5_ch 03 Mechanical Design budynas_SM_ch03

5_ch 03 Mechanical Design budynas_SM_ch03 -...

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18 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (e) ± + ± M B = 0 7 R 1 + 3(400) 3(800) = 0 R 1 =− 171 . 4 lbf ± F y = 0: 171 . 4 400 + R 2 800 = 0 R 2 = 1371 . 4 lbf M 1 171 . 4(4) 685 . 7 lbf · ft M 2 685 . 7 571 . 4(3) 2400 lbf · ft M 3 2400 + 800(3) = 0 checks! (f) Break at A R 1 = V A = 1 2 40(8) = 160 lbf ± + ± M D = 0 12(160) 10 R 2 + 320(5) = 0 R 2 = 352 lbf ± F y = 0 160 + 352 320 + R 3 = 0 R 3 = 128 lbf M 1 = 1 2 160(4) = 320 lbf · in M 2 = 320 1 2 160(4) = 0 checks! (hinge) M 3 = 0 160(2) 320 lbf · in M 4 320 + 192(5) = 640 lbf · in M 5 = 640 128(5) = 0 checks! 40 lbf/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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