28_ch 04 Mechanical Design budynas_SM_ch04

28_ch 04 Mechanical Design budynas_SM_ch04 -...

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Chapter 4 97 4-55 Since θ OA = θ AB T OA (4) GJ = T AB (6) GJ , T OA = 3 2 T AB Also T OA + T AB = 50 T AB ± 3 2 + 1 ² = 50, T AB = 50 2 . 5 = 20 lbf · in Ans. T OA = 3 2 T AB = 3 2 (20) = 30 lbf · in Ans. 4-56 Since θ OA = θ AB , T OA (4) G ( π 32 1 . 5 4 ) = T AB (6) G ( π 32 1 . 75 4 ) , T OA = 0 . 80966 T AB T OA + T AB = 50 0 . 80966 T AB + T AB = 50 T AB = 27 . 63 lbf · in Ans. T OA = 0 . 80966 T AB = 0 . 80966(27 . 63) = 22 . 37 lbf · in Ans. 4-57 F 1 = F 2 T 1 r 1 = T 2 r 2 T 1 1 . 25 = T 2 3 T 2 = 3 1 . 25 T 1 θ 1 + 3 1 . 25 θ 2 = 4 π 180 rad T 1 (48) ( π/ 32)(7 / 8) 4 (11 . 5)(10 6 ) + 3 1 . 25 ³ (3 / 1 . 25) T 1 (48) ( π/ 32)(1 . 25) 4 (11 . 5)(10 6 ) ´ = 4 π 180 T 1 = 403 . 9 lbf · in T 2 = 3 1 . 25 T 1 =
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