Turning time and metal removal rate

Given: D1 = 50 mm, D2 = 44 mm, length 300 mm, v = 30 m/min, f = 0.3 mm/rev, approach and over-run = 5 mm.

(a) Depth of cut

$$d = \frac{D_1 - D_2}{2} = \frac{50-44}{2} = 3\ \text{mm}$$

(b) Spindle speed

The cutting speed is based on the original diameter.

$$N = \frac{1000\,v}{\pi D_1} = \frac{1000\times 30}{\pi\times 50} = 191.0\ \text{rpm}$$

(c) Machining time

Tool travel $L = 300 + 5 = 305$ mm.

$$T = \frac{L}{f\,N} = \frac{305}{0.3\times 191.0} = 5.32\ \text{min}$$

(d) Metal removal rate

Use the mean diameter, $D_{avg} = (50+44)/2 = 47$ mm.

$$\begin{aligned} \text{MRR} &= \pi D_{avg}\, d\, f\, N \\ &= \pi \times 47 \times 3 \times 0.3 \times 191.0 \\ &= 25{,}380\ \text{mm}^3/\text{min} \end{aligned}$$

(Check: volume removed = $\frac{\pi}{4}(50^2-44^2)\times 300 = 132{,}890$ mm³; dividing by the cutting time for 300 mm, 300/(0.3 x 191.0) = 5.24 min, gives about 25,380 mm³/min, which agrees.)

Answer: d = 3 mm, N = 191 rpm, T = 5.32 min, MRR = 25.4 cm³/min.