2011南京理工大学机械原理最后冲刺全真模拟试卷答案.pdf

--- I / 3 / / iI {; - .1... a I WI Pl2 /\ ... Solutions: m a =-(zJ +Z2) Z 10 =-(12xI2) 2 = 120(mm) a cosa a cos a , cos 20° x 120 cosa =---- 130 0.9396926 x 120 - 130 = 0 a = 29.85° . , 2xc . lnva =_. tga + lnva Zc Zs (. ,. ) Xc =_. lnva -lnva 2tga i. , I r I=t ... ------- ??----------------------~: 48 ..... _------- ~~(~~~.#~~)~-~~~~,~~~.o 2~~o~~~mm~m~~~T.ff~x~*~~m.ff m~, ~~~ff~~~#~~~~~o 3 ~~o lfilT~~*T8~1I~~.~lil/J\~*~~p_ , ~~#.W~~~~.~.;ilT~~~~~~M~,~~ ~fttJo 4(1) ~.1tE1fJ p; (2) *m~-tY~mpx;iE~~.IIJ~~; (3) ~tYJJc; (4) ~ItJJc m, jjij~mJJco ~,M: **~ff3~~~,~~=1fJm~~mm~, b 2 p - -- .. .. ==V B3 V B2 + V B3B2 11!tf fft ...L AB ...L !tf fft ? ? ? ? V B2 =cq ·IJUJ =20x 0.05 =1nl/ Is _ ~,uyfFj￡Jl~it1~:. VB3 =Jly. ph3 :lJja]*~ja]~ ,. bll~~-1fJ m~~ I It, I=f ? ~ V =0.72 --51---_-------------.? ...... _------ ------ A ~CDrh LD~C _ (a+b)2+d2-c 2 -17080 UIJ , : .T :n, - arccos ,.., 2d(a+b) () =LDAC 2 - LDAC, =35.33° 2 2 tJjCD ~: LBCD =arccos b +c - (a + d)2 ~ 113.790 900 2bc 180° - LBCD =66.21 ° LBCD rmiD = LBCD =16.21° Md = 28.75N·m n, = !(nmax +n miD ) =1000r/min i. ~ . 2 , ? (2) Q is = nmax - nmiD =0.05 n, I I I=t nmax =1025r/min n min =975r/min ------- ??--------------------------~., 56 M 2 Q , . 3 i. ~ , .. ; I I 1=1 ---6,----1--------...... --------~ M , --- )jtlliML-- )jtlliMo (p 1= p 2=0) I A c .trf1l3lJl(A~f4: :. dl- d2 ~ 4(Pl +P2) ....___---------.:--70---... .... _------- L.J,..~ , f8￥: (a) F=3*6-2*8-1=1, jttft B}~iJd~~fiU$ § mJJl, D}~~!i~~ lit, G, H ff-~~lI~{J~o (b) F=3*5-2*7=1, jttft C }~~~!ifttXlit, E a~jijmtR1SJ~lI ~-1*o = f8￥: (1) ~ b ~:II * tf R1, b)c=5511DD b+d a+c,b a+c-d = 60mm ~b~:II~tf ,bd=35mm lib+ca+d,ba+d-c=20mm :. 20 b d = 35 ~ b ~~:II*, -m~~:II~tfR1, 35 b 55mm --............... lib+a d +c,b c+d -a =50mm :.35 b 50m JYiI2J. b 60mm ~20 b 50mm lJ1mt4J~~1fliffm i. ~ . ; I I ? J::t ------- .... 71 -,-----------