Chap13 10e - chapter 13 of shigley 10th edition PDF

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chapter 13 of shigley 10th edition
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17 / 8 2.125 in 1120 2 2.125 544 3

8 4.375

4.375 in

35 teeth

.

2.125 4.375 / 2 3.25 in

1600 15 / 60 3 mm

.

400 rev/min .

.

 3 15 60  2 112.5 mm

16 4

64 teeth

.

.

64 6

384 mm

.

16 6

96 mm

.

384 96 / 2 240 mm

.

1/ 1/ 3 0.3333 in . 1.25 / 1.25 / 3 0.4167 in 0.0834 in . / / 3 1.047 in . / 2 1.047 / 2 0.523 in . 1 1

2 2

cos

1

/

21/ 3 7 in

7 cos 20 2

/

.

o

6.578 in

.

28 / 3 9.333 in

9.333cos 20 o 8.770 in

/ 3 cos 20o

.

0.984 in

.

/ 1.53 / 0.984 1.55 . See the following figure for a drawing of the gears and the arc lengths.

Shigley’s MED, 10th edition

Chapter 13 Solutions, Page 1/36

1/2

  14 / 6  2  32 / 6  2         2   2  

tan

1

tan

1

14 / 32

23.63o

.

32 /14

o

.

66.37

2.910 in

.

14 / 6 2.333 in 32 / 6 5.333 in

Shigley’s MED, 10th edition

.

Chapter 13 Solutions, Page 2/36

From Table 13-3, 0.3 = 0.3(2.910) = 0.873 in and 10/ = 10/6 = 1.67 0.873 < 1.67 0.873 in .

/

/ 4 0.7854 in

/ cos

0.7854 / cos 30 o

/ tan

o

0.9069 / tan 30 cos

Eq. (13-7): 4 cos 30o

cos

tan

1

tan

/ cos

1.571 in 0.7854 cos 25 o

57 teeth,

2.5 cos tan cos Shigley’s MED, 10th edition

7.854 o cos 30 9.069 o tan 30 2.5 cos30o

0.7380 in

.

3.464 teeth/in

tan 1 (tan 25o / cos 30o ) 28.3o

Table 13-4: 1/ 4 0.250 in . 1.25 / 4 0.3125 in 20 5.774 in 4 cos 30o 36 10.39 in 4 cos 30o

19 teeth,

0.9069 in

20 o,

7.854 mm

. . .

2.5 mm

.

9.069 mm

.

15.71 mm

.

2.887 mm

.

.

Chapter 13 Solutions, Page 3/36

 tan 20o  tan 1  o  cos30  2.5 mm

1.25

22.80o

.

.

1.25 2.5

3.125 mm

.

19 2.887 =54.85 mm

57 2.887

164.6 mm

2 1 2

.

= 1, = 20º, and

Using Eq. (13-11) with

2

sin2 2 1

1 2 2  sin 20 2

o

.

2

sin 2

1 2 2

= 2,

2

1 2 2  sin 2 20o

14.16 teeth

Round up for the minimum integer number of teeth. = 15 teeth Repeating (a) with Repeating (a) with Repeating (a) with

= 3, = 4, = 5,

= 14.98 teeth. Rounding up, = 15.44 teeth. Rounding up, = 15.74 teeth. Rounding up,

= 15 teeth. = 16 teeth. = 16 teeth.

Alternatively, a useful table can be generated to determine the largest gear that can mesh with a specified pinion, and thus also the maximum gear ratio with a specified pinion. The Max column was generated using Eq. (13-12) with = 1, = 20º, and rounding up to the next integer.

13 14 15 16 17 18

16 26 45 101 1309 unlimited

1.23 1.86 3.00 6.31 77.00 unlimited

With this table, we can readily see that gear ratios up to 3 can be obtained with a minimum of 15 teeth, and gear ratios up to 6.31 can be obtained with a minimum of 16 teeth. This is consistent with the results previously obtained.

Shigley’s MED, 10th edition

Chapter 13 Solutions, Page 4/36

Repeating the process shown in the solution to Prob. 13-8, except with = 25º, we obtain the following results. = 10 teeth. For = 2, = 9.43 teeth. Rounding up, = 10 teeth. For = 3, = 9.92 teeth. Rounding up, = 11 teeth. For = 4, = 10.20 teeth. Rounding up, = 10.38 teeth. Rounding up, = 11 teeth. For = 5, For convenient reference, we will also generate the table from Eq. (13-12) for = 25º.

9 10 11 12

13 32 249 unlimited

1.44 3.20 22.64 unlimited

The smallest pinion tooth count that will run with itself is found from Eq. (13-10).

2 3sin 2

1 3sin2

1

21 3sin2 20o 12.32

1 3sin2 20 o

1

13 teeth

.

The smallest pinion that will mesh with a gear ratio of 2 1 2

2

sin

2

21

1 2 2.52

2.5

1 2 2.5  sin 20 14.64 15 teeth o

2

= 2.5, from Eq. (13-11) is

sin 2 1 2 2.5 sin 2 20 o

.

The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-12) is 2

sin 2 4 4 2 sin 2

2

15 2sin 2 20o 4 1 41 45.49

2

2 15 sin 2 20 o 45 teeth

.

The smallest pinion that will mesh with a rack, from Eq. (13-13),

Shigley’s MED, 10th edition

Chapter 13 Solutions, Page 5/36

21 2 2 sin sin 2 20 o 17.097 18 teeth

o

20 ,

30

.

o

From Eq. (13-19),

1

tan

o

o

o

tan 20 / cos 30

22.80

The smallest pinion tooth count that will run with itself, from Eq. (13-21) is

2 cos 3sin 2

1 3sin 2

1

2 1 cos 30 o

1

o

2

3sin 22.80 8.48 9 teeth

1 3sin2 22.80

o

.

The smallest pinion that will mesh with a gear ratio of = 2.5, from Eq. (13-22) is 2 1 cos 30o 2.5 2.5 2 1 2 2.5 sin 2 22.80 o 1 2 2.5  sin 2 22.80o 9.95

10 teeth

.

The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-23) is 2

sin 2 4 cos

4 2 cos2 2 sin 2

10 2 sin 2 22.80o 4 1 cos 2 30o o

4 1 cos 2 30 26.08

2 20 sin 2 22.80

26 teeth

o

.

The smallest pinion that will mesh with a rack, from Eq. (13-24) is

From Eq. (13-19),

Shigley’s MED, 10th edition

2 cos sin 2

2 1 cos 30o

11.53

12 teeth

sin 2 22.80 o

 tan tan 1   cos

  

.

 tan 20o  tan 1  o   cos30 

22.796o

Chapter 13 Solutions, Page 6/36

Program Eq. (13-23) 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 teeth,

= 20 teeth

Note that the given diametral pitch (tooth size) is not relevant to the interference problem.

o  1  tan 1  tan 20  tan tan 27.236o    o  From Eq. (13-19),  cos   cos 45  Program Eq. (13-23) on a computer using a spreadsheet or code, and increment . The ≤ 17.6 teeth. So first value of that can be doubled is = 6 teeth, where = 12 teeth will work. Higher tooth counts will work also, for example 7:14, 8:16, etc.

Use

= 6 teeth,

= 12 teeth

The smallest pinion that will operate with a rack without interference is given by Eq. (1313). 2 sin 2 Setting

= 9 teeth, and solving for ,

= 1 for full depth teeth, sin

1

2

sin

/ tan

Eq. (13-3):

/

tan

1

o

28.126

9

/ cos

Eq. (13-17):

Shigley’s MED, 10th edition

2 1

3 mm

Eq. (13-3): Eq. (13-16):

Eq. (13-19):

1

.

3 / cos 25

o

10.40 mm o

10.40 / tan 25

10.40 /

tan cos

.

.

22.30 mm

3.310 mm

.

.

o

tan

1

tan 20 cos 25o

o

21.88

.

Chapter 13 Solutions, Page 7/36

= = 3.310 (18) = 59.58 mm Eq. (13-2): = = 3.310 (32) = 105.92 mm Eq. (13-2): ______________________________________________________________________________ Sketches of the figures are shown to determine the axial forces by inspection. The axial force of gear 2 on shaft is in the negative -direction. The axial force of gear 3 on shaft is in the positive -direction. The axial force of gear 4 on shaft is in the positive -direction. The axial force of gear 5 on shaft is in the negative -direction.

5

12  16    700 48  36 

77.78 rev/min ccw

2

12 / 12 cos 30 o

1.155 in

3

48 / 12 cos 30o

4.619 in

4

1.155 4.619 2 16 / 8 cos 25o

5

36 / 8 cos 25o

3.586 in

2.887 in

.

.

2.207 in 4.965 in

.

20  8   20  4    40  17   60  51 4 00 47.06 rev/min cw . 51 ______________________________________________________________________________ 6 18  20   3  3     10  38   48  36  304 3 1200 11.84 rev/min cw . 9 304 ______________________________________________________________________________

Shigley’s MED, 10th edition

Chapter 13 Solutions, Page 8/36

    cw about , as viewed from the positive axis    o 12 / 8 cos 23 1.630 in 12 1 540 40 1

40 / 8 cos 23 o

162 rev/min

5.432 in

3.531 in

2

.

.

32 8 in . 4 ______________________________________________________________________________

Applying Eq. (13-30), = ( 2 / 3) ( 4 / 5) = 45. For an exact ratio, we will choose to factor the train value into integers, such that 2 4

/ /

3 5

=9 =5

(1) (2)

Assuming a constant diametral pitch in both stages, the geometry condition to satisfy the in-line requirement of the compound reverted configuration is 2

+

3=

4+

(3)

5

With three equations and four unknowns, one free choice is available. It is necessary that all of the unknowns be integers. We will use a normalized approach to find the minimum free choice to guarantee integers; that is, set the smallest gear of the largest stage to unity, thus 3 = 1. From (1), 2 = 9. From (3), +

3=

9 + 1 = 10 =

Substituting

4=

5

2

5

4

+

5

from (2) gives

10 = 5 5+ 5 = 6 5 = 10 / 6 = 5 / 3

5

To eliminate this fraction, we need to multiply the original free choice by a multiple of 3. In addition, the smallest gear needs to have sufficient teeth to avoid interference. From Eq. (13-11) with = 1, = 20°, and = 9, the minimum number of teeth on the pinion to avoid interference is 17. Therefore, the smallest multiple of 3 greater than 17 is 18. Setting 3 = 18 and repeating the solution of equations (1), (2), and (3) yields

Shigley’s MED, 10th edition

Chapter 13 Solutions, Page 9/36

= 162 teeth = 18 teeth 4 = 150 teeth 5 = 30 teeth ______________________________________________________________________________ 2 3

The solution to Prob. 13-20 applies up to the point of determining the minimum number of teeth to avoid interference. From Eq. (13-11), with = 1, = 25°, and = 9, the minimum number of teeth on the pinion to avoid interference is 11. Therefore, the smallest multiple of 3 greater than 11 is 12. Setting = 12 and repeating the solution of equations (1), (2), and (3) of Prob. 13-20 yields = 108 teeth = 12 teeth 4 = 100 teeth 5 = 20 teeth ______________________________________________________________________________ 2 3

Applying Eq. (13-30), = ( 2 / 3) ( 4 / 5) = 30. For an exact ratio, we will choose to factor the train value into integers, such that / 4/ 2

3 5

=6 =5

(1) (2)

Assuming a constant diametral pitch in both stages, the geometry condition to satisfy the in-line requirement of the compound reverted configuration is 2

+

3=

4+

(3)

5

With three equations and four unknowns, one free choice is available. It is necessary that all of the unknowns be integers. We will use a normalized approach to find the minimum free choice to guarantee integers; that is, set the smallest gear of the largest stage to unity, thus 3 = 1. From (1), 2 = 6. From (3), +

3=

6+1=7=

Substituting

4=

5

2

7=5 5+ 5=7/6

5 5=

4

+

5

from (2) gives 6

5

To eliminate this fraction, we need to multiply the original free choice by a multiple of 6. In addition, the smallest gear needs to have sufficient teeth to avoid interference. From Eq. (13-11) with = 1, = 20°, and = 6, the minimum number of teeth on the pinion to avoid interference is 16. Therefore, the smallest multiple of 6 greater than 16 is 18. = 18 and repeating the solution of equations (1), (2), and (3) yields Setting

Shigley’s MED, 10th edition

Chapter 13 Solutions, Page 10/36

= 108 teeth = 18 teeth 4 = 105 teeth 5 = 21 teeth ______________________________________________________________________________ 2 3

Applying Eq. (13-30), = ( 2 / 3) ( 4 / 5) = 45. For an approximate ratio, we will choose to factor the train value into two equal stages, such that 2

/

3

4

/

5

45

If we choose identical pinions such that interference is avoided, both stages will be identical and the in-line geometry condition will automatically be satisfied. From Eq. 45 , the minimum number of teeth on the pinions (13-11) with = 1, = 20°, and to avoid interference is 17. Setting 3 = 5 = 17, we get 2

17 45 114.04 teeth

4

Rounding to the nearest integer, we obtain 2 3

= =

4= 5=

114 teeth 17 teeth

Checking, the overall train value is = (114 / 17) (114 / 17) = 44.97. ______________________________________________________________________________ = 25 hp, = 2500 rev/min Let ω = 300 rev/min for minimal gear ratio to minimize gear size. 300 2500

Let

2

4

3

5

1 8.333

1 8.333

2

4

3

5

1 2.887

From Eq. (13-11) with = 1, = 20°, and the pinions to avoid interference is 15. Let

2 3

Try

3

=

5

= =

= 2.887, the minimum number of teeth on

= 15 teeth 5 = 2.887(15) = 43.31 teeth 4

= 43 teeth.

Shigley’s MED, 10th edition

Chapter 13 Solutions, Page 11/36

 15   15   43   43  2500   

Too big. Try

3

=

5

304.2

= 44.

 15   15      2500  44   44 

290.55 rev/min

2 = 4 = 15 teeth, 3 = 5 = 44 teeth ______________________________________________________________________________

The planet gears act as keys and the wheel speeds are the same as that of the ring gear. Thus, 900 16 / 48 300 rev/min . 3 0, 300 1 6 0 300 300 6 300 600 rev/min 6 5

6

,

1

.

The wheel spins freely on icy surfaces, leaving no traction for the other wheel. The car is stalled. ______________________________________________________________________________ The motive power is divided equally among four wheels instead of two. Locking the center differential causes 50 percent of the power to be applied to the rear wheels and 50 percent to the front wheels. If one of the rear wheels rests on a slippery surface such as ice, the other rear wheel has no traction. But the front wheels still provide traction, and so you have two-wheel drive. However, if the rear differential is locked, you have 3-wheel drive because the rear-wheel power is now distributed 50-50. ______________________________________________________________________________ Let gear 2 be first, then

=

2=

0. Let gear 6 be last, then

=

6

= –12 rev/min.

20  16  16 30  34  51 16 0 12 51

Shigley’s MED, 10th edition

Chapter 13 Solutions, Page 12/36

12 35 / 51

17.49 rev/min

(negative indicates cw as viewed from the bottom of the figure)

      

.

______________________________________________________________________________ 2=

=

Let gear 2 be first, then

20  16  30  34  16 0 51  16     51   16  1   51  85 16 1 51

0 rev/min. Let gear 6 be last, then

=

6

= 85 rev/min.

16 51 85

85 85

123.9 rev/min

123.9 rev/min ccw . The positive sign indicates the same direction as 6. ______________________________________________________________________________ 2/2 3 4. Since all the gears are meshed, they The geometry condition is 5 / 2 will all have the same diametral pitch. Applying = ,

5

/ (2 )

5

2

=

Let gear 2 be first,

2

2 2

/ (2 )

3

2

4

3

/

12 2 16

4

/

2 12

68 teeth

= 320 rev/min. Let gear 5 be last,

=

5=

. 0 rev/min.

12  16  12  3    16  12  68  17 320

17 0 3 3 320 14

Shigley’s MED, 10th edition

68.57 rev/min

Chapter 13 Solutions, Page 13/36

68.57 rev/min cw . The negative sign indicates opposite of 2. ______________________________________________________________________________ Let

=

2,

then

7=

=

0.

20  16   36     16  30   46  0 10

5

5

0.5217

5

0.5217 10

5

5

5.217 0.5217 5

5

0.5217

5

1 0.5217

5

0

5.217

5.217 1.5217 3.428 turns in same direction ....