hw9.sol.Spring07

# hw9.sol.Spring07 - 1 ME 452: Machine Design II Solution to...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 ME 452: Machine Design II Solution to Homework Set 9 Problem 1 (Problem 14-15, page 761) 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 gearset 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 gearset 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 656, 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 720, can be written as P P P d n V ft / min 12 π = Therefore, the pitch line velocity is given as P 1.417 525 V 194.8 ft/min 12 π × × = = 2 The uncorrected endurance strength for 200 kpsi ut S < , see Equation (6-8), page 274, can be written as 0.5 e ut S S ′ = Therefore, the uncorrected endurance strength is given as 0.5 76 38 kpsi e S ′ = × = The surface factor, see Equation (6-19), page 279 can be written as b a ut k = a S For milled (machined) teeth, see Table (6-2), page 280, the factor a and the exponent b are given as 2.7 and 0.265 = = - a b Therefore, the surface factor is given as 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 717, and Table (13-1), page 676, 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 717 and Figure (14-1), page 717, can be written as 4 = t l x where the value of the distance x , see Equation (14-3), page 717, can be written as 3 2 = Y x P The Lewis form factor for the pinion with 17 teeth, see Table (14-2), page 718, is Y 0.303 = Therefore, the distance x for the pinion is 3 0.303 0.0379 in 2 12 x × = = × Therefore, the tooth thickness for the pinion is 4 0.1875 0.0379 0.1686 in t = × × = 3 The effective diameter for bending, see Equation (6-25), page 281, can be written as ( 29 1/ 2 e d 0.808 = h b where h = F Face width 0.875 in0....
View Full Document

## 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.

### Page1 / 14

hw9.sol.Spring07 - 1 ME 452: Machine Design II Solution to...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online