Tools › Mechanical Design › Brakes, Clutches and Belts › Belt, Clutch and Brake Torque › Example
V-belt drive by Euler's relation, disc clutch by uniform wear and uniform pressure, long-shoe drum brake and band brake with the stop energy
| Tool | Belt, Clutch and Brake Torque |
| Source | Euler, L. (1762). "Remarques sur l'effet du frottement dans l'équilibre." Mémoires de l'Académie Royale des Sciences et Belles-Lettres de Berlin, Vol. 18, pp. 265–278 (the belt relation); Karsh, I., Clement, W., and Hoffman, J. (1966). Tape Recorder Belt Study. Kinelogic Corporation report for JPL contract 950899, NASA contract NAS7-100, Appendix B (the relation in the tool's form); U.S. Navy (1994). Basic Machines, NAVEDTRA 14037, Chapter 8 p. 8-3 (Prony-brake power example). |
| Level | Undergraduate |
Problem
A V belt with a 38° groove runs on 6 in and 12 in pulleys 30 in apart; the driver turns at 1,750 rpm and transmits 5 hp with f = 0.30 and a belt weight of 0.15 lb/ft. Find the wrap angles, the belt length, the belt speed, the centrifugal tension, the tension ratio and the tensions and installation tension that just carry the load. Then, for a two-face disc clutch of 4 and 6 in radii with f = 0.30 limited to 100 lb/in², the actuating force and torque by the uniform-wear and uniform-pressure assumptions; for an internal long shoe on a 10 in drum (pin 4 in from the centre, force arm 8 in, lining 1.5 in wide from 10° to 100°, f = 0.32, 150 lb/in², clockwise) the moments about the pin, the actuating force, the torque and the self-locking friction; for a band wrapped 270° with the same drum (slack-end arm 3 in, lever 12 in) the tensions and lever force; and the energy, time and temperature rise of stopping 2,000 lb·in² from 1,750 rpm with the shoe torque.
Inputs to Enter
- Belt: V belt 2β 38°, D_1 6 in, D_2 12 in, C 30 in, open, 1,750 rpm, 5 hp, f 0.30, w 0.15 lb/ft, least installed tension
- Clutch (select Disc or cone clutch): r_o 6 in, r_i 4 in, 2 faces, f 0.30, p_a 100 lb/in²; cone option α 12°, required torque 1,000 lb-in
- Brake (select Drum or band brake): internal long shoe r 5 in, a 4 in, c 8 in, b 1.5 in, f 0.32, p_a 150 lb/in², θ_1 10°, θ_2 100°, clockwise; band wrap 270°, a_1 0, a_2 3 in, L 12 in
- Stop energy: I 2,000 lb·in², 1,750 rpm to rest, 15 lbm of cast iron at 0.11 Btu/(lb·°F)
Answers
- Belt: sin ψ = 6/60, θ_1 = 168.5°, θ_2 = 191.5°, L = 88.57 in (series 88.57), v = 2,749 ft/min, F_c = m′v² = 9.79 lb, f/sin β = 0.921, e^(f′θ) = 15.03, F_1 − F_2 = P/v = 60.02 lb, F_1 = 74.09 lb, F_2 = 14.06 lb, F_i = 44.08 lb, T_1 = 180.1 lb-in
- Clutch, uniform wear: F = 2π p_a r_i (r_o − r_i) = 5,027 lb, T = f F n (r_o + r_i)/2 = 15,080 lb-in; uniform pressure: F = π p_a (r_o² − r_i²) = 6,283 lb, T = 19,101 lb-in; ratio 0.789; optimum r_i = r_o/√3 = 3.46 in (15,671 lb-in); cone α 12° gives 36,265 / 45,935 lb-in on one face
- Long shoe: θ_a = 90°, M_N = 4,304 lb-in, M_f = 1,409 lb-in (self-energizing), F_a = (M_N − M_f)/c = 361.9 lb, T = f p_a b r² (cos θ_1 − cos θ_2)/sin θ_a = 2,085 lb-in, f_lock = 0.978, pin reaction 902 lb; counter-clockwise F_a = 714.1 lb
- Band: e^(fφ) = 4.518, P_1 = p_a b r = 1,125 lb, P_2 = 249.0 lb, T = 4,380 lb-in, lever force P_2 a_2/L = 62.3 lb; a differential tight-end arm of 1 in would self-lock (a_2 ≤ a_1 e^(fφ) = 4.52 in)
- Stop by the shoe torque: t = I ω_0/T = 0.455 s, E = I ω_0²/2 = 7,249 ft-lb = 9.32 Btu, mean power 28.9 hp (peak 57.9 hp), 6.64 revolutions, ΔT = E/(m c) = 5.6 °F
- Navy Prony brake (test only): net pull 20 lb on a 1 ft pulley at 1,800 rpm = 3.427 hp (the manual prints 3.43 with π = 3.14)
Notes
The tool's default inputs are the belt case; press Submit, then change the Analysis select for the clutch and the brake (the other inputs above are their defaults). The cases are the tool's own illustrative defaults, every closed form checked in test_belt_clutch_brake.php (508 checks) against a scipy integration of the defining integrals and force vectors (References/Belt-Clutch-Brake/checks/beltclutch_check.py); the only printed source number is the Navy manual's Prony-brake example.