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

Answer:

MRR = 0.06 inc^3/ min

Fe = 55 lb

time = 3.28 min

torque  = 13.75 lb -in

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 $fm = f\times z\times N$

                            $= 0.001\times 10\times 250 = 2.5 in/min$

$MRR = w\times d\times fm$

        $= 0.001\times 10\times 250 = 0.06 inc^3/ min$

$over travel = \frac{1}{2} \times (D - \sqrt{D^2 -W^2}$

$over travel = \frac{1}{2} \times (5 - \sqrt{5^2 -2^2} = 0.208 inch$

Le =  L =0.208 = 8.028 inch

$time tm = \frac{Le}{f\times z\times N} = \frac{8.208}{0.001\times 10 \times 250} = 3.28 min$

$e = \frac{Fe\times V}{MRR}$

$Fe\times  v = power = e\times MRR$

$3000,000 \times \frac{0.6}{60} = 300 fr-lb/sec$

$Fe = \frac{300}{\frac{\pi DN}{60}} = \frac{300}{\frac{\pi \frac{5}{12}\times 200}{60}}$

Fe = 55 lb

$torque  = Fe \times r = 55\times 2.5 = 13.75 lb-inch$