4_ch 03 Mechanical Design budynas_SM_ch03

# 4_ch 03 Mechanical Design budynas_SM_ch03 -...

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Chapter 3 17 (b) ± F y = 0 R 0 = 2 + 4(0 . 150) = 2 . 6 kN ± M 0 = 0 M 0 = 2000(0 . 2) + 4000(0 . 150)(0 . 425) = 655 N · m M 1 =− 655 + 2600(0 . 2) 135 N · m M 2 135 + 600(0 . 150) 45 N · m M 3 45 + 1 2 600(0 . 150) = 0 checks! (c) ± M 0 = 0: 10 R 2 6(1000) = 0 R 2 = 600 lbf ± F y = 0: R 1 1000 + 600 = 0 R 1 = 400 lbf M 1 = 400(6) = 2400 lbf · ft M 2 = 2400 600(4) = 0 checks! (d) ± + ± M C = 0 10 R 1 + 2(2000) + 8(1000) = 0 R 1 = 1200 lbf ± F y = 0: 1200 1000 2000 + R 2 = 0 R 2 = 1800 lbf M 1 = 1200(2) = 2400 lbf · ft M 2 = 2400 + 200(6) = 3600 lbf · ft M 3 = 3600 1800(2) = 0 checks! 2000 lbf 1000 lbf R 1 O O M 1 M 2 M 3 R 2 6 ft 2 ft 2 ft
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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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