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Unformatted text preview: SOLUTION {15.3} Known. A pinion with 32 teeth and 8 diametra] pitch meshes with a gear having 55
teeth. Find: Calculate the standard center distance. Schematic and Given Data: Assumptions: 1. The gears are spur gears. 2. The gears have teeth of standard involute profile.
3. The gears mesh along their pitch circles. Analysis:
1. From Eq. (15.3): F = Nl'd = Npidp = Name 2. P = B. N], = 32. Hence. {in = 4.121 in.
3. N3 = :55. Hence, IdE = 3.125 in. rIgrrlp 2 = 6.0625 in. I 4. The center distance, c = Comments:
1. [f the gears did not mesh at the theoretical pitch circles the measured (actual) center distance would not be equal to the sum of the theorelical pitch circle radii of the gears. _ _
2. It should be evident that meshing gears must have the same diametral pitch. SOLUTION [15.4) Known: A spur gear has a size 8 diametral pitch. Find: Calculate the thickness of the spur gear tooth meaeured along the pitch eirete. Schematic and Given Data: Assumption: The gear has teeth of standard involute profile. Analysis:
1. From . “5.5); pP=nt
P = 8. enee, p = 0.392"! in. 2. Tooth thickness. 1: pa? : 1: 0.1963 in. I SOLUTION (15.5) Known: A pair of gears with a known gear ratio and at a specified center distance have
a diametrai pitch of 6. Find: Determine the number of teeth in each gear. Schematic and Given Data: Assumptions: 1. The gears are spur gears. 2. The gears have teeth of standard involute proﬁle.
3. The gears mesh along their pitch circles.
Analysis:. A gear ratio of 2:1 indicatea a 2:1 ratio in the gear diameters. Hence (1% = 2511}.
2. Center distance, c = {dg + dpitz. Hence.
{is + dp = it] in.
3. Substituting dg = up in part 2, we obtain
3dr! 2 lift in. Hence, dp = lﬂi’3 in.
4. Diametral pitch, P = Npi‘tlp, and Np = { lﬂt3}{6] = 2i]. I
5 Since the gear ratio is 2:1, the number of teeth on the gear, N3: 2N};
Therefore, ME = 40. I
Comments:
1. If the gear teeth were not of inrolute proﬁle it would still be possible to have a
constant speed ratio provided it is ensured that the pitch point is stationary.
2. It should be evident that meshing gears must have the We diametral pitch.
3. If the diametral pitch were chosen to be higher, then the number of teeth would be greater on both the pinion and gear {other parameters being kept the same). SOLUTION (15.13] Known: A pair of spur gears of known velocity ratio. center distance. diametrai pitch,
and pressure angle are given. Find: Draw a fullsize layout of the spur gears and label the following: (a) pitch circle, {b} base circle, (c) pressure angle, {d} addendum (for both the pinion and the gear), (e)
dedendttm (for the canton only} Assumption: The gears mesh along their Pitch circles. Schematic and Given Data and Analysis: {e} nectend um
{d} Addendu {h} Base circles inJ Pitch
circles "
Interference docs oxis
ucar addendum should
be. cut back this much. i. \ Gear axis For velocin ratio =45 d3=4dp
Fore: lﬂin.,dl +dP = min.
Therefore, :1 I.+ MP = 20 in.
amid], =4i11.,r_n = Iﬁin. Addendum =—; = 0.2 in. Dedendum = '35 Note: This drawing is not drawn to scale. SOLUTION {15.16)
Known: A gear has a known outside diameter, diametra] pitch and pressure angle. Find: Determine the pitch diameter, the circular pitch, the addendum, the dedendum or.
the gear, and the number of gear teeth. Schematic and Given Data: Assumptions:
I. The gear is a spur gear.
2. The gear has teeth of standard involute proﬁle. Analysis: I. Addendum. a = HF = 0.05 in.
Pitch diameter, {IF = [(outside diameter)  221]
do = 3.0  (EMUUS) = 2.9 in. 2. Module, rn = UP = [mi in. = 1.27 min. From Eq (15.5}, circular pitch, p = nrn.
Therefore, p = 3.99 mm. 3. Dedendutn, d = 1.25 a = 1236.05} = 0.0625 in.
Hence, d = {1.0625 in.
From Eq (15.4), the number of gear teeth, N = dpfm
Therefore, N = 53. There is no prepared solution for Problem 15.21. __,_—__._—_——
SOLUTION (15.22} Known: A pair of standard spur gears of known pressure angle. center distance and
velocity ratio is given. Number of teeth on pinion is specied. Find: Compute the contact ratio using equations in Section 15.3 and compare with
graphical results of Problem 15.21. Schematic and Given Data: Path of contact
nl in. Note: This drawing is not drawn to scale. Assumption: The gears mesh along their pitch circles. Analysis:
1. From Eq. (15.11). rs], = rp cos 4: = Zﬂ cos 20° = 1.379 in.
rag = rg cos it! = 8 cos 20° = 15171n. From Eq (15.8).
"ﬁe .
l'ap(max):‘lfl.379 + lee2 51.1220 =39} m. Tagﬁnax} = «H.5172 + 10.02 sinzZDD = 3.25 in. I {this agrees with graphical snlutien]
2. From Eq. (15.9), (with raw; = 2.2 and Tag = 8.2 from Problem 15.21) CR=J122LL3792 +43.2'="Ee%1._meme 0.590
CR = 1.69 (which is more acCurate than the graphical solution.) I Comment: The contact ratio can be increased by cheesing a greater number of teeth on
the gears andJer increasing the diamelral pitch. (15.25}
Known: A. two stage spur gear speed redueer of specified georrtetrj.r is given. Find: Determine the radial loads applied to the countershaft beatings.
Schematic and Given Data: K Coupled [u EDIbin.
torque meter rotating at Elm
rpm Assumptions: l. The gears mesh aleng their pitch circles. 2. Friction losses in gears and bearings can be neglected (given). 3. The shafts are all parallel. 4. All the gear radial and tangential load is transferred at the pitch point. 5. Bending deﬂection of the eoantershaft is negligible. 6. Gear face width and beating widths are negligible relative to countershaft length.
7'. The gears are rigidly connected to their shafts. Analysis: 35.5 tm 25"= 16.58 1h 29 Ih( gilnl “1) = 35.5515 Hiram“; 9.53 In 1. Fitch diameter of inpui pinion = 3; 2193 = 2 in. 2. For the horizontal plane:
EMA =0:
2101b (2 in.) + 35.55 lb [ID in} — EHUZ in): {1‘
hence, EH = 32.96 lb
EFzﬂ: 2le+35.551h— 32.961h—AH=U
hence, AH = 22.59 lb 3. For the vertical plane:
EMA. = D:
9.33 lb {2 in.)  16.58 lb (1D in.) + BAH in.) = (3!
hence, B... =12.‘25]b
EF=D:  9.33 lh+ 16.5311:  ]2.26 lb + Air = '3
hence, A» = 51]! lb 4. The bearin radial loads are:
Ami = 42159:... 501 E = 23.14 lb, hence, Ami = 23.141]: Brad = «#3195i + 12.26! = 35—171b,henc=2.13m.i = 35. 11' lb Comments: 1. The eﬁ'cet of considering friction losses in the gears and hearings is to reduce the
torque transmitted to the output shaft While keeping the speed ratios the same. A
reduction in torque transmitted will result in lower gear teeth leads and hence
lower radial loads on the conntershaft bearings. The pinion on the ceuntershaft has higher teeth leads than the gear on the
cenntershaft because the pinion has a smaller radius while transmitting the same
torque. The higher teeth load on the pinion leads to a higher radial lead on the
heating closer to the pinienr The effect of choosing a smaller diametral pitch for the gears in the second reduction stage is to provide larger teeth to withstand the higher torques and teeth
leads of the mud stage. If load sharing between teeth is considered, the transfer of gear teeth forces is not
strictly at the pitch point and will lead to different radial forces in the hearings. am (1 . ) Known: A two stage spur gear speed reducer is given which uses a countershaft and
has identical gear pairs in each stage. lGear and shaft geometry is specified such that [lit
input and output shafts are collinear. Shafts and mountings correspond to good
industrial practice but not "high precision". Find: Determine a design of the gears for lIlZIIF‘F cycles with 99% reliability and safer
factor of 1.2. Schematic and Given Data: Load
m rpm
(Mod. shock} ITDUrpm [Light shock) I Decisions:
1. Choose steel for gear material with pinion material 10% harder than the gear
material. 2. Choose standard full depth teeth with pressure angle, 41 = 21]”. 3. Select number of teeth for pinion, N1) = 212}. 4. Choose manufacturing precision between curves C and D in Fi g. 15.24 to estimati
velocity factor, Ks. 5. Choose face width for gears as h = 113. 6. The surface hardness and core hardness for the teeth are equal.
7. The tooth fillet radius is 0.35}? {to enable use of Fig. 15.23{a) to estimate J). Assumptions: 1. The gears are mounted at their theoretical center distance. 2. Friction losses in gears and hearings can be neglected. 3. No load sharing is expected and all tooth toads are transmitted at the pitch point.
4. The operating temperature for the gears is below rso ”F. 5. Surface stress can be estimated by approximating tooth contact region by cylinders.
Surface stress distribution is unaffected by lubricant.
Surface stress due to sliding friction is negligible. ti. T. Design Analysis: 1. With a center distance of 3 in., the 9:1 reduction requires 3:1 reduction by
each gear set.
Hence. dp = 4il1.., d3 = 12 in. 2. was 20pinionteeth,P=24ﬂ.P=5 I 3. Withb=%.b=2.4in. I 4. 5. 6. 1G. Pitch line velocities are, V = {#122111 * 2700' = 232'? fprn (high speed set} and similarly V = 942 fpm (low speed set) With manufacturing precision between curves C and D, from Fig. [5.24.
K1; = 1.? [low speed} and 2.? (high speed} Since FL « K]! is the greatest on the low speed set, we design the gears for
this application. . FIV  Ftﬁ942)
W=m.1ﬂ=m:ﬁ=350lb, with a safety factor of 1.2, F: = 350(1.2[1) = 421} lb Having chosen steel gears with in = 2G0, we ﬁnd hardness needed for surface
fatigue criterion: from Eq. {15.24}: _ 420
on _ BUﬂqX——2_ C“ 4110.12} (1.7111.51n.6> = 8831” psi
where, from Eq. (15.23}, I: sinZﬂoeosZG", 3 = 0.12 2 4
since, from Table 15.431, C1: = 2311!} psi, from Table 15.1, K0 = 1.5. from Table 15.2, Km = 1.6 and K» =1.'?. Equating stress and strength, 33.? ksi = SH = Sr; Cu CR 83.? = [ll=1 Bhn — lﬂ)(1}(i): henoe.B1n = 24? s eeify gear hardness as 251} Bhn, pinion hardness as 275 Ehn o verify that the above solution is adequate for bending fatigue:
from Eq. {15.17}: 420 s
U: 2.46.211} I1'1'3')(153u:1.a)= 14.3?5 psi where. from Fig. 15.22%}, J = [1.24,
and Ky :11 Kc. =1.5, Km =1.6.
from Eq. (15.13): 3:. = 2:5 (l](ﬂ.85){[l.'?l]{.314)(1.4} = 47.233 as = 47,233 psi where. s..* = ifﬂhn} Itsi
CL = Li}, CG = 1135. C3 = £1.71 for .. = % =131s Itsi. It: 2 0.314 from Table 15.3. 111:1, km; = 1.4 from 511.05.19] hence the gears are more than adequate to resist bending fatigue.
Design results: c = E in., h: 2.4 in. Np = 2121, N3 = 61.1 , standard full depth teeth. P: 5, o = 20". steel gears, pinion hardness 2'15 Bhn, gear hardness 250
Him. 11., = 4 in., dg =12 in. Tooth ﬁllet radius = g = 0.07 in. Manufacturing precision between curves C and D in Fig. 15.24. Shafts and
mountings of ordinary good engineering practice. I Comments: 1. Specifying a higher surface hardness and a lower core hardness for the gear teeth
would have resulted in a more balanced factor of safer for bending fatigue and
surface fatigue. 2. By specifying a material of 250 Bhn using design calculations which required a
hardness of 241' Bhn for the pinion1 separate design calculations for the gear were
avoided. The pinion is always more severely stressed than the corresponding gear.
By selecting plnion material 113% harder than the gear an additional factor of
safety is provided. 3. If a larger number of teeth for the pinion were selected, the diametral pitch would
have been larger and a proportionately smaller face width could be selected. These
decisions would result in a higher value of bending stress as well as a higher value
of surface stress thus requiring harder gear material speciﬁcations. S (I . ) Known: A planetary gear train with double planets, two suns and no ring gear is given.
Numbers of teeth on the planets and one of the Suns are speciﬁed. ISine sun is the input
member, the other sun is fixed and the arm is the output member. Find: Determine the inputoutput speed ratio.
Schematic and Given Data:
P] (40 teeth
P2 {31 teeth] Assumption: The gears mesh along their pitch circles.
Analysis:
1. od For a unit clockwise torque applied to SI. forces on the planet—pair are as follows.
(Let 31. P1. etc. represent relative radii of the members): Summing moments to zero: 2Mo=ﬂzF2—S—ll%2l— Summing forces to zero: 29:0 : henee,Fg=Sll (1 g; Therefore, am} torque = FD (arm radius}
i _ £1.
= 31[1 Pzim +13” = ﬁt: @300) = 415333 For 100 ‘ﬁ: efﬁciency, l—I“ : 9.5.1.
SI WA Artﬁtraijr unit
velocity vector We assign unit velocity, as shown in the figure.
let P1, S], em represent relative radii. P2 From known point of zero Ivelocity. we determine planet velocity =  P1 _ P2 From to = Wr: m. = ( PIP—2132).ﬂ = DEL1) 30 — 1314 m5] Sl+Pl 1 30+4o T"
Inputwoutput speed ratio, 9’4 = 4314
ms: Comments: 1. As this problem illustrates, it is not essential to have a ring gear to achieve speed
changes in planetaryr gear trains. Here, the ring {an internal gear) is replaced by a
second sun (an exremal gear} to perform the same function. 2. The use of two suns and two planets allows more ﬂexibility.r in achieving speed
ratio than if a ring gear 1were used because the pairs S 1, F1 and 52. P2 can be
independently chosen as long as 51 + P1 = 52 + P2. set. I NU. Known: A planetary gear train with double planets, two suns, and no ring gear is given. Numbers of teeth on planets and suns are speciﬁed. The arm acts as the input member,
One of the suns is ﬁxed and the other sun acts as the output member. Find: Determine the inputwolltpttt speed ratio. Schematic and Given Data: Pl (101mm P2 (In: teeth}
a: (99 teelJ'I) Dutput S] [100 teeth} Assumption: All gears mesh along their pitch circles. Analysis:
1. Let the center distance be unity; i.e.. .1. ._1 _ .
2(P1+51)._2{l:'12‘.+32}— 1 , Pl _ ~ 31 _ 
Pt +31 —radiusofPl. P1 +51 —radineof51. etc. 2. Rotate input member [arm A) so as to give the unit linear velocity vector
shown; thus . . 3. Using the velocity.r vector method. mout=UJSI=—§f'l—"
[81+P1)
where the pitch line velocity of $1 is
t .. ( .. J
V: P2+52 P1+Sl =_l§’2Pl
( m 1 P2
P2+Sz Thus, PZPl
( P2 ]__(P2P1](SI+PE) mm = (H31_) ‘ _(P2)(S I)
S] + P1
' {102x100} ' 0'0””
and mun"! = +0319? I win Pl {iﬂi}
HUN} 1 {Unit distance} Comment: Increasing the number of teeth on all the gem while keeping the difference
in number of teeth, SI  SE = P2  P1 = 1, will produce an even lower speed ratio: F99!" =£PELPI__)[51 +P_1] 31+ Pl 1 1 we. (szsu ”ﬁﬁi‘j‘m + s: ...
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 Fall '08
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