27_ch 17 Mechanical Design budynas_SM_ch17

27_ch 17 Mechanical Design budynas_SM_ch17 - Decision #1 :...

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446 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The H ±± notation is only necessary because we constructed the first table, which we nor- mally would not do. n fs = K 1 K 2 H ±± tab K s H nom = K 1 K 2 (0 . 618 H ± tab ) K s H nom = 0 . 618[(0 . 0197) K 2 H tab ] = 0 . 0122 K 2 H tab Form a table for a 50 000 h life goal. K 2 H ±± tab Chain # H tab n fs Lub 1 1.0 34.50 120 21.6 0.264 C ± 2 1.7 20.30 120 21.6 0.448 C ± 3 2.5 13.80 120 21.6 0.656 C ± 4 3.3 10.50 120 21.6 0.870 C ± 5 3.9 8.85 120 21.6 1.028 C ± 6 4.6 7.60 120 21.6 1.210 C ± 8 6.0 5.80 80 15.6 1.140 C ± There are two possibilities in the second table with n fs 1 . 1 . (The tables allow for the identification of a longer life one of the outcomes.) We need a figure of merit to help with the choice; costs of sprockets and chains are thus needed, but is more information than we have.
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Unformatted text preview: Decision #1 : #80 Chain (smaller installation) Ans. n f s = . 0122 K 2 H tab = . 0122(8 . 0)(15 . 6) = 1 . 14 O . K . Decision #2 : 8-Strand, No. 80 Ans. Decision #3 : Type C ± Lubrication Ans. Decision #4 : p = 1 . 0 in, C is in midrange of 40 pitches L p = 2 C p + N 1 + N 2 2 + ( N 2 − N 1 ) 2 4 π 2 C / p = 2(40) + 19 + 38 2 + (38 − 19) 2 4 π 2 (40) = 108 . 7 ⇒ 110 even integer Ans . Eq. (17-36): A = N 1 + N 2 2 − L p = 19 + 38 2 − 110 = − 81 . 5 Eq. (17-35): C p = 1 4 81 . 5 + ± 81 . 5 2 − 8 ² 38 − 19 2 π ³ 2 = 40 . 64 C = p ( C / p ) = 1 . 0(40 . 64) = 40 . 64 in (for reference) Ans ....
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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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