Hiper 2

1. SP (Citerna Principal)\n\nP* ΣM1=a\n4x2 + 4x2 + MA=0\nMA = -1.6k\n\nΣFy = -4.4 + RA = 0\nRA = 8 k\n\n2. Equação de Comp.\nS10 + S11X1 + S12X2 = 0\nΣ20 + Σ21X1 + Σ22X2 = 0 1. ΣMA = 0\n25X2 + 287.5x6.25 + 11x12.5 - RBx11 = 0\nRB = 180.39 kN\n\n2. ΣFy = 0\n-287.5 - 25.11 + 180.39 + RA = 0\nRA = 143.11 kN\n\n3. Equação da Compatibilidade grau 2\nS10 + S21X1 + S22X2 = 0\nS10 + S21X1 + S22X2 = 0 1. DMFO (kN)\n\n2m\n2m\n4m\n8m\n258.60\n240.20\n338.50\n42.4\n\n2m\n1.5m\nVirtual (colocando uma carga pontual) S22 = 1 / EI MA M3 + 1 / 6 [M1(2M3 + M4) + M2(M3 + 2M4)] + L3 M1 M3\nS22 = 1 / EI (4 / 3 (A10x1.10) + 1 / EI (4 / 6 [A10 (2xA40 + 2.20) + 2.20 (1.40 + 2x2.20)] +\n1 / EI (3.20 + 2.20 + 2.20) =\nS22 = 41.51 m\nEI\nS12 = ∫M1 M2 ds / EI\n1 / 3 EI\nS12 = 1 / EI (1 / 3 M3) + 1 / EI (M1(2M3 + M4)) + M2 (M3 + 2M4) + 1 / EI (1 / 3 M3)\nS12 = 1 / EI (4 / 3 x2.50 x2.50) + 1 / EI (4 / 6 [1.10(2x2.50 + 1.10) + 2.20(2.50 + 2.10)]) + 1 / EI (3 / 3 (1.40 x 2.20) + X1 = 107.75\nX2 = 90.23\nΣMA = -46x1 + 25x2 + 46x3 + 9.2x6 + 69x9.5 + 34.50 x1.35 + 11x42.5 - 107.75 x4.90 + 26.8 - Rbx11 = 0\nRb = 7.6 kN\nΣFy = -27.50 - 25 - 11 + 7G + 107.75 + 90.23 + RA = 0\nRA = 49.52 kN S20 = 1 / EI (4 / 3 (338.50 x 1.40)) + 4 / 3 (1.40 x 46) + 1 / EI (1 / 6 (1.10(2.338.50 + 258.50) + 2.20(338.50 +\n2.258.60)) + 1 / EI (2.70 (2.258.60 - (-43.70) + 3 / 3 (258.80 + 2.20))\nS20 = 27.61.94 m\nEI\nS11 = 1 / 2 EI (1 / 3 M3 M3) + L3 M4 M3 + 1 / G [M1(2M3 + M4)) + M2 (M3 + 2M4)\nS11 = 1 / EI (4 / 3 x2.50 x2.50) + 1 / EI (4 / 6 [2.50(2 x2.50 + 1.40) + (1.40(2.50 + 2.1.40)]) + 1 / EI (1 / 3 (3 / 6 + 1.40 x 1.40) S10 = ∫ M0 M1 dα E I\n\n\\frac{1}{E I}\n\\frac{1}{E I}\n1.40(338,50 + (2,258,60) - \\frac{4}{3 E I}((2,50)(2,338,50) + 258,60))\n\nS20 = ∫ MOM2 dα E I\n\\frac{1}{E I}\n\\frac{4}{E I}\n338,50\n\\frac{4}{E I} = 42,40\nx2,20 S12 = ∫ M1 M2 dα E I\n\n\\frac{1}{E I} (M1 M3 L) + \\frac{4}{3 E I} (\\frac{L}{4} M M3) + \\frac{1}{E I} (4,4 - 3) + \\frac{4}{3 E I} (\\frac{4}{9}) = S12 = \\frac{48}{E I}\n\n(4, 4, 4) = 48 m\n\n4\n\nuducação do sistema\n\n330,66\n\\frac{424,414 x1}{E I} + \\frac{48 x2}{E I}\n\n234,66\n\\frac{48 x1}{E I} + \\frac{6,44 x2 = 0}{E I}\n\nx1 = -1,95\nx2 = -1,85 S20 = ∫ M0 M2 dα E I\n\n\\frac{4}{E I}\n(x) + \\frac{4}{3 E I}\n4 + \\frac{4}{3 E I}\n(M M3) = S20 = 234,66 m\n\nS11 = \\frac{424,414}{E I}\nS11 = 424,414 m DMF1\n\n4\n\n4\n\n4/3E1\n\n4\n\n4/3E1