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1、The Theory of Machines and Mechanisms Chapter 1. Structural Analysis Main Topics 6.Composition principle of mechanism 机构的组组成原理 6.Replace higher pairs with lower pairs 高副低代 6.Structural analysis 机构的结结构分析 6.Composition principle of mechanism Any mechanism which has a determined motion can be divided i
2、nto two parts: 1.The first part consists of the frame, the driver(s) and the kinematic pair(s) between the frame and driver(s); 2.The second part is the groups of driven links, whose DOF are 0 (called Assur groups) , they connect to each other by kinematic pairs. =+ Assur Group - Assur Group is a ki
3、nd of link group whose DOF is zero and no group can be divided further into smaller 0-DOF groups. Such groups are called Assur Groups in memory of Assurs contribution to this subject. Grade III Assur group Grade II Assur group The grade of mechanism 1.一个机构由几种不同级别的Assur杆组组成，机构的级别取 决于杆组的最高级别； 2.如果组成机构
4、的杆组的最高级别是II级，则机构就是二级机 构,如果杆组的最高级别是III级，则机构就是三级机构； 3.只由主动件和机架构成的机构是一级机构，例如杠杆机构； 4.同一个运动连，当取不同的构件为主动件时，得到的机构的 级别也有可能不同。 For example: 1 is driver: grade III mechanism 7 is driver: grade II mechanism 1 2 3 4 5 6 7 铰链夹紧机构 7.Replace higher pairs with lower pairs (高副低代) Conditions of replacement 1.替代前后机构的自
5、由度不能改变; 2.机构替代前后其瞬时的运动特性 (velocity, acceleration) 保持不变 Methods of replacement 1.Finding the center of the curve at the contact point (找到接触点处构件轮廓线的曲率中心); 2.Draw two circles as revolute pairs at the points and draw a line to connect them (画两个圆作为转 动副，然后用线段将他们连接起来) Replacement types: 1.Permanent replace
6、ment(永久替代): The kinematic properties before and after replacement are completely same 2.Temporary replacement(瞬时替代) Example 1: Both of the links are circles Example 2：One is a circle, the other is a point. Example 3: One is a line, the other is a circle. O1 A O3 1 2 4 O1 A O3 1 2 3 O1 A O3 1 2 3 4 包
7、容面与被包容面对 调 高副低代 2.Temporary replacement（瞬时替代）: 构成高副的两个轮廓是非圆曲线，各处的曲率中心 位置不恒定，引入的杆件长度出处不同； 每一时刻的替代结果不同。 Position 1Position 2 8.Structural analysis - Steps of structural analysis: 1. Delete all redundant constraints and replace higher pairs with lower pairs, 3. Calculate the mechanism DOF correctly and
8、 find the driver. F=3*9-2*13=27-26=1 高副 局部自由度 2. Redraw the kinematic diagram of the mechanism. 3.Disassemble grade II Assur mechanisms begin from the furthest links to the drivers one by one. 4.Every time one Assur mechanism is disassembled, the DOF of the rest mechanism should be equal to its orig
9、inal value. Otherwise, the Assur mechanism is wrong. 12 3 4 5 6 7 8 9 1 9 2 3 4 5 7 8 6 6.If there is no grade II Assur mechanism could be found, then try to disassemble grade III Assur mechanisms as step3 and step4. Until only the driver exist. 5.The grade of the mechanism is equal to the highest g
10、rade of Assur mechanisms disassembled. 二级机构 12 3 4 5 6 7 8 9 9 2 3 4 5 7 1 8 6 1.2 平面连杆机构的结构分析 练习: 1.计算机构的自由度 2.高副低代 3.进行结构组成分析 1.2 平面连杆机构的结构分析 练习: 1.计算机构的自由度 2.高副低代 3.进行结构组成分析 Homework Page 32: 2-23, 2-24 +补充作业 Hello! Anybody home? 补充作业 1.2 平面连杆机构的结构分析 1.计算机构的自由度 2.高副低代 3.进行结构组成分析 补充作业 1.计算机构的自由度 2.高副低代 3.进行结构组成分析
