The document describes different types of gears: 1) Parallel axis gears including spur gears, helical gears, and internal gears 2) Nonparallel co-planar gears including bevel gears, face gears, and conical involute gears 3) Nonparallel non-planar gears including crossed axis helical gears, cylindrical worm gearing, and single enveloping worm gearing. It then provides an example of a simple gear selection procedure to select a suitable gear to mesh with a given 18-tooth pinion operating at 1100 rpm to transmit 1.8 kW of power. The procedure involves selecting the module, pitch diameter, pitch velocity, dynamic factor, transmitted load, and acceptable face width to choose

1. Types of
Gears
1. Parallel axis gears
1.1 Spur gears
1.2 Helical gears
1.3 Internal gears
2. Nonparallel co-planner
gears
2.1 Bevel gears
2.2 Face gears
2.3 Conical involute gearing
3. Nonparallel non-planar
gears
3.1 Crossed axis helical
3.2 Cylindrical worm
gearing
3.3 Single enveloping Worm
Gearing

2. 1. Parallel axis gears
1.1 Spur Gears 1.2 Helical Gears

3. 1. Parallel axis gears
1.2 Double helical gears. 1.3 Internal Gears

4. 2. Nonparallel co-planner gears
2.1 Bevel gears
2.2 Face
gears
2.3 Conical involute
gears

5. 3. Nonparallel non-planar gears
3.1 Crossed axis helical
gears.
3.2 Cylindrical worm gearing
3.3 single enveloping worm gearing.

6. Involute Gear Construction

21. Simple Gear Selection Procedure
Example:
A 20o full depth spur pinion is required to
transmit 1.8 kW at a speed of 1100 rpm. If the
pinion has 18 teeth and is manufactured from
heavy-duty 817M40 steel select a suitable gear.

22. Simple Gear Selection Procedure
1. Select the number of teeth for the pinion and
the gear to give the required gear ratio.
Solution:
pinion has 18 teeth

23. Simple Gear Selection Procedure
2. Select a material.
Solution:
heavy-duty 817M40 steel

24. Simple Gear Selection Procedure
3. Select a module, m
Solution:
Try module =1.

25. Simple Gear Selection Procedure
4.Select the Lewis factor
Solution:
Y18 =0.29327

26. Simple Gear Selection Procedure
5. Calculate the pitch diameter, d = m*N.
Solution:
D= 0.001*18=0.018m
6.Calculate the pitch velocity
Solution:
V =(d/2) * n * (2pi/60)
=0.018/2 * 1100*0.1047=1.037 m/s.
25m/s

27. Simple Gear Selection Procedure
7. Calculate the dynamic factor, Kv =6.1/(6.1+V ).
Solution:
Kv=6.1/(6.1+1.037)=0.8548.
8. Calculate the transmitted load, Wt =Power/V.
Solution:
Wt=1800/1.037=1737 N

28. Simple Gear Selection Procedure
9.Calculate an acceptable face width using the
Lewis formula in the form:
F =Wt / (K*V*m*Y*Ϭp)
Solution:
F=0.0379m=37.9mm.
• According to the catalog the face width should not
exceed than 15mm “1.0 module. Heavy duty
steel 817M40, 655M13. Face width 15 mm.”
Therefore an alternative design needs to be considered

29. Simple Gear Selection Procedure
According to the catalog: “1.5 module. Heavy duty steel
817M40, 655M13.”
Face width 20 mm.
Hence, The gear specification is therefore: NP = 18, m=1.5 mm,
F= 20 mm, 817M40 induction hardened.