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1、精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.2-1 Draw the kinematic diagrams of the mechanisms shown below. 41 BA23scale 3:1B 21AC3C4- 4A1BA1B32C42C3名师归纳总结 - - - - - - -第 1 页,共 28 页精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.2-2 Draw the kinematic dia
2、gram of the mechanism shown below. 1D4AB2D34A1BCC235E56E名师归纳总结 6FF第 2 页,共 28 页- - - - - - -精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.2-3 Draw the kinematic diagram of the mechanism shown below. 6H5GFEA487I1BD3C2G F5H61AE8I4 D3名师归纳总结 7BC第 3 页,共 28 页2- - - - - - -精选学习资料 - -
3、- - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.2-4 Calculate the degree of freedom of the mechanisms shown below. Indicate all points for attention before the calculation of the DOF. 解：I RedundntA32BG5F8a 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； d 滑块C7 与机架 8 间为移动副；F=3n-2PL-Ph=3 7-2 10=1 constr
4、aintD47E 6HJ解：名师归纳总结 I RedundantHD21BA1红线内的构件为重复结构,构成第 4 页,共 28 页虚约束；32去掉以上构件后, C 仍为构件 2、constraintJGC 4 E63、4 的复合铰链；3滑块 5 与机架 6 之间为移动副；KFF= 3n-2PL-Ph =3 52 7=15- - - - - - -精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.2-5 Calculate the degree of freedom of the mechanisms sho
5、wn below. Indicate all points for attention before the calculation of the DOF. 解：3LFM2ONDC6a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；1F=3n-2PL-Ph =3 7 2 9 -2=1BG4E5A7A41B2D385解：1 Link BC is welded to gear 22 A is a compound hinge of gear 4、link 1 、and frame 5.F= 3n-2PL-Ph =3 4 2 5 -1 =1 名师归纳总结 常见错误：认为 B 是复合
6、铰链,而不认为A 是复合铰链；第 5 页,共 28 页- - - - - - -精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.2-6 Calculate the degree of freedom of the mechanisms shown below. Indicate all points for attention before the calculation of the DOF. 解：4CD2E6B1Aa C 为构件 2、3、4 的复合铰链；bC 处有两个转动副和两个移动副；8E 处有一个
7、转动副和两个移动副；3F= 3n-2PL-Ph =3 7 2 10=1留意：E 不是复合铰链！7解：1BAB=CD BC=AD453DC1B2453DC当构件尺寸任意时,构件2作平面复杂运动, 而杆 4 与机架26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AA有相对转动；因此,应当有构件6,并且构件 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
8、6 与构件 4 焊接起来（去掉构件 6）,如左图所示；然而,在运算机构自由度时,应当按一般尺寸情形下进行分析,即：应当按照右图情形来分析机构的自由度；F= 3n-2PL-Ph =3 5 2 7=1 名师归纳总结 - - - - - - -第 6 页,共 28 页精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.2-7 Shown below is the kinematic diagram of an engine mechanism. 1 Calculate the degree of freedom o
9、f the mechanism. Indicate all points for attention before the calculation of the DOF. 2 Carry out the structural analysis for the mechanism. 3 Carry out the structural analysis for the mechanism if link EFG is a driver. Note: During structural analysis, list the assembly order of Assur groups, the t
10、ype 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 mechanism. B DA 1 E 4 2 C38 B 5 F D 7A 1 E 4 G 6 2 H C3 解：（1） F= 3n-2PL-Ph =3 7 2 10=1 8 类型 杆号 5 F 内副 外副 7（2） 当 AB 为原动件时,第一杆组 RRP其次杆组 RRR 2,34,5 G 转C2-3E 6
11、B,移CF,D H 3-8第三杆组 RRP 6,7 转H6-7 G,移H 外副第一杆组 RRP 2,3 转C 2-3 B,移C 3-8其次杆组 RRR 4,5 E F,D第三杆组 RRP6,7 转H6-7 G,移H 7-8（3）当 EFG 为原动件时,名师归纳总结 类型杆号内副外副第 7 页,共 28 页A,E, 移C3-8III 级杆组1,2,3,4B,D, 转CRRP6,7转HG,移H7-8- - - - - - -精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆A8BE4D2C31F57名师归纳总结 类型G6H第 8 页,共 28 页杆号内副外副A,E, 移C
12、3-8III 级杆组 RRP1,2,3,4B,D, 转C6,7转HG,移H7-8- - - - - - -精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.2-8 Carry out the structural analysis for the mechanism a if link 1 is a driver. b if link 5 is a driver. Note: During structural analysis, list the assembly order of Assur group
13、s, 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 mechanism. 解：D 54 EB C21A3 C A6 grade IIa if link 1 is a BAECEABDFE5A1E4 BB2CA1D 3DFCCgrade IV 66grade IIIdriver,Links 2, 3, 4 and 5 constitute a gr
14、ade III Assur group.b if link 5 is a driver. Links 3 and 4 constitute the first RPR Assur group. Links 1 and 2 constitute the second RPR Assur group.A1DE54 B23C6A1B2DCA6DE5A6DE5第 9 页,共 28 页44BB3C3C名师归纳总结 - - - - - - -精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date 2-9 The schematic
15、 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 can t ,please rectify it. 32451Fig2-9 解：不能正常工作；改正如图 或者改成题目 2-3 构件 5、6、7 的连接 名师归纳总结 - -
16、- - - - -第 10 页,共 28 页精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆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 pr
17、operly. If it cant ,please rectify it. EB23C1A54DFig2-10 解：不能正常工作；EB2C4名师归纳总结 1AD第 11 页,共 28 页- - - - - - -精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.3-1 Locate all instant centres of mechanisms for the position shown. 2 2143134n32P23nFCP 23 35D4 P 34 P H1541A P 15nEP P 142
18、P 15P45JP 141P 25O124BP13P14P 12P12 GP 12P 14P 24n名师归纳总结 EP 13 2 14234A P14C P 23 DP13P12第 12 页,共 28 页P12 P 24 G P14 P23 2P 23 V1 P24P 34 B23P 34 1D 234P 24314 13A P23P P43 1BP 14P13 F P 12CP 13P - - - - - - -精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.3-2 For the position s
19、hown of a geared linkage, determine the ratio 3/ 1 of the angular velocity of gear 3 to that of gear 1, using the method of instant centres. 1 BA4E2CP23 5DP36123163 P12 FP 12P 13P 23P 16P 13P 63P16EP 1336解：P13是构件 1 和 3 的瞬心,等速重合点,所以,1L AE = 3LDE, 3/ 1=LAE/L DEEx.3-3 For the position shown of cam mech
20、anism, determine the ratio 2/ 1 of the angular velocity of follower 2 to that of cam 1, using the method of instant centres. 解：A P 23 E（P12）是构件 1 和 2 的瞬心,等速重合点,3 所以,1LOE= 2LAE, 2/ 1=LOE/L AE2名师归纳总结 1EP 12P 13T第 13 页,共 28 页O- - - - - - -精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Dat
21、e Ex.3-4 In the pivot four-bar linkage shown below, 1= -10rad/sec. Using the method of instant centres, a find the velocity of point C for the position shown. b for the position shown, locate the point E on the line BC or its extension which has the minimum velocity among all points on the line BC a
22、nd its extension, and then calculate its velocity. c draw two positions of the crank AB corresponding to VC=0. P 12B1P 23C43EP242P 34D1A P 142F解：P 12BP 23 CC 22LFE；12C13DP 341B 241LAB LFC/L FBb V E=AP 14B 11L AB=V B2= 2LFB,所以,VC= V C2= 2L FC= a VB1= c VC=0 所对应的曲柄 AB 的两个位置 :名师归纳总结 - - - - - - -第 14 页
23、,共 28 页精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.3-5 In the six-bar mechanism shown below, X A=0, Y A=0, X D=450mm, Y D=0, LAB=150mm, L BC=400mm, L DC=350mm, CDE=30 , L DE=150mm, L EF=400mm. The crank AB rotates at a constant speed 10rad/sec. A main program is required to
24、analyze the output motions of the point F. The mechanism will be analyzed for the whole cycle when the driver AB rotates from 0 to 360 with a step size of 5 . 1B2CE5F13AD46解：FOR I=0 TO 360 STEP 5 CALL LINK0, 0, 0, 0, 0, 0, I*PI/180, 10, 0, 150, XB, YB, VBX, VBY , ABX, ABY CALL RRR450, 0, 0, 0, 0, 0,
25、 XB, YB, VBX,VBY , ABX,ABY ,350,400,Q3,W3,E3,Q2,W2,E2 CALL LINK450, 0, 0, 0, 0, 0, Q3-PI/6, W3, E3, 150, XE, YE, VEX, VEY , AEX, AEY CALL RRP1, 0, 400, XE, YE, VEX, VEY, AEX, AEY, Q5, W5, E5 CALL LINKXE,YE,VEX,VEY ,AEX,AEY ,Q5,W5, E5, 400, XF, YF, VFX, VFY, AFX, AFY PRINT I, XF, YF, VFX, VFY , AFX,
26、AFY 名师归纳总结 NEXT I 第 15 页,共 28 页END- - - - - - -精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex3-7 . In the mechanism shown below, XG=Y G=0, X B= - 42, Y B=39, XD=10, YD=75, LBA=23mm, LGF=12mm, L FE=95mm, L EC=69mm, L DC=48mm, EFG=90 . The crank BA rotates at a constant speed of
27、10 rad/sec. A main program is required to analyze the output motions of the point C. The mechanism will be analyzed for the whole cycle when the driver BA rotates from 0 to 360 with a step size of 5 . E 6DB623A451CG名师归纳总结 F第 16 页,共 28 页6FOR I=0 TO 360 STEP 5 CALL CALL LINK-42, 39, 0, 0, 0, 0, I*PI/1
28、80, 10, 0, 23, XA, YA, VAX, VAY, AAX, AAY RPR-1, 0, 0, 0, 0, 0, 0, XA, YA, V AX, V AY, AAX, AAY , 12, QFE, W3, E3 CALL LINK0, 0, 0, 0, 0, 0, QFE+0.12565033, W3, E3, 95.75489544, XE, YE, VEX, VEY , AEX, AEY CALL RRRXE, YE, VEX, VEY , AEX, AEY , 10, 75, 0, 0, 0, 0, 69, 48, QEC, W4, E4, QDC, W5, E5 CAL
29、L LINK10, 75, 0, 0, 0, 0, QDC, W5, E5, 48, XC, YC, VCX, VCY, ACX, ACY NEXT ENDPRINT I, XC, YC, VCX, VCY , ACX, ACY I - - - - - - -精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.3-8 In the six-bar mechanism shown below, X B=0, Y B=0, X F=37.2, Y F=17.5, YC=28.8, L FE=16.8mm, LEC
30、=39.2mm, LCD=20.633mm, L DE=36.4mm, BGA=90 , LBG=9mm, LGA=58mm. The crank FE rotates clockwise at a constant speed of -10 rad/sec. A main program is required to analyze the output motions of the point A. The mechanism will be analyzed for the whole cycle when the driver FE rotates from 360to 0 with
31、a step size of -5 . A名师归纳总结 5CD6第 17 页,共 28 页324F1BEFOR I=360 TO 0 STEP -5 CALL LINK37.2, 17.5, 0, 0, 0, 0, I*PI/180, -10, 0, 16.8, XE, YE, VEX, VEY AEX, AEY CALL RRP-1, 28.8, 39.2, XE, YE, VEX, VEY, AEX, AEY, QEC, W4, E4 , CALL LINKXE, YE, VEX, VEY , AEX, AEY , QEC-0.548, W4, E4, 36.4, XD, YD, VDX,
32、 VDY , ADX, ADY CALL RPR1, 0, 0, 0, 0, 0, 0, XD, YD, VDX, VDY , ADX, ADY , 9, QGA, W2, E2 CALL LINK0, 0, 0, 0, 0, 0, QGA-0.153944664, W2, E2, 58.69412, XA, YA, VAX, VAY, AAX, AAY PRINT NEXT I ENDI, XA, YA, V AX, VAY, AAX, AAY - - - - - - -精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No.
33、Date Ex.4-1 According to link dimensions, determine the type of the pivot four-bar linkages shown below. 9010011070704540 120双曲柄机构 曲柄摇杆机构5010070609070100双摇杆机构 双摇杆机构名师归纳总结 - - - - - - -第 18 页,共 28 页精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.4-3 In an offset slider-crank mech
34、anism ABC, the crank AB is a driver. The maximum pressure angle MAX =30 . Find the stroke H of the slider and the crank acute angle between the two limiting positions. B 3BB 1eAB 230C 1C 3HC 2Ex.4-4 Determine the angular strokes of the rockers AB and CD, respectively, using graphical method. ABCDVB
35、动画名师归纳总结 ABCD第 19 页,共 28 页- - - - - - -精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.4-5 Shown are the two positions, B1C1 and B2C2, of the coupler BC of a revolute four-bar linkage ABCD. The link AB is a driver. The pressure angle at the first position is 0 o. The second posi
36、tion of the mechanism is a toggle position. Design the linkage. Describe briefly the drawing steps. 解：DAB1B2C 1C2The fixed pivot A must be located on the bisector of B1B2. Similarly, D must be located on the bisector of C1C2. The link AB is a driver. The pressure angle a at the first position is 0o.
37、 B1C1 C1DThe second position of the mechanism is a toggle position. A, B2, C2 三点共线名师归纳总结 - - - - - - -第 20 页,共 28 页精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.4-6 In a crank-slider mechanism, two sets of corresponding positions between the slider and a line segment AE on the
38、 crank ABE are known, as shown below. The position C 1 of the slider is its left limiting position. Find the first position B 1 of the revolute B. Describe briefly the drawing steps. 解：C 1C 2B 1AE 1E2作 AC2E1AC2E2,且字母旋向相同,得C2 C2 因 C1 为滑块的极限位置之一,所以 B1 点在 AC1 连线上；作 C1C2 的中垂线与 AC1 交于待定活动铰链点 B 的第一个位置点 B1
39、；名师归纳总结 - - - - - - -第 21 页,共 28 页精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.4-7 In a revolute four-bar linkage ABCD, side link AB is a driver. The positions of the side link CD and a line segment CE on the coupler CBE corresponding to two positions of the linkage are known
40、. The first position of the linkage is also a dead point. Find the second position B2 of the revolute B. Describe briefly the drawing steps. 解：C 2E2E1B1C 1作 A2C1E1DA2AAC2E2,得 A2 点；因 AB 为原动件且机构第一位置为死点,所以 B1 点在 DC1 的延长线上；作 AA 2 的中垂线与 DC1 的延长线交于待定活动铰名师归纳总结 链点 B 的第一个位置点B1；第 22 页,共 28 页- - - - - - -精选学习资料 - - - - - - - - - 学而不思就惘,思而不学就殆Name Class Student No. Date Ex.4-8 In a