Dynamics of Structures 5th Edition Chopra SOLUTIONS MANUAL

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1 Dyamics of Structures 5th Editio Chopra SOLUTIONS MANUAL Full dowload at : Problem.1 CHAPTER A heavy table is supported by flat steel legs (Fig. P.1). Its atural period i lateral vibratio is 0.5 sec. Whe a 50-lb plate is clamped to its surface, the atural period i lateral vibratio is legtheed to 0.75 sec. What are the weight ad the lateral stiffess of the table? T = 0.5 sec T = 0.75 sec Figure P.1 Solutio: Give: m T = π = 0. 5 sec (a) T = π m + 50 g 1. Determie the weight of the table. = sec (b) Taig the ratio of Eq. (b) to Eq. (a) ad squarig the result gives or T m +50 g 1+ = =.5 = T mg = m = 40 lbs mg 0.5. Determie the lateral stiffess of the table. Substitute for m i Eq. (a) ad solve for : 017 Pearso Educatio, Ic., Hoboe, NJ. All rights reserved. This material is protected uder all copyright laws as thebe reproduced, i ay form or by ay meas, without permissio i writig from the publisher.

2 =16π m =16π 40 =16.4lbs i Pearso Educatio, Ic., Hoboe, NJ. All rights reserved. This material is protected uder all copyright laws as thebe reproduced, i ay form or by ay meas, without permissio i writig from the publisher.

3 Problem. A electromaget weighig 400 lb ad suspeded by a sprig havig a stiffess of 100 lb/i. (Fig. P.a) lifts 00 lb of iro scrap (Fig. P.b). Determie the equatio describig the motio whe the electric curret is tured off ad the scrap is dropped (Fig. P.c). Solutio: 1. Determie the atural frequecy. Figure P. = 100 lb i. m = 400 lb sec i ω = = = 9. 8 rads sec m Determie iitial deflectio. Static deflectio due to weight of the iro scrap u(0) = = i. 3. Determie free vibratio. u(t ) = u(0 ) cos ω t = cos (9. 8t ) 017 Pearso Educatio, Ic., Hoboe, NJ. All rights reserved. This material is protected uder all copyright laws as thebe reproduced, i ay form or by ay meas, without permissio i writig from the publisher.

4 Problem.3 A mass m is at rest, partially supported by a sprig ad partially by stops (Fig. P.3). I the positio show, the sprig force is mg/. At time t = 0 the stops are rotated, suddely releasig the mass. Determie the motio of the mass. m u Figure P.3 Solutio: 1. Set up equatio of motio. u+m g/ m g mü u mu&& + u = mg. Solve equatio of motio. u(t ) = A cos ω t + B si ω t + mg At t = 0, u(0) = 0 ad u& (0) = 0 A = mg, B = 0 u(t ) = mg (1 cos ω t ) 017 Pearso Educatio, Ic., Hoboe, NJ. All rights reserved. This material is protected uder all copyright laws as thebe reproduced, i ay form or by ay meas, without permissio i writig from the publisher.

5 Problem.4 The weight of the woode bloc show i Fig. P.4 is 10 lb ad the sprig stiffess is 100 lb/i. A bullet weighig 0.5 lb is fired at a speed of 60 ft/sec ito the bloc ad becomes embedded i the bloc. Determie the resultig motio u(t) of the bloc. m v o Figure P.4 Solutio: u v 0 m m 0 m = = lb sec i m 0 = = lb sec i. 386 = 100 lb i. Coservatio of mometum implies m 0 v 0 = ( m + m 0 ) u& (0) u& (0) = m 0 v 0 =. 857 ft sec = i. sec m + m 0 After the impact the system properties ad iitial coditios are Mass = m + m 0 = lb sec i. Stiffess = = 100 lb i. Natural frequecy: ω = m + m 0 = rads sec Iitial coditios: u(0) = 0, The resultig motio is u&( 0) = i. sec u(t ) = u&( 0 ) si ω t = si (60. 63t ) i. ω 017 Pearso Educatio, Ic., Hoboe, NJ. All rights reserved. This material is protected uder all copyright laws as thebe reproduced, i ay form or by ay meas, without permissio i writig from the publisher.

6 Problem.5 A mass m 1 hags from a sprig ad is i static equilibrium. A secod mass m drops through a height h ad stics to m 1 without reboud (Fig. P.5). Determie the subsequet motio u( t) measured from the static equilibrium positio of m 1 ad. h m m 1 Figure P.5 Solutio: h m f S = u m m 1 m 1 u m g With u measured from the static equilibrium positio of m 1 ad, the equatio of motio after impact is ( m 1 + m ) u&& + u = m g (a) where The geeral solutio is Impose iitial coditios to determie A ad B: u(t ) = A cos ω t + B si ω t + m g (b) u& = gh m g u(0) = 0 A = ω = (c) u&(0) = ω B B = The iitial coditios are m m + m m 1 + m 1 gh Substitutig Eqs. (e) ad (f) i Eq. (b) gives ω (e) (f) m u(0) = 0 u& (0) = gh (d) m1 + m The iitial velocity i Eq. (d) was determied by coservatio of mometum durig impact: m u& = ( m 1 + m ) u&( 0) u(t ) = m g (1 cos ω t ) + gh m si ω t ω m + m Pearso Educatio, Ic., Hoboe, NJ. All rights reserved. This material is protected uder all copyright laws as thebe reproduced, i ay form or by ay meas, without permissio i writig from the publisher.

7 Problem.6 The pacagig for a istrumet ca be modeled as show i Fig. P.6, i which the istrumet of mass m is restraied by sprigs of total stiffess iside a cotaier; m = 10 lb/g ad = 50 lb/i. The cotaier is accidetally dropped from a height of 3 ft above the groud. Assumig that it does ot bouce o cotact, determie the maximum deformatio of the pacagig withi the box ad the maximum acceleratio of the istrumet. u m / / 3 Figure P.6 Solutio: 1. Determie deformatio ad velocity at impact. u(0) = mg = 10 = 0. i. 50 u&( 0 ) = gh = (386 )(36) = i./sec 4. Compute the maximum acceleratio. u&& = ω u = (43.93) (3.8) o o = 7334 i./sec = 18.98g. Determie the atural frequecy. ω = g = w (50)(386) 10 = rad/sec 3. Compute the maximum deformatio. u(t ) = u(0) cos ω t + u&(0) si ω t ω = (0.) cos 316.8t si 316.8t u = [u(0)] + u&(0) o ω = 0. + ( 3.795) = 3.8 i. 017 Pearso Educatio, Ic., Hoboe, NJ. All rights reserved. This material is protected uder all copyright laws as thebe reproduced, i ay form or by ay meas, without permissio i writig from the publisher.

8 Problem.7 Cosider a diver weighig 00 lbs at the ed of a divig board that catilevers out 3 ft. The diver oscillates at a frequecy of Hz. What is the flexural rigidity EI of the divig board? Solutio: Give: m = = lb sec ft f = Hz Determie EI: = 3 EI L 3 = 3 EI 3 3 = EI 9 lb ft 1 1 EI f = = π m π EI = ( 4 π) = 887 lb ft 017 Pearso Educatio, Ic., Hoboe, NJ. All rights reserved. This material is protected uder all copyright laws as thebe reproduced, i ay form or by ay meas, without permissio i writig from the publisher.

9 Problem.8 Show that the motio of a critically damped system due to iitial displacemet u(0) ad iitial velocity u (0) is u(t) = {u(0) + [u (0) + ω u(0)] t} e ω t Solutio: Equatio of motio: Evaluate Eq. (f) at t = 0 : mu&& + cu& + u = 0 (a) u&(0) = ω A 1 + A (1 0 ) Dividig Eq. (a) through by m gives A = u&(0 ) + ω A 1 = u&(0) + ω u(0) (g) u&& + ζω u& + ω u = 0 (b) Substitutig Eqs. (e) ad (g) for A 1 ad A i Eq. (d) gives where ζ = 1. Equatio (b) thus reads u&& + ω u& + ω u = 0 (c) Assume a solutio of the form u(t ) = this solutio ito Eq. (c) yields ( s + ω s + ω ) e st = 0 + ω e st. Substitutig Because e st is ever zero, the quatity withi paretheses must be zero: or s + s = ω s = 0 ω ± ( ω ) 4ω = ω The geeral solutio has the followig form: u(t ) = A e ω t 1 + A t e ω t (double root) where the costats A 1 ad A are to be determied from the iitial coditios: u( 0) ad u&( 0). Evaluate Eq. (d) at t = 0 : u(0) = A 1 A 1 = u(0) Differetiatig Eq. (d) with respect to t gives ω t ω t (d) (e) u&(t ) = ω A 1 e + A (1 ω t ) e (f) u(t ) = {u (0) + [u& (0) + ω u(0) ]t}e ω t (h) 017 Pearso Educatio, Ic., Hoboe, NJ. All rights reserved. This material is protected uder all copyright laws as thebe reproduced, i ay form or by ay meas, without permissio i writig from the publisher.

10 Problem.9 Show that the motio of a overcritically damped system due to iitial displacemet u(0) ad iitial velocity u (0) is u(t) = e ζ ω t A 1 e ω D t + A e ω D t where ω = ω ζ 1 ad D u (0) + ζ + ζ 1 ω u(0) A 1 = ω u (0) + ζ + ζ 1 ω u(0) A = ω D D Solutio: u&(t ) = A ζ ζ 1 ω exp ζ ζ 1 Equatio of motio: ω t 1 mu&& + cu& + u = 0 (a) Dividig Eq. (a) through by m gives u&& + ζω u& + ω u = 0 (b) where ζ > 1. Assume a solutio of the form u(t ) = e st. Substitutig this solutio ito Eq. (b) yields (s + ζω s + ω ) e st = 0 Differetiatig Eq. (c) with respect to t gives + A ζ + ζ 1 ω exp ζ + ζ 1 ω t Evaluate Eq. (e) at t = 0 : u &(0) = or A 1 1 ζ ζ ω A ζ ζ ω = [ u (0) A ζ ζ ] ω A ζ ζ ω 1 (e) Because e st is ever zero, the quatity withi paretheses must be zero: or s + ζω s ω = 0 u&(0) + ζ + ζ 1 ω u(0) + A ω ζ + ζ 1 +ζ + ζ 1 = ζω ± (ζω + s = + ) 4ω u&(0) ζ ζ 1 ω u(0) A = = ζ ± ζ 1 ζ ω 1 ω Substitutig Eq. (f) i Eq. (d) gives The geeral solutio has the followig form: u(t ) = A exp ζ ζ 1 ω t u&(0) + ζ + ζ 1 ω u (0) A 1 = u(0) 1 (c) ζ 1ω + A exp ζ + ζ 1 ω t ζ 1 ω u (0) u& (0) ζ + ζ 1 ω u (0) = ζ where the costats A 1 ad A are to be determied from 1ω the iitial coditios: u( 0) ad u&( 0). u& (0) + ζ + ζ 1 ω u (0) = Evaluate Eq. (c) at t = 0 : ζ 1ω u(0)= A 1 + A A 1 + A =u(0) (d) or (f) (g) 017 Pearso Educatio, Ic., Hoboe, NJ. All rights reserved. This material is protected uder all copyright laws as thebe reproduced, i ay form or by ay meas, without permissio i writig from the publisher.

11 The solutio, Eq. (c), ow reads: u(t ) = e ζω t (A e ω Dt + A e ω D t ) where ω D = ζ 1 ω u& (0) + ζ + ζ 1 ω u (0) A 1 = ω u& (0) + ζ + ζ 1 ω u (0) A = ω D Pearso Educatio, Ic., Hoboe, NJ. All rights reserved. This material is protected uder all copyright laws as they curretly exist. No portio of this material may be reproduced, i ay form or by ay meas, without permissio i writig from the publisher. 10

12 Problem.10 Derive the equatio for the displacemet respose of a viscously damped SDF system due to iitial velocity u (0) for three cases: (a) uderdamped systems; (b) critically damped systems; ad (c) overdamped systems. Plot u(t) u (0)/ω agaist t/t for ζ = 0.1, 1, ad. Solutio: Substitutig A ad B ito Eq. (f) gives u&(0) u(t ) = ω Equatio of motio: 1 ζ u& + ζω u& + ω u = 0 Assume a solutio of the form u( t) = e st Substitutig this solutio ito Eq. (a) yields: Because e st is ever zero s + ζω s + ω = 0 (b) The roots of this characteristic equatio deped o ζ. (a) Uderdamped Systems, ζ<1 The two roots of Eq. (b) are 1 e ζω t sif H ω 1 ζ I K t (g) (a) (b) Critically Damped Systems, ζ = 1 The roots of the characteristic equatio [Eq. (b)] are: s 1 = ω s = ω (h) The geeral solutio is u( t) = A 1 e ω t + A t e ω t (i) s + ζω s + ω I K e st = 0 Determied from the iitial coditios u(0) = 0 ad u&(0) : s 1, = ω F H ζ ± i 1 ζ Hece the geeral solutio is u( t) = A 1 e s1 t + A e s t The geeral solutio is: which after substitutig i Eq. (c) becomes where (c) A 1 = 0 A = u&(0) (j) Substitutig i Eq. (i) gives u(t ) = u&(0) t e ω t () (c) Overdamped Systems, ζ>1 The roots of the characteristic equatio [Eq. (b)] are: s 1, = ω F H ζ ± ζ 1 u(t ) = A s t s t 1 e 1 + A e (m) u(t ) = e ζω t e A e iω D t + A e iω D t j (d) which after substitutig Eq. (l) becomes F u(t ) = A 1 e H 1I ζ + ζ K ω t F + A H 1I e ζ ζ K ω t ω D = ω 1 ζ (e) Determied from the iitial coditios u(0) = 0 ad u&(0) : (l) () Rewrite Eq. (d) i terms of trigoometric fuctios: u( t) = e ζω t (A cosω D t + B si ω D t) (f) Determie A ad B from iitial coditios u(0) = 0 ad u&(0) : u&(0) A = A 1 = ω ζ 1 Substitutig i Eq. () gives A = 0 B = u&(0) u(t ) = u&(0) e ζω t F ω t Ge H ζ 1 e ω t ζ 1 K IJ (o) (p) ω D ω ζ Pearso Educatio, Ic., Hoboe, NJ. All rights reserved. This material is protected uder all copyright laws as they curretly exist. No portio of this material may be reproduced, i ay form or by ay meas, without permissio i writig from the publisher.

13 . u(t) (u(0)) / ω ) (d) Respose Plots Plot Eq. (g) with ζ = 0.1; Eq. (), which is for ζ = 1; ad Eq. (p) with ζ =. ζ = ζ = 1.0 ζ = t/t Pearso Educatio, Ic., Hoboe, NJ. All rights reserved. This material is protected uder all copyright laws as they curretly exist. No portio of this material may be reproduced, i ay form or by ay meas, without permissio i writig from the publisher. 1

14 Problem.11 For a system with dampig ratio ζ, determie the umber of free vibratio cycles required to reduce the displacemet amplitude to 10% of the iitial amplitude; the iitial velocity is zero. Solutio: 1 l F u 1 I G H πζ u j + 1 K J j j 10% l (10) πζ ζ 1 j 10% H 0.1K l 1 I J πζ

15 Problem.1 What is the ratio of successive amplitudes of vibratio if the viscous dampig ratio is ow to be (a) ζ = 0.01, (b) ζ = 0.05, or (c) ζ = 0.5? Solutio: u i = exp πζ 1 ζ u i +1 (a) ζ = 0. 01: (b) ζ = : u i u i + 1 u i u i + 1 = = (c) ζ = 0. 5 : u i u i + 1 = 5. 06

16 Problem.13 The supportig system of the ta of Example.6 is elarged with the objective of icreasig its seismic resistace. The lateral stiffess of the modified system is double that of the origial system. If the dampig coefficiet is uaffected (this may ot be a realistic assumptio), for the modified ta determie (a) the atural period of vibratio T, ad (b) the dampig ratio ζ. Solutio: Give: w = 0.03 ips (empty); m = ip-sec /i. = (8.) = 16.4 ips/i. c = ip-sec/i. (a) T = π m = π (b) ζ = c = m (16. 4) ( ) = 1. 94% = sec =

17 Problem.14 The vertical suspesio system of a automobile is idealized as a viscously damped SDF system. Uder the 3000-lb weight of the car, the suspesio system deflects i. The suspesio is desiged to be critically damped. (a) Calculate the dampig ad stiffess coefficiets of the suspesio. (b) With four 160-lb passegers i the car, what is the effective dampig ratio? (c) Calculate the atural frequecy of damped vibratio for case (b). Solutio: (a) The stiffess coefficiet is 3000 = = 1500 lb/i. The dampig coefficiet is c = c cr = m c = = lb - sec / i. (b) With passegers the weight is w = 3640 lb. The dampig ratio is c ζ = = m = (c) The atural vibratio frequecy for case (b) is ω D = ω 1 ζ 1500 = 1 (0.908) 3640 / 386 = = 5.8 rads / sec

18 Problem.15 The stiffess ad dampig properties of a mass sprig damper system are to be determied by a free vibratio test; the mass is give as m = 0.1 lb-sec /i. I this test, the mass is displaced 1 i. by a hydraulic jac ad the suddely released. At the ed of 0 complete cycles, the time is 3 sec ad the amplitude is 0. i. Determie the stiffess ad dampig coefficiets. Solutio: 1. Determie ζ ad ω. 1 u ζ l 1 1 = l 1 = = 1.8% π j u j +1 π (0) 0. Therefore the assumptio of small dampig implicit i the above equatio is valid. 3 T D = = 0.15 sec ; T 0 T D = 0.15 sec ; π ω = = rads sec Determie stiffess coefficiet. = ω m = ( ) 3. Determie dampig coefficiet. c cr = mω 0.1 = lbs i. = (0.1) ( ) = lb sec i. c = ζ c cr = (8. 377) = lb sec i.

19 Problem.16 A machie weighig 50 lbs is mouted o a supportig system cosistig of four sprigs ad four dampers. The vertical deflectio of the supportig system uder the weight of the machie is measured as 0.8 i. The dampers are desiged to reduce the amplitude of vertical vibratio to oe-eighth of the iitial amplitude after two complete cycles of free vibratio. Fid the followig properties of the system: (a) udamped atural frequecy, (b) dampig ratio, ad (c) damped atural frequecy. Commet o the effect of dampig o the atural frequecy. Solutio: (a) = l F F u 1 I u j + 1 K J 0 G F u I H J u 0 8K I = lbs i. w 50 m = = = lb sec i. g 386 ω = m = rads sec (b) Assumig small dampig, l j π ζ = l (8) () π ζ ζ = This value of ζ may be too large for small dampig assumptio; therefore, we use the exact equatio: l u 1 j π ζ = u j ζ or, l (8) = ( ) π ζ ζ 1 ζ 1 ζ ζ = (1 ζ ) ζ = = = (c) ω D = ω 1 ζ = rads sec Dampig decreases the atural frequecy.

20 Problem.17 Determie the atural vibratio period ad dampig ratio of the alumium frame model (Fig a) from the acceleratio record of its free vibratio show i Fig b. Solutio: Readig values directly from Fig b: Pea 1 31 Time, t i (sec) Pea, u&& i (g) T D = = 0.35 sec 30 1 ζ = l 0. 78g = = 0.36% π (30) 0.50 g

21 Problem.18 Show that the atural vibratio frequecy of the system i Fig. E1.6a is ω = ω (1 w/w cr ) 1/, where ω is the atural vibratio frequecy computed eglectig the actio of gravity, ad w cr is the buclig weight. Solutio: 1. Determie buclig load. L θ w cr w cr ( L θ) = θ w cr = L. Draw free-body diagram ad set up equilibrium equatio. f I L θ w where f S O M O = 0 f I L + f S = w Lθ (a) f I = w L & θ & g f S = θ (b) Substitutig Eq. (b) i Eq. (a) gives w L θ && + ( w L ) θ = 0 (c) g 3. Compute atural frequecy. or w L w L ω = = 1 (w g) L (w g) L ω = ω w 1 wcr (d)

22 Problem.19 A impulsive force applied to the roof slab of the buildig of Example.8 gives it a iitial velocity of 0 i./sec to the right. How far to the right will the slab move? What is the maximum displacemet of the slab o its retur swig to the left? Solutio: For motio of the buildig from left to right, the goverig equatio is mu&& + u = F for which the solutio is u(t ) = A cos ω t + B si ω t u F With iitial velocity of u&( 0) ad iitial displacemet u(0) = 0, the solutio of Eq. (b) is u(t ) = u &( 0 ) si ω t + u F (cos ω t 1) (c) ω u& (t ) = u&(0 ) cos ω t u F ω si ω t At the extreme right, u& (t ) = 0 ; hece from Eq. (d) ta ω t = u &( 0 ) 1 ω u F Substitutig ω = 4 π, u F = 0.15 i. ad 0 i. sec i Eq. (e) gives or ta ω t = π 0.15 = si ω t = ; cos ω t = (a) (b) (d) (e) u& (0 ) = Substitutig i Eq. (c) gives the displacemet to the right: u = 0 ( ) + 4 π 0.15 ( ) = i. After half a cycle of motio the amplitude decreases by u F = 0.15 = 0. 3 i. Maximum displacemet o the retur swig is u = = i.

23 Problem.0 A SDF system cosistig of a weight, sprig, ad frictio device is show i Fig. P.0. This device slips at a force equal to 10% of the weight, ad the atural vibratio period of the system is 0.5 sec. If this system is give a iitial displacemet of i. ad released, what will be the displacemet amplitude after six cycles? I how may cycles will the system come to rest? u F = 0.1w w Figure P.0 Solutio: Give: F = 0.1w, T = 0. 5 sec u F = F = 0.1w = 0.1mg 0.1g 0.1g = ω = ( π T ) 0.1g = (8π) = i. The reductio i displacemet amplitude per cycle is 4u F = i. The displacemet amplitude after 6 cycles is.0 6 (0.44) = = i. Motio stops at the ed of the half cycle for which the displacemet amplitude is less tha u F. Displacemet amplitude at the ed of the 7th cycle is = 0.9 i.; at the ed of the 8th cycle it is = i.; which is less tha u F. Therefore, the motio stops after 8 cycles.

24 Dyamics Problem.1 of Structures 5th Editio Chopra SOLUTIONS MANUAL Full dowload at : Chopra 5th Editio Dyamics of structures ail chopra Dyamics of Structures boo by Ail K. Chopra Dyamics of Structures Ail K. Chopra Dowload eboo dyamics of structures available editios

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