FIGURE 5.1 shows two spur gears that mesh. Gear A 90 teeth Gear B 30 teeth PCD = 90 mm Use the information above and calculate the: 5.1.1 Module of the small gear 5.1.2 Outside diameter of the big gear 5.1.3 PCD of the big gear 5.1.4 Dedendum of the big gear 5.1.5 Centre distance between the two gears (distance Y) 5.1.6 Required indexing for a gear with 33 teeth

NSC Mechanical Technology Automotive Calculations on drill and key sizes: FIGURE 5.1 shows two spur gears

FIGURE 5.1 shows two spur gears that mesh - NSC Mechanical Technology Automotive - Question 5 - 2016 - Paper 1

Question 5

FIGURE 5.1 shows two spur gears that mesh. Gear A 90 teeth Gear B 30 teeth PCD = 90 mm Use the information above and calculate the: 5.1.1 Module of the small ge... show full transcript

Worked Solution & Example Answer:FIGURE 5.1 shows two spur gears that mesh - NSC Mechanical Technology Automotive - Question 5 - 2016 - Paper 1

Step 1

5.1.1 Module of the small gear

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Answer

To find the module of the small gear, we use the formula:

ext{Module} = \frac{PCD}{T}

Substituting the values:

ext{Module} = \frac{90}{30} = 3

Thus, the module of the small gear is 3.

Step 2

5.1.2 Outside diameter of the big gear

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Answer

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

OD = m(T + 2)

Substituting our known values:

OD = 3(30 + 2) = 3 \times 32 = 96 \, mm

Therefore, the outside diameter of the big gear is 96 mm.

Step 3

5.1.3 PCD of the big gear

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Answer

To calculate the PCD of the big gear, we can use:

PCD = m \times T

Substituting in the values:

PCD = 3 \times 90 = 270 \, mm

Thus, the PCD of the big gear is 270 mm.

Step 4

5.1.4 Dedendum of the big gear

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Answer

The dedendum can be calculated with:

Dedendum = 1.25 \times m

Substituting in the module value:

Dedendum = 1.25 \times 3 = 3.75 \, mm

So, the dedendum of the big gear is 3.75 mm.

Step 5

5.1.5 Centre distance between the two gears (distance Y)

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Answer

The centre distance (Y) between two gears can be calculated as:

Y = \frac{PCD_a + PCD_b}{2}

Given the PCD of Gear A is 90 mm and Gear B is 270 mm:

Y = \frac{90 + 270}{2} = \frac{360}{2} = 180 \, mm

Hence, the centre distance between the two gears is 180 mm.

Step 6

5.1.6 Required indexing for a gear with 33 teeth

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Answer

For indexing of a gear, the formula can be applied as follows:

Indexing = \frac{40}{n}

where n is the number of teeth in the new gear. Thus:

Indexing = \frac{40}{33} = 1.21 \, (approx)

Concluding, for one full turn of the crank, you would need approximately 14 holes in a 66-hole plate.

FIGURE 5.1 shows two spur gears that mesh - NSC Mechanical Technology Automotive - Question 5 - 2016 - Paper 1

Worked Solution & Example Answer:FIGURE 5.1 shows two spur gears that mesh - NSC Mechanical Technology Automotive - Question 5 - 2016 - Paper 1

5.1.1 Module of the small gear

5.1.2 Outside diameter of the big gear

5.1.3 PCD of the big gear

5.1.4 Dedendum of the big gear

5.1.5 Centre distance between the two gears (distance Y)

5.1.6 Required indexing for a gear with 33 teeth

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