39_ch 04 Mechanical Design budynas_SM_ch04

# 39_ch 04 Mechanical Design budynas_SM_ch04 - 2" section...

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108 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 4-74 A = π 4 D 2 (1 K 2 ), I = π 64 D 4 (1 K 4 ) = π 64 D 4 (1 K 2 )(1 + K 2 ), k 2 = I A = D 2 16 (1 + K 2 ) From Eq. (4-43) P cr ( π/ 4) D 2 (1 K 2 ) = S y S 2 y l 2 4 π 2 k 2 CE = S y S 2 y l 2 4 π 2 ( D 2 / 16)(1 + K 2 ) CE 4 P cr = π D 2 (1 K 2 ) S y 4 S 2 y l 2 π D 2 (1 K 2 ) π 2 D 2 (1 + K 2 ) CE π D 2 (1 K 2 ) S y = 4 P cr + 4 S 2 y l 2 (1 K 2 ) π (1 + K 2 ) CE D = ± 4 P cr π S y (1 K 2 ) + 4 S 2 y l 2 (1 K 2 ) π (1 + K 2 ) CE π (1 K 2 ) S y ² 1 / 2 = 2 ³ P cr π S y (1 K 2 ) + S y l 2 π 2 CE (1 + K 2 ) ´ 1 / 2 Ans. 4-75 (a) + ± µ M A = 0, 2 . 5(180) 3 3 2 + 1 . 75 2 F BO (1 . 75) = 0 F BO = 297 . 7 lbf Using n d = 5, design for F cr = n d F BO = 5(297 . 7) = 1488 lbf, l = 3 2 + 1 . 75 2 = 3 . 473 ft, S y = 24 kpsi In plane: k = 0 . 2887 h = 0 . 2887", C = 1 . 0 Try 1" × 1
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Unformatted text preview: / 2" section l k = 3 . 473(12) . 2887 = 144 . 4 Â· l k Â¸ 1 = Â· 2 Ï€ 2 (1)(30)(10 6 ) 24(10 3 ) Â¸ 1 / 2 = 157 . 1 Since ( l / k ) 1 > ( l / k ) use Johnson formula P cr = (1) Â· 1 2 Â¸ Â± 24(10 3 ) âˆ’ Â· 24(10 3 ) 2 Ï€ 144 . 4 Â¸ 2 Â· 1 1(30)(10 6 ) Â¸ Â² = 6930 lbf Try 1" Ã— 1 / 4": P cr = 3465 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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