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+{
+ "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": [
+        "<matplotlib.figure.Figure at 0x7fad208>"
+       ]
+      },
+      {
+       "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": {}
+  }
+ ]
+}
+\ No newline at end of file