17_ch 14 Mechanical Design budynas_SM_ch14

17_ch 14 Mechanical Design budynas_SM_ch14 - 099(1 ³ 1 27...

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Chapter 14 365 Eq. (14-16): Eq. (14-42): Eq. (14-23) with m N = 1: Z I = cos 20 sin 20 2 ± 1 . 8 1 . 8 + 1 ² = 0 . 103 Table 14-8: Z E = 191 MPa Strength Grade 1 steel, given H BP = H BG = 200 Fig. 14-2: ( σ FP ) P = ( σ FP ) G = 0 . 533(200) + 88 . 3 = 194 . 9MPa Fig. 14-5: ( σ HP ) P = ( σ HP ) G = 2 . 22(200) + 200 = 644 MPa Fig. 14-15: ( Z N ) P = 1 . 4488(10 8 ) 0 . 023 = 0 . 948 ( Z N ) G = 1 . 4488(10 8 / 1 . 8) 0 . 023 = 0 . 961 Fig. 14-12: H BP / H BG = 1 Z W = 1 Pinion tooth bending ( σ ) P = ± W t K o K v K s 1 bm t K H K B Y J ² P = 458 . 4(1)(1 . 099)(1) ³ 1 18(2 . 5) ´³ 1 . 27(1) 0 . 33 ´ = 43 . 08 MPa Ans. ( S F ) P = ± σ FP σ Y N Y θ Y Z ² P = 194 . 9 43 . 08 ³ 0 . 977 1(0 . 885) ´ = 4 . 99 Ans. Gear tooth bending ( σ ) G = 458 . 4(1)(1 . 099)(1) ³ 1 18(2 . 5) ´³ 1 . 27(1) 0 . 38 ´ = 37 . 42 MPa Ans. ( S F ) G = 194 . 9 37 . 42 ³ 0 . 987 1(0 . 885) ´ = 5 . 81 Ans . Pinion tooth wear ( σ c ) P = µ Z E W t K o K v K s K H d w 1 b Z R Z I · P = 191 458 . 4(1)(1
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Unformatted text preview: . 099)(1) ³ 1 . 27 50(18) ´³ 1 . 103 ´ = 501 . 8 MPa Ans. ( S H ) P = ± σ H P σ c Z N Z W Y θ Y Z ² P = 644 501 . 8 ³ . 948(1) 1(0 . 885) ´ = 1 . 37 Ans. Gear tooth wear ( σ c ) G = ³ ( K s ) G ( K s ) P ´ 1 / 2 ( σ c ) P = ± 1 1 ² 1 / 2 (501 . 8) = 501 . 8 MPa Ans. ( S H ) G = 644 501 . 8 ³ . 961(1) 1(0 . 885) ´ = 1 . 39 Ans. Eq. (14-15): Eq. (14-41):...
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