Chapter 15 15-1
Given: Uncrowned, through-hardened 300 Brinell core and case, Grade 1, N C C = 9 10 rev of pinion pinion at R = 0.999, N P = 20 teeth, N G = 60 teeth, Q = 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 F = S H = 1, K o = 1. v
Mesh
Eq. (15-7): Eq. (15-6):
Eq. (15-5):
Eq. (15-8):
d P = 20 / 6 = 3.333 in, d G = 60 / 6 = 10.000 in v t
= (3 (3.333)(900 / 12) 12) = 785.3 ft/min 2 / 3
B = 0.25(12 – 6) = 0.8255 A = 50 + 56(1 – 0.8255) = 59.77
59.77 785.3 K 59.77
0.8255
1.374
v
v t ,max ,max
2
= [59.77 + (6 – 3)] = 3940 ft/min
Since 785.3 < 3904, K = 1.374 is valid. The size factor for bending is: v
Eq. (15-10):
K s = 0.4867 + 0.2132 / 6 = 0.5222
For one gear straddle-mounted, the load-distribution load -distribution factor is: 2
Eq. (15-11):
K m = 1.10 + 0.0036 (1.25) = 1.106
Eq. (15-15):
(K L ) P = 1.6831(10 ) . = 0.862 9 –0.0323 (K L ) G = 1.6831(10 / 3) = 0.893
Eq. (15-14):
(C L ) P = 3.4822(10 ) . = 1 9 –0.0602 (C L ) G = 3.4822(10 / 3) = 1.069
Eq. (15-19):
K R = 0.50 – 0.25 log(1 – 0.999) = 1.25 (or Table 15-3)
9 –0 0323
9 –0 0602
C R
K R
1.25 1.118
Bending St sat 44(300) 2100 15 300 psi
Fig. 15-13:
0.99 0.99
Eq. (15-4):
( all ) P s
w
t
sat K L S F KT K R
15 300(0.862) 1(1)(1.25)
10 551 psi
Chapter 15, Page 1/20
Eq. (15-3):
W Pt
( all ) P FK x J P Pd Ko K K s K m v
10 551(1.25)(1)(0.249)
6(1)(1.374)(0.5222)(1.106) 690(785.3) H 1 16.4 hp 33 000
Eq. (15-4):
( all )G
690 lbf
15 300(0.893)
10 930 psi 1(1)(1.25) 10 930(1.25)(1)(0.216) t 620 lbf W G 6(1)(1.374)(0.5222)(1.106) 620(785.3) H 2 14.8 hp Ans. 33 000
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): ( c,all ) P ( c,all ) P
sac (CL ) P C H
S H KT C R 125 920(1)(1)
1(1)(1.118)
112 630 psi
For the gear, from Eq. (15-16), B1 0.008 98(300 / 300) 0.008 29 0.000 69 C H 1 0.000 69(3 1) 1.001 38
From Prob. 15-1, (C L ) G = 1.0685. Equation (15-2) thus gives ( c,all )G ( c,all )G
For For stee steel: l:
sac (CL )G C H
S H KT C R 125 920(1.068 920(1.0685)( 5)(1.001 1.001 38)
1(1)(1.118)
120 511 psi
C p 2290 2290 psi psi
Chapter 15, Page 2/20
C s 0.125(1.25) 0.4375 0.593 75
Eq. (15-9): Fig. 15-6:
I = 0.083
Eq. (15-12):
C xc = 2 2
( ) Fd P I W Pt c,all P C p K o K K mCs C xc 2 1.25(3.333)(0.083) 112 630 2290 1(1.374)(1.106)(0.5937)(2) 464 lbf
Eq. (15-1):
v
H 3
464(785.3)
11.0 hp
33 000 2 120 511
W Gt
2290 1(1.374)(1.106)(0.593 75)(2) 1.25(3.333)(0.083)
531 lbf H 4
531(785.3) 33 000
The pinion controls wear:
12.6 hp
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. 1 5-1, 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. d P = 30 / 6 = 5.000 in, d G = 60 / 6 = 10.000 in
Mesh
= (5)(900 / 12) = 1178 ft/min
v t
7
Set N L = 10 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) s t 2250 psi S F KT K R 2(1)(1) w
The velocity factor K 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 v
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 t P I / c FY v
We expect the ratio CI / steel to be CI
( K )CI
steel
v
( K )steel
E CI
E steel
v
In the case of ASTM class 30, from Table A-24(a) ( E CI ) av = (13 + 16.2) / 2 = 14.7 kpsi Then,
( K )CI v
14.7 30
( K )steel 0.7( K ) steel v
v
Our modeling is rough, but it convinces us that (K ) CI < (K ) steel , but we are not sure of the value of (K ) CI . We will use K for steel as a basis for a conservative rating. v
v
Eq. (15-6):
v
v
2 / 3
B = 0.25(12 – 6) = 0.8255 A = 50 + 56(1 – 0.8255) = 59.77
59.77 1178 K 59.77
Eq. (15-5):
0.8255
v
Pinion bending
( all ) P = s
t =
w
1.454
2250 psi
From Prob. 15-1, K x = 1, K m = 1.106, K s = 0.5222 Eq. (15-3):
W Pt
( all ) P FK x J P Pd Ko K K s K m v
2250(1.25)(1)(0.268) 6(1)(1.454)(0.5222)(1.106)
149.6 lbf
Chapter 15, Page 4/20
H 1
149.6(1178) 33 000
5.34 hp
Gear bending J 0.228 WGt W Pt G 149.6 127.3 lbf 0.268 J P 127.3(1178) H 2 4.54 hp 33 000
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 35 355 psi s t c,all 2 w
2
Eq. (15-1):
Fd P I W t c,all C p K o K K mCsC xc v
Fig. 15-6:
I = 0.86
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 2
Thus,
1.25(5.000)(0.086) 35 355 91.6 lbf W 1960 1(1.454)(1.106)(0.59375)(2) t
H 3 H 4
91.6(1178) 33 000
3.27 hp
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
9
Uncrowned, through-hardened to 180 Brinell (core and case), Grade 1, 10 rev of pinion at R = 0.999, N P = z 1 = 22 teeth, N G = z 2 = 24 teeth, Q = 5, m et = 4 mm, v
shaft angle 90°, n 1 = 1800 rev/min, S F = 1, S H
S F
1, J P = Y J 1 = 0.23,
J G = Y J 2 = 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
Eq. (15-7): Eq. (15-6):
d P = d e1 = mz 1 = 4(22) = 88 mm, v et
d G = m et z 2 = 4(24) = 96 mm
–5
= 5.236(10 )(88)(1800) = 8.29 m/s 2 / 3
B = 0.25(12 – 5) = 0.9148 A = 50 + 56(1 – 0.9148) = 54.77 0.9148
Eq. (15-5):
54.77 200(8.29) K 54.77
Eq. (15-10):
K s = Y x = 0.4867 + 0.008 339(4) = 0.520
v
1.663
Eq. (15-11): with K mb = 1 (both straddle-mounted), –6 2 K m = K H = 1 + 5.6(10 )(25 ) = 1.0035 From Fig. 15-8, (C L ) P ( Z NT ) P 3.4822(109 ) 0.0602 1.00 (C L )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 C R Z Z
(uncrowned)
Y Z
1.25 1.118
From Fig. 15-10, C H = Z = 1 w
Eq. (15-9):
Z x = 0.004 92(25) + 0.4375 = 0.560
Wear of Pinion
Fig. 15-12:
H lim = 2.35 H 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 Z W
lim P
S H K Z Z
585.9(1)(1) 1(1)(1.118)
Eq. (15-1):
W H C p t P
524.1 MPa
2
bd e1Z I 1000K A K K H Z x Z xc v
t
The constant 1 000 expresses W in kN.
Chapter 15, Page 6/20
524.1 W 190 t P
Eq. (13-36):
H 3
dn1W t 60 000
2
25(88)(0.066) 1000(1)(1.663)(1.0035)(0.56)(2) 0.591 kN
(88)(1800)(0.591) 60 000
4.90 kW
Wear of Gear
H lim = 585.9 MPa ( H )G
585.9(1.0054)
W Gt W Pt
( H )G
H 4
1(1)(1.118)
526.9 MPa
526.9 0.591 0.594 kN ( H ) P 524.1
(88)(1800)(0.594) 60 000
4.93 kW
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 (Y NT ) P 1.6831(109 ) 0.0323 0.862 ( K L )G (Y NT )G 1.6831[109 (22 / 24)]0.0323 0.864 Fig. 15-13:
F lim = 0.30 H B + 14.48 = 0.30(180) + 14.48 = 68.5 MPa
Eq. (15-13):
K x = Y = 1
Y Z = 1.25, v et = 8.29 m/s, From Prob. 15-5: K A 1, K 1.663, K 1, Y x 0.52, K H 1.0035, YJ 1 0.23 v
Eq. (5-4):
( F ) P
Eq. (5-3):
W Pt
F limY NT S F K Y Z
68.5(0.862) 1(1)(1.25)
( F ) P bmetY YJ 1
47.2 MPa
1000K A K Yx K H v
47.2(25)(4)(1)(0.23) 1000(1)(1.663)(0.52)(1.0035)
1.25 kN
Chapter 15, Page 7/20
H 1
8818001.25 60 000
10.37 kW
Bending of Gear
F lim 68.5 MPa ( F )G W Gt H 2
68.5(0.864)
47.3 MPa 1(1)(1.25) 47.3(25)(4)(1)(0.205)
1000(1)(1.663)(0.52)(1.0035) 8818001.12
1.12 kN
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. ________________________________________________________________________ 15-7
(a)
all all ( S ) F G P G
(SF ) P
(sat K L / KT K R ) P t
(W Pd K o K K s K m / FK x J ) P
v
( sat K L / KT K R )G t
(W Pd K o K K s K m / FK x J )G v
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 ( K L )G J G
(sat ) P
J P J G
m G
where = – 0.0178 or = – 0.0323 as appropriate. This equation is the same as Eq. (14-44). Ans. (b) In bending
all
sat K L FK x J S P K K K K S K K P K K K K 11 s m 11 s m F d o F T R d o
W t
FK x J v
(1)
v
In wear 1/ 2
sacCLCU W t K o K K mCsCxc C p Fd P I 22 S H KT CR 22 v
Chapter 15, Page 8/20
Squaring and solving for W t gives
sac2 CL2CH2 Fd P I W 2 2 2 2 S K C C K K K C C m s xc 22 H T R P 22 o t
(2)
v
Equating the right-hand sides of Eqs. (1) and (2) and canceling terms, and recognizing that C R
K R and P d d P = N P , we obtain
(sac ) 22
S H2 (sat )11( K L )11 K x J11KT Cs Cxc
C p
(C L ) 22
CH2 N P Ks I
SF
t
For equal W in bending and wear 2 H
S
SF
S F
2
S F
1
So we get (sac )G
C p
(sat ) P ( K L ) P J P K x KT Cs Cxc
(C L )G CH
N P IK s
Ans.
(c)
c,all c,all (S H ) P ( S H ) G c P c G Substituting in the right-hand equality gives [ sacC L / (C R KT )]P
C p W t K o K K mCsCxc / ( Fd P I ) P
v
[ sacC LC H / (CR KT )]G
C p W t K o K Km Cs Cxc / ( FdP I ) G v
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
(C L )G (C L ) P
C H
(1)
From Eq. (15-14), C L P 3.4822 N L0.0602 and CL G 3.4822 N L / mG
0.0602
.
Thus,
sac P sac G 1 mG
0.0602
C H sac G mG
0.0602
C H
Ans.
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 (sat )G ( sat ) P ( H B ) 21 (sac )G
J P J G
0.249 0.0323 17 023 psi 3 0.216
mG0.0323 15 300
17 023 2100 44 2290
339 Brinell
15 300(0.862)(0.249)(1)(0.593 25)(2)
1.0685(1) 141 160 psi 141 160 23 600
( H B ) 22
341 0.0602 (sac ) P (sac ) G mG CH ( H B )12
Ans.
Core Case Pinion 300 373 Gear 339 345
345 Brinell
Ans.
141 160(30.0602) 1 150 811 psi
150 811 23 600 341
20(0.086)(0.5222)
373 Brinell
Ans.
Ans.
______________________ _ ___________ ______ __ ______________________________ 15-9
Pinion core
(sat ) P 44(300) 2100 15 300 psi ( all ) P
15 300(0.862)
10 551 psi 1(1)(1.25) 10 551(1.25)(0.249) W t 689.7 lbf 6(1)(1.374)(0.5222)(1.106) Gear core
(sat )G 44(352) 2100 17 588 psi ( all )G
17 588(0.893)
12 565 psi 1(1)(1.25) 12 565(1.25)(0.216) t W 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 ( c,all ) P
150 472(1) 1(1)(1.118)
134 590 W 2290
2
t
134 590 psi
1.25(3.333)(0.086) 1(1.374)(1.106)(0.593 75)(2) 685.8 lbf
Gear case
(sac )G 341(344) 23 620 140 924 psi ( c,all )G
140 924(1.0685)(1) 1(1)(1.118)
134 685 psi
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 W rated 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
(as before)
( all ) P 10 551 psi
(as before)
W t 689.7 lbf
(as before)
Gear core
(sat )G 44(339) 2100 17 016 psi ( all )G
17 016(0.893)
12 156 psi 1(1)(1.25) 12 156(1.25)(0.216) W t 689.3 lbf 6(1)(1.374)(0.5222)(1.106) Pinion case
(sac ) P 341(373) 23 620 150 813 psi ( c,all ) P
150 813(1) 1(1)(1.118)
134 895 W 2290 t
2
134 895 psi
1.25(3.333)(0.086) 1(1.374)(1.106)(0.593 75)(2) 689.0 lbf
Gear case
(sac )G 341(345) 23 620 141 265 psi ( c,all )G
141 265(1.0685)(1) 1(1)(1.118)
135 010 psi
Chapter 15, Page 11/20
135 010 W 2290
2
t
1.25(3.333)(0.086) 1(1.1374)(1.106)(0.593 75)(2) 690.1 lbf
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 = 5 uncrowned. v
Mesh d P 20 / 10 2.000 in, v
t
d P nP
dG 40 / 10 4.000 in
(2)(1200)
628.3 ft/min
12 12 K o 1, S F 1, S H 1
Eq. (15-6):
2 / 3
B = 0.25(12 – 5) = 0.9148 A = 50 + 56(1 – 0.9148) = 54.77 0.9148
Eq. (15-5):
54.77 628.3 K 54.77
Eq. (15-10):
K s = 0.4867 + 0.2132 / 10 = 0.508
Eq. (15-11):
K m = 1.25 + 0.0036(0.71) = 1.252, where K mb = 1.25
Eq. (15-15):
(K L ) P = 1.6831(10 ) = 0.862 9 –0.0323 (K L ) G = 1.6831(10 / 2) = 0.881
Eq. (15-14):
(C L ) P = 3.4822(10 ) = 1.000 9 –0.0602 (C L ) G = 3.4822(10 / 2) = 1.043
1.412
v
2
9 –0.0323
9 –0.0602
9
Analyze for 10 pinion cycles at 0.999 reli ability. Eq. (15-19):
K R = 0.50 – 0.25 log(1 – 0.999) = 1.25 C R
Bending Pinion: Eq. (15-23):
K R
1.25 1.118
(s at ) P = 44(300) + 2100 = 15 300 psi
Chapter 15, Page 12/20
Eq. (15-4):
(s t ) P
Eq. (15-3):
W t
15 300(0.862)
w
1(1)(1.25)
10 551 psi
( s t ) P FK x J P w
Pd Ko K K s K m v
10 551(0.71)(1)(0.241)
10(1)(1.412)(0.508)(1.252) 201(628.3) H 1 3.8 hp 33 000
Gear: Eq. (15-4): Eq. (15-3):
201 lbf
(s at ) G = 15 300 psi 15 300(0.881) (s t ) G 10 783 psi 1(1)(1.25) 10 783(0.71)(1)(0.201) W t 171.4 lbf 10(1)(1.412)(0.508)(1.252) 171.4(628.3) H 2 3.3 hp 33 000 w
Wear Pinion: (C H )G 1,
I 0.078,
C p 2290 psi,
C xc 2
C s 0.125(0.71) 0.4375 0.526 25
Eq. (15-22):
(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):
( ,all ) Fd P I W c P C p K o K K mCs Cxc t
v
112 630 2290 144.0 lbf H 3
2
144(628.3)
0.71(2.000)(0.078) 1(1.412)(1.252)(0.526 25)(2)
2.7 hp
33 000
Gear: (sac )G 125 920 psi ( c,all )
125 920(1.043)(1) 1(1)(1.118)
117 473 W 2290 t
2
117 473 psi
0.71(2.000)(0.078) 1(1.412)(1.252)(0.526 25)(2) 156.6 lbf
Chapter 15, Page 13/20
H 4
156.6(628.3) 33 000
3.0 hp
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 (sac )G
S H2 ( sat ) P ( K L ) P K x J P KT Cs Cxc
C p
(C L )G CH
SF
N P IK s
Substituting (s at ) P from above and the values of the remaining terms from Ex. 15-1, (sac ) G
2290
1.52 10 020(1)(1)(0.216)(1)(0.575)(2)
1.5 114 331 psi 114 331 23 620 1.32(1)
( H B ) 22
341
25(0.065)(0.529)
266 Brinell
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 t his part of th e quest ion would be a function of the extent to which heat-treatment pro cedures were covered in their materials and manufacturing prerequisites, and how qu antitati ve it was. The most important
Chapter 15, Page 14/20
thing is to have the student think a bout 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 c ore 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 • Pinion hardness: ( H B ) 1 , ( H B ) 3 • Gear hardness: ( H B ) 2 , ( H B ) 4 v
First, gather all of the eq uations 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)
st
Tabulated strength
W t PK oK K m K s v
FK x J P
( sat ) P
Associated hardness
Factor of safety
s11S F K T K R
( K L ) P
FK x J G
( sat )G
s21
s21S F K T K R
( K L )G
W t K oK C s Cxc Fd P I
c C p
v
(sac ) P
s12
s12S HK T CR
s 22 = s 12
( sac )G
(C L ) P (C H ) P
s22S HK T CR
(C L )G (C H )G
(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
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
44( H B )11 2100 48( H B )11 5980
(sat1) P n11
Note: S F nd ,
st
v
(sat ) P 2100 44 Bhn (sat ) P 5980 48
Chosen hardness New tabulated strength
s11
W t PK oK K m K s
all
SH
SF
( sat1) P ( K L ) P s11KT K R
44( H B ) 21 2100 48( H B )21 5980
(sat1)G n21
(sat1)G ( K L )G s21KT K R
(sac1)P (CL )P (C H )P s12 KT C R
n12
2
(sac1)G (CL )G (C H )G s22 KT C R
2
n22
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, 2 K a = 1.25, n d = 1, F e = 2 in, A = 850 in (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):
( F W ) max 2 2 7 0.125 2.646 in
Eq. (13-27):
V W (1.5)(1725/12) 677.4 ft/min (7)(1725 / 56) V G 56.45 ft/min 12 L p x N W 0.3927 in
Eq. (13-28):
tan 1
0.3927 o 4.764 (1.5)
Pn pn
Pt
cos Pn
8 cos 4.764
8.028
0.3913 in
(1.5)(1725)
Eq. (15-62):
V s
(b) Eq. (15-38):
f 0.103 exp 0.110(679.8)0.450 0.012 0.0250
12cos4.764
679.8 ft/min
Eq. (15-54): cos n f tan cos 20 0.0250 tan 4.764 0.7563 e cos n f cot cos 20 0.0250 cot 4.764
Ans.
Chapter 15, Page 17/20
33 000nd H o K a
Eq. (15-58):
WGt
Eq. (15-57):
WWt W Gt
VGe
33 000(1)(1)(1.25) 56.45(0.7563)
966 lbf
Ans.
cos n sin f cos cos n cos f sin
cos 20 sin 4 .764 0.025 cos 4.764 966 cos 20 cos 4.764 0.025 sin 4.764 106.4 lbf Ans. (c) Eq. (15-33):
C s = 1190 – 477 log 7.0 = 787
Eq. (15-36):
C m 0.0107 562 56(56) 5145 0.767
Eq. (15-37):
C 0.659exp[0.0011(679.8)] 0.312
Eq. (15-38):
(W t ) all = 787(7) . (2)(0.767)(0.312) = 1787 lbf
v
08
Since WGt (W t )all, the mesh will survive at least 25 000 h. Eq. (15-61): Eq. (15-63):
0.025(966)
W f
0.025 sin 4.764 cos 2 0 cos 4 .764 29.5(679.8) H f 0.608 hp 33 000 106.4(677.4) H W 2.18 hp 33 000 966(56.45) 1.65 hp H G 33 000
The mesh is sufficient
29.5 lbf
Ans.
Pn Pt / cos 8 / cos 4.764o 8.028
pn / 8.028 0.3913 in
G
966 0.3913(0.5)(0.125)
39 500 psi
The stress is high. At the rated horsepower, G
1 1.65
39 500 23 940 psi
acceptable
(d)
Chapter 15, Page 18/20
1.7
2
2
Eq. (15-52):
A min = 43.2(8.5) = 1642 in < 1700 in
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
Eq. (15-50):
CR
Eq. (15-51):
ts 70
3939
0.13 0.568 ft · lbf/(min · in2 · o F) 17 530
88.2o F
Ans. 0.568(1700) ________________________________________________________________________
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.
p x d W F G A H W H G H f N W N G K W C s C m C V G W Gt v
t W
W
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
38.2 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
2430
2430
2430
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 f e 0.948 (P t ) G 1.795 1.979 Pn C -to-C 10.156 t s 177 L 5.25 24.9 5103 G d G 16.71
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
492
563
492
492
2430
2430
2120
2524
2524
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
