3. gear trains - uotechnology.edu.iq · two-step analysis of epicyclic gear trains: the following...

3. GEAR TRAINS

3.1 Introduction:

Sometimes, two or more gears are made to mesh with each other to transmit power from

one shaft to another. Such a combination is called

of the train used depends upon the velocity ratio required and the relative position of the

axes of shafts. A gear train may consist of spur, bevel or spiral gears.

3.2 Types of gear Train:

Depending upon the arrangement of wheels,

3.2.1 Simple Gear Train: A train which each shaft carries only one gear, Figure

It may be noted that when the number of intermediate wheels

follower will be in the opposite direction of the driver as shown in Figure

if the number of intermediate wheels is

follower) will be in the same direction as shown in

Now consider a simple train of wheels with one intermediate wheel

3.2:

Let: N1, N2, N3= speed of driver, intermediate, and follower wheel respectively in (rpm).

T1, T2, T3= Number of teeth on driver, intermediate, and

respectively.

Since the driver gears with the intermediate

N2/N1 = T1/T2

Similarly, as the intermediate wheel gears with the follower

N3/N2 = T2/T3

Multiplying Equation (i) and (ii),

2

1

3

2

1

2

2

3

NNor

TTx

TT

NNx

NN

=

1


one shaft to another. Such a combination is called gear train or train of wheels. The nature


axes of shafts. A gear train may consist of spur, bevel or spiral gears.

Depending upon the arrangement of wheels, there are four different types of gear trains.

A train which each shaft carries only one gear, Figure

Figure 3.1

when the number of intermediate wheels is even, the motion of the

follower will be in the opposite direction of the driver as shown in Figure 12.1

number of intermediate wheels is odd, the motion of both wheels (i.e. driver and

will be in the same direction as shown in Figure 3.1b.

Now consider a simple train of wheels with one intermediate wheel, as shown in Figure

= speed of driver, intermediate, and follower wheel respectively in (rpm).

= Number of teeth on driver, intermediate, and follower wheel

Since the driver gears with the intermediate wheel,

(i)

Similarly, as the intermediate wheel gears with the follower

(ii)

3

1

1

3

TT

NN

=

3 2

Figure 3.2


of wheels. The nature


there are four different types of gear trains.

A train which each shaft carries only one gear, Figure 3.1.

even, the motion of the

12.1a. However,

odd, the motion of both wheels (i.e. driver and

shown in Figure

= speed of driver, intermediate, and follower wheel respectively in (rpm).

follower wheel

In general, the velocity ratio (VR) in simple gear train case can be determined as below:

Similarly, it can be proved that the above equation also holds good, even if there are any

number of intermediate wheels. The velocity ratio, in a simple train of gears, is depends

only on the number of teeth on the last and first gears. The intermediate wheels (idler

gears) decide the direction of follower.

3.2.2 Compound Trains: A train in which each shaft, except the first and last, have

two gears fastened together on the same shaft to operate as an integral part, see Figure

3.3a.

Figure 3.3

Train value (TV):

The train value of the compound gear train in Figure 3.3a is:

( )EDCBFDB

ECA

E

F

C

D

A

B NNandNNTTTTTT

NNx

NNx

NNTV ==== Q

****

15.036*40*5018*20*30

****

−=−===∴FDB

ECA

A

F

TTTTTT

NNTV

The negative sign means that, gear F rotates in opposite direction of A.

In general,

followeronteethofNodriveronteethofNo

driverofSpeedfollowerofSpeedVR

..

==∴

(a) (b)

Driver

Follower

3.2.3 Reverted Gear Train:

axial, the train is known as reverted gear train as shown in Figure

trains are used in automotive transmissions, lathe back

minute- and hour-hand shafts are coaxial).

(SameTTTTrrrr

DCBA

DCBA

+=++=+

where rA, rB, rC, and rD = Radii of the respective gears.

The advantages of compound train over a simple train is that much larger speed reduction

from the first to the last can be obtained with small gears. If a simple train were used to

give a large speed reduction, last gear would have to be large. Usually foe a speed

reduction in excess of 7 to 1 a simple train is not used; instead a compound train or wor

gearing is employed.

Example 3.1: Determine the revolutions per minute and direction of rotation of gear G in

the gear train shown in Figure 3.4.

Solution:

197826.0*197826.0

48*42*46*4042*24*28*26

NNNNTV

AG

A

G

==∴

===

The direction of motion is cw.

gearfirstofSpeedgearlastofSpeedTV ==

When the axes of the first and the last gears are

, the train is known as reverted gear train as shown in Figure 3.3b.

trains are used in automotive transmissions, lathe back gears, and in clocks (where the

hand shafts are coaxial).

)pitchcircularSame

= Radii of the respective gears.


first to the last can be obtained with small gears. If a simple train were used to


a simple train is not used; instead a compound train or wor

Determine the revolutions per minute and direction of rotation of gear G in

Figure 3.4

min/39.2371200*197826

197826.0

r+=

+=

followersonteethofoductPrdriversonteethofoductPr

When the axes of the first and the last gears are co-

b. Reverted gear

gears, and in clocks (where the


first to the last can be obtained with small gears. If a simple train were used to


a simple train is not used; instead a compound train or worm

Determine the revolutions per minute and direction of rotation of gear G in

Example 3.2: Figure 3.5 shows a compound gear train where the individual velocity ratios

for the three pairs of gears are given. Find the train value (

Solution:

(VR3)*(VR2)*(VR1) TV ==

Example 3.3: A conventional automotive transmission is shown diagrammatically in

Figure 3.6. The transmission of power is as follows:

Low gear: Gear 3 shifted to mesh with gear

Second gear: Gear 2 shifted to mesh with gear

High gear: Gear 2 shifted so that clutch teeth on e

teeth on end of gear 1.

Reverse gear: Gear 3 shifted to mesh with gear

A car equipped with this transmission has a differential ratio

diameter of 686 mm. Determine the engine

conditions:

• Low gear and car traveling 28

• High gear and car traveling 80

• Reverse gear and car traveling

shows a compound gear train where the individual velocity ratios

for the three pairs of gears are given. Find the train value (TV).

Figure 3.5

901

4.51*

51*

41

−=

automotive transmission is shown diagrammatically in

. The transmission of power is as follows:

shifted to mesh with gear 6.


shifted so that clutch teeth on end of gear 2 mesh with clutch


A car equipped with this transmission has a differential ratio of 1:3.7 and a tire outside

mm. Determine the engine speed of the car under the following

28 km/h.

80 km/h.

Reverse gear and car traveling 8 km/h.

Figure

shows a compound gear train where the individual velocity ratios

automotive transmission is shown diagrammatically in

mesh with clutch

d a tire outside

speed of the car under the following

Figure 3.6

Figure 3.7

Example 3.4: Figure 3.7 shows a gear train consisting of a pair of same size bevel gears having 16 teeth each, a 4 tooth right-hand worm, and a 40 tooth worm gear. The speed of gear 2 is given as N2 = +200 r/min, which corresponds to counterclockwise about the y-axis. What is the speed and direction of rotation of the worm gear? Solution:

r/min2020010N

104016416

NN

5

2

5

−=−=∴

−=−=

*.

.**

Gear 5 rotates cw 20 r/min about z-axis in a right-handed coordinate.

3.2.4 Epicyclic gear Train (Planetary Gears): These are gear trains in which the

axis of one or more gears moves relative to the frame, i.e. a gear will be so driven that it

not only rotates about its own center but at the same time its center rotates about another

center.

The gear at the center, see Figure 3.8, is called the sun, and gears whose axes move are

called planets.

S is the sun-wheel.

P is a planet wheel which can rotate freely on a pin attached to the arm L.

L is the arm which rotates freely about the axis of S.

A is the annulus having internal teeth.

Two-Step Analysis of Epicyclic Gear Trains:

The following two steps are the same, no matter which member in the train is the driver

and which is the follower.

Step 1: Imagine that the entire gear train is locked together as integral part; then

rotate the entire unit one revolution clockwise.

As a result, every member in the train will rotate (+1) revolution.

Step 2: Now imagine that the gears are again free to turn, and while holding the arm

stationary, rotate the fixed gear one revolution counterclockwise (-1), back to its original

position.

Figure 3.8

Figure 3.10

Figure 3.11

Example 3.5: Figure 3.9 shows the simplest form of Epicyclic gear train. The fixed or sun gear has 80 teeth, and the planet gear has 20 teeth. Find the train value.

Solution:

Step Driver Arm L Fixed S Follower P

1 +1 +1 +1 2 0 -1 +TS/TP

Total +1 0 1+ TS/TP

Example 3.6: Figure 3.10 shows an epicyclic gear train utilizing an annular gear as a sun gear. The sun gear has 80 teeth, and the follower gear has 20 teeth. Find the train value.

Solution:

Example 3.7: Figure 3.11 shows an Epicyclic gear train utilizing an idler gear. The sun gear has 80 teeth, the idler I has 16 teeth, and the follower P has 20 teeth. Find the train value. Solution:

Step Driver Arm L Fixed S Follower P

1 +1 +1 +1 2 0 -1 -TS/TP

Total +1 0 1- TS/TP

Step Driver Arm L Fixed S Idler I Follower P

1 +1 +1 +1 +1 2 0 -1 +TS/TI -TS/TP

Total +1 0 1+ TS/TI 1-TS/TP

Figure 3.9

520801

TT1

1TT1

NNTV

P

SP

S

L

P +=+=+=

+

==∴

320801

TT1

1TT1

NNTV

P

SP

S

L

P −=−=−=

−

==∴

320801

TT1

1TT1

NNTV

P

SP

S

L

P −=−=−=

−

==∴

Figure 3.12

Figure 3.13

Example 3.8: Figure 3.12 shows an Epicyclic gear train in which the wheel D is held stationary by the shaft A and the arm B is rotated at 200 r/min. The wheels E and F are fixed together and rotate freely on the pin carried by the arm. The wheel G is rigidly attached to the shaft C. Find the speed of the shaft C, stating the direction of rotation relative to that of B. The numbers of teeth are as follows: E 20, F 40, and G 30.

Solution: rd + re = rg + rf ∴ Td + Te = Tg +Tf , assuming pitches are equal Td + 20 = 30 +40 ∴ Td = 50 teeth

∴Speed of C = - 466.7 r/min (in opposite direction to B) Example 3.9: In the epicyclic gear train shown in Figure 3.13 the pinion B and the internal wheels E and F are mounted independently on the spindle, O, while C and D form a compound wheel, which rotates on the pin P attached to the arm A. The wheels B, C and D have 15, 30, and 25 teeth, all of the same pitch.

a. If wheel E is fixed, what is the ratio of the speed of F to that of B?

b. If wheel B is fixed, what are the ratios of the speeds of E and F to that of A?

Solution: TE = TB + 2TC = 15 + 60 = 75 teeth

TF = TB + TC + TD = 15 + 30 + 25 = 70 teeth

a. If wheel E is fixed,

Step B C, G F, E A, D

1 +1 +1 +1 +1 2 0 - (TD / TE)*(TF / TG) +TD / TE -1

Total +1 1-(TD / TE)*(TF / TG) 1+ TD/TE 0

minr746620037N

37N

37

1)T / (T*)T / (T-1

NN

BG

GFED

B

G

/.** −=−=−=∴

−==

Figure 3.14

b. If wheel B is fixed,

Example 3.10: In a reverted epicyclic train, the arm A carries two wheels B and C and a compound wheel D and E as shown in Figure 3.14. The number of teeth on wheels B, C and D are 75, 30, and 90 respectively. Find the speed and direction of wheel C when wheel B is fixed and the arm A makes +100 r/min.

Solution:

TB + TE = TC + TD

∴ TE = 30 + 90 – 75 = 45 teeth

Step A E F D,C B

1 +1 +1 +1 +1 +1

2 0 -1 FC

DE

TTTT..

− -TE/TC +TE/TB

Total +1 0 1 FC

DE

TTTT..

− 1- TE/TC 1+ TE/TB

Step A B E F

1 +1 +1 +1 +1

2 0 -1 E

B

TT

+ FC

DB

TTTT..

+

Total +1 0 1E

B

TT

+ 1 FC

DB

TTTT..

+

561

61070

15751

703025751

TT1

TTTT1

NN

B

E

FC

DE

B

F ==+

−=

+

−=

.**

.

.

2833

703025151

TTTT1

NANF

56

75151

TT1

NN

FC

DB

E

B

A

E

=+=+=

=+=+=

**

.

.

Figure 3.15

Step Arm A D, E Wheel B Wheel C

1 +1 +1 +1 +1

2 0 +TB/TE -1 EC

DB

TTTT..

−

Total +1 1+ TB/TE 0 1 EC

DB

TTTT..

−

min/****

.

.

r4001004N4N

4453090751

TTTT1

NN

AC

EC

DB

A

C

−=−=−=∴

−=−=−=

Alternative solution,

Since the arm A makes 100 r/min, ∴ x = 100 r/min

In addition, the wheel B is fixed, ∴ x + y = 0 and y = -x = -100

r/min400453090751100

TTTTyxN

EC

DBC −=

−=+=∴

**

.

.*

Example 3.11: In the train of gear wheels shown in Figure 3.15, the shaft X runs at (-2400 r/min) and drives the bevel wheel B by the pinion A. The compound wheel EC revolves freely on a spindle rigidly attached to B. Wheel C gears with D, and E with F, both wheels D and F being concentric with wheel B. Wheel F is held stationary and wheel K, solid with D, drives L on shaft Z. There are three sets of the wheels CE carried on spindles equally spaced on B. The tooth numbers are: A 18, B 60, C = D = 22, E 23, F 21, K 19, L 64. Find the speed of shaft Z.

Solution:

Wheels A and L are not part of the epicyclic train

and must therefore be treated separately.

Step Arm A D, E Wheel B Wheel C

1 +x +x +x +x

2 0 -y*TB/TE +y EC

DB

TTTTy..*+

Total +x x-y* TB/TE x+ y x EC

DB

TTTTy..*+

Figure 3.16

KKL

KL

AB

AB

N6419N

TT-NAlso,

r/min72024006018N

TTN

*.

)(*.

−==

=−=−=

Example 3.12: In the gear drive shown in Figure 3.16, the casing is fixed. Gear wheel B rotates freely on shaft Q and carries a pin on which rotates the wheels C and D, which are fixed together. The pitch of the teeth is the same for all wheels. Wheel E is fixed on shaft Q. If shaft P rotates at 200 r/min, find the speed and direction of rotation of shaft Q for the two cases: (a) shaft R is fixed, (b) shaft R rotates at 100 r/min in the same direction as P.

Solution:

TE = 28 +16 – 20 = 24 teeth

The wheels A and H are not part of the epicyclic train and must therefore be

treated separately.

Wheel B is the arm of the epicyclic train.

Step B C, E F D, K

1 +1 +1 +1 +1

2 0 +TF/TE -1 E

F

DE

CF

TT

TTTT

−=−**

Total +1 1+TF/TE 0 1 - TF/TE

r/min66272008690N08690N

0869023211

TT1

1TT1

NN

BK

E

FE

F

B

K

.*.*.

.

===∴

=−=−=

−

=

B) to direction in (oppositer/min58196626419NL ..* −=−=∴

HHHG

HG

AB

AB

N45N

2430N

TTN

r/min808032200N

TTN

***

**

−=−=−=

−=−=−=

(a) When R is fixed, a = -80

and a + b = 0 ∴b = -a = 80

∴ NQ = a + b* ED

CF

TTTT.. = -80 + 80*

24162028

** = + 36.7 r/min (in

same direction to P)

(b) When R rotates at 100 r/min, ∴ NG = HN45 *− = 125100

45

−=− * r/min

a = -80

and a + b = -125 ∴b = - 45

NQ = -80 - 45* 2435 = - 145.6 r/min (in opposite direction to P)

Step B C, D E, Q F, G

1 +a +a +a +a

2 0 -b*TF/TD +b*ED

CF

TTTT.. +b

Total +a a- b*TF/TD a + b*ED

CF

TTTT.. a+ b

Figure 3.17

3.2.5 Epicyclics with Two Inputs:

A gear train of this type is shown in Figure 3.17.

Let N1, N2, and No represent the turns of input 1,

input 2, and the output, respectively. By

superposition, the number of turns of the output

turns due to input 1 plus the output turns due to

input 2. This can be expressed in equation form as

follows:

Example 3.13: In Figure 3.17, suppose input 1 turns at 120 r/min ccw, input 2 turns at 360

r/min cw, and the speed and direction of rotation of the output shaft to be determined.

Solution:

In order to evaluate item I in Equation (*), we construct the following table. With input 2

held fixed, B and C are fixed.

Step Driver Arm Fixed C D, E F

1 +1 +1 +1 +1

2 0 -1 + D

C

TT +

FD

EC

TTTT..

Total +1 0 1 + D

C

TT 1+

FD

EC

TTTT..

35

1082436481

1.TT.TT +1

NN

NN FD

EC

Arm

F

fixedheld2Input1

o +=+===

∴

**

(*)**44 344 2144 344 21

II

fixedheld1Input2

o2

I

fixedheld2Input1

o1o N

NNNNNN

+

=

Next, when evaluating item II, we do not construct a table, because with input 1 held

fixed, the rest of the system behaves as an ordinary compound gear train. Hence

Note that the sign is positive because gears F and A turn in the same directions.

Thus the speed of the output shaft is 50 r/min ccw.

125

1082432364820

followersonteethofoductPrdriversonteethofoductPr

NN

NN

A

F

fixedheld1Input2

o +====

****

( ) r/min50125360

35120

NNN

NNNN

2

o2

1

o1o +=−++=

+

= ****

Figure P3.1

Figure P3.2

Figure P3.3

Figure P3.4

Figure P3.5

Problems (Gear Trains):

Q1) Determine the revolutions per minutes

of gear F in Figure P3.1. What is its

direction of rotation when viewed from the

right end? ( Ans: 12.857 rpm, cw when viewed

from the right end) Q2) Determine the number of teeth on

sprocket D in Figure P3.2 if the cable

speed is to be 0.6 m/s approximately. What

is the direction of rotation of A when

viewed from the right end. ( Ans: ccw)

Q3) Determine the speed reduction

between the input and the output shafts in

Figure P3.3. If the input shaft rotates

clockwise when viewed from the right end,

what is the direction of rotation of the

output shaft? ( Ans: cw when viewed from the

right) Q4) Determine the speed of rotation of the

output shaft in Figure P3.4 and its direction

of rotation when viewed from the right. (Ans: 122.727 rpm cw) Q5) Determine the speed and direction of

rotation of the output shaft F in Figure

P3.5. (Ans: 82.63 rpm ccw)

Figure P3.6

Figure P3.7

Figure P3.8

Q6) In the gear train in Figure P3.6, gear A

is fixed, the arm is the driver, and gear D is

the driven member.

a. Determine the speed reduction

for the gear train. If the arm

rotates ccw when viewed from

the right end, what is the

direction of rotation of gear D?

b. Same as (a) except the numbers

of teeth on gear A and D are

interchanged. (Ans:-1/300, +1/301)

Q7) Figure P3.7 shows a gear train

consisting of bevel gears, spur gears, and a

worm and worm gear. The bevel pinion is

mounted on a shaft which is driven by V

belt on pulleys. If pulley 2 rotates 1200

r/min in the direction shown, find the

speed and direction of rotation of gear 9. (Ans: 11.84 rpm cw) Q8) If, in the epicyclic gear train shown in

Figure P3.8, gear A rotates at 1000 r/min

cw, while E rotates at 500 r/min ccw,

determine the speed and direction of

rotation of the annulus D, and of the shaft

F. All gears are of the same pitch, and the

number of teeth in A is 30, in B is 20 and

in E is 80. (Ans: 371.4 rpm ccw, 40 rpm cw)

Q9) In the epicyclic shown in Figure P3.9,

the gear B has 120 teeth externally and 100

internally. The driver A has 20 teeth and

the arm E is connected to the driven shaft.

Gear D has 60 teeth. If gear A revolves at

+ 100 r/min and D revolves at + 27 r/min,

find the speed of the arm E. ( Ans: -0.2916 rpm)

Figure P3.9

Figure P3.10

Figure P3.11

Figure P3.12

Q10) In epicyclic speed reducing gear

shown in Figure P3.10 the input shaft A

runs at 12000 r/min and the annular wheel

B is fixed. Find the speed of the output

shaft Z and the speed of the planet wheels

relative to the spindle on which they are

mounted. The numbers of teeth in the

wheels are: A = 15, C = 41, C1 = 25, B =

81. (Ans: 1217.53 rpm, -3945 rpm)

Q11) In the epicyclic train shown in Figure

P3.11, shaft A rotates at 1000 r/min in a

cw direction while shaft B is driven at 500

r/min in a ccw. Determine the speed and

direction of rotation of the shaft C.

All teeth are of the same module and the

numbers of teeth in the various gears are as

follows: D = 16, E = 24, F = 43, H = 35, K

= 30. (Ans: 2058 rpm cw)

Q12) An epicyclic gear consists of bevel

wheels arranged as shown in Figure P3.12.

The driving pinion A has 20 teeth and

meshes with the wheel B that has 25 teeth.

The wheels B and C are fixed together and

turn freely on the shaft F. The shaft F can

rotate radially about the main axis XX.

The wheel C has 50 teeth and meshes with

wheels D and E, each of which has 60

teeth. Find the speed and direction of E

when A rotates at +200 r/min, (a) if D is

fixed, (b) if D rotates at 100 r/min in the

same direction as A. (Ans: (a) -800 rpm, (b) -

300 rpm)

Figure P3.13

Figure P3.14

Figure P3.15

Q13) In Figure P3.13, the shaft Q is keyed

to gear 4 and turns +1000 r/min. What are

the angular velocities of the shafts R and

M, keyed, respectively, to gears 6 and 7?

The tooth numbers are in parentheses. Find

also the speed of gear 3 relative to arm. All

gears are of the same pitch. ( Ans: 1493.8

rpm, -723.35 rpm, 3.222)

Q14) In the mechanism in Figure P3.14,

gear 2 rotates at 50 r/min in the direction

shown. Determine the speed and direction

of rotation of gear 12.

Q15) The same epicyclic bevel gear train

is shown in Figure P3.6. The numbers of

teeth are A = 75, B = 20, C = 18, and D =

70. Find the speed of the driven shaft if:

1) the driving shaft makes 1000 r/min,

2) the wheel A turns in the same sense

as driving shaft at 400 r/min, the

driving shaft still making 1000

r/min. (Ans: 36 rpm, 421.4 rpm)

Q16) Figure P3.15 shows an epicyclic gear

in which the internally-toothed annulus A

is fixed. Wheels B and G are free to rotate

on the arm J. Wheels F and H are keyed to

shafts D and E respectively. The numbers

of teeth in the gear wheels are as follows:

A = 120 C = 60 H = 50 B = 30 F = 50 G = 30

Figure P3.16

Figure P3.17

Figure P3.18

If D is driven at 80 r/min and C is driven at

50 r/min in the same direction, find the

speed and direction of rotation of the arm J

and of the shaft E. (Ans: +50/3 rpm, -140/3

rpm) Q17) In the planetary gear train shown in

Figure P3.16, gear 2 turns at 500 r/min in

the direction indicated. Determine the

speed and direction of rotation of arm 6 if

gear 5 rotates at 300 r/min in the same

direction as gear 2.

Q18) Figure P3.17 shows part of a gear

train for a vertical milling machine. Power

input is through the pulley and power

output through gear 12. Compound gears 1

and 2, 3 and 4, and 10 and 11 can slide as

shown to give various combinations of

gearing. Determine all of the train values

possible between the pulley and gear 12.

Q19) Figure P3.18 shows an epicyclic gear

in which the wheel P, having 45 teeth 0f 15

mm pitch, is geared with Q through the

intermediate wheel R at the end of the arm

A. When P is rotating at 63 r/min in a

clockwise direction, and A is rotating at 9

r/min, also in a clockwise direction, the

wheel Q is required to rotate at 21 r/min in

an anticlockwise direction.

Find the necessary numbers of teeth in Q

and R, and the P.C.D. of Q. (Ans: 81, 18, 386.75 mm)

3. gear trains - uotechnology.edu.iq · two-step analysis of epicyclic gear trains: the following...

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