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

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

$OD = m(T + 2)$

Substituting the values:

$OD = 3(30 + 2) = 3 \times 32 = 96 \text{ mm}$

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

The Pitch Circle Diameter (PCD) of the big gear can also be calculated as:

$PCD = m \times T$

Substituting values:

$PCD = 3 \times 30 = 90 \text{ mm}$

Hence, the PCD of the big gear is 90 mm.

The dedendum (d) of the big gear can be calculated using:

$d = 1.25 imes m$

Substituting the module value:

$d = 1.25 \times 3 = 3.75 \text{ mm}$

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

The center distance (Y) can be calculated using:

$Y = \frac{PCD_A + PCD_B}{2}$

Where:

• PCD_A = 90 mm (small gear)
• PCD_B = 270 mm (big gear)

Calculating:

$Y = \frac{90 + 270}{2} = 180 \text{ mm}$

Therefore, the center distance between the two gears is 180 mm.

The indexing can be calculated using:

$Index = \frac{40}{n}$

Where n is the number of teeth on the gear.

For a gear with 33 teeth:

$Index = \frac{40}{33} \cdot 2 = \frac{80}{33} \approx 2.42$

Thus, the required indexing for a gear with 33 teeth is approximately 2.42.

The width of the key is given by:

$Width = \frac{Diameter}{4}$

For a diameter of 92 mm:

$Width = \frac{92}{4} = 23 \text{ mm}$

Hence, the width of the key is 23 mm.

The length of the key can be calculated as:

$Length = \frac{Diameter}{2} \cdot 1.5$

Here, Diameter is 92 mm:

$Length = \frac{92}{2} \cdot 1.5 = 69 \text{ mm}$

Thus, the length of the key is 138 mm.

The thickness of the key at the bigger end is:

$Thickness = \frac{Diameter}{6}$

Substituting the diameter:

$Thickness = \frac{92}{6} = 15.33 \text{ mm}$

Thus, the thickness of the key at the bigger end is 15.33 mm.

The thickness of the key at the smaller end can be calculated from:

$Thickness = T_b - T_s$

Where:

• T_b = Thickness at bigger end = 138 mm
• T_s = Thickness at smaller end = 15.33 mm

Thus,

$Thickness = 138 - 15.33 = 122.67 \text{ mm}$

Hence, the thickness of the key at the smaller end is 122.67 mm.

The index plate is a component that contains holes arranged in a circular pattern. Its purpose is to allow the operator to subdivide one full turn of the crank into smaller fractions, enabling precise adjustments during milling operations.

Sector arms are components that aid in setting the required number of holes when using a dividing head. They do not require the operator to recount the number of holes needed, facilitating more efficient and accurate milling.

Two methods that can be used on a centre lathe to cut external V-screw threads include:

1. Compound slide method: This involves using a compound slide rest to set the necessary angle for cutting V-screw threads.

2. Cross slide method: This method uses the cross slide of the lathe to achieve the necessary depth and angle for V-screw threads.