{"id":9180,"date":"2024-01-31T11:15:01","date_gmt":"2024-01-31T16:15:01","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=9180"},"modified":"2026-04-05T20:42:26","modified_gmt":"2026-04-06T01:42:26","slug":"s2eva2023paoii_t2-cable-cuelga-entre-ab","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva30\/s2eva2023paoii_t2-cable-cuelga-entre-ab\/","title":{"rendered":"s2Eva2023PAOII_T2 EDO Cable cuelga entre apoyos A y B"},"content":{"rendered":"\n

Ejercicio<\/strong>: 2Eva2023PAOII_T2 EDO Cable cuelga entre apoyos A y B<\/a><\/p>\n\n\n\n

Literal a<\/h3>\n\n\n\n

La ecuaci\u00f3n diferencial a resolver es:<\/p>\n\n\n \\frac{d^2y}{dx^2} = \\frac{w_0}{T_0} \\Big[ 1+ \\sin \\Big(\\frac{\\pi x}{2l_B} \\Big) \\Big] <\/span>\n\n\n\n

donde w0<\/sub> = 1 000 lbs\/ft, T0<\/sub>. = 0.588345\u00d7106<\/sup>.<\/strong>
dy(0)\/dx = 0 y lB<\/sub>=200 de la gr\u00e1fica presentada.<\/p>\n\n\n \\frac{d^2y}{dx^2} = \\frac{1000}{0.588345\u00d710^6} \\Big[ 1+ \\sin \\Big(\\frac{\\pi x}{2(200)} \\Big) \\Big] <\/span>\n\n\n\n

Para usar Runge Kutta para segunda derivada<\/a>:<\/p>\n\n\n z= y' = f(x,y,z) <\/span>\n\n\n z' = (y')' = 0z + \\frac{1000}{0.588345\u00d710^6} \\Big[ 1+ \\sin \\Big(\\frac{\\pi x}{2(200)} \\Big) \\Big] <\/span>\n\n\n g(x,y,z) = \\frac{1}{0.588345\u00d710^3} \\Big[ 1+ \\sin \\Big(\\frac{\\pi x}{2(200)} \\Big) \\Big] <\/span>\n\n\n\n

los valores iniciales para el ejercicio acorde al enunciado son: x0<\/sub> = 0, y0<\/sub>=0, z0<\/sub> = 0, con h=0.5<\/p>\n\n\n\n

literal b<\/h3>\n\n\n\n

para itera 0<\/p>\n\n\n K1y = h*z = 0.5*0 = 0 <\/span>\n\n\n K1z = (0.5)\\frac{1}{0.588345\u00d710^3} \\Big[ 1+ \\sin \\Big(\\frac{\\pi (0)}{2(200)} \\Big) \\Big] = 0.0008498 <\/span>\n\n\n K2y = h*(z+K1z) = (0.5) (0+0.00084984) = 0.0004249<\/span>\n\n\n K2z = (0.5)\\frac{1}{0.588345\u00d710^3} \\Big[ 1+ \\sin \\Big(\\frac{\\pi (0+0.5)}{2(200)} \\Big) \\Big] =0.0008531 <\/span>\n\n\n y = 0+\\frac{0+0.0004249}{2} = 0.0002124<\/span>\n\n\n z = 0+\\frac{0.0008498+0.0008531}{2} = 0.0008515 <\/span>\n\n\n x = 0 + 0.5 = 0.5 <\/span>\n\n\n\n

...<\/p>\n\n\n\n

Desarrollar dos iteraciones adicionales como tarea.<\/p>\n\n\n\n

Para las primeras iteraciones de un total de 400+1, los valores con Python y en resultados.txt :<\/p>\n\n\n\n

estimado[xi,yi,zi,K1y,K2y,K1z,K2z]\n[0.0000 0.0000e+00 0.0000e+00 0.0000e+00 0.0000e+00 0.0000e+00 0.0000e+00]\n\n[0.5000 2.124603761398499073e-04 8.515101601627471980e-04 0.000000000000000000e+00 4.249207522796998146e-04 8.498415045593996292e-04 8.531788157660947667e-04]\n\n[1.0000 8.515101601627470896e-04 1.706357605799455881e-03 4.257550800813735990e-04 8.523444879644209281e-04 8.531788157660947667e-04 8.565160755073224913e-04]\n\n[1.5000 1.918817981939305653e-03 2.564542259712321911e-03 8.531788028997279406e-04 1.281436840653389295e-03 8.565160755073224913e-04 8.598532323184093513e-04]\n\n[2.0000 3.416052419875068892e-03 3.426063993239661116e-03 1.282271129856160955e-03 1.712197746015365740e-03 8.598532323184093513e-04 8.631902347362692754e-04]\n...<\/code><\/pre>\n\n\n\n
literal c<\/h3>\n\n\n\n
resultado en archivo.txt al ejecutar el algoritmo.<\/p>\n\n\n\n
literal d<\/h3>\n\n\n\n
\"cable<\/figure>\n\n\n\n
Algoritmo con Python<\/h3>\n\n\n
\n# 2Eva_2023PAOII_T2 Cable cuelga entre apoyos A y B\nimport numpy as np\n\ndef rungekutta2_fg(f,g,x0,y0,z0,h,muestras):\n    tamano = muestras + 1\n    estimado = np.zeros(shape=(tamano,3+4),dtype=float)\n\n    # incluye el punto [x0,y0,z0]\n    estimado[0] = [x0,y0,z0,0,0,0,0]\n    xi = x0\n    yi = y0\n    zi = z0\n    for i in range(1,tamano,1):\n        K1y = h * f(xi,yi,zi)\n        K1z = h * g(xi,yi,zi)\n        \n        K2y = h * f(xi+h, yi + K1y, zi + K1z)\n        K2z = h * g(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,K1y,K2y,K1z,K2z]\n    return(estimado)\n\n# PROGRAMA PRUEBA\n# Ref Rodriguez 9.1.1 p335 ejemplo.\n# prueba y'-y-x+(x**2)-1 =0, y(0)=1\n\n# INGRESO\nT0 = 0.588345e6\nLB = 200\nf = lambda x,y,z: z\ng = lambda x,y,z: (1000\/T0)*(1+np.sin(np.pi*x\/(2*LB)))\nx0 = 0\ny0 = 0\nz0 = 0\nh  = 0.5\nmuestras = 401\n\n# PROCEDIMIENTO\npuntosRK2 = rungekutta2_fg(f,g,x0,y0,z0,h,muestras)\nxi = puntosRK2[:,0]\nyiRK2 = puntosRK2[:,1]\n\n# SALIDA\nnp.set_printoptions(precision=4)\nprint('estimado[xi,yi,zi,K1y,K2y,K1z,K2z]')\nprint(puntosRK2)\nnp.savetxt("tablaRk2.txt",puntosRK2)\n\n\n# Gr\u00e1fica\nimport matplotlib.pyplot as plt\n\nplt.plot(xi[0],yiRK2[0],\n         'o',color='r', label ='[x0,y0]')\nplt.plot(xi[1:],yiRK2[1:],\n         color='m',\n         label ='y Runge-Kutta 2 Orden')\n\nplt.title('EDO: Soluci\u00f3n con Runge-Kutta 2do Orden')\nplt.xlabel('x')\nplt.ylabel('y')\nplt.legend()\nplt.grid()\nplt.show()\n<\/pre><\/div>\n\n\n
 <\/p>\n","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2023PAOII_T2 EDO Cable cuelga entre apoyos A y B Literal a La ecuaci\u00f3n diferencial a resolver es: donde w0 = 1 000 lbs\/ft, T0. = 0.588345\u00d7106.dy(0)\/dx = 0 y lB=200 de la gr\u00e1fica presentada. Para usar Runge Kutta para segunda derivada: los valores iniciales para el ejercicio acorde al enunciado son: x0 = 0, […]<\/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":[49],"tags":[58,54],"class_list":["post-9180","post","type-post","status-publish","format-standard","hentry","category-mn-s2eva30","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/9180","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=9180"}],"version-history":[{"count":5,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/9180\/revisions"}],"predecessor-version":[{"id":23887,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/9180\/revisions\/23887"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=9180"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=9180"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=9180"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}