---
title: "Bending stress and safety factor of a spur pinion"
subject: "Machine Design I"
chapter: 9
key: "semester_6/machine_design_i:agma-bending-stress-safety-factor"
source: https://www.bctengineer.com/q/semester_6/machine_design_i/agma-bending-stress-safety-factor
site: BCT Engineering
---

# Bending stress and safety factor of a spur pinion

Machine Design I · Chapter 9: Gear design

- Asked in: year not given, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/machine_design_i/agma-bending-stress-safety-factor
- Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-9

## Question

A spur pinion of module 4 mm with 25 teeth and face width 40 mm transmits 15 kW at 1500 rpm. Use the AGMA bending stress equation sigma = Wt Ko Kv Ks Km Kb/(b m J) with Ko = 1.25, Ks = 1, Km = 1.3, Kb = 1 and J = 0.34. The gears are of quality number Qv = 6, so Kv = ((A + sqrt(200 V))/A)^B with B = 0.25 (12 - Qv)^(2/3) and A = 50 + 56 (1 - B), V in m/s. The pinion is through-hardened Grade 1 steel of 250 HB with allowable bending stress St = 0.533 HB + 88.3 MPa. The life required is 10000 hours, and the life factor is YN = 1.3558 N^(-0.0178). Take KT = KR = 1. Find the bending stress and the factor of safety.

## Short answer

Calculate tangential load Wt = P/V = 15 kW / 7.854 m/s = 1910 N, apply AGMA bending equation with given factors to get sigma = 86.8 MPa. Allowable stress St = 221.6 MPa, life factor YN = 0.939, resulting factor of safety n = 2.40.

## Answer

#### Pitch line velocity and tangential load

$$
d = mz = 4 \times 25 = 100\ \text{mm}, \qquad V = \frac{\pi d n}{60} = \frac{\pi \times 0.1 \times 1500}{60} = 7.854\ \text{m/s}
$$

$$
W_t = \frac{P}{V} = \frac{15\,000}{7.854} = 1910\ \text{N}
$$

#### Dynamic factor ($Q_v = 6$)

$$
\begin{aligned} B &= 0.25\,(12 - 6)^{2/3} = 0.25 \times 3.302 = 0.8255 \\ A &= 50 + 56\,(1 - 0.8255) = 59.77 \\ K_v &= \left(\frac{A + \sqrt{200 V}}{A}\right)^B = \left(\frac{59.77 + \sqrt{1570.8}}{59.77}\right)^{0.8255} = (1.663)^{0.8255} = 1.522 \end{aligned}
$$

#### Bending stress

$$
\sigma = \frac{1910 \times 1.25 \times 1.522 \times 1 \times 1.3 \times 1}{40 \times 4 \times 0.34} = \frac{4724}{54.4} = 86.8\ \text{MPa}
$$

#### Allowable stress and life factor

$$
S_t = 0.533 \times 250 + 88.3 = 221.6\ \text{MPa}
$$

Number of cycles:

$$
N = 1500 \times 60 \times 10\,000 = 9\times10^{8}
$$

$$
Y_N = 1.3558\,(9\times10^8)^{-0.0178} = 0.939
$$

#### Factor of safety

$$
n = \frac{S_t\, Y_N}{K_T\, K_R\, \sigma} = \frac{221.6 \times 0.939}{1 \times 1 \times 86.8} = 2.40
$$

The pinion is safe in bending (n > 1) with a reasonable margin. Surface (contact) strength should be checked separately, because pitting often governs the life of hardened gears.

**Answer: sigma = 86.8 MPa; n = 2.4.**

## Check yourself

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

1. How is W_t found for the 25-tooth, m = 4 pinion at 1500 rpm?
   A. W_t = P / V with V = pi d n / 60 = 7.854 m/s, so 1910 N
   B. W_t = P x V
   C. W_t = P / n
   D. W_t = P / d
   Answer: A. d = 100 mm, V = 7.854 m/s and W_t = 15000 / 7.854 = 1910 N. (Key point: Tangential load)

2. Using the AGMA equation with the given factors, what is the bending stress?
   A. About 86.8 MPa
   B. About 8.7 MPa
   C. About 868 MPa
   D. About 221.6 MPa
   Answer: A. sigma = 1910 x 1.25 x 1.522 x 1 x 1.3 x 1 / (40 x 4 x 0.34) = 86.8 MPa. (Key point: Bending stress)

3. Why is Y_N = 0.939 below 1?
   A. Because N = 9 x 10^8 cycles exceeds 10^7, where Y_N falls with the number of cycles
   B. Because the stress is low
   C. Because the gear is soft
   D. Because the speed is low
   Answer: A. Y_N is 1 at 10^7 cycles and decreases for longer lives, N = 1500 x 60 x 10000 = 9 x 10^8. (Key point: Life factor)

4. What is the factor of safety and conclusion?
   A. n = 221.6 x 0.939 / 86.8 = 2.4, safe in bending
   B. n = 86.8 / 221.6 = 0.39, unsafe
   C. n = 221.6 / 0.939 = 236, safe
   D. n = 1.0 exactly
   Answer: A. n = S_t Y_N / (K_T K_R sigma) = 2.40, above 1, so bending is safe; surface strength should be checked separately. (Key point: Safety factor)

## Related in Machine Design I

- [AGMA bending stress equation and its factors](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aagma-factors-stress-concentration-geometry) — Machine Design I, chapter 9
- [Lewis method module for spur gears](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Alewis-spur-gear-module-design) — Machine Design I, chapter 9
- [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
- [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
- [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

## Related in other subjects

- [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
- [Cantilever 5 m: maximum and 25 mm bending stress](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acantilever-5m-8kn-15kn-bending-stresses) — Strength of Materials, chapter 7
- [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
- [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