13_ch 07 Mechanical Design budynas_SM_ch07

13_ch 07 Mechanical Design budynas_SM_ch07 - However, for...

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190 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Right-hand bearing shoulder The text does not give minimum and maximum shoulder diameters for 03-series bearings (roller). Use D = 1.75 in. r d = 0 . 030 1 . 574 = 0 . 019, D d = 1 . 75 1 . 574 = 1 . 11 From Fig. A-15-9, K t = 2 . 4 From Fig. A-15-8, K ts = 1 . 6 From Fig. 6-20, q = 0 . 65 From Fig. 6-21, q s = 0 . 83 K f = 1 + 0 . 65(2 . 4 1) = 1 . 91 K fs = 1 + 0 . 83(1 . 6 1) = 1 . 50 M = 2178 ± 0 . 453 2 ² = 493 lbf · in Eq. (7-11): 1 n = 16 π (1 . 574 3 ) ³ 4 ± 1 . 91(493) 24 700 ² 2 + 3 ± 1 . 50(2500) 37 500 ² 2 ´ 1 / 2 = 0 . 247, from which n = 4 . 05 Overhanging coupling keyway There is no bending moment, thus Eq. (7-11) reduces to: 1 n = 16 3 K fs T m π d 3 S y = 16 3(1 . 50)(2500) π (1 . 5 3 )(37 500) = 0 . 261 from which n = 3 . 83 (b) One could take pains to model this shaft exactly, using say finite element software.
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Unformatted text preview: However, for the bearings and the gear, the shaft is basically of uniform diameter, 1.875 in. The reductions in diameter at the bearings will change the results insignicantly. Use E = 30(10 6 ) psi . To the left of the load: AB = Fb 6 E Il (3 x 2 + b 2 l 2 ) = 1449(2)(3 x 2 + 2 2 11 2 ) 6(30)(10 6 )( / 64)(1 . 825 4 )(11) = 2 . 4124(10 6 )(3 x 2 117) At x = 0: = 2 . 823(10 4 ) rad At x = 9 in: = 3 . 040(10 4 ) rad...
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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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