Chapter 15 15-1
Given: Uncrowned, through-hardened 300 Brinell core and case, Grade 1, N C = 109 rev of pinion at R = 0.999, N P = 20 teeth, N G = 60 teeth, Q v = 6, P d = 6 teeth/in, shaft angle = 90°, n p = 900 rev/min, J P = 0.249 and J G = 0.216 (Fig. 15-7), F = 1.25 in, S F = S H = 1, K o = 1. Mesh
d P = 20/6 = 3.333 in, d G = 60/6 = 10.000 in
Eq. (15-7):
v t = (3.333)(900/12) = 785.3 ft/min
Eq. (15-6):
B = 0.25(12 – 6)2/3 = 0.8255 A = 50 + 56(1 – 0.8255) = 59.77
Eq. (15-5):
59.77 785.3 K v 59.77
Eq. (15-8):
v t,max = [59.77 + (6 – 3)]2 = 3940 ft/min
0.8255
1.374
Since 785.3 < 3904, K v = 1.374 is valid. The size factor for bending is: Eq. (15-10):
K s = 0.4867 + 0.2132 / 6 = 0.5222
For one gear straddle-mounted, the load-distribution factor is: Eq. (15-11):
K m = 1.10 + 0.0036 (1.25)2 = 1.106
Eq. (15-15):
(K L ) P = 1.6831(109)–0.0323 = 0.862 (K L ) G = 1.6831(109 / 3)–0.0323 = 0.893
Eq. (15-14):
(C L ) P = 3.4822(109)–0.0602 = 1 (C L ) G = 3.4822(109 / 3)–0.0602 = 1.069
Eq. (15-19):
K R = 0.50 – 0.25 log(1 – 0.999) = 1.25 (or Table 15-3) CR K R 1.25 1.118
Bending St sat 44(300) 2100 15 300 psi
Fig. 15-13:
0.99
Eq. (15-4):
( all ) P sw t
sat K L S F KT K R
15 300(0.862) 10 551 psi 1(1)(1.25)
Chapter 15, Page 1/20
Eq. (15-3):
Eq. (15-4):
( all ) P FK x J P Pd K o K v K s K m 10 551(1.25)(1)(0.249) 690 lbf 6(1)(1.374)(0.5222)(1.106) 690(785.3) H1 16.4 hp 33 000
WPt
15 300(0.893) 10 930 psi 1(1)(1.25) 10 930(1.25)(1)(0.216) WGt 620 lbf 6(1)(1.374)(0.5222)(1.106) 620(785.3) H2 14.8 hp Ans. 33 000 ( all )G
The gear controls the bending rating. ________________________________________________________________________ 15-2
Refer to Prob. 15-1 for the gearset specifications. Wear Fig. 15-12:
s ac = 341(300) + 23 620 = 125 920 psi
For the pinion, C H = 1. From Prob. 15-1, C R = 1.118. Thus, from Eq. (15-2): sac (CL ) P CH S H KT C R 125 920(1)(1) ( c,all ) P 112 630 psi 1(1)(1.118) ( c,all ) P
For the gear, from Eq. (15-16), B1 0.008 98(300 / 300) 0.008 29 0.000 69 CH 1 0.000 69(3 1) 1.001 38
From Prob. 15-1, (C L ) G = 1.0685. Equation (15-2) thus gives sac (CL )G CH S H KT C R 125 920(1.0685)(1.001 38) ( c,all )G 120 511 psi 1(1)(1.118) ( c,all )G
For steel:
C p 2290 psi
Chapter 15, Page 2/20
Eq. (15-9):
Cs 0.125(1.25) 0.4375 0.593 75
Fig. 15-6:
I = 0.083
Eq. (15-12):
C xc = 2
Eq. (15-1):
( ) Fd P I W c,all P C p K o K v K mCsC xc
2
t P
1.25(3.333)(0.083) 112 630 2290 1(1.374)(1.106)(0.5937)(2) 464 lbf 464(785.3) 11.0 hp 33 000 2 1.25(3.333)(0.083) 120 511 2290 1(1.374)(1.106)(0.593 75)(2) 531 lbf 531(785.3) 12.6 hp 33 000 2
H3 WGt H4
The pinion controls wear:
H = 11.0 hp
Ans.
The power rating of the mesh, considering the power ratings found in Prob. 15-1, is H = min(16.4, 14.8, 11.0, 12.6) = 11.0 hp Ans. ________________________________________________________________________ 15-3
AGMA 2003-B97 does not fully address cast iron gears. However, approximate comparisons can be useful. This problem is similar to Prob. 15-1, but not identical. We will organize the method. A follow-up could consist of completing Probs. 15-1 and 15-2 with identical pinions, and cast iron gears. Given: Uncrowned, straight teeth, P d = 6 teeth/in, N P = 30 teeth, N G = 60 teeth, ASTM 30 cast iron, material Grade 1, shaft angle 90°, F = 1.25, n P = 900 rev/min, n = 20, one gear straddle-mounted, K o = 1, J P = 0.268, J G = 0.228, S F = 2, S H 2. Mesh
d P = 30/6 = 5.000 in, d G = 60/6 = 10.000 in v t = (5)(900 / 12) = 1178 ft/min
Set N L = 107 cycles for the pinion. For R = 0.99, Table 15-7:
s at = 4500 psi
Chapter 15, Page 3/20
Table 15-5: Eq. (15-4):
s ac = 50 000 psi sat K L 4500(1) swt 2250 psi S F KT K R 2(1)(1)
The velocity factor K v represents stress augmentation due to mislocation of tooth profiles along the pitch surface and the resulting “falling” of teeth into engagement. Equation (5-67) shows that the induced bending moment in a cantilever (tooth) varies directly with E of the tooth material. If only the material varies (cast iron vs. steel) in the same geometry, I is the same. From the Lewis equation of Section 14-1, M K W tP v I /c FY We expect the ratio CI / steel to be CI ( K v )CI ECI steel ( K v )steel Esteel In the case of ASTM class 30, from Table A-24(a) (E CI ) av = (13 + 16.2)/2 = 14.7 kpsi Then,
( K v )CI
14.7 ( K v )steel 0.7( K v )steel 30
Our modeling is rough, but it convinces us that (K v ) CI < (K v ) steel , but we are not sure of the value of (K v ) CI . We will use K v for steel as a basis for a conservative rating. Eq. (15-6):
B = 0.25(12 – 6)2/3 = 0.8255 A = 50 + 56(1 – 0.8255) = 59.77
Eq. (15-5):
59.77 1178 K v 59.77
Pinion bending
0.8255
1.454
( all ) P = s wt = 2250 psi
From Prob. 15-1, K x = 1, K m = 1.106, K s = 0.5222 Eq. (15-3):
( all ) P FK x J P Pd K o K v K s K m 2250(1.25)(1)(0.268) 149.6 lbf 6(1)(1.454)(0.5222)(1.106)
WPt
Chapter 15, Page 4/20
H1
149.6(1178) 5.34 hp 33 000
Gear bending JG 0.228 149.6 127.3 lbf 0.268 JP 127.3(1178) H2 4.54 hp 33 000
WGt WPt
The gear controls in bending fatigue. H = 4.54 hp Ans. ________________________________________________________________________ 15-4
Continuing Prob. 15-3, Table 15-5:
s ac = 50 000 psi 50 000 sw t c,all 35 355 psi 2 2
Eq. (15-1):
Fd P I W c,all C p K o K v K mCsCxc
Fig. 15-6:
I = 0.86
t
From Probs. 15-1 and 15-2: C s = 0.593 75, K s = 0.5222, K m = 1.106, C xc = 2 From Table 14-8:
C p 1960 psi
1.25(5.000)(0.086) 35 355 91.6 lbf Wt 1960 1(1.454)(1.106)(0.59375)(2) 91.6(1178) H3 H 4 3.27 hp 33 000 2
Thus,
Rating Based on results of Probs. 15-3 and 15-4, H = min(5.34, 4.54, 3.27, 3.27) = 3.27 hp
Ans.
The mesh is weakest in wear fatigue. ________________________________________________________________________ 15-5
Uncrowned, through-hardened to 180 Brinell (core and case), Grade 1, 109 rev of pinion at R = 0.999, N P = z 1 = 22 teeth, N G = z 2 = 24 teeth, Q v = 5, m et = 4 mm, shaft angle 90°, n 1 = 1800 rev/min, S F = 1, S H S F 1, J P = Y J1 = 0.23, J G = Y J2 = 0.205, F = b = 25 mm, K o = K A = K T = K = 1 and C p 190 MPa . Chapter 15, Page 5/20
Mesh
d P = d e1 = mz 1 = 4(22) = 88 mm,
d G = m et z 2 = 4(24) = 96 mm
Eq. (15-7):
v et = 5.236(10–5)(88)(1800) = 8.29 m/s
Eq. (15-6):
B = 0.25(12 – 5)2/3 = 0.9148 A = 50 + 56(1 – 0.9148) = 54.77 0.9148
Eq. (15-5): Eq. (15-10):
54.77 200(8.29) Kv 1.663 54.77 K s = Y x = 0.4867 + 0.008 339(4) = 0.520
Eq. (15-11): with K mb = 1 (both straddle-mounted), K m = K H = 1 + 5.6(10–6)(252) = 1.0035 From Fig. 15-8, (CL ) P (Z NT ) P 3.4822(109 ) 0.0602 1.00 (CL )G (Z NT )G 3.4822[109 (22 / 24)]0.0602 1.0054 Eq. (15-12):
C xc = Z xc = 2
Eq. (15-19):
K R = Y Z = 0.50 – 0.25 log (1 – 0.999) = 1.25 CR Z Z YZ 1.25 1.118
(uncrowned)
From Fig. 15-10, C H = Z w = 1 Eq. (15-9):
Z x = 0.004 92(25) + 0.4375 = 0.560
Wear of Pinion Fig. 15-12:
H lim = 2.35H B + 162.89 = 2.35(180) + 162.89 = 585.9 MPa
Fig. 15-6:
I = Z I = 0.066
Eq. (15-2):
( H ) P
( H
) ( Z NT ) P ZW S H K Z Z 585.9(1)(1) 524.1 MPa 1(1)(1.118) lim P
2
bd e1Z I Eq. (15-1): W H C p 1000 K A K v K H Z x Z xc The constant 1000 expresses Wt in kN. t P
Chapter 15, Page 6/20
25(88)(0.066) 524.1 WPt 0.591 kN 190 1000(1)(1.663)(1.0035)(0.56)(2) t dnW (88)(1800)(0.591) 1 H3 4.90 kW 60 000 60 000 2
Eq. (13-36): Wear of Gear
H lim = 585.9 MPa ( H )G
585.9(1.0054) 526.9 MPa 1(1)(1.118)
( H )G 526.9 0.591 0.594 kN ( H ) P 524.1 (88)(1800)(0.594) H4 4.93 kW 60 000
WGt WPt
Thus in wear, the pinion controls the power rating; H = 4.90 kW
Ans.
We will rate the gear set after solving Prob. 15-6. ________________________________________________________________________ 15-6
Refer to Prob. 15-5 for terms not defined below. Bending of Pinion ( K L ) P (YNT ) P 1.6831(109 ) 0.0323 0.862 ( K L )G (YNT )G 1.6831[109 (22 / 24)]0.0323 0.864 Fig. 15-13:
F lim = 0.30H B + 14.48 = 0.30(180) + 14.48 = 68.5 MPa
Eq. (15-13):
Kx = Y = 1
From Prob. 15-5: Y Z = 1.25, v et = 8.29 m/s, K A 1, K v 1.663, K 1, Yx 0.52, K H 1.0035, YJ 1 0.23 Eq. (5-4):
( F ) P
Eq. (5-3):
WPt
F limYNT
68.5(0.862) 47.2 MPa 1(1)(1.25)
S F K YZ ( F ) P bmetY YJ 1
1000K A K vYx K H 47.2(25)(4)(1)(0.23) 1.25 kN 1000(1)(1.663)(0.52)(1.0035)
Chapter 15, Page 7/20
H1
881800 1.25 60 000
10.37 kW
Bending of Gear
F lim 68.5 MPa 68.5(0.864) 47.3 MPa 1(1)(1.25) 47.3(25)(4)(1)(0.205) WGt 1.12 kN 1000(1)(1.663)(0.52)(1.0035) 88 1800 1.12 H2 9.29 kW 60 000 Rating of mesh is H rating = min(10.37, 9.29, 4.90, 4.93) = 4.90 kW Ans. with pinion wear controlling. ________________________________________________________________________ ( F )G
15-7 (a)
(S F ) P all (S F )G all G P (sat K L / KT K R ) P (sat K L / KT K R )G t t (W Pd K o K v K s K m / FK x J ) P (W Pd K o K v K s K m / FK x J )G
All terms cancel except for s at , K L , and J, (s at ) P (K L ) P J P = (s at ) G (K L ) G J G From which (sat )G
(sat ) P ( K L ) P J P J ( sat ) P P mG ( K L )G J G JG
where = – 0.0178 or = – 0.0323 as appropriate. This equation is the same as Eq. (14-44). Ans. (b) In bending sat K L FK x J FK x J W t all S F Pd K o K v K s K m 11 S F KT K R Pd K o K v K s K m 11
(1)
In wear 1/ 2
sacCLCU W t K o K v K mCsCxc C p Fd P I S H KT CR 22 22
Chapter 15, Page 8/20
Squaring and solving for Wt gives s 2 C 2C 2 Fd P I W t 2 ac 2L 2H 2 S H KT CRCP 22 K o K v K mCsCxc 22
(2)
Equating the right-hand sides of Eqs. (1) and (2) and canceling terms, and recognizing that CR K R and P d d P = N P , we obtain (sac )22
S H2 (sat )11( K L )11 K x J11KT CsCxc SF CH2 N P K s I
Cp (CL )22
For equal Wt in bending and wear S H2 SF
SF
2
SF
1
So we get (sac )G
Cp (CL )G CH
(sat ) P ( K L ) P J P K x KT CsCxc N P IK s
Ans.
(c) (S H ) P (S H )G c,all c,all c P c G Substituting in the right-hand equality gives [ sacC L / (C R KT )]P C W K K K C C / ( Fd I ) o v m s xc P p P t
[ sacCLC H / (C R KT )]G C W t K K K C C / ( Fd I ) o v m s xc P p G
Denominators cancel, leaving (s ac ) P (C L ) P = (s ac ) G (C L ) G C H Solving for (s ac ) P gives, ( sac ) P (sac )G
(CL )G CH (CL ) P
(1)
From Eq. (15-14), C L P 3.4822 N L0.0602 and C L G 3.4822 N L / mG Thus, 0.0602 Ans. C H sac G mG0.0602C H sac P sac G 1 mG
0.0602
.
This equation is the transpose of Eq. (14-45).
Chapter 15, Page 9/20
________________________________________________________________________ Core Case Pinion (H B ) 11 (H B ) 12 Gear (H B ) 21 (H B ) 22
15-8
Given (H B ) 11 = 300 Brinell Eq. (15-23):
(s at ) P = 44(300) + 2100 = 15 300 psi J P 0.0323 0.249 0.0323 mG 15 300 17 023 psi 3 JG 0.216 17 023 2100 ( H B ) 21 339 Brinell Ans. 44 2290 15 300(0.862)(0.249)(1)(0.593 25)(2) ( sac )G 1.0685(1) 20(0.086)(0.5222) 141 160 psi 141 160 23 600 ( H B ) 22 345 Brinell Ans. 341 (sac ) P (sac )G mG0.0602CH 141 160(30.0602 ) 1 150 811 psi (sat )G (sat ) P
( H B )12
150 811 23 600 373 Brinell 341
Core Case Pinion 300 373 Gear 339 345
Ans.
Ans.
________________________________________________________________________ 15-9 Pinion core (sat ) P 44(300) 2100 15 300 psi 15 300(0.862) ( all ) P 10 551 psi 1(1)(1.25) 10 551(1.25)(0.249) Wt 689.7 lbf 6(1)(1.374)(0.5222)(1.106) Gear core (sat )G 44(352) 2100 17 588 psi 17 588(0.893) ( all )G 12 565 psi 1(1)(1.25) 12 565(1.25)(0.216) Wt 712.5 lbf 6(1)(1.374)(0.5222)(1.106)
Chapter 15, Page 10/20
Pinion case (sac ) P 341(372) 23 620 150 472 psi 150 472(1) ( c,all ) P 134 590 psi 1(1)(1.118) 1.25(3.333)(0.086) 134 590 Wt 685.8 lbf 2290 1(1.374)(1.106)(0.593 75)(2) 2
Gear case (sac )G 341(344) 23 620 140 924 psi 140 924(1.0685)(1) ( c,all )G 134 685 psi 1(1)(1.118) 2
1.25(3.333)(0.086) 134 685 W 686.8 lbf 2290 1(1.374)(1.106)(0.593 75)(2) t
The rating load would be t Wrated min(689.7, 712.5, 685.8, 686.8) 685.8 lbf
which is slightly less than intended. Pinion core (sat ) P 15 300 psi ( all ) P 10 551 psi
(as before) (as before)
W t 689.7 lbf
(as before)
Gear core (sat )G 44(339) 2100 17 016 psi 17 016(0.893) 12 156 psi ( all )G 1(1)(1.25) 12 156(1.25)(0.216) Wt 689.3 lbf 6(1)(1.374)(0.5222)(1.106) Pinion case (sac ) P 341(373) 23 620 150 813 psi 150 813(1) ( c,all ) P 134 895 psi 1(1)(1.118) 1.25(3.333)(0.086) 134 895 Wt 689.0 lbf 2290 1(1.374)(1.106)(0.593 75)(2) 2
Gear case (sac )G 341(345) 23 620 141 265 psi 141 265(1.0685)(1) ( c,all )G 135 010 psi 1(1)(1.118)
Chapter 15, Page 11/20
1.25(3.333)(0.086) 135 010 Wt 690.1 lbf 2290 1(1.1374)(1.106)(0.593 75)(2) 2
The equations developed within Prob. 15-7 are effective. ________________________________________________________________________ 15-10 The catalog rating is 5.2 hp at 1200 rev/min for a straight bevel gearset. Also given: N P = 20 teeth, N G = 40 teeth, n = 20, F = 0.71 in, J P = 0.241, J G = 0.201, P d = 10 teeth/in, through-hardened to 300 Brinell-General Industrial Service, and Q v = 5 uncrowned. Mesh d P 20 / 10 2.000 in, dG 40 / 10 4.000 in d P nP (2)(1200) vt 628.3 ft/min 12 12 K o 1, S F 1, S H 1 Eq. (15-6):
B = 0.25(12 – 5)2/3 = 0.9148 A = 50 + 56(1 – 0.9148) = 54.77 0.9148
Eq. (15-10):
54.77 628.3 K v 1.412 54.77 K s = 0.4867 + 0.2132/10 = 0.508
Eq. (15-11):
K m = 1.25 + 0.0036(0.71)2 = 1.252, where K mb = 1.25
Eq. (15-15):
(K L ) P = 1.6831(109)–0.0323 = 0.862 (K L ) G = 1.6831(109/2)–0.0323 = 0.881
Eq. (15-14):
(C L ) P = 3.4822(109)–0.0602 = 1.000 (C L ) G = 3.4822(109/2)–0.0602 = 1.043
Eq. (15-5):
Analyze for 109 pinion cycles at 0.999 reliability. Eq. (15-19):
Bending Pinion: Eq. (15-23):
K R = 0.50 – 0.25 log(1 – 0.999) = 1.25 CR K R 1.25 1.118
(s at ) P = 44(300) + 2100 = 15 300 psi
Chapter 15, Page 12/20
15 300(0.862) 10 551 psi 1(1)(1.25) (sw t ) P FK x J P
Eq. (15-4):
(swt ) P
Eq. (15-3):
Wt
Gear:
(s at ) G = 15 300 psi 15 300(0.881) ( sw t ) G 10 783 psi 1(1)(1.25) 10 783(0.71)(1)(0.201) Wt 171.4 lbf 10(1)(1.412)(0.508)(1.252) 171.4(628.3) H2 3.3 hp 33 000
Eq. (15-4): Eq. (15-3):
Pd K o K v K s K m 10 551(0.71)(1)(0.241) 201 lbf 10(1)(1.412)(0.508)(1.252) 201(628.3) H1 3.8 hp 33 000
Wear Pinion: (C H )G 1, I 0.078, C p 2290 psi, Cs 0.125(0.71) 0.4375 0.526 25
Eq. (15-22):
C xc 2
(s ac ) P = 341(300) + 23 620 = 125 920 psi 125 920(1)(1) ( c,all ) P 112 630 psi 1(1)(1.118) 2
Eq. (15-1):
( ) Fd P I W c,all P C p K o K v K mCsC xc t
0.71(2.000)(0.078) 112 630 2290 1(1.412)(1.252)(0.526 25)(2) 144.0 lbf 2
H3
144(628.3) 2.7 hp 33 000
Gear: (sac )G 125 920 psi 125 920(1.043)(1) ( c,all ) 117 473 psi 1(1)(1.118) 117 473 Wt 2290
2
0.71(2.000)(0.078) 1(1.412)(1.252)(0.526 25)(2) 156.6 lbf
Chapter 15, Page 13/20
H4
156.6(628.3) 3.0 hp 33 000
Rating: H = min(3.8, 3.3, 2.7, 3.0) = 2.7 hp Pinion wear controls the power rating. While the basis of the catalog rating is unknown, it is overly optimistic (by a factor of 1.9). ________________________________________________________________________ 15-11 From Ex. 15-1, the core hardness of both the pinion and gear is 180 Brinell. So (H B ) 11 and (H B ) 21 are 180 Brinell and the bending stress numbers are: (sat ) P 44(180) 2100 10 020 psi (sat )G 10 020 psi The contact strength of the gear case, based upon the equation derived in Prob. 15-7, is Cp S H2 (sat ) P ( K L ) P K x J P KT CsCxc (sac )G (CL )G CH S F N P IK s Substituting (s at ) P from above and the values of the remaining terms from Ex. 15-1, 2290 1.52 10 020(1)(1)(0.216)(1)(0.575)(2) 1.32(1) 1.5 25(0.065)(0.529) 114 331 psi 114 331 23 620 266 Brinell (H B ) 22 341
(sac )G
The pinion contact strength is found using the relation from Prob. 15-7: (sac ) P (sac )G mG0.0602CH 114 331(1)0.0602 (1) 114 331 psi 114 331 23 600 ( H B )12 266 Brinell 341 Core Case Pinion 180 266 Gear 180 266 Realization of hardnesses The response of students to this part of the question would be a function of the extent to which heat-treatment procedures were covered in their materials and manufacturing prerequisites, and how quantitative it was. The most important
Chapter 15, Page 14/20
thing is to have the student think about it. The instructor can comment in class when students’ curiosity is heightened. Options that will surface may include: (a) Select a through-hardening steel which will meet or exceed core hardness in the hot-rolled condition, then heat-treating to gain the additional 86 points of Brinell hardness by bath-quenching, then tempering, then generating the teeth in the blank. (b) Flame or induction hardening are possibilities. (c) The hardness goal for the case is sufficiently modest that carburizing and case hardening may be too costly. In this case the material selection will be different. (d)The initial step in a nitriding process brings the core hardness to 33–38 Rockwell C-scale (about 300–350 Brinell), which is too much. ________________________________________________________________________ 15-12 Computer programs will vary. ________________________________________________________________________ 15-13 A design program would ask the user to make the a priori decisions, as indicated in Sec. 15-5, p. 806, of the text. The decision set can be organized as follows: A priori decisions: • Function: H, K o , rpm, m G , temp., N L , R • Design factor: n d (S F = n d , S H nd ) • Tooth system: Involute, Straight Teeth, Crowning, n • Straddling: K mb • Tooth count: N P (N G = m G N P ) Design decisions: • Pitch and Face: P d , F • Quality number: Q v • Pinion hardness: (H B ) 1 , (H B ) 3 • Gear hardness: (H B ) 2 , (H B ) 4 First, gather all of the equations one needs, then arrange them before coding. Find the required hardnesses, express the consequences of the chosen hardnesses, and allow for revisions as appropriate.
Chapter 15, Page 15/20
Pinion Bending
Gear Bending
Pinion Wear
Gear Wear 1/ 2
Load-induced stress (Allowable stress) Tabulated strength Associated hardness
W t PK o K v K m K s st s11 FK x J P ( sat ) P
Factor of safety
( sat )G
s21S F KT K R ( K L )G
W t K o K vCsC xc Fd P I
c Cp
(sac ) P
s12
s 22 = s 12
s12 S H KT CR (CL ) P (CH ) P
( sac )G
s22 S H KT CR (CL )G (CH )G
(sat ) P 2100 44 Bhn (sat ) P 5980 48
(sat )G 2100 44 Bhn (sat )G 5980 48
(sac ) P 23 620 341 Bhn (sac ) P 29 560 363.6
(sac ) P 23 620 341 Bhn (sac ) P 29 560 363.6
(H B ) 11
(H B ) 21
(H B ) 12
(H B ) 22
44( H B )11 2100 (sat1) P 48( H B )11 5980
44( H B ) 21 2100 (sat1)G 48( H B ) 21 5980
341( H B )12 23 620 (sac1) P 363.6( H B )12 29 560
341( H B ) 22 23 620 (sac1)G 363.6( H B )22 29 560
(s ) ( K ) n11 all at1 P L P s11KT K R
(s ) ( K ) n21 at1 G L G s21KT K R
(s ) (C ) (C ) n12 ac1 P L P H P s12 KT CR
Chosen hardness New tabulated strength
s11S F KT K R (K L )P
W t PK o K v K m K s st s21 FK x J G
Note: S F nd , S H
2
n22
(s ) (C ) (C ) ac1 G L G H G s22 KT CR
2
SF
Chapter 15, Page 16/20
15-14 N W = 1, N G = 56, P t = 8 teeth/in, d = 1.5 in, H o = 1hp, n = 20, t a = 70F, K a = 1.25, n d = 1, F e = 2 in, A = 850 in2 (a)
m G = N G /N W = 56, d G = N G /P t = 56/8 = 7.0 in p x = / 8 = 0.3927 in, C = 1.5 + 7 = 8.5 in
Eq. (15-39):
a = p x / = 0.3927 / = 0.125 in
Eq. (15-40):
b = 0.3683 p x = 0.1446 in
Eq. (15-41):
h t = 0.6866 p x = 0.2696 in
Eq. (15-42):
d o = 1.5 + 2(0.125) = 1.75 in
Eq. (15-43):
d r = 3 – 2(0.1446) = 2.711 in
Eq. (15-44):
D t = 7 + 2(0.125) = 7.25 in
Eq. (15-45):
D r = 7 – 2(0.1446) = 6.711 in
Eq. (15-46):
c = 0.1446 – 0.125 = 0.0196 in
Eq. (15-47):
( FW ) max 2 2 7 0.125 2.646 in
Eq. (13-27):
VW (1.5)(1725 /12) 677.4 ft/min (7)(1725 / 56) VG 56.45 ft/min 12 L px NW 0.3927 in
Eq. (13-28):
0.3927
o tan 1 4.764 (1.5)
Pn pn
Eq. (15-62): (b) Eq. (15-38):
Pt 8 8.028 cos cos 4.764
0.3913 in Pn (1.5)(1725) Vs 679.8 ft/min 12 cos 4.764 f 0.103exp 0.110(679.8) 0.450 0.012 0.0250
Eq. (15-54): cos n f tan cos 20 0.0250 tan 4.764 e 0.7563 cos n f cot cos 20 0.0250 cot 4.764
Ans.
Chapter 15, Page 17/20
Eq. (15-58): Eq. (15-57):
33 000nd H o K a 33 000(1)(1)(1.25) 966 lbf VG e 56.45(0.7563) cos n sin f cos WWt WGt cos n cos f sin
WGt
Ans.
cos 20 sin 4.764 0.025cos 4.764 966 cos 20 cos 4.764 0.025sin 4.764 106.4 lbf Ans. (c) Eq. (15-33):
C s = 1190 – 477 log 7.0 = 787
Eq. (15-36):
Cm 0.0107 562 56(56) 5145 0.767
Eq. (15-37):
Cv 0.659 exp[0.0011(679.8)] 0.312
Eq. (15-38):
(Wt) all = 787(7)0.8(2)(0.767)(0.312) = 1787 lbf
Since WGt (W t )all , the mesh will survive at least 25 000 h. Eq. (15-61): Eq. (15-63):
0.025(966) 29.5 lbf 0.025sin 4.764 cos 20 cos 4.764 29.5(679.8) Hf 0.608 hp 33 000 106.4(677.4) HW 2.18 hp 33 000 966(56.45) HG 1.65 hp 33 000
Wf
The mesh is sufficient
Ans.
Pn Pt / cos 8 / cos 4.764o 8.028
pn / 8.028 0.3913 in 966 G 39 500 psi 0.3913(0.5)(0.125) The stress is high. At the rated horsepower,
G
1 39 500 23 940 psi 1.65
acceptable
(d)
Chapter 15, Page 18/20
Eq. (15-52):
A min = 43.2(8.5)1.7 = 1642 in2 < 1700 in2
Eq. (15-49):
H loss = 33 000(1 – 0.7563)(2.18) = 17 530 ft · lbf/min
Assuming a fan exists on the worm shaft, 1725 0.13 0.568 ft · lbf/(min · in 2 · oF) 3939 17 530 Eq. (15-51): ts 70 88.2o F Ans. 0.568(1700) ________________________________________________________________________
Eq. (15-50):
CR
Chapter 15, Page 19/20
15-15 Problem statement values of 25 hp, 1125 rev/min, m G = 10, K a = 1.25, n d = 1.1, n = 20°, t a = 70°F are not referenced in the table. The first four parameters listed in the table were selected as design decisions.
px dW FG A HW HG Hf NW NG KW Cs Cm Cv VG WGt t W
W f e (P t ) G Pn C-to-C ts L
G dG
15-15 1.75 3.60 2.40 2000
15-16 1.75 3.60 1.68 2000
15-17 1.75 3.60 1.43 2000
15-18 1.75 3.60 1.69 2000
15-19 1.75 3.60 2.40 2000
15-20 1.75 4.10 2.25 2000 38.0 36.1 1.85 3 30 50
15-21 1.75 3.60 2.4 2500 FAN 41.2 37.7 3.59 3 30 115
15-22 1.75 3.60 2.4 2600 FAN 41.2 37.7 3.59 3 30 185
38.2 36.2 1.87 3 30
38.2 36.2 1.47 3 30
38.2 36.2 1.97 3 30
38.2 36.2 1.97 3 30 125
38.2 36.2 1.97 3 30 80
607 0.759 0.236 492
854 0.759 0.236 492
1000 0.759 0.236 492
492
492
563
492
492
2430
2430
2430
2430
2430
2120
2524
2524
1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 5103 16.71
1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 7290 16.71
1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 8565 16.71
1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 7247 16.71
1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 5103 16.71
1038 0.0183 0.951 1.571 1.732 11.6 171 6.0 24.98 4158 19.099
1284 0.034 0.913 1.795 1.979 10.156 179.6 5.25 24.9 5301 16.7
1284 0.034 0.913 1.795 1.979 10.156 179.6 5.25 24.9 5301 16.71
Chapter 15, Page 20/20