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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter4-Mechanical Behavior of Materials"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex1-pg101"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#calculate diameterof the bar in all cases\n",
      "import numpy\n",
      "from numpy.linalg import inv\n",
      "sigmayp=350. ##MPa\n",
      "sigma3=0.\n",
      "M=8. ##kN\n",
      "Mt=24. ##kNm\n",
      "N=2.\n",
      "v=0.3\n",
      "\n",
      "## sigma= My/I ==32M/%pid**3\n",
      "## tau= Mt*r/J ==16Mt/%pid**3\n",
      "##sigma1=(16*(M+sqrt(M**2+Mt**2)))/(%pi*d**3)\n",
      "##sigma2=(16*(M-sqrt(M**2+Mt**2)))/(%pi*d**3)\n",
      "\n",
      "##solution a: max principal stress theory\n",
      "##(16*(M+sqrt(M**2+Mt**2)))/(%pi*d**3)=sigmayp/N\n",
      "\n",
      "a=(16.*(M+math.sqrt(M**2.+Mt**2.)))/math.pi\n",
      "print(a)\n",
      "b=sigmayp*10**6./N\n",
      "print(b)\n",
      "d=(a/b)**(1/3.)\n",
      "print'%s %.5f %s'%(\"diameter of the bar in meter is= \",d,\"\")\n",
      "\n",
      "##solution b:max shearing stress theory\n",
      "\n",
      "c=(32.*math.sqrt(M**2.+Mt**2.))/math.pi\n",
      "print(c)\n",
      "d=(c/b)**(1/3.)\n",
      "print'%s %.5f %s'%(\"diameter of the bar in meter is= \",d,\"\")\n",
      "\n",
      "##solution c:max principal strain theory\n",
      "##epsilon1=(sigma1-v(sigma2+sigma3))/E=epsilonyp/N=sigmayp/EN    Or\n",
      "##sigma1-v(sigma2+sigma3)=b    given\n",
      "##sigma1=b+v(sigma2+sigma3)    substituting the values of the sigma 1,2 and 3 we get\n",
      "##(16*(M+sqrt(M**2+Mt**2)-vM-v*sqrt(M**2+Mt**2)))/(%pi*d**3)=b\n",
      "e=(16.*(M+math.sqrt(M**2.+Mt**2.)-v*M-v*math.sqrt(M**2.+Mt**2.)))/math.pi\n",
      "print(e)\n",
      "d=(e/b)**(1/3.)\n",
      "print'%s %.5f %s'%(\"diameter of the bar in meter is= \",d,\"\")\n",
      "\n",
      "##solution d:max energy of distortion theory\n",
      "\n",
      "f=(16.*math.sqrt(4*M**2.+3.*Mt**2.))/math.pi\n",
      "print'%s %.5f %s'%(\"f\",f,\"\")\n",
      "d=(f/b)**(1/3.)\n",
      "print'%s %.5f %s'%(\"diameter of the bar in meter is= \",d,\"\")\n",
      "print(\"all the values are converted into meter\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "169.586448419\n",
        "175000000.0\n",
        "diameter of the bar in meter is=  0.00990 \n",
        "257.685565975\n",
        "diameter of the bar in meter is=  0.01138 \n",
        "118.710513893\n",
        "diameter of the bar in meter is=  0.00879 \n",
        "f 226.85113 \n",
        "diameter of the bar in meter is=  0.01090 \n",
        "all the values are converted into meter\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex3-pg107"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#calculate max principal stress\n",
      "sigmau1=300.  ##MPa\n",
      "sigmau2=700.  ##MPa\n",
      "b=0.105 ##m outer diameter\n",
      "a=0.100 ##m inner diameter\n",
      "\n",
      "##sigma1==(-sigma2)==tau\n",
      "\n",
      "sigma3=0.\n",
      "\n",
      "##Mt=J*tau/r= (%pi*(b**2-a**2))/2*b\n",
      "##Mt=((%pi*(b**4-a**4))/(2*b))*tau         equation a\n",
      "q=(math.pi*(b**4.-a**4.))/(2.*b)\n",
      "\n",
      "##solution a: max principal stress theory\n",
      "tau=sigmau1\n",
      "Mt=q*tau*10**6.\n",
      "print'%s %.2f %s'%(\"max principal stress in Nm is= \",Mt,\"\")\n",
      "\n",
      "##solution b:max shearing stress theory\n",
      "## |sigma1-sigma2|=sigmau1\n",
      "## 2*sigma1==sigmau1==2*tau  or\n",
      "\n",
      "Mt=q*tau*10**6\n",
      "print'%s %.2f %s'%(\"max shearing stress in Nm is= \",Mt,\"\")\n",
      "\n",
      "##solution c:Coulomb- mohr theory\n",
      "##(tau/sigmau1)-(-tau/sigmau2)=1\n",
      "tau=1*((sigmau1*sigmau2)/(sigmau1+sigmau2))\n",
      "print'%s %.2f %s'%(\"tau in MPa is= \",tau,\"\")\n",
      "Mt=q*tau*10**6\n",
      "print'%s %.2f %s'%(\"Coulomb- mohr in Nm is= \",Mt,\"\")\n",
      "\n",
      "\n",
      "\n",
      "\n",
      "\n",
      "\n",
      "\n",
      "\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "max principal stress in Nm is=  96718.98 \n",
        "max shearing stress in Nm is=  96718.98 \n",
        "tau in MPa is=  210.00 \n",
        "Coulomb- mohr in Nm is=  67703.29 \n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex4-pg109"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#calculate laod and intertia\n",
      "a=0.05 ## m\n",
      "Fm=90. ##kN\n",
      "sigmacr=210. ## MPa\n",
      "sigmayp=280. ## MPa\n",
      "\n",
      "##sigmaa=Ma*c/I         equation 1\n",
      "##Ma=0.025*Fa\n",
      "c=0.025\n",
      "I=(a**4.)/12.\n",
      "print'%s %.7f %s'%(\"I\",I,\"\")\n",
      "##sigmaa=((0.025*Fa)*c)/I  substituting the values\n",
      "\n",
      "\n",
      "##sigmam=Fm/A           equation 2\n",
      "sigmam=Fm/(a*a)\n",
      "print'%s %.2f %s'%(\"in kilo Pa is= \",sigmam,\"\")\n",
      "\n",
      "##(((1200*Fa)/sigmacr)+(sigmam/sigmayp))=1\n",
      "Fa=(1.-(sigmam/sigmayp))*(sigmacr/1200.)\n",
      "print'%s %.2f %s'%(\"load Fa in N is= \",Fa,\"\")                         ##wrong ans in textbook\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "I 0.0000005 \n",
        "in kilo Pa is=  36000.00 \n",
        "load Fa in N is=  -22.32 \n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex5-pg110"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#calculate the value of pressure \n",
      "r=0.04 ##m\n",
      "t=5. ##mm\n",
      "sigmae=250. ##MPa\n",
      "sigmay=300. ##MPa\n",
      "\n",
      "##sigmathetamax=(p*r)/t =8*p        max values of tangential stresses\n",
      "##sigmathetamin=((-p/4)*r)/t =-2*p   min values of tangential stresses\n",
      "##sigmazamax=(p*r)/2*t =4*p       axial principal stresses\n",
      "##sigmazmin=((-p/4)*r)/2*t =-p\n",
      "\n",
      "##sigmathetaa=(sigmathetamax-sigmathetamin)/2= 5p    alternating and mean stresses\n",
      "##sigmathetam=(sigmathetamax+sigmathetamin)/2= 3p\n",
      "##sigmaza=(sigmazamax-sigmazmin)/2 =2.5p\n",
      "##sigmazm=(sigmazamax+sigmazmin)/2 =1.5p\n",
      "\n",
      "##sqrt(sigmathetaa**2-sigmathetaa*sigmaza+sigmaza**2)=sigmaea\n",
      "##sqrt(sigmathetam**2-sigmathetam*sigmazm+sigmazm**2)=sigmaem\n",
      "\n",
      "##sqrt(25p**2-12.3p**2+6.25p**2)=sigmaea\n",
      "##sqrt(9p**2-4.5p**2+2.25p**2)=sigmaem           solving this equation we get\n",
      "sigmaea=4.33 ##p\n",
      "sigmaem=2.60 ##p\n",
      "\n",
      "p=1./((sigmaea/sigmae)+(sigmaem/sigmay))\n",
      "print'%s %.2f %s'%(\"the value of p in MPa is= \",p,\"\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the value of p in MPa is=  38.48 \n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex6-pg112"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "import numpy\n",
      "from numpy.linalg import inv\n",
      "#find alternating and mean values of stresses and sigma and cycle\n",
      "a=([[700 ,14 ,0], [14, -350, 0],[0, 0, -350]])\n",
      "print(a)\n",
      "c=([[-660, -7, 0], [-7 ,-350, 0], [0, 0, -350]])\n",
      "print(c)\n",
      "sigmau=2400. ##MPa\n",
      "K=1.\n",
      "sigmae=800. ##MPa\n",
      "Nf=1. ##cycles for SAE\n",
      "Nff=10**3 ##cycles for Gerber\n",
      "Ne=10**8 ##cycles\n",
      "\n",
      "sigmaxa=(700.+660.)/2.\n",
      "print'%s %.2f %s'%(\"alternating and mean values of stresses in MPa is= \",sigmaxa,\"\")\n",
      "sigmaxm=(700-660)/2.\n",
      "print'%s %.2f %s'%(\"alternating and mean values of stresses in MPa is= \",sigmaxm,\"\")\n",
      "sigmaya=(-350+350)/2.\n",
      "print'%s %.2f %s'%(\"alternating and mean values of stresses in MPa is= \",sigmaya,\"\")\n",
      "sigmaym=(-350-350)/2.\n",
      "print'%s %.2f %s'%(\"alternating and mean values of stresses in MPa is= \",sigmaya,\"\")\n",
      "sigmazm=(-350-350)/2.\n",
      "print'%s %.2f %s'%(\"alternating and mean values of stresses in MPa is= \",sigmazm,\"\")\n",
      "tauxya=(14+7)/2.\n",
      "print'%s %.2f %s'%(\"alternating and mean values of stresses in MPa is= \",tauxya,\"\")\n",
      "tauxym=(14-7)/2.\n",
      "print'%s %.2f %s'%(\"alternating and mean values of stresses in MPa is= \",tauxym,\"\")\n",
      "\n",
      "sigmaea=math.sqrt(((sigmaxa-sigmaya)**2.+(sigmaya-sigmaxa)**2.+6.*(tauxya)**2.)/2.)\n",
      "print'%s %.2f %s'%(\"in MPa is =\",sigmae,\"\")\n",
      "sigmaem=math.sqrt(((sigmaxm-sigmaym)**2.+(sigmaym-sigmaxm)**2.+6.*(tauxym)**2.)/2.)\n",
      "print'%s %.2f %s'%(\"in MPa is= \",sigmaem,\"\")\n",
      "\n",
      "##solution a: \n",
      "sigmacr=sigmaea/(1-(sigmaem/2400.))\n",
      "print'%s %.3f %s'%(\"sigmacr\",sigmacr,\"\")\n",
      "b=math.log(sigmau/sigmae)/math.log(1./Ne)\n",
      "print'%s %.3f %s'%(\"b\",b,\"\")\n",
      "\n",
      "Ncr=1.*(sigmacr/2400.)**(1./b)\n",
      "print'%s %.3f %s'%(\"in cycles is= \",Ncr,\"\")\n",
      "\n",
      "##solution b:\n",
      "sigmacr=sigmaea/(1.-(sigmaem/sigmau)**2.)\n",
      "print'%s %.3f %s'%(\"in MPa is= \",sigmacr,\"\")\n",
      "b=math.log(2160./sigmae)/math.log(10**3./Ne)\n",
      "print'%s %.3f %s'%(\"b\",b,\"\")\n",
      "\n",
      "Ncr=Nff*(sigmacr/(0.9*2400))**(-11.587)\n",
      "print'%s %.3e %s'%(\"in cycles is= \",Ncr,\"\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "[[700, 14, 0], [14, -350, 0], [0, 0, -350]]\n",
        "[[-660, -7, 0], [-7, -350, 0], [0, 0, -350]]\n",
        "alternating and mean values of stresses in MPa is=  680.00 \n",
        "alternating and mean values of stresses in MPa is=  20.00 \n",
        "alternating and mean values of stresses in MPa is=  0.00 \n",
        "alternating and mean values of stresses in MPa is=  0.00 \n",
        "alternating and mean values of stresses in MPa is=  -350.00 \n",
        "alternating and mean values of stresses in MPa is=  10.50 \n",
        "alternating and mean values of stresses in MPa is=  3.50 \n",
        "in MPa is = 800.00 \n",
        "in MPa is=  370.05 \n",
        "sigmacr 804.248 \n",
        "b -0.060 \n",
        "in cycles is=  91502934.807 \n",
        "in MPa is=  696.809 \n",
        "b -0.086 \n",
        "in cycles is=  4.934e+08 \n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Ex7-pg115"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#calculate Inertia and deflection of a point in meter and impact factor and pressure and static deflection of the beam and impact facotr\n",
      "W=180. ##N\n",
      "h=0.1 ##m\n",
      "L=1.16 ##m\n",
      "w=0.025 ##m\n",
      "d=0.075 ##m\n",
      "E=200. ##GPa\n",
      "k=180. ##kN/m\n",
      "\n",
      "I=w*d**3.\n",
      "print'%s %.5f %s'%(\"I\",I,\"\")\n",
      "##deltast=(W*L**3)/(48*E*I)       equation 1\n",
      "deltast=(W*L**3*12.)/(48.*E*10**9.*I)\n",
      "print'%s %.5f %s'%(\"deflection of a point in meter is= \",deltast,\"\")\n",
      "\n",
      "##deltastmax=Mc/I                equation 2\n",
      "deltastmax=(W*L*12*0.0375)/(4*I)\n",
      "print'%s %.2f %s'%(\"deflection of a point in Pa is= \",deltastmax,\"\")\n",
      "\n",
      "##solution a:\n",
      "a=1+math.sqrt(1.+((2.*h)/deltast))\n",
      "print'%s %.2f %s'%(\"imapct factor is= \",a,\"\")\n",
      "deltamax=deltast*a\n",
      "print'%s %.2f %s'%(\"in meter is =\",deltastmax,\"\")\n",
      "sigmamax=deltastmax*a\n",
      "print'%s %.2f %s'%(\"in Pa is= \",sigmamax,\"\")\n",
      "\n",
      "##solution b:\n",
      "deltast=deltast+(90./180000.)\n",
      "print'%s %.5f %s'%(\"static deflection of the beam in meter is= \",deltast,\"\")\n",
      "a=1+math.sqrt(1.+((2.*h)/deltast))\n",
      "print'%s %.2f %s'%(\"imapct factor is= \",a,\"\")\n",
      "deltamax=deltast*a\n",
      "print'%s %.3f %s'%(\"in meter is =\",deltamax,\"\")\n",
      "sigmamax=deltastmax*a\n",
      "print'%s %.2f %s'%(\"in Pa is= \",sigmamax,\"\")\n",
      "#round off error\n",
      "\n",
      "\n",
      "\n",
      "\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "I 0.00001 \n",
        "deflection of a point in meter is=  0.00003 \n",
        "deflection of a point in Pa is=  2227200.00 \n",
        "imapct factor is=  78.51 \n",
        "in meter is = 2227200.00 \n",
        "in Pa is=  174848357.07 \n",
        "static deflection of the beam in meter is=  0.00053 \n",
        "imapct factor is=  20.39 \n",
        "in meter is = 0.011 \n",
        "in Pa is=  45415592.13 \n"
       ]
      }
     ],
     "prompt_number": 2
    }
   ],
   "metadata": {}
  }
 ]
}