5_ch 14 Mechanical Design budynas_SM_ch14

5_ch 14 Mechanical Design budynas_SM_ch14 -...

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Chapter 14 353 14-11 d P = 22 6 = 3 . 667 in, d G = 60 6 = 10 in V = π (3 . 667)(1200) 12 = 1152 ft/min Eq. (14-4 b ): K v = 1200 + 1152 1200 = 1 . 96 W t = 63 025(15) 1200(3 . 667 / 2) = 429 . 7 lbf Table 14-8: C p = 2100 ± psi [Note: using Eq. (14-13) can result in wide variation in C p due to wide variation in cast iron properties] Eq. (14-12): r 1 = 3 . 667 sin 20° 2 = 0 . 627 in, r 2 = 10 sin 20° 2 = 1 . 710 in Eq. (14-14): σ C =− C p ² K v W t F cos φ ³ 1 r 1 + 1 r 2 ´µ 1 / 2 =− 2100 ² 1 . 96(429 . 7) 2 cos 20° ³ 1 0 . 627 + 1 1 . 710 ´µ 1 / 2 =− 65 . 6(10 3 ) psi =− 65 . 6 kpsi Ans. 14-12 d P = 16 12 = 1 . 333 in, d G = 48 12 = 4in V = π (1 . 333)(700) 12 = 244 . 3 ft/min Eq. (14-4 b ): K v = 1200 + 244 . 3 1200 = 1 . 204 W t = 63 025(1 . 5) 700(1 . 333 / 2) =
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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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