---
title: "Number of V-belts for a motor drive"
subject: "Machine Design I"
chapter: 8
key: "semester_6/machine_design_i:v-belt-number-of-belts"
source: https://www.bctengineer.com/q/semester_6/machine_design_i/v-belt-number-of-belts
site: BCT Engineering
---

# Number of V-belts for a motor drive

Machine Design I · Chapter 8: Design of belts

- Asked in: year not given, 8 marks
- Video: [V Belts Design Procedure](https://www.youtube.com/watch?v=tPtlwiiSamI) — ROBOMECHTRIX
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/machine_design_i/v-belt-number-of-belts
- Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-8

## Question

A 15 kW, 1440 rpm motor drives a machine at 480 rpm through B-section V-belts. The pitch diameter of the smaller pulley is 140 mm and the centre distance is about 800 mm. The service factor is 1.2. The rated power of one B belt at this speed and pulley size is 3.4 kW; the correction factors are 0.95 for the arc of contact (about 160 degrees) and 1.04 for belt length 2500 mm. Standard B-section pitch lengths are 2240, 2360, 2500, 2650 and 2800 mm. Find the larger pulley diameter, the belt speed, the belt length and actual centre distance, and the number of belts.

## Short answer

Larger pulley D=420 mm. Belt speed v=10.56 m/s. Trial length ≈2504 mm; standard pitch length chosen 2500 mm giving exact centre distance 798 mm. Design power 18 kW, corrected belt rating 3.36 kW → number of belts = 18/3.36 = 5.36 → 6 B-section belts.

## Answer

#### Pulley sizes and speed

$$
D = d\,\frac{n_1}{n_2} = 140 \times \frac{1440}{480} = 420\ \text{mm}
$$

$$
v = \frac{\pi d n_1}{60} = \frac{\pi \times 0.14 \times 1440}{60} = 10.56\ \text{m/s}
$$

(The speed is within the usual limit of 25 m/s for V-belts.)

#### Belt length

Trial centre distance $C = 800$ mm:

$$
L = 2C + \frac{\pi}{2}(D + d) + \frac{(D - d)^2}{4C} = 1600 + 879.6 + 24.5 = 2504\ \text{mm}
$$

Nearest standard pitch length: **2500 mm**.

#### Actual centre distance

$$
C = B + \sqrt{B^2 - \frac{(D - d)^2}{8}}, \qquad B = \frac{L}{4} - \frac{\pi (D + d)}{8}
$$

$$
B = 625 - 219.9 = 405.1\ \text{mm}
$$

$$
C = 405.1 + \sqrt{405.1^2 - \frac{280^2}{8}} = 405.1 + \sqrt{164\,106 - 9800} = 405.1 + 392.8 = 797.9\ \text{mm}
$$

Contact angle on the small pulley:

$$
\theta = 180^\circ - 2\sin^{-1}\frac{D - d}{2C} = 180^\circ - 2\sin^{-1}\frac{280}{1596} = 159.8^\circ
$$

This agrees with the 160 degrees assumed for the factor 0.95.

#### Number of belts

$$
\text{Design power} = 1.2 \times 15 = 18\ \text{kW}
$$

$$
\text{Corrected rating per belt} = 3.4 \times 0.95 \times 1.04 = 3.36\ \text{kW}
$$

$$
\text{Number of belts} = \frac{18}{3.36} = 5.36 \;\Rightarrow\; 6\ \text{belts}
$$

**Answer: D = 420 mm; v = 10.56 m/s; L = 2500 mm; C = 798 mm; 6 B-section belts.**

## Check yourself

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

1. What are the larger pulley diameter and belt speed?
   A. D = 140 x 480/1440 = 46.7 mm; v = 3.5 m/s
   B. D = 420 mm; v = 10.56 m/s
   C. D = 420 mm; v = 3.52 m/s
   D. D = 1260 mm; v = 10.56 m/s
   Answer: B. D = d n1/n2 = 140 x 3 = 420 mm, and v = pi d n1 / 60 = 10.56 m/s using the small pulley and motor speed. (Key point: Pulley and speed)

2. After computing L = 2504 mm from the trial centre distance, what is done next?
   A. Use 2504 mm exactly
   B. Choose the nearest standard pitch length (2500 mm) and recompute the actual centre distance
   C. Round to the largest standard length, 2800 mm
   D. Reduce the pulley size
   Answer: B. Belts come only in standard pitch lengths, so 2500 mm is chosen and the centre distance is recalculated (about 798 mm). (Key point: Standard belt length)

3. How is the corrected power rating per belt obtained?
   A. 3.4 / (0.95 x 1.04) = 3.44 kW
   B. 3.4 x 0.95 x 1.04 = 3.36 kW
   C. 3.4 + 0.95 + 1.04 kW
   D. 3.4 x 1.2 = 4.08 kW
   Answer: B. The rated power is multiplied by the arc-of-contact and belt-length correction factors: 3.36 kW. (Key point: Corrected rating)

4. With design power 1.2 x 15 = 18 kW and 3.36 kW per belt, how many belts?
   A. 5, rounding 5.36 down
   B. 5.36 belts
   C. 6, rounding 5.36 up to a whole number
   D. 18
   Answer: C. 18 / 3.36 = 5.36, and the count must be rounded up to ensure enough capacity: 6 belts. (Key point: Number of belts)

## Related in Machine Design I

- [Roller chain drive: length, speed and factor of safety](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aroller-chain-drive-selection) — Machine Design I, chapter 8
- [Bending stress and safety factor of a spur pinion](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aagma-bending-stress-safety-factor) — Machine Design I, chapter 9
- [Open flat belt power and cross belt length](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aopen-flat-belt-power-and-length) — Machine Design I, chapter 8
- [Diameter of a shaft carrying a spur gear](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Agear-loaded-shaft-diameter) — Machine Design I, chapter 5
- [Shaft diameter for power and bending](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Astatic-shaft-power-bending-diameter) — Machine Design I, chapter 5
- [Weighted decision matrix for drive selection](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aweighted-decision-matrix-drive) — Machine Design I, chapter 1

## Related in other subjects

- [Main dimensions of 15 kW 2810 rpm cage motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Amain-dimensions-15kw-2810rpm) — Electric Machine Design, chapter 4, asked 2×
- [Stator design of 120 kW 750 rpm motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Astator-120kw-750rpm) — Electric Machine Design, chapter 4, asked 2×
- [200 kW 1420 rpm slip ring motor (star)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aslip-ring-200kw-star) — Electric Machine Design, chapter 4, asked 5×
- [250 HP 1410 rpm slip ring motor (kw 0.955)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aslip-ring-250hp-kw0955) — Electric Machine Design, chapter 4, asked 6×