---
title: "Turning time and metal removal rate"
subject: "Workshop Technology"
chapter: 6
key: "semester_2/workshop_technology:lathe-turning-time-mrr-numerical"
source: https://www.bctengineer.com/q/semester_2/workshop_technology/lathe-turning-time-mrr-numerical
site: BCT Engineering
---

# Turning time and metal removal rate

Workshop Technology · Chapter 6: General Machining Processes “Chip-forming”

- Asked in: year not given, 8 marks
- Video: [Process Parameter Of Machining | Cutting Speed | Feed Rate | Depth Of Cut | Material Removal Rate](https://www.youtube.com/watch?v=OZ4fTZK71yw) — TRISHAKTI ACADEMY
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/workshop_technology/lathe-turning-time-mrr-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_2/workshop_technology/chapter-6

## Question

A 50 mm diameter, 300 mm long mild steel bar is turned to 44 mm diameter in a single pass on a lathe at a cutting speed of 30 m/min and feed of 0.3 mm/rev. Take total tool travel as the bar length plus 5 mm for approach and over-run. Find (a) depth of cut, (b) spindle speed, (c) machining time, (d) metal removal rate.

## Short answer

Depth of cut = (50-44)/2 = 3 mm. Spindle speed N = 1000×30/(π×50) ≈ 191 rpm. Tool travel = 300+5 = 305 mm, time = 305/(0.3×191) = 5.32 min. MRR = π × (50+44)/2 × 3 × 0.3 ×191 ≈ 25.4 cm³/min.

## Answer

**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.)

| Quantity | Value |
|---|---|
| Depth of cut | 3 mm |
| N | 191 rpm |
| Time | 5.32 min |
| MRR | 25.4 cm³/min |

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Turning 50 mm to 44 mm diameter gives a depth of cut of
   A. 6 mm
   B. 3 mm
   C. 1.5 mm
   D. 47 mm
   Answer: B. d = (50 - 44)/2 = 3 mm. (Key point: Depth of cut)

2. Spindle speed is calculated using which diameter?
   A. The final diameter 44 mm
   B. The original diameter 50 mm
   C. The mean diameter 47 mm
   D. The depth of cut
   Answer: B. Cutting speed is based on the original diameter: N = 1000 x 30/(pi x 50) = 191 rpm. (Key point: Spindle speed basis)

3. With L = 305 mm, f = 0.3 mm/rev and N = 191 rpm, the machining time is about
   A. 5.32 min
   B. 1.6 min
   C. 53.2 min
   D. 10.6 min
   Answer: A. T = L/(f N) = 305/(0.3 x 191) = 5.32 min. (Key point: Machining time)

4. MRR in turning here is computed as
   A. pi D_avg d f N
   B. D d f
   C. pi D1 f N
   D. d f / N
   Answer: A. MRR = pi x 47 x 3 x 0.3 x 191 = about 25,380 mm^3/min using the mean diameter. (Key point: MRR formula)

## Related in Workshop Technology

- [Drilling time and metal removal rate](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Adrilling-time-numerical) — Workshop Technology, chapter 5
- [Taper turning: set-over and compound angle](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Ataper-turning-setover-numerical) — Workshop Technology, chapter 6
- [Shaper machining time calculation](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Ashaper-machining-time-numerical) — Workshop Technology, chapter 6
- [Thread cutting on lathe and change gears](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Alathe-thread-cutting-change-gears) — Workshop Technology, chapter 6

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