GATE 2024 ME – Question 64
A flat surface of a C60 steel having dimensions of 100 mm (length) × 200 mm (width) is produced by a HSS slab mill cutter. The 8-toothed cutter has 100 mm diameter and 200 mm width. The feed per tooth is 0.1 mm, cutting velocity is 20 m/min and depth of cut is 2 mm. The machining time required to remove the entire stock is _____________ minutes (*rounded off to 2 decimal places*).
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Correct answer: 2.23 to 2.25
Explanation
The spindle speed is $N = \frac{V}{\pi D} = \frac{20\,000}{\pi \times 100} = 63.66$ rpm and the table feed is $f_m = f_tzN = 0.1 \times 8 \times 63.66 = 50.93$ mm/min. The cutter is as wide as the work, so it makes one pass along the 100 mm length. The cutter has to travel an approach distance $\sqrt{d(D - d)} = \sqrt{2 \times 98} = 14$ mm before it is cutting its full depth. So the time is $\frac{100 + 14}{50.93} = 2.24$ minutes.