1_ch 15 Mechanical Design budynas_SM_ch15

# 1_ch 15 Mechanical Design budynas_SM_ch15 - K R = . 50 −...

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Chapter 15 15-1 Given: Uncrowned, through-hardened 300 Brinell core and case, Grade 1, N C = 10 9 rev of pinion at R = 0 . 999, N P = 20 teeth, N G = 60 teeth, Q v = 6, P d = 6 teeth/in, shaft angle 90°, n p = 900 rev/min, J P = 0 . 249 and J G = 0 . 216 (Fig. 15-7), F = 1 . 25 in, S F = S H = 1, K o = 1. Mesh d P = 20 / 6 = 3 . 333 in d G = 60 / 6 = 10 . 000 in Eq. (15-7): v t = π (3 . 333)(900 / 12) = 785 . 3 ft/min Eq. (15-6): B = 0 . 25(12 6) 2 / 3 = 0 . 8255 A = 50 + 56(1 0 . 8255) = 59 . 77 Eq. (15-5): K v = ± 59 . 77 + 785 . 3 59 . 77 ² 0 . 8255 = 1 . 374 Eq. (15-8): v t ,max = [59 . 77 + (6 3)] 2 = 3940 ft/min Since 785 . 3 < 3904, K v = 1 . 374 is valid. The size factor for bending is: Eq. (15-10): K s = 0 . 4867 + 0 . 2132 / 6 = 0 . 5222 For one gear straddle-mounted, the load-distribution factor is: Eq. (15-11): K m = 1 . 10 + 0 . 0036(1 . 25) 2 = 1 . 106 Eq. (15-15): ( K L ) P = 1 . 6831(10 9 ) 0 . 0323 = 0 . 862 ( K L ) G = 1 . 6831(10 9 / 3) 0 . 0323 = 0 . 893 Eq. (15-14): ( C L ) P = 3 . 4822(10 9 ) 0 . 0602 = 1 ( C L ) G = 3 . 4822(10 9 / 3) 0 . 0602 = 1 . 069 Eq. (15-19):
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Unformatted text preview: K R = . 50 − . 25 log(1 − . 999) = 1 . 25 (or Table 15-3) C R = ³ K R = √ 1 . 25 = 1 . 118 Bending Fig. 15-13: . 99 S t = s at = 44(300) + 2100 = 15 300 psi Eq. (15-4): ( σ all ) P = s w t = s at K L S F K T K R = 15 300(0 . 862) 1(1)(1 . 25) = 10 551 psi Eq. (15-3): W t P = ( σ all ) P FK x J P P d K o K v K s K m = 10 551(1 . 25)(1)(0 . 249) 6(1)(1 . 374)(0 . 5222)(1 . 106) = 690 lbf H 1 = 690(785 . 3) 33 000 = 16 . 4 hp Eq. (15-4): ( σ all ) G = 15 300(0 . 893) 1(1)(1 . 25) = 10 930 psi...
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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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