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 AcoshL BsinhL AcosL BsinL w(x L) 0 AsinhL BcoshL AsinL BcosL (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 cosL 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 cosL 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