Hutton - Fundamentals of Finite Element Analysis (523155), страница 72
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Dividing by the mass, we obtainü 2 + 2 u 2 =F (t )m(10.24)where 2 = k/m is the square of the natural circular frequency. Of particularimportance in structural dynamic analysis is the case when external forcing functions exhibit sinusoidal variation in time, since such forces are quite common.Therefore, we consider the case in whichF (t ) = F0 sin f t(10.25)Hutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 200410.2 The Simple Harmonic Oscillatorwhere F0 is the amplitude or maximum value of the force and f is the circularfrequency of the forcing function, or forcing frequency for short.
Equation 10.24becomesF0ü 2 + 2 u 2 =sin f t(10.26)mTo satisfy Equation 10.24 exactly for all values of time, the terms on the left mustcontain a sine function identical to the sine term on the right-hand side. Since thesecond derivative of the sine function is another sine function, we assume asolution in the form u 2 (t ) = U sin f t , where U is a constant to be determined.Differentiating twice and substituting, Equation 10.26 becomesF0−U 2f sin f t + U 2 sin f t =sin f t(10.27)mfrom whichF0 /mU = 2(10.28) − 2fThe particular solution representing response of the simple harmonic oscillatorto a sinusoidally varying force is thenu 2 (t ) =F0 /msin f t 2 − 2f(10.29)The motion represented by Equation 10.29 is most often simply called the forcedresponse and exhibits two important characteristics: (1) the frequency of theforced response is the same as the frequency of the forcing function, and (2) ifthe circular frequency of the forcing function is very near the natural circularfrequency of the system, the denominator in Equation 10.29 becomes very small.The latter is an extremely important observation, as the result is large amplitudeof motion.
In the case f = , Equation 10.29 indicates an infinite amplitude.This condition is known as resonance, and for this reason, the natural circularfrequency of the system is often called the resonant frequency. Mathematically,Equation 10.29 is not a valid solution for the resonant condition (Problem 10.5);however, the correct solution for the resonant condition nevertheless exhibitsunbounded amplitude growth with time.The simple harmonic oscillator just modeled contains no device for energydissipation (damping).
Consequently, the free vibration solution, Equation 10.20,represents motion that continues without end. Physically, such motion is not possible, since all systems contain some type of dissipation mechanism, such asinternal or external friction, air resistance, or devices specifically designed for thepurpose. Similarly, the infinite amplitude indicated for the resonant conditioncannot be attained by a real system because of the presence of damping.
However,relatively large, yet bounded, amplitudes occur at or near the resonant frequency.Hence, the resonant condition is to be avoided if at all possible. As is subsequentlyshown, physical systems actually exhibit multiple natural frequencies, so multiple resonant conditions exist.393Hutton: Fundamentals ofFinite Element Analysis39410. Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural Dynamics10.3 MULTIPLE DEGREES-OF-FREEDOMSYSTEMS113k2U2m2k3U3mFigure 10.4 Aspring-mass systemexhibiting 2 degreesof freedom.Figure 10.4 shows a system of two spring elements having concentrated massesattached at nodes 2 and 3 in the global coordinate system.
As in previous examples, the system is subjected to gravity and the upper spring is attached to a rigidsupport at node 1. Of interest here is the dynamic response of the system of twosprings and two masses when the equilibrium condition is disturbed by someexternal influence and then free to oscillate without external force. We could takethe Newtonian mechanics approach by drawing the appropriate free-body diagrams and applying Newton’s second law of motion to obtain the governingequations. Instead, we take the finite element approach. By now, the procedure ofassembling the system stiffness matrix should be routine.
Following the procedure, we obtain3k −3k0[K ] = −3k 5k −2k (10.30)0−2k 2kas the system stiffness matrix. But what of the mass/inertia matrix? As the massesare concentrated at element nodes, we define the system mass matrix as0 0 0[M] = 0 m 0 (10.31)0 0 mThe equations of motion can be expressed as Ü1 U1 R 1 [M] Ü2 + [K ] U2 = mg U3mgÜ3(10.32)where R1 is the dynamic reaction force at node 1.Invoking the constraint condition U1 = 0, Equation 10.32 become Ü 2m 05k−2kU2mg+=(10.33)0 m−2k2kU3mgÜ 3which is a system of two second-order, linear, ordinary differential equations inthe two unknown system displacements U2 and U3.
As the gravitational forcesindicated by the forcing function represent the static equilibrium condition, theseare neglected and the system of equations rewritten as Ü 2m 05k−2kU20+=(10.34)0 m−2k2kU30Ü 3As a practical matter, most finite element software packages do not includethe structural weight in an analysis problem. Instead, inclusion of the structuralHutton: Fundamentals ofFinite Element Analysis10.
Structural DynamicsText© The McGraw−HillCompanies, 200410.3 Multiple Degrees-of-Freedom Systemsweight is an option that must be selected by the user of the software. Whether toinclude gravitational effects is a judgment made by the analyst based on thespecifics of a given structural geometry and loading.The system of second-order, linear, ordinary, homogeneous differentialequations given by Equation 10.34 represents the free-vibration response of the2 degrees-of-freedom system of Figure 10.4.
As a freely oscillating system, weseek solutions in the form of harmonic motion asU 2 (t ) = A 2 sin(t + )U 3 (t ) = A 3 sin(t + )(10.35)where A2 and A3 are the vibration amplitudes of nodes 2 and 3 (the masses attached to nodes 2 and 3); is an unknown, assumed harmonic circular frequencyof motion; and is the phase angle of such motion. Taking the second derivativeswith respect to time of the assumed solutions and substituting into Equation 10.34results in 0A25k−2kA202 m−sin(t + ) +sin(t + ) =0 m−2k2k0A3A3(10.36)or5k − m 2−2k−2k2k − m 2A2A3sin(t + ) = 00(10.37)Equation 10.37 is a system of two, homogeneous algebraic equations, whichmust be solved for the vibration amplitudes A2 and A3. From linear algebra, asystem of homogeneous algebraic equations has nontrivial solutions if and onlyif the determinant of the coefficient matrix is zero.
Therefore, for nontrivialsolutions, 5k − m 2−2k =0(10.38) −2k2k − m 2 which gives(5k − m 2 )(2k − m 2 ) − 4k 2 = 0(10.39)Equation 10.39 is known as the characteristic equation or frequency equation ofthe physical system. As k and m are known positive constants, Equation 10.39 istreated as a quadratic equation in the unknown 2 and solved by the quadraticformula to obtain two rootsk12 =m(10.40)k22 = 6m395Hutton: Fundamentals ofFinite Element Analysis39610.
Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural Dynamicsor1 =2 =km(10.41)k6mIn mathematical rigor, there are four roots, since the negative values corresponding to Equation 10.41 also satisfy the frequency equation. The negative valuesare rejected because a negative frequency has no physical meaning and use of thenegative values in the assumed solution (Equation 10.35) introduces only a phaseshift and represents the same motion as that corresponding to the positive root.The 2 degrees-of-freedom system of Figure 10.4 is found to have two naturalcircular frequencies of oscillation.
As is customary, the numerically smaller ofthe two is designated as 1 and known as the fundamental frequency. The taskremains to determine the amplitudes A2 and A3 in the assumed solution. For thispurpose, Equation 10.37 is 5k − m 2A20−2k=(10.42)0−2k2k − m 2A3As Equation 10.42 is a set of homogeneous equations, we can find no absolutevalues of the amplitudes. We can, however, obtain information regarding thenumerical relations among the amplitudes as follows. If we substitute 2 = 12 =k/m into either algebraic equation, we obtain A 3 = 2 A 2 , which defines theamplitude ratio A 3 / A 2 = 2 for the first, or fundamental, mode of vibration. Thatis, if the system oscillates at its fundamental frequency 1 , the amplitude ofoscillation of m2 is twice that of m1. (Note that we are unable to calculate theabsolute value of either amplitude; only the ratio can be determined.
The absolutevalues depend on the initial conditions of motion, as is subsequently illustrated.)The displacement equations for the fundamental mode are then(1)U 2 (t ) = A 2 sin(1 t + 1 )(1)(1)U 3 (t ) = A 3 sin(1 + 1 ) = 2 A 2 sin(1 t + 1 )(10.43)where the superscript on the amplitudes is used to indicate that the displacementscorrespond to vibration at the fundamental frequency.Next we substitute the second natural circular frequency 2 = 22 = 6k/minto either equation and obtain the relation A 3 = −0.5 A 2 , which defines the second amplitude ratio as A 3 / A 2 = −0.5.
So, in the second natural mode of vibration, the masses move in opposite directions. The displacements correspondingto the second frequency are then(2)U 2 (t ) = A 2 sin(2 t + 2 )(2)(2)U 3 (t ) = A 3 sin(2 + 2 ) = −0.5 A 2 sin(2 t + 2 )where again the superscript refers to the frequency.(10.44)Hutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 200410.3 Multiple Degrees-of-Freedom Systems397Therefore, the free-vibration response of the 2 degree-of-freedom system isgiven by(1)(2)U 2 (t ) = A 2 sin(1 t + 1 ) + A 2 sin(2 t + 2 )(1)(2)U 3 (t ) = 2 A 2 sin(1 t + 1 ) − 0.5 A 2 sin(2 + 2 )(10.45)and we note the four unknown constants in the solution; specifically, these are the(2)amplitudes A (1)2 , A 2 and the phase angles 1 and 2 .
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