3_ch 08 Mechanical Design budynas_SM_ch08

# 3_ch 08 Mechanical Design budynas_SM_ch08 -...

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206 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 8-7 The force F is perpendicular to the paper. L = 3 1 8 1 4 7 32 = 2 . 406 in T = 2 . 406 F M = ± L 7 32 ² F = ± 2 . 406 7 32 ² F = 2 . 188 F S y = 41 kpsi σ = S y = 32 M π d 3 = 32(2 . 188) F π (0 . 1875) 3 = 41 000 F = 12 . 13 lbf T = 2 . 406(12 . 13) = 29 . 2 lbf · in Ans . (b) Eq. (8-5), 2 α = 60 , l = 1 / 14 = 0 . 0714 in, f = 0 . 075, sec α = 1 . 155, p = 1 / 14 in d m = 7 16 0 . 649 519 ± 1 14 ² = 0 . 3911 in T R = F clamp (0 . 3911) 2 ± Num Den ² Num = 0 . 0714 + π (0 . 075)(0 . 3911)(1 . 155) Den = π (0 . 3911) 0 . 075(0 . 0714)(1 . 155) T = 0 . 028 45 F clamp F clamp = T 0 . 028 45 = 29 . 2 0 . 028 45 = 1030 lbf Ans . (c) The column has one end ﬁxed and the other end pivoted. Base decision on the mean diameter column. Input: C = 1.2, D = 0.391 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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