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-{

- "metadata": {

-  "name": "chapter no.4.ipynb"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter 4:Stresses in Beams"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.1,Page no.130"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "L=5000 #mm #Length of Beam\n",

-      "a=2000 #mm #Length of start of beam to Pt Load\n",

-      "b=3000 #mm #Length of Pt load to end of beam\n",

-      "A=150*250 #m**2 #Area of beam\n",

-      "b=150 #mm #Width of beam\n",

-      "d=250 #mm #Depth of beam\n",

-      "sigma=10#N/mm**2 #stress\n",

-      "l=2000 #m #Load applied from one end\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Moment of Inertia\n",

-      "I=1*12**-1*b*d**3 #m**4\n",

-      "\n",

-      "#Distance from N.A to end\n",

-      "y_max=d*2**-1 #m\n",

-      "\n",

-      "#Section Modulus\n",

-      "Z=1*6**-1*b*d**2 #mm**3\n",

-      "\n",

-      "#Moment Carrying Capacity\n",

-      "M=sigma*Z #N-mm\n",

-      "\n",

-      "#Let w be the Intensity of the Load in N/m,then Max moment\n",

-      "#M_max=w*L**2*8**-1 #N-mm\n",

-      "#After substituting values and further simplifying we get\n",

-      "#M_max=w*25*100*8**-1\n",

-      "\n",

-      "#EQuating it to moment carrying capacity,we get max intensity load\n",

-      "w=M*(25*1000)**-1*8*10**-3\n",

-      "\n",

-      "#Part-2\n",

-      "\n",

-      "#Let P be the concentrated load,then max moment occurs under the load and its value\n",

-      "#M1=P*a*b*L**-1 #N-mm\n",

-      "\n",

-      "#Equting it to moment carrying capacity we get\n",

-      "P=M*1200**-1*10**-3 #N\n",

-      "\n",

-      "#Result\n",

-      "print\"Max Intensity of u.d.l it can carry\",round(w,3),\"KN-m\"\n",

-      "print\"MAx concentrated Load P apllied at 2 m from one end is\",round(P,3),\"KN\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Max Intensity of u.d.l it can carry 5.0 N-mm\n",

-        "MAx concentrated Load P apllied at 2 m from one end is 13.021 KN\n"

-       ]

-      }

-     ],

-     "prompt_number": 9

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.2,Page no.131"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "D=70 #mm #External Diameter\n",

-      "t=8 #mm #Thickness of pipe\n",

-      "L=2500 #mm #span \n",

-      "sigma=150 #N/mm**2 #stress\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Internal Diameter \n",

-      "d=D-2*t #mm\n",

-      "\n",

-      "#M.I Of Pipe\n",

-      "I=pi*64**-1*(D**4-d**4) #mm**4\n",

-      "\n",

-      "y_max=D*2**-1 #mm\n",

-      "Z=I*(y_max)**-1 #mm**3\n",

-      "\n",

-      "#Moment Carrying capacity\n",

-      "M=sigma*Z #N*mm\n",

-      "\n",

-      "#Max moment int the beam occurs at the mid-span and is equal to\n",

-      "#m=P*L*4**-1\n",

-      "\n",

-      "#Equating Max moment to moment carrying capacity we get,\n",

-      "#M=P*2.5*L*4**-1\n",

-      "#After substituting and simplifying we get\n",

-      "P=4*M*(L)**-1*10**-3 #N\n",

-      "\n",

-      "#Result\n",

-      "print\"Max concentrated load that can be applied at the centre of span is\",round(P,3),\"KN\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Max concentrated load that can be applied at the centre of span is 5.22 KN\n"

-       ]

-      }

-     ],

-     "prompt_number": 9

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.3,Page no.132"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "#Flanges Dimension\n",

-      "b1=180 #mm #Width\n",

-      "d1=10 #mm #Thickness\n",

-      "\n",

-      "D=500 #mm #Overall depth\n",

-      "t=8 #mm #Thickness of web\n",

-      "\n",

-      "#Plate Dimensions\n",

-      "b2=240 #mm #Width\n",

-      "t2=12 #mm #Thickness\n",

-      "\n",

-      "sigma=150 #N/mm**2 #Stress\n",

-      "L=3000 #mm #span\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Distance of centroid from bottom fibre\n",

-      "y_bar=(b2*t2*(D+t2*2**-1)+b1*d1*(D-t1*2**-1)+(D-2*t1)*t*D*2**-1+(b1*t1*t1*2**-1))*(b2*t2+b1*d1+b1*d1+(D-2*d1)*t)**-1\n",

-      "\n",

-      "#M.I of section\n",

-      "I=(1*12**-1*b2*t2**3+b2*t2*(D+t2*2**-1-y_bar)**2+1*12**-1*b1*d1**3+b1*d1*(D-t1*2**-1-y_bar)**2+1*12**-1*b1*t1**3+b1*t1*(t1*2**-1-y_bar)**2+1*12**-1*t*(D-2*t1)**3+t*(D-2*t1)*(D*2**-1-y_bar)**2)\n",

-      "\n",

-      "#Section Modulus\n",

-      "Z=I*(y_bar)**-1 #mm**3\n",

-      "\n",

-      "#Moment or Resistance\n",

-      "M=sigma*Z\n",

-      "\n",

-      "#Let Load on Cantilever be w/m Length \n",

-      "#Max M.I produced\n",

-      "#M_max=w*L**2**-1 \n",

-      "\n",

-      "#Now Equating Moment of resistance to Max moment,we get Max load\n",

-      "#4.5*w=M\n",

-      "#After rearranging and further simplifying we get\n",

-      "w=M*4.5**-1*10**3*10**-9\n",

-      "\n",

-      "#Result\n",

-      "print\"Moment of Resistance is\",round(M,2),\"KN-mm\"\n",

-      "print\"Load the section can carry is\",round(w,3),\"KN/m\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Moment of Resistance is 198770121.83 KN-mm\n",

-        "Load the section can carry is 44.171 KN/m\n"

-       ]

-      }

-     ],

-     "prompt_number": 26

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.4,Page no.134"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "#Flange (Top)\n",

-      "b1=80 #mm #Width \n",

-      "t1=40 #mm #Thickness\n",

-      "\n",

-      "#Flange (Bottom)\n",

-      "b2=160 #mm #width\n",

-      "t2=40 #mm #Thickness\n",

-      "\n",

-      "#web\n",

-      "d=120 #mm #Depth\n",

-      "t3=20 #mm #Thickness\n",

-      "\n",

-      "D=200 #mm #Overall Depth\n",

-      "sigma1=30 #N/mm**2 #Tensile stress\n",

-      "sigma2=90 #N/mm**2 #Compressive stress\n",

-      "L=6000 #mm #Span\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Distance of centroid from bottom fibre\n",

-      "y_bar=(b1*t1*(D-t1*2**-1)+d*t3*(d*2**-1+t2)+b2*t2*t2*2**-1)*(b1*t1+d*t3+b2*t2)**-1 #mm\n",

-      "\n",

-      "#Moment of Inertia\n",

-      "I=1*12**-1*b1*t1**3+b1*t1*(D-t1*2**-1-round(y_bar,2))**2+1*12**-1*t3*d**3+t3*d*(d*2**-1+t2-round(y_bar,2))**2+1*12**-1*b2*t2**3+b2*t2*(t2*2**-1-round(y_bar,2))**2\n",

-      "\n",

-      "#Extreme fibre distance of top and bottom fibres are y_t and y_c respectively\n",

-      "\n",

-      "y_t=y_bar #mm\n",

-      "y_c=D-y_bar #mm\n",

-      "\n",

-      "#Moment carrying capacity considering Tensile strength \n",

-      "M1=sigma1*I*y_t**-1*10**-6  #KN-m\n",

-      "\n",

-      "#Moment carrying capacity considering compressive strength \n",

-      "M2=sigma2*I*y_c**-1*10**-6  #KN-m\n",

-      "\n",

-      "#Max Bending moment in simply supported beam 6 m due to u.d.l\n",

-      "#M_max=w*L*10**-3*8**-1\n",

-      "#After simplifying further we get\n",

-      "#M_max=4.5*w\n",

-      "\n",

-      "#Now Equating it to Moment carrying capacity, we get load carrying capacity\n",

-      "w=M1*4.5**-1 #KN/m\n",

-      "\n",

-      "#Result\n",

-      "print\"Max Uniformly Distributed Load is\",round(w,3),\"KN/m\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Max Uniformly Distributed Load is 5.096 KN/m\n"

-       ]

-      }

-     ],

-     "prompt_number": 16

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.5,Page no.136"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "from scipy.integrate import quad\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "#Flanges\n",

-      "b=200 #mm #Width\n",

-      "t=25 #mm #Thickness \n",

-      "\n",

-      "D1=500 #mm #Overall Depth\n",

-      "t2=20 #mm #Thickness of web\n",

-      "\n",

-      "d=450 #mm #Depth of web\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Consider,Element of Thickness \"y\" at Distance \"dy\" from N.A \n",

-      "#Let Bending stress \"sigma_max\"\n",

-      "\n",

-      "#Stress on the element \n",

-      "#sigma=y*(D*2**-1)*sigma_max   ..............(1)\n",

-      "\n",

-      "#Area of Element\n",

-      "#A=b*dy       .................................(2)\n",

-      "\n",

-      "#Force on Element \n",

-      "#F=y*250**-1*sigma_max*b*dy\n",

-      "\n",

-      "#Let M be the Moment of resistance\n",

-      "#M=y*250**-1*sigma_max*b*dy*y\n",

-      "\n",

-      "#Moment of Resistance of top flange be M1\n",

-      "def integrand(y, b, D):\n",

-      "     return b*y**2*D**-1\n",

-      "b=200 \n",

-      "D=250\n",

-      "\n",

-      "X = quad(integrand, 225, 250, args=(b,D))\n",

-      "\n",

-      "Y=2*X[0]\n",

-      "\n",

-      "#M1=Y*sigma\n",

-      "\n",

-      "#Now Moment of Inertia I section is\n",

-      "X=b*D1**3\n",

-      "Y=(b-t2)*d**3\n",

-      "I=(X-Y)*12**-1*10**-8\n",

-      "\n",

-      "#Moment acting on the entire section\n",

-      "#since sigmais the value at y=250\n",

-      "y_max=250\n",

-      "Z=I*10**8*y_max**-1\n",

-      "#M=sigma*Z   \n",

-      "#After Simplifying Further we get\n",

-      "#M2=Z*sigma\n",

-      "\n",

-      "#Percentage Moment resisted by Flanges\n",

-      "P1=2258333.3*(2865833.3)**-1*100\n",

-      "\n",

-      "#Percentage Moment resisted by web\n",

-      "P2=100-P1\n",

-      "\n",

-      "#Result\n",

-      "print\"Percentage Moment resisted by Flanges\",round(P1,2),\"%\"\n",

-      "print\"Percentage Moment resisted by web\",round(P2,2),\"%\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Percentage Moment resisted by Flanges 78.8 %\n",

-        "Percentage Moment resisted by web 21.2 %\n"

-       ]

-      }

-     ],

-     "prompt_number": 38

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.6,Page no.137"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "#Flanges\n",

-      "b1=200 #mm #Width\n",

-      "t1=10 #mm #Thickness\n",

-      "\n",

-      "#Web\n",

-      "d=380 #mm #Depth \n",

-      "t2=8 #mm #Thickness\n",

-      "\n",

-      "D=400 #mm #Overall Depth\n",

-      "sigma=150 #N/mm**2\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Area\n",

-      "A=b1*t1+d*t2+b1*t1 #mm**2\n",

-      "\n",

-      "#Moment of Inertia\n",

-      "I=1*12**-1*(b1*D**3-(b1-t2)*d**3)\n",

-      "\n",

-      "#Bending Moment\n",

-      "M=sigma*I*(D*2**-1)**-1\n",

-      "\n",

-      "#Square Section\n",

-      "\n",

-      "#Let 'a' be the side\n",

-      "a=A**0.5\n",

-      "\n",

-      "#Moment of Resistance of this section\n",

-      "M1=1*6**-1*a*a**2*sigma\n",

-      "\n",

-      "X=M*M1**-1\n",

-      "\n",

-      "#Rectangular section\n",

-      "#Let 'a' be the side and depth be 2*a\n",

-      "\n",

-      "a=(A*2**-1)**0.5\n",

-      "\n",

-      "#Moment of Rectangular secction\n",

-      "M2=1*6**-1*a*(2*a)**2*sigma\n",

-      "\n",

-      "X2=M*M2**-1\n",

-      "\n",

-      "#Circular section\n",

-      "#A=pi*d1**2*4**-1\n",

-      "\n",

-      "d1=(A*4*pi**-1)**0.5\n",

-      "\n",

-      "#Moment of circular section\n",

-      "M3=pi*32**-1*d1**3*sigma\n",

-      "\n",

-      "X3=M*M3**-1\n",

-      "\n",

-      "#Result\n",

-      "print\"Moment of resistance of beam section\",round(M,2),\"mm\"\n",

-      "print\"Moment of resistance of square section\",round(X,2),\"mm\"\n",

-      "print\"Moment of resistance of rectangular section\",round(X2,2),\"mm\"\n",

-      "print\"Moment of resistance of circular section\",round(X3,2),\"mm\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Moment of resistance of beam section 141536000.0 mm\n",

-        "Moment of resistance of square section 9.58 mm\n",

-        "Moment of resistance of rectangular section 6.78 mm\n",

-        "Moment of resistance of circular section 11.33 mm\n"

-       ]

-      }

-     ],

-     "prompt_number": 25

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.7,Page no.139"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "F=12 #KN #Force at End of beam\n",

-      "L=2 #m #span\n",

-      "\n",

-      "#Square section \n",

-      "b=d=200 #mm #Width and depth of beam\n",

-      "\n",

-      "#Rectangular section\n",

-      "b1=150 #mm #Width\n",

-      "d1=300 #mm #Depth\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Max bending Moment\n",

-      "M=F*L*10**6 #N-mm\n",

-      "\n",

-      "#M=sigma*b*d**2\n",

-      "sigma=M*6*(b*d**2)**-1 #N/mm**2\n",

-      "\n",

-      "#Let W be the central concentrated Load in simply supported beam of span L1=3 m\n",

-      "#MAx Moment\n",

-      "#M1=W*L1*4**-1\n",

-      "#After Further simplifying we get\n",

-      "#M1=0.75*10**6 #N-mm\n",

-      "\n",

-      "#The section has  a moment of resistance\n",

-      "M1=sigma*1*6**-1*b1*d1**2\n",

-      "\n",

-      "#Equating it to moment of resistance we get max load W\n",

-      "#0.75*10**6*W=M1\n",

-      "#After Further simplifying we get\n",

-      "W=M1*(0.75*10**6)**-1\n",

-      "\n",

-      "#Result\n",

-      "print\"Minimum Concentrated Load required to brek the beam\",round(W,2),\"KN\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Minimum Concentrated Load required to brek the beam 54.0 KN\n"

-       ]

-      }

-     ],

-     "prompt_number": 30

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.8,Page no.140"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "L=3 #m #span\n",

-      "sigma_t=35 #N/mm**2 #Permissible stress in tension\n",

-      "sigma_c=90 #N/mm**2 #Permissible stress in compression\n",

-      "\n",

-      "#Flanges\n",

-      "t=30 #mm #Thickness\n",

-      "d=250 #mm #Depth\n",

-      "\n",

-      "#Web\n",

-      "t2=25 #mm #Thickness\n",

-      "b=600 #mm #Width\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Let y_bar be the Distance of N.A from Extreme Fibres\n",

-      "y_bar=(t*d*d*2**-1*2+(b-2*t)*t2*t2*2**-1)*(t*d*2+(b-2*t)*t2)**-1\n",

-      "\n",

-      "#Moment of Inertia\n",

-      "I=(1*12**-1*t*d**3+t*d*(d*2**-1-y_bar)**2)*2+1*12**-1*(b-2*t)*t2**3+(b-2*t)*t2*(t2*2**-1-y_bar)**2\n",

-      "\n",

-      "#Part-1\n",

-      "\n",

-      "#If web is in Tension\n",

-      "y_t=y_bar #mm\n",

-      "y_c=d-y_bar #mm\n",

-      "\n",

-      "#Moment carrying caryying capacity From consideration of tensile stress\n",

-      "M=sigma_t*I*(y_bar)**-1 #N-mm\n",

-      "\n",

-      "#Moment carrying caryying capacity From consideration of compressive stress\n",

-      "M1=sigma_c*I*(y_c)**-1 #N-mm\n",

-      "\n",

-      "#If w KN/m is u.d.l in beam,Max bending moment\n",

-      "#M=wl**2*8**-1\n",

-      "#After further simplifyng we get\n",

-      "#M=1.125*w*10**6 N-mm\n",

-      "w=M*(1.125*10**6)**-1 #KN\n",

-      "\n",

-      "#Part-2\n",

-      "\n",

-      "#If web is in compression\n",

-      "y_t2=178.299 #mm\n",

-      "y_c2=71.71 #mm \n",

-      "\n",

-      "#Moment carrying caryying capacity From consideration of tensile stress\n",

-      "M2=sigma_t*I*(y_t2)**-1 #N-mm\n",

-      "\n",

-      "#Moment carrying caryying capacity From consideration of compressive stress\n",

-      "M3=sigma_c*I*(y_c2)**-1 #N-mm\n",

-      "\n",

-      "#Moment of resistance is M2\n",

-      "\n",

-      "#Equating it to bending moment we get\n",

-      "#M2=1.125*10**6*w2\n",

-      "#After further simplifyng we get\n",

-      "w2=M2*(1.125*10**6)**-1\n",

-      "\n",

-      "#Result\n",

-      "print\"Uniformly Distributed Load carrying capacity if:web is in Tension\",round(w,2),\"KN\"\n",

-      "print\"                                               :web is in compression\",round(w2,3),\"KN\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Uniformly Distributed Load carrying capacity if:web is in Tension 73.21 KN\n",

-        "                                               :web is in compression 29.446 KN\n"

-       ]

-      }

-     ],

-     "prompt_number": 26

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.9,Page no.141"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "b1=200 #mm #Width at base\n",

-      "b2=100 #mm #Width at top\n",

-      "\n",

-      "L=8 #m Length\n",

-      "P=500 #N #Load\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Consider a section at y metres from top\n",

-      "\n",

-      "#At this section diameter d is\n",

-      "#d=b2+y*L**-1*(b1-b2)\n",

-      "#After Further simplifying we get\n",

-      "#d=b2+12.5*y #mm\n",

-      "\n",

-      "#Moment of Inertia\n",

-      "#I=pi*64**-1*d**4\n",

-      "\n",

-      "#Section Modulus \n",

-      "#Z=pi*32**-1*(b1+12.5*y)**3\n",

-      "\n",

-      "#Moment \n",

-      "#M=5*10**5*y #N-mm\n",

-      "\n",

-      "#Let sigma be the fibre stress at this section then\n",

-      "#M=sigma*Z\n",

-      "#After sub values in above equation and further simplifying we get\n",

-      "#sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1\n",

-      "\n",

-      "#For sigma to be Max,d(sigma)*(dy)**-1=0\n",

-      "#16*10**6*pi**-1*((b2+12.5*y)**-3+y*(-3)*(b2+12.5*y)**-4*12.5)\n",

-      "#After Further simplifying we get\n",

-      "#b2+12.5*y=37.5*y\n",

-      "#After Further simplifying we get\n",

-      "y=b2*25**-1 #m\n",

-      "\n",

-      "#Stress at this section\n",

-      "sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1\n",

-      "\n",

-      "#Result\n",

-      "print\"Stress at Extreme Fibre is max\",round(y,2),\"m\"\n",

-      "print\"Max stress is\",round(sigma,2),\"N/mm**2\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Stress at Extreme Fibre is max 4.0 m\n",

-        "Max stress is 6.04 N/mm**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 28

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.10,Page no.143"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "H=10 #mm #Height\n",

-      "A1=160*160 #mm**2 #area of square section at bottom\n",

-      "L1=160 #mm #Length of square section at bottom\n",

-      "b1=160 #mm #width of square section at bottom\n",

-      "A2=80*80 #mm**2 #area of square section at top\n",

-      "L2=80 #mm #Length of square section at top\n",

-      "b2=80 #mm #Width of square section at top\n",

-      "P=100 #N #Pull\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Consider a section at distance y from top.\n",

-      "#Let the side of square bar be 'a'\n",

-      "#a=L2+y*(H)**-1*(b1-b2)\n",

-      "#After further simplifying we get\n",

-      "#a=L2+8*y\n",

-      "\n",

-      "#Moment of Inertia\n",

-      "#I=2*1*12**-1*a*(2)**0.5*(a*((2)**0.5)**-1)**3\n",

-      "#After further simplifying we get\n",

-      "#I=a**4*12**-1\n",

-      "\n",

-      "#Section Modulus \n",

-      "#Z=a**4*(12*a*(2)**0.5)**-1\n",

-      "#After further simplifying we get\n",

-      "#Z=2**0.5*a**3*(12)**-1 #mm**3\n",

-      "\n",

-      "#Bending moment at this section=100*y N-mm\n",

-      "#M=100*10**3*y #N-mm\n",

-      "\n",

-      "#But\n",

-      "#M=sigma*Z\n",

-      "#After sub values in above equation we get\n",

-      "#sigma=M*Z**-1\n",

-      "#After further simplifying we get\n",

-      "#sigma=1200*10**3*(2**0.5)**-1*y*((80+80*y)**3)**-1 .......(1)\n",

-      "\n",

-      "#For Max stress df*(dy)**-1=0\n",

-      "#After taking Derivative of above equation we get\n",

-      "#df*(dy)**-1=1200*10**3*(2**0.5)**-1*((80+8*y)**-3+y(-3)*(80+8*y)**-4*8)\n",

-      "#After further simplifying we get\n",

-      "y=80*16**-1 #m\n",

-      "\n",

-      "#Max stress at this level is\n",

-      "sigma=1200*10**3*(2**0.5)**-1*y*((80+8*y)**3)**-1\n",

-      "\n",

-      "#Result\n",

-      "print\"Max Bending stress is Developed at\",round(y,3),\"m\"\n",

-      "print\"Value of Max Bending stress is\",round(sigma,3),\"N/mm**2\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Max Bending stress is Developed at 5.0 m\n",

-        "Value of Max Bending stress is 2.455 N/mm**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 4

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.12,Page no.147"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "b=200 #mm #Width of timber \n",

-      "d=400 #mm #Depth of timber\n",

-      "t=6 #mm #Thickness\n",

-      "b2=200 #mm #width of steel plate\n",

-      "t2=20 #mm #Thickness of steel plate\n",

-      "M=40*10**6 #KN-mm #Moment\n",

-      "#Let E_s*E_t**-1=X\n",

-      "X=20 #Ratio of Modulus of steel to timber\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#let y_bar be the Distance of centroidfrom bottom most fibre\n",

-      "y_bar=(b*d*(b+t)+t2*b2*t*t*2**-1)*(b*d+t2*b2*t)**-1 #mm\n",

-      "\n",

-      "#Moment of Inertia\n",

-      "I=1*12**-1*b*d**3+b*d*(b+t-round(y_bar,3))**2+1*12**-1*t2*b2*t**3+b2*t2*t*(round(y_bar,3)-t*2**-1)**2\n",

-      "\n",

-      "#distance of the top fibre from N-A\n",

-      "y_1=d+t-y_bar #mm\n",

-      "\n",

-      "#Distance of the junction of timber and steel From N-A\n",

-      "y_2=y_bar-t #mm\n",

-      "\n",

-      "#Stress in Timber at the top\n",

-      "Y=M*I**-1*y_1 #N/mm**2\n",

-      "\n",

-      "#Stress in the Timber at the junction point\n",

-      "Z=M*I**-1*y_2\n",

-      "\n",

-      "#Coressponding stress in steel at the junction point\n",

-      "Z2=X*Z #N/mm**2 \n",

-      "\n",

-      "#The stress in Extreme steel fibre \n",

-      "Z3=X*M*I**-1*y_bar\n",

-      "\n",

-      "#Result\n",

-      "print\"Stress in Extreme steel Fibre\",round(Z3,2),\"N/mm**2\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Stress in Extreme steel Fibre 69.67 N/mm**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 23

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.13,Page no.149"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "#Timber size\n",

-      "b=150 #mm #Width\n",

-      "b2=120 #mm \n",

-      "d=300 #mm #Depth\n",

-      "\n",

-      "t=6 #mm #Thickness of steel plate\n",

-      "L=6 #m #Span\n",

-      "\n",

-      "#E_s*E_t**-1=20 \n",

-      "#X=E_s*E_t**-1\n",

-      "X=20 \n",

-      "sigma_timber=8 #N/mm**2  #Stress in timber\n",

-      "sigma_steel=150 #N/mm**2 #Stress in steel plate\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "Y\n",

-      "\n",

-      "#Due to  synnetry cenroid,the neutral axis is half the depth\n",

-      "I=1*12**-1*\n",

-      "\n",

-      "\n",

-      "#Result\n",

-      "print Z"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "153.0\n"

-       ]

-      }

-     ],

-     "prompt_number": 34

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.14,Page no.151"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "L=6000 #mm #Span of beam\n",

-      "W=20*10**3 #N #Load\n",

-      "sigma=8 #N/mm**2 #Stress\n",

-      "b=200 #mm #Width of section\n",

-      "d=300 #mm #Depth of section\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#let x be the distance from left side of beam\n",

-      "\n",

-      "#Bending moment\n",

-      "#M=W*2**-1*x #Nmm   .......(1)\n",

-      "\n",

-      "#But M=sigma*Z   ..........(2)\n",

-      "\n",

-      "#Equating equation 1 and 2 we get\n",

-      "#W*2**-1*x=sigma*Z    ............(3)\n",

-      "\n",

-      "#Section Modulus \n",

-      "#Z=1*6*b*d**2   ...............(4)\n",

-      "\n",

-      "#Equating equation 3 and 4 we get\n",

-      "#b*d**2=3*W*x*sigma**-1 .............(5)\n",

-      "\n",

-      "#Beam of uniform strength of constant depth\n",

-      "#b=3*W*x*(sigma*d**2)    \n",

-      "\n",

-      "#When x=0\n",

-      "b=0\n",

-      "\n",

-      "#When x=L*2**-1\n",

-      "b2=3*W*L*(2*sigma*d**2)**-1 #mm\n",

-      "\n",

-      "#Beam with constant width of 200 mm\n",

-      "\n",

-      "#We have\n",

-      "#d=(3*W*x*(sigma*d)**-1)**0.5\n",

-      "#thus depth varies as (x)**0.5\n",

-      "\n",

-      "#when x=0\n",

-      "d1=0\n",

-      "\n",

-      "#when x=L*2**-1\n",

-      "d2=(2*W*L*(2*sigma*300)**-1)**0.5 #mm\n",

-      "\n",

-      "#Result\n",

-      "print\"Cross section of rectangular beam is:\",round(b2,2),\"mm\"\n",

-      "print\"                                    :\",round(d2,2),\"mm\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Cross section of rectangular beam is: 250.0 mm\n",

-        "                                    : 223.61 mm\n"

-       ]

-      }

-     ],

-     "prompt_number": 22

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.15,Page no.154"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "L=800 #mm #Span\n",

-      "n=5 #number of leaves\n",

-      "b=60 #mm #Width\n",

-      "t=10 #mm #thickness\n",

-      "sigma=250 #N/mm**2 #Stress\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#section Modulus\n",

-      "Z=n*6**-1*b*t**2 #mm**3\n",

-      "\n",

-      "#from the relation\n",

-      "#sigma*Z=M   ...................(1)\n",

-      "#M=P*L*4**-1\n",

-      "#sub values of M in equation 1 we get\n",

-      "P=sigma*Z*4*L**-1*10**-3 #KN #Load\n",

-      "\n",

-      "#Length of Leaves\n",

-      "L1=0.2*L #mm\n",

-      "L2=0.4*L #mm\n",

-      "L3=0.6*L #mm\n",

-      "L4=0.8*L #mm\n",

-      "L5=L #mm\n",

-      "\n",

-      "#Result\n",

-      "print\"Max Load it can take is\",round(P,2),\"KN\"\n",

-      "print\"Length of leaves:L1\",round(L1,2),\"mm\"\n",

-      "print\"                :L2\",round(L2,2),\"mm\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Max Load it can take is 6.25 KN\n",

-        "Length of leaves:L1 160.0 mm\n",

-        "                :L2 320.0 mm\n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.16,Page no.161"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "%matplotlib  inline\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "F=20*10**3 #N #Shear Force\n",

-      "\n",

-      "#Tee section\n",

-      "\n",

-      "#Flange\n",

-      "b=100 #mm #Width\n",

-      "t=12 #mm #Thickness\n",

-      "\n",

-      "#Web\n",

-      "d=88 #mm #Depth\n",

-      "t2=12 #mm #Thicknes\n",

-      "\n",

-      "D=100 #mm #Overall Depth\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Distance of C.G from Top Fibre\n",

-      "y=(b*t*t*2**-1+t2*d*(d*2**-1+t))*(b*t+d*t2)**-1 #mm \n",

-      "\n",

-      "#Moment Of Inertia\n",

-      "I=1*12**-1*b*t**3+b*t*(y-t*2**-1)**2+1*12**-1*t2*d**3+t2*d*(t+d*2**-1-y)**2 #mm**4\n",

-      "\n",

-      "#shear stress at bottom Flange\n",

-      "\n",

-      "#Area above this level\n",

-      "A=b*t #mm**2\n",

-      "\n",

-      "#C.G of this area from N-A\n",

-      "y2=y-t*2**-1\n",

-      "\n",

-      "#Stress at bottom of flange\n",

-      "sigma=F*A*y2*(b*I)**-1 #N/mm**2 \n",

-      "\n",

-      "#sigma2 at same level but in web where width is 12 mm\n",

-      "sigma2=F*A*y2*(t2*I)**-1 #N/mm**2 \n",

-      "\n",

-      "#To find shear stress at N-A\n",

-      "X=t*b*(y-t*2**-1)+t2*(y-t2)*(y-t2)*2**-1 #mm**3\n",

-      "\n",

-      "sigma3=F*X*(t2*I)**-1 #N/mm**2\n",

-      "\n",

-      "#Shear stress at top and bottom fibre is zero\n",

-      "#sigma4 and sigma5 are top and bottom fibre shear stress\n",

-      "sigma4=sigma5=0\n",

-      "\n",

-      "#Result\n",

-      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",

-      "\n",

-      "#Plotting the Shear Force Diagram\n",

-      "\n",

-      "X1=[0,t,t,y,D]\n",

-      "Y1=[sigma4,sigma,sigma2,sigma3,sigma5]\n",

-      "Z1=[0,0,0,0,0]\n",

-      "plt.plot(X1,Y1,X1,Z1)\n",

-      "plt.xlabel(\"Length x in m\")\n",

-      "plt.ylabel(\"Shear Force in kN\")\n",

-      "plt.show()\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The Shear Force and Bending Moment Diagrams are the results\n"

-       ]

-      },

-      {

-       "metadata": {},

-       "output_type": "display_data",

-       "png": 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-       "text": [

-        "<matplotlib.figure.Figure at 0x56a6270>"

-       ]

-      }

-     ],

-     "prompt_number": 15

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.17,Page no.163"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "%matplotlib  inline\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "F=40*10**3 #N #shear Force\n",

-      "\n",

-      "#I-section\n",

-      "\n",

-      "#Flanges\n",

-      "b=80 #mm #Width of flange\n",

-      "t=20 #mm #Thickness\n",

-      "\n",

-      "#Web\n",

-      "d=200 #mm #Depth\n",

-      "t2=20 #mm #Thickness\n",

-      "\n",

-      "#Flange-2\n",

-      "b2=160 #mm #Width\n",

-      "t3=20 #mm #Thickness\n",

-      "\n",

-      "D=240 #mm #Overall Depth\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Distance of N-A from Top Fibre \n",

-      "y=(b*t*t*2**-1+d*t2*(t+d*2**-1)+b2*t3*(t+d+t3*2**-1))*(b*t+d*t2+b2*t3)**-1 #mm\n",

-      "\n",

-      "#Moment of Inertia\n",

-      "I=1*12**-1*b*t**3+b*t*(y-(t*2**-1))**2+1*12**-1*t2*d**3+t2*d*(y-(t+d*2**-1))**2+1*12**-1*b2*t3**3+t3*b2*((d+t+t3*2**-1)-y)**2 #mm**4\n",

-      "\n",

-      "#Shear stress bottom of flange\n",

-      "sigma=F*b*t*(y-t*2**-1)*(b*I)**-1 #N/mm**2\n",

-      "\n",

-      "#At same Level but in web\n",

-      "sigma2=F*b*t*(y-t*2**-1)*(t2*I)**-1 #N/mm**2\n",

-      "\n",

-      "#for shear stress at N.A\n",

-      "X=b*t*(y-t*2**-1)+t2*(y-t)*(y-t)*2**-1 #mm**3\n",

-      "sigma3=F*X*(t2*I)**-1 #N/mm**2\n",

-      "\n",

-      "#Shear stress at bottom of web\n",

-      "\n",

-      "X=b2*t3*((D-y)-t3*2**-1) #mm**3\n",

-      "\n",

-      "#Stress at bottom of web\n",

-      "sigma4=F*X*(t2*I)**-1 #N/mm**2\n",

-      "\n",

-      "#Stress at Lower flange\n",

-      "sigma5=F*X*(b2*I)**-1 #N/mm**2\n",

-      "\n",

-      "#Result\n",

-      "print \"The Shear Force Diagram is the result\"\n",

-      "\n",

-      "#Plotting the Shear Force Diagram\n",

-      "\n",

-      "X1=[0,20,20,140,220,220,240]\n",

-      "Y1=[0,sigma,sigma2,sigma3,sigma4,sigma5,0]\n",

-      "Z1=[0,0,0,0,0,0,0]\n",

-      "plt.plot(X1,Y1,X1,Z1)\n",

-      "plt.xlabel(\"Length in mm\")\n",

-      "plt.ylabel(\"Shear Force in N\")\n",

-      "plt.show()\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The Shear Force Diagram is the result\n"

-       ]

-      },

-      {

-       "metadata": {},

-       "output_type": "display_data",

-       "png": 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OnYL/+i/TF3DLLWZPgXbt7K5KRHyRwsAGJ0+afoBXXjHNQOvXQ1iY3VWJiC8r1gj183MN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\n/w/qZz1xEBCKMwAAAABJRU5ErkJggg==\n",

-       "text": [

-        "<matplotlib.figure.Figure at 0x5614110>"

-       ]

-      }

-     ],

-     "prompt_number": 5

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.18,Page no.164"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "F=30*10**3 #N #Shear Force\n",

-      "\n",

-      "#Channel Section\n",

-      "d=400 #mm #Depth of web \n",

-      "t=10 #mm #THickness of web\n",

-      "t2=15 #mm #Thickness of flange\n",

-      "b=100 #mm #Width of flange\n",

-      "\n",

-      "#Rectangular Welded section\n",

-      "b2=80 #mm #Width\n",

-      "d2=60 #mm #Depth\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Distance of Centroid From Top Fibre\n",

-      "y=(d*t*t*2**-1+2*t2*(b-t)*((b-t)*2**-1+10)+d2*b2*(d2*2**-1+t))*(d*t+2*t2*(b-t)+d2*b2)**-1 #mm\n",

-      "\n",

-      "#Moment Of Inertia of the section about N-A\n",

-      "I=1*12**-1*d*t**3+d*t*(y-t*2**-1)**2+2*(1*12**-1*t2*(b-t)**3+t2*(b-t)*(((b-t)*2**-1+t)-y)**2)+1*12**-1*d2**3*b2+d2*b2*(d2*2**-1+t-y)**2\n",

-      "\n",

-      "#Shear stress at level of weld\n",

-      "sigma=F*d*t*(y-t*2**-1)*((b2+t2+t2)*I)**-1 #N/mm**2\n",

-      "\n",

-      "#Max Shear Stress occurs at Neutral Axis\n",

-      "X=d*t*(y-t*2**-1)+2*t2*(y-t)*(y-t)*2**-1+b2*(y-t)*(y-t)*2**-1\n",

-      "\n",

-      "sigma_max=F*X*((b+t)*I)**-1\n",

-      "\n",

-      "#Result\n",

-      "print\"Shear stress in the weld is\",round(sigma,2),\"N/mm**2\"\n",

-      "print\"Max shear stress is\",round(sigma_max,2),\"N/mm**2\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Shear stress in the weld is 3.62 N/mm**2\n",

-        "Max shear stress is 4.48 N/mm**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 13

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.19,Page no.165"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "#Wooden Section\n",

-      "b=300 #mm #Width\n",

-      "d=300 #mm #Depth\n",

-      "\n",

-      "D=100 #mm #Diameter of Bore\n",

-      "F=10*10**3 #N #Shear Force\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Moment Of Inertia Of Section\n",

-      "I=1*12**-1*b*d**3-pi*64**-1*D**4\n",

-      "\n",

-      "#Shear stress at crown of circle\n",

-      "sigma=F*b*D*(d*2**-1-D*2**-1)*(b*I)**-1\n",

-      "\n",

-      "#Let a*y_bar=X\n",

-      "X=b*d*2**-1*d*4**-1-pi*8**-1*D**2*4*D*2**-1*(3*pi)**-1 #mm**3\n",

-      "\n",

-      "#Shear Stress at Neutral Axis\n",

-      "sigma2=F*X*((b-D)*I)**-1 #N/mm**2\n",

-      "\n",

-      "#Result\n",

-      "print\"Shearing Stress at Crown of Bore\",round(sigma,3),\"N/mm**2\"\n",

-      "print\"Shear Stress at Neutral Axis\",round(sigma2,3),\"N/mm**2\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Shearing Stress at Crown of Bore 0.149 N/mm**2\n",

-        "Shear Stress at Neutral Axis 0.246 N/mm**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 10

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.20,Page no.166"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "#flanges\n",

-      "b=200 #mm #width\n",

-      "t1=25 #mm #Thickness\n",

-      "\n",

-      "#web\n",

-      "d=450 #mm #Depth \n",

-      "t2=20 #mm #thickness\n",

-      "\n",

-      "D=500 #mm #Total Depth of section\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Moment Of Inertia of the section about N-A\n",

-      "I=1*12**-1*b*D**3-1*12**-1*(b-t2)*d**3 #mm**4\n",

-      "\n",

-      "#Consider an element in the web at distance y from y from N-A\n",

-      "#Depth of web section=225-y\n",

-      "\n",

-      "#C.G From N-A\n",

-      "#y2=y+(((D*2**-1-t)-y)*2**-1)\n",

-      "\n",

-      "#ay_bar for section at y\n",

-      "#Let ay_bar be X\n",

-      "#X=X1 be of Flange + X2 be of web above y\n",

-      "#X=b*t1*(D*2**-1-t1*2**-1)+t2*(d-t1)*(d-t1+y)*2**-1\n",

-      "#After Sub values and Further simplifying we get\n",

-      "#X=1187500+10*(225**2-y**2)\n",

-      "\n",

-      "#Shear stress at y\n",

-      "#sigma_y=F*(X)*(t2*I)**-1\n",

-      "\n",

-      "#Shear Force resisted by the Element\n",

-      "#F1=F*X*t2*dy*(t2*I)**-1\n",

-      "\n",

-      "#Shear stress resisted by web \n",

-      "#sigma=2*F*I**-1*(X)*dy\n",

-      "\n",

-      "#After Integrating above equation and further simplifying we get\n",

-      "#sigma=0.9578*F\n",

-      "\n",

-      "sigma=0.9578*100\n",

-      "\n",

-      "#Result\n",

-      "print\"Shear Resisted by web\",round(sigma,2),\"%\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Shear Resisted by web 95.78 %\n"

-       ]

-      }

-     ],

-     "prompt_number": 1

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.21,Page no.167"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "#Wooden Beam\n",

-      "\n",

-      "b=150 #mm #width\n",

-      "d=250 #mm #Depth\n",

-      "\n",

-      "L=5000 #mm #span\n",

-      "m=11.2 #N/mm**2 #Max Bending stress\n",

-      "sigma=0.7 #N/mm**2 #Max shear stress\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Let 'a' be the distance from left support\n",

-      "#Max shear force\n",

-      "#F=R_A=W*(L-a)*L**-1 \n",

-      "\n",

-      "#Max Moment\n",

-      "#M=W*(L-a)*a*L**-1\n",

-      "\n",

-      "#But M=sigma*Z\n",

-      "#W*(L-a)*a*L**-1=m*1*6**-1*b*d**2   .....................(1)\n",

-      "\n",

-      "#In Rectangular Section MAx stress is 1.5 times Avg shear stress\n",

-      "F=sigma*b*d*1.5**-1\n",

-      "\n",

-      "#W*(L-a)*L**-1=F                     .....................(2)\n",

-      "\n",

-      "#Dividing Equation 1 nad 2 we get\n",

-      "a=m*6**-1*b*d**2*1.5*(sigma*b*d)**-1\n",

-      "\n",

-      "#Sub above value in equation 2 we get\n",

-      "W=(L-a)**-1*L*F*10**-3 #KN  \n",

-      "\n",

-      "#Result\n",

-      "print\"Load is\",round(W,2),\"KN\"\n",

-      "print\"Distance from Left support is\",round(a,2),\"mm\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Load is 21.87 KN\n",

-        "Distance from Left support is 1000.0 mm\n"

-       ]

-      }

-     ],

-     "prompt_number": 22

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.22,Page no.168"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "L=1000 #mm #span\n",

-      "\n",

-      "#Rectangular Section\n",

-      "\n",

-      "b=200 #mm #width\n",

-      "d=400 #mm #depth\n",

-      "\n",

-      "sigma=1.5 #N/mm**2 #Shear stress\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#Let AB be the cantilever beam subjected to load W KN at free end\n",

-      "\n",

-      "#MAx shear Force\n",

-      "#F=W*10**3 #KN\n",

-      "\n",

-      "#Since Max shear stress in Rectangular section\n",

-      "#sigma_max=1.5*F*A**-1 \n",

-      "#After sub values and further simplifyng we get\n",

-      "W=1.5*b*d*(1.5*1000)**-1 #KN\n",

-      "\n",

-      "#Moment at fixwed end\n",

-      "M=W*1 #KN-m\n",

-      "y_max=d*2**-1 #mm\n",

-      "\n",

-      "#M.I\n",

-      "I=1*12**-1*b*d**3 #mm**3\n",

-      "\n",

-      "#MAx Stress\n",

-      "sigma_max=M*10**6*I**-1*y_max\n",

-      "\n",

-      "#Result\n",

-      "print\"Concentrated Load is\",round(sigma_max,2),\"N/mm**2\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Concentrated Load is 15.0 N/mm**2\n"

-       ]

-      }

-     ],

-     "prompt_number": 14

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4.4.24,Page no.170"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#Initilization of Variables\n",

-      "\n",

-      "L=4000 #mm #span\n",

-      "\n",

-      "#Rectangular Cross-section\n",

-      "b=100 #mm #Width\n",

-      "d=200 #mm #Thickness\n",

-      "\n",

-      "F_per=10 #N/mm**2 #Max Bending stress\n",

-      "q_max=0.6 #N/mm**2 #Shear stress\n",

-      "\n",

-      "#Calculations\n",

-      "\n",

-      "#If the Load W is in KN/m\n",

-      "\n",

-      "#Max shear Force\n",

-      "#F=w*l*2**-1 #KN\n",

-      "#After substituting values and further simplifying we get\n",

-      "#M=2*w #KN-m\n",

-      "\n",

-      "#Max Load from Consideration of moment\n",

-      "#M=1*6**-1*b*d**2*F_per\n",

-      "#After substituting values and further simplifying we get\n",

-      "w=(1*6**-1*b*d**2*F_per)*(2*10**6)**-1 #KN/m\n",

-      "\n",

-      "#Max Load from Consideration of shear stress\n",

-      "#q_max=1.5*F*(b*d)**-1 #N\n",

-      "#After substituting values and further simplifying we get\n",

-      "F=q_max*(1.5)*b*d #N\n",

-      "\n",

-      "#If w is Max Load in KN/m,then\n",

-      "#2*w*1000=8000\n",

-      "#After Rearranging and Further simplifying we get\n",

-      "w2=8000*(2*1000)**-1 #KN/m\n",

-      "\n",

-      "#Result\n",

-      "print\"Uniformly Distributed Load Beam can carry is\",round(w,2),\"KN/m\""

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Uniformly Distributed Load Beam can carry is 3.33 KN/m\n"

-       ]

-      }

-     ],

-     "prompt_number": 8

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file