Análise das Estruturas - Tabela de Engastamento Perfeito - Método dos Deslocamentos

Esfuerzo de Engastamiento Perfecto \\ \sum 0 \left(\frac{ -p^2 l^2}{8 E_I}\right) = \frac{p^3}{6E_I} \\ \int_0^{1} = \frac{1}{3 E_I} \left(-\frac{p^3}{3 E_I} \right ) = \\ \int_0^{1} = \frac{1}{3 E_I} \left( - \frac{p^4}{8 E_I}\right) = \\ = \frac{p^2l^3}{12E_I} \\ i_{i} = \frac{p^2l^2}{6E_I} \cdots ƩFy = \frac{p l}{2} + R_o - p l = 0 \,.\,\,\, R_o = \frac{p l}{2} \\ ƩM_A = \frac{p l^2}{12} + m_o + \frac{p l}{2} - p \cdot \frac{l}{2} = 0 \,.\,\,\, m_B = \frac{-p l^2}{12} \\ \frac{p^2}{12} \text{(image of a beam with concentrated load)} \\ \frac{p l^2}{12} \uparrow \\ \frac{p l^2}{12} \downarrow \\ \\ 2 \\ m_c \\ \text{(image of a triangular force diagram)} \\ T_x = 1 \\ S.P. \\ E.0 \\ M(x) = -\frac{p l}{l} x - \frac{l}{3} \cdot x = -\frac{p x^3}{6 I} \\ M(l) = -\frac{p l^2}{6} \\ E.1 \\ M(x) = x \\ E.2 \\ M(x) = -1 \\ \int_0^{l} E_I^{-1} \\ \left[-\frac{p x^3}{6 I} \right] \\ dx = \frac{1}{E_I} \left[ \left( \frac{-\left(\frac{p}{6 I}\right)}{0}\right) - \left(\frac{-p}{6 E_I} \cdot l^4\right)\right]_{0}\left[0 - 6 E_I \cdot 0\right] \\ = -\frac{p^4}{30 E_I} \cdots \left\{ x_1 = \frac{1}{30 E_I} p\right) \\ x_2 = x_1 = 0 \\ = \left\lbrace \begin{matrix} = \;\; \frac{p^2}{15} - \frac{p^2}{30} \\ = -\frac{l^2}{24} \end{matrix}\right. \\ 1 - \frac{1}{6}l = \frac{-p^2}{40} \\ x_1 = \frac{3pl}{20} \\ x_2 = \frac{p^2}{30} \\ \sum F_y = \frac{3 p l}{20} + R_o - \frac{p^2}{15} = 0 \\ R_o = -\frac{p^2}{20} \\ \left( \text{(image of a beam with loads)} \\ \\ M = \text{(triangular)} \\ P \cdots \text{(rest of the calculations)} s0x = -\\rho l^4 \n30 E1\n\ns20 = -\\rho l^3 \n24 E1\n\ns11 = \\rho l^2 \n3 E1\n\ns21 = -\\rho l^2 \n2 E1\n\ns22 = l \nE1\n\n\\int\\int\\int \\rho^k \n= 0 \n= \\left \\{ \\begin{array}{l} \n\\frac{l}{3} x_1 + x_2 = 0 \\\\\n\\frac{l^2}{30} (x2) \\\\\n\\frac{l^2}{24} (-x1) + x1 + x2 = 0 \n\\end{array} \\right. \n\n= \\left \\{ \\begin{array}{l} \n1 ^3 x_1 + x_2 = \\frac{\\rho l^2}{15} - \\frac{2}{3} l \\cdot \\frac{3}{20} \\rho l\\\\n\n\\frac{1}{2} l x_1 - x2 = \\frac{\\rho l^2}{24} \\\\ \n\\end{array} \\right. \n\nx2 = -\\frac{\\rho l^2}{30}\n\nx1 = \\frac{3}{20} \\rho l\n\n\\sum F_y = \\frac{3}{20} \\rho l + R_A - \\rho l = 0 \n\\end{array} \nR_A = \\frac{7}{20} \\rho l \n\\sum M_A = M_A + \\frac{3}{20}\\rho l - \\rho l - \\frac{1}{5} = 0 \nM_A = \\frac{\\rho l^2}{20} s0 = \\frac{1}{E1} \\left( \\frac{3}{2} \\frac{\\rho l^2}{0} \\cdot (1 + 2) \\right ) \\] = \\frac{1}{E1} \\left[ -\\frac{\\rho^2 l^2}{24} \\left( \\frac{1}{2} + 2 \\right) \\right ] = - \\frac{5}{48} \\rho^3 \n\ns11 = \\frac{3}{3E1} \n\ns22 = -\\frac{l^2}{2E1} \n\ns20 = \\frac{1}{E1 \\left[ - \\frac{1}{2} \\frac{\\rho l^2}{0.2} ( -1 ) \\right ] = \\frac{\\rho l^2}{8E1} \n\ns22 = \\frac{l}{E1} \n\n\\int\\int\\int = - \\frac{5}{\\rho^3} \n\\=[ \\frac{1}{E1} \\left( \\frac{l}{3} x_1 – \\frac{l^2}{E1} \right) + x_2 = 0 \n\\frac{-l}{3E1} x_1 + x_2 = 0 \n\\sum M_A = \\frac{\\rho^2}{8} + \\frac{M_A}{l} - \\rho l - \\frac{1}{2} = 0 \n\\sum M_R = \\rho^l + R_B - \\frac{\\rho^l}{2} = 0 s0 = \\frac{2 l^2}{\\rho^2 8} \n\\sum M_A = \\sum = R_A + \\frac{l}{2 !} : \\sum_{i=1}^{b} R_E R_1 - P B = P \\cdot R_8 +\\sum = P \\cdot l\n\nR_A = - \\frac{1}{2} \\cdots \\cdots + b : \\cdots - \\cdots \nR = \\rho P \\cdots \\frac{l^2}{2} \nR_B = P + \\frac{l^2}{l}\\] = \\frac{\\rho B}{l = \\frac{1}{3E1}}\n\\int = \\frac{R_M R_A}{10} + r_4 P_B \n=( \\rho^2 \\cdots - b ) \n= P \\sum = F_E = \n{ \n 2 l x_1 - l x_2 = P b c (1 + b / l) \n -1 / 2 l x_1 + l x_2 = -P b c 2 (1 + a / l) \n} \n\n 3 / 2 l x_1 = P b a + P b c_2 / l - P b c^2 / 2 l \n 3 / 2 l x_1 = P b c_2 / 2 + P b^2 c_1 / 2 l \n 3 / 2 l x_1 = P b c (1/2 + b_1 - a / 2 l) \n 3 / 2 l x_1 = P b c (1/2 + 1 - a / 2 l) \n x_1 = 2 P b c / 3 l (3/2 - 3 a / l) \n x_1 = P b c / l (1 - a / l) \n x_1 = P b c / l (l - a) = P b a / l \n\n2 l (P a b)^2 / l^2 - l x_2 = P b c + P a / l^2 \n l x_2 = 2 P a b / l - P a - P b^2 / l \n l x_2 = P a b^2 / l \n l x_2 - P a b / l = P a (b / l - 1) \n x_2 = P a b / l (b - l) = -P a b^2 / l^2 \n\n x_2 = -P a b^2 / l^2 Σ M_B = P a b^2 / l^2 - R_B l + P b - P a b^2 / l^2 = 0 \n MA - R_B l + P b + mb = 0 \n RA = P b + MA + mb / l \n ∑F_y R_A + R_B - P = 0 \n. . . R_B = P - R_A \n Σ M_A = MA = P b^2 / l^2 \n MA + mb = r_A = RA \n- P a b^2 / 6l + P =P b => \n1 / 3 E j J_0 = Ta \cdot \underline{eF_0} \nE / I = -P \cdot J \cdot x_0 = 0 \n= -P b^2 / 6 \n ⇒ x_1 = \lambda I^E = -P I / 6 + m_G \n \nF_1 = = P/3 \cdot I / I + e^{-t_} = I_T \cdot J = l/g m_0 = γ_{E \O \L} \Rightarrow \dots \sum_{\text{r}} i = S + \sum_{E} p_0 ∫ {/e^{ \bar{x}}- / = \frac{ 1-e^4}{6 E^3}\Rightarrow I_J = -\frac{ m}{6m} \} \rightarrow \sum M_A = P_1 = R = 0 \cdots \to M_A = \frac{2ae^2}{3a^2 - m}.dx = α_p \Rightarrow Σ E = \int kij \ right / m_0 \,dx = \right \sqrt{X.B}\cdots \Rightarrow −P^2/I^2 \sum = t^3 - \, x -\\rho x^3 \\over 8EJ + {l \\over EJ} x_1 = 0\nx_1 = {\\rho l^2 \\over 8}\n\\sum M_A: M_A - \\rho l x_1 {x^2 \\over 3} - \\rho x {p \\over 2} = 0 \n\\Rightarrow M_A = {5 \\over 24} \\rho l^2\n\\text{(72\\text{ Marine})}\n\\ f(x) = \\rho p\n\\ p - p_1 + p_2 = {px \\over l}\nE_0 \nM(x) = -\\left(\\rho - {p \\over l}\\right) x^2 {l \\over 3} - p_x {x^2 \\over 2} - \\rho {x^3 \\over 6} + {\\rho x^2 \\over 2} - {p_x^3 \\over 6l}\nE_J: M(x) = 1 \nS_{10} = {1 \\over E} \\left[-\\left( -{\\rho x^2 \\over 6} + {\\rho x^3 \\over 24} \\right) \\right] \n= -{\\rho l^2 \\over 8E} \nJ_{11} = {1 \\over E}\n-\\rho {x^3 \\over 8EJ} + {l \\over EJ} x_1 = 0 \\Rightarrow x_1 = {\\rho l^2 \\over 8} \nM_A = {5 \\rho l^2 \\over 24}\nR_A = {\\rho l \\over 2}\n\\textbf{(5P \\over 24)}\n\\textbf{g \\rho l^2 \\over 2}