gear system

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MECHANICS OF MACHINES – DDA 3043 POWER TRANSMISSION SYSTEM: GEAR SYSTEM 1.1 Introduction to Gear System Mechanical power transmission between shafts can be done in several ways. In most engineering practice, major power transmission used is gear system, belt drive, chain or rope drive. However, among these, gear system is the most efficient. The efficiency can go up to 98%. Gear system is efficient because it can perform high consistency of connection to produce high speed and load transfer with minimal noise of operation. It is analogical with belt drive system because belt drive is a flexible, easy to install way of power transmission mode. However, belt drive’s efficiency depends on the distance between the driver and driven pulley. The efficiency can be affected by belt drive slip, centrifugal effect and creep factor. Gears are used in many machines such as metal cutting machine tools, automobiles, hoists, rolling mill and so on. 1.2 Types of Gear System The function of gear is to transmit mechanical power from one shaft to another shaft with a certain speed ratio. Gear system parts include at least a set of gear that consist of Driver Gear and Driven Gear. Driver Gear is the gear that actuates power 1

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Page 1: Gear System

MECHANICS OF MACHINES – DDA 3043

POWER TRANSMISSION SYSTEM: GEAR SYSTEM

1.1 Introduction to Gear System

Mechanical power transmission between shafts can be done in several ways. In most

engineering practice, major power transmission used is gear system, belt drive, chain or rope

drive. However, among these, gear system is the most efficient. The efficiency can go up to 98%.

Gear system is efficient because it can perform high consistency of connection to produce high

speed and load transfer with minimal noise of operation.

It is analogical with belt drive system because belt drive is a flexible, easy to install way

of power transmission mode. However, belt drive’s efficiency depends on the distance between

the driver and driven pulley. The efficiency can be affected by belt drive slip, centrifugal effect

and creep factor. Gears are used in many machines such as metal cutting machine tools,

automobiles, hoists, rolling mill and so on.

1.2 Types of Gear System

The function of gear is to transmit mechanical power from one shaft to another shaft with

a certain speed ratio. Gear system parts include at least a set of gear that consist of Driver Gear

and Driven Gear. Driver Gear is the gear that actuates power while Driven Gear is the gear that

receives the power. A series of gear set is called Gear Train. Gear can be classified according to

the relative position of the axes of mating gears.

a) Parallel Axes Shaft

The shaft axes between driver and driven gear is parallel to each other. Example of this type of

gear is Spur Gears and Helical Gears.

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Page 2: Gear System

Spur Gear Double Helical Gear

b) Intersecting Axes Shaft

The shaft axes between driver and driven gear is perpendicular to each other. Example of gear is

bevel gears.

Bevel Gear

c) Perpendicular Axes haft

The shaft axes between driver and driven gear are perpendicular to each other and do not

intersect to each other. Example of gear is “Worm Gear” and “Rack and Pinion Gear”.

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Page 3: Gear System

Worm Gear Rack and Pinion Gear

1.3 Relationship between Pitch Diameter and Pitch Circle

Some of the important terminology of gear system is:

Pitch Circle : An imaginary circle which by pure rolling action, would produce the same

motion as the toothed gear wheel.

Circular Pitch : The distance measured along the circumference of the pitch circle from a point

on one tooth to the corresponding point on the adjacent tooth.

Addendum Circle : Circle that limits the top of the teeth.

Tooth Thickness : The width of the tooth measured along the pitch circle

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Page 4: Gear System

For both gear to mate efficiently, the pitch circle of both gear must be the same. Thus;

Thus

Where = Diameter of driver gear

= Diameter of driven gear

= Number of teeth of driver gear

= Number of teeth of driven gear

1.4 Gear Ratio

Consider a gear set below;

When two gear mate efficiently at point A, the velocity, of both gear are the same. Thus;

with

Then from will produce

Where = speed of driver gear

=speed of driven gear

4

Driver gear

Driven gear

Page 5: Gear System

Gear ratio is defined as ratio of speed of driven gear with the speed of driver gear.

Where = angular acceleration of driver gear

=angular acceleration of driven gear

1.5 Gear Train

Combination of gear wheels by means of which motion is transmitted from one shaft to

another shaft is called Gear Train. In simple gear train, each shaft carries one gear only. Some

gear trains consist of three gear that is driver gear, idler gear and driven gear. In Compound Gear

Train, each shaft carries two wheels, except the first and the last.

Simple gear train Compound gear train

The idler gear doesn’t affect the Gear Ratio of a gear system, but only affect the rotation

of the driven gear. When the gear train is complex (consist of many gear sets), it is important for

the designer to identify the rotation of the driver and the final driven gear respectively. However,

there is a simple formula to determine the rotation of each successive gear in a gear train.

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Page 6: Gear System

“For an ODD number of mating gears, the rotation of Driven gear is the SAME as Driver Gear.”

“For an EVEN number of mating gears, the rotation of Driven gear is REVERSE of Driver

Gear.”

Another classification of gear train is called Reverted Gear Train and Epicyclic Gear Train.

1.6 Gear Efficiency

Gear efficiency is defined as the ratio of Output Power from Driven Gear to the Input

Power from Driver Gear. Gear efficiency measures how efficient a gear system is to transmit

power. High value of gear efficiency reflects a more efficient gear system. Power loss in a gear

system may come from sources like friction, slip, backlash and so on.

From Power, , then

Gear Efficiency,

Where = Input power from driver gear

= Output power from driven gear

= Gear ratio

If the , thus the torque at driver gear is;

1.7 Power Transmission in a Gear Train System

In a gear train system, power loss normally happen in the bearing and gear due to friction

and loading imposed on it and also power loss in overcoming shaft inertia. Consider a gear train

consists of two sets of gear reducing arrangement. A motor is attached to the system with is

the moment of inertia of motor shaft, is moment of inertia of middle shaft and is the

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Page 7: Gear System

moment of inertia of hoist which acts as the load of the system. Gear ratio and gear efficiency of

gear set 1-2 is and , between gear set 3-4 is and respectively. Let;

= Torque of motor

= Torque of hoist

= Friction torque at bearing X

Draw free body diagram and using Newton Second Law,

Assume clockwise direction as positive value.

For (A)

……………………………… (1)

For (B)

………………………………. (2)

Since there is gear mating between gear 1 and 2, thus, must include in the analysis its own gear

ratio and gear efficiency, and relate it to the inertia of middle shaft, .

Previously, , thus it follows that

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Hoist

Page 8: Gear System

……………………………….………. (3)

For (C)

……………………………….. (4)

also ………………………………………..(5)

Using power, power transfer to each gear component is;

a) Power transfer by the motor

b) Power at gear 1

c) Power at gear 2

d) Power at gear 3

e) Power at gear 4

f) Power at hoist

g) Overall power transfer efficiency,

Thus if friction torque, effect is neglected,

This concludes that

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Page 9: Gear System

Also;

1.8 Equivalent Moment of Inertia,

Consider a simple gear system as below Figure. In order for the driver gear A to start

rotate, it must have enough torque to overcome its own inertia, first, and then another

additional torque to start accelerate the driver gear B. However, to relate torque with the gear

parameter, inertia term will be taken into account. For a simple gear system, the solution is

straightforward, but when it comes to complex gear train design, it is useful to simplify / group

together all inertia term in the system into a single compact inertia expression. The inertia term

of each moving gear parts will be referred to a single part in the system, normally at motor side.

1. Torque at B to overcome

Refer to gear A side. Use gear ratio,

Thus,

2. Gear efficiency is related to power and thus torque of the mating gears, thus

3. Therefore, torque at A, to accelerate

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Driven gear Driver gear

IB

IA

Page 10: Gear System

4. Therefore total torque at A to accelerate and is

, Or in general form, (referred to motor side)

Thus

The derivation of of this simple gear system can be extended to a double set of gear

reducing problem as in section 1.7. By neglecting the friction torque effect, , thus,

1.9 Gear Train Applications (Solved Problem)

Example 1

A motor is accelerating a 250 kg load with acceleration of 1.2 m/s2 through a gear system as

shown below. The rope that carries the load are encircled on a hoist with diameter 1.2m.Gear for

the hoist’s shaft has 200 teeth, gear for motor shaft has 20 teeth. Gear efficiency is 90%. Mass

and radius of gyration of each shaft is as below;

Mass (kg) Radius of gyration (mm)

Motor shaft 250 100

Hoist shaft 1100 500

Calculate the torque of the motor needed to bring up the load with acceleration 1.2 m/s 2. Neglect

friction effect.

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Page 11: Gear System

Solution

Total torque at motor to bring up load

Where = Torque to overcome equivalent inertia (refer to motor side).

=Torque to accelerate the load through gear system

a) Consider for

From

Thus = Motor shaft inertia

kgm2

= Hoist shaft inertia

kgm2

Gear ratio,

Put into kgm2

Acceleration of hoist,

Thus rad/s

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Dia = 1.2 m

Hoist

Page 12: Gear System

From the gear ratio, angular acceleration of motor, rad/s

Now torque due to equivalent inertia,

Nm.

Then total torque referred to motor side is;

Nm

Example 2

Figure below shows a motor used to accelerate a hoist through two sets of gear reducing

system. Moment of inertia for the motor shaft is 5 kgm2, middle shaft is 40 kgm2 and hoist shaft

is 500 kgm2. Gear ratio for gear set 1 and 2 is 1/3.5 while for gear set 3 and 4 is 1/ 4.5. Gear

efficiency for both gear set is 90%. By neglecting the friction effect, find the total torque

required by the motor to accelerate the load of 6 tones at acceleration of 0.4 m/s2.

250g

Fa

TG

250 kg

r

Hoist

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b) Consider for

From Newton 2nd Law,

Then, torque at hoist

Nm

But due to gear efficiency (since the hoist shaft is connected

to the gear system), torque to accelerate the load,

Nm

Page 13: Gear System

Solution

Given that =5 kgm2, kgm2, =500 kgm2, , ,

Neglect friction effect.

Total torque required for the motor is

Where = Torque to overcome equivalent inertia (refer to motor side).

=Torque to accelerate the load through gear system

a) Consider for

Recall that , but for two set of gear system with friction effect is neglected,

From question, given that, m/s2, thus;

rad/s2

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Diameter = 1.2 m

Hoist

Page 14: Gear System

From gear ratio,

Thus,

rad/s2

Thus,

Nm.

It is known that referred to motor side will be denoted as and is related by

Nm

Thus total torque at motor required is

Nm.

4500g

F1a

4500 kg

Hoist

6000g

6000 kg

F2

a

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b) Consider for

From Newton 2nd Law,

N

N

Resultant torque at hoist

kN

Thus torque at hoist

kN

Page 15: Gear System

Example 3

Figure above shows a motor accelerating a hoist with diameter 0.9m, through two sets of

gear reducing system. Gear ratio for gear 1 and 2 is 1/3.5 while for gear 3 and 4 is 1/ 4.5.

Moment of inertia for the motor shaft is 5 kgm2, middle shaft is 20 kgm2 and hoist shaft is 100

kgm2. The rope that is encircled on the hoist must be capable to lift up a load of 5 tones that is

sliding on a 1 in 50 slope. Friction on the slope is 1000N and the total torque at motor required to

raise the load is 1500N. Use gear efficiency of 90% for both gear set.

If there is friction torque effect on the middle shaft, Nm and at hoist shaft is

Nm. Calculate the acceleration of the load at the above condition.

Solution

For the overall gear ratio,

Total torque required by motor to raise load

Where = Torque to overcome equivalent inertia (refer to motor side).

=Torque to accelerate the load through gear system

Total torque to overcome friction effect.

a) Consider for

Previously,

For double set of gear reducing system,

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Dia = 0.9 mHoist

Page 16: Gear System

kgm2

From , thus

Also from gear ratio, , thus

Thus

Nm

b) Consider for

c) Consider for

Friction effect can be grouped together to form where;

RF 5000g

1F

θ

Mg sinθ

r

F1

hoist

16

From

NThus, torque to accelerate hoist

Nm

Use gear efficiency to relate with

Nm

Page 17: Gear System

Nm

From

Thus m/s2

2.0 Vehicle Dynamics

For a moving vehicle, some of the forces acting on it are;

Friction due to the vehicle’s body (aerodynamic friction),

Forces due to friction from the engine to the wheel such as friction in bearing, shaft,

clutch and gears, .

Forces due the acceleration of the vehicle, which is called tractive force, considering

no slip between the wheel and the road surface.

We can estimate the speed of the moving vehicle by considering the speed of the wheel itself.

Vehicle speed,

Example 1 (Solved Problem)

D/2

wr

FT

R

v

Wheel

Surface

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Page 18: Gear System

Total mass for a two wheeled motorcycle including passenger is 190 kg. The engine produce

torque of 25 Nm at speed of 1800 RPM. Moment of inertia for each wheel is 1.4 kgm2 while for

other rotating parts in the engine is considered as 0.15 kgm2. The wheel’s effective diameter is

610 mm. If the motorcycle is moving on a road with a speed 23 km/hr at second gear, find

(i) Gear ratio for the second gear

(ii) Acceleration at speed 23 km/hr

Assume wind friction is 200 N and gear efficiency is 90%.

Solution

Given that =190 kg, =25 Nm at =1800 PM, =1.4 kgm2, =0.15 kgm2, =610 mm,

If =23 km/hr at 2nd gear, with =200 N, =90%.

1st Step – Draw Free Body Diagram

2nd Step

Total torque at engine,

Where = Torque due to equivalent inertia of rotating parts in the engine.(referred to engine

side).

= Torque to accelerate the wheel.

Engine

Gear system

Wheel

Iint IR

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Page 19: Gear System

3rd Step

Convert all measurement to SI standard.

Engine speed, rad/s

Wheel speed, m/s

4th Step– Determine

In order to find gear ratio for second gear,

(Where )

Thus gear ratio for second gear is

Equivalent moment of inertia is

kgm2

In order to find , use and gear ratio

rad/s2

Thus,

Nm.

5th Step – Determine

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Page 20: Gear System

Thus, total torque at engine

The acceleration at that speed is

By solving the equation for the total torque above, thus

m/s2

m=190 kg

FT

R

a

20

From

Total torque at wheel,

Refer to motor side using gear efficiency

Nm