A2 Mec Geral

A2 - MECÂNICA GERAL\nALUNO: PAULO GABRIEL PEREIRA CASAES - 20103503\n\n1 km + 1,5m + 1,5m + 1\n -\n 50kN\n\nRA = 40kN.m\n\nR1 = (22).(2,5) = 60kN,\n\nx1 = 1/2 (2,5) = 1,25m,\n\nR2 = (36).(2,5) = 45kN → x2 = 1/3 (2,5) = 1/3 (5/2) = 5/6m.\n\nΣM_A = 0 → 40*(-50) + (2)*(6.0) + (60)*(3.75) + (25) + (2/3)*μ_A = 0 → μ_A = 25 kg.\n\nΣF_y = 0 → Ay + 40 + 45 - 50 = 0 → Ay = -55kN.\n\nRA ⇒ Sentido anti-horário\n\nResposta → RA = 55kN e MA = 25,5kN.m. x̄ = 46506188 = 1 32,1 mm\n 35200\n\nȳ = 2(466624) = 75,74 mm\n 35200\n\nPARTE A XC YC XCA YEA\n1 28800 12,0 60 3456000 1728000\n2 6400 186,67 126,66 1194688 938624\n 35200 4650688 2666624 I_x = ΣĪ_x + ΣA dx2\n 7,21 10^9 + 3,06 10^10 → I_x = 3,78 10^10 mm^4\n\nI_y = ΣĪ_y + ΣA dy2\n 1,8 10^9 + 1,845 10 → I_y = 2,025 10^9 mm^4