21_ch 14 Mechanical Design budynas_SM_ch14

21_ch 14 Mechanical Design budynas_SM_ch14 -...

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Chapter 14 369 Fig. 14-5: 0 . 99 ( S c ) 10 7 = 322 H B + 29 100 = 322(232) + 29 100 = 103 804 psi σ c ,all = 103 804(0 . 879) 2(1)(1) = 64 519 psi Eq. (14-16): W t = ± σ c ,all C p ² 2 Fd P I K o K v K s K m C f = ± 64 519 2300 ² 2 ³ 2(2 . 833)(0 . 1205) 1(1 . 472)(1)(1 . 2167)(1) ´ = 300 lbf H = W t V 33 000 = 300(830 . 7) 33 000 = 7 . 55 hp The pinion controls therefore H rated = 7 . 55 hp Ans. 14-23 l = 2 . 25 / P d , x = 3 Y 2 P d t = 4 lx = µ 4 ± 2 . 25 P d ²± 3 Y 2 P d ² = 3 . 674 P d Y d e = 0 . 808 Ft = 0 . 808 µ F ± 3 . 674 P d ² Y = 1 . 5487 µ F Y P d k b = 1 . 5487 F Y / P d 0 . 30 0 . 107 = 0 . 8389 · F Y P d ¸ 0 . 0535 K s = 1 k b = 1 . 192 · F Y P d ¸ 0 . 0535 Ans . 14-24 Y P = 0 . 331, Y G = 0 . 422, J P = 0 . 345, J G = 0 . 410, K o = 1 . 25 . The service conditions
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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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