Mechanical Engineering Design Chapter 14 Solutions

Chapter 14 d 

14-1

N 22   3.667 in P 6

Table 14-2:

Y = 0.331  dn  (3.667)(1200) Eq. (13-34): V    1152 ft/min 12 12 1200  1152 Eq. (14-4b): K v   1.96 1200 H 15 Eq. (13-35) : W t  33 000  33 000  429.7 lbf V 1152 K vW t P 1.96(429.7)(6)     7633 psi  7.63 kpsi Ans. Eq. (14-7): FY 2(0.331) ________________________________________________________________________ 18 N   1.8 in 10 P Table 14-2: Y = 0.309  dn  (1.8)(600) Eq. (13-34): V    282.7 ft/min 12 12 1200  282.7 Eq. (14-4b): K v   1.236 1200 H 2 Eq. (13-35) : W t  33 000  33 000  233.5 lbf V 282.7 K W t P 1.236(233.5)(10) Eq. (14-7):   v   9340 psi  9.34 kpsi Ans. FY 1.0(0.309) ________________________________________________________________________

14-2

d 

14-3

d  mN  1.25(18)  22.5 mm Y = 0.309  dn  (22.5)(103 )(1800) V    2.121 m/s 60 60 6.1  2.121 Kv   1.348 6.1 60 000H 60 000(0.5) Wt    0.2358 kN  235.8 N  dn  (22.5)(1800)

Table 14-2:

Eq. (14-6b): Eq. (13-36):

K vW t 1.348(235.8)   68.6 MPa Ans. FmY 12(1.25)(0.309) ________________________________________________________________________

Eq. (14-8):

 

Chapter 14, Page 1/39

14-4 Table 14-2:

Eq. (14-6b): Eq. (13-36):

d  mN  8(16)  128 mm Y = 0.296  dn  (128)(103 )(150) V    1.0053 m/s 60 60 6.1  1.0053 Kv   1.165 6.1 60 000H 60 000(6) Wt    5.968 kN  5968 N  dn  (128)(150)

K vW t 1.165(5968)   32.6 MPa Ans. FmY 90(8)(0.296) ________________________________________________________________________

Eq. (14-8):

14-5 Table 14-2:

Eq. (14-6b): Eq. (13-36): Eq. (14-8):

 

d  mN  1(16)  16 mm Y = 0.296  dn  (16)(103 )(400) V    0.335 m/s 60 60 6.1  0.335 Kv   1.055 6.1 60 000H 60 000(0.15) Wt    0.4476 kN  447.6 N  dn  (16)(400)

K vW t 1.055(447.6) F    10.6 mm  mY 150(1)(0.296)

From Table 13-2, use F = 11 mm or 12 mm, depending on availability. Ans. ________________________________________________________________________ 14-6 Table 14-2:

Eq. (14-6b): Eq. (13-36):

Eq. (14-8):

d  mN  2(20)  40 mm Y = 0.322  dn  (40)(103 )(200) V    0.419 m/s 60 60 6.1  0.419 Kv   1.069 6.1 60 000H 60 000(0.5) Wt    1.194 kN  1194 N  dn  (40)(200)

F 

K vW t 1.069(1194)   26.4 mm 75(2.0)(0.322)  mY

From Table 13-2, use F = 28 mm. Ans. ________________________________________________________________________

Chapter 14, Page 2/39

14-7

24 N   4.8 in 5 P Table 14-2: Y = 0.337  dn  (4.8)(50) Eq. (13-34): V    62.83 ft/min 12 12 1200  62.83 Eq. (14-4b): K v   1.052 1200 H 6 Eq. (13-35) : W t  33 000  33 000  3151 lbf V 62.83 K vW t P 1.052(3151)(5) F  Eq. (14-7):   2.46 in 20(103 )(0.337) Y d 

Use F = 2.5 in Ans. ________________________________________________________________________ 16 N   4.0 in 4 P Table 14-2: Y = 0.296  dn  (4.0)(400) Eq. (13-34): V    418.9 ft/min 12 12 1200  418.9 Eq. (14-4b): K v   1.349 1200 H 20 Eq. (13-35) : W t  33 000  33 000  1575.6 lbf V 418.9 K W t P 1.349(1575.6)(4) F  v Eq. (14-7):   2.39 in 12(103 )(0.296) Y Use F = 2.5 in Ans. ________________________________________________________________________ d 

14-8

14-9

Try P = 8 which gives d = 18/8 = 2.25 in and Y = 0.309. Eq. (13-34):

Eq. (14-4b): Eq. (13-35): Eq. (14-7):

V 

 dn 12



 (2.25)(600) 12

 353.4 ft/min

1200  353.4  1.295 1200 H 2.5 W t  33 000  33 000  233.4 lbf V 353.4 K W t P 1.295(233.4)(8) F  v   0.783 in 10(103 )(0.309) Y Kv 

Using coarse integer pitches from Table 13-2, the following table is formed.

Chapter 14, Page 3/39

d V Wt Kv 9.000 1413.717 2.178 58.356 6.000 942.478 1.785 87.535 4.500 706.858 1.589 116.713 3.000 471.239 1.393 175.069 2.250 353.429 1.295 233.426 1.800 282.743 1.236 291.782 1.500 235.619 1.196 350.139 1.125 176.715 1.147 466.852

P 2 3 4 6 8 10 12 16

F 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 

 dn

 (36)(103 )(900)

 1.696 m/s 60 60 6.1  1.696 Eq. (14-6b): K v   1.278 6.1 60 000H 60 000(1.5) Eq. (13-36): W t    0.884 kN  884 N  dn  (36)(900) 1.278(884) Eq. (14-8): F   24.4 mm 75(2)(0.309) Using the preferred module sizes from Table 13-2: m 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



d 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

V 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

F Kv Wt 1.139 1768.388 86.917 1.174 1414.711 57.324 1.209 1178.926 40.987 1.278 884.194 24.382 1.417 589.463 12.015 1.556 442.097 7.422 1.695 353.678 5.174 1.834 294.731 3.888 2.112 221.049 2.519 2.391 176.839 1.824 2.669 147.366 1.414 3.225 110.524 0.961 3.781 88.419 0.721 4.476 70.736 0.547 5.450 55.262 0.406 6.562 44.210 0.313 7.953 35.368 0.243

Chapter 14, Page 4/39

1.204(202.6)  1 1   C  2100      F cos 20  0.228 0.684  

1/ 2

 100(103 )

2

 2100  1.204(202.6)   1 1   F     0.669 in 3     100(10 )   cos 20   0.228 0.684 

Use F = 0.75 in Ans. ________________________________________________________________________

d p  5(24)  120 mm, dG  5(48)  240 mm

14-13

V

 (120)(103 )(50)  0.3142 m/s 60

3.05  0.3142  1.103 3.05 60 000H 60(103 ) H Wt    3.183H  dn  (120)(50) where H is in kW and Wt is in kN

Eq. (14-6a):

Kv 

Table 14-8: C p  163 MPa [Note: Using Eq. (14-13) can result in wide variation in C p due to wide variation in cast iron properties]. 120sin 20 240 sin 20 Eq. (14-12): r1   20.52 mm, r2   41.04 mm 2 2 Eq. (14-14):

1.103(3.183) 103  H 690  163  60 cos 20o 

1/ 2

1   1     20.52 41.04  

H  3.94 kW Ans. ________________________________________________________________________

14-14

Eq. (14-6a):

d P  4(20)  80 mm, dG  4(32)  128 mm  (80)(103 )(1000) V   4.189 m/s 60 3.05  4.189 Kv   2.373 3.05 60(10)(103 ) Wt   2.387 kN  2387 N  (80)(1000)

Table 14-8: C p  163 MPa [Note: Using Eq. (14-13) can result in wide variation in C p due to wide variation in cast iron properties.] 80 sin 20 128sin 20 Eq. (14-12): r1   13.68 mm, r2   21.89 mm 2 2 Chapter 14, Page 6/39

1/ 2

 2.373(2387)  1 1      617 MPa Ans.   50 cos 20  13.68 21.89   ________________________________________________________________________

Eq. (14-14):

 C  163 

14-15 The pinion controls the design.

Bending

Y P = 0.303,

Y G = 0.359

17 30  1.417 in, dG   2.500 in 12 12  d Pn  (1.417)(525) V    194.8 ft/min 12 12 1200  194.8 Eq. (14-4b): K v   1.162 1200 Eq. (6-8), p. 282: Se  0.5(76)  38.0 kpsi Eq. (6-19), p. 287: k a = 2.70(76)–0.265 = 0.857 2.25 2.25 l    0.1875 in 12 Pd 3Y 3(0.303) Eq. (14-3): x P   0.0379 in 2P 2(12) dP 

Eq. (b), p. 737: t  Eq. (6-25), p. 289:

4lx 

4(0.1875)(0.0379)  0.1686 in

de  0.808 hb  0.808 0.875(0.1686)  0.310 in 0.107

 0.310  kb   Eq. (6-20), p. 288:  0.996   0.3  kc = kd = ke = 1 Account for one-way bending with k f = 1.66. (See Ex. 14-2.) Eq. (6-18), p. 287:

S e = 0.857(0.996)(1)(1)(1)(1.66)(38.0) = 53.84 kpsi

For stress concentration, find the radius of the root fillet (See Ex. 14-2). rf 

0.300 0.300   0.025 in P 12

From Fig. A-15-6, r r 0.025  f   0.148 d t 0.1686 Approximate D/d = ∞ with D/d = 3; from Fig. A-15-6, K t = 1.68. From Fig. 6-20, with S ut = 76 kpsi and r = 0.025 in, q = 0.62. Eq. (6-32):

K f = 1 + 0.62 (1.68 – 1) = 1.42

Chapter 14, Page 7/39

Se 53.84   16.85 psi 1.42(2.25) K f nd 0.875(0.303)(16 850) FYP all  320.4 lbf Wt   1.162(12) K v Pd 320.4(194.8) W tV   1.89 hp Ans. H  33 000 33 000

 all 

Wear

 1 =  2 = 0.292, E 1 = E 2 = 30(106) psi 1/ 2

Eq. (14-13):

      1 Cp    2   1  0.292    2  30 106      

Eq. (14-12):

r1 

 2285 psi

dP 1.417 sin   sin 20  0.242 in 2 2 d 2.500 r2  G sin   sin 20  0.428 in 2 2 1 1 1 1     6.469 in 1 r1 r2 0.242 0.428

(SC )108  0.4H B  10 kpsi  [0.4(149)  10](103 )  49 600 psi From the discussion and equation developed on the bottom of p. 329,  S  8 49 600  C ,all   C 10   33 067 psi n 2.25

Eq. (6-68), p. 329:

2

 33 067   0.875cos 20  Wt       22.6 lbf  2285   1.162(6.469)  W tV 22.6(194.8) H    0.133 hp Ans. 33 000 33 000 Rating power (pinion controls):

Eq. (14-14):

H 1 = 1.89 hp H 2 = 0.133 hp H all = (min 1.89, 0.133) = 0.133 hp Ans. ________________________________________________________________________ 14-16 See Prob. 14-15 solution for equation numbers.

Chapter 14, Page 8/39

Pinion controls: Bending

Y P = 0.322,

Y G = 0.447

d P = 20/3 = 6.667 in, d G = 100/3 = 33.333 in V   d P n / 12   (6.667)(870) / 12  1519 ft/min K v  (1200  1519) / 1200  2.266 Se  0.5(113)  56.5 kpsi 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 d e  0.808 2.5(0.695)  1.065 in kb  (1.065 / 0.30) 0.107  0.873 kc  kd  ke  1 k f = 1.66 (See Ex. 14-2.) Se  0.771(0.873)(1)(1)(1)(1.66)(56.5)  63.1 kpsi rf  0.300 / 3  0.100 in r r 0.100  f   0.144 d t 0.695 K t = 1.75, q = 0.85, K f = 1.64 Se 63.1  all    25.7 kpsi K f nd 1.64(1.5) FYP all 2.5(0.322)(25 700) Wt   3043 lbf  K v Pd 2.266(3) H  W tV / 33 000  3043(1519) / 33 000  140 hp

Ans.

Wear 1/ 2

Eq. (14-13):

Eq. (14-12):

      1 Cp    2   1  0.292    2  30 106      

 2285 psi

r 1 = (6.667/2) sin 20° = 1.140 in r 2 = (33.333/2) sin 20° = 5.700 in

Eq. (6-68), p. 329:

 C ,all

S C = [0.4(262) – 10](103) = 94 800 psi  SC / nd  94 800 / 1.5  77 400 psi

Chapter 14, Page 9/39

2

  F cos    1 W t   C ,all   C p  K v  1 / r1  1 / r2    2 1  77 400   2.5cos 20          2285   2.266   1 / 1.140  1 / 5.700   1130 lbf W tV 1130(1519) H    52.0 hp Ans. 33 000 33 000 For 108 cycles (revolutions of the pinion), the power based on wear is 52.0 hp. Rating power (pinion controls):

H1  140 hp H 2  52.0 hp H rated  min(140, 52.0)  52.0 hp Ans. ________________________________________________________________________ 14-17 See Prob. 14-15 solution for equation numbers. Given:  = 20°, n = 1145 rev/min, m = 6 mm, F = 75 mm, N P = 16 milled teeth, N G = 30T, S ut = 900 MPa, H B = 260, n d = 3, Y P = 0.296, and Y G = 0.359.

Pinion bending

d P  mN P  6(16)  96 mm dG  6(30)  180 mm  d Pn  (96)(103 )(1145)   5.76 m/s V  60 (60) 6.1  5.76 Kv   1.944 6.1 Se  0.5(900)  450 MPa ka  4.51(900)0.265  0.744 l  2.25m  2.25(6)  13.5 mm x  3Ym / 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 0.107

 24.23   0.884 kb     7.62  kc  k d  k e  1 k f = 1.66 (See Ex. 14-2) S e  0.744(0.884)(1)(1)(1)(1.66)(450)  491.3 MPa

rf  0.300m  0.300(6)  1.8 mm r/d = r f /t = 1.8/12 = 0.15, K t = 1.68, q = 0.86, K f = 1.58

Chapter 14, Page 10/39

 all 

Se 491.3   239.2 MPa K f nd 1.58 1.3

Eq. (14-8):

Wt 

FYm all 75(0.296)(6)(239.2)   16 390 N Kv 1.944

Eq. (13-36):

H 

W t dn 16.39 (96)(1145)   94.3 kW 60 000 60 000

Ans.

Wear: Pinion and gear Eq. (14-12):

r 1 = (96/2) sin 20 = 16.42 mm r 2 = (180/2) sin 20 = 30.78 mm 1/ 2

      1 Eq. (14-13): C p     190 MPa 2   1 0.292     2  207 103       Eq. (6-68), p. 329: S C = 6.89[0.4(260) – 10] = 647.7 MPa 647.7  C ,all  SC / nd   568 MPa 1.3

Eq. (14-14):

  W   C ,all   Cp   

2

t

 F cos   1   K v  1 / r1  1 / r2 

2

Eq. (13-36):

o 1  568   75cos 20          3469 N  190   1.944   1 / 16.42  1 / 30.78  W t dn 3.469 (96)(1145)   20.0 kW H  60 000 60 000

Thus, wear controls the gearset power rating; H = 20.0 kW. Ans. ________________________________________________________________________ N P = 17 teeth, N G = 51 teeth N 17 dP    2.833 in P 6 51 dG   8.500 in 6 V   d P n / 12   (2.833)(1120) / 12  830.7 ft/min

14-18

Eq. (14-4b):

K v = (1200 + 830.7)/1200 = 1.692

Chapter 14, Page 11/39

 all 

Sy nd



90 000  45 000 psi 2

Table 14-2:

Y P = 0.303, Y G = 0.410

Eq. (14-7):

Wt 

Eq. (13-35):

W tV 2686(830.7) H    67.6 hp 33 000 33 000

FYP all 2(0.303)(45 000)   2686 lbf KvP 1.692(6)

Based on yielding in bending, the power is 67.6 hp. (a) Pinion fatigue

Bending Eq. (2-121), p. 41: S ut = 0.5 H B = 0.5(232) = 116 kpsi Eq. (6-8), p. 282: Se  0.5Sut  0.5(116)  58 kpsi Eq. (6-19), p. 287: ka  2.70(116) 0.265  0.766

Table 13-1, p. 696: l 

Eq. (14-3): x 

1 1.25 2.25 2.25     0.375 in 6 Pd Pd Pd

3YP 3(0.303)   0.0758 in 2P 2(6)

Eq. (b), p. 737: t 

4lx 

4(0.375)(0.0758)  0.337 in

Eq. (6-25), p. 289: de  0.808 F t  0.808 2(0.337)  0.663 in

 0.663  Eq. (6-20), p. 288: kb     0.30  kc = kd = ke = 1

0.107

 0.919

Account for one-way bending with k f = 1.66. (See Ex. 14-2.) Eq. (6-18): Se  0.766(0.919)(1)(1)(1)(1.66)(58)  67.8 kpsi For stress concentration, find the radius of the root fillet (See Ex. 14-2). 0.300 0.300 rf    0.050 in P 6 r 0.05 r Fig. A-15-6:  f   0.148 0.338 d t Estimate D/d = ∞ by setting D/d = 3, K t = 1.68.

Chapter 14, Page 12/39

Fig. 6-20, p. 295: q = 0.86 Eq. (6-32), p. 295: K f  1  (0.86)(1.68  1)  1.58

 all 

Se 67.8   21.5 kpsi K f nd 1.58(2)

Wt 

FYP all 2(0.303)(21 500)   1283 lbf K v Pd 1.692(6)

H 

W tV 1283(830.7)   32.3 hp 33 000 33 000

Ans.

(b) Pinion fatigue

Wear 1/ 2

Eq. (14-13):

Eq. (14-12):

Eq. (6-68):

  1 Cp   2 6   2 [(1 - 0.292 ) / 30(10 )] 

 2285 psi

dP 2.833 sin   sin 20o  0.485 in 2 2 dG 8.500 r2  sin   sin 20o  1.454 in 2 2 1 1 1 1   2.750 in     r1 r2  0.485 1.454 r1 

(SC )108  0.4H B  10 kpsi

In terms of gear notation

 C = [0.4(232) – 10]103 = 82 800 psi We will introduce the design factor of n d = 2 and because it is a contact stress apply it to the load Wt by dividing by 2 . (See p. 329.)  82 800  C,all   c    58 548 psi 2 2 Solve Eq. (14-14) for Wt: 2

o  58 548   2 cos 20  W      265 lbf  2285  1.692(2.750)  W tV 265(830.7) H all    6.67 hp Ans. 33 000 33 000 For 108 cycles (turns of pinion), the allowable power is 6.67 hp. (c) Gear fatigue due to bending and wear

t

Chapter 14, Page 13/39

Bending x

Eq. (14-3):

Eq. (b), p. 737: t 

3YG 3(0.4103)   0.1026 in 2P 2(6)

4 lx 

4(0.375)(0.1026)  0.392 in 

Eq. (6-25): de  0.808 F t  0.808 2(0.392)  0.715 in 0.107

 0.715   0.911 Eq. (6-20): kb     0.30  kc = kd = ke = 1 k f = 1.66. (See Ex. 14-2.) Eq. (6-18): Se  0.766(0.911)(1)(1)(1)(1.66)(58)  67.2 kpsi r r 0.050  f   0.128 d t 0.392 Approximate D/d = ∞ by setting D/d = 3 for Fig. A-15-6; K t = 1.80. Fig. 6-20: q = 0.82 Eq. (6-32): K f  1  (0.82)(1.80  1)  1.66

 all 

Se 67.2   20.2 kpsi K f nd 1.66(2)

Wt 

FYP all 2(0.4103)(20 200)   1633 lbf K v Pd 1.692(6)

W tV 1633(830.7)   41.1 hp 33 000 33 000 The gear is thus stronger than the pinion in bending. H all 

Ans.

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, H all = 6.67 hp for 108 revolutions of each. As yet, we have no way to establish S C for 108/3 revolutions. (d)

Pinion bending: H 1 = 32.3 hp Pinion wear: H 2 = 6.67 hp Gear bending: H 3 = 41.1 hp Gear wear: H 4 = 6.67 hp Power rating of the gear set is thus H rated = min(32.3, 6.67, 41.1, 6.67) = 6.67 hp Ans. ________________________________________________________________________ 14-19

d P = 16/6 = 2.667 in, d G = 48/6 = 8 in

Chapter 14, Page 14/39

V 

 (2.667)(300)

 209.4 ft/min 12 33 000(5) Wt   787.8 lbf 209.4

Assuming uniform loading, K o = 1. Eq. (14-28): Qv  6, B  0.25(12  6) 2 / 3  0.8255 A  50  56(1  0.8255)  59.77 0.8255

 59.77  209.4  Eq. (14-27): K v    59.77   Table 14-2: YP  0.296, YG  0.4056 From Eq. (a), Sec. 14-10 with F = 2 in  2 0.296  ( K s ) P  1.192   6  

 1.196

0.0535

 1.088 0.0535

 2 0.4056  ( K s )G  1.192   1.097  6   From Eq. (14-30) with C mc = 1 2  0.0375  0.0125(2)  0.0625 Cp f  10(2.667) C p m  1, Cma  0.093 (Fig. 14 - 11), Ce  1 K m  1  1[0.0625(1)  0.093(1)]  1.156

Assuming constant thickness of the gears → K B = 1 m G = N G /N P = 48/16 = 3 With N (pinion) = 108 cycles and N (gear) = 108/3, Fig. 14-14 provides the relations: (YN ) P  1.3558(108 )0.0178  0.977 (YN )G  1.3558(108 / 3)0.0178  0.996 Fig. 14-6: J P  0.27, J G  0.38 Table 14-10: K R = 0.85 KT = Cf = 1 Eq. (14-23): Table 14-8:

I 

cos 20o sin 20o  3     0.1205 2(1)  3  1

C p  2300 psi

Strength: Grade 1 steel with H BP = H BG = 200

Chapter 14, Page 15/39

Fig. 14-2:

(S t ) P = (S t ) G = 77.3(200) + 12 800 = 28 260 psi

Fig. 14-5:

(S c ) P = (S c ) G = 322(200) + 29 100 = 93 500 psi

Fig. 14-15:

(Z N ) P = 1.4488(108)–0.023 = 0.948 (Z N ) G = 1.4488(108/3)–0.023 = 0.973

Sec. 14-12:

H BP /H BG = 1  C H = 1

Pinion tooth bending Eq. (14-15):

( ) P  W t K o K v K s

Pd K m K B F J

 6   (1.156)(1)   787.8(1)(1.196)(1.088)     2   0.27   13 170 psi Ans.

Eq. (14-41):

 S Y / ( KT K R )  (S F ) P   t N    28 260(0.977) / [(1)(0.85)]   2.47 13 170

Ans.

Gear tooth bending Eq. (14-15): Eq. (14-41):

 6   (1.156)(1)   9433 psi ( )G  787.8(1)(1.196)(1.097)     2   0.38  28 260(0.996) / [(1)(0.85)]  3.51 Ans. ( S F )G  9433

Ans.

Pinion tooth wear 1/ 2

Eq. (14-16):

 K C  ( c ) P  C p  W t K o K v K s m f  d P F I P 

1/ 2

  1.156   1    2300 787.8(1)(1.196)(1.088)     2.667(2)   0.1205     98 760 psi Ans.

Eq. (14-42):  S Z /( KT K R )   93 500(0.948) /[(1)(0.85)]  (S H ) P   c N   1.06   98 760 c   P 

Ans.

Gear tooth wear Chapter 14, Page 16/39

1/ 2

1/ 2

 (K )   1.097  ( c )G   s G  ( c ) P   (98 760)  99 170 psi  1.088   (K s )P  93 500(0.973)(1) /[(1)(0.85)] ( S H )G   1.08 Ans. 99 170

Ans.

The hardness of the pinion and the gear should be increased. ________________________________________________________________________ 14-20

d P = 2.5(20) = 50 mm, d G = 2.5(36) = 90 mm  d PnP  (50)(103 )(100) V    0.2618 m/s 60 60 60(120) Wt   458.4 N  (50)(103 )(100) With no specific information given to indicate otherwise, assume KB = Ko = Y = ZR = 1 Eq. (14-28): Eq. (14-27): Table 14-2:

Q v = 6, B = 0.25(12 – 6)2/3 = 0.8255 A = 50 + 56(1 – 0.8255) = 59.77  59.77  200(0.2618)  Kv    59.77   Y P = 0.322, Y G = 0.3775

0.8255

 1.099

Similar to Eq. (a) of Sec. 14-10 but for SI units: Ks 



1  0.8433 mF Y kb



0.0535

( K s ) P  0.8433  2.5(18) 0.322 

0.0535

( K s )G  0.8433  2.5(18) 0.3775  Cmc  Ce  C pm  1

 1.003

0.0535

=1.007

use 1 use 1

18  0.025  0.011 10(50)  0.247  0.0167(0.709)  0.765(104 )(0.7092 )  0.259 K H  1  1[0.011(1)  0.259(1)]  1.27 F  18 / 25.4  0.709 in, C pf 

Cma

Fig. 14-14:

(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

Chapter 14, Page 17/39

cos 20o sin 20o  1.8     0.103 2(1)  1.8  1 

Eq. (14-23):

ZI 

Table 14-8:

Z E  191 MPa

Strength

Grade 1 steel, H BP = H BG = 200

Fig. 14-2: Fig. 14-5: Fig. 14-15:

(S t ) P = (S t ) G = 0.533(200) + 88.3 = 194.9 MPa (S c ) P = (S c ) G = 2.22(200) + 200 = 644 MPa (Z N ) P = 1.4488(108)–0.023 = 0.948 (Z N )G  1.4488(108 / 1.8)0.023  0.961 H BP / H BG  1  ZW  CH  1

Fig. 14-12:

Pinion tooth bending Eq. (14-15):

 1 KH KB  ( ) P   W t K o K v K s  bmt YJ  P   1  1.27(1)   458.4(1)(1.099)(1)     43.08 MPa 18(2.5)   0.33 

S Y  194.9  0.977  Eq. (14-41) for SI: (S F ) P   t N    4.99 43.08 1(0.885)    Y YZ  P

Ans.

Ans.

Gear tooth bending  1  1.27(1)  ( )G  458.4(1)(1.099)(1)     37.42 MPa 18(2.5)   0.38  194.9  0.987  ( S F )G   5.81 Ans. 37.42 1(0.885) 

Ans.

Pinion tooth wear Eq. (14-16):

 K Z  ( c ) P   Z E W t K o K v K s H R   d w1b Z I  P 

 1.27   1   191 458.4(1)(1.099)(1)     501.8 MPa  50(18)   0.103  S Z Z  644  0.948(1)  Eq. (14-42) for SI: ( S H ) P   c N W    1.37 Ans. 501.8 1(0.885)    c Y YZ  P

Ans.

Gear tooth wear  (K )  ( c )G   s G   (K s )P 

1/ 2

1/ 2

1 ( c ) P    1

(501.8)  501.8 MPa

Ans.

Chapter 14, Page 18/39

644 0.961(1)  1.39 Ans. 501.8 1(0.885) ________________________________________________________________________ ( S H )G 

14-21

Pt  Pn cos  6 cos 30  5.196 teeth/in 48 16 (3.079)  9.238 in dP   3.079 in, dG  5.196 16  (3.079)(300)  241.8 ft/min V  12 0.8255  59.77  241.8  33 000(5) t  682.3 lbf , K v    1.210 W   241.8 59.77  

From Prob. 14-19: YP  0.296, YG  0.4056 ( K s ) P  1.088, ( K s )G  1.097, K B  1 mG  3, (YN ) P  0.977, (YN )G  0.996, K R  0.85 ( St ) P  ( St )G  28 260 psi, CH  1, ( S c ) P  ( S c )G  93 500 psi ( Z N ) P  0.948,

( Z N )G  0.973,

C p  2300 psi

The pressure angle is:  tan 20 

t  tan 1    22.80  cos 30  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. (14-25): 1/ 2

2 2  3.079   9.238    2  0.167   4.2582  Z    0.167   1.419        2   2  3.079 9.238     sin 22.8 2   2  0.9479  2.1852  2.3865  0.7466 Conditions O.K . for use

pN  pn cos  n 

Eq. (14-21):

mN 

 6

1/ 2

cos 20  0.4920 in

pN 0.492   0.6937 0.95Z 0.95(0.7466)

Chapter 14, Page 19/39

Eq. (14-23):

 sin 22.8 cos 22.8   3  I    3  1   0.193 2(0.6937)   

Fig. 14-7:

J P  0.45,

J G  0.54

Fig. 14-8: Corrections are 0.94 and 0.98. J P  0.45(0.94)  0.423, J G  0.54(0.98)  0.529 2  0.0375  0.0125(2)  0.0525 Cmc  1, C pf  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  1.146(1) ( ) P  682.3(1)(1.21)(1.088)   6323 psi   2  0.423 28 260(0.977) / [1(0.85)] (S F ) P   5.14 Ans. 6323

Ans.

Gear tooth bending  5.196  1.146(1) ( )G  682.3(1)(1.21)(1.097)   5097 psi   2  0.529 28 260(0.996) / [1(0.85)] ( S F )G   6.50 Ans. 5097

Ans.

Pinion tooth wear 1/ 2

  1.146   1   ( c ) P  2300 682.3(1)(1.21)(1.088)     3.078(2)   0.193    93 500(0.948) / [(1)(0.85)] (S H ) P   1.54 Ans. 67 700

 67 700 psi

Ans.

Gear tooth wear 1/ 2

1.097  ( c )G   (67 700)  67 980 psi Ans. 1.088  93 500(0.973) /[(1)(0.85)]  1.57 Ans. ( S H )G  67 980 ________________________________________________________________________ 14-22 Given: R = 0.99 at 108 cycles, H B = 232 through-hardening Grade 1, core and case, both gears. N P = 17T, N G = 51T, Table 14-2: Y P = 0.303, Y G = 0.4103 Chapter 14, Page 20/39

Fig. 14-6:

J P = 0.292, J G = 0.396 d P = N P / P = 17 / 6 = 2.833 in, d G = 51 / 6 = 8.500 in.

Pinion bending From Fig. 14-2: 0.99

Fig. 14-14:

( St )107  77.3H B  12 800  77.3(232)  12 800  30 734 psi

Y N = 1.6831(108)–0.0323 = 0.928 V   d P n / 12   (2.833)(1120 / 12)  830.7 ft/min KT  K R  1, S F  2, S H  2 30 734(0.928)  14 261 psi  all  2(1)(1)

Qv  5, B  0.25(12  5)2 / 3  0.9148 A  50  56(1  0.9148)  54.77  54.77  830.7  K v    54.77  

0.9148

 1.472

0.0535

 2 0.303  K s  1.192   1.089  use 1  6   K m  Cm f  1  Cmc (C p f C p m  CmaCe ) Cmc  1 F  0.0375  0.0125F 10d 2   0.0375  0.0125(2)  0.0581 10(2.833) 1

C pf 

C pm

Eq. (14-15):

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)]  1.217 KB  1 FJ P all Wt  K o K v K s Pd K m K B 2(0.292)(14 261)   775 lbf 1(1.472)(1)(6)(1.217)(1) 775(830.7) W tV H    19.5 hp 33 000 33 000

Pinion wear

Chapter 14, Page 21/39

Fig. 14-15: Eq. (14-23): Fig. 14-5:

Z N = 2.466N–0.056 = 2.466(108)–0.056 = 0.879 mG  51 / 17  3 cos 20o sin 20o  3  I     1.205, 2  3  1

CH  1

(Sc )107  322H B  29 100  322(232)  29 100  103 804 psi 103 804(0.879)  c,all   64 519 psi 2(1)(1) 0.99

2

Eq. (14-16):

  Fd P I W   c,all   C p  K o K v K s K mC f   2  64 519   2(2.833)(0.1205)    2300  1(1.472)(1)(1.2167)(1)   300 lbf W tV 300(830.7) H    7.55 hp 33 000 33 000 t

The pinion controls, therefore H rated = 7.55 hp Ans. ________________________________________________________________________ 14-23

l = 2.25/ P d , t 

4 lx 

x = 3Y / 2P d  2.25  3Y  3.674 4 Y    Pd  Pd  2Pd 

 3.674  F Y d e  0.808 F t  0.808 F   Y  1.5487 Pd  Pd  0.107

0.0535  1.5487 F Y / P  F Y  d  kb    0.8389     Pd  0.30    0.0535 F Y  1 Ks  Ans.  1.192   kb  Pd  ________________________________________________________________________

14-24 Y P = 0.331, Y G = 0.422, J P = 0.345, J G = 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 d G = 60 / 4 = 15.000 in

Chapter 14, Page 22/39

V  Pinion bending 0.99

 (5.5)(1145) 12

 1649 ft/min

(St )107  77.3H B  12 800  77.3(250)  12 800  32 125 psi

YN  1.6831[3(109 )]0.0323  0.832 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

 all  P 

0.8255

 59.77  1649  K v    1.534  59.77   K s  1, Cm  1 F Cmc   0.0375  0.0125F 10d 3.25   0.0375  0.0125(3.25)  0.0622 10(5.5) Cma  0.127  0.0158(3.25)  0.093(104 )(3.252 )  0.178

Ce  1

K m  Cm f  1  (1)[0.0622(1)  0.178(1)]  1.240 K B  1,

Eq. (14-15):

KT  1

26 728(3.25)(0.345)  3151 lbf 1.25(1.534)(1)(4)(1.240) 3151(1649) H1   157.5 hp 33 000

W1t 

Gear bending By similar reasoning, W2t  3861 lbf and H 2  192.9 hp Pinion wear mG  60 / 22  2.727

cos 20o sin 20o  2.727     0.1176 2  1  2.727  0.99 (S c )107  322(250)  29 100  109 600 psi I 

(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 ( c,all ) P  1(1)(1)

Chapter 14, Page 23/39

2

  Fd P I W3t   c,all   C  KKKK C  p  o v s m f 2

 79 679   3.25(5.5)(0.1176)       1061 lbf  2300  1.25(1.534)(1)(1.24)(1)  1061(1649) H3   53.0 hp 33 000

Gear wear Similarly,

W4t  1182 lbf ,

H 4  59.0 hp

Rating H rated  min( H1, H 2 , H 3 , H 4 )  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 33 000K o H 33 000(1.25)(40)   1000 lbf Wt  V 1649

Pinion bending: The factor of safety, based on load and stress, is W1t 3151 (S F ) P    3.15 1000 1000 Gear bending based on load and stress ( S F )G 

W2t 3861   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:

n4 

W4t 1182   1.18 1000 1000 Chapter 14, Page 24/39

(SH )G  1.18  1.09

based on stress:

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. ________________________________________________________________________ 14-26 Solution summary from Prob. 14-24: n = 1145 rev/min, K o = 1.25, Grade 1 materials, N P = 22T, N G = 60T, m G = 2.727, Y P = 0.331,Y G = 0.422, J P = 0.345, J G = 0.410, P d = 4T /in, F = 3.25 in, Q v = 6, (N c ) P = 3(109), R = 0.99, K m = 1.240, K T = 1, K B = 1, d P = 5.500 in, d G = 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

Pinion H B : 250 core, 390 case Gear H B : 250 core, 390 case Bending

( all ) P ( all )G W1t W2t

   

26 728 psi 27 546 psi 3151 lbf , 3861 lbf ,

(St ) P  32 125 psi (St )G  32 125 psi H1  157.5 hp H 2  192.9 hp

Wear

  20o ,

I  0.1176,

(Z N ) P  0.727

(Z N )G  0.769, CP  2300 psi (Sc ) P  Sc  322(390)  29 100  154 680 psi 154 680(0.727)  112 450 psi ( c,all ) P  1(1)(1) 154 680(0.769)  118 950 psi ( c,all )G  1(1)(1) 2

 112 450  W3t    (1061)  2113 lbf ,  79 679 

H3 

2113(1649)  105.6 hp 33 000

2

  118 950 W   (1182)  2354 lbf ,  109 600(0.769)  t 4

H4 

2354(1649)  117.6 hp 33 000

Rated power Chapter 14, Page 25/39

H rated = min(157.5, 192.9, 105.6, 117.6) = 105.6 hp

Ans.

Prob. 14-24: H rated = 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

Modification of S t by (Y N ) P = 0.832 produces ( all ) P  45 657 psi,

Similarly for (Y N ) G = 0.859 ( all )G  47 161 psi,

W  4569 lbf , W  5668 lbf , t 1 t 2

and

H1  228 hp H 2  283 hp

From Table 14-8, C p  2300 psi. Also, from Table 14-6: 0.99

(Sc )107  180 000 psi

Modification of S c by Y N produces

( c,all ) P  130 525 psi ( c,all )G  138 069 psi and

W3t  2489 lbf , W4t  2767 lbf ,

H 3  124.3 hp H 4  138.2 hp

Rating H rated = min(228, 283, 124, 138) = 124 hp Ans. ________________________________________________________________________

14-28 Grade 2, 9310 carburized and case-hardened to 285 core and 580 case in Prob. 14-27. Chapter 14, Page 26/39

Summary: Table 14-3:

0.99

(St )107  65 000 psi

( all ) P  53 959 psi ( all )G  55 736 psi and it follows that

W1t  5400 lbf , W2t  6699 lbf ,

H1  270 hp H 2  335 hp

From Table 14-8, C p  2300 psi. Also, from Table 14-6: Sc  225 000 psi

( c,all ) P  181 285 psi ( c,all )G  191 762 psi Consequently,

W3t  4801 lbf , W4t  5337 lbf ,

H 3  240 hp H 4  267 hp

Rating H rated = min(270, 335, 240, 267) = 240 hp. Ans. ________________________________________________________________________ 14-29 Given: n = 1145 rev/min, K o = 1.25, N P = 22T, N G = 60T, m G = 2.727, d P = 2.75 in, d G = 7.5 in, Y P = 0.331,Y G = 0.422, J P = 0.335, J G = 0.405, P = 8T /in, F = 1.625 in, H B = 250, case and core, both gears. C m = 1, F/d P = 0.0591, C f = 0.0419, C pm = 1, C ma = 0.152, C e = 1, K m = 1.1942, K T = 1, K B = 1, K s = 1,V = 824 ft/min, (Y N ) P = 0.8318, (Y N ) G = 0.859, K R = 1, I = 0.117 58 (St )107  32 125 psi ( all ) P  26 668 psi ( all )G  27 546 psi

0.99

and it follows that

W1t  879.3 lbf , W2t  1098 lbf ,

H1  21.97 hp H 2  27.4 hp

For wear

Chapter 14, Page 27/39

W3t  304 lbf , W4t  340 lbf ,

H 3  7.59 hp H 4  8.50 hp

Rating H rated = min(21.97, 27.4, 7.59, 8.50) = 7.59 hp In Prob. 14-24, H rated = 53 hp. Thus, 7.59 1  0.1432  , 53.0 6.98

not

1 8

Ans.

The transmitted load rating is t Wrated  min(879.3, 1098, 304, 340)  304 lbf

In Prob. 14-24 t Wrated  1061 lbf

Thus 304 1 1  0.2865  Ans. , not 1061 3.49 4 ________________________________________________________________________

14-30 S P = S H = 1, P d = 4, J P = 0.345, J G = 0.410, K o = 1.25

Bending Table 14-4:

0.99

(St )107  13 000 psi 13 000(1)  13 000 psi 1(1)(1)  all FJ P 13 000(3.25)(0.345)    1533 lbf K o K v K s Pd K m K B 1.25(1.534)(1)(4)(1.24)(1) 1533(1649)   76.6 hp 33 000  W1t J G / J P  1533(0.410) / 0.345  1822 lbf  H1J G / J P  76.6(0.410) / 0.345  91.0 hp

( all ) P  ( all )G 

W1t H1

W2t H2 Wear Table 14-8: Table 14-7:

C p  1960 psi 0.99

(Sc )107  75 000 psi  ( c,all ) P  ( c,all )G

Chapter 14, Page 28/39

2

 ( )  Fd p I W3t   c,all P   C p  K o K v K s K mC f 2

 75 000  3.25(5.5)(0.1176) W   1295 lbf   1960  1.25(1.534)(1)(1.24)(1) W4t  W3t  1295 lbf 1295(1649) H 4  H3   64.7 hp 33 000 t 3

Rating H rated = 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 S c /S t ratio. Also note that the life is 107 pinion revolutions which is (1/300) of 3(109). Longer life goals require power de-rating. ________________________________________________________________________ 14-31 From Table A-24a, E av = 11.8(106) Mpsi For  = 14.5 and H B = 156

SC 

1.4(81)  51 693 psi 2sin14.5 / [11.8(106 )]

For  = 20 1.4(112)  52 008 psi 2sin 20 / [11.8(106 )] SC  0.32(156)  49.9 kpsi The first two calculations were approximately 4 percent higher. ________________________________________________________________________ SC 

14-32 Programs will vary. ________________________________________________________________________ 14-33

(YN ) P  0.977, (YN )G  0.996 (St ) P  ( St )G  82.3(250)  12 150  32 725 psi 32 725(0.977) ( all ) P   37 615 psi 1(0.85) 37 615(1.5)(0.423) W1t   1558 lbf 1(1.404)(1.043)(8.66)(1.208)(1) 1558(925) H1   43.7 hp 33 000

Chapter 14, Page 29/39

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 ( all )G 

Table 14-6:

0.99

(Sc )107  150 000 psi

 0.948(1)  ( c,allow ) P  150 000    167 294 psi  1(0.85)  2  167 294  1.963(1.5)(0.195)  W3t       2074 lbf  2300   1(1.404)(1.043)  2074(925) H3   58.1 hp 33 000 0.973 ( c,allow )G  (167 294)  171 706 psi 0.948 2  171 706  1.963(1.5)(0.195)  t W4       2167 lbf  2300   1(1.404)(1.052)  2167(925) H4   60.7 hp 33 000 H rated  min(43.7, 56.3, 58.1, 60.7)  43.7 hp Ans. Pinion bending is controlling. ________________________________________________________________________ (YN ) P  1.6831(108 ) 0.0323  0.928

14-34

(YN )G  1.6831(108 / 3.059) 0.0323  0.962 Table 14-3:

S t = 55 000 psi 55 000(0.928)  60 047 psi ( all ) P  1(0.85) 60 047(1.5)(0.423) W1t   2487 lbf 1(1.404)(1.043)(8.66)(1.208)(1) 2487(925) H1   69.7 hp 33 000 0.962 ( all )G  (60 047)  62 247 psi 0.928

Chapter 14, Page 30/39

62 247  0.5346    (2487)  3258 lbf 60 047  0.423  3258 H2  (69.7)  91.3 hp 2487

W2t 

Table 14-6:

S c = 180 000 psi (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 ( c,all ) P  1(0.85) 2

 186 141  1.963(1.5)(0.195)  W      2568 lbf  2300   1(1.404)(1.043)  2568(925) H3   72.0 hp 33 000 0.9358 ( c,all )G  (186 141)  198 169 psi 0.8790 t 3

2

 198 169   1.043  W     (2568)  2886 lbf  186 141   1.052  2886(925) H4   80.9 hp 33 000 H rated  min(69.7, 91.3, 72, 80.9)  69.7 hp t 4

Ans.

Pinion bending controlling ________________________________________________________________________ (Y N ) P = 0.928, (Y N ) G = 0.962 (See Prob. 14-34)

14-35

Table 14-3:

S t = 65 000 psi 65 000(0.928) ( all ) P   70 965 psi 1(0.85) 70 965(1.5)(0.423) W1t   2939 lbf 1(1.404)(1.043)(8.66)(1.208) 2939(925) H1   82.4 hp 33 000 65 000(0.962)  73 565 psi ( all )G  1(0.85) 73 565  0.5346  W2t    (2939)  3850 lbf 70 965  0.423  3850 H2  (82.4)  108 hp 2939

Chapter 14, Page 31/39

Table 14-6:

S c = 225 000 psi ( Z N ) P  0.8790, ( Z N )G  0.9358 225 000(0.879)  232 676 psi ( c,all ) P  1(0.85) 2

 232 676  1.963(1.5)(0.195)  W3t       4013 lbf  2300   1(1.404)(1.043)  4013(925) H3   112.5 hp 33 000 0.9358 ( c,all )G  (232 676)  247 711 psi 0.8790 2

 247 711   1.043  W     (4013)  4509 lbf  232 676   1.052  4509(925) H4   126 hp 33 000 H rated  min(82.4, 108, 112.5, 126)  82.4 hp t 4

Ans.

The bending of the pinion is the controlling factor. ________________________________________________________________________ P = 2 teeth/in, d = 8 in, N = dP = 8 (2) = 16 teeth     F  4 p  4    4    2 P 2  M x  0  10(300) cos 20  4FB cos20

14-36

F B = 750 lbf W t  FB cos 20  750 cos 20  705 lbf n = 2400 / 2 = 1200 rev/min  dn  (8)(1200)   2513 ft/min V  12 12 We will obtain all of the needed factors, roughly in the order presented in the textbook. Fig. 14-2:

S t = 102(300) + 16 400 = 47 000 psi

Fig. 14-5: Fig. 14-6:

S c = 349(300) + 34 300 = 139 000 psi J = 0.27 cos 20o sin 20o  2  I     0.107 2(1)  2  1 C p  2300 psi

Eq. (14-23): Table 14-8:

Assume a typical quality number of 6. Eq. (14-28): B  0.25(12  Qv ) 2 / 3  0.25(12  6) 2 / 3  0.8255

Chapter 14, Page 32/39

A  50  56(1  B)  50  56(1  0.8255)  59.77 B

Eq. (14-27):

A V   59.77  2513  K v       A 59.77    

0.8255

 1.65

To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.296. From Eq. (a), Sec. 14-10, 0.0535

0.0535

F Y   2 0.296  K s  1.192   1.192   1.23   2  P    The load distribution factor is applicable for straddle-mounted gears, which is not the case here since the gear is mounted outboard of the bearings. Lacking anything better, we will use the load distribution factor as a rough estimate.

Eq. (14-31): Eq. (14-32): Eq. (14-33): Fig. 14-11: Eq. (14-35): Eq. (14-30):

C mc = 1 (uncrowned teeth) 2 Cp f   0.0375  0.0125(2 )  0.1196 10(8) C pm = 1.1 C ma = 0.23 (commercial enclosed gear unit) Ce = 1 K m  1  1[0.1196(1.1)  0.23(1)]  1.36

For the stress-cycle factors, we need the desired number of load cycles. Fig. 14-14: Fig. 14-15: Eq. 14-38:

N = 15 000 h (1200 rev/min)(60 min/h) = 1.1 (109) rev Y N = 0.9 Z N = 0.8 K R  0.658  0.0759ln 1  R   0.658  0.0759ln 1  0.95  0.885

With no specific information given to indicate otherwise, assume K o = K B = K T = C f = 1 Tooth bending Eq. (14-15):

Eq. (14-41):

Pd K m K B F J  2   (1.36)(1)   705(1)(1.65)(1.23)   2294 psi   2   0.27 

  W t KoKvK s

 S Y / ( KT K R )  SF   t N    47 000(0.9) / [(1)(0.885)]   20.8 2294

Ans.

Tooth wear Chapter 14, Page 33/39

1/ 2

Eq. (14-16):

 K C   c  C p W t Ko K v K s m f  dPF I  

1/ 2

  1.36   1    2300 705(1)(1.65)(1.23)     8(2 )   0.107     43 750 psi Since gear B is a pinion, C H is not used in Eq. (14-42) (see p. 761), where S c Z N / ( KT K R )

SH 

c

139 000(0.8) / [(1)(0.885)]     2.9 Ans 43 750   ________________________________________________________________________

m = 18.75 mm/tooth, d = 300 mm N = d/m = 300 / 18.75 = 16 teeth F  b  4 p  4  m   4 18.75   236 mm

14-37

M

 0  300(11) cos 20  150 FB cos25 F B = 22.81 kN W t  FB cos 25  22.81cos 25  20.67 kN n = 1800 / 2 = 900 rev/min  dn  (0.300)(900)   14.14 m/s V  60 60 We will obtain all of the needed factors, roughly in the order presented in the textbook. Fig. 14-2: Fig. 14-5: Fig. 14-6: Eq. (14-23): Table 14-8:

x

S t = 0.703(300) + 113 = 324 MPa S c = 2.41(300) + 237 = 960 MPa J = Y J = 0.27 cos 20o sin 20o  5  I  ZI     0.134 2(1)  5  1 Z E  191 MPa

Assume a typical quality number of 6. Eq. (14-28): B  0.25(12  Qv ) 2/ 3  0.25(12  6)2/ 3  0.8255 A  50  56(1  B)  50  56(1  0.8255)  59.77 Eq. (14-27):

 A  200V K v   A 

B

 59.77  200(14.14)       59.77   

0.8255

 1.69

To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.296. Similar to Eq. (a) of Sec. 14-10 but for SI units: Chapter 14, Page 34/39

Ks 



1  0.8433 mF Y kb



0.0535

K s  0.8433 18.75(236) 0.296 

0.0535

 1.28

Convert the diameter and facewidth to inches for use in the load-distribution factor equations. d = 300/25.4 = 11.81 in, F = 236/25.4 = 9.29 in Eq. (14-31): Eq. (14-32): Eq. (14-33): Fig. 14-11: Eq. (14-35): Eq. (14-30):

C mc = 1 (uncrowned teeth) 9.29 C pf   0.0375  0.0125(9.29)  0.1573 10(11.81) C pm = 1.1 C ma = 0.27 (commercial enclosed gear unit) Ce = 1 K m  K H  1  1[0.1573(1.1)  0.27(1)]  1.44

For the stress-cycle factors, we need the desired number of load cycles. N = 12 000 h (900 rev/min)(60 min/h) = 6.48 (108) rev Fig. 14-14: Fig. 14-15: Eq. 14-38:

Y N = 0.9 Z N = 0.85 K R  0.658  0.0759ln 1  R   0.658  0.0759ln 1  0.98  0.955

With no specific information given to indicate otherwise, assume K o = K B = K T = Z R = 1. Tooth bending Eq. (14-15):

  W t KoKvKs

1 KH KB bmt YJ

   (1.44)(1)  1  20 670(1)(1.69)(1.28)     53.9 MPa  236(18.75)   0.27  Eq. (14-41):

 S Y / ( KT K R )  SF   t N    324(0.9) / [(1)(0.955)]   5.66 53.9

Ans.

Tooth wear 1/ 2

Eq. (14-16):

 K Z   c  Z E W t Ko K v K s H R  d w1b Z I  

Chapter 14, Page 35/39

1/ 2

  1.44   1    191  20 670(1)(1.69)(1.28)     300(236)   0.134     498 MPa

Since gear B is a pinion, C H is not used in Eq. (14-42) (see p. 761), where SH 

S c Z N / ( KT K R )

c 960(0.85) / [(1)(0.955)]   1.72 Ans 498 ________________________________________________________________________ 14-38 From the solution to Prob. 13-40, n = 191 rev/min, Wt = 1600 N, d = 125 mm, N = 15 teeth, m = 8.33 mm/tooth. F  b  4 p  4  m   4  8.33  105 mm V 

 dn 60



 (0.125)(191)

 1.25 m/s

60

We will obtain all of the needed factors, roughly in the order presented in the textbook. Table 14-3: Table 14-6: Fig. 14-6: Eq. (14-23): Table 14-8:

S t = 65 kpsi = 448 MPa S c = 225 kpsi = 1550 MPa J = Y J = 0.25 cos 20o sin 20o  2  I  ZI     0.107 2(1)  2  1 Z E  191 MPa

Assume a typical quality number of 6. Eq. (14-28):

Eq. (14-27):

B  0.25(12  Qv ) 2 / 3  0.25(12  6) 2 / 3  0.8255 A  50  56(1  B)  50  56(1  0.8255)  59.77  A  200V K v   A 

B

 59.77  200(1.25)       59.77   

0.8255

 1.21

To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.290. Similar to Eq. (a) of Sec. 14-10 but for SI units: Ks 



1  0.8433 mF Y kb



0.0535

0.0535

K s  0.8433 8.33(105) 0.290   1.17 Convert the diameter and facewidth to inches for use in the load-distribution factor Chapter 14, Page 36/39

equations. d = 125/25.4 = 4.92 in, F = 105/25.4 = 4.13 in Eq. (14-31): C mc = 1 (uncrowned teeth) 4.13 Eq. (14-32): C pf   0.0375  0.0125(4.13)  0.0981 10(4.92) Eq. (14-33): C pm = 1 C ma = 0.32 (open gearing) Fig. 14-11: Eq. (14-35): C e = 1 Eq. (14-30): K m  K H  1  1[0.0981(1)  0.32(1)]  1.42 For the stress-cycle factors, we need the desired number of load cycles. N = 12 000 h (191 rev/min)(60 min/h) = 1.4 (108) rev Fig. 14-14: Y N = 0.95 Fig. 14-15: Z N = 0.88 K R  0.658  0.0759ln 1  R   0.658  0.0759ln 1  0.95  0.885 Eq. 14-38: With no specific information given to indicate otherwise, assume K o = K B = K T = Z R = 1. Tooth bending Eq. (14-15):

  W t Ko K vK s

1 KH KB bmt YJ

   (1.42)(1)  1  1600(1)(1.21)(1.17)     14.7 MPa 105(8.33)   0.25  Since gear is a pinion, C H is not used in Eq. (14-42) (see p. 761), where  S Y / ( KT K R )  SF   t N    448(0.95) / [(1)(0.885)]   32.7 14.7

Ans.

Tooth wear 1/ 2

Eq. (14-16):

 K Z   c  Z E W t Ko K v K s H R  d w1b Z I  

1/ 2

  1.42   1    191 1600(1)(1.21)(1.17)     125(105)   0.107     289 MPa

 S Z / ( KT K R )  SH   c N  c   1550(0.88) / [(1)(0.885)]     5.33 Ans 289   ________________________________________________________________________

Eq. (14-42):

Chapter 14, Page 37/39

14-39 From the solution to Prob. 13-41, n = 2(70) = 140 rev/min, Wt = 180 lbf, d = 5 in N = 15 teeth, P = 3 teeth/in.     F  4 p  4    4    4.2 in P 3  dn  (5)(140) V    183.3 ft/min 12 12

We will obtain all of the needed factors, roughly in the order presented in the textbook. Table 14-3: Table 14-6: Fig. 14-6: Eq. (14-23): Table 14-8:

S t = 65 kpsi S c = 225 kpsi J = 0.25 cos 20o sin 20o  2  I     0.107 2(1)  2  1 C p  2300 psi

Assume a typical quality number of 6. Eq. (14-28):

B  0.25(12  Qv ) 2 / 3  0.25(12  6) 2 / 3  0.8255 A  50  56(1  B)  50  56(1  0.8255)  59.77 B

Eq. (14-27):

A V   59.77  183.3  K v       A 59.77    

0.8255

 1.18

To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.290. From Eq. (a), Sec. 14-10, 0.0535

Eq. (14-31): Eq. (14-32): Eq. (14-33): Fig. 14-11: Eq. (14-35): Eq. (14-30):

F Y   4.2 0.290   1.192  K s  1.192    3  P    C mc = 1 (uncrowned teeth) 4.2 C pf   0.0375  0.0125(4.2)  0.099 10(5) C pm = 1 C ma = 0.32 (Open gearing) Ce = 1 K m  1  1[0.099(1)  0.32(1)]  1.42

0.0535

 1.17

For the stress-cycle factors, we need the desired number of load cycles. N = 14 000 h (140 rev/min)(60 min/h) = 1.2 (108) rev Fig. 14-14: Y N = 0.95 Fig. 14-15: Z N = 0.88 K R  0.658  0.0759ln 1  R   0.658  0.0759ln 1  0.98  0.955 Eq. 14-38: With no specific information given to indicate otherwise, assume K o = K B = K T = C f = 1.

Chapter 14, Page 38/39

Tooth bending Eq. (14-15):

Eq. (14-41):

Pd K m K B F J  3   (1.42)(1)   180(1)(1.18)(1.17)   1010 psi   4.2   0.25 

  W t KoKvK s

 S Y / ( KT K R )  SF   t N    65 000(0.95) / [(1)(0.955)]   64.0 1010

Ans.

Tooth wear 1/ 2

Eq. (14-16):

 K C   c  C p W t Ko K v K s m f  dPF I  

1/ 2

  1.42   1    2300 180(1)(1.18)(1.17)     5(4.2)   0.107     28 800 psi

Since gear B is a pinion, C H is not used in Eq. (14-42) (see p. 761), where  S Z / ( KT K R )  SH   c N  c  

 225 000(0.88) / [(1)(0.955)]     7.28 Ans 28 800   ________________________________________________________________________

Chapter 14, Page 39/39