Ejercicios Modelos Resueltos

5) METODO DE LAS DEFORMACIONES.\n\\[ \\text{...} \\]\n\\[ \\text{Final...} \\]\n\\[ Kab = \\frac{2 E_{11}}{L} = \\frac{27,1 E}{5} = 0,4E \\]\n\\[ Kbc = \\frac{2 E_{12}}{L} = \\frac{2,15 E}{6} = 0,5E \\]\n\\[ Kcd = \\frac{1,5 E_{13}}{L} - \\frac{1,5 E_{12}}{4} = 0,3E \\] 3) cálculo de las reacciones \\( R_b \\) y \\( R_c \\) ecuaciones de Nudos\n\\[ \\Sigma M_b = 0 \\]\n\\[ M_{ba} + M_{bc} + 2 R_b (k_{ab} k_{bc}) + R_c k_{bc} = 0 \\]\n\\[ 3,125 - 3 + 2 R_b (0,4 + 0,3) E + R_c 0,5 E = 0 \\]\n\\[ \\Sigma M_c = 0 \\]\n\\[ M_{cb} + M_{ce} + 2 R_c (k_{bc} k_{cd}) + R_b k_{bc} = 0 \\]\n\\[ 3 - 2,285 + 2 R_c (0,5,0,3) E + R_b 0,5 E = 0 \\]\n\\[ 1,8 R_b + 0,5 R_c = -0,125 \\]\n\\[ 0,5 R_b + 1,6 R_c = -0,715 \\]\n\\[ \\Delta_1 = \\begin{bmatrix} 1,8 & 0,5 \\\\ 0,5 & 1,6 \\end{bmatrix} \\begin{bmatrix} R_b \\\\ R_c \\end{bmatrix} = \\begin{bmatrix} -0,125 \\\\ -0,715 \\end{bmatrix} \\]\n\\[ 2,63 R = 1,6 \\]\n\\[ R_b = 0,594896 \\]\n\\[ R_c = 0,46559 \\] 4) cálculo de los momentos - Ecuaciones de fuerzas\n\\[ M_{ab} = M_{ba} + 2 k_{ab} R_a - k_{ab} R_b \\]\n\\[ M_{ab} = -3,101 \\]\n\\[ -3,125 + 0 + 0,4 E (0,594896) = -3 \\]\n\\[ M_{ab} = 3,173 \\]\n\\[ M_{bc} = M_{cb} + 2 k_{bc} R_b + k_{bc} R_c \\]\n\\[ = -3 \\]\n\\[ M_{bc} = 2,1015 \\]\n\\[ M_{cd} = Q_cD + M_c + M_a - 2,364 \\]\n\\[ Q_{cd} = Q_{ab} + M_c + M_D = 3,256 + 25.64 \\]\n\\[ Q_{dc} = -0,484 \\]\n\\[ Q = 2,494 m \\] MAB = 3.101 + 3.7356 - 1.5x2 \n\nMmax (tramo AB) = 1.55057 \n\nx = 2.4904m \n\ntramo DC \n\nmomento en lame activa P1:\n\nM = 2.864 + 2.516 * 1.5 = 1.21 \n\nE = cE \n\nN = 5 \n\nH6 = 1 (k2) \n\nD) Estimación de los coeficientes de rigidez, K y k^2\n\nK12 = K21 = 2E1 = 2 * 1.5E = 0.75E\n\nK22 = 1.5E\n\nK24 = k42 = 2E2 = 2.41E = 0.5E Moa = Mb = - PL / 8 \n\nM12 = (3x4) / 8 = 1.5 \n\nM21 = PL / 3 = 3x4 / 8 = 1.5\n\nMo = M2.35 = 0 \n\nMoa = - Mb / 12 \n\nM8 = -0.625 \n\nM24 = 0.375 \n\nM2 = 0.46875 \n\nE = [t/m2] M24 = -0.625 + 2.0 * k42 - (-0.46875 / E) \n\nM42 = -0.853375 \n\nSe realiza de M2 = 0 = Ma1 + Ma2 + Ma4 \n\nQ21 = Q12 + Q21 = P1 + P2 = 0.74 \n\nN2 = N21 - N23 = 0 \n\n\n\\Delta = \\begin{bmatrix} 1 & 1 & 1 \\\\ 0 & 1 & -3 \\\\ 0 & 1 -3 \end{bmatrix} \n\n\\Delta 2 = 0.74 N21 = 0,32\nN23 = 0,42\n( > rigidez )\nQ\n1,76\n\n\n0,23\n\n2\n\n1,24\n\nN\n3\n\n0,74\n\nM\n1,85\n0,8\nM = -1,85 + 1,76 * 2 = 1,67\nM = -0,8 * 0,74 = 1,36\n1,67\n1,36\n0,86\n\n0,32\n1\n\n1,47\n\n3,47\n\nZ.H = 0,32+0,42-0,74 = 0\nZ.V = 3 + 2 - 0,23 - 1,76 - 3,47 = 0\nZ.MY = -1,85 = 3,2 + 2,3 + 0,23 + 1,36\n-3,17 + 4 + 0,74 * 4 = 0\n\nDiagrama de cuerpo libre Estructura e muros de despliegles\n1) Determinacion de los coeficientes de rigidez\nK12 = k04 = 2 E I1 = 2 / 3\nK23 = 2 E I2 = 2 . 2.21 E / 5\n\n2) Calculo de los momentos de impulsamiento perfecto M0b\nM12 = M21 = M32 = 0\nM0e32 = 9.12 - 4*9.2 = -8. 3.\n12\nM0b = 4.5 * 2 = 8.3\n3) Ecuaciones de Nudos\nNudo 2\n-8,3 * P2 + 2 L (P2 - A5) E1 + P3 * E1 + 3 * 2/5 E1 = 0\nNudo 3\n8,3 * P4 + 2 (4/5) E1 + P2 * 4/5 E1 + 3 * 2/3 P4 E1 = 0\n1) 2,93 * P2 E1 + 0,8 P3 E1 + 2 * P4 E1 = 8,3\n2) 0,8 * P4 E1 + 2,93 * P3 E1 + 2 * P4 E1 = -8,3\nSigo plantando: \nF = 0\n∑ Pn h n + • + 0,\n\n3) (P-!\nF) = && p =\n\\ S p E / p\n/h = = 0 P.n h + 0 + 0,1 3 \n(2 P K12 + P2K34) + 0 + G (U12 K12 + U13 K13) =\n2 x 3 + 3 (K12 . 2 / 3 E1) + G (U12 2 / 5 E1 + U13 3 / 5 E1) = 0\n\n2 U2E1 + 2 U3E1 + 8 UpE1 = 0\n∆ = 0,8 2,93 = \n=n°I1 = 20,848\nU1 = 3\n ◯ 2\n0,8 2,93 0,8 2,93 = 8,3\n\nZ2 = 4,755 EI1\nE = 2*10^6/m²\nI1 = 0,001 m⁴\nU2 = 7'40\"\nU3 = 5'46\"\nYp = 1'46\"\n\n∆ ∝ Yp h : ∆ = 0,15 cm Mab = m e cb + 2 e a Kab + 4 b k ob + 3 y ab k ab + 2 y ab k ' ob\nMz = 0 + 0 + + l2 k12 = 3 y p k12\nM12 = 4,455.2\n3 + 3(-1.024).2/3 = M12 = 0.922\n\nM21 = 0 + 2 l2 k12 + 0 + 3 y p k12\nM21 = 2.4 455.3 / 2 = 3(-1.024).3/2 = M21 = 3.892\n\nM23 = -8.5 + 2 4.455.1/1 + -3.957.1 = M23 = 3.891\n\nM32 = 3.5 + 2(-3.357).5 + 4.455.4/5 M32 = 6.526\n\nM24 = 0 + 2(-3.357).2/3 + 3(1.024).2/3 M34 = -6.324\n\nM43 = 0 + 0 + 3(-3.351).2/3 = M43 = 4.286\n\nVerifucabion\nΣ Ph = Σ Ao 5\nP = Σ Ms.12 + Ms.6\n\n2 + 0 + 3.894 + 0.42 - 6.52 - 4.29 = 0 / 3\n\n1 = -3.89 9 = 4.5% Qob = q10 - Mob + Mba\nQba = a B M o = 3.894, 0.92/3\nQ12 = 0 - 3.894 / 3 = 1.6\nQ21 = 1.16\nQ22 = 10 - (-3.894 + 6.52)\nQ22 = 9.474\nQ32 = 10 - (-3.894 + 6.52) / 5\nQ32 = 10.536\nQ34 = 0 - (-6.52 + 4.29 / 3)\nQ34 = 3.6\nQ43 = 3.6 9.474\n10.526\n(M12 = 0.92 \nM21 = 3.894\nM23 = 3.891\nM32 = 6.526\nM24 = 6.52\nM43 = 4.29\n\nM23 = 3.894 + 9.474x - 4x² - ... M33 = ... x² - 4.737x + 1.945\n\ndM23 = 9.474 - 2* x = 0 \nX = 2.3565\n\nx1 = 4.228 = (7.737(9.736) - 4.1945)/ 2\n\nx1,2 = 4.228\nx2 = 0.45\nMmax(b) = 7.733\n\ni1,2 = 0.92, 1.6 x\nM1,2 = 0 \nM2,3 = -4.29 + 3.6x M43 = 0 = x = 1.19 m\nN = -Q23\nN21 - 9.474\nN21 - 9.474 / 1\n\nQ22 = -9.474 = Q31 = 1.6 = 3.6\n\nN2 = 0 = Nm21 - 3.6\nN3 = 4.26\n q = 5k/m²\nE = dte\n\n1) Determinacion de los coeficientes de rigidez\nk12 = k23 = 1.35 E I / L - 1.5 E /10\nk25 = 2.1 E / 8 = 0.25\n\nK = 25 = 1.35*E / 6 = 0.2 \nK = 25 0.234 = 0.234\nM21 = -0.0042 \nNo. 25 = 0.234\nm² = 4.227 = (6 - 1.944)\n... 2 ... (m1 = 0.5/3 -> 1) - Omoh (r3,4), m2 = 0.48.\n2.9/m\n(1/s) 2 = 4/09s = 4 = 2 = 0.3.5.\nEqtn de Nu \nN 2\nM14 + M33 + M23 + 2 l( k18 + k29) + 3 y K = 0\n62.5 - 62.5 - 0.234/160 + 2p( 0.32, 0.3, 0.125)\nI + 0 . I j4 = 0.624.\n 0,375 .8 + (0,2341 .0,984) + 3(42 - 0,25 + 6(42 5 - 0,25) = 0\n0,75(42 - 1,5(42 5 = -4,218\n\nΔ =\n1,7 0,75\nΔ1 = 1,9875 Δ1 = -0,234 0,75\n0,75 1,5\nΔ2 = 1,7 -0,234 = -6,981\n\n0,75 -4,218\nU2 = 1,4151\nY25 = -3,52\n\nM12 = 63,5 = 2(1,4151) .0,3 = 63,349\nM23 = 62,5 = 2(1,4151) .0,3 = -61,651\nM32 = 0,2341 2(1,4151) 0,25 + 3(-3,52) -0,25 = -1,698\nM32 = 0,9841 (1,4151) 0,25 + 3(-3,52) -0,95 = -1,3\n\nVerification\nΣH = 0 = 5/8 + 0 - 1,698 -1,3 = 0\n\nQab = Qbc - Mab Mbc\nQba = Qbc + Mbc Mba\nQ12 = 5,10 0 63,349 = 18,665\nQ21 = 5,10 2 63,349 10 = 31,335\nQ23 = 5,10 - 28,651 + 0 = 31,165\nQ32 = 5,10 2 51,651 10 = 18,835\nQ25 = 3 8 - 2 8 = 0\nQ52 = 0 ΣV = 18,665 18,825 + 65,5 - 5,20 - 3 = 0\nΣH = 0\nΣM1 = 5,20 10 - 18,835 20 + 3(11) + 65,5 10 + 1,3 = 0\n\nM12 = 18,665 5 x 2\n= 63,35\n\ndx1 = 18,665 - 5 x\nx = 3,733m\nMmax(t) = 341,828\nx = 3,173\n\nM22 = -18,835 5 x^2 2\n\nMmax(1) = 35,176\n\nQ12 = 18,835 P = 9,36k/m\nK12 = 2E11 0,5EJ\nK23 = 1,55E11 3 = 1,5E11/3\n\nM12 = M21 = 0\nM13 = -3,375\n\nEquations Node 2\n\n-3,375 - 3Q2 + 1,1542(0,5) + 3Q4,2 = 0 - 2(√2) K25 = 0\n\nEquations of displacements\nΣA = 0\nΣ Pn = P = 5\n\n5 .4 + 3Q2 0,5 + 6Q4,12 0,5 = 0\n20 * 1,5Q2 3Q4,2 = 0\n\nΔ = 3 1,5 = 6,75\nΔ1 = 3,375 1,5\n= 40,125\n\nΔ2 = 3 3,375 - 65,0625\n= -9,638 /E1\n\n= M21 = 2(5,9) + 0,5 + 31-9,638) 0,5 = 8,514\n(S23 = -3,375 + 2(5,94)- 1 = 8,514\n\nΣH = 5 + -11,486 -8,514 = 0\n\n\nQ12 = -5\nQ21 = 5 Q23 = -1,662\nQ32 = 7,338\n8.514\n3\n2\n45\n45\n\nΣE V = 9 - 7,338 - 1,662 = 0\nΣE H = 5 - 5 = 0\nΣM2 = 7,338 * 3 + 9,15\n- 11,486 - 5,4 = 0\n\nM12 = 11,486\nM21 = 8,514\nM23 = 8,514\n\nM23 = 8,514 = 3x² + 1,662. x\nQ23 = 1,662 - 3x\nx = 0,554m\nM max = 8,974