22_ch 16 Mechanical Design budynas_SM_ch16

# 22_ch 16 Mechanical Design budynas_SM_ch16 -...

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Chapter 16 417 ω max = T 2 b a = 26 . 771 2690 . 4 21 . 41 = 124 . 41 rad/s Ans . ω min = 117 . 81 rad/s Ans . ¯ ω = 124 . 41 + 117 . 81 2 = 121 . 11 rad/s C s = ω max ω min ( ω max + ω min ) / 2 = 124 . 41 117 . 81 (124 . 41 + 117 . 81) / 2 = 0 . 0545 Ans . E 1 = 1 2 I ω 2 r = 1 2 (110 . 72)(117 . 81) 2 = 768 352 in · lbf E 2 = 1 2 I ω 2 2 = 1 2 (110 . 72)(124 . 41) 2 = 856 854 in · lbf ± E = E 1 E 2 = 768 352 856 854 =− 88 502 in · lbf Eq. (16-64): ± E = C s I ¯ ω 2 = 0 . 0545(110 . 72)(121 . 11) 2 = 88 508 in · lbf, close enough Ans. During the punch T = 63 025 H n H = T L ¯ ω (60 / 2 π ) 63 025 = 1560(121 . 11)(60 / 2 π ) 63 025 = 28 . 6hp The gear train has to be sized for 28.6 hp under shock conditions since the ﬂywheel is on the motor shaft. From Table A-18,
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