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Chapter 13
13-1
dP = 17/8 = 2.125 in
dG = N2
N3
dP = 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.
13-3
NG = 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.
ContactRatio: mc = Lab/pb = 1.53/0.984 = 1.55 Ans.
See the next page for a drawing of the gears and the arc lengths.
334 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
13-5
(a) AO =
__
2.333
2
_2
+
_
5.333
2
_2
_1/2
= 2.910 in Ans.
(b) γ = tan−1(14/32) = 23.63° Ans.
_ = tan−1(32/14) = 66.37° Ans.
(c) dP = 14/6 = 2.333 in,
dG = 32/6 = 5.333in Ans.
(d) From Table 13-3, 0.3AO = 0.873 in and 10/P = 10/6 = 1.67
0.873 < 1.67 ∴ F = 0.873 in Ans.
13-6
(a) pn = π/5 = 0.6283 in
pt = pn/cosψ = 0.6283/cos 30° = 0.7255 in
px = pt/tanψ = 0.7255/tan 30° = 1.25 in
30_
P G
21
3
"
51
3
"
AO
_
_
10.5_
Arc of approach _ 0.87 in Ans.
Arc of recess _ 0.77 in Ans.
Arc of action _ 1.64 in Ans.
Lab _ 1.53 in
10_
O2
O1
14_ 12.6_P
A B
Chapter 13 335
(b) pnb = pn cos φn = 0.6283 cos 20° = 0.590 in Ans.
(c) Pt = Pn cosψ = 5 cos 30° = 4.33 teeth/in
φt = tan−1(tan φn/cosψ) = tan−1(tan 20°/cos 30◦) = 22.8° Ans.
(d) Table 13-4:
a = 1/5 = 0.200 in Ans.
b = 1.25/5 = 0.250 in Ans.
dP = 17
5 cos 30°
= 3.926 in Ans.
dG = 34
5 cos 30°
= 7.852 in Ans.
13-7
φn = 14.5°, Pn = 10 teeth/in
(a) pn = π/10 = 0.3142 in Ans.pt = pn
cosψ
= 0.3142
cos 20°
= 0.3343 in Ans.
px = pt
tanψ
= 0.3343
tan 20°
= 0.9185 in Ans.
(b) Pt = Pn cosψ = 10 cos 20° = 9.397 teeth/in
φt = tan−1
_
tan 14.5°
cos 20°
_
= 15.39° Ans.
(c) a = 1/10 = 0.100 in Ans.
b = 1.25/10 = 0.125 in Ans.
dP = 19
10 cos 20°
= 2.022 in Ans.
dG = 57
10 cos 20°
= 6.066 in Ans.
G
20_
P
336 Solutions Manual • Instructor’s SolutionManual to Accompany Mechanical Engineering Design
13-8 From Ex. 13-1, a 16-tooth spur pinion meshes with a 40-tooth gear, mG = 40/16 = 2.5.
Equations (13-10) through (13-13) apply.
(a) The smallest pinion tooth count that will run with itself is found from Eq. (13-10)
NP ≥ 4k
6 sin2 φ
_
1 +
_
1 + 3 sin2 φ
_
≥ 4(1)
6 sin2 20°
_
1 +
_
1 + 3 sin2 20°
_
≥ 12.32→13 teeth Ans.
(b) Thesmallest pinion that will mesh with a gear ratio of mG = 2.5, from Eq. (13-11) is
NP ≥ 2(1)
[1 + 2(2.5)] sin2 20°

2.5 +
_
2.52 + [1 + 2(2.5)] sin2 20°

≥ 14.64→15 pinion teeth Ans.
(c) The smallest pinion that will mesh with a rack, from Eq. (13-12)
NP ≥ 4k
2 sin2 φ
= 4(1)
2 sin2 20°
≥ 17.097→18 teeth Ans.
(d) The largest gear-tooth count possible to mesh with this pinion, from Eq.(13-13) is
NG ≤
N2P
sin2 φ − 4k2
4k − 2NP sin2 φ
≤ 132 sin2 20° − 4(1)2
4(1) − 2(13) sin2 20°
≤ 16.45→16 teeth Ans.
13-9 From Ex. 13-2, a 20° pressure angle, 30° helix angle, pt = 6 teeth/in pinion with 18 full
depth teeth, and φt = 21.88°.
(a) The smallest tooth count that will mesh with a like gear, from Eq. (13-21), is
NP ≥ 4k cosψ
6 sin2 φt
_
1 +
_
1 + 3 sin2 φt
_
≥ 4(1) cos30°
6 sin2 21.88°
_
1 +
_
1 + 3 sin2 21.88°
_
≥ 9.11→10 teeth Ans.
(b) The smallest pinion-tooth count that will run with a rack, from Eq. (13-23), is
NP ≥ 4k cosψ
2 sin2 φt
≥ 4(1) cos 30◦
2 sin2 21.88°
≥ 12.47→13 teeth Ans.
Chapter 13 337
(c) The largest gear tooth possible, from Eq. (13-24) is
NG ≤
N2P
sin2 φt − 4k2 cos2 ψ
4k cosψ − 2NP sin2 φt
≤ 102 sin2 21.88° − 4(12) cos2...
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