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Chapter_15_Solutions

# Chapter_15_Solutions - Chapter 15 15-1 Given Uncrowned...

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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 = (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): 0.8255 59.77 785.3 1.374 59.77 K v 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): K R = 0 . 50 – 0 . 25 log(1 – 0 . 999) = 1 . 25 (or Table 15-3) 1.25 1.118 R R C K Bending Fig. 15-13: 0.9 9 44(300) 2100 15 300 psi t at S s Eq. (15-4): all 15 300(0.862) ( ) 10 551 psi 1(1)(1.25) at L P t F T R s K s S K K w Chapter 15, Page 1/20

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