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ME 302 DYNAMICS OF ME 302 DYNAMICS OF MACHINERYMACHINERY

GEAR FORCE ANALYSIS

Dr. Sadettin KAPUCU

© 2007 Sadettin Kapucu

2Gaziantep University

GEAR FORCE ANALYSISGEAR FORCE ANALYSIS

Gears are used to transmit force and motion from one shaft to the other.

addendum circle

addendum

dedendum

dedendum circle

clerance

tooth thickness

space width

Circular pitch

top land

face width

face

flank

bottom land

pitch circle

Base circle

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Spur GearSpur Gear In theory, one blank transmit force and motion via the

friction force occurring at the pitch point. There must be no slip between the two blanks at the pitch points. Blanks roll but do not slip with respect to each other. Input /output relationship for circular gear is linear. So:

1

2

3

r2r3

VI23

I23I13I12

Pitch point

Pitch Circle

2 3 2

,

223 * rVI

3

,

323 * rVI

32

323322

r

rr*r*

...

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Spur GearSpur Gear Magnitudes of friction forces are generally small. So, if we want to

transmit larger amounts of forces, friction becomes inadequate and slip occurs. To prevent slip, we make the joint between links 2 and 3 “form closed”. This is obtained by putting teeth around the periphery of the gear blanks. For fitting these details, we need some space. We simply separate the gear blanks apart a bit.

1

2

3

r2

r3I13

Pitch Circle

FG

This separation causes the transmitted force to at an angle called “pressure angle”, denoted by . Pressure angle is standardized;

In imperial system 20In international system 5.18

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Spur GearSpur Gear

1

2

3

r2

r3I13

Pitch CircleFGFr

FGFt

Fr

Ft

Characteristic dimension for the tooth is either the number of teeth on the blank or the length of the portion of the pitch circle within the tooth body.

)inch(diametercirclePitch

teeth#PpitchDiametral d

teeth#

)mm(diametercirclePitchModule

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Helical GearHelical GearTo improve the force carrying capacity of the gears the teeth are cut in a helix. This increase the tooth thickness, so helical gears are stronger. Also they operate with less noise.

Fa

Ft

Fr

FG

Ft

Fa

gear axis of rotation

FaFt

Fr

FG

Force acting normal to the tooth surface, hence it makes an angle of (helix angle) with the gear axis of rotation and with the common tangent.

traG FFFF

tanFF ta

tanFF tr

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Bevel GearBevel GearBevel gears are conical in shape and used to couple the shafts not parallel but intersecting. Point of intersection of shafts is called the apex. Gear force acts as distributed over the whole tooth thickness, but we can assume a resultant single force acting on the mid point of the tooth thickness.

resultant forceacts at the midpoint of the gearthikchness

traG FFFF

sintanFF ta

costanFF tr

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Example 1Example 1The gear train shown in the figure is composed of 6 diametral pitch spur gears and 20 degrees pressure angle. Link 2 is the driving gear, delivering 25 Hp at a CCW speed of 900 rpm. Gear 3 is an idler and gear 4 carries the external load. Draw freebody diagrams of the gears, show all the forces acting and calculate their magnitudes.

2

3

4

18 T

36 T

20 T

Given:o20

6pitchDiametral

HpPower 25

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Example 1Example 12

3

4

18 T

36 T

20 T2

4

18 T

20 T

3

36 T

F’3t

F’3r

F3x

F3yF3t

F3r

F4t

F4r

F2r

F2t

F2x

F2y

F4x

F4y

4

2

x

y+

3

10Gaziantep University

Example 1Example 12

3

4

18 T

36 T

20 T2

18 T

F2r

F2t

F2x

F2y

4

x

y+

From second gear freebody diagram:

txtxx FFFFF 2222 0;0

ryyry FFFFF 2222 0;0

2222222 *0*;0 rFrFM tt

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Example 1Example 12

3

4

18 T

36 T

20 T

3

36 T

F’3t

F’3r

F3x

F3yF3t

F3r

x

y+

From third gear freebody diagram:

trxtrxx FFFFFFF 233233 0;0

trytryy FFFFFFF 423423 0;0

232333323333 **0**;0 rFrFrFrFM tttt

3

12Gaziantep University

Example 1Example 12

3

4

18 T

36 T

20 T

4 20 T F4t

F4rF4x

F4y

4

x

y+

From fourth gear freebody diagram:

txtxx FFFFF 4444 0;0

rytry FFFFF 4444 0;0

24424444 *0*;0 rFrFM tt

Radius of the gears can be calculated following formulas:

d

d

P

teeth#ddiametercirclePitch

diametercirclePitch

teeth#PpitchDiametral

2

"5,16*2

18

*2

##2

222 r

P

teethr

P

teethd

d

"36*2

36

*2

#3

33 r

P

teethr

"67,16*2

20

*2

#4

44 r

P

teethr

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Example 1Example 1Gear forces of the second gear become:

lbrn

PowerF t 14,1167

5,1*900**2

12*25*33000

***2

12**33000

222

lbFFF rtr 8,42420tan*14,1167tan* 222

Speed of the fourth gear is:

rpmnTT

TTn 810900*

36*20

18*36*

*

*2

43

324

Then, gear forces of the fourth gear become:

lbrn

PowerF t 3,1168

67,1*810**2

12*25*33000

***2

12**33000

444

lbFFF rtr 22,42520tan*3,1168tan* 444

Using the equationsunknowns become:

lbF x 14,11672

lbF y 8,4242 inlb.7,17502 lbF x 23,15013 lbF y 18,13703

inlb.03 lbF x 22,4254 lbF y 3,11684 inlb.97,19464

2

4

18 T

20 T

3

36 T

F’3t

F’3r

F3x

F3yF3t

F3r

F4t

F4r

F2r

F2t

F2x

F2y

F4x

F4y

4

4

x

y+