Hutton - Fundamentals of Finite Element Analysis (523155), страница 73

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Evaluation of the constantsis illustrated in a subsequent example.Depending on the reader’s mathematical background, the analysis of the2 degree-of-freedom vibration problem may be recognized as an eigenvalueproblem [1]. The computed natural circular frequencies are the eigenvalues ofthe problem and the amplitude ratios represent the eigenvectors of the problem.Equation 10.45 represents the response of the system in terms of the naturalmodes of vibration. Such a solution is often referred to as being obtained bymodal superposition or simply modal analysis. To represent the complete solution for the system, we use the matrix notation (1) (2)A2A2U 2 (t )=sin(␻1 t + ␾1 ) +sin(␻2 t + ␾2 )(10.46)(1)(2)U 3 (t )2A 2−0.5 A 2which shows that the modes interact to produce the overall motion of the system.EXAMPLE 10.2Given the system of Figure 10.4 with k = 40 lb/in.

and mg = W = 20 lb, determine(a) The natural frequencies of the system.(b) The free response, if the initial conditions areU 2 (t = 0) = 1 in.U 3 (t = 0) = 0.5 in.U̇ 2 (t = 0) = U̇ 3 (t = 0) = 0These initial conditions are specified in reference to the equilibrium position of thesystem, so the computed displacement functions do not include the effect of gravity.■ SolutionPer Equation 10.41, the natural circular frequencies are4040(386.4)== 27.8 rad/sec␻1 =20/g206k6(40)6(40)(386.4)␻2 ==== 68.1 rad/secm20/g20k=mThe free-vibration response is given by Equation 10.35 asU 2 (t ) = AU 3 (t ) =(1)2 sin(27.8t+ ␾1 ) + A(1)2 A 2 sin(27.8t+ ␾1 ) −(2)2 sin(68.1t+ ␾2 )(2)0.5 A 2 sin(68.1t+ ␾2 )Hutton: Fundamentals ofFinite Element Analysis39810.

Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural DynamicsThe amplitudes and phase angles are determined by applying the initial conditions, whichareU 2 (0) = 1 = A(1)2 sinU 3 (0) = 0.5 =(1)2 A 2 sin␾1 + A(2)2 sin␾2␾1 − 0.5 AU̇ 2 (0) = 0 =(1)27.8 A 2 cosU̇ 3 (0) = 0 =(1)2(27.8) A 2 cos␾1 +(2)2 sin␾2(2)68.1 A 2 cos␾2␾1 − 0.5(68.1) A(2)2 cos␾2The initial conditions produce a system of four algebraic equations in the four un(2)knowns A (1)2 , A 2 , ␾1 , ␾2 .

Solution of the equations is not trivial, owing to the presence(2)of the trigonometric functions. Letting P = A (1)2 sin ␾1 and Q = A 2 sin ␾2 , the displacement initial condition equations becomeP+Q=12 P − 0.5 Q = 0.5which are readily solved to obtainP = A(1)2 sin␾1 = 0.4andQ= A(2)2 sin␾2 = 0.6(2)Similarly, setting R = A (1)2 cos ␾1 and S = A 2 sin ␾2 , the initial velocity equations are27.8 R + 68.1S = 02(27.8) R − 0.5(68.1) S = 0representing a homogeneous system in the variables R and S. Nontrivial solutions existonly if the determinant of the coefficient matrix is zero.

In this case, the determinant is notzero, as may easily be verified by direct computation. There are no nontrivial solutions;hence, R = S = 0 . Based on physical argument, the amplitudes cannot be zero, so wemust conclude that cos ␾1 = cos ␾2 = 0 ⇒ ␾1 = ␾2 = ␲/2 . It follows that the sine func(2)tion of the phase angles have unity value; hence, A (1)2 = 0.4 and A 2 = 0.6 . Substitutingthe amplitudes into the general solution form while noting that sin(␻t + ␲/2) = cos ␻t ,the free-vibration response of each mass isU 2 (t ) = 0.4 cos 27.8t + 0.6 cos 68.1tU 3 (t ) = 0.8 cos 27.8t − 0.3 cos 68.1tThe displacement response of each mass is seen to be a combination of motions corresponding to the natural circular frequencies of the system.

Such a phenomenon is characteristic of vibrating structural systems. All the natural modes of vibration participate inthe general motion of a structure.10.3.1 Many-Degrees-of-Freedom SystemsAs illustrated by the system of two springs and masses, there are two naturalfrequencies and two natural modes of vibration. If we extend the analysis toHutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 200410.3 Multiple Degrees-of-Freedom Systems399a system of springs and masses having N degrees of freedom, as depicted inFigure 10.5, and apply the assembly procedure for a finite element analysis, thefinite element equations are of the form[M ]{Ü } + [K ]{U } = {0}(10.47)k1m1where [M ] is the system mass matrix and [K ] is the system stiffness matrix. Todetermine the natural frequencies and mode shapes of the system’s vibrationmodes, we assume, as in the 1 and 2 degrees-of-freedom cases, thatU i (t ) = A i sin(␻t + ␾)(10.48)k2m2Substitution of the assumed solution into the system equations leads to the frequency equation|[K ] − ␻ 2 [M ]| = 0(10.49)which is a polynomial of order N in the variable ␻ 2.

The solution of Equation 10.49results in N natural frequencies ␻ j , which, for structural systems, can be shown tobe real but not necessarily distinct; that is, repeated roots can occur. As discussedmany times, the finite element equations cannot be solved unless boundary conditions are applied so that the equations become inhomogeneous. A similar phenomenon exists when determining the system natural frequencies and modeshapes. If the system is not constrained, rigid body motion is possible and one ormore of the computed natural frequencies has a value of zero. A three-dimensionalsystem has six zero-valued natural frequencies, corresponding to rigid body translation in the three coordinate axes and rigid body rotations about the three coordinate axes.

Therefore, if improperly constrained, a structural system exhibitsrepeated zero roots of the frequency equation.Assuming that constraints are properly applied, the frequencies resultingfrom the solution of Equation 10.49 are substituted, one at a time, into Equation 10.47 and the amplitude ratios (eigenvectors) computed for each naturalmode of vibration. The general solution for each degree of freedom is thenexpressed asU i (t ) =N( j)A i sin(␻ j t + ␾j )k3i = 1, NkNmNFigure 10. 5 Aspring-mass systemexhibiting arbitrarilymany degrees offreedom.(10.50)j=1illustrating that the displacement of each mass is the sum of contributions fromeach of the N natural modes.

Displacement solutions expressed by Equation 10.50 are said to be obtained by modal superposition. We add the independent solutions of the linear differential equations of motion.EXAMPLE 10.3Determine the natural frequencies and modal amplitude vectors for the 3 degrees-offreedom system depicted in Figure 10.6a.Hutton: Fundamentals ofFinite Element Analysis40010. Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural Dynamics1U1 ⫽ 0km2U22km3U3k2m4Figure 10.6 System with3 degrees of freedom forExample 10.3.U4(a)(b)■ SolutionThe finite element model is shown in Figure 10.6b, with node and element numbers asindicated.

Assembly of the global stiffness matrix results ink −k[K ] = 00−k3k−2k00−2k3k−k00 −k kSimilarly, the assembled global mass matrix is0 00 m[M] = 0 00 000m000 0 2mOwing to the constraint U1 = 0, we need consider only the last three equations of motion,given bym000m0   Ü2 0 3k0  Ü3 +  −2k 2m  Ü4 0−2k3k−k   0  U2   0 −k  U3 = 0   0U4kAssuming sinusoidal response as Ui = Ai sin(␻t + ␾), i = 2, 4 and substituting into theequations of motion leads to the frequency equation 3k − ␻2 m −2k0−2k3k − ␻2 m−k0=0−k2k − 2␻ m Hutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 200410.3 Multiple Degrees-of-Freedom SystemsExpanding the determinant and simplifying gives 2 3k 4kk2␻ − 6.5 ␻ + 7.5␻ −=0mmm6which will be treated as a cubic equation in the unknown ␻ 2 .

Setting ␻ 2 = C (k/m) , thefrequency equation becomes(C 3 − 6.5C 2 + 7.5C − 1)km3=0which has the rootsC 1 = 0.1532C 2 = 1.2912C 3 = 5.0556The corresponding natural circular frequencies are␻1 = 0.3914␻2 = 1.1363␻3 = 2.2485kmkmkmTo obtain the amplitude ratios, we substitute the natural circular frequencies into theamplitude equations one at a time while setting (arbitrarily) A (i2) = 1, i = 1, 2, 3 and solvefor the amplitudes A (i3) and A (i4) .

Using ␻1 results in−2k A√(1)2 (1)(1)3k − ␻12 m A 2 − 2k A 3 = 0(1)(1)+ 3k − ␻12 m A 3 − k A 4 = 0 (1)(1)−k A 3 + k − 2␻12 m A 4 = 0Substituting ␻1 = 0.3914 k/m , we obtain2.847 A(1)−2 A 2+(1)2(1)3=0(1)A4=0(1)0.694 A 4=0− 2A(1)2.847 A 3(1)−A 3+−As discussed, the amplitude equations are homogeneous; explicit solutions cannot beobtained. We can, however, determine the amplitude ratios by setting A (1)2 = 1 to obtainA(1)3= 1.4235A(1)4= 2.0511The amplitude vector corresponding to the fundamental mode ␻1 is then represented asA(1)!=A(1)21 1.43252.0511401Hutton: Fundamentals ofFinite Element Analysis40210. Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural Dynamicsand this is the eigenvector corresponding to the eigenvalue ␻1 .

Proceeding identicallywith the values for the other two frequencies, ␻2 and ␻3 , the resulting amplitude vectorsare1A= A(2)0.85442−0.53991!A(3) = A(3)−1.027920.1128!(2)This example illustrates that an N degree-of-freedom system exhibits N naturalmodes of vibration defined by N natural circular frequencies and the corresponding N amplitude vectors (mode shapes). While the examples deal with discretespring-mass systems, where the motions of the masses are easily visualized asrecognizable events, structural systems modeled via finite elements exhibit Nnatural frequencies and N mode shapes, where N is the number of degrees offreedom (displacements in structural systems) represented by the finite elementmodel. Accuracy of the computed frequencies as well as use of the natural modesof vibration to examine response to external forces is delineated in followingsections.10.4 BAR ELEMENTS: CONSISTENTMASS MATRIXIn the preceding discussions of spring-mass systems, the mass (inertia) matrixin each case is a lumped (diagonal) matrix, since each mass is directly attachedto an element node.

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