Método das Forças Exercícios 1

Método das Forças\n\nR: 1) Escala o SP.\n2) Estado 0 (Sólo carga)\n3) Estado 1,2,3,... (Sólo x1, x2)\n4) Deformación x1, x2.\n5) Sistema:\nS1, S10 + S1, S2 + x14 + S3, x2 + ...\n6) Suponer:\nS1 (monto o diagrama final) c) Estado 0 (Sólo carga)\n20 kN/m\n\nd) Estado 1 (Sólo x1)\n\nx1=1\n\ne) Estado 2 (Sólo x2)\nX2=1 A) Deformación (Tabla página 15 - Sussekind)\n\nS10 = M1 • M0\n=- l' • M • M - (1/4) • l' • M • M)\n=-3.6.360 - (4/4 • 3.6.360) = -8100\n\nS11 = M1 • M1\n= l' • M • M + (1/3) • l' • M • M\n= 3 • 6 • 6 + (1/3) • 3 • 6 • 6 = 144\n\nS20 = M2 • M0\n= (1/2) • l' • M • M + (1/3) • l' • M • M\n= (1/2) • 3 • 3 • 360 + (1/3) • 3 • 3 • 360 = 2700\n\nS21 = M2 • M1\n= -(1/2) • l' • M • M - (1/2) • l' • M • M\n= -(1/2) • 3 • 3 • 6 - (1/2) • 3 • 3 • 6 = -54\n\nS22 = M2 • M2\n= (1/3) • l' • M • M + l' • M • M + (1/2) • l' • M • M\n= (1/3) • 3 • 3 • 3 + 3 • 3 • 3 + (1/3) • 3 • 3 • 3 = 45 Sistema\n\nS10 + S11 + S12 · X2 = 0\nS20 + S21 · X1 + S22 · X2 = 0\n\n-8100 + 144 X1 - 54 Y2 = 0\n2700 - 54 X1 + 45 Y2 = 0\n\n144 X1 = 54 Y2 + 8100 ; X1 = 54 Y2 + 8100 / 144\n\nX1 = 0.375 Y2 + 56.25\n2700 - 54(0.375 Y2 + 56.25) + 45 Y2 = 0\n2700 - 20.25 Y2 - 3037.5 + 45 Y2 = 0\n24.75 Y2 + 2700 - 3037.5 = 0\n24.75 Y2 = 337.5 = 0\nX2 = 13.63\nX1 = 61.36 Superposição : E = E0 + X1, E1 + X2 E2\n\nApoio A : M = -360 + 61.36 + 13.63 = 8.16 KN.m\n120 + 6.1,36.(1) + 13.63 = 58.64 kN\n0 + 6.1,36.0 + 13.63 = 1 13.63 kN\nApoio B : M = 0 + 61.36 + 13.63 = 0\n+UV = 0 + 61.36.(1) + 13.63.0 = 61.36 kN\n0 + 61.36 + 13.63.(1) = -13.63 kN E = E0 + X1, E1 + X2 E2\nn0 C: \nM = +360 + 61.36.(−6) + 13.63.(−3) = 32.73 KN.m\nn0 D:\nM = 0 + 61.36.(0) + 13.63.(−3) = 40.89 KN.m\n\n58.64 = -61.36\nX = G.13.36 = 3.068\n\nDEC (KV)\nDEN (KN)