how to solve 15

Question

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 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”

Answer 1

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 2

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 .

Number of teeth for sun gear is S= 60 .

Number of teeth for planetary gear is P = 20 .

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

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 .

Hence 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 .

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

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 .

Hence the speed of the opposite direction from that of the sun gear (planetary gear) is 

150 RPM in anti clockwise direction .

Answer 4

5.2)

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

Number of teeth for gear B is  40 and 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 5

(5.3)

Module of the gear system is m = 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

Hence the addendum of the gear teeth is 1.5 mm .

Answer 6

5.4)

Module of the gear system is m = 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 7

5.5)

Gear A is a input shaft and gear D is out put shaft .

Gear B and gear C are attached to intermediate shaft .

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  = 64.

           Z₂  is number of teeth on gear B = 40 .

           m is module of the gear system is = 1.5 mm .

a = 1.5( 64  + 40 ) / 2  

a = 1.5( 104 ) / 2

a = 1.5 * 52

a = 78 mm

Hence the centre distance X is 78 mm .

Answer 8

5.6)

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 = ( number of teeth of A:number of teeth of 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 .