2_ch 07 Mechanical Design budynas_SM_ch07

2_ch 07 Mechanical Design budynas_SM_ch07 - K f = 1 + ....

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Chapter 7 179 7-2 This problem has to be done by successive trials, since S e is a function of shaft size. The material is SAE 2340 for which S ut = 1226 MPa, S y = 1130 MPa, and H B 368 . Eq. (6-19): k a = 4 . 51(1226) 0 . 265 = 0 . 685 Trial #1 : Choose d r = 22 mm Eq. (6-20): k b = ± 22 7 . 62 ² 0 . 107 = 0 . 893 Eq. (6-18): S e = 0 . 685(0 . 893)(0 . 5)(1226) = 375 MPa d r = d 2 r = 0 . 75 D 2 D / 20 = 0 . 65 D D = d r 0 . 65 = 22 0 . 65 = 33 . 8mm r = D 20 = 33 . 8 20 = 1 . 69 mm Fig. A-15-14: d = d r + 2 r = 22 + 2(1 . 69) = 25 . 4mm d d r = 25 . 4 22 = 1 . 15 r d r = 1 . 69 22 = 0 . 077 K t = 1 . 9 Fig. A-15-15: K ts = 1 . 5 Fig. 6-20: r = 1 . 69 mm, q = 0 . 90 Fig. 6-21: r = 1 . 69 mm, q s = 0 . 97 Eq. (6-32):
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Unformatted text preview: K f = 1 + . 90(1 . 9 − 1) = 1 . 81 K f s = 1 + . 97(1 . 5 − 1) = 1 . 49 We select the DE-ASME Elliptic failure criteria. Eq. (7-12) with d as d r , and M m = T a = 0, d r = 16(2 . 5) π ³ 4 ± 1 . 81(70)(10 3 ) 375 ² 2 + 3 ± 1 . 49(45)(10 3 ) 1130 ² 2 ´ 1 / 2 1 / 3 = 20 . 6 mm...
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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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