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Chapter 15
15-1
Given: Uncrowned, through-hardened 300 Brinell core and case, Grade 1,
N
C
=
10
9
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
=
π
(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):
K
v
=
±
59
.
77
+
√
785
.
3
59
.
77
²
0
.
8255
=
1
.
374
Eq. (15-8):
v
t
,max
=
[59
.
77
+
(6
−
3)]
2
=
3940 ft/min
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(10
9
)
−
0
.
0323
=
0
.
862
(
K
L
)
G
=
1
.
6831(10
9
/
3)
−
0
.
0323
=
0
.
893
Eq. (15-14):
(
C
L
)
P
=
3
.
4822(10
9
)
−
0
.
0602
=
1
(
C
L
)
G
=
3
.
4822(10
9
/
3)
−
0
.
0602
=
1
.
069
Eq. (15-19):
K
R
=
0
.
50
−
0
.
25 log(1
−
0
.
999)
=
1
.
25
(or Table 15-3)
C
R
=
³
K
R
=
√
1
.
25
=
1
.
118
Bending
Fig. 15-13:
0
.
99
S
t
=
s
at
=
44(300)
+
2100
=
15 300 psi
Eq. (15-4):
(
σ
all
)
P
=
s
w
t
=
s
at
K
L
S
F
K
T
K
R
=
15 300(0
.
862)
1(1)(1
.
25)
=
10 551 psi
Eq. (15-3):
W
t
P
=
(
σ
all
)
P
FK
x
J
P
P
d
K
o
K
v
K
s
K
m
=
10 551(1
.
25)(1)(0
.
249)
6(1)(1
.
374)(0
.
5222)(1
.
106)
=
690 lbf
H
1
=
690(785
.
3)
33 000
=
16
.
4hp
Eq. (15-4):
(
σ
all
)
G
=
15 300(0
.
893)
1(1)(1
.
25)
=
10 930 psi

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*Sign up* 380
Solutions Manual
• Instructor’s Solution Manual to Accompany Mechanical Engineering Design
W
t
G
=
10 930(1
.
25)(1)(0
.
216)
6(1)(1
.
374)(0
.
5222)(1
.
106)
=
620 lbf
H
2
=
620(785
.
3)
33 000
=
14
.
8hp
Ans.
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):
(
σ
c
,all
)
P
=
s
ac
(
C
L
)
P
C
H
S
H
K
T
C
R
(
σ
c
,all
)
P
=
125 920(1)(1)
1(1)(1
.
118)
=
112 630 psi
For the gear, from Eq. (15-16),
B
1
=
0
.
008 98(300
/
300)
−
0
.
008 29
=
0
.
000 69
C
H
=
1
+
0
.
000 69(3
−
1)
=
1
.
001 38
And Prob. 15-1, (
C
L
)
G
=
1
.
0685
.
Equation (15-2) thus gives
(
σ
c
,all
)
G
=
s
ac
(
C
L
)
G
C
H
S
H
K
T
C
R
(
σ
c
,all
)
G
=
125 920(1
.
0685)(1
.
001 38)
1(1)(1
.
118)
=
120 511 psi
For steel:
C
p
=
2290
±
psi
Eq. (15-9):
C
s
=
0
.
125(1
.
25)
+
0
.
4375
=
0
.
593 75
Fig. 15-6:
I
=
0
.
083
Eq. (15-12):
C
xc
=
2
Eq. (15-1):
W
t
P
=
²
(
σ
c
,all
)
P
C
p
³
2
Fd
P
I
K
o
K
v
K
m
C
s
C
xc
=
²
112 630
2290
³
2
´
1
.
25(3
.
333)(0
.
083)
1(1
.
374)(1
.
106)(0
.
5937)(2)
µ
=
464 lbf
H
3
=
464(785
.
3)
33 000
=
11
.
0hp
W
t
G
=
²
120 511
2290
³
2
´
1
.
25(3
.
333)(0
.
083)
1(1
.
374)(1
.
106)(0
.
593 75)(2)
µ
=
531 lbf
H
4
=
531(785
.
3)
33 000
=
12
.
6hp

Chapter 15
381
The pinion controls wear:
H
=
11
.
0hp
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
.
0hp
Ans
.
15-3
AGMA 2003-B97 does not fully address cast iron gears, however, approximate compar-
isons can be useful. This problem is similar to Prob. 15-1, but not identical. We will orga-
nize 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,
φ
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
=
π
(5)(900
/
12)
=
1178 ft/min
Set
N
L
=
10
7
cycles for the pinion. For
R
=
0
.
99,
Table 15-7:
s
at
=
4500 psi
Table 15-5:
s
ac
=
50 000 psi
Eq. (15-4):
s
w
t
=
s
at
K
L
S
F
K
T
K
R
=
4500(1)
2(1)(1)
=
2250 psi
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)

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