1-3 Gear Train Application-Example

Mechanics of Machines Gear TrainFadzli’0607

Kuliah Mekanik Mesin BMCM 2723

Application of Gear Train

Gear train can be divided into two major applications:

1. Lifting MachineA simple construction of lifting machine consists of a motor, gear train and drum where load is lifted.

2. Vehicle DynamicsA simple construction of a vehicle consists of an engine, gear train and wheels. Due to tractive force, FT, which acts between the wheel and road, the vehicle experiences linear velocity, where:

v: velocity of vehicle: angular velocity of wheeld: diameter of wheel

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Wheel

Road surface


EXAMPLES OF TRANSMISSION SYSTEM: LIFT

EXAMPLE 1.1

Figure 1.1

A motor accelerates a 250kg mass with acceleration of 1.2ms-2 through a gear system shown

in Figure 1.1. The mass is tied to a rope which is wind to a drum having diameter of 1.2m.

The gear for the drum shaft has 200 teeth and the gear for the motor shaft has 20 teeth. The

contact gear efficiency is 90%. The mass and radius of gyration are respectively as shown:

Mass (kg) Radius of gyration (mm)

Motor shaft 250 100

Drum shaft 1100 500

Find the torque of the motor needed to lift the mass under these conditions. Neglect all

friction.

(a) The mass hangs freely under the drum;

(b) The mass is placed on a plane 10o from the horizontal. Friction under the mass is

1000N while the torque friction is equivalent to fM = 10Nm at the motor shaft and fD

= 200Nm at the drum shaft.

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DRUM

MOTOR

dD = 1.2m

IM

a = 1.2ms-2

ID


ANSWER EXAMPLE 1.1

Given: M = 250kgdD =1.2mt1 = 20t2 = 200mM = 250kgmD = 1100kgkM = 100mmkD = 500mma = 1.2ms-2

G1/2 = 0.9Z = 10NmY = 200Nm

Find: M

(a) Solution:

M1

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M2

According to Newton’s second law:

(b) Solution:

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250kga

F

250g

F

rD


M1

Same as (a):

M2


M3

EXAMPLE 1.2

Figure 1.2 shows a lift is driven by an electric motor. The motor produces the maximum

torque of 1000Nm. The efficiency of each gear contact is 90%. Determine the load M2 to

accelerate the lift at 0.5m/s2 with bring total of load 2000kg (including mass of lift).

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F

250g

a

f 10o

F

rD

250g

250g(sin10)

250g(sin10) 10o250g(cos10)


Movement of lift and load M2 resisted by 500N fix force. The data for gear system are shown

below.

Number of gear teeth, t Moment of inertia, It1 = 80t2 = 100t3 = 20t4 = 40

Drum shaft, ID = 50kgm2

Middle shaft, IT = 20kgm2

Motor shaft, IM = 2kgm2

Figure 1.2

ANSWER EXAMPLE 1.2

Given: M = 2000kgdD =1mt1 = 80t2 = 100t3 = 20

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ID

IT

IM

a a M2


t4 = 40IM = 50kgm2

IT = 20kgm2

ID = 2kgm2

a = 0.5ms-2

G1/2 = 0.9f = 500N

Find: M2

Solution:

M1

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M2


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2000kga f

F1

2000g

M2f a

F2

M2g

F1 F2

rD


EXAMPLE 1.3

Refer to Figure 1.3. A motor is needed to accelerate the drum (diameter = 0.8m) through a

reducer gear. A rope which is wind to a drum is needed to pull a mass, M above a tilt plane at

slope of 1 in 20. The friction between the mass and tilt surface is 1500N. The total of torque

of drum is 3800Nm to accelerate the mass at 0.8ms-2. The gear efficiency is 95% and the

torque friction is equivalent to fM = 10Nm at the motor shaft and fD = 20Nm at the drum

shaft. If the motor speed is 477.5rpm, determine mass, M and power of motor. (Use IM =

9kgm2; ID = 60kgm2; t1 = 25; t2 = 125).

Figure 1.3

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Drum

Motor IM

ID 2

1

M

a = 0.8ms-2


ANSWER EXAMPLE 1.3

Given: dD =0.8mt1 = 25t2 = 125IM = 9kgm2

ID = 60kgm2

f = 1500ND2 = 3800Nma = 0.8ms-2

G1/2 = 0.95Z = 10NmY = 20NmNM = 477.5rpmSlope = 1/20

Find: M; PM

Solution:

M1

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M2


M3

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Mg θ

F a

f 120

θ

FrD

Mg(sin)

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1-3 Gear Train Application-Example