Apostila de Exercícios Método das Forças

El S10 = 6.6 ( -1.20 ) + ( 6.C)( -106.67 ) + ( 6.6)( -1.33 ) = - 3820.0\n\nEl S11 = 6.6 \u00b7 4 / 2 = 144\n\n-3820.0 + X1 . 144 = 0\n\nX1 = 30.00 kN\n\n# diagrama da M.E:\nM = M0 + X1 N1\n\nM0 - 60 kN\n\nM1 = -60 + 30.00 .6 = -30.00 kN\n\nM2 = 40 kN\n\n# diagrama da F.C\n\nArea AB :\nvA = yB + c = 10 kN\n\nArea BC :\nvB = yC + c = -10 kN Método das Forças\n\nP/2001 - (32)\n\n1º passo - diagrama de F.C da liga abaixo: (T, c)\n\n20N\n\n100N\n\n80N/m\n\n2m\n\n6.00\n\n6.00\n\n8.00\n\nG = 4-3-1\n\n40\n\n120\n\n106.67\n\nj -20\n\n20 -40\n\n20\n\n-10\n\n\n0\n\n-Xi + Xi... = 0 Caderno de Exercícios\n\n\" ESTABILIDADE DAS CONSTRUÇÕES \"\n\nMÉTODO DAS FORÇAS\n\nProf . Tati ana Ai ello Método das Forças\n\nAP/99 (3p)\n\n1. Passa a riga abaixaco, pede-se:\na) Parâsel de cargas;\nb) Diagrama de M.F. e F.C.\n\n100kN\n\n20 60 60\n\nx1=1 y1=1\n\n10\n\n10\n\n-\n\n60\n\n40\n\nM1\n\n-\nc\n\n30kN/m\n\nM2\n\n-\nM3\n\nj\n\n20 60 60\n\n-\n\no\n\nx10 + x11 + x21 + x20 = 0\n\nx20 + x11 x21 + xe2.x20 = 0 E1(x10) = (12.12) * (200) + (-180.6) * (10.8) = -17.316\n\nE1(x20) = (6.6) * (40) + (-180.6) * (48) = -56.16\n\nE1 = (182.12) * 9.576\n\nE1 = E1.21 = (20.6) * 10 * 180\n\nE(x20) = (6.6) * 4.72\n\n... +1.316 = X1 * 576 * 180 * (0.40)\n+ 5.616 = X1 * 180 * X2 * 72\n\n-6926.4 = 2.30.X1 - 7.2.X2\n+ 5.616 = 80.X1 + 72.X2\n\nE(x1 + e2 = 0 (X1 + 2.6 KN)/(X2 + 13 KN)\n\n(M1 = -40 kN/m)\nM2 = 160 + (26) - 4 kN/m\nM1 = 220 - 180 (26) * (13) - 70 kN/m\n\nRa = 6 kN\n\nV0 = 30.6 + c - 19 kN\nV1 = 30.6 + c = 71 kN\n\nM = 30.32\n\n0x2 = 30.83 (1.275 w) Método das Forças\n\nAP/2020 - (3p)\n\n3. Resolver a riga abaixaco:\n\ng 6-3-3 -\n\n20 kN/m\n40 kN/m\n20 kN/m\n\n500\n400\n\n400\n\n200\n\n100\n\nx1 = 1\nx2 = 1\n\nR1.28\n\n40 40\n\n60\n\n25\n\n-\n\n40.68\n\n-x1 + x2 + x3 = 0\n\nd0 + x1 d19 + x2 d20 + x3 d13 = 0\n\nd20 + x1 d21 + y3 d22 + ... = 0\n\nd30 + y1 .d31 + ... = 0 S10 = (-250.5).11.75 + (-40.5).105 + (-1.00.4).6 +\n[60.5.573.5.(10.681) + (360.4)(1.506) = -36 087.83\nS20 = [-250.5).3.75 + [5.5)(-40.7) +\n(S60 + 3.9)(5.268) = 8286.50\nS12 = S11(5.5).14.33; 141.67\nS11 = [1.12(8.5).10.69 + [1.8.533] = 646.56\nS22 = (5.5).3.33; 41.63\n3.600.724.3 + X1.646.56 + X0.141.67 = 0\n-8286.50 + X1.141.67 + X2.41.63 = 0\n7906.84 - 41.67.X1 = 31.0.X2 = 0\n8286.50 + 141.67.X1 + 41.63.X2 = 0\n(X2 = 30.18 KN\nX1 = 49.20 KN Barra AB\nVA = VA - 20.K\nVA = c = 50 + 10.62 = 60.62 KN\nVB = -50 + 10.62 = -39.38 KN\nBarra BC\nVB = M + C - 5 - (-4.92) - 9.20 KN\nVC = -9.20 KN\n\n599.01\n(x: 39.38)\n(y: 43.20) + (yz: 40)\n-9.20 - 40\n39.38\n39.38 Método das forças\nPR/2022 - (2)\n4. Resolve a via abaixo:\nGs:6.3 ; 8\n{S10 + X1.S11 + X0.S12 = 0\nS20 + X1.S14 + X0.S13 = 0\nS40 + X1.S15 + Y0.S32 = 0}\nS10(-62.4).5 + (-160.4).67 = 2\n= 256.107\nS20 = (-160.4)(-1) + (-160.4)(-1) = 533.33\nS11 = (6.6).4.72\nS12 = S21(6.4)(1) - 18\nS22 = (-1.6)(-1) 6\n-256.107 + X1.72 - Y2.18 = 0\n1.600.00 = 54X1 + 10X2 = 0\n-96.107 + 18X1 = 0\n(X2 = 71.89 KN.m)\nX1 = 71.89 KN.m Mb = -160 + (53.29) . 6 + (71.29) . (-1) = -70.95 kN.m\nMb = (53.39) . (-1) . 71.29 = 35.49 kN.m\nMc = 71.29 kN.m\n\nVa = 53.33 + 13.33 + C = 60.61 kN\n\nVb = -13.89 kN\n\nVc = -26.67 - 26.67 + C = -53.38 kN