Documento do Scannable em 31 de Mai de 2019 18_10_20

Estabilidade das construções 1 - Montoria\n\nQuestão: Determine as matrizes K^G e R^G da treliça abaixo:\n\nP\n\n30°\n\nEA=1\n\nA1\n\nE1A1\n\n1,90m\n\nE2A2\n\n2,90m\n\n2,10m\n\nE2A2\n\ns\t\n\nNGL=3\n\nG\n\nF1= P.sen30°\n\nF2= P.cos30°\n\nF3= 0\n\ncónd. de eq. (1)\n\n+ (torção do K)\n\nL1 = √(1,90² + 2,00²) = 2,91m\n\nθ1 = arctg(1,90/2,00) = 40,72°\n\ns = sen40,72° = 0,65\n\nc = cos40,72° = 0,76\n\ns² = 0,43\n\nc² = 0,58\n\ncs = 0,48 K^(1) = EA\n\n\t\t\t\t [ 0,49 0,49 -0,57 -0,49 ]\n\t\t\t\t [ 0,49 0,49 -0,49 -0,49 ]\n\t\t\t\t [ -0,57 -0,49 0,57 0,49 ]\n\t\t\t\t [ -0,49 -0,49 0,49 0,42 ]\n\nR^(1) = [ 0,72 0,65 0 0 ]\n\t\t\t\t [ -0,65 0,76 0 0 ]\n\t\t\t\t [ 0 0 -0,65 0,76 ]\n\t\t\t\t [ 0 0 0 0 ]\n\nBanda 2:\n\nL2 = 2m\nθ2 = 0°\n\nc = cos0° = 1\ns = sen0° = 0\n\nc² = 1\ns² = 0\n\nc5 = 0 Banda 3:\n\nL3 = 4,13m\n\nc = cos0° = 1\ns = sen0° = 0\n\nθ3 = 0°\n\nK^(3) = EA\n\n\t\t\t\t [ 1 0 -1 0 ]\n\t\t\t\t [ 0 0 0 0 ]\n\t\t\t\t [ -1 0 1 0 ]\n\t\t\t\t [ 0 0 0 1 ]\n\nR^(3) = [ 1 0 0 0 ]\n\t\t\t\t [ 0 0 0 0 ]\n\t\t\t\t [ 0 0 0 0 ]\n\t\t\t\t [ 0 0 0 0 ] Atenção - Estabilização das construções \n\nQuestão: Determine os esforços da treliça pelo método dos deslocamentos.\n\nAç.: E = 9,1 . 10⁷ t/m³ \n\nelmento \n1 \n2°cm⁴ \n3°cm⁴ \n4°cm² \n5°cm² \n\nSolução:\nDevido à simetria, temos:\nα = arctan(2/2) = 45°, β = arctan(3/2) = 53,13°\n\nObs.: qualquer barra de treliça entre apoios com esforço = zero.\n\nportanto k₂ = k₁, logo deve-se usar k₁, k₀ e k₃. NGL = 3 \n\ncondição de equilíbrio: FₐG = -st \nFₐG = 0 \nF₃G = 0 \n Cálculo do K²:\nElemento 1:\nΘ = 0°\nL₁ = 2 + 1,5 = 3,5m\n\nk⁽²⁾ = 3.10\n\nR₁ = [\n 0 0 0 \n 0 0 0 \n 0 0 0 \n]\n\nElemento 2:\nΘ = 0°\nL₂ = 4,5m\n\nk⁽³⁾ = 5,6.10\n\nR₂ = [\n 1 0 0 \n 0 1 0 \n 0 0 1 \n]\n\nElemento 3:\n2Y\n\nΘ = 360° - α = 360° - 45° = 315°\n\nL₃ = √(2² + 3²) = √13m\n\nE₃ = 9,1.10.30.10⁴\n\nk⁽³⁾ = [\n 0.5 -0.5 0.5 \n -0.5 0.5 -0.5 \n 0.5 -0.5 0.5 \n]\n Elemento 5:\nΘ₅ = β = 53,13°\n\nL₅ = √(2² + 1,5²) = 95m\n\nE₅ = 2,1.10⁷.30.10⁴ / 95\n\nk⁽⁵⁾ = 2,52.10⁴\n\nR₅ = [\n 0.96 0.4 0 \n 0 0.5 0 \n 0 0 0.5 \n]\n\nk₁₁ = k₄₄ + k₄₄ = 0 + 0,64.2,52.10⁴ = 16,928\n\nk₁₂ = k₄₁ + k₄₄ = -0,48.2,52.10⁴ = -12096\n\nk₁₃ = k₁₂ + k₂ = -1648.052.10⁴ = -16128\n\nk₁ = [\n -1628 -12096 -16187 \n -12096 760.207,93 9570,7 \n -16187 9570,7 2764,93 \n]\n\nsolve da equação da equilíbrio; F = k.k⁻¹ f^{(1)}_{xy} = k_{xy} d_{xy} = (1\\;0\\;-1)\\left[ \\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \\end{array} \\right] = \\begin{bmatrix} 0 \\ 0 \\ 0 \\ \\end{bmatrix} + 1,032.10^{-3} \n f^{(1)}_{xy} = R^{T}_{1} f^{(1)}_{xy} \\left[ \\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \\end{array} \\right] = \\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \\end{bmatrix}\n f^{(1)}_{xy} = k^{(4)}_{xy} d_{xy} = 596.4 \\begin{array}{ccc} 1 & -1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \\end{array} = \\begin{bmatrix} -4.563.10^{-4} & 0 \\ -6.052.10^{-4} & -1.039.10^{-3} \\ \\end{bmatrix} \n f^{(2)}_{xy} = k^{(4)}_{R} f^{(1)}_{xy} = \\left[ 1 \\ 0 \\ 0 \\ \\end{bmatrix} \\begin{bmatrix} 7 & -7.5 & 7.5 \\ \\end{array} \\right] \n f^{(3)}_{xy} = R^{T}_{3} f^{(3)}_{xy} = \\left[ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \\end{array} \\right] \\begin{bmatrix} 0.7071 & -0.7071 \\ 0.7071 & 0.7071 \\ \\end{bmatrix} \\left[ 0.5 & -0.5 \\ 0.5 & -0.5 \\ \\end{array} \\right] \n f^{(5)}_{xy} = R^{(5)} f^{(5)}_{xy} = \\begin{bmatrix} 0.6 & 0.8 \\ 0.8 & 0.6 \\ \\end{array} \\right] \\begin{bmatrix} 7.745 \\ 4.923 \\ -3.945 \\ 0 \\ -7.45 \\ 6.24 \\ -6.24 \\end{bmatrix} = \\begin{bmatrix} \\end{bmatrix}