3_ch 07 Mechanical Design budynas_SM_ch07

3_ch 07 Mechanical Design budynas_SM_ch07 - 2 . Examining...

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180 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Trial #2 : Choose d r = 20 . 6mm k b = ± 20 . 6 7 . 62 ² 0 . 107 = 0 . 899 S e = 0 . 685(0 . 899)(0 . 5)(1226) = 377 . 5MPa D = d r 0 . 65 = 20 . 6 0 . 65 = 31 . 7mm r = D 20 = 31 . 7 20 = 1 . 59 mm Figs. A-15-14 and A-15-15: d = d r + 2 r = 20 . 6 + 2(1 . 59) = 23 . 8mm d d r = 23 . 8 20 . 6 = 1 . 16 r d r = 1 . 59 20 . 6 = 0 . 077 We are at the limit of readability of the figures so K t = 1 . 9, K ts = 1 . 5 q = 0 . 9, q s = 0 . 97 K f = 1 . 81 K fs = 1 . 49 Using Eq. (7-12) produces d r = 20 . 5mm . Further iteration produces no change. Decisions : d r = 20 . 5mm D = 20 . 5 0 . 65 = 31 . 5mm , d = 0 . 75(31 . 5) = 23 . 6mm Use D = 32 mm, d = 24 mm, r = 1 . 6mm Ans. 7-3 F cos 20°( d / 2) = T , F = 2 T / ( d cos 20°) = 2(3000) / (6 cos 20°) = 1064 lbf M C = 1064(4) = 4257 lbf · in For sharp fillet radii at the shoulders, from Table 7-1, K t = 2 . 7, and K ts = 2
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Unformatted text preview: 2 . Examining Figs. 6-20 and 6-21, with S ut = 80 kpsi, conservatively estimate q = . 8 and q s = . 9 . These estimates can be checked once a specic llet radius is determined. Eq. (6-32): K f = 1 + (0 . 8)(2 . 7 1) = 2 . 4 K f s = 1 + (0 . 9)(2 . 2 1) = 2 . 1 (a) Static analysis using fatigue stress concentration factors: From Eq. (7-15) with M = M m , T = T m , and M a = T a = 0, max = 32 K f M d 3 2 + 3 16 K f s T d 3 2 1 / 2...
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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