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1、 REFERENCE ANSWER for EXERCISE BOOK of Mechanisms and Machine Theory TRG of Machinery Theory and Design College of Mechanical Engineering Fuzhou University 2003 Name Class Student No. Date 2-1 Draw the kinematic diagrams of the mechanisms shown in Fig2-1. 4scale 3:11BC23A42A3C1BFig2-1(a) 11223344ABC
2、ABCFig2-1(b) Name Class Student No. Date 2-2 Draw the kinematic diagrams of the mechanisms shown in Fig2-2. 12334 45566ABC DEA BCDEFF21Fig2-2(a) Name Class Student No. Date 2-3 Draw the kinematic diagram of the mechanism shown in Fig2-3. 8I7B65HA1G2C3E4 DF27 8I156HAGFC B3ED4Fig2-3 Name Class Student
3、 No. Date 2-4 Calculate the degree of freedom of the mechanisms shown in Fig2-4,and point out what should be paid attention to during the calculation . 解： (a) 左右为对称结构，设左侧为虚约 束。 (b) E 为杆 4、 5、 6 的复合铰链。 (c) 滑 块 7 与机架 8 间为移动副。 F=3n-2PL-Ph=37-210=1 解： (1)红线内的构件为重复结构，构成虚约束。 (2)去掉以上构件后，C 仍为构件 2、3、4 的复合铰链。
4、 (3)滑块 5 与机架 6 之间为移动副。 F= 3n-2PL-Ph =3527=1 ABCDEFGHIJRedundnt constraint23456 78Fig2-4(a) 1BCDEFGHIJKRedundant constraint A23456Fig2-4(b) Name Class Student No. Date 2-5 Calculate the degree of freedom of the mechanisms shown in Fig2-5,and point out what should be paid attention to during the calcu
5、lation . 解： (a) 两个滚子有局部自由度。 (b) 滚子 D 与凸轮 1 之间只能算一 个高副。 F=3n-2PL-Ph =37 29 -2=1 解： (1)杆件 BC 与齿轮 2 焊接在一起。 (2) A 为齿轮 4、杆件 1 和机架 5 的复合 铰链。 F= 3n-2PL-Ph =34 25 -1 =1 常见错误：认为 B 是复合铰链，而不认为 A 是复合铰链。 ABCDEFGLMNO12345678Fig2-5(a) BDA5123 4Fig2-5(b) Name Class Student No. Date 2-6 Calculate the degree of freed
6、om of the mechanisms shown in Fig2-6,and point out what should be paid attention to during the calculation . 解： (a) C 为构件 2、3、4 的复合铰链。 (b) C 处有两个转动副和两个移动 副。 E 处有一个转动副和两个移 动副。 F= 3n-2PL-Ph =37 210=1 注意：E 不是复合铰链！ 解： 当构件尺寸任意时， 构件 2 作平面复杂运动， 而杆 4 与机架间组成移动副， 所以杆 4 仅作平动。因此，构件 2 和构件 4 之间有相对转动。因此，应该有构 件 6，并
7、且构件 4 和 6 之间有转动副，如右图所示。 当 ABCD 且 BCAD 时，杆 2 仅作平动。杆 4 与机架间组成移动副， 所以杆 4 也仅作平动。这样，构件 2 和构件 4 之间就没有相对转动，只有相对 移动。即：构件 4 和构件 6 之间就没有相对转动了，因此，可将构件 6 与构件 4 焊接起来（去掉构件 6） ，如左图所示。 然而，在计算机构自由度时，应该按一般尺寸情况下进行分析，即：应该 按照右图情况来分析机构的自由度。 F= 3n-2PL-Ph =35 27=1 ABCDE1236784Fig2-6(a) ABCDAB=CD BC=ADABCD12345123456Fig2-6(
8、b) Name Class Student No. Date 2-7 The kinematic diagram of an engine mechanism is given in Fig2-7. (1) Calculate the degree of freedom of the mechanism, and point out what should be paid attention to during the calculation. (2) Make the structural analysis for the mechanism. (3) Make the structural
9、 analysis for the mechanism when link EFG is regarded as the driver. Note: During structural analysis, list the assembly order of Assur groups, the type of group, the grade of group, the grade of the mechanism, the link serial numbers, the inner pair and the outer pairs of each group in each mechani
10、sm. 解： （1） F= 3n-2PL-Ph =37 210=1 （2） 当 AB 为原动件时， （3）当 EFG 为原动件时， ABCDEFGH12346578第一杆组RRP RRR2,3 4,5内副E外副B,移C F,D第二杆组类型杆号第三杆组RRP6,7 转H6-7G,移H转C2-33-87-8ABCDEFGH12346578III级杆组 RRP1,2,3,4 6,7内副外副A,E,移C类型杆号 B,D,转C3-8 转HG,移H7-8Name Class Student No. Date 2-8 Make the structural analysis for the mechanis
11、m shown in Fig2-8. (a) When link 1 is regarded as the driver. (b) When link 5 is regarded as the driver. Note: During structural analysis, list the assembly order of Assur groups, the type of group, the grade of group, the grade of the mechanism, the link serial numbers, the inner pair and the outer
12、 pairs of each group in each mechanism. 解： (a) 当 1 为原动件时 杆件 2, 3, 4 和 5 组成一个三级 Assur group. (b) 当 5 为原动件时 杆件 3 和 4 组成第一个 RPR Assur group. 杆件 1 和 2 组成第二个 RPR Assur group BC3 621 AD 45 EFig2-8 23 6A1CDB5 4E23 61 A CBD5 E4ABCDEFABCDEFABCgrade IIgrade IIIgrade IV23 6A1CDB5 4EBD3 6A1524CECADBD53B4A 6CENa
13、me Class Student No. Date 2-9 The schematic diagram of a punch machine designed by someone is shown in Fig2-9. This machine should be able to transform a continuous rotation of gear 1 into a translation of the punch 4. Can the machine work properly? If it cant ,please rectify it. 解：不能正常工作。 改正如图(或者改成
14、题目 2-3 构件 5、6、7 的连接) 1234 5Fig2-9 Name Class Student No. Date 2-10 The schematic diagram of a mechanism designed by someone is shown in Fig2-10. This mechanism should be able to transform a continuous rotation of link1 into an oscillation of link4. Can the mechanism work properly? If it cant ,please rectify it. 解：不能正常工作。 改正后 ADBCE12345Fig2-10 ADBCE124ABCEDABCName Class Student No. Date 3-1 Locate all instant centres of mechanisms in the position shown in Fig3-1. 213 4P14P12P23P34P13P24Fig3-1(a) 1234P12P24)P14P23P34P13Fig3-1(b) 12123P23 PP13OnnFig3-1(c) 2414P(2A 115(P)GB25(P)nP
