1_ch 14 Mechanical Design budynas_SM_ch14

1_ch 14 Mechanical Design budynas_SM_ch14 -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 14 14-1 d = N P = 22 6 = 3 . 667 in Table 14-2: Y = 0 . 331 V = π dn 12 = π (3 . 667)(1200) 12 = 1152 ft/min Eq. (14-4 b ): K v = 1200 + 1152 1200 = 1 . 96 W t = T d / 2 = 63 025 H nd / 2 = 63 025(15) 1200(3 . 667 / 2) = 429 . 7 lbf Eq. (14-7): σ = K v W t P FY = 1 . 96(429 . 7)(6) 2(0 . 331) = 7633 psi = 7 . 63 kpsi Ans. 14-2 d = 16 12 = 1 . 333 in, Y = 0 . 296 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 H nd / 2 = 63 025(1 . 5) 700(1 . 333 / 2) = 202 . 6 lbf Eq. (14-7): σ = K v W t P FY = 1 . 204(202 . 6)(12) 0 . 75(0 . 296) = 13 185 psi = 13 . 2 kpsi Ans. 14-3 d = mN = 1 . 25(18) = 22 . 5mm
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online