budynas SM ch13

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13-1
13-2
13-3
13-4
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• Instructor’s Solution Manual to Accompany Mechanical Engineering Design
13-5
(a)
(b)
(c)
(d)
13-6
(a)
budynas_SM_ch13.qxd 12/04/2006 15:23 Page 326
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(b) Eq. (13-7): cos 0 6283 cos 20° 0 590 in
(c) cos 5 cos 30° 4 33 teeth/in
tan 1(tan cos ) tan 1(tan 20° cos 30 ) 22 8°
(d) Table 13-4:
1 5 0 200 in
1 25 5 0 250 in
17
5 cos 30°
3 926 in
34
5 cos 30°
7 852 in
13-7
19 teeth, 57 teeth, 14 5°, 10 teeth/in
(a) 10 0 3142 in
cos
0 3142
cos 20°
0 3343 in
tan
0 3343
tan 20°
0 9185 in
(b) cos 10 cos 20° 9 397 teeth/in
tan 1
tan 14 5°
cos 20°
15 39°
(c) 1 10 0 100 in
1 25 10 0 125 in
19
10 cos 20°
2 022 in
57
10 cos 20°
6 066 in
20
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328 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
13-8 (a) The smallest pinion tooth count that will run with itself is found from Eq. (13-10)
NP ≥ 2k3 sin2 φ
(
1 +
√
1 + 3 sin2 φ
)
≥ 2(1)
3 sin2 20°
(
1 +
√
1 + 3 sin2 20°
)
≥ 12.32 → 13 teeth Ans.
(b) The smallest 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.
The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-12) is
NG ≤
N 2P sin
2 φ − 4k2
4k − 2NP sin2 φ
≤ 15
2 sin2 20° − 4(1)2
4(1) − 2(15) sin2 20°
≤ 45.49 → 45 teeth Ans.
(c) The smallest pinion that will mesh with a rack, from Eq. (13-13)
NP ≥ 2k
sin2 φ
= 2(1)
sin2 20°
≥ 17.097 → 18 teeth Ans.
13-9 φn = 20°, ψ = 30°, φt = tan−1 (tan 20°/cos 30°) = 22.80°
(a) The smallest pinion tooth count that will run itself is found from Eq. (13-21)
NP ≥ 2k cos ψ3 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 ≤ 10
2 sin2 22.80° − 4(1) cos2 30°
4(1) cos2 30° − 2(20) sin2 22.80°
≤ 26.08 → 26 teeth Ans.
budynas_SM_ch13.qxd 12/04/2006 17:17 Page 328
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(c) The smallest pinion that will mesh with a rack, from Eq. (13-24) is
2(1) cos 30°
sin2 22 80°
11 53 12 teeth
13-10 Pressure Angle: tan 1 tan 20°
cos 30°
22 796°
Program Eq. (13-24) on a computer using a spreadsheet or code and increment The
first value of that can be doubled is 10 teeth, where 26 01 teeth. So
20 teeth will work. Higher tooth counts will work also, for example 11:22, 12:24, etc.
Use 10:20
13-11 Refer to Prob. 13-10 solution. The first value of that can be multiplied by 6 is
11 teeth where 93 6 teeth. So 66 teeth.
Use 11:66
13-12 Begin with the more general relation, Eq. (13-24), for full depth teeth.
2 sin2 4 cos2
4 cos 2 sin2
For a rack, set the denominator to zero
4 cos 2 sin2 0
From which
sin
2 cos
sin 1
2 cos
For 9 teeth and 0 for spur gears,
sin 1
2(1)
9
28 126°
13-13
(a) 3 mm
3 cos 25° 10 4 mm
10 4 tan 25° 22 3 mm
18 32
25 , 20 , 3 mm
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• Instructor’s Solution Manual to Accompany Mechanical Engineering Design
(b)
(c)
13-14 (a)
(b)
(c)
13-15
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13-16
13-17
(a)
(b)
(c)
13-18 (a)
(b)
(c)
13-19 (a)
(b)
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• Instructor’s Solution Manual to Accompany Mechanical Engineering Design
13-20 Let gear 2 be first, then 2 0. Let gear 6 be last, then 6 12 rev/min.
20
30
16
34
16
51
,
(0 ) 16
51
12
12
35 51
17 49 rev/min (negative indicates cw)
13-21 Let gear 2 be first, then 2 180 rev/min. Let gear 6 be last, then 6 0
20
30
16
34
16
51
,
(180 ) 16
51
(0 )
16
35
180 82 29 rev/min
The negative sign indicates opposite 2 82 29 rev/min cw
13-22 5 12 2(16) 2(12) 68 teeth
Let gear 2 be first, 2 320 rev/min. Let gear 5 be last, 5 0
12
16
16
12
12
68
3
17
,
320
17
3
(0 )
3
14
(320) 68 57 rev/min
The negative sign indicates opposite of 2 68 57 rev/min cw
13-23 Let 2 then 7 0
24
18
18
30
36
54
8
15
5
5
8
15
0 5
2 5
8
15 2
5 15
8
(5) 14 375 turns in same direction
13-24 (a) 2 60
2 60 ( in N m, in W)
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So
60 (103)
2
9550 ( in kW, in rev/min)
9550(75)
1800
398 N m
2
2
2
5(17)
2
42 5 mm
So
32
2
398
42 5
9 36 kN
3 3 2(9 36) 18 73 kN in the positive -direction.
See the figure in part (b).
4
4
2
5(51)
2
127 5 mm
4 9 36(127 5) 1193 N m ccw
4 1193 N m cw
Note: The solution is independent of the pressure angle.
9.36
4
4 1193
9.36
3
43
9.36
18.73
23
3
9.36
2
2
398 N•m
32
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• Instructor’s Solution Manual to Accompany Mechanical Engineering Design
13-25
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• Instructor’s Solution Manual to Accompany Mechanical Engineering Design
13-27 Given: 5 teeth/in, 2 18 , 3 45 , 20°, 32 hp, 2
1800 rev/min.
in
63 025(32)
1800
1120 lbf in
18
5
3 600 in
45
5
9 000 in
32
1120
3 6 2
622 lbf
32 622 tan 20° 226 lbf
2 32 622 lbf, 2 32 226 lbf
2 (6222 2262)1 2 662 lbf
Each bearing on shaft has the same radial load of 662 2 331 lbf.
23 32 622 lbf
23 32 226 lbf
3 2 662 lbf
662 2 331 lbf
Each bearing on shaft has the same radial load which is equal to the radial load of bear-
ings, and Thus, all four bearings have the same radial load of 331 lbf. .
13-28 Given: 4 teeth/in, 20 , 20 , 2 900 rev/min.
2
20
4
5 000 in
in
63 025(30)(2)
900
4202 lbf in
32 in ( 2 2) 4202 (5 2) 1681 lbf
32 1681 tan 20 612 lbf
3
2
3
out 23 3
 
2799 lbf•in
3
23
23
3
2
in
32
32
2
2
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The motor mount resists the equivalent forces and torque. The radial force due to torque
4202
14(2) 150 lbf
Forces reverse with rotational 
sense as torque reverses.
The compressive loads at and are absorbed by the base plate, not the bolts. For 32 ,
the tensions in and are
0 1681(4 875 15 25) 2 (15 25) 0 1109 lbf
If 32 reverses, 15.25 in changes to 13.25 in, 4.815 in changes to 2.875 in, and the forces
change direction. For and ,
1681(2 875) 2 1(13 25) 0 1 182 4 lbf
For 32
1681 lbf4.87515.25"1
1
150
14"
150
150
4202 lbf•in150
2 612 lbf
4202 lbf•in
1681 lbf
Equivalent
2
32 1681 lbf
32 612 lbf
Load on 2
due to 3
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• Instructor’s Solution Manual to Accompany Mechanical Engineering Design
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13-29
W i j k
R i j R j
M R W R F T 0
R F i k
R W i j k
i j k i k j 0
F F W
i k i j k
i j k
13-30
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• Instructor’s Solution Manual to Accompany Mechanical Engineering Design
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• Instructor’s Solution Manual to Accompany Mechanical Engineering Design
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Substituting and solving Eq. (1) gives
T = 2404i lbf · in
= −297 5 lbf
= −356 7 lbf
F = F + F + W = 0
Substituting and solving gives
= −344 lbf
= 106 7 lbf
= −297 5 lbf
So
F = −344i − 356 7j − 297 5k lbf
F = 106 7j − 297 5k lbf
13-35 = 8 cos 15° = 7 727 teeth/in
2 = 16 7 727 = 2 07 in
3 = 36 7 727 = 4 66 in
4 = 28 7 727 = 3 62 in
2 =
63 025(7 5)
1720
= 274 8 lbf · in
=
274 8
2 07 2
= 266 lbf
4
4
4
4
34
34
34
22
2
2
32
32
32
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• Instructor’s Solution Manual to Accompany Mechanical Engineering Design
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Next, designate the points of action on gears 4 and 3, respectively, as points and ,
as shown. Position vectors are
1 553 3
2 598 6 5
8 5
Force vectors are
54 1986 748 532
23 1188 500 686
Now, a summation of moments about bearing gives
54 23
The terms for this equation are found to be
54 1412 5961 3086
23 5026 7722 3086
8 5 8 5
When these terms are placed back into the moment equation, the terms, representing
the shaft torque, cancel. The and terms give
3614
8 5
425 lbf
(13 683)
8 5
1610 lbf
Next, we sum the forces to zero.
54 23
Substituting, gives
( 1987 746 532 ) ( 1188 499 686 )
(1610 425 )
Solving gives
1987 1188 1610 1565 lbf
746 499 425 672 lbf
532 686 154 lbf
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• Instructor’s Solution Manual to Accompany Mechanical Engineering Design
13-37
W i j k
R j k R k
M R W R F T 0
R W i j k
R F i j
F i j
F F W R 0
F W F i j k i j
i j k
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Radial F 318 5i 1946j N,
[(318 5)2 ( 1946)2]1 2 1972 N
Thrust 5119 N
13-38 From Prob. 13-37
W 637i 1333j 5119k N
So
48(25)
382 mm
Bearing to take thrust load
M R W R F T 0
R 0 0725i 0 191j
R 0 1075i
The position vectors are in meters.
R W 977 7i 371 1j 25 02k
R F 0 1075 j 0 1075 k
Putting it together and solving
Gives
977 7 N m
F 233j 3450k N, 3460 N
F F W F 0
F (F W ) 637i 1566j 1669k N
Radial F 1566j 1669k N
Or 2289 N (total radial)
F 637i N (thrust)
72.5
191
35
Not to scale
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• Instructor’s Solution Manual to Accompany Mechanical Engineering Design
13-39
W i j k
13-40