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