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  "name": ""
 },
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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 11:Design of Beams and Shafts"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.1 Page No 544"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "sigma_allow = 24    #ksi, stress\n",
      "tou_allow = 14.5      #ksi, allowable stress\n",
      "\n",
      "#Shear and Moment Diagrams\n",
      "V_max = 30         #kip, \n",
      "M_max = 120         #kip, bending moment\n",
      "\n",
      "#Bending Stress\n",
      "S_reqd = (M_max*(10**3))/sigma_allow\n",
      "#Shear Stress\n",
      "d = 17.9   #in\n",
      "tw = 0.315  #inch\n",
      "tou_avg = (V_max)/(d*tw)\n",
      "\n",
      "if tou_avg<14.5:\n",
      "    print\"Use a W18*14\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Use a W18*14\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.2 Page No 545"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "udl = 0.5         #kN/m\n",
      "h_by_a = 1.5      \n",
      "sigma_allow = 9 #MPa, allowable bending stress\n",
      "tou_allow = 0.6 #MPa, allowable shear stress\n",
      "\n",
      "#calculation\n",
      "#Shear and Moment Diagrams\n",
      "import math\n",
      "V_max =20         #kNm\n",
      "M_max =10.67       #kNm\n",
      "#Bending Stress\n",
      "S_reqd = (M_max)/(sigma_allow*1000)\n",
      "c = h_by_a/2.0\n",
      "a_cube = (S_reqd*c*12)/(1.5**3) #S_reqd = I/c\n",
      "a = a_cube**(1/3.0)\n",
      "A = a*h_by_a*a\n",
      "tou_max = (1.5*V_max)/(A*1000)\n",
      "if tou_max>tou_allow:\n",
      "    a_sqr = (3/2.0)*(V_max)/(h_by_a*tou_allow*1000)\n",
      "    a =math.sqrt(a_sqr)\n",
      "else:\n",
      "    print\"not\"\n",
      "\n",
      "#Display\n",
      "print\"The smallest width for the laminated wooden beam = \", round(a,2),\"m\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The smallest width for the laminated wooden beam =  0.18 m\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.3 Page No 546"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "l = 200/1000.0       #m, length\n",
      "t = 30/1000.0        #m\n",
      "sigma_allow = 9     #MPa, shear stress\n",
      "tou_allow = 0.6      #MPa, bending stress\n",
      "V_nail = 1.50        #kN\n",
      "l_bc = 2             #m\n",
      "l_cd = 2            #m\n",
      "\n",
      "#Shear and Moment Diagrams\n",
      "V_max = 20        #kN\n",
      "M_max =10.67         #kNm\n",
      "#Bending Stress\n",
      "y1 = l/2.0\n",
      "A1 = l*t\n",
      "y2 = l+(t/2.0)\n",
      "A2 = t*l\n",
      "y_dash = (y1*A1 + y2*A2)/(A1 + A2)\n",
      "\n",
      "I1 = (t*l**3)/12.0 + (t*l*(y_dash - y1)**2)\n",
      "I2 = (l*t**3)/12.0 + (t*l*(y2 - y_dash)**2)\n",
      "I =I1 + I2\n",
      "\n",
      "c = y_dash\n",
      "sigma = (M_max*c)/(I)\n",
      "flag1 = 0\n",
      "sigma_allow = sigma_allow*1000 #kPa\n",
      "\n",
      "if(sigma<sigma_allow):\n",
      "    flag1 = 1\n",
      "else:\n",
      "    print\"otherwisw not\"\n",
      "\n",
      "#Shear Stress\n",
      "y3 = y_dash/2.0\n",
      "A3 = y_dash*t\n",
      "Q = y3*A3\n",
      "\n",
      "tou = (V_max*Q)/(I*t)\n",
      "tou_allow = tou_allow*1000 #kPa\n",
      "flag2 =0\n",
      "\n",
      "if(tou<tou_allow):\n",
      "    flag2 = 1\n",
      "else:\n",
      "    print\"flag2 is not equal to one\"\n",
      "\n",
      "#Nail Spacing\n",
      "y4a = (l+t-y_dash)\n",
      "y4 = y4a - (t/2.0)\n",
      "A4 = l*t\n",
      "Q4 = y4*A4\n",
      "V_bc = 1.5 #kN\n",
      "V_cd = 1 #kN\n",
      "\n",
      "q_bc = (V_bc*Q4)/I\n",
      "q_cd = (V_cd*Q4)/I\n",
      "\n",
      "s_bc = (V_nail)/(q_bc)\n",
      "s_cd = (V_nail)/(q_cd)\n",
      "\n",
      "chosen_bc = 150 #mm\n",
      "chosen_cd = 250 #mm\n",
      "\n",
      "#Result\n",
      "print'The design is safe in bending and shear.'\n",
      "print'The calculated nail spacing BC ',round(s_bc,2),\"m\"\n",
      "print'The calculated nail spacing CD ',round(s_cd,2),\"m\"\n",
      "print'The chosen nail spacing BC     ',chosen_bc,\"mm\"\n",
      "print'The chosen nail spacing CD     ',chosen_cd,\"mm\"\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "otherwisw not\n",
        "flag2 is not equal to one\n",
        "The design is safe in bending and shear.\n",
        "The calculated nail spacing BC  0.17 m\n",
        "The calculated nail spacing CD  0.26 m\n",
        "The chosen nail spacing BC      150 mm\n",
        "The chosen nail spacing CD      250 mm\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.6 Page No 560"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Find smallest allowable diameter\n",
      "\n",
      "#Given\n",
      "tou_allow = 50*10**6 #MPa, shear stress\n",
      "T = 7.5              #Nm, torque\n",
      "R_ah = 150           #N, horizontal force\n",
      "R_av = 475           #N\n",
      "l_ac = 0.25          #m\n",
      "\n",
      "#Calculation\n",
      "import math\n",
      "mc = R_ah*l_ac\n",
      "m = R_av*l_ac\n",
      "M_c = sqrt(m**2 + mc**2)\n",
      "k = sqrt(M_c**2 + T**2)\n",
      "c1 = (2*k)/(math.pi*tou_allow)\n",
      "c = c1**(1/3.0)\n",
      "d = 2*c*1000\n",
      "\n",
      "#Display\n",
      "print\"The smallest allowable diameter of the shaft =\",round(d,1),\"mm\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The smallest allowable diameter of the shaft = 23.3 mm\n"
       ]
      }
     ],
     "prompt_number": 3
    }
   ],
   "metadata": {}
  }
 ]
}