49958217-Lecture-11-Gear

ENT 253 MECHANICAL DESIGN

Lecture 111/03/2010

GearDr. HAFTIRMAN

MECHANICAL ENGINEERING PROGRAMSCHOOL OF MECHATRONIC ENGINEERING

UniMAP

DR. HAFTIRMAN School of Mechatronic UniMAP

ENT253 MECHANICAL DESIGN I SEM2-2009/2010

1

CO3:CO3:Ability to design and analyzey g ymechanical components such as gears clutches shaft bearing gears, clutches, shaft, bearing, screw and spring.



2

OUTLINEOUTLINE

IntroductionIntroductionType of gearsN l tNomenclatureFundamentalGear ratioInterferenceInterferenceForce analysis



3

INTRODUCTIONINTRODUCTION

• A gear is a component within a transmission deviceg pthat transmits rotational force to another gear ordevice.

• A gear is different from a pulley in that a gear is aA gear is different from a pulley in that a gear is around wheel that has linkages ("teeth" or "cogs") thatmesh with other gear teeth, allowing force to be fullytransferred without slippagetransferred without slippage.

• Depending on their construction and arrangement,geared devices can transmit forces at differentspeeds torques or in a different direction from thespeeds, torques, or in a different direction, from thepower source.


4ENT253 MECHANICAL DESIGN I SEM2-2009/2010

INTRODUCTIONINTRODUCTION

• The most common situation is for a gear to mesh withganother gear, but a gear can mesh with any devicehaving compatible teeth, such as linear moving racks.

• The gear's most important feature is that gears ofThe gear s most important feature is that gears ofunequal sizes (diameters) can be combined to producea mechanical advantage, so that the rotational speedand torque of the second gear are different fromand torque of the second gear are different fromthose of the first. In the context of a particularmachine, the term "gear" also refers to one particulararrangement of gears among other arrangementsarrangement of gears among other arrangements(such as "first gear"). Such arrangements are oftengiven as a ratio, using the number of teeth or geardiameter as unitsdiameter as units.



GearGear

►►Types of GearsTypes of Gears►►Types of GearsTypes of Gears1.1. Spur gearSpur gear22 Helical GearHelical Gear2.2. Helical GearHelical Gear3.3. Bevel GearBevel Gear44 Worm GearWorm Gear4.4. Worm GearWorm Gear



Spur GearSpur Gear

Spur gears have teethSpur gears have teethSpur gears have teeth Spur gears have teeth parallel to the axis of parallel to the axis of rotation and are to rotation and are to transmit motion from transmit motion from one shaft to another one shaft to another

ll l h ftll l h ftparallel shaft.parallel shaft.



Spur GearSpur Gear



Spur gearSpur gear

• Pair of spur gears.PINION(DRIVER) Pair of spur gears.

• The pinion drives the gear.g

GEAR(DRIVEN)



Helical gearHelical gear

Helical gears haveHelical gears have teeth inclined to the axis of rotation.



Bevel gearBevel gear

► Bevel gears have teeth gformed on conical surfaces and are used mostly for transmittingmostly for transmitting motion between intersecting shaft.



Worm gearWorm gear

• The direction ofThe direction of rotation of the worm gear, also called the worm wheel, depends upon the direction of

t ti f throtation of the worm and upon whether the worm teeth are cutworm teeth are cut right-hand or left-hand.



Rack and pinionRack and pinion• A rack is a toothed bar or rod that

can be thought of as a sector gear ca be t oug t o as a secto geawith an infinitely large radius of curvature. Torque can be converted to linear force by meshing a rack with a pinion: the pinion turns; the rack moves in apinion turns; the rack moves in a straight line.

• Racks also feature in the theory of gear geometry, where, for instance the tooth shape of aninstance, the tooth shape of an interchangeable set of gears may be specified for the rack (infinite radius), and the tooth shapes for gears of particular actual radii then g pderived from that.



NomenclatureNomenclature




• Pitch circle is a theoreticalcircle upon which allcircle upon which allcalculations are usually based;its diameter is the pitchdiameter.A i i i th ll f t• A pinion is the smaller of twomating gears.

• The larger is often called gear.• The circular pitch (p) is theThe circular pitch (p) is the

distance, measured on thepitch circle, from a point onone tooth to a correspondingpoint on an adjacent toothpoint on an adjacent tooth.

• The circular pitch is equal tothe sum of the tooth thicknessand the width of space.



NomenclatureNomenclature• The module (m) is the

ratio of the pitch diameterratio of the pitch diameter to the number of teeth.

• The addendum (a) is the radial distance betweenradial distance between the top land and the pitch circle.

• The dedendum (b) is the ( )radial distance from bottom land and the dedendum.

• The whole depth (ht) is the sum of the addendum and the dedendum.



bDdiamterrootDaDdiamteroutsideD

R

o

22

−==+==

2GP dd

C+

=R 2

aaadepthworkinghchbadepthwholeh

k

kt

2=+==+=+==

abcclearancec −=== Ppthicknesstootht

22π

===DR. HAFTIRMAN School of Mechatronic UniMAP



• The clearance circle is a i l h icircular that is tangent to

addendum circle of mating gear.

• The clearance ( c ) is gear• The clearance ( c ) is gear exceeding the sum of addendum and the dedendum.

• The backlash is the amount by ywhich the width of a tooth space exceeds the thickness of the engaging tooth measured on the pitch circlesmeasured on the pitch circles.



FundamentalFundamental

• P = diametral pitch, teeth per inch

• m = module, mm.• d = pitch diameter, mmp ,• N= number of teeth.• p =circular pitch

dNdm =

dNP =

mdp ππ==π=pP in

Pm 1=

mN

p π==πpPmm

Pm 4.25=



Standard modulesStandard modules



Tooth systemsTooth systems

• A tooth system is a standard that specifies the relationshipsA tooth system is a standard that specifies the relationships involving addendum, dedendum, working depth, tooth thickness and pressure angle.



Diametral pitchDiametral pitch



Standard tooth for helical gearsStandard tooth for helical gears



ProblemProblem

Problem 13-1 Problem 13-2A 17-tooth spur pinion has a

diametral pitch of 8 teeth/in, runs at 1120 rev/min, and

A 15-tooth spur pinion has a module of 3mm and runs at a speed of 1600 rev/min. The

drives a gear at a speed of 544 rev/min.

Find the number of teeth on the

driven gear has 60 teeth. Find the speed of the driven gear, the circular pitch, and the th ti l t t tgear and the theoretical

center-to-center distance.theoretical center-to-center distance.

Solutionin

Nd p 125217

===

hPdN

inindnnd

inP

d

PG

P

35)3754(8

375.4)125.2(544

1120

125.28

3

2 ===

===

mmmp

revnG

3

min/40060151600

==

=⎟⎠⎞

⎜⎝⎛=

ππ

inC

teethPdN GG

25.32

)375.4125.2(35)375.4(8

=+

=

===[ ] mmC

p

5.1122

)6015(3=

+=DR. HAFTIRMAN

School of Mechatronic UniMAP



• The pitch-line velocity ωω rrV ==p y2211 ωω rrV ==

1

2

2

1

rr

=ωω



ProblemProblem

13-3.A spur gearset has a module of 4mm and a velocity ratio of 2.80. The pinion has 20 teeth.Find the number of teeth on the driven gear, the pitch diameter, and the theoretical center-to-center distance.SolutionSolution

mmmNdteethNN

GG

PG

224)4)(56(56)80.2(20)80.2(

======

mmdd

C

mmmNd

GP

PP

1522248080)4)(20(

++===

mmC 15222

===DR. HAFTIRMAN School of Mechatronic UniMAP



• Suppose we specify that an 18-tooth pinion is to mesh with a 30-tooth gear and that the diametral pitch of the gear set i t b 2 t th i his to be 2 teeth per inch.

• The pitch diameters of the pinion and gear are

N 18

Ninr

inPN

d

305.4

92

18

1

11

=

===

• The center distance is the suminr

inPN

d

5.7

15230

2

22

=

==

• The center distance is the sum of the pitch radii, inininrr 125.75.421 =+=+DR. HAFTIRMAN

School of Mechatronic UniMAP



• The construct the pitch circles f dii d Thof radii r1 and r2. These are

tangent at P, the pitch point.• Next draw line ab, common

tangent through the pitchtangent, through the pitch point.

• We designate gear 1 as the driver (CCW).( )

• We draw a line cd through point P at an angle Ø to the common tangent ab.

• The line cd has three names; namely the pressure line, the generating line, and the line of actionaction.



Relating base circle to the pressure angle ØRelating base circle to the pressure angle Ø

φcosrr = φcosrrb =DR. HAFTIRMAN School of Mechatronic UniMAP


Pressure angle ØPressure angle Ø



Example 13-1Example 13 1

• A gearset consist of a 16-tooth pinion driving a 40 tooth gear

• SolutionTh i l i h inp 57.1===

ππpinion driving a 40-tooth gear. The diametral pitch (P) is 2, and the addendum and dedendum are 1/P and 1.25/P, respectively The gears are cut

The circular pitchThe pitch diameters of the pinion (dp )and gear (dG) are, respectively,

inP

p 57.12

respectively. The gears are cut using a pressure angle of Ø=20°.

• Compute the circular pitch, the t di t d th dii

p y,

N

inPNdP

40

82

16===

center distance, and the radii of the base circles.

• In mounting these gears, the center distance was incorrectly The center distance

inPNdG 20

240

===

ymade ¼ in larger.

• Compute the new values of the pressure angle and the pitch-circle diameters

indd

c GP 142208

2=

+=

+=

circle diameters.




• Pressure angle 20°, the base- • The velocity ratio does not circle radii is change,

rrB

63208)(

cos

°

= φ dd GG 25.142

4.0 ''

=+

ingearr

inpinionr

B

B

40.920cos220)(

76.320cos28)(

=°=

=°=

ind

indd

P

GG

143.8)357.20(4.0

357.205.284.12

'

''

==

==>=

• Designating d’p and d’

G as the new pitch-circle diameters, the ¼ in increase in the center

• The new pressure angle Ø

2

¼ in increase in the center distance requires that

idd GP 2514

'' + °===

=

−− 56.222/1438

76.3cos2/

)(cos

cos

1'

1' b

b

dpinionr

rr

φ

φ

inGP 25.142

= 2/143.82/Pd



Generation of an involute curveGeneration of an involute curve



Construction of an involuteConstruction of an involute



InvoluteInvolute

• The involute gear profile g pis the most commonly used system for gearingtoday. In an involute gear, y gthe profiles of the teeth are involutes of a circle.

• The involute of a circle isThe involute of a circle is the spiraling curve traced by the end of an imaginary taut stringimaginary taut string unwinding itself from that stationary circle.



InvoluteInvolute

• Equations of an involute of a parametrical define curve are:

22t

∫22

22

''

'''],[

yx

dtyxtxxyxX a

+

+−=

∫

2'2'' dtyxy

yxt

+

+

∫22 ''

],[yx

yyxY a

+−=



Tooth contact nomenclatureTooth contact nomenclature



Point of contact

• Path of Action, ANSI/AGMA 1012-G05

• A point of contact is pany point at which two tooth profiles touch

h heach other.



Plane of action

• Surface of action• The surface of action is the

imaginary surface in which contact occurs between two engaging tooth surfaces. It isengaging tooth surfaces. It is the summation of the paths of action in all sections of the engaging teeth.

• Plane of Action, ANSI/AGMA 1012-G05

• The plane of action is the f f fsurface of action for involute,

parallel axis gears with either spur or helical teeth. It is tangent to the base cylinders.



Line of contactLine of contact

• Line of Contact, ANSI/AGMA 1012-G05

• A line of contact is a line or curve along which two tooth

fsurfaces are tangent to each other.



Zone of action (contact zone)

• Zone of Action, ,ANSI/AGMA 1012-G05

• Zone of action (contact ) f i l tzone) for involute,

parallel-axis gears with either spur or helical teeth, is the rectangular area in the plane of action bounded by the length ofbounded by the length of action and the effective face width.



Path of contactPath of contact

Li f C t t (h li l• Lines of Contact (helical gear), ANSI/AGMA 1012-G05Th th f t t i th• The path of contact is the curve on either tooth surface along which theoretical single pointtheoretical single point contact occurs during the engagement of gears with crowned tooth surfaces or gears that normally engage with only single point contact.



Length of actionLength of action

• Length of Action,Length of Action, ANSI/AGMA 1012-G05

• Length of action is the distance on the line of action through which the point of contact moves during themoves during the action of the tooth profileprofile.



Involute-toothed pinion and rackInvolute toothed pinion and rack

φcosb pp = φcoscb pp

pitchbasethepb =



Internal gear and oinionInternal gear and oinion



Contact ratio (mc)Contact ratio (mc)

pq

m tc =

φcospL

m abc = φp

)( tt qactionofarctheq =DR. HAFTIRMAN School of Mechatronic UniMAP


Problem 13-4Problem 13 4

A 21-tooth spur pinion mates with a 28-tooth gear. The diametral pitch is 3 teeth/in and the pressure angle is 20° M k d i f th h i t th20°. Make a drawing of the gears showing one tooth on each gear.Find the addendum (a), dedendum (b), clearance (c),Find the addendum (a), dedendum (b), clearance (c), circular pitch (p), tooth thickness (t), and base-circle diameters; the lengths of the arc approach, recess, and action; and the base pitch and contact ratioaction; and the base pitch and contact ratio.



ProblemProblem

11

inP

b

inP

a

4167.0325.125.1

333.0311

===

===

770.820cos333.9

333.9328

2

22

=°=

===

b ind

inPN

d

inp

inabcP

0471

0834.03333.04167.03

===

=−=−=ππ 984.020cos

3cos

2

=°⎟⎠⎞

⎜⎝⎛== cb

b

inpp

pitchBaseπφ

inpt

inP

p

523.02047.1

2

047.13

===

===

55153.1

3 ⎠⎝

ab

cb

Lm

ratioContact

inPNd

Pinion

73211

1 ===

55.1984.0

===b

abc P

m

inddP

b 578.620cos720cos3

11 =°=°=DR. HAFTIRMAN School of Mechatronic UniMAP


Gear ratioGear ratio

• The gear ratio is the relationship between the

• This means that for every one l i f h i i hrelationship between the

number of teeth on two gearsthat are meshed or two sprockets connected with a common roller chain or the

revolution of the pinion, the gear has made 1/1.62, or 0.62, revolutions. In practical terms, the gear turns more slowly.common roller chain, or the

circumferences of two pulleysconnected with a drive belt.

• In the picture to the right, the ll (k th

the gear turns more slowly. Suppose the largest gear in the picture has 42 teeth, the gear ratio between the second

d thi d i th 21/42smaller gear (known as the pinion) has 13 teeth, while the second, larger gear (known as the idler gear) has 21 teeth. Th ti i th f

and third gear is thus 21/42 = 1/2, and for every revolution of the smallest gear the largest gear has only turned 0.62/2 =

The gear ratio is therefore 13/21 or 1/1.62 (also written as 1:1.62).

g y0.31 revolution, a total reduction of about 1:3.23.



Gear ratioGear ratio

1st gear 2 97 1• 1st gear 2.97:1• 2nd gear 2.07:1

3 d 1 43 1• 3rd gear 1.43:1• 4th gear 1.00:1• 5th gear 0.84:1• 6th gear 0.56:1

reverse 3.28:1



InterferenceInterference

• The contact pf portions of tooth profiles that are not conjugate is called interference.

• NP= the smallest number of teeth on the pinion without interference.

• k=1 for full depth teeth.• k= 0.8 for stub teeth• Ø = pressure angleØ pressure angle

( )φφ

22 sin311

sin32

++=kN P φsin3




• For pressure angle a 20°, with • For m=4, Ø=20°k=1

( )NP 20sin31120sin3)1(2 2

2 ++=

( )teethN

N

P

P

164.15

20sin)4(21(4420sin)4(21(

)1(2 222

==

°+++°+

=

• If the mating gear has more

• Thus a 16-tooth pinion will mesh with a 64-tooth gear without interference.

• The largest gear with a specified

( )teeth133.1220sin3

==

teethN P 164.15

• If the mating gear has more teeth than the pinion, that is, mG=NG/NP=m more than one, then NP is

• The largest gear with a specified pinion that is interference-free is

φ 222 4sin kNthen NP is

( )φφ

222 sin)21(

sin)21(2 mmmm

kN P ++++

= φφ

2sin244sin

P

PG Nk

kNN

−−

=




• For a 13-tooth pinion with a pressure angle Ø of 20°

teethNG 1645.1620sin)13(2)1(4)1(420sin13

2

222

==°−

−°=

• For a 13-tooth spur pinion, the maximum number of gear teeth. possible without interference is 16.

• The smallest spur pinion that will a rack without interference is• The smallest spur pinion that will a rack without interference is

φ2sin)(2 kN P =

• For a 20° pressure angle full-depth tooth the smallest number of pinion teeth to mesh with a rack is

teethN P 181.1720sin)1(2

2 ==°

=20sin



Straight bevel gearStraight bevel gear

• The pitch angles are defined by the pitch cones meeting at the apex. They are related to the tooth numbers as follows;

G

P

NN

=tan γ

• The virtual number of teeth (N’)P

G

NN

=Γtan

• The virtual number of teeth (N ) and p is the circular pitch.

rbπ2pr

N bπ2'=




• The normal circular pitch pn and is related to the transverse circular pitch pt as follows:

ψcostpp =

• The axial pitch px;

ψcostn pp

p

Th l di t l it h

ψtant

xp

p =

• The normal diametral pitch

tn

PP = nφψ

tancos =

ψcosnPtφ

ψtan

cos




• Figure shows a cylinder cut by an oblique plane ab at an angle to right section.ψ

==>°=DR0ψ

∞==>°=

==>=

R

R

902

0

ψ

ψ

• The virtual number of teeth is related to the actual number:

ψ3cos' NN =



Gear trainGear train

• A gear train is a set or system of gears arranged to transfer rotational torque from one part of a mechanicalsystem to another.system to another.

• Gear trains consists of:• Driving gears - attached to the input shaft • Driven gears/Motor gears - attached to the output shaft • Idler gears - interposed between the driving and driven

gear in order to maintain the direction of the output shaftgear in order to maintain the direction of the output shaft the same as the input shaft or to increase the distance between the drive and driven gears. A compound gear train refers to two or more gears used to transmit motion.train refers to two or more gears used to transmit motion.



Helical gearsHelical gears




• The contact pf portions of tooth profiles that are not conjugate is called interference.

• NP= the smallest number of teeth on the pinion without interference.

• k=1 for full depth teeth.• k= 0.8 for stub teeth• Ø = pressure angleØ pressure angle

( )φφ

22 sin311

sin32

++=kN P



59

φsin3


• For pressure angle a 20°, with • For m=4, Ø=20°k=1

( )NP 20sin31120sin3)1(2 2

2 ++=

( )teethN

N

P

P

164.15

20sin)4(21(4420sin)4(21(

)1(2 222

==

°+++°+

=

• If the mating gear has more

• Thus a 16-tooth pinion will mesh with a 64-tooth gear without interference.

• The largest gear with a specified

( )teeth133.1220sin3

==

teethN P 164.15

• If the mating gear has more teeth than the pinion, that is, mG=NG/NP=m more than one, then NP is

• The largest gear with a specified pinion that is interference-free is

φ 222 4sin kNthen NP is

( )φφ

222 sin)21(

sin)21(2 mmmm

kN P ++++

= φφ

2sin244sin

P

PG Nk

kNN

−−

=



60

Gear-tooth sizeGear tooth size



61


• For a 13-tooth pinion with a pressure angle Ø of 20°

teethNG 1645.1620sin)13(2)1(4)1(420sin13

2

222

==°−

−°=

• For a 13-tooth spur pinion, the maximum number of gear teeth. possible without interference is 16.

• The smallest spur pinion that will a rack without interference is• The smallest spur pinion that will a rack without interference is

φ2sin)(2 kN P =

• For a 20° pressure angle full-depth tooth the smallest number of pinion teeth to mesh with a rack is

teethN P 181.1720sin)1(2

2 ==°

=



62

20sin

Straight bevel gearStraight bevel gear

• The pitch angles are defined by the pitch cones meeting at the apex. They are related to the tooth numbers as follows;

G

P

NN

=tan γ

• The virtual number of teeth (N’)P

G

NN

=Γtan

• The virtual number of teeth (N ) and p is the circular pitch.

rbπ2DR. HAFTIRMAN School of Mechatronic UniMAP


63pr

N bπ2'=


• The normal circular pitch pn and is related to the transverse circular pitch pt as follows:

ψcostpp =ψ

nφ

p

• The axial pitch px;

ψcostn pp

p

ψnp

Th l di t l it h

ψtant

xp

p =

tφtp

xp

• The normal diametral pitch

tn

PP = nφψ

tancos =

tp



64

ψcosnPtφ

ψtan

cos


• Figure shows a cylinder cut by an oblique plane ab at an angle to right section.ψ

==>°=DR0ψ

∞==>°=

==>=

R

R

902

0

ψ

ψ

• The virtual number of teeth is related to the actual number:

ψ3cos' NN =



65


• The pressure angle Øt mN

m G

( )k

nt

cos2costan

tan1⎟⎟⎠

⎞⎜⎜⎝

⎛= −

ψψφ

φ

sin2cos4cos4sin

2

2222

NkkN

N

mN

m

tP

tPP

PG

−−

=

==

φψψφ

( )kN

t

tt

P

20t30,20

sin311sin3cos2 2

2

⎞⎛ °

°=°=

++=

ψφ

φφψ

1202.128022sin)9(230cos)1(430cos)1(480.22sin9

80.22,30,20sin2cos4

2222

N

Nk

G

tn

tP

==°°°

−=

°=°=°=°

φψφφψ

( ) teethNP

t

948.880.22sin3118022i330cos)1(2

80.2230cos20tantan

22

1

==°++°°

=

°=⎟⎠⎞

⎜⎝⎛

°°

= −φ

80223020sin

cos280.22sin)9(230cos)1(4

2

kNt

P

°°°

=

−

φφφψ

( )P 80.22sin3 2 °

.125.1180.22sin30cos)1(2

80.22,30,20

2 teethNP

tn

==°°

=

°=°=°= φψφ



66

Worm gearsWorm gears

• dG= the pitch diameter• NG= the number of teeth• pt= the transverse circular

pitch.p• px= the axial pitch• λ= the lead angle• Ψ = the helix angle• Ψ = the helix angle

pNπ

tGG

pNd =



67


• The pitch diameter of the worm gear should be selected so as to fall into the range

875.0875.0 CC

• C= the center distance

7.10.3CdC

w ≤≤

C the center distance.• L= the lead.• λ= the lead angle

wx

LNpL

λ

=

twdπ

λ =tan



68




69


A gear train is a set or system of gears arranged to transferA gear train is a set or system of gears arranged to transfer rotational torque from one part of a mechanical system to another.

Gear trains consists of:Gear trains consists of:

Driving gears - attached to the input shaft.Driven gears/Motor gears - attached to the output shaft. Idler gears - interposed between the driving and driven gear in order to maintain the direction of the output shaft the same as the input shaft or to increase the distance between the drive and d idriven gears.

A compound gear train refers to two or more gears used to transmit motion.



70




71


⎟⎟⎞

⎜⎜⎛−⎟⎟

⎞⎜⎜⎛−⎟⎟

⎞⎜⎜⎛−= 4315 NNNn

The speed ratio between gear 5 and 1

The minus signs indicate that the pinionand gear rotate in opposite directions⎟⎠

⎜⎝⎟⎠

⎜⎝⎟⎠

⎜⎝ 5421 NNNn and gear rotate in opposite directions



72


• Consider a pinion 2 driving a gear 3.

• The speed of the driven gear is

dN

d= pitch diameter

23

22

3

23 n

dd

nNN

n ==

d= pitch diameter• Gears 2, 3, and 5 are drivers.• Gears 3, 4, and 6 are driven

bNNN members.

• The speed of gear 6 is2

6

5

4

3

3

26 n

NN

NN

NNn −=



73

643

Gear trainGear train• The train value e is • As a rough guideline, a train

value of up to 10 to 1 can be obtained with one pair of gears. A two-stage compound

t i bt i t i

numberstoothdrivenofproductnumberstoothdrivingofproducte =

• For spur gears, e is positive if the last gear rotates in the same sense as the first, and the negative if the last rotates

gear train can obtain a train value of up to 100 to 1.

the negative if the last rotates in the opposite sense.

FL enn =• nL=the speed of the last gear in

the train, nF=the speed of the first.

FL



74

Example 13-4( )N P 20sin)4(21(44

20sin)4(21()1(2 22

2 °+++°+

=

Example 13 4 teethN P 164.15 ==

In a two-stage compound gear train,assign the square root of the overall train value to each stage



75

Example 13-4( )

t thN

N P

16415

20sin)4(21(4420sin)4(21(

)1(2 222 °+++

°+=

Example 13 4 teethN P 164.15 ==



76

A compound reverted gear trainA compound reverted gear train

• This requires the distances between the shafts to be the same for both stages on the train. The distance constraint is

22225432 dddd

+=+

• The diametral pitch relates the diameters and the numbers of teeth, N,

dNP =



77

A compound reverted gear trainA compound reverted gear train

• Replacing all the diameters gives

)2()2()2()2(5432

PN

PN

PN

PN

+=+

• Assuming a constant diametral pitch in both stages and the

)2()2()2()2( PPPP

pitch in both stages, and the geometry condition stated in terms of numbers of teeth:

NNNN 5432 NNNN +=+



78

A planetary gear trainA planetary gear train



79


• Planetary trains always include a sun gear, a planet carrier or arm, and one or more planet gears.

• A planetary train composed af s sun gear 2, an arm or carrier 3, and planet gears 4 and 5. Th l l it f 2S 2 The angular velocity of gear 2 relative to the arm in rev/min is

Sun gear 2

Arm 3

3223 nnn −=

• The velocity of gear 5 relative to the arm is

Planet gears 4 & 5

3223

nnnDR. HAFTIRMAN School of Mechatronic UniMAP


80

Planet gears 4 & 53553 nnn −=


• This equation expresses the ratio of gear 5 to that of gear 2, and both velocities are taken relative to the arm.

• The train value isS 2

32

35

23

53

nnnn

nn

−−

=

• or

Sun gear 2

Arm 3

32

35

nnnn

e−−

= AL

nnnne

−−

=

nF= rev/min of first gearnL= rev/min of last gea

Planet gears 4 & 5

32 nn AF nn



81

Planet gears 4 & 5

Force analysis (Spur gearing)Force analysis (Spur gearing)

• Designate the shafts using letters of the alphabet, a, b, c, etc.

• Figure shows a pinion mounted on shaft a rotating clockwise at n2 rev/min and driving a gear on shaft b at n3

/ irev/min.



82

Force analysis (Spur gearing)Force analysis (Spur gearing)

F32 Force

r = the radial direction.t = the tangential direction.

Th f t d b 2 i t 3

Ta2 Torque

The force exerted by gear 2 against gear 3 as F23. F32 is the force exerted by gear 3against the pinion.

The force of gear 2 against a shaft aor the force of a shaft a against gear 2 is F2a.



83

Fa2 Force

The transmitted loadThe transmitted load

tt FW 32=

The transmitted loadpowertransmitnotdoesF r

32

t FW 32

ωω ⎟⎞

⎜⎛==

dWTH t

22d ωω ⎟

⎠⎜⎝ 2

H = The power transmitted

2

kNdn

HWt π60000

=

ta WdTT22 ==

d= d2 dnV π=



84

2


• Pinion 2 runs at 1750 The pitch diameter of gears 2 and rev/min and transmits 2.5kW to idler gear 3. Theteeth are cut on the 20°

3 aremmmNd

mmmNd125)5.2(5050)5.2(20

33

22

======

teeth are cut on the 20full-depth system andhave a module of m = 2.5

f

The transmitted load

HFW t 60000==mm. Draw a free body

diagram of gear 3 andshow all the forces that

ndFWt

)5.2(600002

23

=

==π

act upon it.• Solution kN546.0

)1750)(50(=π



85

Example 13-7WFF t

tt2343 ==

Example 13 7 kN546.0=kNFF rr 199.02343 ==

kNFF 581.02343 ==

H60000

Idler kNFF 581.02343

kN

ndHFW t

t

546.0))(()5.2(60000

60000

223

==

==π

)1750)(50(π

kWH 5.2=

Pinion

kWH 5.2

kNFF tr

199020tan)546.0(20tan2323 °=°= FF

t23

23 =



86

kN199.0=

kN

F

581.020cos

546.020cos23

=°

=

°

Example 13-7 FFF rtxb )( 43233 =+−=Example 13 7

kN347.0)199.0546.0(

=+−−

FFF tryb )( 43233 =+−=

kN

FFFb

347.0)546.0199.0()( 43233

=−−−

=+=

kNFb

491.0)347.0()347.0( 22

3

=

+=



87

Force analysis-bevel gearingForce analysis bevel gearing

TW =av

t rW =

φ W avr

φtanW

φ av

φtantW

γφ costantr WW =

γφ sintantWW = γ



88

γφ sintanta WW

Force analysis-helical gearingForce analysis helical gearing

ψφ coscos n

tWW =

nr WW φsin=

ψφ sincosWW =ttWW φtan= ψφ sincos na WW =ttr WW φtan

ψtanta WW =

ψφ coscos ntW =



89

Force analysis-worm gearingForce analysis worm gearing

)sincos(cos λλφ fWW z +=

ny WW φsin=

λφ coscos nz WW =

)sincos(cos λλφ fWW n +=

λφ sincos nx WW =

nφλφ sincos nWW

)cossin(cos λλφ fWW nx +=



90

Velocity components in worm gearingVelocity components in worm gearing

λcosW

sV

V =



91

The Lewis bending equationsThe Lewis bending equations

2

23 62 lWFtFtIM t

====>=σ 26)12(/ FttccI>σ

634141116===>=== xxWWlW ttt

σ 62

43/2

4

64

46/ 222 ===>=== xx

ltFltFFt

σ

lt 2/

ltx

tl

xt

4

2/2/

2

=

=



92

l4

The Lewis bending equationsThe Lewis bending equations

pxy

xpF

pW t

==>⎟⎠⎞

⎜⎝⎛

=σ32

32

FpypW t

=σ

, yYp

P

t

== ππ

32xPY

FYPW t

==>=σ



93

Values of the Lewis form factor YValues of the Lewis form factor Y



94

Dynamic effectsDynamic effects

• Kv= the velocity factor • SI units• V = the pitch-line velocity in

ft/min

600 V+ )(1.6

)),(05.3

05.3

profilemildorcutVK

profilecastironcastVKv

+=

+=

)(1200

1200

)),(600

600

profilemildorcutVK

profilecastironcastVK

v

v

+=

+=

)(56.3

56.3

)(1.6

profileshapedorhobbedVK

profilemildorcutK

v

v

+=

=

)(50

501200

profileshapedorhobbedVKv

v

+=

)(56.5

56.5 profilegroundorshavedVKv+

=

t WK t

)(78

78 profilegroundorshavedVKv+

=FY

PWK tv=σ

FmYWK t

v=σ



95

The metric versions

Thank youThank you



96

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