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Exam I Soln - Name ID ME 3403-001/002 TEST 1 Read problems...

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Unformatted text preview: Name: ID: ME 3403-001/002 TEST 1 Read problems carefully and show your calculations in detail in the space provided below each question. 1. For the gear train shown below determine (a) the train value (b) the speed and direction of rotation of the output shaft carrying gear 6 if the input shaft carrying gear 1 rotates at 1725 rev/min clockwise viewed from the positive x-axis. (c) Are there any idler gears in this gear train? If yes, which gear(s) is idler gear(s)? If no, explain why there are no idler gears. (30 points) Ural?“ Einumég IIIIIIEEMflHiHfliHflEI lififlllllllllflllllll lfififlZIIIIIIIIIIIIII (“mm *2 \‘ tumult aha?! Rl {jx'zur‘ 4 a: noui'pui: : _N_‘ m3 N5 _ i ‘8 E. = J. = morsels h input N1 Mg “(9 6’0 4'7- 60 6‘4 9.: "out: = _I_ ”out = .2645 Fey/min (Cow) 1125 eq (‘2) No . All his sews CcniriiomLe in the, cinemas, in Speed ‘F‘fpm (“Put +0 0u+pu+ sin «H . ME 3403-001/002 Test 1 Spring 2008 2. The transmitted load between a spur gear and a pinion is 432 lbf. The pinion has 14 teeth, a 25° pressure angle and a diametral pitch of 6 teeth/in. The gear has 49 teeth. Pinion speed is 2500 rev/min. The face Width is given as 2 in. The load and source are both uniform in nature. A gear quality number of 6 is used. The geometry factors are innion = 0.33 and J gear =0.46. Both the gear and the pinion are made of grade 2 steel and are through-hardened with a hardness of 250 Brinell. Assume Ks and K3 are equal to l and Km = 1.3. Reliability is 0.99. Operating temperature is 200°F. The service life required is 109 pinion revolutions. Use the upper curve in Figure 14—14 for calculating YN. Calculate the factor of safety in bending only for both the gear and the pinion. (Note: Indicate clearly which tables, figures, and equations you are using from the textbook for your calculations.) (30 points) 3W”: wt: “31“”? ' KS“ I “8:“ Km =l'3 I 3P =o'33v38=0c‘~16 ¢ = 15 a Np = “4 RC”? “(3 = lfi iCE‘i’h “P = lSODrcvlmfn, i’L'ékdhlin F=1in. ‘ 9W = 6 , “38 = “89 = 3.50 Bhn ,- grade 2. skel/“throush—karclcned 12: 0.961 ,T: 2.90% , N = \o“ pinion rev . S gench‘na: V: We Koky Ks _P___ «m K3 SF- : 4‘ YN Faun-Hons F‘ J T K1- KR (Agmn) ”~13 N _i_L_q_ l B: OI15( [Du-(v) = O|8155 P: I? = 21¢ dP=jL= 1.33 A: ‘50 + SGCI-o:8155§=§‘1."fl- P dP V: “-92“? = 1T(2|333)(15003 ___ [593‘ lb pflm‘n ‘7- 1:. mass ' Kv= (M W) = [.515 (Ecln. liq—2?, (ti—12) SQI'TW' St: = 102(1503 +l6uoo = 413% psi (Frame. “4—23) RT = i , KR =l (Table IQ 40) -0.01'+% ' in = Lassa (10“) = 0.013% mg: £1 = fl P NP ll} q ~010H‘3 :. [3558 lo _ kg ‘ ( Aim/m» _ (M58? 1 u s: moowmn) T? =08 3 “h SH!) (6) (1630) = “Fisuflepsi SF= M“— =5.o% 1 0.33 2 P 17%.?an U) T%= ssugigs PSi 81:3 = HIQOD(OH5%¥) 2"}:1” 55018.95 (H (1‘) Pn .—. (a Jree’rk/in 4% ME 3403-001/002 Test 1 Spring 2008 3. A proposed two-stage reverted gear reducer utilizes helical gears to provide quiet operation as shown below. The gears have a normal diametral pitch of 6 teeth/in, and a normal pressure angle of 20°. The input shaft is driven clockwise as shown in the figure by a 30 hp, 600 rpm electric motor. The helix angle for gear 1 is wl = 220 and for gear 3 the helix angle is W3 = 30°. The number of teeth on each gear is as follows: N1 = 24 teeth, N2 = 54 teeth, N3 = 22 teeth, N4 = 50 teeth. Gear 1 is a right-hand helical gear and gear 3 is a left-hand helical gear. (40 points) (a)Determine the speed and direction of rotation of shaft B and the output shaft C. (b)Draw free-body diagrams of all four gears and show clearly the directions of all force components acting between the gears and calculate the magnitude of all of these forces. Next to the magnitude of each force indicate in writing the direction of the force based on the coordinate system given below. e/Output shaft C Motor-driven input shaft A Cos {9| Cos 9.9. = 11° 3- q}; ¢t3 = die” = {Tan—4 £9:— :‘i0n-' +0le Cos “’3 005 80 \pH = 30° Rh = Pt). = P11. C05 (l), = (>035 9'1 = 5'56 kefl’lt‘n 1,133 = I?“ COS AP). -.=. 6 Gas 30 = 54‘“; ice-Hahn __ .___ Bi 3.1 _~_ 21. n1 =“B = 166.67!- rev/mh CCLO N1 :1, 5"! 600 NA :21 19;... = .22“ nq=nc =1 'Fh33 Viv/min CLU =+on"l i‘lLile = 2143" = 22a" ME 3403-001/002 Test 1 Spring2008 ' 30 5‘) Fl? 2 wt ____ 3300011 : 33°°°( 3 = 145%.‘13 15F (+% oLitech’on\ | v 878033 I 1 V: MI = M3 = 4,1858 Pc/mm l2. , 19. Fun - 13": +031 (U. -_-: lu5%."13 fig” 21° = 589.05 [HZ (-x direch'onx) a“; = The fin <th = upss‘qs tan ahqgf : 512.53 [L10 (+& oUrec‘Hon) . , F123: .1 “158.93 “3'? (—8 wKCh‘O“) d3: Eli : 11 : 4.13m. dz‘: N2- = £3— ': Qi7lin' 5056 Pt; t 9. a: 1-3 .—_ F43 L 3 2. a. - (w = F ‘0 1&3 ; + = QQS‘BFB) ( T3 qg a a3 Fug“: a} +0111 «)3 = 33m hm 30": Fug" = Fug 113’“th = 33(49 {11;} 33MB. 2: “(-0758 “a? Fat = 3:3th59 ("a ch‘rtcHon) Fag“: #133,55th (PX aUrCch'on) Fad" = tuomw mach-mm) 4 Tar = 57'2-S3 "0-? (-§.— dimch’on) F3} —.- saws MI (+2 Oh‘rech“on) 3309 {be (+‘1 CUNC‘HOA) 1‘13‘35615? (~X dirtch’on) (- a: direch’on) ...
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