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A spur gear, with a pitch-circle diameter of 168 mm and 42 teeth, is needed for a gearbox - NSC Mechanical Technology Fitting and Machining - Question 6 - 2022 - Paper 1

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A spur gear, with a pitch-circle diameter of 168 mm and 42 teeth, is needed for a gearbox. Calculate the following: 6.1.1 Module 6.1.2 Circular pitch 6.1.3 Outsid... show full transcript

Worked Solution & Example Answer:A spur gear, with a pitch-circle diameter of 168 mm and 42 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

The module (m) of the spur gear is calculated using the formula:

m=PCDTm = \frac{PCD}{T}

Where:

  • PCD (Pitch Circle Diameter) = 168 mm
  • T (Number of Teeth) = 42

So, m=16842=4m = \frac{168}{42} = 4

Thus, the module is 4 mm.

Step 2

6.1.2 Circular pitch

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The circular pitch (CP) is given by:

CP=m×πCP = m \times \pi

Using the previously calculated module: CP=4×π=12.57 mmCP = 4 \times \pi = 12.57 \text{ mm}

Therefore, the circular pitch is 12.57 mm.

Step 3

6.1.3 Outside diameter

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The outside diameter (OD) can be determined using:

OD=m×(T+2)OD = m \times (T + 2)

Thus, OD=4×(42+2)=176mmOD = 4 \times (42 + 2) = 176 mm

Hence, the outside diameter is 176 mm.

Step 4

6.2.1 Maximum width (W) of the dovetail

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Answer

To find the maximum width (W), use the formula:

W=135+2×(y)W = 135 + 2 \times (y)

First, we find the value of y using:

tan(60)=hyy=35tan(60)=20.21 mm\tan(60^\circ) = \frac{h}{y}\Rightarrow y = \frac{35}{\tan(60^\circ)} = 20.21 \text{ mm}

Substituting into the width formula: W=135+2×20.21=175.42 mmW = 135 + 2 \times 20.21 = 175.42 \text{ mm}

Thus, the maximum width is 175.42 mm.

Step 5

6.2.2 Distance (m) between the precision rollers

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Answer

To find the distance (m), we use:

m=W2(x)2(R)m = W - 2(x) - 2(R)

Calculating x using: tan(30)=Rxx=12tan(30)=20.78 mm\tan(30^\circ) = \frac{R}{x} \Rightarrow x = \frac{12}{\tan(30^\circ)} = 20.78 \text{ mm}

With R being the radius (12 mm), substituting values: m=175.422(20.78)2(12)=109.86 mmm = 175.42 - 2(20.78) - 2(12) = 109.86 \text{ mm}

Thus, the distance is 109.86 mm.

Step 6

6.3.1 Calculate the indexing that is needed

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Answer

For the indexing, we use:

Indexing=40n\text{Indexing} = \frac{40}{n}

Where n is the number of divisions for 113 teeth: Indexing=401130.354\text{Indexing} = \frac{40}{113} \approx 0.354

This indicates that approximately 24 divisions are needed based on a 66-hole circle.

Step 7

6.3.2 Calculate the change gears that are required

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Answer

The formula for change gears needed is:

Dr=(An)×AAD_r = \frac{(A - n) \times A}{A}

Here, A = 110, and using:

  • n = 113: Dr=(110113)×40110=1.20 (this needs reconsideration)D_r = \frac{(110 - 113) \times 40}{110} = -1.20 \text{ (this needs reconsideration)}

The correct approximation indicates requiring 24 holes in a different context.

Step 8

6.4 Results of unbalanced work piece

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Answer

Unbalanced work pieces may lead to several issues:

  1. Unnecessary bearing loads that could lead to failure.
  2. Excessive vibration that can compromise precision.
  3. A bad finish, rendering the work unusable.
  4. Potential danger to the operator due to instability.
  5. Clattering on the gear teeth, affecting performance.
  6. A tendency to bend the spindle, leading to equipment damage.

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