Camscanner 04-14-2020 15 40 28

2) P=20 KN=20000 N\n(Sada = 150 MPa = 150 N/cm²)\nDAB=?\nDAC=?\n∑FY = 0 => Fab × sin(45°) - P = 0\nFab = 28.2843 N\n∑Fx = 0 => Fac + X × cos(45°) - P = 0 =>\nFac = 20000 N\n(ρ_ad = P → A = P = π/4 d² - P)\nDi = √(4P/π × Sadan)\nDAB = √(4 × Fab) = √(4 × 28.2843) = 15.5 mm\nDNC = √(4 × Fac) = √(42000 / π × 150) = 13.1\n 3) ∑Fx = 0 => O_y × Tac - Tab × cos(60°) = 0\n∑Fy = 0 => O_y × Tac + Tab × sin(60°) - S = 0\n(Sadin = P → A = P = π/4 d² - P)\nD = √(4P/π × Sodin)\nD = (4 × 4 × 25 × 10³) = √(5.26 mm)\nD = (4 × 4 × 7.1 × 10³) = 5.48 mm 1) ∑Fy\n-TAB × cos(45°) + TBC × cos(30°) = 0\n∑Fx\nFAB × sin(45°) + TBC × sin(30°) - P = 0\n-TAB × cos(30°) - TBC × FAB = FAB: IAB + TBC\nTAB × cos(45°) + TBC × P = 0\nFAB = 3800 = 5403 N\nDAB = √(4 × 5403) / π × 16.5 = 6.45 mm\nDRC = √(4 × 4911) / π × 16.5 = 5.83 mm ΣFy = 0\nA_y + B_y - 10 - 18 = 0\nA_y + B_y = 30\nΣFx = 0 -> A_x = 0\n\nΣMA = 0 12,3 + 18,6 + 6,8 - B_y * 11 = 0\n\n38,44 + 12,4 - B_y * 11 = 0\nB_y = 14,35 kN\n\nA_y + B_y - 30 = A_y - 30 - 14,35\nA_y = 15,64 kN B) B.A = 112 16 = 89,6 kN\n\nΣFx = 1 - q - 1 1 2 - 0 3 + 3 m\nA_y\nA_x\n\n89,6\nB\n\n... 3,73 m\n\nΣFy = 0\nA_y + B_y - 89,6 = 0\nA_y + B_y = 89,6\n\n... ΣFx = 0 -> A_x = 0\n\nΣMA = 0 -> 89,6 * 7,47 - B_y * 11,2 = 0\n66,93 = B_y * 11,2\n\nB_y = 83,76 kN\nA_y = 89,6 - 83,16\nA_y = 29,84 kN C) Retângulo\nB.A = 5,8 3 = 73,6 kN\nB.A = 3,3 3 = 46 kN\n\nΣFx = 1 - q - 1 1 2 - 0 3,06 - 4,6 = 0\n\n... ΣFy = 0\nA_y + B_y - 73,6 - 46 = 0\nA_y + B_y = 19,6\n\nΣFx = 0 -> A_y = 0\n\nΣMA = 0 -> 46,3,06 + 73,6,6,4 - 6 - B_y * 1,1 = 0\n\n140,76 + 228,96 = B_y * 9,4\nB_y = 5,2 1 kN\n\nA_y = 11,6 - 52,1 + A_y = 67,5 kN 5)\n∑Fx=0 ∑Fy=0\n0.8FAC - FAB*cos(30°)=0\nFAB=0.8P\n0.6FAC + FAB*cos(30°) - P=0\nFAC=0.9417P\n\n∠CAD=FAC : 180° = 0.9417P\nAAC\nπ/4 x 42\n\nP=29.4 kW\n