{"id":3713,"date":"2019-07-03T07:11:30","date_gmt":"2019-07-03T12:11:30","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=3713"},"modified":"2026-02-26T10:36:16","modified_gmt":"2026-02-26T15:36:16","slug":"s1eva2019ti_t2-catenaria-cable","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s1eva20\/s1eva2019ti_t2-catenaria-cable\/","title":{"rendered":"s1Eva2019TI_T2 Catenaria cable"},"content":{"rendered":"\n

Ejercicio<\/strong>: 1Eva2019TI_T2 Catenaria cable<\/a><\/p>\n\n\n\n

\"Cable<\/figure>\n\n\n\n

Las f\u00f3rmulas con las que se requiere trabajar son:<\/p>\n\n\n y = \\frac{T_A}{w} cosh \\Big( \\frac{w}{T_A}x \\Big) + y_0 - \\frac{T_A}{w}<\/span>\n\n\n\n

\"Cable<\/figure>\n\n\n\n

Donde la altura y del cable est\u00e1 en funci\u00f3n de la distancia x.<\/p>\n\n\n\n

Adem\u00e1s se tiene que:<\/p>\n\n\n cosh(z) = \\frac{e^z+ e^{-z}}{2}<\/span>\n\n\n\n

que sustituyendo la segunda en la primera se convierte en:<\/p>\n\n\n y = \\frac{T_A}{w} \\frac{e^{\\frac{w}{T_A}x}+ e^{-\\frac{w}{T_A}x}}{2} + y_0 - \\frac{T_A}{w}<\/span>\n\n\n\n

y usando los valores del enunciado w=12, y0<\/sub>=6 , y=15, x=50 se convierte en:<\/p>\n\n\n 15 = \\frac{T_A}{12} \\frac{e^{\\frac{12}{T_A}50}+ e^{-\\frac{12}{T_A}50}}{2} + 6 - \\frac{T_A}{12}<\/span>\n\n\n\n

simplificando, para usar el m\u00e9todo de b\u00fasqueda de ra\u00edces:<\/p>\n\n\n \\frac{1}{2}\\frac{T_A}{12} e^{\\frac{12}{T_A}50} + \\frac{1}{2}\\frac{T_A}{12} e^{-\\frac{12}{T_A}50} - \\frac{T_A}{12} - 9 = 0<\/span>\n\n\n\n

cambiando la variable \\frac{12}{T_A}=x <\/span><\/p>\n\n\n \\frac{1}{2x} e^{50x} + \\frac{1}{2x} e^{-50x} - \\frac{1}{x}-9=0 <\/span>\n\n\n\n

la funci\u00f3n a usar para la b\u00fasqueda de ra\u00edces es:<\/p>\n\n\n f(x)=\\frac{1}{2x} e^{50x} + \\frac{1}{2x} e^{-50x} - \\frac{1}{x}-9 <\/span>\n\n\n\n

Para el m\u00e9todo de Newton-Raphson se tiene que:<\/p>\n\n\n x_{i+1} = x_i -\\frac{f(x_i)}{f'(x_i)} <\/span>\n\n\n\n

por lo que se determina:<\/p>\n\n\n f'(x)= - \\frac{1}{2x^2}e^{50x} + \\frac{1}{2x}(50) e^{50x} +<\/span>\n\n\n- \\frac{1}{2x^2} e^{-50x} + \\frac{1}{2x}(-50)e^{-50x} + \\frac{1}{x^2} <\/span>\n\n\n f'(x)= -\\frac{1}{2x^2}[e^{50x}+e^{-50x}] +<\/span>\n\n\n + \\frac{25}{x}[e^{50x}-e^{-50x}] +\\frac{1}{x^2} <\/span>\n\n\n f'(x)= \\Big[\\frac{25}{x} -\\frac{1}{2x^2}\\Big]\\Big[e^{50x}+e^{-50x}\\Big] +\\frac{1}{x^2}<\/span>\n\n\n\n

Con lo que se puede iniciar las iteraciones.<\/p>\n\n\n\n

Por no disponer de valor inicial para TA<\/sub>, considere que el cable colgado no deber\u00eda tener tensi\u00f3n TA<\/sub>=0 N, pues en la forma x=12\/TA<\/sub> se crea una indeterminaci\u00f3n. Si no dispone de alg\u00fan criterio para seleccionar el valor de TA<\/sub> puede iniciar un valor positivo, por ejemplo 120 con lo que el valor de x0<\/sub>=12\/120=0.1<\/p>\n\n\n\n

Iteraci\u00f3n 1<\/h4>\n\n\n f(0.1)=\\frac{1}{2(0.1)} e^{50(0.1)} + \\frac{1}{2(0.1)} e^{-50(0.1)} <\/span>\n\n\n- \\frac{1}{0.1}-9 =723.0994 <\/span>\n\n\n f'(0.1)=\\Big[\\frac{25}{0.1} - \\frac{1}{2(0.1)^2}\\Big]\\Big[e^{50(0.1)}+e^{-50(0.1)}\\Big] +<\/span>\n\n\n+\\frac{1}{(0.1)^2} = 29780.61043 <\/span>\n\n\n x_{1} = 0.1 -\\frac{723.0994}{29780.61043} = 0.07571<\/span>\n\n\n\n

error = | x1<\/sub> - x0<\/sub>| = | 0.07571 - 0.1| = 0.02428<\/p>\n\n\n\n

Iteraci\u00f3n 2<\/h4>\n\n\n f(0.07571)=\\frac{1}{2(0.07571)} e^{50(0.07571)}+<\/span>\n\n\n + \\frac{1}{2(0.07571)} e^{-50(0.07571)} <\/span>\n\n\n - \\frac{1}{0.07571}-9 = 269.0042 <\/span>\n\n\n f'(0.07571)= \\Big[\\frac{25}{0.07571} -\\frac{1}{2(0.07571)^2}\\Big]. <\/span>\n\n\n.\\Big[e^{50(0.07571)}+e^{-50(0.07571)}\\Big] + <\/span>\n\n\n +\\frac{1}{(0.07571)^2} = 10874.0462<\/span>\n\n\n x_{2} = 0.07571 -\\frac{269.0042}{10874.0462} = 0.05098 <\/span>\n\n\n\n

error = | x2<\/sub> - x1<\/sub>| = |0.05098- 0.02428| = 0.02473<\/p>\n\n\n\n

Iteraci\u00f3n 3<\/h4>\n\n\n f(0.05098) = 97.6345 <\/span>\n\n\n f'(0.05098) = 4144.1544 <\/span>\n\n\n x_{3} = 0.0274 <\/span>\n\n\n\n

error = | x3<\/sub> - x2<\/sub>| = |0.05098- 0.0274| = 0.0236<\/p>\n\n\n\n

finalmente despu\u00e9s de varias iteraciones, la ra\u00edz se encuentra en: 0.007124346154337298<\/p>\n\n\n\n

que convitiendo<\/p>\n\n\n T_A = \\frac{12}{x} = \\frac{12}{0.0071243461}<\/span>\n\n\n= 1684.36 N<\/span>\n\n\n\n

\n\n\n\n

Revisi\u00f3n de resultados<\/h4>\n\n\n\n

Usando como base los algoritmos desarrollados en clase:<\/p>\n\n\n\n

['xi', 'xnuevo', 'tramo']\n[0.1    0.0757 0.0243]\n[0.0757 0.051  0.0247]\n[0.051  0.0274 0.0236]\n[0.0274 0.0111 0.0163]\n[0.0111 0.0072 0.0039]\n[7.2176e-03 7.1244e-03 9.3199e-05]\n[7.1244e-03 7.1243e-03 3.8351e-08]\nraiz en:  0.007124346154337298\nTA = 12\/x =  1684.365096815854<\/code><\/pre>\n\n\n\n
\"Cable<\/figure>\n\n\n\n
Algoritmos Python usando el procedimiento de:<\/p>\n\n\n\n
Newton-Raphson<\/p>\n\n\n
\n# 1Eva_IT2019_T2 Catenaria cable\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# INGRESO\na = 0.001\nb = 0.1\nmuestras = 51\n\nx0 = 0.1\ntolera = 0.00001\n\nfx = lambda x: 0.5*(1\/x)*np.exp(50*x) + 0.5*(1\/x)*np.exp(-50*x)-1\/x -9\ndfx = lambda x: -0.5*(1\/(x**2))*(np.exp(50*x)+np.exp(-50*x)) + (25\/x)*(np.exp(50*x)-np.exp(-50*x)) + 1\/(x**2)\n\n# PROCEDIMIENTO\ntabla = []\ntramo = abs(2*tolera)\nxi = x0\nwhile (tramo>=tolera):\n    xnuevo = xi - fx(xi)\/dfx(xi)\n    tramo = abs(xnuevo-xi)\n    tabla.append([xi,xnuevo,tramo])\n    xi = xnuevo\n\ntabla = np.array(tabla)\nn=len(tabla)\n\nTA = 12\/xnuevo\n\n# para la gr\u00e1fica\nxp = np.linspace(a,b,muestras)\nfp = fx(xp)\n\n# SALIDA\nprint(['xi', 'xnuevo', 'tramo'])\nnp.set_printoptions(precision = 4)\nfor i in range(0,n,1):\n    print(tabla[i])\nprint('raiz en: ', xi)\nprint('TA = 12\/x = ', TA)\n\n# Grafica\nplt.plot(xp,fp)\nplt.xlabel('x=12\/TA')\nplt.ylabel('f(x)')\nplt.axhline(0, color = 'green')\nplt.grid()\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 1Eva2019TI_T2 Catenaria cable Las f\u00f3rmulas con las que se requiere trabajar son: Donde la altura y del cable est\u00e1 en funci\u00f3n de la distancia x. Adem\u00e1s se tiene que: que sustituyendo la segunda en la primera se convierte en: y usando los valores del enunciado w=12, y0=6 , y=15, x=50 se convierte en: simplificando, […]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn-ejemplo","format":"standard","meta":{"footnotes":""},"categories":[45],"tags":[58,54],"class_list":["post-3713","post","type-post","status-publish","format-standard","hentry","category-mn-s1eva20","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/3713","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=3713"}],"version-history":[{"count":4,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/3713\/revisions"}],"predecessor-version":[{"id":21793,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/3713\/revisions\/21793"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=3713"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=3713"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=3713"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}