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

6.2)

Simple gear train have the following specifications

circular pitch is (p) 35.5mm .

Pitch circle is (p) = mπ

So the module m = p/π 

m = 35.5/π 

m = 11.30

So module  (m) is 11.3 mm .

Gear speed ratio is 4.5:1 .

So gear ratio is 4.5:1 .

Central distance is 500 mm .

6.2.1)

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 4.5:1  .

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

Then the number of teeth of  ring gear is Z₁ = 4.5x . 

a = ( 4.5x m + x m) / 2

a = m (4.5x  + x) / 2

500 = 11.3 (5.5 x) / 2

11.3 (5.5 x)= 1000

5.5 x = 1000/11.3

5.5x = 88.49

x = 88.49/5.5

x = 16.090

So the number of teeth on  pinion is ≈ 16 .

The number of teeth on  ring gear is 4.5*16.090 = 72.40 .

So the number of teeth on ring gear is ≈ 72 .

 

Answer 2

6.2.2)

Simple gear train have the following specifications

circular pitch is (p) 35.5mm .

Pitch circle is (p) = mπ

So the module m = p/π 

m = 35.5/π 

m = 11.30

So module  (m) is 11.3 mm .

Gear speed ratio is 4.5:1 .

So gear ratio is 4.5:1 .

Central distance is 500 mm .

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 4.5:1  .

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

Then the number of teeth of  ring gear is Z₁ = 4.5x . 

a = ( 4.5x m + x m) / 2

a = m (4.5x  + x) / 2

500 = 11.3 (5.5 x) / 2

11.3 (5.5 x)= 1000

5.5 x = 1000/11.3

5.5x = 88.49

x = 88.49/5.5

x = 16.090

So the number of teeth on  pinion is Z₂ ≈ 16 .

The number of teeth on  ring gear is 4.5*16.090 = 72.40 .

So the number of teeth on ring gear is Z₁ ≈ 72 .

The true  Centre distance can be evaluated using the formula 

a = ( Z₁ m + Z₂ m) / 2  

a = m ( Z₁ + Z₂ ) / 2  

a = 11.3 ( 72 + 16 ) / 2  

a = 11.3 (88 ) / 2  

a = 11.3 * 44

a = 497.2 mm

So the true Centre distance is 497.2 mm .