past year exam 4

Question

1. An epicyclical gear train consists of an annulus A, having 100 teeth, a sun gear B having 40 teeth and a planetary gear C, having 30 teeth that are mounted on arm D. 1.1. if arm D rotated at 30r/min, calculate the speed of sun gear B, when annulus A is fixed. 1.2. Determine the speed of annulus A, when the sun gear B is fixed and arm D rotates at 30 r/min. 2. The following specifications apply to a simple gear train having a pinion A and a gear wheel B: Module=10mm Pressure angle=20 degrees Gear ratio=3:2 PCD of pinion=160mm Calculate the following: 2.1. the number of teeth on each gear wheel 2.2. the addendum and dedendum of the gear teeth 2.3. the outside diameters of both gears 2.4. the pitch circle diameter of the gear wheel 2.5. the total depth of tooth 2.6. the tooth thickness at the pitch circle. 3.1. a simple gear train, comprising of a pinion B with 30 teeth meshing internally with a ring gear A. the centre distances between the shaft is 45mm and the teeth have a module of 1,5mm. Calculate the following: 3.1.1. the number of teeth on the ring gear A 3.1.2. the pitch circle diameter of the ring gear 3.2. A simple gear train consists of a gear and pinion of which the shaft centre is approximately 635mm apart. Assume that the circular pitch is 31,46mm and that gears have a speed ratio 3,3:1 Calculate the following: 3.2.1. the number of teeth on each gear 3.2.2. the actual centre distance between the two shafts 4. The epicyclic gear train consists of an annulus, a sun gear with 60 teeth and three planetary gears each having 20 teeth. The output shaft is connected to the arm carrying the planetary gears, while the input shaft is connected to the sun gear which is rotating at 450r/min in a clockwise direction. Calculate the following: 4.1. the speed and direction of rotation of the output shaft if the annulus is fixed 4.2. the speed and direction of rotation of the opposite direction from that of the sun gear, which is rotating clockwise at 450r/min. 5. A compound gearing system consists of an input gear A having 64 teeth and rotating at 800r/min, an intermediate shaft on which two gears, B and C, are mounted having 40 and 80 teeth respectively and an output gear D rotating at 3200r/min. if this gearing system has a module of 1,5mm, calculate the ffg: 5.1.the rotating speed of gears B and C 5.2. the number of teeth on the output gear D 5.3. the addendum of the gear teeth 5.4. the dedendum of the gear teeth 5.5. the centre distance “X” 5.6. the centre distance “Y” 6.1. a pinion with 40 teeth is meshing with ring gear. The centre distances between the shafts is approximately 75mm and the module of the teeth is 3mm. Determine the pitch-circle diameter of the ring gear and the number of teeth on the ring gear. 6.2. a simple gear train consists of a gear wheel and pinion with centre distance of approximately 500mm. assume the circular pitch is 35,5mm and the gears must have a velocity ration of 4,5:1 Determine the following: 6.2.1. the number of teeth on each gear 6.2.2. the true centre distances between the two shafts

Answer 1

3.2)

Simple gear train have the following specifications

The Centre distances between the shaft is  (a) 635 mm .

Pitch circle is (p) = 31.46 mm

Pitch circle is (p) = mπ

So the module m = p/π 

m = 31.46/π 

m = 10.01

So module  (m) is 10 mm .

Gear ratio is 3.3:1 .

3.2.1)

The number of teeth of gear and pinion can be evaluated using the formula of Centre distance .

a = ( d _g + d _p) / 2    ----------------(1)

Where d _g = gear pitch circle diameter & d _p = pinion pitch circle diameter 

pitch circle diameter  d = z . m   --------------------(2)

From (1) and (2)

a = ( Z₁ m + Z₂ m) / 2  

The Gear ratio (Z₁:Z₂) is 3.3:1  .

 

Let the number of teeth of  pinion is Z₂ = x .

Then the number of teeth of  ring gear is Z₁ = 3.3x . 

a = ( 3.3x m + x m) / 2

a = m (3.3x  + x) / 2

635 = 10 (4.3x) / 2

10 (4.3x) = 1270

4.3x = 1270/10

4.3x = 127

x = 127/4.3

x = 29.49

So the number f teeth of  pinion is ≈ 29 .

The number of teeth of  ring gear is 3.3*29.49 = 97.25 .

So the number f teeth of ring gear is ≈ 97 .

Answer 2

4)

Epicyclical gear train consists of an annulus A, sun gear S and planetary gear P .

Number of teeth for sun gear is S= 60 .

Number of teeth for planetary gear is P = 20 .

So the ratio of speeds of sun gear and planetary gear is 60:20 ⇒ 3:1 .

Input shaft is connected to sun gear .

Output shaft is connected to planetary gear .

4.1)

Sun gear  is rotating with a speed of 450 r/min in a clockwise direction .

The speed of out put shaft ( planetary gear ) can be evaluated by the ratio of speeds .

So the speed of the planetary gear is 450*(1/3)

                                                    = 150 RPM .

The speed of the planetary gear is 150 RPM .

When the sun gear is rotating in clockwise direction then direction of planetary gear is anti clockwise direction .

So the speed of the planetary gear is 150 RPM in anti clockwise direction .

Answer 3

 

 4.2)

Epicyclical gear train consists of an annulus A, sun gear S and planetary gear P .

Number of teeth for sun gear is S= 60 .

Number of teeth for planetary gear is P = 20 .

So the ratio of speeds of sun gear and planetary gear is 60:20 ⇒ 3:1 .

Input shaft is connected to sun gear .

Output shaft is connected to planetary gear .

Sun gear  is rotating with a speed of 450 r/min in a clockwise direction .

So the direction of sun gear is clockwise .

The direction of opposite to the direction of sun gear means anti clock wise direction .

If the direction of sun gear is clock wise then the direction of planetary gear is anti clockwise .

Hence the direction of opposite to the direction of sun gear means simply direction planetary gear .

So the speed of  planetary gear can be evaluated by the ratio of speeds .

So the speed of the planetary gear is 450*(1/3)

                                                    = 150 RPM .

The speed of the planetary gear is 150 RPM .

So the speed of the planetary gear is 150 RPM in anti clockwise direction .

  

Answer 4

5)

A compound gearing system consists of an input gear A having 64 teeth .

Number of teeth for gear B is  40 .

Number of teeth for gear C is 80 .

The gear A is attached to input shaft Which is rotating at 800 r/min .

Gear B and gear C are attached to intermediate shaft .

The gear D is attached to output shaft Which is rotating at 3200 r/min .

Module of the gear system is 1.5 mm .

5.1)

 

The ratio of gear A to gear B 64:40 ⇒ 8:5 .

The speed of gear A is 800 r/min .

Then the speed of gear B is 800*(5/8)

                                           = 500 RPM .

So the speed of gear B is 500 RPM .

Gear B and gear C are attached to intermediate shaft .

Hence thier speeds are same .

So the speed of gear C is 500 RPM .

Answer 5

5.2)

A compound gearing system consists of an input gear A having 64 teeth .

Number of teeth for gear B is  40 .

Number of teeth for gear C is 80 .

The gear A is attached to input shaft Which is rotating at 800 r/min .

Gear B and gear C are attached to intermediate shaft .

The gear D is attached to output shaft Which is rotating at 3200 r/min .

Module of the gear system is 1.5 mm .

 

The ratio of gear A to gear B 64:40 ⇒ 8:5 .

The speed of gear A is 800 r/min .

Then the speed of gear B is 800*(5/8)

                                           = 500 RPM .

So the speed of gear B is 500 RPM .

Gear B and gear C are attached to intermediate shaft .

Hence thier speeds are same .

So the speed of gear C is 500 RPM .

Gear C attached to the gear D so the ratio of speeds of gear C to gear D is 500:3200

                                                                                                        = 5:32

The ratio of speeds of gear C to gear D is 5:32 .

So the number of teeth on gear D is 80*(32/5)

                                                  = 16*32

                                                 = 512

So the number of of teeth on gear D is  512 .

 

Answer 6

5.3)

A compound gearing system consists of an input gear A having 64 teeth .

Number of teeth for gear B is  40 .

Number of teeth for gear C is 80 .

The gear A is attached to input shaft Which is rotating at 800 r/min .

Gear B and gear C are attached to intermediate shaft .

The gear D is attached to output shaft Which is rotating at 3200 r/min .

Module of the gear system is 1.5 mm .

Addendum of the gear teeth can be calculated using the formula  h _a = m .

 h _a = m

h _a =  1.5 mm

So the addendum of the gear teeth is 1.5 mm .

Answer 7

5.4)

A compound gearing system consists of an input gear A having 64 teeth .

Number of teeth for gear B is  40 .

Number of teeth for gear C is 80 .

The gear A is attached to input shaft Which is rotating at 800 r/min .

Gear B and gear C are attached to intermediate shaft .

The gear D is attached to output shaft Which is rotating at 3200 r/min .

Module of the gear system is 1.5 mm .

Dedendum of the gear teeth can be calculated using the formula  h _f = 1.25 m .

 h _f = 1.25 m

 h _f = 1.25 * 1.5 

 h _f = 1.875 mm

So the dedendum of the gear teeth is 1.875 mm .

Answer 8

5.5)

A compound gearing system consists of an input gear A having 64 teeth .

Number of teeth for gear B is  40 .

Number of teeth for gear C is 80 .

The gear A is attached to input shaft Which is rotating at 800 r/min .

Gear B and gear C are attached to intermediate shaft .

The gear D is attached to output shaft Which is rotating at 3200 r/min .

Module of the gear system is 1.5 mm .

The centre distance X is the centre distance between gear A and gear B .

The Centre distance formula X  can be calculated using the  .

a = ( d _g + d _p) / 2    ----------------(1)

Where d _g = pitch circle diameter of gear A & d _p = pitch circle diameter of gear B .

pitch circle diameter  d = z . m   --------------------(2)

From (1) and (2)

a = ( Z₁ m + Z₂ m) / 2  

a =  m( Z₁  + Z₂ ) / 2  

Where Z₁  is number of teeth on gear A .

           Z₂  is number of teeth on gear B .

a = 1.5( 64  + 40 ) / 2  

a = 1.5( 104 ) / 2

a = 1.5 * 52

a = 78 mm

So the centre distance X is 78 mm .

Answer 9

5.6)

A compound gearing system consists of an input gear A having 64 teeth .

Number of teeth for gear B is  40 .

Number of teeth for gear C is 80 .

The gear A is attached to input shaft Which is rotating at 800 r/min .

Gear B and gear C are attached to intermediate shaft .

The gear D is attached to output shaft Which is rotating at 3200 r/min .

Module of the gear system is 1.5 mm .

The centre distance Y is the centre distance between gear C and gear D .

The Centre distance formula Y  can be calculated using the  .

a = ( d _g + d _p) / 2    ----------------(1)

Where d _g = pitch circle diameter of gear C & d _p = pitch circle diameter of gear D .

pitch circle diameter  d = z . m   --------------------(2)

From (1) and (2)

a = ( Z₁ m + Z₂ m) / 2  

a =  m( Z₁  + Z₂ ) / 2  

Where Z₁  is number of teeth on gear C ( Z₁ = 80).

           Z₂  is number of teeth on gear D .

Calculate for Z₂:

The ratio of gear A to gear B 64:40 ⇒ 8:5 .

The speed of gear A is 800 r/min .

Then the speed of gear B is 800*(5/8)

                                           = 500 RPM .

So the speed of gear B is 500 RPM .

Gear B and gear C are attached to intermediate shaft .

Hence thier speeds are same .

So the speed of gear C is 500 RPM .

Gear C attached to the gear D so the ratio of speeds of gear C to gear D is 500:3200

                                                                                                        = 5:32

The ratio of speeds of gear C to gear D is 5:32 .

So the number of teeth on gear D is 80*(32/5)

                                                  = 16*32

                                                 = 512

So the number of of teeth on gear D is  Z₂ = 512 .

Now calculate for centre distance Y :

a =  m( Z₁  + Z₂ ) / 2

a = 1.5( 80  + 512 ) / 2  

a = 1.5( 592) / 2

a = 1.5 * 296

a = 444 mm

So the centre distance Y is 444 mm .

Answer 10

6.1)

Simple gear train have the following specifications

Module of  (m) is 3 mm .

Number of teeth on pinion is 40 .

The centre distance is (a) = 75 mm .

The pitch circle diameter of ring gear can be calculated using the Centre distance formula .

a = ( d _g + d _p) / 2    

Where d _g = gear pitch circle diameter & d _p = pinion pitch circle diameter 

pitch circle diameter of pinion is  d _p = z . m   

d _p = 40*3

d _p = 120 .

SO pitch circle diameter of pinion is  d _p = 120  .

Now calculate pitch circle diameter of ring gear .

a = ( d _g + d _p) / 2  

75 = ( d _g + 120) / 2 

d _g + 120 = 75*2

d _g + 120 = 150

d _g  = 150 - 120 

d _g  = 30 .

So the pitch circle diameter of ring gear is 30 .

Now calculate for number teeths of ring gear :

Number teeth of ring gear can be calculated using the formula  z = d _g/m .

 z =d _g/m

 z = 30/3

z = 10 

So the number of teeth on the ring gear is 10 .