Hutton - Fundamentals of Finite Element Analysis (523155), страница 77
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The free-vibration response of a multiple degrees-of-freedom system is described by Equation 10.95 with {F } = 0 as[M ]{q̈ } + [K ]{q } = {0}(10.99)Assuming that we have solved for the natural circular frequencies and the modalamplitude vectors via the assumed solution form qi (t ) = A i sin(t + ), substitution of a particular frequency i into Equation 10.99 gives!!−i2 [M ] A (i) + [K ] A (i) = 0(10.100)and for any other frequency j!!− 2j [M ] A ( j) + [K ] A ( j) = 0(10.101)Multiplying Equation 10.100 by { A ( j) } T and Equation 10.101 by { A (i) } T gives!T!!T!−i2 A ( j) [M ] A (i) + A ( j) [K ] A (i) = 0(10.102)!!!!TT− 2j A (i) [M ] A ( j) + A (i) [K ] A ( j) = 0(10.103)Subtracting Equation 10.102 from Equation 10.103, we have!T! A ( j) [M ] A (i) i2 − 2j = 0i = j(10.104)In arriving at the result represented by Equation 10.104, we utilize the fact frommatrix algebra that [ A]T [B][C ] = [C ]T [B][ A] , where [ A], [B], [C ] are anythree matrices for which the triple product is defined.
As the two circular frequencies in Equation 10.104 are distinct, we conclude that!T!A ( j) [M ] A (i) = 0i = j(10.105)Hutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText10.7© The McGraw−HillCompanies, 2004Orthogonality of the Principal ModesEquation 10.105 is the mathematical statement of orthogonality of the principalmodes of vibration. The orthogonality property provides a very powerful mathematical technique for decoupling the equations of motion of a multiple degreesof-freedom system.For a system exhibiting P degrees of freedom, we define the modal matrix asa P × P matrix in which the columns are the amplitude vectors for each naturalmode of vibration; that is, (1) ! (2) !![ A] =AA.
. . A ( P)(10.106)and consider the matrix triple product [S] = [ A]T [M ][ A] . Per the orthogonalitycondition, Equation 10.105, each off-diagonal term of the matrix representedby the triple product is zero; hence, the matrix [S] = [ A]T [M ][ A] is a diagonalmatrix. The diagonal (nonzero) terms of the matrix have magnitude!T!Sii = A (i) [M ] A (i)i = 1, P(10.107)As each modal amplitude vector is known only within a constant multiple (recallin earlier examples that we set A (i)1 = 1 arbitrarily), the modal amplitude vectorscan be manipulated such that the diagonal terms described by Equation 10.107can be made to assume any desired numerical value. In particular, if the value isselected as unity, so that!T!Sii = A (i) [M ] A (i) = 1i = 1, P(10.108)then the modal amplitude vectors are said to be orthonormal and the matrix tripleproduct becomes[S] = [ A] T [M ][ A] = [I ](10.109)where [I ] is the P × P identity matrix.Normalizing the modal amplitude vectors per Equation 10.108 is a straightforward procedure, as follows.
Let a specific modal amplitude be represented byEquation 10.98 in which the first term is arbitrarily assigned value of unity. Thecorresponding diagonal term of the modal matrix is thenP P(i)m jk A j A(i)k= Sii = constant(10.110)j=1 k=1If we redefine the terms of the modal amplitude vector so thatA(i)jA(i)A(i)jj=√ =PPSii##(i)m jk A(i)j Aki = 1, P(10.111)j=1 k=1the matrix described by Equation 10.109 is indeed the identity matrix.Having established the orthogonality concept and normalized the modalmatrix, we return to the general problem described by Equation 10.95, in whichthe force vector is no longer assumed to be zero.
For reasons that will become419Hutton: Fundamentals ofFinite Element Analysis42010. Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural Dynamicsapparent, we introduce the change of variables{q } = [ A]{ p}(10.112)where { p} is the column matrix of generalized displacements, which are linearcombinations of the actual nodal displacements {q }, and [A] is the normalizedmodal matrix. Equation 10.95 then becomes[M ][ A] { p̈} + [K ][ A]{ p} = {F }(10.113)Premultiplying by [ A]T , we obtain[ A] T [M ][ A]{ p̈} + [ A] T [K ][ A]{ p} = [ A] T {F }(10.114)Utilizing the orthogonality principle, Equation 10.114 is[I ]{ p̈} + [ A] T [k][ A]{ p} = [ A] T {F }(10.115)TNow we must examine the stiffness effects as represented by [ A] [K ][ A] .
Giventhat [K] is a symmetric matrix, the triple product [ A]T [K ][ A] is also a symmetric matrix. Following the previous development of orthogonality of the principalmodes, the triple product [ A]T [K ][ A] is also easily shown to be a diagonal matrix. The!values of the diagonal terms are found by multiplying Equation 10.100Tby A (i) to obtain!T!!T!−i2 A (i) [M ] A (i) + A (i) [K ] A (i) = 0i = 1, P(10.116)If the modal amplitude vectors have been normalized as described previously,Equation 10.116 is!T!i = 1, PA (i) [K ] A (i) = i2(10.117)hence, the matrix triple product [ A]T [K ][ A] produces a diagonal matrix havingdiagonal terms equal to the squares of the natural circular frequencies of the principal modes of vibration; that is,12 0 · · · 0 0 22. ..
[A]T [K ][A] = .(10.118) ... .. . 0· · ·2PFinally, Equation 10.115 becomes[I ]{ p̈} + [ 2 ]{ p} = [ A] T {F }(10.119)2with matrix [ ] representing the diagonal matrix defined in Equation 10.118.EXAMPLE 10.7Using the data of Example 10.3, normalize the modal matrix and verify that [ A] T [M ][ A] =[I ] and [ A] T [K ][ A] = [ 2 ] .Hutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 200410.7Orthogonality of the Principal Modes■ SolutionFor the first mode, we haveS11 = A!(1) T[M] A= 11.4404 m(1)!m= [1 1.4325 2.0511] 000m00 1 0 1.43252m2.0511so the first modal amplitude vector is normalized by dividing each term by√3.3824 m, which gives the normalized vector asA!(1)√S11 =0.2956 1 = √0.4289m0.6064Applying the same procedure to the modal amplitude vectors for the second and thirdmodes givesA!(2)0.6575 1 = √0.5618m−0.3550A!(3)0.6930 1 = √−0.7124m0.0782and the normalized modal matrix is0.2956 0.65750.69301 [A] = √0.4289 0.5618 −0.7124 m0.6064 −0.3550 0.0782To verify Equation 10.109, we form the triple product0.2956 0.42890.6064m 010.6575 0.5618 −0.3550 0 m[A] [M][A] =m0.6930 −0.7124 0.07820 0 0.2956 0.65750.69301 0× 0.4289 0.5618 −0.7124 = 0 10.6064 −0.3550 0.07820 0T00 2m001as expected.The triple product with respect to the stiffness matrix is0.2956 0.42890.60643 −2 0k[A]T [K ][A] = 0.6575 0.5618 −0.3550 −2 3 −1 m0.6990 −0.7124 0.07820 −1 10.2956 0.65750.6990× 0.4289 0.5618 −0.7124 0.6064 −0.3550 0.0782which evaluates to 20.153200k 1[A]T [K ][A] = 01.29120 = 0m005.05570022000 32421Hutton: Fundamentals ofFinite Element Analysis42210.
Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural Dynamics10.8 HARMONIC RESPONSE USINGMODE SUPERPOSITIONThe orthogonality condition of the principal modes is especially useful in analyzing the steady-state response of finite element models to harmonic forcingfunctions. In this context, a harmonic forcing function is described as F (t ) =F0 sin f t, where F0 is a constant force magnitude and f is a constant circularfrequency of the forcing function.
Prior to applying the mode superpositionmethod, a complete modal analysis must be performed to obtain the natural circular frequencies and normalized modal amplitude vectors (hence, the normalized modal matrix). Using the techniques of the previous section, the equationsof motion for the forced case become[I ]{ p̈} + [ 2 ]{ p} = [ A] T {F }(10.120)Assuming that the structural model under consideration exhibits P total degreesof freedom, Equation 10.120 represents a set of P uncoupled, ordinary differential equations of the formp̈i + i2 pi =P(i)A j Fj (t )i = 1, P(10.121)j=1Observing that the right-hand side is a known linear combination of harmonicforces (these are the so-called generalized forces), the solution to each of theequations is a summation of particular solutions corresponding to each of theharmonic force terms.
By analogy with the procedure used for forced vibrationof a single degree-of-freedom system in Section 10.2, the solutions of Equation 10.121 are given bypi (t ) =(i)PA j Fj0j=1i2 − 2j fsin j f ti = 1, P(10.122)Hence, the generalized displacements pi (t ) are represented by a combinationof independent harmonic motions having frequencies corresponding to theforcing frequencies. Note that, if a forcing frequency is close in value to one ofthe natural frequencies, the denominator term becomes small and the forcedresponse amplitude is large; hence, there are many possibilities for a resonantcondition.The mode superposition method provides mathematical convenience inobtaining the forced response, because the equations of motion become uncoupled and solution is straightforward.
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