18_ch 07 Mechanical Design budynas_SM_ch07

18_ch 07 Mechanical Design budynas_SM_ch07 - gE (1) or d =...

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Chapter 7 195 The students’ initial reaction is that he/she does not know much from the problem state- ment. Then, slowly the realization sets in that they do know some important things that the designer did not. They knew how it failed, where it failed, and that the design wasn’t good enough; it was close, though. Also, a fix at the bearing seat lead-in could transfer the problem to the shoulder fillet, and the problem may not be solved. To many students’ credit, they chose to keep the shaft geometry, and selected a new material to realize about twice the Brinell hardness. 7-14 In Eq. (7-24) set I = π d 4 64 , A = π d 2 4 to obtain ω = ± π l ² 2 ± d 4 ²
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Unformatted text preview: gE (1) or d = 4 l 2 2 r gE (2) (a) From Eq. (1) and Table A-5, = 24 2 1 4 r 386(30)(10 6 ) . 282 = 868 rad/s Ans . (b) From Eq. (2), d = 4(24) 2 (2)(868) 2 . 282 386(30)(10 6 ) = 2 in Ans . (c) From Eq. (2), l = 2 4 d l gE Since d / l is the same regardless of the scale. l = constant = 24(868) = 20 832 = 20 832 12 = 1736 rad/s Ans. Thus the rst critical speed doubles. 7-15 From Prob. 7-14, = 868 rad/s A = . 7854 in 2 , I = . 04909 in 4 , = . 282 lbf/in 3 , E = 30(10 6 ) psi, w = A l = . 7854(0 . 282)(24) = 5 . 316 lbf...
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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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