Método das Forças

hiperestaticidad: 3\n\nSincronismo: 3 equilibrios.\n\nResultados:\n1° Efy = 0: ( - 10(12) - V0 = 0\nV0 = 10 kN\n2° Efx = 0:\n - 40 + 0 + 0 = 0.\nH0 = 40 kN\n3° ΣM0 = 0:\n( - 10(12) + M = 0\nM = 720 kN·m\n\nMEA = 0\nMEB = - 10(6) - V0 = 0\nV0 = 120 kN\n\nMFC = - 10(6) - 12 = - 60\nMFD = - 10(12) / 2 = - 720\n\nMFC = - 10(6/2) + 120(6) - 720 = - 180 Resultados:\n1° Efy = 0: 1 + 12 + V0 = 0\nV0 = 12 kN.\n\n2° Efx = 0: - 40 + 0 + 0 = 0\nH0 = 40 kN\n\nΣM0 = 0:\nMTA = 0\nMFB = - 12 + 6 kN·m\n\nMFC = - 12 - 6 = - 6 kN·m\nMFD = + 12 - 12 kN·m\n\nΣE. END = 0\nV0 = 0 = 20 kN·m.\nV0 = MD = 0\nMD = 10 kN·m. Combina:\n\nδ10 = ?\n\nbarra 1: 3 m\n10\n1/3. 10(30) = 3 m\n\n45°\n1/3 * 180 * 6 = - 1080\n\nbarra 2: 3 m\n\n10\n1/3. 10.8 = 1440\n1/3. 7.41 = 270\n\nbarra 3: 6 m\n\n1 + 1/3. 6 = 720\n\n- 1/3. 10 - 720 - 720 + 160 = - 1620.\n\nδ10 = 140.360 + 60.80 - 4320 - 2880\nδ10 = 47.818 = 0.2394 d1 = \nbarro 1 l=3m\n + 1/2 {1.3\t6 = 36}\nbarro 2 l=3m\n + 3.6 = 108\nbarro 3 l=6m\n*\n\n6\n6 = [6+(2+6+2)]\n12(2m2 + 4) = 504\n\nd11 = 36+108-504 - 000324 = \n\n\n\nd2 = \nbarro 1 l=3m\n - [1/3 \t 6 + 8 - 48]\nbarro 2 l=3m\n + [1/16 { 6(8+16) = 216}]\nbarro 3 l=6m\n - [1/16 { 6(6+12) = 864}]\n\n\nd2 = 1128 - 1188 - 000504\n\nd21 d = 128 +118 - 000504\nE86 = 2.1075 S0 = \n(1)\n\nd1 = [d2\td1 d12 d211 ] {x1}{x2}\n= {0 ; 0}\n\n-28350\ -648\ -1128 = [ 0]\n\n49.810\ -1928\ -d2 = 0\n \n-28350 - 1128 x2 - 2049 x1 = - 14790\n\nx1 = 28350 + 11828 x2 \n648\n\n-1128 28350 - 1128 x2 - 2049 x1 = - 14790\n641\n\n-430350 - 1963.561 x2 + 2049 x1 = - 27980\n\n34.496 x1 = -1490.49350\n\nx2 = 17.409 kN \n\nx1 = 28350 + 1128 (17.408)\n641\n\ny1 = 94.053 kN\n Sussekind - p.921 - v.II\nH=6m\n \nReacción:\nG = EMRA = 0= 1/2\t0.6- 4.230\nG = EMRB = 0 = 1/2 = {3.0 = 0}\n\n-4126\nG = VE = 0\n\n- G H = \n= 14N\n - \nHf = 0 \n\nMAB = 0\nMFB = 0\nMFC = 0\nMFB = 0. *\n\n\nS0 [DMA(0)\n\n(1)\nL G2\norb is 1 H\n\nx1 = \n3\t\n\\ \n\t \n\t( * - 1 H 1 H)\nq2 = 36/4\t\n1.4 L6\t= 8.544 x. 3\n DMF(1)\n\nProof...\n\n(ΣM=0) => -(1/16)\n(ΣM=0) =>\n(ΣM=0) =>\n\nMFA = 0\nMFB = (1 - (1/16))\nMFC = - (1/16)\nMFD = -(1/16)\n\nDMF(2)\n\nProof...\n\n(ΣM=0)\n(ΣM=0)\n(ΣM=0)\n\nMFA = 0\nMFB = - (1/16) + 1\nMFC = 0.5\nMFD = - (1/16)\n\nDMF(3)\n\nProof...\n\n(ΣM=0)\n(ΣM=0)\n(ΣM=0)\n\nMFA = 0\nMFB = 0\nMFC = 0\nMFD = 0 \n\n\n\nCalculations:\n\nb10 = -104\n\nb12 = -64\n\nb16 = -16\n\nb20 = b10\n\nb21 = b12\n\nb23 => -2\nb32 = b20\n\nb25 = -16\nb52 = b20\n\nb54 = b12\n\nb54 => 0\n\nFinal Calculation:\n\nh(β1) = 0\n\nh(β2) = 0\n\nh(β3) = 0\n\nh = ... Regions:\n\nVA = 8\n\nHA = 0\n\nVE = 6\n\nHE = -6/\n\nVA = 6\n\nHA = -6h\n\nVE = 14\n\nHA = -6/\n\nCalculation:\n\nMFA = 0\nMFB = -1.88446 - 12.28\nMFC = ...\n\nQ = .... Diagram\n \n\tEMBc = 60kG - 5.0kN \n\tVs = 150 cm\n\tEFA50 = 0.1 + 27 - (1x1)\n\tVx = 27.5 kN\n\t \n\tC1\n\tMzero = 500\n\n\t \n\tMGA = −500\n\tMGB = 700 - 1.5 × 10-4 = 500\n\tC(F/0) = 100 - 503.5\n\tF = 100kN. \n\n\t \n\tC2 \n\t \n\tC1 = (e.g. 45°)\n\t50°\n\t.5 / 2.5\n\tA\[BM\] = \n\tMFB = −1 =\n\tMGC = 1.5 × 10−3\n\tFMB = 0\n\tMFD = −1/1.5 − 2.60.0