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6_ch 03 Mechanical Design budynas_SM_ch03

6_ch 03 Mechanical Design budynas_SM_ch03 -...

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Chapter 3 19 3-4 (a) q = R 1 x 1 40 x 4 1 + 30 x 8 1 + R 2 x 14 1 60 x 18 1 V = R 1 40 x 4 0 + 30 x 8 0 + R 2 x 14 0 60 x 18 0 (1) M = R 1 x 40 x 4 1 + 30 x 8 1 + R 2 x 14 1 60 x 18 1 (2) for x = 18 + V = 0 and M = 0 Eqs. (1) and (2) give 0 = R 1 40 + 30 + R 2 60 R 1 + R 2 = 70 (3) 0 = R 1 (18) 40(14) + 30(10) + 4 R 2 9 R 1 + 2 R 2 = 130 (4) Solve (3) and (4) simultaneously to get R 1 = − 1 . 43 lbf, R 2 = 71 . 43 lbf. Ans. From Eqs. (1) and (2), at x = 0 + , V = R 1 = − 1 . 43 lbf, M = 0 x = 4 + : V = − 1 . 43 40 = − 41 . 43, M = − 1 . 43 x x = 8 + : V = − 1 . 43 40 + 30 = − 11 . 43 M = − 1 . 43(8) 40(8 4) 1 = − 171 . 44 x = 14 + : V = − 1 . 43 40 + 30 + 71 . 43 = 60 M = − 1 . 43(14) 40(14 4) + 30(14 8) = − 240. x = 18 + : V = 0, M = 0 See curves of V and M in Prob. 3-3 solution. (b) q = R 0 x 1 M 0 x 2 2000 x 0 . 2 1 4000 x 0 . 35 0 + 4000 x 0 . 5 0 V = R 0 M 0 x 1 2000 x 0 . 2 0 4000 x 0 . 35 1 + 4000 x 0 . 5 1 (1) M = R 0 x M 0 2000 x 0 . 2 1 2000 x 0 . 35 2 + 2000 x 0 . 5 2 (2) at x = 0 . 5 + m, V = M = 0, Eqs. (1) and (2) give R 0 2000 4000(0 . 5 0 . 35) = 0 R 1 = 2600 N = 2 . 6 kN Ans. R 0 (0 . 5) M 0 2000(0 . 5 0 . 2) 2000(0 . 5 0 . 35) 2 = 0 with R 0 = 2600 N, M 0 = 655 N · m Ans. With R 0 and M 0 , Eqs. (1) and (2) give the same V and M curves as Prob. 3-3 (note for V , M 0 x 1 has no physical meaning). (c) q = R 1 x 1 1000 x 6 1 + R 2 x 10 1 V = R 1 1000 x 6 0 + R 2 x 10 0 (1) M = R 1
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