--- title: "Tooth force components on a bevel pinion" subject: "Theory of Mechanism and Machine I" chapter: 8 key: "semester_6/theory_of_mechanism_and_machine_i:bevel-gear-tooth-forces-numerical" source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/bevel-gear-tooth-forces-numerical site: BCT Engineering --- # Tooth force components on a bevel pinion Theory of Mechanism and Machine I · Chapter 8: Force Analysis of Mechanisms - Asked in: year not given, 6 marks - Video: [Gear Forces of HELICAL and BEVEL GEARS in Just Over 10 Minutes!](https://www.youtube.com/watch?v=921ToiOynMA) — Less Boring Lectures - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/bevel-gear-tooth-forces-numerical - Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-8 ## Question A straight bevel pinion with 18 teeth and module 5 mm (at the large end), face width 40 mm, meshes with a 54 teeth gear on a shaft at 90 degrees. The pinion transmits 7.5 kW at 900 rpm and the pressure angle is 20 degrees. Considering the forces to act at the mean radius, determine the tangential, radial and axial components of the tooth force on the pinion. ## Short answer At mean radius: tangential force W_t = 2058 N, radial force W_r = 710 N, axial force W_a = 237 N on the pinion. W_t = T/(d_m/2), W_r = W_t tanφ cosγ, W_a = W_t tanφ sinγ with γ = 18.43°. ## Answer #### Force components in a bevel gear The resultant tooth force acts at the mean pitch radius $r_m$ of the face width, normal to the tooth surface. With pitch cone angle $\gamma$ and pressure angle $\phi$: $$ W_t = \frac{T}{r_m}, \qquad W_r = W_t\tan\phi\cos\gamma, \qquad W_a = W_t\tan\phi\sin\gamma $$ The radial force $W_t\tan\phi$ acts normal to the pitch cone; its components are along the radial direction ($\cos\gamma$) and along the axis ($\sin\gamma$). On the pinion the axial force pushes it away from the apex. #### Geometry Pitch cone angle of the pinion: $$ \tan\gamma = \frac{z_1}{z_2} = \frac{18}{54} \Rightarrow \gamma = 18.43^\circ \quad (\sin\gamma = 0.3162,\ \cos\gamma = 0.9487) $$ Large-end pitch diameter: $d = mz_1 = 5 \times 18 = 90$ mm. Mean pitch diameter: $$ d_m = d - b\sin\gamma = 90 - 40 \times 0.3162 = 77.35\ \text{mm} $$ #### Torque $$ \omega = \frac{2\pi \times 900}{60} = 94.248\ \text{rad/s}, \qquad T = \frac{P}{\omega} = \frac{7500}{94.248} = 79.58\ \text{N m} $$ #### Components $$ W_t = \frac{T}{d_m/2} = \frac{79.58}{(77.35/2)\times10^{-3}} = 2058\ \text{N} $$ $$ W_r = W_t\tan20^\circ\cos\gamma = 2058 \times 0.3640 \times 0.9487 = 710\ \text{N} $$ $$ W_a = W_t\tan20^\circ\sin\gamma = 2058 \times 0.3640 \times 0.3162 = 237\ \text{N} $$ The resultant normal force is $W_n = W_t/\cos20^\circ = 2190$ N. On the gear (shaft at 90 degrees) the pinion's radial force becomes its axial force and vice versa: the gear's axial force $= 710$ N and radial force $= 237$ N. **Answer: $W_t = 2058$ N, $W_r = 710$ N, $W_a = 237$ N on the pinion.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For an 18-tooth pinion meshing with a 54-tooth gear at 90 degrees, the pinion pitch cone angle is: A. 18.43 degrees B. 71.57 degrees C. 45 degrees D. 20 degrees Answer: A. tan(gamma) = z1/z2 = 18/54, so gamma = 18.43 degrees. (Key point: Pitch cone angle) 2. Why is the tangential force computed at the mean pitch radius? A. Because torque acts at the large end B. Because the resultant tooth force is taken to act at the mean pitch radius of the face width, d_m = d - b sin(gamma) C. Because the pinion has no face width D. Because the pitch circle is zero at the apex Answer: B. The resultant acts at mean radius; d_m = 90 - 40 x 0.3162 = 77.35 mm. (Key point: Mean diameter) 3. For the pinion, the radial and axial components are: A. W_r = W_t tan(phi) sin(gamma), W_a = W_t tan(phi) cos(gamma) B. W_r = W_t sin(phi), W_a = W_t cos(phi) C. W_r = W_t tan(phi) cos(gamma), W_a = W_t tan(phi) sin(gamma) D. W_r = W_a = W_t tan(phi) Answer: C. The force W_t tan(phi) normal to the pitch cone splits into radial (cos gamma) and axial (sin gamma) parts. (Key point: Radial and axial components) 4. What are the pinion tangential, radial and axial forces? A. 2058 N, 237 N, 710 N B. 2058 N, 710 N, 237 N C. 2190 N, 710 N, 237 N D. 1768 N, 710 N, 237 N Answer: B. W_t = 79.58/0.038675 = 2058 N; W_r = 2058 x 0.364 x 0.9487 = 710 N; W_a = 2058 x 0.364 x 0.3162 = 237 N. (Key point: Numerical result) ## Related in Theory of Mechanism and Machine I - [Straight bevel gear geometry and formative teeth](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Astraight-bevel-gear-geometry-formative-teeth) — Theory of Mechanism and Machine I, chapter 5 - [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 - [Force components on a helical gear tooth](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Ahelical-gear-force-components-axial-thrust) — Theory of Mechanism and Machine I, chapter 8 ## Related in other subjects - [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 - [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 - [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