Solution Manual For Dynamics of Structures, 4th Edition

CHAPTER 2 Problem 2.1 Given: Tn = 2 π m = 0. 5 sec k (a) Tn′ = 2 π m + 50 g = 0. 75 sec k (b) 1. Determine the weight of the table. Taking the ratio of Eq. (b) to Eq. (a) and squaring the result gives 2 ⎛ Tn′ ⎞ m + 50 g ⎜⎜ ⎟⎟ = m ⎝ Tn ⎠ 2 ⇒ 1+ 50 ⎛ 0.75 ⎞ =⎜ ⎟ = 2.25 mg ⎝ 0.5 ⎠ or mg = 50 = 40 lbs 1. 25 2. Determine the lateral stiffness of the table. Substitute for m in Eq. (a) and solve for k: ⎛ 40 ⎞ k =16π 2 m =16π 2 ⎜ ⎟ =16.4lbs in. ⎝ 386 ⎠ © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 1 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.2 1. Determine the natural frequency. m = k = 100 lb in. ωn = k m = 100 400 386 400 386 lb − sec2 in. = 9. 82 rads sec 2. Determine initial deflection. Static deflection due to weight of the iron scrap u( 0 ) = 200 = 2 in. 100 3. Determine free vibration. u ( t ) = u ( 0 ) cos ω nt = 2 cos ( 9. 82 t ) © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 2 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.3 1. Set up equation of motion. ku+mg/2 mü u mg mu&& + ku = mg 2 2. Solve equation of motion. u ( t ) = A cos ω nt + B sin ω nt + mg 2k At t = 0 , u( 0 ) = 0 and u& ( 0 ) = 0 ∴ A = − u(t ) = mg , B = 0 2k mg (1 − cos ω nt ) 2k © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 3 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.4 k u m v0 m0 m = 10 = 0. 0259 lb − sec2 in. 386 m0 = 0. 5 = 1. 3 × 10 −3 lb − sec2 in. 386 k = 100 lb in. Conservation of momentum implies m0 v0 = ( m + m0 ) u& ( 0 ) u& ( 0 ) = m0 v0 = 2. 857 ft sec = 34.29 in. sec m + m0 After the impact the system properties and initial conditions are Mass = m + m0 = 0. 0272 lb − sec2 in. Stiffness = k = 100 lb in. Natural frequency: ωn = k = 60. 63 rads sec m + m0 Initial conditions: u ( 0 ) = 0, u& ( 0 ) = 34. 29 in. sec The resulting motion is u( t ) = u& ( 0 ) ωn sin ω nt = 0. 565 sin ( 60. 63t ) in. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 4 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.5 k m2 f S = ku h m2 m1 m1 u m 2g With u measured from the static equilibrium position of m1 and k, the equation of motion after impact is ( m1 + m2 ) u&& + ku = m2g (a) The general solution is u ( t ) = A cos ω nt + B sin ω nt + ωn = m2 g k k m1 + m2 (b) (c) The initial conditions are u( 0 ) = 0 u& (0) = m2 m1 + m 2 2gh (d) The initial velocity in Eq. (d) was determined by conservation of momentum during impact: m2u&2 = ( m1 + m2 ) u& ( 0 ) where u&2 = 2 gh Impose initial conditions to determine A and B: u( 0 ) = 0 ⇒ A = − m2 g k u& ( 0 ) = ω n B ⇒ B = m2 m1 + m2 (e) 2 gh ωn (f) Substituting Eqs. (e) and (f) in Eq. (b) gives u(t ) = m2 g (1 − cos ω nt ) + k 2 gh ωn m2 sin ω nt m1 + m2 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 5 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.6 1. Determine deformation and velocity at impact. u( 0) = mg 10 = = 0.2 in. k 50 u& ( 0 ) = − 2 gh = − 2( 386 )( 36 ) = − 166.7 in./sec 2. Determine the natural frequency. ωn = kg (50)(386) = = 4393 . rad/sec w 10 3. Compute the maximum deformation. u(t ) = u(0) cos ω n t + u& (0) ωn sin ω n t ⎛ 166.7 ⎞ = ( 0.2) cos 316.8t − ⎜ ⎟ sin 316.8t ⎝ 4393 . ⎠ ⎡ u&( 0) ⎤ uo = [u( 0)]2 + ⎢ ⎥ ⎣ ωn ⎦ 2 = 0.2 2 + ( −3.795) 2 = 38 . in. 4. Compute the maximum acceleration. u&&o = ω n 2 uo = ( 4393 . )2 (38 . ) = 7334 in./sec2 = 18.98g © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 6 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.7 Given: m = 200 = 6. 211 lb − sec2 ft 32. 2 fn = 2 Hz Determine EI: k = fn = 3 EI 3 EI EI = lb ft 3 = 3 3 9 L 1 k 2π m ⇒ 2 = 1 EI 2π 55. 90 ⇒ EI = ( 4 π )2 55. 90 = 8827 lb − ft 2 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 7 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.8 Equation of motion: mu&& + cu& + ku = 0 (a) Dividing Eq. (a) through by m gives u&& + 2 ζω n u& + ω 2n u = 0 (b) where ζ = 1. Equation (b) thus reads u&& + 2 ω n u& + ω 2n u = 0 (c) Assume a solution of the form u ( t ) = e st . Substituting this solution into Eq. (c) yields ( s 2 + 2 ω n s + ω 2n ) e st = 0 Because e st is never zero, the quantity within parentheses must be zero: s 2 + 2 ω n s + ω 2n = 0 or s = − 2ω n ± ( 2 ω n ) 2 − 4 ω 2n 2 = − ωn (double root) The general solution has the following form: u ( t ) = A1 e − ω n t + A2 t e − ω n t (d) where the constants A1 and A2 are to be determined from the initial conditions: u( 0 ) and u& ( 0 ) . Evaluate Eq. (d) at t = 0 : u (0) = A1 ⇒ A1 = u (0) (e) Differentiating Eq. (d) with respect to t gives u& ( t ) = − ω n A1 e − ω n t + A2 (1 − ω n t ) e − ω n t (f) Evaluate Eq. (f) at t = 0 : u& ( 0 ) = − ω n A1 + A2 (1 − 0 ) ∴ A2 = u& ( 0 ) + ω n A1 = u& ( 0 ) + ω n u ( 0 ) (g) Substituting Eqs. (e) and (g) for A1 and A2 in Eq. (d) gives u (t ) = { u (0) + [u& (0) + ω n u (0) ] t} e −ω nt (h) © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 8 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.9 A2 ω n ⎡−ζ + ζ 2 −1 + ζ + ζ 2 −1 ⎤ = ⎢⎣ ⎥⎦ Equation of motion: mu&& + cu& + ku = 0 Dividing Eq. (a) through by m gives u&& + 2 ζω n u& + ω n2 u = 0 or (b) where ζ > 1. A2 = Assume a solution of the form u ( t ) = e st . Substituting this solution into Eq. (b) yields ( s 2 + 2ζω n s + ω n2 ) e st = 0 Because e st is never zero, the quantity within parentheses must be zero: s 2 + 2 ζω n s + ω 2n −2ζω n ± (2ζω n ) 2 − 4ω n2 = 0 u& (0) + ⎛⎜ ζ + ⎝ ⎠ 2 2 ζ −1 ω n (f) Substituting Eq. (f) in Eq. (d) gives u& (0) + ⎛⎜ ζ + ζ 2 − 1 ⎞⎟ω nu (0) ⎝ ⎠ A1 = u (0) − 2 2 ζ −1ω n 2 ζ 2 − 1 ω nu (0) − u& (0) − ⎛⎜ ζ + ζ 2 − 1 ⎞⎟ ω nu (0) ⎝ ⎠ = −u& (0) + ⎛⎜ −ζ + ζ 2 − 1 ⎞⎟ ω nu (0) ⎝ ⎠ = 2 ζ 2 − 1ω n 2 = ⎛⎜ − ζ ± ⎝ ζ 2 −1 ⎞⎟ ω n u (0) 2 ζ 2 −1ω n or s = u& (0) + ⎛⎜ ζ + ζ 2 −1 ⎞⎟ ω n u (0) ⎝ ⎠ (a) (g) ζ 2 −1 ⎞⎟ ω n ⎠ The solution, Eq. (c), now reads: The general solution has the following form: ⎡ ⎤ u (t ) = A1 exp ⎢⎛⎜ −ζ − ζ 2 −1 ⎞⎟ω n t ⎥ ⎠ ⎣⎝ ⎦ ⎡ ⎤ + A2 exp ⎢⎛⎜ −ζ + ζ 2 −1 ⎞⎟ω n t ⎥ ⎠ ⎣⎝ ⎦ u (t ) = e −ζω nt ω ′D = (d) Differentiating Eq. (c) with respect to t gives ⎡ ⎤ + A2 ⎛⎜ −ζ + ζ 2 −1 ⎞⎟ ω n exp⎢⎛⎜ −ζ + ζ 2 −1 ⎞⎟ ω nt ⎥ ⎝ ⎠ ⎠ ⎣⎝ ⎦ −ω ′D t ′ + A2 e ω D t ) ζ2 − 1 ω n −u& (0) + ⎛⎜ −ζ + ζ 2 − 1 ⎞⎟ω n u (0) ⎝ ⎠ A1 = 2ω ′D Evaluate Eq. (c) at t = 0 : ⎡ ⎤ u& (t ) = A1 ⎛⎜ −ζ − ζ 2 −1 ⎞⎟ ω n exp⎢⎛⎜ −ζ − ζ 2 −1 ⎞⎟ ω nt ⎥ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ 1 where (c) where the constants A1 and A2 are to be determined from the initial conditions: u( 0 ) and u& ( 0 ) . u (0) = A1 + A2 ⇒ A1 + A2 =u (0) (A e A2 = u& (0) + ⎛⎜ ζ + ⎝ ζ 2 −1 ⎞⎟ ω n u (0) ⎠ 2ω ′D (e) Evaluate Eq. (e) at t = 0 : u& (0) = A 1 ⎛⎜ −ζ − ζ 2 −1 ⎞⎟ ω n + A 2 ⎛⎜ −ζ + ζ 2 −1 ⎞⎟ ω n ⎠ ⎠ ⎝ ⎝ = [u (0) − A2 ] ⎛⎜ −ζ − ζ 2 −1 ⎞⎟ ω n + A2 ⎛⎜ −ζ + ζ 2 −1 ⎞⎟ ω n ⎝ ⎠ ⎝ ⎠ or © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 9 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.10 The general solution is u(t) = A1 e − ωn t + A2 t e −ω nt Equation of motion: u&& + 2ζω n u& + ω n2 u = 0 (a) u(t) = e A2 = u& (0) (j) st Substituting in Eq. (i) gives Substituting this solution into Eq. (a) yields: u(t ) = u& (0) t e −ω n t FH s2 + 2ζω n s + ω 2n IK e st = 0 The roots of the characteristic equation [Eq. (b)] are: st s 2 + 2ζω n s + ω 2n = 0 (b) The roots of this characteristic equation depend on ζ. FH s1,2 = ω n −ζ ± ζ 2 − 1 IK (l) The general solution is: (a) Underdamped Systems, ζ1 Because e is never zero 1,2 Determined from the initial conditions u(0) = 0 and u& (0) : A1 = 0 Assume a solution of the form (i) (c) F −ζ + ζ −1IK ω t u(t ) = A1e H 2 n F −ζ − ζ −1IK ω t + A 2 eH 2 n (n) Hence the general solution is st Determined from the initial conditions u(0) = 0 and u& (0) : s t u(t) = A1e 1 + A2e 2 which after substituting in Eq. (c) becomes e u(t ) = e −ζω n t A1e iω D t + A2 e −iω D t − A1 = A2 = j (d) (o) ζ 2 −1 2ω n Substituting in Eq. (n) gives where 1− ζ 2 (e) u(t) = e −ζω nt (A cos ω D t + Bsin ω D t) B= ω n 1−ζ 2 0.8 FH IK (g) (b) Critically Damped Systems, ζ = 1 The roots of the characteristic equation [Eq. (b)] are: s2 = −ω n − ω n t ζ 2 −1 IJ K (p) ζ = 0.1 ωD e −ζω n t sin ω n 1 − ζ 2 t −e Plot Eq. (g) with ζ = 0.1; Eq. (k), which is for ζ = 1; and Eq. (p) with ζ = 2. u& (0) Substituting A and B into Eq. (f) gives u& (0) ω n t ζ 2 −1 2 (d) Response Plots (f) Determine A and B from initial conditions u(0) = 0 and u& (0) : s1 = −ω n FG e 2ω ζ − 1 H u& (0) e −ζω n t n Rewrite Eq. (d) in terms of trigonometric functions: A=0 u( t ) = . u(t) ÷ (u(0)) / ωn ) ω D = ωn u( t ) = u& (0) ζ = 1.0 ζ = 2.0 0.4 0 0.25 0.5 0.75 1 1.25 1.5 t/Tn -0.4 -0.8 (h) © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 10 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.11 F GH u1 1 ln j uj +1 I ≈ 2πζ ⇒ 1 ln F 1 I ≈ 2πζ GH 01. JK JK j 10% ∴ j10% ≈ ln (10 ) 2 πζ ≈ 0. 366 ζ © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 11 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.12 ⎛ ⎞ ui ⎜ 2πζ ⎟ = exp ⎜ u i +1 ⎜ 1−ζ 2 ⎟⎟ ⎝ ⎠ (a) ζ = 0. 01: (b) ζ = 0. 05 : (c) ζ = 0. 25 : ui ui + 1 ui ui + 1 ui ui + 1 = 1. 065 = 1. 37 = 5. 06 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 12 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.13 Given: w = 20.03 kips (empty); m = 0.0519 kip-sec2/in. k = 2 (8.2) = 16.4 kips/in. c = 0.0359 kip-sec/in. (a) Tn = 2 π (b) ζ = m 0. 0519 = 2π = 0. 353 sec 16. 4 k c 2 km = 0. 0359 2 (16. 4 ) ( 0. 0519 ) = 0. 0194 = 1. 94% © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 13 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.14 (a) The stiffness coefficient is k= 3000 = 1500 lb/in. 2 The damping coefficient is c = ccr = 2 km c = 2 1500 3000 = 215.9 lb – sec / in. 386 (b) With passengers the weight is w = 3640 lb. The damping ratio is ζ= c 2 km = 215.9 3640 2 1500 386 = 0.908 (c) The natural vibration frequency for case (b) is ω D = ω n 1− ζ 2 1500 1 − (0.908) 2 3640 / 386 = 12.61 × 0.419 = 5.28 rads / sec = © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 14 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.15 1. Determine ζ and ω n . ζ ≈ ⎛ u ⎞ 1 ⎛ 1 ⎞ ln⎜ 1 ⎟ = ln⎜ ⎟ = 0.0128 = 1.28% ⎜ ⎟ 2π j ⎝ u j +1 ⎠ 2π (20) ⎝ 0.2 ⎠ 1 Therefore the assumption of small damping implicit in the above equation is valid. TD = 3 = 0.15 sec ; Tn ≈ TD = 0.15 sec ; 20 ωn = 2π = 41. 89 rads sec 0.15 2. Determine stiffness coefficient. k = ω 2n m = ( 41. 89 )2 0.1 = 175. 5 lbs in. 3. Determine damping coefficient. ccr = 2 mω n = 2 ( 0.1) ( 41. 89 ) = 8. 377 lb − sec in. c = ζ ccr = 0. 0128 ( 8. 377 ) = 0.107 lb − sec in. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 15 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.16 250 = 312. 5 lbs in. 0. 8 (a) k = w m = g = 250 386 = 0. 647 lb − sec2 in. k = 21. 98 rads sec m ωn = (b) Assuming small damping, F u I ≈ 2 jπζ ⇒ GH u JK F u IJ = ln (8) ≈ 2 (2) π ζ ⇒ ζ = 0.165 ln G H u 8K 1 ln j +1 0 0 This value of ζ may be too large for small damping assumption; therefore we use the exact equation: ln F u I = 2 jπ ζ GH u JK 1 − ζ 1 j +1 2 or, ln ( 8) = 2 (2) π ζ 1 − ζ 2 ⇒ ζ 1 − ζ2 = 0.165 ⇒ ζ2 = 0. 027 (1 − ζ2 ) ⇒ ζ = 0. 0267 = 0.163 (c) ω D = ω n 1 − ζ2 = 21. 69 rads sec Damping decreases the natural frequency. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 16 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.17 Reading values directly from Fig. 1.1.4b: Peak 1 31 Time, t i (sec) 0.80 7.84 Peak, u&&i (g) 0.78 0.50 7.84 − 0.80 = 0.235 sec 30 ⎛ 0.78g ⎞ 1 ζ= ln⎜ ⎟ = 0.00236 = 0.236% 2π (30) ⎝ 0.50g ⎠ TD = © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 17 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.18 1. Determine buckling load. wcr θ L k wcr ( L θ ) = k θ wcr = k L 2. Draw free-body diagram and set up equilibrium equation. fI w θ L fS O where ∑ MO = 0 ⇒ fI L + fS = w Lθ fI = w 2 && L θ g fS = k θ Substituting Eq. (b) in Eq. (a) gives w 2 && L θ + (k − w L) θ = 0 g (a) (b) (c) 3. Compute natural frequency. ω n′ = k − wL = ( w g) L2 k wL ⎞ ⎛ ⎜1 − ⎟ 2 k ⎠ ( w g) L ⎝ or ω ′n = ω n 1 − w wcr (d) © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 18 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.19 For motion of the building from left to right, the governing equation is mu&& + ku = − F (a) for which the solution is u ( t ) = A2 cos ω nt + B2 sin ω nt − uF (b) With initial velocity of u& ( 0 ) and initial displacement u( 0 ) = 0 , the solution of Eq. (b) is u(t ) = u& ( 0 ) ωn sin ω nt + uF (cos ω nt − 1) u& ( t ) = u& ( 0 ) cos ω nt − uFω n sin ω nt (c) (d) At the extreme right, u& ( t ) = 0 ; hence from Eq. (d) tan ω nt = u& ( 0 ) 1 ω n uF Substituting ω n = 4 π , uF = 0.15 in. 20 in. sec in Eq. (e) gives tan ω nt = (e) and u& ( 0 ) = 20 1 = 10. 61 4 π 0.15 or sin ω nt = 0. 9956; cos ω nt = 0. 0938 Substituting in Eq. (c) gives the displacement to the right: u = 20 ( 0. 9956 ) + 0.15 ( 0. 0938 − 1) = 1. 449 in. 4π After half a cycle of motion the amplitude decreases by 2 uF = 2 × 0.15 = 0. 3 in. Maximum displacement on the return swing is u = 1. 449 − 0. 3 = 1.149 in. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 19 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problem 2.20 Given: F = 0.1w , Tn = 0. 25 sec 0.1w 0.1mg 0.1g 0.1g F = = = = ( 2 π Tn )2 k k k ω 2n 0.1g = = 0. 061 in . ( 8 π )2 uF = The reduction in displacement amplitude per cycle is 4uF = 0. 244 in. The displacement amplitude after 6 cycles is 2.0 − 6 (0.244) = 2.0 − 1.464 = 0.536 in. Motion stops at the end of the half cycle for which the displacement amplitude is less than uF . Displacement amplitude at the end of the 7th cycle is 0.536 – 0.244 = 0.292 in.; at the end of the 8th cycle it is 0.292 – 0.244 = 0.048 in.; which is less than uF . Therefore, the motion stops after 8 cycles. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: 20 Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.