GEAR FORCES Spur Gears - WordPress.com · Example: The gear train shown in the figure is composed...

GEAR FORCES

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

represented by circular blanks, called “ pitch circles. Pitch circles are tangent to each other at

the “pitch point”. The kinematic requirement is that distance between the axes of the mating

gears should be constant, because shafts carrying the gears are connected to the fixed link by

bearings.

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

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

I23 I13I12

Pitch point

Pitch Circle

2

,

223 * rVI θ= and

3

,

323 * rVI θ=

Then

,

3

2

3,

23

,

32

,

2 ** θθθθr

rrr =⇒=

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. This separation causes the transmitted force to at an angle called “pressure

angle”, denoted by φ . Pressure angle is standardized;

In imperial system o20=φ

In international system o5.18=φ

.

2θ

.

3θ

1

2

3

r2r3I13

Pitch Circle

φ

FG

Gear forces GFv

can be separated into two components tF , tangential force in tangent

direction and rF , radial force in the direction of radius.

For kinematic considerations the relative size of the teeth with respect to the blank is

important. Characteristic dimension for the blank is either the diameter or circumference of

the pitch circle. 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.

diametercirclePitch

teethPpitchDiametral d

#=

teeth

diametercircularPitchPpitchCircular c

#

*π= and

d

cP

Pπ

=

These two are used to Imperial System.

In European, International, German and Turkish standards:

)(#

mmteeth

diametercirclePitchModule =

Module and Diametral pitch number are standardized.

1

2

3

r2r3I13

Pitch CircleFG

Fr

Fr

FG

Ft

Ft

Helical Gears

To 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. Force

acting normal to the tooth surface, hence it makes an angle of ψ (helix angle) with the gear

axis of rotation and φ with the common tangent.

Fa

Ft

Fr

FG

Ft

Fa

gear axis of rotation

ψ

The following relations are evident from figure:

traG FFFFvvvv

++=

ψtanFF ta =

φtanFF tr =

Bevel gears

Bevel 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.

FaFt

Fr

FG

ψ

φ

γ

Γ

resultant forceacts at the midpoint of the gearthikchness

Example: The 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.

Solution:

Given: o20=φ

6=pitchDiametral

HpPower 25=

2

4

18 T

20 T

3

36 T

F2t

F2r

F'3rF'3t

F3t

F3rF4r

F4t

F2x

F2y

τ2

F3x

F3yF4y

F4x

τ4

xy

From second gear freebody diagram:

∑ =⇒=−= txtxx FFFFF 2222 0;0 (1)

∑ =⇒=−= ryyry FFFFF 2222 0;0 (2)

2

3

4

18 T

36 T

20 T

∑ =⇒=−= 2222222 *0*;0 rFrFM tt ττ (3)

From third gear freebody diagram:

∑ +=⇒=++−= trxtrxx FFFFFFF 233233 0;0 (4)

∑ +=⇒=−−= trytryy FFFFFFF 423423 0;0 (5)

∑ −=⇒=+−=232333323333 **0**;0 rFrFrFrFM tttt ττ (6)

From fourth gear freebody diagram:

∑ =⇒=−= txtxx FFFFF 4444 0;0 (7)

∑ =⇒=+−= rytry FFFFF 4444 0;0 (8)

∑ =⇒=+−= 24424444 *0*;0 rFrFM tt ττ (9)

Radius of the gears can be calculated following formulas:

d

dP

teethddiametercirclePitch

diametercirclePitch

teethPpitchDiametral

##2 =⇒=

"5,16*2

18

*2

##2

2

22 ==⇒=⇒= rP

teethr

P

teethd

d

"36*2

36

*2

#3

3

3 ==⇒=⇒ rP

teethr

"67,16*2

20

*2

#4

4

4 ==⇒=⇒ rP

teethr

Gear forces of the second gear become:

lbrn

PowerF t 14,1167

5,1*900**2

12*25*33000

***2

12**33000

22

2 ===ππ

lbFFF rtr 8,42420tan*14,1167tan* 222 ==⇒= φ

Speed of the fourth gear is:

rpmnTT

TTn 810900*

36*20

18*36*

*

*2

42

234 ===

Then, gear forces of the fourth gear become:

lbrn

PowerF t 3,1168

67,1*810**2

12*25*33000

***2

12**33000

44

4 ===ππ


Using the equations (1,2,3,4,5,6,7,8,9) unknowns become:

lbF x 14,11672 =

lbF y 8,4242 =

inlb.7,17502 =τ

lbF x 23,15013 =

lbF y 18,13703 =

inlb.03 =τ

lbF x 22,4254 =

lbF y 3,11684 =

inlb.97,19464 =τ

Example: Consider the gear train

composed of spur gears. Shaft 2 is the input

of the train, and it delivers 6 hp at a speed

of 600 rpm CCW when viewed from

bottom. Calculate the bearing forces at C

and D. Pressure angle is o20=φ .

Solution:

Given: o20=φ

6=pitchDiametral

HpPower 25=

rpmn 6002 =

Top view of the assembly is given below.

Freebody diagram of the gears are given below.

Radius of the gears can be calculated from the definition of diametral pitch;

d

dP

teethddiametercirclePitch

diametercirclePitch

teethPpitchDiametral

##=⇒=

"5,012*2

12

*2

##2

2

22 ==⇒=⇒= rP

teethr

P

teethd

d

"5,112*2

36

*2

#3

3

3 ==⇒=⇒ rP

teethr

Gear forces become:

D

C A

B

12 T12 P, 36 T1"

1"

Y

X

2

z

x

y

F3r

F3t

FCx

FCz

D

FDx

FDz

τ3

z

x

z

F2r

F2t

FAx

FAz

A

BFBx

FBz

lbrn

PowerF t 5,1260

5,0*600**2

12*6*33000

***2

12**33000

22

2 ===ππ


lbFF tt 5,126023 ==

lbFF rr 79,45823 ==

CCWinlbrF t .70,18905,1*5,1260* 233 ===τ

The shaft 3 is in static equilibrium with the gear forces, bearing forces and the torques. All the

forces and all the moments acting should add up to zero. Forces and moments act in 3

dimensional space, so, overall calculations may be difficult. If a body is static, it is seen static

in all directions.

1260,5 lb

FCz FDz

y

z

C D 458,79 lbFCzFDz

y

x

C D

Using the y-z plane view,

∑ = 0yF (2)

∑ =+⇒=−+= lbFFFFFF DzCztDzCzz 5,12600;0 3

lbF

FFFM t

DzDztC 25,630"2

"1*0"2*"1*;0 3

3 ==⇒=+−=∑

lbFlbFF CzDzCz 25,63025,6305,12605,1260 =−=⇒=+

Using the x-y plane view,

∑ =+⇒=−+= lbFFFFFF DxCxrDxCxx 79,4580;0 3

∑ = 0yF (3)

lbF

FFFM rDxDxrC 395,229

"2

"1*0"2*"1*;0 3

3 ==⇒=−=∑

lbFlbFF CxDxCx 395,229395,22979,45879,458 =−=⇒=+

o

CzCxC lbFFF 707,67025,630395,229 2222 ∠=+=+=

o

CxDxD lbFFF 707,67025,630395,229 2222 ∠=+=+=

Example: Gear 2 is rotating CCW at 1000 rpm, actuated by a 5 Hp electric motor. The gears

are all of helical type with 20 degrees of pressure angle and 30 degrees of helical angle,

whose teeth numbers and pitch are shown in the figure. Calculate the gear forces and their

components and draw freebody diagram of each gear with all the force components, showing

the bearing forces and external loads.

3

40 T

4

15 T

2

20 T, 5 P

xy

x

z

Solution:

Given: o20=φ , o20=ψ

5=pitchDiametral , HpPower 5=

rpmn 10002 =

Gear forces come from power.

For second gear;

"25*2

20

*2

#22 ==⇒= r

P

teethr

lbrn

PowerF t 56,157

2*1000**2

12*5*33000

***2

12**33000

22

2 ===ππ


lbFFF rtr 97,9030tan*56,157tan* 222 ==⇒= ψ

For third gear;

rpmnT

Tn 5001000*

40

20* 2

3

2

3 ===

"45*2

40

*2

#33 ==⇒= r

P

teethr

lbrn

PowerF t 56,157

4*500**2

12*5*33000

***2

12**33000'

33

3 ===ππ

lbFFF rtr 35,5720tan*56,157'tan*'' 333 ==⇒= φ

lbFFF rtr 97,9030tan*56,157'tan*'' 333 ==⇒= ψ

x

y

z

F2t

F2r

F2a F3r=F2r

F3a=F2a

F3t=F2t F'3r

F'3a

F'3t

F'3t=F4t

F'3r=F4r

F'3a=F4a


∑ =⇒=−= rxrxx FFFFF 2222 0;0

∑ =⇒=−= tytyy FFFFF 2222 0;0

∑ =⇒=−= azazz FFFFF 2222 0;0

rF

rFM

tz

tzz

*

0*;0

22

2222

=⇒

=−=∑τ

τ

222

222

*

0*;0

rF

rFM

ay

ayy

=⇒

=−−=∑τ

τ

x

y

z

F2t

F2r

F2a

τy2

τz2

F2y

F2x

F2z

2

From the above equations second gear bearing forces and torques become:

lbFlbFlbF zyx 97,90,56,157,35,57 222 ===

zalonginlb

yalonginlb

z

y

+=

−=

.12,315

.94,181

2

2

τ

τ

From third gear freebody diagram:

x

y

z

F3r=F2r

F3a=F2a

F3t=F2t

F'3r

F'3a

F'3t

τy3

τz3

F3y

F3x

F3z

3

∑ −=⇒=+−−= rrxrrxx FFFFFFF 233233 '0';0

ttytyty FFFFFFF 333333 '0';0 +=⇒=+−=∑

∑ +=⇒=++−= aazaazz FFFFFFF 233233 0;0

3332333323 *'*0*'*;0 rFrFrFrFM ttzttzz −=⇒=+−=∑ ττ

3332333323 *'*0*'*;0 rFrFrFrFM aayaayy +−=⇒=−+=∑ ττ

From the above equations and third gear bearing forces and torques become:

lbFlbFlbF zyx 94,181,12,315,0 333 ===

0

0

3

3

=

=

z

y

τ

τ


∑ =⇒=+−= rxrxx FFFFF 3434 '0';0

∑ =⇒=−= tytyy FFFFF 3434 '0';0

∑ =⇒=−= azazz FFFFF 3434 '0';0

434

434

*'

0*';0

rF

rFM

tz

tzz

=⇒

=+−=∑τ

τ

434

434

*'

0*';0

rF

rFM

ay

ayy

=⇒

=−=∑τ

τ

From the above equations fourth gear bearing forces and torques become:

lbFlbFlbF zyx 97,90,56,157,35,57 444 ===

zalonginlb

yalonginlb

z

y

−=

+=

.34,236

.45,136

4

4

τ

τ

y

z

F'3t=F4t

F'3r=F4r

F'3a=F4a

τy4

τz4

F4y

F4x

F4z

4

x

Example: In the figure a simple gear train is

shown, composed of two 10 pitch straight

bevel gears, larger gear having 50 and

smaller 20 teeth, placed to couple two

intersecting shafts at right angles to each

other. Gear 1, rotating CW when viewed

from left is driving gear 2 by delivering it 5

HP at 1000 rpm. The blank thickness of the

smaller gear is 1 inch. Calculate the forces

that bearings at A and B carry, if all the

axial force is carried by the bearing at A.

Pressure angle is o20=φ .

Solution:

Given: o20=φ

10=pitchDiametral

HpPower 5=

rpmn 10001 =

Half cone angle of the first gear;

'2.681

5.2tan 01 ==Γ −

"5,210*2

50

*2

#1

1

1 ==⇒= rP

teethr

"110*2

20

*2

#2

2

2 ==⇒= rP

teethr

"25,05,2"5.011 =−=−= rrav

Forces on the first bevel gear

lbrn

PowerF t 56,157

0,2*1000**2

12*5*33000

***2

12**33000

11

1 ===ππ

lbFFF rtr 25,532,68sin*20tan*56,157sin*tan* 111 ==⇒Γ= φ

lbFFF rtr 256,212,68cos*20tan*56,157cos*tan* 111 ==⇒Γ= φ

1

2

4"

1"

11

1000 rpm

0.5"

Γ

Freebody diagram of the gears are given below.

Fa1

Ft1

Fr1

x

y

z

Ft2=Ft1Fa2=Fr1

Fr2=Fa1

A

BFBx

FBz

FAz

FAx

FAy

τ2

Using the plane views of the second gear,

53,25 lb21,257 lb

FAx

FAy

FBx

4"

1,5"

x

y+

157,54 lb21,257 lb

FAz

FAy

FBz

4"

1,5"

z

y+

∑ =−+= 025,53;0 AzBzz FFF ∑ =−+−= 054,157;0 BzAzz FFF

0257,21;0 =−=∑ Ayy FF 0257,21;0 =−=∑ Ayy FF

∑ =−+= 05,5*25,538,0*257,21"4*;0 AxB FM ∑ =−= 05,5*56,157"4*;0 AzB FM

lbFAz 913,68−= lbFAz 645,216=

lbFAy 257,21= lbFBz 085,59−=

lbFBz 66,15=

lbkjiFA

rrrr645,216257,21913,68 ++−=

lbkiFA

rrr085,5966,15 −=

Example: Gear 2 is rotating CCW at 1000 rpm,

actuated by a 20 HP electric motor. It is delivering

half of its power to gear 3 and other half to gear 4,

whose teeth numbers and pitches are as shown in

the figure. The gears are all of spur type with 20

degrees of pressure angle. Calculate the gear

forces and their components, determine their

directions and draw free body diagram of each

gear with all the force components, showing the

bearing forces and external loads. Assume that the

gears thickness is negligible.

Solution:

Given: o20=φ , 5=pitchDiametral ,

HpPower 20=

3

4

20 T

30 T

2

40 T

F23t

F23r

F32r

F32t

F42rF24r

F24t

F3x

F3y

τ3

F2x

F2yF4y

F4x

τ4

xy

F42t

τ2

25 Tr r'

"*

rP*

teeth#r 4

52

40

2==⇒=

".*

'rP*

teeth#'r 52

52

25

2==⇒=

"*

rP*

teeth#r 2

52

20

233 ==⇒=

"*

rP*

teeth#r 3

52

30

244 ==⇒=


3

2

4

20 T, 5 P

40T

25 T30 T, 5 P

lb.***

**

r*n**

*/Power*F t 56157

410002

121033000

2

12233000

2

32 ===ππ

lb.tan*.Ftan*FF rtr 35572056157323232 ==⇒= φ

lb..***

**

'r*n**

*/Power*F t 10252

5210002

121033000

2

12233000

2

42 ===ππ


From the third gear freebody diagram:

lb.FF tt 651573223 ==

lb.FF rr 35573223 ==

∑ ==⇒=+= lb.FFFF;F txtxx 6515700 233233 (1)

∑ ==⇒=−= lb.FFFF;F ryyry 355700 233233 (2)

∑ ===⇒=−= in.lb.*.r*Fr*F;M tt 1231525615700 323332333 ττ CCW (3)


lb.FFFFFF;F rtxrtxx 3224900 4232242322 =+=⇒=++−=∑ (4)

∑ −=+−=⇒=+−= lb.FFFFFF;F trytryy 7519400 4232242322 (5)

∑ =+=⇒=−−= in.lb.'r*Fr*F'r*Fr*F;M tttt 49126000 42323423222 ττ CCW (6)


lb.FF tt 102524224 ==

lb.FF rr 3557322 ==

∑ ==⇒=−= lb.FFFF;F rxrxx 769100 244244 (7)

lb.FFFF;F tytyy 1025200 44244 ==⇒=−=∑ (8)

∑ ==⇒=−= in.lb.r*Fr*F;M tt 375600 424442444 ττ CCW (9)

Example: Gear 3 is rotating CCW at 1000 rpm, actuated

by a 20 HP electric motor. The gears are all of spur type

with 20 degrees of pressure angle. Calculate the gear

forces and their components, determine their directions

and draw free body diagram of each gear with all the

force components, showing the bearing forces and

external loads. Assume that the gears thickness is

negligible.

Given: o20=φ , 5=pitchDiametral , HpPower 20=

3

4

20 T

30 T

2

40 T

F23t

F23r

F32r

F32t

F42rF24r

F24t

F3x

F3y

τ3

F2x

F2yF4y

F4x

τ4

xy

F42t

τ2

25 Tr r'

"*

rP*

teeth#r 4

52

40

2==⇒=

".*

'rP*

teeth#'r 52

52

25

2==⇒=

"*

rP*

teeth#r 2

52

20

233 ==⇒=

"*

rP*

teeth#r 3

52

30

244 ==⇒=


lb.***

**

r*n**

*Power*F t 25630

210002

122033000

2

1233000

33

23 ===ππ

3

2

4

20 T, 5 P

40T

25 T30 T, 5 P


lb..***

**

'r*n**

*Power*F t 41008

525002

122033000

2

1233000

2

42 ===ππ

lbtan*.Ftan*FF rtr 3672041008424242 ==⇒= φ

From the third gear freebody diagram:

lb.FF tt 256303223 ==

lb.FF rr 42293223 ==

∑ ==⇒=−= lb.FFFF;F txtxx 2563000 233233 (1)

∑ ==⇒=−= lb.FFFF;F ryyry 422900 233323

(2)

∑ ===⇒=−= in.lb.*.r*Fr*F;M tt 5126022563000 323332333 ττ CCW (3)


lb.FFFFFF;F rtxrtxx 2599700 4232242322 =+=⇒=++−=∑ (4)

∑ =+=⇒=−−= lb.FFFFFF;F trytryy 8123700 4232242322 (5)

∑ =−=⇒=+−= in.lb'r*Fr*F'r*Fr*F;M tttt 000 42323423222 ττ (6)


lb.FF tt 410084224 ==

lbFF rr 3674224 ==

∑ ==⇒=−= lbFFFF;F rxrxx 36700 244244 (7)

lb.FFFF;F tytyy 4100800 44244 −=−=⇒=+=∑ (8)

∑ =−=⇒=+= in.lb.r*Fr*F;M tt 2302500 424442444 ττ CW (9)

GEAR FORCES Spur Gears - WordPress.com · Example: The gear train shown in the figure is composed...

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