1/5/013 Egieerig Mechaics Dyaics ad Vibratios Egieerig Mechaics Dyaics & Vibratios Egieerig Mechaics Dyaics & Vibratios Plae Motio of a Rigid Body: Equatios of Motio Motio of a rigid body i plae otio is copletely defied by the resultat ad oet resultat about the ass cetre G of the exteral forces. F a F a M J x x y y G D'lebert's priciple: iertia forces The particle acceleratio we easure fro a fixed set of axes X-Y-Z (Figure (a)) is its absolute acceleratio a. I this case the failiar relatio F a applies Whe we observe the particle fro a ovig syste x-y-z attached to the particle, the particle ecessarily appears to be at rest or i equilibriu. fictitious force -a (so called iertia force) acts o the particle (figure b) 1
1/5/013 Egieerig Mechaics Dyaics & Vibratios Plae Motio of a Rigid Body: D'lebert's priciple D'lebert's priciple: iertia forces D'lebert showed that oe ca trasfor a acceleratig rigid body ito a equivalet static syste by addig the so-called iertia forces - The traslatioal iertia ust act through the ceter of ass ad the rotatioal iertia ca act aywhere. The syste ca the be aalyzed exactly as a static syste. - The iertia forces are see to oppose the otio Egieerig Mechaics Dyaics & Vibratios Sprig-Mass Syste Sall Oscillatios d x x dt Newto s secod law p( t) kx x D lebert s priciple Fx 0 x kx p( t) 0
1/5/013 Egieerig Mechaics Dyaics & Vibratios Sprig-Mass Syste: Gravity Effect t static equilibriu cofiguratio k g st Now the particle is displaced through a distace x o fro its static equilibriu cofiguratio ad released with a velocity v o, the particle will udergo siple haroic otio Fro the free body diagra of the ass at a tie istat t with displaceet x(t) Fv 0 x 0 g k x st x kx 0 Goverig equatio of otio k k x x 0 where Egieerig Mechaics Dyaics & Vibratios Free Vibratios of Sprig-Mass Syste x( t) C si t C cos t 1 Geeral Solutio x is a periodic fuctio ad w is the atural circular frequecy of otio. C 1 ad C are deteried by the iitial coditios: x t C cos t o C1 si v x C1 cos t C si t @ tie t=0; x=x C x @ tie t=0; v=x=v o C 1 o vo v x t t x t 0 ( ) si o cos 3
1/5/013 Egieerig Mechaics Dyaics & Vibratios Siple Haroic Motio vo C1 C x o Displaceet is equivalet to the x copoet of the su of two vectors which rotate with costat agular velocity. v x t t x t 0 ( ) si o cos x x si t v 0 x x 0 aplitude ta 1 v x phase agle period 1 f atural frequecy 0 0 C 1 C Egieerig Mechaics Dyaics & Vibratios Siple Haroic Motio Velocity-tie ad acceleratio-tie curves ca be represeted by sie curves of the sae period as the displaceet-tie curve but differet phase agles. x x si t v x x x a x x x cos t si t si t si t 4
1/5/013 Egieerig Mechaics Dyaics & Vibratios Saple Proble 1 50-kg block oves betwee vertical guides as show. The block is pulled 40 dow fro its equilibriu positio ad released. For each sprig arrageet, deterie a) the period of the vibratio, b) the axiu velocity of the block, ad c) the axiu acceleratio of the block. SOLUTION: For each sprig arrageet, deterie the sprig costat for a sigle equivalet sprig. pply the D lebert s priciple for the haroic otio of a sprig-ass syste. Egieerig Mechaics Dyaics & Vibratios Saple Proble 1 k 4 kn k 6 kn 50kg 1 For equilibriu: Fv 0 x k1x kx 0 x ( k k ) x 0 k k k eq 1 x k x 0 eq 1 10kN Force diagra 4 10 N 4 keq 10 14.14 rad s ec 50 v a v x x t x 0.040 14.14 rad s x cos a x x t si 0.040 14.14 rad s 0.444 s v 0.566 s a 8.00 s 5
1/5/013 Egieerig Mechaics Dyaics & Vibratios Saple Proble 1 k1 4kN k 6kN For equilibriu: Fv 0 x k x 0 eq P k x k ( x x ) k x eq kx x1 k k 1 k1k P k k 1 1 1 1 k k k eq 1 1 1 1 x k x 400N/ 6.93rad s 0kg v x x t v a 0.040 6.93 rad s x cos a x x t si 0.040 6.93 rad s v 0.907 s 0.77 s a 1.90 s Egieerig Mechaics Dyaics & Vibratios Distributed Mass: Rotatioal Iertia J = Mass oet of Iertia about C.G J r d x dx Displaceet diagra J CG Force diagra θ L/ L/ L / J x dx L / ( ) ass / uit legth L ( L) L ML 1 1 1 3 L J CG 1 6
1/5/013 Egieerig Mechaics Dyaics & Vibratios Distributed Mass: Rotatioal Iertia L J y θ = L/ L/ L M Displaceet diagra Iertia Forces at C.G = Force diagra ML 1 Pure Traslatio + ML My + ML Iertia Forces at C.G Moet at due to iertia forces ML ML L ML J 1 3 C.G Pure Rotatio J CG J ML 3 Egieerig Mechaics Dyaics & Vibratios Saple Proble k 1 l l l k For the syste show, =0.4-kg, K 1 =N/ ad K =3N/. Takig the rod o which the ass is fixed as light ad stiff. Deterie a) Natural frequecy of the syste, b) the period of the vibratio. SOLUTION: Select a degree of freedo (sall displaceet). Represet deforatios of sprigs (for elastic forces) ad asses (for iertia forces) i ters of x(t) 7
1/5/013 Egieerig Mechaics Dyaics & Vibratios Saple Proble 1 3 st Gravity Effect 3 st st Displaceet Static equilibriu k st 1 3 Static Forces k st 3 t static equilibriu cofiguratio M 0 st 4 st k1 k g g 9 9 Egieerig Mechaics Dyaics & Vibratios Saple Proble 1 ( ) 3 x t 1 3 st Displaceet 1 For equilibriu: M 0 1 ( ) st k x t 3 3 1 st k1 x( t) l k x( t) st l x ( t)(3 l) g(3 l) 0 3 3 3 3 1 st 1 k1 x( t) k x( t) st x ( t) g 0 3 3 3 3 3 3 k 4k k 4k x t x t g 9 9 9 9 1 1 ( ) ( ) st 0 x t ( ) 3 x t 3 st k 4k 9 9 x( t) st 1 ( ) x( t) 0 k x( t) 3 3 st Equatio of Motio x ( t) forces g 8
1/5/013 Egieerig Mechaics Dyaics & Vibratios Saple Proble 1 ( ) 3 x t ( ) 3 x t static equilibriu cofiguratio l l l 1 k1 x ( t ) k x ( t ) 3 3 For equilibriu: M 0 x( t) x ( t) 1 k1 x( t) l k x( t) l x( t) 3l 0 3 3 1 1 x ( t) k1 k x( t) 0 3 3 3 3 displaceet k eq forces No Gravity Effect Equatio of otio x ( t) k x( t) 0 eq 3x4 9 9 N/ = 14 x10 3 N/ 9 keq 14000 9x0.4 6.36 rad/sec 0.100 sec Egieerig Mechaics Dyaics & Vibratios Saple Proble If distributed ass of the bar is cosidered ML 1 ML l l l 1 k x ( t) 1 x ( t ) k x ( t ) 3 3 For equilibriu: M 0 1 ML k1 x( t) l k x( t) l x( t) 3l 0 3 3 3 x ( t) 3l M 1 4 x( t) k1 k x( t) 0 3 9 9 forces L 3l oe ca also use priciple of virtual work to obtai the equatio of otio 9
1/5/013 Egieerig Mechaics Dyaics & Vibratios Saple Proble 3 k M=10kg a=0.1 b=0.05 SOLUTION: Select a degree of freedo (sall displaceet). Represet deforatios of sprigs (for elastic forces) ad asses (for iertia forces) i ters of θ Derive the equatio of otio of a rectagular block restig o a frictioless surface as show for sall oscillatios i a horizotal plae. Solve the equatio of otio by siplifyig it for M=10Kg, a=0.1, b=0.05, k=10kn/ Deterie a)natural frequecy of the syste, b) the period of the vibratio. Egieerig Mechaics Dyaics & Vibratios Saple Proble 3 bθ θ b/ a/ Displaceets For equilibriu: a 3 ( b ) kbθ b ab 0 M0 0 kb 0 a ab Forces ab a b 1 ( ) ab( a b ) a a b b ab ab kb ( b) 0 ab M 1 M ( a b ) a b M M kb 0 1 4 4 ass / uit area M a J o 3 ( b ) 10
1/5/013 Egieerig Mechaics Dyaics & Vibratios Saple Proble 3 Equatio of otio 0.0416 5 0 k 5 0.0416 4.514 rad/sec 0.56 sec Egieerig Mechaics Dyaics & Vibratios Saple Proble 4 3 3 SOLUTION: Select a degree of freedo (sall displaceet). Represet deforatios of sprigs (for elastic forces) ad asses (for iertia forces) i ters of x What are the differetial equatio of otio about the static equilibriu cofiguratio show ad the atural frequecy of otio of body for sall otio of BC? Neglect iertia effects fro BC. ssue K 1 = 15 N/, K = 0 N/, K 3 = 30 N/ ad W =30N 11
1/5/013 Egieerig Mechaics Dyaics & Vibratios Saple Proble 4 x a x D x D Displaceet diagra Force diagra The cofiguratio show is the static equilibriu ad give that rod BC is ass less (i.e eglect the iertia effect of BC). Here two equilibriu coditios exist i.e Fv 0 ad M c 0 F 0 v x k x k ( x x ) 0 (1) 1 D Egieerig Mechaics Dyaics & Vibratios Saple Proble 4 M c 0 1k x 3 k ( x x ) 0 () 3 D D Fro x k x (), D (3) k 4k3 Force diagra Thus due to iertia less rod BC the -dof proble reduces to 1-dof proble (sice x D depeds purely o x ). Substitute (3) i (1) 4k 3 x k1 k x 0 k 4k3 k1k 4 k3( k1 k) ( k 4k ) 3 3.4 rad/ sec 1
1/5/013 Egieerig Mechaics Dyaics & Vibratios Tutorial Saple proble-3 Proble 5 Rod B is attached to a hige at ad to two sprigs, each of costat k. if h=700, d=300, ad =0 kg, deterie the value of k for which the period of sall oscillatio is (a) 1sec, (b) ifiite. Neglect the ass of the rod ad assue that each sprig ca act i either tesio or copressio. Egieerig Mechaics Dyaics & Vibratios Saple Proble 5 hθ dθ θ Displaceets Forces h ( h) kd ( d) g( h ) 0 kd g 0 h h kd h g h 13
1/5/013 Egieerig Mechaics Dyaics & Vibratios Saple Proble 5 (a) (b) Egieerig Mechaics Dyaics & Vibratios Saple Proble 5 Oe ca also use priciple of iiu potetial eergy to obtai k whe T is ifiite By iiu potetial eergy, We have 1 V g h k d For equilibriu ( cos ) ( ) dv g h d ( si ) kd ( ) 0 θ=0 0 is a equilibriu cofiguratio 14
1/5/013 Egieerig Mechaics Dyaics & Vibratios Saple Proble 5 For stability of Equilibriu: d V d d V d gh k d g( hcos ) kd k >763 N/ gh kd i.e. θ=0 0,Cofiguratio is stable for k >763N/ ( which is sae as that of vibratioal aalysis) 0 Egieerig Mechaics Dyaics & Vibratios Saple Proble 6 k L L M,J Lθ(t) M,J L/θ(t) Lθ(t) L/θ(t) Lθ(t) Lθ(t) k[ L ( t)] M[L/ ( t)] M[L/ ( t)] J ( t) M [ L ( t)] J ( t) Forces [ L ( t)] [ L ( t)] θ(t) Displaceets For equilibriu: M 0 ML ML L L ML 4 1 ML ML kl 0 1 4 5 L ML kl 0 3 15
1/5/013 Egieerig Mechaics Dyaics & Vibratios Daped Free Vibratios ll vibratios are daped to soe degree by forces due to dry frictio, fluid frictio, or iteral frictio. With viscous dapig due to fluid frictio, F a : W k st x cx x x cx kx 0 Substitutig x = e lt ad dividig through by e lt yields the characteristic equatio, c c k 0 c Defie the critical dapig coefficiet such that cc k 0 cc k k Egieerig Mechaics Dyaics & Vibratios Daped Free Vibratios Characteristic equatio, c c k 0 c cc critical dapig coefficiet k Defie dapig ratio c c c c i 1 Light dapig : c < c c c t x e C si t C cos t 1 d d t e C si t C cos t 1 d d Uderdaped Syste 1 daped frequecy d Critical dapig: c = c c t x C 1 C t e - double roots - ovibratory otio Heavy dapig: c > c c t t x C e 1 C e 1 - egative roots - ovibratory otio Critically daped Syste Overdaped Syste 16
1/5/013 Egieerig Mechaics Dyaics & Vibratios Daped Vs. Udaped Free Vibratios Daped t v o x o D o D D x t e x cos t si t vo xo xo D Udaped v x = x cos t o si t o vo xo I the figure use: u=x u(0)=x o u( o )= v o Egieerig Mechaics Dyaics & Vibratios Daped Free Vibratios (logarithic decreet) Fro the two successive peaks x x 1 = e T d 1 TD Note, T D T 1 c ad x l = T x 1 D 1 17
1/5/013 Egieerig Mechaics Dyaics & Vibratios Daped Free Vibratios For u( o )= v o 0,i.e oiitial velocity Egieerig Mechaics Dyaics & Vibratios Saple Proble loaded railroad car weighig 30,000 lb is rollig at a costat velocity v o whe it couples with a sprig ad dashpot buper syste. The recorded displaceet-tie curve of the loaded railroad car after couplig is as show. Deterie (a) the dapig costat, (b) the sprig costat. Source: BJ 18
1/5/013 Egieerig Mechaics Dyaics & Vibratios Saple Proble c 1 D x = e x e D 19