Question

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 mm 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, the approach distance and time taken to complete the required machining task are

• A. 14 mm, 21.4 s
• B. 21 mm, 39.4 s
• C. 14 mm, 18.4 s
• D. 21 mm, 28.9 s

Answer 1

The correct option is A 14 mm, 21.4 s


Cutter diameter,
D = 100 mm
Cutter width = 50 mm that means single pass is sufficient.
Number of teeth,
z = 20, Depth of cut, d = 2 mm
Rotational speed,
N = 1200 rpm,
f = 0.05 mm per tooth
Approach distance,

$A=\sqrt{{\left(\frac{D}{2}\right)}^2-{(\frac{D}{2}-d)}^2}$

$=\sqrt{d(D-d)}$

$=\sqrt{2(100-2)}mm=14mm$

Time for one pass in slab or slot milling

$=\frac{L+A}{fzN}$ for rough milling

$=\frac{L+2A}{fzN}$ for finish milling

As length of approach and over travel are same
$=\frac{L+A+A}{fzN}$

$=\frac{400+14+14}{0.05\times 20\times 1200}min=21.4s$