9_ch 04 Mechanical Design budynas_SM_ch04

9_ch 04 Mechanical Design budynas_SM_ch04 -...

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78 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 4-20 Torque = (600 80)(9 / 2) = 2340 lbf · in ( T 2 T 1 ) 12 2 = T 2 (1 0 . 125)(6) = 2340 T 2 = 2340 6(0 . 875) = 446 lbf, T 1 = 0 . 125(446) = 56 lbf ± M 0 = 12(680) 33(502) + 48 R 2 = 0 R 2 = 33(502) 12(680) 48 = 175 lbf R 1 = 680 502 + 175 = 353 lbf We will treat this as two separate problems and then sum the results. First, consider the 680 lbf load as acting alone. z OA =− Fbx 6 EIl ( x 2 + b 2 l 2 ) ; here b = 36", x = 12", l = 48", F = 680 lbf Also, I = π d 4 64 = π . 5) 4 64 = 0 . 2485 in 4 z A 680(36)(12)(144 + 1296 2304) 6(30)(10 6
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