{"id":7717,"date":"2021-09-15T16:48:23","date_gmt":"2021-09-15T21:48:23","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=7717"},"modified":"2026-04-05T21:07:29","modified_gmt":"2026-04-06T02:07:29","slug":"s3eva2021paoi_t1-tensiones-en-cables-por-carga-variable","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s3eva30\/s3eva2021paoi_t1-tensiones-en-cables-por-carga-variable\/","title":{"rendered":"s3Eva2021PAOI_T1 Tensiones en cables por carga variable"},"content":{"rendered":"\n

Ejercicio<\/strong><\/em>: 3Eva2021PAOI_T1 Tensiones en cables por carga variable<\/a><\/p>\n\n\n\n

Planteamiento del problema<\/h2>\n\n\n\n

Las ecuaciones de equilibrio del sistema corresponden a:<\/p>\n\n\n -T_{CA} \\cos (\\alpha) + T_{CB} \\cos (\\beta) + P \\sin (\\theta) = 0<\/span>\n\n\n T_{CA} \\sin (\\alpha) + T_{CB} \\sin (\\beta) - P \\cos (\\theta) = 0<\/span>\n\n\n\n

se reordenan considerando que P y \u03b8 son valores constantes para cualquier caso<\/p>\n\n\n T_{CA} \\cos (\\alpha) - T_{CB} \\cos (\\beta) = P \\sin (\\theta) <\/span>\n\n\n T_{CA} \\sin (\\alpha) + T_{CB} \\sin (\\beta) = P \\cos (\\theta) <\/span>\n\n\n\n

se convierte a la forma matricial<\/p>\n\n\n \\begin{bmatrix} \\cos (\\alpha) & -\\cos (\\beta) \\\\ \\sin (\\alpha) & \\sin (\\beta) \\end{bmatrix} \\begin{bmatrix} T_{CA} \\\\ T_{CB} \\end{bmatrix} = \\begin{bmatrix} P \\sin (\\theta) \\\\ P \\cos (\\theta) \\end{bmatrix} <\/span>\n\n\n\n

tomando valores por ejemplo:<\/p>\n\n\n\n

\u03b1 = 35\u00b0, \u03b2 = 75\u00b0, P = 400 lb, \u0394\u03b8 = 5\u00b0<\/p>\n\n\n\n

\u03b8 = 75-90 = -15<\/p>\n\n\n \\begin{bmatrix} \\cos (35) & -\\cos (75) \\\\ \\sin (35) & \\sin (75) \\end{bmatrix} \\begin{bmatrix}T_{CA} \\\\ T_{CB} \\end{bmatrix} = \\begin{bmatrix} 400 \\sin (-15) \\\\ 400 \\cos (15) \\end{bmatrix} <\/span>\n\n\n\n

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

matriz aumentada<\/p>\n\n\n \\begin{bmatrix} \\cos (35) & - \\cos (75) & 400 \\sin (-15) \\\\ \\sin (35) & \\sin (75 ) & 400 \\cos (15 ) \\end{bmatrix} <\/span>\n\n\n\n

A = \n[[ 0.81915204 -0.25881905]\n [ 0.57357644  0.96592583]]\nB = \n[-103.52761804  386.37033052]\n\nAB = \n[[ 0.81915204 -0.25881905 -103.52761804]\n [ 0.57357644  0.96592583 386.37033052]]<\/code><\/pre>\n\n\n\n
Pivoteo parcial por filas<\/p>\n\n\n\n
cos(-15\u00b0) tendr\u00eda mayor magnitud que sin(-15\u00b0), por lo que la matriz A se encuentra pivoteada<\/p>\n\n\n\n
Eliminaci\u00f3n hacia adelante<\/p>\n\n\n\n
pivote =  0.81915204\/0.57357644\n[[ 0.81915204 -0.25881905 -103.52761804]\n [ 0.0         1.63830408  655.32162903]]<\/code><\/pre>\n\n\n\n
Sustituci\u00f3n hacia atr\u00e1s<\/p>\n\n\n\n
usando la \u00faltima fila:<\/p>\n\n\n\n
1.63830408 TCB<\/sub> = 655.32162903
TCB<\/sub> = 400<\/p>\n\n\n\n
luego la primera fila:<\/p>\n\n\n\n
0.81915204TCA<\/sub> -0.25881905TCB<\/sub> = -103.52761804<\/p>\n\n\n\n
0.81915204TCA<\/sub> = 0.25881905TCB<\/sub>  -103.52761804<\/p>\n\n\n\n
TCA<\/sub> = 2,392 x10-6<\/sup> \u2248 0<\/p>\n\n\n\n
Se deja como tarea realizar el c\u00e1lculo para:  \u03b8+\u0394\u03b8<\/p>\n\n\n\n
Instrucciones en Python<\/h2>\n\n\n\n
Resultado:<\/p>\n\n\n\n
Resultado: [TCA, TCB], diferencia \n[array([3.46965006e-14, 4.00000000e+02]), -399.99999999999994]<\/code><\/pre>\n\n\n\n
usando el intervalo para \u03b81<\/sub> y \u03b82<\/sub>:<\/p>\n\n\n\n
\"tensiones<\/figure>\n\n\n\n
Algoritmo con Python<\/h2>\n\n\n\n
con las instrucciones:<\/p>\n\n\n
\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# Tema 1\ndef funcion(P,theta,alfa,beta):\n    # ecuaciones\n    A = np.array([[np.cos(alfa), -np.cos(beta)],\n                  [np.sin(alfa),  np.sin(beta)]])\n    B = np.array([P*np.sin(theta), P*np.cos(theta)])\n\n    # usar un algoritmo directo\n    X = np.linalg.solve(A,B)\n    \n    diferencia = X[0]-X[1]\n    return([X,diferencia])    \n\n# INGRESO\nalfa = np.radians(35)\nbeta = np.radians(75)\nP = 400\n\n# PROCEDIMIENTO\ntheta = beta-np.radians(90)\nresultado = funcion(P,theta,alfa, beta)\n\n# SALIDA\nprint("Resultado: [TCA, TCB], diferencia")\nprint(resultado)\n\n# Tema 1b --------------\n# PROCEDIMIENTO\ndtheta = np.radians(5)\ntheta1 = beta-np.radians(90)\ntheta2 = np.radians(90)-alfa\n\ntabla = []\ntheta = theta1\nwhile not(theta>=theta2):\n    X = funcion(P,theta,alfa,beta)[0] # usa vector X\n    tabla.append([theta,X[0],X[1]])\n    theta = theta + dtheta\n    \ntabla = np.array(tabla)\nthetai = np.degrees(tabla[:,0])\nTca = tabla[:,1]\nTcb = tabla[:,2]\n\n# SALIDA\nprint('tetha, TCA, TCB')\nprint(tabla)\n\n# Grafica\nplt.plot(thetai,Tca, label='Tca')\nplt.plot(thetai,Tcb, label='Tcb')\n# plt.axvline(np.degrees(c))\nplt.legend()\nplt.xlabel('theta')\nplt.ylabel('Tensi\u00f3n')\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 3Eva2021PAOI_T1 Tensiones en cables por carga variable Planteamiento del problema Las ecuaciones de equilibrio del sistema corresponden a: se reordenan considerando que P y \u03b8 son valores constantes para cualquier caso se convierte a la forma matricial tomando valores por ejemplo: \u03b1 = 35\u00b0, \u03b2 = 75\u00b0, P = 400 lb, \u0394\u03b8 = 5\u00b0 […]<\/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":[52],"tags":[58,54],"class_list":["post-7717","post","type-post","status-publish","format-standard","hentry","category-mn-s3eva30","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/7717","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=7717"}],"version-history":[{"count":4,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/7717\/revisions"}],"predecessor-version":[{"id":23946,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/7717\/revisions\/23946"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=7717"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=7717"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=7717"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}