Monitoria 2 temp-elast-rec

Estabilidae aas construções 1 - Monitoria 4 (2018.1)\nMétodos das forças: (aprio elástico e temperatura)\nQuestão 1: calcular os esforços da viga abaixo considerando o apoio B elástico e variação de temperatura ao longo da barra.\n\n\n 20 kN/m\n k = 1000 kN/m\n 0,75m 1,5m 4,5m\n |----------------|----|\n | .......30kN......|\n |----------------|\n 30 kN/m\n 4,5m\n\nD GHL = NR - Neq = 3 - 2 = 1\n\n2) 3) SP156C:\n\n4) CC: dL = 0\n\n5) EC: d = d0 + f x\n\n dL =\n\n | | | | P\n |----P-----|\n .......\n d0 + dOL + f\n\n\n\n | f1 |\n | x1 = |\n\n\n\n\n g) Efeito do carregamento (d):\n\nd\u2081 = d\u2081\u00b2 + d\u2082\u00b3 + d\u2083\u00b4 + d\u1d4bc + d\u2081 + d\u2082 + d\u2083\n\n \nd\u2081 = -2.9183\n\t -9.6847 (5-9)\n\t + 9.5 + \u0361l + 9b \n\n\nd\u2082 = d\u2081\u00b3 + d\u2082\u00b2 + d\u2083\u00b4 + d\u2082\n\n \nd\u2082 = \u0017\n\t -\u0017\n\t -8 + 0.5\n\t d\u2082\n \nd\u2083 = -0.260395\n\t d\u0001 = \n\t \n \nd\u2085 = -0.001910\n\nh\n\t 1) Efeito do carregamento (f):\n\nf\u2081 = 2ML/3E = 21.5/5000 = 6.667 * 10\u2074\n \nf\n\t = 6.667 * 10\u2074 + 1.667 * 10\u2074 + 3333.10\u2074\n\n \n\n \n\nh\n\t 8) Solução do item (5):\n\n d = d\u2081 + f.X \n = [0.260395 + j0.001910]\n \n { \n x\u2081 = 23.74 kJm \n x\u2082 = 43.18 KJm\n } * Trecho AB:\n* Trecho B:\n\n 123kN 40kN\n \n 19 kN/m\n 12 kN/m\n 12.5 kN\n 36.95 kN\n 39.01 kN\n \nt\tB \n\n e \n kE = 5m\n A \n \t \n \nR\u2083 = 3.16\n\n* Trecho BC;\n2 \n 16.5\n 90.3\n 90.2\n 19 kN/m\n 13,18 kJm \n 5m\n \n \n \n\ndA = 39.01kN\n 40.99kN