Método das Forças

Ejercicio\n\nJío\n\nSístemo principal\n\nE\n\n\ng = 9.81 m/s²\n\n\n\n\nL = 5 - 1 = 4 m\n\nEMB = D\n\nL1 = 0 1.1. (1 - d) mu = (1 - p)(5 - 1 - 5 m)\n\n\nd = 0\n\nd = 1\n\nd = 0.10\n\nd = 0.3\n\nel 2.\n\n26 + 28 = 81 kN·m\n\n(l = 2 m)\n\np1 + d2 = 0\n 1.2 kN/m 0.6 kN/m\n\n4 m\n\n6 m\n\ng = 9.81 = 0.6 kN/m\n\n10 - 9 = 0\n\nEjercicios de Luz\n\n1kN\n\n4; 2; 4; 24; 0.66 = 3 y\n\nZ\n\nH\n\n S T O Q S S D\nL M M J V S D\n\n1/2\n\n3.6\n\n10 = (1) (4.24-1) + (1)(5.3,6.6,-1) = (-1.96) / (+.216)\n\n1/8\n\n-3.0 + 7.0 = 0.1 kW\n\nm1 = 1\n\n1 = 1/3 0 + \n\nj1 = (1.6-1-1) = 1 + 3.6 + 1.2 kW\n\n1/3\n\n1/1\n\nj2 = (1-1/2)(1-6.6) = +1 kW\n\nj20 = -1\n\n(1.6,3.6,-1) = -7.2 kV\n\n1/3\n\n0.1 =\n\n= +1\n\n0.2 = 0 -5.4 + 2.4 + 12.2 = 0\n-3.8 + 1.2 + 0 = 0\n\nR2 + 18 x2 = 0\n\nVx = -3.8 - 0.26 kN\n\n-15\n\n-7.8 + 12.026 = 1 kN