23_ch 07 Mechanical Design budynas_SM_ch07

23_ch 07 Mechanical Design budynas_SM_ch07 -...

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200 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Modeling the shaft separately using 2 elements gives approximately The spreadsheet can be easily modified to give δ 11 = 9 . 605(10 7 ), δ 12 = δ 21 = 5 . 718(10 7 δ 22 = 5 . 472(10 7 ) y 1 = 1 . 716(10 5 y 2 = 1 . 249(10 5 y 2 1 = 2 . 946(10 10 y 2 2 = 1 . 561(10 10 ± w y = 3 . 316(10 4 ± w y 2 = 5 . 052(10 9 ) ω 1 = ² 386 ³ 3 . 316(10 4 ) 5 . 052(10 9 ) ´ = 5034 rad/s Ans . A finite element model of the exact shaft gives ω 1 = 5340 rad/s. The simple model is 5.7% low. Combination Using Dunkerley’s equation, Eq. (7-32): 1 ω 2 1 = 1 5860 2 + 1 5034 2 3819 rad/s Ans . 7-19 We must not let the basis of the stress concentration factor, as presented, impose a view- point on the designer. Table A-16 shows
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