上海交通大学：《Design and manufacturing（ME371、ME337）》课程教学资源（讲义）Lecture 7 Kinematic Analysis of Mechanisms

e+口 LECTURE 7 ME371/ME337 DESIGN AND MANUFACTURING Kinematic Analysis of Mechanisms Covered:Ch7-in Design of Machinery OUTLINE Some important definitions Vector graphical analysis method Complex vector analytical method 1

1 Kinematic Analysis of Mechanisms LECTURE 7 ME371/ME337 DESIGN AND MANUFACTURING Covered：Ch7- in Design of Machinery OUTLINE Some important definitions 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 yImaginary axis Real axis Some important definitions Velocity-Rate of change of displacement is velocity.Velocity can be linear velocity or angular velocity. First order: d(Re)=ke+R(ej0)-keRoje d de Angular Velocity:@ dt yAImaginary axis Linear Velocity:V= dR R R dt Real axis R 2

2 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 or 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)=ieP+dje+(d+0jeP+RaUej0 d =Re+20e°+iRie-R0ea Angular Acc: a-0-do yAImaginary axis dv R Linear Acc: a== dt Real axis Some important definitions First order: d(Re)=Re+R (ej0)-ke+Roje dt Second order: (R e)=(R-RO)e+20R+Ry d 20Rj eo Tangential Acceleration: perpendicular to the radius of Rj e rotation. Normal Acceleration: R centripetal component at 180 to -RO-en the angle of the original pisiton vector 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 X Y O R  j j e j  Re   j R j e    j e  j Re   2 j R e     2 j Rj e     j Rj e    P 2 2 2 Second order: ( ) ( ) +(2 )        jj j      d Re R R e R Rje dt First order: ( ) ( )= d j jj j j R e Re R e j Re R j e dt             Some important definitions Tangential Acceleration: perpendicular to the radius of rotation. Normal Acceleration: centripetal component at 180 to the angle of the original pisiton vector

Simple cases study A link in pure rotation +0 R孰 Displac- ement R =pero V PA Velocity Vra=poje 1 Ars=pajelo-po'ero +2 Acceleration =A'm+A"m 005 APA Simple cases study ©When point A is moving Displac- Rp=R+R24 RPA ement 可。=可+4 Velocity T+peo（jo) o Graphical solution: VA 4

4 A link in pure rotation Simple cases study Displac￾ement 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 Displac￾ement Velocity RRR P A PA     P A PA j A V V V V pe j           Graphical solution:

Simple cases study +2 When point A is moving APA 02 Rp=R+Rm4 A Displac- ement N PA 可。= 可+了 Velocity V+peio (jo APA Accelerati a。=A,+44 APA on =A-o'per+japer AA Simple cases study-Coriolis Acceleration Position of slider 02 Rp=pelo 2 Velocity of slider V,=De"ig+pe Transmission Slip -X velocity velocity Acceleration: A,=pei0+pe（(io）'+peia+e°+eio Combining terms A,=[p-p0）+i(pa+220]e Coriolis acc.occurs when a body has vasp and w Slip Normal Tangential Coriolis 5

5 When point A is moving Simple cases study Displac￾ement Velocity Accelerati on RRR P A PA     P A PA j A V V V V pe j           2             P A PA j j A AAA A pe j 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 Simple cases study——

Coriolis Acceleration Coriolis acc.occurs or not? 2 B 1/ 3 B B OUTLINE © Some important definitions Vector graphical analysis method Complex vector analytical method 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? (a) (b) (c) (d) OUTLINE Some important definitions Vector graphical analysis method Complex vector analytical method

Velocity Vector graphical analysis method Given linkage configuration and 2.Find @3 and 4 国 We know V and direction of VB and VBA (perpendicular to AB) ® Draw vector triangle.VB VA+VBA 03 VBA Direction Va Direction Va=@;(AB) 02 Vg=@,(O,B). 01 0 Velocity Vector graphical analysis method After finding a and 4,find Vc © VC=VA+VCA Recall that @3 was in the opposite direction as @2 Double Scale 02 7

7 Vector graphical analysis method 13 Given linkage configuration and 2. Find 3 and 4 We know VA and direction of VB and VBA (perpendicular to AB) Draw vector triangle. VB = VA + VBA VBA VB VBA Direction VBA VB VB Direction VA     3 4 4 BA B V AB V OB     Velocity 14 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 Velocity Vector graphical analysis method

Acceleration Vector graphical analysis method Given linkage configuration a2 and 2,3,4.Find AB Ag=(AO2@3A4=(A02)2 AB=AA+ABA ACA 3 03 CA ABA 02 AA Acceleration Vector graphical analysis method Given linkage configuration azand 2,3,4.Find Ac A=(AO2)0 A=(AO2)02 AC=AA+ACA 03 03 02 04 8

8 Vector graphical analysis method Given linkage configuration α2 and 2, 3 , 4, Find AB Acceleration Vector graphical analysis method Given linkage configuration α2 and 2, 3 , 4, Find AC Acceleration

Velocity Vector graphical analysis method The fourbar Crank-slider Ve=VA+VBA VC=VA+VCA Mag:?√？ Dire:√√⊥BA a Aabc is called Velocity image of the link ABC Velocity Vector graphical analysis method Characteristics of Velocity polygon: >Velocity of p equals zero.Velocities from p are absolute velocity. >Edges of velocity image triangle are velocity differences. >Velocity image triangle is similar to a the link ABC.Each point of the link has an image in Velocity image triangle. 9

9 a p A C B Vector graphical analysis method b c VB＝VA+VBA VC＝VA+VCA Mag： ? √ ? Dire：√ √ ⊥BA abc is called Velocity image of the link ABC The fourbar Crank-slider Velocity a p A C B Vector graphical analysis method b c Characteristics of Velocity polygon: Velocity of p equals zero. Velocities from p are absolute velocity. Edges of velocity image triangle are velocity differences. Velocity image triangle is similar to the link ABC. Each point of the link has an image in Velocity image triangle. Velocity

Acceleration Vector graphical analysis method Knowns:and direction of ae ag=aA+a"Ba+a'BA Mag:?√o2laB? A Dire:√√B-→A⊥BA a Select a scale ua m/s2/mm, 0 Draw p'a'from p',let a=uap'a then:ae=μapb' b aBA=μab"b' direction:b”+b' b a aBA=μab'a' Dire: a'→b For ac:ac=a+a"ca+a'cA unsolvable! Mag:?√o2ca-7 Dire:？√C→A⊥CA and:ac=as+a"ce+a'cB unsolvable! Mag:？√o2lcB Dire:？√C→B⊥CB Combine the two eqs: ac=aA+ancA+acA=ae+ance+acB ?√√？ We can find: ac=μap'c'dire:p'→c3 aca=μac"c'dire:c+c acs=μac'c”dire:c→c3 10

10 Vector graphical analysis method 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’ Acceleration aC＝aA + an CA+ at CA ＝ aB + an CB+ at CB and： aC＝ aB + an CB+ at CB unsolvable！ Combine the two eqs： 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’