Vigas Hiperestáticas - Método das Forças

100C\n20C\n10 m\n10 m\n\nSistema Principal:\n\ng=5-3=2\n\ne1=0\n\nEquações compiladas:\n\nd0+d1+F1+d2*Fz=0\n\nd2+d2+Fz=0\n\nCASO 0:\n\n100C\n20C\n\n\n\nCASO 1:\n\nA\n1 1\nC\nF2\n\n\n\n0.1\n0.2\n0.1\n\n\nDMF [kN m]\n\n\n\n10 m\n1 m\n\nMF=0\nRAY=0 (sai)\nRAY=0 kN\n\nM=ME=0\nRBY=0\nRBY=0.2 kN\n\n0\nEFy=0\nRAY=0 kN\nRBY=0 kN\nRCY=0 kN CASO 02\nEFx=0\nRAY=0 kN\nEFy=0\nRAY+RCY+RBY=0\nMC=0\nMC=9\nFCZ=0\nRAY=10-2.0\nRBY=10\nRBY=15\nRBY=0.1 kN\n\nMAMOM de 10=0\nRBY+RAY+RCY+RBY=0\n0-10RBY+R+0=0\nRBY=0.2 kN\n\nDiagrama do caso 1 e 2:\n\nCASO 0:\n\n1000 20C\n\nd0=1/h(ABi-ABs)\n1/h(ABi-ABs)\n\n\nd0=0.0142 rad\n\n(z=0.034 rad)\n\nz=d0=0.034 rad\n\nd1=E1=\n\nd1=1/3 mm+\n\nd2=1 mm=1/3 mm\n\nC1=5*10^-1+5*10^-1.1\n\n1.6 E1=7.700\n Para equação da compatibilidade:\n\nd0+d1+d2*Fz=0\n\nd0+d2*Fz=0\n\n-0.01474+8.58.10^-5 Fz +2.714.10^5 Fz=0\n-0.01421+2.14.10^5 Fz=0\n\n813.08-4.017=0\n0.000-0.034=0\nFz=162.12\nFz=162.54\n\nRestabelecendo a estrutura real:\n\nRAY=10-0.62023=0\n(RAY=162.12)\nRAY=10-162.54=0\nRAY=16.25 kN\n\nMAMO=0\nRAY=10-RBY-30=0\n-lRCY=32.4170+RCY.10+16.25*30=0\n\nRBY=32.14+\n\n\ne=0\n\nRYB0=16.18=0\nRYB=16.19 kN •𝑡 = 20𝑘𝑁/𝑚𝑚\n 𝑬 = 10⁴ 𝑘𝑁/𝑚\n 𝑉 = 7.10⁻⁵ m⁴\n 𝑓 = 0.6m\n 𝑓 = 0.4m\n 𝑑𝑛 = 2.0\n\ndn0 + dn𝑏𝑤4 = 20\n\ndn = 36.𝑢𝑛⁴𝑒𝑒𝑟𝑎𝑠𝑓𝑎𝑡 3.6 = 0\n 𝑓𝑏 = 0.02\n j𝑑𝑜𝑎𝑟\n 𝑑𝑛 = 20 \n𝑑𝑠𝑣𝑎𝑟% = 5\n\n 𝑅𝑟 = 0\n 𝑅𝑗 = 𝑅𝑢\n 𝑓𝑏 = 0\n • 𝑓(\n\n𝑑𝑠𝑣𝑞 = 6\n 𝑓𝑏 = 0\n 𝑅𝑏𝑦 = 0.5 {\n • 𝑓(𝑢 +\n 𝑏𝑙 = 1\n\n𝑓 = 13.5𝑘𝑛\n 𝑳𝑯𝑬𝑒𝑙}\n\n. caso O temperatura.\n jdoact = 𝜅 (𝑎𝑡−𝑎𝑡)\n 𝑛 (𝑎1 + 𝑎2)\n jdoact = 15° (0−50), 𝑃𝑎.\n G.3\n 2\n 0.6\n jdoact = -8.33×10⁴, 78\n jdoact = -0.015\n\n caso carregamento\n\n • 𝑓𝐹𝑥 = 0\n 𝑅𝑎𝑦 = 0\n • 𝑓𝑓𝑦 = 0\n - (20.6) -40 + 𝑅𝑎𝑦 + 𝑅𝑏𝑦 = 0\n 𝑅𝑎𝑦.𝑅𝑏𝑦 = 160\n 𝑅𝑎𝑦 + 60 = 160\n 𝑅𝑏𝑦 = 160 − 60\n 𝑅𝑎𝑦 = 100 𝑘𝑊𝑡\n\n • 𝑓𝑓𝑦 = 0\n 𝑅𝑏𝑦.12 − 40.𝑋 = -20.6, 3 = 0\n 𝑅𝑏𝑦.12 − 12 = 0\n 𝑅𝑏𝑦 = 60 𝑘𝑁𝑡\n\n •\n 𝑄𝑓𝑦 = 40\n\n -20𝑥 + 100 + 20𝑥 = 0\n 𝑄𝑏 = 20𝑥 + 100\n x = 5\n\n. -20𝑥 + 100 + 20𝑥 = 0\n 20𝑥 = 100\n 𝑥 = 5\n\n m = -100 x + 20𝑥\n m = 100 × 20\n m(15) = 100,5 = -10.5\n m(15) = 250𝑘𝑛\n\n Espelhos com batida ideal:\n x = 0 = 𝑓(2)\n j𝑡0 + 𝑑𝑛𝑓 = 𝑑ß\n j𝑡0 + 𝑑𝑛𝑓 = 𝑟𝑒𝑐𝑙𝑢𝑔𝑒\n\n\n\n •\n\n = 𝑑𝑛0 + 𝑑𝑛𝑓 = 0.02\n j𝑑𝑜𝑎𝑟𝑢𝑛 = 1/3 m𝑗 + 1/3 m𝑗𝑛𝑓𝑗' + 1/3 n𝑓\n j𝑑𝑜𝑎𝑟𝑢𝑛 = 1/3.6.3.240 + 1/3.6.3.240 + 1/3.2.15.20\n 1/𝑑𝑠𝑣𝑎𝑠 = 𝑙/6\n\n j𝑑𝑜𝑎𝑟𝑢𝑛 = 1/3\n \n𝑑𝑛 = 36𝑚/𝑒𝑙0\n 𝑑𝑛 = 36𝑚/𝑒𝑙3\n j𝑑𝑛𝑠 = 0,0263.\n\n j𝑑𝑛𝑠 = 0.01\n\n ;\n j𝑑𝑛 = 0.02\n= 0.02\n\n j𝑑𝑛𝑠 + j𝑑𝑛𝑓 = 0,02\n\n depois + 0.02 + 3.6 0.0 𝑛 = 0\n\n Resolvendo o passo real:\n 𝑑𝑡𝑒 = 0\n 𝑅𝑎𝑦 = 0\n 𝑅𝑏𝑦 = 5.28 − 40 + 5𝑦 = 0\n 𝑅𝑏𝑦 = 5.28.6 − 20.6.3 = 0\n 𝑅𝑎𝑦 = 97,36 𝑘𝑁𝑡\n 𝑅𝑏𝑦 = 15,7\n mm de pilares:\n\\n\\n97.36\n\\n\\n97.36\n\\n\\n97.36\n\\n\\n4.863kN\n\\n\\n[one km]\n\nMmax = 2 * 26.97 x km Seja o porto hiperestático, determinar o DMF usando o método das forças.\n\\nSistema principal:\n\nq = 3 + 2 - 3 = 2\n\\n\nCaso (0)\n\nRAx + RDA + 20 + 0 = 0\nRAx + 20 = RDA\nRAx + 20 = 0 Caso (1)\n\\nMA + RAx * 5 + 1 = 0\nMA = -1 kN\n\\n\\nRdx = 0 CASO (Z)\n• Fz\n• EF = 0\nRAY = RDY = 0\nRAX = -0.125 kN\n• ∑MA = 0\nRAY + RDX > 0\n• ∑M = 0 di (d)\nRDX; 8 - 5RDX + 1 = 0\nADY + 5 + 5RDX = 0\nRDX = -1.5 kN\nRDX = 0.2 kN\nRAY = -0.125 kN\nRdx = -1.5\nRdx = 0.2 kN\n• mc = 1 - 1/3 - 1/4 + 1/5\n• [bmr =]\n• 0 - adfn + dzF - 1 = 0\ntotal\ncarga\ncarga\n\nDigitalizado com CamScanner d. o - tec, d - t - le, c, t, ires\nE1 = EBC\nE2 = EIG\nLarg = 5m\nAlt = 5m\nLec = 3m\nL2 = L1. Ic \ns c a?\ncompr\n\nequivalente\nE1A = E31\nE1BC = E51\nE1CB = E21\nL’ = Li. Ic Ei.\nIi\nc a\nL = 5. E1 = 1.667 m\nLac = 8. E1 = 41.6 m\nLec = 5.EI = 2.5 m\n\nDigitalizado com CamScanner EI1 = \n1 m\n= 3.794 × 10^5 rad\nIE1 = E1;\nE1D2 = d2 = 2.67\n\ne\nL = 5m\n\na + 0.1\nn\nk\n= -1/6.\n1.60.1 + 1/3. 1.6.80.1\ne\n= ( -1/2 - 1.667.1.1)\nE1\n= E1.1 + 2/2.66\nE1\nE2D2 = -2.69. 10^5 rad /km \n+ 0 + 0 + 0 + 0 + 0 + 0\nE\n= 91.953 x 10^3 \n\nDigitalizado com CamScanner d1o + d2F1 + d2F2 = 0\nd2o + d2F1 + d2Fz = 0\n{ 6.03 \u00d7 10^{-4} + 1.047 \u00d7 10^{-4}F1 - 2.699 \u00d7 10^{-5}F2 = 0\n3.174 \u00d7 10^{-3} - 2.699 \u00d7 10^{-5}F1 + 9.153 \u00d7 10^{-5}F2 = 0\n6.03 \u00d7 10^{-4} + 1.047 \u00d7 10^{-4}F1 - 2.699 \u00d7 10^{-5}F2.\n22.34 + 3.88FA = Fz\n3.794 \u00d7 10^{-3} - 2.699 \u00d7 10^{-5}F1 + 9.153 \u00d7 10^{-5}(22.34 + 3.88) = 0\n3.794 \u00d7 10^{-3} - 2.699 \u00d7 10^{-5}FA + 2.045 \u00d7 10^{-3} + 3.55 \u00d7 10^{-4}F1 = 0\nFA = 2.28 \u00d7 10^{-3}\nF1 = -15.91\nF2 = -39.40\n\nFz = -39.40\n\n\n\nF1 = -15.91\n\nFz = -39.40\n 20\nRAy = 37.09 kN\n\n25 = 0\nRAy + RBy + 80 = 0\nRAy + RBy = 80\nRAy = 37.09 kN\n\nMa = 8.95 = 0\n8RBy - MA - 10.84 - 20.3 = 0\nMA = 8RBy = 380\n36.68 + 8RBy - 380 = 0\nRBy = 42.31 kN\n\n\n\n\n\n60 - 1(-15.88) + 1(-39.22)\nM = 36.68 km\n\n\n\n\n36.68 - 12.16.3 = 0.192\n