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 Dudley’s 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 Table12.1. See Figure12.1 for a general description of a gear set.

12.1.2 woRking stRess σ

w

Allowable values for σ

w

are given in Table12.2. 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

TABLE12.1

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.

TABLE12.2

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 3to 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 = 24teeth

t = 18teeth

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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