《船舶水动力学》课程教学课件（讲稿）第一章 运动方程及干模态分析 dry hull analysis symmetric

NIVERSITYOSouthampton(1)Equationsofmotion&DryHull AnalysisSymmetricMotions&DistortionsExamplesfrom:Hydroelasticity of Ships,R.E.D.Bishop & W.G.Price, CUP1979Hydrodynamics and HydroelasticityProf.P.Temarel,WUT,July2015EquationsofsymmetricmotionNIVERSITYOSouthampton-Non-uniform EulerbeamAssumptions:LinearAnalysisPlanesectionsremainplane,i.e.shearstrainassumedzeroInfluenceof rotatoryinertiaisneglectedUsing an equilibrium axis system,Oxaxis at still waterline,Oataftend of ship

1 (1) Equations of motion & Dry Hull Analysis Symmetric Motions & Distortions Examples from: Hydroelasticity of Ships, R.E.D. Bishop & W.G. Price, CUP 1979 Hydrodynamics and Hydroelasticity Prof. P. Temarel, WUT, July 2015 2 Equations of symmetric motion – Non-uniform Euler beam Assumptions: Linear Analysis Plane sections remain plane, i.e. shear strain assumed zero Influence of rotatory inertia is neglected. Using an equilibrium axis system, Ox axis at still waterline, O at aft end of ship

Eguations of symmetricmotionNIVERSITYCSouthampton-Non-uniformEulerbeamApplyingNewton's2ndlawforthevertical(i.e.symmetric)motionw(x.t)of theslice:μ(x)Ax w(x,t)=V, -V, + Z(x,t)Axorμ(x)w(x,t)=V'(x,t)+ Z(x,t)(l)sincelimAx →0 ()-V2)/Ax=0V(x,)/ax =V(x,)In this equations,a dot signifies differentiation with respect totime, namelyw(x,1)= Ow(x,1)/at, i(x,1)= a2w(x,1)/at2Equationsof symmetricmotionNIVERSITYOSouthampton-Non-uniformEulerbeamTakingmomentsaboutthelefthandsideoftheslice,i.e.applyingNewton's2nd lawforrotatione(x,t)ofthe slice,anticlockwisetakenaspositive1,(x)Ax 0(x, t)= M, - M, +V Ax +Z(x,t)Ax (Ax/2)or0=M'(x,t)+V(x,t)(2)wherethe higherorderterms (Ax)2hasbeen ignored,V,=V(x,t),neglectingtheeffects of therotatory inertiaresults inzero attheleft handside.Alsolim axr→0 (M,-M2)/Ar = 0M(x,)/ ax = M(x,)resultingintheconventional relationshipforEulerbeamsV=-M'2

2 3 Equations of symmetric motion – Non-uniform Euler beam Applying Newton’s 2nd law for the vertical (i.e. symmetric) motion w(x,t) of the slice: ( ) ( , ) ( , ) x  x x w x t  V1  V 2  Z x t  or (x)w(x,t)  V (x,t) Z(x,t) (1) since )/ ( , )/ ( , ). 1 2 ( 0 lim V V x V x t x V x t x          In this equations, a dot signifies differentiation with respect to time, namely 2 ( , ) / 2 w ( x, t)  w( x, t) / t, w( x, t)   w x t t 4 Equations of symmetric motion – Non-uniform Euler beam ( ) ( ) ( ) x ( x/2) I y x x   x,t  M1  M 2 V1 x  Z x,t   0  M (x,t)V(x,t) (2) ) / ( , ) / ( , ) 1 2 ( 0 lim M M x M x t x M x t x          Taking moments about the left hand side of the slice, i.e. applying Newton’s 2nd law for rotation θ(x,t) of the slice, anticlockwise taken as positive or where the higher order terms (∆x)2 has been ignored, V1≡V(x,t), neglecting the effects of the rotatory inertia results in zero at the left hand side. Also resulting in the conventional relationship for Euler beams V= - M’

Eguations of symmetricmotionNIVERSITYOSouthampton-Non-uniformEulerbeamIn these equations:μ(x):massperunitlength (e.g.tonnes/m),(x):momentof inertia,aboutyaxis,per unit length (e.g.tonnes m?/m)V(x,t): Vertical shear force (e.g.kN)M(x,t): Vertical bending moment(e.g.kN m)Z(x,t): Vertical (or upward) external force, per unit length (e.g.kN/m).You can check the units in eq(1)and (2)and youwill se that theyareconsistent, i.e.eq.(1)in kN/mand eq.(2)inkN.That is because weare working with a slice.The relationship between the vertical displacementw(x,t) and the bendingslope(x,t) isw(x,1)=Ow(x,t)/ax=0(x,t) (3)andtherelationshipbetweenbendingmomentandverticaldisplacementM(x,t)= El(x)aw(x,t)/ax2 =EI(x)w(x,t) : (4)El(x)is referred toas the flexural rigidity,for a ship made of material ofconstantYoung's modulus Eonlythemomentof inertiaI(x)(secondmomentofareaaboutyaxispassingthroughthecentroid,m)varies5along the ship.Equations of symmetricmotionNIVERSITYOSouthampton-Non-uniform Euler beamInanystructurethatexperiencesdistortionsfrictionbetweenthemoleculeswillresult in"resisting"thedistortion,namelydampingofstructural origin.In linear systems damping is assumed proportional to velocity. Hence, wecan rewrite eq.(4),addingtheeffects ofstructural damping asM(x,t)=EI(x)w"(x,t) +β(x)w"(x,t) (5)whereβ(x) represents distributed dampingassociatedwiththe verticalbending moment.Usingeq.(5),eq.(1)becomes:μ(x)i(x,t)+[EI(x)w"(x,t)"+[β(x)w"(x,t)"=Z(x,1)(6)Asyoucan seeeq.(6)is a 4thorderpartial differential equation.Wecan only solve it analyticallyforauniform beam,i.e.wherethecoefficientsdo not varyalong thebeam.b3

3 5 Equations of symmetric motion – Non-uniform Euler beam In these equations: μ(x): mass per unit length (e.g. tonnes/m) Iy(x): moment of inertia, about y axis, per unit length (e.g. tonnes m2/m) V(x,t): Vertical shear force (e.g. kN) M(x,t): Vertical bending moment(e.g. kN m) Z(x,t): Vertical (or upward) external force, per unit length (e.g. kN/m). You can check the units in eq.(1) and (2) and you will se that they are consistent, i.e. eq.(1) in kN/m and eq.(2) in kN. That is because we are working with a slice. w(x,t)  w(x,t)/ x  (x,t) (3) ( , ) ( ) ( , )/ ( ) ( , ) . (4) 2 2 M x t  EI x  w x t x  EI x w x t The relationship between the vertical displacement w(x,t) and the bending slope θ(x,t) is and the relationship between bending moment and vertical displacement EI(x) is referred to as the flexural rigidity; for a ship made of material of constant Young’s modulus E only the moment of inertia I(x) (second moment of area about y axis passing through the centroid, m4) varies along the ship. 6 Equations of symmetric motion – Non-uniform Euler beam M (x,t)  EI(x)w(x,t)  (x) w(x,t) (5) (x)w(x,t) [EI(x)w(x,t)] [(x)w(x,t)]  Z(x,t). (6) In any structure that experiences distortions friction between the molecules will result in “resisting” the distortion, namely damping of structural origin. In linear systems damping is assumed proportional to velocity. Hence, we can rewrite eq.(4), adding the effects of structural damping as where β(x) represents distributed damping associated with the vertical bending moment. Using eq.(5), eq.(1) becomes: As you can see eq.(6) is a 4th order partial differential equation. We can only solve it analytically for a uniform beam, i.e. where the coefficients do not vary along the beam

Dry analysis-Non-uniformNIVERSITYCSouthamptonEulerbeaminvacuoFortheEulerbeaminvacuoexternalforcesareassumedzero,Z(x.t)=0Inadditiontheeffectsofstructuraldampingareneglected,asperclassicalvibrationdefinitionoffreevibration.Accordinglyeq.(6)becomesμ(x)i(x,t)+[EI(x)w'(x,t)]"=0. (7)Theaimofsolvingeq.(7)istoobtainthenatural frequenciesandassociatedprincipalmodeshapesofthisnon-uniformEulerbeam.This implies simpleharmonicmotionforthetimedependenceThesolutionofeq.(7)isobtainedthroughvariableseparation,i.ew(x,t)=w(x)w(t)= w(x)sin@t (8)as wearelookingforsimpleharmonicmotion.Substituting eq.(8)intoeq.(7)weobtain-0μ(x)w(x)+[EI(x)w"(x)"=0. (9)The solution to eq.(9) is of the form:w(x)=AF(x)+BG(x)+CH(x)+DI(x) (10)Dry analysis -Non-uniformNIVERSITYOSouthamptonEulerbeaminvacuowheretheconstantsA,B,CandDdependontheboundaryconditionsattheends of thenon-uniformbeam,i.e.thebowand stern of the beamlikeshipinvacuo.Inthiscase,astherearenoconstraintsattheendsofthebeamtheyaretreated as free,i.e.afree-freebeamForafree-freebeamtherewill bedisplacementandrotation(i.e.bendingslope)ateachend,butthebendingmomentandshearforcewillbezero.ForashipoflengthL(typicallyL=Lpp),withtheaftperpendicularatx=0 and theforward perpendicular at x=L for thebending moment, usingeq.(4),we haveEI(x)w(x) =0atx=0 andx=L(1la)andfortheshearforce,usingeq.(2)wehave[EI(x)w(x)} =00atx=0andx=L(11b)where the negative sign is ofno consequenceto satisfying theboundaryconditions,henceignoredThe functions F(x), G(x), H(x) and I(x) are not of any analytical form for anon-uniformbeam.However,youwill seefortheuniformbeamcasetheytakeanalyticalform.4

4 7 Dry analysis – Non-uniform Euler beam in vacuo For the Euler beam in vacuo external forces are assumed zero, Z(x,t)=0. In addition the effects of structural damping are neglected, as per classical vibration definition of free vibration. Accordingly eq.(6) becomes (x)w(x,t) [EI(x)w(x,t)]  0. (7) . The aim of solving eq.(7) is to obtain the natural frequencies and associated principal mode shapes of this non-uniform Euler beam. This implies simple harmonic motion for the time dependence. The solution of eq.(7) is obtained through variable separation, i.e. w(x,t)  w(x)w(t)  w(x)sin t (8) ( ) ( ) [ ( ) ( )] 0. (9) 2   x w x  EI x w x   w(x)  AF(x)  BG(x)  C H (x)  D I(x) (10) as we are looking for simple harmonic motion. Substituting eq.(8) into eq.(7) we obtain The solution to eq.(9) is of the form: 8 Dry analysis – Non-uniform Euler beam in vacuo where the constants A, B, C and D depend on the boundary conditions at the ends of the non-uniform beam, i.e. the bow and stern of the beamlike ship in vacuo. In this case, as there are no constraints at the ends of the beam they are treated as free, i.e. a free-free beam. For a free-free beam there will be displacement and rotation (i.e. bending slope) at each end, but the bending moment and shear force will be zero. For a ship of length L (typically L=LPP), with the aft perpendicular at x=0 and the forward perpendicular at x=L for the bending moment, using eq.(4), we have EI(x)w(x)  0 at x  0 and x  L (11a) and for the shear force, using eq. (2) we have [EI(x)w(x)]  0 at x  0 and x  L (11b) where the negative sign is of no consequence to satisfying the boundary conditions, hence ignored. The functions F(x), G(x), H(x) and I(x) are not of any analytical form for a non-uniform beam. However, you will see for the uniform beam case they take analytical form

Dry analysis-Non-uniformNIVERSITYCSouthamptonEulerbeaminvacuoTheconstantsA,B,CandDaredeterminedfromsatisfyingegs.(11a,b)-fourequationsintotal.Theconditionforhaving anon-trivial(i.e.non-zero)solutionfortheseconstantsresults inafrequencyorcharacteristicequation,fromwhichonecan determine an infinite number of naturalfrequenciesw, (for r=0, 1, 2, 3,4, etc)andusingeq.(1o)thecorrespondingprincipalmodeshapesw,(x):aswellascorrespondingmodalbendingmomentsM,(x)andshearforcesV(x), using eq.(4) and eq.(2).The solution is carriedout throughanumerical method,forexampleFiniteDifferencemethod.foranon-uniformbeamDry analysis -Uniform EulerJIVERSITYOSouthamptonbeaminvacuoFora uniform beam the properties ofmass perunit length andflexibilityareconstant,i.e.μandEl.Theneq.(9)becomes-0"μ w(x)+EIw""(x)=0. (9a)Denotingx=ou/EI (-x)=x()theneq.(9a)becomesdxUsing a trial solution w(x)=eixand substituting into eq(9b) we obtain -x*=0=(.-x)( +)This equation hasroots =K,,=-,,=ik, a,=-ik, wherei=-1Thegeneral solution ofeq.(9a)then becomesw(x)= Ae + Aze- + A,el + Aye-irwhereA, (i=1,2,3,4)are constants dependingon theboundary conditions ofthebeam.Usingtherelationshipsbetweencomplexexponentialandtrigonometricfunctionsandhyperbolicfunctions,thisgeneralsolutioncanbeexpressedasw(x)=Acoshkx+Bsinhkx+Ccoskx+Dsinkx(12)105

5 9 Dry analysis – Non-uniform Euler beam in vacuo The constants A, B, C and D are determined from satisfying eqs.(11a, b) – four equations in total. The condition for having a non-trivial (i.e. non-zero) solution for these constants results in a frequency or characteristic equation, from which one can determine an infinite number of natural frequencies ωr (for r=0, 1, 2, 3, 4, etc) and using eq.(10) the corresponding principal mode shapes wr (x), as well as corresponding modal bending moments Mr (x) and shear forces Vr (x), using eq.(4) and eq.(2). The solution is carried out through a numerical method, for example Finite Difference method, for a non-uniform beam. 10 Dry analysis – Uniform Euler beam in vacuo For a uniform beam the properties of mass per unit length and flexibility are constant, i.e. μ and EI. Then eq.(9) becomes ( ) ( ) 0. (9a) 2   w x  EI w x  Denoting then eq.(9a) becomes ( ). (9b) d d ( ) ( ) ( ) 0 4 4 4 4 w x x w x w x   w x     Using a trial solution x w x e  ( )  and substituting into eq.(9b) we obtain 0 ( )( ). 4 4 2 2 2 2      .   This equation has roots , , , , where 1. 1   2   3  i 41  i i   The general solution of eq.(9a) then becomes x x i x i x w x A e A e A e A e       1  2  3  4 ( ) where Ai (i=1,2,3,4) are constants depending on the boundary conditions of the beam. Using the relationships between complex exponential and trigonometric functions and hyperbolic functions, this general solution can be expressed as w(x)  Acosh x  Bsinh x  Ccos x  Dsin x (12) / EI 4 2    

Dry analysis-UniformEulerNIVERSITYOSouthamptonbeaminvacuowhich you can see isof the sameformas eq.(10),but withanalytical formsfor the functions F(x), G(x), H(x) and I(x)Theboundaryconditionsofeqs.(11a,b)become,fortheuniformfree-freebeamw(x)=0atx=0 andx=Lw"(x) =0 atx=0 andx=L.Using the derivatives ofeq.(12),wehavew(x)= x (Acosh xx + Bsinh xx -Ccos x - Dsin xx)w"(x)= x(Asinh xx + Bcosh xx + C sin xx - Dcosa)Forafree-free uniformbeamw(x=0)=0=A-CleadingtoA=Cw"(x=0)=0 =B-DleadingtoB=Dthus eliminating two unknowns.Using the boundary conditions at the end x=L(11c)w(x=L)=0=AcoshkL+BsinhkL-AcoskL-BsinkLw"(x=L)=0=AsinhxL+BcoshxL+AsinxL-BcosxL(11d)11Dry analysis-Uniform EulerNIVERSITYOFSouthamptonbeaminvacuoThelast2equationsformasystemof2homogeneousequationsinAandB[(cosh kL - cos kL)(sinh kL - sin L)(11e)(sinh kL +sin kL)(cosh kL- cos kL)RTohaveanon-trivial solution thedeterminantof thecoefficientsofAandBmustbesettozero,i.e(cosh xL - cos xL)(cosh xL - cos xL)-(sinh xL - sin xL)(sinh xL + sin xL)= 0=2-2coshxLcosxLTherefore,thefrequency equation,to obtain thenatural frequencies becomescosh xL cosxL =1.Thisequationhasaninfinitenumberofroots,whichcanbeobtainednumerically(xL)a=O forrigidbodymodes(heaveandpitch)(xL)2 = 4.73,(xL) = 7.853, etc.(KL) EIsothat03=forr=0,1,2,3,4.....L4u126

6 11 Dry analysis – Uniform Euler beam in vacuo which you can see is of the same form as eq.(10), but with analytical forms for the functions F(x), G(x), H(x) and I(x). The boundary conditions of eqs. (11a, b) become, for the uniform free-free beam w(x)  0 at x  0 and x  L w(x)  0 at x  0 and x  L. Using the derivatives of eq.(12), we have ( ) ( cosh sinh cos sin ) 2 w x   A x  B x C x  D x ( ) ( sinh cosh sin cos ) 3 w x   A x  B x  C x  D x For a free-free uniform beam w(x  0)  0  A C leading to A  C w(x  0)  0  B  D leading to B  D thus eliminating two unknowns. Using the boundary conditions at the end x=L w(x  L)  0  AcoshL  BsinhL  AcosL  BsinL w(x  L)  0  AsinhL  BcoshL  AsinL  BcosL (11c) (11d) 12 Dry analysis – Uniform Euler beam in vacuo . The last 2 equations form a system of 2 homogeneous equations in A and B. To have a non-trivial solution the determinant of the coefficients of A and B must be set to zero, i.e.                                0 0 (sinh sin ) (cosh cos ) (cosh cos ) (sinh sin ) B A L L L L L L L L 2 - 2 cosh cos . (cosh cos )(cosh cos ) (sinh sin ) (sinh sin ) 0 L L L L L L L L L L                  Therefore, the frequency equation, to obtain the natural frequencies becomes cosh L cosL 1. This equation has an infinite number of roots, which can be obtained numerically so that ( ) 4.73,( ) 7.853, etc. ( ) 0 for rigid body modes (heave and pitch) 2 3 0,1    L L L    for 0,1,2,3,4,. ( ) 4 4 2  r  EI L L r    (11e)

Dryanalysis-UniformEulerNIVERSITYOFSouthamptonbeaminvacuo1-cosh xLcosxL=0llustrationof201.0E+0EAcosh(bl)cos(bl)Socoalc+cosh(bl)cos(bl)1-cosh(bl)cos(bl)2.5E+00000000588800DE2.5E4035.0E+03L0E+04(xL)2=4.73~3元/2,(xL)3=7.8535元/2,etc.13Dryanalysis-Uniform EulerUNIVERSITYOFSouthamptonbeaminvacuoThe shapeofthedisplacement (as well as, slope,BMand SF) isparticularforeachnaturalfrequency-referredtoasModeshapeThemodeshapeisobtainedfromeq.(12)usingtherelevantboundaryconditions.Forthefree-freeuniformbeamw,(x)= A[cosh(kL),(x / L)+cos(kL),(x/ L)]+ B[sinh(kL),(x/ L)+sin(kL),(x/ L))forr=2.3.4.....ConstantAcanbeobtainedasafunctionofconstantB.fromeithereq.(11c)or(11d),i.e.forr=2.3.4...4 =-B [sinh(kL),-sin(L)][cosh(kL), -cos(kL),1or A=-B[cosh(kL)r-cos(kL),][sinh(kL), + sin(kL),]andthemodeshapeisdefinedforanyvalueofB.e.g.B=1:thisiscallednormalisationForeachmodeshapethereisthecorrespondingBMandSFVr(x)=EI w (x)M(x)=EIw (x)L47

7 13 Dry analysis – Uniform Euler beam in vacuo Illustration of 1- cosh L cosL  0. . ( ) 4.73 3 / 2,( ) 7.853 5 / 2, etc. L 2    L 3    14 The shape of the displacement (as well as, slope, BM and SF) is particular for each natural frequency – referred to as Mode shape. The mode shape is obtained from eq.(12) using the relevant boundary conditions. For the free-free uniform beam: for r=2,3,4,. Constant A can be obtained as a function of constant B, from either eq.(11c) or (11d), i.e. for r=2,3,4,. and the mode shape is defined for any value of B, e.g. B=1; this is called normalisation. For each mode shape there is the corresponding BM and SF Dry analysis – Uniform Euler beam in vacuo . w (x) A[ cosh( L) (x / L) cos( L) (x / L)] B[sinh( L) (x / L) sin( L) (x / L)] r   r   r   r   r [sinh( ) sin( ) ] [cosh( ) cos( ) ] or -B [cosh( ) cos( ) ] [sinh( ) sin( ) ] -B r r r r r r r r L L L L A L L L L A               M (x) EI w (x) r r   V (x) EI w (x) r r  

Dryanalysis-UniformEulerUNIVERSITYOFSouthamptonbeaminvacuoIllustrationofmodeshapesandmodalBMandSFforfree-freebeamThe rigid body modesHeave: wo(x)=1,w o(x)=0=M(x)=Vo(x)Pitch:W;(x)=(1-2x/L), w'(x)=-2/L, M(x)=0=V;(x)Flexiblemodesr=2(mode3),r=3mode4),r=4(mode5）Bending;Free-F0.800.60.8Dryanalysis-UniformEulerUNIVERSITYOFSouthamptonbeaminvacuoIllustrationof modeshapesandmodalBMandSFforfree-freebeamw"(x)proportionaltoBMw""(x)proportionaltoSFBending ; Free-FreeBending ; Free-Fre8

8 15 The rigid body modes Heave: w0(x)= 1, w’0(x)=0=M0(x)=V0(x) Pitch: w1(x)= (1-2x/L), w’1(x)=-2/L, M1(x)=0=V1(x) Flexible modes r=2 (mode3), r=3 (mode4), r=4 (mode5) Dry analysis – Uniform Euler beam in vacuo Illustration of mode shapes and modal BM and SF for free-free beam . Bending ; Free-Free -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0.0 0.2 0.4 0.6 0.8 1.0 x/L displacement mode1 mode2 mode3 mofe4 mode5 16 w’’r (x) proportional to BM w’’’r (x) proportional to SF Dry analysis – Uniform Euler beam in vacuo Illustration of mode shapes and modal BM and SF for free-free beam . Bending ; Free-Free -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0.0 0.2 0.4 0.6 0.8 1.0 x/L d/dx[d/dx(displacement)] mode3 mofe4 mode5 Bending ; Free-Free -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0.0 0.2 0.4 0.6 0.8 1.0 x/L d/dx{d/dx[d/dx(displacement)]} mode3 mofe4 mode5

Dry analysis-Uniform EulerJNIVERSITYOSouthamptonbeaminvacuoSummary of Boundary conditionsfora uniformbeam inbendingBoundary conditionSlidingFixedPinnedFreeDisplacement,C00+0w(x=0or L)Slope,0+0￥00w(x=0orL)Bending MomentFC#0w"(x=0or L)ShearForce,*0w"(x=0or L)Thefile“Bendingmodes"ontheblackboard siteforMSiFcontainsarangeofuniformbeamswithdifferentboundaryconditions,theirnaturalfrequenciesandallmodalcharacteristics17Principal modes and orthogonalityINIVERSITYOSouthamptonconditions-Non-uniform EulerbeamUsingvariableseparationthedisplacements,deflectionsbendingmomentsetcareassumedtovarywithtimeassimpleharmonicfunctions.Thususingtherthprincipalmodewehavew(x,t)=w,(x)sino,tM(x,t)= M,(x)sino,tV(x,t)=V.(x)sino,tforr=0(heave),1(pitch),2(2-nodedistortion),3(3-nodedistortion)etc.Itshould be noted that the rigid body modes have w,=0, for r=0,1 for thenon-uniformbeamin vacuo.Eachprincipalmode satisfieseq.(7),namely-0,μ(x)w,(x)+[EI(x)w(x)"= 0. (13)Multiplyeq.(13)byw.(x)andintegrateoverthelengthLofthenonuniform beam (i.e.ship),where x=0 denotes A.P.and x=L denotes F.P.o, μ(x)w,(x)w,(x) dx=/ [EI(x)w(x)"w,(x) dx. (14)The right-hand-side integral in eq(14) can be evaluated using integrationbyparts,i.e.189

9 17 Summary of Boundary conditions for a uniform beam in bending The file “Bending modes” on the blackboard site for MSiF contains a range of uniform beams with different boundary conditions, their natural frequencies and all modal characteristics. Dry analysis – Uniform Euler beam in vacuo . Boundary condition Fixed Pinned Free Sliding Displacement, 0 0 ≠0 ≠0 Slope, 0 ≠0 ≠0 0 Bending Moment, ≠00 0 ≠0 Shear Force, ≠0 ≠00 0 w(x  0or L) w(x  0or L) w(x  0or L) w(x  0or L) 18 Principal modes and orthogonality conditions– Non-uniform Euler beam w x,t w x t r r ( )  ( ) sin M x,t M x t r r ( )  ( ) sin V x,t V x t r r ( )  ( ) sin Using variable separation the displacements, deflections, bending moments etc are assumed to vary with time as simple harmonic functions. Thus using the rth principal mode we have for r=0 (heave), 1(pitch), 2 (2-node distortion), 3 (3-node distortion) etc. It should be noted that the rigid body modes have ωr =0, for r=0,1 for the non-uniform beam in vacuo. Each principal mode satisfies eq.(7), namely ( ) ( ) [ ( ) ( )] 0. (13) 2 r  x wr x  EI x w r  x   Multiply eq.(13) by ws(x) and integrate over the length L of the non￾uniform beam (i.e. ship), where x=0 denotes A.P. and x=L denotes F.P. ( ) ( ) ( ) [ ( ) ( )] ( ) . (14) 0 0 2 x w x w x dx EI x w x w x dx r s L r s L r        The right-hand-side integral in eq.(14) can be evaluated using integration by parts, i.e

Principal modes and orthogonalityNIVERSITYOSouthamptonconditions-Non-uniformEulerbeam [EI(x)w(x)}"w,(x)dx= ([EI(x)w'(x)w,(x)-1 [EI(x)w(x)w,(x) dx =([EI(x)w(x)w,(x)- ([EI(x)w(x))w,(x) + [EI(x)w(x)/w(x) dx. (15)Thefirst two terms are zero at both x=0 andx=L,as the bending momentandshearforcearezeroattheseends.Henceweendupo,/ μ(x)w,(x)w,(x)dx= EI(x)w(x)w'(x) dx. (16)Eq.(16)remains valid if the subscripts rand s are interchanged,namelyo)f u(x)w,(x)w,(x) dx=T EI(x)w(x)w:(x) dx. (17)Subtractingeq.(17)fromeq.(16)weobtain(o, -02) μ(x)w,(x)w,(x)dx=0. (18)19Principal modes and orthogonalityUNIVERSITYOFSouthamptonconditions-Non-uniformEulerbeamWhen r=s this condition is satisfied,hence the integral term is notzerowhenrstosatisfythisconditiontheintegraltermneedstobezeroTherefore,employingtheKroneckerdeltafunction,thefirstorthogonalityconditionbecomes:u(x)w,(x)w,(x)dx=an 8,(19)wherearsdenotes thegeneralisedmass andOs=1forr=s and Ors=0forrts.Thesecondorthogonality condition is obtainedusing eitherof eqs.(16or17),namely1El(x)w(x)w(x)dx=o,an,=C(20)whereCrsdenotesthegeneralisedstiffness.2010

10 19 Principal modes and orthogonality conditions– Non-uniform Euler beam The first two terms are zero at both x=0 and x=L, as the bending moment and shear force are zero at these ends. Hence we end up ( ) ( ) ( ) ( ) ( ) ( ) . (16) 0 0 2 x w x w x dx EI x w x w x dx r s L r s L r        Eq.(16) remains valid if the subscripts r and s are interchanged, namely ( ) ( ) ( ) ( ) ( ) ( ) . (17) 0 0 2 x w x w x dx EI x w x w x dx r s L r s L s        Subtracting eq.(17) from eq.(16) we obtain ( ) ( ) ( ) ( ) 0. (18) 0 2 2   x wr x ws x dx  L r s        [ ( ) ( )] ( ) . (15) [ ( ) ( )] ( ) [ ( ) ( )] ( ) [ ( ) ( )] ( ) [ ( ) ( )] ( ) [ ( ) ( )] ( ) 0 0 0 0 0 0 EI x w x w x dx EI x w x w x EI x w x w x EI x w x w x dx EI x w x w x EI x w x w x dx r s L L r s L r s r s L L r s r s L                      20 Principal modes and orthogonality conditions– Non-uniform Euler beam When r=s this condition is satisfied, hence the integral term is not zero, when r≠s to satisfy this condition the integral term needs to be zero. Therefore, employing the Kronecker delta function, the first orthogonality condition becomes: ( ) ( ) ( ) (19) 0 r s rs rs L   x w x w x dx  a  where ars denotes the generalised mass and δrs=1 for r=s and δrs=0 for r≠s. The second orthogonality condition is obtained using either of eqs.(16 or 17), namely ( ) ( ) ( ) (20) 2 0 r s r rs rs rs rs L  EI x w x w x dx   a   c  where crs denotes the generalised stiffness