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Solution Manual for Mechanical Vibrations, 6th Edition

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    SOLUTIONS MANUAL MECHANICAL VIBRATIONS SIXTH EDITION SINGIRESU S. RAO University of Miami January 2016

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    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

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    To Lord Sri Venkateswara

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    Contents Preface 1. Fundamentals of Vibration 1-1-1-99 2. Free Vibration of Single-Degree-of-Freedom Systems 2-1-2-184 3. Harmonically Excited Vibration 3-1-3-138 4. Vibration Under General Forcing Conditions 4-1-4-146 5. Two-Degree-of-Freedom Systems 5-1-5-133 6. Multidegree-of-Freedom Systems 6-1-6-152 7. Determination of Natural Frequencies and Mode Shapes ~~ 7-1-7-71 8. Continuous Systems 8-1-8-83 9. Vibration Control 9-1-9-84 10.Vibration Measurement and Applications 10-1 -10-18 11.Numerical Integration Methods in Vibration Analysis ~~ 11-1 -11-50 12.Finite Element Method 12-1-12-96 13.Nonlinear Vibration 13-1-13-52 14.Random Vibration 14-1 -14-28

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    ERRATA (FOR TEXT BOOK) Mechanical Vibrations Sixth Edition, Singiresu S. Rao (January 10, 2016) 1. Page 238 Problems 2.55-2.58 line 3 change k= 10,000 to k= 10,000 N/m 2. Page239 Problem 2.65 Replace (a) by the following: (a) Find whether the trunk of the tree will buckle under the weight of the crown of the tree assuming that the weight of the crown acts as an axial load at the top of the trunk. Assume the height of the trunk as / = I=10m. 3. Page 253 Problem 2.121 Add the following line after part (d): Assume m = 10 kg and k= 10,000 N/m. 4. Page 253 Problem 2.122 ‘Add the following line after part (d): Assume m = 10 kg and k= 10,000 N/m. 5. Page 254 Problem 2.123 Add the following line after part (d): Assume m = 10 kg and k= 10,000 N/m. 6. Page 368 Problem 3.87 line 3 modify the end of the sentence as follows: Replace “respectively.” by “respectively, and & =m/M.”

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    Chapter 1 % . > ) . . S Fundamentals of Vibration a JL ee @ = [ Heed | Head, and neck -. upper torso, upper torso Ah hips and legs spinal 5 = fegs fips d pe spinal PEERS Se 7 | column nn (i legs #~ -- 7 shoulder P 9 legs<:T . (i) Gin) Seat Restraint \ Windshield oN \ 5) Instrument 3 panel . I CHE; — OTR | St EWN tout Impact force ee 1-1

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    unbalanced > ©) unbalanced forces 1 forces 3, E) 1 ! Mass of frame & a Mass of frame, T reciprocating engine + reciprocating x . T engine & ! og -pdamping & fadndatien block x” stiffness of xy : bolts and - elastic pad «damping | block stiffness xp oo damping © - - dl stiffness of soil (@) one degree of i Freedom model (b) Two degree of freedom model ) - od | mass of body mass of body Suspension of wheels damping rear Wheels of 3 0° suspension © wheels (i) 777 (i 1-2

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    [ [emass of passengers % plus seats E) ha <hizdamping © stiffness of seats a a = o- bod. damping Ro _—-ydamping 2 stiffness stiffness of \. yz" of rear suspension Front "w suspension wheels wheels depia NE :- 7damping & stiffness rissa * 0 0 » of rear wheels 7 7’ 7 777 (iii) ®) Radiator sheet metal KN Cente Ne ¥. 7 Bumper Driveline ©) [ Shock absorber 5 3 Ll ) )] Plow 1-3

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    = ® EIN SRN SU Lr. SN z 1k, Kp 24g 0 ~eg TX X ¥ El eg 3 Ky +2ky Ky + keky ~e % F Xx wv At x K -- ep, Keg + *y kg ks Kym Ke(Bgr Ke). tksksrzhblksr kvakg ks pa lakaks PA Kates Ky + Kak +2 K K3 Kg +2 KKsKg pity +i kg Ks thik J 1-4

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    Fo SE & J A’ x |€ T JL re © oo xz @) i + LF Fi Fa b) {te fy From Fig.(@), x= PIE (x2— X01) 9,+42 I CR RI (1 012 2,+4; # vertical force ep pilibrivm from Frg.u(b) : F= F + F, (2> Moment eguilibrium about c’ (Fig.hd): Fp 2, = F, 2, > Solution of Egs. (2) and (3): f= Fla po FA 4) gL, + L, 2 2+ [P} Displacements of springs *, amd tk; are given by oo Bo _Fl oo R_ Fh * 1, (4+ 12) 27 Th, (G12) Displocement of force F can be found using Eps-CsD in Eg. (1: ¢ RE PA 1. SE TH EE 1+, 1, (1,+12) Lita 4p (L442) - _F (2 +1; Ly (6) ISRAEL The epuivalnt Ap2ing constand of the Ayslom in the 1-5

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    - g direction of x, Kg, “ given by Ep. (6): i ke = = = Lr) wk bike D) : NTE Pa eR ce 1-6

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    <3 > 3 Epuiva lence of potential energies gives & z 2 fT EE AC 0) L100) +3 kalo ds) = £ fey =p 4p3= For series springs i» ty and ki: «0 ka LE SE yy = ————————— Kips TR ke ks 123 Kd + Kp fea + Kak Using energy epuivalence, ’ 2 z 2 3 keyot= hk kyo th Kus O +3 ks OF) + ike (eR) ~ Heg = ky + Kizz + R* %g + R" ke fk, ® = K | ®2 %3 ) 2 CT erty +R ( st ®s) ee @ 1) For simply supported beam, 12m Sor load ot middle, + 3 i 54 Ged «= ee _ Wbosro (27) Pap T=— 3 - = 12-36 x10’ Nfm where T= (L2) (er) 5 mt “on . _ om _ 5 9.84 -7 8 = original deflection = = = 200 xto= — 396-8447 xl0 ™ When spring & is added, Keg, = &+ Kg ; mg ) 4 mg @) New deflection= Z% _ 22 ge = ZF = ) d key TF ©” Ts C4 = k+ Hk i Sok = 3% = 37.08 x10 N/m ! (b) New deflection = + . 3 ; k= 2m Lg J? 7 5, k= %,z 12:36 x10 N/m = k+ Kk, ©) New deflection = TZ 4m flection = =24; key ® $F _ 4, % 38 3 Ci k=d k= 402 x10 Nm = +k, Ef 1-7

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    > i») For bar with length L, Youngs modulus E and E) cross-section A, The axial stiffness(x) is given by & = AE a] af ®= 7 when cross -section is solid circular with diameter d, area =A = Td, (2) when cross — section is sguare with side d. ove = Ap = 4 (3) When cross ~section iS hollow circular with mean dio. d and walt thickness t zo-1d, area = TTdt = Td (61d) =o T 4% (CY) } For specified value of k= %& , cross-section areas iced is: fe reguir is A= KL = c (constant (CD) Weight of bar : with solid circular section: c w= rd L= clk with d= RES ©> with hollow ciceul ar section: Wy, = om d L = orm (HAE)L = ok cb (€)) with spuare section: = ok wy = dL = %C | = 4 w, = 2732 Wi 8) ) ™ Ed + The shaft with the hollow ciceular cross-section corresponds fo minimum weight. eye rm 1-8

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    > > @ Stiffress of o cantilever beam under o bend ing 3 wv force of free end = 5 <p) JL For o Apres fled value of *k= %, I= zl -=Cc = constant (©) 3 For a Solid circular section with diameter d5 CL h_ 64 REE 3) I= = =c = 4d = or d = => ( weight of beam = Wy = wd*f —- wk 6c 4 % Jets = 3.5449 {JC “) for a hollow circular section with mean diameter d and woll thickness t=o:1d, weight of beam (wp) is: woz I (do —d) k= pts fey - (4-07) = Li (444) = w dtl = WHE dD —1 = ow REE - igo Le GD For ov Aguas secion with side 4d. weight of the beam (wy) 4: 2 wz= d*4 = [REE = 4.5135 Lic" ) By componing Es: WY, (8) amd (6, The minimum weight 1-9

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    z 3 Spring Sorce is given by F= Soo x + 40 x? (DD) 2 static equilibrium of the rubber mounting (x*) under + the weight of the eledmiomic instrument is given by E= 200 = 8a0 x* + 40 x? 46 2*® 4 900 x" ~ 200 =o (2 The roots of the cubic eppation (2) cam be ford From MATLAB of x*= 02492 » —orlz4é LE 44773 4 @ Thus He sali eguilibriam position of the rubber mounting is given by the real root of Eg-@): x* = 02432 in 4> @) Eguivalent linear Alering constant of vubber mounting ok its shoe eguilibrium position , using CAC keg = | = 860+ 126 x* = goo +1200 (0-24%2) = got 4521 lb/in 52> bY Deflection of rubbeyr mounting corresponding fo the equivalent Linear Apring constant is: BE PE AAT. ©) k= ke T wor4s2t SE 1-10

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    z g @) F(x)= zoox+ 50 x? 4 10 x? a) a dL when the Spring under goes a steady deflection of : x* =0'5 in during the operation of the engine, the force exerted on the pring con be found os g= 200(e'5) + So (05) +10 (o-5)° = 13.75 Lb (2d Equivalent Linear Spring constant at if Ateadly deflection is given by Ey- G7) : Keg = a8 = 200 + 100 2 30 eg d= x=x¥ 2 = 200 + loo (0:8) +30 (o'5) = 253.75 (b/in J 1-11

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