115
12
Gearing
It is not the intention of this chapter to give an in-depth analysis of the design of gear systems. For a
more comprehensive study on the design of gearing see Design Engineer’s Handbook (Chapter 12,
Introduction to Geared Systems) or Dudleys Gear. Only a few brief examples of the primary form
of tooth gearing will be given here; the two forms being considered are:
1. Spur gearing
2. Bevel gearing
12.1 SPUR GEARING
General formula:
=
ω
=E=
T
r
Power
r
Power
V
(12.1)
Lewis formula for strength (American Gear Manufacturers Association):
σ=
φWcos
K.F.m.y
v
(12.2)
where
=W
19.1P
N.D
=−
+
y 0.484
4.24
t6
=
+
K
3.54
3.54 V
v
ϕ = pressure angle (20°)
12.1.1 notation
For gear notation see Table12.1. See Figure12.1 for a general description of a gear set.
12.1.2 woRking stRess σ
w
Allowable values for σ
w
are given in Table12.2. These cover the more common materials used in
gear manufacture.
12.1.3 wiDth of teeth
Using the Lewis factor to nd the wheel proportions involves xing the width in terms of the
circular pitch and substituting the appropriate formula. Generally for slow speeds and where shafts
are inaccurately adjusted, the face width may be 1.25 to 2.5 times the pitch (the average works out
116 Design Engineer's Case Studies and Examples
TABLE12.1
Gear Notation
E = Tangential load on teeth (N)
T = Torque transmitted (Nm)
r = Pitch circle radius (m)
V = Pitch line velocity (m/s)
y = Lewis form factor (See notes)
b = Width of teeth (m)
P = Circular pitch (m)
σ
w
=
Safe working stress (MPa)
D = Pitch circle diameter for wheel (m)
d = Pitch circle diameter for pinion (m)
T = Number of teeth for wheel (m)
t = Number of teeth for pinion (m)
m = Module (m)
A = Addendum = module (m)
B = Dedendum = 1.25 × module (m)
No. of Teeth ‘t
Pitch Point
PCD ‘D’
PCD ‘d’
No. of
Teeth ‘T’
Centre
Distance ‘c’
Wheel
Pinion
FIGURE 12.1 Nomenclature for spur gear set.
TABLE12.2
Lewis Factor Stresses (σ
w
) in MPa
Material
Pitch Circle Velocity (m/min)
30 60 120 180 270 370 550 730
Cast iron 54.4 40.9 32.7 27.2 20.4 16.3 13.9 11.7
Cast steel 136 102 81.7 68.0 51.0 40.8 34.7 29.7
Forged steel 163 122.5 95.3 81.7 68.0 51.0 44.2 37.2
Phosphor bronze 71.5 71.5 44.2 37.4 32.6 27.2 20.8 16.6
Nickel chrome steel
(1544 MPa ult.)
224.6 187.0 150.0 126.0 93.6 75.9
117Gearing
to 3to 4 times the pitch); for high speeds, smooth engagement and high wear, the width may be 6 to
8 times. High ratios result in ner tooth pitches.
Example 12.1
Two spur gears are to have a ratio of 4:3 and a 6 mm module. If the centre distance is to be
approximately 125 mm, calculate:
1. The circular pitch.
2. The number of teeth on each gear.
3. The pitch circle diameters.
4. The exact centre distance.
Solution:
1. The circular pitch:
ρ = πm (12.3)
= 6π
= 18.849 mm
2. The number of teeth on each gear:
From 0.5(D + d) = centre distance
D + d = 125 × 2 (12.4)
Also
D
T
m
d
t
==
=
D
T
6a
nd
6T = D and 6t = d
Substituting in Equation (12.2),
6T + 6t = 250 (12.5)
Also T/t = 4/3 and substitute in Equation (12.3) for T.
6
4
3
t6t 250
hencet 17.86
×+=
=
Both gears must contain a whole number of teeth: therefore to make T a whole number t
must be divisible by 3.
Hence the nearest multiple of 3 is 18.
thereforeT
4
3
18
118 Design Engineer's Case Studies and Examples
T = 24teeth
t = 18teeth
3. The pitch circle diameters:
Wheel pitch circle diameter D = T × module (12.6)
= 24 × 6
= 144 mm
Pinion pitch circle diameter d = t × module (12.7)
= 18 × 6
= 108 mm
4. Exact centre distance:
c = 0.5 × (D + d) (12.8)
= 0.5 × (144 + 108)
= 126.0 mm
Example 12.2
Design a pair of involute spur gears to transmit power of 30 kW at 8.5 rev/s. Gear ratio of 4:1.
Assume a 20 tooth pinion of 6 mm module and a ratio of face width to pitch of 4. The pinion is
forged steel and the wheel is cast steel.
Find the tooth proportions of wheel and pinion.
Solution:
Torque
power
2N
=
π
=
×
π×
×
30 10
28.5
Nm
s
s
3
= 561.72 Nm
Also torque = E × r; here E is the tangential load and
E
pcd
2
561.72 Nm×=
Also
=
π
pcd
t.
p

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