hw8.sol.Spring07

hw8.sol.Spring07 - 1 ME 452: Machine Design II Solution to...

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Unformatted text preview: 1 ME 452: Machine Design II Solution to Homework Set 8. Problem 1 (Problem 13-8, page 701). The gearset in Example 13-1, page 663 is a spur gearset with the pressure angles for the pinion and the gear specified to be 20 = . The specified values of addendum and dedendum, see Table 13-1, 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 (13-10), page 666 as 2 2 2 (1 1 3sin ) 3sin P k N = + + For full depth teeth, see page 666, the factor k is given as 1 k = Substituting the values into Equation (13-10), 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 = 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 (13-11), page 666, is ( 29 2 2 2 2 (1 2 )sin (1 2 )sin P k N m m m m = + + + + For full depth teeth, see page 666, the factor k is given as 1 k = Substituting the values into Equation (13-11), 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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hw8.sol.Spring07 - 1 ME 452: Machine Design II Solution to...

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