Question

A slab-milling operation will take place on a 6061 Aluminum (free machining) 12" long and 2" wide piece of material. A spiral, helical cutter 3" in diameter with 12 teeth will be used. If the feed per tooth (f) is 0.007 inches/tooth and the cutting speed (V) is 444 SFM. Find the machining time.

Answer 1

Answer:

t=0.32 min

Explanation:

Given that

W= 2 in

L= 12 in

D= 3 in

Z= 12 teeth

Speed ,V= 444 SFM

We know that

1 SFM = 0.00508 m/s

So

444 SFM = 2.25 m/s

V= 2.25 m/s

1 m/s= 39.37 in/s

V =88.58 m/s

We know that

$V=\dfrac{\pi DN}{60}$

$N=\dfrac{60V}{\pi D}$

N=\dfrac{60\times 88.58}{\pi \times 3}

N=563.91 RPM

Compulsory approach ,X

$X=\dfrac{1}{2}\left [ D^2-\sqrt{D^2-W^2} \right ]$

$X=\dfrac{1}{2}\left [ 3^2-\sqrt{3^2-2^2} \right ]$

X=3.33 in

Machining time,t

$t=\dfrac{L+X}{fZN}$

$t=\dfrac{12+3.33}{0.007\times 12\times 563.91}$

t=0.32 min