The length, width and thickness of a steel sample are 400 mm, 40 mm and 20 mm, respectively. Its thickness needs to be uniformly reduced by 2 min in a single pass by using horizontal slab milling. The milling cutter (diameter: 100 mm, width: 50 mm) has 20 teeth and rotates at 1200 rpm. The feed per tooth is 0.05 mm. The feed direction is along the length of the sample. If the over travel distance is the same as the approach distance, the approach distance and time taken to complete the required machining task are

The correct option is 14mm, 21.4s.

Let us analyze the parameters provided in the problem and derive the approach distance and the total time taken to complete the machining task step-by-step.

1. Given Data:
- Length of the steel sample, $L=400\text{mm}$
- Width of the steel sample, $w=40\text{mm}$
- Initial thickness of the sample, $t_i=20\text{mm}$
- Uniform reduction in thickness (depth of cut), $d=2\text{mm}$
- Diameter of the milling cutter, $D=100\text{mm}$
- Width of the cutter, $W=50\text{mm}$ (which is greater than the width of the sample, so the cut is completed in a single pass)
- Number of teeth on the cutter, $Z=20$
- Rotational speed of the cutter, $N=1200\text{rpm}$
- Feed per tooth, $f_t=0.05\text{mm/tooth}$
- Over-travel distance, $O_t$ is equal to the approach distance, $A_p$.

2. Calculation of Approach Distance ($A_p$):
In horizontal slab milling, the approach distance $A_p$ is the horizontal distance from the point where the cutter first contacts the workpiece to the point where the cutter axis is directly over the front edge of the workpiece. It is calculated using the formula:

$$A_p=\sqrt{d\left(D-d\right)}$$

Substituting the values of $d=2\text{mm}$ and $D=100\text{mm}$ into the equation:

$$A_p=\sqrt{2\times \left(100-2\right)}=\sqrt{2\times 98}=\sqrt{196}=14\text{mm}$$

Thus, the approach distance is $14\text{mm}$. Since the over-travel distance is the same as the approach distance, we have:
$O_t=14\text{mm}$.

3. Calculation of Feed Rate ($f_m$):
The table feed rate (speed of the workpiece relative to the cutter) in mm/min is given by:

$$f_m=f_t\times Z\times N$$

Substituting the given values:

$$f_m=0.05\times 20\times 1200=1200\text{mm/min}$$

Converting this to mm/second:

$$f_m=\frac{1200}{60}=20\text{mm/s}$$

4. Calculation of Machining Time ($T$):
The total travel length of the cutter to complete the machining task includes the length of the workpiece, the approach distance, and the over-travel distance:

$$L_{\text{total}}=L+A_p+O_t$$

Substituting the values:

$${\mathrm{L<mtext>total</mtext>}}_{}=400+14+14=428\text{mm}$$

The time taken to complete the machining task is given by:

$$T=\frac{{\mathrm{L<mtext>total</mtext>}}_{}}{f_m}=\frac{428}{20}=21.4\text{s}$$

Therefore, the approach distance is 14 mm and the machining time is 21.4 s.