hw13sol.Spring07

hw13sol.Spring07 - ME 452: Machine Design II Solution to...

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1 ME 452: Machine Design II Solution to Homework Set 13 Problem 1 (Problem 16-1, page 851). We will solve the problem for a clockwise rotation of 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 814 and Example 16-2, page 817). (a) The moment of the friction force about the hinge pin, see Equation (16-2), page 815, can be written as 2 1 sin ( cos ) sin a f a fp br M ra d θ θθ =− The moment of the friction force about the hinge pin, by carrying out the above integration, see Equation (16-8), page 816 and Example (16-2), page 817, can be written as () 22 21 2 1 cos cos sin sin sin 2 a f a fp br a Mr ⎡⎤ ⎢⎥ ⎣⎦ The angles 1 and 2 , see Figure for Problem (16-1), page 851 and Figure (16-7), page 814, are given as 12 0 and 120 = ° The angle at which the maximum pressure on the shoe occurs, see Figure 16-6(b), page 814 and Example (16-2), page 817, is given as 90 a = ° The distance a , see Figure (16-7), page 814 and Figure for Problem (16-1), page 851, is given as 5 in aR = = The distance c , see Figure (16-7), page 814 Figure for Problem (16-1), page 851, is given as 2c o s 3 0 cR = ° Therefore, the distance c is given as 25c o s 3 0 i n c =×× ° = 8 . 6 6 Therefore, the moment of the friction force about the hinge pin is given as 0.28 1.5 6 5 6 cos120 cos0 sin 120 sin 0 17.96 in-lb sin90 2 a f a p M p ××× = × − × °− ° − ° = ° The moment of the normal force about the hinge pin, see Equation (16-3), page 815, can be written as
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2 2 1 2 sin sin a N a pbra M d θ = The moment of the normal force about the hinge pin, by carrying out the above integration, see Equation (16-8), page 816 and Example (16-2), page 817, can be written as () 21 2 1 11 sin2 sin 2 4 a N a M θθ ⎡⎤ =− ⎢⎥ ⎣⎦ Therefore, the moment of the normal force about the hinge pin is given as 1.5 6 5 1 120 0 1 sin 240 sin0 56.87 in-lb sin90 2 180 180 4 a N a p Mp ππ ×× × × × ⎛⎞ ° ° = ⎜⎟ ° ⎝⎠ For the self-energizing (right) shoe, the actuating force, see Equation (16-4), page 815, can be written as Nf M M F c = The actuating force is specified to be F = 500 lb. Therefore the actuating force is given as ( ) 56.87 17.96 500 4.49 8.66 aa right right a right pp p == Therefore, the maximum pressure at the self-energizing (right) shoe, is given as 500 111.4 psi 4.49 a right p (b) The braking torque provided by the right shoe, see Equation (16-6), page 815, can be written as ( ) 2 12 (cos cos ) sin a right right a fp b r T = Therefore, the braking torque provided by the right shoe is given as 2 0.28 111.4 1.5 (6) (cos0 cos120 ) 2526.6 in-lb right T × ° ° ° The left shoe is self-deenergizing. Therefore, for the left shoe, the actuation force, see Equation (16-7), page 816, can be written as M M F c + = Therefore the actuation force is given as ( ) 56.87 17.96 500 8.64 8.66 left left a left p +
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3 Therefore, the maximum pressure at the left shoe is given as () 500 57.9 psi 8.64 a left p ==
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This note was uploaded on 02/09/2012 for the course ME 352 taught by Professor Staff during the Fall '08 term at Purdue University.

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hw13sol.Spring07 - ME 452: Machine Design II Solution to...

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