Post on 24-Oct-2014
transcript
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
Mechanics of Machines Gear TrainFadzli’0607
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
Mechanics of Machines Gear TrainFadzli’0607
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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Mechanics of Machines Gear TrainFadzli’0607
M2
According to Newton’s second law:
(b) Solution:
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250kga
F
250g
F
rD
Mechanics of Machines Gear TrainFadzli’0607
M1
Same as (a):
M2
According to Newton’s second law:
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)
Mechanics of Machines Gear TrainFadzli’0607
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
Mechanics of Machines Gear TrainFadzli’0607
t4 = 40IM = 50kgm2
IT = 20kgm2
ID = 2kgm2
a = 0.5ms-2
G1/2 = 0.9f = 500N
Find: M2
Solution:
M1
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Mechanics of Machines Gear TrainFadzli’0607
M2
According to Newton’s second law:
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2000kga f
F1
2000g
M2f a
F2
M2g
F1 F2
rD
Mechanics of Machines Gear TrainFadzli’0607
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
Mechanics of Machines Gear TrainFadzli’0607
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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Mechanics of Machines Gear TrainFadzli’0607
M2
According to Newton’s second law:
M3
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Mg θ
F a
f 120
θ
FrD
Mg(sin)