理论力学运动分析解析法试卷30题.doc

1、第 1 页 共 33 页福州大学 20062007 学年第二学期考试 A 卷课程名称 Mechanisms and Machine Theory 考试日期 2007.07.06考生姓名 学号 专业或类别 题号 一 二 三 四 五 六 七 八 总分题分 32 12 12 8 10 6 10 10 100累分人 签名得分考生注意事项：1、本试卷共 10 页，请查看试卷中是否有缺页。2、考试结束后，考生不得将试卷、答题纸和草稿纸带出考场。教师注意事项：如果整门课程由一个教师评卷的，只需在累分人栏目签名，题首的评卷人栏目可不签名。以下内容命题教师阅读后请删除： 用 B5 纸作为标准试题纸，也可打印在

2、A4 纸上交试卷印刷中心帮助处理。题目不超过方框, 每题必须留有适当空位给学生答题使用。 题目序号统一用一、 1、 （1） ，即第一大标题用 “一、二、”，第一大标题下的题目用“1 、2、”，第 1 标题下的题目用 “（1） 、 （2） 、”。每道大标题下必须加上 题目 “一、二、 ”统一用四号黑体字打印， 其他部分一律用小四号宋体字打印。 教师所填课程名称必须与课表、教学大纲和授课计划吻合，考试形式、考试日期、考生注意事项第 1 条中的空白项必须填写齐全。解析法运动分析考题 30 题（已剔除课本、作业本中的题目！）得分 评卷人人 第 2 页 共 33 页八-1、 In the mechani

3、sm shown below, XE=-10, XA=20.3, YA=0, XC=53.1, YC=0, LAB=11, LCG=4, LDG=63.8, LDE=19.5. The crank AB rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the point E. The mechanism will be analyzed for the whole cycle when the driver AB rotates from

4、0 to 360 with a step size of 5.(10%)得分 评卷人人 第 3 页 共 33 页八-2、 In the mechanism shown below, XF=0, YF=0, XC=42.9, YC=0, XD=-20, LEF=11.8, LBC=5, LAB=63.8, LAD=19.5. The crank FE rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the point D. The mecha

5、nism will be analyzed for the whole cycle when the driver FE rotates from 0 to 360 with a step size of 5.(10%)得分 评卷人人 第 4 页 共 33 页八-3、 In the mechanism shown below, XC=0， YC=0， XA=-30， YA=0, XE=20, LAB=10, LBD=49.6, LDE=28. The crank AB rotates clockwise at a constant speed of -8 rad/sec. A main pro

6、gram is required to analyze the output motions of the point E. The mechanism will be analyzed for the whole cycle when the driver AB rotates from 360 to 0 with a step size of -5.(10%)得分 评卷人人 第 5 页 共 33 页八-4、 In the mechanism shown below, XC=0, YC=0, XA=0, YA=23.5, YE=43.8, LAB=11.8, LCD=47, LDE=19.5

7、. The crank AB rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the point E. The mechanism will be analyzed for the whole cycle when the driver AB rotates from 0 to 360 with a step size of 5.(10%)得分 评卷人人 第 6 页 共 33 页八-5、 In the mechanism shown bel

8、ow, XC=0, YC=0, XB=0, YB=28.8, YE=47, LAB=11, LCG=6.1, LDG=49.7, LDE=30. The crank BA rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the point E. The mechanism will be analyzed for the whole cycle when the driver BA rotates from 0 to 360 with a

9、step size of 5.(10%)得分 评卷人人 第 7 页 共 33 页八-6、 In the mechanism shown below, XB=0, YB=0, XC=0, YC=28.5, YA=-22.6, LCE=8, LEF=6, LDF=50, LAD=27.3. The crank CE rotates clockwise at a constant speed of -8 rad/sec. A main program is required to analyze the output motions of the point A. The mechanism wil

10、l be analyzed for the whole cycle when the driver CE rotates from 360 to 0 with a step size of -5.(10%)得分 评卷人人 第 8 页 共 33 页八-7、 In the mechanism shown below, XG=0, YG=0, XB=-31.6, YB=-8.6, XD=-57.5, YD=22.9, LAB=6.6, LFG=7.4, LEF=58.4, LCE=49.7, LCD=34.6. The crank BA rotates at a constant speed of

11、8 rad/sec. A main program is required to analyze the output motions of the link DC. The mechanism will be analyzed for the whole cycle when the driver BA rotates from 0 to 360 with a step size of 5.(10%)得分 评卷人人 第 9 页 共 33 页八-8、 In the mechanism shown below, XF=0, YF=0, XD=10.6, YD=28.4, XB=28.5, YB=

12、-39.6, LEF=8, LDG=5.7, LCG=59.2, LAB=36.3, LAC=39.9. The crank FE rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the link BA. The mechanism will be analyzed for the whole cycle when the driver FE rotates from 0 to 360 with a step size of 5.(10%)

13、得分 评卷人人 第 10 页 共 33 页八-9、 In the mechanism shown below, XD=0, YD=0, XE=30, YE=7.2, XC=21.7, YC=-8.2, LDG=8, LFG=6, LAF=48.3, LAB=25, LBC=24. The crank DG rotates clockwise at a constant speed of -8 rad/sec. A main program is required to analyze the output motions of the link CB. The mechanism will be analyzed for the whole cycle when the driver DG rotates from 360 to 0 with a step size of -5.(10%)得分 评卷人人