2_ch 03 Mechanical Design budynas_SM_ch03

2_ch 03 Mechanical Design budynas_SM_ch03 -...

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Chapter 3 15 3-2 (a) R A = 2 sin 60 = 1 . 732 kN Ans. R B = 2 sin 30 = 1 kN Ans. (b) S = 0 . 6 m α = tan 1 0 . 6 0 . 4 + 0 . 6 = 30 . 96 R A sin 135 = 800 sin 30 . 96 R A = 1100 N Ans. R O sin 14 . 04 = 800 sin 30 . 96 R O = 377 N Ans. (c) R O = 1 . 2 tan 30 = 2 . 078 kN Ans. R A = 1 . 2 sin 30 = 2 . 4 kN Ans. (d) Step 1: Find R A and R E h = 4 . 5 tan 30 = 7 . 794 m ± + ± M A = 0 9 R E 7 . 794(400 cos 30) 4 . 5(400 sin 30) = 0 R E = 400 N Ans . ± F x = 0 R Ax + 400 cos 30 = 0 R Ax =− 346 . 4N ± F y = 0 R Ay + 400 400 sin 30 = 0 R Ay =− 200 N R A = ² 346 . 4 2 + 200 2 = 400 N Ans. D C h B y E x A 4.5 m 9 m 400 N 3 4 2 30 ° 60 °
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