AA 242B / ME 242B: Mechanical Vibrations (Spring 2016)

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1 AA 242B / ME 242B: Mechanical Vibrations (Spring 206) Solution of Homework #3 Control Tab Figure : Schematic for the control tab. Inadequacy of a static-test A static-test for measuring θ would ideally require the application of a pure moment about the axis A. Such a pure moment would need to be applied through point forces. If these forces are large, they could damage the control tab. If they are small, they would produce small angular displacements that are difficult to measure. Dynamic equilibrium As shown in Fig., the harmonic excitation applied through the driving spring is a vertical displacement of the form a cos(ωt) where a denotes amplitude and ω denotes the angular frequency. Let θ denote the counter-clockwise rotation of the control tab about the axis A. Then, admitting kinematic relations, the equation for dynamic rotational equilibrium is given by I A θ = kθ θ k l 2 θ + k 2 (a cos(ωt) lθ)l

2 which can be rewritten as.2 Natural frequency I A θ + ( kθ + (k + k 2 )l 2) θ = ak 2 l cos(ωt) () The natural frequency of the control tab is obtained in the absence of the supporting and driving springs, i. e. k = k 2 = 0, which yields ω n = kθ I A as expected..3 Resonance frequency θ + k θ I A θ = 0 The resonance frequency of the system is obtained when the driving frequency ω matches the natural frequency of the tab and spring ensemble. From Eq. (), k θ + (k + k 2 )l it follows that ω r = 2, which leads to I A k θ = I A ω 2 r (k + k 2 )l 2 2 Modal Superposition The degrees of freedom for this system have already been indicated in the schematic provided with the problem statement. The derivation of the Lagrange equations governing the free-vibrations of this 3-body oscillator goes as follows. Kinetic energy of the entire system T = 2 m 3 Potential energy of the entire system i= u 2 i V = 2 2 k ( u i+ u i ) 2 Lagrange equations d ( ) T + T V = 0, s =, 2, 3 dt u s u s u s i= 2

3 mü s + k 2 ( u i+ u i )(δ i+,s δ i,s ) = 0, s =, 2, 3 i= where δ denotes the Kronecker delta that is, δ i,j = if and only if i = j. Matrix form Mü + Ku = 0 (2) where u = [u, u 2, u 3 ] T, the mass matrix is diagonal and can be written as M = mi, and the stiffness matrix is given by 0 K = k = k 2 0 Modal superposition The exact solution of Eq. (2) can be obtained using the superposition of the natural vibration modes, Q = [q a, q a2, q a3 ], corresponding to the natural (circular) frequencies Ω = diag([ω, ω 2, ω 3 ]). The natural modes and the squares of the natural frequencies are the solutions of the symmetric eigenvalue problem KQ = MQΩ 2 = mqω 2 where M = mi has been substituted. This shows that (Q, Ω) can be directly obtained from the eigenvectors and eigenvalues of K. Simple algebraic computations lead to k 3k ω = 0, ω 2 = m, ω 3 = m q a =, q a2 = 0, q a3 = 2 3m 2m 6m where the natural modes have been mass-normalized. Note that the first mode corresponds to a rigid body motion of the entire system. Let y = [y, y 2, y 3 ] T denote the vector of modal displacements, u = Qy. Substituting this expansion in Eq. (2) and pre-multiplying the result by Q T gives ÿ + Ω 2 y = 0 The corresponding initial conditions are given by y(0) = Q T Mu(0) = [ 3m, 0, 0] T, ẏ(0) = Q T M u(0) = [0, 0, 0] T The decoupled system can be solved component-wise to obtain y (t) = 3m, y 2 (t) = y 3 (t) = 0 3

4 which by superposition leads to u(t) = Qy(t) = [,, ] T Hence, the system is simply displaced as a rigid body without any velocity or relative motion. Because the initial condition is a rigid body mode displacement (and therefore an eigenvector) and the natural modes (including the rigid body modes) are orthogonal, this is the expected result. 3 Vehicle Dynamics Figure 2: Schematic for the vehicle Degrees of freedom of the entire system vertical displacements y, y 2 of the axles/wheels vertical displacements y A, y B of the points A and B Kinematic constraints y G = by A + ay B a + b θ = y B y A a + b Kinetic energy of the entire system T = 2 I G θ Mẏ2 G + 2 m(ẏ2 + ẏ 2 2) 4

5 Potential energy of the entire system V = Mgy G + mg(y + y 2 ) + 2 k 2(y 2 + y2) k ( (ya y ) 2 + (y B y 2 ) 2) Dissipation function Lagrange equations D = 2 c [ (ẏ A ẏ ) 2 + (ẏ B ẏ 2 ) 2] d dt ( T q s ) + T q s V q s D q s = 0 q s = y q s = y 2 mÿ + (k + k 2 )y + cẏ k y A cẏ A = mg mÿ 2 + (k + k 2 )y 2 + cẏ 2 k y B cẏ B = mg q s = y A Mb 2 + I G (a + b) 2 ÿa + Mab I G b (a + b) 2 ÿb + cẏ A cẏ + k y A k y = Mg a + b q s = y B Ma 2 + I G (a + b) 2 ÿb + Mab I G a (a + b) 2 ÿa + cẏ B cẏ 2 + k y B k y 2 = Mg a + b Matrix form Mÿ + Cẏ + Ky = w where y = [y, y 2, y A, y B ] T, and m m 0 0 M = Mb 2 + I G Mab I G 0 0 (a + b) 2 (a + b) 2 Mab I G Ma 2 + I G 0 0 (a + b) 2 (a + b) 2 C = c

6 K = k + k 2 0 k 0 0 k + k 2 0 k k 0 k 0 0 k 0 k w = m m M b a+b M a a+b g Next, the focus is set on determining the response of the system when the vehicle encounters a pothole. To this end, let the displacements be measured with respect to the state at static equilibrium Ky eq = w Define x = [y y eq, ẏ] T and rewrite the governing equations in first-order form as follows Bẋ + Ax = 0 (3) where B = [ C M M 0 ] [ K 0, A = 0 M Proceeding as in the solution of the previous problem, the response of this system can be obtained using modal superposition. The generalized symmetric eigenvalue problem is given by BQ = AQΛ where Q and Λ can be obtained through a direct computation using MAT- LAB (note that the eigenvectors returned by MATLAB s eig function are not automatically mass-normalized.) For the given parameter values, the solution Λ computed by MATLAB is λ λ 2 λ 3 λ 4 λ 5 λ 6 λ 7 λ 8 = i i i i i i i i The solution Q computed by MATLAB is not reported here for the sake of brevity. Note in Λ above that each even entry is the complex conjugate of its preceding odd entry. Note also that the real part of each λ i is positive, which for a first-order system shows that this system is stable. ] 6

7 Modal superposition consists in searching for the solution x in the form x = Qz that is, in the form of a linear combination of the natural modes stored in the columns of the matrix Q. Hence, substituting x = Qz in Eq. (3) and pre-multiplying by Q T leads to ż + Λz = 0 The initial condition due to the pothole is given by x(0) = [0, 0, h, h, 0, 0, 0, 0] T and therefore z(0) = Q T Bx(0) = i i i i i i i i The decoupled system can now be solved component-wise as follows z j (t) = z j (0) exp( λ j t), j =,..., 8 then grouped into z(t) = [z (t) z 8 (t)] T. One finds that as expected, z = [z z 8 ] T = [z z z 4 z 4 ] T. Then, the response of the system is obtained by modal superposition as x(t) = Qz(t) = 8 8,2 8,2 Q j z j (t) = Q j z j (t) + Q j z j (t) j= where the 2-stepping notation is due here to the alternating storage of Q j and Q j in Q. This response is graphically displayed in Figs. 3 and 4. j= j=2 7

8 Figure 3: Displacements as a function of time Figure 4: Velocities as a function of time 4 Tuned Vibration Absorber 4. Equations of motion Kinetic energy of the entire system T = 2 m u m 2 u 2 2 8

9 Potential energy of the entire system where gravity has been ignored. V = 2 k u k 2(u 2 u ) 2 Lagrange equations d ( ) T + T V = Q s, s =, 2 dt u s u s u s m ü + (k + k 2 )u k 2 u 2 = P cos(ωt) m 2 ü 2 + k 2 u 2 k 2 u = 0 Matrix form where u = [u, u 2 ] T, and Mü + Ku = w [ ] m 0 M = 0 m 2 [ ] k + k K = 2 k 2 k 2 k 2 [ ] P cos(ωt) w = Steady-state forced response Substituting u i = U i cos(ωt), i =, 2, in the above equations leads to [ Ω 2 M + K ] [ ] P U = 0 which yields where U = [ Ω 2 M + K ] [ ] P 0 = [ ] [ k2 Ω 2 m 2 k 2 P k 2 k + k 2 Ω 2 m 0 = P [ k2 Ω 2 m 2 k 2 ] = (k 2 Ω 2 m 2 )(k + k 2 Ω 2 m ) k 2 2 ] 9

10 4.3 Tuned absorber k 2 For the tuned absorber, = k, excitement at resonance implies m 2 m and = (k 2 k 2 )(k + k 2 k ) k 2 2 = k 2 2 U = 0 U 2 = P k 2 which highlights the effect of adding a tuned vibration absorber. 4.4 Amplitude as a function of the driving frequency The desired non-dimensional amplitudes are related to the frequency ratio as follows k U P k 2 U 2 P = k 2 ( r 2 )(k ( r 2 ) + k 2 ) k 2 2 = ( r 2 )( m m 2 ( r 2 ) + ) = ( r 2 )(5 4r 2 ) [ 4( r 2 ) [ k k 2 kr 2 2 m 2 m k2 2 ] [ m ( r 2 ) m 2 These amplitudes are plotted in Figs. 5 and 6. The addition of a tuned absorber has essentially updated the resonance frequency of the system. While the amplitude of mass is zero at the original natural frequency (i. e. when r = ), the system now resonates at r 0.78 and r.28. ] ] 0

11 Figure 5: k U P against frequency ratio Figure 6: k 2 U 2 P against frequency ratio

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