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6.1 A spur gear, with a pitch-circle diameter of 165 mm and 110 teeth, is needed for a gearbox - NSC Mechanical Technology Fitting and Machining - Question 6 - 2022 - Paper 1

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6.1 A spur gear, with a pitch-circle diameter of 165 mm and 110 teeth, is needed for a gearbox. Calculate the following: 6.1.1 Module 6.1.2 Outside diameter FIGU... show full transcript

Worked Solution & Example Answer:6.1 A spur gear, with a pitch-circle diameter of 165 mm and 110 teeth, is needed for a gearbox - NSC Mechanical Technology Fitting and Machining - Question 6 - 2022 - Paper 1

Step 1

6.1.1 Module

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Answer

To calculate the module (m) of a spur gear, we can use the formula:

m=DTm = \frac{D}{T}

Where:

  • DD = pitch-circle diameter (165 mm)
  • TT = number of teeth (110)

Substituting the values gives: m=165110=1.5 mmm = \frac{165}{110} = 1.5 \text{ mm}

Thus, the module of the spur gear is 1.5 mm.

Step 2

6.1.2 Outside diameter

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Answer

The outside diameter (OD) of the spur gear can be calculated using the formula:

OD=D+2mOD = D + 2m

Substituting the values:

  • D=165D = 165 mm
  • m=1.5m = 1.5 mm

We get: OD=165+2×1.5=168 mmOD = 165 + 2 \times 1.5 = 168 \text{ mm}

Therefore, the outside diameter of the spur gear is 168 mm.

Step 3

6.2.1 Maximum width (W) of the dovetail

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Answer

To find the maximum width (W) of the dovetail, we note from the figure:

  • The total width across the two circles is 120 mm
  • The diameter of each circle is 30 mm

The width of the dovetail can be calculated as: W=1202×30=60 mmW = 120 - 2 \times 30 = 60 \text{ mm}

Thus, the maximum width of the dovetail is 60 mm.

Step 4

6.2.2 Distance (m) between the precision rollers

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The distance (m) between the precision rollers can be found from the drawing as:

  • The diameter of the circles is 30 mm, with 11 mm separation between their centers.

Thus: m=30+11=41 mmm = 30 + 11 = 41 \text{ mm}

Therefore, the distance between the precision rollers is 41 mm.

Step 5

6.3.1 Calculate the indexing that is needed

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Answer

The indexing requires calculating the number of divisions to index the spur gear:

The formula to find the indexing (I) is: I=TAI = \frac{T}{A} Where:

  • T=163T = 163 (number of teeth)
  • A=160A = 160 (divisions of the simple indexing)

Substituting: I=1631601.01875I = \frac{163}{160} \approx 1.01875

Thus, about 1.01875 rotations are needed in the dividing head.

Step 6

6.3.2 Calculate the change gears that are required

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Answer

For the change gears required, use the formula for calculating gear ratios:

The required ratio can be derived from: R=TNR = \frac{T}{N} Where NN is the number of turns of the handle that correlate to the index. Assuming a direct proportionality with the given ratio: N=AR=16040=4N = \frac{A}{R} = \frac{160}{40} = 4

Thus, 4 change gears are required.

Step 7

6.4 Name TWO types of balancing methods

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Answer

Two types of balancing methods include:

  1. Static Balancing
  2. Dynamic Balancing

Static balancing focuses on balancing the object at rest. In contrast, dynamic balancing examines forces when the object is in motion.

Step 8

6.5 State TWO advantages of balancing the rotating components of a machine

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Answer

Two advantages of balancing rotating components include:

  1. Reduced Vibration: Balancing minimizes vibrations, leading to smoother operation and reduced wear.
  2. Increased Lifespan: Properly balanced components tend to last longer, which reduces maintenance costs and downtime.

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