Gears are used to transmit mechanical power from one rotating shaft to another. There are several types of gears that are commonly used including spur gears, helical gears, bevel gears, and worm gears. Spur gears have straight teeth that allow for easy engagement and disengagement. This document discusses the design, specification, and selection of spur gears based on failure due to bending stress using the Lewis equation. It provides information on gear terminology, types of gear trains, tooth systems, force analysis, stresses, selection procedures, and wear failure. Examples are also included to demonstrate how to select suitable gears based on given design parameters and constraints.

1. Design of Gears

2. Gears
•
Gears are toothed cylindrical wheels used for
transmitting mechanical power from one rotating
shaft to another. Several types of gears are commonly
used and are available as stock items from original
equipment suppliers worldwide. This chapter
introduces various types of gears and gear
transmission and details the design, specification, and
selection of spur gears, in particular, based on the
consideration of failure due to bending using the
Lewis equation

3. Gears can be divided into several broad classifications.
1. Parallel axis gears:
a. Spur gears.
b. Helical gears, and
c. Internal gears.
2. Nonparallel co-planar gears (intersecting axes):
a. Bevel gears.
b. Face gears, and
c. Conical involute gearing.
3. Nonparallel noncoplanar gears (nonintersecting axes):
a. Crossed axis helical.
b. Cylindrical worm gearing.
c. Single enveloping worm gearing,

4. Spur Gear

5. Helical Gear

6. Bevel Gear

7. Crossed Axis Helical Gear

8. Worm Gear

9. Spur Gear
•
Spur gears are the least expensive of all types for parallel shaft
applications. Their straight teeth allow running engagement or
disengagement using sliding shaft and clutch mechanisms.
Typical applications of spur gears include automatic motor
vehicle gearboxes, machine tool drives, conveyor systems,
electric motor gearboxes. The majority of power gears are
manufactured from hardened and case-hardened steel. Other
materials used include iron, brass, bronze, and polymers such
as polyamide (e.g. nylon)

10. Materials Selection

11. Useful Range of Gear ratios

12. Spur gear schematic showing principal
terminology.

13. Circular pitch
Circular pitch: This is the distance from a point on one tooth to
the corresponding point on the adjacent tooth measured along
the pitch circle.
where p is the circular pitch (mm), m is the module, d is the
pitch diameter (mm), and N is the number of teeth.
Module: This is the ratio of the pitch diameter to the number of
teeth. The unit of the module should be in millimeters (mm).
The module is defined by the ratio of pitch diameter and number
of teeth. Typically the height of a tooth is about 2.25 times
greater than the module.

14. Addendum a: This is the radial distance from the pitch circle to
the outside of the tooth;
Dedendum b: This is the radial distance from the pitch circle to
the bottom land; and

15. Diametral pitch is the ratio of the number of teeth in the gear to the
pitch diameter.
Pressure Angle (φ): is the generating line or line of action in which
the resulting forces actins along this line.

16. Gear Trains
A gear train is one or more pairs of gears operating together to
transmit power. When two gears are in mesh, their pitch circles
roll on each other without slippage. If 𝑟1= pitch radius of gear 1,
𝑟2 = pitch radius of gear 2, 𝜔1 = angular velocity of gear 1,
𝜔2= angular velocity of gear 2, then the pitch line velocity is
given by
The velocity ratio is
It can be defined in any of the following ways:

17. where 𝜔𝑝 and 𝜔𝐺 are the angular velocities of the
pinion and gear respectively (rad/s), 𝑛𝑃 and 𝑛𝐺 are the
rotational speeds of the pinion and gear, respectively
(rpm), 𝑁𝑃 and 𝑁𝐺 are the number of teeth in the pinion
and gear, resp., and 𝑑𝑃and 𝑑𝐺 are the pitch diameter of
the pinion and gear, respectively (mm).
Consider a pinion 1, driving a gear 2. The speed of the
driven gear is

18. Types of Gear Train

19. Example1:
Consider the gear train shown in Figure below,
Calculate the speed of gear 5.

20. Solution:

21. Example 2 :
For the double reduction gear train shown in Figure, if
the input speed is 1750 rpm in a clockwise direction
what is the output speed?

23. Tooth Systems
•
Tooth systems are standards that define the
geometric proportions of gear teeth. Table 8.3
lists the basic tooth dimensions for full depth
teeth with pressure angles of 20 and 25. Table
8.4 lists preferred values for the module, m, and
Table 8.5 lists the preferred standard gear teeth
numbers. The failure of gears can principally be
attributed to tooth breakage, and surface failure

24. Tooth Systems

25. Force Analysis
Figure 1 shows the forces involved for two spur gears in
mesh. The force acting at the pressure angle ∅ can be
subdivided into two components: The tangential
component Ft and the radial component Fr. The radial
component serves no useful purpose. The tangential
component Ft transmits the load from one gear to the
other gear. If Wt is defined as the transmitted load, Wt =
Ft. The transmitted load is related to the power
transmitted through the gears by the following
equation:

26. where Wt = transmitted load (N), P = power (W), and V
= pitch line velocity (m/s).
Alternatively, the pitch line velocity can be defined by
So;
where Wt = transmitted load (kN), H = power (kW),
d = pitch diameter (mm), and n = speed (rpm).

27. Introduction to Gear Stresses
Gears experience two principal types of stresses;
bending stress at the root of the teeth due to the
transmitted load and contact stresses on the flank
of the teeth due to repeated impact, or sustained
contact, of one tooth surface against another.
Bending Stresses: The calculation of bending
stress in gear teeth can be based on the Lewis
formula.

28. where Wt = transmitted load (N), F = face width (m or
mm), m = module (m or mm), and Y = the Lewis form
factor and can be found from Table 8.6.
When teeth mesh, the load is delivered to the teeth with
some degree of impact. The velocity factor is used to
account for this and is given by the Barth equation:

29. Introducing the velocity factor into the Lewis
equation gives;
This equation forms the basis of a simple approach
to the calculation of bending stresses in gears.

30. Lewis Factor ((Y))

31. Example: A 20o full depth spur pinion is to transmit
1.25 kW at 850 rpm. The pinion has 18 teeth.
Determine the Lewis bending stress if the module is 2
and the face width is 25 mm.
Solution
Calculating the pinion pitch diameter: dP = mNP = 2
18 = 36 mm.
Calculating the pitch line velocity,
Calculating the velocity factor:

32. Calculating the transmitted load:
From Table 8.6 for NP = 18, the Lewis form factor Y =
0.29327. The Lewis equation for bending stress gives;

33. Simple Gear Selection Procedure
The Lewis formula, in the form of (𝜎 = 𝑊𝑡 (𝐾𝑣𝐹𝑚𝑌)) can
be used in a provisional spur gear selection procedure for a
given transmission power, input, and output speeds. The
procedure is outlined as follows:
1. Select the number of teeth for the pinion and the gear to
give the required gear ratio (observe the guidelines presented
in Table 8.2 for maximum gear ratios). Note that the min.
number of teeth permissible when using a pressure angle of
20o is 18 (Table 8.3). Use either the standard teeth numbers
as listed in Table 8.5, or as listed in a stock gear catalog.
2. Select a material. This will be limited to those listed in the
stock gear catalogs.

34. 3. Select a module, m from Table 8.4 or as listed in a stock
gear catalog (see Tables 8.7-8.10), which give examples of a
selection of stock gears available).
4. Calculate the pitch diameter, d = mN.
5. Calculate the pitch line velocity, Ensure
this does not exceed the guidelines given in Table 8.2.
6. Calculate the dynamic factor,
7. Calculate the transmitted load, Wt = Power/V.
8. Calculate an acceptable face width using the Lewis
formula in the form

35. The permissible bending stress,𝜎𝑝, can be taken as 𝜎𝑢𝑙𝑡 /factor
of safety, where the factor of safety is set by experience but
may range from 2 to 5. Certain plastics are suitable for use as
gear materials in application where low weight, low friction,
high corrosion resistance and low wear. The strength of plastic
is usually significantly lower than that of metals. Plastics are
often formed using a filler to improve strength, wear, impact
resistance, temperature performance, as well as other
properties.
Values of permissible bending stress for a few gear materials
are listed in Table 8.11.

36. Important Notes:
The design procedure consists of proposing
teeth numbers for the gear and pinion,
selecting a suitable material, selecting a
module, calculating the various parameters
as listed, resulting in a value for the face
width. If the face width is greater than that
available in the stock gear catalog, or if the
pitch line velocity is too high, repeat the
process for a different module. If this does
not provide a sensible solution, try a
different material.

41. Figure (8.11): Permissible bending stresses for various
gear materials

42. 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 from the
limited choice illustrated in Tables 8.7-8.10, specifying the module
and face width based on the Lewis formula.
Solution
Power = 1.8 kW
NP = 18
Try m = 1 mm.
Assuming a hardened material will be used to improve wear
resistance, 𝜎𝑝= 183 MPa.
Y18 = 0.29327
nP = 1100 rpm
dP = mN = 0.001 * 18=0.018 m.

43. This value of face width is greater than the stock value
available as listed in Table 8.7 for a module 1 mm gear.
Therefore, an alternative design needs to be considered.
Trying m=1.5 mm.

44. This is less than the F = 20 mm available for the 1.5 mm
module stock gears listed in Table 8.8. This gear is, therefore,
likely to be acceptable in terms of bending stress capability.
The gear specification is therefore: NP = 18, m = 1.5 mm,
F = 20 mm, 817M40 induction hardened.

45. Example:
A gearbox is required to transmit 18 kW from a shaft rotating at
2650 rpm. The desired output speed is approximately 12,000
rpm. For space limitation and standardization reasons, a double
step-up gearbox is requested with equal ratios. Using the limited
selection of gears presented in Tables 8.8-8.11, select suitable
gears for the gear wheels and pinions. For this case, use 655M13
case-hardened steel gears.
Solution:
The overall ratio =12,000/2650 = 4.528.
First stage ratio = 4.528 = 2.128.
This could be achieved using a gear with 38 teeth and pinion with
18 teeth
(ratio = 38/18 = 2.11).
The gear materials listed in Tables 8.7-8.10 are 817M40 and
655M13 steels.

46. From Table 8.11, the 655M13 is the stronger steel, and this is
selected for this example prior to a more detailed consideration.
For 655M13 case hardened steel gears, the permissible stress
𝜎𝑝 = 345 MPa.
Calculations for gear 1: Y38 = 0.37727, n = 2650 rpm.

47. Note: m = 1.5 gives a face width greater than the catalog value of
20 mm, so try m = 2.
m = 2 gives a face width less than the catalog value of 25 mm, so
OK.
Calculations for pinion 1: Y18 = 0.29327, n = 5594 rpm.
Note: m = 1.5 gives a face width greater than the catalog value of
20 mm, so try m = 2.
m = 2 gives a face width less than the catalog value of 25 mm, so
OK.

48. Calculations for gear 2: Y38 = 0.37727, n = 5594 rpm.
Note: m = 2 gives value for face width lower than catalog
specification, so the design is OK.
Calculations for pinion 2: Y18 = 0.29327, n = 11,810 rpm.
Note: m = 2 gives value for face width lower than catalog
specification, so the design is OK.

49. The overall design for this example is illustrated in Figure below.
Note in this example common shafts, bearings and machine
features have been implemented throughout in order to
minimize the parts inventory.

50. Wear Failure
Gears can fail due to excessive bending stress or wear. Wear
occurs because as the teeth move in and out of contact with
each other, there is accompanying local deformation of the
gear teeth surfaces in the region of contact. The stresses
resulting from the surface deformation are known as contact
stresses. If the stresses are too high, then material failure can
take the form of a loss of material from the surfaces, which is
also known as pitting. Pitting is surface fatigue failure due to
too many repetitions of high contact stresses.
The surface compressive, Hertzian, or contact stress for a gear
can be modeled by:

52. The radii of curvature are given by
where dP and dG are the pitch diameters of the pinion and gear,
respectively.
The velocity factor Kv for cut or milled profile gears is given by:
where V is the pitch line velocity (m/s).
The elastic coefficient Cp can be calculated from Eqn above, or
obtained from Table 9.1.

54. Example:
A speed reducer has a 22-tooth spur pinion made of steel,
driving a 60-tooth gear made of cast iron. The transmitted
power is 10 kW. The pinion speed is 1200 rpm, module 4 mm,
and face width 50 mm. Determine the contact stress.

55. ( - ) sign is Compressive stress

56. AGMA Equations for Bending and Contact
Stress
The calculation of bending and contact stresses in
spur and helical gears can be determined using
standardized methods presented by the AGMA. The
AGMA standards have recently been used and have
therefore been selected for presentation here. The
procedures make extensive use of a series of
geometry and design factors, which can be
determined from design charts and tables.
The AGMA formula for bending stress for spur gears
is:

57. 𝜎 =
𝑊𝑡𝐾𝑎
𝐾𝑣
1
𝐹𝑚
𝐾𝑠𝐾𝑚
𝐽
Where:
𝜎 is the bending stress.
𝑊𝑡 is the transmitted load.
𝐾𝑎 is an application factor (usually taken as 𝐾𝑎=1).
𝐾𝑣 is the dynamic factor (speed factor).
𝐹 is the face width.
𝑚 is the module.
𝐾𝑠 is the size factor (usually taken as 𝐾𝑠=1).
𝐾𝑚 is the load distribution factor (see table 13).
𝐽 is a geometry factor (see table 14).

58. 𝐾𝑣 =
𝐴
𝐴 + 200𝑉
𝐵
Where: 𝐵 =
12−𝑄𝑣
3 4
4
𝐴 = 50 + 56(1 − 𝐵)
Qv is the AGMA quality standard: 3 < 𝑄𝑣 > 12

59. The AGMA equation for pitting (contact) resistance:
𝜎𝑐 = 𝐶𝑝
𝑊𝑡𝐶𝑎
𝐶𝑣
.
𝐶𝑠
𝐹. 𝑑
.
𝐶𝑚. 𝐶𝑓
𝐼
0.5
Where:
𝜎𝑐 is the value of contact stress.
𝐶𝑝 is elastic coefficient (Table 12).
𝐶𝑎 is the application factor (usually taken as 𝐶𝑎 =1).
𝐶𝑣 is the dynamic factor.
𝐶𝑠 is the size factor (usually taken as 𝐶𝑠 =1).
𝑑 is the pitch diameter of the pinion.
𝐶𝑚 is the load distribution factor (Table 13).
𝐶𝑓 is a surface condition factor (𝐶𝑓 =1)
𝐼 is a geometry factor.

60. 𝐶𝑣 =
𝐴
𝐴 + 200𝑉
𝐵
Where: 𝐵 =
12−𝑄𝑣
3 4
4
𝐴 = 50 + 56(1 − 𝐵)
Qv is the AGMA quality standard: 3 < 𝑄𝑣 > 12
𝐼 =
𝑐𝑜𝑠∅𝑠𝑖𝑛∅
2𝑚𝑁
.
𝑚𝐺
𝑚𝐺 + 1
Where: 𝑚𝑁 = 1 for a spur gears.
𝑚𝐺 is the speed ratio, 𝑚𝐺 =
𝑁𝐺
𝑁𝑝

61. Factor of Safety:
The AGMA equation for determining a safe value for the
allowable bending stress is:
𝜎𝑐 𝑎𝑙𝑙 =
𝑆𝑐𝐶𝐿𝐶𝐻
𝐶𝑇𝐶𝑅
Where:
𝑆𝑐 is the AGMA surface fatigue strength (table 17).
𝐶𝐿 is the life factor .
𝐶𝐻 is the hardness factor.
𝐶𝑇 is the temperature factor (𝐶𝑇 =1).
𝐶𝑅 is the reliability factor (table 15).
𝐶𝐿 = 2.466𝑁−0.056
, where: N=number of cycles
𝐶𝐻 = 1 + 𝐴 𝑚𝐺 − 1

62. And
𝐴 = 8.98 ∗ 10−3
𝐻𝐵𝑝
𝐻𝐵𝐺
− 8.29 ∗ 10−3
Where: 𝐻𝐵𝑝 and 𝐻𝐵𝐺 are the Brinell hardness value of the pinion
and gear respectively.
The factor of safety for contact stress is defined as:
𝑛𝑐 =
𝜎𝑐 𝑎𝑙𝑙
𝜎𝑐
The value of factor of safety is (1-2).