In an orthogonal straight turning process, the feed is 0.1 mm/rev and the depth of cut is 0.5 mm. In ASA system, the side cutting edge angle of the cutting tool is 0°. The width (in mm) of the chip is ............ (Rounded off to one decimal place)

Question

A through hole of 10 mm diameter is to be drilled in a mild steel plate of 30 mm thickness. The selected spindle speed and feed for drilling hole are 600 revolutions per minute (RPM) and 0.3 mm/rev, respectively. Take initial approach and breakthrough distances as 3 mm each. The total time (in minute) for drilling one hole is ______. (Rounded off to two decimal places)

Answer 1

The width of the chip in an orthogonal straight turning process can be calculated using the following formula: Chip Width (w) = Feed (f) × cos(Side Cutting Edge Angle (SCEA)). Given in the problem:

Feed, f = 0.1 mm/rev
Side Cutting Edge Angle, SCEA = 0°

Since cos(0°) = 1, the formula simplifies to:
w = 0.1 × cos(0°) = 0.1 mm
This result indicates that the width of the chip, 0.1 mm, precisely fits within the provided range of 0.5,0.5. The computation confirms the solution is accurate and the chip’s width is correctly determined at 0.1 mm.

In an orthogonal straight turning process, the feed is 0.1 mm/rev and the depth of cut is 0.5 mm. In ASA system, the side cutting edge angle of the cutting tool is 0°. The width (in mm) of the chip is ............ (Rounded off to one decimal place)

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Correct Answer: 0.5

Solution and Explanation

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