5_ch 11 Mechanical Design budynas_SM_ch11

5_ch 11 Mechanical Design budynas_SM_ch11 - 272(0 . 342) =...

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Chapter 11 293 11-9 Assume concentrated forces as shown. P z = 8(24) = 192 lbf P y = 8(30) = 240 lbf T = 192(2) = 384 lbf · in ± T x =− 384 + 1 . 5 F cos 20 = 0 F = 384 1 . 5(0 . 940) = 272 lbf ± M z O = 5 . 75 P y + 11 . 5 R y A 14 . 25 F sin 20 = 0 ; thus 5 . 75(240) + 11 . 5 R y A 14 . 25(272)(0 . 342) = 0 R y A =− 4 . 73 lbf ± M y O =− 5 . 75 P z 11 . 5 R z A 14 . 25 F cos 20 = 0 ; thus 5 . 75(192) 11 . 5 R z A 14 . 25(272)(0 . 940) = 0 R z A =− 413 lbf; R A = [( 413) 2 + ( 4 . 73) 2 ] 1 / 2 = 413 lbf ± F z = R z O + P z + R z A + F cos 20 = 0 R z O + 192 413 + 272(0 . 940) = 0 R z O =− 34 . 7 lbf ± F y = R y O + P y + R y A F sin 20 = 0 R y O + 240 4 . 73
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Unformatted text preview: 272(0 . 342) = R y O = 142 lbf R O = [( 34 . 6) 2 + ( 142) 2 ] 1 / 2 = 146 lbf So the reaction at A governs. Reliability Goal: . 92 = . 96 F D = 1 . 2(413) = 496 lbf B O z 11 1 2 " R z O R y O P z P y T F 20 R y A R z A A T y 2 3 4 " x...
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