--- title: "Compound gear train speeds, torques and tooth forces" subject: "Machine Design I" chapter: 9 key: "semester_6/machine_design_i:compound-gear-train-forces" source: https://www.bctengineer.com/q/semester_6/machine_design_i/compound-gear-train-forces site: BCT Engineering --- # Compound gear train speeds, torques and tooth forces Machine Design I · Chapter 9: Gear design - Asked in: year not given, 8 marks - Video: [Calculating Output Torque and Holding Torque for Compound Gears](https://www.youtube.com/watch?v=lWpQ1GZA2s8) — Engineers Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/machine_design_i/compound-gear-train-forces - Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-9 ## Question A 12 kW motor running at 1440 rpm drives a compound reduction gear train of two spur gear pairs, all of module 4 mm and 20 degree pressure angle. Pinion 1 has 20 teeth and meshes with gear 2 of 60 teeth. Gear 3 of 24 teeth is mounted on the same shaft as gear 2 and meshes with gear 4 of 72 teeth. Take 96 percent efficiency for each pair of gears. Find the output speed, the output torque and power, and the tangential and radial tooth forces on the first and the second pair. ## Short answer Output speed = 160 rpm, output torque = 660 N.m, output power = 11.06 kW. First pair: Ft = 1989 N, Fr = 724 N; second pair: Ft = 4775 N, Fr = 1738 N. Forces derived from pitch diameters d=mz and torque after efficiency multipliers. ## Answer ```text Motor ->[1]---[2] shaft 2 20T 60T (2 and 3 fixed together) [3]---[4] -> output 24T 72T ``` #### Speed ratio and output speed $$ \text{Ratio} = \frac{z_2}{z_1}\times\frac{z_4}{z_3} = \frac{60}{20}\times\frac{72}{24} = 3 \times 3 = 9 $$ $$ n_2 = \frac{1440}{3} = 480\ \text{rpm}, \qquad n_4 = \frac{480}{3} = 160\ \text{rpm} $$ #### Torques and power $$ T_1 = \frac{60P}{2\pi n_1} = \frac{60 \times 12\,000}{2\pi \times 1440} = 79.58\ \text{N.m} $$ Each pair multiplies the torque by its ratio and by the efficiency 0.96: $$ \begin{aligned} T_2 &= T_1 \times 3 \times 0.96 = 229.2\ \text{N.m} \\ T_4 &= T_2 \times 3 \times 0.96 = 660.0\ \text{N.m} \end{aligned} $$ $$ P_{out} = 12 \times 0.96^2 = 11.06\ \text{kW} \quad(\text{check: } 660.0 \times 2\pi \times 160/60 = 11.06\ \text{kW}) $$ #### Tooth forces (pitch diameters $d = mz$) **First pair** (pinion 1: $d_1 = 4 \times 20 = 80$ mm): $$ \begin{aligned} F_t &= \frac{T_1}{d_1/2} = \frac{79.58}{0.04} = 1989\ \text{N} \\ F_r &= F_t \tan 20^\circ = 1989 \times 0.364 = 724\ \text{N} \\ F_n &= \frac{F_t}{\cos 20^\circ} = 2117\ \text{N} \end{aligned} $$ **Second pair** (pinion 3: $d_3 = 4 \times 24 = 96$ mm, torque $T_2 = 229.2$ N.m): $$ \begin{aligned} F_t &= \frac{229.2}{0.048} = 4775\ \text{N} \\ F_r &= 4775 \times 0.364 = 1738\ \text{N} \\ F_n &= \frac{4775}{\cos 20^\circ} = 5081\ \text{N} \end{aligned} $$ The tangential force produces the torque; the radial force pushes the gears apart and loads the shaft bearings. **Answer: Output 160 rpm; output torque 660 N.m; output power 11.06 kW; first pair Ft = 1989 N, Fr = 724 N; second pair Ft = 4775 N, Fr = 1738 N.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the output speed of the compound train? A. 1440 x 9 = 12960 rpm B. 1440 / 9 = 160 rpm C. 1440 / 3 = 480 rpm D. 1440 x 3/9 = 480 rpm Answer: B. Overall ratio = (60/20)(72/24) = 9, so n4 = 1440/9 = 160 rpm. (Key point: Output speed) 2. How does the torque change through each gear pair with 96 percent efficiency? A. It is divided by the ratio B. It is multiplied by the ratio and by 0.96 C. It is multiplied by 0.96 only D. It stays constant Answer: B. T2 = 79.58 x 3 x 0.96 = 229.2 N.m, T4 = 229.2 x 3 x 0.96 = 660.1 N.m. (Key point: Output torque) 3. What is the output power? A. 12 kW B. 12 x 0.96 = 11.52 kW C. 12 x 0.96^2 = 11.06 kW D. 12 / 0.96^2 = 13.0 kW Answer: C. Two pairs each with 96 percent efficiency give 12 x 0.96 x 0.96 = 11.06 kW. (Key point: Output power) 4. How are the tangential and radial forces found for the second pair? A. F_t = T1/(d3/2) and F_r = F_t cos 20 B. F_t = T2/(d3/2) = 4775 N and F_r = F_t tan 20 = 1738 N C. F_t = T2/d3 and F_r = F_t sin 20 D. F_t = P / n and F_r = F_t Answer: B. Gear 3 sits on the intermediate shaft carrying torque T2 = 229.2 N.m; F_t = T/(radius) and F_r = F_t tan(pressure angle). (Key point: Tooth forces) ## Related in Machine Design I - [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 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 - [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 - [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 - [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 ## 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 - [Compound gear train speed numerical](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Acompound-gear-train-speed-numerical) — Theory of Mechanism and Machine I, chapter 6 - [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 - [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