12.

Spur Gear Design and Standard proportions

selection
Objectives • American Standard Association (ASA)
• Apply principles learned in Chapter 11 to actual design and selection of spur

•
gear systems.
Calculate forces on teeth of spur gears, including impact forces associated with
• American Gear Manufacturers Association
velocity and clearances. (AGMA)
• Determine allowable force on gear teeth, including the factors necessary due to

•
angle of involute of tooth shape and materials selected for gears.
Design actual gear systems, including specifying materials, manufacturing
• Brown and Sharp
•
accuracy, and other factors necessary for complete spur gear design.
Understand and determine necessary surface hardness of gears to minimize or
• 14 ½ deg; 20 deg; 25 deg pressure angle
prevent surface wear.
• Understand how lubrication can cushion the impact on gearing systems and cool • Full depth and stub tooth systems
them.
• Select standard gears available from stocking manufacturers or distributors.

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Specifications for standard gear

teeth
Forces on spur gear teeth
Item Full depth & pitches Full depth & 14½° full • Ft = Transmitted force
coarser than 20 pitches finer depth
than 20 • Fn = Normal force or separating
Pressure 20° 25° 20° 14½° force
angle
• Fr = Resultant force
Addendum 1.0/P 1.0/P 1.0/P 1/P • θ = pressure angle
(in.)
• Fn = Ft tan θ
Dedendum 1.250/P 1.250/P 1.2/P + 0.002 1.157/P
F
(in.) Fr = t
cos θ
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Example Problem 12-1: Forces on Spur Gear Teeth

Forces on spur gear teeth
• Power, P; P = Tn or T = 63,000 P • 20-tooth, 8 pitch, 1-inch-wide, 20° pinion
63,000 n transmits 5 hp at 1725 rpm to a 60-tooth gear.

• Torque, T = Ft r and r = Dp /2 • Determine driving force, separating force, and

• Combining the above we can write maximum force that would act on mounting
shafts.
2T
Ft =
Dp

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 Example Problem 12-1: Forces on Spur Gear Teeth Example Problem 12-1: Forces on Spur Gear Teeth (cont’d.)
− Find transmitted force:
• 20-tooth, 8 pitch, 1-inch-wide, 20° pinion transmits 5 hp at 1725 rpm to a 60-tooth gear. (12-3)
2T
• Determine driving force, separating force, and maximum force that would act on Ft =
Dp
mounting shafts.
(2-6) (2)183 in-lb
Tn Ft = = 146 lb
P = 2.5 in
63,000
− Find separating force:
63,000P (12-1)
T =
n
Fn = Ft tan θ
(63,000)5
T = = 183 in-lb
1725 Fn = 146 lb tan 20°

Fn = 53 lb

− Find pitch circle: − Find maximum force:

(12-2)
(11-4) Ft
Fr =
Dp =
Np cos θ
Pd
146 lb
20 teeth Fr =
Dp = = 2.5 in cos 20°
8 teeth/in diameter
Fr = 155 lb
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Example Problem 12-2: Surface Speed

Surface Speed
• In previous problem, determine the surface speed:

• Surface speed (Vm) is often referred to as

(12-7)
pitch-line speed Vm = π D n

or
π Dp n
• Vm = ft/min (12-5)

12 Vm =
π Dp n
12
π Dp n
• Vm = m/min -- Metric units Vm = π 2.5 in 1725 rpm
ft
12 in
1000 Vm = 1129 ft/min

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Forces on Gear Tooth

Strength of Gear Teeth
• Lewis form factor method

Figure 14.20 Forces acting

on individual gear tooth.

©1998 McGraw-Hill, Hamrock,

Jacobson and Schmid
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2
 Lewis equation Table 12.1 Lewis form factors (Y)

Sn Y b
Fs =
Pd
• Fs = Allowable dynamic bending force (lb)
• Sn = Allowable stress (lb/in2). Use
endurance limit and account for the fillet as
the stress concentration factor
• b = Face width (in.)
• Y = Lewis form factor (Table 12.1)
• Pd = Diametral pitch
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Example Problem 12-3: Strength of Gear Teeth Example Problem 12-3: Strength of Gear Teeth (cont’d.)
• In Example Problem 12-1, determine the force allowable (Fs) on these teeth if the
− Gear:
pinion is made from an AISI 4140 annealed steel, the mating gear is made from AISI
1137 hot-rolled steel, and long life is desired. Sn = .5 (88 ksi) = 44 ksi
− Pinion:
(Table 12-1)
Sn = .5 Su = .5 (95 ksi) = 47.5 ksi
Y = .421
(12-9)
44,000 (1) .421
Sn b Y Fs =
Fs = 8
Pd
Fs = 2316 lb
− Find Lewis form factor (Y) from Table 12-1, assuming full-depth teeth:

Y = .320

47,500 (1) .320 − Use Fs = 1900 lb for design purposes.

Fs =
8

Fs = 1900 lb

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Classes of Gears Force Transmitted

• Transmitted load depends on the accuracy of the • Transmitted load depends on the accuracy of the
gears gears
• Gear Manufacture • A dynamic load factor is added to take care of
– Casting this.
– Machining
• Forming • Ft = Transmitted force
• Hobbing • Fd = Dynamic force
• Shaping and Planing
– Forming • Commercial 600 + Vm
Fd = Ft
– stamping 600
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3
 Classes of Gears Design Methods
1200 + Vm
• Carefully cut Fd = Ft • Strength of gear tooth should be greater
1200 than the dynamic force; Fs ≥ Fd
• Precision 78 + Vm0.5 • You should also include the factor of safety,
Fd = Ft
78 Nsf
Fs
≥ Fd
50 + Vm0.5 N sf
• Hobbed or shaved Fd = Ft
50
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Table 12.2 Service Factors Face width of Gears

• Relation between the width of gears and the
1 < Nsf < 1.25 Uniform load without shock diametral pitch
1.25 < Nsf < 1.5 Medium shock, frequent starts 8 12.5
<b<
1.5 < Nsf < 1.75 Moderately heavy shock Pd Pd

1.75 < Nsf < 2 Heavy shock

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Example Problem 12-4: Design Methods Example Problem 12-5: Design Methods
• If, in Example Problem 12-1, the gears are commercial grade, determine dynamic load
and, based on force allowable from Example Problem 12-3, would this be an acceptable • Spur gears from the catalog page shown in Figure 12-3 are made from a .2% carbon
design if a factor of safety of 2 were desired? steel with no special heat treatment.
• Use surface speed and force transmitted from Example Problems 12-2 and 12-3. • What factor of safety do they appear to use in this catalog?
– Dynamic load: • Try a 24-tooth at 1800 rpm gear for example purposes.
600 + Vm
Fd = Ft (12-10) • From Appendix 4, an AISI 1020 hot-rolled steel would have .2% carbon with an Sy = 30
600 ksi and Su = 55 ksi.
600 + 1129 − Therefore:
Fd = (146 lb)
600
Fd = 421 lb Sn = .5 Su

Sn = .5 (55 ksi) = 27.5 ksi

– Comparing to force allowable:
− Find Dp :
Fs
≥ Fd (11-4)
N sf
1900 lb Np
≥ 421 Dp =
Pd
2
950 lb > 421 lb Dp =
24
= 2 in
• This design meets the criteria. 12
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 Example Problem 12-5: Design Methods (cont’d) Example Problem 12-5: Design Methods (cont’d)
− Find Vm : − Set Fs = Fd and solve for Ft :

(12-5) ⎛ 600 + Vm ⎞
Fd = Fs = ⎜ ⎟ Ft
π Dp n ⎝ 600 ⎠
Vm =
12
⎛ 600 + 942 ⎞
519 lb = ⎜ ⎟ Ft
⎝ 600 ⎠
ft
Vm = π 2 in (1800 rpm)
12 in Ft = 202 lb
(12-3)
Vm = 942 ft/min
⎛ Dp ⎞
− Find Fs : T = F ⎜ ⎟
⎝ 2 ⎠
(12-9)
⎛ 2 in ⎞
T = 202 lb ⎜ ⎟
Sn b Y ⎝ 2 ⎠
Fs =
Pd
T = 202 in-lb
(from Table 12-1)
(2-6)

Tn 202 (1800)
Y = .302 P = = = 5.8 hp
63,000 63,000

27,500 (¾) .302 or

Fs =
12
(12-6)

Fs = 519 lb Ft Vm 202 (942)

P N RAO 25 P = = P N RAO = 5.8 hp 26
33,000 33,000

Example Problem 12-5: Design Methods (cont’d) Example Problem 12-6: Design Methods

• Pair of commercial-grade spur gears is to transmit 2 hp at a speed of 900 rpm of the

− Compared to catalog:
pinion and 300 rpm for gear.
• If class 30 cast iron is to be used, specify a possible design for this problem.
• The following variables are unknown: Pd, Dp, b, Nt, Nsf, θ.
hp - calculated
Nsf = • As it is impossible to solve for all simultaneously, try the following:
hp - catalog
– Pd = 12, Nt = 48, θ = 14½°, Nsf = 2
– Solve for face width b.
5.8
Nsf = = 1.4
4.14
− Miscellaneous properties:

(11-4)
• Appears to be reasonable value.
Np 48
• Manufacturer may also have reduced its rating for wear purposes as these Dp = = = 4 in
Pd 12
are not hardened gears.
Ng = Np Vr = 48(3) = 144 teeth

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Example Problem 12-6: Design Methods (cont’d.) Example Problem 12-6: Design Methods (cont’d.)
⎛ 600 + Vm ⎞
– Dynamic force Fd = ⎜ ⎟ Ft (12-10)
⎝ 600 ⎠
− Surface speed: ⎛ 600 + 943 ⎞
Fd = ⎜ ⎟ 70
(12-5) ⎝ 600 ⎠
π Dp n Fd = 180 lb
Vm = – Since width b is the unkown:
12
Fs
Vm = π
4 in 900 rpm ≥ Fd
12 in/ft N sf
and
Vm = 943 ft/min
Sn b Y
− Finding force on teeth:
Fs =
Pd (12-8)
(12-6) Sn b Y
= Fd
33,000 hp N sf Pd
Ft =
Vm

33,000 (2) –Class 30 CI; Su = 30 ksi; Sn = .4 Su (.4 is used because cast iron):
Ft =
943 –Sn = 12 ksi

Ft = 70 lb –Y = .344 (Table 12-1)

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 Example Problem 12-6: Design Methods (cont’d.)
− Substituting: • To increase the dynamic beam strength of
(12-9) the gear
Sn b Y
Nsf Pd
= Fd – Increase tooth size by decreasing the
12,000 b .344 diametral pitch
= 180
2 (12)
– Increase face width upto the pitch diameter
b = 1.0 inches
of the pinion
− Check ratio of width to pitch:
(12-14)
– Select material of greater endurance limit
8
< b <
12.5 – Machine tooth profiles more precisely
Pd Pd

8 12.5
– Mount gears more precisely
< 1 <
12 12
– Use proper lubricant and reduce
.66 < 1 < 1.04
contamination
• This is an acceptable design.
• Many other designs are also possible.
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Buckingham Method of Gear Design Fig. 12.4 Expected error in tooth profiles

• It offers greater flexibility

• Expected error is based on different-pitch
teeth
• More conservative design

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Table 12.3 Values of C for e = 0.001 inch Buckingham Method of Gear Design

Material 14 ½ deg 20 deg

Gray iron and Gray iron 800 830 0.05 Vm (b C + Ft )
Fd = Ft +
Gray iron and steel 1,100 1,140 0.05 Vm + (b C + Ft )0.5
Steel and steel 1,600 1,660

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 Example Problem 12-7: Buckingham Method of Gear
Fig. 12.5 Recommended maximum error in gear teeth Design and Expected Error

• A pair of steel gears is made from annealed AISI 3140.

• Each gear has a surface hardness of BHN = 350.
• The pinion is a precision, 16 pitch, 20° involute, with 24 teeth 1 inch wide.
• The gear has 42 teeth.
• To transmit 3 hp at a speed of 3450 rpm for a safety factor of 1.4, is this a
suitable design?
(Appendix 4)

Su = 95 ksi

Sn = .5 Su = .5(95 ksi) = 47.5 ksi

(11-4)

Np 24
Dp = = = 1.5 in
Pd 16

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Example Problem 12-7: Buckingham Method of Gear Example Problem 12-7: Buckingham Method of Gear
Design and Expected Error (cont’d.) Design and Expected Error (cont’d.)
− Find torque: − Find surface speed:

(2-6) (12-5)

Tn π Dp n
P = Vm =
63,000 12

P (63,000) π 1.5 in 3450 rpm

T = Vm =
n 12 ft/in

3 (63,000) Vm = 1355 ft/min

T =
3450
− Find force allowable (Fs):
T = 55 in-lb
(12-9)

− Find force transmitted: Sn b Y

Fs =
Pd
(12-3)
(from Table 12-1)
2T
Ft =
Dp Y = .337
2(55 in-lb) 47,500 (1) .337
Ft = Fs =
1.5 in 16

Ft = 73 lb Fs = 1000
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Example Problem 12-7: Buckingham Method of Gear

Design and Expected Error (cont’d.) Wear strength (Buckingham)
• Expected error from Figure 12-4:
e = .0005
• The value of C from Table 12-3 for steel and steel and 20° involute angle gears is 1660:
C = .0005 (1000) 1660
⎛ 2 Dg ⎞
C = 830
FW = D P b K g ⎜ ⎟
• Solve for dynamic load using Buckingham’s equation: ⎜D + D ⎟
.05Vm (bC + Ft ) ⎝ p g ⎠
Fd = Ft + (12-15)
.05 Vm + (bC + Ft )1/ 2
.05 (1355) (1 • 830 + 73) Fw = tooth wear strength (lb)
Fd = 73 + 1/ 2
.05 (1355) + (1 • 830 + 73)
Fd = 699 lb
Dp = diametral pitch of pinion (in.)
Fs Dg = diametral pitch of gear (in.)
≥ Fd
N sf
1000 b = face width (in.)
≥ 699
1.4
714 > 699 Kg = load stress factor (Appendix 13)

• This meets the criteria.

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 Example Problem 12-8: Wear of Gears Example Problem 12-8: Wear of Gears (cont’d.)
• In prior example problem, verify the surface is suitable for wear considerations. – Substituting into equation 12-16:
For wear use Nss = 1.2
Fw = 1.5 (1) 1.27 (270)
− Wear formula:
(12-16) Fw = 514
Fw = Dp b Q Kg

• This would not be suitable. Try if surfaces each had a BHN = 450.
− Find Q : K g = 470
(from Appendix 13)
(12-17)
Fw = 1.5 (1) (1.27) ( 470)
2 Ng
Q =
Ng + Np
Fw = 895
Fw
2 (42) ≥ Fd
Q =
42 + 24 N sf
895
Q = 1.27 > 699
1 .2
Kg = 270
(from Appendix 13)
746 > 699
• This would now be acceptable if the gear teeth were hardened to a BHN of 450.
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