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Chapter 10
Gears
Transmits
Rotary motionTorques
Why use teeth? Why not frictionrollers?
Teeth need to be specially shaped toallow smooth engagement.
Involute curves are ideal.
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Types of Gears:
Spur Gear Rack and Pinion
Internal Gear Helical Gear
Bevel GearHerrin bone Gear
Miter Gear Worm Gear
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Spur Gear Geometry:
Involute toothUnwind a string from a base circleTrack the path of the string end
Pitch Diameter (d)Size of equivalent friction rollers
Pitch circle
Pitch circle
Line of Centers
Base Circle
Involutetooth profile
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Number of Teeth (N)Must be an integer value
Diametral Pitch (Pd)size of a the gear tooth
d
NPd =
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Pressure Angle ()
Standard values: = 140, 200, 250
Mating gears must have same
pressure angle and diamtral pitch.
Pitch Circle
Line of Centers
Pitch Line
Pressure Angle,
Pressure LinePitch Circle
Base Circle
Base Circle
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Dedendum Circle
Addendum C
Dedendum (b)
ddendum (a)
Face Width (F)
Circular Pitch (p)
Tooth Thickness
Pitch Circle
Flank
Face Fillet Radius
Tip Radius
Pitch Diameter (d)
Others Features
Circular Pitch p = d/NBase circle db=d cos These features are standardized for
interchangeability:
Addendum a=1/PdDedendum b=1.25/PdFace Width F=12/Pd
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Example:A 200, full-depth, involute spur gear
with 18 teeth has a diametral pitch of12. Determine:
Outside (addendum) diameter. Root (dedendum) diameter. Standard face width. Base circle diameter. Circular pitch.
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Line of Centers
Pitch Line
Pressure Angle,
Contact Line
Driven Gear
Driver Gear
(pinion)
123 BaseCircle
Base
Circle
Pitch
Circle
Pitch
Circle
Addendum
Circle
AddendumCircle
Center Distance, c
Mating Spur Gears :
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Center Distance (c)
cexternal gears=r1 + r2
cinternal gears=r2 r1
Example:
Two mating external 4-pitch, spurgears have 14 and 42 teeth. Determinethe center distance.
Pitch circle
Pitch circle
Line of Centers
cexternal
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Contact Ratio (mp) Average number of teeth in
contact at any instant.
b
pp
Zm =
Base pitch (pb):
2
2
1
1 coscos
N
d
N
dpb
==
Length of contact path (Z):
sin)cos()(
sin)cos()(
1
2
1
2
11
2
2
2
2
22
rrar
rrarZ
++
+=
Larger values have smootherload transfer
Recommend 1.4 1.5Example:
Two mating external 4-pitch, spurgears have 18 and 42 teeth. Determine
the contact ratio.
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Interference Gears with too few teeth. Top of one gear digs into
base of the other.
Use table 10.4 to check. = 140 = 200 =250
Number ofPinion teeth
Maximumnumber of
gear teeth
Number ofpinion teeth
Maximumnumber of
gear teeth
Number ofpinion teeth
Maximumnumber of
gear teeth
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Gear Kinematics:
Velocity Ratio (VR)Gear Ratio
2
1
=VR
+ same direction- opposite direction
3:1 or three to one means VR = 3
Commonly written :
1
2
1
2
2
1
N
N
d
dVR ===
2
1
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Example:
A 5-in gear, rotating at 1725 rpm, cw,
meshes with a gear having a pitchdiameter of 20 in. Determine the
velocity of the driven gear.
An 8-pitch, 18 tooth gear rotates at1200 rpm, ccw. It meshes with a gear
with a pitch diameter of 6 in.
Determine the velocity of the drivengear.
A 10-pitch, 16 tooth gear mates with a
gear having a pitch diameter of 6 in.
The driven gear rotates at 300 rpm, cw.Determine the velocity of the driver
gear.
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Pitch Line Velocity (vt)
vt= r11 = r22
watch units:
vt= d n/12
d (in)n (rpm)
vt (fpm)
The pitch line speed determines:
lubrication needed. quality required.
(accuracy & surface finish)
21
1
2
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Gear SelectionNeed to decide:
P Pressure AngleP Suitable diametral pitchP Number of teeth on each gear
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Number of Teeth for
Commercially Available, Stock Gears
32 Diametral Pitch12 16 20 28 36 48 64 80 112
14 18 24 32 40 56 72 96 128
24 Diametral Pitch12 18 24 30 42 54 72 96 144
15 21 27 36 48 60 84 120
20 Diametral Pitch12 16 24 35 50 80 100 160
14 18 25 40 60 84 120 180
15 20 30 45 70 90 140 20016 Diametral Pitch
12 16 24 32 48 64 96 16014 18 28 36 56 72 128 192
15 20 30 40 60 80 144
12 Diametral Pitch12 15 20 28 42 60 84 120 168
13 16 21 30 48 66 96 132 192
14 18 24 36 54 72 108 144 216
10 Diametral Pitch
12 16 24 30 45 55 80 120 20014 18 25 35 48 60 90 140
15 20 28 40 50 70 100 160
8 Diametral Pitch12 16 22 32 44 60 80 11214 18 24 36 48 64 88 120
15 20 28 40 56 72 96 128
6 Diametral Pitch12 16 24 33 48 66 96
14 18 27 36 54 72 10815 21 30 42 60 84 120
5 Diametral Pitch12 16 24 30 45 70 110 160
14 18 25 35 50 80 120 180
15 20 28 40 60 100 140
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Example:A set of gears must be selected that
transmit power from a 5 hp motor at1200 rpm to a grinding wheel at 310
(10) rpm. Select a set of appropriategears.
Example:A set of gears must be selected that
transmit power from a 20 hp engine at
2000 rpm to an air compressor at 660( 10) rpm. These gears must fit into a
housing with a center distance of 6.00in. Select a set of appropriate gears.
Example
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An escalator is being designed to
transport 6000 people per hour.
Assume an average weight of 175 lbper person. Determine the required
power to operate the escalator.
15 ft
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Gear Trains:
Several gear pairs are placed in
series.
Why?
Train Value (TV)out
in
TV
=
TV = (VR)1 (VR)2 (VR)3
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ExampleThe gear train shown is used with an
input speed of 1200 rpm, cw. Giventhe following properties:
N1=24, N2= 48 & Pd = 16
N3=24, N4= 48 & Pd=12
N5=15, N6= 35 & Pd = 10
Determine the output velocity and
center distance of the gear train.
in
1
2
3
4
5
6
out
1 2
3 4
5 6
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ExampleThe gear train shown is used with an
input speed of 1200 rpm, cw. Giventhe following properties:
N1=24 N2= 36 & Pd = 12
D3=2.0 in. & Pd=10 N4= 40
N5=16 & Pd=8 D6= 6.0 inDetermine the output velocity and
center distance of the gear train.
in
out
1
2
3
45
6
4
1
2
63
5
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Gear Train Design Select the appropriate ratios for
each gear pair.
Factor the TV into suitableratios.
Example:Design a gear train with a train value
of +400:1. From interference criteria,no gear should have fewer than 15
teeth and due to size restrictions, no
gear should have more than 75 teeth.
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Worm Gears:
Able to obtain high ratios inlimited space.
Shafts are perpendicular andnon-intersecting
Self-locking 40-50% efficient
Worm Gear Kinematics:
worm
gear
N
NVR ==
2
1
2
1
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Example:Determine the output speed of the gear
drive, when the worm is driven at 1200rpm.
in=1200 rpm1
2
34
5
N2= 18
N3=60
N4=15
N5=45out = ?