Complete_Solutions_for_Homework_2

Complete_Solutions_for_Homework_2 - Homework # 2 —...

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Unformatted text preview: Homework # 2 — Synthesis - Solutions LINKAGE SYNTHESIS QUESTIONS: (*You Must Show All Your Work*) Q1. Function Generation: Assume: 61 = 270 degree, d=1.0 inch 1. Calculate: link lengths a, b, c, all values of 03, and coordinates of joints A and B. b (inch) 1.286 1.719 0.7896 _- __ m _—m 2. Is the fourbar a Grashof linkage? S + L = 0.7896"+1.719"= 2.5086" P + Q =1.286"+1.0"= 2.286" S +L> P+Q NON—GRASHOF LINK! Q2. Motion Generation: Two precision points: P1 and P2 are given as shown in the figure. The angle of rotation of the coupler link is a2 = 20 degree. I7 Case 1: 1. For the L.H.S. Assume: 02 = 150°, B = 10°, 4) = 30 ° Calculate link len ths: a and f a =18.62"; f = 7.3" 2. For the R.H.S. Assume: 94 = 90 °, 5 = 15 0, \V = 150 ° Calculate link lenths: c and g c =10.7"; g =1.95" 3. Calculate coordinates of oints A and B (i.e., XA, YA, X3, Y3) X3 =11.69"; YB =9.03" X04 =11.69"; Y04 = —1.67" XA = 3.68"; YA = 6.35" X02 =19.8"; Y02 = —2.96" 4. Draw the fourbar linkage (to scale) (I = 822'“ b = 8.44" 19 =—8.99"; 0 =18.47" Case 2: 1. For the L.H.S. Assume: B: 40°, (1) = 60°, f = 1 inch Calculate link lengths: a and 92 ax = —0.2"; a. = 4.23"; a = 4.24"; (92 = 92.7” 2. For the R.H.S. Assume: 8 = 30°, w = 120°, g = 1.5 inch Calculate link lengths: c and 64 cx = 0.0593"; c). = 5.186"; c = 5.19";64 = 89.34" 3. Calculate coordinates of oints A and B (i.e., XA, YA, XB, YB) X3 =10.75";YB = 8.7" X04 =10.69";Y04 = 3.52" XA = 9.5";YA =9.13" X02 = 9.7";Y02 = 4.9" 4. Draw the fourbar linkage (to scale) d =1.704";b 21.32" 01 = —54.5°;a3 = ~19.1" -<Q.3— HMKR: Q) P! (as) (045) (>2 (4.0, 27.04) P3 (3.04» 3.23) (X2 = ._ coo Ola 7- 0 52:9!“ (33"5’ : M” 52:62)” 253=7‘i" ‘RHS A: Cook: ovl : .— l-Ol—TSN—I‘GZ 6: $62: 0.qqqg’~I-O C: Cma2_| Dc $05 = —O.I'136 W 7' 2.. 2 2. 0! F2. : V(?Zx— fix) + (1)23— Plj) : ("'§)+(.M’) z figéé W F s. \j 2' a" — ~z-W1-ze "’ = 3.367” 3: @3me + ((333.63) - ( ) ( 2) (Q3) 36§° u-l .- ,I __ .\ ‘ ' 6 .___ «CM a3 (9’3 g. cw. ILL—- => 82: 240.57" 2 P P ’ L5 \ 2mg" 2X m ' (M) 43.9. c 3’ Pavfix «ZLH \21356 “E: ngngz :45 Fsmfi-I=~19‘IS’ C7 = S-vgsz H = (5070(3—4 = O = svx3=o L=g‘m$3=-z‘l‘1 GM=P 3.52:4” ~N=P3I&83= —232 :CQe3-HHK#3(C4WW')= @ A ,3 c -D - 1.02, ——| -o.olb’ o.l736 _ —K .. , - o 0 AA = F 6' H _. -1 Z7§ 0.485 e, A ‘D C I —/-oz 4413!. —o-oIS c, F K H was 4.87: o o E -'.5 NIX _‘ CC _ l— ,_ '2 W'fi : ' F M -l H ) 2m N '2‘32 2:3 O -0 S o O 127 NM 7. o -°~§ WW3 : o -oJ ‘f 2m -on 4.722 -§.718’ 3.836 ’7'“? 5.7M; -3234, 4.979 4'73 ' 3’ 7- II W. 2 \/ WIx + W“j :: ‘74 “’7‘?” 0 Z' = \I 2“ + 2.3 = 8 (a a: 584° /' 4’ = 65 R~H~S. / "'- A A/= CHVZ_‘:_O‘SgI 5:33.);2: 0823 C:Qm°(2-l- 00/52 / 7;: $409: -0474 . E: {Laugh/.5 F: mag—I = 4.90% G’: egg”.qu sza3-::o Kzs;o<3=o . L = g] mgsrgafi - M: P2, S;Sz.—.—I-H ~ N: F3I 5483:. —2«32 :Q'S— kCa—h'rd): : A! —8’ C —D ~O~§3I *0. F, ~G’ — - 0.3 _ ‘ AA 3/ AI 1': CK : Oct 0 qg?’ 0 O o 88% «3‘53! - . - . G’ F’ K H ° '7‘( 0 Olga ° ('73; “93°C? <5 G E "I 6 UN \ L- — CC = M = ‘2‘” l U” = ~ CC -1.” gm N —z.32 SI:3 (c312) U 0 ~06 o o 6o7 +1 '7’ UN = - 0488’ [K UH : o -— Mm 0 ~ 0 g —2 44 => UH ; 2 (04” (4.2 :12: 4qu —o.28l —§7 #18 -3.” 8w: ’2 (H II I 3 5’7 4M; -0.qu Mm ,232 S - 3W )— 9_ II A .. S : ‘ g S 2 27° S. ’ ’4 "(q-3‘3 Ll / I I 0/: 0H ' A,“ = 127° gmflhnahs: A R“fi\'z\:o => A\=Pu'Z| R = (’35, é~'5)-<1*5€ 992) K. = (2.31,; 0.97) ‘Q3— HHK#3 (Com+’J-)‘ g‘: 34‘012 5: ((0037038) :Qg, mum” (Con+>J-): 936.04,) 3.33) F2 (m, 5.04) ; ...
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Complete_Solutions_for_Homework_2 - Homework # 2 —...

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