---
title: "Shaper machining time calculation"
subject: "Workshop Technology"
chapter: 6
key: "semester_2/workshop_technology:shaper-machining-time-numerical"
source: https://www.bctengineer.com/q/semester_2/workshop_technology/shaper-machining-time-numerical
site: BCT Engineering
---

# Shaper machining time calculation

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

- Asked in: year not given, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/workshop_technology/shaper-machining-time-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_2/workshop_technology/chapter-6

## Question

A flat mild steel plate 300 mm long and 160 mm wide is to be machined on a shaper. The cutting speed is 12 m/min, the ratio of cutting time to return time is 3:2, and the feed is 1.5 mm per stroke. Allow 15 mm clearance at each end of the stroke and 10 mm extra on the width for tool approach. Find the number of double strokes per minute and the time to machine one surface in one cut.

## Short answer

Stroke length L = 300+15+15 = 330 mm. Number of double strokes per minute n = v/[L(1+k)] with k=return/cutting = 2/3, v=12000 mm/min → n=21.8. Width to machine = 170 mm, strokes needed = 170/1.5≈114. Machining time = 114/21.8 = 5.23 min.

## Answer

**Given:** plate length 300 mm, width 160 mm; cutting speed v = 12 m/min = 12,000 mm/min; cutting : return time = 3 : 2; feed 1.5 mm per stroke.

#### Stroke length
$$
L = 300 + 15 + 15 = 330\ \text{mm}
$$

#### Double strokes per minute
Let k be the ratio of return time to cutting time: $k = 2/3$.

Time for the cutting stroke = $L/v = 330/12000 = 0.0275$ min.

Time for one double stroke:

$$
t = 0.0275\,(1 + k) = 0.0275\times 1.6667 = 0.04583\ \text{min}
$$

$$
n = \frac{1}{t} = \frac{v}{L(1+k)} = \frac{12000}{330\times 1.6667} = 21.8\ \text{strokes/min}
$$

#### Number of strokes needed
Width to be covered = 160 + 10 = 170 mm.

$$
\text{strokes} = \frac{170}{1.5} = 113.3 \approx 114
$$

#### Machining time
$$
T = \frac{114}{21.8} = 5.23\ \text{min}
$$

| Quantity | Value |
|---|---|
| Stroke length | 330 mm |
| Time per double stroke | 0.0458 min |
| Strokes per minute | 21.8 |
| Strokes required | 114 |
| Time | 5.23 min |

**Answer: n = 21.8 double strokes per minute; machining time = 5.23 min.**

## Check yourself

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

1. The stroke length for a 300 mm plate with 15 mm clearance at each end is
   A. 300 mm
   B. 315 mm
   C. 330 mm
   D. 345 mm
   Answer: C. L = 300 + 15 + 15 = 330 mm. (Key point: Stroke length)

2. With a cutting:return ratio of 3:2 (k = 2/3), the double strokes per minute n = v/(L(1+k)) is about
   A. 36.4
   B. 21.8
   C. 12.0
   D. 43.6
   Answer: B. n = 12000/(330 x 1.6667) = 21.8 strokes/min. (Key point: Double strokes per minute)

3. How many strokes are needed to cover 160 + 10 = 170 mm width at 1.5 mm feed per stroke?
   A. 107
   B. 114
   C. 170
   D. 128
   Answer: B. 170/1.5 = 113.3, rounded up to 114. (Key point: Number of strokes)

4. The machining time is
   A. 5.23 min
   B. 2.5 min
   C. 10.4 min
   D. 114 min
   Answer: A. T = 114/21.8 = 5.23 min. (Key point: Machining time)

## Related in Workshop Technology

- [Shaper types and quick return mechanism](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Ashaper-types-quick-return) — Workshop Technology, chapter 6
- [Turning time and metal removal rate](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Alathe-turning-time-mrr-numerical) — Workshop Technology, chapter 6
- [Drilling time and metal removal rate](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Adrilling-time-numerical) — Workshop Technology, chapter 5
- [Cutting tool materials and carbon steels](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Acutting-tool-materials-carbon-steels) — Workshop Technology, chapter 7
- [Upsetting diameter and force calculation](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Aupsetting-diameter-force-numerical) — Workshop Technology, chapter 9

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