A drilling operation is performed on a steel part using a 12.7 mm diameter twist drill with a point angle of 118 degrees. The hole is a blind hole with a depth of 50 mm. Cutting speed=15 m/min, and feed=0.20 mm/rev.

Determine the following:

a. Rotational speed of the drill
b. Feed rate
c. Approach allowance
d. Cutting time
e. Metal removal rate after the drill bit reaches full diameter.

a. Rotational speed of the drill  = 375.96 rev/min

b. Feed rate  = 75 mm/min

c. Approach allowance  = 3.815 mm

d. Cutting time  = 0.67 minutes

e. Metal removal rate after the drill bit reaches full diameter. = 9525 mm³/min

Explanation:

Here we have

a. N = v/(πD) = 15/(0.0127·π) = 375.96 rev/min

b. Feed rate = fr = Nf = 375.96 × 0.2 = 75 mm/min

c. Approach allowance = tan 118/2 = (12.7/2)/tan 118/2 = 3.815 mm

d. Approach allowance T∞ =L/fr = 50/75 = 0.67 minutes

e. R = 0.25πD²fr = 9525 mm³/min.

answer:

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