文档简介
0F 3,2010 367 0390101003367,at of a , 300072, 66100,6,2009;,2010;0,2010;5,2010 in is ll of of in n a is to of a at or n is he on is as as in on he of of on to on of of be by a is of is by of a he an s an of a of IM,et ,of he be in a he of in to of as to of or a on is et et 。a of of at di et a on to of he as at at or n is o50605060),i o06of is on s a as to n IU。et n to of in as a of by to a et jIn be in a of a !: 竺 里 坐! 兰! 里 !et n a is by a by of et to of of n of of in is to an of is of he of of on of is as as in on A (7)of as in of a of a is in of is to of he of f be in x i he in 1 is is in in in t=mR。l 1,of is at a n be as it is at n is as a is in in he or et Jis no is in is of be by if of 7 on to on of a is is by of a 2 of be as a of at by as in of in of of (1、ne of is at (2、of of is (3)is to (4、(5)A is by a (6、in of in is be on f of is as it on is on in to ,c , 一 一,) of a of it in is oin a is by a as in as of is be he of + ,a R= In 369 at be in (a)b)he is as a of as of in y be by he in e佰:ct of n he be in n of a is It of it ne is is he to be at in v 辨 rN rl r2 r3 。 i : 一f:, s 囝 一 in o a to be s , in on be t is on be rp,in is p)of r1 rp on In of of to rl rp = + , 02+ 2, (1) (2) of of to be + 一 =2+ 01)(pf)360。 3) (4) zl z2 of of n of ri on by sf pf as 。f=R。 as 1 f5) (6) (7) (8) 。1,is to or ri on 9) ri on 量 = R , 【 : ! : 坚 !璺 里 竺 ! ! 苎! 垦! 里 !h R , (10) fk(R。2 of in is an of of rl iS t in to to be rp to be 4 k( 。1 f) (R l i) +( 。1 i) 。2 i) + 。f m。 i。ed m c(f)+c( 一 川)一 c。 +90)一ge as z, 、 l IC S 。 , d + :) g he of of of he of as k(1 k(i+1)11 ( 一l+( 一1 ( 1 i) +( 1 +R 2 f) + f一 。2 0。)一 f+ g=13) of of as Xf)s1 f)tR ( l +(R 1 i) + v 一 R R s +90。)一毫 i=l 0。)+ AR l 0。)一 +90。)一 c 一 )=0,(14) 2 2+R。2(。2 f)i=1 R 2( 2 Yi)t ( 2R 2 f+90。)一毫 c(c(卜 f+90。)一 si f+90o) : n(”9oo f 毫卜s2c。 ( +90。)一戈f= “+Qp r】2、 c。s( 0o): (15) 。 ,J J S O C R 一 + =: 厂 C g F 371 is k c of a in k。c n i=1,2, it be ,i=1is of of he a 1to a he to be do 9) 0),h or q =0 0),h1 5 of an be s a 1is an of be as a he be as a of b ) )be be as to in e1 3,be k be 1 1 1 + + (16) of of a n of of is he to in in n a is to of a y of a is by a he k be by is by is he is he in Z 菪 (a)b) a of =40,z =20 z,=1 0 a8 an of =46,z =25 z,=17,1977 mm 1l 5875 01 0 of is it of of of 1 5875 mm to 372et he be to of a he 2 l m of a m=00 1 6 2 46 kNmma of of a l kNmmof 1=4 0108 2mm on 32 kNAn of is be to a n in 0
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
- 2025年中职(软件与信息服务)软件需求分析阶段测试试题及答案
- 2025年中职会计学(会计教育心理学)试题及答案
- 2025年中职(动物繁殖技术)畜禽人工授精实操阶段测试题及答案
- 2025年大学智能设备运行与维护(智能系统调试)试题及答案
- 2025年大学美术(美术批评)试题及答案
- 2025年高职(应用化工技术)应用化工进阶阶段测试试题及答案
- 2025年中职网络技术(网络设备进阶调试)试题及答案
- 2025年高职第四学年(工程造价咨询)咨询实务阶段测试题及答案
- 2025年中职民俗学(民俗学概论)试题及答案
- 2025年高职铁道运输(铁路客运调度)试题及答案
- 鹤壁供热管理办法
- 01 华为采购管理架构(20P)
- 糖尿病逆转与综合管理案例分享
- 工行信息安全管理办法
- 娱乐场所安全管理规定与措施
- 化学●广西卷丨2024年广西普通高中学业水平选择性考试高考化学真题试卷及答案
- 人卫基础护理学第七版试题及答案
- 烟草物流寄递管理制度
- 被打和解协议书范本
- 《糖尿病合并高血压患者管理指南(2025版)》解读
- 养老院敬老院流动资产管理制度
评论
0/150
提交评论