40_ch 04 Mechanical Design budynas_SM_ch04

40_ch 04 Mechanical Design budynas_SM_ch04 - 3 )(60 2 ) 3...

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Chapter 4 109 Out of plane: k = 0 . 2887(0 . 5) = 0 . 1444 in, C = 1 . 2 l k = 3 . 473(12) 0 . 1444 = 289 Since ( l / k ) 1 < ( l / k ) use Euler equation P cr = 1(0 . 5) 1 . 2( π 2 )(30)(10 6 ) 289 2 = 2127 lbf 1 / 4" increases l / k by 2, ± l k ² 2 by 4, and A by 1 / 2 Try 1" × 3 / 8": k = 0 . 2887(0 . 375) = 0 . 1083 in l k = 385, P cr = 1(0 . 375) 1 . 2( π 2 )(30)(10 6 ) 385 2 = 899 lbf (too low) Use 1" × 1 / 2" Ans. (b) σ b =− P π dl =− 298 π (0 . 5)(0 . 5) =− 379 psi No, bearing stress is not significant. 4-76 This is a design problem with no one distinct solution. 4-77 F = 800 ± π 4 ² (3 2 ) = 5655 lbf, S y = 37 . 5 kpsi P cr = n d F = 3(5655) = 17 000 lbf (a) Assume Euler with C = 1 I = π 64 d 4 = P cr l 2 C π 2 E d = ³ 64 P cr l 2 π 3 CE ´ 1 / 4 = ³ 64(17)(10
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Unformatted text preview: 3 )(60 2 ) 3 (1)(30)(10 6 ) 1 / 4 = 1 . 433 in Use d = 1 . 5 in ; k = d / 4 = . 375 l k = 60 . 375 = 160 l k 1 = 2 2 (1)(30)(10 6 ) 37 . 5(10 3 ) 1 / 2 = 126 use Euler P cr = 2 (30)(10 6 )( / 64)(1 . 5 4 ) 60 2 = 20 440 lbf d = 1 . 5 i n is satisfactory. Ans. (b) d = 64(17)(10 3 )(18 2 ) 3 (1)(30)(10 6 ) 1 / 4 = . 785 in, so use 0 . 875 in...
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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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