上海交通大学：《Design & Manufacturing II and Project》课程教学资源（作业）05_homework kinematic fundamentals solutions

Mechanism Solutions (DOF) Problem 1(10 Points) a)Kinematic diagram(4 points) C(R) Must have links numbered and letters on the joints. ）a) 、 Also label fi joints as (P)or (R)and f joints simply as ) (R) Make sure you understand the nature of each joint.For example.at joint b.you can expect both sliding and rolling,and therefore this will be an f joint. e(R) b)DOF using Gruebler's equation (3 points) 2 F=3m-1)-2-万3 n=8 R) F(R) (R)9 6 f=8 [7 (R)s and 1 (P)] 4 =1 F=3(8-1)-2(8)-1 R) F=21-16-1 Fig.1 F=4 (P) Problem 2 (10 Points) a)Kinematic diagram(4 points) Fig.3 Must have links numbered and letters on the joints. 10 Also label fi joints as(P)or(R)and f2 joints simply as fa. Make sure you understand the nature of each joint. Links 6 and 7,connected at (R)joint h,may be shown in various different ways.One way is shown in Fig.3. Another common way is shown in Fig.4. Note that the (R)joint that connects links 3,4 and 5,is to be counted twice as joint c and d(as explained in the 长 recitation) b)DOF using Gruebler's equation (3 points) F=3(n-10-2f-f n=10 万-13 万-0 F=310-1)-213)-0 F=27-26-0 F=1=

Mechanism Solutions (DOF)

2-21 a)n=4,J1=4,J2=0,M=1 b)n=4,J1=4,J2=0,M=1 c)n=8,J1=10,J2=0,M=1 d)n=4,J1=4,J2=0,M=1 e)n=8,J1=11,J2=0,M=-1(paradox) n=6,J=7,J2=0,M=1(remove the over constraint links 5,7) f)n=4,J1=4,J2=0,M=1 g)n=6,J1=7,J2=0,M=1 h)n=9,J1=12,J2=0,M=0(paradox) n=6,J1=7,J2=0,M=1 (remove the symmetrical parts,link 7,8,9)

2-21 a) n=4, J1=4, J2=0, M=1 b) n=4, J1=4, J2=0, M=1 c) n=8, J1=10, J2=0, M=1 d) n=4, J1=4, J2=0, M=1 e) n=8, J1=11, J2=0, M= -1(paradox) n=6, J1=7, J2=0, M=1 (remove the over constraint links 5, 7) f) n=4, J1=4, J2=0, M=1 g) n=6, J1=7, J2=0, M=1 h) n=9, J1=12, J2=0, M=0 (paradox) n=6, J1=7, J2=0, M=1 (remove the symmetrical parts, link 7, 8, 9)