Hutton - Fundamentals of Finite Element Analysis (523155), страница 71
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A similarapproach can be used to analyze the transient dynamic response of mechanicalstructures. However, in the analysis of structures, an additional tool is available.The tool, known as modal analysis, has its basis in the fact that every mechanical structure exhibits natural modes of vibration (dynamic response) and thesemodes can be readily computed given the elastic and inertia characteristics of thestructure.In this chapter, we introduce the concept of natural modes of vibration via thesimple harmonic oscillator system.
Using the finite element concepts developedin earlier chapters, the simple harmonic oscillator is represented as a finite elementsystem and the basic ideas of natural frequency and natural mode are introduced.The single degree of freedom simple harmonic oscillator is then extended to multiple degrees of freedom, to illustrate the existence of multiple natural frequenciesand vibration modes. From this basis, we proceed to more general dynamic analyses using the finite element method.10.2 THE SIMPLE HARMONIC OSCILLATORThe so-called simple harmonic oscillator is a combination of a linear elasticspring having free length L and a concentrated mass as shown in Figure 10.1a.The mass of the spring is considered negligible.
The system is assumed to besubjected to gravity in the vertical direction, and the upper end of the spring isattached to a rigid support. With the system in equilibrium as in Figure 10.1b, the387Hutton: Fundamentals ofFinite Element Analysis38810. Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural Dynamicsk (␦st ⫹ x)kLLk⫹x⫹x␦stmm(b)(a)m⫹xmg(c)Figure 10.1(a) Simple harmonic oscillator.
(b) Static equilibriumposition. (c) Free-body diagram for arbitraryposition x.gravitational force is in equilibrium with the spring force soFx = 0 = mg − k␦st(10.1)where ␦st is the equilibrium elongation of the spring and x is measured positivedownward from the equilibrium position; that is, when x = 0 , the system is at itsequilibrium position.If, by some action, the mass is displaced from its equilibrium position,the force system becomes unbalanced, as shown by the free-body diagram ofFigure 10.1c. We must apply Newton’s second law to obtain`Fx = ma x = md2 x= mg − k(␦st + x )dt 2(10.2)Incorporating the equilibrium condition expressed by Equation 10.1, Equation 10.2becomesmd2 x+ kx = 0dt 2(10.3)Equation 10.3 is a second-order, linear, ordinary differential equation with constant coefficients.
(And physically, we assume that the coefficients m and k arepositive.) Equation 10.3 is most-often expressed in the formd2 xkd2 x+x=+ 2 x = 0dt 2mdt 2(10.4)The general solution for Equation 10.4 isx (t ) = A sin t + B cos t(10.5)where A and B are the constants of integration. Recall that the solution of asecond-order differential equation requires the specification of two constants todetermine the solution to a specific problem. When the differential equation describes the time response of a mechanical system, the constants of integration aremost-often called the initial conditions.Hutton: Fundamentals ofFinite Element Analysis10.
Structural DynamicsText© The McGraw−HillCompanies, 200410.2 The Simple Harmonic Oscillator389Equation 10.5 shows that the variation of displacement of the mass as a function of time is periodic. Using basic trigonometric identities, Equation 10.5 canbe equivalently expressed asx (t ) = C sin(t + )(10.6)where the constants A and B have been replaced by constants of integration C and. Per Equation 10.6, the mass oscillates sinusoidally at circular frequency andwith constant amplitude C. Phase angle is indicative of position at time 0 sincex (0) = C sin . Also, note that, since x (t ) is measured about the equilibriumposition, the oscillation occurs about that position.
The circular frequency isk=rad/sec(10.7)mand is a constant value determined by the physical characteristics of the system.In this simple case, the natural circular frequency, as it is often called, dependson the spring constant and mass only. Therefore, if the mass is displaced from theequilibrium position and released, the oscillatory motion occurs at a constantfrequency determined by the physical parameters of the system. In the casedescribed, the oscillatory motion is described as free vibration, since the systemis free of all external forces excepting gravitational attraction.Next, we consider the simple harmonic oscillator in the finite element context. From Chapter 2, the stiffness matrix of the spring is (e) 1 −1k=k(10.8)−1 1and the equilibrium equations for the element are 1 −1u1f1k=−1 1u2f2(10.9)f1which is identical to Equation 2.4.
However, the spring element is not in staticequilibrium, so we must examine the nodal forces in detail.Figure 10.2 shows free-body diagrams of the spring element and mass,respectively. The free-body diagrams depict snapshots in time when the systemis in motion and, hence, are dynamic free-body diagrams. As the mass of thespring is considered negligible, Equation 10.9 is valid for the spring element. Forthe mass, we haved2 u 2Fx = ma x = m 2 = mg − f 2dtd2 u 2dt 2f22u2f2(a)(10.10)from which the force on node 2 isf 2 = mg − m1u1(10.11)mg(b)Figure 10.2 Freebody diagrams of (a) aspring and (b) a mass,when treated as partsof a finite elementsystem.Hutton: Fundamentals ofFinite Element Analysis39010. Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural DynamicsSubstituting for f 2 in Equation 10.9 gives 1 −1u1f1k=−1 1u2mg − m ü 2(10.12)where ü 2 = d2 u 2 /dt 2 .
The dynamic effect of the inertia of the attached mass isshown in the second of the two equations represented by Equation 10.12. Equation 10.12 can also be expressed as 0 0ü 11 −1u1f1+k=(10.13)0 mü 2−1 1u2mgwhere we have introduced the mass matrix0 0[m] =0 mand the nodal acceleration matrix{ü} =ü 1ü 2(10.14)(10.15)For the simple harmonic oscillator of Figure 10.1, we have the constraint (boundary) condition u 1 = 0, so the first of Equation 10.13 becomes simply −ku 2 = f 1 ,while the second equation ism ü 2 + ku 2 = mg(10.16)Note that Equation 10.16 is not the same as Equation 10.3. Do the two equationsrepresent the same physical phenomenon? To show that the answer is yes, wesolve Equation 10.16 and compare the results with the solution given in Equation 10.6.Recalling that the solution of any differential equation is the sum of a homogeneous (complementary) solution and a particular solution, both solutions mustbe obtained for Equation 10.16, since the equation is not homogeneous (i.e., theright-hand side is nonzero).
Setting the right-hand side to zero, the form of thehomogeneous equation is the same as that of Equation 10.3, so by analogy,the homogeneous solution isu 2 h (t ) = C sin(t + )(10.17)where , C, and are as previously defined. The particular solution must satisfyEquation 10.16 exactly for all values of time. As the right-hand side is constant,the particular solution must also be constant; hence,u 2 p (t ) =mg= ␦stk(10.18)which represents the static equilibrium solution per Equation 10.1. The completesolution is thenu 2 (t ) = u 2h (t ) + u 2 p (t ) = ␦st + C sin(t + )(10.19)Hutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 200410.2 The Simple Harmonic Oscillator391Equation 10.19 represents a sinusoidal oscillation around the equilibrium position and is, therefore, the same as the solution given in Equation 10.6.
Given thedisplacement of node 2, the reaction force at node 1 is obtained via the constraintequation asf 1 = −ku 2 (t ) = −k(␦st + C sin(t + ))(10.20)Amplitude C and phase angle are determined by application of the initial conditions, as illustrated in the following example.EXAMPLE 10.1A simple harmonic oscillator has k = 25 lb/in.
and mg = 20 lb. The mass is displaceddownward a distance of 1.5 in. from the equilibrium position. The mass is released fromthat position with zero initial velocity at t = 0. Determine (a) the natural circular frequency, (b) the amplitude of the oscillatory motion, and (c) the phase angle of the oscillatory motion.■ SolutionThe natural circular frequency is=k=m25= 21.98 rad/sec20/386.4where, for consistency of units, the mass is obtained from the weight using g = 386.4 in./s2.The given initial conditions areu 2 (t = 0) = ␦st + 1.5 in.u̇ 2 (t = 0) = 0 in./secand the static deflection is ␦st = W / k = 20/25 = 0.8 in. Therefore, we have u 2 (0) =2.3 in.
The motion of node 2 (hence, the mass) is then given by Equation 10.19 asu 2 (t ) = 0.8 + C sin(21.98t + ) in.and the velocity isu̇ 2 (t ) =du 2= 21.98C cos(21.98t + ) in./secdtApplying the initial conditions results in the equationsu 2 (t = 0) = 2.3 = 0.8 + C sin u̇ 2 (t ) = 0 = 21.98C cos The initial velocity equation is satisfied by C = 0 or = /2. If the former is true, theinitial displacement equation cannot be satisfied, so we conclude that = /2. Substituting into the displacement equation then gives the amplitude C as 1.5 in. The completemotion solution isu 2 (t ) = 0.8 + 1.5 sin 21.98t += 0.8 + 1.5 cos(21.98t ) in.2Hutton: Fundamentals ofFinite Element Analysis39210.
Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural Dynamicsindicating that the mass oscillates 1.5 in. above and below the static equilibrium positioncontinuously in time and completes one cycle every 2/21.98 sec. Therefore, the cyclicfrequency isf =21.98== 3.5 cycles/sec (Hz)22The cyclic frequency is often simply referred to as the natural frequency. The time requiredto complete one cycle of motion is known as the period of oscillation, given by =11== 0.286 secf3.510.2.1 Forced Vibrationkm⫹xF(t)Figure 10.3 Simpleharmonic oscillatorsubjected to externalforce F(t).Figure 10.3 shows a simple harmonic oscillator in which the mass is acted on bya time-varying external force F(t). The resulting motion is known as forcedvibration, owing to the presence of the external forcing function.
As the only difference in the applicable free-body diagrams is the external force acting on themass, the finite element form of the system equations can be written directlyfrom Equation 10.13 as 0 01 −1f1ü 1u1+k=(10.21)0 mü 2−1 1u2mg + F (t )While the constraint equation for the reaction force at node 1 is unchanged, thedifferential equation for the motion of node 2 is nowm ü 2 + ku 2 = mg + F (t )(10.22)The complete solution for Equation 10.22 is the sum of the homogeneous solution and two particular solutions, since two nonzero terms are on the right-handside. As we already obtained the homogeneous solution and the particular solution for the mg term, we focus on the particular solution for the external force.The particular solution of interest must satisfym ü 2 + ku 2 = F (t )(10.23)exactly for all values of time.
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