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Problem 6.7 and Solution to it.
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- Central University Of Technology (CUT)
It is a problem in relation with gears, shafts and bearings
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PROBLEM 6,7 PAGE 135A gearbox is required to transmit 6 kW from a shaft rotating at 2200 rpm.
The desired output speed is approximately 300 rpm.
Use the Lewis formula to determine gearbox. And select suitable gears
from the limited selection presented in Tables 6.7-6.10.
To solve this problem, follow procedure elaborated on the section 6,6
pages 124 to 128
Students, please read the procedure.
1
, Gear Selection Procedure
Table 6,2 page 111
GEAR RATIO RANGE PITCH LINE EFFICIENCY
1. Select the number of teeth for the pinion VELOCITY (m/s)
and the gear to give the required gear Spur 1:1 to 6:1 25 98-99%
ratio. Helical 1:1 to 10:1 50 98-99%
▪ Observe the guidelines presented in Table 6.2 Double helical 1:1 to 15:1 150 98-99%
Bevel 1:1 to 4:1 20
for maximum gear ratios.
Worm 5:1 to 75:1 30 20-98%
▪ Note that the minimum number of teeth
Crossed helical 1:1 to 6:1 30
permissible when using a pressure angle of 20o
is 18 or 25o is 12 (Table 6.3).
▪ Use either the standard teeth numbers as listed
in Table 6.5 or as listed in a stock gear
catalogue Table 6.7 to 6,10.
Gear ratios
N1 ……………..…………..…..eq 6,8 pg 116
n2 = − n 1
N2
2. Select a material.
▪ This will be limited to those listed in the Table
6,11 or stock gear catalogues Table 6.7 to 6,10.
▪ The permissible bending stress, p, can be
taken as uts/factor of safety, where the factor of
safety is set by experience but may range from 2
to 5.
▪ Alternatively use values of p as listed for the
appropriate material in a stock gear catalogue.
2
, Gear Selection Procedure
3. Select a module,
▪ Select module based on Table 6.4 or as listed in a
stock gear catalogue Tables 6.7 to 6.10 which give
examples of a selection of stock gears available.
4. Calculate pitch diameter
d=mN ……………................…..….eq 6,1 pg 111
5. Calculate the pitch line velocity.
V = r x n x (2π/60)……………eq 6,21 pg 124
▪ Ensure this does not exceed the guidelines given
in Table 6.2.
Table 6,2
GEAR RATIO RANGE PITCH LINE EFFICIENCY
6. Calculate the dynamic factor, VELOCITY (m/s)
Kv=6.1/(6.1 + V)..……………eq 6,20 pg 123 Spur 1:1 to 6:1 25 98-99%
Helical 1:1 to 10:1 50 98-99%
7. Calculate the transmitted load, Double helical 1:1 to 15:1 150 98-99%
Bevel 1:1 to 4:1 20
Wt=P/V……………………………eq 6,17 pg 123
Worm 5:1 to 75:1 30 20-98%
Crossed helical 1:1 to 6:1 30
8. Calculate the face width using the Lewis formula
F=Wt/(KvmY p) …………………..….eq 6,17 pg 123
3
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