A numerical control drill press drills four 12.0 mm diameter holes at four locations on a flat

aluminum plate in a production work cycle.

The plate is only 10 mm thick, but the drill must travel a full 20 mm vertically at each hole

location to allow for clearance above the plate and breakthrough of the drill on the underside

of the plate. The cutting conditions are the following:

ï‚§ Speed = 0.5 m/s and

ï‚§ Feed = 0.12 mm/rev.

Hole locations are indicated in the following table:

Hole

number

x-coordinate

(mm)

y-coordinate

(mm)

1 50.0 50.0

2 50.0 100.0

3 100.0 100.0

4 100.0 50.0

ï‚§ The drill starts out at point (0,0) and returns to the same position after the work cycle is

completed.

ï‚§ Travel rate of the table in moving from one coordinate position to another is 400 mm/min.

ï‚§ Owing to effects of acceleration and deceleration, and time required for the control system to

achieve final positioning, a time loss of 5 second is experienced at each stopping position of

the table.

ï‚§ Assume that all moves are made so as to minimize the total cycle time. If loading and

unloading the plate take 30 s (total handling time),

Determine the following:

(i) Spindle rotational speed (2 points)

(ii) Feed rate (1 point)

(iii) Time to move between positions (including loss time) (1 point)

(iv) Time for four holes to drill (1 point)

(v) Time required for the work cycle. (1 point)

Hints:

Spindle rotation speed: N = v/(π D)

ï‚§ N = spindle rotational speed (rev/min)

ï‚§ v = cutting speed (m/min, ft/min)

ï‚§ D = milling cutter diameter (m, ft)

Feed rate: f

r = N nt

f

ï‚§ f

r = feed rate (mm/min, in/min)

ï‚§ nt = number of tooth on the milling cutter

ï‚§ f = feed (mm/tooth, inch/tooth

Subject | Business |

Due By (Pacific Time) | 07/27/2015 12:00 am |

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