Chapter_13_Solutions Shigley's Mechanical Engineering Design 9th Edition Solutions Manual

Chapter 13

d P  17 / 8  2.125 in N 1120 dG  2 d P   2.125   4.375 in N3 544

13-1

NG  PdG  8  4.375  35 teeth

Ans.

C   2.125  4.375 / 2  3.25 in

Ans.

______________________________________________________________________________

nG  1600 15 / 60   400 rev/min p   m  3 mm Ans.

13-2

Ans.

C  3 15  60   2  112.5 mm Ans. ______________________________________________________________________________

NG  16  4   64 teeth

13-3

Ans.

dG  NG m  64  6   384 mm

Ans.

d P  N P m  16  6   96 mm

Ans.

C   384  96  / 2  240 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   / P   / 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

Gear Base-Circle:

Ans.

d 2  N 2 / P  28 / 3  9.333 in d 2 b  9.333cos 20  8.770 in

Base pitch:

pb  pc cos    / 3 cos 20  0.984 in

Ans.

Ans.

Contact Ratio: mc  Lab / pb  1.53 / 0.984  1.55 Ans. See the following figure for a drawing of the gears and the arc lengths.

Chapter 13, Page 1/35

______________________________________________________________________________ 13-5 1/2

(a)

 14 / 6  2  32 / 6  2  A0         2   2  

(b)   tan 1 14 / 32   23.63

Ans.

 32 /14   66.37

Ans.

  tan

1

(c) d P  14 / 6  2.333 in dG  32 / 6  5.333 in



 2.910 in

Ans.

Ans. Ans.

Chapter 13, Page 2/35

From Table 13-3, 0.3A 0 = 0.3(2.910) = 0.873 in and 10/P = 10/6 = 1.67 0.873 < 1.67  F  0.873 in Ans. ______________________________________________________________________________ (d)

13-6

(a)

pn   / Pn   / 4  0.7854 in pt  pn / cos  0.7854 / cos 30  0.9069 in px  pt / tan  0.9069 / tan 30  1.571 in pnb  pn cos n  0.7854 cos 25  0.7380 in

(b)

Eq. (13-7):

(c)

pt  Pn cos  4 cos 30  3.464 teeth/in

Ans.

t  tan 1  tan n / cos   tan 1 (tan 25 / cos 30 )  28.3 Ans. Table 13-4: a  1/ 4  0.250 in Ans. b  1.25 / 4  0.3125 in Ans. 20 dP   5.774 in Ans. 4 cos 30 36 dG   10.39 in Ans. 4 cos 30 ______________________________________________________________________________ (d)

13-7

N P  19 teeth, N G  57 teeth, n  20 , mn  2.5 mm

(a)

pn   mn    2.5  7.854 mm

Ans.

pn 7.854   9.069 mm Ans. cos cos 30 pt 9.069 px    15.71 mm Ans. tan tan 30 mn 2.5 mt    2.887 mm Ans. cos cos 30 pt 

(b)

Chapter 13, Page 3/35

 tan 20   22.80   cos 30  

t  tan 1  (c)

a  mn  2.5 mm

Ans.

Ans.

b  1.25mn  1.25  2.5  3.125 mm dP 

Ans.

N  Nmt  19  2.887  =54.85 mm Pt

Ans.

dG  57  2.887   164.6 mm Ans. ______________________________________________________________________________ 13-8

(a)

Using Eq. (13-11) with k = 1, = 20º, and m = 2, NP  



2k m  m 2  1  2m  sin 2  2 1  2m  sin 

 2  1  2  2   sin  20  2 1

2



 2

2

    14.16 teeth

 1  2  2   sin 2 20

Round up for the minimum integer number of teeth. N P = 15 teeth Ans. (b) (c) (d)

Repeating (a) with m = 3, N P = 14.98 teeth. Rounding up, N P = 15 teeth. Ans. Repeating (a) with m = 4, N P = 15.44 teeth. Rounding up, N P = 16 teeth. Ans. Repeating (a) with m = 5, N P = 15.74 teeth. Rounding up, N P = 16 teeth. Ans.

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 N G column was generated using Eq. (13-12) with k = 1, = 20º, and rounding up to the next integer. Min N P 13 14 15 16 17 18

Max N G 16 26 45 101 1309 unlimited

Max m = Max N G / Min N P 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 N P of 15 teeth, and gear ratios up to 6.31 can be obtained with a minimum N P of 16 teeth. This is consistent with the results previously obtained. ______________________________________________________________________________

Chapter 13, Page 4/35

13-9

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

Max N G Max m = Max N G / Min N P Min N P 9 13 1.44 10 32 3.20 11 249 22.64 12 unlimited unlimited ______________________________________________________________________________ 13-10 (a)

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



2k 1  1  3sin 2  3sin 2  2 1

1  3sin 20 2

 12.32

(b)





1  3sin 2 20

 13 teeth



Ans.

The smallest pinion that will mesh with a gear ratio of m G = 2.5, from Eq. (13-11) is NP 



2k m  m 2  1  2m  sin 2  2 1  2m  sin  2 1





2.5  2.52  1  2  2.5   sin 2 20 1  2  2.5   sin 2 20  14.64  15 teeth Ans. 



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

N P2 sin 2   4k 2 4k  2 N P sin 2  152 sin 2 20  4 1

2

4 1  2 15  sin 2 20

 45.49  45 teeth

Ans.

Chapter 13, Page 5/35

(c)

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

2 1 2k  2 sin  sin 2 20  17.097  18 teeth Ans. ______________________________________________________________________________ NP 

13-11 n  20 ,  30

From Eq. (13-19), t  tan 1  tan 20 / cos 30   22.80

(a)

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



2k cos 1  1  3sin 2 t 3sin 2 t 2 1 cos 30

1  3sin 22.80 2



1  3sin 2 22.80

 8.48  9 teeth (b)

 

Ans.

The smallest pinion that will mesh with a gear ratio of m = 2.5, from Eq. (13-22) is 2 1 cos 30 2.5  2.52  1  2  2.5   sin 2 22.80 NP  2  1  2  2.5   sin 22.80  9.95  10 teeth Ans.





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

N P2 sin 2 t  4k 2 cos 2  4k cos  2 N P sin 2 t 102 sin 2 22.80  4 1 cos 2 30

4 1 cos 2 30  2  20  sin 2 22.80

 26.08  26 teeth (c)

Ans.

The smallest pinion that will mesh with a rack, from Eq. (13-24) is  2k cos 2 1 cos 30  NP  sin 2 t sin 2 22.80

 11.53  12 teeth Ans. ______________________________________________________________________________

Chapter 13, Page 6/35

  tan n  1  tan 20    22.796 tan     cos  cos 30     Program Eq. (13-23) on a computer using a spreadsheet or code, and increment N P . The first value of N P that can be doubled is N P = 10 teeth, where N G ≤ 26.01 teeth. So N G = 20 teeth will work. Higher tooth counts will work also, for example 11:22, 12:24, etc.

13-12 From Eq. (13-19),

t  tan 1 

Use N P = 10 teeth, N G = 20 teeth

Ans.

Note that the given diametral pitch (tooth size) is not relevant to the interference problem. ______________________________________________________________________________   tan n  1  tan 20  13-13 From Eq. (13-19),  27.236 t  tan    tan     cos   cos 45  Program Eq. (13-23) on a computer using a spreadsheet or code, and increment N P . The first value of N P that can be doubled is N P = 6 teeth, where N G ≤ 17.6 teeth. So N G = 12 teeth will work. Higher tooth counts will work also, for example 7:14, 8:16, etc. 1

Use N P = 6 teeth, N G = 12 teeth Ans. ______________________________________________________________________________ 13-14 The smallest pinion that will operate with a rack without interference is given by Eq. (1313). 2k NP  sin 2  Setting k = 1 for full depth teeth, N P = 9 teeth, and solving for , 2 1 2k  sin 1  28.126 Ans. NP 9 ______________________________________________________________________________

  sin 1

13-15

(a)

(b)

Eq. (13-3):

pn   mn  3 mm

Eq. (13-16):

pt  pn / cos  3 / cos 25  10.40 mm

Eq. (13-17):

px  pt / tan  10.40 / tan 25  22.30 mm

Eq. (13-3):

mt  pt /   10.40 /   3.310 mm

Ans. 

Ans. Ans.

Ans.

Chapter 13, Page 7/35

Eq. (13-19):

t  tan 1

tan n tan 20  tan 1  21.88  cos cos 25

Ans.

Ans. Eq. (13-2): d p = m t N p = 3.310 (18) = 59.58 mm Eq. (13-2): d G = m t N G = 3.310 (32) = 105.92 mm Ans. ______________________________________________________________________________ (c)

13-16 (a) Sketches of the figures are shown to determine the axial forces by inspection. The axial force of gear 2 on shaft a is in the negative z-direction. The axial force of gear 3 on shaft b is in the positive z-direction. Ans. The axial force of gear 4 on shaft b is in the positive z-direction. The axial force of gear 5 on shaft c is in the negative z-direction. Ans. 12  16     700   77.78 rev/min ccw 48  36 

(b)

nc  n5 

(c)

d P 2  12 / 12 cos 30  1.155 in dG 3

Ans.

   48 / 12 cos 30   4.619 in 

1.155  4.619  2.887 in Ans. 2  16 /  8 cos 25   2.207 in

Cab  dP4





d G 5  36 / 8 cos 25  4.965 in

Cbc  3.586 in Ans. ______________________________________________________________________________ 20  8   20  4     40  17   60  51 4 nd   00   47.06 rev/min cw Ans. 51 ______________________________________________________________________________

13-17

e

6  18   20   3  3      10  38   48   36  304 3 n9  1200   11.84 rev/min cw Ans. 304 ______________________________________________________________________________

13-18

e

Chapter 13, Page 8/35

13-19 (a) (b)

12 1   540   162 rev/min cw about x. Ans. 40 1 d P  12 /  8 cos 23   1.630 in nc 





d G  40 / 8 cos 23  5.432 in

d P  dG  3.531 in 2

Ans.

32  8 in at the large end of the teeth. Ans. 4 ______________________________________________________________________________

(c)

d

13-20 Applying Eq. (13-30), e = (N 2 / N 3 ) (N 4 / N 5 ) = 45. For an exact ratio, we will choose to factor the train value into integers, such that N2 / N3 = 9 N4 / N5 = 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 N2 + N3 = N4 + N5

(3)

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 N 3 = 1. From (1), N 2 = 9. From (3), N 2 + N 3 = 9 + 1 = 10 = N 4 + N 5 Substituting N 4 = 5 N 5 from (2) gives 10 = 5 N 5 + N 5 = 6 N 5 N 5 = 10 / 6 = 5 / 3 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 k = 1, = 20°, and m = 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 N 3 = 18 and repeating the solution of equations (1), (2), and (3) yields N 2 = 162 teeth N 3 = 18 teeth N 4 = 150 teeth N 5 = 30 teeth Ans. ______________________________________________________________________________ Chapter 13, Page 9/35

13-21 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 k = 1, = 25°, and m = 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 N 3 = 12 and repeating the solution of equations (1), (2), and (3) yields N 2 = 108 teeth N 3 = 12 teeth N 4 = 100 teeth N 5 = 20 teeth Ans. ______________________________________________________________________________ 13-22 Applying Eq. (13-30), e = (N 2 / N 3 ) (N 4 / N 5 ) = 30. For an exact ratio, we will choose to factor the train value into integers, such that N2 / N3 = 6 N4 / N5 = 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 N2 + N3 = N4 + N5

(3)

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 N 3 = 1. From (1), N 2 = 6. From (3), N2 + N3 = 6 + 1 = 7 = N4 + N5 Substituting N 4 = 5 N 5 from (2) gives 7 = 5 N5 + N5 = 6 N5 N5 = 7 / 6 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 k = 1, = 20°, and m = 6, the minimum number of teeth on the pinion to avoid interference is 16. Therefore, the smallest multiple of 3 greater than 16 is 18. Setting N 3 = 18 and repeating the solution of equations (1), (2), and (3) yields N 2 = 108 teeth N 3 = 18 teeth N 4 = 105 teeth Ans. N 5 = 21 teeth ______________________________________________________________________________

Chapter 13, Page 10/35

13-23 Applying Eq. (13-30), e = (N 2 / N 3 ) (N 4 / N 5 ) = 45. For an approximate ratio, we will choose to factor the train value into two equal stages, such that

N 2 / N3  N 4 / N5  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. (13-11) with k = 1, = 20°, and m  45 , the minimum number of teeth on the pinions to avoid interference is 17. Setting N 3 = N 5 = 17, we get

N 2  N 4  17 45  114.04 teeth Rounding to the nearest integer, we obtain N 2 = N 4 = 114 teeth N 3 = N 5 = 17 teeth

Ans.

Checking, the overall train value is e = (114 / 17) (114 / 17) = 44.97. ______________________________________________________________________________ 13-24 H = 25 hp,  i = 2500 rev/min Let ω o = 300 rev/min for minimal gear ratio to minimize gear size.

o 300 1   i 2500 8.333 o N N 1   2 4 i 8.333 N 3 N 5 Let

N2 N4 1 1    N3 N5 8.333 2.887

From Eq. (13-11) with k = 1, = 20°, and m = 2.887, the minimum number of teeth on the pinions to avoid interference is 15. Let

N 2 = N 4 = 15 teeth N 3 = N 5 = 2.887(15) = 43.31 teeth

Try N 3 = N 5 = 43 teeth.  15  15 

o      2500   304.2  43  43  Too big. Try N 3 = N 5 = 44.

Chapter 13, Page 11/35

 15   15 

o       2500   290.55 rev/min  44   44  N 2 = N 4 = 15 teeth, N 3 = N 5 = 44 teeth Ans. ______________________________________________________________________________ 13-25 (a) The planet gears act as keys and the wheel speeds are the same as that of the ring gear. Thus, nA  n3  900 16 / 48  300 rev/min Ans. (b)

nF  n5  0, nL  n6 , e  1 n  300 1  6 0  300 300  n6  300 n6  600 rev/min Ans.

The wheel spins freely on icy surfaces, leaving no traction for the other wheel. The car is stalled. Ans. ______________________________________________________________________________ (c)

13-26 (a)

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. ______________________________________________________________________________ (b)

13-27 Let gear 2 be first, then n F = n 2 = 0. Let gear 6 be last, then n L = n 6 = –12 rev/min. 20  16  16 nL  n A    30  34  51 nF  nA 16  0  n A    12  n A 51 12 nA   17.49 rev/min (negative indicates cw) Ans. 35 / 51 ______________________________________________________________________________ e

13-28 Let gear 2 be first, then n F = n 2 = 0 rev/min. Let gear 6 be last, then n L = n 6 = 85 rev/min.

Chapter 13, Page 12/35

20  16  16 nL  n A    30  34  51 nF  nA 16  0  nA   85  nA  51  16  nA    nA  85  51   16  n A  1    85  51  85 nA   123.9 rev/min 16 1 51 e

The positive sign indicates the same direction as n 6 .  nA  123.9 rev/min ccw Ans. ______________________________________________________________________________ 13-29 The geometry condition is d5 / 2  d 2 / 2  d3  d 4 . Since all the gears are meshed, they will all have the same diametral pitch. Applying d = N / P, N5 / (2 P)  N 2 / (2 P)  N3 / P  N 4 / P

N5  N 2  2 N3  2 N 4  12  2 16   2 12   68 teeth

Ans.

Let gear 2 be first, n F = n 2 = 320 rev/min. Let gear 5 be last, n L = n 5 = 0 rev/min. e

12  16  12  3 nL  nA      16  12  68  17 nF  nA

320  nA  nA  

17  0  nA  3

3  320   68.57 rev/min 14

The negative sign indicates opposite of n 2 .  nA  68.57 rev/min cw Ans. ______________________________________________________________________________ 13-30 Let n F = n 2 , then n L = n 7 = 0. e

nL  n5 20  16  36      0.5217  nF  n5 16  30  46 

0  n5  0.5217 10  n5

Chapter 13, Page 13/35

0.5217 10  n5   n5 5.217  0.5217n5  n5  0

n5 1  0.5217   5.217

5.217 1.5217 n5  nb  3.428 turns in same direction ______________________________________________________________________________ n5 

13-31 (a)

  2 n / 60 H  T   2 Tn / 60

(T in N  m, H in W)

 

60 H 103

So

T

So

F32t 

2 n  9550 H / n (H in kW, n in rev/min) 9550  75  Ta   398 N  m 1800 mN 2 5 17  r2    42.5 mm 2 2 Ta 398   9.36 kN r2 42.5

F3b   Fb3  2  9.36   18.73 kN in the positive x-direction. (b)

Ans.

mN 4 5  51   127.5 mm 2 2 Tc 4  9.36 127.5   1193 N  m ccw r4 

T4 c  1193 N  m cw

Ans.

Note: The solution is independent of the pressure angle. ______________________________________________________________________________ Chapter 13, Page 14/35

N N  P 6 d 2  4 in, d 4  4 in, d5  6 in, d 6  24 in d

13-32

 24   24   36  e          1/ 6  24   36   144  nF  n2  1000 rev/min

nL  n6  0 n  nA 0  nA 1 e L   nF  nA 1000  nA 6 nA  200 rev/min Noting that power equals torque times angular velocity, the input torque is T2 

 550 lbf  ft/s   60 s  1 rev  12 in  H 25 hp        1576 lbf  in n2 1000 rev/min  hp   min  2 rad  ft 

For 100 percent gear efficiency, the output power equals the input power, so Tarm 

H 25 hp  550 lbf  ft/s   60 s   1 rev   12 in         7878 lbf  in n A 200 rev/min  hp   min   2 rad   ft 

Next, we’ll confirm the output torque as we work through the force analysis and complete the free body diagrams. Gear 2 1576  788 lbf 2 F32r  788 tan 20  287 lbf

Wt 

Gear 4

FA4  2W t  2  788  1576 lbf

Chapter 13, Page 15/35

Gear 5

Arm

Tout  1576  9   1576  4   7880 lbf  in

Ans.

______________________________________________________________________________ 13-33 Given: m = 12 mm, n P = 1800 rev/min cw, N 2 = 18T, N 3 = 32T, N 4 = 18T, N 5 = 48T Pitch Diameters:

d 2 = 18(12) = 216 mm, d 3 = 32(12) = 384 mm, d 4 = 18(12) = 216 mm, d 5 = 48(12) = 576 mm

Gear 2 From Eq. (13-36), 60 000 150  60 000 H   7.368 kN Wt   dn   216 1800  d   216  Ta 2  Wt  2   7.368    795.7 N  m  2   2  W r  7.368 tan 20  2.682 kN

Gears 3 and 4

 384   216  Wt    7.368 2  2  t W  13.10 kN W r  13.10 tan 20  4.768 kN Ans.

______________________________________________________________________________

Chapter 13, Page 16/35

13-34 Given: P = 5 teeth/in, N 2 = 18T, N 3 = 45T, n  20 , H = 32 hp, n 2 = 1800 rev/min Gear 2 Tin 

63025  32  1800

 1120 lbf  in

18  3.600 in 5 45 dG   9.000 in 5 1120 W32t   622 lbf 3.6 / 2 W32r  622 tan 20  226 lbf dP 

Fat2  W32t  622 lbf,



Fa 2  6222  226 2

Far2  W32r  226 lbf



1/2

 662 lbf

Each bearing on shaft a has the same radial load of R A = R B = 662/2 = 331 lbf. Gear 3 W23t  W32t  622 lbf W23r  W32r  226 lbf

Fb3  Fb 2  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 bearings A and B. Thus, all four bearings have the same radial load of 331 lbf. Ans. ______________________________________________________________________________ 13-35 Given: P = 4 teeth/in,

n  20 ,

N P = 20T,

n 2 = 900 rev/min

N P 20   5.000 in P 4 63025  30  2  Tin   4202 lbf  in 900 W32t  Tin /  d2 / 2   4202 /  5 / 2   1681 lbf d2 

W32r  1681 tan 20  612 lbf

Chapter 13, Page 17/35

The motor mount resists the equivalent forces and torque. The radial force due to torque is Fr 

4202  150 lbf 14  2 

Forces reverse with rotational sense as torque reverses.

The compressive loads at A and D are absorbed by the base plate, not the bolts. For W32t , the tensions in C and D are

M AB  0

1681 4.875  15.25   2 F 15.25  0

F  1109 lbf

Chapter 13, Page 18/35

If W32t reverses, 15.25 in changes to 13.25 in, 4.815 in changes to 2.875 in, and the forces change direction. For A and B,

1681 2.875  2 F1 13.25   0  F1  182.4 lbf For W32r ,

M  612  4.875  11.25 / 2   6426 lbf  in a

14 / 2   11.25 / 2 

F2 

6426  179 lbf 4  8.98 

2

2

 8.98 in

At C and D, the shear forces are: FS 1  153  179  5.625 / 8.98    179  7 / 8.98   2

2

At A and B, the shear forces are: FS 2  153  179  5.625 / 8.98    179  7 / 8.98    145 lbf 2

2

The shear forces are independent of the rotational sense. The bolt tensions and the shear forces for cw rotation are,

Chapter 13, Page 19/35

For ccw rotation,

______________________________________________________________________________ 13-36 (a)

N 2 = N 4 = 15 teeth, P

N d

 d

N 3 = N 5 = 44 teeth

N P

15  2.5 in Ans. 6 44 d3  d5   7.33 in Ans. 6 d2  d4 

(b)

(c)

Vi  V2  V3 

 d 2 n2

  2.5  2500 

 1636 ft/min Ans. 12  d n   2.5   2500 15 / 44   Vo  V4  V5  4 4   558 ft/min 12 12 Input gears: Wti  33000

12



H 33000  25   504.3 lbf  504 lbf Vi 1636

Wri  Wti tan   504.3 tan 20  184 lbf W 504.3  537 lbf Wi  ti  Ans. cos  cos 20

Output gears: H 33000  25 Wto  33000   1478 lbf Vo 558

Ans.

Ans.

Ans.

Ans.

Wro  Wto tan   1478 tan 20  538 lbf Ans. Wto 1478 Wo    1573 lbf Ans.  cos 20 cos 20

(d)

d   2.5  Ti  Wti  2   504.3    630 lbf  in  2   2 

Ans.

Chapter 13, Page 20/35

2

2

 44   44  To  Ti    630    5420 lbf  in Ans.  15   15  ______________________________________________________________________________

(e)

13-37 H  35 hp, ni  1200 rev/min,  =20 N 2  N 4  16 teeth, N3  N5  48 teeth, P  10 teeth/in N 16 (a) nintermediate  n3  n4  2 ni  1200   400 rev/min N3 48 N N 16  16  no  2 4 ni    1200   133.3 rev/min N3 N5 48  48  (b)

P

N d

 d

Ans. Ans.

N P

16  1.6 in Ans. 10 48 d3  d5   4.8 in Ans. 10  d n  1.6 1200   502.7 ft/min Vi  V2  V3  2 2  12 12  d n  1.6  400   167.6 ft/min Vo  V4  V5  4 4  12 12 d2  d4 

(c)

Wti  33000

H 33000  35    2298 lbf lbf Vi 502.7

Wri  Wti tan   2298 tan 20  836.4 lbf W 2298  2445 lbf Wi  ti  Ans. cos  cos 20

Wto  33000

H 33000  35    6891 lbf Vo 167.6

Ans. Ans.

Ans. Ans.

Ans.

Wro  Wto tan   6891tan 20  2508 lbf Ans. Wto 6891 Wo    7333 lbf Ans.  cos 20 cos 20

(d)

d   1.6  Ti  Wti  2   2298    1838 lbf  in  2   2  2

Ans.

2

 48   48  (e) To  Ti    1838    16 540 lbf  in Ans.  16   16  ______________________________________________________________________________

Chapter 13, Page 21/35

13-38 (a)

For

o 2  , from Eq. (13-11), with m = 2, k = 1,   20 i 1 2 1 NP  2  22  1  2  2   sin 2 20  14.16 2  1  2  2   sin 20



So

N P min  15

(b)

P



Ans.

N 15   1.875 teeth/in d 8

Ans.

(c) To transmit the same power with no change in pitch diameters, the speed and transmitted force must remain the same. For A, with  = 20°, W tA = F A cos20° = 300 cos20° = 281.9 lbf For A, with  = 25°, same transmitted load, F A = W tA /cos25° = 281.9/cos25° = 311.0 lbf

Ans.

Summing the torque about the shaft axis, d  d  WtA  A   WtB  B   2   2   d / 2   W  d A   281.9  20   704.75 lbf WtB  WtA A     d B / 2  tA  d B   8  WtB 704.75 FB    777.6 lbf Ans.  cos 25 cos 25 ______________________________________________________________________________

13-39 (a)

For

o 5  , from Eq. (13-11), with m = 5, k = 1,   20 i 1 2 1 NP  5  52  1  2  5   sin 2 25  10.4 1  2  5   sin 2 25



So

N P min  11

(b)

m



Ans.

d 300   27.3 mm/tooth N 11

Ans.

(d) To transmit the same power with no change in pitch diameters, the speed and transmitted force must remain the same.

Chapter 13, Page 22/35

For A, with  = 20°, W tA = F A cos20° = 11 cos20° = 10.33 kN For A, with  = 25°, same transmitted load, F A = W tA /cos25° = 10.33 / cos 25° = 11.40 kN

Ans.

Summing the torque about the shaft axis, d  d  WtA  A   WtB  B   2   2   d / 2   W  d A   11.40  600   22.80 kN WtB  WtA A     d B / 2  tA  d B   300  WtB 22.80 FB    25.16 kN Ans.  cos 25 cos 25 ______________________________________________________________________________

13-40 (a)

Using Eq. (13-11) with k = 1, = 20º, and m = 2, NP  



2k m  m 2  1  2m  sin 2  2 1  2m  sin 

 2  1  2  2   sin  20  2 1

2



 2

2





 1  2  2   sin 2  20   14.16 teeth

Round up for the minimum integer number of teeth. Ans. N F = 15 teeth, N C = 30 teeth d 125   8.33 mm/tooth N 15

(b)

m

(c)

T

(d)

From Eq. (13-36),

H



Wt 



Ans.

2 kW  1000 W   rev   60 s       100 N  m 191 rev/min  kW   2 rad   min 

60 000  2  60 000 H   1.60 kN  1600 N  dn  125 191

Ans.

Or, we could have obtained W t directly from the torque and radius,

Chapter 13, Page 23/35

Wt 

T 100   1600 N d / 2 0.125 / 2

Wr  Wt tan   1600 tan 20  583 N Ans. W 1600 W t   1700 N Ans. cos  cos 20 ______________________________________________________________________________

13-41 (a)

Using Eq. (13-11) with k = 1, = 20º, and m = 2, NP  



2k m  m 2  1  2m  sin 2  2 1  2m  sin 

 2  1  2  2   sin  20  2 1



2

 2

2





 1  2  2   sin 2  20   14.16 teeth

Round up for the minimum integer number of teeth. N C = 15 teeth, N F = 30 teeth Ans. (b)

(c)

(d)

N 30   3 teeth/in d 10

P

Ans.

 550 lbf  ft/s   12 in  rev  60 s  1 hp      70 rev/min  hp   ft  2 rad  min  T  900 lbf  in Ans. T

H



From Eqs. (13-34) and (13-35), V

 dn

 10  70 

 183.3 ft/min 12 H 33000 1 Wt  33000   180 lbf V 183.3 12



Ans.

Wr  Wt tan   180 tan 20  65.5 lbf Ans. W 180 W t   192 lbf Ans. cos  cos 20 ______________________________________________________________________________ N  d   1.30   13-42 (a) Eq. (13-14):   tan 1  P   tan 1  P   tan 1  Ans.   18.5 3.88 N d    G  G

(b)

Eq. (13-34):

V

 dn 12



  2 1.30  600  12

 408.4 ft/min

Ans.

Chapter 13, Page 24/35

(c)

Eq. (13-38):

H  10   33 000    808 lbf V  408.4  Wr  Wt tan  cos   808 tan 20 cos18.5  279 lbf

Eq. (13-38):

Wa  Wt tan  sin   808 tan 20 sin18.5  93.3 lbf

Eq. (13-35):

Wt  33 000

Ans. Ans. Ans.

The tangential and axial forces agree with Prob. 3-74, but the radial force given in Prob. 3-74 is shown here to be incorrect. Ans. ______________________________________________________________________________

Tin  63 025H / n  63025  2.5  / 240  656.5 lbf  in

13-43

W t  T / r  656.5 / 2  328.3 lbf

  tan 1  2 / 4   26.565   tan 1  4 / 2   63.435





a  2  1.5cos 26.565 / 2  2.67 in

W r  328.3 tan 20 cos 26.565  106.9 lbf W a  328.3 tan 20 sin 26.565  53.4 lbf

W = 106.9i – 53.4j + 328.3k lbf R AG = –2i + 5.17j,

R AB = 2.5j

M 4  R AG  W + R AB  FB  T  0 Solving gives R AB  FB  2.5 FBz i  2.5 FBx k R AG  W  1697i  656.6 j  445.9k So

1697i  656.6 j  445.9k    2.5FBz i  2.5 FBxk  Tj  0 FBz  1697 / 2.5  678.8 lbf T  656.6 lbf  in FBx  445.9 / 2.5  178.4 lbf

So 1/ 2

2 2 FB   678.8    178.4    

 702 lbf

Ans.

F A = – (F B + W) = – (–178.8i – 678.8k + 106.9i – 53.4j + 328.3k) = 71.5i + 53.4j + 350.5k

Chapter 13, Page 25/35



FA (radial)  71.52  350.52



1/ 2

 358 lbf

Ans.

FA (thrust)  53.4 lbf Ans. ______________________________________________________________________________ 13-44

d 2  18 /10  1.8 in, d3  30 /10  3.0 in  d2 / 2  1  0.9     tan    30.96 d / 2 1.5    3 

  tan 1 

  180    59.04 9 DE   0.5cos 59.04  0.8197 in 16 t W  25 lbf W r  25 tan 20 cos 59.04  4.681 lbf W a  25 tan 20 sin 59.04  7.803 lbf

W = –4.681i – 7.803j +25k R DG = 0.8197j + 1.25i R DC = –0.625j M D  R DG  W  R DC  FC  T  0 R DG  W  20.49i  31.25 j  5.917k R DC  FC  0.625 FCz i  0.625 FCx k

 20.49i  31.25 j  5.917k    0.625 FCz i  0.625 FCx k   Tj  0 T  31.25 lbf  in

Ans.

FC  9.47i  32.8k lbf



FC  9.47 2  32.82

F  0



1/2

Ans.  34.1 lbf

Ans.

FD  4.79i  7.80 j  57.8k lbf 1/2

2 2 FD (radial)   4.79    57.8    58.0 lbf Ans.   a FD (thrust)  W  7.80 lbf Ans. ______________________________________________________________________________

Chapter 13, Page 26/35

13-45

Pt  Pn cos  4 cos 30  3.464 teeth/in

t  tan 1 dP 

tan n tan 20  tan 1  22.80 cos cos 30

18  5.196 in 3.464

The forces on the shafts will be equal and opposite of the forces transmitted to the gears through the meshing teeth. Pinion (Gear 2) W r  W t tan t  800 tan 22.80  336 lbf W a  W t tan  800 tan 30  462 lbf W  336i  462 j  800k lbf Ans. 1/2

W   336    462   8002    2

Gear 3

2

 983 lbf

Ans.

W  336i  462 j  800k lbf Ans. W  983 lbf Ans. 32 dG   9.238 in 3.464 TG  W t r  800  9.238   7390 lbf  in

______________________________________________________________________________

Chapter 13, Page 27/35

13-46 From Prob. 13-45 solution,

Notice that the idler shaft reaction contains a couple tending to turn the shaft end-overend. Also the idler teeth are bent both ways. Idlers are more severely loaded than other gears, belying their name. Thus, be cautious. ______________________________________________________________________________ 13-47 Gear 3: Pt  Pn cos  7 cos 30  6.062 teeth/in

tan 20  0.4203, t  22.8  cos 30 54 d3   8.908 in 6.062 W t  500 lbf W a  500 tan 30  288.7 lbf W r  500 tan 22.8  210.2 lbf W3  210.2i  288.7 j  500k lbf Ans. Gear 4: 14 d4   2.309 in 6.062 8.908 W t  500  1929 lbf 2.309 W a  1929 tan 30  1114 lbf W r  1929 tan 22.8  811 lbf W4  811i  1114 j  1929k lbf Ans. ______________________________________________________________________________ tan t 

13-48 Pt  6 cos 30  5.196 teeth/in 42 d3   8.083 in 5.196 t  22.8

Chapter 13, Page 28/35

16  3.079 in 5.196 63025  25  T2   916 lbf  in 1720 T 916 Wt    595 lbf r 3.079 / 2 W a  595 tan 30  344 lbf W r  595 tan 22.8  250 lbf d2 

W  344i  250 j  595k lbf

R DC  6i,

R DG  3i  4.04 j

M D  R DC  FC  R DG  W  T  0 R DG  W  2404i  1785 j  2140k

(1)

R DC  FC  6 FCz j  6 FCy k

Substituting and solving Eq. (1) gives

T  2404i lbf  in FCz  297.5 lbf FCy  365.7 lbf

F  FD  FC  W  0 Substituting and solving gives FCx  344 lbf FDy  106.7 lbf FDz  297.5 lbf

FC  344i  356.7 j  297.5k lbf Ans. FD  106.7 j  297.5k lbf Ans. ______________________________________________________________________________

Chapter 13, Page 29/35

13-49

Since the transverse pressure angle is specified, we will assume the given module is also in terms of the transverse orientation.

d 2  mN 2  4 16   64 mm d3  mN3  4  36   144 mm d 4  mN 4  4  28   112 mm 6 kW  1000 W   rev   60 s   35.81 N  m  1600 rev/min  kW   2 rad   min  T 35.81 Wt    1119 N d 2 / 2 0.064 / 2 T

H



Chapter 13, Page 30/35

W r  W t tan t  1119 tan 20  407.3 N

W a  W t tan  1119 tan15  299.8 N F2 a  1119i  407.3j  299.8k N Ans. F3b  1119  407.3 i  1119  407.3 j

 711.7i  711.7 j N Ans. F4c  407.3i  1119 j  299.8k N Ans. ______________________________________________________________________________

13-50

N 14 36   2.021 in, d3   5.196 in  Pn cos 8cos 30 8cos 30 15 45 d4   3.106 in, d5   9.317 in  5cos15 5cos15

d2 

For gears 2 and 3: t  tan 1  tan n / cos   tan 1  tan 20 / cos 30   22.8

For gears 4 and 5: t  tan 1  tan 20 / cos15   20.6

F23t  T2 / r2  1200 /  2.021/ 2   1188 lbf 5.196  1987 lbf 3.106 F23r  F23t tan t  1188 tan 22.8  499 lbf F54t  1188

F54r  1986 tan 20.6  746 lbf F23a  F23t tan  1188 tan 30  686 lbf F54a  1986 tan15  532 lbf

Next, designate the points of action on gears 4 and 3, respectively, as points G and H, as shown. Position vectors are Chapter 13, Page 31/35

R CG  1.553 j  3k R CH  2.598 j  6.5k R CD  8.5k Force vectors are F54  1986i  748 j  532k F23  1188i  500 j  686k FC  FCx i  FCy j FD  FDx i  FDy j  FDz k Now, a summation of moments about bearing C gives

M C  R CG  F54  R CH  F23  R CD  FD  0 The terms for this equation are found to be R CG  F54  1412i  5961j  3086k R CH  F23  5026i  7722 j  3086k R CD  FD  8.5 FDy i  8.5 FDx j

When these terms are placed back into the moment equation, the k terms, representing the shaft torque, cancel. The i and j terms give 3614  425 lbf Ans. 8.5 13683 FDx   1610 lbf Ans. 8.5 FDy  

Next, we sum the forces to zero. F  FC  F54  F23  FD  0 Substituting, gives

 F i  F j   1987i  746 j  532k    1188i  499 j  686k   1610i  425 j  F k   0 x C

y C

z D

Solving gives FCx  1987  1188  1610  1565 lbf F  746  499  425  672 lbf y C z D

Ans.

Ans.

F  532  686  154 lbf Ans. ______________________________________________________________________________

Chapter 13, Page 32/35

VW 

13-51

WW t

 dW nW



  0.100  600 

60 60 H 2000    637 N  VW

  m/s

L  px NW  25 1  25 mm

  tan 1  tan 1

L  dW 25  4.550 lead angle  100 

WW t

W

cos n sin   f cos  V  VS  W   3.152 m/s cos  cos 4.550 In ft/min: V S = 3.28(3.152) = 10.33 ft/s = 620 ft/min Use f = 0.043 from curve A of Fig. 13-42. Then, from the first of Eq. (13-43) W

637  5323 N cos14.5  sin 4.55   0.043cos 4.55 

W y  W sin n  5323sin14.5  1333 N





W z  5323  cos14.5 cos 4.55  0.043 sin 4.55   5119 N The force acting against the worm is W  637i  1333 j  5119k N

Thus, A is the thrust bearing.

Ans.

R AG  0.05 j  0.10k , R AB  0.20k M A  R AG  W  R AB  FB  T  0 R AG  W  122.6i  63.7 j  31.85k R AB  FB  0.2 FBy i  0.2 FBx j

Substituting and solving gives

T  31.85 N  m Ans. FBx  318.5 N, FBy  613 N So

FB  318.5i  613 j N

Ans.

Chapter 13, Page 33/35

2 2 FB   613   318.5     F  FA  W  R B  0

Or

1/2

 691 N radial

FA    W  FB     637i  1333j  5119k  318.5i  613j

 318.5i  1946 j  5119k

Ans.

FAr  318.5i  1946 j N

Radial

1/2

2 2 FAr   318.5    1946    1972 N   a Thrust FA  5119 N ______________________________________________________________________________

13-52 From Prob. 13-51, WG  637i  1333j  5119k N pt  px So

dG 

N G px





48  25 



 382 mm

Bearing D takes the thrust load. M D  R DG  WG  R DC  FC  T  0 R DG  0.0725i  0.191j R DC  0.1075i The position vectors are in meters. R DG  WG  977.7i  371.1j  25.02k R DC  FC  0.1075 FCz j  0.1075 FCy k

Putting it together and solving,

T  977.7 N  m Ans. FC  233 j  3450k N, F  FC  WG  FD  0

FC  3460 N

Ans.

FD    FC  WG   637i  1566 j  1669k N

Ans.

Radial FDr  1566 j  1669k N Or



FDr  15662  1669 2 FDt  637i N



1/ 2

 2289 N (total radial)

(thrust)

Chapter 13, Page 34/35

______________________________________________________________________________ 13-53

VW 

 1.5  600 

 235.7 ft/min 12 33000  0.75  W x  WW t   105.0 lbf 235.7  pt  px   0.3927 in 8 L  0.3927  2   0.7854 in 0.7854  9.46  1.5  105.0 W  515.3 lbf  cos 20 sin 9.46  0.05cos 9.46 W y  515.3sin 20  176.2 lbf W z  515.3  cos 20 cos 9.46  0.05sin 9.46   473.4 lbf

  tan 1



So



W  105i  176 j  473k lbf

Ans.

T  105  0.75  78.8 lbf  in

Ans.

______________________________________________________________________________ 13-54 Computer programs will vary.

Chapter 13, Page 35/35