POWER Transmission Questions- mechanical Principles

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MODULE TITLE : MECHANICAL PRINCIPLES

TOPIC TITLE : POWER TRANSMISSION

LESSON 4 : GEAR TRAINS

MP - 3 - 4

Published by Teesside University Open Learning (Engineering)

School of Science & Engineering

Teesside University

Tees Valley, UK

TS1 3BA

+44 (0)1642 342740

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________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

In many situations the transmission of rotary motion from one part of a

machine to another requires the angular velocity and the torque to be modified

so that the machine can perform its function adequately. Gearboxes, which

often consist of multi-gears, are designed to do this work. FIGURE 1 shows a

typical gearbox.

FIG. 1 Gearbox

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Gears are connected together to form gear trains. So a gear train is a set or

system of gears arranged to transfer rotational torque from one part of a

mechanical system to another by meshing their teeth and turning each other in

a system to generate power and speed.

From our previous study, we already know that the power transmitted may be

considered as a constant (if the mechanical energy loss due to friction is

ignored), which is the product of torque and angular speed. Therefore,

reduction in the angular speed will increase the torque transmitted. Usually,

the rotating speed of the motor shaft is very high whereas the torque it

transmits is small. Then, we require a gear train attached to the shaft of the

electric motor to obtain a large torque and a reduced rotating speed.

Gear trains can be found in many machines in a workshop or factory and at

home when a modification of speed or torque transmitted is required. In a car

the gear trains help the driver to increase and decrease speed. The range of

mechanical devices that are used to modify motion in differing ways, is shown

in FIGURE 2.

FIG. 2

Prime mover

Converter

Load I.C. Engine Turbine Hydraulic Pneumatic Spring Electric

Gears Chain and sprocket Belt and pulley Fluid coupling linkage Cam

Car Train Machine tool Special purpose machinery

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In this lesson, we will analyse systems connected by gears and present an

analytical method for determining velocity ratios, torque, speed and power

relationships and efficiency in gear trains.

________________________________________________________________________________________

YOUR AIMS ________________________________________________________________________________________

After studying this lesson, you should be able to:

• explain selected basic gear terms

• use the gear law to evaluate velocity ratios

• determine the torque and power transmitted through gear trains.

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________________________________________________________________________________________

GEAR CLASSIFICATION ________________________________________________________________________________________

Gears are machine elements that transmit motion by means of successively

engaging teeth, which act like small levers. The teeth, in all types of gearing,

provide a constant velocity ratio between the shafts. We shall consider a

limited selection only which will be sufficient to enable us to analyse the

dynamics of gear trains.

Gears may be classified according to the relative position of the axes of

rotation. The axes may be:

(i) parallel, such as spur gears, parallel helical gears and double-helical

gears, and rack and pinion

(ii) intersecting, such as bevel gears, and worm and gear

(iii) other.

It is necessary to have a quick look at different types of gear, though we only

study spur gears in detail in this lesson.

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SPUR GEARS

Spur gears, shown in FIGURE 3(a), are the simplest and most commonly used

type of toothed gear. The input and output shafts are parallel and the gear

wheels have teeth which are parallel to the shaft axes. The output shaft rotates

in the opposite direction to the input shaft. The smaller wheel is known as the

pinion wheel whilst the larger is termed the gear wheel.

FIGURE 3(b) shows a pinion in contact with an internal gear. The internal

gear is called an annulus or ring gear and is commonly used in planetary gear

trains as we shall encounter later in the topic. The pinion and the ring gears

rotate in the same direction.

(a) (b)

FIG. 3 (a) Spur gears (b) Internal spur gears

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PARALLEL HELICAL GEARS AND DOUBLE-HELICAL GEARS

In parallel helical gears and double-helical gears, the axes of the driving gear

and driven gear are parallel, as shown in FIGURE 4(a). The operation

principles are the same as that of spur gears.

FIG. 4 (a) Parallel helical gears (b) Double-helical gears

(FIG. 4(a) reproduced by courtesy of ‘HPC Gears’ 2006)

(FIG. 4(b) reproduced by courtesy of ‘Hewitt & Topham’ 2006)

(a) (b)

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RACK AND PINION

A set of rack and pinion is shown in FIGURE 5. This converts rotary motion

into linear motion and vice versa. The rack can be thought of as a gear of

infinite radius. In another word, the rack is like a gear whose axis is at infinity.

FIG. 5

RackRack

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BEVEL GEARS

Bevel gears, as shown in FIGURE 6, transmit motion between shafts whose

axes intersect. The teeth are cut on the surfaces of cones.

FIG. 6

OTHER TYPES OF GEARS

FIGURES 7 and 8 shows other types of gears, in which the shafts of the gears

are neither parallel nor intersecting.

A worm and gear, shown in FIGURE 7 opposite, is basically a screw meshing

with a special helical gear. This gear set can have very large velocity ratios (up to

300) and the teeth in contact tend to slide together with high velocities. Frictional

heating and efficiency are of greater concern than with other types of gears.

The typical characteristics of crossed-helical gears, as shown in FIGURE 8, are

skewed shafting, point contact and high sliding. These are ideally suited for

lower speeds and lighter loads.

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FIG. 7 Worm and gear

FIG. 8 Crossed-helical gears

(Reproduced by courtesy of 'HPC Gears' 2006)

Worm

Gear

Worm

Gear

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________________________________________________________________________________________

FRICTION GEARS ________________________________________________________________________________________

Rotary motion can be transferred between shafts by means of cylinders, or

wheels, mounted on the shafts and pressed together so that there is no slip at

the contact.

VELOCITY RATIO

FIGURE 9 shows cylinder A, mounted on the input shaft, driving cylinder B

on the output shaft. If the friction at the contact is sufficiently large, so that

there is no slip between the contacting surfaces, then the surfaces at the point

of contact must have equal velocities. Therefore, the velocity of points on the

surface of A (= rA × ωA m s–1) and the velocity of points on the surface of B (= rB × ωB m s–1), are equal in magnitude.

FIG. 9

A rA

ωB

ωA

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Thus

so the velocity ratio

The negative sign is to indicate that ωB is in the opposite direction to that of ωA.

As rA and rB do not vary, the velocity ratio remains at a constant value.

Rearranging the expression gives

On differentiating both sides with respect to time and remembering that

(the angular acceleration)

we find that

Thus the ratio of the accelerations is the same as that of the velocities. We

have:

Equation (1) is the law of gear motion and can be applied to the evaluation of

velocity ratio and acceleration ratio. You will see in the study of the next

section that it can also be used for working out torque ratio.

r A

= =– – ........................... ω ω

α α

.......... 1( )

α αB A

B A

r = –

α ω

= d dt

ω ωB A

B A

r = –

ω ω

r = –

r rA A B Bω ω=

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Example 1

A small electric motor drives cylinder A of 3 cm diameter which, in turn,

drives output cylinder B of 5 cm diameter. The motor starts from rest and has

an angular acceleration of 5 rad s–2, calculate:

(a) the angular acceleration of cylinder B

(b) the time taken for the angular velocity of cylinder B to reach 21 rad s–1

(d) the angular velocity of cylinder A at this instant.

Solution

(a) α α

B A

r =

⎛ ⎝⎜

⎞ ⎠⎟

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

–

3 2 5 2

3 5

3 55

= – –rad s 2

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(b) Remembering that for constant acceleration

ωB1 = 0 as it starts fom rest and ωB2 is given as 21 rad s–1

then

(d) The velocity of the surface of cylinder A, vA, is the same as vB.

If ωB = 21 rad s–1, then ωA = = –35 rad s–1 at the same

instant.

The usefulness of friction gears is limited as only small torques can be

transmitted without slip occurring at the contact. So toothed gearing is used to

maintain a constant velocity ratio and to transmit a high torque.

– 5 3

21×⎛ ⎝⎜

⎞ ⎠⎟

v rB B B= ×

= ×

0 05 2

0 525

. –m s 1

21 0 3

= + ×

∴ =

t s

ω ω αB B B t2 1= + ×

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________________________________________________________________________________________

TOOTH ACTION ________________________________________________________________________________________

The requirement of a constant velocity ratio between gears A and B means that

as the gears rotate, the shape of the teeth must be such that the common normal

to the tooth surfaces, at the point of contact, must always intersect the line

joining the centres of rotation, OA and OB, at the same point P, called the pitch

point. A circle drawn through the pitch point P and centred at O is known as

the pitch circle. The pitch circles for the gears A and B are shown in

FIGURE 10. Note that they touch at pitch point P.

Note that a 'common normal' is a line at right angles to the tangent made by the

surfaces of the two teeth at the point of contact. This is important as it is the

line of action of the reaction force between the contacting teeth when gear A is

driving gear B against a load. The angle φ shown in FIGURE 10 is known as the pressure angle and is usually 20°.

FIG.10

Driven gear B

Driving gear A

CommonNormal

Ta ng

en t

Pitch circle B

Pitch circle A

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FORCE, TORQUE AND POWER ON GEARS

Consider a simple gear train containing two gears A and B. The force F and

torque T transmitted from gear A to B by the mating gear teeth are shown in

FIGURE 11.

FIG. 11

The force F acting on each tooth has two components Fx and Fy. From

FIGURE 11, we can see that

tan φ = F

F y

........................................... 2( )

Driving torque required from the motor

Bearing support reactions.

Unknown direction and magnitude

Pressure angle φ = 20° Fx Reaction force F

between the teeth equal but opposite

direction on each tooth

Fy RA

OArA

Pitch circle gear B

Pitch circle gear A

Components of F resolved perpendicular and parallel to the line

of centres

Load torque to be overcome

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As none of the parts of the system are accelerating, each part is in static

equilibrium. So for each part:

Sum of the x-components of the forces = 0

Sum of the y-components of the forces = 0

Sum of the moments of those forces about the pivot = 0

For gear B, taking moments about OB

then

Thus, the torque transmitted is

Note that rB is the moment arm for Fx on gear B and that Fy has no moment

about the pivot OB.

Similar analysis of gear A gives,

From equations (3) and (4), we have

Equation (5) shows that the ratio of torques on gears A and B is constant.

r A

= ......................................................... 5( )

T F rA x A= ...................................................... 4( )

T F rB x B= ...................................................... 3( )

F r Tx B B

OB ∑ =

0–

(Since, in general, 20 , then tan 20φ = ° = ° F

F y

== 0.364)

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If the energy loss due to friction is ignored, the power input from gear A

should be equal to the power output from gear B, which is

In practice, we should have

where η is the efficiency of the machine.

NUMBER OF TEETH

A measure of the size of a tooth on a gear is known as the module, m, which is

the ratio of the pitch circle diameter D (in mm) to the number of teeth of a gear

Gears are commonly made to standard values of modules, in the following

ranges:

• 0.2 to 1.0 by increments of 0.1

• 1.0 to 4.0 by increments of 0.25

• 4.0 to 5.0 by increments of 0.5

• then 6, 8, 10, 12, 16, 20 25, 32, 40 and 50.

m D

N = ........................................................... 8( )

η ω ω

= = power at output power at input

– ..B B

A A

T ................... 7( )

ω ωB B A AT T= – ................................................... 6( )

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For two gears to mesh correctly they must have teeth with the same module,

i.e.

where NA and NB are the number of teeth on A and B respectively, and DA and

DB are the respective diameters of the pitch circles. We have, therefore, the

velocity ratio of two mating gears:

Similarly, the magnitude of the acceleration ratio:

GEAR RATIO

The meaning of the term gear ratio can cause a little confusion. It is usually

defined as the ratio of the high speed shaft to the low speed shaft, so that its

value is always greater than 1. For example, a 4:1 reduction gear box means

that the input shaft is the high speed shaft which rotates at 4 times the rate of

the output shaft. Other literature uses terms such as 'velocity ratio' or 'train

value' which are defined as the ratio of the angular velocities of the output

shaft to the input shaft. We will use the latter, giving it the symbol G:

G = angular speed of output shaft angular speedd of input shaft

output

input

= ω ω

α α

N = ................................................................. 11( )

ω ω

N = =– – ................................................... 10( )

m D

N A

= = = ⎛ ⎝⎜

⎞ ⎠⎟

= or .............. 9( )

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Example 2

FIGURE 12 shows a small electric motor with gear A, attached to its output

shaft, which meshes with gear B. The pitch circle diameters of gears A and B

are 3 cm and 5 cm respectively. The motor shaft rotates with a constant

angular speed of 35 rad s–1 whilst gear B is subjected to a load torque TB of

9 Nm. (A load torque is a torque in the opposite direction to the motion of the

shaft.) Calculate:

(a) the angular velocity of the gear B

(b) the power absorbed by the load on gear B and the torque needed on gear

A (assuming that there is no friction on the teeth nor in the support

bearings at A and B)

centres (Fx)

(d) the component of the force along the line of centres (Fy).

FIG. 12

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Solution

(a) As gears are kinematically equivalent to a pair of discs, in contact, with

the same radii as the pitch circles in FIGURE 12, the velocity of wheel or

disc A (vA = rA × ωA) is the same as that for wheel B at the pitch point (vB = rB × ωB).

(b) The power absorbed by the load on gear B is

Since friction losses are being ignored and the system is rotating at

constant angular velocities, the power supplied by the motor must be

equal to that absorbed at the output, i.e. ωBTB = ωATA.

From the question ωA = 35 rad s–1, then the torque needed on gear A is

T T

A B B

ω ω

–

– .

189 35

5 4 Nm

ω B BT = × ( )

9 21

189

–

– W

ω ωB A

B A

r =

= ×

–

– –

3 5

21 rad s 1

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perpendicular to the line of centres Fx is

(d) Applying equation (2)

F Fy x= °

= ×

tan 20

360 0.364

131 N

tan and , thenφ φ= = ° F

F y

F T

rx B

9 0 05

360

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________________________________________________________________________________________

GEAR TRAINS ________________________________________________________________________________________

The term gear train is used to describe a series of gear wheels which transmit

motion from an input shaft to an output shaft or to a component with linear

motion in the case of a rack. We will use the term gear ratio G to mean the

ratio of the output velocity divided by the input velocity.

Gear trains may contain spur, bevel, worm gears, racks and pinions in any

combination. We can broadly classify gear trains by the types shown in

FIGURES 13(a), (b), (c) and (d).

(a) Simple trains are those in which each gear is mounted on a separate shaft.

We will find that gear B does not affect the overall gear ratio so it is

called an idler gear, as shown in FIGURE 13(a).

FIG. 13(a)

A B C

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(b) Compound gear trains are those in which two or more gears share a

common shaft and rotate at the same speed. This shaft is often known as

a Jackshaft. Gears B and C share the common shaft in FIGURE 13(b).

FIG. 13(b)

have a common centre distance as shown in FIGURE 13(c).

Gears A meshing with B have the same centre distance as gears C

meshing with D. This allows the input and output shafts to have a

common centre line. Note that the gear wheels B and C form a compound

gear.

A B

C D

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FIG. 13(c)

(d) Planetary and Epicyclic gear trains are those in which the axis of rotation

of one, or more, gears is mounted on an arm which rotates about the

centre. The rotation of the arm about axis x – x is usually the input

motion. The planet gear B freely rotates about axis y – y which is

mounted on the other end of the arm.

The planet gear meshes with gear D, called the sungear. In the example

shown in FIGURE 13(d), the planet also meshes with an internal gear,

called the annulus, which is not attached to the same shaft as D, but can

rotate about the same axis.

A B

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FIG. 13(d)

Either gear D or gear C can be fixed to the frame, thus preventing rotation, or

used as a second input, the other being used as an output.

For simplicity only one planet gear is shown although two or more may be

needed for dynamic balance of the shaft.

We will take each type of train in turn and evaluate the overall gear ratio in

terms of the number of teeth on each gear.

B CD

Planet, B

Sun, D

Annulus, C

Arm, A

X Y

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THE SIMPLE GEAR TRAIN

In the simple gear train of FIGURE 14(a), gear A is the input driver and we

would like to know the angular velocity and acceleration of gear C. Note the

symbolic representation of the plan view of the train shown in FIGURE 14(b).

FIG. 14

Taking each pair of meshing gears in turn, starting from the input driver A:

Gear A meshes with B

Gear B meshes with C

∴ = ω ω

N –

∴ = ( )ω ω

N – using Equation 10

A B C

Gear A

ωA ωB ωC

Gear B Gear C

(a)

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Noting that the required ratio

We can now replace the velocity ratios by their respective ratios of the number

of teeth:

Since G = , then G = (Note that = G also).

This shows that idler gear B only changes the direction of the output gear

relative to the input.

Example 3

Determine the number of teeth required on each of the three gears A, B and C

shown in FIGURE 14(c), if the output gear is to rotate at an angular velocity of 2/5 of the input velocity and in the same direction. The axes of rotation of the

input driver A and the output gear C are to be 90 mm apart, whilst the

minimum number of teeth on any of the gears is 12 with a module of 2.5

FIG. 14(c)

A B C

90 mm

α α

N A

ω ω

A B

B C

N N

N =

⎛ ⎝⎜

⎞ ⎠⎟

× ⎛ ⎝⎜

⎞ ⎠⎟

= =– –

ω ω

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Solution

The sizes of the gears are restricted by the requirement that the input and

output shaft centres are to be 90 mm apart. This centre distance D, if all three

gear centres are on the same straight line as shown in FIGURE 14(c) (they

don't need to be do they?), is made up by the radius of A + the diameter of B

+ the radius of C.

Now, the radius of A = NA × m/2, the diameter of B = NB × m and the radius of C = NC × m/2, since all three gears must have the same module to mesh correctly.

Thus

or on rearranging

Now let NA be the minimum number, i.e. 12

Since

With m = 2.5 and D = 90, then

90 2 5 2

12 2 30= + +[ ]. N B

N NC A= × = × = 5 2

5 2

12 30

ω ω

N = =

2 5

∴ = N A 12

D m

N N NA B C= + +⎡⎣ ⎤⎦2 2

D mN

mN mNA

B C= + +

2 2

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Multiplying both sides by 2 and clearing the brackets gives

Thus: N B = = 75 5

180 30 5 75

5 180 105

= + +

∴ =

B –

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THE COMPOUND GEAR TRAIN

FIG. 15

When two or more gears are firmly mounted on the same shaft, as are gears

B and C in FIGURE 15, they have the same angular velocity. So, using

equation (10)

Also

But ωC = ωB, so substituting for ωC

ω ωD C

B A

N = × ×

⎛ ⎝⎜

⎞ ⎠⎟

– –

ω ωD C

D C

N = ×– as gears C and D mesh

ω ωB A

B A

N = ×– as gears A and B mesh

A B

ωA ωB

ωD

A B

C D

ωC

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On rearranging

The gear ratio, G, is therefore

and gear D rotates in the same direction as that of gear A.

Note: The expression for G has been arranged so that we can see that the

numerator NA × NC is the product of the number of teeth on the driving gears taken, in sequence, from the input to the output, and the denominator is

the product of the number of teeth on the driven gears taken in sequence,

i.e. –NB × –ND

An 'idler' gear in the train would only alter the direction of rotation of the

output and, as it is both driven by the preceding gear and drives the subsequent

gear, its number of teeth would appear both in the numerator and the

denominator, and hence cancel out.

G = product of teeth on driving gears product oof teeth on driven gears

G N N

N N A C

B D

= × ×

ω ω

D A C

B D A

A C

B D

N N

= ×

∴ =

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Example 4

(a) Sketch the symbolic representation of the gear train shown in

FIGURE 16, with gears 2 and 3 mounted on the same shaft.

FIG. 16

(b) With an input of 100 rad s–1 at gear 1 calculate the angular velocity and

direction of rotation of gear 5. The numbers of teeth on each gear are

shown in parentheses.

calculate the velocity and acceleration of each gear in the train.

1 (42) 3 (67)

4(18) 5 (72)

2 (20)

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Solution

(a) The symbolic representation shows gears 2 and 3 connected to, and

rotating with, a common shaft. It can be seen that gear 4 is an idler gear.

(b) Using the expression

Then

As G is also the ratio between the output and input angular velocities

With ω1 = 100 rad s–1 then ω5 = –195 rad s–1 for gear 5. It is rotating in the opposite direction to gear 1.

G = = ω ω

1 95– .

G N N N

N N N =

× × ( ) × ( ) × ( )

= × ×

( ) ×

2 4 5

3 4

42 67 18 20

– – –

– –118 72

1 95

( ) × ( )

–

– .

G = product of teeth on driving gears product oof teeth on driven gears

Output

3 4

1 2

Input

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easier to work step-by-step for each pair of gears starting from the input at

gear 5.

Remember that the relationship between the acceleration of meshing

gears is the same as that between the velocities, i.e. with gear A as input,

meshing with a gear B,

So, applying these to the gear train in question:

and

Because gears 2 and 3 are attached to, and rotate with, the same shaft

ω ω

α α

2 3

53 7

21 5

= =

–

rad s

ω ω

4 5

72 18

50 4 50 200= = × = × =

– – – –

–

– N

rad s 1

44 5

3 4

4 20 80

18 67

200

× = × =

= = × (

ω ω

– –

– – –

–rad s 2

N ))

= + × =

= × ( ) =

0 269 200 53 7

18 67

80 21 53

. .

– – .

–rad s 1

α rrad s 2–

ω ω

α α

B A

–

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Then,

and

THE REVERTED GEAR TRAIN

A reverted gear train, as shown in FIGURE 17, is a compound gear train

designed so that the centre distance between gears 3 and 4 is the same as the

centre distance between gears 1 and 2, and then the output shaft and the input

shaft can lie on the same centreline. The gear ratio is found in exactly the

same manner as for the compound train.

FIG. 17

1 2

Output shaft

Input shaft

ω ω1 2

1 2

20 42

53 7

0 476 53 7 25 6

= = ×

= × =

– – .

– . . – .

rad s––

–

– .

– . . – .

rad s

α α

1 3

0 476

0 476 21 5 10 24

∴ = × = 22

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Example 5

Two shafts, lying in the same straight line, are geared together through an

intermediate parallel shaft. FIGURE 17 shows this arrangement. The output

shaft is required to rotate at about 0.1 times that of the input, while all of the

gears are to have the same module, m = 3. If the number of teeth on any of

the gears is at least 24, then find suitable numbers of teeth for the gears if we

keep N1 = N3.

Solution

We must first realise that the radius of gear 1 and the radius of gear 2 (i.e. the

centre distance of gears 1 and 2) must equal the sum of the radii of gears 3 and

4 for the centreline of the output shaft to coincide with that of the input.

Thus

Since meshing gears must have the same module m2 = m1 and m4 = m3 and

since m = (i.e. D = mN)

But we are told to use the same module so

m m

N N N N

1 3

1 2 3 4

∴ + = +

m N N

m N N1 1 2

3 3 42 2

+( ) = +( )

D D D D1 2 3 4 2 2 2 2

+ = +

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Since N1 = N3, then N2 must be equal to N4

and

(i) Take N1 = 24 as a trial value

The value of G will then be

(ii) If we take N1 = 25

this gives

We can see that there are many possible options on choosing the teeth numbers

in the gear train under the conditions of N1 = N1 and G ≅ 0.1. You can try others yourself.

25 0 3162

79 06

25 79

= =

= ⎛ ⎝⎜

⎞ ⎠

. .

i.e.

⎟⎟ = 2

0 100144.

G = ⎛ ⎝⎜

⎞ ⎠⎟

= 24 76

0 0997 2

∴ =

∴ = =

24 0 3162

75 9

. .

∴ = × ×

= =

G N N

N N

1 3

2 4

1 2

2 2

0 1

0 1 0 3162

. .

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THE EPICYCLIC GEAR TRAIN

Basically, epicyclic gear trains have two input motions even though one of the

gears is fixed to the frame.

Its angular velocity is zero, but this zero value constitutes one of the input

values. Any of the gears or arms in the train, except the planets, can serve as

an input or an output member.

There are several methods for analysing epicyclic gear trains for output

velocities, but we shall only use one which is based on the relative velocity

equation applied to rotating components.

Using the notation shown in FIGURE 18, we initially determine the angular

velocities of the gears relative to the arm A, here being used as the input

member, i.e. finding the gear velocities with the arm fixed.

Note that at this stage neither gear D nor gear C is considered to be fixed to the

frame.

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FIG. 18

If we write the angular velocity of any gear in the form ω1/2, as for example ωD/A which is ω for gear D in relation to arm A, then, with the arm fixed, gears D, B and C form a compound gear train, for which we know how to evaluate

the angular velocity of each gear in terms of the numbers of teeth on each gear.

Thus, relative to the arm,

ω ω

D A

C A

C B

B D

N N

N /

= ×

( ) × +( ) =– –

D A B

ωD

ωA

ωB

ωC

Input

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But we also know that the angular velocity of D relative to A (ωD/A) is equal to the absolute angular velocity of D (ωD) minus the absolute angular velocity of arm A (ωA).

Similarly

Then

We can generally use the expression in the above form, or we can rearrange it

Expanding the brackets and collecting like terms gives

This gives the characteristic equation for this type of epicyclic gear. In the

form shown, we have D as the output and A and C as inputs.

We can now account for either ωC or ωD being zero, if either of those gears are fixed to the frame.

ω ω ωD C

D A

D C

N = +

⎛ ⎝⎜

⎞ ⎠⎟

1 –

ω ω ω ωD A C

D C A

N – – –= ( )

ω ω

ω ω ω ω

D A

C A

D A

C A

N /

= = –

– –

ω ω ωC A C A/ = –

∴ = /ω ω ωD A D A–

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Example 6

For the simple epicyclic gear train shown in FIGURE 18, the number of teeth,

N, on each gear are as follows

Determine the angular velocity of the output member for an angular velocity of

100 rad s–1 from the input member under each of the following conditions:

(a) Arm A as input, gear C as output, gear D fixed

(b) Gear D as input, arm A as output, gear C fixed

Solution

For all three cases we can use the relative velocity equations.

Relative to the arm

But

Condition (a)

ω ω ω

A C D

= = =

= =

100 0

0 100 100

80 40

rad s 1– , ?

– –

ω ω

ω ω ω ω

D A

C A

D A

C A

N /

= = –

– –

ω ω

D A

C A

N /

= –

N N NB C D= = =20 80 40, ,

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On cross multiplying

Condition (b)

Condition (c)

ω ω ω

D C A

= = =

= =

100 0

100 0 0

80 40

100

rad s 1– ?

– –

– 22

C∴ = – –rad s 1

ω ω ω

ω ω

D A C

= = =

= =

100 0

100

0 80 40

100

rad s 1– ?

–

– – –

– ωω ω

A A

= +

∴ =

100 3

33 1 3

rad s 1–

– – –

– –

–

100 2 100

100 2 200

= ⎡⎣ ⎤⎦

= +

∴ =

C 00

2 150

– –= + rad s 1

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________________________________________________________________________________________

NOTES ________________________________________________________________________________________

...................................................................................................................................................

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. Two spur gears are in mesh, with the pinion having 20 teeth and the

output gear with 32 teeth cut to a module of 8. Determine:

(a) the pitch circle diameter of each gear

(b) the distance between the centres

2. Two parallel shafts are to be separated by a distance of at least 180 mm

and coupled with gears which will give a train value of .

Choose a standard module and determine the number of teeth on each

gear to give the closest value to the given train value.

Module options to be used: 8, 10 and 12.

3. An electric motor drives gear A which meshes with gear B, giving a

velocity ratio of output velocity/input velocity equal to 0.2. The motor

drives gear A with an angular velocity of 100 rad s–1 while producing a

power of 0.5 kW. Gear A has 20 teeth cut to a module of 4. Determine:

• the radius of gear A

• the torque on gear A

• the components (Fx, Fy) of the reaction forces on the teeth in contact

• the load torque on gear B.

1 1 7.

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4. The following table gives the number of teeth on each of the gears

forming the train shown in FIGURE 19. Gears 4 and 6 are subjected to

externally applied torques of T4 = 35 Nm and T6 = 15 Nm in the

directions shown in the figure.

If the input angular velocity of gear 1 is 40 rad s–1, determine:

(a) the angular velocity of gears 4 and 6

(b) the power input to gear 1 required to drive the gear train at constant

velocity.

FIG. 19

2 4

ω1 T6

Gear No. No. of Teeth

1 21

2 45

3 27

4 90

5 27

6 48

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________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

(a) Pinion radius = 4 × 20 = 80 mm ∴ Diameter = 160 mm Output radius = 4 × 32 = 128 mm ∴ Diameter = 256 mm

(b) The centre distance = rA + rB = 80 + 128 = 208 mm

2. Centre distance, and m = i.e. r = N

But N

N G

N N

B A

= =

∴ =

1 1 7.

∴ = +( ) C N N mA B 2

2 2 r N

C r rA B= +

= = = = ω ω

20 32

1 1 6.

m r

r N

= =

∴ =

8 2

2 8

4 for both gears

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with

With:

m = 8

m = 10

m = 12

Selected module would be

m = 12 with NA = 11 and NB = 19

N N

A B= ∴ =

∴ = =

11 1 11 18 7 19

11 19

. ., i.e. , i.e.

0.5579 mmC = × =30 6 180

N N

A B= ∴ =

∴ = =

13 33 13 22 1 22

13 22

. ., i.e. , i.e.

0..591 mmC = × =35 5 175

N N

A B= ∴ =

∴ = =

16 67 17 28 9 29

17 29

. ., i.e. , i.e.

0..586 mmC = × =46 4 184

C G

N m

C G

N m

= +⎛ ⎝⎜

⎞ ⎠⎟

= =

1 1

180 1

1 7

133 33

and .

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Power = torque × angular velocity, i.e. P = T × ω

Free body diagram for gear A

Σ M

F r

x A

× + =

∴ = ×

5 0

5 1000

125

–

A rA

5 Nm

100 rad s–1

Line of centres

∴ = = ×

= NmT P

A Aω

0 5 1000 100

5 .

m r

r m

A A

∴ = × =

= =

2 40

40 0 2

200

mm .

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Since

then

The load torque on gear B can now be found using two methods – this is

useful for checking answers

(a) Free body diagram for gear B

Σ M

T F r

B x B

× =

∴ = ×

125 200

1000

–

tan and

tan

φ φ= = °

= ° ×

F F

y x

0 364. ××

125

45 5. N

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(b) Power absorbed by the load torque TB

with

and

Thus

4. (a) The angular velocity of gear 4

The angular velocity of gear 6

(Note that gears 4 and 5 are acting as idler gears for gear 6.)

ω ω6 1 3 4 5

2 4 5 6 1

= ( ) ( ) ( ) ( )

= ×

N N N N

. . .

– . – . – . –

227 45 48

0 2625 40

10 5

– –

. –

( ) × ( ) ×

= ×

= rad s 1

ω ω4 1 3

2 4 1

21 27 45 90

= ( ) ( )

= ×

( ) × ( ) ×

N N

– . –

– –

. –

14 40

5 6

= rad s 1

P T

B B

= ×

= × =

0 5 1000

0 2 100 20

0 5

–

rad s 1

×× =

1000 20

25 Nm

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(b) Gears 6 and 4, therefore, rotate in the same direction as ω1, i.e. C.C.W., but notice that the torques on gears 4 and 6 are both in the

opposite direction to the motion, i.e. they are load torques and take

power out of the system.

If net power = 0, for no acceleration of any part of the train

The sum of the power in and out is zero.

Therefore, T1 × ω1 = input power, Pi

The input torque would be

Now if either T4 or T6 had been in the opposite direction to that

shown, it would have helped the torque on A to drive the gear train,

i.e. the torque would then input power to the system since it would be

in the same direction as the gear rotation.

P T

i = ×

= ×

∴ =

1 1

8 84

Nm.

P T T

= × + ×

= +

4 4 6 6

35 5 6 15 10 5

196 157 5

ω ω

. .

553 5. W

T T T1 1 4 4 6 6 0× × × =ω ω ω– –

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________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

Examplesof different types of gear pairs have been looked at, along with the

relationship of the number of teeth on a gear to its size and gear module

The simple gear law provides a means of relating angular velocity ω (rad s–1) and acceleration α (rad s–2) to the number of teeth on a gear and gives the gear ratio (G).

There are four classes of gear trains – simple, compound, reverted

compound and epicyclic. The gear ratio for simple and compound trains is

calculated from this simple formula:

The gear ratio for epicyclic gear trains can be obtained by deriving and using

the characteristic equation.

Various forces acting on each gear have been considered and tooth loads and

output torques have been calculated using the fact that, for equilibrium, the

sum of moments is zero.

G = product of no. of teeth on the driving geaars product of no. of teeth on the driven geears

m D

N =⎛

⎝⎜ ⎞ ⎠⎟

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/NOR <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> /SVE <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> >> >> setdistillerparams << /HWResolution [2400 2400] /PageSize [612.000 792.000] >> setpagedevice