Question

A NC drill press is to perform a series of through-hole drilling operations on a 1.75 in thick aluminum plate that is a component in a heat exchanger. Each hole is 3/4 in diameter. There are 100 holes in all, arranged in a 10 by 10 matrix pattern, and the distance between adjacent hole centers (along the square) = 1.5 in. The cutting speed = 300 ft/min, the penetration feed (z-direction) = 0.015 in/rev, and the traverse rate between holes (x-y plane) = 15.0 in/min. Assume that x-y moves are made at a distance of 0.50 in above the work surface, and that this distance must be included in the penetration feed rate for each hole. Also, the rate at which the drill is retracted from each hole is twice the penetration feed rate. The drill has a point angle = 100°. Determine the time required from the beginning of the first hole to the completion of the last hole, assuming the most efficient drilling sequence will be used to accomplish the job.

Answer 1

The total time required is
`T_(total)= 1601 sec`

Step-by-step explanation:

From the question we are told that the

The thickness of the aluminium plate is
`t = 1.75 \ in`

The diameter of the holes is
`d = (3)/(4) \ in`

The number holes is n = 100

The pattern they arranged is = 10 by 10 matrix

The distance between the adjacent holes is
`D = 1.5 \ in`

The cutting speed
`v = 300 \ ft/min =300 *12 = 3600 \ in /min`

The penetration feed is
`P = 0.015 \ in / rev`

The traverse rate is
`T_r = 15.0 \ in /min`

The distance of the x-y moves
`L = 0.50 \ in`

The rate at which the drill id retracted is
`R = 2P`

The point angle of the drill is
`\theta = 100^o`

Generally the cutting speed is mathematically represented as

`v = \pi d N`

Where N is the rotational speed

Substituting values

`3600 = \pi * (3)/(4) N`

=>
`N = (3600)/(0.75 * 3.142 )`

`N = 1527.68 rev /min = (1527.68)/(60) = 25.46 \ rev /sec`

Now the feed rate is mathematically represented as

`Feed \ Rate [F] = P * N`

Substituting values

`F = 0.015 * 25.46`

`= 0.3819 \ in`

Next is to obtain the approach distance

The approach distance is mathematically represented as

`a = 0.5 * d * tan (90 - (\theta )/(2) )`

substituting values

`a = 0.5 * 0.75 * tan ({90 - (100)/(2) })`

`a = 2.5646 \ in`

also we are told that the drill x-y moves is 0.5 in above the work surface

Then the when retracting the drill the total distance moves is

Hence the total drill distance is
`w = 0.5 + 1.75 + 0.3146`

`= 2.5646 \ in`

The time required to drill on hole is

`t_1 = (w )/(F )`

Substituting values

`t_1 = (2.56446)/(0.3819)`

`= 6.71 sce`

The retracting rate of the drill would be

`K = 2P N`

`R = 0.015 * 2 * 25.46`

`= 0.7638`

The time required to move the drill out of one hole is

`t_ 2 = (w)/(K)`

Substituting value

`t_2 = (2.5646)/(0.7638)`

`= 3.3576 \ sec`

The time required to move from one hole to another is

`t_3 = (D)/(T_r)`

`= (1.5)/(0.25)`

`= 6 \ sec`

The time it would take to perform the drilling of 100 holes is mathematically represented as

`t_4 = 100 *t_1`

`= 6.71 *100`

`= 671 \ sec`

Time required for retracting the drill from 100 holes is mathematically represented as

`t_5 = 100 * t_2`

`t_5 = 3.3576 *100`

`= 336 sec`

The time taken to move through the 100 holes is

`t_6 = number of travers time * t_3`

The number of trans verse time is = 99

`t_6 = 99 * 6`

`= 594 \ sec`

The time required the drilling task from the beginning of the first hole to the hole is

`T_(total ) = t_4 + t_5 + t_6`

`= 671 + 336 + 594`

`T_(total)= 1601 sec`