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Chapter 14
Soluzioni degli esercizi aggiuntivi relativi al Capitolo 14, Ruote dentate cilindriche a denti
diritti e a denti elicoidali
14-1
N
22
=
= 3.667 in
P
6
Y = 0.331
d=
Table 14-2:
V =
Eq. (14-4b):
π(3.667)(1200)
πdn
=
= 1152 ft/min
12
12
Kv =
1200 + 1152
= 1.96
1200
Wt =
T
63 025H
63 025(15)
=
=
= 429.7 lbf
d/2
nd/2
1200(3.667/2)
Eq. (14-7):
σ =
14-2
Kv W t P
1.96(429.7)(6)
=
= 7633 psi = 7.63 kpsi Ans.
FY
2(0.331)
16
= 1.333 in, Y = 0.296
12
π(1.333)(700)
V =
= 244.3 ft/min
12
1200 + 244.3
Kv =
= 1.204
1200
d=
Eq. (14-4b):
Wt =
63 025H
63 025(1.5)
=
= 202.6 lbf
nd/2
700(1.333/2)
Eq. (14-7):
σ =
Kv W t P
1.204(202.6)(12)
=
= 13 185 psi = 13.2 kpsi Ans.
FY
0.75(0.296)
14-3
d = m N = 1.25(18) = 22.5 mm, Y = 0.309
π(22.5)(10−3 )(1800)
= 2.121 m/s
60
6.1 + 2.121
= 1.348
Kv =
6.1
60H
60(0.5)(103 )
Wt =
=
= 235.8 N
πdn
π(22.5)(10−3 )(1800)
V =
Eq. (14-6b):
Eq. (14-8):
σ =
Kv W t
1.348(235.8)
=
= 68.6 MPa Ans.
FmY
12(1.25)(0.309)
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Chapter 14
14-4
d = 5(15) = 75 mm, Y = 0.290
π(75)(10−3 )(200)
= 0.7854 m/s
60
Assume steel and apply Eq. (14-6b):
V =
Eq. (14-8):
Kv =
6.1 + 0.7854
= 1.129
6.1
Wt =
60H
60(5)(103 )
=
= 6366 N
πdn
π(75)(10−3 )(200)
σ =
Kv W t
1.129(6366)
=
= 82.6 MPa Ans.
FmY
60(5)(0.290)
14-5
d = 1(16) = 16 mm, Y = 0.296
π(16)(10−3 )(400)
V =
= 0.335 m/s
60
Assume steel and apply Eq. (14-6b):
Kv =
6.1 + 0.335
= 1.055
6.1
Wt =
60H
60(0.15)(103 )
=
= 447.6 N
πdn
π(16)(10−3 )(400)
F=
Eq. (14-8):
Kv W t
1.055(447.6)
=
= 10.6 mm
σ mY
150(1)(0.296)
From Table A-17, use F = 11 mm Ans.
14-6
d = 1.5(17) = 25.5 mm, Y = 0.303
V =
Eq. (14-6b):
π(25.5)(10−3 )(400)
= 0.534 m/s
60
Kv =
6.1 + 0.534
= 1.088
6.1
Wt =
60H
60(0.25)(103 )
=
= 468 N
πdn
π(25.5)(10−3 )(400)
F=
Kv W t
1.088(468)
=
= 14.9 mm
σ mY
75(1.5)(0.303)
Eq. (14-8):
Use F = 15 mm Ans.
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14-7
Eq. (14-4b):
d=
24
= 4.8 in,
5
V =
π(4.8)(50)
= 62.83 ft/min
12
Y = 0.337
Kv =
1200 + 62.83
= 1.052
1200
Wt =
63 025H
63 025(6)
=
= 3151 lbf
nd/2
50(4.8/2)
Eq. (14-7):
F=
Kv W t P
1.052(3151)(5)
=
= 2.46 in
σY
20(103 )(0.337)
d=
16
= 3.2 in,
5
V =
π(3.2)(600)
= 502.7 ft/min
12
Use F = 2.5 in Ans.
14-8
Eq. (14-4b):
Y = 0.296
Kv =
1200 + 502.7
= 1.419
1200
Wt =
63 025(15)
= 984.8 lbf
600(3.2/2)
F=
Eq. (14-7):
Kv W t P
1.419(984.8)(5)
=
= 2.38 in
σY
10(103 )(0.296)
Use F = 2.5 in Ans.
14-9
Try P = 8 which gives d = 18/8 = 2.25 in and Y = 0.309.
V =
Eq. (14-4b):
Eq. (14-7):
π(2.25)(600)
= 353.4 ft/min
12
Kv =
1200 + 353.4
= 1.295
1200
Wt =
63 025(2.5)
= 233.4 lbf
600(2.25/2)
F=
Kv W t P
1.295(233.4)(8)
=
= 0.783 in
σY
10(103 )(0.309)
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363
Chapter 14
Using coarse integer pitches from Table 13-2, the following table is formed.
P
d
V
Kv
Wt
F
2
3
4
6
8
10
12
16
9.000
6.000
4.500
3.000
2.250
1.800
1.500
1.125
1413.717
942.478
706.858
471.239
353.429
282.743
235.619
176.715
2.178
1.785
1.589
1.393
1.295
1.236
1.196
1.147
58.356
87.535
116.713
175.069
233.426
291.782
350.139
466.852
0.082
0.152
0.240
0.473
0.782
1.167
1.627
2.773
Other considerations may dictate the selection. Good candidates are P = 8 ( F = 7/8 in)
and P = 10 ( F = 1.25 in). Ans.
14-10 Try m = 2 mm which gives d = 2(18) = 36 mm and Y = 0.309.
V =
π(36)(10−3 )(900)
= 1.696 m/s
60
6.1 + 1.696
= 1.278
6.1
60(1.5)(103 )
Wt =
= 884 N
π(36)(10−3 )(900)
Kv =
Eq. (14-6b):
F=
Eq. (14-8):
1.278(884)
= 24.4 mm
75(2)(0.309)
Using the preferred module sizes from Table 13-2:
m
d
V
Kv
Wt
F
1.00
1.25
1.50
2.00
3.00
4.00
5.00
6.00
8.00
10.00
12.00
16.00
20.00
25.00
32.00
40.00
50.00
18.0
22.5
27.0
36.0
54.0
72.0
90.0
108.0
144.0
180.0
216.0
288.0
360.0
450.0
576.0
720.0
900.0
0.848
1.060
1.272
1.696
2.545
3.393
4.241
5.089
6.786
8.482
10.179
13.572
16.965
21.206
27.143
33.929
42.412
1.139
1.174
1.209
1.278
1.417
1.556
1.695
1.834
2.112
2.391
2.669
3.225
3.781
4.476
5.450
6.562
7.953
1768.388
1414.711
1178.926
884.194
589.463
442.097
353.678
294.731
221.049
176.839
147.366
110.524
88.419
70.736
55.262
44.210
35.368
86.917
57.324
40.987
24.382
12.015
7.422
5.174
3.888
2.519
1.824
1.414
0.961
0.721
0.547
0.406
0.313
0.243
Other design considerations may dictate the size selection. For the present design,
m = 2 mm ( F = 25 mm) is a good selection. Ans.
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14-11
dP =
V =
Eq. (14-4b):
22
= 3.667 in,
6
dG =
60
= 10 in
6
π(3.667)(1200)
= 1152 ft/min
12
Kv =
1200 + 1152
= 1.96
1200
Wt =
63 025(15)
= 429.7 lbf
1200(3.667/2)
C p = 2100 psi
Table 14-8:
Eq. (14-12):
3.667 sin 20°
10 sin 20°
= 0.627 in, r2 =
= 1.710 in
2
2
1
Kv W t
1 1/2
+
σC = −C p
F cos φ r1 r2
1/2
1
1.96(429.7)
1
= −2100
+
2 cos 20°
0.627 1.710
r1 =
Eq. (14-14):
= −65.6(103 ) psi = −65.6 kpsi Ans.
14-12
dP =
V =
Eq. (14-4b):
16
= 1.333 in,
12
dG =
48
= 4 in
12
π(1.333)(700)
= 244.3 ft/min
12
Kv =
1200 + 244.3
= 1.204
1200
Wt =
63 025(1.5)
= 202.6 lbf
700(1.333/2)
√
C p = 2100 psi
Table 14-8:
Eq. (14-12):
r1 =
1.333 sin 20°
= 0.228 in,
2
Eq. (14-14):
1.202(202.6)
σC = −2100
F cos 20°
F=
2100
100(103 )
2 r2 =
4 sin 20°
= 0.684 in
2
1
1
+
0.228 0.684
1.202(202.6)
cos 20°
1/2
= −100(103 )
1
1
+
0.228 0.684
= 0.668 in
Use F = 0.75 in Ans.
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Chapter 14
14-13
365
24
48
= 4.8 in, dG =
= 9.6 in
5
5
π(4.8)(50)
=
= 62.83 ft/min
12
600 + 62.83
=
= 1.105
600
63 025H
=
= 525.2H
50(4.8/2)
√
= 1960 psi
dP =
V
Kv
Eq. (14-4a):
Wt
Cp
Table 14-8:
Eq. (14-12):
Eq. (14-14):
H = 5.77 hp
4.8 sin 20◦
= 0.821 in, r2 = 2r1 = 1.642 in
2
1/2
1
1.105(525.2H )
1
3
−100(10 ) = −1960
+
2.5 cos 20◦
0.821 1.642
r1 =
Ans.
14-14
d P = 4(20) = 80 mm,
dG = 4(32) = 128 mm
π(80)(10−3 )(1000)
= 4.189 m/s
60
3.05 + 4.189
Kv =
= 2.373
3.05
V =
Eq. (14-6a):
60(10)(103 )
= 2387 N
π(80)(10−3 )(1000)
√
C p = 163 MPa
Wt =
Table 14-8:
Eq. (14-12):
Eq. (14-14):
80 sin 20°
128 sin 20°
= 13.68 mm, r2 =
= 21.89 mm
2
2
1/2
1
2.373(2387)
1
= −617 MPa
σC = −163
+
50 cos 20°
13.68 21.89
r1 =
14-15 The pinion controls the design.
Eq. (7-8):
Y P = 0.303, YG = 0.359
17
30
dP =
= 1.417 in, dG =
= 2.500 in
12
12
πd P n
π(1.417)(525)
V =
=
= 194.8 ft/min
12
12
1200 + 194.8
= 1.162
Kv =
1200
Se = 0.504(76) = 38 300 psi
Eq. (7-18):
ka = 2.70(76) −0.265 = 0.857
Bending
Eq. (14-4b):
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Ans.
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l=
Eq. (14-3):
Eq. (7-24):
Eq. (7-19):
2.25
2.25
=
= 0.1875 in
Pd
12
3Y P
3(0.303)
=
= 0.0379 in
2P
2(12)
t = 4(0.1875)(0.0379) = 0.1686 in
de = 0.808 0.875(0.1686) = 0.310 in
0.310 −0.107
= 0.996
kb =
0.30
x=
kc = kd = ke = 1,
k f1 = 1.65
0.300
= 0.025 in
12
rf
r
0.025
=
=
= 0.148
d
t
0.1686
rf =
Approximate D/d = ∞ with D/d = 3; from Fig. A-15-6, K t = 1.68.
Eq. (7-35):
Kf =
1.68
= 1.323
1.68 − 1
4
2
1+ √
1.68
76
0.025
Miscellaneous-Effects Factor:
k f = k f 1k f 2
Eq. (7-17):
1
= 1.65
1.323
= 1.247
Se = 0.857(0.996)(1)(1)(1)(1.247)(38 300)
= 40 770 psi
σall =
40 770
= 18 120 psi
2.25
Wt =
FY P σall
0.875(0.303)(18 120)
=
K v Pd
1.162(12)
= 345 lbf
H=
Wear
Eq. (14-14):
Eq. (14-12):
345(194.8)
= 2.04 hp Ans.
33 000
ν1 = ν2 = 0.292,
C p = 2300 psi
E 1 = E 2 = 30(106 ) psi
dP
1.417
sin φ =
sin 20° = 0.242 in
2
2
dG
2.500
r2 =
sin φ =
sin 20° = 0.428
2
2
1
1
1
1
+ =
+
= 6.469 in−1
r1 r2
0.242 0.428
r1 =
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Chapter 14
From Eq. (7-68),
(SC ) 108 = 0.4H B − 10 kpsi
= [0.4(149) − 10](103 ) = 49 600 psi
−49 600
= −33 067 psi
σC, all = √
2.25
−33 067 2 0.875 cos 20°
t
= 22.6 lbf
W =
2300
1.162(6.469)
22.6(194.8)
= 0.133 hp Ans.
33 000
Rating power (pinion controls):
H1 = 2.04 hp
H=
H2 = 0.133 hp
Hall = (min 2.04, 0.133) = 0.133 hp
14-16 Pinion controls: Y P = 0.322,
Bending
Ans.
YG = 0.447
d P = 20/3 = 6.667 in,
dG = 33.333 in
V = πd P n/12 = π(6.667)(870)/12 = 1519 ft/min
K v = (1200 + 1519)/1200 = 2.266
Se = 0.504(113) = 56.950 kpsi = 56 950 psi
ka = 2.70(113) −0.265 = 0.771
l = 2.25/Pd = 2.25/3 = 0.75 in
x = 3(0.322)/[2(3)] = 0.161 in
t = 4(0.75)(0.161) = 0.695 in
de = 0.808 2.5(0.695) = 1.065 in
kb = (1.065/0.30) −0.107 = 0.873
kc = kd = ke = 1
r f = 0.300/3 = 0.100 in
rf
r
0.100
=
=
= 0.144
d
t
0.695
From Table A-15-6, K t = 1.75; Eq. (7-35): K f = 1.597
k f 2 = 1/1.597,
k f = k f 1 k f 2 = 1.65/1.597 = 1.033
Se = 0.771(0.873)(1)(1)(1)(1.033)(56 950) = 39 600 psi
σall = 39 600/1.5 = 26 400 psi
Wt =
FY P σall
2.5(0.322)(26 400)
=
= 3126 lbf
K v Pd
2.266(3)
H = W t V /33 000 = 3126(1519)/33 000 = 144 hp Ans.
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Wear
Table 14-8:
Eq. (14-12):
C p = 2300 psi
r1 = (6.667/2) sin 20° = 1.140 in
r2 = (33.333/2) sin 20° = 5.700 in
Eq. (7-68):
SC = [0.4(262) − 10](103 ) = 94 800 psi
√
√
σC, all = −SC / n d = −94.800/ 1.5 = −77 404 psi
1
σC, all 2 F cos φ
t
W =
Cp
K v 1/r1 + 1/r2
1
−77 404 2 2.5 cos 20°
=
2300
2.266
1/1.140 + 1/5.700
= 1115 lbf
H=
WtV
1115(1519)
=
= 51.3 hp Ans.
33 000
33 000
For 108 cycles (revolutions of the pinion), the power based on wear is 51.3 hp.
Rating power–pinion controls
H1 = 144 hp
H2 = 51.3 hp
Hrated = min(144, 51.3) = 51.3 hp Ans.
14-17 Given: φ = 20°, n = 1145 rev/min, m = 6 mm, F = 75 mm, N P = 16 milled teeth,
NG = 30T, Sut = 900 MPa, H B = 260, n d = 3, Y P = 0.296, and YG = 0.359.
Pinion bending
d P = m N P = 6(16) = 96 mm
Eq. (14-6b):
dG = 6(30) = 180 mm
πd P n
π(96)(1145)(10−3 )(12)
V =
=
= 5.76 m/s
12
(12)(60)
6.1 + 5.76
= 1.944
Kv =
6.1
Se = 0.504(900) = 453.6 MPa
a = 4.45,
b = −0.265
ka = 4.51(900) −0.265 = 0.744
l = 2.25m = 2.25(6) = 13.5 mm
x = 3Y m/2 = 3(0.296)6/2 = 2.664 mm
√
t = 4lx = 4(13.5)(2.664) = 12.0 mm
de = 0.808 75(12.0) = 24.23 mm
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Chapter 14
kb =
24.23
7.62
−0.107
= 0.884
kc = kd = ke = 1
r f = 0.300m = 0.300(6) = 1.8 mm
From Fig. A-15-6 for r/d = r f /t = 1.8/12 = 0.15, K t = 1.68.
Kf =
1.68
= 1.537
√ 1 + 2 1.8 [(1.68 − 1)/1.68](139/900)
k f 1 = 1.65
(Gerber failure criterion)
k f 2 = 1/K f = 1/1.537 = 0.651
k f = k f 1 k f 2 = 1.65(0.651) = 1.074
Se = 0.744(0.884)(1)(1)(1)(1.074)(453.6) = 320.4 MPa
Eq. (14-8):
σall =
Se
320.4
=
= 246.5 MPa
nd
1.3
Wt =
FY mσall
75(0.296)(6)(246.5)
=
= 16 890 N
Kv
1.944
H=
16 890(96/2)(1145)
Tn
=
= 97.2 kW
9.55
9.55(106 )
Ans.
Wear: Pinion and gear
r1 = (96/2) sin 20◦ = 16.42 mm
Eq. (14-12):
r2 = (180/2) sin 20◦ = 30.78 mm
Table 14-8:
√
˙ 191 MPa
Cp =
Eq. (7-68):
SC = 6.89[0.4(260) − 10] = 647.7 MPa
Eq. (14-14):
647.7
σC, all = − √
= −568 MPa
1.3
1
σC, all 2 F cos φ
t
W =
Cp
K v 1/r1 + 1/r2
2 1
75 cos 20◦
−568
=
191
1.944
1/16.42 + 1/30.78
= 3433 N
W t dP
3433(96)
=
= 164 784 N · mm = 164.8 N · m
2
2
Tn
164.8(1145)
H=
=
= 19 758.7 W = 19.8 kW Ans.
9.55
9.55
Thus, wear controls the gearset power rating; H = 19.8 kW. Ans.
T =
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14-18 Preliminaries: N P = 17,
NG = 51
dP =
N
17
=
= 2.833 in
Pd
6
51
= 8.500 in
6
V = πd P n/12 = π(2.833)(1120)/12 = 830.7 ft/min
dG =
K v = (1200 + 830.7)/1200 = 1.692
Eq. (14-4b):
σall =
Sy
90 000
=
= 45 000 psi
nd
2
Table 14-2:
Y P = 0.303,
Eq. (14-7):
Wt =
H=
YG = 0.410
FY P σall
2(0.303)(45 000)
=
= 2686 lbf
K v Pd
1.692(6)
Wt V
2686(830.7)
=
= 67.6 hp
33 000
33 000
Based on yielding in bending, the power is 67.6 hp.
(a) Pinion fatigue
Bending
Se = 0.504(232/2) = 58 464 psi
a = 2.70,
Table 13-1:
Eq. (14-3):
l=
b = −0.265,
ka = 2.70(116) −0.265 = 0.766
1
1.25
2.25
2.25
+
=
=
= 0.375 in
Pd
Pd
Pd
6
3(0.303)
3Y P
=
= 0.0758
2Pd
2(6)
√
t = 4lx = 4(0.375)(0.0758) = 0.337 in
√
de = 0.808 Ft = 0.808 2(0.337) = 0.663 in
0.663 −0.107
= 0.919
kb =
0.30
x=
kc = kd = ke = 1. Assess two components contributing to k f . First, based upon
one-way bending and the Gerber failure criterion, k f 1 = 1.65. Second, due to stressconcentration,
rf =
Fig. A-15-6:
0.300
0.300
=
= 0.050 in
Pd
6
rf
r
0.05
=
=
= 0.148
d
t
0.338
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Chapter 14
Estimate D/d = ∞ by setting D/d = 3, K t = 1.68. From Eq. (7-35) and Table 7-8,
Kf =
1.68
= 1.494
√
1 + 2 0.05 [(1.68 − 1)/1.68](4/116)
kf 2 =
1
1
=
= 0.669
Kf
1.494
k f = k f 1 k f 2 = 1.65(0.669) = 1.104
Se = 0.766(0.919)(1)(1)(1)(1.104)(58 464) = 45 436 psi
σall =
Se
45 436
=
= 22 718 psi
nd
2
Wt =
FY P σall
2(0.303)(22 718)
=
= 1356 lbf
K v Pd
1.692(6)
H=
WtV
1356(830.7)
=
= 34.1 hp
33 000
33 000
Ans.
(b) Pinion fatigue
Wear
From Table A-5 for steel:
ν = 0.292, E = 30(106 ) psi
Eq. (14-13) or Table 14-8:
1/2
1
Cp =
=
2285
psi
2π[(1 − 0.2922 )/30(106 )]
In preparation for Eq. (14-14):
Eq. (14-12):
r1 =
dP
2.833
sin φ =
sin 20◦ = 0.485 in
2
2
dG
8.500
sin φ =
sin 20◦ = 1.454 in
r2 =
2
2
1
1
1
1
=
+
+
= 2.750 in
r1 r2
0.485 1.454
Eq. (7-68):
(SC ) 108 = 0.4H B − 10 kpsi
In terms of gear notation
σC = [0.4(232) − 10]103 = 82 800 psi
We√will introduce the design factor of n d = 2 and apply it to the load W t by dividing
by 2.
σc
82 800
σC, all = − √ = − √
= −58 548 psi
2
2
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Solve Eq. (14-14) for W t:
W =
t
Hall =
−58 548
2285
2 2 cos 20◦
= 265 lbf
1.692(2.750)
265(830.7)
= 6.67 hp
33 000
Ans.
For 108 cycles (turns of pinion), the allowable power is 6.67 hp.
(c) Gear fatigue due to bending and wear
Bending
3(0.4103)
3YG
=
= 0.1026 in
2Pd
2(6)
t = 4(0.375)(0.1026) = 0.392 in
de = 0.808 2(0.392) = 0.715 in
0.715 −0.107
kb =
= 0.911
0.30
x=
Eq. (14-3):
kc = kd = ke = 1
rf
r
0.050
=
=
= 0.128
d
t
0.392
Approximate D/d = ∞ by setting D/d = 3 for Fig. A-15-6; K t = 1.80. Use K f =
1.583.
kf 2 =
1
= 0.632, k f = 1.65(0.632) = 1.043
1.583
Se = 0.766(0.911)(1)(1)(1)(1.043)(58 464) = 42 550 psi
σall =
Se
42 550
=
= 21 275 psi
2
2
Wt =
2(0.4103)(21 275)
= 1720 lbf
1.692(6)
Hall =
1720(830.7)
= 43.3 hp
33 000
Ans.
The gear is thus stronger than the pinion in bending.
Wear Since the material of the pinion and the gear are the same, and the contact
stresses are the same, the allowable power transmission of both is the same. Thus,
Hall = 6.67 hp for 108 revolutions of each. As yet, we have no way to establish SC for
108 /3 revolutions.
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Chapter 14
(d) Pinion bending: H1 = 34.1 hp
Pinion wear: H2 = 6.67 hp
Gear bending: H3 = 43.3 hp
Gear wear: H4 = 6.67 hp
Power rating of the gear set is thus
Hrated = min(34.1, 6.67, 43.3, 6.67) = 6.67 hp
Ans.
14-19 d P = 16/6 = 2.667 in, dG = 48/6 = 8 in
π(2.667)(300)
= 209.4 ft/min
12
33 000(5)
Wt =
= 787.8 lbf
209.4
V =
Assuming uniform loading, K o = 1. From Eq. (14-28),
Q v = 6, B = 0.25(12 − 6) 2/3 = 0.8255
A = 50 + 56(1 − 0.8255) = 59.77
Eq. (14-27):
Kv =
From Table 14-2,
0.8255
√
59.77 + 209.4
= 1.196
59.77
N P = 16T, Y P = 0.296
NG = 48T, YG = 0.4056
From Eq. (a), Sec. 14-10 with F = 2 in
√
0.0535
2 0.296
= 1.088
(K s ) P = 1.192
6
√
0.0535
2 0.4056
(K s ) G = 1.192
= 1.097
6
From Eq. (14-30) with Cmc = 1
2
− 0.0375 + 0.0125(2) = 0.0625
10(2.667)
= 1, Cma = 0.093 (Fig. 14-11),
Ce = 1
Cp f =
C pm
K m = 1 + 1[0.0625(1) + 0.093(1)] = 1.156
Assuming constant thickness of the gears → K B = 1
m G = NG /N P = 48/16 = 3
With N (pinion) = 108 cycles and N (gear) = 108 /3, Fig. 14-14 provides the relations:
(Y N ) P = 1.3558(108 ) −0.0178 = 0.977
(Y N ) G = 1.3558(108 /3) −0.0178 = 0.996
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J P = 0.27,
Fig. 14-6:
JG =
˙ 0.38
From Table 14-10 for R = 0.9, K R = 0.85
KT = C f = 1
3
cos 20◦ sin 20◦
= 0.1205
I =
2
3+1
C p = 2300 psi
Eq. (14-23) with m N = 1
Table 14-8:
Strength: Grade 1 steel with H B P = H BG = 200
Fig. 14-2:
(St ) P = (St ) G = 77.3(200) + 12 800 = 28 260 psi
Fig. 14-5:
(Sc ) P = (Sc ) G = 322(200) + 29 100 = 93 500 psi
( Z N ) P = 1.4488(108 ) −0.023 = 0.948
Fig. 14-15:
( Z N ) G = 1.4488(108 /3) −0.023 = 0.973
H B P /H BG = 1
Sec. 14-12:
∴
CH = 1
Pinion tooth bending
Eq. (14-15):
Pd K m K B
6 (1.156)(1)
(σ ) P = W K o K v K s
= 787.8(1)(1.196)(1.088)
F
J
2
0.27
t
= 13 167 psi
Ans.
Factor of safety from Eq. (14-41)
28 260(0.977)/[(1)(0.85)]
St Y N /(K T K R )
(S F ) P =
=
= 2.47 Ans.
σ
13 167
Gear tooth bending
6 (1.156)(1)
= 9433 psi Ans.
(σ ) G = 787.8(1)(1.196)(1.097)
2
0.38
(S F ) G =
Pinion tooth wear
Eq. (14-16):
28 260(0.996)/[(1)(0.85)]
= 3.51 Ans.
9433
K m C f 1/2
(σc ) P = C p W K o K v K s
dP F I P
1/2
1
1.156
= 2300 787.8(1)(1.196)(1.088)
2.667(2)
0.1205
t
= 98 760 psi
Eq. (14-42):
Sc Z N /(K T K R )
(S H ) P =
σc
Ans.
93 500(0.948)/[(1)(0.85)]
= 1.06 Ans.
=
98 760
P
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Chapter 14
Gear tooth wear
(K s ) G
(σc ) G =
(K s ) P
(S H ) G =
1/2
(σc ) P =
1.097
1.088
1/2
(98 760) = 99 170 psi
Ans.
93 500(0.973)(1)/[(1)(0.85)]
= 1.08 Ans.
99 170
The hardness of the pinion and the gear should be increased.
14-20 d P = 2.5(20) = 50 mm,
dG = 2.5(36) = 90 mm
πd P n P
π(50)(10−3 )(100)
=
= 0.2618 m/s
60
60
60(120)
= 458.4 N
Wt =
π(50)(10−3 )(100)
V =
K o = 1,
Eq. (14-28):
Q v = 6,
B = 0.25(12 − 6) 2/3 = 0.8255
A = 50 + 56(1 − 0.8255) = 59.77
√
0.8255
59.77 + 200(0.2618)
= 1.099
Kv =
59.77
Eq. (14-27):
Y P = 0.322,
Table 14-2:
YG = 0.3775
Similar to Eq. (a) of Sec. 14-10 but for SI units:
√ 0.0535
1
= 0.8433 m F Y
kb
√
0.0535
(K s ) P = 0.8433 2.5(18) 0.322
= 1.003 use 1
√
0.0535
> 1 use 1
(K s ) G = 0.8433 2.5(18) 0.3775
Ks =
18
− 0.025 = 0.011
10(50)
= 0.247 + 0.0167(0.709) − 0.765(10−4 )(0.7092 ) = 0.259
Cmc = 1,
F = 18/25.4 = 0.709 in,
C pm = 1,
Cma
Cp f =
Ce = 1
K H = 1 + 1[0.011(1) + 0.259(1)] = 1.27
Eq. (14-40):
Fig. 14-14:
K B = 1,
m G = NG /N P = 36/20 = 1.8
(Y N ) P = 1.3558(108 ) −0.0178 = 0.977
(Y N ) G = 1.3558(108 /1.8) −0.0178 = 0.987
Fig. 14-6:
Eq. (14-38):
(Y J ) P = 0.33,
(Y J ) G = 0.38
Y Z = 0.658 − 0.0759 ln(1 − 0.95) = 0.885
Yθ = Z R = 1
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Eq. (14-23) with m N = 1:
cos 20◦ sin 20◦
ZI =
2
√
Z E = 191 MPa
Table 14-8:
1.8
1.8 + 1
= 0.103
Grade 1 steel, H B P = H BG = 200
Strength
Fig. 14-2:
(σ F P ) P = (σ F P ) G = 0.533(200) + 88.3 = 194.9 MPa
Fig. 14-5:
(σ H P ) P = (σ H P ) G = 2.22(200) + 200 = 644 MPa
( Z N ) P = 1.4488(108 ) −0.023 = 0.948
Fig. 14-15:
( Z N ) G = 1.4488(108 /1.8) −0.023 = 0.961
H B P /H BG = 1
∴
ZW = 1
Pinion tooth bending
1 KH KB
t
(σ ) P = W K o K v K s
bm t Y J
P
1.27(1)
1
= 43.08 MPa Ans.
= 458.4(1)(1.099)(1)
18(2.5)
0.33
0.977
σF P YN
194.9
(S F ) P =
= 4.99 Ans.
=
σ Yθ Y Z P
43.08 1(0.885)
Gear tooth bending
1.27(1)
1
(σ ) G = 458.4(1)(1.099)(1)
= 37.42 MPa Ans.
18(2.5)
0.38
0.987
194.9
(S F ) G =
= 5.81 Ans.
37.42 1(0.885)
Pinion tooth wear
K
Z
H
R
(σc ) P = Z E W t K o K v K s
dw1 b Z I
P
1
1.27
= 501.8 MPa Ans.
= 191 458.4(1)(1.099)(1)
50(18) 0.103
σH P Z N Z W
644 0.948(1)
= 1.37 Ans.
(S H ) P =
=
σc Yθ Y Z P
501.8 1(0.885)
Gear tooth wear
1/2
(K s ) G 1/2
1
(σc ) G =
(σc ) P =
(501.8) = 501.8 MPa
(K s ) P
1
644 0.961(1)
= 1.39 Ans.
(S H ) G =
501.8 1(0.885)
Ans.
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Chapter 14
14-21
Pt = Pn cos ψ = 6 cos 30° = 5.196 teeth/in
16
48
= 3.079 in, dG = (3.079) = 9.238 in
5.196
16
π(3.079)(300)
= 241.8 ft/min
V =
12
0.8255
√
241.8
33
000(5)
59.77
+
Wt =
= 1.210
= 682.3 lbf, K v =
241.8
59.77
dP =
From Prob. 14-19:
Y P = 0.296,
YG = 0.4056
(K s ) P = 1.088,
m G = 3,
(K s ) G = 1.097,
(Y N ) P = 0.977,
(St ) P = (St ) G = 28 260 psi,
( Z N ) P = 0.948,
The pressure angle is:
−1
φt = tan
(Y N ) G = 0.996,
K R = 0.85
C H = 1,
( Z N ) G = 0.973,
tan 20°
cos 30°
KB = 1
(Sc ) P = (Sc ) G = 93 500 psi
√
C p = 2300 psi
= 22.80°
3.079
cos 22.8° = 1.419 in,
2
a = 1/Pn = 1/6 = 0.167 in
(rb ) P =
(rb ) G = 3(rb ) P = 4.258 in
Eq. (14-26):
1/2 1/2
2
2
3.079
9.238
2
2
Z=
+
+ 0.167 − 1.419
+ 0.167 − 4.258
2
2
3.079 9.238
sin 22.8°
−
+
2
2
= 0.9479 + 2.1852 − 2.3865 = 0.7466
Conditions O.K. for use
π
p N = pn cos φn = cos 20° = 0.4920 in
6
Eq. (14-23):
Eq. (14-23):
Fig. 14-7:
pN
0.492
=
= 0.6937
0.95Z
0.95(0.7466)
sin 22.8° cos 22.8°
3
I =
= 0.193
2(0.6937)
3+1
mN =
J P =
˙ 0.45,
JG =
˙ 0.54
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Fig. 14-8: Corrections are 0.94 and 0.98
J P = 0.45(0.94) = 0.423, JG = 0.54(0.98) = 0.529
2
− 0.0375 + 0.0125(2) = 0.0525
Cmc = 1, C p f =
10(3.079)
C pm = 1, Cma = 0.093, Ce = 1
K m = 1 + (1)[0.0525(1) + 0.093(1)] = 1.146
Pinion tooth bending
5.196
(σ ) P = 682.3(1)(1.21)(1.088)
2
(S F ) P =
1.146(1)
= 6323 psi Ans.
0.423
28 260(0.977)/[1(0.85)]
= 5.14 Ans.
6323
Gear tooth bending
5.196
(σ ) G = 682.3(1)(1.21)(1.097)
2
(S F ) G =
1.146(1)
= 5097 psi Ans.
0.529
28 260(0.996)/[1(0.85)]
= 6.50 Ans.
5097
Pinion tooth wear
1/2
1
1.146
(σc ) P = 2300 682.3(1)(1.21)(1.088)
= 67 700 psi Ans.
3.078(2) 0.193
(S H ) P =
93 500(0.948)/[(1)(0.85)]
= 1.54 Ans.
67 700
Gear tooth wear
1.097
(σc ) G =
1.088
(S H ) G =
1/2
(67 700) = 67 980 psi Ans.
93 500(0.973)/[(1)(0.85)]
= 1.57
67 980
Ans.
14-22 Given: R = 0.99 at 108 cycles, H B = 232 through-hardening Grade 1, core and case, both
gears. N P = 17T, NG = 51T, Y P = 0.303, YG = 0.4103, J P = 0.292, JG = 0.396,
d P = 2.833 in, dG = 8.500 in.
Pinion bending
From Fig. 14-2:
0.99 (St ) 107
= 77.3H B + 12 800
= 77.3(232) + 12 800 = 30 734 psi
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Chapter 14
Fig. 14-14:
Y N = 1.6831(108 ) −0.0323 = 0.928
V = πd P n/12 = π(2.833)(1120/12) = 830.7 ft/min
√
K T = K R = 1, S F = 2, S H = 2
σall =
30 734(0.928)
= 14 261 psi
2(1)(1)
Q v = 5,
B = 0.25(12 − 5) 2/3 = 0.9148
A = 50 + 56(1 − 0.9148) = 54.77
0.9148
√
54.77 + 830.7
= 1.472
Kv =
54.77
√
0.0535
2 0.303
K s = 1.192
= 1.089 ⇒
6
use 1
K m = Cm f = 1 + Cmc (C p f C pm + Cma Ce )
Cmc = 1
F
− 0.0375 + 0.0125F
10d
2
=
− 0.0375 + 0.0125(2)
10(2.833)
Cp f =
= 0.0581
C pm = 1
Cma = 0.127 + 0.0158(2) − 0.093(10−4 )(22 ) = 0.1586
Ce = 1
K m = 1 + 1[0.0581(1) + 0.1586(1)] = 1.2167
Kβ = 1
Eq. (14-15):
Wt =
=
H=
F J P σall
K o K v K s Pd K m K β
2(0.292)(14 261)
= 775 lbf
1(1.472)(1)(6)(1.2167)(1)
775(830.7)
WtV
=
= 19.5 hp
33 000
33 000
Pinion wear
Fig. 14-15:
Z N = 2.466N −0.056 = 2.466(108 ) −0.056 = 0.879
MG = 51/17 = 3
sin 20◦ cos 20◦
I =
2
3
3+1
= 1.205,
CH = 1
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0.99 (Sc ) 107
Fig. 14-5:
= 322H B + 29 100
= 322(232) + 29 100 = 103 804 psi
103 804(0.879)
= 64 519 psi
√
2(1)(1)
Fd P I
σc, all 2
t
W =
Cp
Ko Kv Ks Km C f
2 2(2.833)(0.1205)
64 519
=
2300
1(1.472)(1)(1.2167)(1)
σc, all =
Eq. (14-16):
= 300 lbf
WtV
300(830.7)
H=
=
= 7.55 hp
33 000
33 000
The pinion controls therefore Hrated = 7.55 hp
Ans.
4-23
3Y
2Pd
√
3Y
3.674 √
2.25
=
t = 4lx = 4
Y
Pd
2Pd
Pd
√
√
√
F Y
3.674
Y = 1.5487
de = 0.808 Ft = 0.808 F
Pd
Pd
√

−0.107
√ −0.0535
1.5487 F Y/Pd
F Y

kb = 
= 0.8389
0.30
Pd
l = 2.25/Pd ,
x=
1
Ks =
= 1.192
kb
14-24
√ 0.0535
F Y
Pd
Ans.
Y P = 0.331, YG = 0.422, J P = 0.345, JG = 0.410, K o = 1.25. The service conditions
are adequately described by K o . Set S F = S H = 1.
d P = 22/4 = 5.500 in
dG = 60/4 = 15.000 in
V =
π(5.5)(1145)
= 1649 ft/min
12
Pinion bending
0.99 (St ) 107
= 77.3H B + 12 800 = 77.3(250) + 12 800 = 32 125 psi
Y N = 1.6831[3(109 )]−0.0323 = 0.832
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Chapter 14
(σall ) P =
Eq. (14-17):
32 125(0.832)
= 26 728 psi
1(1)(1)
B = 0.25(12 − 6)2/3 = 0.8255
A = 50 + 56(1 − 0.8255) = 59.77
0.8255
√
59.77 + 1649
= 1.534
Kv =
59.77
K s = 1,
F
− 0.0375 + 0.0125F
10d
Cmc =
3.25
− 0.0375 + 0.0125(3.25) = 0.0622
10(5.5)
=
Cma
Ce
Km
Kβ
Cm = 1
= 0.127 + 0.0158(3.25) − 0.093(10−4 )(3.252 ) = 0.178
=1
= Cm f = 1 + (1)[0.0622(1) + 0.178(1)] = 1.240
= 1, K T = 1
W1t =
Eq. (14-15):
26 728(3.25)(0.345)
= 3151 lbf
1.25(1.534)(1)(4)(1.240)
3151(1649)
= 157.5 hp
33 000
By similar reasoning, W2t = 3861 lbf and H2 = 192.9 hp
H1 =
Gear bending
Pinion wear
m G = 60/22 = 2.727
cos 20◦ sin 20◦
I =
2
0.99 (Sc ) 107
2.727
1 + 2.727
= 0.1176
= 322(250) + 29 100 = 109 600 psi
( Z N ) P = 2.466[3(109 )]−0.056 = 0.727
( Z N ) G = 2.466[3(109 )/2.727]−0.056 = 0.769
109 600(0.727)
= 79 679 psi
1(1)(1)
Fd P I
σc, all 2
t
W3 =
Cp
Ko Kv Ks Km C f
2 3.25(5.5)(0.1176)
79 679
= 1061 lbf
=
2300
1.25(1.534)(1)(1.24)(1)
(σc, all ) P =
H3 =
1061(1649)
= 53.0 hp
33 000
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Gear wear
W4t = 1182 lbf,
Similarly,
Rating
H4 = 59.0 hp
Hrated = min( H1 , H2 , H3 , H4 )
= min(157.5, 192.9, 53, 59) = 53 hp Ans.
Note differing capacities. Can these be equalized?
14-25
From Prob. 14-24:
W1t = 3151 lbf,
W2t = 3861 lbf,
W3t = 1061 lbf,
W4t = 1182 lbf
Wt =
33 000K o H
33 000(1.25)(40)
=
= 1000 lbf
V
1649
Pinion bending: The factor of safety, based on load and stress, is
(S F ) P =
W1t
3151
=
= 3.15
1000
1000
Gear bending based on load and stress
W2t
3861
(S F ) G =
=
= 3.86
1000
1000
Pinion wear
based on load:
based on stress:
W3t
1061
=
= 1.06
1000
1000
√
(S H ) P = 1.06 = 1.03
n3 =
Gear wear
based on load:
based on stress:
W4t
1182
=
= 1.18
1000
1000
√
(S H ) G = 1.18 = 1.09
n4 =
Factors of safety are used to assess the relative threat of loss of function 3.15, 3.86, 1.06,
1.18 where the threat is from pinion wear. By comparison, the AGMA safety factors
(S F ) P , (S F ) G , (S H ) P , (S H ) G
are
3.15, 3.86, 1.03, 1.09 or 3.15, 3.86, 1.061/2, 1.181/2
and the threat is again from pinion wear. Depending on the magnitude of the numbers,
using S F and S H as defined by AGMA, does not necessarily lead to the same conclusion
concerning threat. Therefore be cautious.
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Chapter 14
14-26
Solution summary from Prob. 14-24: n = 1145 rev/min, K o = 1.25, Grade 1 materials,
N P = 22T, NG = 60T, m G = 2.727, Y P = 0.331, YG = 0.422, J P = 0.345,
JG = 0.410, Pd = 4T /in, F = 3.25 in, Q v = 6, ( Nc ) P = 3(109 ), R = 0.99
Pinion H B : 250 core, 390 case
Gear H B : 250 core, 390 case
K m = 1.240, K T = 1, K β = 1, d P = 5.500 in, dG = 15.000 in,
V = 1649 ft/min, K v = 1.534, (K s ) P = (K s ) G = 1, (Y N ) P = 0.832,
(Y N ) G = 0.859, K R = 1
Bending
(σall ) P = 26 728 psi
(St ) P = 32 125 psi
(σall ) G = 27 546 psi
(St ) G = 32 125 psi
W1t = 3151 lbf,
H1 = 157.5 hp
W2t = 3861 lbf,
H2 = 192.9 hp
Wear
φ = 20◦ ,
( Z N ) G = 0.769,
I = 0.1176,
( Z N ) P = 0.727,
C P = 2300 psi
(Sc ) P = Sc = 322(390) + 29 100 = 154 680 psi
(σc, all ) P =
154 680(0.727)
= 112 450 psi
1(1)(1)
154 680(0.769)
= 118 950 psi
1(1)(1)
112 450 2
t
(1061) = 2113 lbf,
W3 =
79 679
2
118 950
t
(1182) = 2354 lbf,
W4 =
109 600(0.769)
(σc, all ) G =
H3 =
2113(1649)
= 105.6 hp
33 000
H4 =
2354(1649)
= 117.6 hp
33 000
Rated power
Hrated = min(157.5, 192.9, 105.6, 117.6) = 105.6 hp
Prob. 14-24
Ans.
Hrated = min(157.5, 192.9, 53.0, 59.0) = 53 hp
The rated power approximately doubled.
14-27
The gear and the pinion are 9310 grade 1, carburized and case-hardened to obtain Brinell
285 core and Brinell 580–600 case.
Table 14-3:
0.99 (St ) 107
= 55 000 psi
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Modification of St by (Y N ) P = 0.832 produces
(σall ) P = 45 657 psi,
Similarly for (Y N ) G = 0.859
(σall ) G = 47 161 psi,
W1t = 4569 lbf,
and
H1 = 228 hp
W2t = 5668 lbf,
H2 = 283 hp
From Table 14-8, Cp = 2300 psi. Also, from Table 14-6:
0.99 (Sc ) 107
= 180 000 psi
Modification of Sc by (Y N ) produces
(σc, all ) P = 130 525 psi
(σc, all ) G = 138 069 psi
and
W3t = 2489 lbf,
H3 = 124.3 hp
W4t = 2767 lbf,
H4 = 138.2 hp
Rating
Hrated = min(228, 283, 124, 138) = 124 hp
14-28
Ans.
Grade 2 9310 carburized and case-hardened to 285 core and 580 case in Prob. 14-27.
Summary:
Table 14-3:
0.99 (St ) 107
= 65 000 psi
(σall ) P = 53 959 psi
(σall ) G = 55 736 psi
and it follows that
W1t = 5399.5 lbf,
H1 = 270 hp
H2 = 335 hp
W2 = 6699 lbf,
From Table 14-8, C p = 2300 psi. Also, from Table 14-6:
t
Sc = 225 000 psi
(σc, all ) P = 181 285 psi
(σc, all ) G = 191 762 psi
Consequently,
W3t = 4801 lbf,
H3 = 240 hp
W4t = 5337 lbf,
H4 = 267 hp
Rating
Hrated = min(270, 335, 240, 267) = 240 hp.
Ans.
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Chapter 14
14-29
n = 1145 rev/min, K o = 1.25, N P = 22T, NG = 60T,
dG = 7.5 in, Y P = 0.331, YG = 0.422, J P = 0.335,
F = 1.625 in, H B = 250, case and core, both gears.
C pm = 1,
Cma = 0.152,
Ce = 1,
C f = 0.0419,
K β = 1, K s = 1, V = 824 ft/min, (Y N ) P = 0.8318,
I = 0.117 58
0.99 (St ) 107 = 32 125 psi
(σall ) P = 26 668 psi
(σall ) G = 27 546 psi
and it follows that
m G = 2.727, d P = 2.75 in,
JG = 0.405 , P = 8T /in,
Cm = 1, F/d P = 0.0591,
K m = 1.1942,
K T = 1,
(Y N ) G = 0.859, K R = 1,
W1t = 879.3 lbf,
H1 = 21.97 hp
W2 = 1098 lbf,
H2 = 27.4 hp
W3t = 304 lbf,
H3 = 7.59 hp
W4t = 340 lbf,
H4 = 8.50 hp
t
For wear
Rating
Hrated = min(21.97, 27.4, 7.59, 8.50) = 7.59 hp
In Prob. 14-24, Hrated = 53 hp
Thus
7.59
1
1
= 0.1432 =
, not
Ans.
53.0
6.98
8
The transmitted load rating is
t
= min(879.3, 1098, 304, 340) = 304 lbf
Wrated
In Prob. 14-24
t
Wrated
= 1061 lbf
Thus
304
1
= 0.2865 =
,
1061
3.49
14-30
S P = S H = 1,
Bending
Table 14-4:
Pd = 4,
0.99 (St ) 107
J P = 0.345,
1
not ,
4
JG = 0.410,
Ans.
K o = 1.25
= 13 000 psi
13 000(1)
= 13 000 psi
1(1)(1)
σall F J P
13 000(3.25)(0.345)
W1t =
=
= 1533 lbf
K o K v K s Pd K m K β
1.25(1.534)(1)(4)(1.24)(1)
(σall ) P = (σall ) G =
1533(1649)
= 76.6 hp
33 000
W2t = W1t JG /J P = 1533(0.410)/0.345 = 1822 lbf
H2 = H1 JG /J P = 76.6(0.410)/0.345 = 91.0 hp
H1 =
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Wear
Table 14-8:
Table 14-7:
C p = 1960 psi
= 75 000 psi = (σc, all ) P = (σc, all ) G
FdpI
(σc, all ) P 2
t
W3 =
Cp
Ko Kv Ks Km C f
2
3.25(5.5)(0.1176)
75 000
W3t =
= 1295 lbf
1960
1.25(1.534)(1)(1.24)(1)
0.99 (Sc ) 107
W4t = W3t = 1295 lbf
H4 = H3 =
1295(1649)
= 64.7 hp
33 000
Rating
Hrated = min(76.7, 94.4, 64.7, 64.7) = 64.7 hp
Ans.
Notice that the balance between bending and wear power is improved due to CI’s more
favorable Sc /St ratio. Also note that the life is 107 pinion revolutions which is (1/300) of
3(109). Longer life goals require power derating.
14-31
From Table A-24a, E av = 11.8(106 )
For φ = 14.5◦ and H B = 156
1.4(81)
= 51 693 psi
2 sin 14.5°/[11.8(106 )]
SC =
For φ = 20◦
SC =
1.4(112)
= 52 008 psi
2 sin 20°/[11.8(106 )]
SC = 0.32(156) = 49.9 kpsi
14-32
Programs will vary.
14-33
(Y N ) P = 0.977,
(Y N ) G = 0.996
(St ) P = (St ) G = 82.3(250) + 12 150 = 32 725 psi
(σall ) P =
32 725(0.977)
= 37 615 psi
1(0.85)
W1t =
37 615(1.5)(0.423)
= 1558 lbf
1(1.404)(1.043)(8.66)(1.208)(1)
H1 =
1558(925)
= 43.7 hp
33 000
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Chapter 14
32 725(0.996)
= 38 346 psi
1(0.85)
38 346(1.5)(0.5346)
W2t =
= 2007 lbf
1(1.404)(1.043)(8.66)(1.208)(1)
2007(925)
H2 =
= 56.3 hp
33 000
( Z N ) P = 0.948, ( Z N ) G = 0.973
(σall ) G =
Table 14-6:
0.99 (Sc ) 107
(σc, allow ) P
W3t
H3
(σc, allow ) G
W4t
= 150 000 psi
0.948(1)
= 167 294 psi
= 150 000
1(0.85)
167 294 2 1.963(1.5)(0.195)
= 2074 lbf
=
2300
1(1.404)(1.043)
2074(925)
=
= 58.1 hp
33 000
0.973
=
(167 294) = 171 706 psi
0.948
171 706 2 1.963(1.5)(0.195)
= 2167 lbf
=
2300
1(1.404)(1.052)
2167(925)
= 60.7 hp
33 000
= min(43.7, 56.3, 58.1, 60.7) = 43.7 hp
H4 =
Hrated
Ans.
Pinion bending controlling
14-34
(Y N ) P = 1.6831(108 ) −0.0323 = 0.928
(Y N )G = 1.6831(108 /3.059)−0.0323 = 0.962
Table 14-3:
St = 55 000 psi
(σall ) P =
55 000(0.928)
= 60 047 psi
1(0.85)
W1t =
60 047(1.5)(0.423)
= 2487 lbf
1(1.404)(1.043)(8.66)(1.208)(1)
H1 =
2487(925)
= 69.7 hp
33 000
0.962
(60 047) = 62 247 psi
0.928
62 247 0.5346
t
(2487) = 3258 lbf
W2 =
60 047 0.423
(σall ) G =
H2 =
3258
(69.7) = 91.3 hp
2487
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Sc = 180 000 psi
Table 14-6:
(Z N ) P = 2.466(108 )−0.056 = 0.8790
(Z N )G = 2.466(108 /3.059)−0.056 = 0.9358
180 000(0.8790)
= 186 141 psi
1(0.85)
186 141 2 1.963(1.5)(0.195)
t
= 2568 lbf
W3 =
2300
1(1.404)(1.043)
(σc, all ) P =
2568(925)
= 72.0 hp
33 000
H3 =
0.9358
(186 141) = 198 169 psi
0.8790
198 169 2 1.043
t
(2568) = 2886 lbf
W4 =
186 141
1.052
(σc, all ) G =
H4 =
2886(925)
= 80.9 hp
33 000
Hrated = min(69.7, 91.3, 72, 80.9) = 69.7 hp
Ans.
Pinion bending controlling
14-35
(Y N ) P = 0.928,
Table 14-3:
(Y N ) G = 0.962
(See Prob. 14-34)
St = 65 000 psi
(σall ) P =
65 000(0.928)
= 70 965 psi
1(0.85)
W1t =
70 965(1.5)(0.423)
= 2939 lbf
1(1.404)(1.043)(8.66)(1.208)
H1 =
2939(925)
= 82.4 hp
33 000
65 000(0.962)
= 73 565 psi
1(0.85)
73 565 0.5346
t
(2939) = 3850 lbf
W2 =
70 965 0.423
(σall ) G =
H2 =
3850
(82.4) = 108 hp
2939
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Chapter 14
Table 14-6:
Sc = 225 000 psi
( Z N ) P = 0.8790,
( Z N ) G = 0.9358
225 000(0.879)
= 232 676 psi
1(0.85)
232 676 2 1.963(1.5)(0.195)
t
= 4013 lbf
W3 =
2300
1(1.404)(1.043)
(σc, all ) P =
H3 =
4013(925)
= 112.5 hp
33 000
0.9358
(232 676) = 247 711 psi
0.8790
247 711 2 1.043
t
(4013) = 4509 lbf
W4 =
232 676
1.052
(σc, all ) G =
H4 =
4509(925)
= 126 hp
33 000
Hrated = min(82.4, 108, 112.5, 126) = 82.4 hp
Ans.
The bending of the pinion is the controlling factor.
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