




文档简介
65. 2, 2551997 0 1997 6)002697 F . . 6531, 1 995) is of by in of of of is of on It is a by An on is by To a is a by 0 1997 1. he it to be of on in is by To be in of by on 11, of 2, 3,4. In it is of to be in is he is in is no is in an is in it a In of is to 5of is a to In is of a 8 In a of is of of as of as of is by is by 111. To a is by 2. n of a is as a of of k of is to of to of of of 255 2% S. . is as in of 5 2 To in 4, 5), 7” I” in 1=I ,=I to g,(x, t) 0 P, r = 1, 2, 3 j = 1, . , Y:; = $, + $;,& (1) q=l,., n, s=l,., 12 (7a) of of x is 8&+&,4!: p,r= 1,2,3 V of , is of In q= l,., m s= l,., 12, (7b) is ” is n, is to of m is of of in a of a be in a to a ” of k in a do of be as of is by f 2 mk of k=,=, i q, of a , be (7 7i,j$ j$,j$, 84 M; = f 2 m&, k=,i=l ?“I et s 12 to of +F”+F, (3) + &a + !;,Y;X: + &L.8) y = d is of is of , Q as UI s p”& d V :L$,. = s pkc#&$:; v!- F”=- O I Q!, = ,$, 4 6) + (/?&b;:, + u;&) &,b;:,. (9b) r f to he of by K is a a _ to of z,u= 1,2,3; s,v= l,., 12 is of of By = $?A& + & = $!&,& + $&,&, be 51 i to by of by of of in by , in of a is as of in It be in of of k k in in be to of is a in of A or is 3. he of is , of in of to be C x N, a of is in an An to of C to of of t is to by 3 as g,(x) = - i (10) .=I x) = gj(x, t”) c is a g, g,“). It be a 4 gj is g,n), of c, g,). as j(X) = x)l. (11) In gj a as it to In at of at by In as is by l I is on df/, M = 1, . . . , by in 4. is 1. A is in is to an 0, = 120”, 19 = - 150) to a 0, = 60”, - 30”) a is x =g T 2 E T t - x of T, is to .5 s. is of .6 m is by &, k = 1, 2; i = 1,2 of as of is to = 72 p = 2700 kg is by 1. A 2.0 t t 18.0 f 5 10 15 20 25 30 35 of 2. 258 S. . . 12 22 N) (4 9 9 8 of is at a kg of is 5i=l,.,n, 6 m, at n, of at 6 is of of in x y of 0 mm In by by In of c It of c in as A of c = 50 a It be be of c to if on On in of . 2. by CC of It is .5 t w 3. at of at in 0.6 s P $ 4. in 59 at of at 3. is 4. 5. n a of to of of by on as It of a in is in of by A of in . W. H. . T. . 3. 4. 5. 6. I. a. 9. 11. 12. of in 32, 4331989). E. J. . S. of 5, 3562 (1978). G. . by a 1137 (1979). R. T. 2. . P. 1990). D. A. . S. of 36, 1191990). M. H. . of 2, 2531994). J. H. . of . 16, 3441994). A. A. 1989). S. S. .
温馨提示
- 1. 本站所有资源如无特殊说明,都需要本地电脑安装OFFICE2007和PDF阅读器。图纸软件为CAD,CAXA,PROE,UG,SolidWorks等.压缩文件请下载最新的WinRAR软件解压。
- 2. 本站的文档不包含任何第三方提供的附件图纸等,如果需要附件,请联系上传者。文件的所有权益归上传用户所有。
- 3. 本站RAR压缩包中若带图纸,网页内容里面会有图纸预览,若没有图纸预览就没有图纸。
- 4. 未经权益所有人同意不得将文件中的内容挪作商业或盈利用途。
- 5. 人人文库网仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对用户上传分享的文档内容本身不做任何修改或编辑,并不能对任何下载内容负责。
- 6. 下载文件中如有侵权或不适当内容,请与我们联系,我们立即纠正。
- 7. 本站不保证下载资源的准确性、安全性和完整性, 同时也不承担用户因使用这些下载资源对自己和他人造成任何形式的伤害或损失。
最新文档
- 林地管理办法西藏
- 条幅展板管理办法
- 强捡计量管理办法
- 松原灌溉管理办法
- 施工工区管理办法
- 男性尿失禁的护理方法
- 折弯计件管理办法
- 总部遴选管理办法
- 二零二五年商务居间服务合作协议范本
- 无创血压监测安全操作规范
- 数据标注员基础技能培训手册
- 广东校医考试试题及答案
- 加油站团队管理课件
- GB/T 45760-2025精细陶瓷粉体堆积密度测定松装密度
- 福建省福州市福九联盟2024-2025学年高一下学期7月期末考试数学试卷(含答案)
- 关于水肿的课件
- 企业环境保护工作课件
- 太阳能路灯设计与安装方案
- 2025年高考新课标I卷听力讲评课件-高考英语一轮复习专项
- 石膏固定病人的护理措施
- 2025年湖南省中考语文试卷(含解析)
评论
0/150
提交评论