e国b口 LECTURE 7 Kinematic Analysis of Mechanisms 1日 OUTLINE Some important definitions ©Simple cases study Coriolis acceleration Vector graphical analysis method Complex vector analytical method ME371 Design Manufacturing ll 1
1 Kinematic Analysis of Mechanisms LECTURE 7 ME371 Design & Manufacturing II OUTLINE Some important definitions Simple cases study Coriolis acceleration Vector graphical analysis method Complex vector analytical method
Some important definitions Displacement R=Reio Linear displacement:All particles of a body move in parallel planes and travel by same distance is known as linear displacement Angular displacement:A body rotating about a fixed point in such a way that all particles move in circular path is known as angular displacement axis 2 R Real axis ME371 Design Manufacturing ll Some important definitions Velocity-Rate of change of displacement is velocity.Velocity can be linear velocity of angular velocity. First order: d(Re)=Re+R(e"j0)=Re+ROje d de Angular Velocity:@ dt yImaginary axis Linear Velocity:V=R R Real axis R 2
2 ME371 Design & Manufacturing II Displacement Linear displacement: All particles of a body move in parallel planes and travel by same distance is known as linear displacement Angular displacement: A body rotating about a fixed point in such a way that all particles move in circular path is known as angular displacement Some important definitions j R e R Velocity - Rate of change of displacement is velocity. Velocity can be linear velocity of angular velocity. Some important definitions First order: ( ) ( )= d j jj j j R e Re R e j Re R j e dt Angular Velocity: d dt Linear Velocity: d dt V R
Some important definitions Acceleration-Rate of change of velocity Second order: -Re+20Rj e+Rj ero-ROe Angular Acc:a==do yImaginary axis Linear Acc: a=求=dy R dt Real axis R Simple cases study A link in pure rotation ⑧When point A is moving ME371 Design Manufacturing ll 3
3 Acceleration- Rate of change of velocity Some important definitions 2 2 2 Second order: ( ) ( ) ( ) ( ) = 2 jj j j j j j jj d R e Re R j e RR j e R je j dt Re Rj e Rj e R e Angular Acc: d dt Linear Acc: d a R dt v ME371 Design & Manufacturing II A link in pure rotation When point A is moving Simple cases study
Simple cases study A link in pure rotation +0 Displac- ement Rp=pele RPA N PA Velocity Ve=pojel +2 02 Acceleration APA Apa=pajeo-po'e =A'm+Am A ME371 Design Manufacturing ll Simple cases study +0 When point A is moving Displac- 2 ement Rp=R+Rp4 V. Velocity 可=f4+4 =fa+pe(to）) Graphical solution: VA 4
4 ME371 Design & Manufacturing II A link in pure rotation Simple cases study Displacement Velocity Acceleration j PA pe R j PA p je V 2 j j PA PA PA p je p e t n A A A When point A is moving Simple cases study Displacement Velocity RRR P A PA V pe i V V V i A P A PA Graphical solution:
Simple cases study When point A is moving APA +2 -02 Displac- ement Rp=R+Rm4 APA Velocity 可。=f+ipA A =V,+pe(io) Accelerati on A。=A4+Ap4 AP APA =A-pe+iape AA Coriolis Acceleration Position of slider R=pel Velocity of slider 7。=eigte R Transmission Slip velocity velocity 0 Acceleration: An=peio+pe(io)'+pe”ia+e+pe°i0 Combining terms: 4,=[-po）+i(pa+22o]e Coriolis acc.occurs when a body has vslip and w Slip Normal Tangential Coriolis 5
5 When point A is moving Simple cases study Displacement Velocity Accelerati on RRR P A PA V pe i V V V i A P A PA 2 P A PA i i A AAA A pe i pe Coriolis Acceleration i . R p pe Position of slider Velocity of slider i i V pe i pe p Transmission velocity Slip velocity 2 i i i ii A p pe i pe i pe i pe pe i Acceleration: 2 2 i A p p p ip p e Combining terms: Slip Normal Tangential Coriolis Coriolis acc. occurs when a body has vslip and ω
Coriolis Acceleration Coriolis acc.occurs or not? 2 2 3 3 Vector graphical analysis method © Given linkage configuration &2.Find 3 and 4 We know VA and direction of Ve and VBA (perpendicular to AB) Draw vector triangle.V=or. 03 VBa Direction 3 Va Direction VR=@;(AB) Va=0,(O,B) 02 04 6
6 1 B 2 B 3 1 3 2 B 1 2 3 B 1 2 3 1 B 2 3 B 1 2 3 B 1 2 3 B 1 2 3 Coriolis Acceleration Coriolis acc. occurs or not? Vector graphical analysis method 12 Given linkage configuration & 2. Find 3 and 4 We know VA and direction of VB and VBA (perpendicular to AB) Draw vector triangle. V=r. VBA VB VBA Direction VBA VB VB Direction VA 3 4 4 BA B V AB V OB EX1-Velocity
EX1-Velocity Vector graphical analysis method 入 After finding @3 and 4,find Vc ©Vc=Va+VcA Recall that o3 was in the opposite direction as 2 Double Scale 03 02 0 Vector graphical analysis method Vc=VA+VCA =VB+VCB Mag:？√？√？ A B Dire:?√⊥CA√⊥CB a Aabc is called Velocity image of the link ABC 7
7 EX1-Velocity 13 After finding 3 and 4, find VC VC=VA+VCA Recall that 3 was in the opposite direction as 2 VCA VC VA Double Scale VCA VC Vector graphical analysis method a p A C B Vector graphical analysis method EX2 b c VC＝VA+VCA＝VB+VCB Mag： ? √ ? √ ? Dire： ? √ ⊥CA √ ⊥CB abc is called Velocity image of the link ABC
Vector graphical analysis method Characteristics of Velocity polygon: 1)Velocity of p equals zero.Velocities from p are absolute velocity. A 2)Edges of velocity image triangle are B velocity differences. 3)Velocity image triangle is similar to the link a ABC.Each point of the link has an image in Velocity image triangle. p Vector graphical analysis method EX3-Acc 7入 Knowns:and direction of ae aB=aa+前BA+aA Mag:?√o2aa? A Dire:√√B→A⊥BA Select a scale ua m/s2/mm, p' Draw p'a'from p',let a=uap'a' then:ae=μapb' b aBA=uab”b' direction:b”-→b' b” a agA=μab'a'Dire: a'-b' 8
8 a p A C B Vector graphical analysis method b c Characteristics of Velocity polygon: 1) Velocity of p equals zero. Velocities from p are absolute velocity. 2) Edges of velocity image triangle are velocity differences. 3) Velocity image triangle is similar to the link ABC. Each point of the link has an image in Velocity image triangle. Vector graphical analysis method EX3-Acc b’ B A C then：aB＝μap’ b’ Select a scaleμa m/s2/mm， Draw p’a’ from p’ , let aA＝μap’ a’ b” Knowns： ωand direction of aB aB＝aA + an BA+ at BA at BA＝μab”b’ direction: b” → b’ aBA＝μab’ a’ Dire: a’ →b’ Mag： Dire： ? ⊥BA ? √ √ √ B→A ω2lAB aA aB a’ p’
For ac:ac=a+aca+a'cA unsolvable! Mag: 2ICA 7 Dire: ?√C→A⊥CA and:ac=as+ace+a'cB unsolvable! Mag: √o21cB Dire: √C→B⊥CB Combine the two eqns: ac=aA+anca+a'CA=aB+ance+a'ce √√？ We can find: b ao=μap'c'dire:p'→c' atcA=ac"c'dire:c"c' acB=μac'c"dire:c"→c' Angular Acc:a =a'BAl lA b"b'/AB Dire:CW aBA=(atBA)2+(amBA)2=IAB a2+4=aab acA=(atcA)2+(a"cA)2=ICAa2+4=uaac acB=(atCB)2+(a"CB)2 =ICBa2+4=abc then:a'b'/le=b'c'/lec=a'c/lcA △a'b'c'∽△ABC Denote p'a'b'c'as acc image b C b” C 9
9 aC＝aA + an CA+ at CA ＝ aB + an CB+ at CB and： aC＝ aB + an CB+ at CB unsolvable！ Combine the two eqns： For ac： aC＝aA + an CA+ at CA unsolvable！ We can find: at CA＝μac”’c’ at CB＝μac’c” dire：c”’ → c’ dire：c” → c’ dire：p’ → c’ ? ? √ √ ? √ √ ? √ √ √ √ √ √ B A C Mag： ? Dire： ? √ √ ω2lCA C→A ? ⊥CA Mag： ? Dire： ? √ √ ω2lCB C→B ? ⊥CB b’ b” a’ p’ c”’ c” c’ aC＝μap’ c’ Angular Acc:α＝at BA/ lAB then：a’b’/ lAB＝b’c’/ lBC＝ a’ c’/ lCA Denote p’a’b’c’ as acc image ∴ △a’b’c’∽△ABC aBA＝ (at BA)2+ (an BA)2 aCA＝ (at CA)2+ (an CA)2 aCB＝ (at CB)2+ (an CB)2 ＝μa b”b’ /μl AB Dire：CW b’ b” a’ p’ c”’ c” B c’ A C ＝lCA α2 +ω 4 ＝lCB α2 +ω 4 ＝lAB α2 +ω 4＝μaa’ b’ ＝μa a’ c’ ＝μa b’ c’ α
EX4 Vector graphical analysis method VB3=VB2+VB3B2 Mag: ?√？ Dire: √∥BC Direction of V2:b2→b3o3=μpb3/lCB b3 p 3 03 b2 C Find VBa? EX4 Vector graphical analysis method 7入N aB3=anB3+a'83=aB2+a9382 +akB3B2 Mag:？o23lc?ho2 ak B3B2 2WVB38203 Dire:?B→C√B→A∥BC Found: VB3B2 aB3=μapb3,a'B3B2=μakb3B→C A a3=aB3/lc=μab3”b3'/lgc 01 Bi 3 p 3 C Find Ags? b3 10
10 Vector graphical analysis method EX4 VB3＝VB2+VB3B2 Direction of VB3B2 : b2b3 ω3 = μvpb3 / lCB Mag： Dire： ? √ √ √ ? ∥BC p b2 b3 Find VB3? B 1 3 2 A C ω3 ω1 Vector graphical analysis method EX4 + ak aB3 = a B3B2 n B3+ at B3 = aB2+ ar B3B2 Find AB3? ak B3B2 VB3B2 Mag： Dire： ? ? ω2 3lBC B→C ? √ l1ω21 B→A ? ∥BC 2VB3B2ω3 √ b’ 2 k’ b’ 3 b” 3 p’ aB3 ＝μap’ b3 ’ , α3＝at B3 /lBC＝μab3’’b3’ /lBC ar B3B2 ＝μak’ b3 ’ B → C Found： B 1 3 2 A C ω3 ω1