39_ch 03 Mechanical Design budynas_SM_ch03

# 39_ch 03 Mechanical Design budynas_SM_ch03 - σ = − 92 8...

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52 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design σ max = 7885 296 2 + ± ² 7885 296 2 ³ 2 + 8217 2 = 12 845 psi Ans. 3-47 ´ µ M B z =− 5 . 6(362 . 8) + 1 . 3(92 . 8) + 3 A y = 0 A y = 637 . 0 lbf ´ µ M A z =− 2 . 6(362 . 8) + 1 . 3(92 . 8) + 3 B y = 0 B y = 274 . 2 lbf ´ µ M B y = 0 A z = 5 . 6 3 808 = 1508 . 3 lbf ´ µ M A y = 0 B z = 2 . 6 3 808 = 700 . 3 lbf Torsion: T = 808(1 . 3) = 1050 lbf · in τ = 16(1050) π (1 3 ) = 5348 psi Bending: M p = 92 . 8(1 . 3) = 120 . 6 lbf · in M A = 3 · B 2 y + B 2 z = 3 ¸ 274 . 2 2 + 700 . 3 2 = 2256 lbf · in = M max σ b 32(2256) π (1 3 ) 22 980 psi Axial:
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Unformatted text preview: σ = − 92 . 8 ( π/ 4)1 2 = − 120 psi τ max = ± ² − 22980 − 120 2 ³ 2 + 5348 2 = 12 730 psi Ans. σ max = 22980 − 120 2 + ± ² 22980 − 120 2 ³ 2 + 5348 2 = 24 049 psi Ans. 808 lbf 362.8 lbf 92.8 lbf 3 in 2.6 in 92.8 lbf 1.3 in x y E B y A y B A B z A z P z tens . inAP tens...
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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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