Exemplo Resolvido Metodo das Forças com Temperatura

28/06/18\nQuestão 2 - 11 - 2015.2\n\nDados:\nT c = 20°c\nT s = 40°c\nh = 0.5 m\nA = 5.0 × 10³ kN\nE = 2380 kN/m²\nI = 1.56128 × 10-5 cm⁴\nA = 100 cm²\n\nProblemas:\n a) Reações do vínculo do Estado E1\n b) Diagrama de estados normais, E1\n c) Esforços de j\n d) Reforçantes Yi\n\nDiagrama de Esforços Normais e Tensões 1) Grandeza Hipertatibilidad e Sistemas lineares\nN1=4 j N.E=3 = 6=N.I-N.E=1\nSistemas posibles\n\n2) Estado E0\nPax: Pa.sinθ = 5a.L = 30\nPay: Pa.cosθ = 5a.18 = 40\n\nΣFx= 6/10 = 8/10\n\nΣFAD = HA0 = 0 (1)\nΣFY= VAD - P1 = 0 (2)\nΣMA0= MA0 - P1.4.00 = 0(3) 3) Estado E1\nΣFx=0 HAm = 0 (4)\nΣFy=0 VAm + XA = 0(5)\nΣMA=0 MA1 + X1.800 = 0(6)\n\nα(5) = VAA - X1 = -7\nα(6) = HAm = -800\n\n4) Diagrama de Esfuerzos Seccionais\n\n4.1) P.N.F. 2000d\nE0\n\nA\nC\nO\nB\n\n800\n\n800\n\nE1 4.2) P. E. N.\n\nzo =\n\nNA = VAX = VAx100 - VAxB = 30 - 0.6 -> 70\n\nNE = VAX = 70\nNE' = VAX = 30 - 0 -> 0\n\n5) cálculo de redundantes de\n\ndij = \\int_{0}^{L} (Ni \\alpha Tmdx) + \\int_{0}^{L} (Mi \\alpha 97 dx)\n\n\\int_{0}^{L} + \\int_{0}^{L} (N. \\alpha Tm dx)\n\nd_{10} =\n \n\\left( \\frac{EI}{L} + \\frac{L}{ \\int_{0}^{L} \\alpha T mdx} \\right) + \\left( \\frac{EI}{L} + \\frac{N. \\alpha T m dx}{L} \\right)\n\n+ \\int_{0}^{L} (M_{i} \\alpha T dx)\n 28/06/18\n\nQUESTÃO 2 - M - 2015.2\n\nTm = \\frac{T^{i} + T^{3}}{2} = \\frac{20 + 40}{2} = 30°C\n\n\\tau = \\frac{T^{i} - T^{3}}{h} = \\frac{20 - 40 - 20}{50} = -0.4\n \nEQUAÇÕES AOS DIAGRAMAS\n\nx = 0 x = 1000\n-0.6 = a(1,000) + b\n-0.6 = 0.000a - 0.6 \na = 0\n\n\\left( x \\int_{0}^{L} \\right) = -0.6\n\nN(x|1) = -0.6\n\nx = 0, y = -800\nx = 1000, y = 0\n-800 = a(0) + b + kx - 1800\n0 = a(1000) + b\nb = a(1000) - 800\n\na = \\frac{32,16}{70,5}\n\ny = 5 C_{10} = - \\frac{1}{L} M_{a} (2 M_{u} + M_{b}) - \\frac{L \\cdot M \\cdot \\mu}{E_{F}} + \\int_{0}^{L} \\alpha T dx\n\n+ \\int_{0}^{L} (0.96 \\alpha 30 dx) - \\int_{0}^{L} (0.1 x - 800) \\alpha (C^.* Y) dx\n\nD_{10} = -500.2000(2.800 + 400) - 500.30.0.6\n\n\\frac{6 EI}\n\n-0.6 - 1.7 - 70 x^{10} + 0.8.1.270 - 5.01 x^{10} + 800.0(1.7 - 0.5)x^{10}\n\nD_{10} = 20.10^{-9} - 9.10^{-3} - 0.216 + 1.92 = 384\n\nd_{11} = 20.10^{-9} - 9.10^{-3} - 0.216 + 1.92 - 3.39\n\nL = 3718975000\n\nd_{10} = 8.8.636 - 0.0063 - 0.121 + 1.92 - 3.54 = -11,059\n\nE_{F} = 2380.15663.10^{-5} = 37.187.5.00\n\nE_{F} = 2790.600 = 1423.000 & = 1/3 * Elab * M1 * M2 + Elab * M3 * M4 = 90000.800.500 / 3 * El \n\n + 1000.800.2 / EA = 1000.800.2 / 3.77.187.500 \n\n & = 0.5739 \n\n \\ + 0.5.7367 + 0.10002521 = 0.5739 \n\n Entrai \n\n j11 * X1 = -d10 \n\n X1 = -d10 / j11 = +711.1059 / 0.5739 = 19.352