{"id":1983,"date":"2017-11-15T07:00:29","date_gmt":"2017-11-15T12:00:29","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=1983"},"modified":"2026-04-05T20:21:30","modified_gmt":"2026-04-06T01:21:30","slug":"s2eva2015ti_t2-deflexion-de-mastil","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva20\/s2eva2015ti_t2-deflexion-de-mastil\/","title":{"rendered":"s2Eva2015TI_T2 EDO Deflexi\u00f3n de m\u00e1stil"},"content":{"rendered":"\n

Ejercicio<\/strong><\/em>: 2Eva2015TI_T2 EDO Deflexi\u00f3n de m\u00e1stil<\/a><\/p>\n\n\n \\frac{\\delta ^2 y}{\\delta x^2} = \\frac{f}{2EI} (L-x)^2 <\/span>\n\n\n x= 0, y = 0, \\frac{\\delta y}{\\delta x} = 0<\/span>\n\n\n\n

Ecuaci\u00f3n en forma estandarizada:<\/p>\n\n\n y'' = \\frac{f}{2EI} (L-x)^2<\/span>\n\n\n\n

Sistema de ecuaciones<\/p>\n\n\n y' = z = f(x,y,z)<\/span>\n\n\n z' = \\frac{f}{2EI} (L-x)^2 = g(x,y,z)<\/span>\n\n\n\n

siendo:<\/p>\n\n\n\\frac{f}{2EI} =\\frac{60}{2(1.25 \\times 10 ^{-8}) (0.05)} = 4.8 \\times 10^{-6} <\/span>\n\n\n\n

El m\u00e1stil mide L=30 m y se requieren 30 intervalos, entonces h = 30\/30 = 1.<\/p>\n\n\n\n

Usando muestras = tramos +1 = 31<\/p>\n\n\n\n

Se plantea una soluci\u00f3n usando Runge Kutta de 2do Orden<\/p>\n\n\n f(x,y,z) = z<\/span>\n\n\n g(x,y,z) = (4.8 \\times 10^{-6})(30-x)^2 <\/span>\n\n\n\n

\n\n\n\n

Desarrollo anal\u00edtico<\/h2>\n\n\n\n

Las iteraciones se realizan para llenar los datos de la tabla,<\/p>\n\n\n\n

i<\/th>xi<\/sub><\/th>yi<\/sub><\/th>zi<\/sub><\/th><\/tr><\/thead>
0<\/td>0<\/td>0<\/td>0<\/td><\/tr>
1<\/td>1<\/td>0.00216<\/td>0.00417<\/td><\/tr>
2<\/td>2<\/td>0.00835<\/td>0.00807<\/td><\/tr>
3<\/td>3<\/td>tarea<\/strong><\/em><\/td>...<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

iteraci\u00f3n 1<\/strong><\/p>\n\n\n\n

i = 0 ; x0<\/sub>= 0; y0<\/sub> = 0; z0<\/sub> = 0<\/p>\n\n\n K_{1y} = h f(x_0,y_0, z_0)= 1(0) = 0 <\/span>\n\n\n K_{1z} = h g(x_0,y_0, z_0) = <\/span>\n\n\n= 1(4.8 \\times 10^{-6})(30-0)^2 = 0.00432<\/span>\n\n\n K_{2y} = h f(x_0+h,y_0+K_{1y}, z_0+K_{1z}) = <\/span>\n\n\n= 1(0+0.00432) = 0.00432 <\/span>\n\n\n K_{2z} = h g(x_0+h,y_0+K_{1y}, z_0+K_{1z}) = <\/span>\n\n\n= 1(4.8 \\times 10^{-6})(30-(0+1))^2 = 0.00403<\/span>\n\n\n y_1 = y_0 + \\frac{K_{1y}+K_{2y}}{2} =<\/span>\n\n\n = 0 + \\frac{0+0.00432}{2} = 0.00216<\/span>\n\n\n z_1 = z_0 + \\frac{K_{1z}+K_{2z}}{2} =<\/span>\n\n\n = 0 + \\frac{0.00432+0.00403}{2} = 0.00417<\/span>\n\n\n\n

iteraci\u00f3n 2<\/strong><\/p>\n\n\n\n

i = 2 ; x1<\/sub>= 1; y1<\/sub> = 0.00216; z1<\/sub> = 0.00417<\/p>\n\n\n K_{1y} = h f(x_1,y_1, z_1)= 1(0.00417) = 0.00417 <\/span>\n\n\n K_{1z} = h g(x_1,y_1, z_1) = <\/span>\n\n\n= 1(4.8 \\times 10^{-6})(30-1)^2 = 0.00403<\/span>\n\n\n K_{2y} = h f(x_1+h,y_1+K_{1y}, z_1+K_{1z}) = <\/span>\n\n\n= 1(0.00417+0.00403) = 0.00821 <\/span>\n\n\n K_{2z} = h g(x_1+h,y_1+K_{1y}, z_1+K_{1z}) = <\/span>\n\n\n= 1(4.8 \\times 10^{-6})(30-(1+1))^2 = 0.00376<\/span>\n\n\n y_2 = y_1 + \\frac{K_{1y}+K_{2y}}{2} =<\/span>\n\n\n = 0.00216 + \\frac{0.00417+0.00821}{2} = 0.00835<\/span>\n\n\n z_2 = z_2 + \\frac{K_{1z}+K_{2z}}{2} =<\/span>\n\n\n = 0.00417 + \\frac{0.00403+0.00376}{2} = 0.00807<\/span>\n\n\n\n

iteraci\u00f3n 3<\/strong>
i = 2; x2<\/sub>= 2; y2<\/sub> = 0.00835; z2<\/sub> = 0.00807
tarea<\/em> <\/strong>...<\/p>\n\n\n\n

Para facilitar los c\u00e1lculos se propone usa el algoritmo en Python, como se describe en la siguiente secci\u00f3n.<\/p>\n\n\n\n

\n\n\n\n

Algoritmo en Python<\/h2>\n\n\n\n

Al usar el algoritmo se puede comparar los resultados entre Runge-Kutta de 2do orden y de 4to Orden.<\/p>\n\n\n\n

De los resultados se presenta la siguiente gr\u00e1fica<\/p>\n\n\n\n

\"EDO2EIT2015T2<\/figure>\n\n\n\n

Observe en la gr\u00e1fica la diferencia de escalas entre los ejes.<\/p>\n\n\n\n

Runge Kutta 2do Orden\n[x, y, z]\n[[  0.00000000e+00   0.00000000e+00   0.00000000e+00]\n [  1.00000000e+00   2.16000000e-03   4.17840000e-03]\n [  2.00000000e+00   8.35680000e-03   8.07840000e-03]\n [  3.00000000e+00   1.83168000e-02   1.17096000e-02]\n [  4.00000000e+00   3.17760000e-02   1.50816000e-02]\n [  5.00000000e+00   4.84800000e-02   1.82040000e-02]\n ...\n [  2.90000000e+01   9.29856000e-01   4.32216000e-02]\n [  3.00000000e+01   9.73080000e-01   4.32240000e-02]\n [  3.10000000e+01   1.01630400e+00   4.32264000e-02]]<\/code><\/pre>\n\n\n\n
Runge Kutta 4do Orden\n[x, y, z]\n[[ 0.00000000e+00 0.00000000e+00 0.00000000e+00]\n [ 1.00000000e+00 2.11240000e-03 4.17760000e-03]\n [ 2.00000000e+00 8.26240000e-03 8.07680000e-03]\n [ 3.00000000e+00 1.81764000e-02 1.17072000e-02]\n [ 4.00000000e+00 3.15904000e-02 1.50784000e-02]\n [ 5.00000000e+00 4.82500000e-02 1.82000000e-02]\n ...\n [ 2.90000000e+01 9.28800400e-01 4.31984000e-02]\n [ 3.00000000e+01 9.72000000e-01 4.32000000e-02]\n [ 3.10000000e+01 1.01520040e+00 4.32016000e-02]]\n>>><\/code><\/pre>\n\n\n\n
Las instrucciones en Python obtener los resultados:<\/p>\n\n\n
\n# 2Eva_IT2015_T2 Deflexi\u00f3n de m\u00e1stil\n# soluci\u00f3n propuesta: edelros@espol.edu.ec\nimport numpy as np\n\ndef rungekutta2fg(fx,gx,x0,y0,z0,h,muestras):\n    tamano = muestras + 1\n    estimado = np.zeros(shape=(tamano,3),dtype=float)\n    # incluye el punto [x0,y0]\n    estimado[0] = [x0,y0,z0]\n    xi = x0\n    yi = y0\n    zi = z0\n    for i in range(1,tamano,1):\n        K1y = h * fx(xi,yi,zi)\n        K1z = h * gx(xi,yi,zi)\n        \n        K2y = h * fx(xi+h, yi + K1y, zi + K1z)\n        K2z = h * gx(xi+h, yi + K1y, zi + K1z)\n\n        yi = yi + (K1y+K2y)\/2\n        zi = zi + (K1z+K2z)\/2\n        xi = xi + h\n        \n        estimado[i] = [xi,yi,zi]\n    return(estimado)\n\ndef rungekutta4fg(fx,gx,x0,y0,z0,h,muestras):\n    tamano = muestras + 1\n    estimado = np.zeros(shape=(tamano,3),dtype=float)\n    # incluye el punto [x0,y0]\n    estimado[0] = [x0,y0,z0]\n    xi = x0\n    yi = y0\n    zi = z0\n    for i in range(1,tamano,1):\n        K1y = h * fx(xi,yi,zi)\n        K1z = h * gx(xi,yi,zi)\n        \n        K2y = h * fx(xi+h\/2, yi + K1y\/2, zi + K1z\/2)\n        K2z = h * gx(xi+h\/2, yi + K1y\/2, zi + K1z\/2)\n        \n        K3y = h * fx(xi+h\/2, yi + K2y\/2, zi + K2z\/2)\n        K3z = h * gx(xi+h\/2, yi + K2y\/2, zi + K2z\/2)\n\n        K4y = h * fx(xi+h, yi + K3y, zi + K3z)\n        K4z = h * gx(xi+h, yi + K3y, zi + K3z)\n\n        yi = yi + (K1y+2*K2y+2*K3y+K4y)\/6\n        zi = zi + (K1z+2*K2z+2*K3z+K4z)\/6\n        xi = xi + h\n        \n        estimado[i] = [xi,yi,zi]\n    return(estimado)\n\n# INGRESO\nf = 60\nL = 30\nE = 1.25e8\nI = 0.05\nx0 = 0\ny0 = 0\nz0 = 0\ntramos = 30\n\nfx = lambda x,y,z: z\ngx = lambda x,y,z: (f\/(2*E*I))*(L-x)**2\n\n# PROCEDIMIENTO\nmuestras = tramos + 1\nh = L\/tramos\ntabla2 = rungekutta2fg(fx,gx,x0,y0,z0,h,muestras)\nxi2 = tabla2[:,0]\nyi2 = tabla2[:,1]\nzi2 = tabla2[:,2]\n\ntabla4 = rungekutta4fg(fx,gx,x0,y0,z0,h,muestras)\nxi4 = tabla4[:,0]\nyi4 = tabla4[:,1]\nzi4 = tabla4[:,2]\n\n# SALIDA\nprint('Runge Kutta 2do Orden')\nprint(' [x, y, z]')\nprint(tabla2)\nprint('Runge Kutta 4do Orden')\nprint(' [x, y, z]')\nprint(tabla4)\n\n# GRAFICA\nimport matplotlib.pyplot as plt\nplt.plot(xi2,yi2, label='Runge-Kutta 2do Orden')\nplt.plot(xi4,yi4, label='Runge-Kutta 4do Orden')\nplt.title('Deflexi\u00f3n de m\u00e1stil')\nplt.xlabel('x')\nplt.ylabel('y: deflexi\u00f3n')\nplt.legend()\nplt.grid()\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2015TI_T2 EDO Deflexi\u00f3n de m\u00e1stil Ecuaci\u00f3n en forma estandarizada: Sistema de ecuaciones siendo: El m\u00e1stil mide L=30 m y se requieren 30 intervalos, entonces h = 30\/30 = 1. Usando muestras = tramos +1 = 31 Se plantea una soluci\u00f3n usando Runge Kutta de 2do Orden Desarrollo anal\u00edtico Las iteraciones se realizan para llenar […]<\/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":[48],"tags":[58,54],"class_list":["post-1983","post","type-post","status-publish","format-standard","hentry","category-mn-s2eva20","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1983","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=1983"}],"version-history":[{"count":6,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1983\/revisions"}],"predecessor-version":[{"id":23877,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1983\/revisions\/23877"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=1983"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=1983"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=1983"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}