---
title: "Lewis method module for spur gears"
subject: "Machine Design I"
chapter: 9
key: "semester_6/machine_design_i:lewis-spur-gear-module-design"
source: https://www.bctengineer.com/q/semester_6/machine_design_i/lewis-spur-gear-module-design
site: BCT Engineering
---

# Lewis method module for spur gears

Machine Design I · Chapter 9: Gear design

- Asked in: year not given, 8 marks
- Video: [LEWIS BENDING STRESS at the Teeth of a Gear in Just Over 12 Minutes!](https://www.youtube.com/watch?v=J1i29j43trM) — Less Boring Lectures
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/machine_design_i/lewis-spur-gear-module-design
- Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-9

## Question

A pair of 20 degree full-depth spur gears transmits 18 kW. The pinion has 22 teeth and runs at 1200 rpm, and the gear has 66 teeth. The pinion is steel with allowable static stress 210 MPa and the gear is cast iron with allowable static stress 100 MPa. Take velocity factor Cv = 3/(3 + v) with v in m/s, face width b = 10m (m = module), service factor 1.25 and Lewis form factor y = 0.154 - 0.912/z. Find the module (choose a standard value from 2, 2.5, 3, 4, 5, 6, 8, 10 mm), the face width and the pitch diameters, and check the pinion.

## Short answer

The cast iron gear is weaker (lower σy product). Trial with Lewis equation and velocity factor Cv = 3/(3+v) gives module m = 5 mm, face width b = 50 mm, pinion pitch diameter 110 mm, gear 330 mm. The pinion is safe.

## Answer

#### Form factors

$$
y_1 = 0.154 - \frac{0.912}{22} = 0.1125, \qquad y_2 = 0.154 - \frac{0.912}{66} = 0.1402
$$

Compare $\sigma_o y$:

$$
\text{Pinion: } 210 \times 0.1125 = 23.6, \qquad \text{Gear: } 100 \times 0.1402 = 14.0
$$

The product is smaller for the **gear** (cast iron), so the **gear is the weaker element** and is used for design.

#### Design by trial of module

Load required: $F_t = \dfrac{P \times C_s}{v}$, with $v = \dfrac{\pi\, m\, z_1\, n_1}{60\,000}$ (m in mm). Gear capacity: $F_s = \sigma_{o2}\,C_v\, b\,\pi m\, y_2$ with $b = 10m$.

| m (mm) | v (m/s) | $C_v$ | b (mm) | Required $F_t$ (N) | Capacity $F_s$ (N) | Result |
|---|---|---|---|---|---|---|
| 4 | 5.53 | 0.352 | 40 | $\frac{18\,000 \times 1.25}{5.53} = 4069$ | 2478 | Fails |
| 5 | 6.91 | 0.303 | 50 | $\frac{22\,500}{6.91} = 3255$ | 3332 | Safe |

For example for $m = 5$:

$$
F_s = 100 \times 0.3027 \times 50 \times \pi \times 5 \times 0.1402 = 3332\ \text{N} > 3255\ \text{N}
$$

So the smallest standard module that works is **m = 5 mm**.

#### Dimensions

$$
b = 10m = 50\ \text{mm}, \qquad d_1 = mz_1 = 5 \times 22 = 110\ \text{mm}, \qquad d_2 = 5 \times 66 = 330\ \text{mm}
$$

#### Check of the pinion

$$
F_{s1} = 210 \times 0.3027 \times 50 \times \pi \times 5 \times 0.1125 = 5619\ \text{N} > 3255\ \text{N}
$$

so the pinion is also safe.

**Answer: Module = 5 mm; face width = 50 mm; pinion pitch diameter = 110 mm; gear pitch diameter = 330 mm. The design is controlled by the cast iron gear.**

## Check yourself

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

1. How is the weaker element of the pair identified?
   A. The one with larger diameter
   B. The one with smaller (allowable stress x y); here the cast iron gear (14.0 vs 23.6)
   C. Always the pinion
   D. The one with higher speed
   Answer: B. Compare sigma_o y for pinion and gear; the smaller value is weaker, so the design is based on the cast iron gear. (Key point: Weaker element)

2. Why does m = 4 mm fail?
   A. Required F_t 4069 N exceeds the gear capacity 2478 N
   B. Face width is too large
   C. Velocity factor is above 1
   D. The pinion is too weak
   Answer: A. At m = 4, F_s = 2478 N is less than the required load 4069 N. (Key point: Trial module)

3. Why is m = 5 mm chosen?
   A. It is the largest standard module
   B. It is the smallest standard module whose capacity (3332 N) exceeds the required load (3255 N)
   C. It gives the largest safety margin
   D. It makes the pinion fail
   Answer: B. Modules are tried in increasing order; 5 mm is the first with F_s >= required F_t. (Key point: Chosen module)

4. What are the face width and pitch diameters?
   A. b = 50 mm, d1 = 110 mm, d2 = 330 mm
   B. b = 40 mm, d1 = 88 mm, d2 = 264 mm
   C. b = 50 mm, d1 = 22 mm, d2 = 66 mm
   D. b = 5 mm, d1 = 110 mm, d2 = 330 mm
   Answer: A. b = 10m = 50 mm, d = m z gives 110 mm and 330 mm. (Key point: Dimensions)

## Related in Machine Design I

- [Dynamic and wear load check, spur pair](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Abuckingham-dynamic-wear-check) — Machine Design I, chapter 9
- [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
- [Lewis equation for beam strength of a gear tooth](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Alewis-beam-strength-derivation) — Machine Design I, chapter 9
- [Compound gear train speeds, torques and tooth forces](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Acompound-gear-train-forces) — Machine Design I, chapter 9
- [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
- [Spur gear terminology and standard tooth system](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aspur-gear-terminology) — Machine Design I, chapter 9

## Related in other subjects

- [Spur gear pair dimensions numerical](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Aspur-gear-pair-geometry-numerical) — Theory of Mechanism and Machine I, chapter 4
- [Tooth forces on a spur gear pinion](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Aspur-gear-tooth-forces-numerical) — Theory of Mechanism and Machine I, chapter 8
- [Gearbox shaft design with spur gear](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Agearbox-shaft-design-asme) — Machine Design II, chapter 5
- [Tooth force components on a bevel pinion](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Abevel-gear-tooth-forces-numerical) — Theory of Mechanism and Machine I, chapter 8