燕山大学复试部分机械原理lecture

1、Chapter 3Kinematic Analysis of Mechanisms3.1 Tasks and Methods of Kinematic Analysis3.2 Velocity Analysis by the Method ofInstant Centres(瞬心)P D Fc r e a t e dw i t hp da l v e r s i o nChapter 3Kinematic Analysis of Mechanisms3.1 Tasks and Methods of Kinematic Analysis3.1.1 Tasks: to find position、

2、velocity、andacceleration(度) of driven links according toinput parameters(参数) of driving link(s) andkinematics dimensions of all links.(a)Position:to determine whether all links willinterfere() with each other,to find locus(轨迹)(b)Velocity: to calculate the stored kineticenergy(动能) or power(功率) P.(c )

3、 Acc.：To calculate the dynamic forces,P D Fc r e a t e dw i t hp da l v e r s i o n3.1.2 Methods：（1） Graphical method(图解法)(a) Geometrical(几何学的) method for position(b) Instant(瞬时的) centre method forvelocity(c) Vector(矢量) equation method*（2）Analytical(的) method(a) Closed-loop method*(b) Assur group me

4、thod（3）Experimental(实验) method*P D Fc r e a t e dw i t hp da l v e r s i o nAllkinematicsdimensions(LAD、LAB、LBC、LDC、LDE、ÐCDE、LEF) are known.Draw the kinematic diagram of the mechanismwhen q1=80°.C2BE3511FDA46P D Fc r e a t e dw i t hp da l v e r s i o n(1)Locate A and D according to LAD.C2

5、BE3511FDA46P D Fc r e a t e dw i t hp da l v e r s i o n(2)Locate B according to LAB and q1C2BE3511FDA46P D Fc r e a t e dw i t hp da l v e r s i o n(3)Draw two arcs with the points B and D as thecentres and LBC and LDC as the radii(半径复数). The intersection(交点) of the two arcs is point C.C2BE3511FDA4

6、6P D Fc r e a t e dw i t hp da l v e r s i o n(4) Locate point E according to LDE and ÐCDE.C2BE3511FDA46P D Fc r e a t e dw i t hp da l v e r s i o n(5)Draw an arc with the point E as the centreand LEF as the radius(半径). The intersection of the arc and the horizontal pathway(导路) is point F.C2BE

7、3511FDA46P D Fc r e a t e dw i t hp da l v e r s i o n3.2 Velocity Analysis by the Method of Instant Centres(瞬心)3.2.1 Definition of the Instant Centre(瞬心)1.a pair of coincident(重合) points, theabsolute(绝对) velocities of which are the same, in both magnitude(大小) anddirection.2.relative velocity is zer

8、o.V P23.Instantaneous(瞬时的)VP1centre of relative rotation,2or more briefly the instantP121centre, denoted(标为) as P12 or P21.P D Fc r e a t e dw i t hp da l v e r s i o nSince P12 is the instantaneous(瞬时的) centre of relative rotation, VA2A1AP.VA2A1A2P121P D Fc r e a t e dw i t hp da l v e r s i o nSup

9、pose that the positions of points A and B,the directions of VA2A1 and VB2B1 are known.The position of instant centre P12VA2A1Bis to be located.A2VB2B11P D Fc r e a t e dw i t hp da l v e r s i o nVA2A1AP. VB2B1BP.VA2A1BA2VB2B1P121P D Fc r e a t e dw i t hp da l v e r s i o n3.2.2 Number N of instant

10、 centres:N=K(K-1)/2Note:The frame is included in the number K.Classification(分类) of instant centres：(1) Absolute (绝对) instant centre: one of links is the frame. Its velocity is zero, but its acceleration may not be zero.(2) Relative(相对) instant centre: both links are moving links, velocity of which

11、may not be zero.P D Fc r e a t e dw i t hp da l v e r s i o n3.2.3 Location of the Instant Centre of TwoLinks Connected by a Kinematic Pair(1) Revolute pairIf two links 1 and 2 are connected by a revolute pair, the centre of the revolute pair is obviously(明显地) the instant centre P12 or P21.P1221P D

12、Fc r e a t e dw i t hp da l v e r s i o n(2) Pure-rolling pairThere is no slipping between the twocontacting points A1 and A2, i.e., VA1A2=VA2A1=0. Thus the point of contact A is the instant centre P12 or P21.2A2P121P121P D Fc r e a t e dw i t hp da l v e r s i o n(3) Sliding pairRelative translatio

13、n(平移) is equivalent (等价于) to relative rotation about a point located atinfi(无穷远) in either directionperpendicular(垂直的) to the guideway(导路).Therefore, their instant centre lies at infiineither direction perpendicular to the guideway.P12Attention：The commonnormal(公法线) may pass through any point !21P D

14、 Fc r e a t e dw i t hp da l v e r s i o n(4) Higher pair (rolling & sliding pair)21AP D Fc r e a t e dw i t hp da l v e r s i o n(4) Higher pair (rolling & sliding pair) The direction of relative velocities, VA1A2and VA2A1, between A1 and A2 must be along the common tangent(切线). Otherwise,

15、the two linkswill separate(分离) or interfere().VA1A22n1AnP D Fc r e a t e dw i t hp da l v e r s i o n(4) Higher pair (rolling & sliding pair) The direction of relative velocities, VA1A2and VA2A1, between A1 and A2 must be along the common tangent(切线). Otherwise, the two linkswill separate(分离) or

16、 interfere(So the instant centre P12).or P21 must lie somewhere on the common normal n-n through the point A of contact.VA1A2P12n21AnP D Fc r e a t e dw i t hp da l v e r s i o n(4) Higher pair (rolling & sliding pair)Attention:Instant centre is not located at thepoint of contact or infi！VA1A2P1

17、2n21AnP D Fc r e a t e dw i t hp da l v e r s i o n3.2.4 Theorem(定理) of Three Centres(Aronhold-Kennedy Theorem)321P D Fc r e a t e dw i t hp da l v e r s i o n3.2.4 Theorem(定理) of Three Centres(Aronhold-Kennedy Theorem)Suppose that A is the instant centre P12 of the links 1 and 2. B is the instant c

18、entre P13 of thelinks 1 and 3.Where is P23?Any three linkshave three instant centres: P12, P13, and P23. They must lie on a straight line.32A(P12 )B(P13 )1P D Fc r e a t e dw i t hp da l v e r s i o nLet us consider any point, e.g. point C,outside the line P12P13.3C2B(P13 )A(P12 )1P D Fc r e a t e d

19、w i t hp da l v e r s i o n AC.V C 2C1VC2C13C2B(P13 )A(P12 )1P D Fc r e a t e dw i t hp da l v e r s i o n AC. BC.= V C 2 -V C1.V C 2C1V C 3C1Since V C 2 = V C1+ V C 2C1 ,V C 2C1Similarly, V C3C1 = V C3-V C1.VC3C1VC2C13C2B(P13 )A(P12 )1P D Fc r e a t e dw i t hp da l v e r s i o n AC. BC.= V C 2 -V

20、C1.V C 2C1V C 3C1Since V C 2 = V C1+ V C 2C1 ,V C 2C1Similarly, V C3C1 = V C3 -V C1.Obviously, for any point C outside the line¹ V C3 .P12P13,V C 2C1 ¹ V C 3C1Therefore, V C2.Therefore, thepoint C can not be the instant centre P23 between the links 2 and 3.VC3C1VC2C13C2B(P13 )A(P12 )1P D F

21、c r e a t e dw i t hp da l v e r s i o n AC. BC.= V C 2 -V C1.V C 2C1V C 3C1Since V C 2 = V C1+ V C 2C1 ,V C 2C1Similarly, V C3C1 = V C3 -V C1.Obviously, for any point C outside the line¹ V C3 .P12P13,V C 2C1 ¹ V C 3C1Therefore, V C2.In other words,any point outside the straight line P12P1

22、3 cannot be instant centre P23 .VC3C1VC2C13C2B(P13 )A(P12 )1P D Fc r e a t e dw i t hp da l v e r s i o n3.2.5 Applications of Instant CentresEx.1. The angular velocity w1 of crank 1 is given. For the position shown,(1) locate all instant centres for the mechanismFC2B131DA4P D Fc r e a t e dw i t hp

23、 da l v e r s i o nLocate first the four instant centres of linksconnected by a kinematic pair, i.e.,P14、P12、P23 、and P34. Note: The common normal tomay pass through any point !P23P23FC2)B(1P3121D(P34)A(P14)4P D Fc r e a t e dw i t hp da l v e r s i o nP13 will lie on line P12P23.Since P23 is atinfi

24、perpendicular to the guiding bar 2, theline P12P23 passes through the point P12 and isperpendicular to BF.123P23FPPPC1213232P)B(11231D(P34)A(P14)P134P D Fc r e a t e dw i t hp da l v e r s i o nP13 will lie not only on line P12P23 , but alsoon line P14P34.123P23FPPPC1213232P)B(11214331D(P34)A(P14)P1

25、4P13P344P D Fc r e a t e dw i t hp da l v e r s i o nP13 will lie not only on line P12P23 , but alsoon line P14P34.The intersection E of the two lines P12P23 and P14P34 is the instant center P13.123P23FPPPC1213232P)B(11214331D(P34)A(P14)P14P13P34)E(P413P D Fc r e a t e dw i t hp da l v e r s i o nP

26、D Fc r e a t e dw i t hp d a l v e r s i o nP24 will lie on line P12P14.P24214P23FP12 P24 P14C23B(P12)11D(P34)A(P14)4E(P13 )P D Fc r e a t e dw i t hp d a l v e r s i o nP24 will not only lie on line P12P14 , but also on line P23P34. Line P23P34 passes through point P34 and is perpendicular to BF.2

27、1 4P23FP12 P24 P14C 22 3 43B(P12)11D(P34)A(P14)P23 P24 P344E(P13 )The intersection G of the two lines P12P14 andP23P34 is the instant center P24.G(P24)P23214FP12P24P14C2)B(1P3122341D(P34)A(P14)PPPE(P)423243413P D Fc r e a t e dw i t hp da l v e r s i o n(2) find the ratio w3 /w1.Take advantage of(利用

28、) the frame 4!P34、P13、P14 lie on a straight line.G(P24)P23FC2P)B(13121D(P34)A(P14)E(P13 )4P D Fc r e a t e dw i t hp da l v e r s i o n(2) find the ratio w3 /w1.Take advantage of(利用) frame 4!P34、P13、P14 lie on a straight line.w1LAE=w3LDE，G(P24)P23VE1=VE3,w3=w1LAE/LDE.(in ccw.FCounterClockwise)逆时针C23

29、)B(1P312VVE31E1Measure dimensionD(P34)A(P14)E(P13 )4graphically!P D Fc r e a t e dw i t hp da l v e r s i o n(3) find the velocity VF of point F on link 2G(P24)P23FC23P)B(1312VVE31E1D(P34)A(P14)E(P13 )4P D Fc r e a t e dw i t hp da l v e r s i o n(3) find the velocity VF of point F on link 2Since li

30、nks 2 and 3 are connected by a sliding pair, w2 ? w3. (both in CCW)G(P24)P232FC23)B(1P312VVE31E1D(P34)A(P14)E(P13 )4P D Fc r e a t e dw i t hp da l v e r s i o n(3) find the velocity VF of point F on link 2Since the link 2 rotates relative to the frame 4 about the point P24(G) at this instant, VF= w

31、2LGF.VFGFG(P24)P232V F3FC2P)B(1312VVE31E1D(P34)A(P14)E(P13 )4P D Fc r e a t e dw i t hp da l v e r s i o nIn this mechanism，P34、P14、and P24 areabsolute instant centres， P13、P12、and P23 arerelative instant centres. Although the velocity of point P24 is zero, its acceleration is not zero.G(P24)P232V F

32、3FC2P)B(1312VVE31E1D(P34)A(P14)E(P13 )4P D Fc r e a t e dw i t hp da l v e r s i o nTo find wj if wi is known, we should takeadvantage of(利用) the frame f !We get Pif、Pij、and Pfj.Since B is the instant centre oflinks i and j, VBi= wi LAB= VBj= w j LCB。Note：iffis not the frame，thenA(Pif ) is not the a

33、bsolute instant centre of linki，VBi ¹ w i Ljif_VBi = VBjA(Pif) ia l v e r sB(Pij)C(Pfj)jP D Fc r e a t e dw i t hp di o nEx.3-2In the cam mechanism with translating(平动的) roller(滚子) follower(从动件), the cam is a circular disk.Supposing that the angular velocity w1 of the cam is known, the velocity

34、 V2 of the follower 2 is to be found for the position shown.321OA1P D Fc r e a t e dw i t hp da l v e r s i o nP D Fc r e a t e dw i t hp da l v e r s i o n3The roller has apassive DOF(局部自由度).2The velocity of the follower will not change if the roller is welded(焊接) to the follower 2.1OA1The instant

35、centreP12 must lie somewhere along their common normal n-n through the point3n2ofcontact C.C1OBA)(P121nP D Fc r ea t e dw i t hp da lve r s i o nP D Fc r e a t e dw i t hp da l v e r s i o nP12 must also lie on2313the straight lineconnecting P13 and P23.2nThe intersection B ofP23 P12 P13the two line

36、s P23P13and n-n is the instantcenter P12.CNote: Neither the1centre O of the circleO ( P )nor the contact pointPB13C is the instant centre23(P12 ) A1P12. n1323Since the follower 2is translating(平动), all points on the follower 2 have the same velocity V2.n2P23P12P13C1O( P13)BAP23)(P121nP D Fc r e a t

37、e dw i t hp da l v er s i o n1323Since the follower 2is translating(平动), all points on the follower 2 have the same velocity V2.n2P23P12P13CVV2B1So V2 =VB2 =VB1=w1*LAB.1O( P13)BAP23)(P121nP D Fc r e a t e dw i t hp da l v er s i o nExample 3-3 这是作业中的一Gear 3 rolls on the fixed rack(齿条) 4 without slip

38、ping(打滑). Assuming the velocity V1 of slider(滑块) 1 is known, the velocity VD of the centre D of the gear 3 is to be found.GV 11F2BD3ACP D Fc r e a t e dw i t hp da l v e r s i o nLocate first P12、P23、 P34、and P14. Thegear 3 rolls on the fixed rack 4 without slipping. So the contact point C is their instant centre P34.P14(P23)2GV 11F (P12 )BD3AC (P)a l v e r s i o nP D Fc r e a t e dw i t hp dP13 must lie on line P23P12.1P13P23P14P12GD3P13(P23)2V 11F (P12 )BAC (P)a l v e r s i o nP D Fc r e a t e dw i t hp dP13 must also lie on line P14P34.P34P14 passes through C and is perpen