{ "metadata": { "name": "", "signature": "sha256:996c8013a65a6364550b3850a6d724fb8ccb944aba88d1755142b6b2b68c6a46" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 13: Statically Indeterminate Problems" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.2 page number 693" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given \n", "#First we will solve without the reaction at middle\n", "#Given\n", "import numpy\n", "l_ab = 1.0 #2L in - The length of the beam\n", "F_D = 1.0 #W lb/in - The force distribution \n", "F = F_D*l_ab #WL - The force applied\n", "#Beause of symmetry the moment caliculations can be neglected\n", "#F_Y = 0\n", "R_A = F/2 #wl - The reactive force at A\n", "R_B = F/2 #wl - The reactive force at B\n", "#EI - The flxure rigidity is constant and 1/EI =1 # k\n", "\n", "#part - A\n", "#section 1--1\n", "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.2L distance \n", "M_1 = [0,0,0,0,0,0,0,0,0,0,0]\n", "v = [0,0,0,0,0,0,0,0,0,0,0]\n", "for i in range(10):\n", " v[i] = R_A - F_D*l_1[i] \n", " M_1[i] = R_A*l_1[i] - F_D*(l_1[i]**2)/2\n", "# (EI)y'' = M_1[i] we will integrate M_1[i] twice where variable is l_1[i]\n", "#(EI)y'- \n", "\n", "M_1_intg1 = R_A*(l_1[i]**2)/4 - F_D*(l_1[i]**3)/6 - F_D*(l_ab**3)*l_1[i]/24 #integration of x**n = x**n+1/n+1\n", "#(EI)y- Using end conditions for caliculating constants \n", "\n", "M_1_intg2 = R_A*(l_1[i]**3)/12.0 - F_D*(l_1[i]**4)/24.0 + F_D*(l_ab**3)*l_1[i]/24.0 \n", "#Equations \n", "\n", "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.2L distance \n", "M_1_intg2 = [0,0,0,0,0,0,0,0,0,0,0]\n", "Y = [0,0,0,0,0,0,0,0,0,0,0]\n", "for i in range(10):\n", " M_1_intg2[i] = (l_1[i]**3)/12.0 - (l_1[i]**4)/24.0 - l_1[i]/24.0 # discluding every term for ruling out float values\n", " Y[i] = M_1_intg2[i] #W(l**4)/EI k = 1/EI\n", "Y_min = 16*min(Y)\n", "print \"a) The maximum displacement in y direction is\",16*min(Y),\"W(l**4)/EI \"\n", "print \"a) The maximum deflection occured at\",2*l_1[Y.index(min(Y))],\"L\"\n", "f_bb = 2**3/48.0 #l**3/EI - flexibility coefficient\n", "Reac = - Y_min/f_bb #WL , The reaction at the mid of the bar\n", "print \"The reaction at the mid of the bar\",Reac ,\"WL\"\n", "\n", "#Graphs \n", "Y.extend(Y) #Because of symmetry\n", "import numpy as np\n", "values = Y \n", "y = np.array(values)\n", "t = np.linspace(0,1,22)\n", "poly_coeff = np.polyfit(t, y, 2)\n", "import matplotlib.pyplot as plt\n", "plt.plot(t, y, 'o')\n", "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", "plt.show()\n", "print \"b)The above graph is beam displacement graph\"\n", "print \"b)The minimum occures in the middle from the above graph \"\n", "\n", "\n", "\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "a) The maximum displacement in y direction is -0.208333333333 W(l**4)/EI \n", "a) The maximum deflection occured at 1.0 L\n", "The reaction at the mid of the bar 1.25 WL\n" ] }, { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "b)The above graph is beam displacement graph\n", "b)The minimum occures in the middle from the above graph \n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.3 page number 694 " ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given \n", "#First we will solve without the reaction at middle\n", "#Given\n", "import numpy\n", "l_ab = 1.0 #2L in - The length of the beam\n", "F_D = 1.0 #W lb/in - The force distribution \n", "F = F_D*l_ab #WL - The force applied\n", "#Beause of symmetry the moment caliculations can be neglected\n", "#F_Y = 0\n", "R_A = F/2 #wl - The reactive force at A\n", "R_B = F/2 #wl - The reactive force at B\n", "#EI - The flxure rigidity is constant and 1/EI =1 # k\n", "\n", "#part - A\n", "#section 1--1\n", "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.2L distance \n", "M_1 = [0,0,0,0,0,0,0,0,0,0,0]\n", "v = [0,0,0,0,0,0,0,0,0,0,0]\n", "for i in range(10):\n", " v[i] = R_A - F_D*l_1[i] \n", " M_1[i] = R_A*l_1[i] - F_D*(l_1[i]**2)/2\n", "# (EI)y'' = M_1[i] we will integrate M_1[i] twice where variable is l_1[i]\n", "#(EI)y'- \n", "\n", "M_1_intg1 = R_A*(l_1[i]**2)/4 - F_D*(l_1[i]**3)/6 - F_D*(l_ab**3)*l_1[i]/24 #integration of x**n = x**n+1/n+1\n", "#(EI)y- Using end conditions for caliculating constants \n", "\n", "M_1_intg2 = R_A*(l_1[i]**3)/12.0 - F_D*(l_1[i]**4)/24.0 + F_D*(l_ab**3)*l_1[i]/24.0 \n", "#Equations \n", "\n", "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.2L distance \n", "M_1_intg2 = [0,0,0,0,0,0,0,0,0,0,0]\n", "Y = [0,0,0,0,0,0,0,0,0,0,0]\n", "for i in range(10):\n", " M_1_intg2[i] = (l_1[i]**3)/12.0 - (l_1[i]**4)/24.0 - l_1[i]/24.0 # discluding every term for ruling out float values\n", " Y[i] = M_1_intg2[i] #W(l**4)/EI k = 1/EI\n", "e_1 = 16*min(Y) #WL4/EI - The maximum defection \n", "e_2 = - F_D*((2*l_ab)**3)/24.0 #WL3/EI - The maximum angle\n", "#Caliculating for momentum and force\n", "f_ab = ((2*l_ab)**2)/16.0 #L2/EI \n", "f_bb = ((2*l_ab)**3)/48.0 #L3/EI \n", "f_aa = 2*l_ab/3.0 #L/EI\n", "f_ba = ((l_ab)**2)/4.0 #L2/EI\n", "#F*X = e - Matrix multiplication \n", "#Solving for X\n", "a = np.array([[f_aa,f_ba], [f_ba,f_bb]])\n", "b = np.array([e_2,e_1])\n", "x = np.linalg.solve(a, b)\n", "print \"The reactive moment at A i.e M_A\",x[0],\"WL**2\"\n", "print \"The reactive force at A i.e R_A\",x[1],\"WL\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The reactive moment at A i.e M_A -0.0714285714286 WL**2\n", "The reactive force at A i.e R_A -1.14285714286 WL\n" ] } ], "prompt_number": 26 }, { "cell_type": "code", "collapsed": false, "input": [ "#Given \n" ], "language": "python", "metadata": {}, "outputs": [] } ], "metadata": {} } ] }