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Solutions Manual c to accompany System Dynamics, First Edition by William J Palm III University of Rhode Island Solutions to Problems in Chapter One PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc (”McGrawHill”) and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill c Solutions Manual Copyright 2005 by The McGraw-Hill Companies, Inc 1-1 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 1.1 W = mg = 3(32.2) = 96.6 lb 1.2 m = W/g = 100/9.81 = 10.19 kg W = 100(0.2248) = 22.48 lb m = 10.19(0.06852) = 0.698 slug 1.3 d = (50 + 5/12)(0.3048) = 15.37 m 1.4 n = 1/[60(1.341 × 10−3 )] = 12.43, or approximately 12 bulbs 1.5 5(70 − 32)/9 = 21.1◦ C 1.6 ω = 3000(2π)/60 = 314.16 rad/sec Period P = 2π/ω = 60/3000 − 1/50 sec 1.7 ω = rad/sec Period P = 2π/ω = 2π/5 = 1.257 sec Frequency f = 1/P = 5/2π = 0.796 Hz 1.8 y − x = 5.66 x + y = 10.30 1-2 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 1.9 Ae−1/τ 4.912 = = e2/tau 3.293 Ae−3/τ 4.912 = ln 3.293 τ =5 τ= ln(4.912/3.293) A = 4.912e1/τ = 1.10 y = [0, 7] and t = [0, 4(3)] = [0, 12] 1.11 y = [0, + 12] = [0, 18] Dominant time constant is Thus t = [0, 4(5)] = [0, 20] 1.12 10 sin 3t cos + 10 cos 3t sin = B sin 3t + C cos 3t B = 10 cos = −4.161 C = 10 sin = 9.093 1-3 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 1.13 A sin 6t cos φ + A cos 6t sin φ = −6.062 sin 6t − 3.5 cos 6t A cos φ = −6.062 < A sin φ = −3.5 < Thus φ is in the third quadrant φ = tan−1 3.5 6.062 A= = 0.524 + π = 3.665 rad (6.062)2 + (3.5)2 = 1-4 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 1.14 y˙ = 0.5(7) cos(7t + 4) = 3.5 cos(7t + 4) yă = 0.5(7)2 sin(7t + 4) = 24.5 sin(7t + 4) Velocity amplitude is 3.5 m/s Acceleration amplitude is 24.5 m/s2 1-5 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 1.15 y(t) = A sin(3t + φ) = A sin 3t cos φ + A cos 3t sin φ A sin φ = 0.05 > A cos φ = 0.08 > Thus φ is in the first quadrant φ = tan−1 0.05 0.08 = 0.559 rad Thus y(t) = 0.094 sin(3t + 0.559) The amplitude of the displacement is A= (0.05)2 + (0.08)2 = 0.094 ft For the velocity, y(t) ˙ = 3(0.094) cos(3t + 0.559) = 0.282 cos(3t + 0.559) The amplitude of the velocity is 0.282 ft/sec For the acceleration, yă(t) = 3(0.282) sin(3t + 0.559) = 0.846 sin(3t + 0.559) The amplitude of the acceleration is 0.846 ft/sec2 1-6 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 1.16 y˙ = 5(0.07) sin(5t + φ) Thus y(0) ˙ = 0.35 sin φ = 0.041, and φ = sin−1 0.041 0.35 = 0.117 or π − 0.117 = 3.02 rad Without more information we cannot determine the quadrant of φ 1-7 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 1.17 y(0) = A sin φ = 0.048 > y(0) ˙ = 5A cos φ = 0.062 > Thus φ is in the first quadrant φ = tan−1 A= 0.048 = 1.318 rad (0.062/5) (0.048)2 + (0.062/5)2 = 0.0496 m 1-8 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 1.18 The radian frequency is 2π(10) = 20π The displacement is y = A sin 20πt The velocity and acceleration are y˙ = 20A cos 20t yă = (20)2 A sin 20t Given the acceleration amplitude (20π)2 A = 0.7g we solve for the displacement amplitude A as A= 0.7g = 5.71 × 10−3 ft (20π)2 Thus the amplitude of the velocity is 20πA = 0.359 ft/sec 1-9 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 12.18 We are given that X2 ≤ 0.08/12 ft, ω = 6000(2π)/60 = 628.3 rad/sec, and that the unbalance force amplitude is mu ω = 60 lb Thus mu = 60 60 = = 0.0955 ω 628.3 The first design equation for the absorber is k2 = F 60 = = 9000 lb/ft X2 0.08/12 The second design equation is k2 = ω = 628.3 m2 Thus m2 = k2 = 0.023 slug (628.3)2 12-21 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 12.19 We are given that ω = 200(2π)/60 = 20.94 rad/sec, the amplitude of the unbalance force is mu ω = lb, and that X2 ≤ 1/12 ft Assume that the table legs are rigid The first design equation for the absorber is k2 = ω = 20.94 m2 The second design equation is k2 = F = = 48 lb/ft X2 1/12 m2 = 48 = 0.109 slug (20.94)2 Thus 12-22 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 12.20 The machine mass is m1 = kg a) We are given that ωn = 2π(6) = 12π rad/sec, mu ω = 50 N, ω = 4(2π) = 8π rad/sec, and X2 ≤ 0.1 m The design equation for the absorber is k2 = ω = 8π m2 Thus k2 = 50/0.1 = 500 N/m, and m2 = k2 500 = = 0.792 kg (8π) (8π)2 b) We have that k1 = m1 ωn2 = (12π)2 m1 = 1421m1 = 11368 N/m From (12.3.7), T1 (jω) = where X1 (jω) = F (jω) 11368 − r24 b2 r24 − [1 + (1 + µ)b2 ] r22 + ωn2 8π = = ωn1 12π 0.792 m2 = µ= = 6.336 m1 13 3.168 + = 1.84 + (1 + µ)b2 = b= The amplitude of F (jω) is mu ω , where r2 = ωωn2 = ω/8π Thus X1 = mu 11368 ω − r22 − 1.84r22 + 4 r2 or X1 = mu 11368 ω2 − ω4 4096π ω2 64π 2 ω − 1.84 64π + The plot is shown in the following figure 12-23 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 0.9 0.8 0.6 0.5 u X /m ε (slug−1) 0.7 0.4 0.3 0.2 0.1 0 10 20 30 60 50 40 Forcing frequency ω (rad/s) 70 80 90 100 Figure : for Problem 12.20 12-24 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 12.21 The first printing of the text had incorrect values for some of the frequencies The correct values are as follows The operating speed range is from 1500 to 3000 rpm Without an absorber, excessive vibration was observed at 2100 rpm After the absorber was attached, resonance was observed at 1545 and 2850 rpm An absorber tuned to 2100 rpm requires that 2100(2π) k1 = 219.9 rad/sec = m1 60 k2 = m2 Because we are given that m2 = 5/32.2 = 0.155, we have that k2 = (219.9)2 m2 = 9747.8 Also, (1) k1 = (219.9)2 m1 The characteristic equation of the combined system is given by the denominator of (12.3.1), and is m1 m2 s4 + (m2 k1 + m2 k2 + m1 k2 )s2 + k1 k2 = (1) One of the observed resonances of the combined system is 2850 rpm, or 2850(2π)/60 = 298.45 rad/sec Thus substituting s = 298.45j and relation (1) into (2), along with the values m2 = 0.155 and k2 = 9747, we can determine the value of m1 , which is m1 = 0.4 slug Thus gives k1 = 19, 344 lb/ft With m1 and k1 determined, we now calculate the required values of m2 and k2 Suppose we choose to put the resonances just outside the operating range, say at 1400 and 3100 rpm (146.6 and 324.6 rad/sec) Let λ = s2 , and let λ1 and λ2 denote the desired values of s2 We can factor the characteristic equation (2) as follows: m1 m2 (λ − λ1 )(λ − λ2 ) = m1 m2 λ2 − m1 m2 (λ1 + λ2 )λ + m1 m2 λ1 λ2 = (3) Comparing the coefficients of (3) with (2), we see that (m2 k1 + m2 k2 + m1 k2 = −m1 m2 (λ1 + λ2 ) (4) and m1 m2 λ1 λ2 = k1 k2 (5) We can solve (5) for k2 as follows k2 = m1 m2 λ1 λ2 = Dm2 k1 where D= (6) m1 λ1 λ2 k1 Substitute (6) into (4) and solve for m2 m2 = − m1 (λ1 + λ2 ) + k1 + m1 D D The desired values are λ1 = −(146.6)2 and λ2 = −(324.6)2 These give the absorber values m2 = 0.2706 slug and k2 = 12, 672 lb/ft 12-25 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 12.22 a) From Newtons law, m1 x ă1 = f kx1 + c(x x ) ă2 = −c(x˙ − x˙ ) m2 x Transform these equations with zero initial conditions to obtain (m1 s2 + cs + k)X1 (s) − csX2 (s) = F (s) −csX1 (s) + (m2 s2 + cs)X2 (s) = The solutions obtained with Cramer’s rule are X1 (s) = m2 s2 + cs F (s) D(s) X2 (s) = cs F (s) D(s) where Cramer’s determinant is D(s) = s(m1 m2 s3 + c(m1 + m2 )s2 + km2 s + ck) Define the following parameters: µ= m2 m1 c ζ= √ m1 k1 ω12 = k1 m1 r= ω ω1 Then D(jω) can be written as D(jω) = ωj[−m1 m2 ω j − c(m1 + m2 )ω + km2 ωj + kc] or D(jω) = m21 ω14 r (2ζr)2 [1 − (1 + µ)r ]2 + µ2 r (1 − r )2 The numerator of kX1 (jω)/F (jω) is | − m2 ω + cωj| = m1 rω12 4ζ + µ2 r The numerator of X2 (jω)/F (jω) is |cωj| Thus, kX(jω) = F (jω) and X2 (jω) = F (jω) ω1 4ζ + µ2 r (2ζ)2 [1 − (1 + µ)r ]2 + µ2 r (1 − r )2 (2ζ)2 [1 2ζ − (1 + µ)r ]2 + µ2 r (1 − r )2 12-26 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 12.23 a) From Newton’s law, m1 x ă1 = f kx1 + k2 (x2 − x1 ) + c(x˙ − x˙ ) m2 x ă2 = k2 (x2 x1 ) − c(x˙ − x˙ ) Transform these equations with zero initial conditions to obtain (m1 s2 + cs + k1 + k2 )X1 (s) − (cs + k2 )X2 (s) = F (s) −(cs + k2 )X1 (s) + (m2 s2 + cs + k2 )X2 (s) = The solutions obtained with Cramer’s rule are X1 (s) = m2 s2 + cs + k2 F (s) D(s) X2 (s) = cs + k2 F (s) D(s) where Cramer’s determinant is D(s) = m1 m2 s4 + c(m1 + m2 )s3 + (m2 k1 + m2 k2 + m1 k2 )s2 + ck1 s + k1 k2 Define the following parameters: µ= m2 m1 ω12 = k1 m1 ω22 = k2 m2 ω1 ω r= ω2 ω1 c k2 ζ= √ λ= k1 m1 k1 α= Then D(jω) can be written as D(jω) = m1 m2 ω − (m2 k1 + m2 k2 + m1 k2 )ω + k1 k2 + [ck1 ω − (m1 + m2 )cω ]j or D(jω) = m21 ω [µr − (1 + λ + µλ)r + µα2 ] + [2ζr − 2(1 + µ)r ]j The numerator of k1 X1 (jω)/F (jω) is |k2 − m2 ω + cωj| = k1 [λ − µr + 2ζri] The numerator of k1 X2 (jω)/F (jω) is |cωj + k2 | = k1 |(λ + 2ζrj)| Thus, k1 X(jω) = F (jω) and k1 X2 (jω) = F (jω) (λ − µr )2 + (2ζr)2 [µr − (1 + λ + µλ)r + µα2 ]2 + (2ζr)2 [1 − (1 + µ)r ]2 λ2 + (2ζr)2 [µr − (1 + λ + µλ)r + µα2 ]2 + (2ζr)2 [1 − (1 + µ)r ]2 12-27 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 12.24 The equations of motion are m1 x ă1 = k1 x1 k2 (x1 x2 ) ă2 = k2 (x1 x2 ) m2 x From these equations we can write the modal amplitude equations by substituting x1 = A1 est and x2 = A2 est , and using the given parameter values The result is (10s2 + 30, 000)A1 − 20, 000A2 = −20, 000A1 + (30, 000s2 + 20, 000)A2 = These give the solution s2 + 3000 A1 (1) 2000 The roots are found from Cramer’s determinant of the modal equations, which is A2 = 3s4 + 11, 000s2 + × 106 = The roots are s2 = −3475 and s2 = −192 Substitute s2 = −3475 into (1) to obtain A2 = −0.2375A1 Substitute s2 = −192 into (1) to obtain A2 = 1.404A1 √In the first mode, the masses oscillate in opposite directions with a radian frequency of 3475 The displacement amplitude of mass is 0.2375 times that of mass In √ the second mode, the masses oscillate in the same direction with a radian frequency of 192 The displacement amplitude of mass is 1.404 times that of mass 12-28 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 12.25 The equations of motion are m1 L2 ă1 = −mg1 Lθ1 − kd(dθ1 − dθ2 ) m2 L2 ă2 = m2 gL2 + kd(d1 d2 ) From these equations we can write the modal amplitude equations by substituting θ1 = A1 est and θ2 = A2 est The result is (m1 L2 s2 + m1 gL + kd2 )A1 − kd2 A2 = −kd2 A1 + (m2 L2 s2 + m2 gL + kd2 )A2 = Using the given parameter values and g = 9.81 m/s2 , these equations give the solution A2 = (3.13s2 + 7.13)A1 (1) The roots are found from Cramer’s determinant of the modal equations, which is 2500s4 + 10810s2 + 11 586 = The roots are s2 = −1.96 and s2 = −2.36 Substitute s2 = −1.96 into (1) to obtain A2 = 0.995A1 Substitute s2 = −2.36 into (1) to obtain A2 = −0.257A1 √ In the first mode, the masses oscillate in the same direction with a radian frequency of 1.96 The displacement amplitude of mass is 0.995 times that of mass In√the second mode, the masses oscillate in the opposite direction with a radian frequency of 2.36 The displacement amplitude of mass is 0.257 times that of mass 12-29 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 12.26 The equations of motion are I1 ă1 = k2 (2 ) k1 I2 ă2 = k2 (2 − θ1 ) From these equations we can write the modal amplitude equations by substituting θ1 = A1 est and θ2 = A2 est The result is (I1 s2 + k1 + k2 )A1 − k2 A2 = −k2 A1 + (I2 s2 + k2 )A2 = Using the given parameter values, these equations give A2 = I1 s2 + k1 + k2 s2 + A1 θ1 = k2 (1) The roots are found from Cramer’s determinant of the modal equations, which is 5s4 + 23s2 + = The roots are s2 = −0.134 and s2 = −4.47 Substitute s2 = −0.134 into (1) to obtain A2 = 1.29A1 Substitute s2 = −4.47 into (1) to obtain A2 = −0.157A1 √ In the first mode, the masses oscillate in the same direction with a radian frequency of 0.134 The displacement amplitude of mass is 1.29 times that of mass In√the second mode, the masses oscillate in the opposite direction with a radian frequency of 4.47 The displacement amplitude of mass is 0.157 times that of mass 12-30 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 12.27 The equations of motion are mă x = 2k sin 45 x = 1.41kx mă y = ky 2k sin 45◦ x = −2.41kx In the first mode, the mass oscillates in the x direction with a radian frequency of 1.41k/m In the second mode, the mass oscillates in the y direction with a radian frequency of 2.41k/m 12-31 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 12.28 From (12.4.8): (750)(1350)s4 + 730[(1.95 × 104 )(1.5)2 + 2.3 × 104 (1.1)2 ] + 1350(4.25 × 104 ) s2 + 1.95(2.3) × 108 (2.6) = or s4 + 111.329s2 + 1.166 × 104 = This gives s2 = −99.43 and s2 = −11.9 s = ±9.971j and s = ±3.45j or These correspond to frequencies of 1.587 Hz and 0.549 Hz x 3.95 × 103 5.411 k1 L1 − k2 L2 A1 = = = = 2 A2 θ ms + k1 + k2 730s + 4.25 × 10 s + 58.2192 For mode (s2 = −99.43), x = −0.131 m ahead of the mass center θ For mode (s2 = −11.9), x = 0.1168 m behind the mass center θ 12-32 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 12.29 Referring to Figure 12.4.4, we are given that m1 g = 1000 lb, c1 = 0, and k2 = 1300 lb/in The ride rate should be ke = m1 g 1000 = = 102 ∆ 9.8 lb/in Thus the suspension stiffness should be k1 = k e k2 102(1300) = = 111 lb/in k2 − ke 1300 − 102 12-33 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission 12.30 Define the following (refer to Figure 12.4.2): k1 = rear quarter-car suspension stiffness k2 = front quarter-car suspension stiffness kr = total rear suspension stiffness = 2k1 kf = total front suspension stiffness = 2k2 ker = total rear ride rate (including suspension and tire stiffness) kef = total front ride rate (including suspension and tire stiffness) ke1 =quarter-car rear ride rate ke2 =quarter-car front ride rate kt = individual tire stiffness Referring to the guidelines on page 851, and using equations (4) and (5) of Example 12.4.3 as approximations for the bounce and pitch dynamics, we have kef = 0.7ker ωbounce = 2π 2π ωpitch = 2π 2π (1) kf + kr ≤ 1.3 Hz m (2) kf L21 + kr L22 ≤ 1.3 Hz IG ke1 = (m/4)g ∆1 (4a) ke2 = (m/4)g ∆2 (4b) (3) From the above definitions, kef = 2ke2 ker = 2ke1 (5) Because the tire stiffness is in parallel with the suspension stiffness, ke1 = k1 kt k1 + kt ke2 = k2 kt k2 + kt (6) These relations must be satisfied by k1 , k2 , and kt Using the given values m = 4800/32.2, IG = 1800, L1 = 3.5, and L2 = 2.5, these equations can be rearranged as follows: 0.02019 k1 + k2 ≤ 1.3 (7) 0.0375 24.5k1 + 12.5k2 ≤ 1.3 k1 = k2 = 1200 ∆1 kt kt − 1200 ∆1 1200 ∆ kt kt − 1200 ∆2 (8) (9) (10) 12-34 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission ∆1 = 0.7∆2 (11) The procedure is to select suitable values for kt and ∆2 , solve (9) and (10) for k1 and k2 , and see if (7) and (8) are satisfied Trying ∆2 = 9.8/12 ft, as suggested on page 851, and kt = 1200(12) = 14400 lb/ft, we obtain from (9) and (10) k1 = 2457 lb/ft and k2 = 1636 lb/ft With these values, the left-hand sides of (7) and (8) are 1.29 and 1.07 Thus the requirements have been met 12-35 PROPRIETARY MATERIAL © 2005 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission ...Solutions Manual c to accompany System Dynamics, First Edition by William J Palm III University of Rhode Island Solutions to Problems in Chapter One PROPRIETARY... ln[C(t)/C(0)]/b using C(t)/C(0) = 0.9 and b = 1.2603 × 10−4 The answer is 836 years Thus the organism died 836 years ago (c) Using b = 1.1(1.2603 × 10−4 ) in t = − ln(0.9)/b gives 760 years Using... 10−4 In MATLAB this calculation is t = -log(0.9)/b The answer is 836.0170 years Thus the organism died 836 years ago (c) Using b = 1.1(1.2603 × 10−4 ) in t = − ln(0.9)/b gives 760 years 0.9(1.2603

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