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Unformatted text preview: 1 ME 452: Machine Design II Solution to Homework Set 9 Problem 1 (Problem 14-15, page 781) A steel spur pinion and gear have a diametral pitch of 12 teeth/inch, milled teeth, 17 and 30 teeth respectively, a 20 degree pressure angle, and a pinion speed of 525 rpm. The tooth properties are S ut = 76 kpsi, S y = 42 kpsi, and the Brinell hardness is 149. For a design factor of 2.25, a face width of 0.875 inches, what is the power rating of the gear set? SOLUTION: In general, you should use the AGMA approach for gear analysis, since it is more accurate. But the AGMA approach, at least as summarized in our book, does not give any way to check yield strengths of the gears. This is usually acceptable, because for gears the cycles are so high that fatigue usually dominates. Other than its historical significance, the primary usefulness of the Lewis equation is to check gear tooth yield. Note that ultimate and yield strengths are given in this problem statement, not AGMA fatigue strengths, so the methods of Chapter 6 and the Lewis equation will be used in the solution which follows. Problem 2 of this homework set will focus on the AGMA approach. Since the pinion and the gear of the given spur gear set have the same material properties then the power that the gear set can transmit is governed by the failure of the pinion. The problem is to determine: (i) the power that can be transmitted considering pinion tooth bending failure, and, (ii) the power that can be transmitted considering pinion tooth wear. The power rating of the gear set will then be the minimum of these two power ratings (for the specified design factor of 2.25). Consider bending of the pinion tooth. The diameter of the pinion, see Equation (13-1), page 676, can be written as P P N d P = where P is the diametral pitch; i.e., number of teeth per inch. Therefore, the diameter of the pinion is 17 1.417 in 12 P d = = The pitch line velocity, see Example (14-1), page 740, can be written as P P P d n V ft / min 12 = Therefore, the pitch line velocity is given as P 1.417 525 V 194.8 ft/min 12 = = 2 The uncorrected endurance strength for 200 kpsi ut S < , see Equation (6-8), page 282, can be written as 0.5 e ut S S = Therefore, the uncorrected endurance strength is 0.5 76 38 kpsi e S = = The surface factor, see Equation (6-19), page 287 can be written as b a ut k = a S For milled (machined) teeth, see Table (6-2), page 288, the factor a and the exponent b are given as 2.7 and 0.265 = = - a b Therefore, the surface factor is 0.265 a k 2.7 (76) 0.857- = = Assuming full depth teeth, the sum of the addendum and dedendum, see Figure (14-1), page 737, and Table (13-1), page 696, can be written as 1 1.25 1 1.25 0.1875 in 12 12 l P P = + = + = The tooth thickness, see Equation (b), page 737 and Figure (14-1), page 737, can be written as 4 = t l x where the value of the distance x , see Equation (14-3), page 737, can be written as 3 2 = Y x P The Lewis form factor for the pinion with 17 teeth, see Table (14-2), page 738, is...
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This document was uploaded on 12/28/2011.
- Fall '09
- Machine Design