{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Chapter 5:Axial force, Shear and Bending moment "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example5.2 pagenumber 231"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The X,Y components of reaction force at A is  0 , -410.0 N\n",
      "The X,Y components of reaction force at B is  0 , 670.0 N\n"
     ]
    }
   ],
   "source": [
    "#Given \n",
    "L_ab = 0.4  #mt The total length of the rod\n",
    "M = 200     #N_m - the moment acting on rod\n",
    "l_1 = 0.1   #mt -moment acting point the distance from 'a'\n",
    "R_1 = 100   #N - The Force acting \n",
    "l_2 = 0.2   #mt -R_1 acting point the distance from 'a'\n",
    "R_2 = 160   #N The Force acting \n",
    "l_3 = 0.3   #mt -R_2 acting point the distance from 'a'\n",
    "#caliculations\n",
    "\n",
    "#F_X = 0 forces in x directions \n",
    "R_A_X = 0   # since there are no forces in X-direction \n",
    "R_B_X = 0\n",
    "#M_A = 0 momentum at point a is zero\n",
    "\n",
    "# M + R_1*l_2 + R_2*l_3 = R_B*L_ab *the moment for a force is FxL\n",
    "R_B_Y =  (M + R_1*l_2 + R_2*l_3)/L_ab\n",
    "\n",
    "#M_B= 0 momentum at point b is zero\n",
    "# R_A_Y*L_ab + M - R_1*l_2 - R_2*0.1 = 0\n",
    "\n",
    "R_A_Y = -(M - R_1*l_2 - R_2*0.1)/L_ab\n",
    "    \n",
    "print \"The X,Y components of reaction force at A is \",R_A_X,\",\",R_A_Y,\"N\"\n",
    "print \"The X,Y components of reaction force at B is \",R_B_X,\",\",R_B_Y,\"N\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5.2 page number 233"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The X,Y components of reaction force at A is  0 , -9.0 N\n",
      "The X,Y components of reaction force at B is  0 , 6.0 N\n"
     ]
    }
   ],
   "source": [
    "#Given   \n",
    "P_Max  = 10   #N - the maximum distribution in a triangular distribution\n",
    "L = 3         #mt the total length of force distribution \n",
    "L_X = 5       #mt - the horizantal length of the rod\n",
    "#caliculations \n",
    "\n",
    "F_y = P_Max*L*0.5 #N - The force due to triangular distribition \n",
    "L_com  = 2*L /3   #mt - the resultant force acting as a result of distribution acting position \n",
    "#F_X = 0 forces in x directions\n",
    "R_A_X = 0         # since there are no forces in X-direction\n",
    "R_B_X = 0\n",
    "#M_A = 0 momentum at point a is zero\n",
    "#F_y*L_com - R_B_Y*L_X = 0\n",
    "R_B_Y = F_y*L_com/L_X\n",
    "\n",
    "#M_B= 0 momentum at point b is zero\n",
    "#- R_A_Y*L_X = F_y*(L_X-L )\n",
    "\n",
    "R_A_Y = - F_y*L/L_X \n",
    "print \"The X,Y components of reaction force at A is \",R_A_X,\",\",R_A_Y,\"N\"\n",
    "print \"The X,Y components of reaction force at B is \",R_B_X,\",\",R_B_Y,\"N\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##  Example 5.3 page number 233   "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The X,Y components and resultant of reaction force at A is  4 , 3 , 5.0 N\n",
      "The X,Y components and resultant of reaction force at B is  1 , 1 , 1.41 N\n"
     ]
    }
   ],
   "source": [
    "#given\n",
    "F = 5       #K - force acting on the system\n",
    "tan = (4/3) # the Tan of the angle of force with x axis\n",
    "l_ab = 12   #inch - the total length of ab \n",
    "l = 3       # inch - Distance from 'a'\n",
    "#caliculation\n",
    "F_X = 4 #K\n",
    "F_Y = 3 #k\n",
    "\n",
    "#M_A = 0 momentum at point a is zero\n",
    "# F_X*l- R_B_Y*l_ab = 0 \n",
    "R_B_Y = F_X*l/l_ab\n",
    "\n",
    "#M_B= 0 momentum at point b is zero\n",
    "# R_A_Y*l_ab - F_X*(l_ab - l)\n",
    "R_A_Y = F_X*(l_ab - l)/l_ab\n",
    "    \n",
    "#F_X = 0 forces in x directions\n",
    "R_A_X = F_Y + R_B_Y  \n",
    "R_B_X = R_B_Y      # since the angle is 45 degrees\n",
    "\n",
    "#resultants \n",
    "R_A = pow(R_A_X**2 + R_A_Y**2,0.5)\n",
    "R_B = pow(R_B_X**2 + R_B_Y**2,0.5)\n",
    "\n",
    "print \"The X,Y components and resultant of reaction force at A is \",R_A_X,\",\",R_A_Y,\",\",R_A,\"N\"\n",
    "print \"The X,Y components and resultant of reaction force at B is \",R_B_X,\",\",R_B_Y,\",\",round(R_B,2),\"N\"\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5.4 page number 239"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The force and moment in section a--a are -2.33 KN , -13.644 KN-m\n",
      "The force and moment in section b--b are 6.0 KN , -6.0 KN-m\n"
     ]
    }
   ],
   "source": [
    "#Given   \n",
    "P_Max  = 10   #N - the maximum distribution in a triangular distribution\n",
    "L = 3         #mt the total length of force distribution \n",
    "L_X = 5       #mt - the horizantal length of the rod\n",
    "#caliculations \n",
    "\n",
    "F_y = P_Max*L*0.5 #N - The force due to triangular distribition \n",
    "L_com  = 2*L /3   #mt - the resultant force acting as a result of distribution acting position \n",
    "#F_X = 0 forces in x directions\n",
    "R_A_X = 0         # since there are no forces in X-direction\n",
    "R_B_X = 0\n",
    "#M_A = 0 momentum at point a is zero\n",
    "#F_y*L_com - R_B_Y*L_X = 0\n",
    "R_B_Y = F_y*L_com/L_X\n",
    "\n",
    "#M_B= 0 momentum at point b is zero\n",
    "#- R_A_Y*L_X = F_y*(L_X-L )\n",
    "\n",
    "R_A_Y = - F_y*L/L_X\n",
    "\n",
    "#For a---a section \n",
    "l_a = 2 #mt - a---a section from a \n",
    "l_com_a = 2*l_a/3\n",
    "v_a = R_A_Y + 0.5*l_a*(10.0*2/3) #*(10*2/3) because the maximum moves\n",
    "\n",
    "M_a =  (10.0*0.66)*l_a*(0.33) + R_A_Y*l_a \n",
    "\n",
    "#For b---b section \n",
    "\n",
    "v_b = F_y + R_A_Y #equilabrium conditions\n",
    "M_b =  (F_y + R_A_Y)*(-1)\n",
    "\n",
    "print \"The force and moment in section a--a are\",round(v_a,2),\"KN ,\",M_a,\"KN-m\"\n",
    "print \"The force and moment in section b--b are\",v_b,\"KN ,\",M_b,\"KN-m\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5.5 page number 241"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n"
     ]
    },
    {
     "data": {
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VpWBmZrXHScHMzEqcFMzMrMRJwczMSpwUzMysxEnBzMxKnBTMzKzEScHMzEqc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\nUjAzsxInBTMzK3FSMDOzkv8PlVM9sSzz1iEAAAAASUVORK5CYII=\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x8cd4f60>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "#Given \n",
    "#Lets divide the section into two sections \n",
    "l_ac = 10 # ft -The total length of the rod\n",
    "R = 5     #k - The applies force at c\n",
    "tan = 4/3 # The tan of the angle of the force \n",
    "l_ab = 5  #ft - The distance of applied force from A\n",
    "R_y = 4 #k,downwards X- component of the force\n",
    "R_X = 3 #k Y- component of the force , lets consider only Y direction since we are concentrating on the Shears \n",
    "\n",
    "#F_Y = 0 forces in Y directions\n",
    "#R_A +R_B =  R_y\n",
    "#M_c = 0 making the moment zero at point c \n",
    "#Caliculations \n",
    "# R_A= R_B*(l_ac - l_ab)/(l_ab) so R_A = R_B\n",
    "\n",
    "R_A =  R_y/2 #F_Y = 0\n",
    "R_B =  R_y/2\n",
    "#considering section x--x\n",
    "l_x = 2       #ft - length of section from A\n",
    "v_x = R_A     #k ,F_X = 0 \n",
    "M_x = R_A*l_x #k-ft M_c = 0\n",
    "\n",
    "#considering section at midpoint t--t\n",
    "l_t =  2        #ft - length of section from A\n",
    "v_t = 0         #k ,F_X = 0 \n",
    "M_t = (R_A)*l_t #k-ft M_c = 0\n",
    "\n",
    "##considering section y---y\n",
    "l_y = 2       #ft - length of section from B\n",
    "v_y = - R_B   #k ,F_X = 0 \n",
    "M_y = R_B*l_y #k-ft M_c = 0\n",
    "\n",
    "#Graph\n",
    "%matplotlib inline\n",
    "import math \n",
    "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n",
    "#Drawing of shear and bending moment diagram\n",
    "print \"Given problem is for drawing diagram, this diagram is drawn by step by step manner. \"\n",
    "X = [0,2,4.9999999999999,5,5.00000000000000001,7,10] # For graph precision \n",
    "\n",
    "V = [R_A,v_x,v_x,v_t,v_y,v_y,-R_B];\t\t\t#Shear matrix\n",
    "M = [0,M_x,M_t,M_t,M_t,M_y,0];\t\t\t#Bending moment matrix\n",
    "plot(X,V);\t\t\t#Shear diagram\n",
    "plot(X,M,'r');\t\t\t#Bending moment diagram\n",
    "suptitle( 'Shear and bending moment diagram')\n",
    "xlabel('X axis')\n",
    "ylabel( 'Y axis') ;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5.6 pagenumber 243 "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n"
     ]
    },
    {
     "data": {
      "image/png": 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      "text/plain": [
       "<matplotlib.figure.Figure at 0x8cc3710>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "#Given\n",
    "l = 1                       #L - Length of the cantilever \n",
    "F_app = ((2**0.5))*2        #p - force applies \n",
    "tan = 1                     # The angle of force applied\n",
    "F_app_x = F_app/((2**0.5))  #p The horizantal component of the force , neglected \n",
    "F_app_y = F_app/((2**0.5))  #p  The Vertical component of the force \n",
    "#F_Y = 0 \n",
    "R_A = 1 #p\n",
    "\n",
    "#Considering  section 1-----1\n",
    "l_1 = 0.5       # The length of the section from one end\n",
    "v_1 = R_A       #F_Y = 0\n",
    "M_1 = -R_A*l_1  #MAking moment at section 1 = 0\n",
    "\n",
    "#considering end of cantilever\n",
    "l_2 = 1       # The length of the section from one end\n",
    "v_2 = R_A     #F_Y = 0\n",
    "M_2 = -R_A*l_2#MAking moment at section 1 = 0\n",
    "\n",
    "#Graph\n",
    "%matplotlib inline\n",
    "import math \n",
    "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n",
    "#Drawing of shear and bending moment diagram\n",
    "print \"Given problem is for drawing diagram, this diagram is drawn by step by step manner. \"\n",
    "X = [0,0.5,1] # For graph precision \n",
    "\n",
    "V = [R_A,v_1,v_2 ];\t\t\t#Shear matrix\n",
    "M = [0,M_1,M_2];\t\t\t#Bending moment matrix\n",
    "plot(X,V);\t\t\t        #Shear diagram\n",
    "plot(X,M,'r');\t\t\t    #Bending moment diagram\n",
    "suptitle( 'Shear and bending moment diagram')\n",
    "xlabel('X axis')\n",
    "ylabel( 'Y axis') ;\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5.7 page number 243"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n"
     ]
    },
    {
     "data": {
      "image/png": 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      "text/plain": [
       "<matplotlib.figure.Figure at 0xaaed668>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "#Given\n",
    "l_ab = 3  #L - The total length lets say '3L'\n",
    "R_1 = 1   #p - The force applied at b\n",
    "R_2 = 1   #p - The force applied at c\n",
    "l_ab = 1  #L\n",
    "l_bc = 1  #L \n",
    "\n",
    "#Logical step \n",
    "#Since the system is in symmetry we can avoid moment M = 0 caliculations\n",
    "\n",
    "#F_Y = 0 \n",
    "R_A = (R_1 + R_2)/2\n",
    "R_B = (R_1 + R_2)/2\n",
    "\n",
    "#Lets take '3' sections \n",
    "#Considering section 1-----1 at 0.5L\n",
    "l_1 = 0.5     #L - distance of the section from the A\n",
    "v_1 = R_A     #F_Y = 0 \n",
    "M_1 = R_A*l_1 #MAking moment at section 1 = 0\n",
    "\n",
    "#Considering section 2-----2 at 1L\n",
    "l_2 = 1       #L - distance of the section from the A\n",
    "v_2 = R_A     #F_Y = 0 \n",
    "M_2 = R_A*l_2 #MAking moment at section 2 = 0\n",
    "\n",
    "#Considering section 3-----3 at 1.5L\n",
    "l_3 = 1.5       #L - distance of the section from the A\n",
    "v_3 = 0        #F_Y = 0 \n",
    "M_3 = R_A*l_2  #MAking moment at section 2 = 0 and symmetry \n",
    "\n",
    "#GRAPH\n",
    "#Since the symmetry exists the graphs are also symmetry\n",
    "%matplotlib inline\n",
    "import math \n",
    "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n",
    "#Drawing of shear and bending moment diagram\n",
    "print \"Given problem is for drawing diagram, this diagram is drawn by step by step manner. \"\n",
    "X = [0,0.5,1,1.000000001,1.5,1.999999999999,2,2.5,3] # For graph precision \n",
    "\n",
    "V = [R_A,v_1,v_2,v_3,v_3,v_3,-v_2,-v_1,-R_B];\t\t\t#Shear matrix\n",
    "M = [0,M_1,M_2,1,1,1,M_2,M_1,0];\t\t\t#Bending moment matrix\n",
    "plot(X,V);\t\t\t#Shear diagram\n",
    "plot(X,M);\t\t\t#Bending moment diagram\n",
    "suptitle( 'Shear and bending moment diagram')\n",
    "xlabel('X axis')\n",
    "ylabel( 'Y axis') ;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5.8 page nmber 244"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "0.5\n"
     ]
    },
    {
     "data": {
      "image/png": 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      "text/plain": [
       "<matplotlib.figure.Figure at 0x8f56978>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "image/png": 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      "text/plain": [
       "<matplotlib.figure.Figure at 0x901ada0>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "#Given\n",
    "%matplotlib inline\n",
    "import numpy as np\n",
    "l_ab = 1.0   #L - The length of the beam\n",
    "F_D = 1.0    #W - 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",
    "\n",
    "#considering many sections \n",
    "\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 0.1L 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 #M = 0 in the section\n",
    "print R_A\n",
    "#Graphs\n",
    "import numpy as np\n",
    "values = [0.5,0,-0.5]\n",
    "y = np.array(values)\n",
    "t = np.linspace(0,1,3)\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",
    "\n",
    "import numpy as np\n",
    "values = M_1\n",
    "y = np.array(values)\n",
    "t = np.linspace(0,1,11)\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()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5.9 page number 245"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The force and moment in section a--a are -2.33 KN , -13.644 KN-m\n"
     ]
    }
   ],
   "source": [
    "#Given   \n",
    "P_Max  = 10   #N - the maximum distribution in a triangular distribution\n",
    "L = 3         #mt the total length of force distribution \n",
    "L_X = 5       #mt - the horizantal length of the rod\n",
    "#caliculations \n",
    "\n",
    "F_y = P_Max*L*0.5 #N - The force due to triangular distribition \n",
    "L_com  = 2*L /3   #mt - the resultant force acting as a result of distribution acting position \n",
    "#F_X = 0 forces in x directions\n",
    "R_A_X = 0         # since there are no forces in X-direction\n",
    "R_B_X = 0\n",
    "#M_A = 0 momentum at point a is zero\n",
    "#F_y*L_com - R_B_Y*L_X = 0\n",
    "R_B_Y = F_y*L_com/L_X\n",
    "\n",
    "#M_B= 0 momentum at point b is zero\n",
    "#- R_A_Y*L_X = F_y*(L_X-L )\n",
    "\n",
    "R_A_Y = - F_y*L/L_X\n",
    "\n",
    "#caliculating for some random value\n",
    "#For a---a section \n",
    "l_a = 2 #mt - a---a section from a \n",
    "l_com_a = 2*l_a/3\n",
    "v_a = R_A_Y + 0.5*l_a*(10.0*2/3) #*(10*2/3) because the maximum moves\n",
    "\n",
    "M_a =  (10.0*0.66)*l_a*(0.33) + R_A_Y*l_a\n",
    "\n",
    "print \"The force and moment in section a--a are\",round(v_a,2),\"KN ,\",M_a,\"KN-m\"\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5.13 page number 254"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n"
     ]
    },
    {
     "data": {
      "image/png": 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AKIOy9kMIkBroewAoi7L2Q9gyVkHfA0DZlLEfQoAMQt8DQFmVrR9CgAxC3wNAmZWpH0KA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\nAAC5/H/91y4DSEHt5wAAAABJRU5ErkJggg==\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0xaaedeb8>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "#Given\n",
    "l_ab = 4  #L - The total length lets say '3L'\n",
    "R_1 = 1   #p - The force applied at b\n",
    "R_2 = 1   #p - The force applied at c\n",
    "l_ab = 1  #L\n",
    "l_bc = 3 #L \n",
    "\n",
    "#Logical step \n",
    "#Since the system is in symmetry we can avoid moment M = 0 caliculations\n",
    "\n",
    "#F_Y = 0 \n",
    "R_A = (R_1 + R_2)/2\n",
    "R_B = (R_1 + R_2)/2\n",
    "\n",
    "#Lets take '3' sections \n",
    "#Considering section 1-----1 at 0.5L\n",
    "l_1 = 0.5    #L - distance of the section from the A\n",
    "v_1 = R_A     #F_Y = 0 \n",
    "M_1 = R_A*l_1 #MAking moment at section 1 = 0\n",
    "\n",
    "#Considering section 2-----2 at 1L\n",
    "l_2 = 1       #L - distance of the section from the A\n",
    "v_2 = R_A     #F_Y = 0 \n",
    "M_2 = R_A*l_2 #MAking moment at section 2 = 0\n",
    "\n",
    "#Considering section 3-----3 at 1.5L\n",
    "l_3 = 3       #L - distance of the section from the A\n",
    "v_3 = 0        #F_Y = 0 \n",
    "M_3 = R_A*l_2  #MAking moment at section 2 = 0 and symmetry \n",
    "\n",
    "#GRAPH\n",
    "#Since the symmetry exists the graphs are also symmetry\n",
    "%matplotlib inline\n",
    "import math \n",
    "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n",
    "#Drawing of shear and bending moment diagram\n",
    "print \"Given problem is for drawing diagram, this diagram is drawn by step by step manner. \"\n",
    "X = [0,0.5,1,1.0000001,2,2.9999999999,3,3.5,4] # For graph precision \n",
    "\n",
    "V = [R_A,v_1,v_2,v_3,v_3,v_3,-v_2,-v_1,-R_B];\t\t\t#Shear matrix\n",
    "M = [0,M_1,M_2,M_3,M_3,M_3,M_2,M_1,0];\t\t\t#Bending moment matrix\n",
    "plot(X,V);\t\t\t#Shear diagram\n",
    "plot(X,M);\t\t\t#Bending moment diagram\n",
    "suptitle( 'Shear and bending moment diagram')\n",
    "xlabel('X axis')\n",
    "ylabel( 'Y axis') ;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5.14 page number 255"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "image/png": 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i9NrR9FzUk4iOEVQvXt3pOAHPpgpmUiXzl2RB+wW4xrkIyR9C0xubpmm9qKho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\nAAAAAElFTkSuQmCC\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0xae282e8>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "image/png": 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q8W/kGDN7q6yv+NTMBoZQZlSY2Tgz22Rmn1TR5uD6TneP6g+lwbMGaAHUBZYA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\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0xb34cba8>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "#Given\n",
    "import numpy as np\n",
    "import matplotlib.pyplot as plt\n",
    "l = 1.0 #l - The length of the beam\n",
    "p = 1.0 #W - The total load applied\n",
    "#since it is triangular distribution \n",
    "l_com = 0.66*l#l - The distance of force of action from one end\n",
    "#F_Y = 0\n",
    "#R_A + R_B = p\n",
    "#M_a = 0 Implies that R_B = 2*R_A\n",
    "R_A = p/3.0\n",
    "R_B = 2.0*p/3\n",
    "\n",
    "#Taking Many sections \n",
    "\n",
    "#Section 1----1\n",
    "l = [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 0.1L distance \n",
    "M = [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] = p*(l[i]**2) - p/3.0\n",
    "    M[i] = p*(l[i]**3)/(3.0)- p*l[i]/3.0\n",
    "\n",
    "v[10] = R_B #again concluded Because the value is tearing of  \n",
    "\n",
    "\n",
    "#Graph\n",
    "values = M\n",
    "y = np.array(values)\n",
    "t = np.linspace(0,1,11)\n",
    "poly_coeff = np.polyfit(t, y, 2)\n",
    "\n",
    "plt.plot(t, y, 'o')\n",
    "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n",
    "plt.show()\n",
    "%matplotlib inline\n",
    "values = v\n",
    "y = np.array(values)\n",
    "t = np.linspace(0,1,11)\n",
    "poly_coeff = np.polyfit(t, y, 2)\n",
    "\n",
    "plt.plot(t, y, 'o')\n",
    "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n",
    "plt.show()\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5.16 page number 259"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "0.166666666667\n",
      "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n"
     ]
    },
    {
     "data": {
      "image/png": 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      "text/plain": [
       "<matplotlib.figure.Figure at 0xae30b70>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "#Given\n",
    "import math \n",
    "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n",
    "l = 6.0      #a - length of the rod\n",
    "F = 1.0      #p - force applies in x direction \n",
    "d = 1.0      #a \n",
    "M = 1.0      #pa - torque applies on the rod\n",
    "l_ab = 4.0   #a application of torque point from A\n",
    "#M = 0 implies that\n",
    "R_A =  F/6.0 #p - The reaction at A\n",
    "R_B = - R_A  #F_Y = 0\n",
    "\n",
    "#Caliculations \n",
    "\n",
    "#Taking sections \n",
    "#Section 1---1\n",
    "l_1 = 1 #a - the length of the section \n",
    "M_1 = - R_A*l_1  #M = 0\n",
    "\n",
    "#Section 2---2\n",
    "l_2 = 4 #a - the length of the section \n",
    "M_2 = - R_A*l_2  #M = 0\n",
    "\n",
    "l_4 = 2 #a - the length of the section \n",
    "M_4 = 1/3.0 #pa #M = 0 '-M' because there is moment couple in between\n",
    "\n",
    "\n",
    "#Section 3---3\n",
    "l_3 = 1 #a - the length of the section \n",
    "M_3 = 1/6.0#pa M = 0 '-M' because there is moment couple in between\n",
    "print R_A\n",
    "\n",
    "#GRAPH\n",
    "#Since the symmetry exists the graphs are also symmetry\n",
    "%matplotlib inline\n",
    "#Drawing of shear and bending moment diagram\n",
    "print \"Given problem is for drawing diagram, this diagram is drawn by step by step manner. \"\n",
    "X = [0,1,4,4.00001,5,6] # For graph precision \n",
    "M = [0,M_1,M_2,M_4,M_3,0];\t\t\t#Bending moment matrix\n",
    "plot(X,M);\t\t\t#Bending moment diagram\n",
    "suptitle( 'Shear and bending moment diagram')\n",
    "xlabel('X axis')\n",
    "ylabel( 'Y axis') ;\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
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