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**Unformatted text preview: **1 ME 452: Machine Design II Solution to Homework Set 13 Problem 1. The figure below shows an internal rim=type brake having an inside rim diameter of 12 inches and a dimension R = 5 inches. The shoes have a face width of 1.5 inches and are both actuated by a force of 500 lb. The mean coefficient of friction is 0.28. (a) Calculate the maximum pressure and indicate the shoe on which it occurs. (b) Calculate the braking torque effected by each shoe, and find the total braking torque. (c) Calculate the resulting hinge-pin reaction forces. SOLUTION: There are two shoes interacting with the brake drum. Each generates friction and applies torque to the drum, but one shoe will be self-energizing and the other will not be. This will be true regardless of the drum rotation direction, because of the symmetry of the geometry. The total torque on the drum will be the same regardless of drum rotation direction, so we are free to choose a clockwise rotation for the drum. For a clockwise rotation of the drum, the right shoe is self-energizing, because for the right shoe the friction force acting on the shoe causes a clockwise moment about its hinge pin, which is in the same direction as the applied force, F. In this way, the friction helps apply the brake (see Figure 16-7, page 834 and Example 16-2, page 837). The other shoe will therefore not be self energizing. 2 (a) The moment of the friction force about the hinge pin, see Equation (16-2), page 835, can be written as 2 1 sin ( cos ) sin a f a fp br M r a d θ θ θ θ θ θ =- ∫ The moment of the friction force about the hinge pin, by carrying out the above integration, see Equation (16-8), page 836 and Example (16-2), page 837, can be written as ( 29 ( 29 2 2 2 1 2 1 cos cos sin sin sin 2 a f a fp br a M r θ θ θ θ θ =---- The angles 1 θ and 2 θ , see Figure (16-7), page 834, are 1 2 0 and 120 θ θ = ° = ° The angle at which the maximum pressure on the shoe occurs, see Figure 16-6(b), page 834 and Example (16-2), page 837, is 90 a θ = ° The distance a , see Figure (16-7), page 834, is 5 in a R = = The distance c , see Figure (16-7), page 834, is 2 cos30 c R = ° Therefore, the distance c is 2 5 cos30 in c = × × ° = 8.66 Therefore, the moment of the friction force about the hinge pin is ( 29 ( 29 2 2 0.28 1.5 6 5 6 cos120 cos0 sin 120 sin 0 17.96 in-lb sin 90 2 a f a p M p × × × = × - × ° - ° - ° - ° = ° The moment of the normal force about the hinge pin, see Equation (16-3), page 835, is 2 1 2 sin sin a N a p bra M d θ θ θ θ θ = ∫ The moment of the normal force about the hinge pin, by carrying out the above integration, see Equation (16-8), page 836 and Example (16-2), page 837, is 3 ( 29 ( 29 2 1 2 1 1 1 sin 2 sin 2 sin 2 4 a N a p bra M θ θ θ θ θ =--- Therefore, the moment of the normal force about the hinge pin is ( 29 1.5 6 5 1 120 1 sin 240 sin 0 56.87 in-lb sin 90 2 180 180 4 a N a p M p π π...

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