Atividade Prática Princípios de Mecânica e Resistência dos Materiais

RU: 21391801\na = 103 mm\nA = 2°\nNF = 45 N\n\nΣMA = 0\n45.10^3 - Fr * cos 2° * 0.065 = 0\n45.45 - Fr * 0.064936 = 0\nFr = 4.545\n\n0.064906\nFr = 69.97 N RU: 21391801\nρ = 59 kg = 578.6 N\na = 2 m\n\nΣMA = 0\n-ρ * (xA + 3) + FB * (xA + 3 + 2 * xB) = 0\n-578.6 * (1.0.15 + 3) + FB * (0.15 + 3 + 2 + 0.10) = 0\n1822.16 + 5.25 FB = 0\nFB = 1822.26 / 5.25\nFB ≠ 347.36 N → substituting in Equation 2\n\nFB = 0.15 * WB\n347.16 = 0.15 WB\nWB = 347.16 / 0.15\nWB = 2314.4 N/m\n\nΣFy = 0\nFA - 620 - FB = 0\nFA - 620 + 347.36\nFA = 272.84 N → substituting in Equation 1\nFA = 0.225 WB\nFA = 272.84 = 0.225 WB\nWA = 272.84 / 0.225\nWA = 1212.6 N/m RU: 21391801\na = 2 mm\nb = 2 mm\n\nX̄ = 1 mm\nX̄2 = a + b / 2 = 2 + 2 / 2 = 4 / 2 = 3 mm\nA1 = 2.2 ⇒ A1 = 4 mm²\nA2 = 2.2 ⇒ A2 = 4 mm²\n\nX̄1 = a / 2 = 2 / 2 = 1 mm\nX̄2 = X̄A * A1 = 3.4 ⇒ ĀX2 = 12 mm\n\nȲ1 = b / 2 = 2 / 2 = 1 mm\nȲ2 = a / 2 = 2 / 2 = 1 mm\n\nĀȲ1 = A. Ȳ1 = 4 mm\nĀȲA = 4 mm\n\nX̄ = ΣA * X̄A / ΣA = 4 * 4 / 8 = 36 / 8\nȲ = ΣA * ȲA / ΣA = 2 * 4 / 8 = 8 / 8 I_y3 = 2.23 \\ 12 = 2.8 \\Rightarrow \\bar{I_y} = 0.67 mm^4 \nI_y2 = 2.8^2 \\ 12 \n \\Rightarrow \\bar{I_y2} = 0.67 mm^4 \n\\bar{I_y} = \\bar{I_{y1}} + A \\cdot d x^2 \nI_y3 = 0.67 + 7 \\cdot (2 - 1)^2 \nI_y = 0.67 + 4 \n\\bar{I_y} = 4.67 mm^4 \n\\bar{I_{y2}} = \\bar{I_{y1}} + A \\cdot d x^2 \nI_{y1} = 0.67 + 4 \\cdot (3 - 2)^2 \nI_y = 0.67 + 4 \n\\bar{I_y2} = 4.67 mm^2 \n\\bar{I_y} = \\bar{I_y1} + \\bar{I_y2} \n4.67 + 4.67 \n\\bar{I_y} = 9.34 mm^4\n RU = 2291801 \nF1 = 16 kN \nF2 = 9 kN \na = 2 m \n\\Sigma M_A = 0 \n-F2 \\cdot a - F1 \\cdot a - 5.3 \\cdot a - F1 \\cdot a - E_y \\cdot a = 0 \n-9 \\cdot 2 - 9 \\cdot 2 - 5.3 \\cdot 2 - 2 + E_y \\cdot 4 = 0 \n-18 - 36 - 30 - 8 + E_y \\cdot 8 = 0 \nE_y = 92/8 \n\\Rightarrow E_y = 11.5 kN\n RU = 2291801 \nF1 = 15 kN \nF2 = 10 kN \n\\Sigma M_B = 0 \n-A.y \\cdot (3 + a + 2) + 10 \\cdot (2.2) - 1.5 \\cdot (3 - 1) = 0 \nF_y = 7 + 20 - 6 = 0 \nA_y = 34/7 \nA_y = 4.8573 kN \n\\Sigma F_y = 0 \nA_y + F_a - F1 - F2 - F_by = 0 \n24.8573 - 10 - 3.5 - 1.5 + B_y = 0 \nB_y = 7.1429 kN \n\\Sigma V_c = 0 \nA_y - F_a - V_c = 0 \n24.8573 - 10 - V_c = 0 \nV_c = -5.1429 kN \n\\Sigma H_c = 0 \n-A_y - 5 + F_a + 2 + M_c = 0 \n-24.8573 - 5 + 10 + M_c = 0 \nM_c = 42.2855 kNm \nM_D = 10 kNm RU = 2191801\n w = 3 kN/m\n M = 10 kN·m\n ΣFy = 0 => Ay - 3.2 = 0\n Ay = 6 kN\n ΣFx = 0 => Ax = 0\n ΣMA = 0 + MA - 3.(2).(2/2) - 10 = 0\n MA - 3.2 - 10 = 0\n MA = -6 - 10 = 0\n MA = 16 kN/m\n ΣFy = 0 => Ay - Fy - v = 0\n hence, Fy = 3x\n 6 - 3x - v = 0\n v = 6 - 3x kN\n ΣM = 0 => -MA - Fy·x + Fx·x/2 + M = 0\n -16 - 6x + 3x²/2 + M = 0\n -16 - 6x + 3x²/2 + M = 0\n M = 3x²/2 + 6x + 16 kN RU = 2191801\n a = 11 x 15 = 165 mm\n b = 21° RU = 2191801\n D = 21 mm\n v = 0.3\n E = 39 MPa\n y = 3:1000 and y = 0.003\n G = Jc = 39 => 13000 MPa\n G = (13000 = E)\n 2(1.3 + 0.2)\n E = 33800 MPa\n A = π/4 d² => A = 314π\n 2 => A = 346185 MPa\n TE = ρA => 39 = P\n 346185\n P = 13501.245 N