26_ch 04 Mechanical Design budynas_SM_ch04

# 26_ch 04 Mechanical Design budynas_SM_ch04 -...

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Chapter 4 95 δ B = u F = 1 EI ± a 0 Fx ( x ) dx + (3 / 5) F (3 / 5) l AE + (4 / 5) Fa (4 a / 5) l JG + 1 EI ± l 0 4 5 F ¯ x ² 4 5 ¯ x ³ d ¯ x + 1 EI ± l 0 3 5 Fa ² 3 5 a ³ d ¯ x = Fa 3 3 EI + 9 25 ² Fl AE ³ + 16 25 ² Fa 2 l JG ³ + 16 75 ² Fl 3 EI ³ + 9 25 ² Fa 2 l EI ³ I = π 64 d 4 , J = 2 I , A = π 4 d 2 δ B = 64 Fa 3 3 E π d 4 + 9 25 ² 4 Fl π d 2 E ³ + 16 25 ² 32 Fa 2 l π d 4 G ³ + 16 75 ² 64 Fl 3 E π d 4 ³ + 9 25 ² 64 Fa 2 l E π d 4 ³ = 4 F 75 π Ed 4 ² 400 a 3 + 27 ld 2 + 384 a 2 l E G + 256 l 3 + 432 a 2 l ³ Ans. 4-51 The force applied to the copper and steel wire assembly is F c + F s = 250 lbf Since δ c = δ s F c L 3( π/ 4)(0 . 0801) 2 (17 . 2)(10 6 ) = F s L ( π/ 4)(0 . 0625) 2 (30)(10 6 ) F c = 2 . 825 F s 3 . 825 F s = 250 F s = 65 . 36 lbf, F c = 2 . 825 F s = 184 . 64 lbf σ c = 184 . 64 3( π/ 4)(0 . 0801) 2 = 12 200 psi = 12 . 2 kpsi Ans. σ s = 65 . 36 ( π/ 4)(0 . 0625 2 ) = 21 300 psi = 21 . 3 kpsi Ans. 4-52 (a) Bolt stress σ b = 0 . 9 ( 85 ) = 76 . 5 kpsi Ans.
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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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