22_ch 13 Mechanical Design budynas_SM_ch13

# 22_ch 13 Mechanical Design budynas_SM_ch13 - = 45 5 cos...

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354 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design d 2 = 16 / 7 . 727 = 2 . 07 in d 3 = 36 / 7 . 727 = 4 . 66 in d 4 = 28 / 7 . 727 = 3 . 62 in T 2 = 63 025(7 . 5) 1720 = 274 . 8 lbf · in W t = 274 . 8 2 . 07 / 2 = 266 lbf W r = 266 tan 20° = 96 . 8 lbf W a = 266 tan 15° = 71 . 3 lbf F 2 a =− 266 i 96 . 8 j 71 . 3 k lbf Ans . F 3 b = (266 96 . 8) i (266 96 . 8) j = 169 i 169 j lbf Ans . F 4 c = 96 . 8 i + 266 j + 71 . 3 k lbf Ans . 13-40 d 2 = N P n cos ψ = 14 8 cos 30° = 2 . 021 in, d 3 = 36 8 cos 30° = 5 . 196 in d 4 = 15 5 cos 15° = 3 . 106 in, d 5
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Unformatted text preview: = 45 5 cos 15° = 9 . 317 in C x y z b F t 23 F r 23 F a 23 F t 54 F a 54 F r 54 D G H 3" 2" 3 2.6" R 1.55" R 4 3 1" 2 F y D F x D F x C F y C F z D y F r 43 F x b 3 F y b 3 F a 23 F r 23 F t 23 F t 43 F a 43 3 F b 3 z x b y F t c 4 F r c 4 F a c 4 4 F a 34 F r 34 F t 34 z x c...
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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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