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Example 1(Taken from Roslan, Che’ Abas, Yunus (2001), UTM) 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 sha...

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Example 1(Taken from Roslan, Che’ Abas, Yunus (2001), UTM) 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) 250 1100

Motor shaft Hoist shaft

Radius of gyration (mm) 100 500

Calculate the torque of the motor needed to bring up the load with acceleration 1.2 m/s 2. Neglect friction effect. Dia = 1.2 m Hoist

Figure 13: Gear system attached to hoist Solution Total torque at motor to bring up load Ttotal  TM 1  TM 2 Where TM 1 = Torque to overcome equivalent inertia (refer to motor side). TM 2 =Torque to accelerate the load through gear system a) Consider for TM 1 I n2 I equiv  I M  G G From Thus

I M = Motor shaft inertia

I M  mk 2  2500.1  2.5 kgm2 IG = Hoist shaft inertia 2 I G  11000.5  275 kgm2 2

N1 20   0.1 N 200 2 Gear ratio,  2750.12  I equiv  2.5     5.55 0.9   Put into kgm2 n

aG   G rG Acceleration of hoist, 1.2 2 G  0.6 rad/s2 Thus

m 

From the gear ratio, angular acceleration of motor, T  I equiv m Now torque due to equivalent inertia, M 1 TM 1  5.55(20)  111.1 Nm.

G n



2  20 0.1 rad/s2

b) Consider for TM 2

F  250 g 250a F  ma From Newton 2nd Law, F  250g  a   2509.81  1.2 F  2752.5 N

Then, torque at hoist TG  Fr  2752.50.6  1651.5

Nm

But due to gear efficiency (since the hoist shaft is connected to the gear system), torque to accelerate the load, T n 1651.50.1 TM 2  G 1 / 2   183.5  G ,1 / 2 0.9

Nm

TG Hoist r

Then total torque referred to motor side is; Ttotal  TM 1  TM 2

a

F

Ttotal  111.1  183.5

Ttotal  294.6

Nm

250 kg

250g

Figure 14: Free body diagram of hoist and load

Example 2(Taken from Roslan, Che’ Abas, Yunus, (2001), UTM) Figure 12 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.

Hoist

Diameter = 1.2 m

Figure 15: Loading system on gear

Solution n1 / 2  1 n 1 I 3.5 , 3 / 4 4.5 ,  G  0.9 Given that I M =5 kgm2, I T  40 kgm2, G =500 kgm2, Neglect friction effect. Total torque required for the motor is Ttotal  TM 1  TM 2 Where TM 1 = Torque to overcome equivalent inertia (refer to motor side). TM 2 =Torque to accelerate the load through gear system a) Consider for TM 1 T  I 2 m 2 of gear system with friction effect is neglected, Recall that MI1 (n equiv I, but ) (n for) 2two (n3 / set 4) I equiv  I m  T 1 / 2  G 1 / 2  G ,1 / 2 ( G ,1 / 2 )( G ,3 / 4 ) I equiv

40( 1 ) 2 500( 1 ) 2 ( 1 ) 2 3.5  3.5 4.5  11.116 Nm  5 0.9 (0.9)(0.9)

From question, a 0.4 given that, G  G  rG 0.6

aG  0.4

m/s2, thus;

 G  0.6667 rad/s2  G  G T 1 1 1      From gear ratio,  m  T  m 4.5 3.5 15.75 Thus, Thus,

 m  15.75 G  m  15.750.6667  10.5 rad/s2 Tm1  I equiv m  11.11610.5 TM 1  116.72 Nm. b) Consider for TM 2 in Figure 13; F  ma From Newton 2nd Law, 

Hoist

4500 g  F1  4500a a

a

F1

F1  4500g  a   45009.41  42345 N

F2

4500 kg

F2  6000 g  6000a F2  6000a  g   61260 N

6000 kg

Resultant torque at hoist FR  F2  F1  18.915 kN 4500g

6000g

Figure 16: Loading FBD

Thus torque at hoist TG  FR rG  18.9150.6  11.349

kN

T to motor side will be denoted as TM 2 and is related by It is known thatTG nG1 /referred 2 n3 / 4 TM 2 

 G ,1 / 2 G ,3 / 4

TM 2 

  

1 11.349 1 3.5 4.5  889.6 0.90.9 Nm

Thus total torque at motor required is Ttotal  TM 1  TM 2 Ttotal  116.72  889.6

Ttotal  1006.32

Nm.

Example 3(Taken from Roslan, Che’ Abas, Yunus, (2001), UTM) Hoist Dia = 0.9 m

Figure 17: Gear with inclined loading

Figure 14 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 T  800 Nm. Calculate the effect on the middle shaft, TX  150 Nm and at hoist shaft is y acceleration of the load at the above condition.

Solution For the overall gear ratio,1 1 1 n0  n1 / 2 n3 / 4    3.5 4.5 15.75 Total torque required by motor to raise load Ttotal  TM 1  TM 2  TM 3 Where TM 1 = Torque to overcome equivalent inertia (refer to motor side). TM 2 =Torque to accelerate the load through gear system TM 3  Total torque to overcome friction effect. a) Consider for TM 1 T  I equiv m Previously, M 1 I reducing I ( n ) 2 ( n3 / 4 ) 2 (n ) 2 system, For double set of gear I equiv  I m  T 1 / 2  G 1 / 2  G ,1 / 2 ( G ,1 / 2 )( G ,3 / 4 ) I equiv

20( 1 ) 2 100( 1 ) 2 ( 1 ) 2 3.5  3.5 4.5  7.312  5 0.9 (0.9)(0.9) kgm2

From

a   G rG

G  , thus,

a a  rG 0.45

G 1  a    m  15.75 G  15.75   35a  15 . 75 0 . 45   m Also from gear ratio, , thus, Thus TM 1  7.31235a   255.92a Nm b) Consider for TM 2 as in Figure 15; From

 F  ma, F1  FR  mg sin   ma F1  5000a  1000  50009.81 1

F1

 50

FR

Mg sinθ

5000g

θ

F1  5000a  1981 N Thus, torque to accelerate hoist TG  F1r  5000a  19810.45 TG  2250a  891.45

F1

Use gear efficiency to relate T n n TM 2  G 1 / 2 3 / 4

hoist

TM 3

Friction effect Tcan n be grouped T n ntogether to form TM 3  X 1 / 2  Y 1 / 2 3 / 4

 G ,1 / 2

TM 3

From

with TM 2

TM 2  176.4a  69.876 Nm

Figure 18: inclined loading

Consider for

TG

 G ,1 / 2 G ,3 / 4

r

c)

Nm

TM 3

where;

 G ,1 / 2 G ,3 / 4

 1  1   1  150    800 3.5  4.5  3.5     110.33   0.9 0.9 2 Nm

Ttotal  TM 1  TM 2  TM 3

1500  255.92a   176.4a  69.876  110.33

Thus a  3.1 m/s2

Tutorial 1-GEAR SYSTEM

1.

The axes of two parallel shafts are to be 600mm apart approximately, and have to be connected by spur gear, having a circular pitch of 30 mm. If gear A rotate at 200 rpm and gear B rotate at 600 rpm, find the number of teeth on each gear.

600 mm

2.

Figure below showed a motor used to accelerate a hoist through a set of gear system. Gear for the hoist’s shaft has 200 teeth and gear for motor shaft has 20 teeth. Gear efficiency is 90 %. Moment of inertia for the motor shaft is 2.5 kgm² and hoist shaft is 275 kgm². The rope that carries a 250 kg load are encircle on hoist with diameter 1.2 m. By neglecting the friction, find a) gear ratio, b) equivalent moment inertia for a gear system, c) the total torque required by the motor to accelerate the load at acceleration of 1.0 m/s².

Motor

Set of gear Hoist

Rope Load

3

The diagram above shows a gear train composed of three gears. Gear A revolves at 60 revs/min in a clockwise direction.

a)

GEAR A 20 TEETH What is the output in revolutions per minute at Gear C?

b)

In what direction does Gear C revolve ?

GEAR B 60 TEETH

GEAR C 10 TEETH

[ 120 rpm, clockwise ]

4

Spur gears are arranged as shown below. Driver gear is rotating at 200 rpm in clockwise direction. Calculate;

(i)

Rotational speed, N of gear D and

( ii ) Gear ratio, n of the gear train system

(…..)

5

FIGURE Q1(b) shows a motor used to accelerate a hoist through a set of gear system. Gear for the hoist’s shaft has 100 teeth and gear for motor shaft has 10 teeth. Gear efficiency is 90 %. Moment of inertia for the motor shaft is 2 kgm² and hoist shaft is 200 kgm². Motor shaft attached by a support bearing with friction, Tx is 9 Nm. The rope that carries a 147 kg load are encircled on hoist with radius 0.5 m. By neglecting the gear friction, find (i)

Gear ratio n,

( ii )

Equivalent moment inertia, Iequi for a gear system and

( iii ) The total torque, Ttotal required by the motor to accelerate the load at acceleration

Set of gear

Motor Tx

Hoist

Rope Load...