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A Engineering_Mechanics_of_Solids_by_Popov_E_P/Chapter1.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter10.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter11.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter12.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter2.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter4.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter5.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter6.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter7.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter8.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter9.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/charpter_3.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/charpter_3_1.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/Untitled.png A Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/cap2.png A Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/cap3.png A Microwave_Engineering_by_G._S._Raghuvanshi/Ch1.ipynb A Microwave_Engineering_by_G._S._Raghuvanshi/Ch10.ipynb A Microwave_Engineering_by_G._S._Raghuvanshi/Ch11.ipynb A Microwave_Engineering_by_G._S._Raghuvanshi/Ch12.ipynb A Microwave_Engineering_by_G._S._Raghuvanshi/Ch2.ipynb A Microwave_Engineering_by_G._S._Raghuvanshi/Ch3.ipynb A Microwave_Engineering_by_G._S._Raghuvanshi/Ch4.ipynb A Microwave_Engineering_by_G._S._Raghuvanshi/Ch5.ipynb A Microwave_Engineering_by_G._S._Raghuvanshi/Ch6.ipynb A Microwave_Engineering_by_G._S._Raghuvanshi/Ch7.ipynb A Microwave_Engineering_by_G._S._Raghuvanshi/Ch8.ipynb A Microwave_Engineering_by_G._S._Raghuvanshi/Ch9.ipynb A Microwave_Engineering_by_G._S._Raghuvanshi/screenshots/12refCoeff.png A Microwave_Engineering_by_G._S._Raghuvanshi/screenshots/1RefCoeff.png A Microwave_Engineering_by_G._S._Raghuvanshi/screenshots/6axialPhasVel.png
author: Trupti Kini 2016-01-11 23:30:05 +0600
committer: Trupti Kini 2016-01-11 23:30:05 +0600
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tree: 5a9269d094b4a98947094c7c23cee3dbb3ae6c68 /Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter5.ipynb
parent: e1dbc4e45afcfcd26275010a9c0008b5d7f7e5ea (diff)
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+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:71148b069e338fd0ffabcfe7f6bd37d7ab47477a297f58633926ede4f66ab53c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Chapter 5:Axial force, Shear and Bending moment "
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example5.2 pagenumber 231"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#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\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "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"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2 page number 233"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#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"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "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"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "\n",
+      "Example 5.3 page number 233\n",
+      "\n",
+      "\n",
+      "\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#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"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "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"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4 page number 239"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#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\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "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"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5 page number 241"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#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') ;"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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gZlZyTgRmZiXnRGBmVnJOBGZmJedEYGZWck4EZmYl50RgZlZyTgRmZiXnRGBm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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x87a5ac8>"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6 pagenumber 243 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#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"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": [
+        "<matplotlib.figure.Figure at 0x8154208>"
+       ]
+      }
+     ],
+     "prompt_number": 128
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7 page number 243"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#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') ;"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": [
+        "<matplotlib.figure.Figure at 0x8154240>"
+       ]
+      }
+     ],
+     "prompt_number": 127
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8 page nmber 244"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given \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()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0.5\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": [
+        "<matplotlib.figure.Figure at 0x87884e0>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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FRKLQEbWPILVNKrcsiK6h8cMtCI2B9YWmN4TmlbTNPcDfgLwwc5SrMW+P4fpT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\nFQQREQlRQRAREQD+H2Gz7ECR6pNEAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x9e8f588>"
+       ]
+      }
+     ],
+     "prompt_number": 111
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9 page number 245"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#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"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The force and moment in section a--a are -2.33 KN , -13.644 KN-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 112
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.13 page number 254"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#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') ;"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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VtAAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x9e8f1d0>"
+       ]
+      }
+     ],
+     "prompt_number": 126
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.14 page number 255"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\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",
+      "import numpy as np\n",
+      "values = M\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()\n",
+      "import numpy as np\n",
+      "values = v\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()\n",
+      "\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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dGnl6UHPpVdyiYGMKpVCF8ArM7zWfaydey/lVzueupncVuk1S0hIGDlxIevrf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\nY/L8f6G/NIrkiT5JAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x80c3f98>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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ZdY50jZFysGNhZteXHYPFZvaxmZ0ZRJ11rTq/E2XtuphZkZn1j2R9kVTN70dK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\nbKRBM81FAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0xa487c18>"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.16 page number 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\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",
+      "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 = [0,1,4,4.00001,5,6] # 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"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0.166666666667\n",
+        "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": [
+        "<matplotlib.figure.Figure at 0x9e65940>"
+       ]
+      }
+     ],
+     "prompt_number": 150
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
author	Trupti Kini	2016-01-11 23:30:05 +0600
committer	Trupti Kini	2016-01-11 23:30:05 +0600
commit	8fa3a985b0772b28f8e6e6f5aec18bca1c0f7332 (patch)
tree	5a9269d094b4a98947094c7c23cee3dbb3ae6c68 /Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter5.ipynb
parent	e1dbc4e45afcfcd26275010a9c0008b5d7f7e5ea (diff)
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