T
g
1627
ft lbf
⋅
=
T
g
19524
in lbf
⋅
=
T
g
N
g
N
p
T
p
⋅
=
4.
We could have calculated the torque on the gear shaft directly without finding the gear shaft speed,
T
g
1627
ft lbf
⋅
=
T
g
19524
in lbf
⋅
=
T
g
P
ϖ
g
=
g
403.509
rpm
=
g
N
p
N
g
p
⋅
=
3.
For the gear shaft
2.
The gear shaft will rotate at a lower speed, which is determined by the gear ratio.
(The speed will be decreased i
proportion to the ratio and the torque will be increased by the reciprocal of the ratio).
T
p
656.5
ft lbf
⋅
=
T
p
7878
in lbf
⋅
=
T
p
P
p
=
1.
For the pinion shaft
Solution:
See Mathcad file P1114.
Assumptions:
There is no loss of power in the gear mesh (100% efficiency).
P
125
hp
⋅
=
Transmitted power
N
g
57
=
gear
p
1000
rpm
⋅
=
Pinion speed
N
p
23
=
pinion
Tooth numbers:
Given:
rpm
2
π
⋅
rad
⋅
min
1

⋅
=
Units:
Statement:
If the gearset in Problem 113 transmits 125 HP at 1000 pinion rpm, find the torque on each
shaft.
PROBLEM 1114
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View Full DocumentPitchline velocity (fpm)
V
t
p
d
(
29
d p
d
(
29
n
⋅
2
=
Transmitted load (lbf)
W
t
p
d
(
29
H
V
t
p
d
(
29
=
3.
Set the application factor,
K
a
K
a
1
=
4.
Write the equation for the dynamic load factor,
K
v
B
0.25 12
Q
v

(
29
0.6667
⋅
=
B
0.52
=
A
50
56 1
B

(
)
⋅
+
=
A
76.878
=
K
v
p
d
(
29
A
A
V
t
p
d
(
29
min
ft
⋅
+
B
=
5.
Tentatively choose the mounting factor,
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 Spring '11
 Staff
 Machine Design, gear, Pinion, face width

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