shi20396_ch15

# shi20396_ch15 - 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 = 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 = Ã 59 . 77 + 785 . 3 59 . 77 ! 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): K R = 0 . 50 0 . 25 log(1 0 . 999) = 1 . 25 (or Table 15-3) C R = p K R = 1 . 25 = 1 . 118 Bending Fig. 15-13: 0 . 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 . 4hp Eq. (15-4): ( σ all ) G = 15 300(0 . 893) 1(1)(1 . 25) = 10 930 psi

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Chapter 15 391 W t G = 10 930(1 . 25)(1)(0 . 216) 6(1)(1 . 374)(0 . 5222)(1 . 106) = 620 lbf H 2 = 620(785 . 3) 33 000 = 14 . 8hp Ans. The gear controls the bending rating. 15-2 Refer to Prob. 15-1 for the gearset speciFcations. Wear ±ig. 15-12: s ac = 341(300) + 23 620 = 125 920 psi ±or the pinion, C H = 1 . ±rom Prob. 15-1, C R = 1 . 118. Thus, from Eq. (15-2): ( σ c ,all ) P = s ac ( C L ) P C H S H K T C R ( σ c ,all ) P = 125 920(1)(1) 1(1)(1 . 118) = 112 630 psi ±or the gear, from Eq. (15-16), B 1 = 0 . 008 98(300 / 300) 0 . 008 29 = 0 . 000 69 C H = 1 + 0 . 000 69(3 1) = 1 . 001 38 And Prob. 15-1, ( C L ) G = 1 . 0685 . Equation (15-2) thus gives ( σ c ,all ) G = s ac ( C L ) G C H S H K T C R ( σ c ,all ) G = 125 920(1 . 0685)(1 . 001 38) 1(1)(1 . 118) = 120 511 psi ±or steel: C p = 2290 p psi Eq. (15-9): C s = 0 . 125(1 . 25) + 0 . 4375 = 0 . 593 75 ±ig. 15-6: I = 0 . 083 Eq. (15-12): C xc = 2 Eq. (15-1): W t P = µ ( σ c ,all ) P C p 2 Fd P I K o K v K m C s C xc = µ 112 630 2290 2 · 1 . 25(3 . 333)(0 . 083) 1(1 . 374)(1 . 106)(0 . 5937)(2) ¸ = 464 lbf H 3 = 464(785 . 3) 33 000 = 11 . 0hp W t G = µ 120 511 2290 2 · 1 . 25(3 . 333)(0 . 083) 1(1 . 374)(1 . 106)(0 . 593 75)(2) ¸ = 531 lbf H 4 = 531(785 . 3) 33 000 = 12 . 6hp
392 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The pinion controls wear: H = 11 . 0hp 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 . 0hp Ans . 15-3 AGMA 2003-B97 does not fully address cast iron gears, however, approximate compar- isons can be useful. This problem is similar to Prob. 15-1, but not identical. We will orga- nize 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, φ n = 20 , one gear straddle-mounted, K o = 1, J P = 0 . 268, J G = 0 . 228, S F = 2, S H = 2 . Mesh d P = 30 / 6 = 5 . 000 in d G = 60 / 6 = 10 . 000 in v t = π (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 Table 15-5: s ac = 50 000 psi Eq. (15-4): s w t = s at K L S F K T K R = 4500(1) 2(1)(1) = 2250 psi The velocity factor K v represents stress augmentation due to mislocation of tooth pro±les along the pitch surface and the resulting “falling” of teeth into engagement. Equation (5-67)

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## This note was uploaded on 04/30/2008 for the course MAE 4900 taught by Professor Khanna during the Fall '06 term at Missouri (Mizzou).

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shi20396_ch15 - Chapter 15 15-1 Given: Uncrowned,...

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