shi20396_ch18

# shi20396_ch18 - Chapter 18 Comment: This chapter, when...

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Chapter 18 Comment : This chapter, when taught immediately after Chapter 7, has the advantage of im- mediately applying the fatigue information acquired in Chapter 7. We have often done it ourselves. However, the disadvantage is that many of the items attached to the shaft have to be explained sufFciently so that the in±uence on the shaft is understood. It is the in- structor’s call as to the best way to achieve course objectives. This chapter is a nice note upon which to Fnish a study of machine elements. A very popular Frst design task in the capstone design course is the design of a speed-reducer, in which shafts, and many other elements, interplay. 18-1 The Frst objective of the problem is to move from shaft attachments to in±uences on the shaft. The second objective is to “see” the diameter of a uniform shaft that will satisfy de- ±ection and distortion constraints. (a) d P + (80 / 16) d P 2 = 12 in d P = 4 . 000 in W t = 63 025(50) 1200(4 / 2) = 1313 lbf W r = W t tan 25° = 1313 tan 25° = 612 lbf W = W t cos 25° = 1313 cos 25° = 1449 lbf T = W t ( d / 2) = 1313(4 / 2) = 2626 lbf · in Reactions R A , R B , and load W are all in the same plane. R A = 1449(2 / 11) = 263 lbf R B = 1449(9 / 11) = 1186 lbf M max = R A (9) = 1449(2 / 11)(9) = 2371 lbf · in Ans . 2" 9" Components in xyz 238.7 111.3 A B z x y 1074.3 612 1313 500.7 M O M max 5 2371 lbf•in 9" 01 1 " 11" 9" 2" w R A R B

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Chapter 18 465 (b) Using n d = 2 and Eq. (18-1) d A = ¯ ¯ ¯ ¯ 32 n d Fb ( b 2 l 2 ) 3 π El θ all ¯ ¯ ¯ ¯ 1 / 4 = ¯ ¯ ¯ ¯ 32(2)(1449)(2)(2 2 11 2 ) 3 π (30)(10 6 )(11)(0 . 001) ¯ ¯ ¯ ¯ 1 / 4 = 1 . 625 in A design factor of 2 means that the slope goal is 0.001 / 2 or 0.0005. Eq. (18-2): d B = ¯ ¯ ¯ ¯ 32 n d Fa ( l 2 a 2 ) 3 π θ all ¯ ¯ ¯ ¯ 1 / 4 = ¯ ¯ ¯ ¯ 32(2)(1449)(9)(11 2 9 2 ) 3 π 6 . 001) ¯ ¯ ¯ ¯ 1 / 4 = 1 . 810 in The diameter of a uniform shaft should equal or exceed 1.810 in. Ans. 18-2 This will be solved using a deterministic approach with n d = 2. However, the reader may wish to explore the stochastic approach given in Sec. 7-17. Table A-20: S ut = 68 kpsi and S y = 37 . 5 kpsi Eq. (7-8): S 0 e = 0 . 504(68) = 34 . 27 kpsi Eq. (7-18): k a = 2 . 70(68) 0 . 265 = 0 . 883 Assume a shaft diameter of 1.8 in. Eq. (7-19): k b = µ 1 . 8 0 . 30 0 . 107 = 0 . 826 k c = k d = k f = 1 From Table 7-7 for R = 0 . 999, k e = 0 . 753. Eq. (7-17): S e = 0 . 883(0 . 826)(1)(1)(1)(0 . 753)(34 . 27) = 18 . 8 kpsi From p. 444, K t = 2 . 14, K ts = 2 . 62 With r = 0 . 02 in, Figs. 7-20 and 7-21 give q = 0 . 60 and q s = 0 . 77, respectively. Eq. (7-31): K f = 1 + 0 . 60(2 . 14 1) = 1 . 68 K fs = 1 + 0 . 77(2 . 62 1) = 2 . 25 (a) DE-elliptic from Eq. (18-21), d = 16(2) π " 4 µ 1 . 68(2371) 18 800 2 + 3 µ 2 . 25(2626) 37 500 2 # 1 / 2 1 / 3 = 1 . 725 in Ans .
466 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (b) DE-Gerber from Eq. (18-16), d = 16 ( 2 )( 1 . 68 )( 2371 ) π( 18 800 ) 1 + " 1 + 3 µ 2 . 25 ( 2626 )( 18 800 ) 1 . 68 ( 2371 )( 68 000 ) 2 # 1 / 2 1 / 3 = 1 . 687 in Ans. From Prob. 18-1, de±ection controls d = 1 . 81 in 18-3 It is useful to provide a cylindrical roller bearing as the heavily-loaded bearing and a ball bearing at the other end to control the axial ±oat, so that the roller grooves are not subject to thrust hunting. Pro²le keyways capture their key. A small shoulder can locate the pinion, and a shaft collar (or a light press ²t) can capture the pinion. The key transmits the torque in either case. The student should:

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## This note was uploaded on 04/30/2008 for the course MAE 4900 taught by Professor Khanna during the Fall '06 term at Missouri (Mizzou).

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shi20396_ch18 - Chapter 18 Comment: This chapter, when...

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