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

1)

Epicyclical gear train consists of an annulus A, having A = 100 teeth

Number of teeth for sun gear is S= 40 .

Number of teeth for planetary gear is P = 30 .

1.1)

Speed of arm D is  30 r/min .

When annulus A is fixed , we have calculate speed of sun gear .

Let us consider the case where planetary gear is driven by sun gear 

The ratio of revolutions can be given by S/(A+S) .

                                                      ⇒40/(100+40)

                                                      ⇒ 40/140

                                                      ⇒ 0.2857 

which when calculated, {(1 / 0.2857 = 3.5}, gives us a ratio of just a hair over 1 : 3.5

This means that to get one complete rotation on the output side, the input side needs to make  three and half rotations.

Speed of the arm is 30 r/min .

So the speed of the sun gear is 30 * 0.2857

                                             = 8.5714 RPM .

The speed of the sun gear is 8.5714 RPM .

 

Answer 2

 

 1.2)

Epicyclical gear train consists of an annulus A, having A = 100 teeth

Number of teeth for sun gear is S= 40 .

Number of teeth for planetary gear is P = 30 .

Speed of arm D is  30 r/min .

When sun S is fixed , we have calculate speed of annulus gear .

Let us consider the case where planetary gear is driven by annulus gear 

The ratio of revolutions can be given by (A+S) / S .

                                                      ⇒ (100+40) /40

                                                      ⇒ 140/40

                                                      ⇒ 3.5 

Gives us a ratio  3.5 : 1

This means that to get three and half complete rotation on the output side, the input side needs to make  one complete rotation .

Speed of the arm is 30 r/min .

So the speed of the sun gear is 30 * 3.5

                                             = 105 RPM .

The speed of the sun gear is 105 RPM . 

 

Answer 3

2)

Simple gear train have the following specifications

Module of pinion (m) is 10 mm .

Pressure angle of pinion is 20° .

Gear ratio is 3:2 .

PCD of pinion (d) is 160 mm .

2.1)

Number teeth of pinion can be calculated using the formula  z = d/m .

z = d/m

z = 160 mm / 10 mm

z = 16 teeth .

So the number of teeth on pinion gear is 16 teeth .

Given the wheel to pinion gear ratio is 3:2 .

so  number of teeth on wheel gear is (16*3)/2

                                                          = 48/2

                                                         = 24 teeth 

So the number of teeth on wheel gear is 24 teeth .

 

Answer 4

 

 2.2)

Simple gear train have the following specifications

Module of pinion (m) is 10 mm .

Pressure angle of pinion is 20° .

Gear ratio is 3:2 .

PCD of pinion (d) is 160 mm .

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

 h _a = m

h _a =  10 mm

So the addendum of the gear teeth is 10 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 * 10 

 h _f = 12.5 mm

So the dedendum of the gear teeth is 12.5 mm . 

 

Answer 5

2.3)

Simple gear train have the following specifications

Module of pinion (m) is 10 mm .

Pressure angle of pinion is 20° .

Gear ratio is 3:2 .

PCD of pinion (d) is 160 mm .

Number teeth of pinion can be calculated using the formula  z = d/m .

z = 160 mm / 10 mm

z = 16 teeth .

So the number of teeth on pinion gear is 16 teeth .

Given the wheel to pinion gear ratio is 3:2 .

so  number of teeth on wheel gear is (16*3)/2

                                                          = 48/2

                                                         = 24 teeth 

So the number of teeth on wheel gear is 24 teeth .

The outside diameter of the pinion gear can be calculated using the formula D _o = (z + 2) x m .

D _o = (16 + 2) x 10

D _o = 18 x 10

D _o = 180

The outside diameter of the pinion gear is 180 mm .

The outside diameter of the wheel gear can be calculated using the formula D _o = (z + 2) x m .

D _o = (24 + 2) x 10

D _o = 26 x 10

D _o = 260

The outside diameter of the pinion gear is 260 mm .

 

Answer 6

2.4)

Simple gear train have the following specifications

Module  (m) is 10 mm .

Pressure angle of pinion is 20° .

Gear ratio is 3:2 .

PCD of pinion (d) is 160 mm .

Number teeth of pinion can be calculated using the formula  z = d/m .

z = 160 mm / 10 mm

z = 16 teeth .

So the number of teeth on pinion gear is 16 teeth .

Given the wheel to pinion gear ratio is 3:2 .

so  number of teeth on wheel gear is (16*3)/2

                                                          = 48/2

                                                         = 24 teeth 

So the number of teeth on wheel gear is 24 teeth .

The pitch circle diameter of the gear wheel is can be calculated using the formula d _g  = z . m .

d _g  = z . m

d _g  = 24*10

d _g  =  240 mm

So the pitch circle diameter of the gear wheel is 240 mm .

 

Answer 7

2.5)

Simple gear train have the following specifications

Module  (m) is 10 mm .

Pressure angle of pinion is 20° .

Gear ratio is 3:2 .

PCD of pinion (d) is 160 mm .

Total depth or whole depth tooth can be calculated using the formula  h = 2.25 m .

h = 2.25 m

h = 2.25 * 10

h = 22.5 mm

So the total depth or whole depth tooth is 22.5 mm .

Answer 8

2.6)

Simple gear train have the following specifications

Module  (m) is 10 mm .

Pressure angle of pinion is 20° .

Gear ratio is 3:2 .

PCD of pinion (d) is 160 mm .

The tooth thickness at the pitch circle can be calculated using the formula  ctt = p/2 .

Where p is  the circular pitch and p = m.π .

So the tooth thickness at the pitch circle is  ctt =  m.π  /2 .

 ctt =  m.π  /2

 ctt =  10.π  /2 

 ctt =  10 * 1.5707

 ctt =  15.707 mm .

So the tooth thickness at the pitch circle is 15.7 mm .

Answer 9

3.1)

Simple gear train have the following specifications

The number of teeth on pinion gear  (Z₁) is 30 .

The Centre distances between the shaft is  (a) 45 mm .

Module  (m) is 1.5 mm .

3.1.1)

The number of teeth on ring gear can be calculated using the Centre distance formula .

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  

a =  m( Z₁  + Z₂ ) / 2  

Where Z₁  is number of teeth on pinion gear .

           Z₂  is number of teeth on ring gear .

45  =  1.5 ( 30  + Z₂ ) / 2  

1.5 ( 30  + Z₂ ) = 90

( 30  + Z₂ ) = 90/1.5

30  + Z₂  = 60

Z₂  = 60 - 30

Z₂  = 30 

So the number  of teeth on ring gear  (Z₂) is 30 .

Answer 10

3.1.2)

Simple gear train have the following specifications

The number of teeth on pinion gear  (Z₁) is 30 .

The Centre distances between the shaft is  (a) 45 mm .

Module  (m) is 1.5 mm .

Pitch circle diameter of ring gear d =Z₂ . m . 

So first we need to find the  number of teeth on ring gear Z_2 .

Calculate for  Z₂  :

The number of teeth on ring gear can be calculated using the Centre distance formula .

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  

a =  m( Z₁  + Z₂ ) / 2  

Where Z₁  is number of teeth on pinion gear .

           Z₂  is number of teeth on ring gear .

45  =  1.5 ( 30  + Z₂ ) / 2  

1.5 ( 30  + Z₂ ) = 90

( 30  + Z₂ ) = 90/1.5

30  + Z₂  = 60

Z₂  = 60 - 30

Z₂  = 30 

So the number  of teeth on ring gear  (Z₂) is 30 .

The Pitch circle diameter of ring gear d =Z₂ . m .

d = 30*1.5

d = 45 mm

The Pitch circle diameter of ring gear is 45 mm .