A spur gear has 51 teeth and a module of 3 - NSC Mechanical Technology Fitting and Machining - Question 6 - 2020 - Paper 1

The cutting depth of a spur gear can be calculated using the formula:

$\text{Cutting Depth} = \frac{2}{3} \times \text{Module}$

With the given module of 3, we have:

$\text{Cutting Depth} = \frac{2}{3} \times 3 = 2 \text{ mm}$

Therefore, the cutting depth of the gear is 2.5 mm.

For simple indexing, we need to consider the number of teeth in the gear and how many divisions are required:

The formula for simple indexing is:

$\text{Simple Indexing} = \frac{N}{N_0}$

Where N is the total number of teeth and $N_0$ is the number of divisions.

For this gear:

• Total teeth (N) = 51
• Using a common division method, we find: 51 full turns yield 0 full turns and 40 holes on the 51-hole circle.

Thus, the required simple indexing is 51/8 = 6.375.

For differential indexing, using 80 divisions is ideal as it provides better accuracy for the milling process. This is calculated based on the formula:

$\text{Differential Indexing} = \frac{80 \text{ divisions}}{N}$

This means the differential indexing results in 40 divisions.

To find the change gears for the process, we use:

$\text{Driver} = 40 \text{ and } \text{Driven} = 40 \text{ when using 80 teeth.}$

From the calculations, the gearing ratio can be reduced as follows:

• With Driver as 40 and Driven as 80, the required change gears would be 32, 48, 72, or 80.

The index plate rotates in the opposite direction to the index crank handle. Hence, if the crank handle turns, the index plate will turn counter-clockwise.

To calculate distance X across the rollers, we consider:

$X = Y - 2(AC + r)$

Where Y is the vertical distance and AC can be calculated as follows:

• For triangle ABC:

Using trigonometric functions: $\tan(30^{\circ}) = \frac{BC}{AC}$ Solving for AC: $AC = \frac{BC}{\tan(30^{\circ})}$ Here, BC is 12.5 and we find: $AC = \frac{12.5}{\tan(30^{\circ})} = 21.65 mm$

Now replacing the values into the formula for calculating X leads to:

• Substitute Y, X, and any additional constants to find the final X = 123.04 mm.

1. Prevent unnecessary bearing loads: Balancing ensures that the workpiece maintains stability during rotation, which minimizes uneven wear on the bearings.

2. Ensure safety of the worker: An unbalanced workpiece can cause vibrations that may lead to loss of control, posing safety risks during the machining process.