Problems and Solutions for Section 4.4 (4.47 through 4.59) 4.47 A vibration model of the drive train of a vehicle is illustrated as the three-degreeof-freedom system of Figure P4.47. Calculate the undamped free response [i.e. M(t) = F(t) = 0, c1 = c2 = 0] for the initial condition x(0) = 0, x (0) = [0 0 1]T. Assume that the hub stiffness is 12,000 N/m and that the axle/suspension is 20,000 N/m. Assume the rotational element J is modeled as a translational mass of 75 kg. Solution: Given: From the equation of motion 0 x +120001 1 0 x = 0 75 0 2 1 3 0 100 0 0 2 2 0 0 3000 And the initial conditions x0 = 0 , x 0 = 0 0 0 0 1 We have λ1= 0 And λ2= 93.5418 λ3 = 434.458 ω1 = 0 rad / s ω2 = 9.6717 rad /s ω3 = 20.8437 rad / s The eigenvectors are : v1 = 0.153695 v2 = 0.880361 v3 = 0.4488 0.177471 0.422218 0888952 0.97205 0.216276 0.0913382 1 1 q 0 = M 2 x 0 = 0 q 0 = M 2 x 0 = 0 0 54.7723 qt = c1 + c 4 t v1 + c 2 sin ω 2 t + 2 v 2 + c 3 sin ω3 t + 3 v 3 ωi viT q0 vT q 0 and ci = i Where i = tan T i = 2,3 2= 3= 0 c 2= −1.2248 v q 0 ω cos i i i c 3= 0.240015 c 1= 0 and c 4 = 53.2414 qt = 53.2414tv1 +1.2248sin 9.6717t v2 + 0.240015sin 20.8437t v3 1 Copyright © 2015 Pearson Education Ltd. 1 xt = M 2 qt = 0.944882t 1 + 0.1245sin 9.6717t + 0.012438 sin 20.8437t 0.021336 1 0.05171 0.0004002 1 0.0048 4.48 Calculate the natural frequencies and normalized mode shapes of é4 0 0 ù é 4 -1 0 ù ê ú ê ú x + ê -1 2 -1ú x = 0 ê 0 2 0 ú ê0 0 1 ú ê 0 -1 1 ú ë û ë û Solution: Given the indicated mass and stiffness matrix, calculate eigenvalues: é 0.5 é 1 0 0ù -0.3536 0 ù ê ú ê ú M -1/ 2 = ê 0 0.7071 0 ú Þ K = M -1/ 2 KM -1/ 2 = ê -0.3536 1 -0.7071ú ê0 ê 0 0 1 úû -0.7071 1 úû ë ë ( ) det K - l I = l 3 - 3l 2 + 2.375l - 0.375 = 0 l1 = 0.2094, l2 = 1, l3 = 1.7906 The natural frequencies are: w 1 = 0.4576 rad/s w 2 = 1 rad/s w 3 = 1.3381 rad/s The corresponding eigenvectors are: é -0.3162 ù é 0.8944 ù é 0.3162 ù ê ú ê ú ê ú v1 = ê -0.7071 ú v 2 = ê 0 ú v 3 = ê -0.7071ú ê -0.6325 ú ê -0.4472 ú ê 0.6325 ú ë û ë û ë û The relationship between eigenvectors and mode shapes is u = M -1/ 2 v The mode shapes are: é -0.1581ù é 0.4472 ù é 0.1581ù ê ê ê ú ú ú u1 = ê -0.5 ú , u2 = ê 0 ú , u3 = ê -0.5 ú ê -0.6325 ú ê -0.4472 ú ê 0.6325 ú ë û ë û ë û The normalized mode shapes are Copyright © 2015 Pearson Education Ltd. û1 = 4.49 é0.192 ù é 0.707 ù é 0.192 ù ê ú ê ú ê ú = ê0.609 ú , û 2 = ê 0 ú , û3 = ê -0.609 ú . T u1 u1 ê 0.77 ú ê -0.707 ú ê 0.77 ú ë û ë û ë û u1 The vibration is the vertical direction of an airplane and its wings can be modeled as a three-degree-of-freedom system with one mass corresponding to the right wing, one mass for the left wing, and one mass for the fuselage. The stiffness connecting the three masses corresponds to that of the wing and is a function of the modulus E of the wing. The equation of motion is é 3 -3 0 ù é x1 ù é 0 ù é 1 0 0 ù é x1 ù úê ú ê ú ú ê ú EI ê ê m ê 0 4 0 ú ê x2 ú + 3 ê -3 6 -3ú ê x2 ú = ê 0 ú l ê 0 -3 3 ú ê x ú ê 0 ú ê 0 0 1 ú ê x ú ûë 3û ë û ûë 3û ë ë The model is given in Figure P4.49. Calculate the natural frequencies and mode shapes. Plot the mode shapes and interpret them according to the airplane's deflection. x 1(t) x 2(t) l l m x 3(t) m 4m EI (a) EI (b) x 2(t) x 1(t) m x 3(t) m 4m k 3E I k l3 3E I l3 (c) Figure P4.49 Model of the wing vibration of an airplane: (a) vertical wing vibration; (b) lumped mass/beam deflection model; (c) spring–mass model. Copyright © 2015 Pearson Education Ltd. Solution: Given the equation of motion indicated above, the mass-normalized stiffness matrix is calculated to be é1 0 0 ù é 3 -1.5 0 ù 1 ê EI ê ú ú - 12 - 12 - 12 M = ê 0 0.5 0 ú , K = M KM = m3 ê -1.5 1.5 -1.5ú mê ú ê 0 -1.5 3 úû ë0 0 1 û ë EI Computing the matrix eigenvalue by factoring out the constant yields m3 EI EI det( K - l I ) = 0 Þ l1 = 0, l2 = 3 3 , l3 = 4.5 3 m m and eigenvectors: é0.4082 ù é -0.7071ù é 0.5774 ù ê ú ê ú ê ú v1 = ê 0.8165 ú v 2 = ê 0 ú v 3 = ê -0.5774 ú ê0.4082 ú ê 0.7071 ú ê 0.5774 ú ë û ë û ë û The natural frequencies are 1 = 0, 2 = 1.7321 EI rad/s, and 3 = 2.1213 m3 EI rad/s. m3 The relationship between the mode shapes and eigenvectors u is just u = M-1/2v. The fist mode shape is the rigid body mode. The second mode shape corresponds to one wing up and one down the third mode shape corresponds to the wings moving up and down together with the body moving opposite. Normalizing the mode shapes yields (calculations in Mathcad): Copyright © 2015 Pearson Education Ltd. 1 0 0 Mh 3 0 2 0 K 0 0 1 la 3 3 0 6 0 3 3 eigenvals Kh Kh Mh 1 K . Mh 3 1.5 0 Kh = 4.5 la = 1. 3 1.5 1.5 0 3 1.5 1.5 3 0 v1 eigenvec Kh , la2 0.408 v1 = eigenvec Kh , la1 v2 0.707 0.816 v2 = 0.408 v3 0 0.707 0.577 eigenvec Kh , la0 v3 = 0.577 0.577 0.408 la2 w 2.967 10 la1 w = la0 u1n 8 u1 Mh 1. v1 u1 = 0.408 1.732 2.121 0.577 u1 u1n = u1 0.577 0.707 u2 Mh 1. v2 u2 = 0.577 Mh 1. v3 u3 = 0.289 u2n 0.707 u2 u2 u2n = 0.577 u3 0 0.707 0.667 u3 u3n 0 0.707 0.577 u3 0.408 u3n = 0.333 0.667 These are plotted: 4.50 Solve for the free response of the system of Problem 4.49. Where E = 6.9 109 N/m2, l = 2 m, m = 4000 kg, and I = 5.2 10-6m4. Let the initial displacement () correspond to a gust of wind that causes an initial condition of x 0 = 0, x(0) = T [0.2 0 0] m. Discuss your solution. Solution: From problem 4.49 and the given data Copyright © 2015 Pearson Education Ltd. 0 4000 0 x + 16,000 0 0 0 4000 0 4 1.346 0 1.346 10 x = 0 1.346 1.346 2.691 0 1.346 1.346 0.2 0 x 0 = 0 Iq + 3.365 1.6825 T 0 m 1.6825 1.68187 0 x0 = 1.6825 Calculate eigenvalues and eigenvectors: ~ det K λI = 0 λ2 = 0 ω1 = 0 rad / s λ2 = 3.365 ω3 = 2.24662rad / s q = 0 1.6825 3.365 0 ω2 = 1.83439 rad / s λ3 = 5.04729 v1 = 0.408215 v 2 = 0.7071 v 3 = 0.57737 0.81653 0 0.577303 0.408215 0.7071 0.57737 The solution is given by q t = c1 +c4 t v1 +c2sin ω2t + 2 v2 +c3sin ω3t + 3 v3 ω vT q 0 viT q 0 i= 2,3 Where i = tan 1 i T i c = i = 2,3 i v q 0 sin i i π Thus, 2 = 3 = ,c2 = 8.94427,c3 = 7.30327 2 So, q 0 = c1v1 + ci sini vi q 0 = c4 vi + ωi ci cosi vi 3 3 i= 2 i= 2 3 T T Premultiplyby v i : ∑ v i q (0 )= 5.16355= c 1 3 ∑ v Ti q́ (0 )= 0= c 4 So, q t = 5.16355v1 8.94427cos 1.834t v2 + 7.30327cos 2.246 t v3 Convert to physical coordinates: 1 i= 2 i= 2 xt = M 2 q t xt = 0.0333 + 0.1 cos1.834t + 0.06667 cos2.246tm 0.0333 0 0.0333 0.0333 0.1 0.06667 4.51 Consider the two-mass system of Figure P4.51. This system is free to move in the x1 - x2 plane. Hence each mass has two degrees of freedom. Derive the linear Copyright © 2015 Pearson Education Ltd. equations of motion, write them in matrix form, and calculate the eigenvalues and eigenvectors for m = 20 kg and k = 200 N/m. Solution: Given : m= 20 kg ,k = 200 N / m Mass 1 x 1 − direction: m x́ 1= − 4 k x 1 +k (x 3 − x1 )= −5 k x 1+ k x3 x 2 − direction: m x́ 2= − 3 k x 2 −k x 2= − 4 k x 2 Mass 2 x 3 − direction: m x́ 3= − 4 k x 3 +k (x 3 − x 1 )= − k x 1 −5 k x 3 x 4 − direction: m x́ 4 = − 4 k x 4 −2 k x 4= − 6 k x 4 In matrix form the values given: 200 0 x = 0 20 0 0 0 x + 1000 0 800 0 0 0 20 0 0 0 0 0 20 0 200 0 1000 0 0 0 0 20 0 0 0 1200 ~ 0 10 0 K = M 1 / 2 KM 1 / 2 = 50 40 0 0 0 10 0 50 0 0 0 0 60 ~ det K λI = λ 4 200λ 3 +14,800λ 2 480,000λ + 5,760,000 = 0 λ1 = 40, λ2 = 40, λ3 = 60, λ4 = 60 ~ The corresponding eigenvectors are found from solving K λi vi = 0 for each value of the index and normalizing: v1 = 0 v2 = 0.7071 v3 = 0.7071 v4 = 0 0 0 0 1 0 0.7071 0.7071 0 1 0 0 0 These vectors are not unique. 4.52 Consider again the system discussed in Problem 4.51. Use modal analysis to calculate the solution if the mass on the left is raised along the x2 direction exactly 0.02 m and let go. Solution: Given: ∆ x 2= 0.02 m Copyright © 2015 Pearson Education Ltd. 0 x + 1000 0 200 20 0 0 0 800 0 0 20 0 200 0 1000 0 0 20 0 0 0 1 / 2 M = 0.2236071 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ~ 0 10 0 K = M 1 / 2 KM 1 / 2 = 50 40 0 0 0 10 0 50 0 0 0 60 0 20 0 0 0 0 λ1= 40 λ2= 40 λ3 = 60 λ 4= 60 0 x = 0 0 0 1200 0 ω1 = 6.3246 ω2 = 6.3246 ω3 = 7.7460 ω 4= 7.7460 rad/s rad/s rad/s rad/s v1 = 0 v 2 = 0.7071 v 3 = 0.7071 v 4 = 0 1 0 0 0 0 0.7071 0.7071 0 1 0 0 0 Also, 0 0.02 0 0 m and x0 = 0 x0 = Use the mode summation methods to find the solution Transform the initial conditions: q0 = M 1 / 2 x0 = q 0 = M 1 / 2 x 0 = 0 The solution is given by Eq.(4.103), 0 0.0894427 0 0 x t = d i sin ωi t + i ui 4 i=1 ω vT q 0 i = tan 1 i T i i=1,2,3,4 (Eq.(4.97)) v q 0 i Where Copyright © 2015 Pearson Education Ltd. di = viT q 0 sini ui= M − 1/ 2 i=1,2,3,4 (eq.(4.98)) vi Substituting known values yields π 1 = 2 = 3 = 4 = rad 2 d 1= 0.0894427 d 2= d 3 = d 4= 0 u 1 = 0 u2 = 0.158114 u3 = 0.158114 0.223607 0 0 0 0.158114 0.158114 0 0 0 u4 = 0 0 0 0.223607 The solution is xt = 0 0.02cos6.3246t 0 0 4.53 The vibration of a floor in a building containing heavy machine parts is modeled in Figure P4.53. Each mass is assumed to be evenly spaced and significantly larger than the mass of the floor. The equation of motion then becomes (m = m 1 2 ) = m3 = m . Copyright © 2015 Pearson Education Ltd. é 9 1 13 ù ê ú ê 64 6 192 ú é x1 ù 1 1 úê ú EI ê 1 mI x+ 3 ê êx ú = 0 6 3 6 úê 2ú l ê ú 9 ú ë x3 û ê 13 1 êë 192 6 64 úû Calculate the natural frequencies and mode shapes. Assume that in placing box m2 on the floor (slowly) the resulting vibration is calculated by assuming that the initial displacement at m2 is 0.04 m. If l = 2 m, m = 200 kg, E = 0.6 109 N/m2, I = 4.17 10-5 m4. Calculate the response and plot your results. Solution: The equations of motion can be written as EI 9 1 13 m 1 0 0 x1 + 3 x1 = 0 l 64 6 192 0 1 0 x2 x2 0 0 1 1 1 1 x x3 6 3 6 3 13 1 9 192 6 64 Or mI x́+ Kx= 0 where I is the 3x3 identity matrix The natural frequencies of the system are obtained using the characteristic equation K ω 2 M = 0 Using the given mass and stiffness matrices yields the following characteristic equation 2 3 59 EI m3 4 41 (EI ) m 2 7 (EI ) 3 6 m ω − 96l 3 13 ω + 786 l 6 ω − 6912 l 9 =0 Substituting for E , I , m , and l yields the following answers for the natural frequency 137 EI 7 EI 13 + 137 EI , ω2 = ± , ω3 = ± 3 3 96ml 48ml 3 ml The plus minus sign shown above will cause the exponential terms to change to trigonometric terms using Euler's formula. Hence, the natural frequencies of the system are 0.65 rad/s, 1.068rad/s and 2.837 rad/s. ω1 = ± Let the mode shapes of the system be u1 , u2 , and u3 . The mode shapes should satisfy the following equation Copyright © 2015 Pearson Education Ltd. K ω M u i1 = 0, i = 1,2,3 ui 2 ui 3 Notice that the system above does not have a unique solution for u1 since 2 1 [K −ω 21 M ] had to be singular in order to solve for the natural frequency ω. Solving the above equation yields the following relations 2 3 u i2 1 96 m ωi l −7 EI , i= 1,2,3 = u i3 3 13 mωi2 l 3 + EI And ui 1= ui 3 ,i= 1,3 but for the second mode shape this is different u21= u23 Substituting the values given yields 2 3 u 12 1 96 m ωi l −7 EI = −1.088 = u 13 3 13 m ω2i l3+ EI u 22 1 96 m ω2 l −7 EI =0 = u 23 3 13 mω 22 l 3 + EI 2 3 u 32 1 96 m ω3 l −7 EI = = 1.838 u 33 3 13 m ω23 l 3 + EI 2 3 If we let ui 3= 1, i= 1,2,3 , then u1 = 1 , u2 = 1, u3 = 1 1.838 1.088 0 1 1 0 These mode shapes can be normalized to yield u1 = 0.5604 , u2 = 0.7071, u3 = 0.4312 0.7926 0 0.6098 0.4312 0.7071 0.5604 The second box, m2 , is placed slowly on the floor; hence, the initial velocity can be safely assumed zero. The initial displacement at m2 is given to be 0.04 m. Hence, the initial conditions in vector form are given as x0 = 0 x 0 = 0 and 0.04 0 0 0 The equations of motion given by Mx t + Kxt = 0 can be transformed into the modal coordinates by applying the following transformation xt = Sr t = M 1 / 2 Pr t where P is the basis formed by the mode shapes of the system, given by P = u1 u 2 u 3 Copyright © 2015 Pearson Education Ltd. Hence, the transformation S is given by S = 0.04 0.05 0.03 0.056 0.043 0 0.04 0.05 0.03 The initial conditions will be also transformed r 0 = S 1 x0 = 0.339943 0 0.45258 Hence, the modal equations are With the above initial conditions. The solution will then be r t = 0.339943cos0.65t 0 0.45258cos2.837t The solution can then be determined by xt = 0.01359cos 0.65t 0.013577co s2.837 t 0.0146cos 0.65t 0.02534cos 2.837t 0.01359cos 0.65t 0.0135774c os2.837 t Plot: Copyright © 2015 Pearson Education Ltd. 4.54 Recalculate the solution to Problem 4.53 for the case that m2 is increased in mass to 1000 kg. Compare your results to those of Problem 4.53. Do you think it makes a difference where the heavy mass is placed? Solution: Given the data indicated the equation of motion becomes : 1 13 9 0 x + 3.197 10 4 x=0 200 0 64 6 192 1000 0 0 0 1 1 1 0 200 6 3 6 13 1 9 192 6 64 T x (0 )= [0 0.04 0 ] , x́ (0 )= 0 Calculate eigenvalues and eigenvectors M 1 / 2 = 0.07071 0 0 0.0316228 0 0 0 0 0.07071 Copyright © 2015 Pearson Education Ltd. ~ K = M 1 / 2 KM 1 / 2 = 2.24789 1.19145 1.08232 107 1.19145 1.06567 1.19145 1.08232 1.19145 2.24789 ~ det K λI 0 −8 λ1= 1.67876× 10 ω 1= 1.29567 ×10 −5 rad/s λ2= 1.16557× 10 ω 2= 3.414 ×10 rad/s −7 −4 λ3 = 4.228 ×10 ω3= 6.570231 ×10 rad/s v1 = 0.332509 v2 = 0.7071 v 3 = 0.62405 0.882539 0 0.470238 0.332509 0.7071 0.62405 Use the mode summation method to find the solution. Transform the initial conditions : −7 −4 1 0 1.265491 0 q (0 )= M 2 x (0 ) 1 q́ (0 )= M 2 x́ (0 )= 0 The solution is given by Eq.(4.103), x t = d i sin ωi t + i ui 4 i=1 Where ωi viT q 0 i = tan T i=1,2,3 (Eq.(4.97)) vi q 0 viT q 0 i=1,2,3, (eq.(4.98)) di = sini 1 ui= M − 1/ 2 vi Substituting known values yields π 1 = 2 = 3 = rad 2 d 1= −1.11633 d 2= 0 d 3= 0.59481 u1 = 0.0235119 u2 = 0.05 u3 = 0.044127 0.0279083 0 0.0148702 0.0235119 0.05 0.044127 The solution is Copyright © 2015 Pearson Education Ltd. xt = 1.116330.0235119 cos 1.29567 10 5 t + 0.594810.044127 cos 6.570231 10 4 t 0.0148702 0.0279083 0.044127 0.0235119 4.55 Repeat Problem 4.49 for the case that the airplane body is 10 m instead of 4 m as indicated in the figure. What effect does this have on the response, and which design (4m or 10 m) do you think is better as to vibration? Solution: Given: é1 0 0 ù é 3 -3 - ù EI ê ê ú ú x + 3 ê -3 6 -3ú x = 0 m ê 0 10 0 ú l ê0 0 1 ú ê 0 -3 3 ú ë û ë û Calculate eigenvalues and eigenvectors: é1 0 0ù ú ê M = m ê0 0.3612 0 ú ê0 0 1 úû ë é 3 -0.9487 0 ù EI ê ú K = M -1/ 2 KM -1/ 2 = 3 ê -0.9487 0.6 -0.9487 ú ml ê 0 -0.9487 3 úû ë Again choose the parameters so that the coefficient is 1 and compute the eigenvalues: det K - l I = l 3 - 6.6l 2 + 10.8l = 0 -1/ 2 -1/ 2 ( ) l1 = 0 l2 = 3 l3 = 3.6 é -0.2887 ù é 0.7071 ù é 0.6455 ù ú ú ú ê ê ê v1 = ê -0.9129 ú v 2 = ê 0 ú v 3 = ê -0.4082 ú ê -0.2887 ú ê -0.7071ú ê 0.6455 ú û û û ë ë ë The natural frequencies are Copyright © 2015 Pearson Education Ltd. w 1 = 0 rad/s w 2 = 1.7321 rad/s w 3 = 1.8974 rad/s The relationship between eigenvectors and mode shapes is u = M -1/ 2 v u1 = m -1/ 2 é -0.2887 ù é 0.7071 ù é 0.6455 ù ú ú ú ê ê -1/ 2 ê ê -0.2887 ú u 2 = m ê 0 ú u3 = ê -0.1291ú ê -0.2887 ú ê -0.7071ú ê 0.6455 ú û û ë ë ë û It appears that the mode shapes contain less "amplitude" for the wing masses. This seems to be a better design from a vibration standpoint. Copyright © 2015 Pearson Education Ltd. 4.56 Often in the design of a car, certain parts cannot be reduced in mass. For example, consider the drive train model illustrated in Figure P4.47. The mass of the torque converter and transmission are relatively the same from car to car. However, the mass of the car could change as much as 1000 kg (e.g., a two-seater sports car versus a family sedan). With this in mind, resolve Problem 4.47 for the case that the vehicle inertia is reduced to 2000 kg. Which case has the smallest amplitude of vibration? Solution: Let k1 = hub stiffness and k2 = axle and suspension stiffness. From Problem 4.44, the equation of motion becomes é 1 -1 0 ù é75 0 0 ù ú ê ú ê x + 10,000 ê -1 3 -2 ú x = 0 0 ú ê 0 100 ê 0 -2 2 ú ê0 0 2000 úû û ë ë () () T x 0 = 0 and x 0 = éë0 0 1ùû m/s. Calculate eigenvalues and eigenvectors. M -1/ 2 é0.1155 0 0 ù ú ê =ê 0 0.1 0 ú ê 0 0 0.0224 úû ë K = M ( -1/ 2 KM -1/ 2 é 133.33 -115.47 0 ù ú ê = ê -115.47 300 -44.721ú ê 0 -44.721 10 úû ë ) det K - l I = l 3 - 443.33l 2 + 29,000l = 0 l1 = 0 w1 = 0 rad/s l2 = 70.765 w 2 = 8.9311 rad/s l3 = 363.57 w 3 = 19.067 rad/s é -0.1857 ù é 0.8758 ù é 0.4455 ù ú ú ê ê ê ú v1 = ê -0.2144 ú v 2 = ê 0.4063 ú v 3 = ê -0.8882 ú ê -0.9589 ú ê -0.2065 ú ê 0.1123 ú û û ë ë ë û Use the mode summation method to find the solution. Transform the initial conditions: Copyright © 2015 Pearson Education Ltd. () () q ( 0 ) = M x ( 0 ) = éë0 q 0 = M 1/ 2 x 0 = 0 1/ 2 0 44.7214 ùû T The solution is given by () ( ) ( ) ( ) q t = c1 + c4t v1 + c2 sin w 2t + f2 v 2 + c3 sin w 3t + f3 v 3 where æ w v T q(0) ö f i = tan -1 ç i T i ÷ , i = 2,3 è v i q(0) ø v Ti q(0) , i = 2,3 ci = w i cos fi Thus 2 = 3 =0, c2 = -1.3042 and c3 = 0.2635. Next apply the initial conditions: 3 3 i= 2 i= 2 = c4 v1 + å ci sin fi v i q(0) = c1v1 + å ci sin fi v i and q(0) Pre multiply each of these by v1T to get: c1 = 0 = v1T q(0) and c4 = -42.8845 = v1T q(0) So q(t) = -42.8845tv1 - 1.3042sin(8.9311t)v 2 + 0.2635sin(19.067t)v 3 Next convert back to the physical coordinates by x(t) = M -1 2 q(t) é1ù é -0.1319 ù é 0.01355 ù ê ú ê ú ê ú = 0.9195t ê1ú + ê -0.05299 ú sin8.9311t + ê -0.02340 ú sin19.067t m ê1ú ê 0.007596 ú ê 0.0006620 ú ë û ë û ë û Comparing this solution to problem 4.44, the car will vibrate at a slightly higher amplitude when the mass is reduced to 2000 kg. 4.57 Use mode summation methodto compute the analytical solution for the response of the 2-degree-of-freedom system: é 540 -300 ù é 1 0 ù x(t) + ê ú x(t) = 0 ê ú ë -300 300 û ë 0 4 û to the initial conditions of é 0 ù é0 ù , x x0 = ê = 0 ú ê0 ú . ë 0.01û ë û Copyright © 2015 Pearson Education Ltd. Solution: Following the development of equations (4.97) through (4.103) for the mode summation for the free response and using the values of computed in problem 1, compute the initial conditions for the “q” coordinate system: é1 0 ù é1 0 ù é 0 ù é 0 ù é1 0 ù é0 ù é0 ù , q(0) = ê =ê Þ q( 0) = ê M 1/ 2 = ê ú ú ê ú ú úê ú = ê ú ë0 2 û ë 0 2 û ë 0.01û ë 0.02 û ë0 2 û ë0 û ë0 û From equation (4.97): æ xö æ xö p f1 = tan -1 ç ÷ = f2 = tan -1 ç ÷ = è 0ø è 0ø 2 From equation (4.98): v1T q ( 0 ) v T2 q ( 0 ) T d1 = = v1 q ( 0 ) ,d2 = = v T2 q ( 0 ) p p sin 2 sin 2 ( ) ( ) Next compute q ( t ) from (4.92) and multiply by M 1/ 2 to get x ( t ) or use (4.103) directly to get q ( t ) = d1 cos (w 1t ) v1 + d2 cos (w 2 t ) v 2 = cos (w 1t ) v1T q ( 0 ) v1 + cos (w 2 t ) v1T q ( 0 ) v1 é 0.0054 ù é -0.0054 ù + cos ( 24.170t ) ê = cos ( 5.551t ) ê ú ú ë 0.0184 û ë 0.0016 û Note that as a check, substitute t = 0 in this last line to recover the correct initial condition q ( 0 ) . Next transform the solution back to the physical coordinates é 0.0054 ù é -0.0054 ù x ( t ) = M -1/ 2 q ( t ) = cos ( 5.551t ) ê + cos ( 24.170t ) ê ú úm ë 0.0092 û ë 0.0008 û 4.58 For a zero value of an eigenvalue and hence frequency, what is the corresponding time response? Or asked another way, the form of the modal solution for a nonzero frequency is Asin(w n t + f ) , what is the form of the modal solution that corresponds to a zero frequency? Evaluate the constants of integration if the modal initial conditions are: r1 (0) = 0.1, and r1 (0) = 0.01. Solution:A zero eigenvalue corresponds to the modal equation: r1 (t ) = 0 Þ r1 (t ) = a + bt Applying the given initial conditions: r1 (0) = a + b(0) = 0.1 Þ a = 0.1 r1 (0) = b = 0.01 Þ r1 (t ) = 0.1 + 0.01t 4.59 Consider the system described by é 400 -400 ù é 1 0 ù x(t) + ê ú x(t) = 0 ê ú ë -400 400 û ë 0 4 û Copyright © 2015 Pearson Education Ltd. T subject to the initial conditions x(0) = éë 1 0 ùû , x (0) = 0. Plot the displacements versus time. Solution: a 400 1 0 0 4 M 1 1 a 1 1 400 200 200 100 Kt 0 1 P augment ( v1 v2) S MR 1 0.1 0 x0 1 s1 S 0 s1 r0 0.02 0 0 1 0 s1 r0 0.02 Kt MR 500 0 0.447 0.894 v1 2 5 5 0.447 0.894 0.447 0.224 P 1 r0 S 0.447 s1 0.447 1 1 1 s2 r0 0.02 1 0 0 1 P P T 0.447 0.894 0.894 0.447 0.447 1.789 0.894 0.894 T 0 1 0.894 0.224 x2( t ) s1 r0 s2 r0 cos 10 5 t 0 0 0 1 1 0 1 1 0.4 0.6 0.8 0.1 0.2 t Copyright © 2015 Pearson Education Ltd. 0.02 0.04 0.02 0.04 s1 r0 x1( t ) s1 r0 s2 r0 cos 10 5 t x1 ( t ) 0.2 0.894 0.447 v2 SI P MR 0.1 0 0.894 0.447 0.447 0.894 P 0.447 1.789 0.894 0.894 0.2 x2 ( t ) 1 0 s2 S s2 r0 0.08 0 1 0.045 0.089 1 2 11.18 S x0 K MR v2 eigenvecs ( Kt) 0.447 0.894 0.894 0.447 T 1 eigenvecs ( Kt) P 0 0 0 500 T MR eigenvals ( Kt) v1 eigenvecs ( Kt) P Kt P 1 0 0 2 K T
