19_ch 14 Mechanical Design budynas_SM_ch14

19_ch 14 Mechanical Design budynas_SM_ch14 - 67 980 = 1 ....

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Chapter 14 367 Fig. 14-8: Corrections are 0.94 and 0.98 J P = 0 . 45(0 . 94) = 0 . 423, J G = 0 . 54(0 . 98) = 0 . 529 C mc = 1, C pf = 2 10(3 . 079) 0 . 0375 + 0 . 0125(2) = 0 . 0525 C pm = 1, C ma = 0 . 093, C e = 1 K m = 1 + (1)[0 . 0525(1) + 0 . 093(1)] = 1 . 146 Pinion tooth bending ( σ ) P = 682 . 3(1)(1 . 21)(1 . 088) ± 5 . 196 2 ²³ 1 . 146(1) 0 . 423 ´ = 6323 psi Ans. ( S F ) P = 28 260(0 . 977) / [1(0 . 85)] 6323 = 5 . 14 Ans. Gear tooth bending ( σ ) G = 682 . 3(1)(1 . 21)(1 . 097) ± 5 . 196 2 ²³ 1 . 146(1) 0 . 529 ´ = 5097 psi Ans. ( S F ) G = 28 260(0 . 996) / [1(0 . 85)] 5097 = 6 . 50 Ans. Pinion tooth wear ( σ c ) P = 2300 µ 682 . 3(1)(1 . 21)(1 . 088) ³ 1 . 146 3 . 078(2) ´± 1 0 . 193 ²¶ 1 / 2 = 67 700 psi Ans. ( S H ) P = 93 500(0 . 948) / [(1)(0 . 85)] 67 700 = 1 . 54 Ans. Gear tooth wear ( σ c ) G = ³ 1 . 097 1 . 088 ´ 1 / 2 (67 700) = 67 980 psi Ans. ( S H ) G = 93 500(0 . 973) / [(1)(0 . 85)]
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Unformatted text preview: 67 980 = 1 . 57 Ans . 14-22 Given: N P = 17 T , N G = 51 T , R = . 99 at 10 8 cycles, H B = 232 through-hardening Grade 1, core and case, both gears. Table 14-2: Y P = 0.303, Y G = 0.4103 Fig. 14-6: J P = 0.292, J G = 0.396 d P = N P / P = 17 / 6 = 2.833 in, d G = 51 / 6 = 8.5 in Pinion bending From Fig. 14-2: . 99 ( S t ) 10 7 = 77 . 3 H B + 12 800 = 77 . 3(232) + 12 800 = 30 734 psi...
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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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