Chapter 13 13-1 d P = 17/8 = 2.125 in d G = N 2 N 3 d P = 1120 544 (2.125) = 4.375 in N G = Pd G = 8(4.375) = 35 teeth Ans. C = (2.125 + 4.375) /2 = 3.25 in Ans. 13-2 n G = 1600(15/60) = 400 rev/min Ans. p = π m = 3π mm Ans. C = [3(15 + 60)]/2 = 112.5 mm Ans. 13-3 N G = 20(2.80) = 56 teeth Ans. d G = N G m = 56(4) = 224 mm Ans. d P = N P m = 20(4) = 80 mm Ans. C = (224 + 80) /2 = 152 mm Ans. 13-4 Mesh: a = 1/ P = 1/3 = 0.3333 in Ans. b = 1.25/ P = 1.25/3 = 0.4167 in Ans. c = b − a = 0.0834 in Ans. p = π/ P = π/3 = 1.047 in Ans. t = p/2 = 1.047/2 = 0.523 in Ans. Pinion Base-Circle: d 1 = N 1 / P = 21/3 = 7 in d 1b = 7 cos 20° = 6.578 in Ans. Gear Base-Circle: d 2 = N 2 / P = 28/3 = 9.333 in d 2b = 9.333 cos 20° = 8.770 in Ans. Base pitch: p b = p c cos φ = ( π/3) cos 20° = 0.984 in Ans. Contact Ratio: m c = L ab / p b = 1.53/0.984 = 1.55 Ans. See the next page for a drawing of the gears and the arc lengths.
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FIRST PAGES
Chapter 13
13-1
dP = 17/8 = 2.125 in
dG =N2
N3dP =
1120
544(2.125) = 4.375 in
NG = PdG = 8(4.375) = 35 teeth Ans.
C = (2.125 + 4.375)/2 = 3.25 in Ans.
13-2
nG = 1600(15/60) = 400 rev/min Ans.
p = πm = 3π mm Ans.
C = [3(15 + 60)]/2 = 112.5 mm Ans.
13-3NG = 20(2.80) = 56 teeth Ans.
dG = NGm = 56(4) = 224 mm Ans.
dP = NPm = 20(4) = 80 mm Ans.
C = (224 + 80)/2 = 152 mm Ans.
13-4 Mesh: a = 1/P = 1/3 = 0.3333 in Ans.
b = 1.25/P = 1.25/3 = 0.4167 in Ans.
c = b − a = 0.0834 in Ans.
p = π/P = π/3 = 1.047 in Ans.
t = p/2 = 1.047/2 = 0.523 in Ans.
Pinion Base-Circle: d1 = N1/P = 21/3 = 7 in
d1b = 7 cos 20° = 6.578 in Ans.
Gear Base-Circle: d2 = N2/P = 28/3 = 9.333 in
d2b = 9.333 cos 20° = 8.770 in Ans.
Base pitch: pb = pc cos φ = (π/3) cos 20° = 0.984 in Ans.
Contact Ratio: mc = Lab/pb = 1.53/0.984 = 1.55 Ans.
See the next page for a drawing of the gears and the arc lengths.
(a) The smallest pinion tooth count that will run itself is found from Eq. (13-21)
NP ≥ 2k cos ψ
3 sin2 φt
(1 +
√1 + 3 sin2 φt
)
≥ 2(1) cos 30°
3 sin2 22.80°
(1 +
√1 + 3 sin2 22.80°
)
≥ 8.48 → 9 teeth Ans.
(b) The smallest pinion that will mesh with a gear ratio of m = 2.5, from Eq. (13-22) is
NP ≥ 2(1) cos 30°
[1 + 2(2.5)] sin2 22.80°
{2.5 +
√2.52 + [1 + 2(2.5)] sin2 22.80°
}
≥ 9.95 → 10 teeth Ans.
The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-23) is
NG ≤ 102 sin2 22.80° − 4(1) cos2 30°
4(1) cos2 30° − 2(20) sin2 22.80°
≤ 26.08 → 26 teeth Ans.
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Chapter 13 329
(c) The smallest pinion that will mesh with a rack, from Eq. (13-24) is
NP ≥2(1) cos 30°
sin2 22.80°
≥ 11.53 → 12 teeth Ans.
13-10 Pressure Angle: φt = tan−1(
tan 20°
cos 30°
)= 22.796°
Program Eq. (13-24) on a computer using a spreadsheet or code and increment NP . Thefirst value of NP that can be doubled is NP = 10 teeth, where NG ≤ 26.01 teeth. So NG =20 teeth will work. Higher tooth counts will work also, for example 11:22, 12:24, etc.
Use 10:20 Ans.
13-11 Refer to Prob. 13-10 solution. The first value of NP that can be multiplied by 6 isNP = 11 teeth where NG ≤ 93.6 teeth. So NG = 66 teeth.
Use 11:66 Ans.
13-12 Begin with the more general relation, Eq. (13-24), for full depth teeth.
NG =N 2
P sin2 φt − 4 cos2 ψ
4 cos ψ − 2NP sin2 φt
For a rack, set the denominator to zero4 cos ψ − 2NP sin2 φt = 0
13-14 (a) The axial force of 2 on shaft a is in the negative direction. The axial force of 3 onshaft b is in the positive direction of z. Ans.
The axial force of gear 4 on shaft b is in the positive z-direction. The axial force ofgear 5 on shaft c is in the negative z-direction. Ans.
(b) nc = n5 =14
54
(16
36
)(900) = +103.7 rev/min ccw Ans.
(c) dP2 = 14/(10 cos 30°) = 1.6166 in
dG3 = 54/(10 cos 30°) = 6.2354 in
Cab =1.6166 + 6.2354
2= 3.926 in Ans.
dP4 = 16/(6 cos 25°) = 2.9423 in
dG5 = 36/(6 cos 25°) = 6.6203 in
Cbc = 4.781 in Ans.
13-15 e =20
40
(8
17
)(20
60
)=
4
51
nd =4
51(600) = 47.06 rev/min cw Ans.
5
4
c
bz
a
3
z
2
b
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Chapter 13 331
13-16
e =6
10
(18
38
)(20
48
)(3
36
)=
3
304
na =3
304(1200) = 11.84 rev/min cw Ans.
13-17
(a) nc =12
40·
1
1(540) = 162 rev/min cw about x . Ans.
(b) dP = 12/(8 cos 23°) = 1.630 in
dG = 40/(8 cos 23°) = 5.432 in
dP + dG
2= 3.531 in Ans.
(c) d =32
4= 8 in at the large end of the teeth. Ans.
13-18 (a) The planet gears act as keys and the wheel speeds are the same as that of the ring gear.Thus
n A = n3 = 1200(17/54) = 377.8 rev/min Ans.
(b) nF = n5 = 0, nL = n6, e = −1
−1 =n6 − 377.8
0 − 377.8
377.8 = n6 − 377.8
n6 = 755.6 rev/min Ans.
Alternatively, the velocity of the center of gear 4 is v4c ∝ N6n3 . The velocity of theleft edge of gear 4 is zero since the left wheel is resting on the ground. Thus, the ve-locity of the right edge of gear 4 is 2v4c ∝ 2N6n3. This velocity, divided by the radiusof gear 6 ∝ N6, is angular velocity of gear 6–the speed of wheel 6.
∴ n6 =2N6n3
N6= 2n3 = 2(377.8) = 755.6 rev/min Ans.
(c) The wheel spins freely on icy surfaces, leaving no traction for the other wheel. Thecar is stalled. Ans.
13-19 (a) The motive power is divided equally among four wheels instead of two.
(b) Locking the center differential causes 50 percent of the power to be applied to therear wheels and 50 percent to the front wheels. If one of the rear wheels, rests ona slippery surface such as ice, the other rear wheel has no traction. But the frontwheels still provide traction, and so you have two-wheel drive. However, if the reardifferential is locked, you have 3-wheel drive because the rear-wheel power is nowdistributed 50-50.
Each bearing on shaft a has the same radial load of RA = RB = 662/2 = 331 lbf.
Gear 3
W t23 = W t
32 = 622 lbf
Wr23 = Wr
32 = 226 lbf
Fb3 = Fb2 = 662 lbf
RC = RD = 662/2 = 331 lbf
Each bearing on shaft b has the same radial load which is equal to the radial load of bear-ings, A and B. Thus, all four bearings have the same radial load of 331 lbf. Ans.
Notice that the idler shaft reaction contains a couple tending to turn the shaft end-over-end. Also the idler teeth are bent both ways. Idlers are more severely loaded than othergears, belying their name. Thus be cautious.