23_ch 16 Mechanical Design budynas_SM_ch16

23_ch 16 Mechanical Design budynas_SM_ch16 - T L = 1300(12)...

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418 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design where l is the rim width as shown in Table A-18. The specific weight of cast iron is γ = 0 . 260 lbf · in 3 , therefore the volume of cast iron is V = W γ = 189 . 1 0 . 260 = 727 . 3in 3 Thus 188 . 5 l = 727 . 3 l = 727 . 3 188 . 5 = 3 . 86 in wide Proportions can be varied. 16-30 Prob. 16-29 solution has I for the motor shaft flywheel as I = 110 . 72 in · lbf · s 2 /rad A flywheel located on the crank shaft needs an inertia of 10 2 I (Prob. 16-26, rule 2) I = 10 2 (110 . 72) = 11 072 in · lbf · s 2 /rad A 100-fold inertia increase. On the other hand, the gear train has to transmit 3 hp under shock conditions. Stating the problem is most of the solution. Satisfy yourself that on the crankshaft:
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Unformatted text preview: T L = 1300(12) = 15 600 lbf in T r = 10(168 . 07) = 1680 . 7 lbf in r = 117 . 81 / 10 = 11 . 781 rad/s s = 125 . 66 / 10 = 12 . 566 rad/s a = 21 . 41(100) = 2141 b = 2690 . 35(10) = 26903 . 5 T M = 2141 c + 26 903 . 5 lbf in T 2 = 1680 . 6 15 600 1680 . 5 15 600 T 2 19 The root is 10(26 . 67) = 266 . 7 lbf in = 121 . 11 / 10 = 12 . 111 rad/s C s = . 0549 (same) max = 121 . 11 / 10 = 12 . 111 rad/s Ans . min = 117 . 81 / 10 = 11 . 781 rad/s Ans . E 1 , E 2 , E and peak power are the same. From Table A-18 W = 8 gI d 2 o + d 2 i = 8(386)(11 072) d 2 o + d 2 i...
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