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 break down the solution step-by-step:

1. Identify the given parameters:
Length of the steel sample, L = 400 mm
Width of the steel sample, w = 40 mm
Initial thickness of the steel sample, t = 20 mm
Depth of cut (thickness reduction), d = 2 mm
Diameter of the milling cutter, D = 100 mm
Number of teeth on the cutter, z = 20
Rotational speed of the cutter, N = 1200 rpm
Feed per tooth, f_t = 0.05 mm/tooth

2. Calculate the Approach Distance (e):
For slab milling, the approach distance e required for the cutter to fully engage with the workpiece is given by the formula:
$e=\sqrt{d(D-d)}$
Substituting the given values:
$e=\sqrt{2(100-2)}$
$e=\sqrt{2\times 98}=\sqrt{196}=14mm$
Therefore, the approach distance is 14 mm.

3. Calculate the Total Table Travel Length (L_total):
Since the over travel distance is the same as the approach distance, we have:
Over travel distance, o = 14 mm
The total length of table travel during machining is:
$L_{\mathrm{total}}=L+e+o$
Substituting the values:
$L_{\mathrm{total}}=400+14+14=428mm$

4. Calculate the Feed Rate (v_f):
The table feed rate can be determined using the feed per tooth, number of teeth, and rotational speed:
$v_f=f_t\times z\times N$
Substituting the values:
$v_f=0.05mm/tooth\times 20teeth\times 1200rpm$
$v_f=1200mm/min$
To convert the feed speed to millimeters per second (mm/s):
$v_f=\frac{1200}{60}=20mm/s$

5. Calculate the Machining Time (T):
The total time taken to complete the machining pass is:
$T=\frac{L_{\mathrm{total}}}{v_f}$
Substituting the values:
$T=\frac{428}{20}=21.4s$
Thus, the approach distance is 14 mm and the time taken is 21.4 s.