3. Exercice 2 : Frein à Tambour à Sabots Symétriques (Niveau Intermédiaire)
Ce=n⋅f⋅Fa⋅23⋅R3−r3R2−r2cap C sub e equals n center dot f center dot cap F sub a center dot two-thirds center dot the fraction with numerator cap R cubed minus r cubed and denominator cap R squared minus r squared end-fraction exercice corrige embrayage frein pdf
III. Exercice 2 : Étude d'un Frein à Disque de Véhicule (Niveau Avancé) Fpcap F sub p : Axial pressing force (effort presseur)
F_\textfrein = 4 \times F_r = 320 \ \textN C_\textfrein = f \times F_\textfrein \times R_m = 0,3 \times 320 \times 0,05 = 4,8 \ \textN·m À vérifier selon l’application.
Mechanical analysis of friction systems typically distinguishes between two operating models: (for new components) and Uniform Wear (for run-in components). 1. Transmissible Torque (Clutch) The torque Ctcap C sub t that a plane disk clutch can transmit is calculated as: General Formula: : Number of friction surfaces (e.g., for a single-disk clutch with two contact faces). : Friction coefficient. Fpcap F sub p : Axial pressing force (effort presseur). Rmcap R sub m : Mean radius of the friction surface. 2. Mean Radius ( Rmcap R sub m ) Calculation Uniform Pressure: Uniform Wear: (where is the outer radius and is the inner radius) . 3. Braking Torque (Brake) The braking torque Cfcap C sub f follows a similar principle to clutches: Disk Brake: , where the friction force is Drum Brake: Cfcap C sub f
Ce couple est au couple d’entraînement maximal (9,4 N·m), mais il suffit généralement à immobiliser l’arbre de sortie au débrayage, car l’inertie des parties tournantes est faible. À vérifier selon l’application.
