14_ch 04 Mechanical Design budynas_SM_ch04

14_ch 04 Mechanical Design budynas_SM_ch04 -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: budynas_SM_ch04.qxd 11/28/2006 20:50 FIRST PAGES Page 83 83 Chapter 4 Designating the slope constraint as ξ , we then have ξ = |θ L | = 1 6E I l Fi bi bi2 − l 2 2 1/2 Setting I = π d 4 /64 and solving for d 32 d= 3π El ξ Fi bi bi − l 2 2 1/4 2 1/2 For the LH bearing, E = 30 Mpsi, ξ = 0.001, b1 = 12, b2 = 6, and l = 16. The result is d L = 1.31 in. Using a similar flip beam procedure, we get d R = 1.36 in for the RH bearing. So use d = 1 3/8 in Ans. 4-27 I= π (1.3754 ) = 0.17546 in4 . For the xy plane, use yBC of Table A-9-6 64 100(4)(16 − 8) y= [82 + 42 − 2(16)8] = −1.115(10−3 ) in 6(30)(106 )(0.17546)(16) For the xz plane use yAB z= 300(6)(8) [82 + 62 − 162 ] = −4.445(10−3 ) in 6(30)(106 )(0.17546)(16) δ = ( −1.115j − 4.445k)(10−3 ) in |δ | = 4.583(10−3 ) in Ans. 4-28 32n dL = 3π El ξ = Fi bi bi − l 2 2 1/4 2 1/2 32(1.5) [3.5(150)(1502 − 2502 )]2 3π (207)(109 )(250)0.001 + [2.7(75)(752 − 2502 )]2 1/2 33 1/4 (10 ) = 39.2 mm dR = 32(1.5) [3.5(100)(1002 − 2502 )]2 3π (207)109 (250)0.001 + [2.7(175)(1752 − 2502 )]2 1/2 = 39.1 mm Choose d ≥ 39.2 mm Ans. 4-29 From Table A-9-8 we have yL = MB x 2 ( x + 3a 2 − 6al + 2l 2 ) 6E I l dy L MB = (3x 2 + 3a 2 − 6al + 2l 2 ) dx 6E I l 33 (10 ) 1/4 ...
View Full Document

This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

Ask a homework question - tutors are online