---
title: "Dynamic and wear load check, spur pair"
subject: "Machine Design I"
chapter: 9
key: "semester_6/machine_design_i:buckingham-dynamic-wear-check"
source: https://www.bctengineer.com/q/semester_6/machine_design_i/buckingham-dynamic-wear-check
site: BCT Engineering
---

# Dynamic and wear load check, spur pair

Machine Design I · Chapter 9: Gear design

- Asked in: year not given, 8 marks
- Video: [Design of Spur gear problems | DME |Design procedure for spur gears in Machine design | DME 2](https://www.youtube.com/watch?v=WBRA4qeeJHg) — Mahesh Gadwantikar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/machine_design_i/buckingham-dynamic-wear-check
- Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-9

## Question

A steel spur pinion of 20 teeth and module 6 mm, face width 60 mm, meshes with a steel gear of 60 teeth (20 degree full depth) and transmits 10 kW at 900 rpm of the pinion. The tooth error is 0.015 mm and E = 200 GPa for both gears. Take endurance strength of the teeth as 400 MPa, y = 0.154 - 0.912/z, the deformation factor C = 0.111 e/(1/Ep + 1/Eg) in N/mm (e in mm, E in N/mm^2), and the load-stress factor K = 1.1 N/mm^2. Find the dynamic load by Buckingham's equation, the beam strength, the limiting wear load, and state whether the design is safe.

## Short answer

Dynamic load Fd = 7.92 kN, beam strength Fs = 49.0 kN, wear load Fw = 11.9 kN. Both Fs and Fw exceed Fd, so the design is safe; wear governs.

## Answer

Data: $z_1 = 20$, $z_2 = 60$, $m = 6$ mm, $b = 60$ mm, $P = 10$ kW, $n_1 = 900$ rpm, $e = 0.015$ mm.

#### Velocity and transmitted load

$$
d_1 = mz_1 = 120\ \text{mm}, \qquad v = \frac{\pi d_1 n_1}{60\,000} = \frac{\pi \times 120 \times 900}{60\,000} = 5.655\ \text{m/s}
$$

$$
F_t = \frac{P}{v} = \frac{10\,000}{5.655} = 1768\ \text{N}
$$

#### Dynamic load (Buckingham)

Deformation factor:

$$
C = \frac{0.111\, e}{\frac{1}{E_p} + \frac{1}{E_g}} = \frac{0.111 \times 0.015}{2/200\,000} = 166.5\ \text{N/mm}
$$

$$
F_d = F_t + \frac{21 v\,(bC + F_t)}{21 v + \sqrt{bC + F_t}}
$$

$$
\begin{aligned} bC + F_t &= 60 \times 166.5 + 1768 = 11\,758\ \text{N} \\ \sqrt{11\,758} &= 108.4, \qquad 21v = 118.8 \\ F_d &= 1768 + \frac{118.8 \times 11\,758}{118.8 + 108.4} = 1768 + 6147 = 7915\ \text{N} \end{aligned}
$$

#### Beam strength

$$
y = 0.154 - \frac{0.912}{20} = 0.1084
$$

$$
F_s = \sigma_e\, b\, \pi m\, y = 400 \times 60 \times \pi \times 6 \times 0.1084 = 49\,040\ \text{N}
$$

(The weaker is the pinion, as both are steel and the pinion has smaller $y$.)

#### Wear load

$$
Q = \frac{2 z_2}{z_1 + z_2} = \frac{2 \times 60}{80} = 1.5
$$

$$
F_w = d_1\, b\, Q\, K = 120 \times 60 \times 1.5 \times 1.1 = 11\,880\ \text{N}
$$

#### Check

| Quantity | Value (N) | Ratio to $F_d$ |
|---|---|---|
| Dynamic load $F_d$ | 7915 | 1.0 |
| Beam strength $F_s$ | 49 040 | 6.2 |
| Wear load $F_w$ | 11 880 | 1.5 |

Both $F_s > F_d$ and $F_w > F_d$, so the design is safe against bending and wear. Wear is the controlling criterion (factor 1.5 only).

**Answer: Fd = 7.92 kN; Fs = 49.0 kN; Fw = 11.9 kN; the design is safe (wear governs).**

## Check yourself

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

1. How is the tangential load found?
   A. F_t = P / v = 10000 / 5.655 = 1768 N
   B. F_t = P x v
   C. F_t = P / n
   D. F_t = P / d1
   Answer: A. v = pi d1 n1 / 60000 = 5.655 m/s and F_t = power / pitch line velocity. (Key point: Transmitted load)

2. What does the deformation factor C = 0.111 e / (1/Ep + 1/Eg) depend on, and what is its value?
   A. Tooth error and moduli of elasticity; 166.5 N/mm
   B. Tooth width and speed; 11.1 N/mm
   C. Pressure angle only; 20 N/mm
   D. Module only; 6 N/mm
   Answer: A. C = 0.111 x 0.015 / (2/200000) = 166.5 N/mm; it grows with error e and stiffness. (Key point: Deformation factor)

3. What is the limiting wear load?
   A. F_w = 120 x 60 x 1.5 x 1.1 = 11880 N, with Q = 2 z2 / (z1 + z2)
   B. F_w = 400 x 60 x pi x 6 x 0.1084
   C. F_w = 1768 N
   D. F_w = 120 x 60 x 1.1 = 7920 N
   Answer: A. Buckingham: F_w = d1 b Q K, with Q = 120/80 = 1.5. (Key point: Wear load)

4. Which conclusion is correct?
   A. Bending governs because F_s is largest
   B. Wear governs because F_w = 11.9 kN is only 1.5 times F_d = 7.9 kN; both exceed F_d so design is safe
   C. The design fails since F_d is greater than F_t
   D. The design is unsafe because F_w < F_s
   Answer: B. Both F_s (49.0 kN) and F_w (11.9 kN) exceed F_d (7.9 kN); wear has the smallest margin so it controls. (Key point: Governing criterion)

## Related in Machine Design I

- [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
- [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
- [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
- [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

## 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
- [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
- [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