5_ch 15 Mechanical Design budynas_SM_ch15

# 5_ch 15 Mechanical Design budynas_SM_ch15 -...

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Unformatted text preview: budynas_SM_ch15.qxd 12/05/2006 17:42 Page 383 FIRST PAGES Chapter 15 Thus, W= t 35 355 1960 2 383 1.25(5.000)(0.086) = 91.6 lbf 1(1.454)(1.106)(0.59375)(2) 91.6(1178) = 3.27 hp Ans. 33 000 Rating Based on results of Probs. 15-3 and 15-4, H = min(5.34, 4.54, 3.27, 3.27) = 3.27 hp Ans. The mesh is weakest in wear fatigue. H3 = H4 = 15-5 Uncrowned, through-hardened to 180 Brinell (core and case), Grade 1, 109 rev of pinion at R = 0.999, N P = z 1 = 22 teeth, NG = z 2 = 24 teeth, Q v = 5, m et = 4 mm, shaft angle √ 90°, n 1 = 1800 rev/min, S F = 1, S H = S F = 1, J P = Y J 1 √ 0.23, JG = Y J 2 = = 0.205, F = b = 25 mm, K o = K A = K T = K θ = 1 and C p = 190 MPa . Mesh Eq. (15-7): Eq. (15-6): Eq. (15-5): Eq. (15-10): Eq. (15-11) with d P = de1 = mz 1 = 4(22) = 88 mm dG = m et z 2 = 4(24) = 96 mm vet = 5.236(10−5 )(88)(1800) = 8.29 m/s B = 0.25(12 − 5) 2/3 = 0.9148 A = 50 + 56(1 − 0.9148) = 54.77 √ 0.9148 54.77 + 200(8.29) Kv = = 1.663 54.77 K s = Yx = 0.4867 + 0.008 339(4) = 0.520 K mb = 1 (both straddle-mounted), K m = K Hβ = 1 + 5.6(10−6 )(252 ) = 1.0035 From Fig. 15-8, Eq. (15-12): Eq. (15-19): From Fig. 15-10, Eq. (15-9): Wear of Pinion Fig. 15-12: Fig. 15-6: Eq. (15-2): Eq. (15-1): ( C L ) P = ( Z N T ) P = 3.4822(109 ) −0.0602 = 1.00 ( C L ) G = ( Z N T ) G = 3.4822[109 (22/24)]−0.0602 = 1.0054 C xc = Z xc = 2 (uncrowned) K R = Y Z = 0.50 − 0.25 log (1 − 0.999) = 1.25 √ C R = Z Z = Y Z = 1.25 = 1.118 C H = Zw = 1 Z x = 0.004 92(25) + 0.4375 = 0.560 σ H lim = 2.35 H B + 162.89 = 2.35(180) + 162.89 = 585.9 MPa I = Z I = 0.066 ( σ H lim ) P ( Z N T ) P Z W (σ H ) P = SH K θ Z Z 585.9(1)(1) =√ = 524.1 MPa 1(1)(1.118) t WP = σH Cp 2 bde1 Z I 1000 K A K v K Hβ Z x Z xc ...
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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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