Question

A face milling operation is to be performed on cast iron. Tool diameter is 5 inches, w = 2 inches, part length = 8 inches, depth of cut = 0.012 inches, f = 0.001 inch/tooth, and N = 250 RPM. Find: MRR, power, time, and torque.

Answer 1

MRR = 0.06 inc^3/ min

Fe = 55 lb

time = 3.28 min

torque = 13.75 lb -in

Step-by-step explanation:

Given data:

D = 5 inch

`f = 0.001 inch/ tooth`

Part length = 8 inch

N =250 RPM

let assume Z = 10

e =25J/mm^3 or 30,000 ft lb/inc^2

machine feed
`fm = f* z* N`

`= 0.001* 10* 250 = 2.5 in/min`

`MRR = w* d* fm`

`= 0.001* 10* 250 = 0.06 inc^3/ min`

`over travel = (1)/(2) * (D - √(D^2 -W^2)`

`over travel = (1)/(2) * (5 - √(5^2 -2^2) = 0.208 inch`

Le = L =0.208 = 8.028 inch

`time tm = (Le)/(f* z* N) = (8.208)/(0.001* 10 * 250) = 3.28 min`

`e = (Fe* V)/(MRR)`

`Fe*  v = power = e* MRR`

`3000,000 * (0.6)/(60) = 300 fr-lb/sec`

`Fe = (300)/((\pi DN)/(60)) = (300)/((\pi (5)/(12)* 200)/(60))`

Fe = 55 lb

`torque  = Fe * r = 55* 2.5 = 13.75 lb-inch`