{
 "metadata": {
  "name": "",
  "signature": "sha256:32b849ce2b6f77e54947c1d30daf4cb67054a775f92b165f8c389aea69b70715"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter6-Torision;Including Non-Circular Sections"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex3-pg187"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## initialization o variables\n",
      "#find the shear force on each rivet and suitable diameter of rivets\n",
      "p=5. ##cm\n",
      "D=10. ##cm\n",
      "d=2. ##mm\n",
      "T= 10. ##kgm\n",
      "ss= 785. ##kg/cm^2\n",
      "## calculations\n",
      "P= 2.*T/(math.pi*D**2)\n",
      "P=P*5*100.\n",
      "## results\n",
      "print'%s %.2f %s'%('Force per rivet is ',P,' kg')\n",
      "print'%s %.2f %s'%('\\n The diameter of rivet, using a permissible stress of',ss,' kg/cm^2 = 0.227 cm')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Force per rivet is  31.83  kg\n",
        "\n",
        " The diameter of rivet, using a permissible stress of 785.00  kg/cm^2 = 0.227 cm\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex4-pg189"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## initialization of variables\n",
      "#what is maximum torque\n",
      "D=5. ##cm\n",
      "Y=3500. ##kg/cm^2\n",
      "\n",
      "##part (a)\n",
      "Ta=350. ##kg-m\n",
      "tau=Y/2.\n",
      "Ip=Ta*D*100./(2.*tau)\n",
      "d1=Ip*32./math.pi\n",
      "d1=(D**4-d1)**(1/4.)\n",
      "\n",
      "##part (b)\n",
      "Tb= 700. ##kg-m\n",
      "Ip=Tb*D*100./(2.*tau)\n",
      "d2=Ip*32./math.pi\n",
      "d2=(D**4-d2)\n",
      "T=tau*math.pi*(D**4)*2./(32.*D)\n",
      "## results\n",
      "print'%s %.2f %s'%('The maximum diameter corresponding to the case a is ',d1,' cm')\n",
      "print'%s %.2f %s'%('\\n Since the daimeter for the case (b) is coming out to be negative, \\n The maximum torque transmitted is ',T/100,' kg-m')\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The maximum diameter corresponding to the case a is  3.28  cm\n",
        "\n",
        " Since the daimeter for the case (b) is coming out to be negative, \n",
        " The maximum torque transmitted is  429.51  kg-m\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex5-pg190"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## initialization of variables\n",
      "#find the weight of students\n",
      "A=3 ##cm^2\n",
      "E= 2*10**6 ##kg/cm^2\n",
      "nu= 0.25\n",
      "l= 60. ##m\n",
      "L=150. ##cm\n",
      "d=0.5 ##cm\n",
      "dd=10 ##cm\n",
      "D=180 ##cm\n",
      "##calculations\n",
      "K=(l*100./(A*E))+(L*D/2.*D*32.*2.*(1.+nu)/(E*math.pi*dd**4*2))\n",
      "P=d/K\n",
      "## results\n",
      "print'%s %.2f %s'%('The weight of the students that entered the length is',P,' kg')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The weight of the students that entered the length is 196.31  kg\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex6-pg192"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## initialization of variables\n",
      "#find the stress in the cable and the shaft\n",
      "## linked to 6_5\n",
      "A=3. ##cm^2\n",
      "E= 2.*10**6 ##kg/cm^2\n",
      "nu= 0.25\n",
      "l= 60. ##m\n",
      "L=150. ##cm\n",
      "d=0.5 ##cm\n",
      "dd=10 ##cm\n",
      "D=180. ##cm\n",
      "##calculations\n",
      "K=(l*100./(A*E))+(L*D/2.*D*32.*2.*(1+nu)/(E*math.pi*dd**4*2.))\n",
      "P=d/K\n",
      "Ts=P/A\n",
      "fs=dd*D*P*32./(math.pi*4.*dd**4)\n",
      "\n",
      "## results\n",
      "print'%s %.2f %s'%('The tensile stress is ',Ts,'kg/cm^2')\n",
      "print'%s %.2f %s'%('\\n Maximum shear stress is ',fs,' kg/cm^2')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The tensile stress is  65.44 kg/cm^2\n",
        "\n",
        " Maximum shear stress is  89.98  kg/cm^2\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex7-pg192"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## initialization of variables\n",
      "#find the value of x and length of the suspension arm and D\n",
      "F=500. ##kg\n",
      "k=25. ##kg/cm\n",
      "dd=15. ##cm\n",
      "ss=3500. ##kg/cm^2\n",
      "L=2. ##m\n",
      "G=8.*10**5 ##kg/cm^2\n",
      "## calculations\n",
      "x=math.sqrt(math.pi*G/(25.*L*32.*100.))\n",
      "d=x*16.*(F+dd*k)/(ss*math.pi)\n",
      "x2=x*d**2.\n",
      "## results\n",
      "print'%s %.2f %s'%('d=',d,' cm')\n",
      "print'%s %.2f %s'%('\\n x=',x,' cm')\n",
      "\n",
      "## Text: not exact\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "d= 5.05  cm\n",
        "\n",
        " x= 3.96  cm\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex11-pg201"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "##initialization of variables\n",
      "#find the horse power\n",
      "d=5. ##cm\n",
      "rpm1=300. ##rpm\n",
      "rpm2=30000. ##rpm\n",
      "s=1000. ##kg/cm^2\n",
      "##calcuations\n",
      "T=(d/2.)*math.pi*10**2*s/32.\n",
      "hp1= 2*math.pi*rpm1*T/4500.\n",
      "hp2=hp1*100.\n",
      "## results\n",
      "print'%s %.2f %s %.2f %s'%('Horse power at 300 rpm and 30000 rpm are respecively ',hp1/10,''and  '',hp2/10,' h.p.')\n",
      "\n",
      "print('wrong/approximate answers in the text')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Horse power at 300 rpm and 30000 rpm are respecively  1028.08  102808.38  h.p.\n",
        "wrong/approximate answers in the text\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex12-pg200"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## initialization of variables\n",
      "#selct solid transmits and diameter required find\n",
      "hp=300. ##h.p.\n",
      "N1=30. ##rpm\n",
      "N2=30000. ##rpm\n",
      "fs=600. ##kg/cm^2\n",
      "## calculations\n",
      "T1=4500.*hp*100./(2.*math.pi*N1)\n",
      "T2=T1/1000.\n",
      "D1=16.*T1/(math.pi*fs)\n",
      "D1=D1**(1/3.)\n",
      "D2=16.*T2/(math.pi*fs)\n",
      "D2=D2**(1/3.)\n",
      "## results\n",
      "print'%s %.2f %s %.2f %s '%('Diameters required are ',D1,''and '',D2,'cm')\n",
      "\n",
      "print('wrong calculations in the text')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Diameters required are  18.25  1.83 cm \n",
        "wrong calculations in the text\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex13-pg202"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## initialization of variables\n",
      "#find the strain energy per unit volume stored and the energy if the lenght 5cm \n",
      "d=10. ##cm\n",
      "t =1. ##mm\n",
      "T= 100. ##kg-m\n",
      "L=5. ##m\n",
      "G=8*10**5 ##kg/cm^2\n",
      "\n",
      "##calculations\n",
      "r=d/2.\n",
      "fs=T*r*100./(r**2*2*math.pi*L*t*10**-1)\n",
      "U=fs**2/(2*G)\n",
      "U1=U*(math.pi*L*100.)\n",
      "## results\n",
      "print'%s %.2f %s'%('Energy per unit volume = ',U,' kg-cm/cm^3')\n",
      "print'%s %.2f %s'%('\\n Total strain energy= ',U1,' kg-cm')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Energy per unit volume =  0.25  kg-cm/cm^3\n",
        "\n",
        " Total strain energy=  397.89  kg-cm\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex14-pg206"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "##initialization of variables\n",
      "#find the actual deflection of closely coiled comperssion spring\n",
      "D=10. ##cm\n",
      "d= 1. ##cm\n",
      "n=20.\n",
      "P=60. ##kg\n",
      "G=8*10**5 ##kg/cm^2\n",
      "##calculations\n",
      "n=n-0.75*2\n",
      "delta=P*n*math.pi*D**3*32./(4.*math.pi*G)\n",
      "## results\n",
      "print'%s %.2f %s'%('The deflection is ',delta,' cm')\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The deflection is  11.10  cm\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex15-pg206"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "##initialization of variables\n",
      "#find the shear stress in the wire\n",
      "## linked to 6_14\n",
      "D=10. ##cm\n",
      "d= 1. ##cm\n",
      "n=20.\n",
      "P=60. ##kg\n",
      "G=8*10**5 ##kg/cm^2\n",
      "## calculations\n",
      "m=D/d\n",
      "fs=8.*P*D/(d**3*math.pi)\n",
      "fs1=fs*(1+0.615/m+3./(4.*m-4.))\n",
      "## results\n",
      "print'%s %.2f %s %.2f %s '%('The shear stress with and without correction facor are \\n respectively ',fs,''and  '',fs1,'kg/cm^2')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The shear stress with and without correction facor are \n",
        " respectively  1527.89  1749.18 kg/cm^2 \n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex16-pg214"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "##initialization of variables\n",
      "#find their angles of twist and maximum stresses\n",
      "\n",
      "## circle\n",
      "D=1. ##unit diameter\n",
      "Ip=D**4/32.\n",
      "Zp=D**3/16.\n",
      "\n",
      "##Square\n",
      "s=math.sqrt(math.pi/4.)*D\n",
      "Is=0.886*D**4/32.\n",
      "Zs=0.7383*D**3/16.\n",
      "\n",
      "##Rectangle\n",
      "a=math.sqrt(math.pi/2.)*D\n",
      "b=math.sqrt(math.pi/8.)*D\n",
      "Ir=0.719*D**4/32.\n",
      "Zr=0.616*D**3/16.\n",
      "\n",
      "## Trianle\n",
      "t=math.sqrt(math.pi/math.sqrt(3))*D\n",
      "It=0.725*D**4/32.\n",
      "Zt=0.622*D**3/16.\n",
      "\n",
      "##Ellipse\n",
      "A=D/math.sqrt(2)\n",
      "B=D/math.sqrt(8)\n",
      "Ie=A**3*B**3/(A**2+B**2)\n",
      "Ze=A*B**2/2.\n",
      "\n",
      "##Normalization\n",
      "Is=Is/Ip\n",
      "Ie=Ie/Ip\n",
      "It=It/Ip\n",
      "Ir=Ir/Ip\n",
      "\n",
      "Zs=Zs/Zp\n",
      "Ze=Ze/Zp\n",
      "Zt=Zt/Zp\n",
      "Zr=Zr/Zp\n",
      "Ip=1.\n",
      "Zp=1.\n",
      "##results\n",
      "print'%s %.2f %s %.2f %s %.2f %s %.2f %s %.2f %s '%('Z:: Circle:Square:Ellipse:Triangle:Rectangle = ',Zp,'' and '',Zs,'' and '',Ze,'' and '',Zt,' 'and '',Zr,'')\n",
      "\n",
      "print'%s %.2f %s %.2f %s %.2f %s %.2f %s %.2f %s '%('\\n I:: Circle:Square:Ellipse:Triangle:Rectangle = ',Ip,'' and  '',Is,'' and '',Ie,'' and '',It,'' and '',Ir,'')\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Z:: Circle:Square:Ellipse:Triangle:Rectangle =  1.00  0.74  0.71  0.62  0.62  \n",
        "\n",
        " I:: Circle:Square:Ellipse:Triangle:Rectangle =  1.00  0.89  0.80  0.72  0.72  \n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex17-pg215"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "##initialization of variables\n",
      "#find the maximum torque that can be transmitted without yeilding\n",
      "\n",
      "yp=2450. ##kg/cm^2\n",
      "d=0.4 ##cm\n",
      "ys=4200. ##kg/cm^2\n",
      "sa=1.6 ##mm\n",
      "sb=7. ##mm\n",
      "## calculations\n",
      "sa=sa/10.\n",
      "sb=sb/10.\n",
      "T1=yp*math.pi*d**3/16.\n",
      "T2=ys*0.303*sa**2*sb\n",
      "## results\n",
      "print'%s %.2f %s'%('The maximum torque that can be transitted by the screw-driver is ',T2,' kg-cm')\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The maximum torque that can be transitted by the screw-driver is  22.80  kg-cm\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex18-pg216"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "##initialization of variables\n",
      "#find the maximum stress in the walls \n",
      "b=5. ##cm\n",
      "h=10. ##cm\n",
      "tL=3. ##mm\n",
      "tl=1.5 ##mm\n",
      "T=100. ##kg-cm\n",
      "## calculations\n",
      "tl=tl/10.\n",
      "fs=T*100./(2.*b*h*tl)\n",
      "## results\n",
      "print'%s %.2f %s'%('The maximum stress is ',fs,' kg/cm^2')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The maximum stress is  666.67  kg/cm^2\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex19-pg216"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "##initialization of variables\n",
      "#compare its strength with hollow circular shaft of same thickness\n",
      "b=5. ##cm\n",
      "h=10. ##cm\n",
      "tL=3. ##mm\n",
      "tl=1.5 ##mm\n",
      "T=100. ##kg-cm\n",
      "## calculations\n",
      "D=2.*(b+h)/math.pi\n",
      "AR=b*h\n",
      "AC=math.pi*D**2/4.\n",
      "r=AC/AR\n",
      "## results\n",
      "print'%s %.2f %s'%('The ratio is 1:',r,'')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The ratio is 1: 1.43 \n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex20-pg217"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "##initialization of variables\n",
      "#find the value of maximum shear stress and angle of twist\n",
      "G=8*10**5 ##kg/cm^2\n",
      "##part (a)\n",
      "T =20. ##kg-m\n",
      "t1=0.9 ##cm\n",
      "t2=0.5 ##cm\n",
      "b1=6.8 ##cm\n",
      "b2=14.2 ##cm\n",
      "I0=1/3.*(2.*b1*t1**3+b2*t2**3)\n",
      "Zt=I0/max(t1,t2)\n",
      "fs=T*100./Zt\n",
      "Phi=T*100./(G*I0)\n",
      "print('part (a)')\n",
      "print'%s %.2f %s %.5f %s '%('\\n The maximum shear stress and twist rate are respectively \\n ',fs,' kg/cm^2'and'',Phi, ' radians/cm ')\n",
      "\n",
      "##part (b)\n",
      "t1=1. ##cm\n",
      "t2=1. ##cm\n",
      "b1=10. ##cm\n",
      "b2=9. ##cm\n",
      "I0=1/3.*(b1*t1**3+b2*t2**3)\n",
      "Zt=I0/max(t1,t2)\n",
      "fs=T*100./Zt\n",
      "Phi=T*100/(G*I0)\n",
      "print('\\n part (b)')\n",
      "print'%s %.2f %s %.5f %s '%('\\n The maximum shear stress and twist rate are respectively \\n ',fs,'kg/cm^2'and '',Phi,'radians/cm ')\n",
      "\n",
      "##part (c)\n",
      "t1=0.76 ##cm \n",
      "t2=0.48 ##cm\n",
      "b1=8. ##cm\n",
      "b2=14.04 ##cm\n",
      "I0=(1/3.)*(2*b1*t1**3+b2*t2**3)\n",
      "Zt=I0/max(t1,t2)\n",
      "fs=T*100./Zt\n",
      "Phi=T*100./(G*I0)\n",
      "print('\\n part (c)')\n",
      "print'%s %.2f %s %.5f %s ' %('\\n The maximum shear stress and twist rate are respectively \\n ',fs,' kg/cm^2'and '',Phi,'radians/cm ')\n",
      "\n",
      "##part(d)\n",
      "t=1 ##cm \n",
      "b=19 ##cm\n",
      "I0=1/3.*t**3*b\n",
      "Zt=I0/t\n",
      "fs=T*100./Zt\n",
      "Phi=T*100./(G*I0)\n",
      "print('\\n part (d)')\n",
      "print'%s %.2f %s %.5f %s '%('\\n The maximum shear stress and twist rate are respectively \\n ',fs,' kg/cm^2'and '',Phi,'radians/cm ')\n",
      "\n",
      "print('Twist rate: answers differ by a scale of 10. wrong answers in the text')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "part (a)\n",
        "\n",
        " The maximum shear stress and twist rate are respectively \n",
        "  461.96  0.00064  radians/cm  \n",
        "\n",
        " part (b)\n",
        "\n",
        " The maximum shear stress and twist rate are respectively \n",
        "  315.79  0.00039 radians/cm  \n",
        "\n",
        " part (c)\n",
        "\n",
        " The maximum shear stress and twist rate are respectively \n",
        "  531.70  0.00087 radians/cm  \n",
        "\n",
        " part (d)\n",
        "\n",
        " The maximum shear stress and twist rate are respectively \n",
        "  315.79  0.00039 radians/cm  \n",
        "Twist rate: answers differ by a scale of 10. wrong answers in the text\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex21-pg221"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## initialization of variables\n",
      "#caclualte Maximum +ve residual stress occurs at r and'radius\n",
      "D=5. ##cm\n",
      "d=2. ##cm\n",
      "t_y=3000. ##kg/cm^2\n",
      "## calculations\n",
      "R=D/2. \n",
      "r=d/2. \n",
      "Tep=2.*math.pi*R**3*t_y/3.-math.pi*r**3.*t_y/6.\n",
      "t_er=2*Tep/(math.pi*R**3)\n",
      "t_er1=t_er*r/R\n",
      "prs=t_y-t_er1\n",
      "nrs=t_er-t_y\n",
      "## results\n",
      "print'%s %.2f %s %.2f %s '%('Maximum +ve residual stress occurs at ',r,' cm' and'radius and is equal to \\n ',prs,' kg/cm^2')\n",
      "print'%s %.2f %s %.2f %s '%('\\n Maximum -ve residual stress occurs at ',R,' cm'and' radius and is equal to \\n ',-nrs,' kg/cm^2')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Maximum +ve residual stress occurs at  1.00 radius and is equal to \n",
        "  1425.60  kg/cm^2 \n",
        "\n",
        " Maximum -ve residual stress occurs at  2.50  radius and is equal to \n",
        "  -936.00  kg/cm^2 \n"
       ]
      }
     ],
     "prompt_number": 17
    }
   ],
   "metadata": {}
  }
 ]
}