Chapter 2: Structure Analysis of Mechanisms

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1 Chapter 2: Structure Analysis of Mechanisms
(第二章 机构的结构分析)

2 Contents (内容)： Goals (机构结构分析的内容及目的)
Contents and goals of structural analysis of mechanisms (机构结构分析的内容及目的) Contents (内容)： 1）Composition of Mechanism（机构的组成） 2）Conditions for a mechanism to have a determined motion(机构具有确定运动的条件) 3）Degree of freedom(机构的自由度) 4）Composition principle of mechanisms(机构的组成 原理) Goals (目的)： To study conditions for a mechanism to move and to have a determined motion (研究机构在何种条件下可动，具备何种条件时具有确定的相对运动).

3 2-1 Composition of mechanism
(机构的组成 ) 1 Links(构件) Unitbody can move independently(独立运动的单元体). Consisting of one or more elements(由单个或多个零件构成). 2 Degree of freedom of links(DOF,构件的自由度)： The number of independent parameters in a mechanism with respect to the frame (构件所具有的独立运动的数目). A xA yA zA X Y Z O S A X Y O yA xA S

4 3 Constraint(约束)： 4 Kinematic pairs 1 2 (a) (b) n (c) Y X O
To impose restrictions on independent motion (对独立运动的限制). Constraint number=lost DOF number(构件失去的自由度) 4 Kinematic pairs (运动副)： Two links contact directly and have relative motion connection(两个构件直接接触，并能产生 一定相对运动的连接). 1 2 X Y (a) O (b) n (c)

5 Types(运动副分类)： 1）For the type of constraints(按约束类型分)： 引入n个约束的运动 副称为n级副。

6 2）For the contact type between links
(根据构件间接触形式分)： 2 1 A higher pair(高副)： Two pair-elements of a kinematic pair have a point or a line contact(点、线接触的 运动副).

7 Two pair-elements of a kinematic pair have a surface contact(面接触的运动副).
1 2 A lower pair(低副)： Two pair-elements of a kinematic pair have a surface contact(面接触的运动副).

8 2 1 1 2 A lower pair(低副)： (a) (b) A revolute pair(转动副)： Y DOF=1
X Y o (a) A revolute pair(转动副)： DOF=1 constraint number : in plane 2 on space 5 A sliding/prismatic pair(移动副)： DOF=1 constraint number : in plane 2 on space 5 1 2 X Y o (b)

9 DOF=2 constraint number： A planar higher pair (平面高副)： A A gear pair
in plane 1 on space 4 t n 1 2 V12 A (c) A gear pair （齿轮副）： in Plane ： A lower pair：DOF=1， constraint number =2 A higher pair：DOF=2， constraint number =1

10 5 Kinematic chain(运动链)：
When a number of links are connected by means of kinematic pairs, the resulting mobile system is a kinematic chain(构件通过运动副联接而构成的相对可动的系统称为运动链). 1）Open chains(开式链)： One or more links in a kinematic chain have only one pair-element, the kinematic chain will be an open chain. 2）Closed chains(闭式链)： Every link in a kinematic chain has at least two pair-elements, all links form a closed chain.

11 6 Mechanisms(机构)： A kinematic chain in which at least one link has been “grounded”,or attached, to the frame of regerence (which itself may be in motion) and other links have a determined motion (在运动链中，固定某一构件为机架，且其它构件都具有确定运动，这种运动链称为机构). Type of links in mechanisms(机构中构件的种类)： 1）frame(机架) 2）driving links(原动件) 3）driven links(从动件)

12 Composition of mechanism(机构的组成)：
Rigid connection (刚性连接) Kinematic pair connection (运动副连接) Element (零件) Links (构件) Kinematic chain (运动链) Fix a link (固 定 一 个 构 件) A mechanism (机构)

13 2.2 The simple kinematic diagram of mechanism
(机构运动简图 ) Concept A simple kinematic diagram is to omit complicated shapes of a link and indicate the relative motion of all links. In the kinematic diagram, links and kinematic pairs should be represented by simple and specified symbols in proportion. (抛开构件的复杂外形，按一定比例用简单 线条表示构件，用规定符号表示运动副，能表明各构件间相对运动关系的简单图形)。

14 Representation of a kinematic pair and a link (运动副及构件的表示方法)
2 1 A sliding pair (移动副) 1 2 A revolute pair (转动副)

15 A Cam pair (凸轮副) t n 1 2 A A gear pair (齿轮副)

16 2.Links(构件) A fixed link (固定件) The same link (同一构件)

17 Two-pairs link (两副构件) Three-pairs link (三副构件)

18 2.2.2 Procedures for drawing the kinematic diagram of a mechanism(机构运动简图的画法)
1)Run the mechanism slowly, study carefully the structure of the mechanism. Determine the types of all kinematic pairs. Analyze the transmission route from the driving link to the driven link and determine the types of all kinematic pairs. Then stop the mechanism at a suitable position for drawing. Steps： 2)A plane paralleled to these planes is chosen as a drawing plane. Sometimes, a local view may be drawn to clarify the structure.

19 3)Select a suitable scale μl and then draw the schematic diagram of the mechanism. Firstly,draw all fixed pair elements or the pair elements on the frame. Be careful about the relative positions between these fixed pair elements. Then begin to draw the moving links(draw the drivers first and then the driven links according to the route of motion tranmission). 4)Calculate the degree of freedom of the mechanism according to the schematic diagram, and check if it is the same as that of the actual mechanism.

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21 画出图示机构简图

22 2.3 Degree of Freedom of planar mechanisms
(平面机构自由度的计算) 2.3.1 Formula of Degree of Freedom of planar mechanisms (平面机构自由度的计算公式) The number of independent parameters needed to define uniquely the locations of all moving links within a mechanism with respect to the frame is called degree of freedom of a mechanism(机构相对于机架所具有的独立运动的数目). If a mechanism contains K links and n=K-1 moving links will have a total DOF of 3n. Since each planar lower pair eliminates 2 DOF and each planar higher pair eliminates 1 DOF, the DOF of a planar mechanism with n moving links, PL lower pairs, PH higher pairs can be calculated as F=3n-2PL-PH

23 2.3.2 Conditions for a mechanism to have a determined motion(机构具有确定运动的条件)
If F=0，it is a truss and can’t move； If F>0，it can move . Conditions for a mechanism to move：F>0 If the number of driving links is less(greater) than the DOF of the mechanism, some driven links will not have determined motion. If F=m，there must be m driving links to ensure other links having a determined motion.

24 Example 1 n=3, PL=4, PH=0, DOF=Number of the driving links
Condition for a mechanism to have a determined motion DOF=Number of the driving links Calculate the DOF of the following mechanisms: Example 1 Solution： D C A B n=3, PL=4, PH=0, 4 3 1 2

25 n=4, PL=5, PH=0, n=3, PL=4, PH=0, Example 2 Example 3 Solution： C 3 D
B E n=4, PL=5, PH=0, 4 3 1 2 5 Example 3 Solution： C A B n=3, PL=4, PH=0, 1 2 3 4

26 Kinematic diagram---运动简图
1.Words: Link---构件 Kinematic diagram---运动简图 Lower pair---低副 Revolute pair --转动副 Sliding pair --移动副 Higher pair---高副

27 F= Number of driving links
2.Formula of DOF of a planar mechanism (平面机构自由度的计算公式) F=3n-2PL-PH 3.Condition for a mechanism to have a determined motion（机构具有确定运动的条件） F= Number of driving links

28 Example1：Calculate DOF of the four-bar linkage(四杆机构).
2.4 Calculation of DOF and Points for Attention （机构自由度的计算及注意事项） 2.4.1 Calculation of DOF Example1：Calculate DOF of the four-bar linkage(四杆机构). Solution： n=3, PL=4, PH=0

29 Example2: Calculate DOF of the disc saw（圆盘锯）.
B F 1 2 5 4 3 6 8 7 Solution： n=7, PL=6, PH=0

30 2.4.2 Points for Attention 1. Compound hinge(复合铰链) 两个以上构件在同一处以转动副形式连接。
More than two links are hinged at one point by revolute pairs. 两个以上构件在同一处以转动副形式连接。 If m links are hinged at one point, the number of the revolute pairs is m-1. 2 revolute pairs 3 revolute pairs

31 Example2: Calculate DOF of the disc saw.
n=7, Solution： A E D C B F 1 2 5 4 3 6 8 7 PL=10, PH=0

32 机构中构件所具有的与整个机构运动无关的自由度。
Example 3: Calculate DOF of the cam mechanism（凸轮机构）. Solution： roller (滚子) follower (从动件) cam (凸轮) w 2 4 3 n=3, PL=3, PH=1 1 2. Passive DOF(局部自由度) The DOF of links within a mechanism has no effect on the whole mechanism’s motion. 机构中构件所具有的与整个机构运动无关的自由度。

33 w 1 2 3 4 1 2 3 w We can delete the Passive DOF by welding the roller to the follower before the calculation of DOF of a mechanism. n=2, PL=2, PH=1

34 3. Redundant constraints(虚约束）
Notes ：only at geometric center of the roller coincides with its turning center, the DOF is passive DOF(只有在小滚子处，且小滚子几何中心与转动中心重合，才是局部自由度) No Yes 3. Redundant constraints(虚约束） Ineffective constraints for a mechanism motion (对机构的运动不起作用的约束)

35 = AB CD EF n=3, PL=4, PH=0, F=3×3－2×4－0=1 = n=4, PL=6, PH=0,
If 5 bar is added 4 3 1 C D B A E F 2 5 AB CD EF = = AB CD EF n=4, PL=6, PH=0, F=3×4－2×6－0=0

36 1）Track coincidence(轨迹重合)
When redundant constraints occur in a mechanism, they should not be counted during the calculation(机构中有虚约束时，应将虚约束去掉). Redundant constraints occur in many situations as described below： 1）Track coincidence(轨迹重合) geometrical condition(要注意几何条件，有时需证明) B D C A 4 3 2 1 AB=BC=BD AC⊥AD 5 n=4, PL=6, PH=0, So In fact, the mechanism can move.F=0 because of the redundant constraints.

37 F=3×3－2×4－0=1 需证明当D点滑块拿掉后，仍有AC⊥AD， 即 证明： 则 即 将D点滑块去掉后,n=3,PL=4,PH=0, D
5 B D C A 4 AB=BC=BD AC⊥AD 证明： 即 将D点滑块去掉后,n=3,PL=4,PH=0, 则 F=3×3－2×4－0=1

38 2）When two links are connected by more than one parallel sliding pair or more than one revolute pair whose axes coincide, only one of the sliding pair of revolute pair must be counted during the calculation. Others are redundant constraints and must not be counted(两构件组成多个移动副且导路平行，或组成多个转动副且轴线重合，则只考虑一处约束，其他看做虚约束) When two links are connected by more than one higher pair andcommon normal lines coincide, only one of the higher pairs can be counted during the calculation. Others are redundant constraints and should not be counted. (若两构件组成多个高副且公法线重合，则也只考虑一处约束)。As shown in fig. bellow

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40 3）Redundant constraints include symmetrical structure or duplicated structure which have no effect on mechanism motion(对机构运动不起作用的结构对称部分所带来的约束，也存在虚约束). 1 4 3 2 n=3, PL=3, PH=2, So

41 n=8, PL=11, PH=1, F=3×8－2×11－1=1 = 1) B D C AB CD EF
Example 1: Calculate the DOF of the mechanisms 1) C D B A E F AB CD EF = 1 5 4 3 2 6 7 8 9 P C R n=8, PL=11, PH=1, Solution： F=3×8－2×11－1=1

42 = n=6, PL=8, PH=1, F=3×6－2×8－1=1 2) AB CD A B C D 1 6 5 4 3 2 7 R P
Solution： F=3×6－2×8－1=1

43 3) 5 4 3 2 1 n=4, PL=4, PH=2, So Solution： F=3×4－2×4－2=2

44 n=6, PL=8, PH=1 1) n=5, PL=7, PH=1, So F=3×5－2×7－1=0
Example2: Try to judge mechanism below having determined motion or not. If not, how to revise? 1) n=6, PL=8, PH=1 A B C D E F G H 1 2 5 6 4 3 n=5, PL=7, PH=1, So Solution： F=3×5－2×7－1=0 The mechanism can’t move.

45 n=6, PL=8, PH=1 n=6, PL=8, PH=1， So F=3×6－2×8－1=1
If point C is revised as it is shown below n=6, PL=8, PH=1 A B C D E F G 1 2 5 6 4 3 7 H n=6, PL=8, PH=1， So F=3×6－2×8－1=1

46 2) n=3, PL=4, PH=1， F=3×3－2×4－1=0 So n=4, PL=5, PH=1， F=3×4－2×5－1=1
Solution： n=3, PL=4, PH=1， A D C B F E 4 3 2 1 F=3×3－2×4－1=0 The mechanism can’t move. point D is revised as it is shown on the left diagram A D C B F E 5 4 3 2 1 So n=4, PL=5, PH=1， F=3×4－2×5－1=1

47 2.4.1Substitute Equivalent Linkage Instead of Higher Pair(高副低代)
2.4 Lower pair replacing higher pair，structure analysis and composing principle of planar mechanism (平面机构的高副低代、结构分析与组成原理) 2.4.1Substitute Equivalent Linkage Instead of Higher Pair(高副低代) 1.Concept: On certain conditions, higher pairs in the mechanism are equivalent substituted by virtual lower pairs, this method is called Substitute Equivalent Linkage Instead of Higher Pair (将机构中的高副根据一定的条 件用虚拟的低副来等效替代，这种方法为高副低代). 2.Conditions: The number of DOF and instantaneous motion between substitution and post-substitution must be same( 代替前后机构的自由度数和瞬时运动必相同). 3.Means: A higher pair can be substituted by a mechanism and two lower pairs(用一个构件两个低副代替一个高副).

48 2.4.2 Structural analysis and the composition principle of planar mechanisms(平面机构的结构分析及机构组成原理)
1. Structural analysis of planar mechanisms Basic group(杆组) If the DOF of each group is zero and no group can be divided further into two or more zero-DOF groups, then such groups are called Basic groups, shorter from groups (不可再拆的自由度为零的构件组称为基本杆组，简称杆组).

49 F=3n-2PL 3 2 6 4 … Grade Ⅱ group( Ⅱ级杆组) Grade Ⅲ group(Ⅲ级杆组)
Grade Ⅱ group( Ⅱ级杆组) Grade Ⅲ group(Ⅲ级杆组) … Grade Ⅱ group( Ⅱ级杆组) Grade Ⅲ group(Ⅲ级杆组) The grade of a mechanism is defined as the highest grade of the group in the mechanism(机构的级别由机构中最高杆组的级别来定).

50 5 4 5 1 4 3 6 3 2 2 6 2 4 1 3 1 5 6 2.Structural analysis(平面机构的结构分析)
1）1 link as driver 2 1 6 5 4 3 Grade Ⅲ mechanism 2）5 link as driver 2 1 6 5 4 3 Grade Ⅱ mechanism 3.Composition principle (组成原理) A mechanism=the frame+drivers+groups