Vibration 2-1 MENG331

Transcription

1 Vibraio MENG33

2 Roos of Char. Eq. of DOF m,c,k sysem for λ o he splae λ, ζ ± ζ FIG..5

3 Dampig raios of commo maerials 3

4 4 T d T d / si cos B B e d d ζ ˆ ˆ d T N e B e B ζ ζ d T T w w e e e B e B ˆ ˆ ζ ζ ζ ζ Aoher dampig coefficie: Log decreme δ Sec.6.3

5 Log decreme δ v.s. ζ Whe ˆ ˆ e ζ T d Eq..85 δ πζ ζ Eq..86 δ πζ 5

6 Eample. Free Vibraio Respose Due o Impac A cailever beam carries a mass M a he free ed as show i he figure. A mass m falls from a heigh h o o he mass M ad adheres o i wihou reboudig. Deermie he resulig rasverse vibraio of he beam. 6

7 Soluio o Eample. Usig he priciple of coservaio of momeum: & mv m M m M v m m m & gh E. The iiial codiios of he problem ca be saed: m M mg m, & gh k M m m E. Thus he resulig free rasverse vibraio of he beam ca be epressed as: Acos φ 7

8 8 where 3 a 3 / m M l EI m M k A φ & & Soluio o Eample.

9 Eample. Shock Absorber for a Moorcycle A uderdamped shock absorber is o be desiged for a moorcycle of mass kg show i Fig.a. Whe he shock absorber is subjeced o a iiial verical velociy due o a road bump, he resulig displacemeime curve is o be as idicaed i Fig.b. Fid he ecessary siffess ad dampig cosas of he shock absorber if he damped period of vibraio is o be s ad he ampliude is o be reduced o oefourh i oe half cycle i.e.,.5 /4. Also fid he miimum iiial velociy ha leads o a maimum displaceme of 5 mm. 9

10 Eample. Soluio Approach: We use he equaio for he logarihmic decreme i erms of he dampig raio, equaio for he damped period of vibraio, ime correspodig o maimum displaceme for a uderdamped sysem, ad evelope passig hrough he maimum pois of a uderdamped sysem.

11 Eample. Soluio / 4, / 4 Sice.5.5, Hece he logarihmic decreme becomes δ l πζ.776 E. l 6 ζ From which ζ ca be foud as.437. The damped period of vibraio give by s. Hece, τ d π d π.437 π ζ / rad/s

12 c c Eample. Soluio The criical dampig cosa ca be obaied: m N s/m Thus he dampig cosa is give by: c ζc N s/m c ad he siffess by: k m N/m The displaceme of he mass will aai is ma value a ime, give by This gives: or si si d d si ζ siπ.949 π From Problem sec.949

13 Eample. Soluio 3 The evelope passig hrough he ma pois is: Sice 5mm,.5 Xe as & Xe ζ Xe ζ.437 X.455 m ζ ζ si d ζ si d Xe E The velociy of mass ca be obaied by differeiaig he displaceme: Whe, & & X X ζ d cos d m/s d E.3.437

14 Desig Cosideraios Usig he aalysis so far o guide he selecio of compoes. 4

15 Wha is Desig Desig has may defiiios Here we cosider choosig favorable parameers as desig The sprig eamples provide a mehod of desigig vibraio sysems The siffess sprig formulas offer addiioal desig soluios 5

16 Eample Mass kg < m < 3kg ad k > N/m For a possible frequecy rage of 8.6 rad/s < < rad/s For iiial codiios:, v < 3 mm/s Choose a viscous damper c so vibraio respose is always < 5 mm 6

17 Soluio: Wrie dow for zero iiial displaceme Look for ma ampliude Occurs a ime of firs peak T ma Compue he ampliude a T ma Compue dampig raioζ for AT ma.5 7

18 v e ζ si d 4 d 43 Ampliude wors case happes a smalles d 8.6 rad/s wors case happes a ma v 3 mm/s Wih ad v fied a hese values, ivesigae how varies wih ζ Firs peak is highes ad occurs a d d e ζ cos d ζ e ζ si d d Solve for T ma T m a d ζ a d ζ d ζ Sub T ma io :A m ζ T m v ζ e ζ ζ a ζ ζ ζ sia ζ A m ζ v e ζ ζ a ζ ζ 8

19 To keep he ma value less he.5 m solve A ma ζ.5 ζ.8 Usig he upper limi o he mass m 3 kg yields c m ζ kg/s FYI, ζ yields A ma v 37 mm 9 hp://

20 Eample Wha happes o a good desig whe some oe chages he parameers? Car suspesio sysem. How does ζ chage wih mass? Give ζ, m 36 kg, Δ.5 m, compue c, k. k m k 36, mg kδ k mg Δ mg mδ rad/s.5 k N/m ζ c m kg/s

21 Now add 9 kg of passegers ad luggage. Wha happes? m kg Δ mg k ζ c c cr m g Δ m.9.7 rad/s So some oscillaio resuls a a lower frequecy.

22 Mechaicalelecrical aalogies

23 Mechaicalelecrical aalogies 3

24 Numerical Simulaio Solvig differeial equaios by umerical iegraio Euler, RugeKua, ec. Available i Malab, Mahemaica ad Fora or i MS Visual C Use hese o eamie oliear vibraio problems ha do o have aalyical epressios for soluios 4

25 Euler or Tage mehod udo he derivaive d i d lim Δ i i Δ Δ i i 5

26 6 a a a i i i i i i i i i i Δ Δ Δ,,, solve & Udo he derivaive

27 Eample solve d/d3, umerically a 3, ake Δ M M M 7

28 8 Aalyical Soluio e A Ae Ae Ae Ae 3 so ha 3, 3 3 λ λ λ λ λ &

29 For Δ.5.5 i A. i Δ i Δ Time sep is oo large o capure he respose 9

30 For Δ.5 i.. i 3 i Δ.5 i A. i Δ i Δ 3

31 3 Numerical soluio of he d order equaio of vibraio:, Le m k m c k c m & & & & &&

32 3 Wrie as a s order vecor equaio i i i i i i A A m c m k A A Δ Δ,, where, &

33 33 Usually use RugeKua Mos use firs order form Ofe picks Δ Works for oliear equaios oo Δ Δ v f v v v v f i i i i i i i,, && Check Malab [,y]ode3 fu, spa, y

34 Respose o Harmoic Eciaio Forced Vibraio Rao Chap 3 Iroduces he impora cocep of resoace 34

35 Harmoic Eciaio of Udamped Sysems Cosider he usual sprig mass damper sysem wih applied force FF cos is he drivig frequecy F is he magiude of he applied force We akec o sar wih k m F c 35

36 Free Body Diagram y c m F F c& c? k k N F F k k mg mg 36

37 37 Equaios of moio m k m F f f F k m, / where cos cos && && Normalized force

38 Liear ohomogeous ODE: Soluio is sum of homogeous ad paricular soluio The paricular soluio assumes form of forcig fucio physically he ipu wis p X cos 38

39 39 Subsiue io he equaio of moio: yields: solvig cos cos cos f X f X X p p & & / / / s k F m k F X δ

40 Thus he paricular soluio has he form: p f cos 4

41 Add paricular ad homogeeous soluios o ge geeral soluio: A si A cos homogeeous paricular f cos A ad A are cosas of iegraio. 4

42 Could use alerae form: Asi φ homogeeous paricular f cos A ad φ are cosas of iegraio. The relaioship bewee he wo homogeeous forms is give i Usig he form A si A cos homogeeous makes he algebra less complicaed. 4

43 43 f f v v A v A f A A f A f A f A A cos cos si si si cos cos cos si & Apply he iiial codiios o evaluae he cosas

44 Compariso of free ad forced respose Sum of wo harmoic erms of differe frequecies Free respose has ampliude ad phase effeced by forcig fucio Our soluio is o defied for w w because i produces divisio by. 44 The ma ampliude of he forced respose ca be epressed as: X 3. δ s

45 where he quaiy X / δ s is called he magificaio facor, amplificaio facor, or ampliude raio. The variaio of he ampliude raio wih he frequecy raio is show i he Figure. The respose of he sysem ca be ideified o be of hree ypes from he figure. X / δ s r / 45

46 Case. Whe < / <, he deomiaor i Eq.3. is posiive ad he respose is give by Eq.3.5 wihou chage. The harmoic respose of he sysem is i phase wih eeral force, show i figure. I/P X δ s 3. p X cos 3.5 O/P 46

47 / Case. Whe >, he deomiaor i Eq.3. is egaive ad he seadysae soluio ca be epressed as p X cos where he ampliude, δ s X 3. I/P O/P The variaios are show i figure. Ou of phase

48 Toal Respose free vibraio & forced vibraio The oal respose of he sysem, Eq.3.7 or Eq.3.9, ca also be epressed as δ s Acos φ cos; for < FF cos

49 ad δ s Acos φ cos; for >

50 Respose for m kg, k N/m, F N, 5 v.m/s ad. m. v si. f. cos. f. cos ime Noe he obvious presece of wo harmoic sigals 5

51 5 Case 3: Wha happes whe is ear? Beaig m F m F si si / cos cos / & ε m F ε ε si si / π ε π τ b Period of beaig ε b Beaig freq ε π π

52 Bea frequecy.5 f si Time s The bea frequecy is half ha of he blue lie: bea 5

53 Resoace p subsiue io eq.ad solve for A X si X f si A cos X grows wih ou boud f si 53

54 Resoace v f si.. cos... si.. 5 b b

55 Eample : Compue ad plo he respose for m kg, k N/m,,v. m/s, F3 N,. f F m f Equaio for he geeral sol k m 3 N kg N/m kg.si.3 N/kg,.3 N/kg rad rad/s, v / s m/s rad/s he yields: 3 cos rad/s. m 3 m cos 55

56 Eample Give zero iiial codiios a harmoic ipu of Hz wih N magiude ad k N/m, ad measured respose ampliude of.m, compue he mass of he sysem. f rig ideiy cosπ cos for zero iiial codiios f si 4 43 si. m f. / m m π. m.45 kg 56

57 Harmoic Eciaio of Damped Sysems Eedig resoace ad respose calculaio o damped sysems Sec 3.4 &

58 58 Harmoic eciaio of damped sysems & && & && phase shif ow icludes a cos cos cos θ ζ X f F k c m p

59 59 B A B A A B B A X B A s s p s s p s s s s s s p θ si cos cos si a, si cos && & Le p have he form: where: Take derivaives:

60 6 /, ime.specifically for for all si cos s s s s s s s s s s B A f B A B A B f A B A ζ ζ π ζ ζ SUBSTITE THESE VALUES INTO EQUATIONS OF MOTION:

61 Wrie as a mari equaio: ζ ζ A s B s f A s f ζ B s ζ f ζ 6

62 Subsiue he values of A s ad B s io p : p X [ k m c ] f cos ζ ζ Ae si d φ homogeeous or rasie soluio F X / 3.8 θ a X cos θ paricular or seady sae soluio ζ a Add homogeeous ad paricular o ge oal soluio: c k m θ 6 Noe: ha A ad φ will o have he same values as before, as ges large, rasie dies ou

63 Thigs o oice Ifζ, udamped equaios resul Seady sae soluio prevails for large Ofe we igore he rasie erm how large is ζ, how log is? Coefficies of rasie erms cosas of iegraio are effeced by he iiial codiios AND he forcig fucio 63

64 Eample: rad/s, 5 rad/s, ζ., F N, m kg, ad he iiial codiios.5 m ad v. Compare A ad φ for forced ad uforced case: X.33,θ.3 Ae. si.999 φ.33cos5.3 64

65 v.ae. si9.999 φ 9.999Ae. cos9.999 φ.665si5.3 applyig he iial codiios : A.59.5, φ The umbers i are obaied by usig he free respose values 65

66 Proceedig wih igorig he rasie Always check o make sure he rasie is o sigifica For eample, rasies are very impora i earhquakes However, i may machie applicaios rasies may be igored 66

67 Magiude: X f ζ No dimesioal Form: Phase: Frequecy raio: Xk F X f θ a ζr r r r ζr

68 Magiude plo r,... X r, ζ r. ζ. r 6 X r,. X r,.5 4 X r,.7 X r, r 68

69 Magiude plo Fig

70 7 X δ s r ζ r. For a udamped sysem ζ, Eq.3.3 reduces o Eq.3., ad M as r.. Ay amou of dampig ζ > reduces he magificaio facor M for all values of he forcig frequecy. 3. For ay specified value of r, a higher value of dampig reduces he value of M. 4. I he degeerae case of a cosa force whe r, he value of M. X δ s

71 5. The reducio i M i he presece of dampig is very sigifica a or ear resoace. 6. The ampliude of forced vibraio becomes smaller wih icreasig values of he forcig frequecy ha is, M as r. 7. For < ζ <, he maimum value of M occurs whe r ζ or ζ X δ s r ζ r which ca be see o be lower ha he udamped aural frequecy ad he damped frequecy d ζ.

72 8. The maimum value of X whe r ζ is give by: X 3.33 δ s ζ ζ ma ad he value of X a by X δ s dm ζ Useful o measure dampig raio For ζ, whe r. For ζ > dr, he graph of M moooically decreases wih icreasig values of r..77 7

73 Phase plo The followig characerisics of he phase agle ca be observed from figure ad Eq.3.3 as follows: 73

74 θ a ζr r. For a udamped sysem ζ, Eq. 3.3 shows ha he phase agle is for < r < ad 8 for r >. This implies ha he eciaio ad respose are i phase for < r < ad ou of phase for r > whe ζ.. For ζ > ad < r <, he phase agle is give by < Φ < 9, implyig ha he respose lags he eciaio. 3. For ζ > ad r >, he phase agle is give by 9 < Φ < 8, implyig ha he respose leads he eciaio. c k m a 74

75 4. For ζ > ad r, he phase agle is give by Φ 9, implyig ha he phase differece bewee he eciaio ad he respose is For ζ > ad large values of r, he phase agle approaches 8, implyig ha he respose ad he eciaio are ou of phase. 75

76 Phase plo. r. ζ. r. ζ θ r, ζ aa. Φ r π aa. Φ r r r 4 θ r,. θ r,.5 θ r,.7 θ r,.5 z r 3 π π/.5.5 r 76

77 Log plo of magiude X r, ζ r. ζ. r r,... X r,. X r,.5 X r,. X r,.5 X r, r 77

78 Some oes o resoace Resoace is close o r For ζ, r defies resoace As ζ grows resoace moves r < The eac value of r, ca be foud from differeiaig he magiude Resoace occurs a φ π/ 78

79 79 ma peak / ζ ζ ζ ζ ζ < < F Xk r r r dr d F Xk dr d Compue ma peak by differeiaig:

80 ζ.,... Log plo of peak ampliude value r ζ. ζ A ζ A ζ 3. ζ. ζ ζ Depedece of peak locaio o dampig raio r ζ ζ 8

81 Qualiy facor ad badwidh For values of dampig ζ <.5, we ca ake X δ s small ma X δ Power absorbed by damper: ΔW πcx s ζ From figure, R ad R is he badwidh of he sysem. Se X / δ s Q /, hece, r X δ s ma ζ ζ Q Q ζr ζ

82 8 Solvig he equaio for small values of, or ζ ζ r r ζ 3.44 Thus, we obaied 3.46 ζ Q Usig he relaio, 3.4, ζ ζ R r R r R R ζ Qualiy facor ad badwidh

83 Qualiy facor ad badwidh Q ζ 3.46 X / δ s Q / 83 Ref: Tebook, p.34

84 Qualiy facor Q i he s plae ad Freq Resp && && ζ & & Q Q ζ Tuig fork Quarz wach Caesium beam 5 for aomic clock 84

85 Respose of a Damped Sysem Uder The equaio of moio becomes i m && c& k F e Assumig he paricular soluio p Xe i Subsiuig, F X k m ic F F e i X k m c F i k m c k m c Ref: Tebook, Sec 3.5

86 F X F e i iy F Ae Usig he relaio, iφ iφ / [ ] k m c φ c a k m Hece, he seadysae soluio becomes, e p F i φ / [ ] k m c e

87 87 Respose of a Damped Sysem Uder Frequecy Respose The comple frequecy respose is give by: kx H i 3.54 F r iζ The absolue value becomes, where H i r H i e iφ ζr φ a r Thus, he seadysae soluio becomes, F i φ p H i e k F F e i

88 88 If, he correspodig seadysae soluio is give by he imagiary par of Eq.3.53 If, he correspodig seadysae soluio is give by he real par of Eq.3.53 [ ] 3.59 Re Re cos / φ φ i i p e i H k F e i H k F c m k F F F cos F F si [ ] 3.6 Im si / φ φ i p e i H k F c m k F

89 89 Comple Vecor Represeaio of Harmoic Moio Differeiaig Eq.3.58 wih respec o ime, The various erms of he equaio of moio ca be represeed i he figure, 3.6 Acceleraio Velociy e i H k F i i e i H k F i p i p p i p φ φ && &

90 Comple Vecor Represeaio of Harmoic Moio 9 m && c& k F e i 3.47