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ID: ME 3403001/002 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 xaxis. (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 3403001/002 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 throughhardened 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
FaunHons F‘ J T K1 KR
(Agmn)
”~13 N _i_L_q_ l
B: OI15( [Du(v) = O8155 P: I? = 21¢ dP=jL= 1.33
A: ‘50 + SGCIo:8155§=§‘1."ﬂ P dP
V: “92“? = 1T(2333)(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 3403001/002 Test 1
Spring 2008 3. A proposed twostage 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 righthand helical gear and gear 3
is a lefthand helical gear. (40 points) (a)Determine the speed and direction of rotation of shaft B and the output shaft C.
(b)Draw freebody 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 Motordriven
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‘“; iceHahn
__ .___ 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 3403001/002 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: 13 .—_ 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 machmm)
4 Tar = 57'2S3 "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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 Spring '08
 Nersersov

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