Đề thi cuối kì 2 - Môn: Cơ kĩ thuật

lire nhu tren Hinh 1. Bi~t lire Fi nlim trong m~t phling Oyz, lire F2 nlim trong khong gian va c6 cac t9a de;\ g6c nhu tren hinh ( 0<90°). a) Bieu di~n cac vector lire tren du6i d!IJlg vector Descartes (theo cac vector dan vi T,],k), b) Tim vector hqp lire FR (du6i d1mg vector Descartes) cua hai lire Fi va F2 ; c) Tim de;\ 16-n ctia vector FR. z Hinh 1 Cau 2: (1,5 diem) z F=200N r<--75 mm .,,..,,. 1 I ¢1 Hinh2 Tay quay ABC duqc dung d~ quay tr11c OA nhu tren Hinh 2. Tf\lc OA nilm d9c theo tr11c Oy, AB d6ng thai vuong g6c v6i Oy va BC. Lire F = 200 N tac d11ng t~i C va vuong g6c v6i tay dm BC. Bi~t 0 = 60° va ¢ = 30°. a) Bieu di~n vector lire F du6i d1mg vector Descartes; b) Tim d~ng vector Descartes ctia moment do lire F gay ra quanh tam O. Cau 3: (1,5.di~m) Beam AB is subjected to the loads and constraints as shown in Figure 3. Determine reactions at A and B. Cau 4: (2,0 di~m) Figure 3. Ca du ctia mc;\t may ep Ion duqc mo hinh h6a nhu tren Hinh 4. Mc;\t lire F = 300 N tac d11ng t~i B theo phuang vuong g6c v6i tay dm BC. B6 qua ma sat, hay: a) Ve sa d6 giai ph6ng,lien ket (FBD) cho tay c!m ABC; b) Tim phan lire lien ket t~i A; S6 hi~u: BMI/QT-PE>BCL-RE>TV Trang: 1/2 y c) Tim lµc ep do con trugt C ep len Ion. B 0,3 m A~ 02m , Hinh 4 Hinh 5 Cau 5: (2.5 di€m) Cho he hai banh rang an khcrp nhu tren Hinh 5. Bi~t R1 = 400 mm, R2 = OA = 300 mm. Tl!i thoi di€m khao sat nhu tren hinh ve, tay quay OA va banh rang I dang quay nguqc chi€u kim d6ng h6 quanh ch6t ban IS c6 djnh t(li O vcri v~ t6c g6c tmmg ung la % A =IO rad/s va w, = 5 rad/s, hay xac djnh: a) V~n t6c cila di€m A; b) V~n t6c cila di€m B; C) Tam v~n t6c tuc thoi cua banh rang 2; d) V~n t6e g6e cila banh rling 2. Cau 6: (I di€m) The disk shown in Figure 6, which has a mass of 20 kg, is subjected to the couple moment of M = 80 N.m. The radius of gyration of the disk about its mass center O is k0 = 0,2 m. lfthe disk starts from rest, determine its angular velocity when it has made I 0 revolutions. Figure6 - HET- Ghi chu: Can b9 coi thi kh6ng au(J'c giai thich ai thi. Chuan dau ra cua hoc ohfi n (v~ ki~n thuc) [CDR I. I]: Thu g.;m duqc m9t he h,rc ve m9t tam thu gQn, phan tieh duqc cac thanh ph/in phan lµc lien k€t (k€ ca lire ma sat) va thi€t l~p duqc cac ohuong trinh can bin!! eua ehftt di€m, cua v~t ri1n, he v~t rlin. (CDR 1.2]: Tinh duqc cac thong so d9ng hQe (vi tri, v~ toe, gia toe) cua v~t rlin, di€m thm;,c v~t rlin tron!! chuv€n d9ng song phi1ng. (CDR 1.3] : Su d11ng duqc phuang ph~p luqng va ph!l'ang phap nl\ng luqng d€ xac djnh duqc thong so d9ng hQc cua v~t ran trong chuv€n d9ng song ohin!! ducri tac d11ng cila lµc. (CDR 2.1]: Phan tich va mo hinh h6a duqc so co M tinh djnh trong ca khi bi1ng cac mo hinh tinh tuan!! ung. (CDR 2.2]: Phan tich va mo hinh h6a duqc so ca cau phang trong ca khi bi1ng cac mo hinh tinh tUOD !! uni!. fCDR 3.11: DQc hieu cac tai li~u ca hQc ban!! tie1rn Anh. Noi dun2 kiem tra Cau I, 2, 3, 4 Cau 5, 6 Cau6 Cau 3,4 Cau4, 5 Cau 3,6 Ngay IO thang 07 n1im 2020 Thong qua bl} mon /::r (Icy va ghi ro h9 ten) S6 hi~u: BMI /QT-PDBCL-RDTV T 61.M~ T,>.,:, 4':ang: 2/2 DAP AN DE THI CUCH KY HK2-NH 19-20 TRVONG O~I HQC SV P~M KY THUA T THANH PHO HO CHI MINH Mon: Ca ky thu~t ... ...... ........................................... . KHOA CO KHi CHE TAO MAY B() MON co so THIET.KE MAY Ma n;ion hQ~: ENME 130620 ······················ ................. . 08 s6/Ma de: 01 ......... ........... DA c6 03 trang. Cau 1: 1,50 d a) • F, =(200{153)1-(200{:!} N 76,9]-184,6k N 0,75 d • cos0=-Jl-cos2 (60°)-cos2 (120°) = (0 <90°) 2 • F2 = (350)(cos60" )1 + (350)(cosl20°)} + (350{ ~} N =175t-175}+247,5f N b) t. = 1151 -98,1} + 62,9f 0,50 d • N c) 0,25 d F• = + (-98,1) 2 +62,92 = 210,2 N Cau 2: 1,50 d • F = (Fcos¢)i +0]-(Fsin¢)k = 173,2i -IOOk N 0,50d • i;, 10 =OC=(l50cos60")i +(100+75)}+(150sin60")k mm 0,50 d • i'c,o = 75i + 175] + 129,9k mm • M0 = i'c,o x F = -I 7500i +30000}-303 I0,9k = N.mm 0,50 d Cau3: 1,50 d S<1d6GPLK: 0,50 d y 2000Nlm~t L A, A . . - - • -, • 1- I B X 600N.ml-c- Ay 4m •--2m ....... Phu<1ng trinh can bing cho d§m AB: • F. = (0,5)(2000 N/m)(6 m) = 6000 N 0,75 d • _LF, =0 A, -N8 sin30° = 0 Ax -0,5N8 = 0 (I) • L FY= 0 AY -F,1 + N8 cos30° = 0 AY -6000+ = 0 (2) • _LMA = 0 (F.)(2 m) + 600 N.m-(N8 cos30°)(6 m) = 0 =0 (3) S6 hi~u: BM 1/QT-POBCL-ROTV Trang: 1/3 Giai ra cac uhan I!!£ lien k~t: 0,25 d Ax=l212,4 N· ' AY =3900 N; NB =2424,9 N Cau4: 2,00d a) SO' d6 GPLK cho tay dm ABC: 1,00d F B 0,3 m 0 0,2m LF()A 45° I X c , l c,. b) • LMc = 0 (F)(0,5 m)-(F0A)(0,2 m) = 0 0,50 d =750N c) Xet can bing clia tay dim ABC theo phucmg x: 0,50 d • LFx =0 SO' d6 GPLK cho con tr!!Q'.! C: y~ Lxcx Cy C Fe Xet can bing clia con trugt theo phucmg x: • LFx =0~Fc =Cx =318,2N Cau 5: 2,50 d a) • VA =(OA)(%A)=(0,3m)(!0rad/s)=3m/s 0,75 d • Phucmg, chi~u nhu hinh ve b) • V8 = (R1)(m1) = (0,4 m)(5 rad/s) = 2 mis 0,75 d • Phucmg, chi~u nhu hinh ve c) Tam van t6c tfrc thin cua banh rang 2: 0,50 d S6 hi~u: BM 1/QT-PE>BCL-RE>TV Trang: 2/3 d) 10) -~-~ 2- -rA IIC r8/IC rA IIC - r8/IC = 0,1 m 0)2 = IO rad/s ( chi€u nhu hinh ve) Cau 6: Dong nang: • T, = 0 (Ca h~ dung yen) I 2 2 2 • T2 = -(mk0 )(J) = 0,4 (J) 2 Cong cua ngoai hrc: Sa d6 GPLK (FBD) Iu1_2 = M.0 = (80 N.m)[(l0)(21r) rad]= 1600Jr J Iu1-2 =5026,55 1 Ap dung dinh ly dong nang: T, + Iu,_2 = T2 (J) = 112,1 rad/s Ghi chu: Sinh vien c6 Mi giai khac dap an nhung dung vdn c6 didm t6i da. -HET- 0,50 d 1,00 d 0,50 d 0,25 d 0,25 d S6 hi~u: BMI/QT-POBCL-ROTV Trang: 3/3