Question

A slot of 25 mm × 25 mm is to be milled in a workpiece of 300 mm length using a side and face milling cutter of diameter 100 mm, width 25 mm and having 20 teeth.

For a depth of cut 5 mm, feed per tooth 0.1 mm, cutting speed 35 m/min and approach and over-travel distance of 5 mm each, the time required for milling the slot is ______ minutes (round off to one decimal place).

Answer 1

Concept:

Cutting speed of the milling cutter is given by, V = πDN m/min

Where, D = Diameter of cutter, N = speed in rpm

In Face milling:

Length of cut (L) = Workpiece length + Approach + Over travel + Radius of tool

Time required for milling (single pass):

\(t = \frac{{Length\;of\;cut\;\left( L \right)}}{{Table\;feed}}\)

Table feed is given by, F = f × Z × N

Where f is the feed per tooth, Z is the number of tooth and N is the r.p.m. of the cutter

Calculation:

Given, cutter speed = 35 m/min, length of the workpiece = 300 mm, Approach and over travel = 5 mm, Diameter of the cutter = 100 mm, ∴ radius of the tool = 50 mm

Therefore,                         

35 = π × 0.100 × N

N= 111.4084 rpm             

Length of cut (L) = Workpiece length + Approach + Over travel + Radius of tool

L = 300 + 5 + 5 + 50 = 360 mm

Time required for milling (single pass)

t = \(\frac{{Length\;of\;cut\;\left( L \right)}}{{f\; \times \;Z\; \times \;N}} = \frac{{360}}{{0.1 \times 20 \times 111.4084}}\)

∴ t = 1.615 min per pass

Since depth of cut is 5 mm, therefore for 25 mm of cut, 5 passes are required.

Total machining time = time per pass × number of pass

t = 1.6157 × 5 = 8.078 min