Questões - Resistência dos Materiais - 2023-1

BC V_{BC} = -\int 10 \, dx = -10x + C \quad x = 5,5 \quad V = 0 -10 \cdot 5,5 + C = 0 \quad C = 55 V_{BC} = -10x + 55 M_{BC} = \int -10x + 55 \, dx = -5x^2 + 55x + C \quad x = 5,5 \quad M = 0 -5 \cdot 5,5^2 + 55 \cdot 5,5 + C = 0 \quad C = -151,25 M_{BC} = -5x^2 + 55x -151,25 c) V_{AB} = 0 -\frac{15x^2}{2} + 27,1875x = M_{AB} V_{AC} = -15x + 27,1875 x = \frac{-27,1875}{15} \quad x = 1,8125m \quad V = 0\, kN d) M_{max P} = \frac{-15 \cdot 1,8125^2}{2} + 27,1875 \cdot 1,8125 M_{max P} = 24,6387 \, kNm V_{AB} = \int 15 \, dx = -15x + C \quad x = 0 \quad V = AY V_{AB} = -15x + 27,1875 M_{AB} = \int -15x + 27,1875 \, dx M_{AB} = \frac{-15x^2}{2} + 27,1875x M_{med N} = \frac{10 \cdot 1,5^2}{2} \quad M_{med N} = -11,25 kNm c) CORTANTE Ay - 15\cdot 4 = -32,8125 27,1875 15 -32,8125 V (kN) x (m) MOMENTO 1,8125m -11,25 24,6387 M (kNm) x (m) Mmax = 24,6387 kNm σ = \frac{M . \overline{y}}{I} σ = \frac{24,6387.10^3 N . cm . 10^2 . 13 cm}{10527,1093 cm^4} σ = 3042,65 \frac{N}{cm^2} σ = 30,4265 MPa \sigma_{máx} \sigma_{máx} C = -30,4265 MPa y = 26 cm \sigma_{máx} T = 30,4265 MPa y = 0 cm Y̅ = \frac{x̄}{2} = 13 cm X̅ = 10 cm I_x = \frac{16.20^3 . 2 + 22,8 . 1^3}{12} + 20,16.2 . (10)^4 + 1.22,8. (10 - 10)^2 I_x = 10527,1093 cm^4 I_y = \frac{1,6.20^3 . 2 + 22,8.1^3}{12} + 20,16 . 2 . (10-10)^4 + +1.22,8. (10-10)^2 I_y = 2135,23 cm4 Vmax = 32,8125 kN τ = \frac{V . Q}{I . t} Q = 1,1 . 4,5 . 7 + 1,6 . 20.23,6 Q = 820,18 cm^3 τ = \frac{32,8125.10^3 N . 820,18 cm^3}{10527,123 cm^4 . 1 cm} τ = 2556,4625 \frac{N}{cm^2} τ = 25,564625 MPa \tau_{máx} y̅ = 43 cm 5\nE)\nMAB = -7,5x^2 + 27,1875x\nMBC = -5x^2 + 55x - 151,25\nΘ = \int M dx\nV = \int Θ dx\nΘAB = \frac{-7,5x^3}{3} + \frac{27,1875x^2}{2} + C_1\nVAB = \frac{-7,5x^4}{12} + \frac{27,1875x^3}{6} + C_1x + C_2\nΘBC = \frac{-5x^3}{3} + \frac{55x^2}{2} - 151,25x + C_3\nVBC = \frac{-5x^4}{12} + \frac{55x^3}{6} - \frac{151,25x^2}{2} + C_3x + C_4\nb)\nVAB|_{x=0} = 0\nVAB|_{x=4} = 0\nVBC|_{x=4} = 0\nΘAB|_{x=4} = ΘBC|_{x=4}\nc)\nVAB|_{x=0} \to \frac{-7,5 \cdot 0^4}{12} + \frac{27,1875 \cdot 0^3}{6} + C_1 \cdot 0 + C_2 = 0\n\text{C}_2 = 0 VAB|_{x=4}\n\frac{-7,5 \cdot 4^4}{12} + \frac{27,1875 \cdot 4^3}{6} + C_1 \cdot 4 = 0\n\text{C}_1 = -32,5\nΘAB|_{x=4} = ΘBC|_{x=4}\n\frac{-7,5 \cdot 4^3}{3} + \frac{27,1875 \cdot 4^2}{2} - 32,5 = \frac{-5 \cdot 4^3}{3} + \frac{55 \cdot 4^2}{2} - 151,25 \cdot 4 + C_3\n\text{C}_3 = 296,6\nVBC|_{x=4} = 0\n\frac{-5 \cdot 4^4}{12} + \frac{55 \cdot 4^3}{6} - \frac{151,25 \cdot 4^2}{2} + 296,6 \cdot 4 + C_4 = 0\n\text{C}_4 = -457,6\nE)\nX=1\n\frac{-7,5 \cdot 1^4}{12} + \frac{27,1875 \cdot 1^3}{6} - 32,5 \cdot 1 = \frac{\nu}{EI}\n\frac{\nu}{EI} = \frac{-28,59375}{EI}\n\frac{\nu}{EI} = -5,4323389 \cdot 10^{-3} m\n\nu = -5,4323389 mm\nE = 506 Pa\nI = 10527,23 cm^4