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

t=0.32 min

Step-by-step 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=(\pi DN)/(60)`

`N=(60V)/(\pi D)`

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

N=563.91 RPM

Compulsory approach ,X

`X=(1)/(2)\left [ D^2-√(D^2-W^2) \right ]`

`X=(1)/(2)\left [ 3^2-√(3^2-2^2) \right ]`

X=3.33 in

Machining time,t

`t=(L+X)/(fZN)`

`t=(12+3.33)/(0.007* 12* 563.91)`

t=0.32 min