numerical investigation of galloping instabilities in z-shaped … · 2016-06-09 ·...

Research ArticleNumerical Investigation of Galloping Instabilities inZ-Shaped Profiles

Ignacio Gomez1 Miguel Chavez1 Gustavo Alonso2 and Eusebio Valero1

1 ETSI Aeronauticos Universidad Politecnica de Madrid 28040 Madrid Spain2 IDRUPM ETSI Aeronauticos Universidad Politecnica de Madrid 28040 Madrid Spain

Correspondence should be addressed to Gustavo Alonso gustavoalonsoupmes

Received 12 March 2014 Accepted 29 May 2014 Published 25 June 2014

Academic Editor Angel Sanz-Andres

Copyright copy 2014 Ignacio Gomez et alThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Aeroelastic effects are relatively common in the design of modern civil constructions such as office blocks airport terminalbuildings and factories Typical flexible structures exposed to the action of wind are shading devices normally slats or louvers Atypical cross-section for such elements is a Z-shaped profilemade out of a central web and two-sidewings Galloping instabilities areoften determined in practice using theGlauert-DenHartog criterionThis criterion relies on accurate predictions of the dependenceof the aerodynamic force coefficients with the angle of attack The results of a parametric analysis based on a numerical analysisand performed on different Z-shaped louvers to determine translational galloping instability regions are presented in this paperThese numerical analysis results have been validated with a parametric analysis of Z-shaped profiles based on static wind tunneltests In order to perform this validation the DLR TAU Code which is a standard code within the European aeronautical industryhas been used This study highlights the focus on the numerical prediction of the effect of galloping which is shown in a visibleway through stability maps Comparisons between numerical and experimental data are presented with respect to various meshesand turbulence models

1 Introduction

Aeroelastic phenomena are becomingmore andmore impor-tant from the point of view of its potential relevance inmodern structural design It is well known that bluff bodiesin cross-flow are subject to typical aeroelastic phenomenalike vortex shedding translational and torsional gallopingand even flutter Some of these phenomena can even appearcoupled occasionally Galloping is a typical instability offlexible lightly damped structures Under certain conditionsthese structures may have large amplitude normal to windoscillations at much lower frequencies than those of vortexshedding found in the von Karman vortex street

Theoretical foundations of galloping are well establishedand understood [1] very often supported by experimentslike in Parkinson and Smith [2] Novak [3] Novak [4] andSimiu and Scanlan [5] As it is well known galloping can beexplained by taking into account that although the incidentvelocity 119880 is uniform and constant because of the lateraloscillation of the body in a body reference system the totalvelocity changes both magnitude and direction with time

Therefore the structure angle of attack also changes withtime hence the aerodynamic forces acting on the body

In the simplestmodel of galloping (one degree of freedommodel) it is assumed that the two-dimensional body (119909-119911plane) whosemass per unit length is119898 is elasticallymountedon a support with a damping coefficient 120577 and a stiffness1198981205962

119899

(where 120596119899is the angular natural frequency) This structure

at rest is oriented at given angle of attack 1205720with respect to

the incident flow Assuming the structure oscillating along 119911-axis direction within an uniform flow with velocity 119880 therelative velocity between the fluid and the body is 119880

119903=

[1198802+ (d119911d119905)2]12 and the angle of attack due to oscillation

is 120572 = (d119911d119905)119880 Therefore drag 119889(120572) and lift 119897(120572) are asfollows

119889 (120572) =1

21205881198881198802

119903119888119889(120572)

119897 (120572) =1

21205881198881198802

119903119888119897(120572)

(1)

Hindawi Publishing Corporatione Scientific World JournalVolume 2014 Article ID 363274 14 pageshttpdxdoiorg1011552014363274

2 The Scientific World Journal

In these expressions 120588 stands for the fluid density 119888 fora characteristic length normal to incident flow 119888

119889for the

drag coefficient and 119888119897for the lift coefficient The projection

of those forces in 119911-axis direction is 119891119911(120572) = minus119889(120572) sin120572 minus

119897(120572) cos120572 On the other hand the equation of the movementof the body is

119898[d2119911d1199052

+ 2120577120596119899

d119911d119905

+ 1205962

119899119911] = 119891

119911(120572) (2)

Considering the movement to be quasisteady and assum-ing the angles of attack due to oscillation to be small enough(120572 ≪ 1) from the expression (2) the Glauert-Den Hartogcriterion for galloping instability is derived In effect if theaerodynamic force (proportional in this case to d119911d119905) isconsidered as a contribution to the total damping of thesystem (aerodynamic damping) the total damping coefficientis

120577119879= 120577 +

120588119880119888

4119898120596119899

(d119888119897

d120572+ 119888119889)

10038161003816100381610038161003816100381610038161003816120572=0

(3)

and therefore the oscillation will be stable if 120577119879

gt 0 andunstable if 120577

119879lt 0 As the mechanical damping 120577 is generally

positive instability will only occur if (d119888119897d120572 + 119888

119889) lt 0

expression known as Glauert-Den Hartog criterion which isa necessary condition for galloping instability The sufficientcondition is 120577

119879lt 0 or according to (3)

(d119888119897

d120572+ 119888119889)

10038161003816100381610038161003816100381610038161003816120572=0

lt minus4119898120577120596

119899

120588119880119888 (4)

Later in a number of papers Novak [3 4] extendedthe analysis of transverse galloping to the three-dimensionalcase Other researchers have investigated other interestingphenomena like the influence of turbulence [6ndash10] or thehysteresis phenomenon [11 12] in transverse galloping Inthe last decades besides theoretical work large efforts havebeen devoted to experimentally study the galloping featuresof many bodies having different cross-sections Althoughmost of the effort in galloping oscillation research has beenconcentrated in bodies with square or rectangular cross-sections prismatic bodies with other cross-sectional shapeshave been also considered [13ndash18]

In the last years some research on galloping has beencarried out at IDRUPM because of the above-mentionedinterest on the use of large louvers for sun-shading in buildingfacades amongst other applications A systematic parametricanalysis of simple cross-section two-dimensional bodies hasbeen accomplished The geometries analysed up to now areisosceles triangular cross-sections (the varying parameterbeing the main vertex angle 120573 [19ndash21] and ellipses [22]as well as biconvex and rhomboidal cross-sections [23] andthe varying parameter in these last cases being the relativethickness of the cross-section 120591 In all cases the unstableregions in the 120573-120572 plane (isosceles triangles) and in the 120591-120572plane (elliptical biconvex and rhomboidal bodies) where 120572stands for the wind angle of incidence were determined)

Z-shaped cross-sections are not rare in modern archi-tecture Sun-shading louvers are examples where Z-shaped

cross-sections may be found in slender structures susceptibleto gallop This type of Z-shaped cross-sections has also beenexperimentally analysed [24]

The validity of the numerical methods in the study of thiskind of problemswould have an important added value sinceit would allow for predicting the aerodynamic propertiesof different shapes and profiles without performing themore expensive experimental tests However most of thenumerical studies encountered are mainly focused on under-standing the physical mechanism involved in flow separationof bluff bodies such as circular rectangular square andtrapezoidal cylinders in a particular configuration In thisline different high-fidelity numerical simulations have beenperformed using direct numerical simulation (DNS) largeEddy simulation (LES) Reynolds-averaged Navier-Stokes(RANS) or URANS models Examples of these studies oncylinders of rectangular sections are provided by Tamura etal [25] Hayashi and Ohya [26] and Rodi [27] in computingthe mean aerodynamic coefficients using DNS Hirano et al[28] performed an LES simulation of the flow showing goodaccuracy compared to the experimental results for meandrag lift and the Strouhal number or Oka and Ishihara[29] who also include an accurate method for estimating theaerodynamic coefficients using a systematic elongation of thespanwise length Several RANS computations can be foundin Bosch and Rodi [30] using the 119896-120576 turbulence model andBao et al [31] who tried to use one-equation models such asSpalart-Allmaras RANS models provide considerable savingin computational effort but normally have problems to getaccuracy in flows with high pressure gradients

Apart from the square and circular cross-sections trape-zoidal and triangular sections have also been analysed withdetails by several researchers Lee [32] Cheng and Liu [33]and Oka and Ishihara [29] and Bao et al [31] have studiedthe developing of the recirculation region vortex sheddingphenomena and its interaction with the separated shear layerThese detailed works draw conclusions about the variation ofthe Strouhal number with the Reynolds number and aspectratio

Currently these studies have been focused on the predic-tion of drag lift Strouhal number and flow features in thedownstream region of the profile for one particular config-uration showing that precise numerical computations canprovide valuable information of the flow solutions in goodagreement with experimental data However a thoroughanalysis of galloping stability requires the study of a largenumber of parameters (geometrical and physical) whichmakes previous computations though accurate nonpracti-cal In this line Robertson et al [34] made a numericalstudy of rotational and transverse galloping rectangularbodies using two-dimensional spectral high-order methodsVery good qualitative results were obtained showing thepossibility to use numerical methods in the prediction anddiscussion of this problem The work was performed at verylow Reynolds numbers (250) far beyond current industrialconfigurations where turbulence effects can be of importanceand although the separation point is almost independent ofthe Reynolds number in rectangular structures it is not sofor the reattachment length and base pressure

The Scientific World Journal 3

w

s

ap

k

h

U

120572

120593

Figure 1 Z-shaped profile parameters

To the authorsrsquo knowledge a complete and critical studyof the application of current ldquostate of the artrdquo numericalmethods to the analysis of galloping instabilities at highReynolds numbers has not been performed yet

In this paper the transverse galloping characteristicsof two-dimensional bodies having Z-shaped cross-sections(Figure 1) are analysed numerically The overall aerodynamicforces and pressure distributions on the body walls as longas the stability maps are computed and compared withreferenced experimental test cases The study is a criticaldemonstration of the feasibility of using standard industrialsolvers in the prediction of galloping So contrary to otherauthors we are more interested in global results and compu-tational efficiency than the detailed analysis of one particularconfiguration After a first evaluation of different modelsthe numerical solution is solved with a 2D RANS-SST tur-bulence model which provides a good compromise betweenaccuracy and computational effort Detailed analysis of somecritical configurations where numerical and experimentalresults show some disagreement is also included In orderto clarify the influence of the geometry on the transversegalloping fourteen different Z-shaped cross-sections havebeen considered

The paper is organized as follows In Section 2 after thedefinition of the geometry and flow condition of the problema short introduction of the DLR TAU Code is used in thecomputations and a critical evaluation of different turbulencemodels and grid sensitivity are performed Section 3 showsthe aerodynamic coefficients stabilitymaps andpressure dis-tributions of the whole range of configurations with differentaspect ratios A comparison between RANS and URANS forsome critical configuration is also included Finally Section 4extracts the conclusions from the study

2 Numerical Analysis and Validation Tests

21 Problem Description The Z-shaped louvers are made outa thick central body (web) with two thin plates attached to itsextremes (wings) The geometry of the Z-shaped louvers isdefined by some geometrical parameters shown in Figure 1All the parameters are in function of the chord of the slats119888 the slat angle 120593 and 119899 The angle 120593 is defined by the anglebetween the central body and the wings 119899 is an integer usedto classify the different the aspect ratios studied Two different

Table 1 Case matrix of the numerical study

119899 = 1 119899 = 2 119899 = 3 119899 = 4 119899 = 5 119899 = 6 119899 = 7

120593 = 45∘

120593 = 90∘

families of louvers identified by the angle 120593 and for each oneof them different geometries were considered namely 120593 =

90∘ with 119899 = 1 2 7 and 120593 = 45

∘ with 119899 = 1 2 7 Thewhole set of tested louver shapes is shown in Figure 2

These configurations have been experimentally studiedat IDRUPM Alonso et al [24] The test facility is a two-dimensional open circuit wind tunnel where the test chamberdimensions are 015m width 090m high and 120m longThe aerodynamic loads weremeasured with a six-componentstrain-gauge balance Additional details of the wind tunnelchamber and the experimental results can be found in [24]The flow conditions were an inlet velocity of 25ms aReynolds number of 17 sdot 105 with the chord 119888 = 01m asreference length and 4 of turbulence intensity The anglesof attack used for the numerical study vary from 0∘ to 180∘with increments of 5∘

The parameters used in the present study are defined tomatch the experimental results the Reynolds number Re isdefined on the basis of the free stream velocity and the chordof the slats as reference length giving a value of 17 sdot 105 Thedrag and lift coefficients are defined as 119888

119889= 2119889120588119880

2

infin119888 119888119897=

21198971205881198802

infin119888 respectively 119889 and 119897 are the drag an lift forces acting

on the Z-profile respectively and 120588 is the flow density Thepressure coefficient is defined as 119888

119901= 2(119901 minus 119901

infin)1205881198802

infin with

119901infin

being the pressure at the far fieldThe case matrix of the numerical study is given in Table 1

22 DLR TAU Solver The industrial DLR TAU Code [35]is used in the computations The DLR TAU Code solvesthe compressible Reynolds-averaged Navier-Stokes (RANS)equations on unstructured hybrid grids employing a second-order finite volume discretization The solver uses an edge-based dual-cell approach that is a vertex-centred schemeThe inviscid terms are computed employing either a second-order central scheme with the standard Jameson-Schmidt-Turkel (JST) numerical dissipation model or a variety ofupwind schemes using linear reconstruction for second-order accuracy and a scalar or matrix artificial dissipationViscous and turbulent terms are discretized with a centralsecond-order scheme

The turbulence models implemented within the TAUcode include linear as well as nonlinear eddy viscositymodelsspanning to both one and two equation model families

Two different turbulence models have been tested inthe computations the Spalart-Allmaras original (SAO) [36]yielding highly satisfactory results for a wide range of appli-cations while being numerically robust and the two equa-tion standard 119896-120596 model and its variation the SST (shear-stress-transport) 119896-120596 model [37] which combines severaldesirable elements of 119896-120596 and 119896-120576 models The SST (shear-stress-transport) 119896-120596 model uses Wilcoxrsquos 119896-120596 with a good


n = 1

2

3

4

5

6

7

(a)

n = 1

2

3

4

5

6

7

(b)

Figure 2 Set of tested louver Z-shaped profiles with 120593 = 90∘ (a) and 120593 = 45

∘ (b) with 119899 = 1 2 7

Table 2 Characteristics of validation grids

120572 = 60∘ Number of elements Number of nodes Residual Conver iterations Comp time (min)

Coarse 12923 16642 56119890 minus 5 50000 46Medium 28397 35632 95119890 minus 6 100000 116Fine 57408 70276 38119890 minus 6 200000 672Extra fine 99562 119552 22119890 minus 6 300000 1434

behaviour in modelling the near solid walls and the standard119896-120576 model appropriate for near boundary layer edges andfree-shear layers This turbulence model is appropriate foradverse pressure gradients and separating flow

23 Mesh Validation Figure 3 shows the computationaldomain and a sketch of a typical mesh used in the presentstudy A 2D hybrid mesh allows creating a good resolu-tion boundary layer zone with quadrilateral and triangularelements for the rest of the domain A circular domain witha radius of 100-chords (100119888) is employed which ensures thatfar field boundary condition does not affect the numericalresults

Four different grids (see Table 2) of the same configura-tion (120593 = 45

∘ and 119899 = 5 angle of attack120572 = 60∘) with different

densities have been created to check the mesh convergence ofthe numerical results The four meshes (Figure 4) have beensolved with the Spalart-Allmaras model turbulence

No-slip boundary condition is imposed at the wall ofthe profile To solve correctly the boundary conditions it isimportant to define the 119910

+ carefully (119910+ = 119906+Δ119910]) 119910+

depends on the wall normal direction near the surface of theprofile (Δ119910) the viscosity (]) and the friction velocity (119906+)In this study Δ119910 has been fixed to 5 sdot 10minus5 m which gives avalue of 119910+ around the unity for all the cases

Local time step multigrid acceleration and LU-SGSsemi-implicit smoother with a Courant-Friedrich-Levynumber CFL = 15 are used in the computations of thesteady states Moreover the Jameson-Schmidt-Turkel (JST)numerical dissipation model has been used with 119896

2and 119896

4

coefficients fixed in 23 and 164 respectively (values bydefault in DLR TAU Code) These values have been provento give a good numerical stability of the solution

The DLR-TAU solver uses a preconditioner for the low-mach number problem This preconditioner reduces thestiffness of the problem by rendering the eigenvalues ofthe Jacobian matrix of the same magnitude acting on thetemporal derivative of the unknowns A modification of thewell-known Choi-Merkle and Turkel preconditioner schemeis implemented in TAU (see Turkel et al [38] for details)

The preconditioner is only activated for steady solutionsIn case of unsteady problems it should be deactivated torecover the temporal accuracy A lower time step than usualis considered to reduce numerical errors associated with thelow-mach number problem

The simulations have been computed in an 8 processorsmulticore cluster Each node is an Intel(R) Xeon(R) CPUE5620 240GHz with 24Gb of RAM The computationaltime for each simulation is shown in Table 2

The aerodynamic coefficients as a function of the numberof elements obtained with these four validation meshes


1050minus5

10

5

0

minus5

minus10

X

Z

X

Z

0minus002minus004

002

0

minus002

minus004

minus006minus008 002

Figure 3 Computational circular domain and hybrid mesh used in the present study

are shown in Figure 5 The numerical results converge tothe experimental value for this configuration With thecurrent results the fine mesh (see Table 2) which guaranteea good compromise between mesh independence and thecomputational effort has been chosen to perform the batteryof the simulations of this study

24 Turbulence Model Validation In order to compare theeffect of the turbulencemodel a complete lift and drag curvesare obtained for the configuration 120593 = 45

∘ and 119899 = 5 Theangle of attack ranged from 0∘ to 180∘ with an increment of2∘ making a total of 46 angles of attack computed

As previously mentioned the Navier-Stokes equationshave been solved with the Spalart-Allmaras original (SAO)and the 119896-120596 SST (shear-stress-transport) turbulence modelsIn addition to those RANS computations the problem hasalso been solved with the Euler equations

The reason to solve the problem with Euler equations isthe fact that in this kind of profiles the detachment of theflow is geometry induced (and almost independent of theReynolds number) instead of a pressure induced separationwhich requires a better modelling of the boundary layerA mesh (Figure 6) without refinement for the boundarylayer created with 40604 triangle elements has been usedin this computation The Euler solution has the advantageof requiring coarse meshes since the boundary layer is notsolved which obviously will affect the computational costand taking into account the large number of simulations to

Table 3 Comparison between the computational cost of Euler andRANS calculations for 120572 = 60

∘ Similar values are obtained in otherconfigurations

Number ofelements Residual Conver

iterationsComp time

(min)Euler 12923 35119890 minus 4 200000 160Fine-SA 57408 38119890 minus 6 200000 672Fine-119896-120596 SST 57408 10119890 minus 8 200000 770

carry out the choice of the turbulence model is balancedbetween the accuracy of the results and the computationaltime The aim of this comparison is to check the capacity ofthe turbulence model in obtaining the base pressure in thevortex detached zone behind the profile which have a strongimpact on the final forces

Figure 8 shows a comparison between the three simula-tions It is observed that the Euler results underpredict thestall area which confirms the fact that even in this kind ofconfigurations it is necessary to account for the viscosity inthe prediction of the detachmentTherefore it is necessary touse Navier-Stokes with turbulence model even if it means anincrease in the computational time as observed in Table 3

An additional analysis of the simulations shows goodagreement with the experimental results However there aretwo areas where the turbulence models and the experimentsshow largest discrepancies (see Figure 8)The 119888

119897max and the 119888119889


minus005 0 005

X

004

002

0

minus002

minus004

Z

minus005 0 005

X

minus005 0 005

X

minus005 0 005

X

004

002

0

minus002

minus004

Z

004

002

0

minus002

minus004

Z

004

002

0

minus002

minus004

Z

Figure 4 Validation grids with different refinement (coarse medium fine and extrafine)

066

064

062

06

058

056

00E + 00 20E + 04 40E + 04 60E + 04 80E + 04 10E + 05

Number of elements20E + 04 40E + 04 60E + 04 80E + 04

Number of elements

c l

Resid

ual

CoarseMediumFine

Extrafine CoarseMedium

FineExtrafineExperimental

5E minus 054E minus 05

3E minus 05

2E minus 05

1E minus 05

Figure 5 Convergence of the aerodynamic coefficients and residual in function of the number of elements

at high angle of attack zones are better predicted by the 119896-120596SST The 119896-120596 SST model estimates with more accuracy thebase pressure value giving overall better results Finally the119896-120596 SST has been selected to solve all the configurations Amore detailed study of these two areas is shown in Section 32

As an additional validation it has been checked thepossibility that the wind tunnel blockage could substantially

modify the numerical results To this end the dimensions ofthe wind tunnel test chamber which is 015m width 090mhigh and 120m long have been reproduced In this casedifferentmeshes for the following angles of attack 0∘ 20∘ 40∘60∘ 80∘ 100∘ 120∘ 140∘ 160∘ and 180∘ have been created

The results in Figure 7 show no difference with and with-out tunnel blockage consideration This conclusion allows


001

minus002

0

minus001

minus002

minus003

minus004

X

Z

Figure 6 Euler mesh without refinement for the boundary layer

1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

2

18

16

14

12

1

08

06

04

02

120572

Spalart-Allmaras fine meshSpalart-Allmaras wind tunnel

Experimental

c l c d

0 20 40 60 80 100 120 140 160 180

120572

20 40 60 80 100 120 140 160 180


Experimental

Figure 7 Checking the possibility of wind tunnel blockage simulation in the experimental results with Spalart-Allmaras

120572

0 20 40 60 80 100 120 140 160 180

120572

20 40 60 80 100 120 140 160 180

1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

c l

minus12

Euler equationsSpalart-Allmaras Experimental


2

18

16

14

12

1

08

06

04

02

c d

24

22

k-120596 SST k-120596 SST

Figure 8 Validation model turbulence in medium grid 119899 = 5 120593 = 45∘ with Euler equations Spalart-Allmaras and 119896-120596 SST turbulence

model


minus04minus06minus08minus1

minus12

2181614121

08060402

22

0minus02


minus12

2181614121

08060402

22

0minus02


minus12

2181614121

08060402

22

0minus02

minus14

c dcl

c dcl

c dcl

0 15 30 45 60 75 90 105 120 135 150 165 180

120572

180

180

0 15 30 45 60 75 90 105 120 135 150 165

120572

0 15 30 45 60 75 90 105 120 135 150 165

120572

n = 1 n = 3

n = 5 n = 7

Experimental cdExperimental cl



minus12

2181614121

080604020

minus02c dcl

0 15 30 45 60 75 90 105 120 135 150 165 180

120572

k-120596 SST cd

k-120596 SST cl

k-120596 SST cd

k-120596 SST cl

Figure 9 Lift and drag coefficient integral over the surface profile of each configuration (119899 = 1 3 5 and 7) for 120572 = 45∘

using the mesh with circle domain for all the rest of configu-rations which facilitates the numerical study in terms of gridgeneration

3 Numerical Results for GallopingStability Analysis

After completion of the problem definition mesh andturbulence model validation the simulation of the gallopinginstability maps was computed using the fine mesh and a 119896-120596SST RANS turbulence model

The computations are ldquosteadyrdquo which means that someflow unsteady features that can appear in the solution arenumerically dampedWe will see in our analysis that in mostof the cases the numerical steady solutions give enough res-olution for standard industrial problems however unsteadydetailed computations can improve the global estimations incertain configurations where as we will see the unsteadyfeatures are more important

As previously mentioned the angle of attack is variedfrom 0∘ to 180∘ with an increment of 2∘ making a total of 46angles of attack for each configuration

31 Aerodynamic Coefficients As a preliminary resultFigure 9 shows the lift and drag coefficient integrated overthe surface profile for four representative configurations

(119899 = 1 119899 = 3 119899 = 5 and 119899 = 7) of the 120593 = 45∘ family profiles

The numerical results agree well with the experimentalvalues even in the critical 119888

119897max and in the 119888119889values at high

angles of attack regions This is not the case for the 120593 = 90∘

family profiles In Figure 10 the lift and drag coefficientsfor configurations 119899 = 1 3 5 and 7 for 120593 = 90

∘ areshown In these cases some differences in the areas of 119888

119897maxand the 119888

119889values at high angles of attack are numerically

underestimatedComparison of the pressure distribution at 120593 = 90

∘

and 119899 = 4 provides an explanation of the reasons of thesediscrepancies Figure 11 sketches the values of 119888

119901as a function

of the coordinates 119883 and 119885 at 120572 = 75∘ in the region of

119888119897max It is shown that on the upstream part of the profilenumerical and experimental results agree accordingly but asexpected in the downstream region the detached flow showsdifferences between the base pressures These differencesare the cause of the discrepancies between the aerodynamicforces

The main reason of this inaccuracy is due to the strongunsteady flow which appears in that region and it is notproperly solved by a steady RANS solver A better resolutionshould be obtained by performing an unsteady simulationwith a time step small enough to solve these structures As wewill see in Section 32 these structures have a strong impacton the evaluation of the global coefficients




k-120596 SST cdk-120596 SST cl


120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

minus04minus06minus08

2181614121

08060402

22

0minus02

c dcl

minus04minus06

2181614121

08060402

22

0minus02

c dcl


minus12

2181614121

080604020

minus02c dcl


2181614121

08060402

22

0minus02

c dcl

n = 1 n = 3

n = 5 n = 7

Figure 10 Lift and drag coefficient integrate over the surface profile of each configuration (119899 = 1 3 5 and 7) for 120572 = 90∘

minus2

minus15

minus1

minus05

0

05

1

minus002

minus002

minus001

X

c p

cp

TAU k-120596 fine meshExperimental

Z

002

002

001 minus2 minus1 0

0

0 1

ExperimentalTAU k-120596 fine mesh

Figure 11 Value of pressure coefficient in function of the both coordinates119883 and 119885 for 120572 = 75∘ in configuration 119899 = 4 120593 = 90

∘

For demonstration purposes Figure 12 also shows the 119888119901

values as a function of the coordinates 119883 and 119885 in this caseat 120572 = 100

∘ where a good agreement between numerical andexperimental lift is obtained As expected the numerical andexperimental pressure values are in good consonance

However we will see in Section 33 that these discrep-ancies do not have a strong influence in the stability maps

since they are more defined by the global behavior of thecoefficients than the detailed solutions

32 Unsteady Simulation (URANS) The URANS numericalmethod is used to improve the prediction and explain thediscrepancies in the two areas where a lack of accuracy hasappeared the 119888

119897max and the 119888119889 values at high angles of attack It



minus2

minus15

minus1

minus05

0

05

1

c p

minus002

Z

002

0

cp

minus2 minus1 0 1minus002 minus001

X

0020010



∘

is important to see if by improving the numerical resolutionof the solution it is possible to obtain additional details ofthe downstream flow structures which eventually producechanges on the average pressure distribution over the profile

The URANS computations use a dual-time steppingscheme A Backward difference formula (BDF) is used to dis-cretize the global time and a fourth-order Runge Kutta withmultigrid acceleration scheme being employed to obtain asteady state in the fictitious pseudotime Taking into accountthat the frequency of the oscillation is unknown a priori inorder to define the global time-step of the computations firsta characteristic time scale is defined based on the convectivevelocity and the length of the profile (119905

119888asymp 119871119880

infin) The first

estimation of the time-step is taken as Δ119905119888= 119905119888100 With

this time step an unsteady oscillation of 100Hz frequencyis computed Finally a more detailed simulation has beencomputed taking 100 steps per cycle giving a global time stepof (with the CFL = 1) Δ119905

119888= 10minus5

The unsteady computation is initialized with the steadysolution and evolved until statistical convergence whichrequires a total number of 6sdot104 time steps to get a simulationtime of 06 sThemean values of this unsteady simulation arecompared with the results obtained with the RANS solver

Figure 13 shows the comparison between the aerody-namic coefficients obtained with the RANS and URANSsimulations and the experimental results As expected the119888119897max and the 119888

119889values at high angles of attack have greatly

improved compared to the experimental results The rest ofthe angles of attack where unsteady effects are negligiblekeep the same accuracy of steady computations By com-paring the mean pressure distribution (Figure 14) at angle ofattack120572 = 165

∘ it is observed how theURANS resultsmodifythemeanbase pressure to provide a better accuracy comparedto experimental data on 119888

119897and 119888119889 An additional improvement

should be possible if complex numerical models like LES orDES are used but at the cost of increasing even more thecomputational effort

Finally for demonstration purposes Figure 15 repro-duces the pressure contours of the numerical solution at

Table 4 Comparison between RANS URANS and experimentalvalues of 119888

119897 119888119889and the computational time for the angle of attack

120572 = 80∘

120572 = 80∘

119888119897

119888119889

Comp time (h)Experimental 114 07155 mdashSteady 0755 0584 11Unsteady 1108 0728 238

different time steps in a temporal period of lift The flowfollows a typical von Karman vortex street with a maindetachment frequency of 76Hz the detachment appears inboth wings of the profile alternately

In terms of computational effort Table 4 shows the resultsof 119888119897and 119888

119889and the computational time for the angle of

attack 120572 = 80∘ between the steady and unsteady simulations

The unsteady simulation takes around 24 times the steadycomputational time this is a difference to be taken intoaccount

33 Stability Maps The stability parameter (119867 = d119888119897d120572 +

119888119889) following the Glauert-Den Hartog criterion and the

numerical computations have been calculated with a centraldifference approximation for the first derivative Despitethe detailed numerical analysis which shows that for someangles of attack better accuracy is obtained with unsteadysimulations in the computation of the stability maps onlyldquoindustrialrdquo steady solutions are considered for comparison

In Figure 16 the numerical and experimental values of119867 as a function of the parameter 119899 are compared for bothconfigurations studied 120593 = 45

∘ and 120593 = 90∘ Inside the

shadow areas the stability parameter (119867) value is negativein other words the Glauert-Den Hartog criterion is satisfiedand galloping instability appears Consider

120593 = 45∘ (5)

As it can be observed a good agreement between theexperimental and numerical results is obtained even for the


Steady k-120596 SSTUnsteady k-120596 SST

Experimental

60 75 90 105 120 135 150 165 180120572

90 105 120 135 150 165 180120572


Experimental

2

18

16

14

12

1

08

06

04

22

24

26

minus04

minus06

minus08

12

1

08

06

04

02

0

minus02

c l c d

Figure 13 Aerodynamic coefficients comparison between the RANS and URANS simulation with the experimental results in configuration119899 = 4 120593 = 90

∘

XX

XX

XXXXXXXXXXXXXXXXXX

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

XX

XX

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

X

XXXXXXXXXXXXXXXXXXXXX

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

X


X

minus2

minus15

minus1

minus05

0

05

1

c p

minus25

minus002 minus001

X

0020010 minus2 minus15 minus1 minus05 0 05 1

cp

minus25

003

002

001

0

minus001

minus002

minus003

Z

Unsteady TAU k-120596 fine meshSteady TAU k-120596 fine mesh

Experimental XUnsteady TAU k-120596 fine meshSteady TAU k-120596 fine mesh

Experimental

Figure 14 Pressure coefficient comparison between RANS and URANS at angle of attack 120572 = 165∘ configuration 119899 = 4 120593 = 90

∘

critical regions of 119888119897max and high angles of attackNonetheless

some differences are observed in the family profiles 120593 = 45∘

a good agreement in the area where the instability appears (inthe range of angles of attack between 120572 = 40

∘ and 120572 = 60∘) is

found As depicted in Figure 9 this area corresponds to theregion of maximum lift and close to the minimum drag andit is related to the two-peak structure shown for higher valuesof 119899 Indeed for lower values of 119899 the lift curve shows a largeflat zone close to its maximum and galloping instability doesnot show up For 119899 = 3 or higher a first peak of maximumlift appears around 120572 = 40

∘ followed by a strong decayand a second maximum close to 120572sim60∘ndash70∘ This effect ismore accused as long as 119899 is increased It is precisely thisstructure which defines the galloping instability At highervalues of 119899sim6-7 the two peaks separate giving place to a smalllift flat region between them this eventually can produce astable area between these two branches Although the globalmagnitudes arewell predicted this particular detail is notwell

captured by the numericalmethodmainly because of the lackof resolution in the Δ120572 of the predicted lift curves Consider

120593 = 90∘ (6)

Despite some numerical inaccuracies observed in thenumerical computations of this case the instability regionsof the steady numerical solutions are well captured and agreewith the experimental results

Two main regions of instability appear for this configura-tion a region at low angles of attack (between120572 = 20

∘ and120572 =

40∘) for 119899 values ranging from 2 to 6 and a second area for

with 120572 = 90∘ndash110∘ for all 119899The first region is related to a peak

of lift at around 120572 = 20∘ Although the numerical results are

somehow inaccurate and this maximum lift is overpredicted(making necessary unsteady computation to improve theresults) this inaccuracy has almost no effect on the instabilityregions since the global behavior is well capturedThe second


clcd

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

14

12

1

08

06

0505

051

0515

052

0525

t (s)

120572 = 80∘

04

02

0

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

c lc

d

(a)

3

2

1

0

minus1

01 011 012 013 014

120572 = 165∘

Pres

sure

100800

100200

99600

99000

98400

t (s)

cl

cd

04

02

0

Z

04

02

0

Z

04

02

0Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Pres

sure

100800

100200

99600

99000

98400

c lc

d

(b)

Figure 15 Pressure field of a period at angle of attack 120572 = 80∘ (a) and 165∘ (b)

region (120572 = 90∘ndash110∘) shows a similar behavior to the

case 120593 = 45∘ The presence of two maxima in the lift co-

efficient in a region of growing drag is the cause of gallop-ing The region is slightly wider for larger values 119899 effectwhich is well reproduced by the numerical and experimentaldata

An additional and small region appears at high anglesof attack (120572sim160∘ndash170∘) This region is predicted by bothnumerical and experimental results but slightly shifted tolower 120572 in the numerical results The reason to that is thelack of numerical precision in this areaThe problemhas been

discussed in the previous section and unsteady numericalcomputations are able to improve these results

Even with these differences only with the numericalresults we are able to design our geometry in terms of thestability region Within the shadow region the necessarycondition for transverse galloping is satisfied so this designgeometry is appropriated to extract energy through move-ment caused by this phenomenon Otherwise if the goal is toavoid this phenomenon the geometry will be designed witha configuration which is outside of the transverse gallopinginstability zone


7

6

5

4

3

2

10 0

7

6

5

4

3

2

120 40 60 8080 100 120 140 160 180 20 40 60 100 120 140 160 180

120593 = 45∘

120593 = 90∘

n n

120572 120572

ExperimentalTAU k120596-SST


Figure 16 Stability map of 120593 = 45∘ and 120593 = 90

∘ family profiles as a function of 119899 and 120572

4 Conclusions

The use of wind tunnel tests for the evaluation of the aero-dynamic coefficients can be sometimes difficult and costlyThis paper investigates the suitability of numerical models forthe prediction of galloping instability of bodies with differentcross-sections A numerical method based on the DLRTAU Code has been used The method has been validatedcomparing the results with experimental measurements fora cross-section which is very interesting from the civilconstruction point of view the Z-profile The Glauert-DenHartog criterion has been used to determine the gallopingregions of instability with respect to the incident wind angleof attack and cross-section geometrical parameters

This paper reports the results obtained in a systematicanalysis of the galloping instability performed on a family ofZ-shaped cross-sections This profile is made out of a centralweb and two sidewingsThe analysis determines underwhichinterval of angles of attacks and geometrical parameters andthis type of profiles become unstable From the computedaerodynamic coefficients the 119867(120593 120572 119899) = d119888

119897d120572 + 119888

119889

functions have been determined and from them stabilitymaps in the 119899-120572 plane have been plotted

According to the results the family of profiles with betterperformance against the apparition of transverse gallopinginstabilities is120593 = 45

∘Within this family instabilities can stillappear at angles of attack between 40∘ and 60∘ contrary toprofiles with smaller aspect ratio for which this phenomenondoes not appear On the contrary if the goal is to induce thisphenomenon to extract energy from the movement [39] thesuitable profiles are with120593 = 90

∘ because they aremore proneto gallop

Good accuracy in the prediction of the galloping insta-bility with the numerical method (DLR TAU Code) has beenobtained

The methodology employed allows important savings incomputational effort

It has been observed that the RANS simulation is goodenough to predict the necessary condition for the occurrenceof the transverse galloping phenomenon that is the Glauert-Den Hartog criterion is satisfied However better accuracycan be obtained (at a higher computational costs) in the areawhere some discrepancies still appear

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J P den Hartog Mechanical Vibrations McGraw-Hill NewYork NY USA 4th edition 1956

[2] G Parkinson and J Smith ldquoThe square prism as an aeroelasticnon-linear oscillatorrdquo The Quarterly Journal of Mechanics andApplied Mathematics vol 17 no 2 pp 225ndash239 1964

[3] MNovak ldquoAeroelastic galloping of prismatic bodiesrdquo Journal ofEngineering Mechanics Division vol 96 no 1 pp 115ndash142 1969

[4] M Novak ldquoGalloping oscillations of prismatic structuresrdquoJournal of Engineering Mechanics Division vol 98 no 1 pp 27ndash46 1972

[5] E Simiu and R H ScanlanWind Effects on Structures Funda-mentals and Applications to Design John Wiley amp Sons NewYork NY USA 1996


[6] M Novak and H Tanaka ldquoEffect of turbulence on gallopinginstabilityrdquo Journal of the Engineering Mechanics Division vol100 no 1 pp 27ndash47 1974

[7] Q S Li J Q Fang and A P Jeary ldquoEvaluation of 2D coupledgalloping oscillations of slender structuresrdquo Computers andStructures vol 66 no 5 pp 513ndash523 1998

[8] C Ziller and H Ruscheweyh ldquoA new approach for determiningthe onset velocity of galloping instability taking into accountthe nonlinearity of the aerodynamic damping characteristicrdquoJournal of Wind Engineering and Industrial Aerodynamics vol69-71 pp 303ndash314 1997

[9] P Hemon F Santi B Schnoerringer and J WojciechowskildquoInfluence of free-stream turbulence on themovement-inducedvibrations of an elongated rectangular cylinder in cross-flowrdquoJournal of Wind Engineering and Industrial Aerodynamics vol89 no 14-15 pp 1383ndash1395 2001

[10] P Hemon and F Santi ldquoOn the aeroelastic behaviour of rectan-gular cylinders in cross-flowrdquo Journal of Fluids and Structuresvol 16 no 7 pp 855ndash889 2002

[11] S C Luo Y T Chew and Y T Ng ldquoHysteresis phenomenon inthe galloping oscillation of a square cylinderrdquo Journal of Fluidsand Structures vol 18 no 1 pp 103ndash118 2003

[12] A Barrero-Gil A Sanz-Andres and G Alonso ldquoHysteresis intransverse galloping the role of the inflection pointsrdquo Journal ofFluids and Structures vol 25 no 6 pp 1007ndash1020 2009

[13] H RuscheweyhMHortmanns andC Schnakenberg ldquoVortex-excited vibrations and galloping of slender elementsrdquo Journal ofWind Engineering and Industrial Aerodynamics vol 65 no 1ndash3pp 347ndash352 1996

[14] H Kawai ldquoEffect of corner modifications on aeroelastic insta-bilities of tall buildingsrdquo Journal of Wind Engineering andIndustrial Aerodynamics vol 74ndash76 pp 719ndash729 1998

[15] S C Luo Y T Chew T S Lee and M G Yazdani ldquoStability totranslational galloping vibration of cylinders at different meanangles of attackrdquo Journal of Sound and Vibration vol 215 no 5pp 1183ndash1194 1998

[16] A S Richardson ldquoPredicting galloping amplitudesrdquo ASCEJournal of Engineering Mechanics vol 114 no 4 pp 716ndash7231988

[17] R D Blevins Flow-Induced Vibrations Krieger Malabar FlaUSA 2nd edition 1990

[18] E Naudascher and D Rockwell Flow-Induced Vibrations AnEngineering Guide AA Balkema RotterdamTheNetherlands1994

[19] G Alonso J Meseguer and I Perez-Grande ldquoGalloping insta-bilities of two-dimensional triangular cross-section bodiesrdquoExperiments in Fluids vol 38 no 6 pp 789ndash795 2005

[20] G Alonso J Meseguer and I Perez-Grande ldquoGalloping stabil-ity of triangular cross-sectional bodies a systematic approachrdquoJournal of Wind Engineering and Industrial Aerodynamics vol95 no 9ndash11 pp 928ndash940 2007

[21] G Alonso and J Meseguer ldquoA parametric study of the gallopingstability of two-dimensional triangular cross-section bodiesrdquoJournal ofWind Engineering amp Industrial Aerodynamics vol 94no 4 pp 241ndash253 2006

[22] G Alonso J Meseguer A Sanz-Andres and E Valero ldquoOn thegalloping instability of two-dimensional bodies having ellipticalcross-sectionsrdquo Journal of Wind Engineering and IndustrialAerodynamics vol 98 no 8-9 pp 438ndash448 2010

[23] G Alonso E Valero and J Meseguer ldquoAn analysis on thedependence on cross section geometry of galloping stability of

two-dimensional bodies having either biconvex or rhomboidalcross sectionsrdquo European Journal of Mechanics B vol 28 no 2pp 328ndash334 2009

[24] G Alonso I Perez-Grande and J Meseguer ldquoGalloping insta-bilities ofZ-shaped shading louversrdquo Indoor and Built Environ-ment In press

[25] T Tamura I Ohta and K Kuwahara ldquoOn the reliability of two-dimensional simulation for unsteady flows around a cylinder-type structurerdquo Journal of Wind Engineering and IndustrialAerodynamics vol 35 no 1ndash3 pp 275ndash298 1990

[26] K Hayashi and Y Ohya ldquoA numerical study of the flow arounda rectangular cylinder with critical depthrdquo in Proceedings of the16th National Symposium on Wind Engineering pp 179ndash1842000 (Japanese)

[27] W Rodi ldquoComparison of LES and RANS calculations of theflow around bluff bodiesrdquo Journal of Wind Engineering andIndustrial Aerodynamics vol 69ndash71 pp 55ndash75 1997

[28] HHirano AMaruoka and SWatanabe ldquoCalculations of aero-dynamic properties of rectangular cylinder with slendernessratio of 21 under various angles of attackrdquo Journal of StructuralEngineering A vol 48 pp 971ndash978 2002

[29] S Oka and T Ishihara ldquoNumerical study of aerodynamiccharacteristics of a square prism in a uniform flowrdquo Journal ofWindEngineering and Industrial Aerodynamics vol 97 no 11-12pp 548ndash559 2009

[30] G Bosch and W Rodi ldquoSimulation of vortex shedding past asquare cylinder with different turbulencemodelsrdquo InternationalJournal Numerical Methods in Fluids vol 28 pp 601ndash616 1998

[31] Y Bao D Zhou C Huang Q Wu and X Chen ldquoNumericalprediction of aerodynamic characteristics of prismatic cylinderby finite element method with Spalart-Allmaras turbulencemodelrdquo Computers and Structures vol 89 no 3-4 pp 325ndash3282011

[32] T S Lee ldquoEarly stages of an impulsively started unsteadylaminar flow past tapered trapezoidal cylindersrdquo InternationalJournal for Numerical Methods in Fluids vol 26 no 10 pp 1181ndash1203 1998

[33] M Cheng and G R Liu ldquoEffects of afterbody shape on flowaround prismatic cylindersrdquo Journal of Wind Engineering ampIndustrial Aerodynamics vol 84 no 2 pp 181ndash196 2000

[34] I Robertson L Li S J Sherwin and PW Bearman ldquoA numer-ical study of rotational and transverse galloping rectangularbodiesrdquo Journal of Fluids and Structures vol 17 no 5 pp 681ndash699 2003

[35] DLR Institute TAU-Code User Guide Release 201120 2011[36] P R Spalart and S R Allmaras ldquoA one-equation turbulence

model for aerodynamic flowsrdquo AIAA Paper 92-0439 1992[37] F R Menter ldquoZonal two equation k-120596 turbulence models for

aerodynamic flowsrdquo AIAA Paper 93-2906 1993[38] E Turkel R Radespiel and N Kroll ldquoAssessment of pre-

conditioning methods for multidimensional aerodynamicsrdquoComputers and Fluids vol 26 no 6 pp 613ndash634 1997

[39] G Alonso A Sanz-Andres and A Barrero-Gil ldquoEnergyharvesting from transverse gallopingrdquo Journal of Sound andVibration vol 329 no 14 pp 2873ndash2883 2010


In these expressions 120588 stands for the fluid density 119888 fora characteristic length normal to incident flow 119888

119889for the

drag coefficient and 119888119897for the lift coefficient The projection

of those forces in 119911-axis direction is 119891119911(120572) = minus119889(120572) sin120572 minus

119897(120572) cos120572 On the other hand the equation of the movementof the body is

119898[d2119911d1199052

+ 2120577120596119899

d119911d119905

+ 1205962

119899119911] = 119891

119911(120572) (2)

Considering the movement to be quasisteady and assum-ing the angles of attack due to oscillation to be small enough(120572 ≪ 1) from the expression (2) the Glauert-Den Hartogcriterion for galloping instability is derived In effect if theaerodynamic force (proportional in this case to d119911d119905) isconsidered as a contribution to the total damping of thesystem (aerodynamic damping) the total damping coefficientis

120577119879= 120577 +

120588119880119888

4119898120596119899

(d119888119897

d120572+ 119888119889)

10038161003816100381610038161003816100381610038161003816120572=0

(3)

and therefore the oscillation will be stable if 120577119879

gt 0 andunstable if 120577

119879lt 0 As the mechanical damping 120577 is generally

positive instability will only occur if (d119888119897d120572 + 119888

119889) lt 0

expression known as Glauert-Den Hartog criterion which isa necessary condition for galloping instability The sufficientcondition is 120577

119879lt 0 or according to (3)

(d119888119897

d120572+ 119888119889)

10038161003816100381610038161003816100381610038161003816120572=0

lt minus4119898120577120596

119899

120588119880119888 (4)

Later in a number of papers Novak [3 4] extendedthe analysis of transverse galloping to the three-dimensionalcase Other researchers have investigated other interestingphenomena like the influence of turbulence [6ndash10] or thehysteresis phenomenon [11 12] in transverse galloping Inthe last decades besides theoretical work large efforts havebeen devoted to experimentally study the galloping featuresof many bodies having different cross-sections Althoughmost of the effort in galloping oscillation research has beenconcentrated in bodies with square or rectangular cross-sections prismatic bodies with other cross-sectional shapeshave been also considered [13ndash18]

In the last years some research on galloping has beencarried out at IDRUPM because of the above-mentionedinterest on the use of large louvers for sun-shading in buildingfacades amongst other applications A systematic parametricanalysis of simple cross-section two-dimensional bodies hasbeen accomplished The geometries analysed up to now areisosceles triangular cross-sections (the varying parameterbeing the main vertex angle 120573 [19ndash21] and ellipses [22]as well as biconvex and rhomboidal cross-sections [23] andthe varying parameter in these last cases being the relativethickness of the cross-section 120591 In all cases the unstableregions in the 120573-120572 plane (isosceles triangles) and in the 120591-120572plane (elliptical biconvex and rhomboidal bodies) where 120572stands for the wind angle of incidence were determined)

Z-shaped cross-sections are not rare in modern archi-tecture Sun-shading louvers are examples where Z-shaped

cross-sections may be found in slender structures susceptibleto gallop This type of Z-shaped cross-sections has also beenexperimentally analysed [24]

The validity of the numerical methods in the study of thiskind of problemswould have an important added value sinceit would allow for predicting the aerodynamic propertiesof different shapes and profiles without performing themore expensive experimental tests However most of thenumerical studies encountered are mainly focused on under-standing the physical mechanism involved in flow separationof bluff bodies such as circular rectangular square andtrapezoidal cylinders in a particular configuration In thisline different high-fidelity numerical simulations have beenperformed using direct numerical simulation (DNS) largeEddy simulation (LES) Reynolds-averaged Navier-Stokes(RANS) or URANS models Examples of these studies oncylinders of rectangular sections are provided by Tamura etal [25] Hayashi and Ohya [26] and Rodi [27] in computingthe mean aerodynamic coefficients using DNS Hirano et al[28] performed an LES simulation of the flow showing goodaccuracy compared to the experimental results for meandrag lift and the Strouhal number or Oka and Ishihara[29] who also include an accurate method for estimating theaerodynamic coefficients using a systematic elongation of thespanwise length Several RANS computations can be foundin Bosch and Rodi [30] using the 119896-120576 turbulence model andBao et al [31] who tried to use one-equation models such asSpalart-Allmaras RANS models provide considerable savingin computational effort but normally have problems to getaccuracy in flows with high pressure gradients

Apart from the square and circular cross-sections trape-zoidal and triangular sections have also been analysed withdetails by several researchers Lee [32] Cheng and Liu [33]and Oka and Ishihara [29] and Bao et al [31] have studiedthe developing of the recirculation region vortex sheddingphenomena and its interaction with the separated shear layerThese detailed works draw conclusions about the variation ofthe Strouhal number with the Reynolds number and aspectratio

Currently these studies have been focused on the predic-tion of drag lift Strouhal number and flow features in thedownstream region of the profile for one particular config-uration showing that precise numerical computations canprovide valuable information of the flow solutions in goodagreement with experimental data However a thoroughanalysis of galloping stability requires the study of a largenumber of parameters (geometrical and physical) whichmakes previous computations though accurate nonpracti-cal In this line Robertson et al [34] made a numericalstudy of rotational and transverse galloping rectangularbodies using two-dimensional spectral high-order methodsVery good qualitative results were obtained showing thepossibility to use numerical methods in the prediction anddiscussion of this problem The work was performed at verylow Reynolds numbers (250) far beyond current industrialconfigurations where turbulence effects can be of importanceand although the separation point is almost independent ofthe Reynolds number in rectangular structures it is not sofor the reattachment length and base pressure


w

s

ap

k

h

U

120572

120593








119899 = 1 119899 = 2 119899 = 3 119899 = 4 119899 = 5 119899 = 6 119899 = 7

120593 = 45∘

120593 = 90∘


90∘ with 119899 = 1 2 7 and 120593 = 45




119889= 2119889120588119880

2

infin119888 119888119897=

21198971205881198802



119901= 2(119901 minus 119901

infin)1205881198802

infin with

119901infin






n = 1

2

3

4

5

6

7

(a)

n = 1

2

3

4

5

6

7

(b)


∘ (b) with 119899 = 1 2 7










+ carefully (119910+ = 119906+Δ119910]) 119910+



2and 119896

4







1050minus5

10

5

0

minus5

minus10

X

Z

X

Z

0minus002minus004

002

0

minus002

minus004











iterationsComp time





119897max and the 119888119889


minus005 0 005

X

004

002

0

minus002

minus004

Z

minus005 0 005

X

minus005 0 005

X

minus005 0 005

X

004

002

0

minus002

minus004

Z

004

002

0

minus002

minus004

Z

004

002

0

minus002

minus004

Z


066

064

062

06

058

056

00E + 00 20E + 04 40E + 04 60E + 04 80E + 04 10E + 05


Number of elements

c l

Resid

ual

CoarseMediumFine




3E minus 05

2E minus 05

1E minus 05







001

minus002

0

minus001

minus002

minus003

minus004

X

Z


1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

2

18

16

14

12

1

08

06

04

02

120572


Experimental

c l c d

0 20 40 60 80 100 120 140 160 180

120572

20 40 60 80 100 120 140 160 180


Experimental


120572

0 20 40 60 80 100 120 140 160 180

120572

20 40 60 80 100 120 140 160 180

1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

c l

minus12



2

18

16

14

12

1

08

06

04

02

c d

24

22

k-120596 SST k-120596 SST


model



minus12

2181614121

08060402

22

0minus02


minus12

2181614121

08060402

22

0minus02


minus12

2181614121

08060402

22

0minus02

minus14

c dcl

c dcl

c dcl

0 15 30 45 60 75 90 105 120 135 150 165 180

120572

180

180

0 15 30 45 60 75 90 105 120 135 150 165

120572

0 15 30 45 60 75 90 105 120 135 150 165

120572

n = 1 n = 3

n = 5 n = 7




minus12

2181614121

080604020

minus02c dcl

0 15 30 45 60 75 90 105 120 135 150 165 180

120572

k-120596 SST cd

k-120596 SST cl

k-120596 SST cd

k-120596 SST cl














119897maxand the 119888



∘


119901as a function









120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180


2181614121

08060402

22

0minus02

c dcl

minus04minus06

2181614121

08060402

22

0minus02

c dcl


minus12

2181614121

080604020

minus02c dcl


2181614121

08060402

22

0minus02

c dcl

n = 1 n = 3

n = 5 n = 7


minus2

minus15

minus1

minus05

0

05

1

minus002

minus002

minus001

X

c p

cp


Z

002

002

001 minus2 minus1 0

0

0 1



∘










minus2

minus15

minus1

minus05

0

05

1

c p

minus002

Z

002

0

cp


X

0020010



∘



119888asymp 119871119880

infin) The first



119888= 10minus5











120572 = 80∘

120572 = 80∘

119888119897

119888119889













120593 = 45∘ (5)




Experimental

60 75 90 105 120 135 150 165 180120572

90 105 120 135 150 165 180120572


Experimental

2

18

16

14

12

1

08

06

04

22

24

26

minus04

minus06

minus08

12

1

08

06

04

02

0

minus02

c l c d


∘

XX

XX

XXXXXXXXXXXXXXXXXX


XX

XX


X



X


X

minus2

minus15

minus1

minus05

0

05

1

c p

minus25

minus002 minus001

X


cp

minus25

003

002

001

0

minus001

minus002

minus003

Z



Experimental


∘




∘ and 120572 = 60∘) is




120593 = 90∘ (6)



∘ and120572 =






clcd

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

14

12

1

08

06

0505

051

0515

052

0525

t (s)

120572 = 80∘

04

02

0

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

c lc

d

(a)

3

2

1

0

minus1

01 011 012 013 014

120572 = 165∘

Pres

sure

100800

100200

99600

99000

98400

t (s)

cl

cd

04

02

0

Z

04

02

0

Z

04

02

0Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Pres

sure

100800

100200

99600

99000

98400

c lc

d

(b)









7

6

5

4

3

2

10 0

7

6

5

4

3

2

120 40 60 8080 100 120 140 160 180 20 40 60 100 120 140 160 180

120593 = 45∘

120593 = 90∘

n n

120572 120572





4 Conclusions



119897d120572 + 119888

119889










References











































w

s

ap

k

h

U

120572

120593








119899 = 1 119899 = 2 119899 = 3 119899 = 4 119899 = 5 119899 = 6 119899 = 7

120593 = 45∘

120593 = 90∘


90∘ with 119899 = 1 2 7 and 120593 = 45




119889= 2119889120588119880

2

infin119888 119888119897=

21198971205881198802



119901= 2(119901 minus 119901

infin)1205881198802

infin with

119901infin






n = 1

2

3

4

5

6

7

(a)

n = 1

2

3

4

5

6

7

(b)


∘ (b) with 119899 = 1 2 7










+ carefully (119910+ = 119906+Δ119910]) 119910+



2and 119896

4







1050minus5

10

5

0

minus5

minus10

X

Z

X

Z

0minus002minus004

002

0

minus002

minus004











iterationsComp time





119897max and the 119888119889


minus005 0 005

X

004

002

0

minus002

minus004

Z

minus005 0 005

X

minus005 0 005

X

minus005 0 005

X

004

002

0

minus002

minus004

Z

004

002

0

minus002

minus004

Z

004

002

0

minus002

minus004

Z


066

064

062

06

058

056

00E + 00 20E + 04 40E + 04 60E + 04 80E + 04 10E + 05


Number of elements

c l

Resid

ual

CoarseMediumFine




3E minus 05

2E minus 05

1E minus 05







001

minus002

0

minus001

minus002

minus003

minus004

X

Z


1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

2

18

16

14

12

1

08

06

04

02

120572


Experimental

c l c d

0 20 40 60 80 100 120 140 160 180

120572

20 40 60 80 100 120 140 160 180


Experimental


120572

0 20 40 60 80 100 120 140 160 180

120572

20 40 60 80 100 120 140 160 180

1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

c l

minus12



2

18

16

14

12

1

08

06

04

02

c d

24

22

k-120596 SST k-120596 SST


model



minus12

2181614121

08060402

22

0minus02


minus12

2181614121

08060402

22

0minus02


minus12

2181614121

08060402

22

0minus02

minus14

c dcl

c dcl

c dcl

0 15 30 45 60 75 90 105 120 135 150 165 180

120572

180

180

0 15 30 45 60 75 90 105 120 135 150 165

120572

0 15 30 45 60 75 90 105 120 135 150 165

120572

n = 1 n = 3

n = 5 n = 7




minus12

2181614121

080604020

minus02c dcl

0 15 30 45 60 75 90 105 120 135 150 165 180

120572

k-120596 SST cd

k-120596 SST cl

k-120596 SST cd

k-120596 SST cl














119897maxand the 119888



∘


119901as a function









120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180


2181614121

08060402

22

0minus02

c dcl

minus04minus06

2181614121

08060402

22

0minus02

c dcl


minus12

2181614121

080604020

minus02c dcl


2181614121

08060402

22

0minus02

c dcl

n = 1 n = 3

n = 5 n = 7


minus2

minus15

minus1

minus05

0

05

1

minus002

minus002

minus001

X

c p

cp


Z

002

002

001 minus2 minus1 0

0

0 1



∘










minus2

minus15

minus1

minus05

0

05

1

c p

minus002

Z

002

0

cp


X

0020010



∘



119888asymp 119871119880

infin) The first



119888= 10minus5











120572 = 80∘

120572 = 80∘

119888119897

119888119889













120593 = 45∘ (5)




Experimental

60 75 90 105 120 135 150 165 180120572

90 105 120 135 150 165 180120572


Experimental

2

18

16

14

12

1

08

06

04

22

24

26

minus04

minus06

minus08

12

1

08

06

04

02

0

minus02

c l c d


∘

XX

XX

XXXXXXXXXXXXXXXXXX


XX

XX


X



X


X

minus2

minus15

minus1

minus05

0

05

1

c p

minus25

minus002 minus001

X


cp

minus25

003

002

001

0

minus001

minus002

minus003

Z



Experimental


∘




∘ and 120572 = 60∘) is




120593 = 90∘ (6)



∘ and120572 =






clcd

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

14

12

1

08

06

0505

051

0515

052

0525

t (s)

120572 = 80∘

04

02

0

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

c lc

d

(a)

3

2

1

0

minus1

01 011 012 013 014

120572 = 165∘

Pres

sure

100800

100200

99600

99000

98400

t (s)

cl

cd

04

02

0

Z

04

02

0

Z

04

02

0Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Pres

sure

100800

100200

99600

99000

98400

c lc

d

(b)









7

6

5

4

3

2

10 0

7

6

5

4

3

2

120 40 60 8080 100 120 140 160 180 20 40 60 100 120 140 160 180

120593 = 45∘

120593 = 90∘

n n

120572 120572





4 Conclusions



119897d120572 + 119888

119889










References











































n = 1

2

3

4

5

6

7

(a)

n = 1

2

3

4

5

6

7

(b)


∘ (b) with 119899 = 1 2 7










+ carefully (119910+ = 119906+Δ119910]) 119910+



2and 119896

4







1050minus5

10

5

0

minus5

minus10

X

Z

X

Z

0minus002minus004

002

0

minus002

minus004











iterationsComp time





119897max and the 119888119889


minus005 0 005

X

004

002

0

minus002

minus004

Z

minus005 0 005

X

minus005 0 005

X

minus005 0 005

X

004

002

0

minus002

minus004

Z

004

002

0

minus002

minus004

Z

004

002

0

minus002

minus004

Z


066

064

062

06

058

056

00E + 00 20E + 04 40E + 04 60E + 04 80E + 04 10E + 05


Number of elements

c l

Resid

ual

CoarseMediumFine




3E minus 05

2E minus 05

1E minus 05







001

minus002

0

minus001

minus002

minus003

minus004

X

Z


1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

2

18

16

14

12

1

08

06

04

02

120572


Experimental

c l c d

0 20 40 60 80 100 120 140 160 180

120572

20 40 60 80 100 120 140 160 180


Experimental


120572

0 20 40 60 80 100 120 140 160 180

120572

20 40 60 80 100 120 140 160 180

1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

c l

minus12



2

18

16

14

12

1

08

06

04

02

c d

24

22

k-120596 SST k-120596 SST


model



minus12

2181614121

08060402

22

0minus02


minus12

2181614121

08060402

22

0minus02


minus12

2181614121

08060402

22

0minus02

minus14

c dcl

c dcl

c dcl

0 15 30 45 60 75 90 105 120 135 150 165 180

120572

180

180

0 15 30 45 60 75 90 105 120 135 150 165

120572

0 15 30 45 60 75 90 105 120 135 150 165

120572

n = 1 n = 3

n = 5 n = 7




minus12

2181614121

080604020

minus02c dcl

0 15 30 45 60 75 90 105 120 135 150 165 180

120572

k-120596 SST cd

k-120596 SST cl

k-120596 SST cd

k-120596 SST cl














119897maxand the 119888



∘


119901as a function









120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180


2181614121

08060402

22

0minus02

c dcl

minus04minus06

2181614121

08060402

22

0minus02

c dcl


minus12

2181614121

080604020

minus02c dcl


2181614121

08060402

22

0minus02

c dcl

n = 1 n = 3

n = 5 n = 7


minus2

minus15

minus1

minus05

0

05

1

minus002

minus002

minus001

X

c p

cp


Z

002

002

001 minus2 minus1 0

0

0 1



∘










minus2

minus15

minus1

minus05

0

05

1

c p

minus002

Z

002

0

cp


X

0020010



∘



119888asymp 119871119880

infin) The first



119888= 10minus5











120572 = 80∘

120572 = 80∘

119888119897

119888119889













120593 = 45∘ (5)




Experimental

60 75 90 105 120 135 150 165 180120572

90 105 120 135 150 165 180120572


Experimental

2

18

16

14

12

1

08

06

04

22

24

26

minus04

minus06

minus08

12

1

08

06

04

02

0

minus02

c l c d


∘

XX

XX

XXXXXXXXXXXXXXXXXX


XX

XX


X



X


X

minus2

minus15

minus1

minus05

0

05

1

c p

minus25

minus002 minus001

X


cp

minus25

003

002

001

0

minus001

minus002

minus003

Z



Experimental


∘




∘ and 120572 = 60∘) is




120593 = 90∘ (6)



∘ and120572 =






clcd

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

14

12

1

08

06

0505

051

0515

052

0525

t (s)

120572 = 80∘

04

02

0

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

c lc

d

(a)

3

2

1

0

minus1

01 011 012 013 014

120572 = 165∘

Pres

sure

100800

100200

99600

99000

98400

t (s)

cl

cd

04

02

0

Z

04

02

0

Z

04

02

0Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Pres

sure

100800

100200

99600

99000

98400

c lc

d

(b)









7

6

5

4

3

2

10 0

7

6

5

4

3

2

120 40 60 8080 100 120 140 160 180 20 40 60 100 120 140 160 180

120593 = 45∘

120593 = 90∘

n n

120572 120572





4 Conclusions



119897d120572 + 119888

119889










References











































1050minus5

10

5

0

minus5

minus10

X

Z

X

Z

0minus002minus004

002

0

minus002

minus004











iterationsComp time





119897max and the 119888119889


minus005 0 005

X

004

002

0

minus002

minus004

Z

minus005 0 005

X

minus005 0 005

X

minus005 0 005

X

004

002

0

minus002

minus004

Z

004

002

0

minus002

minus004

Z

004

002

0

minus002

minus004

Z


066

064

062

06

058

056

00E + 00 20E + 04 40E + 04 60E + 04 80E + 04 10E + 05


Number of elements

c l

Resid

ual

CoarseMediumFine




3E minus 05

2E minus 05

1E minus 05







001

minus002

0

minus001

minus002

minus003

minus004

X

Z


1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

2

18

16

14

12

1

08

06

04

02

120572


Experimental

c l c d

0 20 40 60 80 100 120 140 160 180

120572

20 40 60 80 100 120 140 160 180


Experimental


120572

0 20 40 60 80 100 120 140 160 180

120572

20 40 60 80 100 120 140 160 180

1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

c l

minus12



2

18

16

14

12

1

08

06

04

02

c d

24

22

k-120596 SST k-120596 SST


model



minus12

2181614121

08060402

22

0minus02


minus12

2181614121

08060402

22

0minus02


minus12

2181614121

08060402

22

0minus02

minus14

c dcl

c dcl

c dcl

0 15 30 45 60 75 90 105 120 135 150 165 180

120572

180

180

0 15 30 45 60 75 90 105 120 135 150 165

120572

0 15 30 45 60 75 90 105 120 135 150 165

120572

n = 1 n = 3

n = 5 n = 7




minus12

2181614121

080604020

minus02c dcl

0 15 30 45 60 75 90 105 120 135 150 165 180

120572

k-120596 SST cd

k-120596 SST cl

k-120596 SST cd

k-120596 SST cl














119897maxand the 119888



∘


119901as a function









120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180


2181614121

08060402

22

0minus02

c dcl

minus04minus06

2181614121

08060402

22

0minus02

c dcl


minus12

2181614121

080604020

minus02c dcl


2181614121

08060402

22

0minus02

c dcl

n = 1 n = 3

n = 5 n = 7


minus2

minus15

minus1

minus05

0

05

1

minus002

minus002

minus001

X

c p

cp


Z

002

002

001 minus2 minus1 0

0

0 1



∘










minus2

minus15

minus1

minus05

0

05

1

c p

minus002

Z

002

0

cp


X

0020010



∘



119888asymp 119871119880

infin) The first



119888= 10minus5











120572 = 80∘

120572 = 80∘

119888119897

119888119889













120593 = 45∘ (5)




Experimental

60 75 90 105 120 135 150 165 180120572

90 105 120 135 150 165 180120572


Experimental

2

18

16

14

12

1

08

06

04

22

24

26

minus04

minus06

minus08

12

1

08

06

04

02

0

minus02

c l c d


∘

XX

XX

XXXXXXXXXXXXXXXXXX


XX

XX


X



X


X

minus2

minus15

minus1

minus05

0

05

1

c p

minus25

minus002 minus001

X


cp

minus25

003

002

001

0

minus001

minus002

minus003

Z



Experimental


∘




∘ and 120572 = 60∘) is




120593 = 90∘ (6)



∘ and120572 =






clcd

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

14

12

1

08

06

0505

051

0515

052

0525

t (s)

120572 = 80∘

04

02

0

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

c lc

d

(a)

3

2

1

0

minus1

01 011 012 013 014

120572 = 165∘

Pres

sure

100800

100200

99600

99000

98400

t (s)

cl

cd

04

02

0

Z

04

02

0

Z

04

02

0Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Pres

sure

100800

100200

99600

99000

98400

c lc

d

(b)









7

6

5

4

3

2

10 0

7

6

5

4

3

2

120 40 60 8080 100 120 140 160 180 20 40 60 100 120 140 160 180

120593 = 45∘

120593 = 90∘

n n

120572 120572





4 Conclusions



119897d120572 + 119888

119889










References











































minus005 0 005

X

004

002

0

minus002

minus004

Z

minus005 0 005

X

minus005 0 005

X

minus005 0 005

X

004

002

0

minus002

minus004

Z

004

002

0

minus002

minus004

Z

004

002

0

minus002

minus004

Z


066

064

062

06

058

056

00E + 00 20E + 04 40E + 04 60E + 04 80E + 04 10E + 05


Number of elements

c l

Resid

ual

CoarseMediumFine




3E minus 05

2E minus 05

1E minus 05







001

minus002

0

minus001

minus002

minus003

minus004

X

Z


1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

2

18

16

14

12

1

08

06

04

02

120572


Experimental

c l c d

0 20 40 60 80 100 120 140 160 180

120572

20 40 60 80 100 120 140 160 180


Experimental


120572

0 20 40 60 80 100 120 140 160 180

120572

20 40 60 80 100 120 140 160 180

1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

c l

minus12



2

18

16

14

12

1

08

06

04

02

c d

24

22

k-120596 SST k-120596 SST


model



minus12

2181614121

08060402

22

0minus02


minus12

2181614121

08060402

22

0minus02


minus12

2181614121

08060402

22

0minus02

minus14

c dcl

c dcl

c dcl

0 15 30 45 60 75 90 105 120 135 150 165 180

120572

180

180

0 15 30 45 60 75 90 105 120 135 150 165

120572

0 15 30 45 60 75 90 105 120 135 150 165

120572

n = 1 n = 3

n = 5 n = 7




minus12

2181614121

080604020

minus02c dcl

0 15 30 45 60 75 90 105 120 135 150 165 180

120572

k-120596 SST cd

k-120596 SST cl

k-120596 SST cd

k-120596 SST cl














119897maxand the 119888



∘


119901as a function









120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180


2181614121

08060402

22

0minus02

c dcl

minus04minus06

2181614121

08060402

22

0minus02

c dcl


minus12

2181614121

080604020

minus02c dcl


2181614121

08060402

22

0minus02

c dcl

n = 1 n = 3

n = 5 n = 7


minus2

minus15

minus1

minus05

0

05

1

minus002

minus002

minus001

X

c p

cp


Z

002

002

001 minus2 minus1 0

0

0 1



∘










minus2

minus15

minus1

minus05

0

05

1

c p

minus002

Z

002

0

cp


X

0020010



∘



119888asymp 119871119880

infin) The first



119888= 10minus5











120572 = 80∘

120572 = 80∘

119888119897

119888119889













120593 = 45∘ (5)




Experimental

60 75 90 105 120 135 150 165 180120572

90 105 120 135 150 165 180120572


Experimental

2

18

16

14

12

1

08

06

04

22

24

26

minus04

minus06

minus08

12

1

08

06

04

02

0

minus02

c l c d


∘

XX

XX

XXXXXXXXXXXXXXXXXX


XX

XX


X



X


X

minus2

minus15

minus1

minus05

0

05

1

c p

minus25

minus002 minus001

X


cp

minus25

003

002

001

0

minus001

minus002

minus003

Z



Experimental


∘




∘ and 120572 = 60∘) is




120593 = 90∘ (6)



∘ and120572 =






clcd

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

14

12

1

08

06

0505

051

0515

052

0525

t (s)

120572 = 80∘

04

02

0

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

c lc

d

(a)

3

2

1

0

minus1

01 011 012 013 014

120572 = 165∘

Pres

sure

100800

100200

99600

99000

98400

t (s)

cl

cd

04

02

0

Z

04

02

0

Z

04

02

0Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Pres

sure

100800

100200

99600

99000

98400

c lc

d

(b)









7

6

5

4

3

2

10 0

7

6

5

4

3

2

120 40 60 8080 100 120 140 160 180 20 40 60 100 120 140 160 180

120593 = 45∘

120593 = 90∘

n n

120572 120572





4 Conclusions



119897d120572 + 119888

119889










References











































001

minus002

0

minus001

minus002

minus003

minus004

X

Z


1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

2

18

16

14

12

1

08

06

04

02

120572


Experimental

c l c d

0 20 40 60 80 100 120 140 160 180

120572

20 40 60 80 100 120 140 160 180


Experimental


120572

0 20 40 60 80 100 120 140 160 180

120572

20 40 60 80 100 120 140 160 180

1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

c l

minus12



2

18

16

14

12

1

08

06

04

02

c d

24

22

k-120596 SST k-120596 SST


model



minus12

2181614121

08060402

22

0minus02


minus12

2181614121

08060402

22

0minus02


minus12

2181614121

08060402

22

0minus02

minus14

c dcl

c dcl

c dcl

0 15 30 45 60 75 90 105 120 135 150 165 180

120572

180

180

0 15 30 45 60 75 90 105 120 135 150 165

120572

0 15 30 45 60 75 90 105 120 135 150 165

120572

n = 1 n = 3

n = 5 n = 7




minus12

2181614121

080604020

minus02c dcl

0 15 30 45 60 75 90 105 120 135 150 165 180

120572

k-120596 SST cd

k-120596 SST cl

k-120596 SST cd

k-120596 SST cl














119897maxand the 119888



∘


119901as a function









120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180


2181614121

08060402

22

0minus02

c dcl

minus04minus06

2181614121

08060402

22

0minus02

c dcl


minus12

2181614121

080604020

minus02c dcl


2181614121

08060402

22

0minus02

c dcl

n = 1 n = 3

n = 5 n = 7


minus2

minus15

minus1

minus05

0

05

1

minus002

minus002

minus001

X

c p

cp


Z

002

002

001 minus2 minus1 0

0

0 1



∘










minus2

minus15

minus1

minus05

0

05

1

c p

minus002

Z

002

0

cp


X

0020010



∘



119888asymp 119871119880

infin) The first



119888= 10minus5











120572 = 80∘

120572 = 80∘

119888119897

119888119889













120593 = 45∘ (5)




Experimental

60 75 90 105 120 135 150 165 180120572

90 105 120 135 150 165 180120572


Experimental

2

18

16

14

12

1

08

06

04

22

24

26

minus04

minus06

minus08

12

1

08

06

04

02

0

minus02

c l c d


∘

XX

XX

XXXXXXXXXXXXXXXXXX


XX

XX


X



X


X

minus2

minus15

minus1

minus05

0

05

1

c p

minus25

minus002 minus001

X


cp

minus25

003

002

001

0

minus001

minus002

minus003

Z



Experimental


∘




∘ and 120572 = 60∘) is




120593 = 90∘ (6)



∘ and120572 =






clcd

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

14

12

1

08

06

0505

051

0515

052

0525

t (s)

120572 = 80∘

04

02

0

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

c lc

d

(a)

3

2

1

0

minus1

01 011 012 013 014

120572 = 165∘

Pres

sure

100800

100200

99600

99000

98400

t (s)

cl

cd

04

02

0

Z

04

02

0

Z

04

02

0Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Pres

sure

100800

100200

99600

99000

98400

c lc

d

(b)









7

6

5

4

3

2

10 0

7

6

5

4

3

2

120 40 60 8080 100 120 140 160 180 20 40 60 100 120 140 160 180

120593 = 45∘

120593 = 90∘

n n

120572 120572





4 Conclusions



119897d120572 + 119888

119889










References












































minus12

2181614121

08060402

22

0minus02


minus12

2181614121

08060402

22

0minus02


minus12

2181614121

08060402

22

0minus02

minus14

c dcl

c dcl

c dcl

0 15 30 45 60 75 90 105 120 135 150 165 180

120572

180

180

0 15 30 45 60 75 90 105 120 135 150 165

120572

0 15 30 45 60 75 90 105 120 135 150 165

120572

n = 1 n = 3

n = 5 n = 7




minus12

2181614121

080604020

minus02c dcl

0 15 30 45 60 75 90 105 120 135 150 165 180

120572

k-120596 SST cd

k-120596 SST cl

k-120596 SST cd

k-120596 SST cl














119897maxand the 119888



∘


119901as a function









120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180


2181614121

08060402

22

0minus02

c dcl

minus04minus06

2181614121

08060402

22

0minus02

c dcl


minus12

2181614121

080604020

minus02c dcl


2181614121

08060402

22

0minus02

c dcl

n = 1 n = 3

n = 5 n = 7


minus2

minus15

minus1

minus05

0

05

1

minus002

minus002

minus001

X

c p

cp


Z

002

002

001 minus2 minus1 0

0

0 1



∘










minus2

minus15

minus1

minus05

0

05

1

c p

minus002

Z

002

0

cp


X

0020010



∘



119888asymp 119871119880

infin) The first



119888= 10minus5











120572 = 80∘

120572 = 80∘

119888119897

119888119889













120593 = 45∘ (5)




Experimental

60 75 90 105 120 135 150 165 180120572

90 105 120 135 150 165 180120572


Experimental

2

18

16

14

12

1

08

06

04

22

24

26

minus04

minus06

minus08

12

1

08

06

04

02

0

minus02

c l c d


∘

XX

XX

XXXXXXXXXXXXXXXXXX


XX

XX


X



X


X

minus2

minus15

minus1

minus05

0

05

1

c p

minus25

minus002 minus001

X


cp

minus25

003

002

001

0

minus001

minus002

minus003

Z



Experimental


∘




∘ and 120572 = 60∘) is




120593 = 90∘ (6)



∘ and120572 =






clcd

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

14

12

1

08

06

0505

051

0515

052

0525

t (s)

120572 = 80∘

04

02

0

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

c lc

d

(a)

3

2

1

0

minus1

01 011 012 013 014

120572 = 165∘

Pres

sure

100800

100200

99600

99000

98400

t (s)

cl

cd

04

02

0

Z

04

02

0

Z

04

02

0Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Pres

sure

100800

100200

99600

99000

98400

c lc

d

(b)









7

6

5

4

3

2

10 0

7

6

5

4

3

2

120 40 60 8080 100 120 140 160 180 20 40 60 100 120 140 160 180

120593 = 45∘

120593 = 90∘

n n

120572 120572





4 Conclusions



119897d120572 + 119888

119889










References















































120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180

120572

0 20 40 60 80 100 120 140 160 180


2181614121

08060402

22

0minus02

c dcl

minus04minus06

2181614121

08060402

22

0minus02

c dcl


minus12

2181614121

080604020

minus02c dcl


2181614121

08060402

22

0minus02

c dcl

n = 1 n = 3

n = 5 n = 7


minus2

minus15

minus1

minus05

0

05

1

minus002

minus002

minus001

X

c p

cp


Z

002

002

001 minus2 minus1 0

0

0 1



∘










minus2

minus15

minus1

minus05

0

05

1

c p

minus002

Z

002

0

cp


X

0020010



∘



119888asymp 119871119880

infin) The first



119888= 10minus5











120572 = 80∘

120572 = 80∘

119888119897

119888119889













120593 = 45∘ (5)




Experimental

60 75 90 105 120 135 150 165 180120572

90 105 120 135 150 165 180120572


Experimental

2

18

16

14

12

1

08

06

04

22

24

26

minus04

minus06

minus08

12

1

08

06

04

02

0

minus02

c l c d


∘

XX

XX

XXXXXXXXXXXXXXXXXX


XX

XX


X



X


X

minus2

minus15

minus1

minus05

0

05

1

c p

minus25

minus002 minus001

X


cp

minus25

003

002

001

0

minus001

minus002

minus003

Z



Experimental


∘




∘ and 120572 = 60∘) is




120593 = 90∘ (6)



∘ and120572 =






clcd

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

14

12

1

08

06

0505

051

0515

052

0525

t (s)

120572 = 80∘

04

02

0

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

c lc

d

(a)

3

2

1

0

minus1

01 011 012 013 014

120572 = 165∘

Pres

sure

100800

100200

99600

99000

98400

t (s)

cl

cd

04

02

0

Z

04

02

0

Z

04

02

0Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Pres

sure

100800

100200

99600

99000

98400

c lc

d

(b)









7

6

5

4

3

2

10 0

7

6

5

4

3

2

120 40 60 8080 100 120 140 160 180 20 40 60 100 120 140 160 180

120593 = 45∘

120593 = 90∘

n n

120572 120572





4 Conclusions



119897d120572 + 119888

119889










References












































minus2

minus15

minus1

minus05

0

05

1

c p

minus002

Z

002

0

cp


X

0020010



∘



119888asymp 119871119880

infin) The first



119888= 10minus5











120572 = 80∘

120572 = 80∘

119888119897

119888119889













120593 = 45∘ (5)




Experimental

60 75 90 105 120 135 150 165 180120572

90 105 120 135 150 165 180120572


Experimental

2

18

16

14

12

1

08

06

04

22

24

26

minus04

minus06

minus08

12

1

08

06

04

02

0

minus02

c l c d


∘

XX

XX

XXXXXXXXXXXXXXXXXX


XX

XX


X



X


X

minus2

minus15

minus1

minus05

0

05

1

c p

minus25

minus002 minus001

X


cp

minus25

003

002

001

0

minus001

minus002

minus003

Z



Experimental


∘




∘ and 120572 = 60∘) is




120593 = 90∘ (6)



∘ and120572 =






clcd

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

14

12

1

08

06

0505

051

0515

052

0525

t (s)

120572 = 80∘

04

02

0

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

c lc

d

(a)

3

2

1

0

minus1

01 011 012 013 014

120572 = 165∘

Pres

sure

100800

100200

99600

99000

98400

t (s)

cl

cd

04

02

0

Z

04

02

0

Z

04

02

0Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Pres

sure

100800

100200

99600

99000

98400

c lc

d

(b)









7

6

5

4

3

2

10 0

7

6

5

4

3

2

120 40 60 8080 100 120 140 160 180 20 40 60 100 120 140 160 180

120593 = 45∘

120593 = 90∘

n n

120572 120572





4 Conclusions



119897d120572 + 119888

119889










References












































Experimental

60 75 90 105 120 135 150 165 180120572

90 105 120 135 150 165 180120572


Experimental

2

18

16

14

12

1

08

06

04

22

24

26

minus04

minus06

minus08

12

1

08

06

04

02

0

minus02

c l c d


∘

XX

XX

XXXXXXXXXXXXXXXXXX


XX

XX


X



X


X

minus2

minus15

minus1

minus05

0

05

1

c p

minus25

minus002 minus001

X


cp

minus25

003

002

001

0

minus001

minus002

minus003

Z



Experimental


∘




∘ and 120572 = 60∘) is




120593 = 90∘ (6)



∘ and120572 =






clcd

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

14

12

1

08

06

0505

051

0515

052

0525

t (s)

120572 = 80∘

04

02

0

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

c lc

d

(a)

3

2

1

0

minus1

01 011 012 013 014

120572 = 165∘

Pres

sure

100800

100200

99600

99000

98400

t (s)

cl

cd

04

02

0

Z

04

02

0

Z

04

02

0Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Pres

sure

100800

100200

99600

99000

98400

c lc

d

(b)









7

6

5

4

3

2

10 0

7

6

5

4

3

2

120 40 60 8080 100 120 140 160 180 20 40 60 100 120 140 160 180

120593 = 45∘

120593 = 90∘

n n

120572 120572





4 Conclusions



119897d120572 + 119888

119889










References











































clcd

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

Zve

loci

ty

minus25

minus15

minus5

5

15

25

35

14

12

1

08

06

0505

051

0515

052

0525

t (s)

120572 = 80∘

04

02

0

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

c lc

d

(a)

3

2

1

0

minus1

01 011 012 013 014

120572 = 165∘

Pres

sure

100800

100200

99600

99000

98400

t (s)

cl

cd

04

02

0

Z

04

02

0

Z

04

02

0Z

04

02

0

Z

04

02

0

Z

04

02

0

Z

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

minus02 0 02

X

Pres

sure

100800

100200

99600

99000

98400

c lc

d

(b)









7

6

5

4

3

2

10 0

7

6

5

4

3

2

120 40 60 8080 100 120 140 160 180 20 40 60 100 120 140 160 180

120593 = 45∘

120593 = 90∘

n n

120572 120572





4 Conclusions



119897d120572 + 119888

119889










References











































7

6

5

4

3

2

10 0

7

6

5

4

3

2

120 40 60 8080 100 120 140 160 180 20 40 60 100 120 140 160 180

120593 = 45∘

120593 = 90∘

n n

120572 120572





4 Conclusions



119897d120572 + 119888

119889










References










































numerical investigation of galloping instabilities in z-shaped … · 2016-06-09 ·...

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