25_ch 17 Mechanical Design budynas_SM_ch17

# 25_ch 17 Mechanical Design budynas_SM_ch17 - n f s = 1 ....

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444 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Evaluate K 1 and K 2 Eq. (17-38): H d = H nom K s n d Eq. (17-37): H a = K 1 K 2 H tab Equate H d to H a and solve for H tab : H tab = K s n d H nom K 1 K 2 Table 17-22: K 1 = 1 Table 17-23: K 2 = 1, 1.7, 2.5, 3.3 for 1 through 4 strands H ± tab = 1 . 5 ( 1 . 1 )( 25 ) ( 1 ) K 2 = 41 . 25 K 2 Prepare a table to help with the design decisions: Chain Lub. Strands K 2 H ± tab No. H tab n fs Type 1 1.0 41.3 100 59.4 1.58 B 2 1.7 24.3 80 31.0 1.40 B 3 2.5 16.5 80 31.0 2.07 B 4 3.3 12.5 60 13.3 1.17 B Design Decisions We need a ﬁgure of merit to help with the choice. If the best was 4 strands of No. 60 chain, then Decision #1 and #2 : Choose four strand No. 60 roller chain with
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Unformatted text preview: n f s = 1 . 17. n f s = K 1 K 2 H tab K s H nom = 1(3 . 3)(13 . 3) 1 . 5(25) = 1 . 17 Decision #3 : Choose Type B lubrication Analysis: Table 17-20: H tab = 13 . 3 hp Table 17-19: p = . 75 in Try C = 30 in in Eq. (17-34): L p = 2 C p + N 1 + N 2 2 + ( N 2 N 1 ) 2 4 2 C / p = 2(30 / . 75) + 17 + 84 2 + (84 17) 2 4 2 (30 / . 75) = 133 . 3 134 From Eq. (17-35) with p = . 75 in, C = 30 . 26 in . Decision #4 : Choose C = 30 . 26 in ....
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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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