to zero that is when120596119899= 2120587119873119879 where 119873 is an integer larger than 0 This says
that the structure will experience a resonance when the natu-ral frequency is an integral multiple of the forcing frequency
16 Mathematical Problems in Engineering
as would be expected In the previous numerical simulationsthis condition has been observed in several cases Figures12(a) and 13(b) For instance the structural frequency ofFigure 13(b) is 2120587119879 = 212058705 = 1257 rads The naturalfrequency of the structure is radic119896119898 = radic10000001600 =
25 rads With119873 = 2 the structural frequency is close to thenatural frequency The result is what is seen in Figure 13(b)The two masses vibrate at the same frequency of 2120587119879 whichis half of the natural frequency The structural frequencyremains at that level which is lock-in and so does thefrequency of the free mass response
Thus the impact model successfully predicts the exis-tence of a lock-in condition and also predicts that it can occurwhen the frequency of the structure and freemass impacts aresynchronized together at a multiple of the natural frequencyof the structure The question of the precise conditionsnecessary for the frequency of the impacts to occur at amultiple of the natural frequency still remains open
6 Summary and Conclusions
This paper analyzes the vibrations caused by repeated impactsof a flow-driven freemass onto a second structure Conserva-tion of linearmomentum and the coefficient of restitution areused to characterize the nature of the impacts between thetwo masses This model is referred to as the impact modelTheproperties of the impactmodel were investigated throughnumerical simulations The mass of the free mass is one ofthe factors determining the vibration characteristics sinceit is related to the momentum of each impact This modelis proposed to explain the nature of ice-induced vibrationswithout requiring a microscopic model of the mechanism ofice failure The magnitudes of the vibrations are governedin part by the coefficient of restitution which when usedto explain IIVs encapsulates all ice properties mentionedin other IIV models In addition to the ice properties thecoefficient of restitution also governs the interaction betweenthe structure and the free mass The free mass is used torepresent the ice when modelling IIVs The last propertyof the impact model is that of the free mass area whichinfluences the dissipation of the system momenta throughthe drag force A free mass with a small area shows the samebehavior of a mass driven by lower flow velocity
Therefore through the impact model two factors whichcontrol the characteristics of IIVs have been proposed theice area and the coefficient of restitution By changing thesize of the contact area between the ice and structure thedriving force of the ice can be reduced as if the ice weredriven by lower flow velocities Furthermore by changingfactors related to the coefficient of restitution such as thesurface roughness or shape of the structure IIVs can bereduced to avoid the lock-in conditionwhich occurswhen thefrequencies of the structure and ice impacts are synchronizedat a multiple of the natural frequency of the structure Inaddition to the inclusion of the effect of flow the newapproach to ice forces through the impact model permitsthe answering of questions which originate from the weaktheoretical points of existing IIV models
Particular strengths of the model are that it qualitativelymatches observed results and in particular that a mechanismof ice failure is not required in order to perform simulationsand predictions It also proposes a simple mechanism toexplain the potentially destructive resonance lock-in condi-tions
References
[1] J P Dempsey ldquoResearch trends in ice mechanicsrdquo InternationalJournal of Solids and Structures vol 37 no 1-2 pp 131ndash153 2000
[2] K A Blenkarn ldquoMeasurement and analysis of ice forces oncook inlet structuresrdquo in Proceedings of the Offshore TechnologyConference (OTC rsquo70) pp 365ndash378 1970
[3] Y Toyama T Sensu M Minami and N Yashima ldquoModel testson ice-induced self-excited vibration of cylindrical structuresrdquoin Proceedings of the 7th International Conference on Port andOcean Engineering under Arctic Conditions vol 2 pp 834ndash8441983
[4] M Ranta and R Raty ldquoOn the analytic solution of ice-inducedvibrations in a marine pile structurerdquo in Proceedings of the 7thInternational Conference on Port and Ocean Engineering underArctic Conditions vol 2 pp 901ndash908 1983
[5] A Barker G Timco H Gravesen and P Voslashlund ldquoIce loadingon Danish wind turbinesmdashpart 1 dynamic model testsrdquo ColdRegions Science and Technology vol 41 no 1 pp 1ndash23 2005
[6] D S Sodhi ldquoIce-induced vibrations of structuresrdquo SpecialReport 89-5 International Association for Hydraulic ResearchWorking Group on Ice Forces 1989
[7] D S Sodhi and C E Morris ldquoCharacteristic frequency of forcevariations in continuous crushing of sheet ice against rigidcylindrical structuresrdquo Cold Regions Science and Technologyvol 12 no 1 pp 1ndash12 1986
[8] D S Sodhi C E Morris and G F N Cox ldquoDynamic analysisof failure modes on ice sheets encountering sloping structuresrdquoin Proceedings of the 6th International Conference on OffshoreMechanics and Arctic Engineering vol 4 pp 281ndash284 1987
[9] H R Peyton ldquoSea ice forcesrdquo in Proceedings of the Conferenceon Ice Pressure against Structures 1966
[10] C R Neill ldquoDynamic ice forces on piers and piles an assess-ment of design guidelines in the light of recent researchrdquoCanadian Journal of Civil Engineering vol 3 no 2 pp 305ndash3411976
[11] G Matlock W Dawkins and J Panak ldquoAnalytical model forice-structure interactionrdquo Journal of the Engineering MechanicsDivision pp 1083ndash11092 1971
[12] M M Maattanen ldquoIce-induced vibrations of structures self-excitationrdquo Special Report 89-5 International Association forHydraulic Research Working Group on Ice Forces 1989
[13] M M Maattanen ldquoNumerical simulation of ice-induced vibra-tions in offshore structuresrdquo in Proceedings of the 14th NordicSeminar on Computational Mechanics 2001 Keynote lecture
[14] J Xu and L Wang ldquoIce force oscillator model and its numericalsolutionsrdquo in Proceedings of the 7th International Conference onOffshore Machanics and Arctic Engineering pp 171ndash176 1988
[15] G TsinkerMarine Structures Engineering Chapman-Hall NewYork NY USA 1st edition 1995
[16] G Huang and P Liu ldquoA dynamic model for ice-inducedvibration of structuresrdquo Journal of OffshoreMechanics andArcticEngineering vol 131 no 1 Article ID 011501 6 pages 2009
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
Mathematical Problems in Engineering 5
4 Numerical Simulations
41 Linearized Drag Force For the numerical simulationsthere were two primary concerns how to identify the instantof an impact and how to define the impact condition Sincethe basis of this repeated impact model is the exchangeof momenta through impacts the time and condition ofthe impact need to be defined as precisely as possibleThe solution chosen to address this was to use a timestepping algorithm In the beginning 119888
1 1198882 1198883 and 119888
4are
calculated with the initial conditions By substituting 1198881 1198882
1198883 and 119888
4into (24) the responses of the free mass and
structure are obtained Based on the calculated response therelative displacement between the free mass and structure iscompared to the impact condition at every time step Theimpact condition is theoretically given by
|119909 (119905) minus 119911 (119905)| = 0 (25)
The actual numerical simulation however cannot achieveenough precision to capture the exact impact conditionTherefore a tolerance for the impact condition is requiredto allow for numerical round-off errors of the numericalsimulation
The tolerance should be considered carefully because toosmall tolerance has no meaning and a large tolerance makesimpacts occur more often than they should In the algorithmthe tolerance for the impact condition is set flexibly by thefollowing equation
Δ =1
10min (|120575119909| |120575119911|) (26)
where Δ is the impact condition such that |119909(119905) minus 119911(119905)| le Δ
and 120575119909 = (119905)Δ119905 and 120575119911 = (119905)Δ119905 Furthermore Δ119905 is thetime stepThe flexible impact condition hasmerit over a fixedone When the responses of the two masses are very smallthe fixed impact condition cannot capture some impactsTheflexible impact condition however is determined accordingto the size of the responses therefore the impact conditionis able to capture impacts whether the responses are slow orfast
After the comparison of the relative displacement andthe impact condition if the relative displacement is largerthan the impact condition the same values of 119888
1 1198882 1198883 and
1198884will be forwarded to the next time step Otherwise the
relative displacement is equal to or less than the impactcondition and thus an impact takes place Once the algorithmcatches the impact 119888
1 1198882 1198883 and 119888
4are calculated based on
the values from the previous time step and new conditionscorresponding to the impact as explained in the previoussection In addition the algorithm is performed on twodifferent time scales one is global time and the other is localtime The global time is the overall time starting with thebeginning of the algorithm and the local time counts timeduration between impacts thus after each impact the localtime is reset to 0 while the global time increases continuouslyWith 0 local time new responses of the system based on thenew 1198881 1198882 1198883 and 119888
4begin and evolve until next impact This
process keeps repeating until the system reaches the steadystate
The steady-state condition is also implemented in thealgorithm to avoid infinitesimal impacts which occur at theend of the responses The steady-state condition involvesdisplacements velocities and accelerations The algorithmverifies at every step whether or not both responses of the iceand structure satisfy the steady-state conditions The steady-state conditions are given as
(i) max(|119911(119905)| |119909(119905)|) lt 00001m(ii) max(|(119905)| |(119905)|) lt 00001ms(iii) max(|(119905)| |(119905)|) lt 00001ms2
The value of 00001 was chosen through trial and error for thesteady-state condition When the system reaches the steadystate the structure is constantly forced by the free mass andso the equation of motion in the steady state is expressed as
119898 + 119888 + 119896119909 = minus119887 ( minus ) (27)
Since the system is in the steady state
119896119909 = 119887 (28)
In the steady state condition the twomasses collide andmovetogether therefore the responses are given by
119909 (119905) = 119911 (119905) =119887
119896 (29)
which is a constant (steady state) displacement since isthe (assumed) constant velocity of the flow The algorithmreplaces (24) with (28) when the steady-state condition issatisfied
Figure 2 is plotted using the same structural propertiesin Table 1 which are obtained from [27] The two free massproperties mass 119872 and area 119860 are arbitrarily chosen as1600 kg which is the same as the structural mass and 1m2respectively The reason for this is as follows If the free massis much larger than the structural mass then its displacementand velocity would be hardly affected by the collisionwith thestructure and in fact repeated impacts would not occur Onthe other hand if the free mass is much smaller than that ofthe structure then the structure itself would hardly be affectedby the impacts Thus the most meaningful dynamics willoccur when the two masses are relatively the same order ofmagnitude For simplicity we begin our analysis by choosingthem to be the same and later in the paper we investigate theeffect of different choices of for the free mass
The linearization factor | minus | is set as 01ms becauseit is the maximum difference between the free mass velocityand flow velocity which is set as 01ms Therefore the valueof 119887 is
119887 =1
2120588119862119863119860 | minus | = 4995 kgs (30)
where 120588 is the density of water and119862119863 the drag coefficient is
1The coefficient of restitution another important parameteris assumed to be 1 which implies no energy dissipationthrough impact In addition the initial displacement and
6 Mathematical Problems in Engineering
0
minus02
minus04
minus06
minus08
minus10 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall Responses
Disp
lace
men
t (m
)
Time (s)
0
minus0002
minus0004
minus0006
186 188 190 192 194 196 198 200
StructureIce
(b) Responses from 185 to 200 seconds
Disp
lace
men
t (m
)
Time (s)
StructureIce
0
380 385 390 395 400
times10minus4
minus1
minus2
minus3
(c) Responses from 380 to 400 seconds
Figure 2 Responses of the ice and structure with a linearized drag force
Table 1 Parameters for simulations
Parameter Model structureℎ (mm) 41119863 (mm) 76119898 (kg) 1600119888 (kNsm) 02119896 (MNm) 1ℓ (m) 4493120590119888(MPa) 14
119909max (mm) 20
velocity of the free mass are minus001m and 001ms respec-tively and 0m and 0ms for the structure Figure 2 shows theresponse based on these parameters
Since the motivation for this model was to attempt toexplain IIV without analyzing specific ice-failure mecha-nisms we attempt to compare our results to some measuredIIV data There is very little IIV data in the open literatureespecially for displacement measurements Only laboratorymeasurements are available from [28 29] for displacementmeasurements Since the structural data for their tests arenot presented it is inadequate to compare their experimentaldata to Figure 2 directly but general vibration characteristicscan be compared Structural vibrations of IIVs becomequasi-static vibrations at low ice velocities and steady-statevibrations at medium ice velocities [30] although there isno quantitative definition of low and high ice velocities Aquasi-static vibration is defined by Karna [30] as a transientresponse which is followed by its maximum response atthe peak ice force and is not amplified by the dynamicsof the structure Figure 3(b) represents typical quasi-static
Mathematical Problems in Engineering 7
(a) [28]
1
0
2 4 6 8 10
Stru
ctur
aldi
spla
cem
ent(
cm)
Ice velocity = 76mms Time (s)
(b) [29]
2 4 6 8 10
1
0Stru
ctur
aldi
spla
cem
ent(
cm)
Time (s)Ice velocity = 395mms
(c) [29]
Figure 3 Measured structural responses from the literature
vibrations at ice velocity 76mms and Figure 3(c) shows thecharacteristic of a steady-state vibration at free mass velocity395mms Figure 3(a) however presents the displacementresponse although it is plotted at a lower free mass velocitythan Figure 3(b) This demonstrates that the vibration char-acteristics of IIVs are not determined by the velocity of theice alone but by involving all parameters
As shown in Figure 3(a) actual IIVs display multiplevibration characteristics from quasi-static to steady-statevibrations even at the same ice velocity contrary to thecharacteristic failure frequency model which predicts onetype of characteristic vibration at one ice velocity In thisrespect the proposed impact model possesses an advantageover the characteristic failure frequency model In an impactmodel the velocity of the ice varies from time to time whichis more realistic because part of the ice (free mass) contactingthe structure is deformed as the vibrations proceed Figure 2clearly shows multiple vibration characteristics Figure 2(b)displays quasi-static vibrations similar to Figure 3(b) From380 to 400 seconds the vibrations become stable and steadystate shown in Figure 2(c) which is close to the vibrationcharacteristic of Figure 3(c) The vibration characteristics ofthe impact model however cannot be clearly distinguishedfor the entire duration rather they show a mixture of thecharacteristics Therefore it is reasonable to conclude thatthe vibration characteristics of the impact IIV model aregoverned by the momenta of the ice and structure at the timeof impact
The weakness of modelling IIVs as repeated impacts asproposed in this paper model is how to model the relevantproperties of the ice The definitions of the ice mass and areain the impact IIVmodel imply that themass and area directlyaffect the impact or interactionwith the structure Among thetwo ice properties the influence of the ice area 119860 is directlyrelated to the velocity of the flowbecause of the relationwith 119887given by 119887 = (12)120588119862
119863119860|minus | It is inappropriate to simulate
the linearized model with large flow velocities since then thelinearized model would not be applicable
The mass of the free mass plays an important role in thismodel In order to understand its role it will be investigatedby conducting several numerical simulations Figures 4 5and 6 are plotted with the same parameters of Figure 2 exceptwith differentmasses which are proportional to the structuralmass 50 and 150 percent of 1600 kg so that the structuralmass can be used as a benchmark Including Figure 2 whichis plotted with the free mass of 100 percent of the structuralmass the responses with three different free masses arecompared At the beginning of the responses the heaviermass transfers higher momentum thus impacts happen lessfrequently but with largermagnitudes On the other hand thelighter mass allows the structure to approach the steady-statevibration in relatively short time In Figure 6 the magnitudesof the structural responses are not too different but theheavier mass induces more quasi-static vibrations for longertime Therefore it can be concluded that the free mass hasa greater effect on determining the vibration characteristicsthan on the magnitude of the structural response
Another important factor in this model is the coefficientof restitution The coefficient of restitution also requiresexperimental research to be defined quantitatively Theconducted numerical simulations can only show how thecoefficient of restitution influences the responses of thesystem Figure 2 is plotted with the coefficient of restitution119890 = 1 which is an unrealistic case Figures 7 8 and 9are simulated based on the same parameters of Figure 2 butwith different values of 119890 It becomes obvious that the factorthat most influences the magnitudes of the vibrations is thecoefficient of restitution rather than the velocities of the freemass The system with 119890 = 01 approaches the steady stateafter the first impact The system with 119890 = 05 not onlyapproaches the steady-state vibration faster but also showslower structural magnitudes than the system with 119890 = 1Figure 2 Figure 8 indicates that the system with 119890 = 05
loses the majority of its momentum after approximately54 seconds This suggests that vibrations due to repeatedimpacts can be controlled by manipulating the coefficient
8 Mathematical Problems in Engineering
002
0
minus002
minus004
minus006
minus008
minus01
minus0120 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with119872 = 800
0 50 100 150 200 250 300 350 400
02
0
minus02
minus04
minus06
minus08
minus1
minus12
minus14
Time (s)
Disp
lace
men
t (m
)
StructureIce
(b) Overall responses with119872= 2400
Figure 4 Responses with different ice masses 1
186 188 190 192 194 196 198 200
Disp
lace
men
t (m
)
StructureIce
Time (s)
0
minus2
minus4
minus6
times10minus4
(a) Responses with119872 = 800 from 185 to 200 seconds
0
minus0002
minus0004
minus0006
minus0008
186 188 190 192 194 196 198 200
StructureIce
Time (s)
Disp
lace
men
t (m
)
(b) Responses with119872= 2400 from 185 to 200 seconds
Figure 5 Responses with different ice masses 2
of restitution via the properties of the contacting area Forinstance the magnitudes of vibrations can be significantlyreduced by increasing the surface roughness of the structure
42 Quadratic Drag Force In the previous section themodelwas developed based on the linearized drag force whichassumes small relative velocity between the velocities of thefree mass and structure The linearized model howeveris definitely vulnerable to two factors namely when theinterval between impacts exceeds several seconds and whenthe velocity of the flow is no longer small Figure 10 clearlyshows the error between the models based on the linearized
drag and quadratic drag as time passes The displacement oflinearized equations (3) and (6) is compared with numer-ically solved equation (3) by using the built-in ordinarydifferential equation solver ofMathCadThe difference growsas time goes by but at high flow velocity the differencedecreases in Figure 10(b) This is because of the linearizationfactor | minus | which calibrates the difference between thelinear and quadratic models The solution however to thequadratic drag force needs to be developed for a more precisesimulation
A numerical solution is needed to solve (3) but insteadof using the built-in ordinary differential equation solver
Mathematical Problems in Engineering 9
StructureIce
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
Time (s)
1
0
minus1
minus2
minus3
minus4
times10minus4
(a) Responses with119872 = 800 from 385 to 400 seconds
StructureIce
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
0
minus2
minus4
minus6
times10minus4
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 6 Responses with different ice masses 3
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Overall responses with 119890 = 01
0
minus005
minus01
minus015
minus02
minus025
minus03
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400Time (s)
StructureIce
(b) Overall responses with 119890 = 05
Figure 7 Responses with different coefficients of restitution 1
of MathCad a simple numerical solution is developed for afaster algorithm From (3)
119872 = minus1
2120588119862119863119860 ( minus ) | minus | = minus119861 ( minus ) | minus | (31)
or
=119889
119889119905= minus
119861
119872( minus ) | minus | (32)
where
119861 =1
2120588119862119863119860 (33)
Using 119894 as a counter for the a number of time steps thedisplacement of (34) can be expressed as
119911119894= 119911119894minus1
+119889119911119894minus1
119889119905Δ119905 = 119911
119894minus1+ 119894minus1Δ119905 (34)
in which Δ119905 is previously defined as a time step z is given by
119894= 119894minus1
+119889119894minus1
119889119905Δ119905 =
119894minus1+ 119894minus1Δ119905 (35)
By substituting (32) into (35)
119894= 119894minus1
+ [minus119861
119872(119894minus1
minus )1003816100381610038161003816119894minus1 minus
1003816100381610038161003816] Δ119905(36)
10 Mathematical Problems in Engineering
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 2 4 6 8 10 12 14
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Responses with 119890 = 01 from 0 to 15 seconds
Disp
lace
men
t (m
)
Time (s)
StructureIce
46 48 50 52 54 56 58 60
2
0
minus2
minus4
minus6
minus8
times10minus4
(b) Responses with 119890 = 05 from 45 to 60 seconds
Figure 8 Responses with different coefficients of restitution 2
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus001386 388 390 392 394 396 398 400
Time (s)
Disp
lace
men
t (m
)
(a) Responses with 119890 = 01 from 385 to 400 seconds
StructureIce
Time (s)
Disp
lace
men
t (m
)
398 3985 399 3995 400
times10minus6
6
5
4
3
(b) Responses with 119890 = 05 from 398 to 400 seconds
Figure 9 Responses with different coefficients of restitution 3
Given the initial displacement and velocity of the free massthe displacement can be numerically solved by using (34)and (36) as simultaneous equationsThe simulation results of(34) and (36) are also plotted in Figure 10 At the high flowvelocity Figure 10(b) (34) and (36) predict the displacementclose enough to the one evaluated with the built-in ordinarydifferential equation solver In the algorithm of the numericalsimulation (34) and (36) replace (6) and (7)
Using the same parameters as those used in Figure 2but with a different flow velocity the responses of themodel with the quadratic drag force are compared withthe one with the linearized drag force in Figure 11 At high
flow velocities over 1ms the quadratic drag force modelapproaches steady-state vibration more quickly than thelinearized drag force model because at high flow velocity thequadratic drag induces more damping while it transfers moreforces than the linearized drag In Figure 12 the quadraticdrag force model exhibits the characteristic of the steady-state vibration while the linearized drag force model stilldisplays quasi-static vibration As velocities increase whichcorresponds to flowvelocities increasing vibrations approachthe steady state The linearized drag force model howeverdoes not show a steady-state vibration until 400 secondsin Figure 12 The linearization factor allows the linearized
Mathematical Problems in Engineering 11
0 100 200 300 400 500 600
100
80
60
40
20
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(a) Flow velocity 01ms
0 100 200 300 400 500 600
1000
800
600
400
200
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(b) Flow velocity 1ms
Figure 10 Comparison of the linearized and quadratic Drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
Disp
lace
men
t (m
)
(a) Overall responses with the quadratic drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
0
minus02
minus04
minus06
minus08
minus1
Disp
lace
men
t (m
)
(b) Overall responses with the linear drag
Figure 11 Responses with the quadratic and linear drags 1
drag force model to follow the quadratic drag force modelin the beginning but when the system is in a series of lowvelocity impacts the linearization factor causes a differenceThe linearized drag force model therefore is inappropriateto simulate the repeated impact model at high flow veloci-ties
Since the free mass area 119860 plays a similar role as theflow velocity in the model the model with wide area is alsosimulated with the quadratic drag force The influence of thedrag force on the response of the free mass increases as thesize of the area increases because the area is related to the dragforce by (1) 119891
119863= minus(12)120588119862
119863119860( minus )
2 Figure 13 is plottedwith the area 119860 = 2m2 the flow velocity = 01ms and
the rest of parameters are the same as the parameters usedin Figure 11 The flow velocity of Figure 13 is identical to thatof Figure 2 but the responses quickly approach the steady-state vibrations as if they are driven by high flow velocityFigure 13(b) also shows a similar vibration characteristic toFigure 12(a) because both flow velocity and area influence thedrag force in the impact model Therefore the area can beanother controlling factor of the vibrations Since the areais related to the size of the contacting area of the structurereducing the contacting area has a similar effect to that ofreducing the flow velocity
In the previous section the roles of the free mass andthe coefficient of restitution have been defined by changing
12 Mathematical Problems in EngineeringD
ispla
cem
ent (
m)
395 396 397 398 399 400Time (s)
StructureIce
8
6
4
2
0
times10minus4
(a) Responses with the quadratic drag from 395 to 400 secondsD
ispla
cem
ent (
m)
386 388 390 392 394 396 398 400Time (s)
StructureIce
0
minus0005
(b) Responses with the linear drag from 385 to 400 seconds
Figure 12 Responses with the quadratic and linear drags 2
005
0
minus005
minus01
minus015
minus02
minus025
minus03
minus035
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
(a) Overall responses with 119860 = 2
Disp
lace
men
t (m
)
395 396 397 398 399 400Time (s)
StructureIce
1
0
minus1
times10minus5
(b) Responses with 119860 = 2 from 395 to 400 seconds
Figure 13 Responses with 119860 = 2m2
relevant parameters To verify if those definitions still holdwith the quadratic drag force the impact model with thequadratic drag is plotted with change in free mass and thecoefficient of restitution Figures 14 15 and 16 are plottedon the same parameters used in Figures 2 4(a) and 4(b)respectively except that they are based on the quadratic dragforce The responses of Figures 14 15 and 16 prove thatthe defined role of the free mass holds with the quadraticdrag force The responses of the system with heavier freemass Figure 16(b) is still unstable while the system with thefree mass 119872 = 800 kg Figure 15(b) is in the steady-statevibration The quadratic drag force however induces more
drag force thus the system responses tend to be stabilizedfaster than those with the linearized drag forces
Figures 17 and 18 are plotted on the same conditions withFigures 7(a) and 7(b) respectively but with the quadraticdrag forceThe coefficient of restitution also governs themag-nitudes of responses in the impact model with the quadraticdrag force however the system with the quadratic drag forceis more sensitive to change in the coefficient of restitutionthan the one with the linearized drag force The system with119890 = 05 Figure 18(b) loses the momentum at approximately43 seconds under the influence of the quadratic drag forcewhile the system with the linearized drag force and 119890 = 05
Mathematical Problems in Engineering 13
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
01
0
minus01
minus02
minus03
minus04
minus05
minus06
minus07
(a) Overall responses with119872 = 1600
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
StructureIce
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 1600 from 385 to 400 seconds
Figure 14 Responses with119872 = 1600
0
minus001
minus002
minus003
minus004
minus005
minus006
minus0070 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall responses with119872 = 800
395 396 397 398 399 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
0
5
minus5
times10minus6
(b) Responses with119872 = 800 from 395 to 400 seconds
Figure 15 Responses with119872 = 800
Figure 8(b) loses the majority of its momentum at approxi-mately 54 seconds This is again because the quadratic dragforce dissipates energy more quickly than the linearized dragforceHowever it is demonstrated in Figures 17 and 18 that thecoefficient of restitution plays the same role in both quadraticand linearized drag forces
5 Resonance Conditions
One of the main objectives of the analysis aims to find aresonance condition (which is often referred to as lock-in
when investigating IIVs) The structural vibration frequencyincreases as the free mass velocity increases but when lock-in occurs the structural frequency remains at the naturalfrequency even with increasing free mass velocity [6] In theimpact model however the velocities of the free mass cannotbe the primary lock-in condition because the velocities of thefree mass are not fixed
One possible interpretation of modelling IIVs with theimpact model is that it considers the forces to be a seriesof discrete impulses as shown in Figure 19(a) while otherIIV models consider them as continuous periodic forces
14 Mathematical Problems in Engineering
0 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
(a) Overall responses with119872= 2400
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 16 Responses with119872 = 2400
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
(a) Overall responses with 119890 = 01
StructureIce
0 2 4 6 8 10 12 14
0002
0
minus0002
minus0004
minus0006
minus0008
Disp
lace
men
t (m
)
Time (s)
(b) Responses with 119890 = 01 from 0 to 15 seconds
Figure 17 Responses with 119890 = 01
In fact the impact IIV model proposes a way to calculate thestrength of each impulse and the times between impulses andthese depend on the momenta of the ice and structure aswell as the coefficient of restitution for the collision Undercertain conditions it is possible for the discrete impulsesof Figure 19(a) to approach the regular periodic series ofimpulses of Figure 19(b) which is suspected to cause the lock-in condition
This view of the impact IIV model can be analyzed byusing a Fourier transform [31] There are several definitions
of the Fourier transform (FT) that may be used and in thispaper the chosen version of the FT is given by
119865 (120596) = int
infin
minusinfin
119891 (119905) 119890minus119894120596119905
119889119905 (37)
The corresponding inverse Fourier transform is then given by
119891 (119905) =1
2120587int
infin
minusinfin
119865 (120596) 119890119894120596119905119889120596 (38)
Mathematical Problems in Engineering 15
005
0
minus005
minus01
minus015
minus02
minus0250 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with 119890 = 05
Time (s)
Disp
lace
men
t (m
)
StructureIce
36 38 40 42 44 46 48 50
4
2
0
minus2
minus4
times10minus4
(b) Responses with 119890 = 05 from 35 to 50 seconds
Figure 18 Responses with 119890 = 05
f(t)
fj
tj t
(a) Random impulses
f(t)
f
T 2T 3Tt
pT
(b) Uniform impulses
Figure 19 Ice forces as a series of impulses
The discrete ice forces of Figure 19(a) can be expressed asa series of impulses as
119891 (119905) =
infin
sum
119895=1
119891119895120575119863(119905 minus 119905119895) (39)
where 120575119863indicates the Dirac delta function and 119891
119895indicates
the strength of each impact If the sizes and intervals ofthe impulses are uniform as shown in Figure 19(b) (39) isrewritten as
119891 (119905) =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (40)
In reality the repeated impacts will not all have the samestrength or occur at a fixed interval however it is sufficient forthe situation of Figure 19(a) to approach that of Figure 19(b)for this analysis to be relevant
Equation (40) is often called a comb functionTheFouriertransform of the comb function is given [32] by
119865 (120596) = 2120587
infin
sum
119873=1
119891120575119863(120596 minus
2120587119873
119879) (41)
For a typical single-degree of freedom forced vibrationsproblem
+ 2120577120596119899 + 1205962
119899119909 =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (42)
Taking Fourier transforms of both sides and rearranging give
119883 (120596) = 2120587suminfin
119873=1119891120575119863 (120596 minus 2120587119873119879)
(minus1205962 + 2120577120596119899120596119894 + 1205962
119899)
(43)
Taking the inverse Fourier transforms to find the structureresponse in time gives
119909 (119905) = 119891
infin
sum
119873=1
119890119894(2120587119873119879)119905
(minus4120587211987321198792 + 119894120577120596119899 (4120587119873119879) + 1205962
119899) (44)
Equation (44) experiences a resonance when the denom-inator of any of the terms in the sum goes to zero that is when120596119899= 2120587119873119879 where 119873 is an integer larger than 0 This says
that the structure will experience a resonance when the natu-ral frequency is an integral multiple of the forcing frequency
16 Mathematical Problems in Engineering
as would be expected In the previous numerical simulationsthis condition has been observed in several cases Figures12(a) and 13(b) For instance the structural frequency ofFigure 13(b) is 2120587119879 = 212058705 = 1257 rads The naturalfrequency of the structure is radic119896119898 = radic10000001600 =
25 rads With119873 = 2 the structural frequency is close to thenatural frequency The result is what is seen in Figure 13(b)The two masses vibrate at the same frequency of 2120587119879 whichis half of the natural frequency The structural frequencyremains at that level which is lock-in and so does thefrequency of the free mass response
Thus the impact model successfully predicts the exis-tence of a lock-in condition and also predicts that it can occurwhen the frequency of the structure and freemass impacts aresynchronized together at a multiple of the natural frequencyof the structure The question of the precise conditionsnecessary for the frequency of the impacts to occur at amultiple of the natural frequency still remains open
6 Summary and Conclusions
This paper analyzes the vibrations caused by repeated impactsof a flow-driven freemass onto a second structure Conserva-tion of linearmomentum and the coefficient of restitution areused to characterize the nature of the impacts between thetwo masses This model is referred to as the impact modelTheproperties of the impactmodel were investigated throughnumerical simulations The mass of the free mass is one ofthe factors determining the vibration characteristics sinceit is related to the momentum of each impact This modelis proposed to explain the nature of ice-induced vibrationswithout requiring a microscopic model of the mechanism ofice failure The magnitudes of the vibrations are governedin part by the coefficient of restitution which when usedto explain IIVs encapsulates all ice properties mentionedin other IIV models In addition to the ice properties thecoefficient of restitution also governs the interaction betweenthe structure and the free mass The free mass is used torepresent the ice when modelling IIVs The last propertyof the impact model is that of the free mass area whichinfluences the dissipation of the system momenta throughthe drag force A free mass with a small area shows the samebehavior of a mass driven by lower flow velocity
Therefore through the impact model two factors whichcontrol the characteristics of IIVs have been proposed theice area and the coefficient of restitution By changing thesize of the contact area between the ice and structure thedriving force of the ice can be reduced as if the ice weredriven by lower flow velocities Furthermore by changingfactors related to the coefficient of restitution such as thesurface roughness or shape of the structure IIVs can bereduced to avoid the lock-in conditionwhich occurswhen thefrequencies of the structure and ice impacts are synchronizedat a multiple of the natural frequency of the structure Inaddition to the inclusion of the effect of flow the newapproach to ice forces through the impact model permitsthe answering of questions which originate from the weaktheoretical points of existing IIV models
Particular strengths of the model are that it qualitativelymatches observed results and in particular that a mechanismof ice failure is not required in order to perform simulationsand predictions It also proposes a simple mechanism toexplain the potentially destructive resonance lock-in condi-tions
References
[1] J P Dempsey ldquoResearch trends in ice mechanicsrdquo InternationalJournal of Solids and Structures vol 37 no 1-2 pp 131ndash153 2000
[2] K A Blenkarn ldquoMeasurement and analysis of ice forces oncook inlet structuresrdquo in Proceedings of the Offshore TechnologyConference (OTC rsquo70) pp 365ndash378 1970
[3] Y Toyama T Sensu M Minami and N Yashima ldquoModel testson ice-induced self-excited vibration of cylindrical structuresrdquoin Proceedings of the 7th International Conference on Port andOcean Engineering under Arctic Conditions vol 2 pp 834ndash8441983
[4] M Ranta and R Raty ldquoOn the analytic solution of ice-inducedvibrations in a marine pile structurerdquo in Proceedings of the 7thInternational Conference on Port and Ocean Engineering underArctic Conditions vol 2 pp 901ndash908 1983
[5] A Barker G Timco H Gravesen and P Voslashlund ldquoIce loadingon Danish wind turbinesmdashpart 1 dynamic model testsrdquo ColdRegions Science and Technology vol 41 no 1 pp 1ndash23 2005
[6] D S Sodhi ldquoIce-induced vibrations of structuresrdquo SpecialReport 89-5 International Association for Hydraulic ResearchWorking Group on Ice Forces 1989
[7] D S Sodhi and C E Morris ldquoCharacteristic frequency of forcevariations in continuous crushing of sheet ice against rigidcylindrical structuresrdquo Cold Regions Science and Technologyvol 12 no 1 pp 1ndash12 1986
[8] D S Sodhi C E Morris and G F N Cox ldquoDynamic analysisof failure modes on ice sheets encountering sloping structuresrdquoin Proceedings of the 6th International Conference on OffshoreMechanics and Arctic Engineering vol 4 pp 281ndash284 1987
[9] H R Peyton ldquoSea ice forcesrdquo in Proceedings of the Conferenceon Ice Pressure against Structures 1966
[10] C R Neill ldquoDynamic ice forces on piers and piles an assess-ment of design guidelines in the light of recent researchrdquoCanadian Journal of Civil Engineering vol 3 no 2 pp 305ndash3411976
[11] G Matlock W Dawkins and J Panak ldquoAnalytical model forice-structure interactionrdquo Journal of the Engineering MechanicsDivision pp 1083ndash11092 1971
[12] M M Maattanen ldquoIce-induced vibrations of structures self-excitationrdquo Special Report 89-5 International Association forHydraulic Research Working Group on Ice Forces 1989
[13] M M Maattanen ldquoNumerical simulation of ice-induced vibra-tions in offshore structuresrdquo in Proceedings of the 14th NordicSeminar on Computational Mechanics 2001 Keynote lecture
[14] J Xu and L Wang ldquoIce force oscillator model and its numericalsolutionsrdquo in Proceedings of the 7th International Conference onOffshore Machanics and Arctic Engineering pp 171ndash176 1988
[15] G TsinkerMarine Structures Engineering Chapman-Hall NewYork NY USA 1st edition 1995
[16] G Huang and P Liu ldquoA dynamic model for ice-inducedvibration of structuresrdquo Journal of OffshoreMechanics andArcticEngineering vol 131 no 1 Article ID 011501 6 pages 2009
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
6 Mathematical Problems in Engineering
0
minus02
minus04
minus06
minus08
minus10 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall Responses
Disp
lace
men
t (m
)
Time (s)
0
minus0002
minus0004
minus0006
186 188 190 192 194 196 198 200
StructureIce
(b) Responses from 185 to 200 seconds
Disp
lace
men
t (m
)
Time (s)
StructureIce
0
380 385 390 395 400
times10minus4
minus1
minus2
minus3
(c) Responses from 380 to 400 seconds
Figure 2 Responses of the ice and structure with a linearized drag force
Table 1 Parameters for simulations
Parameter Model structureℎ (mm) 41119863 (mm) 76119898 (kg) 1600119888 (kNsm) 02119896 (MNm) 1ℓ (m) 4493120590119888(MPa) 14
119909max (mm) 20
velocity of the free mass are minus001m and 001ms respec-tively and 0m and 0ms for the structure Figure 2 shows theresponse based on these parameters
Since the motivation for this model was to attempt toexplain IIV without analyzing specific ice-failure mecha-nisms we attempt to compare our results to some measuredIIV data There is very little IIV data in the open literatureespecially for displacement measurements Only laboratorymeasurements are available from [28 29] for displacementmeasurements Since the structural data for their tests arenot presented it is inadequate to compare their experimentaldata to Figure 2 directly but general vibration characteristicscan be compared Structural vibrations of IIVs becomequasi-static vibrations at low ice velocities and steady-statevibrations at medium ice velocities [30] although there isno quantitative definition of low and high ice velocities Aquasi-static vibration is defined by Karna [30] as a transientresponse which is followed by its maximum response atthe peak ice force and is not amplified by the dynamicsof the structure Figure 3(b) represents typical quasi-static
Mathematical Problems in Engineering 7
(a) [28]
1
0
2 4 6 8 10
Stru
ctur
aldi
spla
cem
ent(
cm)
Ice velocity = 76mms Time (s)
(b) [29]
2 4 6 8 10
1
0Stru
ctur
aldi
spla
cem
ent(
cm)
Time (s)Ice velocity = 395mms
(c) [29]
Figure 3 Measured structural responses from the literature
vibrations at ice velocity 76mms and Figure 3(c) shows thecharacteristic of a steady-state vibration at free mass velocity395mms Figure 3(a) however presents the displacementresponse although it is plotted at a lower free mass velocitythan Figure 3(b) This demonstrates that the vibration char-acteristics of IIVs are not determined by the velocity of theice alone but by involving all parameters
As shown in Figure 3(a) actual IIVs display multiplevibration characteristics from quasi-static to steady-statevibrations even at the same ice velocity contrary to thecharacteristic failure frequency model which predicts onetype of characteristic vibration at one ice velocity In thisrespect the proposed impact model possesses an advantageover the characteristic failure frequency model In an impactmodel the velocity of the ice varies from time to time whichis more realistic because part of the ice (free mass) contactingthe structure is deformed as the vibrations proceed Figure 2clearly shows multiple vibration characteristics Figure 2(b)displays quasi-static vibrations similar to Figure 3(b) From380 to 400 seconds the vibrations become stable and steadystate shown in Figure 2(c) which is close to the vibrationcharacteristic of Figure 3(c) The vibration characteristics ofthe impact model however cannot be clearly distinguishedfor the entire duration rather they show a mixture of thecharacteristics Therefore it is reasonable to conclude thatthe vibration characteristics of the impact IIV model aregoverned by the momenta of the ice and structure at the timeof impact
The weakness of modelling IIVs as repeated impacts asproposed in this paper model is how to model the relevantproperties of the ice The definitions of the ice mass and areain the impact IIVmodel imply that themass and area directlyaffect the impact or interactionwith the structure Among thetwo ice properties the influence of the ice area 119860 is directlyrelated to the velocity of the flowbecause of the relationwith 119887given by 119887 = (12)120588119862
119863119860|minus | It is inappropriate to simulate
the linearized model with large flow velocities since then thelinearized model would not be applicable
The mass of the free mass plays an important role in thismodel In order to understand its role it will be investigatedby conducting several numerical simulations Figures 4 5and 6 are plotted with the same parameters of Figure 2 exceptwith differentmasses which are proportional to the structuralmass 50 and 150 percent of 1600 kg so that the structuralmass can be used as a benchmark Including Figure 2 whichis plotted with the free mass of 100 percent of the structuralmass the responses with three different free masses arecompared At the beginning of the responses the heaviermass transfers higher momentum thus impacts happen lessfrequently but with largermagnitudes On the other hand thelighter mass allows the structure to approach the steady-statevibration in relatively short time In Figure 6 the magnitudesof the structural responses are not too different but theheavier mass induces more quasi-static vibrations for longertime Therefore it can be concluded that the free mass hasa greater effect on determining the vibration characteristicsthan on the magnitude of the structural response
Another important factor in this model is the coefficientof restitution The coefficient of restitution also requiresexperimental research to be defined quantitatively Theconducted numerical simulations can only show how thecoefficient of restitution influences the responses of thesystem Figure 2 is plotted with the coefficient of restitution119890 = 1 which is an unrealistic case Figures 7 8 and 9are simulated based on the same parameters of Figure 2 butwith different values of 119890 It becomes obvious that the factorthat most influences the magnitudes of the vibrations is thecoefficient of restitution rather than the velocities of the freemass The system with 119890 = 01 approaches the steady stateafter the first impact The system with 119890 = 05 not onlyapproaches the steady-state vibration faster but also showslower structural magnitudes than the system with 119890 = 1Figure 2 Figure 8 indicates that the system with 119890 = 05
loses the majority of its momentum after approximately54 seconds This suggests that vibrations due to repeatedimpacts can be controlled by manipulating the coefficient
8 Mathematical Problems in Engineering
002
0
minus002
minus004
minus006
minus008
minus01
minus0120 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with119872 = 800
0 50 100 150 200 250 300 350 400
02
0
minus02
minus04
minus06
minus08
minus1
minus12
minus14
Time (s)
Disp
lace
men
t (m
)
StructureIce
(b) Overall responses with119872= 2400
Figure 4 Responses with different ice masses 1
186 188 190 192 194 196 198 200
Disp
lace
men
t (m
)
StructureIce
Time (s)
0
minus2
minus4
minus6
times10minus4
(a) Responses with119872 = 800 from 185 to 200 seconds
0
minus0002
minus0004
minus0006
minus0008
186 188 190 192 194 196 198 200
StructureIce
Time (s)
Disp
lace
men
t (m
)
(b) Responses with119872= 2400 from 185 to 200 seconds
Figure 5 Responses with different ice masses 2
of restitution via the properties of the contacting area Forinstance the magnitudes of vibrations can be significantlyreduced by increasing the surface roughness of the structure
42 Quadratic Drag Force In the previous section themodelwas developed based on the linearized drag force whichassumes small relative velocity between the velocities of thefree mass and structure The linearized model howeveris definitely vulnerable to two factors namely when theinterval between impacts exceeds several seconds and whenthe velocity of the flow is no longer small Figure 10 clearlyshows the error between the models based on the linearized
drag and quadratic drag as time passes The displacement oflinearized equations (3) and (6) is compared with numer-ically solved equation (3) by using the built-in ordinarydifferential equation solver ofMathCadThe difference growsas time goes by but at high flow velocity the differencedecreases in Figure 10(b) This is because of the linearizationfactor | minus | which calibrates the difference between thelinear and quadratic models The solution however to thequadratic drag force needs to be developed for a more precisesimulation
A numerical solution is needed to solve (3) but insteadof using the built-in ordinary differential equation solver
Mathematical Problems in Engineering 9
StructureIce
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
Time (s)
1
0
minus1
minus2
minus3
minus4
times10minus4
(a) Responses with119872 = 800 from 385 to 400 seconds
StructureIce
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
0
minus2
minus4
minus6
times10minus4
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 6 Responses with different ice masses 3
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Overall responses with 119890 = 01
0
minus005
minus01
minus015
minus02
minus025
minus03
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400Time (s)
StructureIce
(b) Overall responses with 119890 = 05
Figure 7 Responses with different coefficients of restitution 1
of MathCad a simple numerical solution is developed for afaster algorithm From (3)
119872 = minus1
2120588119862119863119860 ( minus ) | minus | = minus119861 ( minus ) | minus | (31)
or
=119889
119889119905= minus
119861
119872( minus ) | minus | (32)
where
119861 =1
2120588119862119863119860 (33)
Using 119894 as a counter for the a number of time steps thedisplacement of (34) can be expressed as
119911119894= 119911119894minus1
+119889119911119894minus1
119889119905Δ119905 = 119911
119894minus1+ 119894minus1Δ119905 (34)
in which Δ119905 is previously defined as a time step z is given by
119894= 119894minus1
+119889119894minus1
119889119905Δ119905 =
119894minus1+ 119894minus1Δ119905 (35)
By substituting (32) into (35)
119894= 119894minus1
+ [minus119861
119872(119894minus1
minus )1003816100381610038161003816119894minus1 minus
1003816100381610038161003816] Δ119905(36)
10 Mathematical Problems in Engineering
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 2 4 6 8 10 12 14
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Responses with 119890 = 01 from 0 to 15 seconds
Disp
lace
men
t (m
)
Time (s)
StructureIce
46 48 50 52 54 56 58 60
2
0
minus2
minus4
minus6
minus8
times10minus4
(b) Responses with 119890 = 05 from 45 to 60 seconds
Figure 8 Responses with different coefficients of restitution 2
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus001386 388 390 392 394 396 398 400
Time (s)
Disp
lace
men
t (m
)
(a) Responses with 119890 = 01 from 385 to 400 seconds
StructureIce
Time (s)
Disp
lace
men
t (m
)
398 3985 399 3995 400
times10minus6
6
5
4
3
(b) Responses with 119890 = 05 from 398 to 400 seconds
Figure 9 Responses with different coefficients of restitution 3
Given the initial displacement and velocity of the free massthe displacement can be numerically solved by using (34)and (36) as simultaneous equationsThe simulation results of(34) and (36) are also plotted in Figure 10 At the high flowvelocity Figure 10(b) (34) and (36) predict the displacementclose enough to the one evaluated with the built-in ordinarydifferential equation solver In the algorithm of the numericalsimulation (34) and (36) replace (6) and (7)
Using the same parameters as those used in Figure 2but with a different flow velocity the responses of themodel with the quadratic drag force are compared withthe one with the linearized drag force in Figure 11 At high
flow velocities over 1ms the quadratic drag force modelapproaches steady-state vibration more quickly than thelinearized drag force model because at high flow velocity thequadratic drag induces more damping while it transfers moreforces than the linearized drag In Figure 12 the quadraticdrag force model exhibits the characteristic of the steady-state vibration while the linearized drag force model stilldisplays quasi-static vibration As velocities increase whichcorresponds to flowvelocities increasing vibrations approachthe steady state The linearized drag force model howeverdoes not show a steady-state vibration until 400 secondsin Figure 12 The linearization factor allows the linearized
Mathematical Problems in Engineering 11
0 100 200 300 400 500 600
100
80
60
40
20
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(a) Flow velocity 01ms
0 100 200 300 400 500 600
1000
800
600
400
200
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(b) Flow velocity 1ms
Figure 10 Comparison of the linearized and quadratic Drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
Disp
lace
men
t (m
)
(a) Overall responses with the quadratic drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
0
minus02
minus04
minus06
minus08
minus1
Disp
lace
men
t (m
)
(b) Overall responses with the linear drag
Figure 11 Responses with the quadratic and linear drags 1
drag force model to follow the quadratic drag force modelin the beginning but when the system is in a series of lowvelocity impacts the linearization factor causes a differenceThe linearized drag force model therefore is inappropriateto simulate the repeated impact model at high flow veloci-ties
Since the free mass area 119860 plays a similar role as theflow velocity in the model the model with wide area is alsosimulated with the quadratic drag force The influence of thedrag force on the response of the free mass increases as thesize of the area increases because the area is related to the dragforce by (1) 119891
119863= minus(12)120588119862
119863119860( minus )
2 Figure 13 is plottedwith the area 119860 = 2m2 the flow velocity = 01ms and
the rest of parameters are the same as the parameters usedin Figure 11 The flow velocity of Figure 13 is identical to thatof Figure 2 but the responses quickly approach the steady-state vibrations as if they are driven by high flow velocityFigure 13(b) also shows a similar vibration characteristic toFigure 12(a) because both flow velocity and area influence thedrag force in the impact model Therefore the area can beanother controlling factor of the vibrations Since the areais related to the size of the contacting area of the structurereducing the contacting area has a similar effect to that ofreducing the flow velocity
In the previous section the roles of the free mass andthe coefficient of restitution have been defined by changing
12 Mathematical Problems in EngineeringD
ispla
cem
ent (
m)
395 396 397 398 399 400Time (s)
StructureIce
8
6
4
2
0
times10minus4
(a) Responses with the quadratic drag from 395 to 400 secondsD
ispla
cem
ent (
m)
386 388 390 392 394 396 398 400Time (s)
StructureIce
0
minus0005
(b) Responses with the linear drag from 385 to 400 seconds
Figure 12 Responses with the quadratic and linear drags 2
005
0
minus005
minus01
minus015
minus02
minus025
minus03
minus035
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
(a) Overall responses with 119860 = 2
Disp
lace
men
t (m
)
395 396 397 398 399 400Time (s)
StructureIce
1
0
minus1
times10minus5
(b) Responses with 119860 = 2 from 395 to 400 seconds
Figure 13 Responses with 119860 = 2m2
relevant parameters To verify if those definitions still holdwith the quadratic drag force the impact model with thequadratic drag is plotted with change in free mass and thecoefficient of restitution Figures 14 15 and 16 are plottedon the same parameters used in Figures 2 4(a) and 4(b)respectively except that they are based on the quadratic dragforce The responses of Figures 14 15 and 16 prove thatthe defined role of the free mass holds with the quadraticdrag force The responses of the system with heavier freemass Figure 16(b) is still unstable while the system with thefree mass 119872 = 800 kg Figure 15(b) is in the steady-statevibration The quadratic drag force however induces more
drag force thus the system responses tend to be stabilizedfaster than those with the linearized drag forces
Figures 17 and 18 are plotted on the same conditions withFigures 7(a) and 7(b) respectively but with the quadraticdrag forceThe coefficient of restitution also governs themag-nitudes of responses in the impact model with the quadraticdrag force however the system with the quadratic drag forceis more sensitive to change in the coefficient of restitutionthan the one with the linearized drag force The system with119890 = 05 Figure 18(b) loses the momentum at approximately43 seconds under the influence of the quadratic drag forcewhile the system with the linearized drag force and 119890 = 05
Mathematical Problems in Engineering 13
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
01
0
minus01
minus02
minus03
minus04
minus05
minus06
minus07
(a) Overall responses with119872 = 1600
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
StructureIce
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 1600 from 385 to 400 seconds
Figure 14 Responses with119872 = 1600
0
minus001
minus002
minus003
minus004
minus005
minus006
minus0070 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall responses with119872 = 800
395 396 397 398 399 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
0
5
minus5
times10minus6
(b) Responses with119872 = 800 from 395 to 400 seconds
Figure 15 Responses with119872 = 800
Figure 8(b) loses the majority of its momentum at approxi-mately 54 seconds This is again because the quadratic dragforce dissipates energy more quickly than the linearized dragforceHowever it is demonstrated in Figures 17 and 18 that thecoefficient of restitution plays the same role in both quadraticand linearized drag forces
5 Resonance Conditions
One of the main objectives of the analysis aims to find aresonance condition (which is often referred to as lock-in
when investigating IIVs) The structural vibration frequencyincreases as the free mass velocity increases but when lock-in occurs the structural frequency remains at the naturalfrequency even with increasing free mass velocity [6] In theimpact model however the velocities of the free mass cannotbe the primary lock-in condition because the velocities of thefree mass are not fixed
One possible interpretation of modelling IIVs with theimpact model is that it considers the forces to be a seriesof discrete impulses as shown in Figure 19(a) while otherIIV models consider them as continuous periodic forces
14 Mathematical Problems in Engineering
0 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
(a) Overall responses with119872= 2400
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 16 Responses with119872 = 2400
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
(a) Overall responses with 119890 = 01
StructureIce
0 2 4 6 8 10 12 14
0002
0
minus0002
minus0004
minus0006
minus0008
Disp
lace
men
t (m
)
Time (s)
(b) Responses with 119890 = 01 from 0 to 15 seconds
Figure 17 Responses with 119890 = 01
In fact the impact IIV model proposes a way to calculate thestrength of each impulse and the times between impulses andthese depend on the momenta of the ice and structure aswell as the coefficient of restitution for the collision Undercertain conditions it is possible for the discrete impulsesof Figure 19(a) to approach the regular periodic series ofimpulses of Figure 19(b) which is suspected to cause the lock-in condition
This view of the impact IIV model can be analyzed byusing a Fourier transform [31] There are several definitions
of the Fourier transform (FT) that may be used and in thispaper the chosen version of the FT is given by
119865 (120596) = int
infin
minusinfin
119891 (119905) 119890minus119894120596119905
119889119905 (37)
The corresponding inverse Fourier transform is then given by
119891 (119905) =1
2120587int
infin
minusinfin
119865 (120596) 119890119894120596119905119889120596 (38)
Mathematical Problems in Engineering 15
005
0
minus005
minus01
minus015
minus02
minus0250 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with 119890 = 05
Time (s)
Disp
lace
men
t (m
)
StructureIce
36 38 40 42 44 46 48 50
4
2
0
minus2
minus4
times10minus4
(b) Responses with 119890 = 05 from 35 to 50 seconds
Figure 18 Responses with 119890 = 05
f(t)
fj
tj t
(a) Random impulses
f(t)
f
T 2T 3Tt
pT
(b) Uniform impulses
Figure 19 Ice forces as a series of impulses
The discrete ice forces of Figure 19(a) can be expressed asa series of impulses as
119891 (119905) =
infin
sum
119895=1
119891119895120575119863(119905 minus 119905119895) (39)
where 120575119863indicates the Dirac delta function and 119891
119895indicates
the strength of each impact If the sizes and intervals ofthe impulses are uniform as shown in Figure 19(b) (39) isrewritten as
119891 (119905) =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (40)
In reality the repeated impacts will not all have the samestrength or occur at a fixed interval however it is sufficient forthe situation of Figure 19(a) to approach that of Figure 19(b)for this analysis to be relevant
Equation (40) is often called a comb functionTheFouriertransform of the comb function is given [32] by
119865 (120596) = 2120587
infin
sum
119873=1
119891120575119863(120596 minus
2120587119873
119879) (41)
For a typical single-degree of freedom forced vibrationsproblem
+ 2120577120596119899 + 1205962
119899119909 =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (42)
Taking Fourier transforms of both sides and rearranging give
119883 (120596) = 2120587suminfin
119873=1119891120575119863 (120596 minus 2120587119873119879)
(minus1205962 + 2120577120596119899120596119894 + 1205962
119899)
(43)
Taking the inverse Fourier transforms to find the structureresponse in time gives
119909 (119905) = 119891
infin
sum
119873=1
119890119894(2120587119873119879)119905
(minus4120587211987321198792 + 119894120577120596119899 (4120587119873119879) + 1205962
119899) (44)
Equation (44) experiences a resonance when the denom-inator of any of the terms in the sum goes to zero that is when120596119899= 2120587119873119879 where 119873 is an integer larger than 0 This says
that the structure will experience a resonance when the natu-ral frequency is an integral multiple of the forcing frequency
16 Mathematical Problems in Engineering
as would be expected In the previous numerical simulationsthis condition has been observed in several cases Figures12(a) and 13(b) For instance the structural frequency ofFigure 13(b) is 2120587119879 = 212058705 = 1257 rads The naturalfrequency of the structure is radic119896119898 = radic10000001600 =
25 rads With119873 = 2 the structural frequency is close to thenatural frequency The result is what is seen in Figure 13(b)The two masses vibrate at the same frequency of 2120587119879 whichis half of the natural frequency The structural frequencyremains at that level which is lock-in and so does thefrequency of the free mass response
Thus the impact model successfully predicts the exis-tence of a lock-in condition and also predicts that it can occurwhen the frequency of the structure and freemass impacts aresynchronized together at a multiple of the natural frequencyof the structure The question of the precise conditionsnecessary for the frequency of the impacts to occur at amultiple of the natural frequency still remains open
6 Summary and Conclusions
This paper analyzes the vibrations caused by repeated impactsof a flow-driven freemass onto a second structure Conserva-tion of linearmomentum and the coefficient of restitution areused to characterize the nature of the impacts between thetwo masses This model is referred to as the impact modelTheproperties of the impactmodel were investigated throughnumerical simulations The mass of the free mass is one ofthe factors determining the vibration characteristics sinceit is related to the momentum of each impact This modelis proposed to explain the nature of ice-induced vibrationswithout requiring a microscopic model of the mechanism ofice failure The magnitudes of the vibrations are governedin part by the coefficient of restitution which when usedto explain IIVs encapsulates all ice properties mentionedin other IIV models In addition to the ice properties thecoefficient of restitution also governs the interaction betweenthe structure and the free mass The free mass is used torepresent the ice when modelling IIVs The last propertyof the impact model is that of the free mass area whichinfluences the dissipation of the system momenta throughthe drag force A free mass with a small area shows the samebehavior of a mass driven by lower flow velocity
Therefore through the impact model two factors whichcontrol the characteristics of IIVs have been proposed theice area and the coefficient of restitution By changing thesize of the contact area between the ice and structure thedriving force of the ice can be reduced as if the ice weredriven by lower flow velocities Furthermore by changingfactors related to the coefficient of restitution such as thesurface roughness or shape of the structure IIVs can bereduced to avoid the lock-in conditionwhich occurswhen thefrequencies of the structure and ice impacts are synchronizedat a multiple of the natural frequency of the structure Inaddition to the inclusion of the effect of flow the newapproach to ice forces through the impact model permitsthe answering of questions which originate from the weaktheoretical points of existing IIV models
Particular strengths of the model are that it qualitativelymatches observed results and in particular that a mechanismof ice failure is not required in order to perform simulationsand predictions It also proposes a simple mechanism toexplain the potentially destructive resonance lock-in condi-tions
References
[1] J P Dempsey ldquoResearch trends in ice mechanicsrdquo InternationalJournal of Solids and Structures vol 37 no 1-2 pp 131ndash153 2000
[2] K A Blenkarn ldquoMeasurement and analysis of ice forces oncook inlet structuresrdquo in Proceedings of the Offshore TechnologyConference (OTC rsquo70) pp 365ndash378 1970
[3] Y Toyama T Sensu M Minami and N Yashima ldquoModel testson ice-induced self-excited vibration of cylindrical structuresrdquoin Proceedings of the 7th International Conference on Port andOcean Engineering under Arctic Conditions vol 2 pp 834ndash8441983
[4] M Ranta and R Raty ldquoOn the analytic solution of ice-inducedvibrations in a marine pile structurerdquo in Proceedings of the 7thInternational Conference on Port and Ocean Engineering underArctic Conditions vol 2 pp 901ndash908 1983
[5] A Barker G Timco H Gravesen and P Voslashlund ldquoIce loadingon Danish wind turbinesmdashpart 1 dynamic model testsrdquo ColdRegions Science and Technology vol 41 no 1 pp 1ndash23 2005
[6] D S Sodhi ldquoIce-induced vibrations of structuresrdquo SpecialReport 89-5 International Association for Hydraulic ResearchWorking Group on Ice Forces 1989
[7] D S Sodhi and C E Morris ldquoCharacteristic frequency of forcevariations in continuous crushing of sheet ice against rigidcylindrical structuresrdquo Cold Regions Science and Technologyvol 12 no 1 pp 1ndash12 1986
[8] D S Sodhi C E Morris and G F N Cox ldquoDynamic analysisof failure modes on ice sheets encountering sloping structuresrdquoin Proceedings of the 6th International Conference on OffshoreMechanics and Arctic Engineering vol 4 pp 281ndash284 1987
[9] H R Peyton ldquoSea ice forcesrdquo in Proceedings of the Conferenceon Ice Pressure against Structures 1966
[10] C R Neill ldquoDynamic ice forces on piers and piles an assess-ment of design guidelines in the light of recent researchrdquoCanadian Journal of Civil Engineering vol 3 no 2 pp 305ndash3411976
[11] G Matlock W Dawkins and J Panak ldquoAnalytical model forice-structure interactionrdquo Journal of the Engineering MechanicsDivision pp 1083ndash11092 1971
[12] M M Maattanen ldquoIce-induced vibrations of structures self-excitationrdquo Special Report 89-5 International Association forHydraulic Research Working Group on Ice Forces 1989
[13] M M Maattanen ldquoNumerical simulation of ice-induced vibra-tions in offshore structuresrdquo in Proceedings of the 14th NordicSeminar on Computational Mechanics 2001 Keynote lecture
[14] J Xu and L Wang ldquoIce force oscillator model and its numericalsolutionsrdquo in Proceedings of the 7th International Conference onOffshore Machanics and Arctic Engineering pp 171ndash176 1988
[15] G TsinkerMarine Structures Engineering Chapman-Hall NewYork NY USA 1st edition 1995
[16] G Huang and P Liu ldquoA dynamic model for ice-inducedvibration of structuresrdquo Journal of OffshoreMechanics andArcticEngineering vol 131 no 1 Article ID 011501 6 pages 2009
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
Mathematical Problems in Engineering 7
(a) [28]
1
0
2 4 6 8 10
Stru
ctur
aldi
spla
cem
ent(
cm)
Ice velocity = 76mms Time (s)
(b) [29]
2 4 6 8 10
1
0Stru
ctur
aldi
spla
cem
ent(
cm)
Time (s)Ice velocity = 395mms
(c) [29]
Figure 3 Measured structural responses from the literature
vibrations at ice velocity 76mms and Figure 3(c) shows thecharacteristic of a steady-state vibration at free mass velocity395mms Figure 3(a) however presents the displacementresponse although it is plotted at a lower free mass velocitythan Figure 3(b) This demonstrates that the vibration char-acteristics of IIVs are not determined by the velocity of theice alone but by involving all parameters
As shown in Figure 3(a) actual IIVs display multiplevibration characteristics from quasi-static to steady-statevibrations even at the same ice velocity contrary to thecharacteristic failure frequency model which predicts onetype of characteristic vibration at one ice velocity In thisrespect the proposed impact model possesses an advantageover the characteristic failure frequency model In an impactmodel the velocity of the ice varies from time to time whichis more realistic because part of the ice (free mass) contactingthe structure is deformed as the vibrations proceed Figure 2clearly shows multiple vibration characteristics Figure 2(b)displays quasi-static vibrations similar to Figure 3(b) From380 to 400 seconds the vibrations become stable and steadystate shown in Figure 2(c) which is close to the vibrationcharacteristic of Figure 3(c) The vibration characteristics ofthe impact model however cannot be clearly distinguishedfor the entire duration rather they show a mixture of thecharacteristics Therefore it is reasonable to conclude thatthe vibration characteristics of the impact IIV model aregoverned by the momenta of the ice and structure at the timeof impact
The weakness of modelling IIVs as repeated impacts asproposed in this paper model is how to model the relevantproperties of the ice The definitions of the ice mass and areain the impact IIVmodel imply that themass and area directlyaffect the impact or interactionwith the structure Among thetwo ice properties the influence of the ice area 119860 is directlyrelated to the velocity of the flowbecause of the relationwith 119887given by 119887 = (12)120588119862
119863119860|minus | It is inappropriate to simulate
the linearized model with large flow velocities since then thelinearized model would not be applicable
The mass of the free mass plays an important role in thismodel In order to understand its role it will be investigatedby conducting several numerical simulations Figures 4 5and 6 are plotted with the same parameters of Figure 2 exceptwith differentmasses which are proportional to the structuralmass 50 and 150 percent of 1600 kg so that the structuralmass can be used as a benchmark Including Figure 2 whichis plotted with the free mass of 100 percent of the structuralmass the responses with three different free masses arecompared At the beginning of the responses the heaviermass transfers higher momentum thus impacts happen lessfrequently but with largermagnitudes On the other hand thelighter mass allows the structure to approach the steady-statevibration in relatively short time In Figure 6 the magnitudesof the structural responses are not too different but theheavier mass induces more quasi-static vibrations for longertime Therefore it can be concluded that the free mass hasa greater effect on determining the vibration characteristicsthan on the magnitude of the structural response
Another important factor in this model is the coefficientof restitution The coefficient of restitution also requiresexperimental research to be defined quantitatively Theconducted numerical simulations can only show how thecoefficient of restitution influences the responses of thesystem Figure 2 is plotted with the coefficient of restitution119890 = 1 which is an unrealistic case Figures 7 8 and 9are simulated based on the same parameters of Figure 2 butwith different values of 119890 It becomes obvious that the factorthat most influences the magnitudes of the vibrations is thecoefficient of restitution rather than the velocities of the freemass The system with 119890 = 01 approaches the steady stateafter the first impact The system with 119890 = 05 not onlyapproaches the steady-state vibration faster but also showslower structural magnitudes than the system with 119890 = 1Figure 2 Figure 8 indicates that the system with 119890 = 05
loses the majority of its momentum after approximately54 seconds This suggests that vibrations due to repeatedimpacts can be controlled by manipulating the coefficient
8 Mathematical Problems in Engineering
002
0
minus002
minus004
minus006
minus008
minus01
minus0120 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with119872 = 800
0 50 100 150 200 250 300 350 400
02
0
minus02
minus04
minus06
minus08
minus1
minus12
minus14
Time (s)
Disp
lace
men
t (m
)
StructureIce
(b) Overall responses with119872= 2400
Figure 4 Responses with different ice masses 1
186 188 190 192 194 196 198 200
Disp
lace
men
t (m
)
StructureIce
Time (s)
0
minus2
minus4
minus6
times10minus4
(a) Responses with119872 = 800 from 185 to 200 seconds
0
minus0002
minus0004
minus0006
minus0008
186 188 190 192 194 196 198 200
StructureIce
Time (s)
Disp
lace
men
t (m
)
(b) Responses with119872= 2400 from 185 to 200 seconds
Figure 5 Responses with different ice masses 2
of restitution via the properties of the contacting area Forinstance the magnitudes of vibrations can be significantlyreduced by increasing the surface roughness of the structure
42 Quadratic Drag Force In the previous section themodelwas developed based on the linearized drag force whichassumes small relative velocity between the velocities of thefree mass and structure The linearized model howeveris definitely vulnerable to two factors namely when theinterval between impacts exceeds several seconds and whenthe velocity of the flow is no longer small Figure 10 clearlyshows the error between the models based on the linearized
drag and quadratic drag as time passes The displacement oflinearized equations (3) and (6) is compared with numer-ically solved equation (3) by using the built-in ordinarydifferential equation solver ofMathCadThe difference growsas time goes by but at high flow velocity the differencedecreases in Figure 10(b) This is because of the linearizationfactor | minus | which calibrates the difference between thelinear and quadratic models The solution however to thequadratic drag force needs to be developed for a more precisesimulation
A numerical solution is needed to solve (3) but insteadof using the built-in ordinary differential equation solver
Mathematical Problems in Engineering 9
StructureIce
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
Time (s)
1
0
minus1
minus2
minus3
minus4
times10minus4
(a) Responses with119872 = 800 from 385 to 400 seconds
StructureIce
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
0
minus2
minus4
minus6
times10minus4
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 6 Responses with different ice masses 3
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Overall responses with 119890 = 01
0
minus005
minus01
minus015
minus02
minus025
minus03
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400Time (s)
StructureIce
(b) Overall responses with 119890 = 05
Figure 7 Responses with different coefficients of restitution 1
of MathCad a simple numerical solution is developed for afaster algorithm From (3)
119872 = minus1
2120588119862119863119860 ( minus ) | minus | = minus119861 ( minus ) | minus | (31)
or
=119889
119889119905= minus
119861
119872( minus ) | minus | (32)
where
119861 =1
2120588119862119863119860 (33)
Using 119894 as a counter for the a number of time steps thedisplacement of (34) can be expressed as
119911119894= 119911119894minus1
+119889119911119894minus1
119889119905Δ119905 = 119911
119894minus1+ 119894minus1Δ119905 (34)
in which Δ119905 is previously defined as a time step z is given by
119894= 119894minus1
+119889119894minus1
119889119905Δ119905 =
119894minus1+ 119894minus1Δ119905 (35)
By substituting (32) into (35)
119894= 119894minus1
+ [minus119861
119872(119894minus1
minus )1003816100381610038161003816119894minus1 minus
1003816100381610038161003816] Δ119905(36)
10 Mathematical Problems in Engineering
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 2 4 6 8 10 12 14
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Responses with 119890 = 01 from 0 to 15 seconds
Disp
lace
men
t (m
)
Time (s)
StructureIce
46 48 50 52 54 56 58 60
2
0
minus2
minus4
minus6
minus8
times10minus4
(b) Responses with 119890 = 05 from 45 to 60 seconds
Figure 8 Responses with different coefficients of restitution 2
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus001386 388 390 392 394 396 398 400
Time (s)
Disp
lace
men
t (m
)
(a) Responses with 119890 = 01 from 385 to 400 seconds
StructureIce
Time (s)
Disp
lace
men
t (m
)
398 3985 399 3995 400
times10minus6
6
5
4
3
(b) Responses with 119890 = 05 from 398 to 400 seconds
Figure 9 Responses with different coefficients of restitution 3
Given the initial displacement and velocity of the free massthe displacement can be numerically solved by using (34)and (36) as simultaneous equationsThe simulation results of(34) and (36) are also plotted in Figure 10 At the high flowvelocity Figure 10(b) (34) and (36) predict the displacementclose enough to the one evaluated with the built-in ordinarydifferential equation solver In the algorithm of the numericalsimulation (34) and (36) replace (6) and (7)
Using the same parameters as those used in Figure 2but with a different flow velocity the responses of themodel with the quadratic drag force are compared withthe one with the linearized drag force in Figure 11 At high
flow velocities over 1ms the quadratic drag force modelapproaches steady-state vibration more quickly than thelinearized drag force model because at high flow velocity thequadratic drag induces more damping while it transfers moreforces than the linearized drag In Figure 12 the quadraticdrag force model exhibits the characteristic of the steady-state vibration while the linearized drag force model stilldisplays quasi-static vibration As velocities increase whichcorresponds to flowvelocities increasing vibrations approachthe steady state The linearized drag force model howeverdoes not show a steady-state vibration until 400 secondsin Figure 12 The linearization factor allows the linearized
Mathematical Problems in Engineering 11
0 100 200 300 400 500 600
100
80
60
40
20
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(a) Flow velocity 01ms
0 100 200 300 400 500 600
1000
800
600
400
200
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(b) Flow velocity 1ms
Figure 10 Comparison of the linearized and quadratic Drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
Disp
lace
men
t (m
)
(a) Overall responses with the quadratic drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
0
minus02
minus04
minus06
minus08
minus1
Disp
lace
men
t (m
)
(b) Overall responses with the linear drag
Figure 11 Responses with the quadratic and linear drags 1
drag force model to follow the quadratic drag force modelin the beginning but when the system is in a series of lowvelocity impacts the linearization factor causes a differenceThe linearized drag force model therefore is inappropriateto simulate the repeated impact model at high flow veloci-ties
Since the free mass area 119860 plays a similar role as theflow velocity in the model the model with wide area is alsosimulated with the quadratic drag force The influence of thedrag force on the response of the free mass increases as thesize of the area increases because the area is related to the dragforce by (1) 119891
119863= minus(12)120588119862
119863119860( minus )
2 Figure 13 is plottedwith the area 119860 = 2m2 the flow velocity = 01ms and
the rest of parameters are the same as the parameters usedin Figure 11 The flow velocity of Figure 13 is identical to thatof Figure 2 but the responses quickly approach the steady-state vibrations as if they are driven by high flow velocityFigure 13(b) also shows a similar vibration characteristic toFigure 12(a) because both flow velocity and area influence thedrag force in the impact model Therefore the area can beanother controlling factor of the vibrations Since the areais related to the size of the contacting area of the structurereducing the contacting area has a similar effect to that ofreducing the flow velocity
In the previous section the roles of the free mass andthe coefficient of restitution have been defined by changing
12 Mathematical Problems in EngineeringD
ispla
cem
ent (
m)
395 396 397 398 399 400Time (s)
StructureIce
8
6
4
2
0
times10minus4
(a) Responses with the quadratic drag from 395 to 400 secondsD
ispla
cem
ent (
m)
386 388 390 392 394 396 398 400Time (s)
StructureIce
0
minus0005
(b) Responses with the linear drag from 385 to 400 seconds
Figure 12 Responses with the quadratic and linear drags 2
005
0
minus005
minus01
minus015
minus02
minus025
minus03
minus035
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
(a) Overall responses with 119860 = 2
Disp
lace
men
t (m
)
395 396 397 398 399 400Time (s)
StructureIce
1
0
minus1
times10minus5
(b) Responses with 119860 = 2 from 395 to 400 seconds
Figure 13 Responses with 119860 = 2m2
relevant parameters To verify if those definitions still holdwith the quadratic drag force the impact model with thequadratic drag is plotted with change in free mass and thecoefficient of restitution Figures 14 15 and 16 are plottedon the same parameters used in Figures 2 4(a) and 4(b)respectively except that they are based on the quadratic dragforce The responses of Figures 14 15 and 16 prove thatthe defined role of the free mass holds with the quadraticdrag force The responses of the system with heavier freemass Figure 16(b) is still unstable while the system with thefree mass 119872 = 800 kg Figure 15(b) is in the steady-statevibration The quadratic drag force however induces more
drag force thus the system responses tend to be stabilizedfaster than those with the linearized drag forces
Figures 17 and 18 are plotted on the same conditions withFigures 7(a) and 7(b) respectively but with the quadraticdrag forceThe coefficient of restitution also governs themag-nitudes of responses in the impact model with the quadraticdrag force however the system with the quadratic drag forceis more sensitive to change in the coefficient of restitutionthan the one with the linearized drag force The system with119890 = 05 Figure 18(b) loses the momentum at approximately43 seconds under the influence of the quadratic drag forcewhile the system with the linearized drag force and 119890 = 05
Mathematical Problems in Engineering 13
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
01
0
minus01
minus02
minus03
minus04
minus05
minus06
minus07
(a) Overall responses with119872 = 1600
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
StructureIce
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 1600 from 385 to 400 seconds
Figure 14 Responses with119872 = 1600
0
minus001
minus002
minus003
minus004
minus005
minus006
minus0070 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall responses with119872 = 800
395 396 397 398 399 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
0
5
minus5
times10minus6
(b) Responses with119872 = 800 from 395 to 400 seconds
Figure 15 Responses with119872 = 800
Figure 8(b) loses the majority of its momentum at approxi-mately 54 seconds This is again because the quadratic dragforce dissipates energy more quickly than the linearized dragforceHowever it is demonstrated in Figures 17 and 18 that thecoefficient of restitution plays the same role in both quadraticand linearized drag forces
5 Resonance Conditions
One of the main objectives of the analysis aims to find aresonance condition (which is often referred to as lock-in
when investigating IIVs) The structural vibration frequencyincreases as the free mass velocity increases but when lock-in occurs the structural frequency remains at the naturalfrequency even with increasing free mass velocity [6] In theimpact model however the velocities of the free mass cannotbe the primary lock-in condition because the velocities of thefree mass are not fixed
One possible interpretation of modelling IIVs with theimpact model is that it considers the forces to be a seriesof discrete impulses as shown in Figure 19(a) while otherIIV models consider them as continuous periodic forces
14 Mathematical Problems in Engineering
0 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
(a) Overall responses with119872= 2400
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 16 Responses with119872 = 2400
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
(a) Overall responses with 119890 = 01
StructureIce
0 2 4 6 8 10 12 14
0002
0
minus0002
minus0004
minus0006
minus0008
Disp
lace
men
t (m
)
Time (s)
(b) Responses with 119890 = 01 from 0 to 15 seconds
Figure 17 Responses with 119890 = 01
In fact the impact IIV model proposes a way to calculate thestrength of each impulse and the times between impulses andthese depend on the momenta of the ice and structure aswell as the coefficient of restitution for the collision Undercertain conditions it is possible for the discrete impulsesof Figure 19(a) to approach the regular periodic series ofimpulses of Figure 19(b) which is suspected to cause the lock-in condition
This view of the impact IIV model can be analyzed byusing a Fourier transform [31] There are several definitions
of the Fourier transform (FT) that may be used and in thispaper the chosen version of the FT is given by
119865 (120596) = int
infin
minusinfin
119891 (119905) 119890minus119894120596119905
119889119905 (37)
The corresponding inverse Fourier transform is then given by
119891 (119905) =1
2120587int
infin
minusinfin
119865 (120596) 119890119894120596119905119889120596 (38)
Mathematical Problems in Engineering 15
005
0
minus005
minus01
minus015
minus02
minus0250 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with 119890 = 05
Time (s)
Disp
lace
men
t (m
)
StructureIce
36 38 40 42 44 46 48 50
4
2
0
minus2
minus4
times10minus4
(b) Responses with 119890 = 05 from 35 to 50 seconds
Figure 18 Responses with 119890 = 05
f(t)
fj
tj t
(a) Random impulses
f(t)
f
T 2T 3Tt
pT
(b) Uniform impulses
Figure 19 Ice forces as a series of impulses
The discrete ice forces of Figure 19(a) can be expressed asa series of impulses as
119891 (119905) =
infin
sum
119895=1
119891119895120575119863(119905 minus 119905119895) (39)
where 120575119863indicates the Dirac delta function and 119891
119895indicates
the strength of each impact If the sizes and intervals ofthe impulses are uniform as shown in Figure 19(b) (39) isrewritten as
119891 (119905) =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (40)
In reality the repeated impacts will not all have the samestrength or occur at a fixed interval however it is sufficient forthe situation of Figure 19(a) to approach that of Figure 19(b)for this analysis to be relevant
Equation (40) is often called a comb functionTheFouriertransform of the comb function is given [32] by
119865 (120596) = 2120587
infin
sum
119873=1
119891120575119863(120596 minus
2120587119873
119879) (41)
For a typical single-degree of freedom forced vibrationsproblem
+ 2120577120596119899 + 1205962
119899119909 =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (42)
Taking Fourier transforms of both sides and rearranging give
119883 (120596) = 2120587suminfin
119873=1119891120575119863 (120596 minus 2120587119873119879)
(minus1205962 + 2120577120596119899120596119894 + 1205962
119899)
(43)
Taking the inverse Fourier transforms to find the structureresponse in time gives
119909 (119905) = 119891
infin
sum
119873=1
119890119894(2120587119873119879)119905
(minus4120587211987321198792 + 119894120577120596119899 (4120587119873119879) + 1205962
119899) (44)
Equation (44) experiences a resonance when the denom-inator of any of the terms in the sum goes to zero that is when120596119899= 2120587119873119879 where 119873 is an integer larger than 0 This says
that the structure will experience a resonance when the natu-ral frequency is an integral multiple of the forcing frequency
16 Mathematical Problems in Engineering
as would be expected In the previous numerical simulationsthis condition has been observed in several cases Figures12(a) and 13(b) For instance the structural frequency ofFigure 13(b) is 2120587119879 = 212058705 = 1257 rads The naturalfrequency of the structure is radic119896119898 = radic10000001600 =
25 rads With119873 = 2 the structural frequency is close to thenatural frequency The result is what is seen in Figure 13(b)The two masses vibrate at the same frequency of 2120587119879 whichis half of the natural frequency The structural frequencyremains at that level which is lock-in and so does thefrequency of the free mass response
Thus the impact model successfully predicts the exis-tence of a lock-in condition and also predicts that it can occurwhen the frequency of the structure and freemass impacts aresynchronized together at a multiple of the natural frequencyof the structure The question of the precise conditionsnecessary for the frequency of the impacts to occur at amultiple of the natural frequency still remains open
6 Summary and Conclusions
This paper analyzes the vibrations caused by repeated impactsof a flow-driven freemass onto a second structure Conserva-tion of linearmomentum and the coefficient of restitution areused to characterize the nature of the impacts between thetwo masses This model is referred to as the impact modelTheproperties of the impactmodel were investigated throughnumerical simulations The mass of the free mass is one ofthe factors determining the vibration characteristics sinceit is related to the momentum of each impact This modelis proposed to explain the nature of ice-induced vibrationswithout requiring a microscopic model of the mechanism ofice failure The magnitudes of the vibrations are governedin part by the coefficient of restitution which when usedto explain IIVs encapsulates all ice properties mentionedin other IIV models In addition to the ice properties thecoefficient of restitution also governs the interaction betweenthe structure and the free mass The free mass is used torepresent the ice when modelling IIVs The last propertyof the impact model is that of the free mass area whichinfluences the dissipation of the system momenta throughthe drag force A free mass with a small area shows the samebehavior of a mass driven by lower flow velocity
Therefore through the impact model two factors whichcontrol the characteristics of IIVs have been proposed theice area and the coefficient of restitution By changing thesize of the contact area between the ice and structure thedriving force of the ice can be reduced as if the ice weredriven by lower flow velocities Furthermore by changingfactors related to the coefficient of restitution such as thesurface roughness or shape of the structure IIVs can bereduced to avoid the lock-in conditionwhich occurswhen thefrequencies of the structure and ice impacts are synchronizedat a multiple of the natural frequency of the structure Inaddition to the inclusion of the effect of flow the newapproach to ice forces through the impact model permitsthe answering of questions which originate from the weaktheoretical points of existing IIV models
Particular strengths of the model are that it qualitativelymatches observed results and in particular that a mechanismof ice failure is not required in order to perform simulationsand predictions It also proposes a simple mechanism toexplain the potentially destructive resonance lock-in condi-tions
References
[1] J P Dempsey ldquoResearch trends in ice mechanicsrdquo InternationalJournal of Solids and Structures vol 37 no 1-2 pp 131ndash153 2000
[2] K A Blenkarn ldquoMeasurement and analysis of ice forces oncook inlet structuresrdquo in Proceedings of the Offshore TechnologyConference (OTC rsquo70) pp 365ndash378 1970
[3] Y Toyama T Sensu M Minami and N Yashima ldquoModel testson ice-induced self-excited vibration of cylindrical structuresrdquoin Proceedings of the 7th International Conference on Port andOcean Engineering under Arctic Conditions vol 2 pp 834ndash8441983
[4] M Ranta and R Raty ldquoOn the analytic solution of ice-inducedvibrations in a marine pile structurerdquo in Proceedings of the 7thInternational Conference on Port and Ocean Engineering underArctic Conditions vol 2 pp 901ndash908 1983
[5] A Barker G Timco H Gravesen and P Voslashlund ldquoIce loadingon Danish wind turbinesmdashpart 1 dynamic model testsrdquo ColdRegions Science and Technology vol 41 no 1 pp 1ndash23 2005
[6] D S Sodhi ldquoIce-induced vibrations of structuresrdquo SpecialReport 89-5 International Association for Hydraulic ResearchWorking Group on Ice Forces 1989
[7] D S Sodhi and C E Morris ldquoCharacteristic frequency of forcevariations in continuous crushing of sheet ice against rigidcylindrical structuresrdquo Cold Regions Science and Technologyvol 12 no 1 pp 1ndash12 1986
[8] D S Sodhi C E Morris and G F N Cox ldquoDynamic analysisof failure modes on ice sheets encountering sloping structuresrdquoin Proceedings of the 6th International Conference on OffshoreMechanics and Arctic Engineering vol 4 pp 281ndash284 1987
[9] H R Peyton ldquoSea ice forcesrdquo in Proceedings of the Conferenceon Ice Pressure against Structures 1966
[10] C R Neill ldquoDynamic ice forces on piers and piles an assess-ment of design guidelines in the light of recent researchrdquoCanadian Journal of Civil Engineering vol 3 no 2 pp 305ndash3411976
[11] G Matlock W Dawkins and J Panak ldquoAnalytical model forice-structure interactionrdquo Journal of the Engineering MechanicsDivision pp 1083ndash11092 1971
[12] M M Maattanen ldquoIce-induced vibrations of structures self-excitationrdquo Special Report 89-5 International Association forHydraulic Research Working Group on Ice Forces 1989
[13] M M Maattanen ldquoNumerical simulation of ice-induced vibra-tions in offshore structuresrdquo in Proceedings of the 14th NordicSeminar on Computational Mechanics 2001 Keynote lecture
[14] J Xu and L Wang ldquoIce force oscillator model and its numericalsolutionsrdquo in Proceedings of the 7th International Conference onOffshore Machanics and Arctic Engineering pp 171ndash176 1988
[15] G TsinkerMarine Structures Engineering Chapman-Hall NewYork NY USA 1st edition 1995
[16] G Huang and P Liu ldquoA dynamic model for ice-inducedvibration of structuresrdquo Journal of OffshoreMechanics andArcticEngineering vol 131 no 1 Article ID 011501 6 pages 2009
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
8 Mathematical Problems in Engineering
002
0
minus002
minus004
minus006
minus008
minus01
minus0120 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with119872 = 800
0 50 100 150 200 250 300 350 400
02
0
minus02
minus04
minus06
minus08
minus1
minus12
minus14
Time (s)
Disp
lace
men
t (m
)
StructureIce
(b) Overall responses with119872= 2400
Figure 4 Responses with different ice masses 1
186 188 190 192 194 196 198 200
Disp
lace
men
t (m
)
StructureIce
Time (s)
0
minus2
minus4
minus6
times10minus4
(a) Responses with119872 = 800 from 185 to 200 seconds
0
minus0002
minus0004
minus0006
minus0008
186 188 190 192 194 196 198 200
StructureIce
Time (s)
Disp
lace
men
t (m
)
(b) Responses with119872= 2400 from 185 to 200 seconds
Figure 5 Responses with different ice masses 2
of restitution via the properties of the contacting area Forinstance the magnitudes of vibrations can be significantlyreduced by increasing the surface roughness of the structure
42 Quadratic Drag Force In the previous section themodelwas developed based on the linearized drag force whichassumes small relative velocity between the velocities of thefree mass and structure The linearized model howeveris definitely vulnerable to two factors namely when theinterval between impacts exceeds several seconds and whenthe velocity of the flow is no longer small Figure 10 clearlyshows the error between the models based on the linearized
drag and quadratic drag as time passes The displacement oflinearized equations (3) and (6) is compared with numer-ically solved equation (3) by using the built-in ordinarydifferential equation solver ofMathCadThe difference growsas time goes by but at high flow velocity the differencedecreases in Figure 10(b) This is because of the linearizationfactor | minus | which calibrates the difference between thelinear and quadratic models The solution however to thequadratic drag force needs to be developed for a more precisesimulation
A numerical solution is needed to solve (3) but insteadof using the built-in ordinary differential equation solver
Mathematical Problems in Engineering 9
StructureIce
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
Time (s)
1
0
minus1
minus2
minus3
minus4
times10minus4
(a) Responses with119872 = 800 from 385 to 400 seconds
StructureIce
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
0
minus2
minus4
minus6
times10minus4
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 6 Responses with different ice masses 3
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Overall responses with 119890 = 01
0
minus005
minus01
minus015
minus02
minus025
minus03
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400Time (s)
StructureIce
(b) Overall responses with 119890 = 05
Figure 7 Responses with different coefficients of restitution 1
of MathCad a simple numerical solution is developed for afaster algorithm From (3)
119872 = minus1
2120588119862119863119860 ( minus ) | minus | = minus119861 ( minus ) | minus | (31)
or
=119889
119889119905= minus
119861
119872( minus ) | minus | (32)
where
119861 =1
2120588119862119863119860 (33)
Using 119894 as a counter for the a number of time steps thedisplacement of (34) can be expressed as
119911119894= 119911119894minus1
+119889119911119894minus1
119889119905Δ119905 = 119911
119894minus1+ 119894minus1Δ119905 (34)
in which Δ119905 is previously defined as a time step z is given by
119894= 119894minus1
+119889119894minus1
119889119905Δ119905 =
119894minus1+ 119894minus1Δ119905 (35)
By substituting (32) into (35)
119894= 119894minus1
+ [minus119861
119872(119894minus1
minus )1003816100381610038161003816119894minus1 minus
1003816100381610038161003816] Δ119905(36)
10 Mathematical Problems in Engineering
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 2 4 6 8 10 12 14
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Responses with 119890 = 01 from 0 to 15 seconds
Disp
lace
men
t (m
)
Time (s)
StructureIce
46 48 50 52 54 56 58 60
2
0
minus2
minus4
minus6
minus8
times10minus4
(b) Responses with 119890 = 05 from 45 to 60 seconds
Figure 8 Responses with different coefficients of restitution 2
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus001386 388 390 392 394 396 398 400
Time (s)
Disp
lace
men
t (m
)
(a) Responses with 119890 = 01 from 385 to 400 seconds
StructureIce
Time (s)
Disp
lace
men
t (m
)
398 3985 399 3995 400
times10minus6
6
5
4
3
(b) Responses with 119890 = 05 from 398 to 400 seconds
Figure 9 Responses with different coefficients of restitution 3
Given the initial displacement and velocity of the free massthe displacement can be numerically solved by using (34)and (36) as simultaneous equationsThe simulation results of(34) and (36) are also plotted in Figure 10 At the high flowvelocity Figure 10(b) (34) and (36) predict the displacementclose enough to the one evaluated with the built-in ordinarydifferential equation solver In the algorithm of the numericalsimulation (34) and (36) replace (6) and (7)
Using the same parameters as those used in Figure 2but with a different flow velocity the responses of themodel with the quadratic drag force are compared withthe one with the linearized drag force in Figure 11 At high
flow velocities over 1ms the quadratic drag force modelapproaches steady-state vibration more quickly than thelinearized drag force model because at high flow velocity thequadratic drag induces more damping while it transfers moreforces than the linearized drag In Figure 12 the quadraticdrag force model exhibits the characteristic of the steady-state vibration while the linearized drag force model stilldisplays quasi-static vibration As velocities increase whichcorresponds to flowvelocities increasing vibrations approachthe steady state The linearized drag force model howeverdoes not show a steady-state vibration until 400 secondsin Figure 12 The linearization factor allows the linearized
Mathematical Problems in Engineering 11
0 100 200 300 400 500 600
100
80
60
40
20
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(a) Flow velocity 01ms
0 100 200 300 400 500 600
1000
800
600
400
200
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(b) Flow velocity 1ms
Figure 10 Comparison of the linearized and quadratic Drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
Disp
lace
men
t (m
)
(a) Overall responses with the quadratic drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
0
minus02
minus04
minus06
minus08
minus1
Disp
lace
men
t (m
)
(b) Overall responses with the linear drag
Figure 11 Responses with the quadratic and linear drags 1
drag force model to follow the quadratic drag force modelin the beginning but when the system is in a series of lowvelocity impacts the linearization factor causes a differenceThe linearized drag force model therefore is inappropriateto simulate the repeated impact model at high flow veloci-ties
Since the free mass area 119860 plays a similar role as theflow velocity in the model the model with wide area is alsosimulated with the quadratic drag force The influence of thedrag force on the response of the free mass increases as thesize of the area increases because the area is related to the dragforce by (1) 119891
119863= minus(12)120588119862
119863119860( minus )
2 Figure 13 is plottedwith the area 119860 = 2m2 the flow velocity = 01ms and
the rest of parameters are the same as the parameters usedin Figure 11 The flow velocity of Figure 13 is identical to thatof Figure 2 but the responses quickly approach the steady-state vibrations as if they are driven by high flow velocityFigure 13(b) also shows a similar vibration characteristic toFigure 12(a) because both flow velocity and area influence thedrag force in the impact model Therefore the area can beanother controlling factor of the vibrations Since the areais related to the size of the contacting area of the structurereducing the contacting area has a similar effect to that ofreducing the flow velocity
In the previous section the roles of the free mass andthe coefficient of restitution have been defined by changing
12 Mathematical Problems in EngineeringD
ispla
cem
ent (
m)
395 396 397 398 399 400Time (s)
StructureIce
8
6
4
2
0
times10minus4
(a) Responses with the quadratic drag from 395 to 400 secondsD
ispla
cem
ent (
m)
386 388 390 392 394 396 398 400Time (s)
StructureIce
0
minus0005
(b) Responses with the linear drag from 385 to 400 seconds
Figure 12 Responses with the quadratic and linear drags 2
005
0
minus005
minus01
minus015
minus02
minus025
minus03
minus035
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
(a) Overall responses with 119860 = 2
Disp
lace
men
t (m
)
395 396 397 398 399 400Time (s)
StructureIce
1
0
minus1
times10minus5
(b) Responses with 119860 = 2 from 395 to 400 seconds
Figure 13 Responses with 119860 = 2m2
relevant parameters To verify if those definitions still holdwith the quadratic drag force the impact model with thequadratic drag is plotted with change in free mass and thecoefficient of restitution Figures 14 15 and 16 are plottedon the same parameters used in Figures 2 4(a) and 4(b)respectively except that they are based on the quadratic dragforce The responses of Figures 14 15 and 16 prove thatthe defined role of the free mass holds with the quadraticdrag force The responses of the system with heavier freemass Figure 16(b) is still unstable while the system with thefree mass 119872 = 800 kg Figure 15(b) is in the steady-statevibration The quadratic drag force however induces more
drag force thus the system responses tend to be stabilizedfaster than those with the linearized drag forces
Figures 17 and 18 are plotted on the same conditions withFigures 7(a) and 7(b) respectively but with the quadraticdrag forceThe coefficient of restitution also governs themag-nitudes of responses in the impact model with the quadraticdrag force however the system with the quadratic drag forceis more sensitive to change in the coefficient of restitutionthan the one with the linearized drag force The system with119890 = 05 Figure 18(b) loses the momentum at approximately43 seconds under the influence of the quadratic drag forcewhile the system with the linearized drag force and 119890 = 05
Mathematical Problems in Engineering 13
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
01
0
minus01
minus02
minus03
minus04
minus05
minus06
minus07
(a) Overall responses with119872 = 1600
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
StructureIce
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 1600 from 385 to 400 seconds
Figure 14 Responses with119872 = 1600
0
minus001
minus002
minus003
minus004
minus005
minus006
minus0070 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall responses with119872 = 800
395 396 397 398 399 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
0
5
minus5
times10minus6
(b) Responses with119872 = 800 from 395 to 400 seconds
Figure 15 Responses with119872 = 800
Figure 8(b) loses the majority of its momentum at approxi-mately 54 seconds This is again because the quadratic dragforce dissipates energy more quickly than the linearized dragforceHowever it is demonstrated in Figures 17 and 18 that thecoefficient of restitution plays the same role in both quadraticand linearized drag forces
5 Resonance Conditions
One of the main objectives of the analysis aims to find aresonance condition (which is often referred to as lock-in
when investigating IIVs) The structural vibration frequencyincreases as the free mass velocity increases but when lock-in occurs the structural frequency remains at the naturalfrequency even with increasing free mass velocity [6] In theimpact model however the velocities of the free mass cannotbe the primary lock-in condition because the velocities of thefree mass are not fixed
One possible interpretation of modelling IIVs with theimpact model is that it considers the forces to be a seriesof discrete impulses as shown in Figure 19(a) while otherIIV models consider them as continuous periodic forces
14 Mathematical Problems in Engineering
0 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
(a) Overall responses with119872= 2400
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 16 Responses with119872 = 2400
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
(a) Overall responses with 119890 = 01
StructureIce
0 2 4 6 8 10 12 14
0002
0
minus0002
minus0004
minus0006
minus0008
Disp
lace
men
t (m
)
Time (s)
(b) Responses with 119890 = 01 from 0 to 15 seconds
Figure 17 Responses with 119890 = 01
In fact the impact IIV model proposes a way to calculate thestrength of each impulse and the times between impulses andthese depend on the momenta of the ice and structure aswell as the coefficient of restitution for the collision Undercertain conditions it is possible for the discrete impulsesof Figure 19(a) to approach the regular periodic series ofimpulses of Figure 19(b) which is suspected to cause the lock-in condition
This view of the impact IIV model can be analyzed byusing a Fourier transform [31] There are several definitions
of the Fourier transform (FT) that may be used and in thispaper the chosen version of the FT is given by
119865 (120596) = int
infin
minusinfin
119891 (119905) 119890minus119894120596119905
119889119905 (37)
The corresponding inverse Fourier transform is then given by
119891 (119905) =1
2120587int
infin
minusinfin
119865 (120596) 119890119894120596119905119889120596 (38)
Mathematical Problems in Engineering 15
005
0
minus005
minus01
minus015
minus02
minus0250 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with 119890 = 05
Time (s)
Disp
lace
men
t (m
)
StructureIce
36 38 40 42 44 46 48 50
4
2
0
minus2
minus4
times10minus4
(b) Responses with 119890 = 05 from 35 to 50 seconds
Figure 18 Responses with 119890 = 05
f(t)
fj
tj t
(a) Random impulses
f(t)
f
T 2T 3Tt
pT
(b) Uniform impulses
Figure 19 Ice forces as a series of impulses
The discrete ice forces of Figure 19(a) can be expressed asa series of impulses as
119891 (119905) =
infin
sum
119895=1
119891119895120575119863(119905 minus 119905119895) (39)
where 120575119863indicates the Dirac delta function and 119891
119895indicates
the strength of each impact If the sizes and intervals ofthe impulses are uniform as shown in Figure 19(b) (39) isrewritten as
119891 (119905) =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (40)
In reality the repeated impacts will not all have the samestrength or occur at a fixed interval however it is sufficient forthe situation of Figure 19(a) to approach that of Figure 19(b)for this analysis to be relevant
Equation (40) is often called a comb functionTheFouriertransform of the comb function is given [32] by
119865 (120596) = 2120587
infin
sum
119873=1
119891120575119863(120596 minus
2120587119873
119879) (41)
For a typical single-degree of freedom forced vibrationsproblem
+ 2120577120596119899 + 1205962
119899119909 =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (42)
Taking Fourier transforms of both sides and rearranging give
119883 (120596) = 2120587suminfin
119873=1119891120575119863 (120596 minus 2120587119873119879)
(minus1205962 + 2120577120596119899120596119894 + 1205962
119899)
(43)
Taking the inverse Fourier transforms to find the structureresponse in time gives
119909 (119905) = 119891
infin
sum
119873=1
119890119894(2120587119873119879)119905
(minus4120587211987321198792 + 119894120577120596119899 (4120587119873119879) + 1205962
119899) (44)
Equation (44) experiences a resonance when the denom-inator of any of the terms in the sum goes to zero that is when120596119899= 2120587119873119879 where 119873 is an integer larger than 0 This says
that the structure will experience a resonance when the natu-ral frequency is an integral multiple of the forcing frequency
16 Mathematical Problems in Engineering
as would be expected In the previous numerical simulationsthis condition has been observed in several cases Figures12(a) and 13(b) For instance the structural frequency ofFigure 13(b) is 2120587119879 = 212058705 = 1257 rads The naturalfrequency of the structure is radic119896119898 = radic10000001600 =
25 rads With119873 = 2 the structural frequency is close to thenatural frequency The result is what is seen in Figure 13(b)The two masses vibrate at the same frequency of 2120587119879 whichis half of the natural frequency The structural frequencyremains at that level which is lock-in and so does thefrequency of the free mass response
Thus the impact model successfully predicts the exis-tence of a lock-in condition and also predicts that it can occurwhen the frequency of the structure and freemass impacts aresynchronized together at a multiple of the natural frequencyof the structure The question of the precise conditionsnecessary for the frequency of the impacts to occur at amultiple of the natural frequency still remains open
6 Summary and Conclusions
This paper analyzes the vibrations caused by repeated impactsof a flow-driven freemass onto a second structure Conserva-tion of linearmomentum and the coefficient of restitution areused to characterize the nature of the impacts between thetwo masses This model is referred to as the impact modelTheproperties of the impactmodel were investigated throughnumerical simulations The mass of the free mass is one ofthe factors determining the vibration characteristics sinceit is related to the momentum of each impact This modelis proposed to explain the nature of ice-induced vibrationswithout requiring a microscopic model of the mechanism ofice failure The magnitudes of the vibrations are governedin part by the coefficient of restitution which when usedto explain IIVs encapsulates all ice properties mentionedin other IIV models In addition to the ice properties thecoefficient of restitution also governs the interaction betweenthe structure and the free mass The free mass is used torepresent the ice when modelling IIVs The last propertyof the impact model is that of the free mass area whichinfluences the dissipation of the system momenta throughthe drag force A free mass with a small area shows the samebehavior of a mass driven by lower flow velocity
Therefore through the impact model two factors whichcontrol the characteristics of IIVs have been proposed theice area and the coefficient of restitution By changing thesize of the contact area between the ice and structure thedriving force of the ice can be reduced as if the ice weredriven by lower flow velocities Furthermore by changingfactors related to the coefficient of restitution such as thesurface roughness or shape of the structure IIVs can bereduced to avoid the lock-in conditionwhich occurswhen thefrequencies of the structure and ice impacts are synchronizedat a multiple of the natural frequency of the structure Inaddition to the inclusion of the effect of flow the newapproach to ice forces through the impact model permitsthe answering of questions which originate from the weaktheoretical points of existing IIV models
Particular strengths of the model are that it qualitativelymatches observed results and in particular that a mechanismof ice failure is not required in order to perform simulationsand predictions It also proposes a simple mechanism toexplain the potentially destructive resonance lock-in condi-tions
References
[1] J P Dempsey ldquoResearch trends in ice mechanicsrdquo InternationalJournal of Solids and Structures vol 37 no 1-2 pp 131ndash153 2000
[2] K A Blenkarn ldquoMeasurement and analysis of ice forces oncook inlet structuresrdquo in Proceedings of the Offshore TechnologyConference (OTC rsquo70) pp 365ndash378 1970
[3] Y Toyama T Sensu M Minami and N Yashima ldquoModel testson ice-induced self-excited vibration of cylindrical structuresrdquoin Proceedings of the 7th International Conference on Port andOcean Engineering under Arctic Conditions vol 2 pp 834ndash8441983
[4] M Ranta and R Raty ldquoOn the analytic solution of ice-inducedvibrations in a marine pile structurerdquo in Proceedings of the 7thInternational Conference on Port and Ocean Engineering underArctic Conditions vol 2 pp 901ndash908 1983
[5] A Barker G Timco H Gravesen and P Voslashlund ldquoIce loadingon Danish wind turbinesmdashpart 1 dynamic model testsrdquo ColdRegions Science and Technology vol 41 no 1 pp 1ndash23 2005
[6] D S Sodhi ldquoIce-induced vibrations of structuresrdquo SpecialReport 89-5 International Association for Hydraulic ResearchWorking Group on Ice Forces 1989
[7] D S Sodhi and C E Morris ldquoCharacteristic frequency of forcevariations in continuous crushing of sheet ice against rigidcylindrical structuresrdquo Cold Regions Science and Technologyvol 12 no 1 pp 1ndash12 1986
[8] D S Sodhi C E Morris and G F N Cox ldquoDynamic analysisof failure modes on ice sheets encountering sloping structuresrdquoin Proceedings of the 6th International Conference on OffshoreMechanics and Arctic Engineering vol 4 pp 281ndash284 1987
[9] H R Peyton ldquoSea ice forcesrdquo in Proceedings of the Conferenceon Ice Pressure against Structures 1966
[10] C R Neill ldquoDynamic ice forces on piers and piles an assess-ment of design guidelines in the light of recent researchrdquoCanadian Journal of Civil Engineering vol 3 no 2 pp 305ndash3411976
[11] G Matlock W Dawkins and J Panak ldquoAnalytical model forice-structure interactionrdquo Journal of the Engineering MechanicsDivision pp 1083ndash11092 1971
[12] M M Maattanen ldquoIce-induced vibrations of structures self-excitationrdquo Special Report 89-5 International Association forHydraulic Research Working Group on Ice Forces 1989
[13] M M Maattanen ldquoNumerical simulation of ice-induced vibra-tions in offshore structuresrdquo in Proceedings of the 14th NordicSeminar on Computational Mechanics 2001 Keynote lecture
[14] J Xu and L Wang ldquoIce force oscillator model and its numericalsolutionsrdquo in Proceedings of the 7th International Conference onOffshore Machanics and Arctic Engineering pp 171ndash176 1988
[15] G TsinkerMarine Structures Engineering Chapman-Hall NewYork NY USA 1st edition 1995
[16] G Huang and P Liu ldquoA dynamic model for ice-inducedvibration of structuresrdquo Journal of OffshoreMechanics andArcticEngineering vol 131 no 1 Article ID 011501 6 pages 2009
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
Mathematical Problems in Engineering 9
StructureIce
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
Time (s)
1
0
minus1
minus2
minus3
minus4
times10minus4
(a) Responses with119872 = 800 from 385 to 400 seconds
StructureIce
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
0
minus2
minus4
minus6
times10minus4
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 6 Responses with different ice masses 3
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Overall responses with 119890 = 01
0
minus005
minus01
minus015
minus02
minus025
minus03
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400Time (s)
StructureIce
(b) Overall responses with 119890 = 05
Figure 7 Responses with different coefficients of restitution 1
of MathCad a simple numerical solution is developed for afaster algorithm From (3)
119872 = minus1
2120588119862119863119860 ( minus ) | minus | = minus119861 ( minus ) | minus | (31)
or
=119889
119889119905= minus
119861
119872( minus ) | minus | (32)
where
119861 =1
2120588119862119863119860 (33)
Using 119894 as a counter for the a number of time steps thedisplacement of (34) can be expressed as
119911119894= 119911119894minus1
+119889119911119894minus1
119889119905Δ119905 = 119911
119894minus1+ 119894minus1Δ119905 (34)
in which Δ119905 is previously defined as a time step z is given by
119894= 119894minus1
+119889119894minus1
119889119905Δ119905 =
119894minus1+ 119894minus1Δ119905 (35)
By substituting (32) into (35)
119894= 119894minus1
+ [minus119861
119872(119894minus1
minus )1003816100381610038161003816119894minus1 minus
1003816100381610038161003816] Δ119905(36)
10 Mathematical Problems in Engineering
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 2 4 6 8 10 12 14
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Responses with 119890 = 01 from 0 to 15 seconds
Disp
lace
men
t (m
)
Time (s)
StructureIce
46 48 50 52 54 56 58 60
2
0
minus2
minus4
minus6
minus8
times10minus4
(b) Responses with 119890 = 05 from 45 to 60 seconds
Figure 8 Responses with different coefficients of restitution 2
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus001386 388 390 392 394 396 398 400
Time (s)
Disp
lace
men
t (m
)
(a) Responses with 119890 = 01 from 385 to 400 seconds
StructureIce
Time (s)
Disp
lace
men
t (m
)
398 3985 399 3995 400
times10minus6
6
5
4
3
(b) Responses with 119890 = 05 from 398 to 400 seconds
Figure 9 Responses with different coefficients of restitution 3
Given the initial displacement and velocity of the free massthe displacement can be numerically solved by using (34)and (36) as simultaneous equationsThe simulation results of(34) and (36) are also plotted in Figure 10 At the high flowvelocity Figure 10(b) (34) and (36) predict the displacementclose enough to the one evaluated with the built-in ordinarydifferential equation solver In the algorithm of the numericalsimulation (34) and (36) replace (6) and (7)
Using the same parameters as those used in Figure 2but with a different flow velocity the responses of themodel with the quadratic drag force are compared withthe one with the linearized drag force in Figure 11 At high
flow velocities over 1ms the quadratic drag force modelapproaches steady-state vibration more quickly than thelinearized drag force model because at high flow velocity thequadratic drag induces more damping while it transfers moreforces than the linearized drag In Figure 12 the quadraticdrag force model exhibits the characteristic of the steady-state vibration while the linearized drag force model stilldisplays quasi-static vibration As velocities increase whichcorresponds to flowvelocities increasing vibrations approachthe steady state The linearized drag force model howeverdoes not show a steady-state vibration until 400 secondsin Figure 12 The linearization factor allows the linearized
Mathematical Problems in Engineering 11
0 100 200 300 400 500 600
100
80
60
40
20
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(a) Flow velocity 01ms
0 100 200 300 400 500 600
1000
800
600
400
200
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(b) Flow velocity 1ms
Figure 10 Comparison of the linearized and quadratic Drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
Disp
lace
men
t (m
)
(a) Overall responses with the quadratic drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
0
minus02
minus04
minus06
minus08
minus1
Disp
lace
men
t (m
)
(b) Overall responses with the linear drag
Figure 11 Responses with the quadratic and linear drags 1
drag force model to follow the quadratic drag force modelin the beginning but when the system is in a series of lowvelocity impacts the linearization factor causes a differenceThe linearized drag force model therefore is inappropriateto simulate the repeated impact model at high flow veloci-ties
Since the free mass area 119860 plays a similar role as theflow velocity in the model the model with wide area is alsosimulated with the quadratic drag force The influence of thedrag force on the response of the free mass increases as thesize of the area increases because the area is related to the dragforce by (1) 119891
119863= minus(12)120588119862
119863119860( minus )
2 Figure 13 is plottedwith the area 119860 = 2m2 the flow velocity = 01ms and
the rest of parameters are the same as the parameters usedin Figure 11 The flow velocity of Figure 13 is identical to thatof Figure 2 but the responses quickly approach the steady-state vibrations as if they are driven by high flow velocityFigure 13(b) also shows a similar vibration characteristic toFigure 12(a) because both flow velocity and area influence thedrag force in the impact model Therefore the area can beanother controlling factor of the vibrations Since the areais related to the size of the contacting area of the structurereducing the contacting area has a similar effect to that ofreducing the flow velocity
In the previous section the roles of the free mass andthe coefficient of restitution have been defined by changing
12 Mathematical Problems in EngineeringD
ispla
cem
ent (
m)
395 396 397 398 399 400Time (s)
StructureIce
8
6
4
2
0
times10minus4
(a) Responses with the quadratic drag from 395 to 400 secondsD
ispla
cem
ent (
m)
386 388 390 392 394 396 398 400Time (s)
StructureIce
0
minus0005
(b) Responses with the linear drag from 385 to 400 seconds
Figure 12 Responses with the quadratic and linear drags 2
005
0
minus005
minus01
minus015
minus02
minus025
minus03
minus035
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
(a) Overall responses with 119860 = 2
Disp
lace
men
t (m
)
395 396 397 398 399 400Time (s)
StructureIce
1
0
minus1
times10minus5
(b) Responses with 119860 = 2 from 395 to 400 seconds
Figure 13 Responses with 119860 = 2m2
relevant parameters To verify if those definitions still holdwith the quadratic drag force the impact model with thequadratic drag is plotted with change in free mass and thecoefficient of restitution Figures 14 15 and 16 are plottedon the same parameters used in Figures 2 4(a) and 4(b)respectively except that they are based on the quadratic dragforce The responses of Figures 14 15 and 16 prove thatthe defined role of the free mass holds with the quadraticdrag force The responses of the system with heavier freemass Figure 16(b) is still unstable while the system with thefree mass 119872 = 800 kg Figure 15(b) is in the steady-statevibration The quadratic drag force however induces more
drag force thus the system responses tend to be stabilizedfaster than those with the linearized drag forces
Figures 17 and 18 are plotted on the same conditions withFigures 7(a) and 7(b) respectively but with the quadraticdrag forceThe coefficient of restitution also governs themag-nitudes of responses in the impact model with the quadraticdrag force however the system with the quadratic drag forceis more sensitive to change in the coefficient of restitutionthan the one with the linearized drag force The system with119890 = 05 Figure 18(b) loses the momentum at approximately43 seconds under the influence of the quadratic drag forcewhile the system with the linearized drag force and 119890 = 05
Mathematical Problems in Engineering 13
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
01
0
minus01
minus02
minus03
minus04
minus05
minus06
minus07
(a) Overall responses with119872 = 1600
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
StructureIce
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 1600 from 385 to 400 seconds
Figure 14 Responses with119872 = 1600
0
minus001
minus002
minus003
minus004
minus005
minus006
minus0070 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall responses with119872 = 800
395 396 397 398 399 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
0
5
minus5
times10minus6
(b) Responses with119872 = 800 from 395 to 400 seconds
Figure 15 Responses with119872 = 800
Figure 8(b) loses the majority of its momentum at approxi-mately 54 seconds This is again because the quadratic dragforce dissipates energy more quickly than the linearized dragforceHowever it is demonstrated in Figures 17 and 18 that thecoefficient of restitution plays the same role in both quadraticand linearized drag forces
5 Resonance Conditions
One of the main objectives of the analysis aims to find aresonance condition (which is often referred to as lock-in
when investigating IIVs) The structural vibration frequencyincreases as the free mass velocity increases but when lock-in occurs the structural frequency remains at the naturalfrequency even with increasing free mass velocity [6] In theimpact model however the velocities of the free mass cannotbe the primary lock-in condition because the velocities of thefree mass are not fixed
One possible interpretation of modelling IIVs with theimpact model is that it considers the forces to be a seriesof discrete impulses as shown in Figure 19(a) while otherIIV models consider them as continuous periodic forces
14 Mathematical Problems in Engineering
0 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
(a) Overall responses with119872= 2400
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 16 Responses with119872 = 2400
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
(a) Overall responses with 119890 = 01
StructureIce
0 2 4 6 8 10 12 14
0002
0
minus0002
minus0004
minus0006
minus0008
Disp
lace
men
t (m
)
Time (s)
(b) Responses with 119890 = 01 from 0 to 15 seconds
Figure 17 Responses with 119890 = 01
In fact the impact IIV model proposes a way to calculate thestrength of each impulse and the times between impulses andthese depend on the momenta of the ice and structure aswell as the coefficient of restitution for the collision Undercertain conditions it is possible for the discrete impulsesof Figure 19(a) to approach the regular periodic series ofimpulses of Figure 19(b) which is suspected to cause the lock-in condition
This view of the impact IIV model can be analyzed byusing a Fourier transform [31] There are several definitions
of the Fourier transform (FT) that may be used and in thispaper the chosen version of the FT is given by
119865 (120596) = int
infin
minusinfin
119891 (119905) 119890minus119894120596119905
119889119905 (37)
The corresponding inverse Fourier transform is then given by
119891 (119905) =1
2120587int
infin
minusinfin
119865 (120596) 119890119894120596119905119889120596 (38)
Mathematical Problems in Engineering 15
005
0
minus005
minus01
minus015
minus02
minus0250 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with 119890 = 05
Time (s)
Disp
lace
men
t (m
)
StructureIce
36 38 40 42 44 46 48 50
4
2
0
minus2
minus4
times10minus4
(b) Responses with 119890 = 05 from 35 to 50 seconds
Figure 18 Responses with 119890 = 05
f(t)
fj
tj t
(a) Random impulses
f(t)
f
T 2T 3Tt
pT
(b) Uniform impulses
Figure 19 Ice forces as a series of impulses
The discrete ice forces of Figure 19(a) can be expressed asa series of impulses as
119891 (119905) =
infin
sum
119895=1
119891119895120575119863(119905 minus 119905119895) (39)
where 120575119863indicates the Dirac delta function and 119891
119895indicates
the strength of each impact If the sizes and intervals ofthe impulses are uniform as shown in Figure 19(b) (39) isrewritten as
119891 (119905) =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (40)
In reality the repeated impacts will not all have the samestrength or occur at a fixed interval however it is sufficient forthe situation of Figure 19(a) to approach that of Figure 19(b)for this analysis to be relevant
Equation (40) is often called a comb functionTheFouriertransform of the comb function is given [32] by
119865 (120596) = 2120587
infin
sum
119873=1
119891120575119863(120596 minus
2120587119873
119879) (41)
For a typical single-degree of freedom forced vibrationsproblem
+ 2120577120596119899 + 1205962
119899119909 =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (42)
Taking Fourier transforms of both sides and rearranging give
119883 (120596) = 2120587suminfin
119873=1119891120575119863 (120596 minus 2120587119873119879)
(minus1205962 + 2120577120596119899120596119894 + 1205962
119899)
(43)
Taking the inverse Fourier transforms to find the structureresponse in time gives
119909 (119905) = 119891
infin
sum
119873=1
119890119894(2120587119873119879)119905
(minus4120587211987321198792 + 119894120577120596119899 (4120587119873119879) + 1205962
119899) (44)
Equation (44) experiences a resonance when the denom-inator of any of the terms in the sum goes to zero that is when120596119899= 2120587119873119879 where 119873 is an integer larger than 0 This says
that the structure will experience a resonance when the natu-ral frequency is an integral multiple of the forcing frequency
16 Mathematical Problems in Engineering
as would be expected In the previous numerical simulationsthis condition has been observed in several cases Figures12(a) and 13(b) For instance the structural frequency ofFigure 13(b) is 2120587119879 = 212058705 = 1257 rads The naturalfrequency of the structure is radic119896119898 = radic10000001600 =
25 rads With119873 = 2 the structural frequency is close to thenatural frequency The result is what is seen in Figure 13(b)The two masses vibrate at the same frequency of 2120587119879 whichis half of the natural frequency The structural frequencyremains at that level which is lock-in and so does thefrequency of the free mass response
Thus the impact model successfully predicts the exis-tence of a lock-in condition and also predicts that it can occurwhen the frequency of the structure and freemass impacts aresynchronized together at a multiple of the natural frequencyof the structure The question of the precise conditionsnecessary for the frequency of the impacts to occur at amultiple of the natural frequency still remains open
6 Summary and Conclusions
This paper analyzes the vibrations caused by repeated impactsof a flow-driven freemass onto a second structure Conserva-tion of linearmomentum and the coefficient of restitution areused to characterize the nature of the impacts between thetwo masses This model is referred to as the impact modelTheproperties of the impactmodel were investigated throughnumerical simulations The mass of the free mass is one ofthe factors determining the vibration characteristics sinceit is related to the momentum of each impact This modelis proposed to explain the nature of ice-induced vibrationswithout requiring a microscopic model of the mechanism ofice failure The magnitudes of the vibrations are governedin part by the coefficient of restitution which when usedto explain IIVs encapsulates all ice properties mentionedin other IIV models In addition to the ice properties thecoefficient of restitution also governs the interaction betweenthe structure and the free mass The free mass is used torepresent the ice when modelling IIVs The last propertyof the impact model is that of the free mass area whichinfluences the dissipation of the system momenta throughthe drag force A free mass with a small area shows the samebehavior of a mass driven by lower flow velocity
Therefore through the impact model two factors whichcontrol the characteristics of IIVs have been proposed theice area and the coefficient of restitution By changing thesize of the contact area between the ice and structure thedriving force of the ice can be reduced as if the ice weredriven by lower flow velocities Furthermore by changingfactors related to the coefficient of restitution such as thesurface roughness or shape of the structure IIVs can bereduced to avoid the lock-in conditionwhich occurswhen thefrequencies of the structure and ice impacts are synchronizedat a multiple of the natural frequency of the structure Inaddition to the inclusion of the effect of flow the newapproach to ice forces through the impact model permitsthe answering of questions which originate from the weaktheoretical points of existing IIV models
Particular strengths of the model are that it qualitativelymatches observed results and in particular that a mechanismof ice failure is not required in order to perform simulationsand predictions It also proposes a simple mechanism toexplain the potentially destructive resonance lock-in condi-tions
References
[1] J P Dempsey ldquoResearch trends in ice mechanicsrdquo InternationalJournal of Solids and Structures vol 37 no 1-2 pp 131ndash153 2000
[2] K A Blenkarn ldquoMeasurement and analysis of ice forces oncook inlet structuresrdquo in Proceedings of the Offshore TechnologyConference (OTC rsquo70) pp 365ndash378 1970
[3] Y Toyama T Sensu M Minami and N Yashima ldquoModel testson ice-induced self-excited vibration of cylindrical structuresrdquoin Proceedings of the 7th International Conference on Port andOcean Engineering under Arctic Conditions vol 2 pp 834ndash8441983
[4] M Ranta and R Raty ldquoOn the analytic solution of ice-inducedvibrations in a marine pile structurerdquo in Proceedings of the 7thInternational Conference on Port and Ocean Engineering underArctic Conditions vol 2 pp 901ndash908 1983
[5] A Barker G Timco H Gravesen and P Voslashlund ldquoIce loadingon Danish wind turbinesmdashpart 1 dynamic model testsrdquo ColdRegions Science and Technology vol 41 no 1 pp 1ndash23 2005
[6] D S Sodhi ldquoIce-induced vibrations of structuresrdquo SpecialReport 89-5 International Association for Hydraulic ResearchWorking Group on Ice Forces 1989
[7] D S Sodhi and C E Morris ldquoCharacteristic frequency of forcevariations in continuous crushing of sheet ice against rigidcylindrical structuresrdquo Cold Regions Science and Technologyvol 12 no 1 pp 1ndash12 1986
[8] D S Sodhi C E Morris and G F N Cox ldquoDynamic analysisof failure modes on ice sheets encountering sloping structuresrdquoin Proceedings of the 6th International Conference on OffshoreMechanics and Arctic Engineering vol 4 pp 281ndash284 1987
[9] H R Peyton ldquoSea ice forcesrdquo in Proceedings of the Conferenceon Ice Pressure against Structures 1966
[10] C R Neill ldquoDynamic ice forces on piers and piles an assess-ment of design guidelines in the light of recent researchrdquoCanadian Journal of Civil Engineering vol 3 no 2 pp 305ndash3411976
[11] G Matlock W Dawkins and J Panak ldquoAnalytical model forice-structure interactionrdquo Journal of the Engineering MechanicsDivision pp 1083ndash11092 1971
[12] M M Maattanen ldquoIce-induced vibrations of structures self-excitationrdquo Special Report 89-5 International Association forHydraulic Research Working Group on Ice Forces 1989
[13] M M Maattanen ldquoNumerical simulation of ice-induced vibra-tions in offshore structuresrdquo in Proceedings of the 14th NordicSeminar on Computational Mechanics 2001 Keynote lecture
[14] J Xu and L Wang ldquoIce force oscillator model and its numericalsolutionsrdquo in Proceedings of the 7th International Conference onOffshore Machanics and Arctic Engineering pp 171ndash176 1988
[15] G TsinkerMarine Structures Engineering Chapman-Hall NewYork NY USA 1st edition 1995
[16] G Huang and P Liu ldquoA dynamic model for ice-inducedvibration of structuresrdquo Journal of OffshoreMechanics andArcticEngineering vol 131 no 1 Article ID 011501 6 pages 2009
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
10 Mathematical Problems in Engineering
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 2 4 6 8 10 12 14
Disp
lace
men
t (m
)
Time (s)
StructureIce
(a) Responses with 119890 = 01 from 0 to 15 seconds
Disp
lace
men
t (m
)
Time (s)
StructureIce
46 48 50 52 54 56 58 60
2
0
minus2
minus4
minus6
minus8
times10minus4
(b) Responses with 119890 = 05 from 45 to 60 seconds
Figure 8 Responses with different coefficients of restitution 2
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus001386 388 390 392 394 396 398 400
Time (s)
Disp
lace
men
t (m
)
(a) Responses with 119890 = 01 from 385 to 400 seconds
StructureIce
Time (s)
Disp
lace
men
t (m
)
398 3985 399 3995 400
times10minus6
6
5
4
3
(b) Responses with 119890 = 05 from 398 to 400 seconds
Figure 9 Responses with different coefficients of restitution 3
Given the initial displacement and velocity of the free massthe displacement can be numerically solved by using (34)and (36) as simultaneous equationsThe simulation results of(34) and (36) are also plotted in Figure 10 At the high flowvelocity Figure 10(b) (34) and (36) predict the displacementclose enough to the one evaluated with the built-in ordinarydifferential equation solver In the algorithm of the numericalsimulation (34) and (36) replace (6) and (7)
Using the same parameters as those used in Figure 2but with a different flow velocity the responses of themodel with the quadratic drag force are compared withthe one with the linearized drag force in Figure 11 At high
flow velocities over 1ms the quadratic drag force modelapproaches steady-state vibration more quickly than thelinearized drag force model because at high flow velocity thequadratic drag induces more damping while it transfers moreforces than the linearized drag In Figure 12 the quadraticdrag force model exhibits the characteristic of the steady-state vibration while the linearized drag force model stilldisplays quasi-static vibration As velocities increase whichcorresponds to flowvelocities increasing vibrations approachthe steady state The linearized drag force model howeverdoes not show a steady-state vibration until 400 secondsin Figure 12 The linearization factor allows the linearized
Mathematical Problems in Engineering 11
0 100 200 300 400 500 600
100
80
60
40
20
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(a) Flow velocity 01ms
0 100 200 300 400 500 600
1000
800
600
400
200
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(b) Flow velocity 1ms
Figure 10 Comparison of the linearized and quadratic Drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
Disp
lace
men
t (m
)
(a) Overall responses with the quadratic drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
0
minus02
minus04
minus06
minus08
minus1
Disp
lace
men
t (m
)
(b) Overall responses with the linear drag
Figure 11 Responses with the quadratic and linear drags 1
drag force model to follow the quadratic drag force modelin the beginning but when the system is in a series of lowvelocity impacts the linearization factor causes a differenceThe linearized drag force model therefore is inappropriateto simulate the repeated impact model at high flow veloci-ties
Since the free mass area 119860 plays a similar role as theflow velocity in the model the model with wide area is alsosimulated with the quadratic drag force The influence of thedrag force on the response of the free mass increases as thesize of the area increases because the area is related to the dragforce by (1) 119891
119863= minus(12)120588119862
119863119860( minus )
2 Figure 13 is plottedwith the area 119860 = 2m2 the flow velocity = 01ms and
the rest of parameters are the same as the parameters usedin Figure 11 The flow velocity of Figure 13 is identical to thatof Figure 2 but the responses quickly approach the steady-state vibrations as if they are driven by high flow velocityFigure 13(b) also shows a similar vibration characteristic toFigure 12(a) because both flow velocity and area influence thedrag force in the impact model Therefore the area can beanother controlling factor of the vibrations Since the areais related to the size of the contacting area of the structurereducing the contacting area has a similar effect to that ofreducing the flow velocity
In the previous section the roles of the free mass andthe coefficient of restitution have been defined by changing
12 Mathematical Problems in EngineeringD
ispla
cem
ent (
m)
395 396 397 398 399 400Time (s)
StructureIce
8
6
4
2
0
times10minus4
(a) Responses with the quadratic drag from 395 to 400 secondsD
ispla
cem
ent (
m)
386 388 390 392 394 396 398 400Time (s)
StructureIce
0
minus0005
(b) Responses with the linear drag from 385 to 400 seconds
Figure 12 Responses with the quadratic and linear drags 2
005
0
minus005
minus01
minus015
minus02
minus025
minus03
minus035
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
(a) Overall responses with 119860 = 2
Disp
lace
men
t (m
)
395 396 397 398 399 400Time (s)
StructureIce
1
0
minus1
times10minus5
(b) Responses with 119860 = 2 from 395 to 400 seconds
Figure 13 Responses with 119860 = 2m2
relevant parameters To verify if those definitions still holdwith the quadratic drag force the impact model with thequadratic drag is plotted with change in free mass and thecoefficient of restitution Figures 14 15 and 16 are plottedon the same parameters used in Figures 2 4(a) and 4(b)respectively except that they are based on the quadratic dragforce The responses of Figures 14 15 and 16 prove thatthe defined role of the free mass holds with the quadraticdrag force The responses of the system with heavier freemass Figure 16(b) is still unstable while the system with thefree mass 119872 = 800 kg Figure 15(b) is in the steady-statevibration The quadratic drag force however induces more
drag force thus the system responses tend to be stabilizedfaster than those with the linearized drag forces
Figures 17 and 18 are plotted on the same conditions withFigures 7(a) and 7(b) respectively but with the quadraticdrag forceThe coefficient of restitution also governs themag-nitudes of responses in the impact model with the quadraticdrag force however the system with the quadratic drag forceis more sensitive to change in the coefficient of restitutionthan the one with the linearized drag force The system with119890 = 05 Figure 18(b) loses the momentum at approximately43 seconds under the influence of the quadratic drag forcewhile the system with the linearized drag force and 119890 = 05
Mathematical Problems in Engineering 13
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
01
0
minus01
minus02
minus03
minus04
minus05
minus06
minus07
(a) Overall responses with119872 = 1600
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
StructureIce
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 1600 from 385 to 400 seconds
Figure 14 Responses with119872 = 1600
0
minus001
minus002
minus003
minus004
minus005
minus006
minus0070 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall responses with119872 = 800
395 396 397 398 399 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
0
5
minus5
times10minus6
(b) Responses with119872 = 800 from 395 to 400 seconds
Figure 15 Responses with119872 = 800
Figure 8(b) loses the majority of its momentum at approxi-mately 54 seconds This is again because the quadratic dragforce dissipates energy more quickly than the linearized dragforceHowever it is demonstrated in Figures 17 and 18 that thecoefficient of restitution plays the same role in both quadraticand linearized drag forces
5 Resonance Conditions
One of the main objectives of the analysis aims to find aresonance condition (which is often referred to as lock-in
when investigating IIVs) The structural vibration frequencyincreases as the free mass velocity increases but when lock-in occurs the structural frequency remains at the naturalfrequency even with increasing free mass velocity [6] In theimpact model however the velocities of the free mass cannotbe the primary lock-in condition because the velocities of thefree mass are not fixed
One possible interpretation of modelling IIVs with theimpact model is that it considers the forces to be a seriesof discrete impulses as shown in Figure 19(a) while otherIIV models consider them as continuous periodic forces
14 Mathematical Problems in Engineering
0 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
(a) Overall responses with119872= 2400
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 16 Responses with119872 = 2400
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
(a) Overall responses with 119890 = 01
StructureIce
0 2 4 6 8 10 12 14
0002
0
minus0002
minus0004
minus0006
minus0008
Disp
lace
men
t (m
)
Time (s)
(b) Responses with 119890 = 01 from 0 to 15 seconds
Figure 17 Responses with 119890 = 01
In fact the impact IIV model proposes a way to calculate thestrength of each impulse and the times between impulses andthese depend on the momenta of the ice and structure aswell as the coefficient of restitution for the collision Undercertain conditions it is possible for the discrete impulsesof Figure 19(a) to approach the regular periodic series ofimpulses of Figure 19(b) which is suspected to cause the lock-in condition
This view of the impact IIV model can be analyzed byusing a Fourier transform [31] There are several definitions
of the Fourier transform (FT) that may be used and in thispaper the chosen version of the FT is given by
119865 (120596) = int
infin
minusinfin
119891 (119905) 119890minus119894120596119905
119889119905 (37)
The corresponding inverse Fourier transform is then given by
119891 (119905) =1
2120587int
infin
minusinfin
119865 (120596) 119890119894120596119905119889120596 (38)
Mathematical Problems in Engineering 15
005
0
minus005
minus01
minus015
minus02
minus0250 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with 119890 = 05
Time (s)
Disp
lace
men
t (m
)
StructureIce
36 38 40 42 44 46 48 50
4
2
0
minus2
minus4
times10minus4
(b) Responses with 119890 = 05 from 35 to 50 seconds
Figure 18 Responses with 119890 = 05
f(t)
fj
tj t
(a) Random impulses
f(t)
f
T 2T 3Tt
pT
(b) Uniform impulses
Figure 19 Ice forces as a series of impulses
The discrete ice forces of Figure 19(a) can be expressed asa series of impulses as
119891 (119905) =
infin
sum
119895=1
119891119895120575119863(119905 minus 119905119895) (39)
where 120575119863indicates the Dirac delta function and 119891
119895indicates
the strength of each impact If the sizes and intervals ofthe impulses are uniform as shown in Figure 19(b) (39) isrewritten as
119891 (119905) =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (40)
In reality the repeated impacts will not all have the samestrength or occur at a fixed interval however it is sufficient forthe situation of Figure 19(a) to approach that of Figure 19(b)for this analysis to be relevant
Equation (40) is often called a comb functionTheFouriertransform of the comb function is given [32] by
119865 (120596) = 2120587
infin
sum
119873=1
119891120575119863(120596 minus
2120587119873
119879) (41)
For a typical single-degree of freedom forced vibrationsproblem
+ 2120577120596119899 + 1205962
119899119909 =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (42)
Taking Fourier transforms of both sides and rearranging give
119883 (120596) = 2120587suminfin
119873=1119891120575119863 (120596 minus 2120587119873119879)
(minus1205962 + 2120577120596119899120596119894 + 1205962
119899)
(43)
Taking the inverse Fourier transforms to find the structureresponse in time gives
119909 (119905) = 119891
infin
sum
119873=1
119890119894(2120587119873119879)119905
(minus4120587211987321198792 + 119894120577120596119899 (4120587119873119879) + 1205962
119899) (44)
Equation (44) experiences a resonance when the denom-inator of any of the terms in the sum goes to zero that is when120596119899= 2120587119873119879 where 119873 is an integer larger than 0 This says
that the structure will experience a resonance when the natu-ral frequency is an integral multiple of the forcing frequency
16 Mathematical Problems in Engineering
as would be expected In the previous numerical simulationsthis condition has been observed in several cases Figures12(a) and 13(b) For instance the structural frequency ofFigure 13(b) is 2120587119879 = 212058705 = 1257 rads The naturalfrequency of the structure is radic119896119898 = radic10000001600 =
25 rads With119873 = 2 the structural frequency is close to thenatural frequency The result is what is seen in Figure 13(b)The two masses vibrate at the same frequency of 2120587119879 whichis half of the natural frequency The structural frequencyremains at that level which is lock-in and so does thefrequency of the free mass response
Thus the impact model successfully predicts the exis-tence of a lock-in condition and also predicts that it can occurwhen the frequency of the structure and freemass impacts aresynchronized together at a multiple of the natural frequencyof the structure The question of the precise conditionsnecessary for the frequency of the impacts to occur at amultiple of the natural frequency still remains open
6 Summary and Conclusions
This paper analyzes the vibrations caused by repeated impactsof a flow-driven freemass onto a second structure Conserva-tion of linearmomentum and the coefficient of restitution areused to characterize the nature of the impacts between thetwo masses This model is referred to as the impact modelTheproperties of the impactmodel were investigated throughnumerical simulations The mass of the free mass is one ofthe factors determining the vibration characteristics sinceit is related to the momentum of each impact This modelis proposed to explain the nature of ice-induced vibrationswithout requiring a microscopic model of the mechanism ofice failure The magnitudes of the vibrations are governedin part by the coefficient of restitution which when usedto explain IIVs encapsulates all ice properties mentionedin other IIV models In addition to the ice properties thecoefficient of restitution also governs the interaction betweenthe structure and the free mass The free mass is used torepresent the ice when modelling IIVs The last propertyof the impact model is that of the free mass area whichinfluences the dissipation of the system momenta throughthe drag force A free mass with a small area shows the samebehavior of a mass driven by lower flow velocity
Therefore through the impact model two factors whichcontrol the characteristics of IIVs have been proposed theice area and the coefficient of restitution By changing thesize of the contact area between the ice and structure thedriving force of the ice can be reduced as if the ice weredriven by lower flow velocities Furthermore by changingfactors related to the coefficient of restitution such as thesurface roughness or shape of the structure IIVs can bereduced to avoid the lock-in conditionwhich occurswhen thefrequencies of the structure and ice impacts are synchronizedat a multiple of the natural frequency of the structure Inaddition to the inclusion of the effect of flow the newapproach to ice forces through the impact model permitsthe answering of questions which originate from the weaktheoretical points of existing IIV models
Particular strengths of the model are that it qualitativelymatches observed results and in particular that a mechanismof ice failure is not required in order to perform simulationsand predictions It also proposes a simple mechanism toexplain the potentially destructive resonance lock-in condi-tions
References
[1] J P Dempsey ldquoResearch trends in ice mechanicsrdquo InternationalJournal of Solids and Structures vol 37 no 1-2 pp 131ndash153 2000
[2] K A Blenkarn ldquoMeasurement and analysis of ice forces oncook inlet structuresrdquo in Proceedings of the Offshore TechnologyConference (OTC rsquo70) pp 365ndash378 1970
[3] Y Toyama T Sensu M Minami and N Yashima ldquoModel testson ice-induced self-excited vibration of cylindrical structuresrdquoin Proceedings of the 7th International Conference on Port andOcean Engineering under Arctic Conditions vol 2 pp 834ndash8441983
[4] M Ranta and R Raty ldquoOn the analytic solution of ice-inducedvibrations in a marine pile structurerdquo in Proceedings of the 7thInternational Conference on Port and Ocean Engineering underArctic Conditions vol 2 pp 901ndash908 1983
[5] A Barker G Timco H Gravesen and P Voslashlund ldquoIce loadingon Danish wind turbinesmdashpart 1 dynamic model testsrdquo ColdRegions Science and Technology vol 41 no 1 pp 1ndash23 2005
[6] D S Sodhi ldquoIce-induced vibrations of structuresrdquo SpecialReport 89-5 International Association for Hydraulic ResearchWorking Group on Ice Forces 1989
[7] D S Sodhi and C E Morris ldquoCharacteristic frequency of forcevariations in continuous crushing of sheet ice against rigidcylindrical structuresrdquo Cold Regions Science and Technologyvol 12 no 1 pp 1ndash12 1986
[8] D S Sodhi C E Morris and G F N Cox ldquoDynamic analysisof failure modes on ice sheets encountering sloping structuresrdquoin Proceedings of the 6th International Conference on OffshoreMechanics and Arctic Engineering vol 4 pp 281ndash284 1987
[9] H R Peyton ldquoSea ice forcesrdquo in Proceedings of the Conferenceon Ice Pressure against Structures 1966
[10] C R Neill ldquoDynamic ice forces on piers and piles an assess-ment of design guidelines in the light of recent researchrdquoCanadian Journal of Civil Engineering vol 3 no 2 pp 305ndash3411976
[11] G Matlock W Dawkins and J Panak ldquoAnalytical model forice-structure interactionrdquo Journal of the Engineering MechanicsDivision pp 1083ndash11092 1971
[12] M M Maattanen ldquoIce-induced vibrations of structures self-excitationrdquo Special Report 89-5 International Association forHydraulic Research Working Group on Ice Forces 1989
[13] M M Maattanen ldquoNumerical simulation of ice-induced vibra-tions in offshore structuresrdquo in Proceedings of the 14th NordicSeminar on Computational Mechanics 2001 Keynote lecture
[14] J Xu and L Wang ldquoIce force oscillator model and its numericalsolutionsrdquo in Proceedings of the 7th International Conference onOffshore Machanics and Arctic Engineering pp 171ndash176 1988
[15] G TsinkerMarine Structures Engineering Chapman-Hall NewYork NY USA 1st edition 1995
[16] G Huang and P Liu ldquoA dynamic model for ice-inducedvibration of structuresrdquo Journal of OffshoreMechanics andArcticEngineering vol 131 no 1 Article ID 011501 6 pages 2009
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
Mathematical Problems in Engineering 11
0 100 200 300 400 500 600
100
80
60
40
20
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(a) Flow velocity 01ms
0 100 200 300 400 500 600
1000
800
600
400
200
0
Time (s)
Disp
lace
men
t (m
)
OdeLinear dragQuadratic drag
(b) Flow velocity 1ms
Figure 10 Comparison of the linearized and quadratic Drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
Disp
lace
men
t (m
)
(a) Overall responses with the quadratic drag
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
0
minus02
minus04
minus06
minus08
minus1
Disp
lace
men
t (m
)
(b) Overall responses with the linear drag
Figure 11 Responses with the quadratic and linear drags 1
drag force model to follow the quadratic drag force modelin the beginning but when the system is in a series of lowvelocity impacts the linearization factor causes a differenceThe linearized drag force model therefore is inappropriateto simulate the repeated impact model at high flow veloci-ties
Since the free mass area 119860 plays a similar role as theflow velocity in the model the model with wide area is alsosimulated with the quadratic drag force The influence of thedrag force on the response of the free mass increases as thesize of the area increases because the area is related to the dragforce by (1) 119891
119863= minus(12)120588119862
119863119860( minus )
2 Figure 13 is plottedwith the area 119860 = 2m2 the flow velocity = 01ms and
the rest of parameters are the same as the parameters usedin Figure 11 The flow velocity of Figure 13 is identical to thatof Figure 2 but the responses quickly approach the steady-state vibrations as if they are driven by high flow velocityFigure 13(b) also shows a similar vibration characteristic toFigure 12(a) because both flow velocity and area influence thedrag force in the impact model Therefore the area can beanother controlling factor of the vibrations Since the areais related to the size of the contacting area of the structurereducing the contacting area has a similar effect to that ofreducing the flow velocity
In the previous section the roles of the free mass andthe coefficient of restitution have been defined by changing
12 Mathematical Problems in EngineeringD
ispla
cem
ent (
m)
395 396 397 398 399 400Time (s)
StructureIce
8
6
4
2
0
times10minus4
(a) Responses with the quadratic drag from 395 to 400 secondsD
ispla
cem
ent (
m)
386 388 390 392 394 396 398 400Time (s)
StructureIce
0
minus0005
(b) Responses with the linear drag from 385 to 400 seconds
Figure 12 Responses with the quadratic and linear drags 2
005
0
minus005
minus01
minus015
minus02
minus025
minus03
minus035
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
(a) Overall responses with 119860 = 2
Disp
lace
men
t (m
)
395 396 397 398 399 400Time (s)
StructureIce
1
0
minus1
times10minus5
(b) Responses with 119860 = 2 from 395 to 400 seconds
Figure 13 Responses with 119860 = 2m2
relevant parameters To verify if those definitions still holdwith the quadratic drag force the impact model with thequadratic drag is plotted with change in free mass and thecoefficient of restitution Figures 14 15 and 16 are plottedon the same parameters used in Figures 2 4(a) and 4(b)respectively except that they are based on the quadratic dragforce The responses of Figures 14 15 and 16 prove thatthe defined role of the free mass holds with the quadraticdrag force The responses of the system with heavier freemass Figure 16(b) is still unstable while the system with thefree mass 119872 = 800 kg Figure 15(b) is in the steady-statevibration The quadratic drag force however induces more
drag force thus the system responses tend to be stabilizedfaster than those with the linearized drag forces
Figures 17 and 18 are plotted on the same conditions withFigures 7(a) and 7(b) respectively but with the quadraticdrag forceThe coefficient of restitution also governs themag-nitudes of responses in the impact model with the quadraticdrag force however the system with the quadratic drag forceis more sensitive to change in the coefficient of restitutionthan the one with the linearized drag force The system with119890 = 05 Figure 18(b) loses the momentum at approximately43 seconds under the influence of the quadratic drag forcewhile the system with the linearized drag force and 119890 = 05
Mathematical Problems in Engineering 13
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
01
0
minus01
minus02
minus03
minus04
minus05
minus06
minus07
(a) Overall responses with119872 = 1600
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
StructureIce
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 1600 from 385 to 400 seconds
Figure 14 Responses with119872 = 1600
0
minus001
minus002
minus003
minus004
minus005
minus006
minus0070 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall responses with119872 = 800
395 396 397 398 399 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
0
5
minus5
times10minus6
(b) Responses with119872 = 800 from 395 to 400 seconds
Figure 15 Responses with119872 = 800
Figure 8(b) loses the majority of its momentum at approxi-mately 54 seconds This is again because the quadratic dragforce dissipates energy more quickly than the linearized dragforceHowever it is demonstrated in Figures 17 and 18 that thecoefficient of restitution plays the same role in both quadraticand linearized drag forces
5 Resonance Conditions
One of the main objectives of the analysis aims to find aresonance condition (which is often referred to as lock-in
when investigating IIVs) The structural vibration frequencyincreases as the free mass velocity increases but when lock-in occurs the structural frequency remains at the naturalfrequency even with increasing free mass velocity [6] In theimpact model however the velocities of the free mass cannotbe the primary lock-in condition because the velocities of thefree mass are not fixed
One possible interpretation of modelling IIVs with theimpact model is that it considers the forces to be a seriesof discrete impulses as shown in Figure 19(a) while otherIIV models consider them as continuous periodic forces
14 Mathematical Problems in Engineering
0 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
(a) Overall responses with119872= 2400
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 16 Responses with119872 = 2400
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
(a) Overall responses with 119890 = 01
StructureIce
0 2 4 6 8 10 12 14
0002
0
minus0002
minus0004
minus0006
minus0008
Disp
lace
men
t (m
)
Time (s)
(b) Responses with 119890 = 01 from 0 to 15 seconds
Figure 17 Responses with 119890 = 01
In fact the impact IIV model proposes a way to calculate thestrength of each impulse and the times between impulses andthese depend on the momenta of the ice and structure aswell as the coefficient of restitution for the collision Undercertain conditions it is possible for the discrete impulsesof Figure 19(a) to approach the regular periodic series ofimpulses of Figure 19(b) which is suspected to cause the lock-in condition
This view of the impact IIV model can be analyzed byusing a Fourier transform [31] There are several definitions
of the Fourier transform (FT) that may be used and in thispaper the chosen version of the FT is given by
119865 (120596) = int
infin
minusinfin
119891 (119905) 119890minus119894120596119905
119889119905 (37)
The corresponding inverse Fourier transform is then given by
119891 (119905) =1
2120587int
infin
minusinfin
119865 (120596) 119890119894120596119905119889120596 (38)
Mathematical Problems in Engineering 15
005
0
minus005
minus01
minus015
minus02
minus0250 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with 119890 = 05
Time (s)
Disp
lace
men
t (m
)
StructureIce
36 38 40 42 44 46 48 50
4
2
0
minus2
minus4
times10minus4
(b) Responses with 119890 = 05 from 35 to 50 seconds
Figure 18 Responses with 119890 = 05
f(t)
fj
tj t
(a) Random impulses
f(t)
f
T 2T 3Tt
pT
(b) Uniform impulses
Figure 19 Ice forces as a series of impulses
The discrete ice forces of Figure 19(a) can be expressed asa series of impulses as
119891 (119905) =
infin
sum
119895=1
119891119895120575119863(119905 minus 119905119895) (39)
where 120575119863indicates the Dirac delta function and 119891
119895indicates
the strength of each impact If the sizes and intervals ofthe impulses are uniform as shown in Figure 19(b) (39) isrewritten as
119891 (119905) =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (40)
In reality the repeated impacts will not all have the samestrength or occur at a fixed interval however it is sufficient forthe situation of Figure 19(a) to approach that of Figure 19(b)for this analysis to be relevant
Equation (40) is often called a comb functionTheFouriertransform of the comb function is given [32] by
119865 (120596) = 2120587
infin
sum
119873=1
119891120575119863(120596 minus
2120587119873
119879) (41)
For a typical single-degree of freedom forced vibrationsproblem
+ 2120577120596119899 + 1205962
119899119909 =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (42)
Taking Fourier transforms of both sides and rearranging give
119883 (120596) = 2120587suminfin
119873=1119891120575119863 (120596 minus 2120587119873119879)
(minus1205962 + 2120577120596119899120596119894 + 1205962
119899)
(43)
Taking the inverse Fourier transforms to find the structureresponse in time gives
119909 (119905) = 119891
infin
sum
119873=1
119890119894(2120587119873119879)119905
(minus4120587211987321198792 + 119894120577120596119899 (4120587119873119879) + 1205962
119899) (44)
Equation (44) experiences a resonance when the denom-inator of any of the terms in the sum goes to zero that is when120596119899= 2120587119873119879 where 119873 is an integer larger than 0 This says
that the structure will experience a resonance when the natu-ral frequency is an integral multiple of the forcing frequency
16 Mathematical Problems in Engineering
as would be expected In the previous numerical simulationsthis condition has been observed in several cases Figures12(a) and 13(b) For instance the structural frequency ofFigure 13(b) is 2120587119879 = 212058705 = 1257 rads The naturalfrequency of the structure is radic119896119898 = radic10000001600 =
25 rads With119873 = 2 the structural frequency is close to thenatural frequency The result is what is seen in Figure 13(b)The two masses vibrate at the same frequency of 2120587119879 whichis half of the natural frequency The structural frequencyremains at that level which is lock-in and so does thefrequency of the free mass response
Thus the impact model successfully predicts the exis-tence of a lock-in condition and also predicts that it can occurwhen the frequency of the structure and freemass impacts aresynchronized together at a multiple of the natural frequencyof the structure The question of the precise conditionsnecessary for the frequency of the impacts to occur at amultiple of the natural frequency still remains open
6 Summary and Conclusions
This paper analyzes the vibrations caused by repeated impactsof a flow-driven freemass onto a second structure Conserva-tion of linearmomentum and the coefficient of restitution areused to characterize the nature of the impacts between thetwo masses This model is referred to as the impact modelTheproperties of the impactmodel were investigated throughnumerical simulations The mass of the free mass is one ofthe factors determining the vibration characteristics sinceit is related to the momentum of each impact This modelis proposed to explain the nature of ice-induced vibrationswithout requiring a microscopic model of the mechanism ofice failure The magnitudes of the vibrations are governedin part by the coefficient of restitution which when usedto explain IIVs encapsulates all ice properties mentionedin other IIV models In addition to the ice properties thecoefficient of restitution also governs the interaction betweenthe structure and the free mass The free mass is used torepresent the ice when modelling IIVs The last propertyof the impact model is that of the free mass area whichinfluences the dissipation of the system momenta throughthe drag force A free mass with a small area shows the samebehavior of a mass driven by lower flow velocity
Therefore through the impact model two factors whichcontrol the characteristics of IIVs have been proposed theice area and the coefficient of restitution By changing thesize of the contact area between the ice and structure thedriving force of the ice can be reduced as if the ice weredriven by lower flow velocities Furthermore by changingfactors related to the coefficient of restitution such as thesurface roughness or shape of the structure IIVs can bereduced to avoid the lock-in conditionwhich occurswhen thefrequencies of the structure and ice impacts are synchronizedat a multiple of the natural frequency of the structure Inaddition to the inclusion of the effect of flow the newapproach to ice forces through the impact model permitsthe answering of questions which originate from the weaktheoretical points of existing IIV models
Particular strengths of the model are that it qualitativelymatches observed results and in particular that a mechanismof ice failure is not required in order to perform simulationsand predictions It also proposes a simple mechanism toexplain the potentially destructive resonance lock-in condi-tions
References
[1] J P Dempsey ldquoResearch trends in ice mechanicsrdquo InternationalJournal of Solids and Structures vol 37 no 1-2 pp 131ndash153 2000
[2] K A Blenkarn ldquoMeasurement and analysis of ice forces oncook inlet structuresrdquo in Proceedings of the Offshore TechnologyConference (OTC rsquo70) pp 365ndash378 1970
[3] Y Toyama T Sensu M Minami and N Yashima ldquoModel testson ice-induced self-excited vibration of cylindrical structuresrdquoin Proceedings of the 7th International Conference on Port andOcean Engineering under Arctic Conditions vol 2 pp 834ndash8441983
[4] M Ranta and R Raty ldquoOn the analytic solution of ice-inducedvibrations in a marine pile structurerdquo in Proceedings of the 7thInternational Conference on Port and Ocean Engineering underArctic Conditions vol 2 pp 901ndash908 1983
[5] A Barker G Timco H Gravesen and P Voslashlund ldquoIce loadingon Danish wind turbinesmdashpart 1 dynamic model testsrdquo ColdRegions Science and Technology vol 41 no 1 pp 1ndash23 2005
[6] D S Sodhi ldquoIce-induced vibrations of structuresrdquo SpecialReport 89-5 International Association for Hydraulic ResearchWorking Group on Ice Forces 1989
[7] D S Sodhi and C E Morris ldquoCharacteristic frequency of forcevariations in continuous crushing of sheet ice against rigidcylindrical structuresrdquo Cold Regions Science and Technologyvol 12 no 1 pp 1ndash12 1986
[8] D S Sodhi C E Morris and G F N Cox ldquoDynamic analysisof failure modes on ice sheets encountering sloping structuresrdquoin Proceedings of the 6th International Conference on OffshoreMechanics and Arctic Engineering vol 4 pp 281ndash284 1987
[9] H R Peyton ldquoSea ice forcesrdquo in Proceedings of the Conferenceon Ice Pressure against Structures 1966
[10] C R Neill ldquoDynamic ice forces on piers and piles an assess-ment of design guidelines in the light of recent researchrdquoCanadian Journal of Civil Engineering vol 3 no 2 pp 305ndash3411976
[11] G Matlock W Dawkins and J Panak ldquoAnalytical model forice-structure interactionrdquo Journal of the Engineering MechanicsDivision pp 1083ndash11092 1971
[12] M M Maattanen ldquoIce-induced vibrations of structures self-excitationrdquo Special Report 89-5 International Association forHydraulic Research Working Group on Ice Forces 1989
[13] M M Maattanen ldquoNumerical simulation of ice-induced vibra-tions in offshore structuresrdquo in Proceedings of the 14th NordicSeminar on Computational Mechanics 2001 Keynote lecture
[14] J Xu and L Wang ldquoIce force oscillator model and its numericalsolutionsrdquo in Proceedings of the 7th International Conference onOffshore Machanics and Arctic Engineering pp 171ndash176 1988
[15] G TsinkerMarine Structures Engineering Chapman-Hall NewYork NY USA 1st edition 1995
[16] G Huang and P Liu ldquoA dynamic model for ice-inducedvibration of structuresrdquo Journal of OffshoreMechanics andArcticEngineering vol 131 no 1 Article ID 011501 6 pages 2009
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
12 Mathematical Problems in EngineeringD
ispla
cem
ent (
m)
395 396 397 398 399 400Time (s)
StructureIce
8
6
4
2
0
times10minus4
(a) Responses with the quadratic drag from 395 to 400 secondsD
ispla
cem
ent (
m)
386 388 390 392 394 396 398 400Time (s)
StructureIce
0
minus0005
(b) Responses with the linear drag from 385 to 400 seconds
Figure 12 Responses with the quadratic and linear drags 2
005
0
minus005
minus01
minus015
minus02
minus025
minus03
minus035
Disp
lace
men
t (m
)
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
(a) Overall responses with 119860 = 2
Disp
lace
men
t (m
)
395 396 397 398 399 400Time (s)
StructureIce
1
0
minus1
times10minus5
(b) Responses with 119860 = 2 from 395 to 400 seconds
Figure 13 Responses with 119860 = 2m2
relevant parameters To verify if those definitions still holdwith the quadratic drag force the impact model with thequadratic drag is plotted with change in free mass and thecoefficient of restitution Figures 14 15 and 16 are plottedon the same parameters used in Figures 2 4(a) and 4(b)respectively except that they are based on the quadratic dragforce The responses of Figures 14 15 and 16 prove thatthe defined role of the free mass holds with the quadraticdrag force The responses of the system with heavier freemass Figure 16(b) is still unstable while the system with thefree mass 119872 = 800 kg Figure 15(b) is in the steady-statevibration The quadratic drag force however induces more
drag force thus the system responses tend to be stabilizedfaster than those with the linearized drag forces
Figures 17 and 18 are plotted on the same conditions withFigures 7(a) and 7(b) respectively but with the quadraticdrag forceThe coefficient of restitution also governs themag-nitudes of responses in the impact model with the quadraticdrag force however the system with the quadratic drag forceis more sensitive to change in the coefficient of restitutionthan the one with the linearized drag force The system with119890 = 05 Figure 18(b) loses the momentum at approximately43 seconds under the influence of the quadratic drag forcewhile the system with the linearized drag force and 119890 = 05
Mathematical Problems in Engineering 13
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
01
0
minus01
minus02
minus03
minus04
minus05
minus06
minus07
(a) Overall responses with119872 = 1600
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
StructureIce
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 1600 from 385 to 400 seconds
Figure 14 Responses with119872 = 1600
0
minus001
minus002
minus003
minus004
minus005
minus006
minus0070 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall responses with119872 = 800
395 396 397 398 399 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
0
5
minus5
times10minus6
(b) Responses with119872 = 800 from 395 to 400 seconds
Figure 15 Responses with119872 = 800
Figure 8(b) loses the majority of its momentum at approxi-mately 54 seconds This is again because the quadratic dragforce dissipates energy more quickly than the linearized dragforceHowever it is demonstrated in Figures 17 and 18 that thecoefficient of restitution plays the same role in both quadraticand linearized drag forces
5 Resonance Conditions
One of the main objectives of the analysis aims to find aresonance condition (which is often referred to as lock-in
when investigating IIVs) The structural vibration frequencyincreases as the free mass velocity increases but when lock-in occurs the structural frequency remains at the naturalfrequency even with increasing free mass velocity [6] In theimpact model however the velocities of the free mass cannotbe the primary lock-in condition because the velocities of thefree mass are not fixed
One possible interpretation of modelling IIVs with theimpact model is that it considers the forces to be a seriesof discrete impulses as shown in Figure 19(a) while otherIIV models consider them as continuous periodic forces
14 Mathematical Problems in Engineering
0 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
(a) Overall responses with119872= 2400
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 16 Responses with119872 = 2400
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
(a) Overall responses with 119890 = 01
StructureIce
0 2 4 6 8 10 12 14
0002
0
minus0002
minus0004
minus0006
minus0008
Disp
lace
men
t (m
)
Time (s)
(b) Responses with 119890 = 01 from 0 to 15 seconds
Figure 17 Responses with 119890 = 01
In fact the impact IIV model proposes a way to calculate thestrength of each impulse and the times between impulses andthese depend on the momenta of the ice and structure aswell as the coefficient of restitution for the collision Undercertain conditions it is possible for the discrete impulsesof Figure 19(a) to approach the regular periodic series ofimpulses of Figure 19(b) which is suspected to cause the lock-in condition
This view of the impact IIV model can be analyzed byusing a Fourier transform [31] There are several definitions
of the Fourier transform (FT) that may be used and in thispaper the chosen version of the FT is given by
119865 (120596) = int
infin
minusinfin
119891 (119905) 119890minus119894120596119905
119889119905 (37)
The corresponding inverse Fourier transform is then given by
119891 (119905) =1
2120587int
infin
minusinfin
119865 (120596) 119890119894120596119905119889120596 (38)
Mathematical Problems in Engineering 15
005
0
minus005
minus01
minus015
minus02
minus0250 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with 119890 = 05
Time (s)
Disp
lace
men
t (m
)
StructureIce
36 38 40 42 44 46 48 50
4
2
0
minus2
minus4
times10minus4
(b) Responses with 119890 = 05 from 35 to 50 seconds
Figure 18 Responses with 119890 = 05
f(t)
fj
tj t
(a) Random impulses
f(t)
f
T 2T 3Tt
pT
(b) Uniform impulses
Figure 19 Ice forces as a series of impulses
The discrete ice forces of Figure 19(a) can be expressed asa series of impulses as
119891 (119905) =
infin
sum
119895=1
119891119895120575119863(119905 minus 119905119895) (39)
where 120575119863indicates the Dirac delta function and 119891
119895indicates
the strength of each impact If the sizes and intervals ofthe impulses are uniform as shown in Figure 19(b) (39) isrewritten as
119891 (119905) =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (40)
In reality the repeated impacts will not all have the samestrength or occur at a fixed interval however it is sufficient forthe situation of Figure 19(a) to approach that of Figure 19(b)for this analysis to be relevant
Equation (40) is often called a comb functionTheFouriertransform of the comb function is given [32] by
119865 (120596) = 2120587
infin
sum
119873=1
119891120575119863(120596 minus
2120587119873
119879) (41)
For a typical single-degree of freedom forced vibrationsproblem
+ 2120577120596119899 + 1205962
119899119909 =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (42)
Taking Fourier transforms of both sides and rearranging give
119883 (120596) = 2120587suminfin
119873=1119891120575119863 (120596 minus 2120587119873119879)
(minus1205962 + 2120577120596119899120596119894 + 1205962
119899)
(43)
Taking the inverse Fourier transforms to find the structureresponse in time gives
119909 (119905) = 119891
infin
sum
119873=1
119890119894(2120587119873119879)119905
(minus4120587211987321198792 + 119894120577120596119899 (4120587119873119879) + 1205962
119899) (44)
Equation (44) experiences a resonance when the denom-inator of any of the terms in the sum goes to zero that is when120596119899= 2120587119873119879 where 119873 is an integer larger than 0 This says
that the structure will experience a resonance when the natu-ral frequency is an integral multiple of the forcing frequency
16 Mathematical Problems in Engineering
as would be expected In the previous numerical simulationsthis condition has been observed in several cases Figures12(a) and 13(b) For instance the structural frequency ofFigure 13(b) is 2120587119879 = 212058705 = 1257 rads The naturalfrequency of the structure is radic119896119898 = radic10000001600 =
25 rads With119873 = 2 the structural frequency is close to thenatural frequency The result is what is seen in Figure 13(b)The two masses vibrate at the same frequency of 2120587119879 whichis half of the natural frequency The structural frequencyremains at that level which is lock-in and so does thefrequency of the free mass response
Thus the impact model successfully predicts the exis-tence of a lock-in condition and also predicts that it can occurwhen the frequency of the structure and freemass impacts aresynchronized together at a multiple of the natural frequencyof the structure The question of the precise conditionsnecessary for the frequency of the impacts to occur at amultiple of the natural frequency still remains open
6 Summary and Conclusions
This paper analyzes the vibrations caused by repeated impactsof a flow-driven freemass onto a second structure Conserva-tion of linearmomentum and the coefficient of restitution areused to characterize the nature of the impacts between thetwo masses This model is referred to as the impact modelTheproperties of the impactmodel were investigated throughnumerical simulations The mass of the free mass is one ofthe factors determining the vibration characteristics sinceit is related to the momentum of each impact This modelis proposed to explain the nature of ice-induced vibrationswithout requiring a microscopic model of the mechanism ofice failure The magnitudes of the vibrations are governedin part by the coefficient of restitution which when usedto explain IIVs encapsulates all ice properties mentionedin other IIV models In addition to the ice properties thecoefficient of restitution also governs the interaction betweenthe structure and the free mass The free mass is used torepresent the ice when modelling IIVs The last propertyof the impact model is that of the free mass area whichinfluences the dissipation of the system momenta throughthe drag force A free mass with a small area shows the samebehavior of a mass driven by lower flow velocity
Therefore through the impact model two factors whichcontrol the characteristics of IIVs have been proposed theice area and the coefficient of restitution By changing thesize of the contact area between the ice and structure thedriving force of the ice can be reduced as if the ice weredriven by lower flow velocities Furthermore by changingfactors related to the coefficient of restitution such as thesurface roughness or shape of the structure IIVs can bereduced to avoid the lock-in conditionwhich occurswhen thefrequencies of the structure and ice impacts are synchronizedat a multiple of the natural frequency of the structure Inaddition to the inclusion of the effect of flow the newapproach to ice forces through the impact model permitsthe answering of questions which originate from the weaktheoretical points of existing IIV models
Particular strengths of the model are that it qualitativelymatches observed results and in particular that a mechanismof ice failure is not required in order to perform simulationsand predictions It also proposes a simple mechanism toexplain the potentially destructive resonance lock-in condi-tions
References
[1] J P Dempsey ldquoResearch trends in ice mechanicsrdquo InternationalJournal of Solids and Structures vol 37 no 1-2 pp 131ndash153 2000
[2] K A Blenkarn ldquoMeasurement and analysis of ice forces oncook inlet structuresrdquo in Proceedings of the Offshore TechnologyConference (OTC rsquo70) pp 365ndash378 1970
[3] Y Toyama T Sensu M Minami and N Yashima ldquoModel testson ice-induced self-excited vibration of cylindrical structuresrdquoin Proceedings of the 7th International Conference on Port andOcean Engineering under Arctic Conditions vol 2 pp 834ndash8441983
[4] M Ranta and R Raty ldquoOn the analytic solution of ice-inducedvibrations in a marine pile structurerdquo in Proceedings of the 7thInternational Conference on Port and Ocean Engineering underArctic Conditions vol 2 pp 901ndash908 1983
[5] A Barker G Timco H Gravesen and P Voslashlund ldquoIce loadingon Danish wind turbinesmdashpart 1 dynamic model testsrdquo ColdRegions Science and Technology vol 41 no 1 pp 1ndash23 2005
[6] D S Sodhi ldquoIce-induced vibrations of structuresrdquo SpecialReport 89-5 International Association for Hydraulic ResearchWorking Group on Ice Forces 1989
[7] D S Sodhi and C E Morris ldquoCharacteristic frequency of forcevariations in continuous crushing of sheet ice against rigidcylindrical structuresrdquo Cold Regions Science and Technologyvol 12 no 1 pp 1ndash12 1986
[8] D S Sodhi C E Morris and G F N Cox ldquoDynamic analysisof failure modes on ice sheets encountering sloping structuresrdquoin Proceedings of the 6th International Conference on OffshoreMechanics and Arctic Engineering vol 4 pp 281ndash284 1987
[9] H R Peyton ldquoSea ice forcesrdquo in Proceedings of the Conferenceon Ice Pressure against Structures 1966
[10] C R Neill ldquoDynamic ice forces on piers and piles an assess-ment of design guidelines in the light of recent researchrdquoCanadian Journal of Civil Engineering vol 3 no 2 pp 305ndash3411976
[11] G Matlock W Dawkins and J Panak ldquoAnalytical model forice-structure interactionrdquo Journal of the Engineering MechanicsDivision pp 1083ndash11092 1971
[12] M M Maattanen ldquoIce-induced vibrations of structures self-excitationrdquo Special Report 89-5 International Association forHydraulic Research Working Group on Ice Forces 1989
[13] M M Maattanen ldquoNumerical simulation of ice-induced vibra-tions in offshore structuresrdquo in Proceedings of the 14th NordicSeminar on Computational Mechanics 2001 Keynote lecture
[14] J Xu and L Wang ldquoIce force oscillator model and its numericalsolutionsrdquo in Proceedings of the 7th International Conference onOffshore Machanics and Arctic Engineering pp 171ndash176 1988
[15] G TsinkerMarine Structures Engineering Chapman-Hall NewYork NY USA 1st edition 1995
[16] G Huang and P Liu ldquoA dynamic model for ice-inducedvibration of structuresrdquo Journal of OffshoreMechanics andArcticEngineering vol 131 no 1 Article ID 011501 6 pages 2009
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
Mathematical Problems in Engineering 13
0 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
01
0
minus01
minus02
minus03
minus04
minus05
minus06
minus07
(a) Overall responses with119872 = 1600
Disp
lace
men
t (m
)
386 388 390 392 394 396 398 400Time (s)
StructureIce
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 1600 from 385 to 400 seconds
Figure 14 Responses with119872 = 1600
0
minus001
minus002
minus003
minus004
minus005
minus006
minus0070 50 100 150 200 250 300 350 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
(a) Overall responses with119872 = 800
395 396 397 398 399 400
StructureIce
Time (s)
Disp
lace
men
t (m
)
0
5
minus5
times10minus6
(b) Responses with119872 = 800 from 395 to 400 seconds
Figure 15 Responses with119872 = 800
Figure 8(b) loses the majority of its momentum at approxi-mately 54 seconds This is again because the quadratic dragforce dissipates energy more quickly than the linearized dragforceHowever it is demonstrated in Figures 17 and 18 that thecoefficient of restitution plays the same role in both quadraticand linearized drag forces
5 Resonance Conditions
One of the main objectives of the analysis aims to find aresonance condition (which is often referred to as lock-in
when investigating IIVs) The structural vibration frequencyincreases as the free mass velocity increases but when lock-in occurs the structural frequency remains at the naturalfrequency even with increasing free mass velocity [6] In theimpact model however the velocities of the free mass cannotbe the primary lock-in condition because the velocities of thefree mass are not fixed
One possible interpretation of modelling IIVs with theimpact model is that it considers the forces to be a seriesof discrete impulses as shown in Figure 19(a) while otherIIV models consider them as continuous periodic forces
14 Mathematical Problems in Engineering
0 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
(a) Overall responses with119872= 2400
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 16 Responses with119872 = 2400
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
(a) Overall responses with 119890 = 01
StructureIce
0 2 4 6 8 10 12 14
0002
0
minus0002
minus0004
minus0006
minus0008
Disp
lace
men
t (m
)
Time (s)
(b) Responses with 119890 = 01 from 0 to 15 seconds
Figure 17 Responses with 119890 = 01
In fact the impact IIV model proposes a way to calculate thestrength of each impulse and the times between impulses andthese depend on the momenta of the ice and structure aswell as the coefficient of restitution for the collision Undercertain conditions it is possible for the discrete impulsesof Figure 19(a) to approach the regular periodic series ofimpulses of Figure 19(b) which is suspected to cause the lock-in condition
This view of the impact IIV model can be analyzed byusing a Fourier transform [31] There are several definitions
of the Fourier transform (FT) that may be used and in thispaper the chosen version of the FT is given by
119865 (120596) = int
infin
minusinfin
119891 (119905) 119890minus119894120596119905
119889119905 (37)
The corresponding inverse Fourier transform is then given by
119891 (119905) =1
2120587int
infin
minusinfin
119865 (120596) 119890119894120596119905119889120596 (38)
Mathematical Problems in Engineering 15
005
0
minus005
minus01
minus015
minus02
minus0250 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with 119890 = 05
Time (s)
Disp
lace
men
t (m
)
StructureIce
36 38 40 42 44 46 48 50
4
2
0
minus2
minus4
times10minus4
(b) Responses with 119890 = 05 from 35 to 50 seconds
Figure 18 Responses with 119890 = 05
f(t)
fj
tj t
(a) Random impulses
f(t)
f
T 2T 3Tt
pT
(b) Uniform impulses
Figure 19 Ice forces as a series of impulses
The discrete ice forces of Figure 19(a) can be expressed asa series of impulses as
119891 (119905) =
infin
sum
119895=1
119891119895120575119863(119905 minus 119905119895) (39)
where 120575119863indicates the Dirac delta function and 119891
119895indicates
the strength of each impact If the sizes and intervals ofthe impulses are uniform as shown in Figure 19(b) (39) isrewritten as
119891 (119905) =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (40)
In reality the repeated impacts will not all have the samestrength or occur at a fixed interval however it is sufficient forthe situation of Figure 19(a) to approach that of Figure 19(b)for this analysis to be relevant
Equation (40) is often called a comb functionTheFouriertransform of the comb function is given [32] by
119865 (120596) = 2120587
infin
sum
119873=1
119891120575119863(120596 minus
2120587119873
119879) (41)
For a typical single-degree of freedom forced vibrationsproblem
+ 2120577120596119899 + 1205962
119899119909 =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (42)
Taking Fourier transforms of both sides and rearranging give
119883 (120596) = 2120587suminfin
119873=1119891120575119863 (120596 minus 2120587119873119879)
(minus1205962 + 2120577120596119899120596119894 + 1205962
119899)
(43)
Taking the inverse Fourier transforms to find the structureresponse in time gives
119909 (119905) = 119891
infin
sum
119873=1
119890119894(2120587119873119879)119905
(minus4120587211987321198792 + 119894120577120596119899 (4120587119873119879) + 1205962
119899) (44)
Equation (44) experiences a resonance when the denom-inator of any of the terms in the sum goes to zero that is when120596119899= 2120587119873119879 where 119873 is an integer larger than 0 This says
that the structure will experience a resonance when the natu-ral frequency is an integral multiple of the forcing frequency
16 Mathematical Problems in Engineering
as would be expected In the previous numerical simulationsthis condition has been observed in several cases Figures12(a) and 13(b) For instance the structural frequency ofFigure 13(b) is 2120587119879 = 212058705 = 1257 rads The naturalfrequency of the structure is radic119896119898 = radic10000001600 =
25 rads With119873 = 2 the structural frequency is close to thenatural frequency The result is what is seen in Figure 13(b)The two masses vibrate at the same frequency of 2120587119879 whichis half of the natural frequency The structural frequencyremains at that level which is lock-in and so does thefrequency of the free mass response
Thus the impact model successfully predicts the exis-tence of a lock-in condition and also predicts that it can occurwhen the frequency of the structure and freemass impacts aresynchronized together at a multiple of the natural frequencyof the structure The question of the precise conditionsnecessary for the frequency of the impacts to occur at amultiple of the natural frequency still remains open
6 Summary and Conclusions
This paper analyzes the vibrations caused by repeated impactsof a flow-driven freemass onto a second structure Conserva-tion of linearmomentum and the coefficient of restitution areused to characterize the nature of the impacts between thetwo masses This model is referred to as the impact modelTheproperties of the impactmodel were investigated throughnumerical simulations The mass of the free mass is one ofthe factors determining the vibration characteristics sinceit is related to the momentum of each impact This modelis proposed to explain the nature of ice-induced vibrationswithout requiring a microscopic model of the mechanism ofice failure The magnitudes of the vibrations are governedin part by the coefficient of restitution which when usedto explain IIVs encapsulates all ice properties mentionedin other IIV models In addition to the ice properties thecoefficient of restitution also governs the interaction betweenthe structure and the free mass The free mass is used torepresent the ice when modelling IIVs The last propertyof the impact model is that of the free mass area whichinfluences the dissipation of the system momenta throughthe drag force A free mass with a small area shows the samebehavior of a mass driven by lower flow velocity
Therefore through the impact model two factors whichcontrol the characteristics of IIVs have been proposed theice area and the coefficient of restitution By changing thesize of the contact area between the ice and structure thedriving force of the ice can be reduced as if the ice weredriven by lower flow velocities Furthermore by changingfactors related to the coefficient of restitution such as thesurface roughness or shape of the structure IIVs can bereduced to avoid the lock-in conditionwhich occurswhen thefrequencies of the structure and ice impacts are synchronizedat a multiple of the natural frequency of the structure Inaddition to the inclusion of the effect of flow the newapproach to ice forces through the impact model permitsthe answering of questions which originate from the weaktheoretical points of existing IIV models
Particular strengths of the model are that it qualitativelymatches observed results and in particular that a mechanismof ice failure is not required in order to perform simulationsand predictions It also proposes a simple mechanism toexplain the potentially destructive resonance lock-in condi-tions
References
[1] J P Dempsey ldquoResearch trends in ice mechanicsrdquo InternationalJournal of Solids and Structures vol 37 no 1-2 pp 131ndash153 2000
[2] K A Blenkarn ldquoMeasurement and analysis of ice forces oncook inlet structuresrdquo in Proceedings of the Offshore TechnologyConference (OTC rsquo70) pp 365ndash378 1970
[3] Y Toyama T Sensu M Minami and N Yashima ldquoModel testson ice-induced self-excited vibration of cylindrical structuresrdquoin Proceedings of the 7th International Conference on Port andOcean Engineering under Arctic Conditions vol 2 pp 834ndash8441983
[4] M Ranta and R Raty ldquoOn the analytic solution of ice-inducedvibrations in a marine pile structurerdquo in Proceedings of the 7thInternational Conference on Port and Ocean Engineering underArctic Conditions vol 2 pp 901ndash908 1983
[5] A Barker G Timco H Gravesen and P Voslashlund ldquoIce loadingon Danish wind turbinesmdashpart 1 dynamic model testsrdquo ColdRegions Science and Technology vol 41 no 1 pp 1ndash23 2005
[6] D S Sodhi ldquoIce-induced vibrations of structuresrdquo SpecialReport 89-5 International Association for Hydraulic ResearchWorking Group on Ice Forces 1989
[7] D S Sodhi and C E Morris ldquoCharacteristic frequency of forcevariations in continuous crushing of sheet ice against rigidcylindrical structuresrdquo Cold Regions Science and Technologyvol 12 no 1 pp 1ndash12 1986
[8] D S Sodhi C E Morris and G F N Cox ldquoDynamic analysisof failure modes on ice sheets encountering sloping structuresrdquoin Proceedings of the 6th International Conference on OffshoreMechanics and Arctic Engineering vol 4 pp 281ndash284 1987
[9] H R Peyton ldquoSea ice forcesrdquo in Proceedings of the Conferenceon Ice Pressure against Structures 1966
[10] C R Neill ldquoDynamic ice forces on piers and piles an assess-ment of design guidelines in the light of recent researchrdquoCanadian Journal of Civil Engineering vol 3 no 2 pp 305ndash3411976
[11] G Matlock W Dawkins and J Panak ldquoAnalytical model forice-structure interactionrdquo Journal of the Engineering MechanicsDivision pp 1083ndash11092 1971
[12] M M Maattanen ldquoIce-induced vibrations of structures self-excitationrdquo Special Report 89-5 International Association forHydraulic Research Working Group on Ice Forces 1989
[13] M M Maattanen ldquoNumerical simulation of ice-induced vibra-tions in offshore structuresrdquo in Proceedings of the 14th NordicSeminar on Computational Mechanics 2001 Keynote lecture
[14] J Xu and L Wang ldquoIce force oscillator model and its numericalsolutionsrdquo in Proceedings of the 7th International Conference onOffshore Machanics and Arctic Engineering pp 171ndash176 1988
[15] G TsinkerMarine Structures Engineering Chapman-Hall NewYork NY USA 1st edition 1995
[16] G Huang and P Liu ldquoA dynamic model for ice-inducedvibration of structuresrdquo Journal of OffshoreMechanics andArcticEngineering vol 131 no 1 Article ID 011501 6 pages 2009
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
14 Mathematical Problems in Engineering
0 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
02
0
minus02
minus04
minus06
minus08
(a) Overall responses with119872= 2400
386 388 390 392 394 396 398 400
Disp
lace
men
t (m
)
StructureIce
Time (s)
4
2
0
minus2
minus4
times10minus5
(b) Responses with119872 = 2400 from 385 to 400 seconds
Figure 16 Responses with119872 = 2400
StructureIce
0002
0
minus0002
minus0004
minus0006
minus0008
minus0010 50 100 150 200 250 300 350 400
Disp
lace
men
t (m
)
Time (s)
(a) Overall responses with 119890 = 01
StructureIce
0 2 4 6 8 10 12 14
0002
0
minus0002
minus0004
minus0006
minus0008
Disp
lace
men
t (m
)
Time (s)
(b) Responses with 119890 = 01 from 0 to 15 seconds
Figure 17 Responses with 119890 = 01
In fact the impact IIV model proposes a way to calculate thestrength of each impulse and the times between impulses andthese depend on the momenta of the ice and structure aswell as the coefficient of restitution for the collision Undercertain conditions it is possible for the discrete impulsesof Figure 19(a) to approach the regular periodic series ofimpulses of Figure 19(b) which is suspected to cause the lock-in condition
This view of the impact IIV model can be analyzed byusing a Fourier transform [31] There are several definitions
of the Fourier transform (FT) that may be used and in thispaper the chosen version of the FT is given by
119865 (120596) = int
infin
minusinfin
119891 (119905) 119890minus119894120596119905
119889119905 (37)
The corresponding inverse Fourier transform is then given by
119891 (119905) =1
2120587int
infin
minusinfin
119865 (120596) 119890119894120596119905119889120596 (38)
Mathematical Problems in Engineering 15
005
0
minus005
minus01
minus015
minus02
minus0250 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with 119890 = 05
Time (s)
Disp
lace
men
t (m
)
StructureIce
36 38 40 42 44 46 48 50
4
2
0
minus2
minus4
times10minus4
(b) Responses with 119890 = 05 from 35 to 50 seconds
Figure 18 Responses with 119890 = 05
f(t)
fj
tj t
(a) Random impulses
f(t)
f
T 2T 3Tt
pT
(b) Uniform impulses
Figure 19 Ice forces as a series of impulses
The discrete ice forces of Figure 19(a) can be expressed asa series of impulses as
119891 (119905) =
infin
sum
119895=1
119891119895120575119863(119905 minus 119905119895) (39)
where 120575119863indicates the Dirac delta function and 119891
119895indicates
the strength of each impact If the sizes and intervals ofthe impulses are uniform as shown in Figure 19(b) (39) isrewritten as
119891 (119905) =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (40)
In reality the repeated impacts will not all have the samestrength or occur at a fixed interval however it is sufficient forthe situation of Figure 19(a) to approach that of Figure 19(b)for this analysis to be relevant
Equation (40) is often called a comb functionTheFouriertransform of the comb function is given [32] by
119865 (120596) = 2120587
infin
sum
119873=1
119891120575119863(120596 minus
2120587119873
119879) (41)
For a typical single-degree of freedom forced vibrationsproblem
+ 2120577120596119899 + 1205962
119899119909 =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (42)
Taking Fourier transforms of both sides and rearranging give
119883 (120596) = 2120587suminfin
119873=1119891120575119863 (120596 minus 2120587119873119879)
(minus1205962 + 2120577120596119899120596119894 + 1205962
119899)
(43)
Taking the inverse Fourier transforms to find the structureresponse in time gives
119909 (119905) = 119891
infin
sum
119873=1
119890119894(2120587119873119879)119905
(minus4120587211987321198792 + 119894120577120596119899 (4120587119873119879) + 1205962
119899) (44)
Equation (44) experiences a resonance when the denom-inator of any of the terms in the sum goes to zero that is when120596119899= 2120587119873119879 where 119873 is an integer larger than 0 This says
that the structure will experience a resonance when the natu-ral frequency is an integral multiple of the forcing frequency
16 Mathematical Problems in Engineering
as would be expected In the previous numerical simulationsthis condition has been observed in several cases Figures12(a) and 13(b) For instance the structural frequency ofFigure 13(b) is 2120587119879 = 212058705 = 1257 rads The naturalfrequency of the structure is radic119896119898 = radic10000001600 =
25 rads With119873 = 2 the structural frequency is close to thenatural frequency The result is what is seen in Figure 13(b)The two masses vibrate at the same frequency of 2120587119879 whichis half of the natural frequency The structural frequencyremains at that level which is lock-in and so does thefrequency of the free mass response
Thus the impact model successfully predicts the exis-tence of a lock-in condition and also predicts that it can occurwhen the frequency of the structure and freemass impacts aresynchronized together at a multiple of the natural frequencyof the structure The question of the precise conditionsnecessary for the frequency of the impacts to occur at amultiple of the natural frequency still remains open
6 Summary and Conclusions
This paper analyzes the vibrations caused by repeated impactsof a flow-driven freemass onto a second structure Conserva-tion of linearmomentum and the coefficient of restitution areused to characterize the nature of the impacts between thetwo masses This model is referred to as the impact modelTheproperties of the impactmodel were investigated throughnumerical simulations The mass of the free mass is one ofthe factors determining the vibration characteristics sinceit is related to the momentum of each impact This modelis proposed to explain the nature of ice-induced vibrationswithout requiring a microscopic model of the mechanism ofice failure The magnitudes of the vibrations are governedin part by the coefficient of restitution which when usedto explain IIVs encapsulates all ice properties mentionedin other IIV models In addition to the ice properties thecoefficient of restitution also governs the interaction betweenthe structure and the free mass The free mass is used torepresent the ice when modelling IIVs The last propertyof the impact model is that of the free mass area whichinfluences the dissipation of the system momenta throughthe drag force A free mass with a small area shows the samebehavior of a mass driven by lower flow velocity
Therefore through the impact model two factors whichcontrol the characteristics of IIVs have been proposed theice area and the coefficient of restitution By changing thesize of the contact area between the ice and structure thedriving force of the ice can be reduced as if the ice weredriven by lower flow velocities Furthermore by changingfactors related to the coefficient of restitution such as thesurface roughness or shape of the structure IIVs can bereduced to avoid the lock-in conditionwhich occurswhen thefrequencies of the structure and ice impacts are synchronizedat a multiple of the natural frequency of the structure Inaddition to the inclusion of the effect of flow the newapproach to ice forces through the impact model permitsthe answering of questions which originate from the weaktheoretical points of existing IIV models
Particular strengths of the model are that it qualitativelymatches observed results and in particular that a mechanismof ice failure is not required in order to perform simulationsand predictions It also proposes a simple mechanism toexplain the potentially destructive resonance lock-in condi-tions
References
[1] J P Dempsey ldquoResearch trends in ice mechanicsrdquo InternationalJournal of Solids and Structures vol 37 no 1-2 pp 131ndash153 2000
[2] K A Blenkarn ldquoMeasurement and analysis of ice forces oncook inlet structuresrdquo in Proceedings of the Offshore TechnologyConference (OTC rsquo70) pp 365ndash378 1970
[3] Y Toyama T Sensu M Minami and N Yashima ldquoModel testson ice-induced self-excited vibration of cylindrical structuresrdquoin Proceedings of the 7th International Conference on Port andOcean Engineering under Arctic Conditions vol 2 pp 834ndash8441983
[4] M Ranta and R Raty ldquoOn the analytic solution of ice-inducedvibrations in a marine pile structurerdquo in Proceedings of the 7thInternational Conference on Port and Ocean Engineering underArctic Conditions vol 2 pp 901ndash908 1983
[5] A Barker G Timco H Gravesen and P Voslashlund ldquoIce loadingon Danish wind turbinesmdashpart 1 dynamic model testsrdquo ColdRegions Science and Technology vol 41 no 1 pp 1ndash23 2005
[6] D S Sodhi ldquoIce-induced vibrations of structuresrdquo SpecialReport 89-5 International Association for Hydraulic ResearchWorking Group on Ice Forces 1989
[7] D S Sodhi and C E Morris ldquoCharacteristic frequency of forcevariations in continuous crushing of sheet ice against rigidcylindrical structuresrdquo Cold Regions Science and Technologyvol 12 no 1 pp 1ndash12 1986
[8] D S Sodhi C E Morris and G F N Cox ldquoDynamic analysisof failure modes on ice sheets encountering sloping structuresrdquoin Proceedings of the 6th International Conference on OffshoreMechanics and Arctic Engineering vol 4 pp 281ndash284 1987
[9] H R Peyton ldquoSea ice forcesrdquo in Proceedings of the Conferenceon Ice Pressure against Structures 1966
[10] C R Neill ldquoDynamic ice forces on piers and piles an assess-ment of design guidelines in the light of recent researchrdquoCanadian Journal of Civil Engineering vol 3 no 2 pp 305ndash3411976
[11] G Matlock W Dawkins and J Panak ldquoAnalytical model forice-structure interactionrdquo Journal of the Engineering MechanicsDivision pp 1083ndash11092 1971
[12] M M Maattanen ldquoIce-induced vibrations of structures self-excitationrdquo Special Report 89-5 International Association forHydraulic Research Working Group on Ice Forces 1989
[13] M M Maattanen ldquoNumerical simulation of ice-induced vibra-tions in offshore structuresrdquo in Proceedings of the 14th NordicSeminar on Computational Mechanics 2001 Keynote lecture
[14] J Xu and L Wang ldquoIce force oscillator model and its numericalsolutionsrdquo in Proceedings of the 7th International Conference onOffshore Machanics and Arctic Engineering pp 171ndash176 1988
[15] G TsinkerMarine Structures Engineering Chapman-Hall NewYork NY USA 1st edition 1995
[16] G Huang and P Liu ldquoA dynamic model for ice-inducedvibration of structuresrdquo Journal of OffshoreMechanics andArcticEngineering vol 131 no 1 Article ID 011501 6 pages 2009
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
Mathematical Problems in Engineering 15
005
0
minus005
minus01
minus015
minus02
minus0250 50 100 150 200 250 300 350 400
Time (s)
Disp
lace
men
t (m
)
StructureIce
(a) Overall responses with 119890 = 05
Time (s)
Disp
lace
men
t (m
)
StructureIce
36 38 40 42 44 46 48 50
4
2
0
minus2
minus4
times10minus4
(b) Responses with 119890 = 05 from 35 to 50 seconds
Figure 18 Responses with 119890 = 05
f(t)
fj
tj t
(a) Random impulses
f(t)
f
T 2T 3Tt
pT
(b) Uniform impulses
Figure 19 Ice forces as a series of impulses
The discrete ice forces of Figure 19(a) can be expressed asa series of impulses as
119891 (119905) =
infin
sum
119895=1
119891119895120575119863(119905 minus 119905119895) (39)
where 120575119863indicates the Dirac delta function and 119891
119895indicates
the strength of each impact If the sizes and intervals ofthe impulses are uniform as shown in Figure 19(b) (39) isrewritten as
119891 (119905) =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (40)
In reality the repeated impacts will not all have the samestrength or occur at a fixed interval however it is sufficient forthe situation of Figure 19(a) to approach that of Figure 19(b)for this analysis to be relevant
Equation (40) is often called a comb functionTheFouriertransform of the comb function is given [32] by
119865 (120596) = 2120587
infin
sum
119873=1
119891120575119863(120596 minus
2120587119873
119879) (41)
For a typical single-degree of freedom forced vibrationsproblem
+ 2120577120596119899 + 1205962
119899119909 =
infin
sum
119901=1
119891120575119863(119905 minus 119901119879) (42)
Taking Fourier transforms of both sides and rearranging give
119883 (120596) = 2120587suminfin
119873=1119891120575119863 (120596 minus 2120587119873119879)
(minus1205962 + 2120577120596119899120596119894 + 1205962
119899)
(43)
Taking the inverse Fourier transforms to find the structureresponse in time gives
119909 (119905) = 119891
infin
sum
119873=1
119890119894(2120587119873119879)119905
(minus4120587211987321198792 + 119894120577120596119899 (4120587119873119879) + 1205962
119899) (44)
Equation (44) experiences a resonance when the denom-inator of any of the terms in the sum goes to zero that is when120596119899= 2120587119873119879 where 119873 is an integer larger than 0 This says
that the structure will experience a resonance when the natu-ral frequency is an integral multiple of the forcing frequency
16 Mathematical Problems in Engineering
as would be expected In the previous numerical simulationsthis condition has been observed in several cases Figures12(a) and 13(b) For instance the structural frequency ofFigure 13(b) is 2120587119879 = 212058705 = 1257 rads The naturalfrequency of the structure is radic119896119898 = radic10000001600 =
25 rads With119873 = 2 the structural frequency is close to thenatural frequency The result is what is seen in Figure 13(b)The two masses vibrate at the same frequency of 2120587119879 whichis half of the natural frequency The structural frequencyremains at that level which is lock-in and so does thefrequency of the free mass response
Thus the impact model successfully predicts the exis-tence of a lock-in condition and also predicts that it can occurwhen the frequency of the structure and freemass impacts aresynchronized together at a multiple of the natural frequencyof the structure The question of the precise conditionsnecessary for the frequency of the impacts to occur at amultiple of the natural frequency still remains open
6 Summary and Conclusions
This paper analyzes the vibrations caused by repeated impactsof a flow-driven freemass onto a second structure Conserva-tion of linearmomentum and the coefficient of restitution areused to characterize the nature of the impacts between thetwo masses This model is referred to as the impact modelTheproperties of the impactmodel were investigated throughnumerical simulations The mass of the free mass is one ofthe factors determining the vibration characteristics sinceit is related to the momentum of each impact This modelis proposed to explain the nature of ice-induced vibrationswithout requiring a microscopic model of the mechanism ofice failure The magnitudes of the vibrations are governedin part by the coefficient of restitution which when usedto explain IIVs encapsulates all ice properties mentionedin other IIV models In addition to the ice properties thecoefficient of restitution also governs the interaction betweenthe structure and the free mass The free mass is used torepresent the ice when modelling IIVs The last propertyof the impact model is that of the free mass area whichinfluences the dissipation of the system momenta throughthe drag force A free mass with a small area shows the samebehavior of a mass driven by lower flow velocity
Therefore through the impact model two factors whichcontrol the characteristics of IIVs have been proposed theice area and the coefficient of restitution By changing thesize of the contact area between the ice and structure thedriving force of the ice can be reduced as if the ice weredriven by lower flow velocities Furthermore by changingfactors related to the coefficient of restitution such as thesurface roughness or shape of the structure IIVs can bereduced to avoid the lock-in conditionwhich occurswhen thefrequencies of the structure and ice impacts are synchronizedat a multiple of the natural frequency of the structure Inaddition to the inclusion of the effect of flow the newapproach to ice forces through the impact model permitsthe answering of questions which originate from the weaktheoretical points of existing IIV models
Particular strengths of the model are that it qualitativelymatches observed results and in particular that a mechanismof ice failure is not required in order to perform simulationsand predictions It also proposes a simple mechanism toexplain the potentially destructive resonance lock-in condi-tions
References
[1] J P Dempsey ldquoResearch trends in ice mechanicsrdquo InternationalJournal of Solids and Structures vol 37 no 1-2 pp 131ndash153 2000
[2] K A Blenkarn ldquoMeasurement and analysis of ice forces oncook inlet structuresrdquo in Proceedings of the Offshore TechnologyConference (OTC rsquo70) pp 365ndash378 1970
[3] Y Toyama T Sensu M Minami and N Yashima ldquoModel testson ice-induced self-excited vibration of cylindrical structuresrdquoin Proceedings of the 7th International Conference on Port andOcean Engineering under Arctic Conditions vol 2 pp 834ndash8441983
[4] M Ranta and R Raty ldquoOn the analytic solution of ice-inducedvibrations in a marine pile structurerdquo in Proceedings of the 7thInternational Conference on Port and Ocean Engineering underArctic Conditions vol 2 pp 901ndash908 1983
[5] A Barker G Timco H Gravesen and P Voslashlund ldquoIce loadingon Danish wind turbinesmdashpart 1 dynamic model testsrdquo ColdRegions Science and Technology vol 41 no 1 pp 1ndash23 2005
[6] D S Sodhi ldquoIce-induced vibrations of structuresrdquo SpecialReport 89-5 International Association for Hydraulic ResearchWorking Group on Ice Forces 1989
[7] D S Sodhi and C E Morris ldquoCharacteristic frequency of forcevariations in continuous crushing of sheet ice against rigidcylindrical structuresrdquo Cold Regions Science and Technologyvol 12 no 1 pp 1ndash12 1986
[8] D S Sodhi C E Morris and G F N Cox ldquoDynamic analysisof failure modes on ice sheets encountering sloping structuresrdquoin Proceedings of the 6th International Conference on OffshoreMechanics and Arctic Engineering vol 4 pp 281ndash284 1987
[9] H R Peyton ldquoSea ice forcesrdquo in Proceedings of the Conferenceon Ice Pressure against Structures 1966
[10] C R Neill ldquoDynamic ice forces on piers and piles an assess-ment of design guidelines in the light of recent researchrdquoCanadian Journal of Civil Engineering vol 3 no 2 pp 305ndash3411976
[11] G Matlock W Dawkins and J Panak ldquoAnalytical model forice-structure interactionrdquo Journal of the Engineering MechanicsDivision pp 1083ndash11092 1971
[12] M M Maattanen ldquoIce-induced vibrations of structures self-excitationrdquo Special Report 89-5 International Association forHydraulic Research Working Group on Ice Forces 1989
[13] M M Maattanen ldquoNumerical simulation of ice-induced vibra-tions in offshore structuresrdquo in Proceedings of the 14th NordicSeminar on Computational Mechanics 2001 Keynote lecture
[14] J Xu and L Wang ldquoIce force oscillator model and its numericalsolutionsrdquo in Proceedings of the 7th International Conference onOffshore Machanics and Arctic Engineering pp 171ndash176 1988
[15] G TsinkerMarine Structures Engineering Chapman-Hall NewYork NY USA 1st edition 1995
[16] G Huang and P Liu ldquoA dynamic model for ice-inducedvibration of structuresrdquo Journal of OffshoreMechanics andArcticEngineering vol 131 no 1 Article ID 011501 6 pages 2009
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
16 Mathematical Problems in Engineering
as would be expected In the previous numerical simulationsthis condition has been observed in several cases Figures12(a) and 13(b) For instance the structural frequency ofFigure 13(b) is 2120587119879 = 212058705 = 1257 rads The naturalfrequency of the structure is radic119896119898 = radic10000001600 =
25 rads With119873 = 2 the structural frequency is close to thenatural frequency The result is what is seen in Figure 13(b)The two masses vibrate at the same frequency of 2120587119879 whichis half of the natural frequency The structural frequencyremains at that level which is lock-in and so does thefrequency of the free mass response
Thus the impact model successfully predicts the exis-tence of a lock-in condition and also predicts that it can occurwhen the frequency of the structure and freemass impacts aresynchronized together at a multiple of the natural frequencyof the structure The question of the precise conditionsnecessary for the frequency of the impacts to occur at amultiple of the natural frequency still remains open
6 Summary and Conclusions
This paper analyzes the vibrations caused by repeated impactsof a flow-driven freemass onto a second structure Conserva-tion of linearmomentum and the coefficient of restitution areused to characterize the nature of the impacts between thetwo masses This model is referred to as the impact modelTheproperties of the impactmodel were investigated throughnumerical simulations The mass of the free mass is one ofthe factors determining the vibration characteristics sinceit is related to the momentum of each impact This modelis proposed to explain the nature of ice-induced vibrationswithout requiring a microscopic model of the mechanism ofice failure The magnitudes of the vibrations are governedin part by the coefficient of restitution which when usedto explain IIVs encapsulates all ice properties mentionedin other IIV models In addition to the ice properties thecoefficient of restitution also governs the interaction betweenthe structure and the free mass The free mass is used torepresent the ice when modelling IIVs The last propertyof the impact model is that of the free mass area whichinfluences the dissipation of the system momenta throughthe drag force A free mass with a small area shows the samebehavior of a mass driven by lower flow velocity
Therefore through the impact model two factors whichcontrol the characteristics of IIVs have been proposed theice area and the coefficient of restitution By changing thesize of the contact area between the ice and structure thedriving force of the ice can be reduced as if the ice weredriven by lower flow velocities Furthermore by changingfactors related to the coefficient of restitution such as thesurface roughness or shape of the structure IIVs can bereduced to avoid the lock-in conditionwhich occurswhen thefrequencies of the structure and ice impacts are synchronizedat a multiple of the natural frequency of the structure Inaddition to the inclusion of the effect of flow the newapproach to ice forces through the impact model permitsthe answering of questions which originate from the weaktheoretical points of existing IIV models
Particular strengths of the model are that it qualitativelymatches observed results and in particular that a mechanismof ice failure is not required in order to perform simulationsand predictions It also proposes a simple mechanism toexplain the potentially destructive resonance lock-in condi-tions
References
[1] J P Dempsey ldquoResearch trends in ice mechanicsrdquo InternationalJournal of Solids and Structures vol 37 no 1-2 pp 131ndash153 2000
[2] K A Blenkarn ldquoMeasurement and analysis of ice forces oncook inlet structuresrdquo in Proceedings of the Offshore TechnologyConference (OTC rsquo70) pp 365ndash378 1970
[3] Y Toyama T Sensu M Minami and N Yashima ldquoModel testson ice-induced self-excited vibration of cylindrical structuresrdquoin Proceedings of the 7th International Conference on Port andOcean Engineering under Arctic Conditions vol 2 pp 834ndash8441983
[4] M Ranta and R Raty ldquoOn the analytic solution of ice-inducedvibrations in a marine pile structurerdquo in Proceedings of the 7thInternational Conference on Port and Ocean Engineering underArctic Conditions vol 2 pp 901ndash908 1983
[5] A Barker G Timco H Gravesen and P Voslashlund ldquoIce loadingon Danish wind turbinesmdashpart 1 dynamic model testsrdquo ColdRegions Science and Technology vol 41 no 1 pp 1ndash23 2005
[6] D S Sodhi ldquoIce-induced vibrations of structuresrdquo SpecialReport 89-5 International Association for Hydraulic ResearchWorking Group on Ice Forces 1989
[7] D S Sodhi and C E Morris ldquoCharacteristic frequency of forcevariations in continuous crushing of sheet ice against rigidcylindrical structuresrdquo Cold Regions Science and Technologyvol 12 no 1 pp 1ndash12 1986
[8] D S Sodhi C E Morris and G F N Cox ldquoDynamic analysisof failure modes on ice sheets encountering sloping structuresrdquoin Proceedings of the 6th International Conference on OffshoreMechanics and Arctic Engineering vol 4 pp 281ndash284 1987
[9] H R Peyton ldquoSea ice forcesrdquo in Proceedings of the Conferenceon Ice Pressure against Structures 1966
[10] C R Neill ldquoDynamic ice forces on piers and piles an assess-ment of design guidelines in the light of recent researchrdquoCanadian Journal of Civil Engineering vol 3 no 2 pp 305ndash3411976
[11] G Matlock W Dawkins and J Panak ldquoAnalytical model forice-structure interactionrdquo Journal of the Engineering MechanicsDivision pp 1083ndash11092 1971
[12] M M Maattanen ldquoIce-induced vibrations of structures self-excitationrdquo Special Report 89-5 International Association forHydraulic Research Working Group on Ice Forces 1989
[13] M M Maattanen ldquoNumerical simulation of ice-induced vibra-tions in offshore structuresrdquo in Proceedings of the 14th NordicSeminar on Computational Mechanics 2001 Keynote lecture
[14] J Xu and L Wang ldquoIce force oscillator model and its numericalsolutionsrdquo in Proceedings of the 7th International Conference onOffshore Machanics and Arctic Engineering pp 171ndash176 1988
[15] G TsinkerMarine Structures Engineering Chapman-Hall NewYork NY USA 1st edition 1995
[16] G Huang and P Liu ldquoA dynamic model for ice-inducedvibration of structuresrdquo Journal of OffshoreMechanics andArcticEngineering vol 131 no 1 Article ID 011501 6 pages 2009
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
Mathematical Problems in Engineering 17
[17] R Frederking and J Schwarz ldquoModel tests of ice forces on fixedand oscillating conesrdquoCold Regions Science and Technology vol6 no 1 pp 61ndash72 1982
[18] K Hirayama and I Obara ldquoIce forces on inclined structuresrdquoin Proceedings of the 5th International Offshore Mechanics andArctic Engineering pp 515ndash520 1986
[19] Q Yue and X Bi ldquoFull-scale test and analysis of dynamic inter-action between ice sheet and conical structurerdquo inProceedings ofthe 14th International Association for Hydraulic Research (IAHR)Symposium on Ice vol 2 1998
[20] Q Yue and X Bi ldquoIce-induced jacket structure vibrations inBohai Seardquo Journal of Cold Regions Engineering vol 14 no 2pp 81ndash92 2000
[21] Y Qu Q Yue X Bi and T Karna ldquoA random ice forcemodel for narrow conical structuresrdquo Cold Regions Science andTechnology vol 45 no 3 pp 148ndash157 2006
[22] X Liu G Li R Oberlies and Q Yue ldquoResearch on short-termdynamic ice cases for dynamic analysis of ice-resistant jacketplatform in the Bohai Gulfrdquo Marine Structures vol 22 no 3pp 457ndash479 2009
[23] X Liu G Li Q Yue and R Oberlies ldquoAcceleration-orienteddesign optimization of ice-resistant jacket platforms in theBohai GulfrdquoOcean Engineering vol 36 no 17-18 pp 1295ndash13022009
[24] B Munson D Young and T Okiishi Fundamentals of FluidMechanics John Wiley amp Sons Ames Iowa USA 4th edition2002
[25] S Salapaka M V Salapaka M Dahleh and I Mezic ldquoComplexdynamics in repeated impact oscillatorsrdquo in Proceedings of the37th IEEE Conference on Decision and Control (CDC rsquo98) pp2053ndash2058 December 1998
[26] A Bedford and W Fowler Engineering Mechanics DynamicsAddison Wesley Longman Inc Menlo Park Calif USA 2ndedition 1999
[27] T Karna and R Turunen ldquoDynamic response of narrowstructures to ice crushingrdquoCold Regions Science andTechnologyvol 17 no 2 pp 173ndash187 1989
[28] E Eranti F D Hayes M M Maattanen and T T SoongldquoDynamic ice-structure interaction analysis for narrow verticalstructuresrdquo in Proceedings of the 6th International Conference onPort and Ocean Engineering under Arctic Conditions vol 1 pp472ndash479 1981
[29] L Y Shih ldquoAnalysis of ice-induced vibrations on a flexiblestructurerdquoAppliedMathematicalModelling vol 15 no 11-12 pp632ndash638 1991
[30] T K Karna ldquoSteady-state vibrations of offshore structuresrdquoHydrotechnical Construction vol 28 no 9 pp 446ndash453 1994
[31] CW de SilvaVibration Fundamentals and Practice CRCPressLLC Boca Raton Fla USA 1st edition 1999
[32] H J Weaver Applications of Discrete and Continuous FourierAnalysis John Wiley amp Sons New York NY USA 1st edition1983
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
Submit your manuscripts athttpwwwhindawicom
OperationsResearch
Advances in
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Abstract and Applied Analysis
ISRN Applied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
International Journal of
Combinatorics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal of Function Spaces and Applications
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Geometry
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Discrete Dynamicsin Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2013
Advances in
Mathematical Physics
ISRN Algebra
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ProbabilityandStatistics
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Mathematical Analysis
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Journal ofApplied Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Advances in
DecisionSciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
Stochastic AnalysisInternational Journal of
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013
The Scientific World Journal
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2013
ISRN Discrete Mathematics
Hindawi Publishing Corporationhttpwwwhindawicom
DifferentialEquations
International Journal of
Volume 2013
Related Documents
