6_ch 15 Mechanical Design budynas_SM_ch15

6_ch 15 Mechanical Design budynas_SM_ch15 - K = 1, Y x = ....

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384 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The constant 1000 expresses W t in kN W t P = ± 524 . 1 190 ² 2 ³ 25(88)(0 . 066) 1000(1)(1 . 663)(1 . 0035)(0 . 56)(2) ´ = 0 . 591 kN Eq. (13-36): H 3 = π dn W t 60 000 = π (88)1800(0 . 591) 60 000 = 4 . 90 kW Wear of Gear σ H lim = 585 . 9 MPa ( σ H ) G = 585 . 9(1 . 0054) 1(1)(1 . 118) = 526 . 9 MPa W t G = W t P ( σ H ) G ( σ H ) P = 0 . 591 ± 526 . 9 524 . 1 ² = 0 . 594 kN H 4 = π (88)1800(0 . 594) 60 000 = 4 . 93 kW Thus in wear, the pinion controls the power rating; H = 4 . 90 kW Ans. We will rate the gear set after solving Prob. 15-6. 15-6 Refer to Prob. 15-5 for terms not defined below. Bending of Pinion Eq. (15-15): ( K L ) P = ( Y NT ) P = 1 . 6831(10 9 ) 0 . 0323 = 0 . 862 ( K L ) G = ( Y NT ) G = 1 . 6831[10 9 (22 / 24)] 0 . 0323 = 0 . 864 Fig. 15-13: σ F lim = 0 . 30 H B + 14 . 48 = 0 . 30(180) + 14 . 48 = 68 . 5 MPa Eq. (15-13): K x = Y β = 1 From Prob. 15-5: Y Z = 1 . 25, v et = 8 . 29 m/s K A = 1, K v = 1 . 663,
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Unformatted text preview: K = 1, Y x = . 52, K H = 1 . 0035, Y J 1 = . 23 Eq. (5-4): ( F ) P = F lim Y NT S F K Y Z = 68 . 5(0 . 862) 1(1)(1 . 25) = 47 . 2 MPa Eq. (5-3): W t p = ( F ) P bm et Y Y J 1 1000 K A K v Y x K H = 47 . 2(25)(4)(1)(0 . 23) 1000(1)(1 . 663)(0 . 52)(1 . 0035) = 1 . 25 kN H 1 = (88)1800(1 . 25) 60 000 = 10 . 37 kW Bending of Gear F lim = 68 . 5 MPa ( F ) G = 68 . 5(0 . 864) 1(1)(1 . 25) = 47 . 3 MPa W t G = 47 . 3(25)(4)(1)(0 . 205) 1000(1)(1 . 663)(0 . 52)(1 . 0035) = 1 . 12 kN H 2 = (88)1800(1 . 12) 60 000 = 9 . 29 kW...
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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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