A slot of 25 mm x 25 mm Is lo be milled in a workpiece of 300 mm length using a side and face milling cutter of diameter 100 mm, width 25 mm and having 20 teeth. For a depth of cut 5 mm, feed per tooth 0.1 mm, cutting speed 35 m/min and approach and over travel distance of 5 mm each, the time required for milling the slot is_______ minutes (round off to one decimal place)

The correct answer is 8.1

1. Identify the given parameters:
- Dimensions of the slot to be milled: 25 mm depth �� 25 mm width
- Length of the workpiece (L_w) = 300 mm
- Cutter diameter (D) = 100 mm
- Cutter width = 25 mm
- Number of teeth (Z) = 20
- Depth of cut per pass (d) = 5 mm
- Feed per tooth (f) = 0.1 mm/tooth
- Cutting speed (V) = 35 m/min
- Approach distance (A) = 5 mm
- Over travel distance (O) = 5 mm
- Radius of the tool (R) = D / 2 = 100 / 2 = 50 mm

2. Calculate the rotational speed of the cutter (N):
The relation between cutting speed and rotational speed is given by:

$$V=\pi \times D\times N$$
Here, cutting speed V must be in m/min, and diameter D must be converted to meters (100 mm = 0.1 m):

$$35=\pi \times 0.1\times N$$
Solving for N:

$$N=\frac{35}{\pi \times 0.1}\approx 111.4084\text{rpm}$$

3. Calculate the total length of travel (L):
According to the provided solution, the total travel length for a pass is the sum of the workpiece length, the approach distance, the over travel distance, and the tool radius:

$$L=L_w+\text{Approach}+\text{Over travel}+\text{Radius of tool}$$
$$L=300+5+5+50=360\text{mm}$$

4. Calculate the time required for a single pass (t_pass):
The feed rate of the table is:

$$F=f\times Z\times N$$
The time required for one pass is:

$$t_{\text{pass}}=\frac{L}{f\times Z\times N}$$
Substituting the values:

$$t_{\text{pass}}=\frac{360}{0.1\times 20\times 111.4084}\approx 1.6157\text{minutes}$$

5. Determine the number of passes and total machining time:
The total depth of the slot is 25 mm, and the depth of cut per pass is 5 mm.

$$\text{Number of passes}=\frac{\text{Total depth}}{\text{Depth of cut per pass}}=\frac{25}{5}=5\text{passes}$$
Thus, the total machining time is:

$$t_{\text{total}}=1.6157\times 5=8.078\text{minutes}$$
Rounding off to one decimal place, we get 8.1 minutes.