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 RR CK Bending Fig. 15-13: 0.9 9 44(300) 2100 15 300 psi ta t Ss Eq. (15-4): all 15 300(0.862) ( ) 10 551 psi 1(1)(1.25) at L Pt FTR sK s SKK w Chapter 15, Page 1/20
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Eq. (15-3): all () t Px P P do sm FKJ W PKK KK v 1 10 551(1.25)(1)(0.249) 690 lbf 6(1)(1.374)(0.5222)(1.106) 690(785.3) 16.4 hp 33 000 H  Eq. (15-4): all 15 300(0.893) ( ) 10 930 psi 1(1)(1.25) G 2 10 930(1.25)(1)(0.216) 620 lbf 6(1)(1.374)(0.5222)(1.106) 620(785.3) 14.8 hp . 33 000 t G W HA n s The gear controls the bending rating. ________________________________________________________________________ 15-2 Refer to Prob. 15-1 for the gearset specifications. Wear Fig. 15-12: s ac = 341(300) + 23 620 = 125 920 psi For the pinion, C H = 1 . From Prob. 15-1, C R = 1 . 118. Thus, from Eq. (15-2): ,all ,all 125 920(1)(1) ( ) 112 630 psi 1(1)(1.118) ac L P H cP HTR sC C SKC For the gear, from Eq. (15-16), 1 0.008 98(300 / 300) 0.008 29 0.000 69 1 0.000 69(3 1) 1.001 38 H B C     From Prob. 15-1, ( C L ) G = 1 . 0685 . Equation (15-2) thus gives ,all ,all 125 920(1.0685)(1.001 38) ( ) 120 511 psi ac L G H cG For steel: 2290 psi p C Chapter 15, Page 2/20
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Eq. (15-9): 0.125(1.25) 0.4375 0.593 75 s C  Fig. 15-6: I = 0 . 083 Eq. (15-12): C xc = 2 Eq. (15-1): 2 ,all () cP t P P po m s Fd I W CK K K C C    v x c 2 3 2 4 112 630 1.25(3.333)(0.083) 2290 1(1.374)(1.106)(0.5937)(2) 464 lbf 464(785.3) 11.0 hp 33 000 120 511 1.25(3.333)(0.083) 2290 1(1.374)(1.106)(0.593 75)(2) 531 lbf 531(785.3) 3 t G H W H  12.6 hp 3 000 The pinion controls wear: H = 11 . 0 hp Ans. The power rating of the mesh, considering the power ratings found in Prob. 15-1, is H = min(16 . 4, 14 . 8, 11 . 0, 12 . 6) = 11 . 0 hp Ans. ________________________________________________________________________ 15-3 AGMA 2003-B97 does not fully address cast iron gears. However, approximate comparisons can be useful. This problem is similar to Prob. 15-1, but not identical. We will organize the method. A follow-up could consist of completing Probs. 15-1 and 15-2 with identical pinions, and cast iron gears. Given: Uncrowned, straight teeth, P d = 6 teeth/in, N P = 30 teeth, N G = 60 teeth, ASTM 30 cast iron, material Grade 1, shaft angle 90°, F = 1 . 25, n P = 900 rev/min, n = 20 , one gear straddle-mounted, K o = 1, J P = 0 . 268, J G = 0 . 228, S F = 2, 2. H S Mesh d P = 30 / 6 = 5 . 000 in, d G = 60 / 6 = 10 . 000 in v t = (5)(900 / 12) = 1178 ft/min Set N L = 10 7 cycles for the pinion. For R = 0 . 99, Table 15-7: s at = 4500 psi Chapter 15, Page 3/20
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Table 15-5: s ac = 50 000 psi Eq. (15-4): 4500(1) 2250 psi 2(1)(1) at L t FTR sK s SKK  w The velocity factor K v represents stress augmentation due to mislocation of tooth profiles along the pitch surface and the resulting “falling” of teeth into engagement. Equation (5-67) shows that the induced bending moment in a
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This note was uploaded on 11/20/2011 for the course MCE 325 taught by Professor Ahm during the Spring '11 term at American Dubai.

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Chapter_15_Solutions - Chapter 15 15-1 Given: Uncrowned,...

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