Método das Forças Exercícios 3

S_{10} = M_1 \\times M_0 = \\frac{1}{3} \\cdot l \\cdot M_b \\cdot M_m = \\frac{1}{3} \\cdot 3.90 \\cdot \\Delta = 90.0\nS_{11} = M_1 \\times M_1 = l \\cdot M_1 + \\frac{1}{3} \\cdot l \\cdot M_0 \\cdot M_b = 3.\\Delta \\cdot 3.\\Delta \\cdot \\frac{1}{3} = 4.0\nS_{20} = M_2 \\times M_0 = \\frac{2}{3} \\cdot l \\cdot M_b \\cdot M_m = \\frac{2}{3} \\cdot 3.90 \\cdot 3 = 540.0\nS_{21} = M_2 \\times M_1 = \\frac{1}{2} \\cdot 3.3.1 + \\frac{1}{2} \\cdot 3.3.1 = 9.0\nS_{22} = M_2 \\times M_2 = \\frac{1}{3} \\cdot l \\cdot M_b \\cdot M_m + 0. \\cdot M_m + \\frac{1}{3} \\cdot l \\cdot M_8 = \\frac{1}{3} \\cdot 3.3.3 + 3.3.3 + \\frac{1}{3} \\cdot 3.3.3 = 450. Solução:\na) Grau de hiperestaticidade total\ng = g_e + g_i\nq_e = I - e - R; q_e = 5 - 3 - 0 = 2\nX_1 + X_2\n\nb) Sistema Principal (S.P)\n\nl' = l \\cdot \\frac{J_c}{J}\nJ_c = menor de toda a estrutura\nJ = J da barra\n\nl_1' = 3. \\cdot \\frac{1}{1} = 3\nl_2' = 6. \\cdot \\frac{1}{2} = 3\nl_3 = 3. \\cdot \\frac{1}{1} = 3 c) Estado 0 (só cargas)\n+1 = \\sum F_y = 0\nV_A + V_B = 0\nV_A = -V_B\n\\Sigma M_A = 0 + 7 + 7\n-6U_B + A = 0\nX_A \\cdot V_A = 1\nX_A = \\Delta\n\nd) Estado \\Delta (só X_1)\n\n+1 = \\sum F_y = 0\nV_A + V_B = 0\nV_A = -V_B = -\\frac{\\Delta}{g} - 0.166\n\ne) Estado 2 (só X_2)\nV_A = 0\n\\Delta X_2 = \\Delta g) Sistema.\nS10 + S11 + S4 + S22X2 = 0\nS20 + S21X1 + S22X2 = 0\n90 + 4X1 + 9X2 = 0\n540 + 9X1 + 45X2 = 0\n4X1 = -90 - 9X2: X1 = -22.5 - 2.25X2\n540 + 9(-22.5 - 2.25X2) + 45X2 = 0\n540 - 20.25 - 20.25X2 + 45X2 = 0\n337.5 + 24.75X2 = 0: X2 = -13.64\nX1 = -22.5 - 2.25(-13.64)\nX1 = -22.5 + 30.69 = 8.19\n\nh) Superposição:\nX1 = -22.5 + 30.69 = 8.19\nE = E0 + X1E1 + X2E2\napoio A: M0 = 0 + 8.19(-1) + (-13.64) = 0 = 8.19\n+ U: 60 + 8.19(-0.96) + (-13.64) = 0 = 58.64\n+ H: 0 + 8.19(0) + (-13.64) = -A = 13.64\n\napoio B; + U = 60 + 8.19(0.466) + (-13.64) = 6.136 = 6.136\n+ H = 0 + 8.19(-1) + (-13.64) = -1 = -13.63\nnó C: M01 = 0 + 8.19(-1) + (-13.64) = -3 = 32.73\nnó D: M03 = 0 + 8.19(0) + (-13.64) + 3 = 40.89 + ΣzFy = 0: VA + VB = 120\nΣFy ≡ MRA = -6VB + 20(-6.3) + 8.16 → -6VB = 368.16\nVB = 61.36 kN\nVA = 58.64 kN\n\nM_D = -13.64(-3) + 8.16 = -40.89\n(i) (ii)\nMc = -13.64(6.43) + 8.16 = -32.73