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We want to machine by turning this part on a CNC lathe with uncoated carbide tools in optimal condition. A long rod of Ø15 mm is used. Cite the sources and then calculate: 1) The depth of cut to produce Ø14mm in two equal passes. 2) Cutting speed (m/min) and feed (mm/rev) for turning the Ø15mm rod. 3) The part rotation speed (rpm) and tool feed rate (mm/min) for CNC programming.

Respuesta :

Answer:

1) The initial diameter of the rod is 15 mm

the final diameter should be 14 mm in two-pass

the total reduction of 1mm in two passes

so in one pass, it should be 1/2 = 0.5 mm

Depth of cut in circular cross-section is equal in both sides in one pass

so, Depth of cut = 0.5/2 = 0.25 mm

b) IN turning operation we usually take spindle speed is 20,000 rpm

then velocity = (pi * D * 20,000)/1000 = (3.14 * 15 * 20) = 942 m/min.

feed = cutting velocity / spindle speed = (942*1000)/ (20000) = 47.1 mm/rev

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