5.2 GRAPHICAL VELOCITY ANALYSIS Polygon Method

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1 ME 352 GRHICL VELCITY NLYSIS 52 GRHICL VELCITY NLYSIS olygon Mehod Velociy analyi form he hear of kinemaic and dynamic of mechanical yem Velociy analyi i uually performed following a poiion analyi; ie, he poiion and orienaion of all he link in a mechanim are aumed known In hi coure we concenrae on one analyical and wo graphical mehod for planar mechanim Velociy polygon i a graphical pencil-and-paper approach for deermining unknown velociie of a ingle degree-of-freedom mechanim The mehod require conrucing a velociy loop equaion (a polygon) graphically polygon may have hree or more edge depending on he number of velociy vecor in he equaion For a vecor loop equaion, he polygon mehod i he graphical procedure of olving wo algebraic equaion in wo unknown The velociy polygon mehod i demonraed for everal commonly ued mechanim Tool: ruler, righ angle, proracor, drawing compa Four-bar For a known four-bar mechanim, in a given configuraion and for a known angular velociy of he crank, ω 2, we wan o deermine ω 3 and ω 4 In hi example we aume ω 2 i CCW For he poiion vecor loop equaion R 2 +R R 4 R 4 he velociy equaion i V +V V (a) R R R R 4 Since vecor R 2, R, and R 2 have conan lengh, heir correponding velociy vecor are angenial; ie, V +V V (b) or, ω 2R 2 +ω 3R ω 4R 2 (b) The unknown are ω 3 and ω 4 ince all vecor axe are known Velociy polygon 1 Nex o he diagram of he four-bar, elec a poin in a convenien poiion a he reference for zero velociie Name hi poin (origin of velociie) R 2 Compue he magniude of V a R 2 ω 2 From V conruc vecor V perpendicular o R 2 by roaing V R 2 90 o in he direcion of ω 2 3 From draw a line perpendicular o R V mu reide on hi line E Nikraveh 5-4

2 ME 352 GRHICL VELCITY NLYSIS R 4 From draw a line perpendicular o R 4 V mu reide on hi line V R 5 Conruc vecor V and V 6 Deermine he magniude of V from he polygon Compue ω 3 = V / L Deermine he direcion of ω 3 In hi example i i CW ince R mu roae 90 o CW V o line up wih V V V V 7 Deermine he magniude of V from he polygon Compue ω 4 = V / L 4 Deermine he direcion of ω 4 In hi example i i CCW ince R 4 mu roae 90 o CCW o line up wih V Secondary equaion() To deermine he velociy of a econdary poin, uch a a coupler poin, we refer o he poiion expreion and he correponding velociy expreion: R 2 = R 2 + R y R R R 2 V = V + V = ω 2R 2 + ω 3R Since he angular velociie are already known, V and V are conruced We add hee wo vecor graphically o deermine V x V V V E Nikraveh 5-5

3 ME 352 GRHICL VELCITY NLYSIS Example F-V-1 four-bar mechanim ha he following conan daa: L 2 = 10, L = 40, L 4 = 30, L 4 = 30, L = 18, β 3 = 75 o The crank angle i a θ 2 = 170 o wih an angular velociy of ω 2 = 10 rad/ec CCW The velociy polygon i conruced and he following velociie are deermined from he polygon: ω 3 = V rad/ec, CCW L 40 V ω 4 = V rad/ec, CCW L 4 30 V V V econd polygon provide he velociy of poin a V = 13 in he direcion hown V V Slider-crank (inverion 1) Thi lider-crank mechanim in he given configuraion ha a known angular velociy of he crank, ω 2 We wan o deermine ω 3 and he velociy of he lider block In hi example we aume ω 2 i CCW The poiion vecor loop equaion i: R 2 +R R 2 The velociy (loop) equaion i expreed a R R V +V V We noe ha V and V are angenial and V i of lip ype (along he axi of R 2 ) Therefore he velociy equaion can be expreed a ω 2R 2 +ω 3R V Velociy polygon 1 Nex o he diagram of he mechanim, elec a poin for he origin of velociie 2 Compue he magniude of V a R 2 ω 2 From conruc vecor V perpendicular o R 2 by roaing R 2 90 o in he direcion of ω 2 R V 3 From draw a line perpendicular o R V mu reide on hi line 4 From draw a line parallel o he axi of he lider; ie, parallel o R 2 V mu reide on hi line E Nikraveh 5-6

4 ME 352 GRHICL VELCITY NLYSIS 5 Conruc vecor V and V, conidering heir ign in he velociy equaion 6 Deermine he magniude of V from he polygon R Compue ω 3 = V / R Deermine he direcion of ω 3, which i CW in hi example 7 Deermine he magniude of V from he polygon The direcion of hi vecor indicae ha he lider block i moving o he lef V V V V Example SC-V-1 For a lider-crank mechanim, he following lengh are given: L 2 = 15, L = 30 The crank angle i θ 2 = 120 o, and ω 2 = 1 rad/ec CW The mechanim i drawn for θ 2 = 120 o For he given angular velociy, he velociy polygon i conruced The following velociie are deermined from he polygon: ω 3 28 rad/ec, CW; V 94 o he righ V V V Slider-crank (inverion 2) For hi lider-crank mechanim (inverion 2), in he given configuraion and for a known angular velociy of he crank, ω 2, conruc he velociy polygon Then deermine ω 4 and he velociy of he lider block ume ω 2 i CW The poiion vecor loop equaion i R 2 R 4 R 4 Since R 2 i a roaing fixedlengh vecor, V 2 i angenial However, R 4 i a variable-lengh, variable-angle vecor, and herefore V 4 conain boh angenial and lip componen Hence he velociy (loop) equaion i wrien a V 2 V 4 V 4 = ω 2R 2 V 4 ω 4R 4 Noe: In hi velociy equaion we do no drop he index of he non-moving poin, and Dropping hem may caue confuion R 2 R 4 R 4 E Nikraveh 5-7

5 ME 352 GRHICL VELCITY NLYSIS Velociy polygon 1 Selec a poin for he origin of velociie 2 Conruc vecor V 2 3 From he end of V 2 draw a line parallel o R 4 R 2 V 4 hould reide on hi line V 2 4 From draw a line perpendicular o he axi of he lider; ie, perpendicular o R 2 V 4 hould reide on hi line R 4 5 Conruc vecor V 4 and V 4, conidering heir ign in he velociy equaion 6 Deermine he magniude of V 4 Compue V 2 ω 4 = V 4 / R 4 Deermine he direcion of ω 4 In hi example ω 4 i CW 7 Deermine he magniude of V 4 from he polygon V 4 V 2 V 4 Secondary equaion() Deermine he velociy of a econdary poin, uch on link 4, where L 4 = R 4 i a known conan R There are wo poible way o deermine V : (a) We can poiion wih repec o he ground poin a R 2 = R 2 + R The correponding R 2 R 2 velociy expreion i V = V + V, where V conain boh angenial and lip componen Compuing he angenial componen require he magniude of R, which mu be deermined baed on L 4 R 4 The lip componen of V mu be V (a) baed on he lip componen of R 4 I hould be obviou ha we have made a imple problem unnecearily difficul! (b) We can poiion wih repec o he ground poin a R 2 = R 4 + R 4 The correponding velociy expreion i V = V 4 = ω 4R4 a hown R 2 R 4 R 4 on he figure baed on a CW direcion of ω 4 (b) Example SC-V-2 Conider he following lengh for a lider-crank, inverion 2: L 2 = 10, L 4 = 20, L 4 = 40 The crank angle i in θ 2 = 125 o orienaion The angular velociy of he E Nikraveh 5-8

6 ME 352 GRHICL VELCITY NLYSIS crank i ω 2 = 1 rad/ec CW The mechanim i drawn for he given crank angle The velociy polygon i conruced and he following value are obained: ω 3 = ω 4 3 rad/ec CW; V 4 61 in he direcion hown The velociy of poin i compued a V = (40)(03) 12, in he direcion hown Q: I he lider moving away from? V V 4 V 4 V 2 Slider-crank (inverion 2 - offe) For hi offe lider-crank mechanim he poiion vecor equaion i R 2 R R 4 R 4 The correponding velociy equaion i expreed a V 2 V V 4 V 4 r, ω 2R 2 ω 4R ω 4R 4 V 4 Unlike he velociy equaion for he four-bar or he andard lider-crank, hi velociy equaion conain four vecor Since hree of he vecor conain he unknown, conrucing hi polygon may no appear eay However, we can combine wo of he vecor and reduce he number of vecor in he polygon from four o hree: ω 2 R 2 ω 4 ( R + R 4 ) V 4 For a known angular velociy ω 2, we fir conruc V 2 R 4 for V 4 angenial velociy V 4, and anoher axi parallel o R 4 for V 4 R R R θ 4 R 4 V 4 V 4 V 2 ω 2R 2 ω 4R 4 V 4 can be decompoed ino i componen V V 4 V V 4 V 2 We add an axi perpendicular o o complee he polygon Then he and V 4 Slider-crank (inverion 3) For hi lider-crank mechanim (inverion 3), in he given configuraion and for a known angular velociy of he crank, ω 2, conruc he velociy polygon Then deermine ω 4 and he velociy of he lider block ume ω 2 i CW The poiion vecor loop equaion i wrien a: R 2 + R R 4 The correponding velociy equaion i V 2 + V 4 + V 4 = ω 2R 2 + V 4 + ω 3R4 Velociy polygon 1 Selec a poin for he origin of velociie 2 Conruc vecor V 2 R 2 R 4 R 4 E Nikraveh 5-9

7 ME 352 GRHICL VELCITY NLYSIS 3 From he end of V 2 draw a line parallel o R 4 V 4 hould reide on hi line 4 From draw a line perpendicular o he axi of he lider; ie, perpendicular o R V 4 on hi line mu reide R 2 V 2 R 4 5 Conruc vecor V 4 and V 4, conidering heir ign in he velociy equaion V 2 V 2 V 4 V 4 6 Deermine he magniude of V 4 Compue ω 3 = ω 4 = V 4 / R 4 Deermine he direcion of ω 3 In hi example, i i CW 7 Deermine he magniude of V 4 Secondary poin Deermine he velociy of poin on link 3, where R = L 3 i a known conan oin can be poiioned wih repec o he ground poin a R 2 = R 2 + R The correponding velociy expreion i V = V + V = ω 2R 2 + ω 3R Since boh angular velociie are known, V can be conruced graphically R 2 V R R 4 V 2 V Example SC-V-3 The following lengh are provided for a lidercrank (inverion 3) mechanim: L 2 = 10, L 4 = 15, L 5 The crank angle in he hown configuraion i θ 2 = 30 o The angular velociy of he crank i ω 2 = 1 rad/ec CCW The velociy polygon i conruced and he following velociie are deermined from he polygon: E Nikraveh 5-10

8 ME 352 GRHICL VELCITY NLYSIS ω 3 = ω 4 46 rad/ec CW, V 4 9 in he direcion hown econd polygon provide he velociy of a V = 11 in he direcion hown V V 4 V V 4 V 4 V 4 Exercie In hee exercie ake direc meauremen from he figure for link lengh and he magniude of velociy vecor Conruc velociy polygon o deermine he unknown Exercie 1 4 are example of four-bar mechanim ume known value and direcion for ω 2 Deermine ω 3, ω 4, and V Exercie 5 8 are example of lider-crank mechanim ume known value and direcion for ω 2 For 5 and 6 deermine ω 3, ω 4, and he velociy of he lider block For 7 and 8 deermine ω 3, ω 4, and V 5 6 E Nikraveh 5-11

9 ME 352 GRHICL VELCITY NLYSIS For hi ix-bar mechanim ω 2 i given Deermine ω 5, velociy of, and he velociy of he lider block Q (5) (6) 10 For hi ix-bar mechanim ω 2 i given Deermine ω 5 and he velociy of he lider block (6) (6) (5) E Nikraveh 5-12