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Unformatted text preview: 1 ME 452: Machine Design II Solution to Homework Set 8. Problem 1 (Problem 138, page 701). The gearset in Example 131, page 663 is a spur gearset with the pressure angles for the pinion and the gear specified to be 20 φ = ° . The specified values of addendum and dedendum, see Table 131, page 676, show that the gears have full depth teeth. (a) The smallest number of teeth on the pinion that will allow the pinion to run in mesh with a similar gear without interference is given by Equation (1310), page 666 as 2 2 2 (1 1 3sin ) 3sin P k N φ φ = + + For full depth teeth, see page 666, the factor k is given as 1 k = Substituting the values into Equation (1310), the smallest number of teeth on the pinion is 2 2 2 1 (1 1 3 sin 20 ) 12.3 3 sin 20 P N × = + + × ° = × ° Since it is not possible to have12.3 teeth then 13 teeth represents the least number of teeth on the pinion. Therefore, a spur gearset with full depth teeth, pressure angle of 20 φ = ° and 13 teeth on both the pinion and the gear is free of interference. (b) The smallest number of teeth on the pinion for meshing with a gear to obtain a gear ratio of 2.5 G m = without interference, see Equation (1311), page 666, is ( 29 2 2 2 2 (1 2 )sin (1 2 )sin P k N m m m m φ φ = + + + + For full depth teeth, see page 666, the factor k is given as 1 k = Substituting the values into Equation (1311), the smallest number of teeth on the pinion is ( 29 2 2 2 2 1 2.5 (2.5) (1 2 2.5) sin 20 14.6 (1 2 2.5) sin 20 P N × = + + + × × ° = + × × ° 2 Since it is not possible to have 14.6 teeth then 15 teeth represents the smallest number of teeth Since it is not possible to have 14....
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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.
 Fall '08
 Staff
 Machine Design

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