15_ch 11 Mechanical Design budynas_SM_ch11

# 15_ch 11 Mechanical Design budynas_SM_ch11 - 9 = 11 48 Now...

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Chapter 11 303 Point B x B = 0 . 02 + 4 . 439[ln(1 / 0 . 99)] 1 / 1 . 483 = 0 . 220 turns log x B = log 0 . 220 =− 0 . 658 F B = F D ± x D x B ² 1 / 3 = 495 . 6 ± 540 0 . 220 ² 1 / 3 = 6685 lbf Note: Example 11-3 used Eq. (11-7). Whereas, here we basically used Eq. (11-6). log F B = log(6685) = 3 . 825 K D = 6685 3 (0 . 220) = 65 . 7(10 9 ) lbf 3 · turns (as it should) Point A F A = F B = C 10 = 6685 lbf log C 10 = log(6685) = 3 . 825 x A = 1 log x A = log(1) = 0 K 10 = F 3 A x A = C 3 10 (1) = 6685 3 = 299(10 9 ) lbf 3 · turns Note that K D / K 10 = 65 . 7(10 9 ) / [299(10 9 )] = 0 . 220, which is x B . This is worth knowing since K 10 = K D x B log K 10 = log[299(10
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Unformatted text preview: 9 )] = 11 . 48 Now C 10 = 6685 lbf = 29 . 748 kN, which is required for a reliability goal of 0.99. If we select an angular contact 02-40 mm ball bearing, then C 10 = 31 . 9 kN = 7169 lbf . 0.1 ± 1 ± 0.658 1 10 1 10 2 2 10 2 2 10 3 495.6 6685 3 10 4 4 10 3 3 x log x F A D B log F 540...
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