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+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:c3ab3016805321c8a28748b1937c77b2125357dfc661b20013df0624feaa950e"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter 16: Bending of open and closed thin-walled beams \n"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 16.1 Pg. No.456"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# Fig 16.6 Dimensions\n",

+      "from __future__ import division\n",

+      "import math\n",

+      "\n",

+      "#variable declaration\n",

+      "l=300            #length of I section (mm)\n",

+      "b=200            #width of I section beam (mm)\n",

+      "w1=25            #width of center section of I beam (mm)\n",

+      "w2=20            #width of upper and lower section of I beam (mm)\n",

+      "M=100*10**6      #moment applied in vertical plane (N*mm)\n",

+      "\n",

+      "#second moment of area Ixx\n",

+      "# Ixx=b*l^3/12\n",

+      "Ixx=b*l**3/12-(b-w1)*(l-w2-w2)**3/12\n",

+      "print \"\\nSecond moment of area of I section beam = %5.3e mm^4\"%(Ixx)\n",

+      "\n",

+      "#sigma_z=My/I  reference 16.9\n",

+      "# @ \n",

+      "y=150\n",

+      "sigma_z=M*y/Ixx\n",

+      "print \"\\ndirect stress at the top of the I section (y=150) = %3.2f N/mm^2 (compression)\"%(sigma_z)\n",

+      "\n",

+      "# @\n",

+      "y=-150\n",

+      "sigma_z=M*y/Ixx\n",

+      "print \"\\ndirect stress at the bottom of the I section (y=-150) = %3.2f N/mm^2 (tension)\"%(sigma_z)\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        "Second moment of area of I section beam = 1.937e+08 mm^4\n",

+        "\n",

+        "direct stress at the top of the I section (y=150) = 77.45 N/mm^2 (compression)\n",

+        "\n",

+        "direct stress at the bottom of the I section (y=-150) = -77.45 N/mm^2 (tension)\n"

+       ]

+      }

+     ],

+     "prompt_number": 13

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 16.2 Pg. No.457"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# Fig 16.6 reference\n",

+      "from __future__ import division\n",

+      "import math\n",

+      "\n",

+      "#variable declaration\n",

+      "l=300            #length of I section (mm)\n",

+      "b=200            #width of I section beam (mm)\n",

+      "w1=25            #width of center section of I beam (mm)\n",

+      "w2=20            #width of upper and lower section of I beam (mm)\n",

+      "M=100*10**6      #moment applied in vertical plane (N*mm)\n",

+      "\n",

+      "# second moment of area\n",

+      "# Iyy=wb^3/12\n",

+      "Iyy=2*20*200**3/12+260*25**3/12\n",

+      "print \"\\nSecond moment of area of I section beam = %5.3e mm^4\"%(Iyy)\n",

+      "\n",

+      "#sigma_z=Mx/I  \n",

+      "# @ \n",

+      "x=100\n",

+      "sigma_z=M*x/Iyy\n",

+      "print \"\\ndirect stress at the top of the I section (y=150) = %3.2f N/mm^2 (compression)\"%(sigma_z)\n",

+      "\n",

+      "# @\n",

+      "x=-100\n",

+      "sigma_z=M*x/Iyy\n",

+      "print \"\\ndirect stress at the bottom of the I section (y=-150) = %3.2f N/mm^2 (tension)\"%(sigma_z)\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        "Second moment of area of I section beam = 2.701e+07 mm^4\n",

+        "\n",

+        "direct stress at the top of the I section (y=150) = 370.30 N/mm^2 (compression)\n",

+        "\n",

+        "direct stress at the bottom of the I section (y=-150) = -370.30 N/mm^2 (tension)\n"

+       ]

+      }

+     ],

+     "prompt_number": 17

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 16.3 Pg.No.458 "

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "from __future__ import division\n",

+      "import math\n",

+      "\n",

+      "#variable declaration\n",

+      "l=300            #length of I section (mm)\n",

+      "b=200            #width of I section beam (mm)\n",

+      "w1=25            #width of center section of I beam (mm)\n",

+      "w2=20            #width of upper and lower section of I beam (mm)\n",

+      "M=100*10**6      #moment applied in vertical plane (N*mm)\n",

+      "theta=30         #angle at which bending moment is applied(degree)\n",

+      "\n",

+      "Mx=M*math.cos(math.radians(30))\n",

+      "My=M*math.sin(math.radians(30))\n",

+      "\n",

+      "# sigma_z=Mx/Ixx*y+My/Iyy*x\n",

+      "# @top left hand corner\n",

+      "y=150\n",

+      "x=-100\n",

+      "sigma_z=Mx/Ixx*y-My/Iyy*x  \n",

+      "print \"\\ndirect stress at the top left hand corner = %3.1f N/mm^2 (tension)\"%(sigma_z)\n",

+      "\n",

+      "# @ top right hand corner\n",

+      "x=100\n",

+      "y=150\n",

+      "sigma_z=Mx/Ixx*y-My/Iyy*x  \n",

+      "print \"\\ndirect stress at the top right hand corner = %3.1f N/mm^2 (compression)\"%(sigma_z)\n",

+      "\n",

+      "alpha=math.atan(My*Ixx/Mx/Iyy)\n",

+      "print \"\\ninclination = %3.1f degree\\n\"%(alpha*180/math.pi)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        "direct stress at the top left hand corner = 252.2 N/mm^2 (tension)\n",

+        "\n",

+        "direct stress at the top right hand corner = -118.1 N/mm^2 (compression)\n",

+        "\n",

+        "inclination = 76.4 degree\n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 38

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 16.4 Pg.No.466"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# reference Fig 16.13\n",

+      "import math\n",

+      "from __future__ import division\n",

+      "#variable declaration\n",

+      "l=80           #length of one section (mm)\n",

+      "w=8            #thickness of each section (mm)\n",

+      "b1=40          #width of one section (mm)\n",

+      "b2=80          #width of other section (mm)\n",

+      "Mx=1500*10*3   #bending moment (N.mm)\n",

+      "My=0\n",

+      "\n",

+      "#centroid\n",

+      "# in this example C is taken at top surface x axis aligned to AB and y axis is \n",

+      "# aligned to E surface \n",

+      "y_bar=((b1+b2)*w*w/2+l*w*(w+l/2))/(l*w+(b1+b2)*w)\n",

+      "print \"\\ny coordinate of centroid = %3.1f mm\"%(y_bar)\n",

+      "\n",

+      "x_bar=((b1+b2)*w*((b1+b2)/2-b1+w/2)+l*w*(w/2))/(l*w+(b1+b2)*w)\n",

+      "print \"\\nx coordinate of centroid = %3.1f mm\"%(x_bar)\n",

+      "\n",

+      "#second area of moment\n",

+      "#IB=IC+Ab^2              IB=moment about any point,IC=moment about centroid, b=distace between both points\n",

+      "Ixx=(b1+b2)*w**3/12+(b1+b2)*w*(y_bar-w/2)**2+w*l**3/12+l*w*(w+l/2-y_bar)**2\n",

+      "print \"\\nsecond moment of area about x axis = %3.2e mm^4\"%(Ixx)\n",

+      "\n",

+      "Iyy=w*(b1+b2)**3/12+(b1+b2)*w*((b1+b2)/2-b1+w/2-x_bar)**2+l*w**3/12+l*w*(x_bar-w/2)**2\n",

+      "print \"\\nsecond moment of area about y axis = %3.2e mm^4\"%(Iyy)\n",

+      "\n",

+      "Ixy=(b1+b2)*w*(y_bar-w/2)*((b1+b2)/2-b1+w/2-x_bar)+l*w*(w+l/2-y_bar)*(x_bar-w/2)\n",

+      "print \"\\nsecond moment of area about x and y axis = %3.2e mm^4\"%(Ixy)\n",

+      "\n",

+      "Mx=15*10**5\n",

+      "def f(x,y):\n",

+      "        return Mx*(Iyy*y-Ixy*x)/(Ixx*Iyy-Ixy**2)\n",

+      "sigma_z_max= f(-8,-66.4)\n",

+      "print \"\\nmaximum direct shear stress = %3.0f N/mm^2 (compressive)\\n\"%(sigma_z_max)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "\n",

+        "y coordinate of centroid = 21.6 mm\n",

+        "\n",

+        "x coordinate of centroid = 16.0 mm\n",

+        "\n",

+        "second moment of area about x axis = 1.09e+06 mm^4\n",

+        "\n",

+        "second moment of area about y axis = 1.31e+06 mm^4\n",

+        "\n",

+        "second moment of area about x and y axis = 3.38e+05 mm^4\n",

+        "\n",

+        "maximum direct shear stress = -96 N/mm^2 (compressive)\n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 62

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [],

+     "language": "python",

+     "metadata": {},

+     "outputs": []

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
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