4_ch 08 Mechanical Design budynas_SM_ch08

4_ch 08 Mechanical Design budynas_SM_ch08 -...

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Chapter 8 207 Eq. (8-5): T = 0 . 5417( F / 2) ± 0 . 1667 + π (0 . 15)(0 . 5417)(1 . 033) π (0 . 5417) 0 . 15(0 . 1667)(1 . 033) ² = 0 . 0696 F Eq. (8-6): T c = 0 . 15(7 / 16)( F / 2) = 0 . 032 81 F T total = (0 . 0696 + 0 . 0328) F = 0 . 1024 F F = 16 . 5 0 . 1024 = 161 lbf Ans . 8-9 d m = 40 3 = 37 mm, l = 2(6) = 12 mm From Eq. (8-1) and Eq. (8-6) T R = 10(37) 2 ± 12 + π (0 . 10)(37) π (37) 0 . 10(12) ² + 10(0 . 15)(60) 2 = 38 . 0 + 45 = 83 . 0N · m Since n = V / l = 48 / 12 = 4 rev/s ω = 2 π n = 2 π (4) = 8 π rad/s so the power is H = T ω = 83 . 0(8 π ) = 2086 W Ans . 8-10 (a) d m = 36 3 = 33 mm, l = p = 6mm From Eqs. (8-1) and (8-6) T = 33 F 2 ± 6 + π (0 . 14)(33) π (33) 0 . 14(6) ² + 0 . 09(90) F 2 = (3 . 292 + 4 . 050) F = 7 . 34 F N · m ω = 2 π n = 2 π (1) = 2 π rad/s H = T ω T = H ω = 3000 2 π = 477 N
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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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